From Classical to Quantum Shannon
TheoryMark M. Wilde
Hearne Institute for Theoretical Physics
Department of Physics and Astronomy
Center for Computation and Technology
Louisiana State University
Baton Rouge, Louisiana 70803, USA
December 5, 2021

2
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Contents

I

Introduction

17

1 Concepts in Quantum Shannon Theory
1.1 Overview of the Quantum Theory . . . . . . . . . . . . . . . . . . . . . . . .
1.2 The Emergence of Quantum Shannon Theory . . . . . . . . . . . . . . . . .

19
22
26

2 Classical Shannon Theory
2.1 Data Compression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Channel Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

41
41
50
64

II

67

The Quantum Theory

3 The Noiseless Quantum Theory69
3.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
3.2 Quantum Bits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
3.3 Reversible Evolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76
3.4 Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83
3.5 Composite Quantum Systems . . . . . . . . . . . . . . . . . . . . . . . . . . 88
3.6 Entanglement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96
3.7 Summary and Extensions to Qudit States . . . . . . . . . . . . . . . . . . . 107
3.8 Schmidt Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113
3.9 History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
4 The Noisy Quantum Theory
117
4.1 Noisy Quantum States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118
4.2 Measurement in the Noisy Quantum Theory . . . . . . . . . . . . . . . . . . 130
4.3 Composite Noisy Quantum Systems . . . . . . . . . . . . . . . . . . . . . . . 133
4.4 Quantum Evolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145
4.5 Interpretations of Quantum Channels . . . . . . . . . . . . . . . . . . . . . . 156
4.6 Quantum Channels are All Encompassing . . . . . . . . . . . . . . . . . . . 158
3

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4.7
4.8
4.9

Examples of Quantum Channels . . . . . . . . . . . . . . . . . . . . . . . . . 170
Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

5 The Purified Quantum Theory
181
5.1 Purification . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182
5.2 Isometric Evolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185
5.3 Coherent Quantum Instrument . . . . . . . . . . . . . . . . . . . . . . . . . 196
5.4 Coherent Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197
5.5 History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 198

III

Unit Quantum Protocols

201

6 Three Unit Quantum Protocols
203
6.1 Non-Local Unit Resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204
6.2 Protocols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
6.3 Optimality of the Three Unit Protocols . . . . . . . . . . . . . . . . . . . . . 215
6.4 Extensions for Quantum Shannon Theory . . . . . . . . . . . . . . . . . . . 217
6.5 Three Unit Qudit Protocols . . . . . . . . . . . . . . . . . . . . . . . . . . . 218
6.6 History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 223
7 Coherent Protocols
225
7.1 Definition of Coherent Communication . . . . . . . . . . . . . . . . . . . . . 227
7.2 Implementations of a Coherent Bit Channel . . . . . . . . . . . . . . . . . . 228
7.3 Coherent Dense Coding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230
7.4 Coherent Teleportation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
7.5 Coherent Communication Identity . . . . . . . . . . . . . . . . . . . . . . . . 233
7.6 History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 234
8 Unit Resource Capacity Region
235
8.1 The Unit Resource Achievable Region . . . . . . . . . . . . . . . . . . . . . . 235
8.2 The Direct Coding Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 239
8.3 The Converse Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 240
8.4 History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 244

IV

Tools of Quantum Shannon Theory

245

9 Distance Measures
247
9.1 Trace Distance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 248
9.2 Fidelity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261
9.3 Relations between Trace Distance and Fidelity . . . . . . . . . . . . . . . . . 271
9.4 Gentle Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275
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9.5
9.6
9.7

5

Fidelity of a Quantum Channel . . . . . . . . . . . . . . . . . . . . . . . . . 278
The Hilbert–Schmidt Distance Measure . . . . . . . . . . . . . . . . . . . . . 282
History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 283

10 Classical Information and Entropy
285
10.1 Entropy of a Random Variable . . . . . . . . . . . . . . . . . . . . . . . . . . 286
10.2 Conditional Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291
10.3 Joint Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292
10.4 Mutual Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293
10.5 Relative Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 294
10.6 Conditional Mutual Information . . . . . . . . . . . . . . . . . . . . . . . . . 295
10.7 Entropy Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296
10.8 Near Saturation of Entropy Inequalities . . . . . . . . . . . . . . . . . . . . . 308
10.9 Classical Information from Quantum Systems . . . . . . . . . . . . . . . . . 314
10.10History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 316
11 Quantum Information and Entropy
317
11.1 Quantum Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 318
11.2 Joint Quantum Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322
11.3 Potential yet Unsatisfactory Definitions of Conditional Quantum Entropy . . 326
11.4 Conditional Quantum Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . 327
11.5 Coherent Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330
11.6 Quantum Mutual Information . . . . . . . . . . . . . . . . . . . . . . . . . . 332
11.7 Conditional Quantum Mutual Information . . . . . . . . . . . . . . . . . . . 335
11.8 Quantum Relative Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337
11.9 Quantum Entropy Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . 343
11.10Continuity of Quantum Entropy . . . . . . . . . . . . . . . . . . . . . . . . . 355
11.11History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 361
12 Quantum Entropy Inequalities and Recoverability363
12.1 Recoverability Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363
12.2 Schatten Norms and Complex Interpolation . . . . . . . . . . . . . . . . . . 364
12.3 Petz Recovery Map . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 369
12.4 Rényi Information Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . 371
12.5 Proof of the Recoverability Theorem . . . . . . . . . . . . . . . . . . . . . . 376
12.6 Refinements of Quantum Entropy Inequalities . . . . . . . . . . . . . . . . . 378
12.7 History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 382
13 The Information of Quantum Channels
385
13.1 Mutual Information of a Classical Channel . . . . . . . . . . . . . . . . . . . 386
13.2 Private Information of a Wiretap Channel . . . . . . . . . . . . . . . . . . . 394
13.3 Holevo Information of a Quantum Channel . . . . . . . . . . . . . . . . . . . 398
13.4 Mutual Information of a Quantum Channel . . . . . . . . . . . . . . . . . . 404
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13.5 Coherent Information of a Quantum Channel . . . . . . . . . . . . . . . . .
13.6 Private Information of a Quantum Channel . . . . . . . . . . . . . . . . . . .
13.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13.8 History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . .

408
413
419
420

14 Classical Typicality421
14.1 An Example of Typicality . . . . . . . . . . . . . . . . . . . . . . . . . . . . 422
14.2 Weak Typicality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423
14.3 Properties of the Typical Set . . . . . . . . . . . . . . . . . . . . . . . . . . . 425
14.4 Application: Data Compression . . . . . . . . . . . . . . . . . . . . . . . . . 427
14.5 Weak Joint Typicality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429
14.6 Weak Conditional Typicality . . . . . . . . . . . . . . . . . . . . . . . . . . . 432
14.7 Strong Typicality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 435
14.8 Strong Joint Typicality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 444
14.9 Strong Conditional Typicality . . . . . . . . . . . . . . . . . . . . . . . . . . 446
14.10Application: Channel Capacity Theorem . . . . . . . . . . . . . . . . . . . . 452
14.11Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 456
14.12History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 457
15 Quantum Typicality459
15.1 The Typical Subspace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 460
15.2 Conditional Quantum Typicality . . . . . . . . . . . . . . . . . . . . . . . . 469
15.3 The Method of Types for Quantum Systems . . . . . . . . . . . . . . . . . . 479
15.4 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481
15.5 History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 481
16 The Packing Lemma
483
16.1 Introductory Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484
16.2 The Setting of the Packing Lemma . . . . . . . . . . . . . . . . . . . . . . . 484
16.3 Statement of the Packing Lemma . . . . . . . . . . . . . . . . . . . . . . . . 486
16.4 Proof of the Packing Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . 488
16.5 Derandomization and Expurgation . . . . . . . . . . . . . . . . . . . . . . . 494
16.6 Sequential Decoding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496
16.7 History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 500
17 The Covering Lemma
501
17.1 Introductory Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502
17.2 Setting and Statement of the Covering Lemma . . . . . . . . . . . . . . . . . 504
17.3 Operator Chernoff Bound . . . . . . . . . . . . . . . . . . . . . . . . . . . . 505
17.4 Proof of the Covering Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . 510
17.5 History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 516
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V

Noiseless Quantum Shannon Theory

7517

18 Schumacher Compression
519
18.1 The Information-Processing Task . . . . . . . . . . . . . . . . . . . . . . . . 520
18.2 The Quantum Data-Compression Theorem . . . . . . . . . . . . . . . . . . . 522
18.3 Quantum Compression Example . . . . . . . . . . . . . . . . . . . . . . . . . 526
18.4 Variations on the Schumacher Theme . . . . . . . . . . . . . . . . . . . . . . 527
18.5 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 529
18.6 History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 529
19 Entanglement Manipulation
531
19.1 Sketch of Entanglement Manipulation . . . . . . . . . . . . . . . . . . . . . . 532
19.2 LOCC and Relative Entropy of Entanglement . . . . . . . . . . . . . . . . . 536
19.3 Entanglement Manipulation Task . . . . . . . . . . . . . . . . . . . . . . . . 538
19.4 The Entanglement Manipulation Theorem . . . . . . . . . . . . . . . . . . . 539
19.5 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 548
19.6 History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 549

VI

Noisy Quantum Shannon Theory551

20 Classical Communication
555
20.1 Naive Approach: Product Measurements . . . . . . . . . . . . . . . . . . . . 557
20.2 The Information-Processing Task . . . . . . . . . . . . . . . . . . . . . . . . 559
20.3 The Classical Capacity Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 562
20.4 Examples of Channels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 569
20.5 Superadditivity of the Holevo Information . . . . . . . . . . . . . . . . . . . 578
20.6 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 581
20.7 History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 582
21 Entanglement-Assisted Classical Communication
585
21.1 The Information-Processing Task . . . . . . . . . . . . . . . . . . . . . . . . 587
21.2 A Preliminary Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 588
21.3 Entanglement-Assisted Capacity Theorem . . . . . . . . . . . . . . . . . . . 591
21.4 The Direct Coding Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 592
21.5 The Converse Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 600
21.6 Examples of Channels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 606
21.7 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 610
21.8 History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 611
22 Coherent Communication with Noisy Resources613
22.1 Entanglement-Assisted Quantum Communication . . . . . . . . . . . . . . . 614
22.2 Quantum Communication . . . . . . . . . . . . . . . . . . . . . . . . . . . . 619
22.3 Noisy Super-Dense Coding . . . . . . . . . . . . . . . . . . . . . . . . . . . . 620
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22.4 State Transfer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22.5 Trade-off Coding . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22.6 Concluding Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22.7 History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . .

623
626
634
635

23 Private Classical Communication
637
23.1 The Information-Processing Task . . . . . . . . . . . . . . . . . . . . . . . . 638
23.2 The Private Classical Capacity Theorem . . . . . . . . . . . . . . . . . . . . 640
23.3 The Direct Coding Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 641
23.4 The Converse Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 649
23.5 Discussion of Private Classical Capacity . . . . . . . . . . . . . . . . . . . . 650
23.6 History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 653
24 Quantum Communication
655
24.1 The Information-Processing Task . . . . . . . . . . . . . . . . . . . . . . . . 656
24.2 No-Cloning and Quantum Communication . . . . . . . . . . . . . . . . . . . 658
24.3 The Quantum Capacity Theorem . . . . . . . . . . . . . . . . . . . . . . . . 659
24.4 The Direct Coding Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 659
24.5 Converse Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 666
24.6 An Interlude with Quantum Stabilizer Codes . . . . . . . . . . . . . . . . . . 669
24.7 Example Channels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 675
24.8 Discussion of Quantum Capacity . . . . . . . . . . . . . . . . . . . . . . . . 678
24.9 Entanglement Distillation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 684
24.10History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 687
25 Trading Resources for Communication
691
25.1 The Information-Processing Task . . . . . . . . . . . . . . . . . . . . . . . . 692
25.2 The Quantum Dynamic Capacity Theorem . . . . . . . . . . . . . . . . . . . 694
25.3 The Direct Coding Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 699
25.4 The Converse Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 700
25.5 Examples of Channels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 711
25.6 History and Further Reading . . . . . . . . . . . . . . . . . . . . . . . . . . . 727
26 Summary and Outlook
729
26.1 Unit Protocols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 730
26.2 Noiseless Quantum Shannon Theory . . . . . . . . . . . . . . . . . . . . . . 731
26.3 Noisy Quantum Shannon Theory . . . . . . . . . . . . . . . . . . . . . . . . 731
26.4 Protocols Not Covered In This Book . . . . . . . . . . . . . . . . . . . . . . 734
26.5 Network Quantum Shannon Theory . . . . . . . . . . . . . . . . . . . . . . . 735
26.6 Future Directions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 736
A Supplementary Results

737

B Unique Linear Extension of a Quantum Physical Evolution741

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For Students
Prerequisites for understanding the content in this book are a solid background in probability
theory and linear algebra. If you are new to information theory, then there should be enoughbackground in this book to get you up to speed (Chapters 2, 10, 13, and 14). However, classics
on information theory such as Cover and Thomas (2006) and MacKay (2003) could be helpful
as a reference. If you are new to quantum mechanics, then there should be enough material
in this book (Part II) to give you the background necessary for understanding quantum
Shannon theory. The book of Nielsen and Chuang (2000) (sometimes affectionately known as“Mike and Ike”) has become the standard starting point for students in quantum information
science and might be helpful as well. Some of the content in that book is available in thedissertation of Nielsen (1998). If you are familiar with Shannon’s information theory (at thelevel of Cover and Thomas (2006), for example), then the present book should be a helpful
entry point into the field of quantum Shannon theory. We build on intuition developedclassically to help in establishing schemes for communication over quantum channels. If you
are familiar with quantum mechanics, it might still be worthwhile to review Part II becausesome content there might not be part of a standard course on quantum mechanics.
The aim of this book is to develop “from the ground up” many of the major, exciting,pre- and post-millennium developments in the general area of study known as quantum
Shannon theory. As such, we spend a significant amount of time on quantum mechanics
for quantum information theory (Part II), we give a careful study of the important unit
protocols of teleportation, super-dense coding, and entanglement distribution (Part III),and we develop many of the tools necessary for understanding information transmission orcompression (Part IV). Parts V and VI are the culmination of this book, where all of thetools developed come into play for understanding many of the important results in quantum
Shannon theory.

For Instructors
This book could be useful for self-learning or as a reference, but one of the main goals
is for it to be employed as an instructional aid for the classroom. To aid instructors indesigning a course to suit their own needs, a draft, pre-publication copy of this book is
9

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available under a Creative Commons Attribution-NonCommercial-ShareAlike license. Thismeans that you can modify and redistribute this draft, pre-publication copy as you wish, aslong as you attribute the author, you do not use it for commercial purposes, and you sharea modification or derivative work under the same license (seehttp://creativecommons.org/licenses/by-nc-sa/3.0/
for a readable summary of the terms of the license). These requirements can be waived if
you obtain permission from the present author. By releasing the draft, pre-publication copyof the book under this license, I expect and encourage instructors to modify it for their own
needs. This will allow for the addition of new exercises, new developments in the theory, andthe latest open problems. It might also be a helpful starting point for a book on a related
topic, such as network quantum Shannon theory.
I used an earlier version of this book in a one-semester course on quantum Shannon
theory at McGill University during Winter semester 2011 (in many parts of the USA, thissemester is typically called “Spring semester”). We almost went through the entire book,but it might also be possible to spread the content over two semesters instead. Here is theorder in which we proceeded:
1. Introduction in Part I.
2. Quantum mechanics in Part II.
3. Unit protocols in Part III.
4. Chapter 9 on distance measures, Chapter 10 on classical information and entropy, andChapter 11 on quantum information and entropy.
5. The first part of Chapter 14 on classical typicality and Shannon compression.
6. The first part of Chapter 15 on quantum typicality.
7. Chapter 18 on Schumacher compression.
8. Back to Chapters 14 and 15 for the method of types.
9. Chapter 19 on entanglement concentration.
10. Chapter 20 on classical communication.
11. Chapter 21 on entanglement-assisted classical communication.
12. The final explosion of results in Chapter 22 (one of which is a route to proving theachievability part of the quantum capacity theorem).
c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

CONTENTS

11

The above order is just a particular order that suited the needs for the class at McGill,but other orders are of course possible. One could sacrifice the last part of Part III on theunit resource capacity region if there is no desire to cover the quantum dynamic capacity
theorem. One could also focus on going from classical communication to private classicalcommunication to quantum communication in order to develop some more intuition behindthe quantum capacity theorem. I later did this when teaching the course at LSU in Fall 2013.
But just recently in Fall 2015, I went back to the ordering above while including lectures
devoted to the CHSH game and the new results in Chapter 12.

Other Sources
There are many other sources to obtain a background in quantum Shannon theory. The
standard reference has become the book of Nielsen and Chuang (2000), but it does notfeature any of the post-millennium results in quantum Shannon theory. Other excellent
books that cover some aspects of quantum Shannon theory are (Hayashi, 2006; Holevo,
2002a, 2012; Watrous, 2015). Patrick Hayden has had a significant hand as a collaborative
guide for many PhD and Masters’ theses in quantum Shannon theory, during his time as a
postdoctoral fellow at the California Institute of Technology and as a professor at McGill
University. These include the theses of Yard (2005), Abeyesinghe (2006), Savov (2008, 2012),
Dupuis (2010), and Dutil (2011). All of these theses are excellent references. Hayden also
had a strong influence over the present author during the development of the first edition ofthis book.

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12

c 2019CONTENTSMark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

Preface to the Second Edition
It has now been some years since I completed the first draft of the first edition of this book.
In this time, I have learned much from many collaborators and I am grateful to them. During
the past few years, Mario Berta, Nilanjana Datta, Saikat Guha, and Andreas Winter havestrongly shaped my thinking about quantum information theory, and Mario and Nilanjanain particular have influenced my technical writing style, which is reflected in the new editionof the book. Also, the chance to work with them and others has led me to new research
directions in quantum information theory that I never would have imagined on my own.
I am also thankful to Todd Brun, Paul Cuff, Ludovico Lami, Ciara Morgan, and Giannicola Scarpa for using the book as the main text in their graduate courses on quantum
information theory and for feedback. One can try as much as possible to avoid typos in abook, but inevitably, they seem to show up in unexpected places. I am grateful to many
people for pointing out typos or errors and for suggesting how to fix them, including Todd
Brun, Giulio Chiribella, Paul Cuff, Dawei (David) Ding, Will Matthews, Milan Mosonyi,
David Reeb, and Marco Tomamichel. I also thank Corsin Pfister for helpful discussions
about unique linear extensions of quantum physical evolutions. I am grateful to David
Tranah and the editorial staff at Cambridge University Press for their help with publishing
the second edition.
So what’s new in the second edition? Suffice it to say that every page of the book has
been rewritten and there are over 100 pages of new material! I formulated many thoughts
about revising during Fall 2013 while teaching a graduate course on quantum information
at LSU, and I then formulated many more thoughts and made the actual changes during
Fall 2015 (when teaching it again). In that regard, I am thankful to both the Departmentof Physics and Astronomy and the Center for Computation and Technology at LSU for providing a great environment and support. I also thank the graduate students at LSU whogave feedback during and after lectures. There are many little changes throughout thatwill probably go unnoticed. For example, I have come to prefer writing a quantum stateshared between Alice and Bob as ρAB rather than ρAB (i.e., with system labels as subscripts
rather than superscripts). Admittedly, several collaborators influenced me here, but there
are a few good reasons for this convention: the phrase “state of a quantum system” suggests that the state ρ should be “resting on” the systems AB, the often used partial trace
TrA {ρAB } looks better than TrA {ρAB }, and the notation ρAB is more consistent with thestandard notation pX for a probability distribution corresponding to a random variable X.
OK, that’s perhaps minor. Major changes include the addition of many new exercises, a13

14

CONTENTS

detailed discussion of Bell’s theorem, the CHSH game, and Tsirelson’s theorem, the axiomatic approach to quantum channels, a proof of the Choi–Kraus theorem, a definition ofunital and adjoint maps, a discussion of states, channels, and measurements all as quantum
channels, the equivalence of purifications, the adjoint map in terms of isometric extension,
the definition of the diamond norm and its interpretation, how a measurement achieves the
fidelity, how the Hilbert–Schmidt distance is not monotone with respect to channels, moredetailed definitions of classical and quantum relative entropies, new continuity bounds forclassical and quantum entropies, refinements of classical entropy inequalities, streamlined
proofs of data processing inequalities using relative entropy, the equivalence of quantum entropy inequalities like strong subadditivity and monotonicity of relative entropy, Chapter 12
on recoverability, modified proofs of additivity of channel information quantities, sequential
decoding for classical communication, simpler proofs of the Schumacher compression theorem, a complete rewrite of Chapter 19, alternate proofs for the achievability part of the HSW
theorem, a proof for the classical capacity of the erasure channel, simpler converse proofs forthe entanglement-assisted capacity theorem, a revised proof of the trade-off coding resource
inequality, a revised proof of the hashing bound, a simplified converse proof of the quantum
dynamic capacity theorem, a completely revised discussion of the importance of the quantum dynamic capacity formula, and the addition of many new references that have been
influential in recent years. Minor changes include improved presentations of many theorems
and definitions throughout.
I am most grateful to my family for all of their support and encouragement throughout
my life, including my mother, father, sister, and brother and all of my surrounding family
members. I am still indebted to my wife Christabelle and her family for warmth and love.
Christabelle has been an unending source of support and love for me. I dedicate this second
edition to my nephews David and Matthew.
Mark M. Wilde
Baton Rouge, Louisiana, USA
December 2015

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Preface to the First Edition
I began working on this book in the summer of 2008 in Los Angeles, with much time to sparein the final months of dissertation writing. I had a strong determination to review quantum
Shannon theory, a beautiful area of quantum information science that Igor Devetak had
taught me three years earlier at USC in fall 2005. I was carefully studying a manuscript
entitled “Principles of Quantum Information Theory,” a text that Igor had initiated incollaboration with Patrick Hayden and Andreas Winter. I read this manuscript many times,
and many parts of it I understood well, though other parts I did not.
After a few weeks of reading and rereading, I decided “if I can write it out myself fromscratch, perhaps I would then understand it!”, and thus began the writing of the chapters on
the packing lemma, the covering lemma, and quantum typicality. I knew that Igor’s (nowformer) students Min-Hsiu Hsieh and Zhicheng Luo knew the topic well because they had
already written several quality research papers with him, so I requested if they could meetwith me weekly for an hour to review the fundamentals. They kindly agreed and helped mequite a bit in understanding the packing and covering techniques.
Not much later, after graduating, I began collaborating with Min-Hsiu on a research
project that Igor had suggested to the both of us: “find the triple trade-off capacity formulasof a quantum channel.” This was perhaps the best starting point for me to learn quantum
Shannon theory because proving this theorem required an understanding of most everythingthat had already been accomplished in the area. After a month of effort, I continued towork with Min-Hsiu on this project while joining Andreas Winter’s Singapore group for a
two-month visit. As I learned more, I added more to the notes, and they continued to grow.
After landing a job in the DC area for January 2009, I realized that I had almost enoughmaterial for teaching a course, and so I contacted local universities in the area to see if theywould be interested. Can Korman, formerly chair of the Electrical Engineering Department
at George Washington University, was excited about the possibility. His enthusiasm wasenough to keep me going on the notes, and so I continued to refine and add to them in myspare time in preparing for teaching. Unfortunately (or perhaps fortunately?), the course
ended up being canceled. This was disheartening to me, but in the mean time, I had
contacted Patrick Hayden to see if he would be interested in having me join his group at
McGill University for postdoctoral studies. Patrick Hayden and David Avis then offered mea postdoctoral fellowship, and I moved to Montréal in October 2009.
After joining, I learned a lot by collaborating and discussing with Patrick and his groupmembers. Patrick offered me the opportunity to teach his graduate class on quantum Shan15

16

CONTENTS

non theory while he was away on sabbatical, and this encouraged me further to persist with
the notes.
I am grateful to everyone mentioned above for encouraging and supporting me during
this project, and I am also grateful to everyone who provided feedback during the course ofwriting up. In this regard, I am especially grateful to Dave Touchette for detailed feedback on
all of the chapters in the book. Dave’s careful reading and spotting of errors has immensely
improved the quality of the book. I am grateful to my father, Gregory E. Wilde, Sr., for
feedback on earlier chapters and for advice and love throughout. I thank Ivan Savov forencouraging me, for feedback, and for believing that this is an important scholarly work.
I also thank Constance Caramanolis, Raza-Ali Kazmi, John M. Schanck, Bilal Shaw, andAnna Vershynina for valuable feedback. I am grateful to Min-Hsiu Hsieh for the many
research topics we have worked on together that have enhanced my knowledge of quantum
Shannon theory. I thank Michael Nielsen and Victor Shoup for advice on Creative Commonslicensing and Kurt Jacobs for advice on book publishing. I am grateful to Sarah Payne andDavid Tranah of Cambridge University Press for their extensive feedback on the manuscript
and their outstanding support throughout the publication process. I acknowledge funding
from the MDEIE (Quebec) PSR-SIIRI international collaboration grant.
I am indebted to my mentors who took me on as a student during my career. Todd
Brun was a wonderful PhD supervisor—helpful, friendly, and encouraging of creativity andoriginal pursuit. Igor Devetak taught me quantum Shannon theory in fall 2005 and helped
me once per week during his office hours. He also invited me to join Todd’s and his group,and more recently, Igor provided much encouragement and “big-picture” feedback during thewriting of this book. Bart Kosko shaped me as a scholar during my early years at USC andprovided helpful advice regarding the book project. Patrick Hayden has been an immense
bedrock of support at McGill. His knowledge of quantum information and many other areas
is unsurpassed, and he has been kind, inviting, and helpful during my time at McGill. I
am also grateful to Patrick for giving me the opportunity to teach at McGill and for advice
throughout the development of this book.
I thank my mother, father, sister, and brother and all of my surrounding family members
for being a source of love and support. Finally, I am indebted to my wife Christabelle and herfamily for warmth and love. I dedicate this book to the memory of my grandparents Joseph
and Rose McMahon, and Norbert Jay and Mary Wilde. Lux aeterna luceat eis, Domine.
Mark M. Wilde
Montreal, Quebec, Canada
June 2011

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicensePart I
Introduction

17

CHAPTER 1

Concepts in Quantum Shannon
TheoryIn these first few chapters, our aim is to establish a firm grounding so that we can addresssome fundamental questions regarding information transmission over quantum channels.
This area of study has become known as “quantum Shannon theory” in the broader quantum information community, in order to distinguish this topic from other areas of study inquantum information science. In this text, we will use the terms “quantum Shannon theory” and “quantum information theory” somewhat interchangeably. We will begin by briefly
overviewing several fundamental aspects of the quantum theory. Our study of the quantum
theory, in this chapter and future ones, will be at an abstract level, without giving preference
to any particular physical system such as a spin- 12 particle or a photon. This approach will
be more beneficial for the purposes of our study, but, here and there, we will make somereference to actual physical systems to ground us in reality.
You may be wondering, what is quantum Shannon theory and why do we name thisarea of study as such? In short, quantum Shannon theory is the study of the ultimate
capability of noisy physical systems, governed by the laws of quantum mechanics, to preserve
information and correlations. Quantum information theorists have chosen the name quantum
Shannon theory to honor Claude Shannon, who single-handedly founded the field of classicalinformation theory, with a groundbreaking paper (Shannon, 1948). In particular, the namerefers to the asymptotic theory of quantum information, which is the main topic of study
in this book. Information theorists since Shannon have dubbed him the “Einstein of theinformation age.”1 The name quantum Shannon theory is fit to capture this area of study
because we often use quantum versions of Shannon’s ideas to prove some of the main theorems
in quantum Shannon theory.
We prefer the name “quantum Shannon theory” over such names as “quantum information science” or just “quantum information.” These other names are too broad, encompassing
subjects as diverse as quantum computation, quantum algorithms, quantum complexity the1

It is worthwhile to look up “Claude Shannon—Father of the Information Age” on YouTube and watch
several reknowned information theorists speak with awe about “the founding father” of information theory.

19

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CHAPTER 1. CONCEPTS IN QUANTUM SHANNON THEORY

ory, quantum communication complexity, entanglement theory, quantum key distribution,
quantum error correction, and even the experimental implementation of quantum protocols.
Quantum Shannon theory does overlap with some of the aforementioned subjects, such as
quantum computation, entanglement theory, quantum key distribution, and quantum error
correction, but the name “quantum Shannon theory” should evoke a certain paradigm forquantum communication with which the reader will become intimately familiar after some
exposure to the topics in this book. For example, it is necessary for us to discuss quantum
gates (a topic in quantum computing) because quantum Shannon-theoretic protocols exploit
them to achieve certain information-processing tasks. Also, in Chapter 23, we are interested in the ultimate limitation on the ability of a noisy quantum communication channel
to transmit private information (information that is secret from any third party besides theintended receiver). This topic connects quantum Shannon theory with quantum key distribution because the private information capacity of a noisy quantum channel is strongly
related to the task of using the quantum channel to distribute a secret key. As a final connection, one of the most important theorems of quantum Shannon theory is the quantum
capacity theorem. This theorem determines the ultimate rate at which a sender can reliablytransmit quantum information over a quantum channel to a receiver. The result provided by
the quantum capacity theorem is closely related to the theory of quantum error correction,
but the mathematical techniques used in quantum Shannon theory and in quantum error
correction are so different that these subjects merit different courses of study.
Quantum Shannon theory intersects two of the great sciences of the twentieth century: thequantum theory and information theory. It was really only a matter of time before physicists,
mathematicians, computer scientists, and engineers began to consider the convergence of thetwo subjects because the quantum theory was essentially established by 1926 and information
theory by 1948. This convergence has sparked what we may call the “quantum information
revolution” or what some refer to as the “second quantum revolution” (Dowling and Milburn,2003) (with the first revolution being the discovery of the quantum theory).
The fundamental components of the quantum theory are a set of postulates that govern
phenomena on the scale of atoms. Uncertainty is at the heart of the quantum theory—“quantum uncertainty” or “Heisenberg uncertainty” is not due to our lack or loss of information or due to imprecise measurement capability, but rather, it is a fundamental uncertainty
inherent in nature itself. The discovery of the quantum theory came about as a total shockto the physics community, shaking the foundations of scientific knowledge. Perhaps it isfor this reason that every introductory quantum mechanics course delves into its history indetail and celebrates the founding fathers of the quantum theory. In this book, we do notdiscuss the history of the quantum theory in much detail and instead refer to several great
introductory books for these details (Bohm, 1989; Sakurai, 1994; Griffiths, 1995; Feynman,
1998). Physicists such as Planck, Einstein, Bohr, de Broglie, Born, Heisenberg, Schrödinger,
Pauli, Dirac, and von Neumann contributed to the foundations of the quantum theory in the
1920s and 1930s. We introduce the quantum theory by briefly commenting on its history
and major underlying concepts.
Information theory is the second great foundational science for quantum Shannon theory.
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21

In some sense, it could be viewed as merely an application of probability theory. Its aim isto quantify the ultimate compressibility of information and the ultimate ability for a sender
to transmit information reliably to a receiver. It relies upon probability theory because“classical” uncertainty, arising from our lack of total information about any given scenario,
is ubiquitous throughout all information-processing tasks. The uncertainty in classical information theory is the kind that is present in the flipping of a coin or the shuffle of a deckof cards, the uncertainty due to imprecise knowledge. “Quantum” uncertainty is inherent
in nature itself and is perhaps not as intuitive as the uncertainty that classical information
theory measures. We later expand further on these differing kinds of uncertainty, and Chapter 4 shows how a theory of quantum information captures both kinds of uncertainty within
one formalism.2
The history of classical information theory began with Claude Shannon. Shannon’scontribution is heralded as one of the single greatest contributions to modern science because
he established the field in his seminal paper (Shannon, 1948). In this paper, he coined theessential terminology, and he stated and justified the main mathematical definitions andthe two fundamental theorems of information theory. Many successors have contributed to
information theory, but most, if not all, of the follow-up contributions employ Shannon’sline of thinking in some form. In quantum Shannon theory, we will notice that many of
Shannon’s original ideas are present, though they take a particular “quantum” form.
One of the major assumptions in both classical information theory and quantum Shannon
theory is that local computation is free but communication is expensive. In particular, forthe classical case, we assume that each party has unbounded computation available. For thequantum case, we assume that each party has a fault-tolerant quantum computer available
at his or her local station and the power of each quantum computer is unbounded. We also
assume that both communication and a shared resource are expensive, and for this reason,
we keep track of these resources in a resource count. Sometimes, however, we might say thatclassical communication is free in order to simplify a scenario. A simplification like this one
can lead to greater insights that might not be possible without making such an assumption.
We should first study and understand the postulates of the quantum theory in order tostudy quantum Shannon theory properly. Your heart may sink when you learn that theNobel Prize-winning physicist Richard Feynman is famously quoted as saying, “I think I
can safely say that nobody understands quantum mechanics.” We should take the liberty ofclarifying Feynman’s statement. Of course, Feynman does not intend to suggest that no one
knows how to work with the quantum theory. Many well-abled physicists are employed tospend their days exploiting the laws of the quantum theory to do fantastic things, such as
the trapping of ions in a vacuum or applying the quantum tunneling effect in a transistor toprocess a single electron. I am hoping that you will give me the license to interpret Feynman’sstatement. I think he means that it is very difficult for us to understand the quantum theory
intuitively because we do not experience the phenomena that it predicts. If we were the sizeof atoms and we experienced the laws of quantum theory on a daily basis, then perhaps the2

Von Neumann established the density operator formalism in his 1932 book on the quantum theory. This
mathematical framework captures both kinds of uncertainty (von Neumann, 1996).
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CHAPTER 1. CONCEPTS IN QUANTUM SHANNON THEORY

quantum theory would be as intuitive to us as Newton’s law of universal gravitation.3 Thus,in this sense, I would agree with Feynman—nobody can really understand the quantum
theory because it is not part of our everyday experiences. Nevertheless, our aim in this book
is to work with the laws of quantum theory so that we may begin to gather insights about
what the theory predicts. Only by exposure to and practice with its postulates can we reallygain an intuition for its predictions. It is best to imagine that the world in our everyday
life does incorporate the postulates of quantum mechanics, because, indeed, as many, many
experiments have confirmed, it does!
We delve into the history of the convergence of the quantum theory and information
theory in some detail in this introductory chapter because this convergence does have aninteresting history and is relevant to the topic of this book. The purpose of this historicalreview is not only to become familiar with the field itself but also to glimpse into the mindsof the founders of the field so that we may see the types of questions that are important tothink about when tackling new, unsolved problems.4 Many of the most important results
come about from asking simple, yet profound, questions and exploring the possibilities.
We first briefly review the history and the fundamental concepts of the quantum theory
before delving into the convergence of the quantum theory and information theory. We buildon these discussions by introducing some of the initial fundamental contributions to quantum
Shannon theory. The final part of this chapter ends by posing some of the questions to whichquantum Shannon theory provides answers.

1.1 Overview of the Quantum Theory
1.1.1

Brief History of the Quantum Theory

A physicist living around 1890 would have been well pleased with the progress of physics,
but perhaps frustrated at the seeming lack of open research problems. It seemed as though
the Newtonian laws of mechanics, Maxwell’s theory of electromagnetism, and Boltzmann’s
theory of statistical mechanics explained most natural phenomena. In fact, Max Planck, oneof the founding fathers of the quantum theory, was searching for an area of study in 1874
and his advisor gave him the following guidance:“In this field [of physics], almost everything is already discovered, and all thatremains is to fill a few holes.”3

Of course, Newton’s law of universal gravitation was a revolutionary breakthrough because the phenomenon of gravity is not entirely intuitive when a student first learns it. However, we do experience thegravitational law in our daily lives, and I would argue that this phenomenon is much more intuitive than,say, the phenomenon of quantum entanglement.
4
Another way to discover good questions is to attend parties that well-established professors hold. The
story goes that Oxford physicist David Deutsch attended a 1981 party at the Austin, Texas house of reknowned physicist John Archibald Wheeler, in which many attendees discussed the foundations of computing(Mullins, 2001). Deutsch claims that he could immediately see that the quantum theory would give an improvement for computation. A few years later, in 1985, he published an algorithm that was the first instanceof a quantum speed-up over the fastest classical algorithm (Deutsch, 1985).
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1.1. OVERVIEW OF THE QUANTUM THEORY23

Two Clouds
Fortunately, Planck did not heed this advice and instead began his physics studies. Not
everyone agreed with Planck’s former advisor. Lord Kelvin stated in his famous April 1900
lecture that “two clouds” surrounded the “beauty and clearness of theory” (Kelvin, 1901).
The first cloud was the failure of Michelson and Morley to detect a change in the speed oflight as predicted by an “ether theory,” and the second cloud was the ultraviolet catastrophe,
the classical prediction that a blackbody emits radiation with an infinite intensity at highultraviolet frequencies. Also in 1900, Planck started the quantum revolution that began
to clear the second cloud. He assumed that light comes in discrete bundles of energy andused this idea to produce a formula that correctly predicts the spectrum of blackbody radiation (Planck, 1901). A great cartoon lampoon of the ultraviolet catastrophe shows Planckcalmly sitting fireside with a classical physicist whose face is burning to bits because of theintense ultraviolet radiation that his classical theory predicts the fire is emitting (McEvoy
and Zarate, 2004). A few years later, Einstein (1905) contributed a paper that helped tofurther clear the second cloud (he also cleared the first cloud with his other 1905 paper on
special relativity). He assumed that Planck was right and showed that the postulate thatlight arrives in “quanta” (now known as the photon theory) provides a simple explanation
for the photoelectric effect, the phenomenon in which electromagnetic radiation beyond acertain threshold frequency impinging on a metallic surface induces a current in that metal.
These two explanations of Planck and Einstein fueled a theoretical revolution in physics
that some now call the first quantum revolution (Dowling and Milburn, 2003). Some years
later, de Broglie (1924) postulated that every element of matter, whether an atom, electron,or photon, has both particle-like behavior and wave-like behavior. Just two years later,
Schrödinger (1926) used the de Broglie idea to formulate a wave equation, now known as
Schrödinger’s equation, that governs the evolution of a closed quantum-mechanical system.
His formalism later became known as wave mechanics and was popular among physicists
because it appealed to notions with which they were already familiar. Meanwhile, Heisenberg(1925) formulated an “alternate” quantum theory called matrix mechanics. His theory usedmatrices and linear algebra, mathematics with which many physicists at the time were notreadily familiar. For this reason, Schrödinger’s wave mechanics was more popular than
Heisenberg’s matrix mechanics. In 1930, Paul Dirac published a textbook (now in its fourth
edition and reprinted 16 times) that unified the formalisms of Schrödinger and Heisenberg,
showing that they were actually equivalent (Dirac, 1982). In a later edition, he introduced
the now ubiquitous “Dirac notation” for quantum theory that we will employ in this book.
After the publication of Dirac’s textbook, the quantum theory then stood on firm mathematical grounding and the basic theory had been established. We thus end our historical
overview at this point and move on to the fundamental concepts of the quantum theory.

1.1.2

Fundamental Concepts of the Quantum Theory

Quantum theory, as applied in quantum information theory, really has only a few importantconcepts. We review each of these aspects of quantum theory briefly in this section. Somec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

24

CHAPTER 1. CONCEPTS IN QUANTUM SHANNON THEORYof these phenomena are uniquely “quantum” but others do occur in the classical theory. In
short, these concepts are as follows:5
1. indeterminism,
2. interference,
3. uncertainty,
4. superposition,
5. entanglement.
The quantum theory is indeterministic because the theory makes predictions about probabilities of events only. This aspect of quantum theory is in contrast with a deterministicclassical theory such as that predicted by the Newtonian laws. In the Newtonian system, it
is possible to predict, with certainty, the trajectories of all objects involved in an interaction
if one knows only the initial positions and velocities of all the objects. This deterministic
view of reality even led some to believe in determinism from a philosophical point of view.
For instance, the mathematician Pierre-Simon Laplace once stated that a supreme intellect,
colloquially known as “Laplace’s demon,” could predict all future events from present andpast events:“We may regard the present state of the universe as the effect of its past andthe cause of its future. An intellect which at a certain moment would know allforces that set nature in motion, and all positions of all items of which nature iscomposed, if this intellect were also vast enough to submit these data to analysis,
it would embrace in a single formula the movements of the greatest bodies of theuniverse and those of the tiniest atom; for such an intellect nothing would be
uncertain and the future just like the past would be present before its eyes.”
The application of Laplace’s statement to atoms is fundamentally incorrect, but we canforgive him because the quantum theory had not yet been established in his time. Many
have extrapolated from Laplace’s statement to argue the invalidity of human free will. We
leave such debates to philosophers.6
In reality, we never can possess full information about the positions and velocities ofevery object in any given physical system. Incorporating probability theory then allows
us to make predictions about the probabilities of events and, with some modifications, theclassical theory becomes an indeterministic theory. Thus, indeterminism is not a uniqueaspect of the quantum theory but merely a feature of it. But this feature is so crucial to thequantum theory that we list it among the fundamental concepts.
5

I have used Todd A. Brun’s list from his lecture notes (Brun).
John Archibald Wheeler may disagree with this approach. He once said, “Philosophy is too important
to be left to the philosophers” (Misner et al., 2009).
6

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1.1. OVERVIEW OF THE QUANTUM THEORY

25

Interference is another feature of the quantum theory. It is also present in any classicalwave theory—constructive interference occurs when the crest of one wave meets the crest ofanother, producing a stronger wave, while destructive interference occurs when the crest of
one wave meets the trough of another, canceling out each other. In any classical wave theory,a wave occurs as a result of many particles in a particular medium coherently displacing one
another, as in an ocean surface wave or a sound pressure wave, or as a result of coherent
oscillating electric and magnetic fields, as in an electromagnetic wave. The strange aspectof interference in the quantum theory is that even a single “particle” such as an electron
can exhibit wave-like features, as in the famous double slit experiment (see, e.g., Greene(1999) for a history of these experiments). This quantum interference is what contributes
wave–particle duality to every fundamental component of matter.
Uncertainty is at the heart of the quantum theory. Uncertainty in the quantum theory
is fundamentally different from uncertainty in the classical theory (discussed in the former
paragraph about an indeterministic classical theory). The archetypal example of uncertainty
in the quantum theory occurs for a single particle. This particle has two complementary variables: its position and its momentum. The uncertainty principle states that it is impossible
to know both the particle’s position and momentum to arbitrary accuracy. This principle
even calls into question the meaning of the word “know” in the previous sentence in thecontext of quantum theory. We might say that we can only know that which we measure,and thus, we can only know the position of a particle after performing a precise measurementthat determines it. If we follow with a precise measurement of its momentum, we lose all
information about the position of the particle after learning its momentum. In quantum information science, the BB84 protocol for quantum key distribution exploits the uncertainty
principle and statistical analysis to determine the presence of an eavesdropper on a quantum
communication channel by encoding information into two complementary variables (Bennett
and Brassard, 1984).
The superposition principle states that a quantum particle can be in a linear combination
state, or superposed state, of any two other allowable states. This principle is a result of thelinearity of quantum theory. Schrodinger’s wave equation is a linear differential equation,
meaning that the linear combination αψ +βφ is a solution of the equation if ψ and φ are both
solutions of the equation. We say that the solution αψ +βφ is a coherent superposition of thetwo solutions. The superposition principle has dramatic consequences for the interpretationof the quantum theory—it gives rise to the notion that a particle can somehow “be in one
location and another” at the same time. There are different interpretations of the meaningof the superposition principle, but we do not highlight them here. We merely choose to
use the technical language that the particle is in a superposition of both locations. The
loss of a superposition can occur through the interaction of a particle with its environment.
Maintaining an arbitrary superposition of quantum states is one of the central goals of aquantum communication protocol.
The last, and perhaps most striking, quantum feature that we highlight here is entanglement. There is no true classical analog of entanglement. The closest analog of entanglement
might be a secret key that two parties possess, but even this analogy does not come close.
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Entanglement refers to the strong quantum correlations that two or more quantum particles can possess. The correlations in quantum entanglement are stronger than any classicalcorrelations in a precise, technical sense. Schrödinger (1935) first coined the term “entanglement” after observing some of its strange properties and consequences. Einstein, Podolsky,
and Rosen then presented an apparent paradox involving entanglement that raised concerns
over the completeness of the quantum theory (Einstein et al., 1935). That is, they suggested
that the seemingly strange properties of entanglement called the uncertainty principle into
question (and thus the completeness of the quantum theory) and furthermore suggested
that there might be some “local hidden-variable” theory that could explain the results ofexperiments. It took about 30 years to resolve this paradox, but John Bell did so by presenting a simple inequality, now known as a Bell inequality (Bell, 1964). He showed that any
two-particle classical correlations that satisfy the assumptions of the “local hidden-variable
theory” of Einstein, Podolsky, and Rosen must be less than a certain amount. He thenshowed how the correlations of two entangled quantum particles can violate this inequality,
and thus, entanglement has no explanation in terms of classical correlations but is instead a
uniquely quantum phenomenon. Experimentalists later verified that two entangled quantum
particles can violate Bell’s inequality (Aspect et al., 1981).
In quantum information science, the non-classical correlations in entanglement play afundamental role in many protocols. For example, entanglement is the enabling resource inteleportation, a protocol that disembodies a quantum state in one location and reproduces
it in another. We will see many other examples of entanglement throughout this book.
Entanglement theory concerns methods for quantifying the amount of entanglement
present not only in a two-particle state but also in a multiparticle state. A large bodyof literature exists that investigates entanglement theory (Horodecki et al., 2009), but weonly address aspects of it that are relevant in our study of quantum Shannon theory.
The above five features capture the essence of the quantum theory, but we will see more
aspects of it as we progress through our overview in Chapters 3, 4, and 5.

1.2 The Emergence of Quantum Shannon Theory
In the previous section, we discussed several unique quantum phenomena such as superposition and entanglement, but it is not clear what kind of information these unique quantum
phenomena represent. Is it possible to find a convergence of the quantum theory and Shannon’s information theory, and if so, what is the convergence?

1.2.1

The Shannon Information Bit

A fundamental contribution of Shannon is the notion of a bit as a measure of information.
Typically, when we think of a bit, we think of a two-valued quantity that can be in the state“off” or the state “on.” We represent this bit with a binary number that can be “0” or “1.”We also associate a physical representation with a bit—this physical representation can bewhether a light switch is off or on, whether a transistor allows current to flow or not, whether
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a large number of magnetic spins point in one direction or another, the list going on and on.
These are all physical notions of a bit.
Shannon’s notion of a bit is quite different from these physical notions, and we motivate
his notion with the example of a fair coin. Without flipping the coin, we have no idea what
the result of a coin flip will be—our best guess at the result is to guess randomly. If someone
else learns the result of a random coin flip, we can ask this person the question: What was
the result? We then learn one bit of information.
Though it may seem obvious, it is important to stress that we do not learn any (or not
as much) information if we do not ask the right question. This point becomes even more
important in the quantum case. Suppose that the coin is not fair—without loss of generality,
suppose the probability of “heads” is greater than the probability of “tails.” In this case, wewould not be as surprised to learn that the result of a coin flip is “heads.” We may say inthis case that we would learn less than one bit of information if we were to ask someone theresult of the coin flip.
The Shannon binary entropy is a measure of information. Given a probability distribution(p, 1 − p) for a binary random variable, its Shannon binary entropy ish2 (p) ≡ −p log p − (1 − p) log(1 − p),(1.1)

where here and throughout the book (unless stated explicitly otherwise), the logarithm istaken base two. The Shannon binary entropy measures information in units of bits. We will
discuss it in more detail in the next chapter and in Chapter 10.
The Shannon bit, or Shannon binary entropy, is a measure of the surprise upon learning
the outcome of a random binary experiment. Thus, the Shannon bit has a completely
different interpretation from that of the physical bit. The outcome of the coin flip resides ina physical bit, but it is the information associated with the random nature of the physical
bit that we would like to measure. It is this notion of a bit that is important in information
theory.

1.2.2

A Measure of Quantum Information

The above section discusses Shannon’s notion of a bit as a measure of information. A natural
question is whether there is an analogous measure of quantum information, but before we
can even ask that question, we might first wonder: What is quantum information? As in theclassical case, there is a physical notion of quantum information. A quantum state always
resides “in” a physical system. Perhaps another way of stating this idea is that every physicalsystem is in some quantum state. The physical notion of a quantum bit, or qubit for short(pronounced “cue · bit”), is a two-level quantum system. Examples of two-level quantum
systems are the spin of the electron, the polarization of a photon, or an atom with a groundstate and an excited state. The physical notion of a qubit is straightforward to understandonce we have a grasp of the quantum theory.
A more pressing question for us in this book is to understand an informational notionof a qubit, as in the Shannon sense. In the classical case, we quantify information by thec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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amount of knowledge we gain after learning the answer to a probabilistic question. In thequantum world, what knowledge can we have of a quantum state?
Sometimes we may know the exact quantum state of a physical system because we prepared the quantum system in a certain way. For example, we may prepare an electron inits “spin-up in the z direction” state, where |↑z i denotes this state. If we prepare the statein this way, we know for certain that the state is indeed |↑z i and no other state. Thus, we
do not gain any information, or equivalently, there is no removal of uncertainty if someone
else tells us that the state is |↑z i. We may say that this state has zero qubits of quantum
information, where the term “qubit” now refers to a measure of the quantum information ofa state.
In the quantum world, we also have the option of measuring this state in the x direction.
The postulates of quantum theory, given in Chapter 3, predict that the state will then be|↑x i or |↓x i with equal probability after measuring in the x direction. One interpretationof this aspect of quantum theory is that the system does not have any definite state in
the x direction: in fact there is maximal uncertainty about its x direction, if we knowthat the physical system has a definite z direction. This behavior is one manifestation of
the Heisenberg uncertainty principle. So before performing the measurement, we have noknowledge of the resulting state and we gain one Shannon bit of information after learning
the result of the measurement. If we use Shannon’s notion of entropy and perform an x
measurement, this classical measure loses some of its capability here to capture our knowledgeof the state of the system. It is inadequate to capture our knowledge of the state because
we actually prepared it ourselves and know with certainty that it is in the state |↑z i. With
these different notions of information gain, which one is the most appropriate for the quantumcase?
It turns out that the first way of thinking is the one that is most useful for quantifying
quantum information. If someone tells us the definite quantum state of a particular physicalsystem and this state is indeed the true state, then we have complete knowledge of the state
and thus do not learn more “qubits” of quantum information from this point onward. Thisline of thinking is perhaps similar in one sense to the classical world, but different from theclassical world, in the sense of the case presented in the previous paragraph.
Now suppose that a friend (let us call him “Bob”) randomly prepares quantum states as aprobabilistic ensemble. Suppose Bob prepares |↑z i or |↓z i with equal probability. With only
this probabilistic knowledge, we acquire one bit of information if Bob reveals which state
he prepared. We could also perform a quantum measurement on the system to determine
what state Bob prepared (we discuss quantum measurements in detail in Chapter 3). One
reasonable measurement to perform is a measurement in the z direction. The result of
the measurement determines which state Bob actually prepared because both states in theensembles are states with definite z direction. The result of this measurement thus gives
us one bit of information—the same amount that we would learn if Bob informed us whichstate he prepared. It seems that most of this logic is similar to the classical case—i.e., theresult of the measurement only gave us one Shannon bit of information.
Another measurement to perform is a measurement in the x direction. If the actual
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state prepared is |↑z i, then the quantum theory predicts that the state becomes |↑x i or |↓x i
with equal probability. Similarly, if the actual state prepared is |↓z i, then the quantum
theory predicts that the state again becomes |↑x i or |↓x i with equal probability. Calculating
probabilities, the resulting state is |↑x i with probability 1/2 and |↓x i with probability 1/2.
So the Shannon bit content of learning the result is again one bit, but we arrived at thisconclusion in a much different fashion from the scenario in which we measured in the z
direction. How can we quantify the quantum information of this ensemble? We claim for
now that this ensemble contains one qubit of quantum information and this result derives
from either the measurement in the z direction or the measurement in the x direction forthis particular ensemble.
Let us consider one final example that perhaps gives more insight into how we might
quantify quantum information. Suppose Bob prepares |↑z i or |↑x i with equal probability.
The first state is spin-up in the z direction and the second is spin-up in the x direction. If Bob
reveals which state he prepared, then we learn one Shannon bit of information. But suppose
now that we would like to learn the prepared state on our own, without the help of our friend
Bob. One possibility is to perform a measurement in the z direction. If the state prepared is|↑z i, then we learn this result with probability 1/2. But if the state prepared is |↑x i, then thequantum theory predicts that the state becomes |↑z i or |↓z i with equal probability (while welearn what the new state is). Thus, quantum theory predicts that the act of measuring thisensemble inevitably disturbs the state some of the time. Also, there is no way that we canlearn with certainty whether the prepared state is |↑z i or |↑x i. Using a measurement in thez direction, the resulting state is |↑z i with probability 3/4 and |↓z i with probability 1/4. We
learn less than one Shannon bit of information from this ensemble because the probability
distribution becomes skewed when we perform this particular measurement.
The probabilities resulting from the measurement in the z direction are the same thatwould result from an ensemble where Bob prepares |↑z i with probability 3/4 and |↓z i withprobability 1/4 and we perform a measurement in the z direction. The actual Shannon
entropy of the distribution (3/4, 1/4) is about 0.81 bits, confirming our intuition that welearn approximately less than one bit. A similar, symmetric analysis holds to show that we
gain 0.81 bits of information when we perform a measurement in the x direction.
We have more knowledge of the system in question if we gain less information fromperforming measurements on it. In the quantum theory, we learn less about a system if we
perform a measurement on it that does not disturb it too much. Is there a measurement that
we can perform in which we learn the least amount of information? Recall that learning theleast amount of information is ideal because it has the interpretation that we require fewer
questions on average to learn the result of a random experiment. Indeed, it turns out that ameasurement in the x + z direction reveals the least amount of information. Avoiding details
for now, this measurement returns a state that we label |↑x+z i with probability cos2 (π/8)
and a state |↓x+z i with probability sin2 (π/8). This measurement has the desirable effect thatit causes the least amount of disturbance to the original states in the ensemble. The entropyof the distribution resulting from the measurement is about 0.6 bits and is less than the one
bit that we learn if Bob reveals the state. The entropy ≈ 0.6 is also the least amount ofc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 1. CONCEPTS IN QUANTUM SHANNON THEORY

information among all possible sharp measurements that we may perform on the ensemble.
We claim that this ensemble contains ≈ 0.6 qubits of quantum information.
We can determine the ultimate compressibility of classical data with Shannon’s source
coding theorem (we overview this technique in the next chapter). Is there a similar way that
we can determine the ultimate compressibility of quantum information? This question was
one of the early and profitable ones for quantum Shannon theory and the answer is affirmative. The technique for quantum compression is called Schumacher compression, named
after Benjamin Schumacher. Schumacher used ideas similar to that of Shannon—he createdthe notion of a quantum information source that emits random physical qubits, and he invoked the law of large numbers to show that there is a so-called typical subspace where mostof the quantum information really resides. This line of thought is similar to that which wewill discuss in the overview of data compression in the next chapter. The size of the typicalsubspace for most quantum information sources is exponentially smaller than the size of thespace in which the emitted physical qubits resides. Thus, one can “quantum compress” thequantum information to this subspace without losing much. Schumacher’s quantum source
coding theorem then quantifies, in an operational sense, the amount of actual quantum information that the ensemble contains. The amount of actual quantum information corresponds
to the number of qubits, in the informational sense, that the ensemble contains. It is thismeasure that is equivalent to the “optimal measurement” one that we suggested in the previous paragraph. We will study this idea in more detail later when we introduce the quantum
theory and a rigorous notion of a quantum information source.
Some of the techniques of quantum Shannon theory are the direct quantum analog of thetechniques from classical information theory. We use the law of large numbers and the notionof the typical subspace, but we require generalizations of measures from the classical world todetermine how “close” two different quantum states are. One measure, the fidelity, has theoperational interpretation that it gives the probability that one quantum state would pass atest for being another. The trace distance is another distance measure that is perhaps moresimilar to a classical distance measure—its classical analog is a measure of the closenessof two probability distributions. The techniques in quantum Shannon theory also residefirmly in the quantum theory and have no true classical analog for some cases. Some of thetechniques will seem similar to those in the classical world, but the answer to some of thefundamental questions in quantum Shannon theory are rather different from some of theanswers in the classical world. It is the purpose of this book to explore the answers to thefundamental questions of quantum Shannon theory, and we now begin to ask what kinds oftasks we can perform.

1.2.3

Operational Tasks in Quantum Shannon Theory

Quantum Shannon theory has several resources that two parties can exploit in a quantum
information-processing task. Perhaps the most natural quantum resource is a noiseless qubit
channel. We can think of this resource as some medium through which a physical qubit can
travel without being affected by any noise. One example of a noiseless qubit channel could
be the free space through which a photon travels, where it ideally does not interact with any
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other particles along the way to its destination.7
A noiseless classical bit channel is a special case of a noiseless qubit channel because
we can always encode classical information into quantum states. For the example of aphoton, we can say that horizontal polarization corresponds to a “0” and vertical polarization
corresponds to a “1.” We refer to the dynamic resource of a noiseless classical bit channel asa cbit, in order to distinguish it from the noiseless qubit channel.
Perhaps the most intriguing resource that two parties can share is noiseless entanglement.
Any entanglement resource is a static resource because it is one that they share. Examples
of static resources in the classical world are an information source that we would like tocompress or a common secret key that two parties may possess. We actually have a wayof measuring entanglement that we discuss later on, and for this reason, we can say that asender and receiver have bits of entanglement or ebits.
Entanglement turns out to be a useful resource in many quantum communication tasks.
One example where it is useful is in the teleportation protocol, where a sender and receiveruse one ebit and two classical bits to transmit one qubit faithfully. This protocol is an
example of the extraordinary power of noiseless entanglement. The name “teleportation”
is really appropriate for this protocol because the physical qubit vanishes from the sender’sstation and appears at the receiver’s station after the receiver obtains the two transmittedclassical bits. We will see later on that a noiseless qubit channel can generate the othertwo noiseless resources, but it is impossible for each of the other two noiseless resources togenerate the noiseless qubit channel. In this sense, the noiseless qubit channel is the strongestof the three unit resources.
The first quantum information-processing task that we have discussed is Schumacher
compression. The goal of this task is to use as few noiseless qubit channels as possible inorder to transmit the output of a quantum information source reliably. After we understandSchumacher compression in a technical sense, the main focus of this book is to determine
what quantum information-processing tasks a sender and receiver can accomplish with theuse of a noisy quantum channel. The first and perhaps simplest task is to determine howmuch classical information a sender can transmit reliably to a receiver, by using a noisy
quantum channel a large number of times. This task is known as HSW coding, named afterits discoverers Holevo, Schumacher, and Westmoreland. The HSW coding theorem is one
quantum generalization of Shannon’s channel coding theorem (the latter overviewed in thenext chapter). We can also assume that a sender and receiver share some amount of noiseless entanglement prior to communication. They can then use this noiseless entanglementin addition to a large number of uses of a noisy quantum channel. This task is known asentanglement-assisted classical communication over a noisy quantum channel. The capacity
theorem corresponding to this task again highlights one of the marvelous features of entanglement. It shows that entanglement gives a boost to the amount of noiseless classicalcommunication we can generate using a noisy quantum channel—the classical capacity is
generally higher with entanglement assistance than without it.
7

We should be careful to note here that this is not actually a perfect channel because even empty spacecan be noisy in quantum mechanics, but nevertheless, it is a simple physical example to imagine.
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32

CHAPTER 1. CONCEPTS IN QUANTUM SHANNON THEORYOne of the most important theorems for quantum Shannon theory is the quantum channel
capacity theorem. Any proof of a capacity theorem consists of two parts: one part establishes
a lower bound on the capacity and the other part establishes an upper bound. If the twobounds coincide, then we have a characterization of the capacity in terms of these bounds.
The lower bound on the quantum capacity is colloquially known as the LSD coding theorem,8and it gives a characterization of the highest rate at which a sender can transmit quantum
information reliably over a noisy quantum channel so that a receiver can recover it perfectly.
The rate is generally lower than the classical capacity because it is more difficult to keep
quantum information intact. As we have said before, it is possible to encode classical information into quantum states, but this classical encoding is only a special case of a quantum
state. In order to preserve quantum information, we have to be able to preserve arbitrary
quantum states, not merely a classical encoding within a quantum state.
The pinnacle of this book is in Chapter 24 where we finally reach our study of thequantum capacity theorem. All efforts and technical developments in preceding chaptershave this goal in mind.9 Our first coding theorem in the dynamic setting is the HSW
coding theorem. A rigorous study of this coding theorem lays an important foundation—an
understanding of the structure of a code for reliable communication over a noisy quantum
channel. The method for the HSW coding theorem applies to the “entanglement-assistedclassical capacity theorem,” which is one building block for other protocols in quantum
Shannon theory. We then develop a more complex coding structure for sending private
classical information over a noisy quantum channel. In private coding, we are concerned withcoding in such a way that the intended receiver can learn the transmitted message perfectly,
but a third-party eavesdropper cannot learn anything about what the sender transmits to theintended receiver. This study of the private classical capacity may seem like a detour at first,
but it is closely linked with our ultimate aim. The coding structure developed for sending
private information proves to be indispensable for understanding the structure of a quantum
code. There are strong connections between the goals of keeping classical information private
and keeping quantum information coherent. In the private coding scenario, the goal is to
avoid leaking any information to an eavesdropper so that she cannot learn anything about thetransmission. In the quantum coding scenario, we can think of quantum noise as resulting
from the environment learning about the transmitted quantum information and this actof learning disturbs the quantum information. This effect is related to the information–
disturbance trade-off that is fundamental in quantum information theory. If the environment
learns something about the state being transmitted, there is inevitably some sort of noisy
disturbance that affects the quantum state. Thus, we can see a correspondence between
private coding and quantum coding. In quantum coding, the goal is to avoid leaking any
information to the environment because the avoidance of such a leak implies that there is no8

The LSD coding theorem does not refer to the synthetic crystalline compound, lysergic acid diethylamide(which one may potentially use as a hallucinogenic drug), but refers rather to Lloyd (1997), Shor (2002b),
and Devetak (2005), all of whom gave separate proofs of the lower bound on the quantum capacity withincreasing standards of rigor.
9
One goal of this book is to unravel the mathematical machinery behind Devetak’s proof of the quantum
channel coding theorem (Devetak, 2005).
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disturbance to the transmitted state. So the role of the environment in quantum coding issimilar to the role of the eavesdropper in private coding, and the goal in both scenarios is todecouple either the environment or eavesdropper from the picture. It is then no coincidence
that private codes and quantum codes have a similar structure. In fact, we can say that thequantum code inherits its structure from that of the private code.10
We also consider “trade-off” problems in addition to discussing the quantum capacity
theorem. Chapter 22 is another high point of the book, featuring a whole host of results thatemerge by combining several of the ideas from previous chapters. The most appealing aspectof this chapter is that we can construct virtually all of the protocols in quantum Shannon
theory from just one idea in Chapter 21. Also, Chapter 22 provides partial answers to many
practical questions concerning information transmission over noisy quantum channels. Some
example questions are as follows:• How much quantum and classical information can a noisy quantum channel transmit?
• An entanglement-assisted noisy quantum channel can transmit more classical information than an unassisted one, but how much entanglement is really necessary?
• Does noiseless classical communication help in transmitting quantum information reliably over a noisy quantum channel?
• How much entanglement can a noisy quantum channel generate when aided by classicalcommunication?
• How much quantum information can a noisy quantum channel communicate when
aided by entanglement?
These are examples of trade-off problems because they involve a noisy quantum channel
and either the consumption or generation of a noiseless resource. For every combinationof the generation or consumption of a noiseless resource, there is a corresponding coding
theorem that states what rates are achievable (and in some cases optimal). Some of these
trade-off questions admit interesting answers, but some of them do not. Our final aim inthese trade-off questions is to determine the full triple trade-off solution where we study
the optimal ways of combining all three unit resources (classical communication, quantum
communication, and entanglement) with a noisy quantum channel.
The coding theorems for a noisy quantum channel are just as important as (if not more
important than) Shannon’s classical coding theorems because they determine the ultimate
capabilities of information processing in a world where the postulates of quantum theory
apply. It is thought that quantum theory is the ultimate theory underpinning all physicalphenomena, and any theory of gravity will have to incorporate the quantum theory in some
10

There are other methods of formulating quantum codes using random subspaces (Shor, 2002b; Hayden
et al., 2008a,b; Klesse, 2008), but we prefer the approach of Devetak because we learn about other aspectsof quantum Shannon theory, such as the private capacity, along the way to proving the quantum capacity
theorem.
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CHAPTER 1. CONCEPTS IN QUANTUM SHANNON THEORY

fashion. Thus, it is reasonable that we should be focusing our efforts now on a full Shannon theory of quantum information processing in order to determine the tasks that thesesystems can accomplish. In many physical situations, some of the assumptions of quantum
Shannon theory may not be justified (such as an independent and identically distributed
quantum channel), but nevertheless, it provides an ideal setting in which we can determine
the capabilities of these physical systems.

1.2.4

History of Quantum Shannon Theory

We conclude this introductory chapter by giving a brief overview of the problems that researchers were thinking about that ultimately led to the development of quantum Shannon
theory.
The 1970s—The first researchers in quantum information theory were concerned withtransmitting classical data by optical means. They were ultimately led to a quantum formulation because they wanted to transmit classical information by means of a coherent laser.
Coherent states are special quantum states that a coherent laser ideally emits. Glauber provided a full quantum-mechanical theory of coherent states in two seminal papers (Glauber,
1963a,b), for which he shared the Nobel Prize in 2005 (Glauber, 2005). The first researchersof quantum information theory were Helstrom, Gordon, Stratonovich, and Holevo. Gordon(1964) first conjectured an important bound for our ability to access classical informationfrom a quantum system and Levitin (1969) stated it without proof. Holevo (1973a,b) later
provided a proof that the bound holds. This important bound is now known as the Holevo
bound, and it is useful in proving converse theorems (theorems concerning optimality) in
quantum Shannon theory. The simplest (yet rough) statement of the Holevo bound statesthat it is not possible to transmit more than one classical bit of information using a noiseless qubit channel, while at the same time being able to decode it reliably—i.e., we get one
cbit per qubit. Helstrom (1976) developed a full theory of quantum detection and quantum
estimation and published a textbook that discusses this theory. Fannes (1973) contributed
a useful continuity property of the entropy that is also useful in proving converse theorems
in quantum Shannon theory. Wiesner also used the uncertainty principle to devise a notionof “quantum money” in 1970, but unfortunately, his work was not accepted upon its initialsubmission. This work was way ahead of its time, and it was only until much later that
it was accepted (Wiesner, 1983). Wiesner’s ideas paved the way for the BB84 protocol for
quantum key distribution. Fundamental entropy inequalities, such as the strong subadditivity of quantum entropy (Lieb and Ruskai, 1973a,b) and the monotonicity of quantum relative
entropy (Lindblad, 1975), were proved during this time as well. These entropy inequalities
generalize the Holevo bound and are foundational for establishing optimality theorems inquantum Shannon theory.
Park (1970) produced one of the simplest, yet most profound, results that is crucial toquantum information science (Wootters and Zurek (1982) and Dieks (1982) also proved thisresult many years later). He proved the no-cloning theorem, showing that the postulates of
the quantum theory imply the impossibility of universally cloning quantum states. Given an
arbitrary unknown quantum state, it is impossible to build a device that can copy this state.
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This result has deep implications for the processing of quantum information and shows astrong divide between information processing in the quantum world and that in the classicalworld. We will prove this theorem in Chapter 3 and use it time and again in our reasoning.
For a history of the no-cloning theorem, see Ortigoso (2018).
The 1980s—The 1980s witnessed only a few advances in quantum information theory
because just a handful of researchers thought about the possibilities of linking quantum
theory with information-theoretic ideas. The Nobel Prize-winning physicist Richard Feynman published an interesting 1982 article that was one of the first to discuss computing
with quantum-mechanical systems (Feynman, 1982). His interest was in using a quantum
computer to simulate quantum-mechanical systems—he figured there should be a speed-up
over a classical simulation if we instead use one quantum system to simulate another. This
work is less quantum Shannon theory than it is quantum computing, but it is still a landmark because Feynman began to think about exploiting the actual quantum information ina physical system, rather than just using quantum systems to process classical information
as the researchers in the 1970s suggested.
The work of Wiesner on conjugate coding inspired an IBM physicist named Charles
Bennett. Bennett and Brassard (1984) published a groundbreaking paper that detailed thefirst quantum communication protocol: the BB84 protocol. This protocol shows how asender and a receiver can exploit a quantum channel to establish a secret key. The securityof this protocol, roughly speaking, relies on the uncertainty principle. If any eavesdropper
tries to learn about the random quantum data that they use to establish the secret key, thisact of learning inevitably disturbs the transmitted quantum data and the two parties candiscover this disturbance by noticing the change in the statistics of random sample data. The
secret key generation capacity of a noisy quantum channel is inextricably linked to the BB84
protocol, and we study this capacity problem in detail when we study the ability of quantum
channels to communicate private information. Interestingly, the physics community largely
ignored the BB84 paper when Bennett and Brassard first published it, likely because theypresented it at an engineering conference and the merging of physics and information had
not yet taken effect.
The 1990s—The 1990s were a time of much increased activity in quantum information
science, perhaps some of the most exciting years with many seminal results. One of the firstmajor results was from Ekert. He published a different way for performing quantum key distribution, this time relying on the strong correlations of entanglement (Ekert, 1991). He wasunaware of the BB84 protocol when he was working on his entanglement-based quantum key
distribution. The physics community embraced this result and a short time later, Ekert andBennett and Brassard became aware of each other’s respective works (Bennett et al., 1992a).
Bennett, Brassard, and Mermin later showed a sense in which these two seemingly different
schemes are equivalent (Bennett et al., 1992b). Bennett later developed the B92 protocol for
quantum key distribution using any two non-orthogonal quantum states (Bennett, 1992).
Two of the most profound results that later impacted quantum Shannon theory appearedin the early 1990s. First, Bennett and Wiesner (1992) devised the super-dense coding protocol. This protocol consumes one noiseless ebit of entanglement and one noiseless qubitc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 1. CONCEPTS IN QUANTUM SHANNON THEORY

channel to simulate two noiseless classical bit channels. Let us compare this result to that ofHolevo. Holevo’s bound states that we can reliably send only one classical bit per qubit, but
the super-dense coding protocol states that we can double this rate if we consume entanglement as well. Thus, entanglement is the enabler in this protocol that boosts the classical rate
beyond that possible with a noiseless qubit channel alone. The next year, Bennett and someother coauthors reversed the operations in the super-dense coding protocol to devise a protocol that has more profound implications. They devised the teleportation protocol (Bennett
et al., 1993)—this protocol consumes two classical bit channels and one ebit to transmit aqubit from a sender to a receiver. Right now, without any technical development yet, it may
be unclear how the qubit gets from the sender to the receiver. The original authors described
it as the “disembodied transport of a quantum state.” Suffice it for now to say that it is
the unique properties of entanglement (in particular, the ebit) that enable this disembodied
transport to occur. Yet again, it is entanglement that is the resource that enables this protocol, but let us be careful not to overstate the role of entanglement. Entanglement alone
does not suffice for implementing quantum teleportation. These protocols show that it is theunique combination of entanglement and quantum communication or entanglement and classical communication that yields these results. These two noiseless protocols are cornerstonesof quantum Shannon theory, originally suggesting that there are interesting ways of combining the resources of classical communication, quantum communication, and entanglement toformulate uniquely quantum protocols and leading the way to more exotic protocols thatcombine the different noiseless resources with noisy resources. Simple questions concerningthese protocols lead to quantum Shannon-theoretic protocols. In super-dense coding, howmuch classical information can Alice send if the quantum channel becomes noisy? What if
the entanglement is noisy? In teleportation, how much quantum information can Alice send
if the classical channel is noisy? What if the entanglement is noisy? Researchers addressed
these questions quite a bit after the original super-dense coding and teleportation protocols
were available, and we discuss these important questions in this book.
The year 1994 was a landmark for quantum information science. Shor (1994) published
his algorithm that factors a number in polynomial time—this algorithm gives an exponential
speed-up over the best known classical algorithm. We cannot overstate the importance of
this algorithm for the field. Its major application is to break RSA encryption (Rivest et al.,1978) because the security of that encryption algorithm relies on the computational difficultyof factoring a large number. This breakthrough generated wide interest in the idea of aquantum computer and started the quest to build one and study its capabilities.
Initially, much skepticism met the idea of building a practical quantum computer (Landauer, 1995; Unruh, 1995). Some experts thought that it would be impossible to overcome
errors that inevitably occur during quantum interactions, due to the coupling of a quantum system with its environment. Shor met this challenge by devising the first quantum
error-correcting code (Shor, 1995) and a scheme for fault-tolerant quantum computation(Shor, 1996). His paper on quantum error correction is the one most relevant for quantum
Shannon theory. At the end of this paper, he posed the idea of the quantum capacity of anoisy quantum channel as the highest rate at which a sender and receiver can maintain thec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

1.2. THE EMERGENCE OF QUANTUM SHANNON THEORY

37

fidelity of a quantum state when it is sent over a large number of uses of the noisy channel.
This open problem set the main task for researchers interested in quantum Shannon theory.
A flurry of theoretical activity then ensued in quantum error correction (Calderbank andShor, 1996; Steane, 1996; Laflamme et al., 1996; Gottesman, 1996, 1997; Calderbank et al.,1997, 1998) and fault-tolerant quantum computation (Aharonov and Ben-Or, 1997; Kitaev,
1997; Preskill, 1998; Knill et al., 1998). These two areas are now important subfields withinquantum information science, but we do not focus on them in any detail in this book.
Schumacher published a critical paper in 1995 as well (Schumacher, 1995) (we discussed
some of his contributions in the previous section). This paper gave the first informational
notion of a qubit, and it even established the now ubiquitous term “qubit.” He proved thequantum analog of Shannon’s source coding theorem, giving the ultimate compressibility ofquantum information. He used the notion of a typical subspace as an analogy of Shannon’s
typical set. This notion of a typical subspace proves to be one of the most crucial ideas
for constructing codes in quantum Shannon theory, just as the notion of a typical set is so
crucial for Shannon’s information theory.
Not much later, several researchers began investigating the capacity of a noisy quantum
channel for sending classical information (Hausladen et al., 1996). Holevo (1998) and Schumacher and Westmoreland (1997) independently proved that the Holevo information of aquantum channel is an achievable rate for classical communication over it. They appealed
to Schumacher’s notion of a typical subspace and constructed channel codes for sendingclassical information. The proof looks somewhat similar to the proof of Shannon’s channel
coding theorem (discussed in the next chapter) after taking a few steps away from it. The
proof of the converse theorem proceeds somewhat analogously to that of Shannon’s theorem,with the exception that one of the steps uses Holevo’s bound from 1973. In hindsight, itis perhaps somewhat surprising that it took over 20 years between the appearance of theproof of Holevo’s bound (the main step in the converse proof) and the appearance of a directcoding theorem for sending classical information.
The quantum capacity theorem is perhaps one of the most fundamental theorems ofquantum Shannon theory. Initial work by several researchers provided some insight into thequantum capacity theorem (Bennett et al., 1996b,c, 1997; Schumacher and Westmoreland,1998), and a series of papers established an upper bound on the quantum capacity (Schumacher, 1996; Schumacher and Nielsen, 1996; Barnum et al., 1998, 2000). For the lower
bound, Lloyd (1997) was the first to construct an idea for a proof, but it turns out that his
proof was more of a heuristic argument. Shor (2002b) then followed with another proof ofthe lower bound, and some of Shor’s ideas appeared much later in a full publication (Hayden
et al., 2008b). Devetak (2005) and Cai et al. (2004) independently solved the private capacity
theorem at approximately the same time (with the publication of the CWY paper appearing
a year after Devetak’s arXiv post). Devetak took the proof of the private capacity theorem
a step further and showed how to apply its techniques to construct a quantum code that
achieves a good lower bound on the quantum capacity, while also providing an alternate,cleaner proof of the converse theorem (Devetak, 2005). It is Devetak’s technique that we
mainly explore in this book because it provides some insight into the coding structure (howc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 1. CONCEPTS IN QUANTUM SHANNON THEORY

ever, we also explore a different technique via the entanglement-assisted classical capacity
theorem).
The 2000s—In recent years, we have had many advancements in quantum Shannon
theory (technically some of the above contributions were in the 2000s, but we did not want
to break the continuity of the history of the quantum capacity theorem). One major result
was the proof of the entanglement-assisted classical capacity theorem—it is the noisy versionof the super-dense coding protocol where the quantum channel is noisy (Bennett et al.,1999, 2002; Holevo, 2002b). This theorem assumes that Alice and Bob share unlimited
entanglement and they exploit the entanglement and the noisy quantum channel to sendclassical information.
A few fantastic results have arisen in recent years. Horodecki, Oppenheim, and Wintershowed the existence of a state-merging protocol (Horodecki et al., 2005, 2007). This protocol
gives the minimum rate at which Alice and Bob consume noiseless qubit channels in orderfor Alice to send her share of a quantum state to Bob. This rate is the conditional quantum
entropy—the protocol thus gives an operational interpretation to this entropic quantity.
What was most fascinating about this result is that the conditional quantum entropy can be
negative in quantum Shannon theory. Prior to their work, no one really understood what it
meant for the conditional quantum entropy to become negative (Wehrl, 1978; Horodecki andHorodecki, 1994; Cerf and Adami, 1997), but this state-merging result gave a compelling
operational interpretation. A negative rate implies that Alice and Bob gain the ability for
future quantum communication, instead of consuming quantum communication as when therate is positive.
Another fantastic result came from (Smith and Yard, 2008). Suppose we have two noisy
quantum channels and each of them individually has zero capacity to transmit quantum
information. One would expect intuitively that the “joint quantum capacity” (when using
them together) would also have zero ability to transmit quantum information. But this result is not generally the case in the quantum world. It is possible for some particular noisy
quantum channels with no individual quantum capacity to have a non-zero joint quantum
capacity. It is not clear yet how we might practically take advantage of such a “superactivation” effect, but the result is nonetheless fascinating, counterintuitive, and not yet fully
understood.
The latter part of the 2000s saw the unification of quantum Shannon theory. The resource
inequality framework was the first step because it unified many previously known results
into one formalism (Devetak et al., 2004, 2008). Devetak, Harrow, and Winter provided afamily tree for quantum Shannon theory and showed how to relate the different protocols
in the tree to one another. We will go into the theory of resource inequalities in somedetail throughout this book because it provides a tremendous conceptual simplification when
considering coding theorems in quantum Shannon theory. In fact, the last chapter of thisbook contains a concise summary of many of the major quantum Shannon-theoretic protocols
in the language of resource inequalities. Abeyesinghe et al. (2009) published a work showing
a sense in which the mother protocol of the family tree can generate the father protocol. We
have seen unification efforts in the form of triple trade-off coding theorems (Abeyesinghe andc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

1.2. THE EMERGENCE OF QUANTUM SHANNON THEORY

39

Hayden, 2003; Hsieh and Wilde, 2010a,b). These theorems give the optimal combination ofclassical communication, quantum communication, entanglement, and an asymptotic noisy
resource for achieving a variety of quantum information-processing tasks.
We have also witnessed the emergence of a study of network quantum Shannon theory.
Some authors have tackled the quantum broadcasting paradigm (Guha and Shapiro, 2007;
Guha et al., 2007; Dupuis et al., 2010; Yard et al., 2011), where one sender transmits tomultiple receivers. A multiple-access quantum channel has many senders and one receiver.
Some of the same authors (and others) have tackled multiple-access communication (Winter,
2001; Yard, 2005; Yen and Shapiro, 2005; Yard et al., 2005, 2008; Hsieh et al., 2008a; Czekaj
and Horodecki, 2009). This network quantum Shannon theory should become increasingly
important as we get closer to the ultimate goal of a quantum Internet.
Quantum Shannon theory has now established itself as an important and distinct fieldof study. The next few chapters discuss the concepts that will prepare us for tackling someof the major results in quantum Shannon theory.

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c 2019CHAPTER 1. CONCEPTS IN QUANTUM SHANNON THEORYMark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 2Classical Shannon Theory
We cannot overstate the importance of Shannon’s contribution to modern science. His
introduction of the field of information theory and his solutions to its two main theorems
demonstrate that his ideas on communication were far beyond the other prevailing ideas inthis domain around 1948.
In this chapter, our aim is to discuss Shannon’s two main contributions in a descriptive
fashion. The goal of this high-level discussion is to build up the intuition for the problem
domain of information theory and to understand the main concepts before we delve into
the analogous quantum information-theoretic ideas. We avoid going into deep technicaldetail in this chapter, leaving such details for later chapters where we formally prove bothclassical and quantum Shannon-theoretic coding theorems. We do use some mathematics
from probability theory (namely, the law of large numbers).
We will be delving into the technical details of this chapter’s material in later chapters(specifically, Chapters 10, 13, and 14). Once you have reached later chapters that develop
some more technical details, it might be helpful to turn back to this chapter to get an overall
flavor for the motivation of the development.

2.1 Data Compression
We first discuss the problem of data compression. Those who are familiar with the Internet
have used several popular data formats such as JPEG, MPEG, ZIP, GIF, etc. All of these fileformats have corresponding algorithms for compressing the output of an information source.
A first glance at the compression problem might lead one to believe that it is not possible
to compress the output of the information source to an arbitrarily small size, and Shannon
proved that this is the case. This result is the content of Shannon’s first noiseless coding
theorem.
41

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CHAPTER 2. CLASSICAL SHANNON THEORY

2.1.1

An Example of Data Compression

We begin with a simple example that illustrates the concept of an information source. We
then develop a scheme for coding this source so that it requires fewer bits to represent its
output faithfully.
Suppose that Alice is a sender and Bob is a receiver. Suppose further that a noiselessbit channel connects Alice to Bob—a noiseless bit channel is one that transmits information
perfectly from sender to receiver, e.g., Bob receives “0” if Alice transmits “0” and Bob
receives “1” if Alice transmits “1.” Alice and Bob would like to minimize the number oftimes that they use this noiseless channel because it is expensive to use it.
Alice would like to use the noiseless channel to communicate information to Bob. Supposethat an information source randomly chooses from four symbols {a, b, c, d} and selects themwith a skewed probability distribution:
Pr{a} = 1/2,
Pr{b} = 1/8,
Pr{c} = 1/4,
Pr{d} = 1/8.

(2.1)
(2.2)
(2.3)
(2.4)

So it is clear that the symbol a is the most likely one, c the next likely, and both b andd are least likely. We make the additional assumption that the information source chooses
each symbol independently of all previous ones and chooses each with the same probability
distribution above. After the information source makes a selection, it gives the symbol to
Alice for coding.
A noiseless bit channel accepts only bits as input—it does not accept the symbols a, b, c,d as input. So, Alice has to encode her information into bits. Alice could use the following
coding scheme:a → 00, b → 01, c → 10, d → 11,(2.5)
where each binary representation of a letter is a codeword. How do we measure the performance of a particular coding scheme? The expected length of a codeword is one way tomeasure performance. For the above example, the expected length is equal to two bits. Thismeasure reveals a problem with the above scheme—the scheme does not take advantage ofthe skewed nature of the distribution of the information source because each codeword has
the same length.
One might instead consider a scheme that uses shorter codewords for symbols that are
more likely and longer codewords for symbols that are less likely.1 Then the expected length1

Such coding schemes are common. Samuel F. B. Morse employed this idea in his popular Morse code.
Also, in the movie The Diving Bell and the Butterfly, a writer becomes paralyzed with “locked-in” syndrome
so that he can only blink his left eye. An assistant then develops a “blinking code” where she reads a list ofletters in French, beginning with the most commonly used letter and ending with the least commonly usedletter. The writer blinks when she says the letter he wishes and they finish an entire book using this coding
scheme.
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2.1. DATA COMPRESSION

43

of a codeword with such a scheme should be shorter than that in the former scheme. The
following coding scheme gives an improvement in the expected length of a codeword:a → 0, b → 110, c → 10, d → 111.

(2.6)

This scheme has the advantage that any coded sequence is uniquely decodable. For example,
suppose that Bob obtains the following sequence:0011010111010100010.

(2.7)

Bob can parse the above sequence as
0 0 110 10 111 0 10 10 0 0 10,(2.8)

and determine that Alice transmitted the message
aabcdaccaac.

(2.9)

We can calculate the expected length of this coding scheme as follows:
1
1
1
7
1
(1) + (3) + (2) + (3) = .
2
8
4
8
4

(2.10)

This scheme is thus more efficient because its expected length is 7/4 bits as opposed to two
bits. It is a variable-length code because the number of bits in each codeword depends on
the source symbol.

2.1.2

A Measure of Information

The above scheme suggests a way to measure information. Consider the probability distribution in (2.1)–(2.4). Would we be more surprised to learn that the information source
produced the symbol a or to learn that it produced the symbol d? The answer is d because
the source is less likely to produce it. Let X denote a random variable with distribution
given in (2.1)–(2.4). One measure of the surprise of symbol x ∈ {a, b, c, d} is1= − log (pX (x)) ,(2.11)
i(x) ≡ log
pX (x)
where the logarithm is base two—this convention implies the units of this measure are bits.
This measure of surprise has the desirable property that it is higher for lower probability
events and lower for higher probability events. Here, we take after Shannon, and we namei(x) the information content or surprisal of the symbol x. Observe that the length ofeach codeword in the coding scheme in (2.6) is equal to the information content of its
corresponding symbol.
The information content has another desirable property called additivity. Suppose thatthe information source produces two symbols x1 and x2 , with corresponding random variables
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CHAPTER 2. CLASSICAL SHANNON THEORY

X1 and X2 . The probability for this event is pX1 X2 (x1 , x2 ) and the joint distribution factors
as pX1 (x1 )pX2 (x2 ) if we assume the source is memoryless—that it produces each symbol
independently. The information content of the two symbols x1 and x2 is additive becausei(x1 , x2 ) = − log (pX1 X2 (x1 , x2 ))= − log (pX1 (x1 )pX2 (x2 ))= − log (pX1 (x1 )) − log (pX2 (x2 ))= i(x1 ) + i(x2 ).

(2.12)(2.13)(2.14)(2.15)

In general, additivity is a desirable property for any information measure. We will return tothe issue of additivity in many different contexts in this book (especially in Chapter 13).
The expected information content of the information source isXX
pX (x)i(x) = −pX (x) log (pX (x)) .
(2.16)xx

The above quantity is so important in information theory that we give it a name: the entropyof the information source. The reason for its importance is that the entropy and variationsof it appear as the answer to many questions in information theory. For example, in theabove coding scheme, the expected length of a codeword is the entropy of the information
source because
1 1
1 1
1 1
1
1
− log − log − log − log
2
2 8
8 4
4 8
8
1
1
1
1
= (1) + (3) + (2) + (3)(2.17)
2
8
4
8
7
= .
(2.18)
4
It is no coincidence that we chose the particular coding scheme in (2.6). The effectiveness ofthe scheme in this example is related to the structure of the information source—the numberof symbols is a power of two and the probability of each symbol is the reciprocal of a powerof two.

2.1.3

Shannon’s Source Coding Theorem

The next question to ask is whether there is any other scheme that can achieve a better
compression rate than the scheme in (2.6). This question is the one that Shannon asked in
his first coding theorem. To answer this question, we consider a more general information
source and introduce a notion of Shannon, the idea of the set of typical sequences.
We can represent a more general information source with a random variable X whose
realizations x are letters in an alphabet X . Let pX (x) be the probability mass function
associated with random variable X, so that the probability of realization x is pX (x). Let
H(X) denote the entropy of the information source:
X
H(X) ≡ −pX (x) log (pX (x)) .
(2.19)
x∈X
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2.1. DATA COMPRESSION

45

The entropy H(X) is also the entropy of the random variable X. Another way of writing itis H(p), but we use the more common notation H(X) throughout this book.
The information content i(X) of random variable X isi(X) ≡ − log (pX (X)) ,(2.20)and is itself a random variable. There is nothing wrong mathematically here with having
random variable X as the argument to the density function pX , though this expression may
seem self-referential at a first glance. This way of thinking turns out to be useful later.
Again, the expected information content of X is equal to the entropy:
EX {− log (pX (X))} = H(X).

(2.21)

Exercise 2.1.1 Show that the entropy of a uniform random variable is equal to log |X |,where |X | is the size of the variable’s alphabet.
We now turn to source coding the above information source. We could associate a binary
codeword for each symbol x as we did in the scheme in (2.6). But this scheme may lose someefficiency if the size of our alphabet is not a power of two or if the probabilities are not areciprocal of a power of two as they are in our nice example. Shannon’s breakthrough idea
was to let the source emit a large number of realizations and then code the emitted data asa large block, instead of coding each symbol as the above example does. This technique iscalled block coding. Shannon’s other insight was to allow for a slight error in the compression
scheme, but to show that this error vanishes as the block size becomes arbitrarily large. To
make the block coding scheme more clear, Shannon suggests to let the source emit thefollowing sequence:xn ≡ x1 x2 · · · xn ,(2.22)
where n is a large number that denotes the size of the block of emitted data and xi , for alli = 1, . . . , n, denotes the ith emitted symbol. Let X n denote the random variable associatedwith the sequence xn , and let Xi be the random variable for the ith symbol xi . Figure 2.1
depicts Shannon’s idea for a classical source code.
An important assumption regarding this information source is that it is independent andidentically distributed (i.i.d.). The i.i.d. assumption means that each random variable Xi
has the same distribution as random variable X, and we use the index i merely to track
to which symbol xi the random variable Xi corresponds. Under the i.i.d. assumption, theprobability of any given emitted sequence xn factors as
pX n (xn ) = pX1 ,X2 ,...,Xn (x1 , x2 , . . . , xn )= pX1 (x1 )pX2 (x2 ) · · · pXn (xn )= pX (x1 )pX (x2 ) · · · pX (xn )nY=pX (xi ).

(2.23)(2.24)(2.25)(2.26)

i=1

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CHAPTER 2. CLASSICAL SHANNON THEORY

Figure 2.1: This figure depicts Shannon’s idea for a classical source code. The information source emits along sequence xn to Alice. She encodes this sequence as a block with an encoder E and produces a codeword
whose length is less than that of the original sequence xn (indicated by fewer lines coming out of the encoder
E). She transmits the codeword over noiseless bit channels (each indicated by “id” which stands for theidentity bit channel) and Bob receives it. Bob decodes the transmitted codeword with a decoder D andproduces the original sequence that Alice transmitted, only if their chosen code is good, in the sense thatthe code has a small probability of error.

The above rule from probability theory results in a remarkable simplification of the mathematics. Suppose that we now label the letters in the alphabet X as a1 , . . . , a|X | in order todistinguish the letters from the realizations. Let N (ai |xn ) denote the number of occurrencesof the letter ai in the sequence xn (where i = 1, . . . , |X |). As an example, consider thesequence in (2.9). The quantities N (ai |xn ) for this example are
N (a|xn ) = 5,
N (b|xn ) = 1,
N (c|xn ) = 4,
N (d|xn ) = 1.

(2.27)(2.28)(2.29)(2.30)

We can rewrite the result in (2.26) asn

pX n (x ) =nYpX (xi ) =

i=1

|X |
Y

n

pX (ai )N (ai |x ) .

(2.31)

i=1

Keep in mind that we are allowing the length n of the emitted sequence to be extremely
large, so that it is much larger than the alphabet size |X |:n  |X | .

(2.32)

The formula on the right in (2.31) is much simpler than the formula in (2.26) because it has
fewer iterations of multiplications. There is a sense in which the i.i.d. assumption allows us
to permute the sequence xn as
xn → a1 · · · a1 a2 · · · a2 · · · a|X | · · · a|X | ,| {z } | {z } | {z }N (a1 |xn ) N (a2 |xn )c 2019(2.33)

N (a|X | |xn )Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

2.1. DATA COMPRESSION

47

because the probability calculation is invariant under this permutation. We introduce theabove way of thinking right now because it turns out to be useful later when we develop
some ideas in quantum Shannon theory (specifically in Section 14.9). Thus, the formula on
the right in (2.31) characterizes the probability of any given sequence xn .
The above discussion applies to a particular sequence xn that the information source
emits. Now, we would like to analyze the behavior of a random sequence X n that thesource emits, and this distinction between the realization xn and the random variable X n
is important. In particular, let us consider the sample average of the information content
of the random sequence X n (divide the information content of X n by n to get the sample
average):−1log (pX n (X n )) .
n(2.34)

It may seem strange at first glance that X n , the argument of the probability mass function
pX n is itself a random variable, but this type of expression is perfectly well defined mathematically. (This self-referencing type of expression is similar to (2.20), which we used to
calculate the entropy.) For reasons that will become clear shortly, we call the above quantity
the sample entropy of the random sequence X n .
Suppose now that we use the function N (ai |•) to calculate the number of appearancesof the letter ai in the random sequence X n . We write the desired quantity as N (ai |X n ) andnote that it is also a random variable, whose random nature derives from that of X n . We
can reduce the expression in (2.34) to the following one with some algebra and the resultin (2.31):


|X |
Y
1
1
n
− log (pX n (X n )) = − log  pX (ai )N (ai |X ) nn
i=1(2.35)

|X |

1X
n 
log pX (ai )N (ai |X )
=−
n i=1
=−

|X |
X
N (ai |X n )i=1nlog (pX (ai )) .

(2.36)(2.37)

We stress again that the above quantity is random.
Is there any way that we can determine the behavior of the above sample entropy when nbecomes large? Probability theory gives us a way. The expression N (ai |X n )/n represents anempirical distribution for the letters ai in the alphabet X . As n becomes large, one form of
the law of large numbers states that it is overwhelmingly likely that a random sequence has
its empirical distribution N (ai |X n )/n close to the true distribution pX (ai ), and conversely,
it is highly unlikely that a random sequence does not satisfy this property. Thus, a random
emitted sequence X n is highly likely to satisfy the following condition for all δ > 0 as n
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CHAPTER 2. CLASSICAL SHANNON THEORY

becomes large:

lim Pr

n→∞




|X |

X
1
pX (ai ) log− log (pX n (X n )) −
 n
i=1



1
pX (ai )


≤δ




= 1.

(2.38)P |The quantity − |X
i=1 pX (ai ) log (pX (ai )) is none other than the entropy H(X) so that theabove expression is equivalent to the following one for all δ > 0:1nlim Pr − log (pX n (X )) − H(X) ≤ δ = 1.
(2.39)n→∞n
Another way of stating this property is as follows:
It is highly likely that the information source emits a sequence whose sample
entropy is close to the true entropy, and conversely, it is highly unlikely that the
information source emits a sequence that does not satisfy this property.2
Now we consider a particular realization xn of the random sequence X n . We name aparticular sequence xn a typical sequence if its sample entropy is close to the true entropy
H(X) and the set of all typical sequences is the typical set. Fortunately for data compression,
the set of typical sequences is not too large. In Chapter 14 on typical sequences, we provethat the size of this set is much smaller than the set of all sequences. We accept it for now(and prove later) that the size of the typical set is ≈ 2nH(X) , whereas the size of the set of
all sequences is equal to |X |n . We can rewrite the size of the set of all sequences as|X |n = 2n log|X | .

(2.40)

Comparing the size of the typical set to the size of the set of all sequences, the typical set isexponentially smaller than the set of all sequences whenever the random variable is not equal
to the uniform random variable. Figure 2.2 illustrates this concept.We summarize these two
crucial properties of the typical set and give another that we prove later:
Property 2.1.1 (Unit Probability) The probability that an emitted sequence is typicalapproaches one as n becomes large. Another way of stating this property is that the typicalset has almost all of the probability.
Property 2.1.2 (Exponentially Smaller Cardinality) The size of the typical set is ≈2nH(X) and is exponentially smaller than the size 2n log|X | of the set of all sequences wheneverrandom variable X is not uniform.
2

Do not fall into the trap of thinking “The possible sequences that the source emits are typical sequences.”That line of reasoning is quantitatively far from the truth. In fact, what we can show is much different because
the set of typical sequences is much smaller than the set of all possible sequences.
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Figure 2.2: This figure indicates that the typical set is much smaller (exponentially smaller) than the setof all sequences. The typical set is roughly the same size as the set of all sequences only when the entropy
H(X) of the random variable X is equal to log |X |—implying that the distribution of random variable X is
uniform.

Property 2.1.3 (Equipartition) The probability of a particular typical sequence is roughly
uniform ≈ 2−nH(X) . (The probability 2−nH(X) is easy to calculate if we accept that the typicalset has all of the probability, its size is 2nH(X) , and the distribution over typical sequences is
uniform.)
These three properties together are collectively known as the asymptotic equipartition
theorem. The word “asymptotic” applies because the theorem exploits the asymptotic limit
when n is large and the word “equipartition” refers to the third property above.
With the above notions of a typical set under our belt, a strategy for compressing information should now be clear. The strategy is to compress only the typical sequences that thesource emits. We simply need to establish a one-to-one encoding function that maps fromthe set of typical sequences (size 2nH(X) ) to the set of all binary strings of length nH(X)(this set also has size 2nH(X) ). If the source emits an atypical sequence, we declare an error.
This coding scheme is reliable in the asymptotic limit because the probability of an error
event vanishes as n becomes large, due to the unit probability property in the asymptotic
equipartition theorem. We measure the rate of this block coding scheme as follows:
compression rate ≡# of noiseless channel bits.
# of source symbols(2.41)For the case of Shannon compression, the number of noiseless channel bits is equal to nH(X)and the number of source symbols is equal to n. Thus, the compression rate is equal to theentropy H(X).
One may then wonder whether this rate of data compression is the best that we can
do—whether this rate is optimal (we could achieve a lower rate of compression if it were not
optimal). In fact, the above rate is the optimal rate at which we can compress information,and this is the content of Shannon’s data compression theorem. We hold off on a formal
proof of optimality for now and delay it until we reach Chapter 18. We just mention forc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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now that this data compression protocol gives an operational interpretation to the Shannon
entropy H(X) because it appears as the optimal rate of data compression.
The above discussion highlights the common approach in information theory for establishing a coding theorem. Proving a coding theorem has two parts—traditionally called the
direct coding theorem and the converse theorem. First, we give a coding scheme that canachieve a given rate for an information-processing task. This first part includes a direct
construction of a coding scheme, hence the name direct coding theorem. The statement of
the direct coding theorem for the above task is as follows:
If the rate of compression is greater than the entropy of the source, then there
exists a coding scheme that can achieve lossless data compression in the sense that
it is possible to make the probability of error for incorrectly decoding arbitrarily
small.
The second task is to prove that the rate from the direct coding theorem is optimal—that
we cannot do any better than the suggested rate. We traditionally call this part the converse
theorem because it corresponds to the converse of the above statement:
If there exists a coding scheme that can achieve lossless data compression with
arbitrarily small probability of decoding error, then the rate of compression isgreater than the entropy of the source.
The techniques used in proving each part of the coding theorem are completely different.
For most coding theorems in information theory, we can prove the direct coding theorem
by appealing to the ideas of typical sequences and large block sizes. That this techniquegives a good coding scheme is directly related to the asymptotic equipartition properties
that govern the behavior of random sequences of data as the length of the sequence becomeslarge. The proof of a converse theorem relies on information inequalities that give tightbounds on the entropic quantities appearing in the coding constructions. We spend sometime with information inequalities in Chapter 10 to build up our ability to prove converse
theorems.
Sometimes, in the course of proving a direct coding theorem, one may think to havefound the optimal rate for a given information-processing task. Without a matching converse
theorem, it is not generally clear that the suggested rate is optimal. So, always prove converse
theorems!

2.2 Channel Capacity
The next issue that we overview is the transmission of information over a noisy classicalchannel. We begin with a standard example—transmitting a single bit of information overa noisy bit-flip channel.
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Figure 2.3: This figure depicts the action of the bit-flip channel. It preserves the input bit with probability
1 − p and flips it with probability p.

2.2.1

An Example of an Error Correction CodeWe again have our protagonists, Alice and Bob, as respective sender and receiver. Thistime, however, we assume that a noisy classical channel connects them, so that informationtransfer is not reliable. Alice and Bob realize that a noisy channel is not as expensive
as a noiseless one, but it still is expensive for them to use. For this reason, they would
like to maximize the amount of information that Alice can communicate reliably to Bob,
where reliable communication implies that there is a negligible probability of error whentransmitting this information.
The simplest example of a noisy classical channel is a bit-flip channel, with the technicalname binary symmetric channel. This channel flips the input bit with probability p andleaves it unchanged with probability 1 − p. Figure 2.3 depicts the action of the bit-flipchannel. Alice and Bob are allowed to use the channel multiple times, and in so doing, we
assume that the channel behaves independently from one use to the next and behaves in thesame random way as described above. For this reason, we describe the multiple uses of thechannel as i.i.d. channels. This assumption will be helpful when we go to the asymptotic
regime of a large number of uses of the channel.
Suppose that Alice and Bob just use the channel as is—Alice just sends plain bits toBob. This scheme works reliably only if the probability of bit-flip error vanishes. So, Aliceand Bob could invest their best efforts into engineering the physical channel to make itreliable. But, generally, it is not possible to engineer a classical channel this way for physicalor logistical reasons. For example, Alice and Bob may only have local computers at theirends and may not have access to the physical channel because the telephone company may
control the channel.
Alice and Bob can employ a “systems engineering” solution to this problem rather thanan engineering of the physical channel. They can redundantly encode information in a way
such that Bob can have a higher probability of determining what Alice is sending, effectively
reducing the level of noise on the channel. A simple example of this systems engineeringsolution is the three-bit majority vote code. Alice and Bob employ the following encoding:0 → 000,

1 → 111,(2.42)

where both “000” and “111” are codewords. Alice transmits the codeword “000” with threec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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Channel Output Probability000(1 − p)3
001, 010, 100p(1 − p)2
011, 110, 101
p2 (1 − p)111p3Table 2.1: The first column gives the eight possible outputs of the noisy bit-flip channel when Alice encodesa “0” with the majority vote code. The second column gives the corresponding probability of Bob receiving
the particular outputs.

independent uses of the noisy channel if she really wants to communicate a “0” to Bob
and she transmits the codeword “111” if she wants to send a “1” to him. The physical orchannel bits are the actual bits that she transmits over the noisy channel, and the logicalor information bits are those that she intends for Bob to receive. In our example, “0” is alogical bit and “000” corresponds to the physical bits.
The rate of this scheme is 1/3 because it encodes one information bit. The term “rate” isperhaps a misnomer for coding scenarios that do not involve sending bits in a time sequence
over a channel. We may just as well use the majority vote code to store one bit in a memory
device that may be unreliable. Perhaps a more universal term is efficiency. Nevertheless,we follow convention and use the term rate throughout this book.
Of course, the noisy bit-flip channel does not always transmit these codewords without
error. So how does Bob decode in the case of error? He simply takes a majority vote todetermine the transmitted message—he decodes as “0” if the number of zeros in the codeword
he receives is greater than the number of ones.
We now analyze the performance of this simple “systems engineering” solution. Table 2.1
enumerates the probability of receiving every possible sequence of three bits, assuming that
Alice transmits a “0” by encoding it as “000.” The probability of no error is (1 − p)3 , theprobability of a single-bit error is 3p(1−p)2 , the probability of a double-bit error is 3p2 (1−p),and the probability of a total failure is p3 . The majority vote solution can “correct” for noerror and it corrects for all single-bit errors, but it has no ability to correct for double-bit
and triple-bit errors. In fact, it actually incorrectly decodes these latter two scenarios by“correcting” “011”, “110”, or “101” to “111” and decoding “111” as a “1.” Thus, these latter
two outcomes are errors because the code has no ability to correct them. We can employ
similar arguments as above to the case in which Alice transmits a “1” to Bob using the
majority vote code.
When does this majority vote scheme perform better than no coding at all? It is exactly
when the probability of error with the majority vote code is less than p, the probability oferror with no coding. Letting e denote the event that an error occurs, the probability oferror is equal to the following quantity:
Pr{e} = Pr{e|0} Pr{0} + Pr{e|1} Pr{1}.

(2.43)

Our analysis above suggests that the conditional probabilities Pr{e|0} and Pr{e|1} are equal
for the majority vote code because of the symmetry in the noisy bit-flip channel. This resultc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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implies that the probability of error is
Pr{e} = 3p2 (1 − p) + p3
= 3p2 − 2p3 ,(2.44)(2.45)

because Pr{0}+Pr{1} = 1. We consider the following inequality to determine if the majority
vote code reduces the probability of error:
3p2 − 2p3 < p.

(2.46)

0 < 2p3 − 3p2 + p∴ 0 < p (2p − 1) (p − 1) .

(2.47)(2.48)

This inequality simplifies as

The only values of p that satisfy the above inequality are 0 < p < 1/2. Thus, the majority
vote code reduces the probability of error only when 0 < p < 1/2, i.e., when the noise on
the channel is not too much. Too much noise has the effect of causing the codewords to flip
too often, throwing off Bob’s decoder.
The majority vote code gives a way for Alice and Bob to reduce the probability of error
during their communication, but unfortunately, there is still a non-zero probability for the
noisy channel to disrupt their communication. Is there any way that they can achieve reliablecommunication by reducing the probability of error to zero?
One simple approach to achieve this goal is to exploit the majority vote idea a second
time. They can concatenate two instances of the majority vote code to produce a code witha larger number of physical bits. Concatenation consists of using one code as an “inner”
code and another as an “outer” code. There is no real need for us to distinguish between the
inner and outer code in this case because we use the same code for both the inner and outercode. The concatenation scheme for our case first encodes the message i, where i ∈ {0, 1},
using the majority vote code. Let us label the codewords as follows:0̄ ≡ 000,1̄ ≡ 111.

(2.49)For the second layer of the concatenation, we encode 0̄ and 1̄ with the majority vote code
again:
0̄ → 0̄0̄0̄,
1̄ → 1̄1̄1̄.
(2.50)
Thus, the overall encoding of the concatenated scheme is as follows:
0 → 000 000 000,

1 → 111 111 111.

(2.51)The rate of the concatenated code is 1/9 and smaller than the original rate of 1/3. A simple
application of the above performance analysis for the majority vote code shows that thisconcatenation scheme reduces the probability of error as follows:
3[Pr{e}]2 − 2[Pr{e}]3 = O(p4 ).
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The error probability Pr{e} is in (2.45) and O(p4 ) indicates that the leading order term of
the left-hand side is the fourth power in p.
The concatenated scheme achieves a lower probability of error at the cost of using more
physical bits in the code. Recall that our goal is to achieve reliable communication, where
there is no probability of error. A first guess for achieving reliable communication is tocontinue concatenating. If we concatenate again, the probability of error reduces to O(p6 ),and the rate drops to 1/27. We can continue indefinitely with concatenating to make theprobability of error arbitrarily small and achieve reliable communication, but the problem isthat the rate approaches zero as the probability of error becomes arbitrarily small.
The above example seems to show that there is a trade-off between the rate of theencoding scheme and the desired order of error probability. Is there a way that we can code
information for a noisy channel while maintaining a good rate of communication?

2.2.2

Shannon’s Channel Coding Theorem

Shannon’s second breakthrough coding theorem provides an affirmative answer to the above
question. This answer came as a complete shock to communication researchers in 1948.
Furthermore, the techniques that Shannon used in demonstrating this fact were rarely usedby engineers at the time. We give a broad overview of Shannon’s main idea and techniquesthat he used to prove his second important theorem—the noisy channel coding theorem.

2.2.3

General Model for a Channel CodeWe first generalize some of the ideas in the above example. We still have Alice trying
to communicate with Bob, but this time, she wants to be able to transmit a larger setof messages with asymptotically perfect reliability, rather than merely sending “0” or “1.”Suppose that she selects messages from a message set [M ] that consists of M messages:[M ] ≡ {1, . . . , M } .

(2.53)

Suppose furthermore that Alice chooses a particular message m with uniform probability
from the set [M ]. This assumption of a uniform distribution for Alice’s messages indicatesthat we do not really care much about the content of the actual message that she is transmitting. We just assume total ignorance of her message because we only really care about
her ability to send any message reliably. The message set [M ] requires log(M ) bits to represent it, where the logarithm is again base two. This number becomes important when we
calculate the rate of a channel code.
The next aspect of the model that we need to generalize is the noisy channel that connects Alice to Bob. We used the bit-flip channel before, but this channel is not generalenough for our purposes. A simple way to extend the channel model is to represent it asa conditional probability distribution involving an input random variable X and an output
random variable Y :N :pY |X (y|x).
(2.54)
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Figure 2.4: This figure depicts Shannon’s idea for a classical channel code. Alice chooses a message mfrom a message set [M ] ≡ {1, . . . , M }. She encodes the message m with an encoding operation E. This
encoding operation assigns a codeword xn to the message m and inputs the codeword xn to a large numberof i.i.d. uses of a noisy channel N . The noisy channel randomly corrupts the codeword xn to a sequence y n .
Bob receives the corrupted sequence y n and performs a decoding operation D to estimate the codeword xn .
This estimate of the codeword xn then produces an estimate m̂ of the message that Alice transmitted. A
reliable code has the property that Bob can decode each message m ∈ [M ] with a vanishing probability oferror when the block length n becomes large.

We use the symbol N to represent this more general channel model. One assumption that
we make about random variables X and Y is that they are discrete, but the respective sizesof their outcome sets do not have to match. The other assumption that we make concerning
the noisy channel is that it is i.i.d. Let X n ≡ X1 X2 · · · Xn and Y n ≡ Y1 Y2 · · · Yn be therandom variables associated with respective sequences xn ≡ x1 x2 · · · xn and y n ≡ y1 y2 · · · yn .
If Alice inputs the sequence xn to the n inputs of n respective uses of the noisy channel, apossible output sequence may be y n . The i.i.d. assumption allows us to factor the conditional
probability of the output sequence y n :pY n |X n (y n |xn ) = pY1 |X1 (y1 |x1 )pY2 |X2 (y2 |x2 ) · · · pYn |Xn (yn |xn )= pY |X (y1 |x1 )pY |X (y2 |x2 ) · · · pY |X (yn |xn )nY=
pY |X (yi |xi ).

(2.55)(2.56)(2.57)

i=1

The technical name of this more general channel model is a discrete memoryless channel.
A coding scheme or code translates all of Alice’s messages into codewords that can be
input to n i.i.d. uses of the noisy channel. For example, suppose that Alice selects a message
m to encode. We can write the codeword corresponding to message m as xn (m) because theinput to the channel is some codeword that depends on m.
The last part of the model involves Bob receiving the corrupted codeword y n over thechannel and determining a potential codeword xn with which it should be associated. We
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do not get into any details just yet for this last decoding part—imagine for now that it
operates similarly to the majority vote code example. Figure 2.4 displays Shannon’s modelof communication that we have described.
We calculate the rate of a given coding scheme as follows:
rate ≡# of message bits.
# of channel uses

(2.58)

In our model, the rate of a given coding scheme isR=

1
log(M ),
n(2.59)

where log(M ) is the number of bits needed to represent any message in the message set [M ]and n is the number of channel uses. The capacity of a noisy channel is the highest rate at
which it can communicate information reliably.
We also need a way to determine the performance of any given code. Here, we list several
measures of performance. Let C ≡ {xn (m)}m∈[M ] represent a code that Alice and Bob choose,where xn (m) denotes each codeword corresponding to the message m. Let pe (m, C) denotethe probability of error when Alice transmits a message m ∈ [M ] using the code C. We
denote the average probability of error asM

1 X
pe (m, C).
p̄e (C) ≡M m=1(2.60)The maximal probability of error isp∗e (C) ≡ max pe (m, C).

(2.61)m∈[M ]

Our ultimate aim is to make the maximal probability of error p∗e (C) arbitrarily small, but
the average probability of error p̄e (C) is important in the analysis. These two performancemeasures are related—the average probability of error is small if the maximal probabilityof error is. Perhaps surprisingly, the maximal probability is small for at least half of themessages if the average probability of error is. We make this statement more quantitativein the following exercise.
Exercise 2.2.1 Let ε ∈ [0, 1/2] and let pe (m, C) denote the probability of error when Alicetransmits a message m ∈ [M ] using the code C. Use Markov’s inequality to prove that thefollowing upper bound on the average probability of error:
1 X
pe (m, C) ≤ ε
M m(2.62)

implies the following upper bound for at least half of the messages m:pe (m, C) ≤ 2ε.
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You may have wondered why we use the random sequence X n to model the inputs to thechannel. We have already stated that Alice’s message is a uniform random variable, and thecodewords in any coding scheme directly depend on the message to be sent. For example, inthe majority vote code, the channel inputs are always “000” whenever the intended message
is “0” and similarly for the channel inputs “111” and the message “1”. So why is there aneed to overcomplicate things by modeling the channel inputs as the random variable X nwhen it seems like each codeword is a deterministic function of the intended message? We
are not yet ready to answer this question but will return to it shortly.
We should also stress an important point before proceeding with Shannon’s ingenious
scheme for proving the existence of reliable codes for a noisy channel. In the above model,
we described essentially two “layers of randomness”:
1. The first layer of randomness is the uniform random variable associated with Alice’s
choice of a message.
2. The second layer of randomness is the noisy channel. The output of the channel isa random variable because we cannot always predict the output of the channel withcertainty.
It is not possible to “play around” with these two layers of randomness. The randomvariable associated with Alice’s message is fixed as a uniform random variable because we
assume ignorance of Alice’s message. The conditional probability distribution of the noisy
channel is also fixed. We are assuming that Alice and Bob can learn the conditional probability distribution associated with the noisy channel by estimating it. Alternatively, we
may assume that a third party has knowledge of the conditional probability distribution andinforms Alice and Bob of it in some way. Regardless of how they obtain the knowledge ofthe distribution, we assume that they both know it and that it is fixed.

2.2.4

Proof Sketch of Shannon’s Channel Coding Theorem

We are now ready to present an overview of Shannon’s technique for proving the existenceof a code that can achieve the capacity of a given noisy channel. Some of the methods
that Shannon uses in his outline of a proof are similar to those in the first coding theorem.
We again use the channel a large number of times so that the law of large numbers fromprobability theory comes into play and allow for a small probability of error that vanishes as
the number of channel uses becomes large. If the notion of typical sequences is so importantin the first coding theorem, we might suspect that it should be important in the noisy channel
coding theorem as well. The typical set captures a certain notion of efficiency because it isa small set when compared to the set of all sequences, but it is the set that has almost allof the probability. Thus, we should expect this efficiency to come into play somehow in thechannel coding theorem.
The aspect of Shannon’s technique for proving the noisy channel coding theorem that isdifferent from the other ideas in the first theorem is the idea of random coding. Shannon’sc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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technique adds a third layer of randomness to the model given above (recall that the firsttwo are Alice’s random message and the random nature of the noisy channel).
The third layer of randomness is to choose the codewords themselves in a random fashionaccording to a random variable X, where we choose each letter xi of a given codeword xn
independently according to the distribution pX (xi ). It is for this reason that we model thechannel inputs as a random variable. We can then write each codeword as a random variable
X n (m). The probability distribution for choosing a particular codeword xn (m) is
Pr {X n (m) = xn (m)} = pX1 ,X2 ,...,Xn (x1 (m), x2 (m), . . . , xn (m))= pX (x1 (m))pX (x2 (m)) · · · pX (xn (m))
nY=pX (xi (m)).

(2.64)(2.65)(2.66)

i=1

The important result to notice is that the probability for a given codeword factors because wechoose the code in an i.i.d. fashion, and perhaps more importantly, the distribution of eachcodeword has no explicit dependence on the message m with which it is associated. Thatis, the probability distribution of the first codeword is exactly the same as the probability
distribution of all of the other codewords. The code C itself becomes a random variable inthis scheme for choosing a code randomly. We now let C refer to the random variable that
represents a random code, and we let C0 represent any particular deterministic code. The
probability of choosing a particular code C0 = {xn (m)}m∈[M ] ispC (C0 ) =

M Y
n
Y

pX (xi (m)),(2.67)

m=1 i=1

and this probability distribution again has no explicit dependence on each message m in thecode C0 .
Choosing the codewords in a random way allows for a dramatic simplification in themathematical analysis of the probability of error. One of Shannon’s breakthrough ideas wasto analyze the expectation of the average probability of error, where the expectation is withrespect to the random code C, rather than analyzing the average probability of error itself.
The expectation of the average probability of error is
EC {p̄e (C)} .

(2.68)

This expectation is much simpler to analyze because of the random way that we choose thecode. Consider that
)
(
M
1 X
pe (m, C) .
(2.69)
EC {p̄e (C)} = EC
M m=1Using linearity of the expectation, we can exchange the expectation with the sum so thatM

EC {p̄e (C)} =c 20191 X
EC {pe (m, C)} .
M m=1(2.70)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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Now, the expectation of the probability of error for a particular message m does not actuallydepend on the message m because the distribution of each random codeword X n (m) does not
explicitly depend on m. This line of reasoning leads to the dramatic simplification because
EC {pe (m, C)} is then the same for all messages. So we can then say that
EC {pe (m, C)} = EC {pe (1, C)} .

(2.71)(We could have equivalently chosen any message instead of the first.) We then have thatM

1 X
EC {pe (1, C)}
EC {p̄e (C)} =M m=1= EC {pe (1, C)} ,(2.72)(2.73)

where the last step follows because the quantity EC {pe (1, C)} has no dependence on m.
We now only have to determine the expectation of the probability of error for one message
instead of determining the expectation of the average error probability of the whole set. This
simplification follows because random coding results in the equality of these two quantities.
Shannon then determined a way to obtain a bound on the expectation of the average
probability of error (we soon discuss this technique briefly) so that
EC {p̄e (C)} ≤ ε,

(2.74)

where ε is some number ∈ (0, 1) that we can make arbitrarily small by letting the block
size n become arbitrarily large. If it is possible to obtain a bound on the expectation of
the average probability of error, then surely there exists some deterministic code C0 whose
average probability of error meets this same bound:p̄e (C0 ) ≤ ε.

(2.75)

If it were not so, then the original bound on the expectation would not be possible. This stepis the derandomization step of Shannon’s proof. Ultimately, we require a deterministic codewith a high rate and arbitrarily small probability of error and this step shows the existenceof such a code. The random coding technique is only useful for simplifying the mathematicsof the proof.
The last step of the proof is the expurgation step. It is an application of the result of
Exercise 2.2.1. Recall that our goal is to show the existence of a high-rate code that has low
maximal probability of error. But so far we only have a bound on the average probabilityof error. In the expurgation step, we simply throw out the half of the codewords with theworst probability of error. Throwing out the worse half of the codewords reduces the numberof messages by a factor of two, but only has a negligible impact on the rate of the code.
Consider that the number of messages is 2nR where R is the rate of the code. Thus, the1number of messages is 2n(R− n ) after throwing out the worse half of the codewords, and therate R − n1 is asymptotically equal to the rate R. After throwing out the worse half of thec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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Figure 2.5: This figure depicts the notion of a conditionally typical set. Associated to every input sequencexn is a conditionally typical set consisting of the likely output sequences. The size of this conditionallytypical set is ≈ 2nH(Y |X) . It is exponentially smaller than the set of all output sequences whenever theconditional random variable is not uniform.

codewords, the result of Exercise 2.2.1 shows that the following bound then applies to the
maximal probability of error:p∗e (C0 ) ≤ 2ε.
(2.76)
This last expurgation step ends the analysis of the probability of error.
We now discuss the size of the code that Alice and Bob employ. Recall that the rate of
the code is R = log(M )/n. It is convenient to define the size M of the message set [M ] interms of the rate R. When we do so, the size of the message set isM = 2nR .

(2.77)

What is peculiar about the message set size when defined this way is that it grows exponentially with the number of channel uses. But recall that any given code exploits n channel
uses to send M messages. So when we take the limit as the number of channel uses tends
to infinity, we are implying that there exists a sequence of codes whose message set size is
M = 2nR and number of channel uses is n. We are focused on keeping the rate of the code
constant and use the limit n → ∞ to make the probability of error vanish for a certain fixed
rate R.
What is the maximal rate at which Alice can communicate to Bob reliably? We needto determine the number of distinguishable messages that Alice can reliably send to Bob,
and we require the notion of conditional typicality to do so. Consider that Alice chooses
codewords randomly according to random variable X with probability distribution pX (x).
By the asymptotic equipartition theorem, it is highly likely that each of the codewords
that Alice chooses is a typical sequence with sample entropy close to H(X). In the coding
scheme, Alice transmits a particular codeword xn over the noisy channel and Bob receives
a random sequence Y n . The random sequence Y n is a random variable that depends on xn
through the conditional probability distribution pY |X (y|x). We would like a way to determine
the number of possible output sequences that are likely to correspond to a particular input
sequence xn . A useful entropic quantity for this situation is the conditional entropy H(Y |X),
the technical details of which we leave for Chapter 10. For now, just think of this conditional
entropy as measuring the uncertainty of a random variable Y when one already knows thec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

2.2. CHANNEL CAPACITY

61

Figure 2.6: This figure depicts the packing argument that Shannon used. The channel induces a conditionally typical set corresponding to each codeword xn (i) where i ∈ {1, . . . , M }. The size of each conditionallytypical output set is 2nH(Y |X) . The size of the typical set of all output sequences is 2nH(Y ) . These sizes
suggest that we can divide the output typical set into M conditionally typical sets and be able to distinguishM ≈ 2nH(Y ) /2nH(Y |X) messages without error.

value of the random variable X. The conditional entropy H(Y |X) is always less than theentropy H(Y ) unless X and Y are independent. This inequality holds because knowledgeof a correlated random variable X does not increase the uncertainty about Y . It turns
out that there is a notion of conditional typicality (depicted in Figure 2.5), similar to thenotion of typicality, and a similar asymptotic equipartition theorem holds for conditionallytypical sequences (more details in Section 14.9). This theorem also has three important
properties. For each input sequence xn , there is a corresponding conditionally typical set
with the following properties:
1. It has almost all of the probability—it is highly likely that a random channel output
sequence is conditionally typical given a particular input sequence.
2. Its size is ≈ 2nH(Y |X) .
3. The probability of each conditionally typical sequence y n , given knowledge of the input
sequence xn , is ≈ 2−nH(Y |X) .
If we disregard knowledge of the input sequence used to generate an output sequence,the probability distribution that generates the output sequences isX
pY (y) =pY |X (y|x)pX (x).
(2.78)x

We can think that this probability distribution is the one that generates all the possible
output sequences. The likely output sequences are in an output typical set of size 2nH(Y ) .
We are now in a position to describe the structure of a random code and the size of themessage set. Alice generates 2nR codewords according to the distribution pX (x) and suppose
for now that Bob has knowledge of the code after Alice generates it. Suppose Alice sends
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CHAPTER 2. CLASSICAL SHANNON THEORY

one of the codewords over the channel. Bob is ignorant of the transmitted codeword, so fromhis point of view, the output sequences are generated according to the distribution pY (y).
Bob then employs typical sequence decoding. He first determines if the output sequence y n
is in the typical output set of size 2nH(Y ) . If not, he declares an error. The probability of thistype of error is small by the asymptotic equipartition theorem. If the output sequence y n
is in the output typical set, he uses his knowledge of the code to determine a conditionallytypical set of size 2nH(Y |X) to which the output sequence belongs. If he decodes an output
sequence y n to the wrong conditionally typical set, then an error occurs. This last type oferror suggests how they might structure the code in order to prevent this type of error fromhappening. If they structure the code so that the output conditionally typical sets do not
overlap too much, then Bob should be able to decode each output sequence y n to a uniqueinput sequence xn with high probability. This line of reasoning suggests that they should
divide the set of output typical sequences into M sets of conditionally typical output sets,
each of size 2nH(Y |X) . Thus, if they set the number of messages M = 2nR as follows:2nR ≈2nH(Y )= 2n(H(Y )−H(Y |X)) ,2nH(Y |X)

(2.79)

then our intuition is that Bob should be able to decode correctly with high probability. Suchan argument is a “packing” argument because it shows how to pack information into thespace of all output sequences. Figure 2.6 gives a visual depiction of the packing argument.
It turns out that this intuition is correct—Alice can reliably send information to Bob if thequantity H(Y ) − H(Y |X) bounds the rate R:
R < H(Y ) − H(Y |X).

(2.80)

A rate less than H(Y ) − H(Y |X) ensures that we can make the expectation of the average
probability of error as small as we would like. We then employ the derandomization andexpurgation steps, discussed before, in order to show that there exists a code whose maximal
probability of error vanishes as the number n of channel uses tends to infinity.
The entropic quantity H(Y ) − H(Y |X) deserves special attention because it is anotherimportant entropic quantity in information theory. It is the mutual information between
random variables X and Y and we denote it asI(X; Y ) ≡ H(Y ) − H(Y |X).

(2.81)

It is important because it arises as the limiting rate of reliable communication. We will
discuss its properties in more detail throughout this book.
There is one final step that we can take to strengthen the above coding scheme. We
mentioned before that there are three layers of randomness in the coding construction: Alice’s
uniform choice of a message, the noisy channel, and Shannon’s random coding scheme. The
first two layers of randomness we do not have control over. But we actually do have control
over the last layer of randomness. Alice chooses the code according to the distribution pX (x).
She can choose the code according to any distribution that she would like. If she chooses it
according to pX (x), the resulting rate of the code is the mutual information I(X; Y ). We willc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

2.2. CHANNEL CAPACITY

63

prove later on that the mutual information I(X; Y ) is a concave function of the distributionpX (x) when the conditional distribution pY |X (y|x) is fixed. Concavity implies that there isa distribution p∗X (x) that maximizes the mutual information. Thus, Alice should choose an
optimal distribution p∗X (x) when she randomly generates the code, and this choice gives thelargest possible rate of communication that they could have. This largest possible rate is
the capacity of the channel and we denote it asC(N ) ≡ max I(X; Y ).

(2.82)pX (x)

Our discussion here is just an overview of Shannon’s channel capacity theorem. In Section 14.10, we give a full proof of this theorem after having developed some technical tools
needed for a formal proof.
We clarify one more point. In the discussion of the operation of the code, we mentionedthat Alice and Bob both have knowledge of the code. Well, how can Bob know the code if anoisy channel connects Alice to Bob? One solution to this problem is to assume that Aliceand Bob have unbounded computation on their local ends. Thus, for a given code that uses
the channel n times, they can both compute the above optimization problem and generate“test” codes randomly until they determine the best possible code to employ for n channel
uses. They then both end up with the unique, best possible code for n uses of the given
channel. This scheme might be impractical, but nevertheless, it provides a justification forboth of them to have knowledge of the code that they use. Another solution to this problem
is simply to allow them to meet before going their separate ways in order to coordinate on
the choice of code.
We have said before that the capacity C(N ) is the maximal rate at which Alice and Bob
can communicate. But in our discussion above, we did not prove optimality—we only proveda direct coding theorem for the channel capacity theorem. It took quite some time and effort
to develop this elaborate coding procedure—along the way, we repeatedly invoked one of thepowerful tools from probability theory, the law of large numbers. It perhaps seems intuitivethat typical sequence coding and decoding should lead to optimal code constructions. Typicalsequences exhibit some kind of asymptotic efficiency by being the most likely to occur, but inthe general case, their cardinality is exponentially smaller than the set of all sequences. But isthis intuition about typical sequence coding correct? Is it possible that some other scheme forcoding might beat this elaborate scheme that Shannon devised? Without a converse theorem
that proves optimality, we would never know! If you recall from our previous discussion in
Section 2.1.3 about coding theorems, we stressed how important it is to prove a converse
theorem that matches the rate that the direct coding theorem suggests is optimal. For now,
we delay the proof of the converse theorem because the tools for proving it are much different
from the tools we described in this section. For now, accept that the formula in (2.82) is
indeed the optimal rate at which two parties can communicate and we will prove this resultin a later chapter.
We end the description of Shannon’s channel coding theorem by summarizing the statements of the direct coding theorem and the converse theorem. The statement of the directcoding theorem is as follows:c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 2. CLASSICAL SHANNON THEORY

If the rate of communication is less than the channel capacity, then it is possible
for Alice to communicate reliably to Bob, in the sense that a sequence of codes
exists whose maximal probability of error vanishes as the number of channel uses
tends to infinity.
The statement of the converse theorem is as follows:
If a reliable sequence of codes exists, then the rate of this sequence of codes isless than the channel capacity.
Another way of stating the converse proves to be useful later on:
If the rate of a coding scheme is greater than the channel capacity, then a reliable
code does not exist, in the sense that the error probability of the coding scheme
is bounded away from zero.

2.3 Summary
A general communication scenario involves one sender and one receiver. In the classicalsetting, we discussed two information-processing tasks that they can perform. The first task
was data compression or source coding, and we assumed that the sender and receiver are
linked together by a noiseless classical bit channel that they can use a large number of times.
We can think of this noiseless classical bit channel as a noiseless dynamic resource that thetwo parties share. The resource is dynamic because we assume that there is some physicalmedium through which the physical carrier of information travels in order to get from thesender to the receiver. It was our aim to count the number of times they would have to use
the noiseless resource in order to send information reliably. The result of Shannon’s source
coding theorem is that the entropy gives the minimum rate at which they have to use thenoiseless resource. The second task we discussed was channel coding and we assumed thatthe sender and receiver are linked together by a noisy classical channel that they can use alarge number of times. This noisy classical channel is a noisy dynamic resource that theyshare. We can think of this information-processing task as a simulation task, where the goal
is to simulate a noiseless dynamic resource by using a noisy dynamic resource in a redundantway. This redundancy is what allows Alice to communicate reliably to Bob, and reliablecommunication implies that they have effectively simulated a noiseless resource. We again
had a resource count for this case, where we counted n as the number of times they use thenoisy resource and nC is the number of noiseless bit channels they simulate (where C is thecapacity of the channel). This notion of resource counting may not seem so important forthe classical case, but it becomes much more important for the quantum case.
We now conclude our overview of Shannon’s information theory. The main points to takehome from this overview are the ideas that Shannon employed for constructing source andchannel codes. We let the information source emit a large sequence of data, or similarly,
we use the channel a large number of times so that we can invoke the law of large numbers
c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License2.3. SUMMARY

65

from probability theory. The result is that we can show vanishing error for both schemes bytaking a limit. In Chapter 14, we develop the theory of typical sequences in detail, proving
many of the results taken for granted in this overview.
In hindsight, Shannon’s methods for proving the two coding theorems are merely a tour
de force for one idea from probability theory: the law of large numbers. Perhaps, thisviewpoint undermines the contribution of Shannon, until we recall that no one else had
even come close to devising these methods for data compression and channel coding. The
theoretical development of Shannon is one of the most important contributions to modern
science because his theorems determine the ultimate rate at which we can compress andcommunicate classical information.

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c 2019CHAPTER 2. CLASSICAL SHANNON THEORYMark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicensePart IIThe Quantum Theory

67

CHAPTER 3

The Noiseless Quantum Theory
The simplest quantum system is the physical quantum bit or qubit. The qubit is a two-level
quantum system—example qubit systems are the spin of an electron, the polarization of aphoton, or a two-level atom with a ground state and an excited state. We do not worry too
much about physical implementations in this chapter, but instead focus on the mathematicalpostulates of the quantum theory and operations that we can perform on qubits. From qubits
we progress to a study of physical qudits. Qudits are quantum systems that have d levels
and are an important generalization of qubits. Again, we do not discuss physical realizationsof qudits.
Noise can affect quantum systems, and we must understand methods of modeling noise in
the quantum theory because our ultimate aim is to construct schemes for protecting quantum
systems against the detrimental effects of noise. In Chapter 1, we remarked on the different
types of noise that occur in nature. The first, and perhaps more easily comprehensible typeof noise, is that which is due to our lack of information about a given scenario. We observe
this type of noise in a casino, with every shuffle of cards or toss of dice. These events are
random, and the random variables of probability theory model them because the outcomesare unpredictable. This noise is the same as that in all classical information-processing
systems.
On the other hand, the quantum theory features a fundamentally different type of noise.
Quantum noise is inherent in nature and is not due to our lack of information, but is due
rather to nature itself. An example of this type of noise is the “Heisenberg noise” that
results from the uncertainty principle. If we know the momentum of a given particle fromperforming a precise measurement of it, then we know absolutely nothing about its position—a measurement of its position gives a random result. Similarly, if we know the rectilinear
polarization of a photon by precisely measuring it, then a future measurement of its diagonal
polarization will give a random result. It is important to keep the distinction clear between
these two types of noise.
We explore the postulates of the quantum theory in this chapter, by paying particularattention to qubits. These postulates apply to a closed quantum system that is isolated
from everything else in the universe. We label this first chapter “The Noiseless Quantum69

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CHAPTER 3. THE NOISELESS QUANTUM THEORY

Theory” because closed quantum systems do not interact with their surroundings and arethus not subject to corruption and information loss. Interaction with surrounding systems
can lead to loss of information in the sense of the classical noise that we described above.
Closed quantum systems do undergo a certain type of quantum noise, such as that from theuncertainty principle and the act of measurement, because they are subject to the postulatesof the quantum theory. The name “noiseless quantum theory” thus indicates the closed, ideal
nature of the quantum systems discussed.
This chapter introduces the four postulates of the quantum theory. The mathematical
tools of the quantum theory rely on the fundamentals of linear algebra—vectors and matricesof complex numbers. It may seem strange at first that we need to incorporate the machineryof linear algebra in order to describe a physical system in the quantum theory, but it turns outthat this description uses the simplest set of mathematical tools to predict the phenomenathat a quantum system exhibits. A hallmark of the quantum theory is that certain operations
do not commute with one another, and matrices are the simplest mathematical objects that
capture this idea of non-commutativity.

3.1 Overview
We first briefly overview how information is processed with quantum systems. This usuallyconsists of three steps: state preparation, quantum operations, and measurement. Statepreparation is the initialization of a quantum system to some beginning state, depending on
what operation we would like a quantum system to execute. There could be some classicalcontrol device that initializes the state of the quantum system. Observe that the input system
for this step is a classical system, and the output system is quantum. After initializing thestate of the quantum system, we perform some quantum operations that evolve its state. This
stage is where we can take advantage of quantum effects for enhanced information-processing
abilities. Both the input and output systems of this step are quantum. Finally, we need some
way of reading out the result of the computation, and we can do so with a measurement.
The input system for this step is quantum, and the output is classical. Figure 3.1 depicts
all of these steps. In a quantum communication protocol, spatially separated parties may
execute different parts of these steps, and we are interested in keeping track of the non-local
resources needed to implement a communication protocol. Section 3.2 describes quantum
states (and thus state preparation), Section 3.3 describes the noiseless evolution of quantum
states, and Section 3.4 describes “read out” or measurement. For now, we assume that wecan perform all of these steps perfectly and later chapters discuss how to incorporate theeffects of noise.

3.2 Quantum Bits
The simplest quantum system is a two-state system: a physical qubit. Let |0i denote one
possible state of the system. The left vertical bar and the right angle bracket indicate that wec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License3.2. QUANTUM BITS

71

Figure 3.1: All of the steps in a typical noiseless quantum information processing protocol. A classicalcontrol (depicted by the thick black line on the left) initializes the state of a quantum system. The quantum
system then evolves according to some unitary operation (described in Section 3.3). The final step is ameasurement that reads out some classical data m from the quantum system.

are using the Dirac notation to represent this state. The Dirac notation has some advantages
for performing calculations in the quantum theory, and we highlight some of these advantages
as we progress through our development. Let |1i denote another possible state of the qubit.
We can encode a classical bit or cbit into a qubit with the following mapping:0 → |0i,

1 → |1i.

(3.1)

So far, nothing in our description above distinguishes a classical bit from a qubit, except
for the funny vertical bar and angle bracket that we place around the bit values. However,the quantum theory predicts that the above states are not the only possible states of aqubit. Arbitrary superpositions (linear combinations) of the above two states are possible
as well because the quantum theory is a linear theory. Suffice it to say that the linearityof the quantum theory results from the linearity of Schrödinger’s equation that governs theevolution of quantum systems.1 A general noiseless qubit can be in the following state:|ψi ≡ α|0i + β|1i,(3.2)

where the coefficients α and β are arbitrary complex numbers with unit norm: |α|2 +|β|2 = 1.
The coefficients α and β are probability amplitudes—they are not probabilities themselves,
but they do allow us to calculate probabilities. The unit-norm constraint leads to the Bornrule (the probabilistic interpretation) of the quantum theory, and we speak more on thisconstraint and probability amplitudes when we introduce the measurement postulate.
The possibility of superposition states indicates that we cannot represent the states |0i
and |1i with the Boolean algebra of the respective classical bits 0 and 1 because Boolean
algebra does not allow for superposition states. We instead require the mathematics of linear
algebra to describe these states. It is beneficial at first to define a vector representation of
the states |0i and |1i:
 
 
1
0
|0i ≡
,
|1i ≡
.
(3.3)
0
1
The |0i and |1i states are called “kets” in the language of the Dirac notation, and it is bestat first to think of them merely as column vectors. The superposition state in (3.2) then has
1

We will not present Schrödinger’s equation in this book, but instead focus on a “quantum information”
presentation of the quantum theory. Griffith’s book on quantum mechanics introduces the quantum theory
from the Schrödinger equation if you are interested (Griffiths, 1995).
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72

CHAPTER 3. THE NOISELESS QUANTUM THEORYa representation as the following two-dimensional vector: α|ψi = α|0i + β|1i =
.
β

(3.4)

The representation of quantum states with vectors is helpful in understanding some of themathematics that underpins the theory, but it turns out to be much more useful for our
purposes to work directly with the Dirac notation. We give the vector representation for
now, but later on, we will exclusively employ the Dirac notation.
The Bloch sphere, depicted in Figure 3.2, gives a valuable way to visualize a qubit.
Consider any two qubits that are equivalent up to a differing global phase. For example,
these two qubits could be|ψ0 i ≡ |ψi,
|ψ1 i ≡ eiχ |ψi,
(3.5)
where 0 ≤ χ < 2π. These two qubits are physically equivalent because they give thesame physical results when we measure them (more on this point when we introduce themeasurement postulate in Section 3.4). Suppose that the probability amplitudes α and βhave the following representations as complex numbers:
α = r0 eiϕ0 ,β = r1 eiϕ1 .

(3.6)

We can factor out the phase eiϕ0 from both coefficients α and β, and we still have a statethat is physically equivalent to the state in (3.2):|ψi ≡ r0 |0i + r1 ei(ϕ1 −ϕ0 ) |1i,

(3.7)

where we redefine |ψi to represent the state because of the equivalence mentioned in (3.5).
Let ϕ ≡ ϕ1 − ϕ0 , where 0 ≤ ϕ < 2π. Recall that the unit-norm constraint requires |r0 |2 +|r1 |2 = 1. We can thus parametrize the values of r0 and r1 in terms of one parameter θ sothat
r0 = cos(θ/2),
r1 = sin(θ/2).
(3.8)
The parameter θ varies between 0 and π. This range of θ and the factor of two give a uniquerepresentation of the qubit. One may think to have θ vary between 0 and 2π and omit thefactor of two, but this parametrization would not uniquely characterize the qubit in termsof the parameters θ and ϕ. The parametrization in terms of θ and ϕ gives the Bloch sphere
representation of the qubit in (3.2):|ψi ≡ cos(θ/2)|0i + sin(θ/2)eiϕ |1i.

(3.9)

In linear algebra, column vectors are not the only type of vectors—row vectors are usefulas well. Is there an equivalent of a row vector in Dirac notation? The Dirac notation provides
an entity called a “bra,” that has a representation as a row vector. The bras corresponding
to the kets |0i and |1i are as follows:
h0| ≡ 1 0 ,h1| ≡ 0 1 ,(3.10)
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73

Figure 3.2: The Bloch sphere representation of a qubit. Any qubit |ψi admits a representation in termsof two angles θ and ϕ where 0 ≤ θ ≤ π and 0 ≤ ϕ < 2π. The state of any qubit in terms of these angles is|ψi = cos(θ/2)|0i + eiϕ sin(θ/2)|1i.

and are the matrix conjugate transpose of the kets |0i and |1i:h0| = (|0i)† ,h1| = (|1i)† .

(3.11)We require the conjugate transpose operation (as opposed to just the transpose) because themathematical representation of a general quantum state can have complex entries.
The bras do not represent quantum states, but are helpful in calculating probability
amplitudes. For our example qubit in (3.2), suppose that we would like to determine theprobability amplitude that the state is |0i. We can combine the state in (3.2) with the bra
h0| as follows:h0||ψi = h0| (α|0i + β|1i)= αh0||0i + βh0||1i
 
 

 1

 0
=α 1 0+β 1 001=α·1+β·0= α.

(3.12)(3.13)(3.14)(3.15)(3.16)

The above calculation may seem as if it is merely an exercise in linear algebra, with a“glorified” Dirac notation, but it is a standard calculation in the quantum theory. A quantity
like h0||ψi occurs so often in the quantum theory that we abbreviate it ash0|ψi ≡ h0||ψi,c 2019(3.17)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 3. THE NOISELESS QUANTUM THEORY

and the above notation is known as a “braket.”2 The physical interpretation of the quantityh0|ψi is that it is the probability amplitude for being in the state |0i, and likewise, thequantity h1|ψi is the probability amplitude for being in the state |1i. We can also determinethat the amplitude h1|0i (for the state |0i to be in the state |1i) and the amplitude h0|1i
are both equal to zero. These two states are orthogonal states because they have no overlap.
The amplitudes h0|0i and h1|1i are both equal to one by following a similar calculation.
Our next task may seem like a frivolous exercise, but we would like to determine theamplitude for any state |ψi to be in the state |ψi, i.e., to be itself. Following the above
method, this amplitude is hψ|ψi and we calculate it ashψ|ψi = (h0|α∗ + h1|β ∗ ) (α|0i + β|1i)= α∗ α h0|0i + β ∗ α h1|0i + α∗ β h0|1i + β ∗ β h1|1i= |α|2 + |β|2
= 1,

(3.18)(3.19)(3.20)(3.21)

where we have used the orthogonality relations of h0|0i, h1|0i, h0|1i, and h1|1i, and theunit-norm constraint. We also write this in terms of the Euclidean norm of |ψi asp(3.22)
k|ψik2 ≡ hψ|ψi = 1.
We come back to the unit-norm constraint in our discussion of quantum measurement, but
for now, we have shown that any quantum state has a unit amplitude for being itself.
The states |0i and |1i are a particular basis for a qubit that we call the computational
basis. The computational basis is the standard basis that we employ in quantum computationand communication, but other bases are important as well. Consider that the following two
vectors form an orthonormal basis:
 


1
1
1
1
√
√
,
.
(3.23)
2 1
2 −1
The above alternate basis is so important in quantum information theory that we define a
Dirac notation shorthand for it, and we can also define the basis in terms of the computational
basis:|0i − |1i|0i + |1i|+i ≡ √,|−i ≡ √.
(3.24)
2
2
The common names for this alternate basis are the “+/−” basis, the Hadamard basis, orthe diagonal basis. It is preferable for us to use the Dirac notation, but we are using thevector representation as an aid for now.
Exercise 3.2.1 Determine the Bloch sphere angles θ and ϕ for the states |+i and |−i.
2

It is for this (somewhat silly) reason that Dirac decided to use the names “bra” and “ket,” because
putting them together gives a “braket.” The names in the notation may be silly, but the notation itself has
persisted over time because this way of representing quantum states turns out to be useful. We will avoid
the use of the terms “bra” and “ket” as much as we can, only resorting to these terms if necessary.
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75

What is the amplitude that the state in (3.2) is in the state |+i? What is the amplitude
that it is in the state |−i? These are questions to which the quantum theory provides simple
answers. We employ the bra h+| and calculate the amplitude h+|ψi ash+|ψi = h+| (α|0i + β|1i)= α h+|0i + β h+|1i
α+β
= √ .
2

(3.25)(3.26)(3.27)The result follows by employing the definition in (3.24) and doing similar linear algebraic
calculations as the example in (3.16). We can also calculate the amplitude h−|ψi as
α−βh−|ψi = √ .
2(3.28)

The above calculation follows from similar manipulations.
The +/− basis is a complete orthonormal basis, meaning that we can represent any qubit
state in terms of the two basis states |+i and |−i. Indeed, the above probability amplitude
calculations and the fact that the +/− basis is complete imply that we can represent thequbit in (3.2) as the following superposition state:




α−β
α+β
√
√
|+i +
|−i .
(3.29)|ψi =2
2
The above representation is an alternate one if we would like to “see” the qubit state represented in the +/− basis. We can substitute the equalities in (3.27) and (3.28) to represent
the state |ψi as|ψi = h+|ψi |+i + h−|ψi |−i.
(3.30)
The amplitudes h+|ψi and h−|ψi are both scalar quantities so that the above quantity isequal to the following one:
|ψi = |+ih+|ψi + |−ih−|ψi.
(3.31)The order of the multiplication in the terms |+ih+|ψi and |−ih−|ψi does not matter, i.e.,
the following equality holds:|+i (h+|ψi) = (|+ih+|) |ψi,

(3.32)and the same for |−i h−|ψi. The quantity on the left is a ket multiplied by an amplitude,
whereas the quantity on the right is a linear operator multiplying a ket, but linear algebra
tells us that these two quantities are equal. The operators |+ih+| and |−ih−| are special
operators—they are rank-one projection operators, meaning that they project onto a onedimensional subspace. Using linearity, we have the following equality:|ψi = (|+ih+| + |−ih−|) |ψi.
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CHAPTER 3. THE NOISELESS QUANTUM THEORY

The above equation indicates a seemingly trivial, but important point—the operator |+ih+|+|−ih−| is equal to the identity operator and we can write
I = |+ih+| + |−ih−|,(3.34)

where I stands for the identity operator. This relation is known as the completeness relationor the resolution of the identity. Given any orthonormal basis, we can always construct aresolution of the identity by summing over the rank-one projection operators formed fromeach of the orthonormal basis states. For example, the computational basis states give
another way to form a resolution of the identity operator:
I = |0ih0| + |1ih1|.

(3.35)

This simple trick provides a way to find the representation of a quantum state in any basis.

3.3 Reversible Evolution
Physical systems evolve as time progresses. The application of a magnetic field to an electron
can change its spin and pulsing an atom with a laser can excite one of its electrons from aground state to an excited state. These are only a couple of ways in which physical systems
can change.
The Schrödinger equation governs the evolution of a closed quantum system. In this book,we will not even state the Schrödinger equation, but we will instead focus on an important
implication of it. The evolution of a closed quantum system is reversible if we do not learn
anything about the state of the system (that is, if we do not measure it). Reversibility implies
that we can determine the input state of an evolution given the output state and knowledgeof the evolution. An example of a single-qubit reversible operation is a NOT gate:|0i → |1i,

|1i → |0i.

(3.36)In the classical world, we would say that the NOT gate merely flips the value of the input
classical bit. In the quantum world, the NOT gate flips the basis states |0i and |1i. The
NOT gate is reversible because we can simply apply the NOT gate again to recover theoriginal input state—the NOT gate is its own inverse.
In general, a closed quantum system evolves according to a unitary operator U . Unitaryevolution implies reversibility because a unitary operator always possesses an inverse—its
inverse is merely U † , the conjugate transpose. This property gives the relations:U † U = U U † = I.

(3.37)The unitary property also ensures that evolution preserves the unit-norm constraint (animportant requirement for a physical state that we discuss in Section 3.4). Consider applying
the unitary operator U to the example qubit state in (3.2): U |ψi. Figure 3.3 depicts aquantum circuit diagram for unitary evolution.
c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

3.3. REVERSIBLE EVOLUTION

77

Figure 3.3: A quantum circuit diagram that depicts the evolution of a quantum state |ψi according to aunitary operator U .

The bra that is dual to the above state is hψ|U † (we again apply the conjugate transpose
operation to get the bra). We showed in (3.18)–(3.21) that every quantum state should havea unit amplitude for being itself. This relation holds for the state U |ψi because the operator
U is unitary:hψ|U † U |ψi = hψ|I|ψi = hψ|ψi = 1.
(3.38)
The assumption that a vector always has a unit amplitude for being itself is one of the
crucial assumptions of the quantum theory, and the above reasoning demonstrates thatunitary evolution complements this assumption.
Exercise 3.3.1 A linear operator T is norm preserving if kT |ψik2 = k|ψik2 holds for all
quantum states |ψi (unit vectors), where the Euclidean norm is defined in (3.22). Prove thatan operator T is an isometry (satisfying T † T = I) if and only if it is norm preserving. Hint:
For showing the “only-if ” part, consider using the polarization identity:hψ|φi =

3.3.1

1
k|ψi + |φik22 − k|ψi − |φik22 + ik|ψi + i|φik22 − ik|ψi − i|φik22 .
4(3.39)

Matrix Representations of Operators

We now explore some properties of the NOT gate. Let X denote the operator corresponding
to a NOT gate. The action of X on the computational basis states is as follows:
X|ii = |i ⊕ 1i ,(3.40)

where i = {0, 1} and ⊕ denotes binary addition. Suppose the NOT gate acts on a superposition state:
X (α|0i + β|1i) .
(3.41)
By linearity of the quantum theory, the X operator distributes so that the above expression
is equal to the following one:αX|0i + βX|1i = α|1i + β|0i.

(3.42)

Indeed, the NOT gate X merely flips the basis states of any quantum state when represented
in the computational basis.
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78

CHAPTER 3. THE NOISELESS QUANTUM THEORY

We can determine a matrix representation for the operator X by using the bras h0| andh1|. Consider the relations in (3.40). Let us combine the relations with the bra h0|:
h0|X|0i = h0|1i = 0,

h0|X|1i = h0|0i = 1.

(3.43)

h1|X|1i = h1|0i = 0.

(3.44)

Likewise, we can combine with the bra h1|:
h1|X|0i = h1|1i = 1,

We can place these entries in a matrix to give a matrix representation of the operator X:
h0|X|0i h0|X|1i,(3.45)
h1|X|0i h1|X|1i
where we order the rows according to the bras and order the columns according to the kets.
We then say that


0 1
X=
,
(3.46)
1 0
and adopt the convention that the symbol X refers to both the operator X and its matrix
representation (this is an abuse of notation, but it should be clear from the context when X
refers to an operator and when it refers to the matrix representation of the operator).
Let us now observe some uniquely quantum behavior. We would like to consider theaction of the NOT operator X on the +/− basis. First, let us consider what happens√if we
operate on the |+i state with the X operator. Recall that the state |+i = (|0i + |1i) / 2 sothat
|0i + |1i
|1i + |0i
X|0i + X|1i√√
X|+i = X= √= |+i.
(3.47)=
2
2
2
The above development shows that the state |+i is a special state with respect to the NOT
operator X—it is an eigenstate of X with eigenvalue one. An eigenstate of an operator is one
that is invariant under the action of the operator. The coefficient in front of the eigenstateis the eigenvalue corresponding to the eigenstate. Under a unitary evolution, the coefficientin front of the eigenstate is just a complex phase, but this global phase has no effect on
the observations resulting from a measurement of the state because two quantum states areequivalent up to a differing global phase.
Now, let us consider
the action of the NOT operator X on the state |−i. Recall that√|−i = (|0i − |1i) / 2. Calculating similarly, we get that
|0i − |1i
X|0i − X|1i
|1i − |0i√√
X|−i = X== √= −|−i.
(3.48)
2
2
2
So the state |−i is also an eigenstate of the operator X, but its eigenvalue is −1.
We can find a matrix representation of the X operator in the +/− basis as well: 
h+|X|+i h+|X |−i
1 0=.
(3.49)
h−|X|+i h−|X|−i
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3.3. REVERSIBLE EVOLUTION

79This representation demonstrates that the X operator is diagonal with respect to the +/−basis, and therefore, the +/− basis is an eigenbasis for the X operator. It is always handy
to know the eigenbasis of a unitary operator U because this eigenbasis gives the states thatare invariant under an evolution according to U .
Let Z denote the operator that flips states in the +/− basis:Z|+i → |−i,Z|−i → |+i.

(3.50)

Using an analysis similar to that which we did for the X operator, we can find a matrix
representation of the Z operator in the +/− basis: 
h+|Z|+i h+|Z |−i
0 1=.
(3.51)
h−|Z|+i h−|Z|−i1 0Interestingly, the matrix representation for the Z operator in the +/− basis is the same asthat for the X operator in the computational basis. For this reason, we call the Z operator
the phase-flip operator.3
We expect the following steps to hold because the quantum theory is a linear theory:|−i + |+i|+i + |−iZ|+i + Z|−i|+i + |−i
√
√
√
√
=
=
,
(3.52)
=
Z
2
2
2
2




|+i − |−iZ|+i − Z|−i|−i − |+i|+i − |−i
√
√
√
√
Z
=
=
=−
.
(3.53)
2
2
2
2√√
The above steps demonstrate that the states (|+i + |−i) / 2 and (|+i − |−i) / 2 are botheigenstates of the Z operators. These states are none other than the respective computational
basis states |0i and |1i, by inspecting the definitions in (3.24). Thus, a matrix representationof the Z operator in the computational basis is h0|Z|0i h0|Z|1i
1 0=,(3.54)
h1|Z|0i h1|Z|1i
0 −1and is a diagonalization of the operator Z. So, the behavior of the Z operator in thecomputational basis is the same as the behavior of the X operator in the +/− basis.

3.3.2

Commutators and Anticommutators

The commutator [A, B] of two operators A and B is as follows:[A, B] ≡ AB − BA.

(3.55)

Two operators commute if and only if their commutator is equal to zero.
The anticommutator {A, B} of two operators A and B is as follows:{A, B} ≡ AB + BA.

(3.56)

We say that two operators anticommute if their anticommutator is equal to zero.
Exercise 3.3.2 Find a matrix representation for [X, Z] in the basis {|0i, |1i}.
3

A more appropriate name might be the “bit flip in the +/− basis operator,” but this name is too long,so we stick with the term “phase flip.”c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

80

3.3.3

CHAPTER 3. THE NOISELESS QUANTUM THEORY

The Pauli Matrices

The convention in quantum theory is to take the computational basis as the standard basis
for representing physical qubits. The standard matrix representation for the above two
operators is as follows when we choose the computational basis as the standard basis:




0 1
1 0
X≡
, Z≡
.
(3.57)
1 0
0 −1
The identity operator I has the following representation in any basis:


1 0
I≡
.
0 1

(3.58)

Another operator, the Y operator, is a useful one to consider as well. The Y operator has
the following matrix representation in the computational basis:


0 −i
Y ≡
.
(3.59)
i 0
It is easy to check that Y = iXZ, and for this reason, we can think of the Y operator
as a combined bit and phase flip. The four matrices I, X, Y , and Z are special for the
manipulation of physical qubits and are known as the Pauli matrices.
Exercise 3.3.3 Show that the Pauli matrices are all Hermitian, unitary, they square to theidentity, and their eigenvalues are ±1.
Exercise 3.3.4 Represent the eigenstates of the Y operator in the computational basis.
Exercise 3.3.5 Show that the Pauli matrices either commute or anticommute.
Exercise 3.3.6 Let us label the Pauli matrices as σ0 ≡ I, σ1 ≡ X, σ2 ≡ Y , and σ3 ≡ Z.
Show that Tr {σi σj } = 2δij for all i, j ∈ {0, . . . , 3}, where Tr denotes the trace of a matrix,
defined as the sum of the entries along the diagonal (see also Definition 4.1.1).

3.3.4

Hadamard Gate

Another important unitary operator is the transformation that takes the computational basis
to the +/− basis. This transformation is the Hadamard transformation:|0i → |+i,

|1i → |−i.

(3.60)

Using the above relations, we can represent the Hadamard transformation as the following
operator:
H ≡ |+ih0| + |−ih1|.
(3.61)
It is straightforward to check that the above operator implements the transformation in (3.60).
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3.3. REVERSIBLE EVOLUTION81

Now consider a generalization of the above construction. Suppose that one orthonormal
basis is {|ψi i}i∈{0,1} and another is {|φi i}i∈{0,1} where the index i merely indexes the statesin each orthonormal basis. Then the unitary operator that takes states in the first basis tostates in the second basis isX|φi ihψi |.
(3.62)
i=0,1

Exercise 3.3.7 Show that the Hadamard operator H has the following matrix representationin the computational basis:


1
1 1
.
(3.63)
H=√
2 1 −1
Exercise 3.3.8 Show that the Hadamard operator is its own inverse by employing the above
matrix representation and by using its operator form in (3.61).
Exercise 3.3.9 If the Hadamard gate is its own inverse, then it takes the states |+i and|−i to the respective states |0i and |1i and we can represent it as the following operator:
H = |0ih+| + |1ih−|. Show that |0ih+| + |1ih−| = |+ih0| + |−ih1|.
Exercise 3.3.10 Show that HXH = Z and that HZH = X.

3.3.5

Rotation Operators

We end this section on the evolution of quantum states by discussing “rotation evolutions”
and by giving a more complete picture of the Bloch sphere. The rotation operators RX (φ),
RY (φ), RZ (φ) are functions of the respective Pauli operators X, Y , Z whereRX (φ) ≡ exp {iXφ/2} ,RY (φ) ≡ exp {iY φ/2} ,RZ (φ) ≡ exp {iZφ/2} ,(3.64)and φ is some angle such that 0 ≤ φ < 2π. How do we determine a function of an operator?
The standard way is to represent the operator in its diagonal basis and apply the function
to the non-zero eigenvalues of the operator. For example, the diagonal representation of the
X operator is
X = |+ih+| − |−ih−|.
(3.65)
Applying the function exp {iXφ/2} to the non-zero eigenvalues of X gives
RX (φ) = exp {iφ/2} |+ih+| + exp {−iφ/2} |−ih−|.

(3.66)

This is a special case of the following more general convention that we follow throughout
this book:
Definition 3.3.1 (Function of a Hermitian
operator) Suppose that a Hermitian operP
ator A has a spectral decomposition A = i ai |iihi| for some orthonormal basis {|ii}. Thenthe operator f (A) for some function f is defined as follows:Xf (A) ≡f (ai )|iihi|,(3.67)
i:ai ∈Dom(f )

where Dom(f ) denotes the domain of the function f .
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82

CHAPTER 3. THE NOISELESS QUANTUM THEORY

Figure 3.4: This figure provides more labels for states on the Bloch sphere. The Z axis has its points on
the sphere as eigenstates of the Pauli Z operator, the X axis has eigenstates of the Pauli X operator, andthe Y axis has eigenstates of the Pauli Y operator. The rotation operators RX (φ), RY (φ), and RZ (φ) rotate
a state on the sphere by an angle φ about the respective X, Y , and Z axis.

Exercise 3.3.11 Show that the rotation operators RX (φ), RY (φ), RZ (φ) are equal to thefollowing expressions:
RX (φ) = cos(φ/2)I + i sin(φ/2)X,RY (φ) = cos(φ/2)I + i sin(φ/2)Y,RZ (φ) = cos(φ/2)I + i sin(φ/2)Z,(3.68)(3.69)(3.70)


by using the facts that cos(φ/2) = 21 eiφ/2 + e−iφ/2 and sin(φ/2) = 2i1 eiφ/2 − e−iφ/2 .
Figure 3.4 provides a more detailed picture of the Bloch sphere since we have now established the Pauli operators and their eigenstates. The computational basis states are theeigenstates of the Z operator and are the north and south poles on the Bloch sphere. The+/− basis states are the eigenstates of the X operator and the calculation from Exercise 3.2.1
shows that they are the “east and west poles” of the Bloch sphere. We leave it as anotherexercise to show that the Y eigenstates are the other poles along the equator of the Bloch
sphere.

Exercise 3.3.12 Determine the Bloch sphere angles θ and ϕ for the eigenstates of the Pauli
Y operator.
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3.4. MEASUREMENT

83

Figure 3.5: This figure depicts our diagram of a quantum measurement. Thin lines denote quantum
information and thick lines denote classical information. The result of the measurement is to output aclassical variable m according to a probability distribution governed by the Born rule of the quantum theory.

3.4 Measurement
Measurement is another type of evolution that a quantum system can undergo. It is anevolution that allows us to retrieve classical information from a quantum state and thus is
the way that we can “read out” information. Suppose that we would like to learn something
about the quantum state |ψi in (3.2). Nature prevents us from learning anything about
the probability amplitudes α and β if we have only one quantum measurement that we canperform on one copy of the state. Nature only allows us to measure observables. Observables
are physical variables such as the position or momentum of a particle. In the quantum theory,
we represent observables as Hermitian operators in part because their eigenvalues are real
numbers and every measuring device outputs a real number. Examples of qubit observables
that we can measure are the Pauli operators X, Y , and Z.
Suppose that we measure the Z operator. This measurement is called a “measurement in
the computational basis” or a “measurement of the Z observable” because we are measuring
the eigenvalues of the Z operator. The measurement postulate of the quantum theory, also
known as the Born rule, states that the system reduces to the state |0i with probability |α|2
and reduces to the state |1i with probability |β|2 . That is, the resulting probabilities are
the squares of the probability amplitudes. After the measurement, our measuring apparatus
tells us whether the state reduced to |0i or |1i—it returns +1 if the resulting state is |0i
and returns −1 if the resulting state is |1i. These returned values are the eigenvalues of theZ operator. The measurement postulate is the aspect of the quantum theory that makes it
probabilistic or “jumpy” and is part of the “strangeness” of the quantum theory. Figure 3.5
depicts the notation for a measurement that we will use in diagrams throughout this book.
What is the result if we measure the state |ψi in the +/− basis? Consider that we
can represent |ψi as a superposition of the |+i and |−i states, as given in (3.29). The
measurement postulate then states that a measurement of the X operator gives the state|+i with probability |α + β|2 /2 and the state |−i with probability |α − β|2 /2. Quantum
interference is now coming into play because the amplitudes α and β interfere with eachother. So this effect plays an important role in quantum information theory.
In some cases, the basis states |0i and |1i may not represent the spin states of an electron,but may represent the location of an electron. So, a way to interpret this measurementc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

84

CHAPTER 3. THE NOISELESS QUANTUM THEORY

Quantum State
Probability of |+i Probability of |−iSuperposition state|α + β|2 /2|α − β|2 /2
Probabilistic description 1/2
1/2
Table 3.1: This table summarizes the differences in probabilities for a quantum state in a superposition
α|0i + β|1i and a classical state that is a probabilistic mixture of |0i and |1i.

postulate is that the electron “jumps into” one location or another depending on the outcomeof the measurement. But what is the state of the electron before the measurement? We will
just say in this book that it is in a superposed, indefinite, or unsharp state, rather thantrying to pin down a philosophical interpretation. Some might say that the electron is in“two different locations at the same time.”
Also, we should stress that we cannot interpret the measurement postulate as meaningthat the state is in |0i or |1i with respective probabilities |α|2 and |β|2 before the measurement
occurs, because this latter interpretation is physically different from what we described above
and is also completely classical. The superposition state α|0i + β|1i gives fundamentallydifferent behavior from the probabilistic description of a state that is in |0i or |1i withrespective probabilities |α|2 and |β|2 . Suppose that we have the two different descriptions ofa state (superposition and probabilistic) and measure the Z operator. We get the same result
for both cases—the resulting state is |0i or |1i with respective probabilities |α|2 and |β|2 .
But now suppose that we measure the X operator. The superposed state gives theresult from before—we get the state |+i with probability |α + β|2 /2 and the state |−i withprobability |α − β|2 /2. The probabilistic description gives a much different result. Supposethat the state is |0i. We know that |0i is a uniform superposition of |+i and |−i:|0i =|+i + |−i√.
2(3.71)

So the state collapses to |+i or |−i with equal probability in this case. If the state is |1i,then it collapses again to |+i or |−i with equal probabilities. Summing up these probabilities, it follows
that a measurement of the X operator gives the state |+i with probability22|α| + |β| /2 = 1/2 and gives the state |−i with the same probability. These results arefundamentally different from those in which the state is the superposition state |ψi, and experiment after experiment has supported the predictions of the quantum theory. Table 3.1
summarizes the results that we just discussed.
Now we consider a “Stern–Gerlach”-like argument to illustrate another example of fundamental quantum behavior (Gerlach and Stern, 1922). The Stern–Gerlach experiment wasa crucial one for determining the “strange” behavior of quantum spin states. Suppose that
we prepare the state |0i. If we measure this state in the Z basis, the result is that we always
obtain the state |0i because the prepared state is a definite Z eigenstate. Suppose now that
we measure the X operator. The state |0i is equal to a uniform superposition of |+i and|−i. The measurement postulate then states that we get the state |+i or |−i with equal
probability after performing this measurement. If we then measure the Z operator again, thec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

3.4. MEASUREMENT

85

result is completely random. The Z measurement result is |0i or |1i with equal probability
if the result of the X measurement is |+i and the same outcome occurs if the result of the X
measurement is |−i. This argument demonstrates that the measurement of the X operator“throws off” the measurement of the Z operator. The Stern–Gerlach experiment was one of
the earliest to validate the predictions of the quantum theory.

3.4.1

Probability, Expectation, and Variance of an Operator

We have an alternate, more formal way of stating the measurement postulate that turns
out to be more useful for a general quantum system. Suppose that we are measuring
the Z operator. The diagonal representation of this operator is
Z = |0ih0| − |1ih1|.

(3.72)

Π0 ≡ |0ih0|.

(3.73)

Consider the Hermitian operator
It is a projection operator because applying it twice has the same effect as applying it
once: Π20 = Π0 . It projects onto the subspace spanned by the single vector |0i. A similarline of analysis applies to the projection operator
Π1 ≡ |1ih1|.

(3.74)

So we can represent the Z operator as Π0 − Π1 . Performing a measurement of the Z operator
is equivalent to asking the question: Is the state |0i or |1i? Consider the quantity hψ|Π0 |ψi:hψ|Π0 |ψi = hψ|0i h0|ψi = α∗ α = |α|2 .

(3.75)

A similar analysis demonstrates thathψ|Π1 |ψi = |β|2 .

(3.76)

These two quantities then give the probability that the state reduces to |0i or |1i.
A more general way of expressing a measurement of the Z basis is to say that we havea set {Πi }i∈{0,1} of measurement operators that determine the outcome probabilities. These
measurement operators also determine the state that results after the measurement. If themeasurement result is +1, then the resulting state is
Π |ψi
p 0
= |0i,
hψ|Π0 |ψi

(3.77)

αwhere we implicitly ignore the irrelevant global phase factor |α|. If the measurement resultis −1, then the resulting state is

Π |ψi
p 1
= |1i,
hψ|Π1 |ψic 2019(3.78)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

86

CHAPTER 3. THE NOISELESS QUANTUM THEORYpβ
where we again implicitly ignore the irrelevant global phase factor |β|. Dividing by hψ|Πi |ψi
for i = 0, 1 ensures that the state resulting after measurement corresponds to a physical state(a unit vector).
We can also measure any orthonormal basis in this way—this type of projective measurement is called a von Neumann measurement. For any orthonormal basis {|φi i}i∈{0,1} ,the measurement operators are {|φi ihφi |}i∈{0,1} , and the state reduces to |φi ihφi |ψi/ |hφi |ψi|with probability hψ|φi i hφi |ψi = |hφi |ψi|2 .
Exercise 3.4.1 Determine the set of measurement operators corresponding to a measurement of the X observable.
We might want to determine the expectation of the measurement result when measuring
the Z operator. The probability of getting the +1 value corresponding to the |0i stateis |α|2 and the probability of getting the −1 value corresponding to the −1 eigenstate is|β|2 . Standard probability theory then gives us a way to calculate the expected value of ameasurement of the Z operator when the state is |ψi:
E [Z] = |α|2 (1) + |β|2 (−1) = |α|2 − |β|2 .

(3.79)

We can formulate an alternate way to write this expectation, by making use of the Diracnotation:
E [Z] = |α|2 (1) + |β|2 (−1)= hψ|Π0 |ψi + hψ|Π1 |ψi (−1)= hψ|Π0 − Π1 |ψi= hψ|Z|ψi.

(3.80)(3.81)(3.82)(3.83)

It is common for physicists to denote the expectation as
hZi ≡ hψ|Z|ψi,(3.84)

when it is understood that the expectation is with respect to the state |ψi. This type of
expression is a general one and the next exercise asks you to show that it works for the X
and Y operators as well.
Exercise 3.4.2 Show that the expressions hψ|X|ψi and hψ|Y |ψi give the respective expectations E [X] and E [Y ] when measuring the state |ψi in the respective X and Y basis.
We also might want to determine the variance of the measurement of the Z operator.
Standard probability theory again gives that 
Var [Z] = E Z 2 − E [Z]2 .
(3.85)
Physicists denote the standard deviation of the measurement of the Z operator as∆Z ≡ (Z − hZi)2c 20191/2,(3.86)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

3.4. MEASUREMENT

87

and thus the variance is equal to (∆Z)2 . Physicists often refer to ∆Z as the uncertainty ofthe observable Z when the state is |ψi.
In order to calculate the variance Var[Z], we really just need the second moment E [Z 2 ]because we already have the expectation E [Z]: 
E Z 2 = |α|2 (1)2 + |β|2 (−1)2 = |α|2 + |β|2 .
(3.87)
We can again calculate this quantity with the Dirac notation. The quantity hψ|Z 2 |ψi is thesame as E [Z 2 ] and the next exercise asks you for a proof.
Exercise 3.4.3 Show that E [X 2 ] = hψ|X 2 |ψi, E [Y 2 ] = hψ|Y 2 |ψi, and E [Z 2 ] = hψ|Z 2 |ψi.

3.4.2

The Uncertainty Principle

The uncertainty principle is a fundamental feature of the quantum theory. In the case of
qubits, one instance of the uncertainty principle gives a lower bound on the product of theuncertainty of the Z operator and the uncertainty of the X operator:∆Z∆X ≥

1|hψ| [Z, X] |ψi| .
2(3.88)

We can prove this principle using the postulates of the quantum theory. Let us define theoperators Z0 ≡ Z − hZi and X0 ≡ X − hXi. First, consider that
∆Z∆X = hψ|Z02 |ψi1/2 hψ|X02 |ψi1/2 ≥ |hψ|Z0 X0 |ψi| .

(3.89)

The above step follows by applying the Cauchy–Schwarz inequality to the vectors X0 |ψi andZ0 |ψi. For any operator A, we define its real part Re {A} as Re {A} ≡ (A + A† )/2, and its
imaginary part Im {A} as Im {A} ≡ (A − A† )/2i, so that A = Re {A} + i Im {A} . So thereal and imaginary parts of the operator Z0 X0 are{Z0 , X0 }Z0 X0 + X0 Z0≡,
2
2
Z0 X0 − X0 Z0[Z0 , X0 ]Im {Z0 X0 } =
≡
,
2i
2i
Re {Z0 X0 } =(3.90)(3.91)

where {Z0 , X0 } is the anticommutator of Z0 and X0 and [Z0 , X0 ] is the commutator of thetwo operators. We can then express the quantity |hψ|Z0 X0 |ψi| in terms of the real andimaginary parts of Z0 X0 :|hψ|Z0 X0 |ψi| = |hψ| Re {Z0 X0 } |ψi + ihψ| Im {Z0 X0 } |ψi|
≥ |hψ| Im {Z0 X0 } |ψi|
= |hψ| [Z0 , X0 ] |ψi| /2= |hψ| [Z, X] |ψi| /2.
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The first equality follows by substitution. The first inequality follows because the magnitudeof any complex number is greater than the magnitude of its imaginary part. The second
equality follows by substitution with (3.91). Finally, the third equality follows from the resultof Exercise 3.4.4 below. We worked out the above derivation for particular observables acting
on qubit states, but note that it holds for general observables and quantum states.
The commutator of the operators Z and X arises in the lower bound, and thus, thenon-commutativity of the operators Z and X is the fundamental reason that there is anuncertainty principle for them. Also, there is no uncertainty principle for any two operatorsthat commute with each other.
It is worthwhile to interpret the uncertainty principle in (3.88), which really receives aninterpretation after conducting a large number of independent experiments of two different
kinds. In the first kind of experiment, one prepares the state |ψi and measures the Z
observable. After repeating this experiment independently many times, one can calculate an
estimate of the standard deviation ∆Z, which becomes closer and closer to the true standard
deviation ∆Z as the number of independent experiments becomes large. In the second kindof experiment, one prepares the state |ψi and measures the X observable. After repeating
many times, one can calculate an estimate of ∆X. The uncertainty principle then states thatthe product of the estimates (for a large number of independent experiments) is boundedfrom below by the expectation of the commutator: 12 |hψ| [X, Z] |ψi|.
Exercise 3.4.4 Show that [Z0 , X0 ] = [Z, X] and that [Z, X] = 2iY .
Exercise 3.4.5 The uncertainty principle in (3.88) has the property that the lower bound has
a dependence on the state |ψi. Find a state |ψi for which the lower bound on the uncertainty
product ∆X∆Z vanishes.4

3.5 Composite Quantum Systems
A single physical qubit is an interesting physical system that exhibits uniquely quantum
phenomena, but it is not particularly useful on its own (just as a single classical bit is notparticularly useful for classical communication or computation). We can only perform interesting quantum information-processing tasks when we combine qubits together. Therefore,
we should have a way to describe their behavior when they combine to form a composite
quantum system.
Consider two classical bits c0 and c1 . In order to describe bit operations on the pair ofcbits, we write them as an ordered pair (c1 , c0 ). The space of all possible bit values is the
Cartesian product Z2 × Z2 of two copies of the set Z2 ≡ {0, 1}:
Z2 × Z2 ≡ {(0, 0) , (0, 1) , (1, 0) , (1, 1)} .

(3.96)

4

Do not be alarmed by the result of this exercise! The usual formulation of the uncertainty principle only
gives a lower bound on the uncertainty product. This lower bound never vanishes for the case of position andmomentum observables because the commutator of these two observables is equal to the identity operator
multiplied by i, but it can vanish for the operators given in the exercise.
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3.5. COMPOSITE QUANTUM SYSTEMS

89

Typically, we make the abbreviation c1 c0 ≡ (c1 , c0 ) when representing cbit states.
We can represent the state of two cbits with particular states of qubits. For example, we
can represent the two-cbit state 00 using the following mapping:
00 → |0i|0i.

(3.97)

Many times, we make the abbreviation |00i ≡ |0i|0i when representing two-cbit states with
qubits. Any two-cbit state c1 c0 has the following representation as a two-qubit state:c1 c0 → |c1 c0 i .

(3.98)

The above qubit states are not the only possible states that can occur in the quantum
theory. By the superposition principle, any possible linear combination of the set of two-cbit
states is a possible two-qubit state:|ξi ≡ α|00i + β|01i + γ|10i + δ|11i.

(3.99)The unit-norm condition |α|2 + |β|2 + |γ|2 + |δ|2 = 1 again must hold for the two-qubit
state to correspond to a physical quantum state. It is now clear that the Cartesian productis not sufficient for representing two-qubit quantum states because it does not allow for
linear combinations of states (just as the mathematics of Boolean algebra is not sufficient torepresent single-qubit states).
We again turn to linear algebra to determine a representation that suffices. The tensor product is a mathematical operation that suffices to give a representation of two-qubit
quantum states. Suppose we have two two-dimensional vectors:




a1
a2
,
.
(3.100)b1
b2
The tensor product of these two vectors is
  
 

a
a
a
2
1
2

 

 a1 b 2   a1 b 2 
a1
a2
.

 =
⊗
≡

b1
b2
a2   b 1 a2 
b1
b2
b1 b2

(3.101)

One can understand this operation as taking the vector on the right and stacking two copies
of it together, while multiplying each copy by the corresponding number in the first vector.
Recall, from (3.3), the vector representation of the single-qubit states |0i and |1i. Using
these vector representations and the above definition of the tensor product, the two-qubit
basis states have the following vector representations:
 
 
 
 
1
0
0
0
 0 
 1 
 0 
 0 

 
 
 
|00i = (3.102)
 0  , |01i =  0  , |10i =  1  , |11i =  0  .
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CHAPTER 3. THE NOISELESS QUANTUM THEORY

A simple way to remember these representations is that the bits inside the ket index theelement equal to one in the vector. For example, the vector representation of |01i has a
one as its second element because 01 is the second index for the two-bit strings. The vector
representation of the superposition state in (3.99) is
 
α
 β 
 .
(3.103) γ δThere are actually many different ways that we can write two-qubit states, and we list
all of these right now. Physicists have developed many shorthands, and it is important toknow each of them because they often appear in the literature (this book even uses different
notations depending on the context). We may use any of the following two-qubit notations
if the two qubits are local to one party and only one party is involved in a protocol:
α|0i ⊗ |0i + β|0i ⊗ |1i + γ|1i ⊗ |0i + δ|1i ⊗ |1i,α|0i|0i + β|0i|1i + γ|1i|0i + δ|1i|1i,
α|00i + β|01i + γ|10i + δ|11i.

(3.104)(3.105)(3.106)

We can put labels on the qubits if two or more parties, such as A and B, are involved
α|0iA ⊗ |0iB + β|0iA ⊗ |1iB + γ|1iA ⊗ |0iB + δ|1iA ⊗ |1iB ,α|0iA |0iB + β|0iA |1iB + γ|1iA |0iB + δ|1iA |1iB ,α|00iAB + β|01iAB + γ|10iAB + δ|11iAB .

(3.107)(3.108)(3.109)This second scenario is different from the first scenario because two spatially separated parties
share the two-qubit state. If the state has quantum correlations, then it can be valuable as acommunication resource. We go into more detail on this topic in Section 3.6, which discusses
entanglement.

3.5.1

Evolution of Composite Systems

The postulate on unitary evolution extends to the two-qubit scenario as well. First, let us
establish that the tensor product A ⊗ B of two operators A and B is 
a11 a12b11 b12A⊗B ≡⊗(3.110)a21 a22b21 b22 b11 b12b11 b12a12
 a11 b21 b22
 b21 b22  ≡(3.111)b11 b12b11 b12 
a21
a22
b21 b22b21 b22
a11 b11 a11 b12 a12 b11 a12 b12
 a11 b21 a11 b22 a12 b21 a12 b22 =(3.112)
 a21 b11 a21 b12 a22 b11 a22 b12  .
a21 b21 a21 b22 a22 b21 a22 b22
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3.5. COMPOSITE QUANTUM SYSTEMS

91

Figure 3.6: This figure depicts circuits for the example two-qubit unitaries X1 I2 , I1 X2 , and X1 X2 .

The tensor-product operation for matrices is similar to what we did for vectors, but now weare stacking copies of the matrix on the right both vertically and horizontally, and multiplying
each copy by the corresponding number in the first matrix.
Consider the two-qubit state in (3.99). We can perform a NOT gate on the first qubit
so that it changes to α|10i + β|11i + γ|00i + δ|01i. We can alternatively flip its second
qubit: α|01i + β|00i + γ|11i + δ|10i, or flip both at the same time: α|11i + β|10i + γ|01i +δ|00i. Figure 3.6 depicts quantum circuit representations of these operations. These are allreversible operations because applying them again gives the original state in (3.99). In thefirst case, we did nothing to the second qubit, and in the second case, we did nothing to thefirst qubit. The identity operator acts on the qubits that have nothing happen to them.
Let us label the first qubit as “1” and the second qubit as “2.” We can then label theoperator for the first operation as X1 I2 because this operator flips the first qubit and doesnothing (applies the identity) to the second qubit. We can also label the operators for thesecond and third operations respectively as I1 X2 and X1 X2 . The matrix representation of
the operator X1 I2 is the tensor product of the matrix representation of X with the matrix
representation of I—this relation similarly holds for the operators I1 X2 and X1 X2 . Weshow that it holds for the operator X1 I2 and ask you to verify the other two cases. We
can use the two-qubit computational basis to get a matrix representation for the two-qubit
operator X1 I2 :
h00|X1 I2 |00i h00|X1 I2 |01i h00|X1 I2 |10i h00|X1 I2 |11i
 h01|X1 I2 |00i h01|X1 I2 |01i h01|X1 I2 |10i h01|X1 I2 |11i  h10|X1 I2 |00i h10|X1 I2 |01i h10|X1 I2 |10i h10|X1 I2 |11i 
h11|X1 I2 |00i h11|X1 I2 |01i h11|X1 I2 |10i h11|X1 I2 |11i 

0 0 1 0h00|10i h00|11i h00|00i h00|01i
 h01|10i h01|11i h01|00i h01|01i   0 0 0 1  = h10|10i h10|11i h10|00i h10|01i  =  1 0 0 0  . (3.113)
0 1 0 0h11|10i h11|11i h11|00i h11|01i
This last matrix is equal to the tensor product X ⊗ I by inspecting the definition of thetensor product for matrices in (3.110).
Exercise 3.5.1 Show that the matrix representation of the operator I1 X2 is equal to thetensor product I ⊗ X. Show the same for X1 X2 and X ⊗ X.
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3.5.2

Probability Amplitudes for Composite Systems

We relied on the orthogonality of the two-qubit computational basis states for evaluating
amplitudes such as h00|10i or h00|00i in the above matrix representation. It turns out that
there is another way to evaluate these amplitudes that relies only on the orthogonality of thesingle-qubit computational basis states. Suppose that we have four single-qubit states |φ0 i,|φ1 i, |ψ0 i, |ψ1 i, and we make the following two-qubit states from them:|φ0 i ⊗ |ψ0 i,|φ1 i ⊗ |ψ1 i.

(3.114)

We may represent these states equally well as follows:|φ0 , ψ0 i ,|φ1 , ψ1 i ,(3.115)

because the Dirac notation is versatile (virtually anything can go inside a ket as long as its
meaning is not ambiguous). The bra hφ1 , ψ1 | is dual to the ket |φ1 , ψ1 i, and we can use it tocalculate the following amplitude:
hφ1 , ψ1 |φ0 , ψ0 i .

(3.116)This amplitude is equal to the multiplication of the single-qubit amplitudes:
hφ1 , ψ1 |φ0 , ψ0 i = hφ1 |φ0 i hψ1 |ψ0 i .

(3.117)

Exercise 3.5.2 Verify that the amplitudes {hij|kli}i,j,k,l∈{0,1} are respectively equal to theamplitudes {hi|ki hj|li}i,j,k,l∈{0,1} . By linearity, this exercise justifies the relation in (3.117)(at least for two-qubit states).

3.5.3

Controlled Gates

An important two-qubit unitary evolution is the controlled-NOT (CNOT) gate. We consider
its classical version first. The classical gate acts on two cbits. It does nothing if the first bit
is equal to zero, and flips the second bit if the first bit is equal to one:
00 → 00,

01 → 01,

10 → 11,

11 → 10.

(3.118)We turn this gate into a quantum gate5 by demanding that it act in the same way on the
two-qubit computational basis states:|00i → |00i,|01i → |01i,|10i → |11i,|11i → |10i.

(3.119)By linearity, this behavior carries over to superposition states as well:
α|00i + β|01i + γ|10i + δ|11i

CNOT
−−−−→

α|00i + β|01i + γ|11i + δ|10i.

(3.120)5

There are other terms for the action of turning a classical operation into a quantum one. Some examples
are “making it coherent,” “coherifying,” or the quantum gate is a “coherification” of the classical one. The
term “coherify” is not a proper English word, but we will use it regardless at certain points.
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3.5. COMPOSITE QUANTUM SYSTEMS

93

Figure 3.7: Circuit diagrams that we use for (a) a CNOT gate and (b) a controlled-U gate.

A useful operator representation of the CNOT gate is
CNOT ≡ |0ih0| ⊗ I + |1ih1| ⊗ X.

(3.121)

The above representation truly captures the coherent quantum nature of the CNOT gate. In
the classical CNOT gate, we can say that it is a conditional gate, in the sense that the gateapplies to the second bit conditioned on the value of the first bit. In the quantum CNOT
gate, the second operation is controlled on the basis state of the first qubit (hence the choiceof the name “controlled-NOT”). That is, the gate acts on superpositions of quantum statesand maintains these superpositions, shuffling the probability amplitudes around while it does
so. The one case in which the gate has no effect is when the first qubit is prepared in thestate |0i and the state of the second qubit is arbitrary.
A controlled-U gate is similar to the CNOT gate in (3.121). It simply applies theunitary U (assumed to be a single-qubit unitary) to the second qubit, controlled on thefirst qubit:
controlled-U ≡ |0ih0| ⊗ I + |1ih1| ⊗ U.
(3.122)The control qubit could alternatively be controlled with respect to any orthonormal basis{|φ0 i, |φ1 i}:|φ0 ihφ0 | ⊗ I + |φ1 ihφ1 | ⊗ U.
(3.123)Figure 3.7 depicts the circuit diagrams for a controlled-NOT and controlled-U operation.
Exercise 3.5.3 Verify that the matrix representation of the CNOT gate in the computational
basis is

1 0 0 0
 0 1 0 0 
(3.124)
 0 0 0 1 .
0 0 1 0Exercise 3.5.4 Consider applying Hadamards to the first and second qubits before and after
a CNOT acts on them. Show that this gate is equivalent to a CNOT in the +/− basis (recall
that the Z operator flips the +/− basis):
H1 H2 CNOT H1 H2 = |+ih+| ⊗ I + |−ih−| ⊗ Z.

(3.125)

Exercise 3.5.5 Show that two CNOT gates with the same control qubit commute.
Exercise 3.5.6 Show that two CNOT gates with the same target qubit commute.
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3.5.4

CHAPTER 3. THE NOISELESS QUANTUM THEORYThe No-Cloning Theorem

The no-cloning theorem is one of the simplest results in the quantum theory, yet it has someof the most profound consequences. It states that it is impossible to build a universal copierof quantum states. A universal copier would be a device that could copy any arbitrary
quantum state that is input to it. It may be surprising at first to hear that copying quantum
information is impossible because copying classical information is ubiquitous.
We now give a simple proof of the no-cloning theorem. Suppose for a contradiction that
there is a two-qubit unitary operator U acting as a universal copier of quantum information.
That is, if we input an arbitrary state |ψi = α|0i+β|1i as the first qubit and input an ancillaqubit |0i as the second qubit, such a device should “write” the first qubit to the second qubit
slot as follows:
U |ψi|0i = |ψi|ψi= (α|0i + β|1i) (α|0i + β|1i)= α2 |0i|0i + αβ|0i|1i + αβ|1i|0i + β 2 |1i|1i.

(3.126)(3.127)(3.128)The copier is universal, meaning that it copies an arbitrary state. In particular, it also copies
the states |0i and |1i:
U |0i|0i = |0i|0i,

U |1i|0i = |1i|1i.

(3.129)Linearity of the quantum theory then implies that the unitary operator acts on a superposition α|0i + β|1i as follows:
U (α|0i + β|1i) |0i = α|0i|0i + β|1i|1i.

(3.130)However, the consequence in (3.128) contradicts the consequence in (3.130) because these
two expressions do not have to be equal for all α and β:∃α, β : α2 |0i|0i + αβ|0i|1i + αβ|1i|0i + β 2 |1i|1i =6 α|0i|0i + β|1i|1i.

(3.131)Thus, linearity of the quantum theory contradicts the existence of a universal quantum
copier.
We would like to stress that this proof does not mean that it is impossible to copy certain
quantum states—it only implies the impossibility of a universal copier. Observe that (3.131)
is satisfied for α = 1, β = 0 or α = 0, β = 1, so that we can copy unknown classical states
prepared in the basis |0i, |1i (or any other orthonormal basis for that matter).
Another proof of the no-cloning theorem arrives at a contradiction by exploiting unitarityof quantum evolutions. Let us again suppose that a universal copier U exists. Consider two
arbitrary states |ψi and |φi. If a universal copier U exists, then it performs the following
copying operation for both states:
U |ψi|0i = |ψi|ψi,
c 2019

U |φi|0i = |φi |φi.

(3.132)

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3.5. COMPOSITE QUANTUM SYSTEMS

95

Consider the probability amplitude hψ|hψ||φi|φi:hψ|hψ||φi|φi = hψ|φi hψ|φi = hψ|φi2 .

(3.133)

The following relation for hψ|hψ||φi|φi holds as well by using the results in (3.132) and theunitarity property U † U = I:hψ|hψ||φi|φi = hψ|h0|U † U |φi|0i= hψ|h0||φi|0i= hψ|φi h0|0i= hψ|φi .

(3.134)(3.135)(3.136)(3.137)

As a consequence, we find thathψ|hψ||φi|φi = hψ|φi2 = hψ|φi ,(3.138)by employing the above two results. The equality hψ|φi2 = hψ|φi holds for exactly twocases, hψ|φi = 1 and hψ|φi = 0. The first case holds only when the two states are the samestate and the second case holds when the two states are orthogonal to each other. Thus, it
is impossible to copy quantum information in any other case because we would contradict
unitarity.
The no-cloning theorem has several applications in quantum information processing.
First, it underlies the security of the quantum key distribution protocol because it ensuresthat an attacker cannot copy the quantum states that two parties use to establish a secretkey. It finds application in quantum Shannon theory because we can use it to reason about
the quantum capacity of a certain quantum channel known as the erasure channel. We will
return to this point in Chapter 24.
Exercise 3.5.7 Suppose that two states |ψi and |ψ ⊥ i are orthogonal: hψ|ψ ⊥ i = 0. Construct
a two-qubit unitary that can copy the states, i.e., find a unitary U that acts as follows:
U |ψi|0i = |ψi|ψi, U |ψ ⊥ i|0i = |ψ ⊥ i|ψ ⊥ i.
Exercise 3.5.8 (No-Deletion Theorem) Somewhat related to the no-cloning theorem,
there is a no-deletion theorem. Suppose that two copies of a quantum state |ψi are available,
and the goal is to delete one of these states by a unitary interaction. That is, there should
exist a universal quantum deleter U that has the following action on the two copies of |ψi
and an ancilla state |Ai, regardless of the input state |ψi:
U |ψi|ψi |Ai = |ψi|0i |A0 i ,(3.139)where |A0 i is another state. Show that this is impossible.
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CHAPTER 3. THE NOISELESS QUANTUM THEORY

3.5.5

Measurement of Composite Systems

The measurement postulate also extends to composite quantum systems. Suppose againthat we have the two-qubit quantum state in (3.99). By a straightforward analogy with thesingle-qubit case, we can determine the following probability amplitudes:h00|ξi = α,h01|ξi = β,h10|ξi = γ,h11|ξi = δ.

(3.140)Π11 ≡ |11ih11|,(3.141)

We can also define the following projection operators:
Π00 ≡ |00ih00|,

Π01 ≡ |01ih01|,Π10 ≡ |10ih10|,

and apply the Born rule to determine the probabilities for each result:hξ| Π00 |ξi = |α|2 ,hξ| Π01 |ξi = |β|2 ,hξ| Π10 |ξi = |γ|2 ,hξ| Π11 |ξi = |δ|2 .

(3.142)

Suppose that we wish to perform a measurement of the Z operator on the first qubit
only. What is the set of projection operators that describes this measurement? The answer
is similar to what we found for the evolution of a composite system. We apply the identity operator to the second qubit because no measurement occurs on it. Thus, the set ofmeasurement operators is{Π0 ⊗ I, Π1 ⊗ I} ,(3.143)where the definition of Π0 and Π1 is in (3.73)–(3.74). The state reduces toα|00i + β|01i(Π ⊗ I) |ξi
p 0
,
= q
hξ| (Π0 ⊗ I) |ξi|α|2 + |β|2

(3.144)

with probability hξ| (Π0 ⊗ I) |ξi = |α|2 + |β|2 , and reduces to(Π ⊗ I) |ξi
γ|10i + δ|11i
p 1
= q
,
hξ| (Π1 ⊗ I) |ξi|γ|2 + |δ|2

(3.145)pp
with probability hξ| (Π1 ⊗ I) |ξi = |γ|2 +|δ|2 . Normalizing by hξ| (Π0 ⊗ I) |ξi and hξ| (Π1 ⊗ I) |ξi
again ensures that the resulting vector corresponds to a physical state.

3.6 Entanglement
Composite quantum systems give rise to a uniquely quantum phenomenon: entanglement.
Schrödinger first observed that two or more quantum systems can be entangled and coined
the term after noticing some of the bizarre consequences of this phenomenon.6
6

Schrödinger actually used the German word “Verschränkung” to describe the phenomenon, which literally translates as “little parts that, though far from one another, always keep the exact same distance fromeach other.” The one-word English translation is “entanglement.” Einstein described the “Verschränkung” asa “spukhafte Fernwirkung,” most closely translated as “long-distance ghostly effect” or the more commonlystated “spooky action at a distance.”
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97

We first consider a simple, unentangled state that two parties, Alice and Bob, may share,in order to see how an unentangled state contrasts with an entangled state. Suppose thatthey share the state|0iA |0iB ,(3.146)
where Alice has the qubit in system A and Bob has the qubit in system B. Alice candefinitely say that her qubit is in the state |0iA and Bob can definitely say that his qubit isin the state |0iB . There is nothing really too strange about this scenario.
Now, consider the composite quantum state |Φ+ iAB :1
Φ+ AB ≡ √ (|0iA |0iB + |1iA |1iB ) .
2(3.147)

Alice again has possession of the first qubit in system A and Bob has possession of the second
qubit in system B. But now, it is not clear from the above description how to determine theindividual state of Alice or the individual state of Bob. The above state is really a uniform
superposition of the joint state |0iA |0iB and the joint state |1iA |1iB , and it is not possible
to describe either Alice’s or Bob’s individual state in the noiseless quantum theory. We also
cannot describe the entangled state |Φ+ iAB as a product state of the form |φiA |ψiB , for any
states |φiA or |ψiB . This leads to the following general definition:
Definition 3.6.1 (Pure-State Entanglement) A pure bipartite state |ψiAB is entangled
if it cannot be written as a product state |φiA ⊗ |ϕiB for any choices of states |φiA and |ϕiB .
Exercise 3.6.1 Show that the entangled state |Φ+ iAB has the following representation in
the +/− basis:1(3.148)
Φ+ AB = √ (|+iA |+iB + |−iA |−iB ) .
2
Figure 3.8 gives a graphical depiction of entanglement. We use this depiction often
throughout this book. Alice and Bob must receive the entanglement in some way, and the
diagram indicates that some source distributes the entangled pair to them. It indicates that
Alice and Bob are spatially separated and they possess the entangled state after some time.
If they share the entangled state in (3.147), we say that they share one bit of entanglement,or one ebit. The term “ebit” implies that there is some way to quantify entanglement andwe will make this notion clear in Chapter 19.

3.6.1

Entanglement as a Resource

In this book, we are interested in the use of entanglement as a resource. Much of this bookconcerns the theory of quantum information-processing resources, and we have a standard
notation for the theory of resources. Let us represent the resource of a shared ebit as[qq] ,
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CHAPTER 3. THE NOISELESS QUANTUM THEORY

Figure 3.8: We use the above diagram to depict entanglement shared between two parties A and B. The
diagram indicates that a source location creates the entanglement and distributes one system to√A and theother system to B. The standard unit of entanglement is the ebit |Φ+ iAB ≡ (|00iAB + |11iAB )/ 2.

meaning that the ebit is a noiseless, quantum resource shared between two parties. Square
brackets indicate a noiseless resource, the letter q indicates a quantum resource, and the twocopies of the letter q indicate a two-party resource.
Our first example of the use of entanglement is its role in generating shared randomness.
We define one bit of shared randomness as the following probability distribution for two
binary random variables XA and XB :1
pXA ,XB (xA , xB ) = δ(xA , xB ),2(3.150)where δ is the Kronecker delta function. Suppose Alice possesses random variable XA andBob possesses random variable XB . Thus, with probability 1/2, they either both have a zeroor they both have a one. We represent the resource of one bit of shared randomness as[cc] ,(3.151)indicating that a bit of shared randomness is a noiseless, classical resource shared between
two parties.
Now suppose that Alice and Bob share an ebit and they decide that they will eachmeasure their qubits in the computational basis. Without loss of generality, suppose that
Alice performs a measurement first. Thus, Alice performs a measurement of the ZA operator,
meaning that she measures ZA ⊗ IB (she cannot perform anything on Bob’s qubit because
they are spatially separated). The projection operators for this measurement are given in(3.143), and they project the joint state. Just before Alice looks at her measurement result,she does not know the outcome, and we can describe the system as being in the following
ensemble of states:
1,2
1|1iA |1iB with probability .
2|0iA |0iB with probability(3.152)(3.153)The interesting thing about the above ensemble is that Bob’s result is already determinedeven before he measures, just after Alice’s measurement occurs. Suppose that Alice knows
the result of her measurement is |0iA . When Bob measures his system, he obtains the statec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License3.6. ENTANGLEMENT

99|0iB with probability one and Alice knows that he has measured this result. Additionally,
Bob knows that Alice’s state is |0iA if he obtains |0iB . The same results hold if Alice knowsthat the result of her measurement is |1iA . Thus, this protocol is a method for them togenerate one bit of shared randomness as defined in (3.150).
We can phrase the above protocol as the following resource inequality:[qq] ≥ [cc] .

(3.154)The interpretation of the above resource inequality is that there exists a protocol which
generates the resource on the right by consuming the resource on the left and using only local
operations, and for this reason, the resource on the left is a stronger resource than the oneon the right. The theory of resource inequalities plays a prominent role in this book and isa useful shorthand for expressing quantum protocols.
A natural question is to wonder if there exists a protocol to generate entanglement exclusively from shared randomness. It is not possible to do so and one reason justifying thisinequivalence of resources is another type of inequality (different from the resource inequality mentioned above), called a Bell’s inequality. In short, Bell’s theorem places an upperbound on the correlations present in any two classical systems. Entanglement violates thisinequality, showing that it has no known classical equivalent. Thus, entanglement is a strictly
stronger resource than shared randomness and the resource inequality in (3.154) only holds
in the given direction.
Shared randomness is a resource in classical information theory, and may be useful in somescenarios, but it is actually a rather weak resource. Surely, generating shared randomness
is not the only use of entanglement. It turns out that we can construct far more exotic
protocols such as the teleportation protocol or the super-dense coding protocol by combining
the resource of entanglement with other resources. We discuss these protocols in Chapter 6.
Exercise 3.6.2 Use the representation of the ebit in Exercise 3.6.1 to show that Alice andBob can measure the X operator to generate shared randomness. This ability to obtain sharedrandomness by both parties measuring in either the Z or X basis is the foundation for an
entanglement-based secret key distribution protocol.
Exercise 3.6.3 (Cloning Implies Signaling) Prove that if a universal quantum cloner
were to exist, then it would be possible for Alice to signal to Bob faster than the speed of lightby exploiting only the ebit state |Φ+ iAB shared between them and no communication. Thatis, show the existence of a protocol that would allow for this. (Hint: One possibility is for
Alice to measure the X or Z Pauli operator locally on her share of the ebit, and then for Bob
to exploit the universal quantum cloner. Consider the representation of the ebit in (3.147)
and (3.148). Note that there could be a variety of answers to this question because quantum
theory becomes effectively nonlinear if we assume the existence of a cloner!)

3.6.2

Entanglement in the CHSH GameOne of the simplest means for demonstrating the power of entanglement is with a two-player
game known as the CHSH game (after Clauser, Horne, Shimony, and Holt), which is ac 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 3. THE NOISELESS QUANTUM THEORY

Figure 3.9: A depiction of the CHSH game. The referee distributes the bits x and y to Alice and Bob in
the first round. In the second round, Alice and Bob return the bits a and b to the referee.

particular variation of the original setup in Bell’s theorem. We first present the rules of thegame, and then we find an upper bound on the probability that players operating according
to a classical strategy can win. We finally leave it as an exercise to show that players sharing
a maximally entangled Bell state |Φ+ i can have an approximately 10% higher chance of
winning the game using a quantum strategy. This result, known as Bell’s theorem, represents
one of the most striking separations between classical and quantum physics.
The players of the game are Alice and Bob, who are spatially separated from each otherfrom the time that the game starts until it is over. The game begins with a referee selecting
two bits x and y uniformly at random. The referee then sends x to Alice and y to Bob.
Alice and Bob are not allowed to communicate with each other in any way at this point.
Alice sends back to the referee a bit a, and Bob sends back a bit b. Since they are spatiallyseparated, Alice’s response bit a cannot depend on Bob’s input bit y, and similarly, Bob’s
response bit b cannot depend on Alice’s input bit x. After receiving the response bits a andb, the referee determines if the AND of x and y is equal to the exclusive OR of a and b. If
so, then Alice and Bob win the game. That is, the winning condition isx ∧ y = a ⊕ b.

(3.155)Figure 3.9 depicts the CHSH game.
We need to figure out an expression for the winning probability of the CHSH game.
Let V (x, y, a, b) denote the following indicator function for whether they win in a particularinstance of the game:
1 if x ∧ y = a ⊕ b
V (x, y, a, b) =.
(3.156)0else
There is a conditional probability distribution pAB|XY (a, b|x, y), which corresponds to theparticular strategy that Alice and Bob employ. Since the inputs x and y are chosen uniformly
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Figure 3.10: Various reductions of a classical strategy in the CHSH game: (i) an unconstrained strategy,(ii) strategy resulting from demanding that the parameter λ is independent of the input bits x and y, and(iii) further demanding that Alice and Bob’s actions are independent and that they do not have access toeach other’s input bits.

at random and each take on two possible values, the distribution pXY (x, y) for x and y is asfollows:
pXY (x, y) = 1/4.
(3.157)
So an expression for the winning probability of the CHSH game is
1 X
V (x, y, a, b)pAB|XY (a, b|x, y).
4 a,b,x,y(3.158)In order to calculate this winning probability for a classical or quantum strategy, weneed to understand the distribution pAB|XY (a, b|x, y) further. In order to do so, we needa way for describing the strategy that Alice and Bob employ. For this purpose, we willassume that there is a random variable Λ taking values λ, which describes either a classicalor quantum strategy, and its values could be all of the entries in a matrix and even taking
on continuous values. Using the law of total probability, we can expand the conditional
probability pAB|XY (a, b|x, y) as follows:Z
pAB|XY (a, b|x, y) = dλ pAB|ΛXY (a, b|λ, x, y) pΛ|XY (λ|x, y),(3.159)where pΛ|XY (λ|x, y) is a conditional probability distribution. Decomposing the distribution
pAB|XY (a, b|x, y) in this way leads to the depiction of their strategy given in Figure 3.10(i).
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CHAPTER 3. THE NOISELESS QUANTUM THEORYClassical Strategies
Let us suppose that they act according to a classical strategy. What is the most generalform of such a strategy? Looking at the picture in Figure 3.10(i), there are a few aspects of
it which are not consistent with our understanding of how the game works.
In a classical strategy, the random variable Λ corresponds to classical correlations that
Alice and Bob can share before the game begins. They could meet beforehand and select avalue λ of Λ at random. According to the specification of the game, the input bits x andy for Alice and Bob are chosen independently at random, and so the random variable Λcannot depend on the bits x and y. So the conditional distribution pΛ|XY (λ|x, y) simplifies
as follows:pΛ|XY (λ|x, y) = pΛ (λ),(3.160)and Figure 3.10(ii) reflects this constraint.
Next, Alice and Bob are spatially separated and acting independently, so that the distribution pAB|ΛXY (a, b|λ, x, y) factors as follows:pAB|ΛXY (a, b|λ, x, y) = pA|ΛXY (a|λ, x, y) pB|ΛXY (b|λ, x, y).

(3.161)

But we also said that Alice’s strategy cannot depend on Bob’s input bit y and neither can
Bob’s strategy depend on Alice’s input x, because they are spatially separated. However,their strategies could depend on the random variable Λ, which they are allowed to share
before the game begins. All of this implies that the conditional distribution describing theirstrategy should factor as follows:pAB|ΛXY (a, b|λ, x, y) = pA|ΛX (a|λ, x) pB|ΛY (b|λ, y),(3.162)and Figure 3.10(iii) reflects this change. Now Figure 3.10(iii) depicts the most generalclassical strategy that Alice and Bob could employ if Λ corresponds to a random variable
that Alice and Bob are both allowed to access before the game begins.
Putting everything together, the conditional distribution pAB|XY (a, b|x, y) for a classicalstrategy takes the following form:
Z
pAB|XY (a, b|x, y) = dλ pA|ΛX (a|λ, x) pB|ΛY (b|λ, y) pΛ (λ),(3.163)and we can now consider optimizing the winning probability in (3.158) with respect to allclassical strategies. Consider that any stochastic map pA|ΛX (a|λ, x) can be simulated byapplying a deterministic binary-valued function f (a|λ, x, n) to a local random variable N
taking values labeled by n. That is, we can always find a random variable N such thatZ
pA|ΛX (a|λ, x) = dn f (a|λ, x, n) pN (n).
(3.164)The same is true for the stochastic map pB|ΛY (b|λ, y); i.e., there is a random variable M suchthat
Z
pB|ΛY (b|λ, y) = dm g(b|λ, y, m) pM (m),(3.165)c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License3.6. ENTANGLEMENT

103

where g is a deterministic binary-valued function. So this implies that
pAB|XY (a, b|x, y)Z= dλ pA|ΛX (a|λ, x) pB|ΛY (b|λ, y) pΛ (λ)
Z
 Z

Z
= dλ
dn f (a|λ, x, n) pN (n)dm g(b|λ, y, m) pM (m) pΛ (λ)
Z Z Z=dλ dn dm f (a|λ, x, n) g(b|λ, y, m) pΛ (λ) pN (n) pM (m).

(3.166)(3.167)(3.168)

By inspecting the last line above, it is clear that we could then have the shared randomvariable Λ subsume the local random variables N and M , allowing us to write any conditional
distribution pAB|XY (a, b|x, y) for a classical strategy as follows:Z
pAB|XY (a, b|x, y) =dλ f 0 (a|λ, x) g 0 (b|λ, y) pΛ (λ),(3.169)where f 0 and g 0 are deterministic binary-valued functions (related to f and g). Substituting
this expression into the winning probability expression in (3.158), we find that
1 X
V (x, y, a, b)pAB|XY (a, b|x, y)
4 a,b,x,yZ
1 X
V (x, y, a, b) dλ f 0 (a|λ, x) g 0 (b|λ, y) pΛ (λ)=
4 a,b,x,y"#Z
X
1= dλ pΛ (λ)
V (x, y, a, b) f 0 (a|λ, x) g 0 (b|λ, y)
4 a,b,x,y
1 X≤
V (x, y, a, b) f 0 (a|λ∗ , x) g 0 (b|λ∗ , y).
4 a,b,x,y(3.170)(3.171)(3.172)In the second equality, we just exchanged the integral over λ with the sum. In the inequality
in the last step, we used the fact that the average is always less than the maximum. That is,there is always a particular value λ∗ that leads to the same or a higher winning probability
than when averaging over all values of λ. As a consequence of the above development, we
see that it suffices to consider deterministic strategies of Alice and Bob when analyzing thewinning probability.
Since we now know that deterministic strategies are optimal among all classical strategies,
let us focus on these. A deterministic strategy would have Alice select a bit ax conditionedon the bit x that she receives, and similarly, Bob would select a bit by conditioned on y. The
following table presents the winning conditions for the four different values of x and y withc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 3. THE NOISELESS QUANTUM THEORY

this deterministic strategy:
x y
0 0
0 1
1 0
1 1

x∧y
0
0
0
1

= ax ⊕ b y= a0 ⊕ b 0= a0 ⊕ b 1= a1 ⊕ b 0= a1 ⊕ b 1.

(3.173)

However, we can observe that it is impossible for them to always win. If we add the entries
in the column x ∧ y, the binary sum is equal to one, while if we add the entries in the column= ax ⊕ by , the binary sum is equal to zero. Thus, it is impossible for all of these equations
to be satisfied. At most, only three out of four of them can be satisfied, so that the maximal
winning probability with a classical deterministic strategy pAB|XY (a, b|x, y) is at most 3/4:
3
1 X
V (x, y, a, b)pAB|XY (a, b|x, y) ≤ .
4 a,b,x,y4(3.174)

We can then see that a strategy for them to achieve this upper bound is for Alice and Bob
always to return a = 0 and b = 0 no matter the values of x and y.
Quantum Strategies
What does a quantum strategy of Alice and Bob look like? Here the parameter λ cancorrespond to a shared quantum state |φiAB . Alice and Bob perform local measurements
depending on the values of the inputs x and y that they receive. We can write
xP Alice’s
(x)
(x)
(x)
dependent measurement as {Πa } where for each x, Πa is a projector and a Πa = I.
(y)Similarly, we can write Bob’s y-dependent measurement as {Πb }. Then we instead employ
the Born rule to determine the conditional probability distribution pAB|XY (a, b|x, y):(y)pAB|XY (a, b|x, y) = hφ|AB Π(x)a ⊗ Πb |φiAB ,(3.175)so that the winning probability with a particular quantum strategy is as follows:
1 X(y)
V (x, y, a, b)hφ|AB Π(x)a ⊗ Πb |φiAB .
4 a,b,x,y(3.176)

Interestingly, if Alice and Bob share a maximally entangled state, they can achieve ahigher winning probability than if they share classical correlations only. This is one demonstration of the power of entanglement, and we leave it as an exercise to prove that thefollowing quantum strategy achieves a winning probability of cos2 (π/8) ≈ 0.85 in the CHSH
game.
Exercise 3.6.4 Suppose that Alice and Bob share a maximally entangled state |Φ+ i. Show
that the following strategy has a winning probability of cos2 (π/8). If Alice receives x = 0
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105

from the referee, then she performs a measurement of Pauli Z on her system and returns themeasurement outcome as “a” after identifying a = 0 with the measurement outcome +1 anda = 1 with the measurement outcome −1. (The same convention is applied to the following
scenarios.) If she receives x = 1, then she performs a measurement of Pauli X and returns
the outcome √as “a.” If Bob receives y = 0 from the referee, then he performs a measurementof (X + Z) / 2 on his system and returns the outcome√as b. If Bob receives y = 1 from thereferee, then he performs a measurement of (Z − X) / 2 and returns the outcome as b.
Maximum Quantum Winning Probability
Given that classical strategies cannot win with probability any larger than 3/4, it is natural
to wonder if there is a bound on the winning probability of a quantum strategy. It turns outthat cos2 (π/8) is the maximum probability with which Alice and Bob can win the CHSH
game using a quantum strategy, a result known as Tsirelson’s bound. To establish this result,
let us go back to the CHSH game. Conditioned on the inputs x and y being equal to 00, 01,
or 10, we know that Alice and Bob win if they report back the same results. The probability
for this to happen with a given quantum strategy is
(x)

(y)

(x)

(y)

(3.177)(x)(y)(3.178)hφ|AB Π0 ⊗ Π0 |φiAB + hφ|AB Π1 ⊗ Π1 |φiAB ,and the probability for it not to happen is
(x)

(y)

hφ|AB Π0 ⊗ Π1 |φiAB + hφ|AB Π1 ⊗ Π0 |φiAB .

So, conditioned on x and y being equal to 00, 01, or 10, the probability of winning minus
the probability of losing is
hφ|AB A(x) ⊗ B (y) |φiAB ,(3.179)where we define the observables A(x) and B (y) as follows:
A(x) ≡ Π0 − Π1 ,

(x)

(x)

(3.180)(y)(y)
B (y) ≡ Π0 − Π1 .

(3.181)

If x and y are both equal to one, then Alice and Bob should report back different results, andsimilar to the above, one can work out that the probability of winning minus the probabilityof losing is equal to− hφ|AB A(1) ⊗ B (1) |φiAB .
(3.182)Thus, when averaging over all values of the input bits, the probability of winning minus theprobability of losing is equal to1
hφ|AB CAB |φiAB ,(3.183)4
where CAB is the CHSH operator, defined as
CAB ≡ A(0) ⊗ B (0) + A(0) ⊗ B (1) + A(1) ⊗ B (0) − A(1) ⊗ B (1) .
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CHAPTER 3. THE NOISELESS QUANTUM THEORY

It is a simple exercise to check that

 

2
CAB
= 4IAB − A(0) , A(1) ⊗ B (0) , B (1) .

(3.185)The infinity norm kRk∞ of an operator R is equal to its largest singular value. It obeys thefollowing relations:kcRk∞ = |c| kRk∞ ,kRSk∞ ≤ kRk∞ kSk∞ ,kR + Sk∞ ≤ kRk∞ + kSk∞ ,(3.186)(3.187)(3.188)where c ∈ C and S is another operator. Using these, we find that
 (0) (1)   (0) (1) 
2⊗ B ,B
CAB
=
4I
−
A ,A
AB
∞
∞
 (0) (1)   (0) (1)⊗ B ,B≤ 4 kIAB k∞ + A , A
∞




= 4 + A(0) , A(1) ∞ B (0) , B (1) ∞≤ 4 + 2 · 2 = 8,
implying that
kCAB k∞ ≤(3.189)(3.190)(3.191)(3.192)√√
8 = 2 2.

(3.193)Given this and the expression in (3.183),
the probability of winning minus the probability√of losing can never be larger than 2/2 for any quantum strategy. Combined with thefact that the probability of winning summed with the probability of losing is equal to one,
we find√ that the winning probability of any quantum strategy can never be larger than
1/2 + 2/4 = cos2 (π/8).

3.6.3

The Bell States

There are other useful entangled states besides the standard ebit. Suppose that Alice performs a ZA operation on her share of the ebit |Φ+ iAB . Then the resulting state is1
Φ− AB ≡ √ (|00iAB − |11iAB ) .
2(3.194)

Similarly, if Alice performs an X operator or a Y operator, the global state transforms to
the following respective states (up to a global phase):1
Ψ+ AB ≡ √ (|01iAB + |10iAB ) ,21
Ψ− AB ≡ √ (|01iAB − |10iAB ) .
2
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The states |Φ+ iAB , |Φ− iAB , |Ψ+ iAB , and |Ψ− iAB are known as the Bell states and are themost important entangled states for a two-qubit system. They form an orthonormal basis,
called the Bell basis, for a two-qubit space. We can also label the Bell states as|Φzx iAB ≡ ZAz XAx Φ+ AB ,(3.197)

where the two-bit binary number zx indicates whether Alice applies IA , ZA , XA , or ZA XA .
Then the states |Φ00 iAB , |Φ01 iAB , |Φ10 iAB , and |Φ11 iAB are in correspondence with therespective states |Φ+ iAB , |Ψ+ iAB , |Φ− iAB , and |Ψ− iAB .
Exercise 3.6.5 Show that the Bell states form an orthonormal basis:hΦz1 x1 |Φz2 x2 i = δz1 ,z2 δx1 ,x2 .
Exercise 3.6.6 Show that the following identities hold:
1|00iAB = √Φ+ AB + Φ− AB ,21
Ψ+ AB + Ψ− AB ,|01iAB = √2
1|10iAB = √Ψ+ AB − Ψ− AB ,21
Φ+ AB − Φ− AB .
|11iAB = √2(3.198)(3.199)(3.200)(3.201)(3.202)

Exercise 3.6.7 Show that the following identities hold by using the relation in (3.197):
1(3.203)
Φ+ AB = √ (|++iAB + |−−iAB ) ,21
Φ− AB = √ (|−+iAB + |+−iAB ) ,(3.204)2
1
Ψ+ AB = √ (|++iAB − |−−iAB ) ,(3.205)2
1
Ψ− AB = √ (|−+iAB − |+−iAB ) .
(3.206)
2
Entanglement is one of the most useful resources in quantum computing, quantum communication, and in the setting of quantum Shannon theory that we explore in this book. Our
goal in this book is merely to study entanglement as a resource, but there are many otheraspects of entanglement that one can study, such as measures of entanglement, multiparty
entanglement, and generalized Bell’s inequalities (Horodecki et al., 2009).

3.7 Summary and Extensions to Qudit States
We now end our overview of the noiseless quantum theory by summarizing its main postulates
in terms of quantum states that are on d-dimensional systems. Such states are called qudit
states, in analogy with the name “qubit” for two-dimensional quantum systems.
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3.7.1

CHAPTER 3. THE NOISELESS QUANTUM THEORY

QuditsA qudit state |ψi is an arbitrary superposition of some set of orthonormal basis states{|ji}j∈{0,...,d−1} for a d-dimensional quantum system:
|ψi ≡

d−1
X

αj |ji.

(3.207)j=0

The amplitudes αj obey the normalization condition

3.7.2

Pd−1

2
j=0 |αj | = 1.

Unitary Evolution

The first postulate of the quantum theory is that we can perform a unitary (reversible)
evolution U on this state. The resulting state is U |ψi, meaning that we apply the operator
U to the state |ψi.
One example of a unitary evolution is the cyclic shift operator X(x) that acts on theorthonormal states {|ji}j∈{0,...,d−1} as follows:
X(x)|ji = |x ⊕ ji ,(3.208)where ⊕ is a cyclic addition operator, meaning that the result of the addition is (x + j) mod (d).
Notice that the X Pauli operator has a similar behavior on the qubit computational basis
states because
X|ii = |i ⊕ 1i ,(3.209)for i ∈ {0, 1}. Therefore, the operator X(x) is a qudit generalization of the X Pauli operator.
Exercise 3.7.1 Show that the inverse of X(x) is X(−x).
Exercise 3.7.2 Show that the matrix representation X(x) of the X(x) operator, with respect
to the standard basis {|ji}, is a matrix with elements [X(x)]i,j = δi,j⊕x .
Another example of a unitary evolution is the phase operator Z(z). It applies a statedependent phase to a basis state. It acts as follows on the qudit computational basis states{|ji}j∈{0,...,d−1} :Z(z)|ji = exp {i2πzj/d} |ji .
(3.210)This operator is the qudit analog of the Pauli Z operator. The d2 operators {X(x)Z(z)}x,z∈{0,...,d−1}
are known as the Heisenberg–Weyl operators.
Exercise 3.7.3 Show that Z(1) is equivalent to the Pauli Z operator for the case that thedimension d = 2.
Exercise 3.7.4 Show that the inverse of Z(z) is Z(−z).
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Exercise 3.7.5 Show that the matrix representation of the phase operator Z(z), with respect
to the standard basis {|ji}, is[Z(z)]j,k = exp {i2πzj/d} δj,k .

(3.211)

In particular, this result implies that the Z(z) operator has a diagonal matrix representation
with respect to the qudit computational basis states {|ji}j∈{0,...,d−1} . Thus, the qudit computational basis states {|ji}j∈{0,...,d−1} are eigenstates of the phase operator Z(z) (similar to thequbit computational basis states being eigenstates of the Pauli Z operator). The eigenvalue
corresponding to the eigenstate |ji is exp {i2πzj/d}.
Exercise 3.7.6 Show that the eigenstates |eli of the cyclic shift operator X(1) are the Fourieretransformed states |li, whered−1

1 X
exp {i2πlj/d} |ji,|eli ≡ √
d j=0(3.212)and l is an integer in the set {0, . . . , d − 1}. Show that the eigenvalue corresponding to thestate |eli is exp {−i2πl/d}. Conclude that these states are also eigenstates of the operator
X(x), but the corresponding eigenvalues are exp {−i2πlx/d}.
Exercise 3.7.7 Show that the +/− basis states are a special case of the states in (3.212)
when d = 2.
Exercise 3.7.8 The Fourier transform operator F is a qudit
analog of the Hadamard H.
Pd−1
jihj|, where the states |eji
We define it to take Z eigenstates to X eigenstates: F ≡ j=0 |eare defined in (3.212). It performs the following transformation on the qudit computational
basis states:
d−1
1 X
exp {i2πjk/d} |ki.
(3.213)|ji → √
d k=0
Show that the following relations hold for the Fourier transform operator F : F X(x)F † =Z(x), F Z(z)F † = X(−z).
Exercise 3.7.9 Show that the commutation relations of the cyclic shift operator X(x) andthe phase operator Z(z) are as follows:X(x1 )Z(z1 )X(x2 )Z(z2 ) =exp {2πi (z1 x2 − x1 z2 ) /d} X(x2 )Z(z2 )X(x1 )Z(z1 ). (3.214)
You can get this result by first showing that
X(x)Z(z) = exp {−2πizx/d} Z(z)X(x).
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3.7.3

CHAPTER 3. THE NOISELESS QUANTUM THEORY

Measurement of Qudits

Measurement of qudits is similar to measurement of qubits. Suppose that we have somestate |ψi. Suppose further that we would like to measure some Hermitian operator A withthe following diagonalization:
XA=
f (j)Πj ,(3.216)jP

where Πj Πk = Πj δj,k , and j Πj = I. A measurement of the operator A then returns theresult j with the following probability:p(j) = hψ|Πj |ψi,

(3.217)

Π |ψi
pj
.
p(j)

(3.218)and the resulting state is

The calculation of the expectation of the operator A is similar to how we calculate in thequbit case:XX
E [A] =
f (j)hψ|Πj |ψi = hψ|f (j)Πj |ψi = hψ|A|ψi.
(3.219)jj

We give two quick examples of qudit operators that we might like to measure. The
operators X(1) and Z(1) are not completely analogous to the respective Pauli X and Pauli
Z operators because X(1) and Z(1) are not Hermitian. Thus, we cannot directly measurethese operators. Instead, we construct operators that are essentially equivalent to “measuring
the operators” X(1) and Z(1). Let us first consider the Z(1) operator. Its eigenstates are
the qudit computational basis states {|ji}j∈{0,...,d−1} . We can form the operator MZ(1) as
MZ(1) ≡

d−1
X

j|jihj|.

(3.220)j=0Measuring this operator is equivalent to measuring in the qudit computational basis. The
expectation of this operator for a qudit |ψi in the state in (3.207) is
E MZ(1) = hψ|MZ(1) |ψi(3.221)
d−1
X

d−1
d−1
X
X
∗
=
hj |αj 0
j|jihj|αj 00 |j 00 i
j=0
j 0 =0j 00 =0

=

0

d−1
X

jαj∗0 αj 00 hj 0 |ji hj|j 00 i

(3.222)(3.223)

j 0 ,j,j 00 =0

=

d−1
X

j |αj |2 .

(3.224)j=0

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111

Similarly, we can construct an operator MX(1) for “measuring the operator X(1)” by using
the eigenstates |jiX of the X(1) operator:
MX(1) ≡

d−1
X

j|e
jihe
j|.

(3.225)j=0

We leave it as an exercise to determine the expectation when measuring the MX(1) operator.
Exercise 3.7.10 Suppose the qudit is in the state |ψi in (3.207). Show that the expectationof the MX(1) operator is
2

d−1
d−1
X

 1X
αj 0 exp {−i2πj 0 j/d} .
j
E MX(1) =d j=0 j 0 =0(3.226)

Hint: First show that we can represent the state |ψi in the X(1) eigenbasis as follows:
d−1
X
1
√
|ψi =
d
l=0

3.7.4

d−1
X

!
αj exp {−i2πlj/d} |eli.

(3.227)j=0

Composite Systems of Qudits

We can define a system of multiple qudits again by employing the tensor product. A general
two-qudit state on systems A and B has the following form:|ξiAB ≡d−1Xαj,k |jiA |kiB .

(3.228)j,k=0

Evolution of two-qudit states is similar as before. Suppose Alice applies a unitary UA to her
qudit. The result is as follows:(UA ⊗ IB ) |ξiAB = (UA ⊗ IB )d−1
Xαj,k |jiA |kiB

(3.229)

j,k=0

=

d−1
X

αj,k (UA |jiA ) |kiB ,(3.230)j,k=0which follows by linearity. Bob applying a local unitary UB has a similar form. The application of some global unitary UAB results in the state
UAB |ξiAB .
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112

CHAPTER 3. THE NOISELESS QUANTUM THEORYThe Qudit Bell States
Two-qudit states can be entangled as well. The maximally entangled qudit state is as follows:d−1

1 X|ΦiAB ≡ √|iiA |iiB .
d i=0(3.232)

When Alice possesses the first qudit and Bob possesses the second qudit and they are also
separated in space, the above state is a resource known as an edit (pronounced “ee · dit”).
It is useful in the qudit versions of the teleportation protocol and the super-dense coding
protocol discussed in Chapter 6. Throughout the book, we often find it convenient to makeuse of the unnormalized maximally entangled vector:|ΓiAB ≡d−1
X

|iiA |iiB .

(3.233)i=0Consider applying the operator X(x)Z(z) to Alice’s share of the maximally entangledstate |ΦiAB . We use the following notation:|Φx,z iAB ≡ (XA (x)ZA (z) ⊗ IB ) |ΦiAB .

(3.234)The d2 states {|Φx,z iAB }d−1x,z=0 are known as the qudit Bell states and are important in qudit
quantum protocols and in quantum Shannon theory. Exercise 3.7.11 asks you to verify thatthese states form a complete, orthonormal basis. Thus, one can measure two qudits in thequdit Bell basis. Similar to the qubit case, it is straightforward to see that the qudit state
can generate a dit of shared randomness by extending the arguments in Section 3.6.1.
Exercise 3.7.11 Show that the set of states {|Φx,z iAB }d−1x,z=0 forms a complete, orthonormal
basis:d−1XhΦx1 ,z1 |Φx2 ,z2 i = δx1 ,x2 δz1 ,z2 ,(3.235)|Φx,z ihΦx,z |AB = IAB .

(3.236)x,z=0

Exercise 3.7.12 (Transpose Trick) Show that the following “transpose trick” or “ricochet” property holds for a maximally entangled state |ΦiAB (as defined in (3.232)) and any
d × d matrix M :
(MA ⊗ IB ) |ΦiAB = IA ⊗ MBT |ΦiAB ,(3.237)
where M T is the transpose of the operator M with respect to the basis {|iiB } from (3.232).
The implication is that some local action of Alice on |ΦiAB is equivalent to Bob performing
the transpose of this action on his share of the state. Of course, the same equality is true forthe unnormalized maximally entangled vector |ΓiAB from (3.233):

(MA ⊗ IB ) |ΓiAB = IA ⊗ MBT |ΓiAB .
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3.8. SCHMIDT DECOMPOSITION

113

3.8 Schmidt Decomposition
The Schmidt decomposition is one of the most important tools for analyzing bipartite purestates in quantum information theory, showing that it is possible to decompose any pure
bipartite state as a superposition of coordinated orthonormal states. It is a consequenceof the well known singular value decomposition theorem from linear algebra. We state thisresult formally as the following theorem:
Theorem 3.8.1 (Schmidt decomposition) Suppose that we have a bipartite pure state,|ψiAB ∈ HA ⊗ HB ,(3.238)where HA and HB are finite-dimensional Hilbert spaces, not necessarily of the same dimension, and k|ψiAB k2 = 1. Then it is possible to express this state as follows:|ψiAB ≡d−1X

λi |iiA |iiB ,(3.239)i=0

P
where the amplitudes λi are real, strictly positive, and normalized so that i λ2i = 1, thestates {|iiA } form an orthonormal basis for system A, and the states {|iiB } form an orthonormal basis for the system B. The vector [λi ]i∈{0,...,d−1} is called the vector of Schmidt
coefficients. The Schmidt rank d of a bipartite state is equal to the number of Schmidt
coefficients λi in its Schmidt decomposition and satisfiesd ≤ min {dim(HA ), dim(HB )} .

(3.240)

Proof. This is essentially a restatement of the singular value decomposition of a matrix.
Consider an arbitrary bipartite pure state |ψiAB ∈ HA ⊗ HB . Let dA ≡ dim(HA ) anddB ≡ dim(HB ). We can express |ψiAB as follows:|ψiAB =dX
A −1 dXB −1
j=0αj,k |jiA |kiB ,(3.241)k=0

for some amplitudes αj,k and some orthonormal bases {|jiA } and {|kiB } on the respective
systems A and B. Let us write the matrix formed by the coefficients αj,k as some dA × dB
matrix G where[G]j,k = αj,k .
(3.242)Since every matrix has a singular value decomposition, we can write G as
G = U ΛV,

(3.243)where U is a dA × dA unitary matrix, V is a dB × dB unitary matrix, and Λ is a dA × dB
matrix with d real, strictly positive numbers λi along the diagonal and zeros elsewhere. Let
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CHAPTER 3. THE NOISELESS QUANTUM THEORY

us write the matrix elements of U as uj,i and those of V as vi,k . The above matrix equationis then equivalent to the following set of equations:
αj,k =d−1
Xuj,i λi vi,k .

(3.244)i=0

Let us make this substitution into the expression for the state in (3.241):!
dX
d−1
A −1 dX
B −1
X|ψiAB =
uj,i λi vi,k |jiA |kiB .
j=0

k=0

(3.245)i=0

Readjusting some terms by exploiting the properties of the tensor product, we find that
!
!
dX
dX
d−1
A −1
B −1
X
|ψiAB =
λiuj,i |jiA ⊗vi,k |kiB(3.246)
i=0

=

d−1
X

j=0

λi |iiA |iiB ,k=0(3.247)i=0

P
where we define the orthonormal basis on the A systemas|ii≡
j uj,i |jiA and we defineAP
the orthonormal basis on the B system as |iiB ≡ k vi,k |kiB . This final step completes theproof of the theorem, but Exercise 3.8.1 asks you to verify that the set of states {|iiA } form
an orthonormal basis (the proof for the set of states {|iiB } is similar).
The statement of Theorem 3.8.1 is rather remarkable after pausing to think about it
further. For example, the Hilbert space HA of Alice could be a qubit Hilbert space ofdimension two, and the Hilbert space HB of Bob could be of dimension one billion (or someother large number). Then, in spite of this large dimension for Bob’s Hilbert space, if weknow that the state of systems A and B is a pure state, then it is always possible to find a
two-dimensional subspace of HB which along with HA suffices to represent the state. So all
those extra degrees of freedom are unnecessary in this example. Often in quantum Shannon
theory, we are optimizing certain functions over pure states. In such cases, the Schmidt
decomposition theorem is helpful in limiting the size of the space we have to consider in such
optimization problems.
Remark 3.8.1 The Schmidt decomposition applies not only to bipartite systems but to any
number of systems where we can make a bipartite cut of the systems. For example, supposethat there is a state |φiABCDE on systems ABCDE. We could say that AB are part of one
system and CDE are part of another system and write a Schmidt decomposition for thisstate as follows:
Xp
pY (y)|yiAB |yiCDE ,(3.248)|φiABCDE =ywhere {|yiAB } is an orthonormal basis for the joint system AB and {|yiCDE } is an orthonormal basis for the joint system CDE.
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3.9. HISTORY AND FURTHER READING

115

Exercise 3.8.1 Verify that the set of states {|iiA } from the proof of Theorem 3.8.1 forms
an orthonormal basis by exploiting the unitarity of the matrix U .
Exercise 3.8.2 Prove that the Schmidt decomposition gives a way to identify if a pure stateis entangled or product. In particular, prove that a pure bipartite state is entangled if andonly if it has more than one Schmidt coefficient. First, suppose that a pure bipartite state|φiAB has only one Schmidt coefficient. Prove that its maximum overlap with a product stateis equal to one:max |hϕ|A ⊗ hψ|B |φiAB |2 = 1.
(3.249)|ϕiA ,|ψiB

Now, suppose that there is more than one Schmidt coefficient for a state |φiAB . Prove that
this state’s maximum overlap with a product state is strictly less than one (and thus it cannot
be written as a product state):
max |hϕ|A ⊗ hψ|B |φiAB |2 < 1.

|ϕiA ,|ψiB(3.250)(Hint: Use the Schmidt! Use the Schwarz! (as in Cauchy–Schwarz...) )3.9 History and Further Reading
There are many great books on quantum mechanics that outline the mathematical background. The books of Bohm (1989), Sakurai (1994), and Nielsen and Chuang (2000) areamong these. The ideas for the resource inequality formalism first appeared in the popular
article (Bennett, 1995) and another of Bennett’s papers (Bennett, 2004). The no-deletion
theorem is in (Pati and Braunstein, 2000). The review article of the Horodecki family isa helpful reference on the study of quantum entanglement (Horodecki et al., 2009). Ourpresentation of the CHSH game and its analysis follows the approach detailed in (Scarani,2013). The bound on the maximum quantum winning probability of the CHSH game wasestablished in (Tsirelson, 1980).

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c 2019CHAPTER 3. THE NOISELESS QUANTUM THEORYMark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 4

The Noisy Quantum Theory
In general, we may not know for certain whether we possess a particular quantum state.
Instead, we may only have a probabilistic description of an ensemble of quantum states.
This chapter re-establishes the foundations of the quantum theory so that they incorporatea lack of complete information about a quantum system. The density operator formalism isa powerful mathematical tool for describing this scenario. This chapter also establishes how
to model the noisy evolution of a quantum system, and we explore models of noisy quantum
channels that are generalizations of the noisy classical channel discussed in Section 2.2.3 ofChapter 2.
You might have noticed that the development in the previous chapter relied on thepremise that the possessor of a quantum system has perfect knowledge of the state of agiven system. For instance, we assumed that Alice knows that she possesses a qubit in thestate |ψi where|ψi = α|0i + β|1i,(4.1)
for some α, β ∈ C such that |α|2 + |β|2 = 1. Also, in some examples, we assumed that
Alice and Bob know that they share an ebit |Φ+ iAB . We even assumed perfect knowledgeof a unitary evolution or a particular measurement that a possessor of a quantum state may
apply to it.
This assumption of perfect, definite knowledge of a quantum state is a difficult one tojustify in practice. In reality, it is challenging to prepare, evolve, or measure a quantum state
exactly as we wish. Slight errors may occur in the preparation, evolution, or measurement
due to imprecise devices or to coupling with other degrees of freedom outside of the systemthat we are controlling. An example of such imprecision can occur in the coupling of two
photons at a beamsplitter. We may not be able to tune the reflectivity of the beamsplitter
exactly or may not have the timing of the arrival of the photons exactly set. The noiselessquantum theory as we presented it in the previous section cannot handle such imprecisions.
In this chapter, we relax the assumption of perfect knowledge of the preparation, evolution, or measurement of quantum states and develop a noisy quantum theory that incorporates an imprecise knowledge of these states. The noisy quantum theory fuses probability
theory and the quantum theory into a single formalism.
117

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CHAPTER 4. THE NOISY QUANTUM THEORY

We proceed with the development of the noisy quantum theory in the following order:
1. We first present the density operator formalism, which gives a representation for anoisy, imprecise quantum state.
2. We then discuss a general form of measurements and the effect of them on our description of a noisy quantum state.
3. We proceed to composite noisy systems, which admit a particular form, and we discuss
several possible states of composite noisy systems including product states, separable
states, classical–quantum states, entangled states, and arbitrary states.
4. Next, we consider the Kraus representation of a quantum channel, which gives a way
to describe noisy evolution, and we discuss important examples of noisy quantum
channels. We also stress how every operation we have discussed so far, including
preparations and measurements, can be viewed as quantum channels.

4.1 Noisy Quantum States
We generally may not have perfect knowledge of a prepared quantum state. Suppose a third
party, Bob, prepares a state for us and only gives us a probabilistic description of it. Thatis, we might only know that Bob selects the state |ψx i with a certain probability pX (x). Our
description of the state is then as an ensemble E of quantum states whereE ≡ {pX (x), |ψx i}x∈X .

(4.2)

In the above, X is a random variable with distribution pX (x). Each realization x of randomvariable X belongs to an alphabet X . Thus, the realization x merely acts as an index,
meaning that the quantum state is |ψx i with probability pX (x). We also assume that eachstate |ψx i is a d-dimensional qudit state.
A simple example is the following ensemble: {{1/3, |1i} , {2/3, |3i}}. The states |1i
and |3i are in a four-dimensional Hilbert space with basis states {|0i, |1i, |2i, |3i}. Theinterpretation of this ensemble is that the state is |1i with probability 1/3 and the stateis |3i with probability 2/3.

4.1.1

The Density Operator

Suppose now that we have the ability to perform a perfect, projective measurement of asystem with ensembleP
description E in (4.2). Let Πj be the elements of this projective
measurement so that j Πj = I, and let J be the random variable corresponding to themeasurement outcome j. Let us suppose at first, without loss of generality, that the state in
the ensemble is |ψx i for some x ∈ X . Then the Born rule of the noiseless quantum theoryc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

4.1. NOISY QUANTUM STATES

119

states that the conditional probability pJ|X (j|x) of obtaining measurement result j (giventhat the state is |ψx i) is equal to
pJ|X (j|x) = hψx |Πj |ψx i,

(4.3)

p
and the post-measurement state is Πj |ψx i/ pJ|X (j|x). However, we would also like to know
the unconditional probability pJ (j) of obtaining measurement result j for the ensemble description E. By the law of total probability, the unconditional probability pJ (j) isX
pJ (j) =pJ|X (j|x)pX (x)
(4.4)
x∈X

=

X

hψx |Πj |ψx ipX (x).

(4.5)

x∈X

At this point, it is helpful for us to introduce the trace of a square operator, which will
be used extensively throughout this book.
Definition 4.1.1 (Trace) The trace Tr {A} of a square operator A acting on a Hilbert
space H is defined as follows:X
Tr {A} ≡hi|A|ii,(4.6)
iwhere {|ii} is some complete, orthonormal basis for H.
Observe that the trace operation is linear. It is also independent of which orthonormal
basis we choose because
X
Tr {A} =hi|A|ii
(4.7)
i
!
X
X
=
hi|A
|φj ihφj | |ii
i

(4.8)

j

X
=
hi|A|φj ihφj |ii

(4.9)

i,j

X
=
hφj |iihi|A|φj i(4.10)

i,j

!
X
X
=
hφj |
|iihi| A|φj ij=(4.11)iX
hφj |A|φj i.

(4.12)

jIn the above, {|φj i} is P
some other orthonormal
basis for H and we made use of the comP
pleteness relation: I = j |φj ihφj | = i |iihi|.
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CHAPTER 4. THE NOISY QUANTUM THEORY

Exercise 4.1.1 Prove that the trace is cyclic. That is, for three operators A, B, and C, thefollowing relation holds Tr{ABC} = Tr{CAB} = Tr{BCA}.
Returning to (4.5), we can then show the following useful property:!
Xhψx |Πj |ψx i = hψx ||iihi| Πj |ψx i(4.13)i=Xhψx |ii hi|Πj |ψx i(4.14)

hi|Πj |ψx i hψx |ii(4.15)

i=

X
i= Tr {Πj |ψx ihψx |} .
(4.16)P
The first equality uses the completeness relation i |iihi| = I. Thus, we continue with thedevelopment in (4.5) and show thatX
Tr {Πj |ψx ihψx |} pX (x)(4.17)pJ (j) =x∈X(= Tr Πj)X

pX (x)|ψx ihψx | .

(4.18)x∈X

We can rewrite the last equation as follows:pJ (j) = Tr {Πj ρ} ,(4.19)

introducing ρ as the density operator corresponding to the ensemble E:
Definition 4.1.2 (Density Operator) The density operator ρ corresponding to an ensemble E ≡ {pX (x), |ψx i}x∈X is defined asXρ≡pX (x)|ψx ihψx |.
(4.20)x∈X

The operator ρ as defined above is known as the density operator because it is thequantum generalization of a probability density function. Throughout this book, we often
use the symbols ρ, σ, τ , π, and ω to denote density operators.
We sometimes refer to the density operator as the expected density operator because there
is a sense in which we are taking the expectation over all of the states in the ensemble inorder to obtain the density operator. We can equivalently write the density operator as
follows:
ρ = EX {|ψX ihψX |} ,(4.21)
where the expectation is with respect to the random variable X. Note that we are careful
to use the notation |ψX i instead of the notation |ψx i for the state inside of the expectation
because the state |ψX i is a random quantum state, random with respect to a classical randomvariable X.
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4.1. NOISY QUANTUM STATES

121

Exercise 4.1.2 Suppose the ensemble has a degenerate probability distribution, say pX (0) =1 and pX (x) = 0 for all x 6= 0. What is the density operator of this degenerate ensemble?
Exercise 4.1.3 Prove the following equality:
Tr {A} = hΓ|RS IR ⊗ AS |ΓiRS ,(4.22)

where A is a square operator acting on a Hilbert space HS , IR is the identity operator acting
on a Hilbert space HR isomorphic to HS and |ΓiRS is the unnormalized maximally entangled
vector from (3.233). This gives an alternate formula for the trace of a square operator A.
Exercise 4.1.4 Prove that Tr{f (G† G)} = Tr{f (GG† )}, where G is any operator (not necessarily Hermitian) and f is any function. (Recall the convention for a function of an
operator given in Definition 3.3.1.)
Properties of the Density Operator
What are the properties that a given density operator corresponding to an ensemble satisfies?
Let us consider taking the trace of ρ:()
X
Tr {ρ} = Tr
pX (x)|ψx ihψx |(4.23)x∈X=

X

pX (x) Tr {|ψx ihψx |}(4.24)pX (x) hψx |ψx i(4.25)pX (x)(4.26)

x∈X

=

X
x∈X

=

X
x∈X

= 1.

(4.27)

The above development shows that every density operator corresponding to an ensemble has
unit trace.
Let us consider taking the conjugate transpose of the density operator ρ:
!†
X

ρ† =

pX (x)|ψx ihψx |(4.28)pX (x) (|ψx ihψx |)†(4.29)pX (x)|ψx ihψx |(4.30)

x∈X

=

X
x∈X

=

X
x∈X

= ρ.
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CHAPTER 4. THE NOISY QUANTUM THEORY

Every density operator is thus a Hermitian operator as well because the conjugate transposeof ρ is ρ.
Every density operator is furthermore positive semi-definite, meaning thathϕ|ρ|ϕi ≥ 0∀ |ϕi.

(4.32)

We write ρ ≥ 0 to indicate that an operator is positive semi-definite. A proof for nonnegativity of any density operator ρ is as follows:!
X
hϕ|ρ|ϕi = hϕ|pX (x)|ψx ihψx | |ϕi(4.33)x∈X=

X

pX (x) hϕ|ψx i hψx |ϕi(4.34)pX (x) |hϕ|ψx i|2 ≥ 0.

(4.35)

x∈X

=

X
x∈X

The inequality follows because each pX (x) is a probability and is therefore non-negative.
Ensembles and the Density Operator
Every ensemble has a unique density operator, but the opposite does not necessarily hold:every density operator does not correspond to a unique ensemble and could correspond to
many ensembles. However, there are restrictions on which ensembles can realize a given
density operator and there is a relation between them. We return to this question in Section 5.1.2, after we have developed more tools.
Exercise 4.1.5 Show that the ensembles {{1/2, |0i} , {1/2, |1i}} and {{1/2, |+i} , {1/2, |−i}}have the same density operator.
This last result has profound implications for the predictions of the quantum theory
because it is possible for two or more completely different ensembles to have the same probabilities for measurement results. It also has important implications for quantum Shannon
theory as well.
By the spectral theorem, it follows that every density operator ρ has a spectral decomposition in terms of eigenstates {|φx i}x∈{0,...,d−1} because every ρ is Hermitian:
ρ=

d−1
X

λx |φx ihφx |,(4.36)

x=0

where the coefficients λx are the eigenvalues.
Exercise 4.1.6 Show that the coefficients λx are probabilities using the facts that Tr {ρ} = 1
and ρ ≥ 0.
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Thus, given any density operator ρ, we can define a “canonical” ensemble {λx , |φx i}
corresponding to it. Note that this ensemble is not unique: if λx = λx0 for x 6= x0 , then the
choice of eigenvectors corresponding to these eigenvalues is not unique. The fact that an
ensemble can correspond to a density operator is so important for quantum Shannon theory
that we see this idea arise again and again throughout this book. Any ensemble arising fromthe spectral theorem is the most “efficient” ensemble, in a sense, and we will explore thisidea more in Chapter 18 on quantum data compression.
Density Operator as the State
We can also refer to the density operator as the state of a given quantum system because it ispossible to use it to calculate probabilities for any measurement performed on that system.
We can make these calculations without having an ensemble description—all we need is thedensity operator. The noisy quantum theory also subsumes the noiseless quantum theory
because any state |ψi has a corresponding density operator |ψihψ| in the noisy quantum
theory, and all calculations with this density operator in the noisy quantum theory give thesame results as using the state |ψi in the noiseless quantum theory. For these reasons, wewill say that the state of a given quantum system is a density operator.
Definition 4.1.3 (Density Operator as the State) The state of a quantum system isgiven by a density operator ρ, which is a positive semi-definite operator with trace equal toone. Let D(H) denote the set of all density operators acting on a Hilbert space H.
One of the most important states is the maximally mixed state π:
Definition 4.1.4 (Maximally Mixed State) The maximally mixed state π is the density
operator corresponding to a uniform ensemble of orthogonal states d1 , |xi , where d is thedimensionality of the Hilbert space. The maximally mixed state π is then equal toI
1X|xihx| = .
(4.37)π≡d x∈XdExercise 4.1.7 Show that π is the density operator of the ensemble that chooses |0i, |1i,|+i, |−i with equal probability.
Exercise 4.1.8 (Convexity) Show that the set of density operators acting on a given
Hilbert space is a convex set. That is, if λ ∈ [0, 1] and ρ and σ are density operators,
then λρ + (1 − λ)σ is a density operator.
Definition 4.1.5 (Purity) The purity P (ρ) of a density operator ρ is equal to
P (ρ) ≡ Tr ρ† ρ = Tr ρ2 .

(4.38)The purity is one particular measure of the noisiness of a quantum state. The purity ofa pure state is equal to one, and the purity of a mixed state is strictly less than one, as thefollowing exercise asks you to verify.
Exercise 4.1.9 Prove that the purity of a density operator ρ is equal to one if and only if
ρ is a pure state, such that it can be written as ρ = |ψihψ| for some unit vector ψ.
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124

CHAPTER 4. THE NOISY QUANTUM THEORYThe Density Operator on the Bloch Sphere
Consider that the following pure qubit state|ψi ≡ cos(θ/2)|0i + eiϕ sin(θ/2)|1i
has the following density operator representation:|ψihψ| = cos(θ/2)|0i + eiϕ sin(θ/2)|1i cos(θ/2)h0| + e−iϕ sin(θ/2)h1|(4.39)(4.40)= cos2 (θ/2)|0ih0| + e−iϕ sin(θ/2) cos(θ/2)|0ih1|+ eiϕ sin(θ/2) cos(θ/2)|1ih0| + sin2 (θ/2)|1ih1|.

(4.41)The matrix representation, or density matrix, of this density operator with respect to thecomputational basis is as follows:
cos2 (θ/2)
e−iϕ sin(θ/2) cos(θ/2).
(4.42)
eiϕ sin(θ/2) cos(θ/2)
sin2 (θ/2)
Using trigonometric identities, it follows that the density matrix is equal to the following
matrix:
1
1 + cos(θ)
sin(θ) (cos (ϕ) − i sin(ϕ)).
(4.43)
1 − cos(θ)
2 sin(θ) (cos(ϕ) + i sin (ϕ))
We can further exploit the Pauli matrices, defined in Section 3.3.3, to represent the density
matrix as follows:1(I + rx X + ry Y + rz Z) ,(4.44)
2
where rx = sin(θ) cos(ϕ), ry = sin(θ) sin(ϕ), and rz = cos(θ). The coefficients rx , ry , and rz
are none other than the Cartesian coordinate representation of the angles θ and ϕ, and they
thus correspond to a unit vector.
More generally, the formula in (4.44) can represent an arbitrary qubit density operator
where the coefficients rx , ry , and rz do not necessarily correspond to a unit vector, but rather
a vector r such that krk2 ≤ 1. Consider that the density matrix in (4.44) is as follows:
1 1 + rz rx − iry.
(4.45)
2 rx + iry 1 − rz
The above matrix corresponds to a valid density matrix because it has unit trace, it is
Hermitian, and it is non-negative (the next exercise asks you to verify these facts). This
alternate representation of the density matrix as a vector in the Bloch sphere is useful for
visualizing noisy qubit processes in the noisy quantum theory.
Exercise 4.1.10 Show that the matrix in (4.45) has unit trace, is Hermitian, and is nonnegative for all r such that krk2 ≤ 1. It thus corresponds to any valid density matrix.
Exercise 4.1.11 Show that we can compute the Bloch sphere coordinates rx , ry , and rz withthe respective formulas Tr {Xρ}, Tr {Y ρ}, and Tr {Zρ} using the representation in (4.45)
and the result of Exercise 3.3.6.
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Figure 4.1: The mixing process by which we can generate an “ensemble of ensembles.” First choose arealization x according to distribution pX (x). Then choose a realization y according to the conditional
distribution pY |X (y|x). Finally, choose
P a state |ψx,y i according to the realizations x and y. This leads to an
ensemble {pX (x), ρx } where ρx ≡ y pY |X (y|x)|ψx,y ihψx,y |.

Exercise 4.1.12 Show that the eigenvalues of a general qubit density operator with density
matrix representation in (4.45) are as follows: 21 (1 ± krk2 ) .
Exercise 4.1.13 Show that a mixture of pure states |ψj i each with Bloch
P vector rj andprobability p(j) gives a density matrix with the Bloch vector r where r = j p(j)rj .

4.1.2

An Ensemble of Ensembles

The most general ensemble that we can construct is an ensemble of ensembles, i.e., anensemble F of density operators whereF ≡ {pX (x), ρx } .

(4.46)The ensemble F essentially has two layers of randomization. The first layer is from the distribution pX (x). Each density operator ρx in F arises from an ensemble pY |X (y|x), |ψx,y i .
The conditional distribution pY |X (y|x) represents the second layer of randomization. Each ρx
is a density operator with respect to the above ensemble:
Xρx ≡pY |X (y|x)|ψx,y ihψx,y |.
(4.47)y

The ensemble F has its own density operator ρ whereXXρ≡pY |X (y|x)pX (x)|ψx,y ihψx,y | =pX (x)ρx .
x,y(4.48)x

The density operator ρ is the density operator from the perspective of someone who does
not possess x. Figure 4.1 displays the process by which we can select the ensemble F.

4.1.3

Noiseless Evolution of an Ensemble

Quantum states can evolve in a noiseless fashion either according to a unitary operator ora measurement. In this section, we determine the noiseless evolution of an ensemble andits corresponding density operator. We also show how density operators evolve under aquantum measurement.
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CHAPTER 4. THE NOISY QUANTUM THEORY

Noiseless Unitary Evolution of a Noisy StateWe first consider noiseless evolution according to some unitary U . Suppose we have theensemble E in (4.2) with density operator ρ. Suppose without loss of generality that thestate is |ψx i. Then the evolution postulate of the noiseless quantum theory gives that thestate after the unitary evolution is as follows: U |ψx i. This result implies that the evolution
leads to a new ensemble
EU ≡ {pX (x), U |ψx i}x∈X .
(4.49)The density operator of the evolved ensemble is!
X

pX (x)U |ψx ihψx |U † = U

x∈X

X

pX (x)|ψx ihψx | U †x∈X†= U ρU .

(4.50)(4.51)

Thus, the above relation shows that we can keep track of the evolution of the density
operator ρ, rather than worrying about keeping track of the evolution of every state in
the ensemble E. It suffices to keep track of only the density operator evolution because thisoperator is sufficient to determine probabilities when performing any measurement on thesystem.
Noiseless Measurement of a Noisy State
In a similar fashion, we can analyze the result of a measurement on a system with ensemble
description E in (4.2). P
Suppose that we perform a projective measurement with projection
operators {Πj }j where j Πj = I. The main result of this section is that two things happen
after a measurement occurs. First, as shown in the development preceding (4.19), we receive
the outcome j with probability pJ (j) = Tr{Πj ρ}. Second, if the outcome of the measurement
is j, then the state evolves as follows:ρ −→Πj ρΠj.
pJ (j)(4.52)To see the above, let us suppose that the state in the ensemble E is |ψx i. Then thenoiseless quantum theory predicts that the probability of obtaining outcome j conditionedon the index x is
pJ|X (j|x) = hψx |Πj |ψx i,(4.53)and the resulting state isΠ |ψ ip j x.
pJ|X (j|x)(4.54)

Supposing that we receive outcome j, then we have a new ensemble:()Πj |ψx i.
Ej ≡ pX|J (x|j), p
pJ|X (j|x)(4.55)

x∈X

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4.1. NOISY QUANTUM STATES127

The density operator for this ensemble is
X

pX|J (x|j)

x∈X

= Πj

X pX|J (x|j)
x∈X

= Πj
=
=

pJ|X (j|x)!
|ψx ihψx | Πj

X pJ|X (j|x)pX (x)x∈XΠjΠj |ψx ihψx |Πj
pJ|X (j|x)

P

pJ|X (j|x)pJ (j)(4.56)!

|ψx ihψx | Πjx∈X pX (x)|ψx ihψx |ΠjpJ (j)
Πj ρΠj.
pJ (j)(4.57)(4.58)(4.59)The second equality follows from applying the Bayes rule: pX|J (x|j) = pJ|X (j|x)pX (x)/pJ (j).

4.1.4

Probability Theory as a Special Case

It may help to build some intuition for the noisy quantum theory by showing how it contains
probability theory as a special case. Indeed, we should expect this containment of probability
theory within the noisy quantum theory to hold if the noisy quantum theory is making
probabilistic predictions about the physical world.
Let us again begin with an ensemble of quantum states, but this time, let us pick thestates in the ensemble to be special states, where they are all orthogonal to one another. If
the states in the ensemble are all orthogonal to one another, then they are essentially classicalstates because there is a measurement that distinguishes them from one another. So, let us
pick the ensemble to be {pX (x), |xi}x∈X where the states {|xi}x∈X form an orthonormal basis
for a Hilbert space of dimension |X |. These states are classical because a measurement withthe following projection operators can distinguish them:{|xihx|}x∈X .

(4.60)The generalization of a probability distribution to the quantum world is the density
operator:pX (x) ↔ ρ.
(4.61)The reason for this is that we can use the density operator to calculate expectations andmoments of observables. Furthermore, a probability distribution can be encoded as a density
operator that is diagonal with respect to a known orthonormal basis, as follows:X

pX (x)|xihx|.

(4.62)

x∈X
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CHAPTER 4. THE NOISY QUANTUM THEORY

The generalization of a random variable is an observable. For example, let us consider
the following observable:
X
X≡
x|xihx|,(4.63)
x∈X

analogous to the observable in (3.220). We perform the following calculation to determine
the expectation of the observable X:
Eρ [X] = Tr {Xρ} .

(4.64)

Explicitly calculating this quantity, we find that it is consistent with the formula for the
expectation of random variable X with probability distribution pX (x):()
X
X
Tr {Xρ} = Tr
x|xihx|pX (x0 )|x0 ihx0 |(4.65)
x0 ∈X

x∈X

=

X

2

x pX (x0 ) |hx|x0 i|(4.66)

x,x0 ∈X=

X

x pX (x).

(4.67)x∈X

Another useful notion in probability theory is the notion of an indicator random variable IA (X). We define the indicator function IA (x) for a set A as follows:
1 : x∈A
IA (x) ≡.
(4.68)
0 : x∈/A
The expectation E [IA (X)] of the indicator random variable IA (X) is
X
pX (x) ≡ pX (A),E [IA (X)] =(4.69)

x∈A

where pX (A) represents the probability of the set A. In the quantum theory, we can define
an indicator observable IA (X):X
IA (X) ≡|xihx|.
(4.70)x∈A

It has eigenvalues equal to one for all eigenvectors with labels x in the set A, and it has zero
eigenvalues for those eigenvectors with labels outside of A. It is straightforward to show thatthe expectation Tr {IA (X)ρ} of the indicator observable IA (X) is pX (A).
You may have noticed that the indicator observable is also a projection operator. So,
according to the postulates of the quantum theory, we can perform a measurement withelements:{IA (X), IAc (X) ≡ I − IA (X)} .
(4.71)The result of such a projective measurement is to project onto the subspace given by IA (X)with probability pX (A) and to project onto the complementary subspace given by IAc (X)with probability 1 − pX (A).
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129

We highlight the connection between the noisy quantum theory and probability theory
with two more examples. First, suppose that we have two disjoint sets A and B. Then theprobability of their union is the sum of the probabilities of the individual sets:
Pr {A ∪ B} = Pr {A} + Pr {B} ,(4.72)and the probability of the complementary set (A ∪ B)c = Ac ∩ B c is equal to 1 − Pr {A} −Pr {B}. We can perform the analogous calculation in the noisy quantum theory. Let us
consider two projection operators
X
X
Π(A) ≡|xihx|, Π(B) ≡|xihx|.
(4.73)x∈Ax∈B

The sum of these projection operators gives a projection onto the union set A ∪ B:X
Π (A ∪ B) ≡|xihx| = Π(A) + Π(B).

(4.74)x∈A∪B

Exercise 4.1.14 Show that Tr {Π(A ∪ B)ρ} = Pr{A} + Pr{B} whenever the projectors
Π(A) and Π(B) satisfy Π(A)Π(B) = 0 and the density operator ρ is diagonal in the same
basis as Π(A) and Π(B).
We can also consider intersections of sets. Suppose that we have two sets A and B.
The intersection of these two sets consists of all the elements that are common to both sets.
There is an associated probability Pr {A ∩ B} with the intersection. We can again formulate
this idea in the noisy quantum theory. Consider the projection operators in (4.73). The
multiplication of these two projectors gives a projector onto the intersection of the two
spaces:Π (A ∩ B) = Π(A)Π(B).
(4.75)
Exercise 4.1.15 Show that Tr {Π(A)Π(B)ρ} = Pr {A ∩ B} whenever the density operator
ρ is diagonal in the same basis as Π(A) and Π(B).
Such ideas and connections to the classical world are crucial for understanding quantum
Shannon theory. Many times, we will be thinking about unions of disjoint subspaces and it is
helpful to make the analogy with a union of disjoint sets. Also, in Chapter 17 on the coveringlemma, we will use projection operators to remove some of the support of an operator, andthis operation is analogous to taking intersections of sets.
Despite the fact that there is a strong connection for classical states, some of this intuition
breaks down by considering the non-orthogonality of quantum states. For example, consider
the case of the projectors Π0 ≡ |0ih0| and Π+ ≡ |+ih+|. The two subspaces onto which these
operators project do not intersect, yet we know that the projectors have some overlap because
their corresponding states are non-orthogonal. One analogy of the intersection operation is
to sandwich one operator with another. For example, we can form the operators
Π0 Π+ Π0 ,c 2019Π+ Π0 Π+ .

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CHAPTER 4. THE NOISY QUANTUM THEORY

If the two projectors were to commute, then this ordering would not matter, and the resulting
operator would be a projector onto the intersection of the two subspaces. But this is not thecase for our example here, and the resulting operators are quite different.
Exercise 4.1.16 (Union Bound) Prove a union bound for commuting projectors Π1 andΠ2 where 0 ≤ Π1 , Π2 ≤ I and for an arbitrary density operator ρ (not necessarily diagonalin the same basis as Π1 and Π2 ):Tr {(I − Π1 Π2 ) ρ} ≤ Tr {(I − Π1 ) ρ} + Tr {(I − Π2 ) ρ} .

(4.77)

4.2 Measurement in the Noisy Quantum Theory
We have described measurement in the quantum theory using a set of projectors that form aresolutionof the identity. For example, the set {Πj }j of projectors that satisfy the conditionPΠ=I
form a valid projective quantum measurement.
j j
There is an alternate description of quantum measurements that follows from allowing
the system of interest to interact unitarily with a probe system that we measure after theinteraction occurs. So suppose that the system of interest is in a state |ψiS and that theprobe is in a state |0iP , so that the overall state before anything happens is as follows:|ψiS ⊗ |0iP .

(4.78)

Let {|0iP , |1iP , . . . , |d − 1iP } be an orthonormal basis for the probe system (assuming that it
has dimension d). Now suppose that the system and the probe interact according to a unitary
USP , and then we perform a measurement of the probe system, described by measurement
operators {|jihj|P }. The probability to obtain outcome j is†
pJ (j) = hψ|S ⊗ h0|P USP (IS ⊗ |jihj|P ) (USP |ψiS ⊗ |0iP ) ,(4.79)and the post-measurement state upon obtaining outcome j is1p
(IS ⊗ |jihj|P ) (USP |ψiS ⊗ |0iP ) .
pJ (j)(4.80)

We can rewrite the expressions above in a different way. Let us expand the unitary
operator USP in the orthonormal basis of the probe system P as follows:
USP =

X

MSj,k ⊗ |jihk|P ,(4.81)

j,k

where {MSj,k } is a set of operators. Up to a permutation of the S and P systems and using
the mathematics of the tensor product (described in Section 3.5.1), this is the same as writingc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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131

the unitary USP as follows:MS0,0 M 1,0
 S
 ..
 .

MS0,1
MS1,1
..
.



···
···
...

MSd−1,0 MSd−1,1 · · ·
MS0,d−1MS1,d−1 

.
..

.
d−1,d−1MS(4.82)

This set {MSj,k } needs to satisfy some constraints corresponding to the unitarity of USP . In
particular, consider the following operator:
X j,0MS ⊗ |jih0|P ,(4.83)
jwhich corresponds to the first column of operator-valued entries in USP , as illustrated in†(4.82). In what follows, we employ the shorthand MSj ≡ MSj,0 . From the fact that USP
USP =ISP = IS ⊗ IP , we deduce that the following equality must hold
!
!
X j0†
X j
IS ⊗ |0ih0|P =MS ⊗ |0ihj 0 |PMS ⊗ |jih0|P(4.84)
j0

=

X

j
j0†

MS MSj ⊗ |0i hj 0 |ji h0|P

(4.85)

MSj† MSj ⊗ |0ih0|P ,(4.86)

j 0 ,j=X
jwhere the last line follows from the fact that we chose an orthonormal basis in the representation of USP in (4.81). So the above equality implies that the following condition
holds
X j† jMS MS = IS .
(4.87)
j

Plugging (4.81) into (4.79) and (4.80), a short calculation (similar to the above one)
reveals that they simplify as follows:pJ (j) = hψ|Mj† Mj |ψi,
Mj |ψiS ⊗ |jiP1pp
(IS ⊗ |jihj|P ) (USP |ψiS ⊗ |0iP ) =.
pJ (j)pJ (j)(4.88)(4.89)

Since the system and the probe are in a pure product state (and thus independent of eachother) after the measurement occurs, we can discard the probe
p system and deduce that the
post-measurement state of the system S is simply Mj |ψiS / pJ (j).
Motivated by the above development, we allow for an alternate notion of quantum measurement, saying that it consists of a set of measurement operators {Mj }j that satisfy thefollowing completeness condition:X †Mj Mj = I.
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CHAPTER 4. THE NOISY QUANTUM THEORY

Observe from the above development that this is the only constraint that the operators{Mj } need to satisfy. This constraint is a consequence of unitarity, but can be viewed
as a generalization of the completeness relation for a set of projectors that constitute a
projective quantum measurement. Given a set of measurement operators of the above form,
the probability for obtaining outcome j when measuring a state |ψi is
pJ (j) ≡ hψ|Mj† Mj |ψi,

(4.91)and the post-measurement state when we receive outcome j is
M |ψi
pj
.
pJ (j)(4.92)

Suppose that we instead have an ensemble {pX (x), |ψx i} with density operator ρ. We
can carry out an analysis similar to that which led to (4.59) to conclude that the probabilitypJ (j) for obtaining outcome j ispJ (j) ≡ Tr{Mj† Mj ρ},(4.93)and the post-measurement state when we measure result j is
Mj ρMj†.
pJ (j)(4.94)The expression pJ (j) = Tr{Mj† Mj ρ} is a reformulation of the Born rule.

4.2.1

POVM Formalism

Sometimes, we simply may not care about the post-measurement state of a quantum measurement, but instead we only care about the probability for obtaining a particular outcome.
For example, this situation arises in the transmission of classical data over a quantum channel. In this situation, we are merely concerned with minimizing the error probabilities of
the classical transmission. The receiver does not care about the post-measurement state because he no longer needs it in the quantum information-processing protocol. We can specify
a measurement of this sort by a POVM, defined as follows:
Definition 4.2.1 (POVM) A positive operator-valued measure (POVM) is a set {Λj }j of
operators that satisfy non-negativity and completeness:X∀j : Λj ≥ 0,
Λj = I.
(4.95)
j

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if the state is some pure state |ψi. The probability for obtaining outcome j is
Tr {Λj ρ} ,(4.97)

if the state is in a mixed state described by some density operator ρ. This is anotherreformulation of the Born rule.
Exercise 4.2.1 Consider the following five “Chrysler” states:|ek i ≡ cos(2πk/5)|0i + sin(2πk/5)|1i,(4.98)

where k ∈ {0, . . . , 4}. These states are the “Chrysler” states because they form a pentagon
on the XZ-planeof the Bloch sphere. Show that the following set of operators forms a valid
POVM: 52 |ek ihek | .
Exercise 4.2.2 Suppose we have an ensemble {pX (x), ρx } of density operators and a POVM
with elements {Λx } that should identify the states ρx with high probability, i.e., we would like
Tr {Λx ρx } to be as high as possible. The expected success probability of the POVM is thenXpX (x) Tr {Λx ρx } .
(4.99)
x

Suppose that there exists some operator τ such thatτ ≥ pX (x)ρx ,(4.100)where the condition τ ≥ pX (x)ρx is the same as τ − pX (x)ρx ≥ 0 (i.e., that the operator
τ − pX (x)ρx is a positive semi-definite operator). Show that Tr {τ } is an upper bound on
the expected success probability of the POVM. After doing so, consider the case of encoding
n bits into a d-dimensional subspace. By choosing states uniformly at random (in the caseof the ensemble {2−n , ρi }i∈{0,1}n ), show that the expected success probability is bounded above
by d 2−n . Thus, it is not possible to store more than n classical bits in n qubits and have aperfect success probability of retrieval.

4.3 Composite Noisy Quantum Systems
We are again interested in the behavior of two or more quantum systems when we jointhem together. Some of the most exotic, truly “quantum” behavior occurs in joint quantum
systems, and we observe a marked departure from the classical world.

4.3.1

Independent Ensembles

Let us first suppose that we have two independent ensembles for quantum systems A and
B. The first quantum system belongs to Alice and the second quantum system belongs to
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CHAPTER 4. THE NOISY QUANTUM THEORY

Bob, and they may or may not be spatially separated. Let {pX (x), |ψx i} be the ensemble
for the system A and let {pY (y), |φy i} be the ensemble for the system B. Suppose for now
that the state on system A is |ψx i for some x and the state on system B is |φy i for somey. Then, using the composite system postulate of the noiseless quantum theory, the jointstate for a given x and y is |ψx i ⊗ |φy i. The density operator for the joint quantum systemis the expectation of the states |ψx i ⊗ |φy i with respect to the random variables X and Y
that describe the individual ensembles:
EX,Y {(|ψX i ⊗ |φY i) (hψX | ⊗ hφY |)} .

(4.101)

The above expression is equal to the following one:
EX,Y {|ψX ihψX | ⊗ |φY ihφY |} ,(4.102)

because (|ψx i ⊗ |φy i) (hψx | ⊗ hφy |) = |ψx ihψx | ⊗ |φy ihφy |. We then explicitly write out the
expectation as a sum over probabilities:
X
pX (x)pY (y)|ψx ihψx | ⊗ |φy ihφy |.
(4.103)x,y

We can distribute the probabilities and the sum because the tensor product obeys a distributive property:
XX
pY (y)|φy ihφy |.
(4.104)pX (x)|ψx ihψx | ⊗xyThe density operator for this ensemble admits the following simple form:ρ ⊗ σ,(4.105)P
P
where ρ = x pX (x)|ψx ihψx | is the density operator of the X ensemble and σ = y pY (y)|φy ihφy |is the density operator of the Y ensemble. We can say that Alice’s local density operator isρ and Bob’s local density operator is σ. The overall state is a tensor product of these two
density operators.
Definition 4.3.1 (Product State) A density operator is a product state if it is equal to atensor product of two or more density operators.
We should expect the density operator to factor as it does above because we assumed thatthe ensembles are independent. There is nothing much that distinguishes this situation fromthe classical world, except for the fact that the states in each respective ensemble may be
non-orthogonal to other states in the same ensemble. But even here, there is some equivalentdescription of each ensemble in terms of an orthonormal basis so that there is really not too
much difference between this description and a joint probability distribution that factors as
two independent distributions.
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4.3. COMPOSITE NOISY QUANTUM SYSTEMS

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Exercise 4.3.1 Show that the purity P (ρA ) is equal to the following expression:
P (ρA ) = Tr {(ρA ⊗ ρA0 ) FAA0 } .

(4.106)

where system A0 has a Hilbert space structure isomorphic to that of system A and FAA0 is
the swap operator that has the following action on kets in A and A0 :∀x, y

FAA0 |xiA |yiA0 = |yiA |xiA0 .

(4.107)(One can in fact show more generally that Tr {f (ρA )} = Tr {(f (ρA ) ⊗ IA0 ) FAA0 } for any
function f on the operators in system A.)

4.3.2

Separable States

Let us now consider two systems A and B whose corresponding ensembles are correlated ina classical way. We describe this correlated ensemble as the joint ensemble{pX (x), |ψx i ⊗ |φx i} .

(4.108)It is straightforward to verify that the density operator of this correlated ensemble has thefollowing form:XpX (x)|ψx ihψx | ⊗ |φx ihφx |.
(4.109)EX {(|ψX i ⊗ |φX i) (hψX | ⊗ hφX |)} =x

By ignoring Bob’s system, Alice’s local density operator is of the formXpX (x)|ψx ihψx |,EX {|ψX ihψX |} =(4.110)x

and similarly, Bob’s local density operator is
EX {|φX ihφX |} =X

pX (x)|φx ihφx |.

(4.111)x

States of the form in (4.109) can be generated by a classical procedure. A third party
generates a symbol x according to the probability distribution pX (x) and sends the symbol
x to both Alice and Bob. Alice then prepares the state |ψx i and Bob prepares the state |φx i.
If they then discard the symbol x, the state of their systems is given by (4.109).
We can generalize this classical preparation procedure one step further, using an idea
similar to the “ensemble of ensembles” idea in Section 4.1.2. Let us suppose that we first
generate a random variable Z according to some distribution pZ (z). We then generate twoother ensembles, conditioned on the value of the random variable Z. Let {pX|Z (x|z), |ψx,z i}be the first ensemble and let {pY |Z (y|z), |φy,z i} be the second ensemble, where the randomvariables X and Y are independent when conditioned on Z. Let us label the density operatorsof the first and second ensembles when conditioned on a particular realization z by ρz and σz ,respectively. It is then straightforward to verify that the density operator of an ensemble
created from this classical preparation procedure has the following form:
X
EX,Y,Z {(|ψX,Z i ⊗ |φY,Z i) (hψX,Z | ⊗ hφY,Z |)} =pZ (z)ρz ⊗ σz .
(4.112)z
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CHAPTER 4. THE NOISY QUANTUM THEORY

Exercise 4.3.2 By ignoring Bob’s system, we can determine Alice’s local density operator.
Show that
X
EX,Y,Z {|ψX,Z ihψX,Z |} =pZ (z)ρz ,(4.113)z

so that the above expression is the density operator for Alice. It similarly follows that the
local density operator for Bob isEX,Y,Z {|φY,Z ihφY,Z |} =X

pZ (z)σz .

(4.114)z

Exercise 4.3.3 Show that we can always write a state of the form in (4.112) as a convex
combination of pure product states:X

pW (w)|φw ihφw | ⊗ |ψw ihψw |,(4.115)w

by manipulating the general form in (4.112).
As a consequence of Exercise 4.3.3, we see that any state of the form in (4.112) can be
written as a convex combination of pure product states. Such states are called separable
states, defined formally as follows:
Definition 4.3.2 (Separable State) A bipartite density operator σAB is a separable state
if it can be written in the following form:
σAB =

X

pX (x)|ψx ihψx |A ⊗ |φx ihφx |B(4.116)xfor some probability distribution pX (x) and sets {|ψx iA } and {|φx iB } of pure states.
The term “separable” implies that there is no quantum entanglement in the above state,i.e., there is a completely classical procedure that prepares the above state. In fact, thisleads to the definition of entanglement for a general bipartite density operator:
Definition 4.3.3 (Entangled State) A bipartite density operator ρAB is entangled if it is
not separable.
Exercise 4.3.4 (Convexity) Show that the set of separable states acting on a given tensorproduct Hilbert space is a convex set. That is, if λ ∈ [0, 1] and ρAB and σAB are separable
states, then λρAB + (1 − λ)σAB is a separable state.
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4.3. COMPOSITE NOISY QUANTUM SYSTEMS

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Separable States and the CHSH GameOne motivation for Definitions 4.3.2 and 4.3.3 was already given above: for a separable
state, there is a classical procedure that can be used to prepare it. Thus, for an entangledstate, there is no such procedure. That is, a non-classical (quantum) interaction between
the systems is necessary to prepare an entangled state.
Another related motivation is that separable states admit an explanation in terms of aclassical strategy for the CHSH game, discussed in Section 3.6.2. Recall from (3.163) thatclassical strategies pAB|XY (a, b|x, y) are of the following form:
Z
pAB|XY (a, b|x, y) = dλ pΛ (λ) pA|ΛX (a|λ, x) pB|ΛY (b|λ, y).
(4.117)
If we allow for a continuous index λ for a separable state, then we can write such a state as
follows:Z(4.118)σAB = dλ pΛ (λ) |ψλ ihψλ |A ⊗ |φλ ihφλ |B .
(x)

(y)

Recall that in a general quantum strategy, there are measurements {Πa } and {Πb }, giving
output bits a and b based on the input bits x and y and leading to the following strategy:(y)pAB|XY (a, b|x, y) = Tr{(Π(x)⊗ Πb )σAB }
 a
Z

(y)
(x)
= Tr (Πa ⊗ Πb )dλ pΛ (λ) |ψλ ihψλ |A ⊗ |φλ ihφλ |B
Z
n
o
(y)
= dλ pΛ (λ) Tr Π(x)
|ψ
ihψ
|
⊗
Π
|φ
ihφ
|
λ
λ
A
λ
λ
B
a
b
Z
(y)
= dλ pΛ (λ) hψλ |A Π(x)a |ψλ iA hφλ |B Πb |φλ iB .

(4.119)(4.120)(4.121)(4.122)

By picking the probability distributions pA|ΛX (a|λ, x) and pB|ΛY (b|λ, y) in (4.117) as follows:
pA|ΛX (a|λ, x) = hψλ |A Π(x)a |ψλ iA ,(y)pB|ΛY (b|λ, y) = hφλ |B Πb |φλ iB ,(4.123)(4.124)

we see that there is a classical strategy that can simulate any quantum strategy which uses
separable states in the CHSH game. Thus, the winning probability of quantum strategies
involving separable states are subject to the classical bound of 3/4 derived in Section 3.6.2.
In this sense, such strategies are effectively classical.

4.3.3

Local Density Operators and Partial Trace

A First Example
Consider the entangled Bell state |Φ+ iAB shared on systems A and B. In the above analyses,
we determined a local density operator description for both Alice and Bob. Now, we arec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 4. THE NOISY QUANTUM THEORY

curious if it is possible to determine such a local density operator description for Alice and
Bob with respect to the state |Φ+ iAB or more general ones.
As a first approach to this issue, recall that the density operator description arises fromits usefulness in determining the probabilities of the outcomes of a particular measurement.
We say that the density operator is “the state” of the system because it is a mathematicalrepresentation that allows us to compute the probabilities resulting from a physical measurement. So, if we would like to determine a “local density operator,” such a local density
operator should predict the result of a local measurement.
Let us consider a local POVM {Λj }j that Alice can perform on her system. The global
measurement operators for this local measurement are {ΛjA ⊗ IB }j because nothing (theidentity) happens to Bob’s system. The probability of obtaining outcome j when performing
this measurement on the state |Φ+ iAB is1

Φ+ AB ΛjA ⊗ IB Φ+ AB =1X
hkk|AB ΛjA ⊗ IB |lliAB
2 k,l=0(4.125)

1

1X=
hk|A ΛjA |liA hk|liB
2 k,l=01h0|A ΛjA |0iA + h1|A ΛjA |1iA
2



1
Tr ΛjA |0ih0|A + Tr ΛjA |1ih1|A
=
2 

j 1
= Tr ΛA (|0ih0|A + |1ih1|A )2 j= Tr ΛA πA .
=(4.126)(4.127)(4.128)(4.129)(4.130)

The above steps follow by applying the rules of taking the inner product with respect totensor product operators. The last line follows by recalling the definition of the maximally
mixed state π in (4.37), where π here is a qubit maximally mixed state.
The above calculation demonstrates that we can predict the result of any local “Alice”
measurement using the density operator π. Therefore, it is reasonable to say that Alice’s
local density operator is π, and we even go as far to say that her local state is π. A symmetriccalculation shows that Bob’s local state is also π.
This result concerning their local density operators may seem strange at first. The
following global state gives equivalent predictions for local measurements:
πA ⊗ π B .

(4.131)

Can we then conclude that an equivalent representation of the global state is the above
state? Absolutely not. The global state |Φ+ iAB and the above state give drastically different predictions for global measurements. Exercise 4.3.6 below asks you to determine theprobabilities for measuring the global operator ZA ⊗ ZB when the global state is |Φ+ iAB orπA ⊗ πB , and the result is that the predictions are rather different.
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4.3. COMPOSITE NOISY QUANTUM SYSTEMS

139

Exercise 4.3.5 Show that the projection operators corresponding to a measurement of theobservable ZA ⊗ ZB are as follows:1(IA ⊗ IB + ZA ⊗ ZB ) = |00ih00|AB + |11ih11|AB ,21
Πodd ≡ (IA ⊗ IB − ZA ⊗ ZB ) = |01ih01|AB + |10ih10|AB .
2Πeven ≡

(4.132)(4.133)This measurement is a parity measurement, where the measurement operator Πeven coherently
measures even parity and the measurement operator Πodd measures odd parity.
Exercise 4.3.6 Show that a parity measurement (defined in the previous exercise) of thestate |Φ+ iAB returns an even parity result with probability one, and a parity measurementof the state πA ⊗ πB returns even or odd parity with equal probability. Thus, despite the factthat these states have the same local description, their global behavior is very different. Show
that the same is true for the phase parity measurement, given by1(IA ⊗ IB + XA ⊗ XB ) ,2
1(IA ⊗ IB − XA ⊗ XB ) .
ΠX
odd ≡
2

ΠX
even ≡(4.134)(4.135)

Exercise 4.3.7 Show that the maximally correlated state ΦAB , whereΦAB =1(|00ih00|AB + |11ih11|AB ) ,2(4.136)gives results for local measurements that are the same as those for the maximally entangledstate |Φ+ iAB . Show that the above parity measurements can distinguish these states.
Partial Trace
In general, we would like to determine a local density operator that predicts the outcomesof all local measurements. The general method for determining a local density operator isto employ the partial trace operation, which we motivate and define here, as a generalizationof the example discussed at the beginning of Section 4.3.3.
Suppose that Alice and Bob share a bipartite state ρAB and that Alice performs a local
measurement on her system, described by a POVM {ΛjA }. Then the overall POVM on the
joint system is {ΛjA ⊗ IB } because we are assuming that Bob is not doing anything to his
system. According to the Born rule, the probability for Alice to receive outcome j after
performing the measurement is given by the following expression:pJ (j) = Tr{(ΛjA ⊗ IB )ρAB }.

(4.137)

In order to evaluate the trace, we can choose any orthonormal basis that we wish (seeDefinition 4.1.1 and subsequent statements). Taking {|kiA } as an orthonormal basis forc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 4. THE NOISY QUANTUM THEORY

Alice’s Hilbert space and {|liB } as an orthonormal basis for Bob’s Hilbert space, the set{|kiA ⊗ |liB } constitutes an orthonormal basis for the tensor product of their Hilbert spaces.
So we can evaluate (4.137) as follows:
Tr{(ΛjA ⊗ IB )ρAB } =X(hk|A ⊗ hl|B ) (ΛjA ⊗ IB )ρAB (|kiA ⊗ |liB )(4.138)
hk|A (IA ⊗ hl|B ) (ΛjA ⊗ IB )ρAB (IA ⊗ |liB ) |kiA

(4.139)hk|A ΛjA (IA ⊗ hl|B ) ρAB (IA ⊗ |liB ) |kiA

(4.140)

k,l

=

X

=

X

k,l

k,l

"
=

X

#
X

hk|A ΛjAk(IA ⊗ hl|B ) ρAB (IA ⊗ |liB ) |kiA .

(4.141)l

The first equality follows from the definition of the trace in Definition 4.1.1 and using theorthonormal basis {|kiA ⊗ |liB }. The second equality follows because|kiA ⊗ |liB = (IA ⊗ |liB ) |kiA .

(4.142)The third equality follows because(IA ⊗ hl|B ) (ΛjA ⊗ IB ) = ΛjA (IA ⊗ hl|B ) .

(4.143)

The fourth equality follows by bringing the sum over l inside. Using the definition of the
trace in Definition 4.1.1 and the fact that {|kiA } is an orthonormal basis for Alice’s Hilbert
space, we can rewrite (4.141) as("Tr ΛjA#)X(IA ⊗ hl|B ) ρAB (IA ⊗ |liB ).

(4.144)l

Our final step is to define the partial trace operation as follows:
Definition 4.3.4 (Partial Trace) Let XAB be a square operator acting on a tensor product
Hilbert space HA ⊗ HB , and let {|liB } be an orthonormal basis for HB . Then the partial
trace over the Hilbert space HB is defined as follows:
TrB {XAB } ≡X(IA ⊗ hl|B ) XAB (IA ⊗ |liB ) .

(4.145)l

For simplicity, we often suppress the identity operators IA and write this as follows:
TrB {XAB } ≡X

hl|B XAB |liB .

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4.3. COMPOSITE NOISY QUANTUM SYSTEMS141

For the same reason that the definition of the trace is invariant under the choice of an
orthonormal basis, the same is true for the partial trace operation. We can also observefrom the above definition that the partial trace is a linear operation. Continuing with ourdevelopment above, we can define a local operator ρA , using the partial trace, as follows:ρA = TrB {ρAB }.

(4.147)This then allows us to arrive at a rewriting of (4.144) as Tr{ΛjA ρA }, which allows us toconclude thatpJ (j) = Tr{(ΛjA ⊗ IB )ρAB } = Tr{ΛjA ρA }.
(4.148)Thus, from the operator ρA , we can predict the outcomes of local measurements that Alice
performs on her system. Also important here is that the global picture, in which we havea density operator ρAB and a measurement of the form {ΛjA ⊗ IB }, is consistent with thelocal picture, in which the measurement is written as {ΛjA } and the operator ρA is used to
calculate the probabilities pJ (j). The operator ρA is itself a density operator, called the
local or reduced density operator, and the next exercise asks you to verify that it is indeed adensity operator.
Exercise 4.3.8 (Local Density Operator) Let ρAB be a density operator acting on a bipartite Hilbert space. Prove that ρA = TrB {ρAB } is a density operator, meaning that it ispositive semi-definite and has trace equal to one.
In conclusion, given a density operator ρAB describing the joint state held by Alice andBob, we can always calculate a local density operator ρA , which describes the local state ofAlice if Bob’s system is inaccessible to her.
There is an alternate way of describing partial trace, of which it is helpful to be aware.
For a simple state of the form
|xihx|A ⊗ |yihy|B ,(4.149)with |xiA and |yiB each unit vectors, the partial trace has the following action:
TrB {|xihx|A ⊗ |yihy|B } = |xihx|A Tr {|yihy|B } = |xihx|A ,(4.150)where we “trace out” the second system to determine the local density operator for the
first. If the partial trace acts on a tensor product of rank-one operators (not necessarily
corresponding to a state)|x1 ihx2 |A ⊗ |y1 ihy2 |B ,(4.151)
its action is as follows:
TrB {|x1 ihx2 |A ⊗ |y1 ihy2 |B } = |x1 ihx2 |A Tr {|y1 ihy2 |B }= |x1 ihx2 |A hy2 |y1 i .

(4.152)(4.153)

In fact, an alternate way of defining the partial trace is as above and to extend it by linearity.
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CHAPTER 4. THE NOISY QUANTUM THEORY

Exercise 4.3.9 Show that the two notions of the partial trace operation are consistent. Thatis, show thatX
TrB {|x1 ihx2 |A ⊗ |y1 ihy2 |B } =hi|B (|x1 ihx2 |A ⊗ |y1 ihy2 |B ) |iiB
(4.154)i= |x1 ihx2 |A hy2 |y1 i ,(4.155)for some orthonormal basis {|iiB } on Bob’s system.
It can be helpful to see the alternate notion of partial trace worked out in detail. The
most general density operator on two systems A and B is some operator ρAB that is positive
semi-definite with unit trace. We can obtain the local density operator ρA from ρAB bytracing out the B system:
ρA = TrB {ρAB } .
(4.156)
In more detail, let us expand an arbitrary density operator ρAB with an orthonormal basis{|iiA ⊗ |jiB }i,j for the bipartite (two-party) state:XρAB =λi,j,k,l (|iiA ⊗ |jiB )(hk|A ⊗ hl|B ).
(4.157)i,j,k,lThe coefficients λi,j,k,l are the matrix elements of ρAB with respect to the basis {|iiA ⊗|jiB }i,j ,and they are subject to the constraint of non-negativity and unit trace for ρAB . We canrewrite the above operator as
X
λi,j,k,l |iihk|A ⊗ |jihl|B .
(4.158)ρAB =i,j,k,lWe can now evaluate the partial trace:)(Xλi,j,k,l |iihk|A ⊗ |jihl|B
ρA = TrB(4.159)i,j,k,l=X

λi,j,k,l TrB {|iihk|A ⊗ |jihl|B }(4.160)λi,j,k,l |iihk|A Tr {|jihl|B }(4.161)

λi,j,k,l |iihk|A hj|li(4.162)λi,j,k,j |iihk|A(4.163)

i,j,k,l=X
i,j,k,l

=

X

=

X

i,j,k,l

i,j,k

!
=

X X
i,k

λi,j,k,j|iihk|A .

(4.164)j

The second equality exploits the linearity of the partial trace operation.
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4.3. COMPOSITE NOISY QUANTUM SYSTEMS

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Exercise 4.3.10 Verify that the partial trace of a product state gives one of the density
operators in the product state:
TrB {ρA ⊗ σB } = ρA .
(4.165)This result is consistent with the observation near (4.105).
Exercise 4.3.11 Verify that the partial trace of a separable state gives the result in (4.113):)(X
X(4.166)pZ (z)ρzA .
TrB
pZ (z)ρzA ⊗ σBz =zz

Exercise 4.3.12 Consider the following density operator that embeds a joint probability distribution pX,Y (x, y) in a bipartite quantum state:Xρ=pX,Y (x, y)|xihx| ⊗ |yihy|,(4.167)x,ywhere the set of states {|xi}x and {|yi}y each form an orthonormal basis. Show that, in
this case,Ptracing out the second system is the same as taking the marginal distributionpX (x) = y pX,Y (x, y) of the joint distribution pX,Y (x, y). That is, we are left with a density
operator of the formXpX (x)|xihx|.
(4.168)x

Keep in mind that the partial trace is a generalization of the marginalization because it
handles more exotic quantum states besides the above “classical” state.
Exercise 4.3.13 Show that the two partial traces in any order on a bipartite system areequivalent to a full trace:
Tr {ρAB } = TrA {TrB {ρAB }} = TrB {TrA {ρAB }} .

(4.169)Exercise 4.3.14 Verify that Alice’s local density operator does not change if Bob performs
a unitary operator or a measurement in which he does not inform her of the measurement
result.
Exercise 4.3.15 Prove that the partial trace operation obeys a cyclicity relation with respect
to operators that act exclusively on the system over which we trace. That is, let XAB be asquare operator acting on the tensor-product Hilbert space HA ⊗ HB , and let YB , ZB and WB
be square operators acting on the Hilbert space HB . Prove that
TrB {XAB YB ZB WB } = TrB {WB XAB YB ZB }= TrB {ZB WB XAB YB }= TrB {YB ZB WB XAB }.

(4.170)(4.171)(4.172)In the above, it is implicit that YB = IA ⊗ YB , etc.
Exercise 4.3.16 Recall that the purity of a density operator ρA is equal to Tr {ρ2A }. Supposethat ρA = TrB {ΦAB }, where ΦAB is a maximally entangled state. Prove that the purity isequal to the inverse of the dimension of the A system.
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4.3.4

CHAPTER 4. THE NOISY QUANTUM THEORYClassical–Quantum Ensemble

We end our overview of composite noisy quantum systems by discussing one last type of
joint ensemble: the classical–quantum ensemble. This ensemble is a generalization of the“ensemble of ensembles” from before.
Let us consider the following ensemble of density operators:{pX (x), ρxA }x∈X .

(4.173)The intuition here is that Alice prepares a quantum system in the state ρxA with probabilitypX (x). She then passes this ensemble to Bob, and it is Bob’s task to learn about it. He canlearn about the ensemble if Alice prepares a large number of them in the same way.
There is generally a loss of the information in the random variable X once Alice has
prepared this ensemble. It is easier for Bob to learn about the distribution of the randomvariable X if each density operator ρxA is a pure state |xihx| where the states {|xi}x∈X form
an orthonormal basis. The resulting density operator would beXρA =pX (x)|xihx|A .
(4.174)x∈X

Bob could then perform a measurement with measurement operators {|xihx|}x∈X , and learn
about the distribution pX (x) with a large number of measurements.
In the general case, the density operators {ρxA }x∈X do not correspond to pure states,
much less orthonormal ones, and it is more difficult for Bob to learn about random variable
X. The density operator of the ensemble isXρA =pX (x)ρxA ,(4.175)x∈X

and the information about the distribution of random variable X becomes “mixed in” withthe “mixedness” of the density operators ρx . There is then no measurement that Bob canperform on ρ that allows him to directly learn about the probability distribution of randomvariable X.
One solution to this issue is for Alice to prepare the following classical–quantum ensemble:{pX (x), |xihx|X ⊗ ρxA }x∈X ,(4.176)

where we label the first system as X and the second as A. She simply correlates a state |xi
with each density operator ρxA , where the states {|xi}x∈X form an orthonormal basis. We
call this ensemble a “classical–quantum” ensemble because the first system is classical andthe second system is quantum. This then leads to the notion of a classical–quantum state
ρXA defined as follows:
Definition 4.3.5 (Classical–Quantum State) The density operator corresponding to aclassical–quantum ensemble {pX (x), |xihx|X ⊗ ρxA }x∈X , as discussed above, is called a classical–
quantum state and takes the following form:XρXA ≡pX (x)|xihx|X ⊗ ρxA .
(4.177)x∈X
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145

It is a particular kind of separable state of systems X and A, in which the individual statesof the X system are perfectly distinguishable and thus classical.
The “enlarged” ensemble in (4.176) lets Bob easily learn about random variable X whileat the same time he can learn about the ensemble that Alice prepares. Bob can learn about
the distribution of random variable X by performing a local measurement of the system X.
He also can learn about the states ρx by performing a measurement on A and combining theresult of this measurement with the result of the first measurement. The next exercises ask
you to verify these statements.
Exercise 4.3.17 Show that a local measurement of system X reproduces the probability
distribution pX (x). Use local measurement operators {|xihx|}x∈X to show that pX (x) =Tr {ρXA (|xihx|X ⊗ IA )}.
Exercise 4.3.18 Show that performing a measurement with measurement operators {ΛjA }on system A is the same as performing a measurement of the ensemble in (4.173). That is,show that Tr{ρA ΛjA } = Tr{ρXA (IX ⊗ ΛjA )}, where ρA is defined in (4.175).
Exercise 4.3.19 (Lack of Convexity) Prove that the set of classical–quantum states is
not a convex set. That is, show that there exists a classical-quantum state ρXA and anotherσXA and λ ∈ [0, 1], such that λρXA + (1 − λ)σXA is not a classical–quantum state.

4.4 Quantum Evolutions
The evolution of a quantum state is never perfect. In this section, we begin by discussing
the most general approach to understanding quantum evolutions: the axiomatic approach.
This powerful approach starts with three physically reasonable axioms that should holdfor any quantum evolution and from there we deduce a set of mathematical constraints
that any quantum evolution should satisfy (this is known as the Choi–Kraus theorem).
Throughout the book, we will refer to quantum evolutions satisfying these constraints asquantum channels. We then show how noise resulting from the loss of information about aquantum system or from lack of access to an environment system is equivalent to what we
find from the Choi–Kraus theorem. We finally discuss how every operation we have discussed
so far, including preparations and measurements, can be viewed as a quantum channel, andwe follow by giving several important examples of quantum channels.

4.4.1

Axiomatic Approach to Quantum Evolutions

We now discuss a powerful approach to understanding quantum physical evolutions called
the axiomatic approach. Here we make three physically reasonable assumptions that any
quantum evolution should satisfy and then prove that these axioms imply mathematicalconstraints on the form of any quantum physical evolution.
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146

CHAPTER 4. THE NOISY QUANTUM THEORYAll of the constraints we impose are motivated by the reasonable requirement for the output of the evolution to be a quantum state (density operator) if the input to the evolution
is a quantum state (density operator). An important assumption to clarify at the outset isthat we are viewing a quantum physical evolution as a “black box,” meaning that Alice canprepare any state that she wishes before the evolution begins, including pure states or mixed
states. Critically, we even allow her to input one share of an entangled state. This is a standard assumption in quantum information theory, but one could certainly question whether
this assumption is reasonable. If we do accept this criterion as physically reasonable, thenthe Choi–Kraus representation theorem for quantum evolutions follows as a consequence.
Notation 4.4.1 (Density Operators and Linear Operators) Let D(H) denote the spaceof density operators acting on a Hilbert space H, let L(H) denote the space of square linear operators acting on H, and let L(HA , HB ) denote the space of linear operators taking a
Hilbert space HA to a Hilbert space HB .
Throughout this development, we let N denote a map which takes density operators in
D(HA ) to those in D(HB ). In general, the respective input and output Hilbert spaces HA
and HB need not be the same. Implicitly, we have already stated a first physically reasonablerequirement that we impose on N , namely, that N (ρA ) ∈ D(HB ) if ρA ∈ D(HA ). Extending
this requirement, we demand that N should be convex linear when acting on D(HA ):N (λρA + (1 − λ)σA ) = λN (ρA ) + (1 − λ)N (σA ),(4.178)where ρA , σA ∈ D(HA ) and λ ∈ [0, 1].
The physical interpretation of this convex-linearity requirement is in terms of repeated experiments. Suppose a large number of experiments are conducted in which identical quantum
systems are prepared in the state ρA for a fraction λ of the experiments and in the state σAfor the other fraction 1 − λ of the experiments. Suppose further that it is not revealed whichstates are prepared for which experiments. Before you are allowed to perform measurements
on each system, the evolution N is applied to each of the systems. The density operator
characterizing the state of each system for these experiments is then N (λρA + (1 − λ)σA ).
You are then allowed to perform a measurement on each system, which after a large numberof experiments allow you to infer that the density operator is N (λρA + (1 − λ)σA ). Now, in
principle, it could have been revealed which fraction of the experiments had the state ρA prepared and which fraction had σA prepared. In this case, the density operator describing theρA experiments would be N (ρA ) and that describing the σA experiments would be N (σA ).
So, it is reasonable to expect that the statistics observed in your measurement outcomes in
the first scenario would be consistent with those observed in the second scenario, and this is
the physical statement that the requirement (4.178) makes.
Now, it is mathematically convenient to extend the domain and range of the quantum
channel to apply not only to density operators but to all linear operators. To this end, it ise of any quantum evolution N defined as above
possible to find a unique linear extension N(originally defined exclusively by its action on density operators and satisfying convex linearity). See Appendix B for a full development of this idea. Thus, it is reasonable to associate
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4.4. QUANTUM EVOLUTIONS

147

e to the quantum physical evolution N mathematically, andthis unique linear extension N
in what follows (and for the rest of the book), we simply identify a physical evolution N
e , and this is what we call a quantum channel. For thesewith its unique linear extension N
reasons, we now impose that any quantum channel N is linear:
Criterion 4.4.1 (Linearity) A quantum channel N is a linear map:
N (αXA + βYA ) = αN (XA ) + βN (YA ),(4.179)where XA , YA ∈ L(HA ) and α, β ∈ C.
We have already demanded that quantum physical evolutions should take density operators to density operators. Combining with linearity (in particular, scale invariance) implies
that quantum channels should preserve the class of positive semi-definite operators. Thatis, they should be positive maps, as defined below:
Definition 4.4.1 (Positive Map) A linear map M : L(HA ) → L(HB ) is positive if
M(XA ) is positive semi-definite for all positive semi-definite XA ∈ L(HA ).
If we were dealing with classical systems, then positivity would be sufficient to describe
the class of physical evolutions. However, above we argued that we are working in the“black box” picture of quantum physical evolutions, and here, in principle, we allow for
Alice to prepare the input system A to be one share of an arbitrary two-party state ρRA ∈D(HR ⊗HA ), where R is a reference system of arbitrary size. So this means that the evolution
consisting of the identity acting on the reference system R and the map N acting on system
A should take ρRA to a density operator on systems R and B. Let idR ⊗NA→B denote thisevolution, where idR denotes the identity superoperator acting on the system R.
How do we describe the evolution idR ⊗NA→B mathematically? Let XRA be an arbitrary
operator acting on HR ⊗ HA , and let {|iiR } be an orthonormal basis for HR . Then we can
expand XRA with respect to this basis as follows:X
XRA =|iihj|R ⊗ XAi,j ,(4.180)i,jand the action of idR ⊗NA→B on XRA (for linear N ) is defined as follows:!
Xi,j(idR ⊗NA→B ) (XRA ) = (idR ⊗NA→B )|iihj|R ⊗ XA

(4.181)

i,j=

X(idR ⊗NA→B ) |iihj|R ⊗ XAi,j(4.182)

i,j

=

X

=

X

idR (|iihj|R ) ⊗ NA→B XAi,j(4.183)i,j|iihj|R ⊗ NA→B XAi,j .

(4.184)i,j

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CHAPTER 4. THE NOISY QUANTUM THEORY

That is, the identity superoperator idR has no effect on the R system. The above development
leads to the notion of a linear map being completely positive and our next criterion for any
quantum physical evolution:
Definition 4.4.2 (Completely Positive Map) A linear map M : L(HA ) → L(HB ) iscompletely positive if idR ⊗M is a positive map for a reference system R of arbitrary size.
Criterion 4.4.2 (Complete Positivity) A quantum channel is a completely positive map.
There is one last requirement that we impose for quantum physical evolutions, known
as trace preservation. This requirement again stems from the reasonable constraint that
N should map density operators to density operators. That is, it should be the case that
Tr{ρA } = Tr{N (ρA )} = 1 for all input density operators ρA . However, now that we haveargued for linearity of every quantum physical evolution, trace preservation on density operators combined with linearity implies that quantum channels are trace preserving on theset of all operators. This is due to the fact that there are sets of density operators that forma basis for L(HA ). Indeed, one such basis of density operators is as follows:
ρx,y
A =





|xihx|A
1
(|xi
+
|yi
A
A ) (hx|A + hy|A )2
1(|xiA + i|yiA ) (hx|A − ihy|A )2

if x = y
if x < y .
if x > y(4.185)

Consider that for all x, y such that x < y, the following holds
1 x,x 1 y,yy,x− i ρA − ρA − ρA ,|xihy|A =
2
2
2
2




1 x,x 1 y,y
1 x,x 1 y,yx,yy,x|yihx|A = ρA − ρA − ρA + i ρA − ρA − ρA ,
2
2
2
2


1 x,x 1 y,yρx,yA − ρA − ρA

(4.186)(4.187)so that we can represent any operator XA as a linear combination of density operators fromthe set {ρx,yA }. This leads to our final criterion for quantum channels:
Criterion 4.4.3 (Trace Preservation) A quantum channel is trace preserving, in the sensethat Tr{XA } = Tr{N (XA )} for all XA ∈ L(HA ).
Definition 4.4.3 (Quantum Channel) A quantum channel is a linear, completely positive, trace preserving map, corresponding to a quantum physical evolution.
Criteria 4.4.1, 4.4.2, and 4.4.3 detailed above lead naturally to the Choi–Kraus representation theorem, which states that a map satisfies all three criteria if and only if it takes aparticular form according to a Choi–Kraus decomposition:c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

4.4. QUANTUM EVOLUTIONS

149

Theorem 4.4.1 (Choi–Kraus) A map N : L(HA ) → L(HB ) (denoted also by NA→B ) is
linear, completely positive, and trace-preserving if and only if it has a Choi–Kraus decomposition as follows:d−1
X
NA→B (XA ) =Vl XA Vl† ,(4.188)l=0where XA ∈ L(HA ), Vl ∈ L(HA , HB ) for all l ∈ {0, . . . , d − 1},d−1
XVl† Vl = IA ,(4.189)l=0

and d need not be any larger than dim(HA ) dim(HB ).
Before we delve into a proof, it is helpful to give a sketch. There is an easier part anda more challenging part of the proof. For the more challenging part, a helpful tool is anoperator called the Choi operator:
Definition 4.4.4 (Choi Operator) Let HR and HA be isomorphic Hilbert spaces, and let{|iiR } and {|iiA } be orthonormal bases for HR and HA , respectively. Let HB be some other
Hilbert space, and let N : L(HA ) → L(HB ) be a linear map (written also as NA→B ). The
Choi operator corresponding to NA→B and the bases {|iiR } and {|iiA } is defined as thefollowing operator:(idR ⊗NA→B ) (|ΓihΓ|RA ) =

dX
A −1

|iihj|R ⊗ NA→B (|iihj|A ),(4.190)i,j=0where dA ≡ dim(HA ) and |ΓiRA is an unnormalized maximally entangled vector, as definedin (3.233):
dX
A −1|ΓiRA ≡
|iiR ⊗ |iiA .
(4.191)i=0

The rank of the Choi operator is called the Choi rank.
If NA→B is a completely positive map, then the Choi operator is positive semi-definite.
This follows as a direct consequence of Definition 4.4.2 and the fact that |ΓihΓ|RA is positive
semi-definite. The converse is true as well, and Exercise 4.4.1 asks you to verify this. The
converse is in some sense a much more powerful statement. Definition 4.4.2 suggests that wewould have to check a seemingly infinite number of cases in order to verify whether a given
linear map is completely positive, but the converse statement establishes that we need tocheck only one condition: whether the Choi operator is positive semi-definite.
Why else is the Choi operator a useful tool? One other important reason is that it
encodes how a quantum channel acts on any possible input operator XA , and thus specifies
the channel completely. Consider that we can expand the Choi operator as a matrix ofc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 4. THE NOISY QUANTUM THEORY

matrices (of total size dA dB × dA dB ) in the following way, by exploiting properties of thetensor product:
NA→B (|0ih0|A )NA→B (|0ih1|A )···NA→B (|0ihdA − 1|A ) NA→B (|1ih0|A )NA→B (|1ih1|A )···NA→B (|1ihdA − 1|A )  . (4.192)
..
..
..
..


.
.
.
.
NA→B (|dA − 1ih0|A ) NA→B (|dA − 1ih1|A ) · · · NA→B (|dA − 1ihdA − 1|A )So if we would like to figure out how the channel NA→B acts on an input operator
XA , weP
can first expand XA with respect to the orthonormal basis {|iiA } as XA = i,j xi,j |iihj|Aand then apply the channel, using linearity:
!
X
X
i,j
NA→B (XA ) = NA→B
x |iihj|A =xi,j NA→B (|iihj|A ).
(4.193)i,ji,jSo the procedure is to expand XA as above, multiple the (i, j) coefficient xi,j with the (i, j)
entry in the Choi operator, and then sum these operators over all indices i and j.
Proof of Theorem 4.4.1. We first prove the easier “if-part” of the theorem. So let us
suppose that NA→B has the form in (4.188) and that the condition in (4.189) holds as well.
Then NA→B is clearly a linear map. It is completely positive because (idR ⊗NA→B )(XRA ) ≥ 0
if XRA ≥ 0 when NA→B has the form in (4.188), and this holds for a reference system R ofarbitrary size. That is, consider from (4.184) that {IR ⊗ Vl } is a set of Kraus operators forthe extended channel idR ⊗NA→B and thus
(idR ⊗NA→B )(XRA ) =d−1
X(IR ⊗ Vl )XRA (IR ⊗ Vl† ).

(4.194)l=0

We know that (IR ⊗ Vl )XRA (IR ⊗ Vl )† ≥ 0 for all l when XRA ≥ 0, and the same is true forthe sum. Trace preservation follows because
( d−1
)
X
†
Tr {NA→B (XA )} = Tr
Vl XA Vl
(4.195)
l=0

= Tr

( d−1
X

)
†

Vl Vl XA

(4.196)l=0= Tr {XA } ,(4.197)

where the second line follows from linearity and cyclicity of trace and the last line follows
from the condition in (4.189).
We now prove the more difficult “only-if” part. Let dA ≡ dim(HA ) and dB ≡ dim(HB ).
Consider that we can diagonalize the Choi operator as given in Definition 4.4.4, because it
is positive semi-definite:d−1
X
NA→B (|ΓihΓ|RA ) =|φl ihφl |RB ,(4.198)l=0c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

4.4. QUANTUM EVOLUTIONS

151

where d ≤ dA dB is the Choi rank of the map NA→B . (This decomposition does not necessarily
have to be such that the vectors {|φl iRB } are orthonormal, but keep in mind that there isalways a choice such that d ≤ dA dB .) Consider by inspecting (4.190) that(hi|R ⊗ IB ) (NA→B (|ΓihΓ|RA )) (|jiR ⊗ IB ) = NA→B (|iihj|) .

(4.199)Now, consider that for any bipartite vector |φiRB , we can expand it in terms of an
orthonormal basis {|jiB } and the basis {|iiR } given above:|φiRB =dX
A −1 dXB −1
i=0

αij |iiR ⊗ |jiB .

(4.200)j=0

Let VA→B denote the following linear operator:
VA→B ≡dX
A −1 dX
B −1
i=0αi,j |jiB hi|A ,(4.201)j=0

where {|iiA } is the orthonormal basis given above. Then we see that(IR ⊗ VA→B ) |ΓiRA =dX
A −1 dX
B −1
i=0

=

j=0

dX
A −1 dX
B −1
i=0|kiR ⊗ |kiA

(4.202)k=0

dX
A −1 dX
B −1 dX
A −1
i=0

=

αi,j |jiB hi|A

j=0

dX
A −1

αi,j |kiR ⊗ |jiB hi|kiA(4.203)k=0

αij |iiR ⊗ |jiB

(4.204)j=0= |φiRB .

(4.205)So this means that for all bipartite vectors |φiRB , we can find a linear operator VA→B suchthat (IR ⊗ VA→B ) |ΓiRA = |φiRB . Consider also thathi|R |φiRB = hi|R (IR ⊗ VA→B ) |ΓiRA= VA→B |iiA .

(4.206)(4.207)

Applying this to our case of interest, for each l, we can write|φl iRB = IR ⊗ (Vl )A→B |ΓiRA ,(4.208)where (Vl )A→B is some linear operator of the form in (4.201). After making this observation,c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 4. THE NOISY QUANTUM THEORY

we realize that it is possible to write
NA→B (|iihj|) = (hi|R ⊗ IB ) (NA→B (|ΓihΓ|RA )) (|jiR ⊗ IB )= (hi|R ⊗ IB )d−1
X

|φl ihφl |RB (|jiR ⊗ IB )(4.209)(4.210)

l=0

=

=

d−1
X
l=0
d−1
X

[(hi|R ⊗ IB ) |φl iRB ] [hφl |RB (|jiR ⊗ IB )](4.211)

Vl |iihj|A Vl† .

(4.212)l=0

By linearity of the map NA→B , exploiting the above result, and the development in (4.193),
it follows that the action of NA→B on any input operator XA can be written as follows:
NA→B (XA ) =d−1
X

V l XA V l † .

(4.213)l=0

To prove the condition in (4.189), let us begin by exploiting the fact that the map NA→B
is trace preserving, so that
Tr {NA→B (|iihj|A )} = Tr {|iihj|A } = δij .

(4.214)for all operators {|iihj|A }i,j . But consider also that()
X†
Tr {NA→B (|iihj|A )} = Tr
Vl (|iihj|A ) Vl

(4.215)

l

= Tr

(
X

)
†

Vl Vl (|iihj|A )(4.216)Vl† Vl |iiA .

(4.217)l= hj|AX
lThus, in order to have consistency with (4.214), we require that hj|Aequivalently, for (4.189) to hold.

†l Vl Vl |iiA = δi,j , orP

Remark 4.4.1 If the decomposition in (4.198) is a spectral decomposition, then it follows
that the Kraus operators {Vl } are orthogonal with respect to the Hilbert–Schmidt inner product:nono
Tr Vl† Vk = Tr Vl† Vl δl,k .
(4.218)This follows from the fact that
δl,k hφl |φl i = hφl |φk ih i= hΓ|RB IR ⊗ Vl† Vk
|ΓiRB
B
n
o
= Tr Vl† Vk ,c 2019(4.219)(4.220)(4.221)

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4.4. QUANTUM EVOLUTIONS

153

where in the third line we have applied the result of Exercise 4.1.3.
Exercise 4.4.1 Prove that a linear map N is completely positive if its corresponding Choi
operator, as defined in Definition 4.4.4, is a positive semi-definite operator. (Hint: Use thefact that any positive semi-definite operator can be diagonalized, the fact that idR ⊗N islinear, and use something similar to (4.202)–(4.205)).

4.4.2

Unique Specification of a Quantum Channel

We emphasize again that any linear map N : L(HA ) → L(HB ) is specified completely by its
action NA→B (|iihj|A ) on an operator of the form |iihj|A where {|iiA } is some orthonormal
basis. Thus, two linear maps NA→B and MA→B are equal if they have the same effect on all
operators of the form |iihj|:
NA→B = MA→B

⇔

∀i, j

NA→B (|iihj|A ) = MA→B (|iihj|A ) .

(4.222)

As a consequence, there is an interesting way to test whether two quantum channels areequal to each other. Let us now consider a maximally entangled qudit state |ΦiRA whered−1

1 X|iiR |iiA ,|ΦiRA = √d i=0(4.223)and d is the dimension of each system R and A. The density operator ΦRA corresponding
to |ΦiRA is as follows:
d−1
1X
ΦRA =
|iihj|R ⊗ |iihj|A .
(4.224)
d i,j=0
Let us now send the A system of ΦRA through a quantum channel N :d−1(idR ⊗NA→B ) (ΦRA ) =1X|iihj|R ⊗ NA→B (|iihj|A ).
d i,j=0(4.225)

The resulting state completely characterizes the quantum channel N because the following map translates between the state in (4.225) and the operators NA→B (|iihj|A ) in (4.222):
dhi|R (idR ⊗NA→B ) (ΦRA ) |jiR = NA→B (|iihj|A ) .

(4.226)

Thus, we can completely characterize a quantum channel by determining the quantum state
resulting from sending one share of a maximally entangled state through it, and the following
condition is necessary and sufficient for any two quantum channels to be equal:
N =M⇔(idR ⊗NA→B ) (ΦRA ) = (idR ⊗MA→B ) (ΦRA ) .

(4.227)

It is equivalent to the condition in (4.222).
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4.4.3

CHAPTER 4. THE NOISY QUANTUM THEORY

Serial Concatenation of Quantum Channels

A quantum state may undergo not just one type of quantum evolution—it can of course
undergo one quantum channel followed by another quantum channel. Let N : L(HA ) →L(HB ) denote a first quantum channel and let M : L(HB ) → L(HC ) denote a second
quantum channel. Suppose that the Kraus operators of N are {Nk } and the Kraus operatorsof M are {Mk }. It is straightforward to define the serial concatenation MB→C ◦ NA→B ofthese two quantum channels. Consider that the output of the first channel is
X
NA→B (ρA ) ≡(4.228)
Nk ρA Nk† ,kfor some input density operator ρA ∈ D(HA ). The output of the serially concatenatedchannel MB→C ◦ NA→B is then
X
X
(MB→C ◦ NA→B ) (ρA ) =Mk NA→B (ρ)Mk† =Mk Nk0 ρA Nk†0 Mk† .
(4.229)k,k0k

It is clear that the Kraus operators of the serially concatenated channel MB→C ◦ NA→B
are {Mk Nk0 }k,k0 . Serial concatenation of channels has an obvious generalization to a serial
concatenation of more than two channels.

4.4.4

Parallel Concatenation of Quantum Channels

We can also use two channels in parallel. That is, suppose that we send a system A througha channel N : L(HA ) → L(HC ) and a system B through a channel M : L(HB ) → L(HD ).
Suppose further that the Kraus operators of NA→C are {Nk } and those for MB→D are{Mk0 }. Then the parallel concatenation of the two channels is equal to the following serial
concatenation:
NA→C ⊗ MB→D = (NA→C ⊗ idD )(idA ⊗MB→D ),(4.230)or equivalently
NA→C ⊗ MB→D = (idC ⊗MB→D )(NA→C ⊗ idB ).

(4.231)

Intuitively, if Alice is conducting a local action and Bob is as well, the order in which they
conduct their actions does not matter for determining the final output state. We havealready discussed that a set of Kraus operators for NA→C ⊗ idD is {Nk ⊗ ID } and a set for
idA ⊗MB→D is {IA ⊗Mk0 }, so that it is straightforward to verify that a set of Kraus operators
for NA→C ⊗ MB→D is {Nk ⊗ Mk0 }. The parallel concatenated channel NA→C ⊗ MB→D thus
has the following action on an input density operator ρAB ∈ D(HA ⊗ HB ):X(NA→C ⊗ MB→D )(ρAB ) =(Nk ⊗ Mk0 ) (ρAB ) (Nk ⊗ Mk0 )† ,(4.232)k,k0

Parallel concatenation of channels also has an obvious generalization to more than two
channels.
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4.4. QUANTUM EVOLUTIONS

4.4.5

155

Unital Maps and Adjoints of Quantum Channels

Recall that the adjoint G† of a linear operator G is defined as the unique linear operator
satisfying the following set of equations:hy, Gxi = hG† y, xi,(4.233)P
for all vectors x and y, and with hz, wi = i zi∗ wi defined as the inner product between
vectors z and w.
As an extension of this idea, we can define an inner product for operators:
Definition 4.4.5 (Hilbert–Schmidt Inner Product) The Hilbert–Schmidt inner product between two operators C, D ∈ L(H) is defined as follows:hC, Di ≡ Tr{C † D}.

(4.234)This then allows us to define the adjoint N † of a linear map N in a way similar to (4.233):
Definition 4.4.6 (Adjoint Map) Let N : L(HA ) → L(HB ) be a linear map. The adjoint
N † : L(HB ) → L(HA ) of a linear map N is the unique linear map satisfying the following
set of equations:hY, N (X)i = hN † (Y ), Xi,(4.235)for all X ∈ L(HA ) and Y ∈ L(HB ).
Another important class of linear maps are unital maps, defined as follows:
Definition 4.4.7 (Unital Map) A linear map N : L(HA ) → L(HB ) is unital if it preserves the identity operator, in the sense that N (IA ) = IB .
Given the notion of an adjoint map, it is natural to inquire what is the adjoint of aquantum channel, and furthermore, what is an interpretation of it. So let us now supposethat N : L(HA ) → L(HB ) is a quantum channel with a set {Vl } of Kraus operators satisfyingP†l Vl Vl = IA . Then we compute
)
(
)
(
X
X
hY, N (X)i = Tr Y †Vl XVl† = Tr
Vl† Y † Vl X
(4.236)

= Tr

l

 X



l

!†
Vl† Y Vl

X





l

=

*
X

+
Vl† Y Vl , X,(4.237)lwhere the second equality is from linearity and cyclicity of trace and the last is from thedefinition of the Hilbert–Schmidt inner product. Thus, the adjoint N † of any quantum
channel N is given byX †N † (Y ) =Vl Y Vl .
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CHAPTER 4. THE NOISY QUANTUM THEORY

The adjoint N † is completely positive, as one can verify by applying Exercise 4.4.1. Furthermore, the adjoint N † is unital becauseX †X †N † (IB ) =(4.239)Vl IB Vl =Vl Vl = IA .
ll

We summarize these results as follows:
Proposition 4.4.1 The adjoint N † : L(HB ) → L(HA ) of a quantum channel N : L(HA ) →L(HB ) is a completely positive, unital map.
What is an interpretation of the adjoint of a quantum channel? It provides a connection
from the Schrödinger picture of quantum physics, in which the focus is on the evolutionof states, to the Heisenberg picture, in which the focus is on the evolution of observables
or measurement operators. To see this, let {ΛjB } be a POVM, ρA be a density operator,and N : L(HA ) → L(HB ) be a quantum channel. Suppose that we prepare the state ρA ,apply the channel N , and then perform the measurement {ΛjB }. The probability of getting
outcome j from the measurement is given by the Born rule:pJ (j) = Tr{ΛjB N (ρA )} = Tr{N † (ΛjB )ρA },(4.240)where the second equality follows because N † is the adjoint of N . This latter expressionis what corresponds to the Heisenberg picture. Here, the interpretation is that each measurement operator ΛjB “evolves backwards” to become N † (ΛjB ) and then the measurement{N † (ΛjB )} is performed on the state ρA . We should verify that the set {N † (ΛjB )} indeed
constitutes a measurement. Consider that each N † (ΛjB ) is positive semi-definite, given thatthe adjoint is a completely positive map, and that!
XX
N † (ΛjB ) = N †ΛjB = N † (IB ) = IA ,(4.241)jjwhere the equalities are following because N † is linear and unital. The interpretation of
the measurement {N † (ΛjB )} is that it is the physical procedure corresponding to applying
the channel N and then performing the measurement {ΛjB }, which is of course a valid
measurement procedure.

4.5 Interpretations of Quantum Channels
We now detail two interpretations of quantum channels that are consistent with the Choi–
Kraus theorem (Theorem 4.4.1). The first is that we can interpret the noise occurring ina quantum channel as the loss of a measurement outcome, and the second is that we caninterpret noise as being due to a unitary interaction with an environment to which we do
not have access.
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4.5. INTERPRETATIONS OF QUANTUM CHANNELS

157

Figure 4.2: The diagram on the left depicts a quantum channel NA→B that takes a quantum system A toa quantum system B. This quantum channel has an interpretation in terms of the diagram on the right, inwhich some third party performs a measurement on the input system and does not inform the receiver ofthe measurement outcome.

4.5.1

Noisy Evolution as the Loss of a Measurement Outcome

We can interpret the noise resulting from a quantum channel as arising from the loss of ameasurement outcome (see Figure 4.2). Suppose that the state of a system is described by adensity operator ρ andPthat we then perform a measurement with a set {Mk } of measurement†
operators for whichk Mk Mk = I. The probability of obtaining outcome k from themeasurement is given by the Born rule: pK (k) = Tr{Mk† Mk ρ}, and the post-measurement
state is Mk ρMk† /pK (k), as discussed at the end of Section 4.2. Let us now suppose that welose track of the measurement outcome, or equivalently, someone else measures the systemand does not inform us of the measurement outcome. The resulting ensemble description isthennopK (k), Mk ρMk† /pK (k) .
(4.242)k

The density operator corresponding to this ensemble is thenXkpK (k)Mk ρMk† X=
Mk ρMk† .
pK (k)
k(4.243)P
We can thus write this evolution as a quantum channel N (ρ) where N (ρ) = k Mk ρMk† .
The measurement operators are playing the role of Kraus operators in this evolution.

4.5.2

Noisy Evolution from a Unitary Interaction

There is another perspective on quantum noise that is helpful to consider. It is equivalent
to the perspective given in Chapter 5 when we discuss isometric evolution. Suppose that aquantum system A begins in the state ρA and that there is an environment system E in apure state |0iE . So the initial state of the joint system AE is ρA ⊗ |0ih0|E . Suppose thatthese two systems interact according to some unitary operator UAE acting on both systems
A and E. If we only have access to the system A after the interaction, then we calculate thestate σA of this system by taking the partial trace over the environment E:
no†
σA = TrE UAE (ρA ⊗ |0ih0|E ) UAE.
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158

CHAPTER 4. THE NOISY QUANTUM THEORYThis evolution is equivalent to that of a completely positive, trace-preserving map with Kraus
operators {Bi ≡ (IA ⊗ hi|E ) UAE (IA ⊗ |0iE )}i . This follows because we can take the partial
trace with respect to an orthonormal basis {|iiE } for the environment:no†
TrE UAE (ρA ⊗ |0ih0|E ) UAE
X†=(IA ⊗ hi|E ) UAE (ρA ⊗ |0ih0|E ) UAE(IA ⊗ |iiE )(4.245)i=

X†(IA ⊗ hi|E ) UAE (IA ⊗ |0iE ) (ρA ) (IA ⊗ h0|E ) UAE(IA ⊗ |iiE )(4.246)

Bi ρA Bi† .

(4.247)i=

X
i

The first equality follows from Definition 4.3.4 for partial trace. The second equality follows
becauseρA ⊗ |0ih0|E = (IA ⊗ |0iE ) (ρA ) (IA ⊗ h0|E ) .
(4.248)P †That the operators {Bi } are a legitimate set of Kraus operators satisfying i Bi Bi = IAfollows from the unitarity of UAE and the orthonormality of the basis {|iiE }:X †X†
Bi Bi =
(IA ⊗ h0|E ) UAE(IA ⊗ |iiE ) (IA ⊗ hi|E ) UAE (IA ⊗ |0iE )(4.249)ii!
†= (IA ⊗ h0|E ) UAE

IA ⊗

X

|iihi|E

UAE (IA ⊗ |0iE )i†= (IA ⊗ h0|E ) UAE UAE (IA ⊗ |0iE )= (IA ⊗ h0|E ) IA ⊗ IE (IA ⊗ |0iE )= IA .

(4.250)(4.251)(4.252)(4.253)

4.6 Quantum Channels are All Encompassing
In this section, we show how everything we have considered so far can be viewed as a quantum
channel. This includes physical evolutions as we have discussed so far, but additionally (andperhaps surprisingly) density operators, discarding of systems, and quantum measurements.
From this perspective, one could argue that that there really is just a single underlying
postulate of quantum physics, that everything we consider in the theory is just a quantum
channel of some sort.

4.6.1

Preparation and Appending Channels

The preparation of a system A in a state ρA ∈ D(HA ) is a particular type of quantum
channel,Pwith trivial input Hilbert space C and output Hilbert space HA . Let ρA = x pX (x)|xihx|Abe a spectral decomposition of ρA . Then the Kraus operators of this channel are {Nx ≡c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

4.6. QUANTUM CHANNELS ARE ALL ENCOMPASSING

159

p
pX (x)|xiA }, and we can easily verify that these are legitimate Kraus operators by calculating
 p
 X
X
X p
†
Nx Nx =pX (x)hx|A
pX (x)|xiA =pX (x) = 1,
(4.254)xx

x

so that the completeness relation holds, given that the number 1 is the identity for the trivial
Hilbert space C. Considering that the number 1 is also the only density operator in D(C),we can view this channel as mapping the trivial density operator 1 to a density operatorρA ∈ D(HA ). It is thus a preparation channel.
Definition 4.6.1 (Preparation Channel) A preparation channel PA ≡ PC→A prepares aquantum system A in a given state ρA ∈ D(HA ).
This leads to a related channel, called an appending channel:
Definition 4.6.2 (Appending Channel) An appending channel is the parallel concatenation of the identity channel and a preparation channel.
Thus, an appending channel has the following action on a system B in the state σB :(PA ⊗ idB ) (σB ) = ρA ⊗ σB .

(4.255)p
The Kraus operators of such an appending channel are then { pX (x)|xiA ⊗ IB }.

4.6.2

Trace-out and Discarding Channels

In some sense, the opposite of preparation is discarding. So suppose that we completely
discard the contents of a quantum system A. The channel that does so is called a trace-out
channel TrA , and its action is to map any density operator ρA ∈ D(HA ) to the trivial density
operator 1. The Kraus operators of the trace-out channel are {Nx ≡ hx|A }, where {|xiA } issome orthonormal basis for the system A. These Kraus operators satisfy the completeness
relation because
X
X
Nx† Nx =|xihx|A = IA .
(4.256)xxThis channel is in direct correspondence with the trace operation, given in Definition 4.1.1.
Now suppose that we have two systems A and B, and we would like to discard system A
only. The channel that does so is a discarding channel, which is the parallel concatenationof the trace-out channel TrA and the identity channel idB . It has the following action on adensity operator ρAB ∈ D(HA ⊗ HB ):X(TrA ⊗ idB ) (ρAB ) =(hx|A ⊗ IB ) ρAB (|xiA ⊗ IB ) = TrA {ρAB },(4.257)x

where we have taken the Kraus operators of TrA ⊗ idB to be {hx|A ⊗IB }. Clearly, this channel
is in direct correspondence with the partial trace operation, given in Definition 4.3.4.
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4.6.3

CHAPTER 4. THE NOISY QUANTUM THEORY

Unitary and Isometric Channels

Unitary evolution is a special kind of quantum channel in which there is a single Kraus
operator U ∈ L(H), satisfying U U † = U † U = IH . Unitary channels are thus completely
positive, trace-preserving, and unital. Let ρ ∈ D(H). Under the action of a unitary channel
U, this state evolves asU(ρ) = U ρU † ,(4.258)where U(ρ) ∈ D(H). Our convention henceforth is to denote a unitary channel by U and aunitary operator by U .
There is a related, but more general kind of quantum channel called an isometric quantum
channel. Before defining it, we need to define the notion of a linear isometry:
Definition 4.6.3 (Isometry) Let H and H0 be Hilbert spaces such that dim(H) ≤ dim(H0 ).
An isometry V is a linear map from H to H0 such that V † V = IH . Equivalently, an isometry
V is a linear, norm-preserving operator, in the sense that k|ψik2 = kV |ψik2 for all |ψi ∈ H.
An isometry is a generalization of a unitary, because it maps between spaces of different
dimensions and is thus generally rectangular and need not satisfy V V † = IH0 . Rather, itsatisfies V V † = ΠH0 , where ΠH0 is some projection onto H0 , because(V V † )(V V † ) = V (V † V )V † = V IH V † = V V † .

(4.259)In later chapters, we repeatedly use the notion of an isometry.
We can now define an isometric channel:
Definition 4.6.4 (Isometric Channel) A channel V : L(H) → L(H0 ) is an isometricchannel if there exists a linear isometry V : H → H0 such that
V(X) = V XV † ,(4.260)for X ∈ L(H).
Isometric channels are completely positive and trace-preserving. Furthermore, as in thecase of unitary channels, there is just a single Kraus operator V satisfying V † V = IH .
Reversing Unitary and Isometric Channels
Suppose that we would like to reverse the action of a unitary channel U. It is easy to do so:
the adjoint map U † is a unitary channel, and by performing it after U, we get(U † ◦ U)(X) = U † U XU † U = X,

(4.261)for X ∈ L(H).
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4.6. QUANTUM CHANNELS ARE ALL ENCOMPASSING

161

If we would like to reverse the action of an isometric channel V, we need to be a bit more
careful. In this case, the adjoint map V † is not a channel, because it is not trace-preserving.
Consider that
Tr{V † (Y )} = Tr{V † Y V } = Tr{V V † Y }= Tr{ΠH0 Y } ≤ Tr{Y },(4.262)(4.263)for Y ∈ L(H0 ) and where the projection ΠH0 ≡ V V † .
However, it is possible to construct a reversal channel R for any isometric channel V in
the following way:R(Y ) ≡ V † (Y ) + Tr{(IH0 − ΠH0 ) Y }σ,(4.264)
where σ ∈ D(H). One can verify that the map R is completely positive, and it is tracepreserving because
Tr{R(Y )} = Tr{ V † (Y ) + Tr{(IH0 − ΠH0 ) Y }σ }(4.265)= Tr{V † (Y )} + Tr{(IH0 − ΠH0 ) Y } Tr{σ}= Tr{ΠH0 Y } + Tr{(IH0 − ΠH0 ) Y }= Tr{Y }.

(4.266)(4.267)(4.268)Furthermore, it perfectly reverses the action of the isometric channel V because(R ◦ V)(X) = V † (V(X)) + Tr{(IH0 − ΠH0 ) V(X)}σ= V † V XV † V + Tr{ IH0 − V V † V XV † }σ
= X + Tr{V XV † } − Tr{V V † V XV † } σ= X + Tr{V † V X} − Tr{V † V V † V X} σ
= X + [Tr{X} − Tr{X}] σ= X,(4.269)(4.270)(4.271)(4.272)(4.273)(4.274)for X ∈ L(H).

4.6.4Classical-to-Classical Channels

It is natural to expect that classical channels are special cases of quantum channels, andindeed, this is the case. To see this, fix an input probability distribution pX (x) and a classicalchannel pY |X (y|x). Fix an orthonormal basis {|xi} corresponding to the input letters and an
orthonormal basis {|yi} corresponding to the output letters. We can then encode the input
probability distribution pX (x) as a density operator ρ of the following form:Xρ=pX (x)|xihx|.
(4.275)x

Let N be a quantum channel with the following Kraus operatorsonq
pY |X (y|x)|yihx|.

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162

CHAPTER 4. THE NOISY QUANTUM THEORY(The fact that these are legitimate Kraus operators follows directly from the fact that
pY |X (y|x) is a conditional probability distribution.) The quantum channel then has the following action on the input ρ:
!
q
X
Xq
0
0
0
pY |X (y|x)|yihx|
pX (x )|x ihx |pY |X (y|x)|xihy|(4.277)N (ρ) =
x0

x,y

=

X

2

pY |X (y|x)pX (x0 ) |hx0 |xi| |yihy|(4.278)

pY |X (y|x)pX (x)|yihy|(4.279)x,y,x0

=

X
x,y

!
=

X X
y

pY |X (y|x)pX (x) |yihy|.

(4.280)x

Thus, the evolution is the same that a noisy classical channel pY |X (y|x) would enact on aprobability distribution pX (x) by taking it topY (y) =X

pY |X (y|x)pX (x)(4.281)x

at the output.
Since a noiseless classical channel has pY |X (y|x) = δx,y , we are led to the following
definition:
Definition 4.6.5 (Noiseless Classical Channel) Let {|xi} be an orthonormal basis for a
Hilbert space H. A noiseless classical channel has the following action on a density operatorρ ∈ D(H):X|xihx|ρ|xihx|.
(4.282)ρ→xThat is, it removes the off-diagonal elements of ρ when represented as a matrix with respect
to the basis {|xi}.

4.6.5

Classical-to-Quantum Channels

Classical-to-quantum channels, or classical–quantum channels for short, are channels whichtake classical systems to quantum systems. They thus go one step beyond both classicalto-classical channels and preparation channels. More generally, they make a given quantum
system classical and then prepare a quantum state, as discussed in the following definition:
Definition 4.6.6 (Classical–Quantum Channel) A classical–quantum channel first measures the input state in a particular orthonormal basis and outputs a density operator conditioned on the result of the measurement. Given an orthonormal basis {|kiA } and a set
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4.6. QUANTUM CHANNELS ARE ALL ENCOMPASSING

163

Figure 4.3: This figure illustrates the internal workings of a classical–quantum channel. It first measures the
input state in some basis {|ki} and outputs a quantum state σk conditioned on the measurement outcome.

of states {σBk }, each of which is in D(HB ), a classical–quantum channel has the following
action on an input density operator ρA ∈ D(HA ):ρA →Xhk|A ρA |kiA σBk .

(4.283)k

Let us see how this comes about, using the definition above. The classical–quantum
channel first measures the input state ρA in the basis {|kiA }. Given that the result of themeasurement is k, the post measurement state is|kihk|ρA |kihk|.
hk|ρA |ki

(4.284)The channel then correlates a density operator σBk with the post-measurement state k:|kihk|ρA |kihk|⊗ σBk .
hk|ρA |kiThis action leads to an ensemble:
|kihk|ρA |kihk|khk|ρA |ki,
⊗ σB ,
hk|ρA |ki

(4.285)(4.286)and the density operator of the ensemble isX
X|kihk|ρA |kihk|hk|ρA |ki
⊗ σBk =
|kihk|ρA |kihk| ⊗ σBk .
hk|ρ
|ki
A
k
k

(4.287)The channel then only outputs the system on the right (tracing out the first system) so thatthe resulting channel is as given in (4.283).
Exercise 4.6.1 What is a set of Kraus operators for a classical–quantum channel?
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4.6.6

CHAPTER 4. THE NOISY QUANTUM THEORY

Quantum-to-Classical Channels (Measurement Channels)

Quantum-to-classical, or quantum–classical channels for short, are in some sense the opposite
of classical–quantum channels. They take a quantum system to a classical one, and as such,they are in direct correspondence with measurements. So sometimes they are referred to asmeasurement channels. They also represent a way of generalizing classical channels different
from classical–quantum channels.
Definition 4.6.7 (Quantum–Classical Channels) Let {|xiX } be an orthonormal basis
for a Hilbert space HX , and let {ΛxA } be a POVM acting on the system A. A quantum–
classical channel has the following action on an input density operator ρA ∈ D(HA ):X
ρA →Tr{ΛxA ρA }|xihx|X .
(4.288)x

We should verify that this is indeed a quantum channel, by determining
its Kraus opP
erators. Consider that the trace operation can be written as Tr{·} = j hj|A · |jiA , where{|jiA } is some orthonormal basis for HA . Then we can rewrite (4.288) as
X
X np
p o
(4.289)ΛxA ρA ΛxA |xihx|XTr{ΛxA ρA }|xihx|X =
Tr
x

x

=

X

=

X

hj|A

p

ΛxA ρAp

ΛxA |jiA |xihx|X(4.290)pp
ΛxA ρA ΛxA |jiA hx|X .

(4.291)x,j|xiX hj|Ax,j

So this development
p x reveals that a set of Kraus operators for the channel in (4.288) is{Nx,j ≡ |xiX hj|A ΛA }. Let us verify the completeness relation for them:Xp
X †p
Nx,j Nx,j =ΛxA |jiA hx|X |xiX hj|A ΛxA(4.292)
x,j

x,j

Xp
p
ΛxA |jiA hj|A ΛxA=(4.293)x,j=X

ΛxA = IA ,(4.294)xwhere the last equality follows because {ΛxA } is a POVM.

4.6.7

Entanglement-Breaking Channels

An important class of channels is the set of entanglement-breaking channels, and we will seethat both quantum–classical and classical–quantum channels are special cases of them.
Definition 4.6.8 (Entanglement-Breaking Channel) An entanglement-breaking chanEBnel N EB : L(HA ) → L(HB ) is defined by the property that the channel idR ⊗NA→B
takes
any state ρRA to a separable state, where R is a reference system of arbitrary size.
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4.6. QUANTUM CHANNELS ARE ALL ENCOMPASSING

165

Fortunately, we do not need to checkthis property for all possible ρRA . In fact, itEB
suffices to check whether idR ⊗NA→B (ΦRA ) is a separable state, where ΦRA is a maximallyentangled state, as defined in (3.232).
Exercise 4.6.2 Prove that a quantum channel NA→B is entanglement-breaking if (idR ⊗NA→B ) (ΦRA )is a separable state, where ΦRA is a maximally entangled state. (Hint: You can use a trick
similar to that which you used to solve Exercise 4.4.1. Alternatively, you can inspect theproof of Theorem 4.6.1 below.)
Exercise 4.6.3 Show that both a classical–quantum channel and a quantum–classical channel are entanglement-breaking—i.e., if we input the A system of a bipartite state ρRA toeither of these channels, then the resulting state on systems RB is separable.
We can prove a more general structural theorem regarding entanglement-breaking channels by exploiting its definition.
Theorem 4.6.1 A channel is entanglement-breaking if and only if it has a Kraus representation with Kraus operators that are unit rank.
Proof. We first prove the “if-part” of the theorem. Suppose that the Kraus operators of aquantum channel NA→B are{Nz ≡ |ξz iB hϕz |A }.
(4.295)
Without loss of generality, we can take each |ξz iB to be a unit vector, simply by rescaling
the corresponding |ϕz iA . In order for this set to be a legitimate set of Kraus operators, thefollowing condition should holdXXX
|ϕz ihϕz |A .
(4.296)|ϕz iA hξz |B |ξz iB hϕz |A =Nz† Nz =
IA =
z

z

z

Now consider when such a channel acts on one share of a general bipartite state ρRA ∈D(HA ⊗ HB ):X(IR ⊗ |ξz iB hϕz |A ) ρRA (IR ⊗ |ϕz iA hξz |B )(4.297)(idR ⊗NA→B )(ρRA ) =z=

X

TrA {|ϕz ihϕz |A ρRA } ⊗ |ξz ihξz |B(4.298)pZ (z)ρzR ⊗ |ξz ihξz |B ,(4.299)z=

Xz

where in the last line we define the state ρzR ≡ TrA {|ϕz ihϕz |A ρRA }/pZ (z) and the probability
distribution pZ from pZ (z) = Tr{|ϕz ihϕz |A ρRA } (the fact that pZ is a probability distribution
follows from (4.296)). Consider now that the density operator in the last line above isseparable. Since ρRA is arbitrary, the “if-part” of the theorem follows.
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CHAPTER 4. THE NOISY QUANTUM THEORY

We now prove the “only-if” part. Consider that the output of an entanglement-breaking
channel N EB acting on one share of a maximally entangled state ΦRA is as follows:XEB
idR ⊗NA→B(ΦRA ) =pZ (z)|φz ihφz |R ⊗ |ψz ihψz |B ,(4.300)z

where pZ is a probability distribution and {|φz iR } and {|ψz iB } are sets of pure states. This
holds because the output of a channel is a separable state (the channel “breaks” entanglement), and it is always possible to find a representation of the separable state with pure states(see Exercise 4.3.3). Now consider constructing a quantum channel M with the following
unit-rank Kraus operators:
np
o
∗
Nz ≡
d pZ (z)|ψz iB hφz |A ,(4.301)z

where d is the Schmidt rank of the maximally entangled state ΦRA and |φ∗z iA is the state |φz iA
with all of its elements conjugated with respect to the bases defined from ΦP
RA . We should
first verify that these Kraus operators form a valid channel, by checking that z Nz† Nz = IA :X

Nz† Nz =X

d pZ (z)|φ∗z iA hψz |ψz iB hφ∗z |A(4.302)X(4.303)zz=dpZ (z)|φ∗z ihφ∗z |A .

z

Consider that
TrB




EB
idR ⊗NA→B(ΦRA ) = πR
= TrB

(4.304)(X)
pZ (z)|φz ihφz |R ⊗ |ψz ihψz |B(4.305)z=

XpZ (z)|φz ihφz |R ,(4.306)z

where πR is the maximally mixed state. Thus, it follows that M is a valid quantum channel
becausedXpZ (z)|φz ihφz |R = d πR = IR = (IA )∗(4.307)

z

=d

X
z

c 2019pZ (z)|φ∗z ihφ∗z |A =X

Nz† Nz .

(4.308)z

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4.6. QUANTUM CHANNELS ARE ALL ENCOMPASSING167

Now let us consider the action of the channel M on the maximally entangled state:(idR ⊗MA→B ) (ΦRA )pp
1X=|iihj|R ⊗ d pZ (z)|ψz iB hφ∗z |A |iihj|A |φ∗z iA hψz |B d pZ (z)d z,i,jX=pZ (z) |iihj|R ⊗ hφ∗z |ii hj|φ∗z i |ψz ihψz |B(4.309)(4.310)(4.311)

z,i,j

=

X

pZ (z) |ii hj|φ∗z i hφ∗z |ii hj|R ⊗ |ψz ihψz |B(4.312)pZ (z) |φz ihφz |R ⊗ |ψz ihψz |B .

(4.313)

z,i,j

=

X
z

P
The last equality follows from recognizing i,j |ii hj| · |iihj| as the transpose operation (withrespect to the bases from ΦRA ) and noting that the transpose is equivalent to conjugationEB
and MA→B on thefor a Hermitian operator |φz ihφz |. Finally, since the action of both NA→B
maximally entangled state is the same, we can conclude that the two channels are equal (seeSection 4.4.2). Thus, M is a representation of the channel with unit-rank Kraus operators.
The proof of the above theorem leads to the following important corollary:EB
is a serial concatenation of a
Corollary 4.6.1 An entanglement-breaking channel NA→B
EB
quantum–classical channel MA→Z with a classical–quantum channel PZ→B , i.e., NA→B=
PZ→B ◦ MA→Z . That is, every entanglement-breaking channel can be written as a measurement followed by a preparation.

Proof. Due to the above theorem, we can take the Kraus operators for an entanglementEB
breaking channel NA→B
to be as in (4.295), with {|ξz iB } a set of unit vectors and {|ϕz iA }satisfying (4.296). Let {|ziZ } be an orthonormal basis for a Hilbert space HZ . Then takethe quantum–classical channel MA→Z to be
X
MA→Z (ρA ) =Tr{|ϕz ihϕz |A ρA }|zihz|Z(4.314)z

and the classical–quantum channel PZ→B to be
X
PZ→B (σZ ) =hz|σZ |ziZ |ξz ihξz |B .

(4.315)zEB
One can then verify that NA→B
= PZ→B ◦ MA→Z .

4.6.8

Quantum Instruments

The description of a quantum channel with Kraus operators gives the most general evolutionthat a quantum state can undergo. We may want to specialize this definition somewhat for
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CHAPTER 4. THE NOISY QUANTUM THEORY

another scenario. Suppose that we would like to determine the most general evolution where
the input is a quantum system and the output consists of both a quantum system and aclassical system. Such a scenario may arise in a case where Alice is trying to transmit bothclassical and quantum information, and Bob exploits a quantum instrument to decode bothkinds of information. A quantum instrument gives such an evolution with a hybrid output.
Definition 4.6.9 (Trace Non-Increasing Map) A linear map M is trace non-increasing
if Tr{M(X)} ≤ Tr{X} for all positive semi-definite X ∈ L(H), with H a Hilbert space.
Definition 4.6.10 (Quantum Instrument) A quantum instrument consists of aPcollection {Ej } of completely positive, trace non-increasing maps such that the sum map j Ej istrace preserving. Let {|ji} be an orthonormal basis for a Hilbert space HJ . The action ofa quantum instrument on a density operator ρ ∈ D(H) is the following quantum channel,
which features a quantum and classical output:Xρ→
Ej (ρ) ⊗ |jihj|J .
(4.316)j

Let us see one way in which this definition comes about. Recall that we may view a noisy
quantum channel as arising from the forgetting of a measurement outcome, as in (4.243). Let
us now suppose that some third party performs a measurement with two outcomes j and k,
but does not give us access to the measurement outcome j. Suppose that the measurement
operators for this two-outcome measurement are {Mj,k }j,k . Let us first suppose that thethird party performs the measurement on a quantum system with density operator ρ andgives us both of the measurement outcomes. The post-measurement state in such a scenariois†
Mj,k ρMj,k,(4.317)
pJ,K (j, k)where the joint distribution of outcomes j and k is†
pJ,K (j, k) = Tr{Mj,kMj,k ρ}.

(4.318)

We can calculate the marginal distributions pJ (j) and pK (k) according to the law of total
probability:
XX†
pJ (j) =pJ,K (j, k) =Tr{Mj,kMj,k ρ},(4.319)pK (k) =

k

k

X

X

j

pJ,K (j, k) =†Tr{Mj,kMj,k ρ}.

(4.320)j

Suppose the measuring device also places the classical outcomes in classical registers J andK, so that the post-measurement state is†
Mj,k ρMj,k⊗ |jihj|J ⊗ |kihk|K ,pJ,K (j, k)c 2019(4.321)

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4.6. QUANTUM CHANNELS ARE ALL ENCOMPASSING

169

Figure 4.4: The figure on the left illustrates a quantum instrument, a general noisy evolution that produces
both a quantum and classical output. The figure on the right illustrates the internal workings of a quantum
instrument, showing that it results from having only partial access to a measurement outcome.

where the sets {|ji} and {|ki} form respective orthonormal bases. Such an operation ispossible physically, and we could retrieve the classical information at some later point byperforming a complete projective measurement of the registers J and K. If we would like todetermine the Kraus map for the overall quantum evolution, we simply take the expectation
over all measurement outcomes j and k:!
†
X
Mj,k ρMj,k⊗ |jihj|J ⊗ |kihk|K
pJ,K (j, k)
pJ,K (j, k)j,kX†Mj,k ρMj,k=⊗ |jihj|J ⊗ |kihk|K . (4.322)j,kLet us now suppose that we do not have access to the measurement result k. This lackof access is equivalent to lacking access to classical register K. To determine the resulting
state, we should trace out the classical register K. Our map then becomesX†
Mj,k ρMj,k⊗ |jihj|J .
(4.323)j,k

The above map corresponds to a quantum instrument, and is a general noisy quantum
evolution that produces both a quantum output and a classical output. Figure 4.4 depicts aquantum instrument.
We can rewrite the above map more explicitly as follows:
!
X X
X
†
Mj,k ρMj,k ⊗ |jihj|J =Ej (ρ) ⊗ |jihj|J ,(4.324)jjkwhere we define
Ej (ρ) ≡X†
Mj,k ρMj,k.

(4.325)k

Each j-dependent map Ej (ρ) is a completely positive trace-non-increasing map because
Tr {Ej (ρ)} ≤ 1. In fact, by examining the definition of Ej (ρ) and comparing to (4.319),
it holds that
Tr {Ej (ρ)} = pJ (j).
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CHAPTER 4. THE NOISY QUANTUM THEORY

It is important to note that the probability pJ (j) is dependent on the density operator ρthat is input to the instrument. We can determine the quantum output of the instrument
by tracing over the classical register J. The resulting quantum output is then
(
)
X
X
TrJ
Ej (ρ) ⊗ |jihj|J =Ej (ρ).
(4.327)jj

The above “sum map” is a trace-preserving map because
(
)
X
X
X
Tr
Ej (ρ) =Tr {Ej (ρ)} =pJ (j) = 1,jj(4.328)jwhere the last equality follows because the marginal probabilities pJ (j) sum to one. The
above points that we have mentioned are the most salient for the quantum instrument. We
will exploit this type of evolution when we require a device that outputs both a classical andquantum system.
We should stress that a quantum instrument is more general than applying a mixture of
CPTP maps to a quantum state. Suppose that we apply a mixture {Nj } of CPTP maps toa quantum state ρ, chosen according to a distribution pJ (j). The resulting expected state is
as follows:X
pJ (j)|jihj|J ⊗ Nj (ρ).
(4.329)j

The probabilities pJ (j) here are independent of the state ρ that is input to the mixtureof CPTP maps, but this is not generally the case for a quantum instrument. There, theprobabilities pJ (j) can depend on the state ρ that is input—it may be beneficial then to
write these probabilities as pJ (j|ρ) because there is an implicit conditioning on the statethat is input to the instrument.

4.7 Examples of Quantum Channels
This section discusses some of the most important examples of quantum channels that wewill consider in this book. Throughout, we will be considering the information-carrying
ability of these various channels. They will provide some useful, “hands on” insight intoquantum Shannon theory.

4.7.1

Noisy Evolution from a Random Unitary

Perhaps the simplest example of a quantum channel is the quantum bit-flip channel, which
has the following action on a qubit density operator ρ:
pXρX † + (1 − p)ρ.
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4.7. EXAMPLES OF QUANTUM CHANNELS

171

The above density operator is more “mixed” than the original density operator and we will
make this statement more precise in Chapter 10, when we study entropy. The evolution†
ρ → pXρX√+ (1 − p)ρ is clearly a legitimate quantum channel. Here the Kraus operators√
are { pX, 1 − pI} and it is clear that they satisfy the completeness relation.
A generalization of the above discussion is to consider some ensemble of unitaries (arandom unitary) {p(k), Uk } that we can apply to a density operator ρ, resulting in thefollowing output density operator:
X

p(k)Uk ρUk† .

(4.331)

k

4.7.2

Dephasing Channels

We have already given the example of a noisy quantum bit-flip channel in Section 4.7.1.
Another important example is a bit flip in the conjugate basis, or equivalently, a phase-flipchannel. This channel acts as follows on any given density operator:
ρ → (1 − p)ρ + pZρZ.

(4.332)

It is also known as a dephasing channel.
For p = 1/2, the action of the dephasing channel on a given quantum state is equivalent
to the action of measuring the qubit in the computational basis and forgetting the result of
the measurement. We make this idea more clear with an example. First, suppose that wehave a qubit|ψi = α|0i + β|1i,(4.333)and we measure it in the computational basis. Then the postulates of quantum theory statethat the qubit becomes |0i with probability |α|2 and it becomes |1i with probability |β|2 .
Suppose that we forget the measurement outcome, or alternatively, that we do not haveaccess to it. Then our best description of the qubit is with the following ensemble:|α|2 , |0i , |β|2 , |1i.

(4.334)The density operator of this ensemble is|α|2 |0ih0| + |β|2 |1ih1|.

(4.335)

Now let us check if the dephasing channel gives the same behavior as the forgetful measurement above. We can consider the qubit as being an ensemble {1, |ψi}, i.e., the state iscertain to be |ψi. The density operator of the ensemble is then ρ whereρ = |α|2 |0ih0| + αβ ∗ |0ih1| + α∗ β|1ih0| + |β|2 |1ih1|.
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CHAPTER 4. THE NOISY QUANTUM THEORY

If we act on the density operator ρ with the dephasing channel with p = 1/2, then it preserves
the density operator with probability 1/2 and phase flips the qubit with probability 1/2:
11
ρ + ZρZ22
1=|α|2 |0ih0| + αβ ∗ |0ih1| + α∗ β|1ih0| + |β|2 |1ih1| +2
1|α|2 |0ih0| − αβ ∗ |0ih1| − α∗ β|1ih0| + |β|2 |1ih1|
2= |α|2 |0ih0| + |β|2 |1ih1|.

(4.337)(4.338)The dephasing channel eliminates the off-diagonal terms of the density operator when represented with respect to the computational basis. The resulting density operator description is
the same as what we found for the forgetful measurement. It is also equivalent to a classicalchannel, as given in Definition 4.6.5.
Exercise 4.7.1 Verify that the action of the dephasing channel on the Bloch vector is1(I + rx X + ry Y + rz Z) →2
1
(I + (1 − 2p)rx X + (1 − 2p)ry Y + rz Z) , (4.339)
2
so that the channel preserves any component of the Bloch vector in the Z direction, whileshrinking any component in the X or Y direction.

4.7.3

Pauli Channels

A Pauli channel is a generalization of the above dephasing channel and the bit-flip channel.
It simply applies a random Pauli operator according to a probability distribution. The map
for a qubit Pauli channel is1Xρ→p(i, j)Z i X j ρX j Z i .
(4.340)i,j=0

The generalization of this channel to qudits is straightforward. We simply replace the Pauli
operators with the Heisenberg–Weyl operators. The Pauli qudit channel is
ρ→

d−1
X

p(i, j)Z(i)X(j)ρX † (j)Z † (i).

(4.341)i,j=0These channels have been prominent in the study of quantum key distribution.
Exercise 4.7.2 We can write a Pauli channel asρ → pI ρ + pX XρX + pY Y ρY + pZ ZρZ.
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4.7. EXAMPLES OF QUANTUM CHANNELS

173

Verify that the action of the Pauli channel on the Bloch vector is(rx , ry , rz ) →((pI + pX − pY − pZ ) rx , (pI + pY − pX − pZ ) ry , (pI + pZ − pX − pY ) rz ) . (4.343)

4.7.4

Depolarizing Channels

The depolarizing channel is a “worst-case scenario” channel. It assumes that we completely
lose the input qubit with some probability, i.e., it replaces the lost qubit with the maximally
mixed state. The map for the depolarizing channel isρ → (1 − p)ρ + pπ,(4.344)

where π is the maximally mixed state: π = I/2.
Most of the time, this channel is too pessimistic. Usually, we can learn something about
the physical nature of the channel by some estimation process. We should only consider
using the depolarizing channel as a model if we have little to no information about theactual physical channel.
Exercise 4.7.3 (Pauli Twirl) Show that randomly applying the Pauli operators I, X, Y ,Z with uniform probability to any density operator gives the maximally mixed state:
1
1
1
1
ρ + XρX + Y ρY + ZρZ = π.
4444(4.345)(Hint: Represent the density operator as ρ = (I + rx X + ry Y + rz Z) /2 and apply the commutation rules of the Pauli operators.) This is known as the “twirling” operation.
Exercise 4.7.4 Show that we can rewrite the depolarizing channel as the following Paulichannel:111ρ → (1 − 3p/4) ρ + p
XρX + Y ρY + ZρZ .
(4.346)4
4
4
Exercise 4.7.5 Show that the action of a depolarizing channel on the Bloch vector is(rx , ry , rz ) → ((1 − p)rx , (1 − p)ry , (1 − p)rz ) .

(4.347)

Thus, it uniformly shrinks the Bloch vector to become closer to the maximally mixed state.
The generalization of the depolarizing channel to qudits is again straightforward. It is
the same as the map in (4.344), with the exception that the density operators ρ and π arequdit density operators.
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CHAPTER 4. THE NOISY QUANTUM THEORY

Exercise 4.7.6 (Qudit Twirl) Show that randomly applying the Heisenberg–Weyl operators{X(i)Z(j)}i,j∈{0,...,d−1}(4.348)with uniform probability to any qudit density operator gives the maximally mixed state π:d−1

1 X
X(i)Z(j)ρZ † (j)X † (i) = π.
d2 i,j=0(4.349)(Hint: You can do the full calculation, or you can decompose this channel into the composition of two completely dephasing channels where the first is a dephasing in the computational
basis and the next is a dephasing in the conjugate basis).

4.7.5

Amplitude Damping Channels

The amplitude damping channel is an approximation to a noisy evolution that occurs in many
physical systems ranging from optical systems to chains of spin-1/2 particles to spontaneousemission of a photon from an atom.
In order to motivate this channel, we give a physical interpretation to our computational
basis states. Let us think of the |0i state as the ground state of a two-level atom and let us
think of the state |1i as the excited state of the atom. Spontaneous emission is a processthat tends to decay the atom from its excited state to its ground state, even if the atom is in
a superposition of the ground and excited states. Let the parameter γ denote the probabilityof decay so that 0 ≤ γ ≤ 1. One Kraus operator that captures the decaying behavior is√(4.350)A0 = γ|0ih1|.
The operator A0 annihilates the ground state:
A0 |0ih0|A†0 = 0,(4.351)and it decays the excited state to the ground state:
A0 |1ih1|A†0 = γ|0ih0|.

(4.352)†The Kraus operator A0 alone does not specify a physical map because AP
0 A0 = γ|1ih1| (recall
that the Kraus operators of any channel should satisfy the condition k A†k Ak = I). We
can satisfy this condition by choosing another operator A1 such thatA†1 A1 = I − A†0 A0 = |0ih0| + (1 − γ) |1ih1|.
The following choice of A1 satisfies the above condition:p
A1 ≡ |0ih0| + 1 − γ|1ih1|.

(4.353)(4.354)Thus, the operators A0 and A1 are valid Kraus operators for the amplitude damping channel.
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4.7. EXAMPLES OF QUANTUM CHANNELS

175

Exercise 4.7.7 Consider a single-qubit density operator with the following matrix representation with respect to the computational basis:


1−p η
ρ=
,
(4.355)η∗pwhere 0 ≤ p ≤ 1 and η is some complex number. Show that applying the amplitude damping
channel with parameter γ to a qubit with the above density operator gives a density operator
with the following matrix representation:√1√− (1 − γ) p
1 − γη
.
(4.356)
1 − γη ∗(1 − γ) p
Exercise 4.7.8 Show that the amplitude damping channel obeys a composition rule. Consider an amplitude damping channel N1 with transmission parameter (1 − γ1 ) and consider
another amplitude damping channel N2 with transmission parameter (1 − γ2 ). Show that thecomposition channel N2 ◦ N1 is an amplitude damping channel with transmission parameter(1 − γ1 ) (1 − γ2 ). (Note that the transmission parameter is equal to one minus the damping
parameter.)

4.7.6

Erasure Channels

The erasure channel is another important channel in quantum Shannon theory. It admits a
simple model and is amenable to relatively straightforward analysis when we later discuss
its capacity. The erasure channel can serve as a simplified model of photon loss in opticalsystems.
We first recall the classical definition of an erasure channel. A classical erasure channel
either transmits a bit with some probability 1 − ε or replaces it with an erasure symbol ewith some probability ε. The output alphabet contains one more symbol than the input
alphabet, namely, the erasure symbol e.
The generalization of the classical erasure channel to the quantum world is straightforward. It implements the following map:
ρ → (1 − ε) ρ + ε|eihe|,(4.357)

where |ei is some state that is not in the input Hilbert space, and thus is orthogonal to it.
The output space of the erasure channel is larger than its input space by one dimension.
The interpretation of the quantum erasure channel is similar to that for the classical erasurechannel. It transmits a qubit with probability 1 − ε and “erases” it (replaces it with an
orthogonal erasure state) with probability ε.
Exercise 4.7.9√Show that the following operators
are Kraus√√ operators for the quantum erasure channel: { 1 − ε(|0iB h0|A + |1iB h1|A ), ε|eiB h0|A , ε|eiB h1|A }.
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CHAPTER 4. THE NOISY QUANTUM THEORY

Figure 4.5: The figure on the left depicts a general operation, a conditional quantum encoder, that takes aclassical system to a quantum system. The figure on the right depicts the inner workings of the conditional
quantum encoder.

At the receiving end of the channel, a simple measurement can determine whether an
erasure has occurred. We perform a measurement with measurement operators {Πin , |eihe|},where Πin is the projector onto the input Hilbert space. This measurement has the benefit ofdetecting no more information than necessary. It merely detects whether an erasure occurs,
and thus preserves the quantum information at the input if an erasure does not occur.

4.7.7

Conditional Quantum Channels

We end this chapter by considering one final type of evolution. A conditional quantummencoder EM A→B , or conditional quantum channel , is a collection {EA→B}m of CPTP maps.
Its inputs are a classical system M and a quantum system A and its output is a quantum
system B. A conditional quantum encoder can function as an encoder of both classical andquantum information.
A classical–quantum state ρM A , whereXp(m)|mihm|M ⊗ ρm
(4.358)
ρM A ≡
A,m

can act as an input to a conditional quantum encoder EM A→B . The action of the conditional
quantum encoder EM A→B on the classical–quantum state ρM A is as follows:
(
)
X
m
m
EM A→B (ρM A ) = TrM
p(m)|mihm|M ⊗ EA→B (ρA ) .
(4.359)m

Figure 4.5 depicts the behavior of the conditional quantum encoder.
It is actually possible to write any quantum channel as a conditional quantum encoderwhen its input is a classical–quantum state. Indeed, consider any quantum channel NXA→B
that has input systems X and A and output system B. Suppose the Kraus decompositionof this channel is as follows:
X
NXA→B (ρ) ≡Aj ρA†j .
(4.360)j

Suppose now that the input to the channel is the following classical–quantum state:XσXA ≡pX (x)|xihx|X ⊗ ρxA .
(4.361)x
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4.7. EXAMPLES OF QUANTUM CHANNELS

177

Then the channel NXA→B acts as follows on the classical–quantum state σXA :X
NXA→B (σXA ) =Aj (pX (x)|xihx|X ⊗ ρxA ) A†j .

(4.362)j,xConsider that a classical–quantum state admits the following matrix representation by exploiting the tensor product:X(4.363)pX (x)|xihx|X ⊗ ρxAx∈X

pX (x1 )ρxA1
0
···
0
..


0
pX (x2 )ρxA2
.


=

..
...


.
0
x|X |
0
···
0 pX (x|X | )ρA
M=
pX (x)ρx .

(4.364)(4.365)x∈X

It is possible to specify a matrix representation for each Kraus operator Aj in terms of |X |block matrices:
Aj = Aj,1 Aj,2 · · · Aj,|X | .
(4.366)
Each operator Aj (pX (x)|xihx|X ⊗ ρxA ) A†j in the sum in (4.362) then takes the following form:
Aj (pX (x)|xihx|X ⊗ ρxA ) A†j= Aj,1 Aj,2 · · ·=X(4.367) † pX (x1 )ρxA1 0 · · ·
0
Aj,1
† 
.
.



.
.

.
0
.
  Aj,2 
Aj,|X | 
  .. ..
..
 . 

.
0
.
x|X |
A†j,|X |0· · · 0 pX (x|X | )ρA

pX (x)Aj,x ρxA A†j,x .

(4.368)(4.369)x∈|X |

We can write the overall map as follows:
NXA→B (σXA ) =

XX
j

=
=

pX (x)Aj,x ρxA A†j,x(4.370)x∈X

X

pX (x)X

Aj,x ρxA A†j,x(4.371)

x∈X

j

X

x
pX (x)NA→B(ρxA ),(4.372)X(4.373)x∈Xx
where we define each map NA→B
as follows:
x
NA→B(ρxA ) =Aj,x ρxA A†j,x .

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CHAPTER 4. THE NOISY QUANTUM THEORY

Thus, the action of any quantum channel on a classical–quantum state is the same as theaction of the conditional quantum encoder.
P
Exercise 4.7.10 Show that the condition j A†j Aj = I implies the |X | conditions:X †∀x ∈ X :Aj,x Aj,x = I.
(4.374)j

4.8 Summary
We give a brief summary of the main results in this chapter. We derived all of these resultsfrom the noiseless quantum theory and an ensemble viewpoint. An alternate viewpoint is
to say that the density operator is the state of the system and then give the postulates ofquantum mechanics in terms of the density operator. Regardless of which viewpoint you
consider as more fundamental, they are consistent with each other.
The density operator ρ for an ensemble {pX (x), |ψx i} is the following expectation:Xρ=pX (x)|ψx ihψx |.
(4.375)x

The evolution of the density operator according to a unitary operator U is
ρ → U ρU † .
A measurement of the state according to a measurement {Mj } whereto the following post-measurement state:
Mj ρMj†,ρ→pJ (j)
where the probability pJ (j) for obtaining outcome j isno†pJ (j) = Tr Mj Mj ρ .

(4.376)†
j Mj Mj = I leadsP(4.377)(4.378)The most general noisy evolution that a quantum state can undergo is according to a completely positive, trace-preserving map N (ρ) that we can write as follows:X
N (ρ) =
Aj ρA†j ,(4.379)jP
where j A†j Aj = I. A special case of this evolution is a quantum instrument. A quantum
instrument has a quantum input and a classical and quantum output. The most general way
to represent a quantum instrument is as follows:Xρ→
Ej (ρ) ⊗ |jihj|J ,(4.380)j
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4.9. HISTORY AND FURTHER READING

where each map Ej is a completely positive, trace-non-increasing map, whereX
Ej (ρ) =Aj,k ρA†j,k ,179(4.381)k

and†j,k Aj,k Aj,k = I, so that the overall map is trace-preserving.

P

4.9 History and Further Reading
Nielsen and Chuang (2000) have given an excellent introduction to noisy quantum channels. Werner (1989) defined what it means for a multiparty quantum state to be entangled. Horodecki et al. (2003) introduced entanglement-breaking channels and proved several
properties of them (e.g., the proof of Theorem 4.6.1). Davies and Lewis (1970) introduced
the quantum instrument formalism, and Ozawa (1984) developed it further. Grassl et al.
(1997) introduced the quantum erasure channel and constructed some simple quantum errorcorrecting codes for it. A discussion of the conditional quantum channel appears in Yard
(2005).

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180

c 2019CHAPTER 4. THE NOISY QUANTUM THEORYMark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 5

The Purified Quantum Theory
The final chapter of our development of the quantum theory gives perhaps the most powerful viewpoint, by providing a mathematical tool, the purification theorem, which offers acompletely different way of thinking about noise in quantum systems. This theorem statesthat our lack of information about a set of quantum states can be thought of as arising fromentanglement with another system to which we do not have access. The system to which
we do not have access is known as a purifying system. In this purified view of the quantum
theory, noisy evolution arises from the interaction of a quantum system with its environment.
The interaction of a quantum system with its environment leads to correlations between thequantum system and its environment, and this interaction leads to a loss of information
because we cannot access the environment. The environment is thus the purification of theoutput of the noisy quantum channel.
In Chapter 3, we introduced the noiseless quantum theory. The noiseless quantum theory
is a useful theory to learn so that we can begin to grasp an intuition for some uniquelyquantum behavior, but it is an idealized model of quantum information processing. In
Chapter 4, we introduced the noisy quantum theory as a generalization of the noiselessquantum theory. The noisy quantum theory can describe the behavior of imperfect quantum
systems that are subject to noise.
In this chapter, we actually show that we can view the noisy quantum theory as a
special case of the noiseless quantum theory. This relation may seem strange at first, but
the purification theorem allows us to make this connection. The quantum theory that we
present in this chapter is a noiseless quantum theory, but we name it the purified quantum
theory, in order to distinguish it from the description of the noiseless quantum theory in
Chapter 3.
The purified quantum theory shows that it is possible to view noise as resulting fromentanglement of a system with another system. We have actually seen a glimpse of thisphenomenon in the previous chapter when we introduced the notion of the local density
operator, but we did not highlight it in detail there. The example was the maximallyentangled Bell state |Φ+ iAB . This state is a pure state on the two systems A and B, but
the local density operator of Alice is the maximally mixed state πA . We saw that the local
181

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CHAPTER 5. THE PURIFIED QUANTUM THEORY

Figure 5.1: This diagram depicts a purification |ψiRA of a density operator ρA . The above diagram
indicates that the reference system R is generally entangled with the system A. An interpretation of thepurification theorem is that the noise inherent in a density operator ρA is due to entanglement with areference system R.

density operator is a mathematical object that allows us to make all the predictions about
any local measurement or evolution. We also have seen that a density operator arises froman ensemble, but there is also the reverse interpretation, that an ensemble corresponds toa convex decomposition of any density operator. There is a sense in which we can view
this local density operator as arising from an ensemble where we choose the states |0i and|1i with equal probability 1/2. The purification idea goes as far as to say that the noisy
ensemble for Alice with density operator πA arises from the entanglement of her system with
Bob’s. We explore this idea in more detail in this final chapter on the quantum theory.

5.1 Purification
Suppose we are given a density operator ρA on a system A. Every such density operator has
a purification, as defined below and depicted in Figure 5.1:
Definition 5.1.1 (Purification) A purification of a density operator ρA ∈ D(HA ) is apure bipartite state |ψiRA ∈ HR ⊗ HA on a reference system R and the original system A,
with the property that the reduced state on system A is equal to ρA :
ρA = TrR {|ψihψ|RA } .
Suppose that a spectral decomposition for the density operator ρA is as follows:XρA =pX (x)|xihx|A .

(5.1)

(5.2)

x

We claim that the following state |ψiRA is a purification of ρA :Xp
pX (x)|xiR |xiA ,|ψiRA ≡

(5.3)

xwhere the set {|xiR }x of vectors is some set of orthonormal vectors for the reference system
R. The next exercise asks you to verify this claim.
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5.1. PURIFICATION

183

Exercise 5.1.1 Show that the state |ψiRA , as defined in (5.3), is a purification of the density
operator ρA , with a spectral decomposition as given in (5.2).
√
Exercise 5.1.2 (Canonical purification) Let ρA be a density operator and let ρA be its√ √unique positive semi-definite square root (i.e., ρA = ρA ρA .) We define the canonicalpurification of ρA as follows:√(5.4)(IR ⊗ ρA ) |ΓiRA ,where |ΓiRA is the unnormalized maximally entangled vector from (3.233). Show that (5.4)
is a purification of ρA .

5.1.1

Interpretation of Purifications

The purification idea has an interesting physical interpretation: we can think of the noisinessinherent in a particular quantum system as being due to entanglement with some externalreference system to which we do not have access. That is, we can think that the density
operator ρA arises from the entanglement of the system A with the reference system R andfrom our lack of access to the system R.
Stated another way, the purification idea gives us a fundamentally different way to interpret noise. The interpretation is that any noise on a local system is due to entanglementwith another system to which we do not have access. This interpretation extends to thenoise from a noisy quantum channel. We can view this noise as arising from the interactionof the system that we possess with an external environment over which we have no control.
The global state |ψiRA is a pure state, but a reduced state ρA is not a pure state ingeneral because we trace over the reference system to obtain it. A reduced state ρA is pure
if and only if the global state |ψiRA is a pure product state.

5.1.2

Equivalence of Purifications

Theorem 5.1.1 below states that there is an equivalence relation between all purifications of agiven density operator ρA . It is a consequence of the Schmidt decomposition (Theorem 3.8.1).
Before stating it, recall the definition of an isometry from Definition 4.6.3.
Theorem 5.1.1 All purifications of a density operator are related by an isometry acting on
the purifying system. That is, let ρA be a density operator, and let |ψiR1 A and |ϕiR2 A be
purifications of ρA , such that dim(HR1 ) ≤ dim(HR2 ). Then there exists an isometry UR1 →R2
such that|ϕiR2 A = (UR1 →R2 ⊗ IA ) |ψiR1 A .
(5.5)
Proof. Let us first suppose that the eigenvalues of ρA are distinct, so that a unique spectral
decomposition of ρA is as follows:XρA =pX (x)|xihx|A .
(5.6)
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CHAPTER 5. THE PURIFIED QUANTUM THEORY

Then a Schmidt decomposition of |ϕiR2 A necessarily has the form
|ϕiR2 A =Xp
pX (x)|ϕx iR2 |xiA ,

(5.7)

x

where {|ϕx iR2 } is an orthonormal basis for the R2 system, and similarly, the Schmidt decomposition of |ψiR1 A necessarily has the form
|ψiR1 A =Xp
pX (x)|ψx iR1 |xiA .

(5.8)

x

(If it were not the case then we could not have TrR2 {|ϕihϕ|R2 A } = TrR1 {|ψihψ|R1 A } = ρA , asgiven in the statement of the theorem.) Given the above, we can take the isometry UR1 →R2
to be
XUR1 →R2 =|ϕx iR2 hψx |R1 ,(5.9)
xwhich is an isometry because U † U = IR1 . If the eigenvalues of ρA are not distinct, thenthere is more freedom in the Schmidt decompositions, but here we are free to choose themas above, and then the development is the same.
This theorem leads to a way of relating all convex decompositions of a given density
operator, addressing a question raised in Section 4.1.1:
Corollary 5.1.1 Let two convex decompositions of a density operator ρ be as follows:
ρ=

d
X

0

pX (x)|ψx ihψx | =dXpY (y)|φy ihφy |,(5.10)

y=1

x=1

where d0 ≤ d. Then there exists an isometry U such thatXpppX (x)|ψx i =Ux,y pY (y)|φy i.

(5.11)y

Proof. Let {|xiR } be an orthonormal basis for a purification system, with a number ofstates equal to max {d, d0 }. Then a purification for the first decomposition is as follows:|ψiRA ≡Xp
pX (x)|xiR ⊗ |ψx iA ,(5.12)xand a purification of the second decomposition is|φiRA ≡Xp
pY (y)|yiR ⊗ |φy iA .

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5.2. ISOMETRIC EVOLUTION

185

From Theorem 5.1.1, we know that there exists an isometry UR such that |ψiRA = (UR ⊗ IA ) |φiRA .
Then consider thatXpppX (x)|ψx iA =pX (x0 )hx|R |x0 iR ⊗ |ψx0 iA = (hx|R ⊗ IA ) |ψiRA(5.14)x0= (hx|R UR ⊗ IA ) |φiRA =Xp

pY (y)hx|R UR |yiR |φy iA(5.15)yXp
pY (y)Ux,y |φy iA ,=(5.16)ywhere in the last step we have defined Ux,y = hx|R UR |yiR .
Exercise 5.1.3 Find a purification of the following classical–quantum state:XpX (x)|xihx|X ⊗ ρxA .

(5.17)x

Exercise 5.1.4 Let {pX (x), ρxA } be an ensemble of density operators. SupposePthat |ψ x iRA
is a purification of ρxA . The expected density operator of the ensemble is ρA ≡ x pX (x)ρxA .
Find a purification of ρA .

5.1.3

Extension of a Quantum State

We can also define an extension of a quantum state ρA :Definition 5.1.2 (Extension) An extension of a density operator ρA ∈ D(HA ) is a density
operator ΩRA ∈ D(HR ⊗ HA ) such that ρA = TrR {ΩRA } .
This notion can be useful, but keep in mind that we can always find a purification |ψiR0 RA
of the extension ΩRA .

5.2 Isometric Evolution
A quantum channel admits a purification as well. We motivate this idea with a simple
example.

5.2.1

Example: Isometric Extension of the Bit-Flip Channel

Consider the bit-flip channel from (4.330)—it applies the identity operator with some probability 1 − p and applies the bit-flip Pauli operator X with probability p. Suppose that we
input a qubit system A in the state |ψi to this channel. The ensemble corresponding to thestate at the output has the following form:{{1 − p, |ψi} , {p, X|ψi}} ,c 2019(5.18)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 5. THE PURIFIED QUANTUM THEORY

and the density operator of the resulting state is(1 − p)|ψihψ| + pX|ψihψ|X.

(5.19)

The following state is a purification of the above density operator (you should quickly checkthat this relation holds):p√
1 − p|ψiA |0iE + pX|ψiA |1iE .

(5.20)We label the original system as A and label the purification system as E. In this context,
we can view the purification system as the environment of the channel.
There is another way for interpreting the dynamics of the above bit-flip channel. Insteadof determining the ensemble for the channel and then purifying, we can say that the channel
directly implements the following map from the system A to the larger joint system AE:
p√(5.21)|ψiA → 1 − p|ψiA |0iE + pX|ψiA |1iE .
We see that any p ∈ (0, 1), i.e., any amount of noise in the channel, can lead to entanglementof the input system with the environment E. We then obtain the noisy dynamics of thechannel by discarding (tracing out) the environment system E.
Exercise 5.2.1 Find two input states for which the map in (5.21) does not lead to entanglement between systems A and E.
The map in (5.21) is an isometric extension of the bit-flip channel. Let us label it as
UA→AE where the notation indicates that the input system is A and the output system is
AE. As discussed around Definition 4.6.3, an isometry is similar to a unitary operator but
different because it maps states in one Hilbert space (for an input system) to states in alarger Hilbert space (which could be for a joint system). It generally does not admit asquare matrix representation, but instead admits a rectangular matrix representation. The
matrix representation of the isometric operation in (5.21) consists of the following matrix
elements:


  √
1−p
0
h0|A h0|E UA→AE |0iA h0|A h0|E UA→AE |1iA√
 h0|A h1|E UA→AE |0iA h0|A h1|E UA→AE |1iA  0
=√ p .
(5.22)
 h1|A h0|E UA→AE |0iA h1|A h0|E UA→AE |1iA  
0
1−p 
√
p
0
h1|A h1|E UA→AE |0iA h1|A h1|E UA→AE |1iA
There is no reason that we have to choose the environment states as we did in (5.21). We
could have chosen the environment states to be any orthonormal basis—isometric behavior
only requires that the states on the environment be distinguishable. This is related to thefact that all purifications are related by an isometry acting on the purifying system (seeTheorem 5.1.1).
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5.2. ISOMETRIC EVOLUTION

187

An Isometry is Part of a Unitary on a Larger System
We can view the dynamics in (5.21) as an interaction between an initially pure environment
and the qubit state |ψi. So, an equivalent way to implement an isometric mapping is witha two-step procedure. We first assume that the environment of the channel is in a purestate |0iE before the interaction begins. The joint state of the qubit |ψi and the environmentis|ψiA |0iE .
(5.23)
These two systems then interact according to a unitary operator VAE . We can specify twocolumns of the unitary operator (we make this more clear in a bit) by means of the isometricmapping in (5.21):
p√(5.24)
VAE |ψiA |0iE = 1 − p|ψiA |0iE + pX|ψiA |1iE .
In order to specify the full unitary VAE , we must also specify how the map behaves when
the initial state of the qubit and the environment is|ψiA |1iE .
We choose the mapping to be as follows so that the overall interaction is unitary:p√
VAE |ψiA |1iE = p|ψiA |0iE − 1 − pX|ψiA |1iE .

(5.25)(5.26)

Exercise 5.2.2 Check that the operator VAE , defined by (5.24) and (5.26), is unitary bydetermining its action on the computational basis {|0iA |0iE , |0iA |1iE , |1iA |0iE , |1iA |1iE } andshowing that all of the outputs for each of these inputs form an orthonormal basis.
Exercise 5.2.3 Verify that the matrix representation of the full unitary operator VAE , defined by (5.24) and (5.26), is

 √
√
1−p
p
0
0
√
√

− 1−p 
0
0
,

√ p
√
(5.27)


0
0
1
−
p
p
√
√
p
− 1−p
0
0
by considering the matrix elements hi|A hj|E V |kiA |liE .
Complementary Channel
We may not only be interested in the receiver’s output of the quantum channel. We may also
be interested in determining the environment’s output from the channel. This idea becomesincreasingly important as we proceed in our study of quantum Shannon theory. We should
consider all parties in a quantum protocol, and the purified quantum theory allows us to doso. We consider the environment as one of the parties in a quantum protocol because theenvironment could also be receiving some quantum information from the sender.
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CHAPTER 5. THE PURIFIED QUANTUM THEORY

We can obtain the environment’s output from the quantum channel simply by tracingout every system besides the environment. The map from the sender to the environment is
known as a complementary channel. In our example of the isometric extension of the bit-flipchannel in (5.21), we can check that the environment receives the following output state if
the channel input is |ψiA :
 p
o
np
√
√
1 − p|ψiA |0iE + pX|ψiA |1iE
1 − phψ|A h0|E + phψ|A Xh1|E
TrA
n
o
p
= TrA (1 − p)|ψihψ|A ⊗ |0ih0|E + p(1 − p)X|ψihψ|A ⊗ |1ih0|E
np
o
+ TrA
p (1 − p)|ψihψ|A X ⊗ |0ih1|E + pX|ψihψ|A X ⊗ |1ih1|E(5.28)p= (1 − p)|0ih0|E + p (1 − p)hψ|X|ψi|1ih0|Ep(5.29)+ p(1 − p)hψ|X|ψi|0ih1|E + p|1ih1|Ep= (1 − p)|0ih0|E + p (1 − p)hψ|X|ψi (|1ih0|E + |0ih1|E ) + p|1ih1|E(5.30)
p(5.31)= (1 − p)|0ih0|E + p (1 − p)2 Re {α∗ β} (|1ih0|E + |0ih1|E ) + p|1ih1|E ,where in the last line we assume that the qubit |ψi ≡ α|0i + β|1i.
It is helpful to examine several cases of the above example. Consider the case in which
the noise parameter p = 0 or p = 1. In this case, the environment receives one of therespective states |0i or |1i. Therefore, in these cases, the environment does not receive anyof the quantum information about the state |ψi transmitted down the channel—it does notlearn anything about the probability amplitudes α or β. This viewpoint is a completely
different way to see that the channel is truly noiseless in these cases. A channel is noiselessif the environment of the channel does not learn anything about the states that we transmit
through it, i.e., if the channel does not leak quantum information to the environment. Now
let us consider p
the case in which p ∈ (0, 1). As p approaches 1/2 from either above or below,
the amplitude p(1 − p) of the off-diagonal terms is a monotonic function that reaches its
peak at 1/2. Thus, at the peak 1/2, the off-diagonal terms are the strongest, implying thatthe environment is generally “stealing” much of the coherence from the original quantum
state |ψi.
Exercise 5.2.4 Show that the receiver’s output density operator for a bit-flip channel with
p = 1/2 is the same as what the environment obtains.

5.2.2

Isometric Extension of a Quantum Channel

We now give a general definition for an isometric extension of a quantum channel:
Definition 5.2.1 (Isometric Extension) Let HA and HB be Hilbert spaces, and let N :L(HA ) → L(HB ) be a quantum channel. Let HE be a Hilbert space with dimension nosmaller than the Choi rank of the channel N . An isometric extension or Stinespring dilation
U : HA → HB ⊗ HE of the channel N is a linear isometry such that
TrE {U XA U † } = NA→B (XA ),c 2019(5.32)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

5.2. ISOMETRIC EVOLUTION

189

N
Figure 5.2: This figure depicts an isometric extension UA→BEof a quantum channel NA→B . The extension
N
UA→BE
includes the inaccessible environment on system E as a “receiver” of quantum information. Ignoring
the environment E gives the quantum channel NA→B .

for XA ∈ L(HA ). The fact that U is an isometry is equivalent to the following conditions:U † U = IA ,U U † = ΠBE ,(5.33)

where ΠBE is a projection of the tensor-product Hilbert space HB ⊗ HE .
Notation 5.2.1 We often write a channel N : L(HA ) → L(HB ) as NA→B in order toindicate the input and output systems explicitly. Similarly, we often write an isometricNin order to indicate its association with N
extension U : HA → HB ⊗ HE of N as UA→BE
explicitly, as well the fact that it accepts an input system A and has output systems B and E.
The system E is often referred to as an “environment” system. Finally, there is a quantumNN
channel UA→BE
associated to an isometric extension UA→BE, which is defined by
N
UA→BE(XA ) = U XA U † ,(5.34)

Nis a quantum channel with a single Kraus operator U
for XA ∈ L(HA ). Note that UA→BE†
given that U U = IA .

We can think of an isometric extension of a quantum channel as a purification of thatchannel: the environment system E is analogous to the purification system from Section 5.1
because we trace over it to get back the original channel. An isometric extension extends theoriginal channel because it produces the evolution of the quantum channel NA→B if we trace
out the environment system E. It also behaves as an isometry—it is analogous to a rectangular matrix that behaves somewhat like a unitary operator. The matrix representation of
an isometry is a rectangular matrix formed from selecting only a few of the columns froma unitary matrix. The property U † U = IA indicates that the isometry behaves analogously
to a unitary operator, because we can determine an inverse operation simply by taking its
conjugate transpose. The property U U † = ΠBE distinguishes an isometric operation from a
unitary one. It states that the isometry takes states in the input system A to a particularsubspace of the joint system BE. The projector ΠBE projects onto the subspace where theisometry takes input quantum states. Figure 5.2 depicts a quantum circuit for an isometricextension.
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CHAPTER 5. THE PURIFIED QUANTUM THEORY

Isometric Extension from Kraus Operators
It is possible to determine an isometric extension of a quantum channel directly from a setof Kraus operators. Consider a quantum channel NA→B with the following Kraus representation:
X
NA→B (ρA ) =Nj ρA Nj† .
(5.35)
j

An isometric extension of the channel NA→B is the following linear map:
N
UA→BE
≡

X

Nj ⊗ |jiE .

(5.36)

j

It is straightforward to verify that the above map is an isometry:
!
U


N †

X

UN =

Nk† ⊗ hk|E=

Nj ⊗ |jiE(5.37)

j

k

X

!
X

Nk† Nj hk|ji(5.38)

Nk† Nk(5.39)

k,j

=

X
k

= IA .

(5.40)The last equality follows from the completeness condition of the Kraus operators. As a†consequence, we get that U N U N is a projector on the joint system BE, which follows by
the same reasoning given in (4.259). Finally, we should verify that U N is an extension of N .
N
to an arbitrary density operator ρA gives the following map:
Applying the channel UA→BE
†
N
UA→BE
(ρA ) ≡ U N ρA U N!
X=Nj ⊗ |jiE ρA(5.41)
!
X

j

=

X

Nk† ⊗ hk|E

(5.42)k

Nj ρA Nk† ⊗ |jihk|E ,(5.43)

j,k

and tracing out the environment system gives back the original quantum channel NA→B :X N
TrE UA→BE(ρA ) =
Nj ρA Nj† = NA→B (ρA ).

(5.44)j

Exercise 5.2.5 Show that all isometric extensions of a quantum channel are equivalent up
to an isometry on the environment system (this is similar to the result of Theorem 5.1.1).
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191

Exercise 5.2.6 Show that an isometric extension of the erasure channel is
√
N
UA→BE
= 1 − ε(|0iB h0|A + |1iB h1|A ) ⊗ |eiE√√
+ ε|eiB h0|A ⊗ |0iE + ε|eiB h1|A ⊗ |1iE√√
= 1 − εIA→B ⊗ |eiE + εIA→E ⊗ |eiB . (5.45)
Exercise 5.2.7 Determine the resulting state when Alice inputs an arbitrary pure state |ψiinto an isometric extension of the erasure channel. Verify that Bob and Eve receive the sameensemble (they have the same local density operator) when the erasure probability ε = 1/2.
Exercise 5.2.8 Show that the matrix representation of an isometric extension of the erasurechannel is
 


N
N
0
0
h0|B h0|E UA→BE|0iA h0|B h0|E UA→BE
|1iA
N
N

 h0|B h1|E UA→BE0|0iA h0|B h1|E UA→BE|1iA   √ 0NN
 h0|B he|E UA→BE |0iA h0|B he|E UA→BE |1iA   1 − ε

0


 
N
N


 h1|B h0|E UA→BE00|0iA h1|B h0|E UA→BE |1iA  


N
N
=
 h1|B h1|E UA→BE
.
|1i
|0i
h1|
h1|
U
0
0
(5.46)
A
A
B
E
A→BE
 


√
N
 h1|B he|E U N



|1i
|0i
h1|
he|
U
1
−
ε
0
A 
A
B
E A→BE
A→BEN  √ε he|B h0|E U N

|0i
he|
h0|
U
|1i
0
A
B
E
A
A→BE
A→BE
 


√
N
N
 he|B h1|E UA→BE |0iA he|B h1|E UA→BE |1iA  
0
ε 
N
N
he|B he|E UA→BE|0iA he|B he|E UA→BE
|1iA
0
0
N
of
Exercise 5.2.9 Show that the matrix representation of an isometric extension UA→BE
the amplitude damping channel is
 


√
N
N
0
h0|B h0|E UA→BE|0iA h0|B h0|E UA→BE
|1iA
γ
N
N

 h0|B h1|E UA→BE|0iA h0|B h1|E UA→BE|1iA 0
= 1.
(5.47)
NN
 h1|B h0|E UA→BE |0iA h1|B h0|E UA→BE |1iA   0

0
√
N
N
|1iA
|0iA h1|B h1|E UA→BE
h1|B h1|E UA→BE
1−γ
0

Exercise 5.2.10 Consider a full unitary VAE→BE such that
TrE V (ρA ⊗ |0ih0|E ) V †(5.48)

gives the amplitude damping channel. Show that a matrix representation of V ish0|B h0|E V |0iA |0iE h0|B h0|E V |0iA |1iE h0|B h0|E V |1iA |0iE h0|B h0|E V |1iA |1iE
 h0|B h1|E V |0iA |0iE h0|B h1|E V |0iA |1iE h0|B h1|E V |1iA |0iE h0|B h1|E V |1iA |1iE  h1|B h0|E V |0iA |0iE h1|B h0|E V |0iA |1iE h1|B h0|E V |1iA |0iE h1|B h0|E V |1iA |1iE 
h1|B h1|E V |0iA |0iE h1|B h1|E V |0iA |1iE h1|B h1|E V |1iA |0iE h1|B h1|E V |1iA |1iE


√
√
γ
0
0 − 1−γ 1000 
 . (5.49)
=

 0
0
0
1
√
√
0
γ
1−γ 0
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CHAPTER 5. THE PURIFIED QUANTUM THEORY

Exercise 5.2.11 Consider the full unitary operator for the amplitude damping channel fromthe previous exercise. Show that the density operator(5.50)TrB V (ρA ⊗ |0ih0|E ) V †that Eve receives has the following matrix representation:



√ ∗
γp
1−p η
γη
√
if
ρA =
.
γη 1 − γpη∗p(5.51)

By comparing with (4.356), observe that the output to Eve is the bit flip of the output of anamplitude damping channel with damping parameter 1 − γ.
Complementary Channel
In the purified quantum theory, it is useful to consider all parties that are participating
in a given protocol. One such party is the environment of the channel, even if it is notnecessarily an active participant in a protocol. However, in a cryptographic setting, in somesense the environment is active, and we associate it with an eavesdropper, thus personifying
it as “Eve.”N
For any quantum channel NA→B , there exists an isometric extension UA→BEof thatcchannel. The complementary channel NA→E is a quantum channel from the sender to theenvironment, formally defined as follows:
Definition 5.2.2 (Complementary Channel) Let N : L(HA ) → L(HB ) be a quantum
channel, and let U : HA → HB ⊗ HE be an isometric extension of the channel N . The
complementary channel N c : L(HA ) → L(HE ) of N , associated with U , is defined as follows:
N c (XA ) = TrB U XA U † ,(5.52)
for XA ∈ L(HA ).
That is, we obtain a complementary channel by tracing out Bob’s system B from theoutput of an isometric extension. It captures the noise that Eve “sees” by having her system
coupled to Bob’s system.
Exercise 5.2.12 Show that Eve’s density operator (the output of a complementary channel)
is of the following form:Xρ→
Tr{Ni ρNj† }|iihj|,(5.53)
i,j

if we take an isometric extension of the channel to be of the form in (5.36).
The complementary channel is unique only up to an isometry acting on Eve’s system.
It inherits this property from the fact that an isometric extension of a quantum channel isunique only up to isometries acting on Eve’s system. For all practical purposes, this lackof uniqueness does not affect our study of the noise that Eve sees because the measures ofnoise in Chapter 11 are invariant with respect to isometries acting on Eve’s system.
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5.2. ISOMETRIC EVOLUTION

5.2.3

193

Further Examples of Isometric Extensions

Generalized Dephasing Channels
A generalized dephasing channel is one that preserves states diagonal in some preferredorthonormal basis {|xi}, but it can add arbitrary phases to the off-diagonal elements of adensity operator represented in this basis. An isometric extension of a generalized dephasing
channel acts as follows on the basis {|xi}:
ND
UA→BE|xiA = |xiB |ϕx iE ,(5.54)

where |ϕx iE is some state for the environment (these states need not be mutually orthogonal).
Thus, we can represent the isometry as follows:
X
ND
UA→BE
≡
|xiB |ϕx iE hx|A ,(5.55)xand its action on a density operator ρ is† X
U ND ρ U ND =hx|ρ|x0 i |xihx0 |B ⊗ |ϕx ihϕx0 |E .

(5.56)

x,x0

Tracing out the environment gives the action of the channel ND to the receiverX
ND (ρ) =hx|ρ|x0 ihϕx0 |ϕx i |xihx0 |B ,(5.57)

x,x0

where we observe that this channel preserves the diagonal components {|xihx|} of ρ, but
it multiplies the d (d − 1) off-diagonal elements of ρ by arbitrary phases, depending on thed (d − 1) overlaps hϕx0 |ϕx i of the environment states (where x 6= x0 ). Tracing out the receiver
gives the action of the complementary channel NDc to the environmentXhx|ρ|xi |ϕx ihϕx |E .
(5.58)NDc (ρ) =x

Observe that the channel to the environment is entanglement-breaking. That is, the actionof the channel is the same as first performing a complete projective measurement in the basis{|xi} and preparing a state |ϕx iE conditioned on the outcome of the measurement (it is aclassical–quantum channel, as discussed in Section 4.6.7). Additionally, the receiver Bob cansimulate the action of this channel to the receiver by performing the same actions on thestate that he receives.
Exercise 5.2.13 Explicitly show that the following qubit dephasing channel is a special caseof a generalized dephasing channel:
ρ → (1 − p)ρ + pZρZ.
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CHAPTER 5. THE PURIFIED QUANTUM THEORY

Quantum Hadamard Channels
Quantum Hadamard channels are those whose complements are entanglement-breaking, andso generalized dephasing channels are a subclass of quantum Hadamard channels. We canwrite the output of a quantum Hadamard channel as the Hadamard product (element-wise
multiplication) of a representation of the input density operator with another operator.
c
To discuss how this comes about, suppose that the complementary channel NA→Eof achannel NA→B is entanglement-breaking. Then, using the fact that its Kraus operators|ξi iE hζi |A are unit rank (see Theorem 4.6.1) and the construction in (5.36) for an isometriccextension, we can write an isometric extension U N for N c asXcc †
U N ρA U N=|ξi iE hζi |A ρA |ζj iA hξj |E ⊗ |iiB hj|B
(5.60)i,j=X

hζi |A ρA |ζj iA |ξi iE hξj |E ⊗ |iiB hj|B .

(5.61)

i,j

The sets {|ξi iE } and {|ζi iA } each do not necessarily consist of orthonormal states, but theset {|iiB } does because it is the environment of the complementary channel. Tracing overthe system E gives the original channel from system A to B:
X
HNA→B(ρA ) =
hζi |A ρA |ζj iA hξj |ξi iE |iiB hj|B .
(5.62)
i,jLet Σ denote the matrix with elements [Σ]i,j = hζi |A ρA |ζj iA , a representation of the input
state ρ, and let Γ denote the matrix with elements [Γ]i,j = hξi |ξj iE . Then, from (5.62), it isclear that the output of the channel is the Hadamard product ∗ of Σ and Γ† with respect tothe basis {|iiB }:HNA→B(ρ) = Σ ∗ Γ† .
(5.63)For this reason, such a channel is known as a Hadamard channel.
Hadamard channels are degradable, as introduced in the following definition:cDefinition 5.2.3 (Degradable Channel) Let NA→B be a quantum channel, and let NA→E
denote a complementary channel for NA→B . The channel NA→B is degradable if there exists
a degrading channel DB→E such thatc
DB→E (NA→B (XA )) = NA→E
(XA ),(5.64)

for all XA ∈ L(HA ).
To see that a quantum Hadamard channel is degradable, let Bob perform a complete
projective measurement of his state in the basis {|iiB } and prepare the state |ξi iE conditioned on the outcome of the measurement. This procedure simulates the complementarycchannel NA→E
and also implies that the degrading channel DB→E is entanglement-breaking.
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5.2. ISOMETRIC EVOLUTION

195

To be more precise, the Kraus operators of the degrading channel DB→E are {|ξi iE hi|B } sothat
XH(σA )) =DB→E (NA→B|ξi iE hi|B NA→B (σA )|iiB hξi |E(5.65)i=Xhζi |A σA |ζi iA |ξi i hξi |E ,(5.66)

iH
demonstrating that this degrading channel simulates the complementary channel NA→E.
Note that we can view this degrading channel as the composition of two channels: a first1
channel DB→Y
performs the complete projective measurement, leading to a classical variable
Y , and a second channel DY2 →E performs the state preparation, conditioned on the value of1
the classical variable Y . We can therefore write DB→E = DY2 →E ◦ DB→Y
. This particularform of the channel has implications for its quantum capacity (see Chapter 24) and its moregeneral capacities (see Chapter 25). Observe that a generalized dephasing channel from theprevious section is a quantum Hadamard channel because the channel to its environment isentanglement-breaking.

5.2.4

Isometric Extension and Adjoint of a Quantum Channel

Recall the notion of an adjoint of a quantum channel from Section 4.4.5. Here we show analternate way of representing an adjoint of a quantum channel using an isometric extensionof it.
Proposition 5.2.1 Let N : L(HA ) → L(HB ) be a quantum channel and let U : HA →HB ⊗ HE be an isometric extension of it. Then the adjoint map N † : L(HB ) → L(HA ) canbe written as follows:N † (YB ) = U † (YB ⊗ IE )U,(5.67)
for YB ∈ L(HB ).
Proof. We can see this by using the definition of the adjoint map (Definition 4.4.6), thedefinition of an isometric extension (Definition 5.2.1), and the definition of partial trace(Definition 4.3.4). Consider from the definition of the adjoint map that N † is such thathYB , N (XA )i = N † (YB ), XA ,(5.68)

for all XA ∈ L(HA ) and YB ∈ L(HB ). ThenhYB , N (XA )i = Tr{YB† N (XA )}(5.70)

†= Tr{ U † (YB ⊗ IE )U XA }(5.73)†= U (YB ⊗ IE ) U, XA .
c 2019(5.69)= Tr{YB† TrE {U XA U † }}= Tr{(YB† ⊗ IE )U XA U † }= Tr{U † (YB† ⊗ IE )U XA }(5.71)(5.72)(5.74)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 5. THE PURIFIED QUANTUM THEORY

The second equality is from the definition of an isometric extension. The third equality
follows by applying the definition of partial trace. The fourth uses cyclicity of trace. Since wehave shown that hYB , N (XA )i = U † (YB ⊗ IE ) U, XA for all XA ∈ L(HA ) and YB ∈ L(HB ),the statement in (5.67) follows.
We can verify the formula in (5.67) in a different way. Suppose that we have a Kraus
representation of the channel N as follows:
N (XA ) =X

Vl XA Vl† ,(5.75)lwhere Vl ∈ L(HA , HB ) for all l andis then as given in (5.36):†
l Vl Vl = IA . An isometric extension U for this channel

P

U=

X

Vl ⊗ |liE ,(5.76)lwhere {|liE } is some orthonormal basis. We can then explicitly compute the formula in(5.67) as follows:!

!
U † (YB ⊗ IE )U =X

Vl† ⊗ hl|E

Vl0 ⊗ |l0 iE(5.77)Vl† YB Vl = N † (YB ),(5.78)(YB ⊗ IE )

l0

l

=

X

X

Vl† YB Vl0 hl|l0 iE =

l,l0

X
l

where the last equality follows from what we calculated before in (4.238).

5.3 Coherent Quantum Instrument
It is useful to consider an isometric extension of a quantum instrument (we discussed quantum instruments in Section 4.6.8). This viewpoint is important when we recall that a quantum instrument is the most general map from a quantum system to a quantum system anda classical system.
Recall from Section 4.6.8 that a quantum instrument acts as follows on an input ρA ∈D(HA ):X j
ρA →EA→B (ρA ) ⊗ |jihj|J ,(5.79)jj
where each EA→B
is a completely positive trace-non-increasing (CPTNI) map that has thefollowing form:
X
j
†
EA→B
(ρA ) =
Mj,k ρA Mj,k,(5.80)ksuch thatc 2019P†
k Mj,k Mj,k ≤ I for all j.

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5.4. COHERENT MEASUREMENT197

We now describe a particular coherent evolution that implements the above transformation when we trace over certain degrees of freedom. A pure extension of each CPTNI map
Ej is as follows:
X
Ej
UA→BE
≡
Mj,k ⊗ |kiE ,(5.81)kwhere the operator Mj,k acts on the input system A and the environment system E is large
enough to accomodate all of the CPTNI maps Ej . That is, if the first map E1 has states{|1iE , . . . , |d1 iE }, then the second map E2 has states {|d1 + 1iE , . . . , |d1 + d2 iE } so that thestates on E are orthogonal for all the different maps Ej that are part of the instrument. We
can embed this pure extension into the evolution in (5.79) as follows:
ρA →

X

E

j
UA→BE
(ρA ) ⊗ |jihj|J ,(5.82)

j
E

E

E

j
j
j
where UA→BE(ρA ) = UA→BE(ρA )(UA→BE)† . This evolution is not quite fully coherent, but asimple modification of it does make it fully coherent:
X E
j
UA→BE
⊗ |jiJ ⊗ |jiEJ .
(5.83)

j

The full action of the coherent instrument is then as follows:
†
 E
X E
j
j0
⊗ |jihj 0 |J ⊗ |jihj 0 |EJ
ρA →
UA→BE
ρA UA→BE(5.84)

j,j 0

=

X

Mj,k ρA Mj†0 ,k0 ⊗ |kihk 0 |E ⊗ |jihj 0 |J ⊗ |jihj 0 |EJ .

(5.85)

j,k,j 0 ,k0

One can then check that tracing over the environmental degrees of freedom E and EJ reproduces the action of the quantum instrument in (5.79).

5.4 Coherent Measurement
We end this chapter by discussing a coherent measurement. This last section, combinedwith the notion of an isometric extension of a quantum channel, shows that it is sufficient todescribe all of the quantum theory in the so-called “traditionalist” way by using only unitaryevolutions and von Neumann (complete projective) measurements.
P
Suppose that we have a set of measurement operators {Mj }j such that j Mj† Mj = I.
In the noisy quantum theory, we found that the post-measurement state of a measurement
on a quantum system S with density operator ρ is
Mj ρMj†,pJ (j)
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CHAPTER 5. THE PURIFIED QUANTUM THEORY

where the measurement outcome j occurs with probabilitynopJ (j) = Tr Mj† Mj ρ .

(5.87)

We would like a way to perform the above measurement on system S in a coherent fashion.
The isometry in (5.36) gives a hint for how we can structure such a coherent measurement.
We can build the coherent measurement as the following isometry:
X j
US→SS 0 ≡MS ⊗ |jiS 0 .
(5.88)
jAppying this isometry to a density operator ρS gives the following state:
US→SS 0 (ρS ) = US→SS 0 ρS (US→SS 0 )†X j
0=MS ρS (MSj )† ⊗ |jihj 0 |S 0 .

(5.89)(5.90)

j,j 0

We can then apply a complete projective measurement with projection operators {|jihj|}jto the system S 0 , which gives the following post-measurement state:(IS ⊗ |jihj|S 0 )(US→SS 0 (ρS ))(IS ⊗ |jihj|S 0 )Tr {(IS ⊗ |jihj|S 0 )(US→SS 0 (ρS ))}=MSj ρS (MSj )†
⊗ |jihj|S 0 . (5.91)
Tr (MSj )† MSj ρS

The result is then the same as that in (5.86). In fact, this is the same as the way in which
Section 4.2 motivated an alternate description of quantum measurements.

Exercise 5.4.1 Suppose that there is a set of density operators ρkS and a POVM ΛkS that
identifies these states with high probability, in the sense that∀k
Tr ΛkS ρkS ≥ 1 − ε,(5.92)
where ε ∈ (0, 1). Construct a coherent measurement US→SS 0 and show that the coherent
measurement has a high probability of success in the sense that|hφk |RS hk|S 0 US→SS 0 |φk iRS | ≥ 1 − ε,(5.93)

where each |φk iRS is a purification of ρk .

5.5 History and Further Reading
The purified view of quantum mechanics has long been part of quantum information theory(e.g., see Nielsen and Chuang (2000) or Yard (2005)). Early work of Stinespring (1955)
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5.5. HISTORY AND FURTHER READING

199

showed that every linear CPTP map can be realized as a linear isometry with an output ona larger Hilbert space and followed by a partial trace. Giovannetti and Fazio (2005) discussed
some of the observations about the amplitude damping channel that appear in our exercises.
Devetak and Shor (2005) introduced generalized dephasing channels in the context of tradeoff coding and they also introduced the notion of a degradable quantum channel. King et al.
(2007) studied the quantum Hadamard channels. Coherent instruments and measurements
appeared in (Devetak and Winter, 2004; Devetak, 2005; Hsieh et al., 2008a) as part of thedecoder used in several quantum coding theorems. We exploit them in Chapters 24 and 25.

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200

c 2019CHAPTER 5. THE PURIFIED QUANTUM THEORYMark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicensePart III
Unit Quantum Protocols

201

CHAPTER 6

Three Unit Quantum Protocols
This chapter begins our first exciting application of the postulates of the quantum theory toquantum communication. We study the fundamental, unit quantum communication protocols. These protocols involve a single sender Alice and a single receiver Bob. The protocols
are ideal and noiseless because we assume that Alice and Bob can exploit perfect classicalcommunication, perfect quantum communication, and perfect entanglement. At the end of
this chapter, we suggest how to incorporate imperfections into these protocols for later study.
Alice and Bob may wish to perform one of several quantum information-processing tasks,
such as the transmission of classical information, quantum information, or entanglement.
Several fundamental protocols make use of these resources:
1. We will see that noiseless entanglement is an important resource in quantum Shannon
theory because it enables Alice and Bob to perform other protocols that are not possible
with classical resources only. We will present a simple, idealized protocol for generating
entanglement, named entanglement distribution.
2. Alice may wish to communicate classical information to Bob. A trivial method, named
elementary coding, is a simple way of doing so and we discuss it briefly.
3. A more interesting technique for transmitting classical information is super-dense coding. It exploits a noiseless qubit channel and shared entanglement to transmit moreclassical information than would be possible with a noiseless qubit channel alone.
4. Finally, Alice may wish to transmit quantum information to Bob. A trivial method
for her to do so is to exploit a noiseless qubit channel. However, it is useful to haveother ways for transmitting quantum information because such a resource is difficult
to engineer in practice. An alternative, surprising method for transmitting quantum
information is quantum teleportation. The teleportation protocol exploits classicalcommunication and shared entanglement to transmit quantum information.
Each of these protocols is a fundamental unit protocol and provides a foundation for
asking further questions in quantum Shannon theory. In fact, the discovery of these latter
203

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CHAPTER 6. THREE UNIT QUANTUM PROTOCOLS

two protocols was the stimulus for much of the original research in quantum Shannon theory.
One could take each of these protocols and ask about its performance if one or more of theresources involved is noisy rather than noiseless. Later chapters of this book explore manyof these possibilities.
This chapter introduces the technique of resource counting, which is of practical importance because it quantifies the communication cost of achieving a certain task. We include
only non-local resources in a resource count—non-local resources include classical or quantum
communication or shared entanglement.
It is important to minimize the use of certain resources, such as noiseless entanglement
or a noiseless qubit channel, in a given protocol because they are expensive. Given a certain
implementation of a quantum information-processing task, we may wonder if there is a wayof implementing it that consumes fewer resources. A proof that a given protocol is the bestthat we can hope to do is an optimality proof (also known as a converse proof, as discussed
in Section 2.1.3). We argue, based on good physical grounds, that the protocols in thischapter are the best implementations of the desired quantum information-processing task.
Chapter 25 gives information-theoretic proofs of optimality.

6.1 Non-Local Unit Resources
We first briefly define what we mean by a noiseless qubit channel, a noiseless classical bitchannel, and noiseless entanglement. Each of these resources is a non-local, unit resource.
A resource is non-local if two spatially separated parties share it or if one party uses it tocommunicate to another. We say that a resource is unit if it comes in some “gold standard”
form, such as qubits, classical bits, or entangled bits. It is important to establish thesedefinitions so that we can check whether a given protocol is truly simulating one of theseresources.
A noiseless qubit channel is any mechanism that implements the following map:|iiA → |iiB ,(6.1)

extended linearly to arbitrary state vectors and where i ∈ {0, 1}, {|0iA , |1iA } is some preferred orthonormal basis on Alice’s system, and {|0iB , |1iB } is some preferred orthonormal
basis on Bob’s system. The bases do not have to be the same, but it must be clear whichbasis each party is using. The above map is linear so that it preserves arbitrary superposition states (it preserves any qubit). For example, the map acts as follows on a superposition
state:
α|0iA + β|1iA → α|0iB + β|1iB .
(6.2)
We can also write it as the following isometry:
1
X|iiB hi|A .

(6.3)

i=0
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6.1. NON-LOCAL UNIT RESOURCES

205

Any information-processing protocol that implements the above map simulates a noiselessqubit channel. We label the communication resource of a noiseless qubit channel as follows:[q → q] ,(6.4)

where the notation indicates one forward use of a noiseless qubit channel.
A noiseless classical bit channel is any mechanism that implements the following map:
|iihi|A → |iihi|B ,|iihj|A → 0 for i 6= j,

(6.5)
(6.6)

extended linearly to density operators and where i, j ∈ {0, 1} and the orthonormal bases
are again arbitrary. This channel maintains the diagonal elements of a density operator in
the basis {|0iA , |1iA }, but it eliminates the off-diagonal elements. We can write it as thefollowing linear map acting on a density operator ρA :
ρA →

1
X

|iiB hi|A ρA |iiA hi|B .

(6.7)

i=0

The form above is consistent with Definition 4.6.5 for noiseless classical channels. This
resource is weaker than a noiseless qubit channel because it does not require Alice and Bob
to maintain arbitrary superposition states—it merely transfers classical information. Alicecan use the above channel to transmit classical information to Bob. She can prepare eitherof the classical states |0ih0| or |1ih1|, send it through the classical channel, and Bob performs
a computational basis measurement to determine the message Alice transmits. We denotethe communication resource of a noiseless classical bit channel as follows:[c → c] ,(6.8)

where the notation indicates one forward use of a noiseless classical bit channel.
We can study other ways of transmitting classical information. For example, supposethat Alice flips a fair coin that chooses the state |0iA or |1iA with equal probability. The
resulting state is the following density operator:
1(|0ih0|A + |1ih1|A )
2

(6.9)

Suppose that she sends the above state through a noiseless classical channel. The resulting
density operator for Bob is as follows:1(|0ih0|B + |1ih1|B ) .
2(6.10)

The above classical bit channel map does not preserve off-diagonal elements of a density
operator. Suppose instead that Alice prepares a superposition state|0iA + |1iA√.
2
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206

CHAPTER 6. THREE UNIT QUANTUM PROTOCOLSThe density operator corresponding to this state is
1(|0ih0|A + |0ih1|A + |1ih0|A + |1ih1|A ) .
2(6.12)

Suppose Alice then transmits this state through the above classical channel. The classicalchannel eliminates all the off-diagonal elements of the density operator and the resulting
state for Bob is as follows:1(|0ih0|B + |1ih1|B ) .
(6.13)
2
Thus, it is impossible for a noiseless classical channel to simulate a noiseless qubit channel
because it cannot maintain arbitrary superposition states. However, it is possible for anoiseless qubit channel to simulate a noiseless classical bit channel, and we denote this factwith the following resource inequality:[q → q] ≥ [c → c] .

(6.14)

Noiseless quantum communication is therefore a stronger resource than noiseless classicalcommunication.
Exercise 6.1.1 Show that the noisy dephasing channel in (4.332) with p = 1/2 is equal toa noiseless classical bit channel.
The final resource that we consider is shared entanglement. The ebit is our “gold standard” resource for pure bipartite (two-party) entanglement, and we will make this point
more clear operationally in Chapter 19. An ebit is the following state of two qubits:1
Φ+ AB ≡ √ (|00iAB + |11iAB ) ,2(6.15)

where Alice possesses the first qubit and Bob possesses the second.
Below, we show how a noiseless qubit channel can generate a noiseless ebit through a simple protocol named entanglement distribution. However, an ebit cannot simulate a noiselessqubit channel (for reasons which we explain later). Therefore, noiseless quantum communication is the strongest of all three resources, and entanglement and classical communication
are in some sense “orthogonal” to one another because neither can simulate the other.

6.2 Protocols
6.2.1

Entanglement Distribution

The entanglement distribution protocol is the most basic of the three unit protocols. It
exploits one use of a noiseless qubit channel to establish one shared noiseless ebit. It consistsof the following two steps:
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6.2. PROTOCOLS

207

Figure 6.1: This figure depicts a protocol for entanglement distribution. Alice performs local operations(the Hadamard and CNOT) and consumes one use of a noiseless qubit channel to generate one noiseless ebit
|Φ+ iAB shared with Bob.

1. Alice prepares a Bell state locally in her laboratory. She prepares two qubits in thestate |0iA |0iA0 , where we label the first qubit as A and the second qubit as A0 . She
performs a Hadamard gate on qubit A to produce the following state:|0iA + |1iA√
|0iA0 .
(6.16)
2
She then performs a CNOT gate with qubit A as the source qubit and qubit A0 as thetarget qubit. The state becomes the following Bell state:
Φ+ AA0 =|00iAA0 + |11iAA0√.
2(6.17)

2. She sends qubit A0 to Bob with one use of a noiseless qubit channel. Alice and Bob
then share the ebit |Φ+ iAB .
Figure 6.1 depicts the entanglement distribution protocol.
The following resource inequality quantifies the non-local resources consumed or generated in the above protocol:[q → q] ≥ [qq] ,(6.18)
where [q → q] denotes one forward use of a noiseless qubit channel and [qq] denotes a shared,
noiseless ebit. The meaning of the resource inequality is that there exists a protocol that
consumes the resource on the left in order to generate the resource on the right. The best
analogy is to think of a resource inequality as a “chemical reaction”-like formula, where theprotocol is like a chemical reaction that transforms one resource into another.
There are several subtleties to notice about the above protocol and its corresponding
resource inequality:c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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1. We are careful with the language when describing the resource state. We described
the state |Φ+ i as a Bell state in the first step because it is a local state in Alice’s
laboratory. We only used the term “ebit” to describe the state after the second step,when the state becomes a non-local resource shared between Alice and Bob.
2. The resource count involves non-local resources only—we do not factor any local operations, such as the Hadamard gate or the CNOT gate, into the resource count. Thisline of thinking is different from the theory of computation, where it is of utmost importance to minimize the number of steps involved in a computation. In this book, weare developing a theory of quantum communication and thus count non-local resourcesonly.
3. We are assuming that it is possible to perform all local operations perfectly. Thisline of thinking is another departure from practical concerns that one might have in
fault-tolerant quantum computation, the study of the propagation of errors in quantum
operations. Performing a CNOT gate is a highly non-trivial task at the current stage ofexperimental development in quantum computation, with most implementations being
far from perfect. Nevertheless, we proceed forward with this communication-theoretic
line of thinking.
The following exercises outline classical information-processing tasks that are analogous
to the task of entanglement distribution.
Exercise 6.2.1 Outline a protocol for shared randomness distribution. Suppose that Aliceand Bob have available one use of a noiseless classical bit channel. Give a method for themto implement the following resource inequality:[c → c] ≥ [cc] ,(6.19)

where [c → c] denotes one forward use of a noiseless classical bit channel and [cc] denotes ashared, non-local bit of shared randomness.
Exercise 6.2.2 Consider three parties Alice, Bob, and Eve and suppose that a noiselessprivate channel connects Alice to Bob. Privacy here implies that Eve does not learn anything
about the information that traverses the private channel—Eve’s probability distribution is
independent of Alice and Bob’s:
pA,B,E (a, b, e) = pA (a)pB|A (b|a)pE (e).

(6.20)For a noiseless private bit channel, pB|A (b|a) = δb,a . A noiseless secret key corresponds tothe following distribution:
1(6.21)
pA,B,E (a, b, e) = δb,a pE (e),2where 21 implies that the key is equal to “0” or “1” with equal probability, δb,a implies aperfectly correlated secret key, and the factoring of the distribution pA,B,E (a, b, e) implies thec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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secrecy of the key (Eve’s information is independent of Alice and Bob’s). The difference
between a noiseless private bit channel and a noiseless secret key is that the private channel
is a dynamic resource while the secret key is a shared, static resource. Show that it ispossible to upgrade the protocol for shared randomness distribution to a protocol for secretkey distribution, if Alice and Bob share a noiseless private bit channel. That is, show thatthey can achieve the following resource inequality:[c → c]priv ≥ [cc]priv ,(6.22)

where [c → c]priv denotes one forward use of a noiseless private bit channel and [cc]priv denotes
one bit of shared, noiseless secret key.
Entanglement and Quantum Communication
Can entanglement enable two parties to communicate quantum information? It is natural towonder if there is a protocol corresponding to the following resource inequality:?

[qq] ≥ [q → q] .

(6.23)

Unfortunately, it is physically impossible to construct a protocol that implements the above
resource inequality. The argument against such a protocol arises from the theory of relativity.
Specifically, the theory of relativity prohibits information transfer or signaling at a speed
greater than the speed of light. Suppose that two parties share noiseless entanglement
over a large distance. That resource is a static resource, possessing only shared quantum
correlations. If a protocol were to exist that implements the above resource inequality, it
would imply that two parties could communicate quantum information faster than the speedof light, because they would be exploiting the entanglement for the instantaneous transfer
of quantum information.
The entanglement distribution resource inequality is only “one-way,” as in (6.18). Quantum communication is therefore strictly stronger than shared entanglement when no othernon-local resources are available.

6.2.2

Elementary Coding

We can also send classical information using a noiseless qubit channel. A simple protocol fordoing so is elementary coding. This protocol consists of the following steps:
1. Alice prepares either |0i or |1i, depending on the classical bit that she would like tosend.
2. She transmits this state over the noiseless qubit channel, and Bob receives the qubit.
3. Bob performs a measurement in the computational basis to determine the classical bitthat Alice transmitted.
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Elementary coding succeeds without error because Bob’s measurement can always distinguish the classical states |0i and |1i. The following resource inequality applies to elementarycoding:[q → q] ≥ [c → c] .
(6.24)
Again, we are only counting non-local resources in the resource count—we do not count thestate preparation at the beginning or the measurement at the end.
If no other resources are available for consumption, the above resource inequality is
optimal—one cannot do better than to transmit one classical bit of information per use ofa noiseless qubit channel. This result may be a bit frustrating at first, because it may seemthat we could exploit the continuous degrees of freedom in the probability amplitudes ofa qubit state for encoding more than one classical bit per qubit. Unfortunately, there is
no way that we can access the information in the continuous degrees of freedom using any
measurement scheme. The result of Exercise 4.2.2 demonstrates the optimality of the above
protocol, and it holds as well by invoking the Holevo bound from Chapter 11.

6.2.3

Quantum Super-Dense Coding

We now outline a protocol named super-dense coding. It is named as such because it has thestriking property that noiseless entanglement can double the classical communication abilityof a noiseless qubit channel. It consists of three steps:
1. Suppose that Alice and Bob share an ebit |Φ+ iAB . Alice applies one of four unitary
operations {I, X, Z, XZ} to her share of the above state. The state becomes one of
the following four Bell states (up to a global phase), depending on the message that
Alice chooses:
Φ+ AB ,

Φ− AB ,

Ψ+ AB ,

Ψ− AB .

(6.25)The definitions of these Bell states are in (3.194)–(3.196).
2. She transmits her qubit to Bob with one use of a noiseless qubit channel.
3. Bob performs a Bell measurement (a measurement in the basis {|Φ+ iAB , |Φ− iAB ,|Ψ+ iAB , |Ψ− iAB }) to distinguish the four states perfectly—he can distinguish the states
because they are all orthogonal to each other.
Thus, Alice can transmit two classical bits (corresponding to the four messages) if she
shares a noiseless ebit with Bob and uses a noiseless qubit channel. Figure 6.2 depicts the
protocol for quantum super-dense coding.
The super-dense coding protocol realizes the following resource inequality:[qq] + [q → q] ≥ 2 [c → c] .

(6.26)

Notice again that the resource inequality counts the use of non-local resources only—we do
not count the local operations at the beginning of the protocol or the Bell measurement at
the end of the protocol.
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Figure 6.2: This figure depicts the dense coding protocol. Alice and Bob share an ebit before the protocol
begins. Alice would like to transmit two classical bits x1 x2 to Bob. She performs a Pauli rotation conditionedon her two classical bits and sends her share of the ebit over a noiseless qubit channel. Bob can then recover
the two classical bits by performing a Bell measurement.

Also, notice that we could have implemented two noiseless classical bit channels with two
instances of elementary coding:
2 [q → q] ≥ 2 [c → c] .

(6.27)

However, this method is not as powerful as the super-dense coding protocol—in super-dense
coding, we consume the weaker resource of an ebit to help transmit two classical bits, insteadof consuming the stronger resource of an extra noiseless qubit channel.
The super-dense coding protocol also transmits the classical bits privately. Supposea third party intercepts the qubit that Alice transmits. There is no measurement thatthe third party can perform to determine which message Alice transmits because the local
density operator of all of the Bell states is the same and equal to the maximally mixed
state πA (the information for the eavesdropper is the same irrespective of each message that
Alice transmits). The privacy of the protocol is due to Alice and Bob sharing maximal
entanglement. We exploit this aspect of the super-dense coding protocol when we “make it
coherent” in Chapter 7.

6.2.4

Quantum Teleportation

Perhaps the most striking protocol in noiseless quantum communication is the quantum
teleportation protocol. The protocol destroys the quantum state of a qubit in one location
and recreates it on a qubit at a distant location, with the help of shared entanglement. Thus,
the name “teleportation” corresponds well to the mechanism that occurs.
The teleportation protocol is actually a flipped version of the super-dense coding protocol, in the sense that Alice and Bob merely “swap their equipment.” The first step inunderstanding teleportation is to perform a few algebraic steps using the tricks of the tensor
product and the Bell state substitutions from Exercise 3.6.6. Consider a qubit |ψiA0 thatc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 6. THREE UNIT QUANTUM PROTOCOLS

Alice possesses, where|ψiA0 ≡ α|0iA0 + β|1iA0 .

(6.28)

Suppose she shares an ebit |Φ+ iAB with Bob. The joint state of the systems A0 , A, and B
is as follows:|ψiA0 Φ+ AB .
(6.29)Let us first explicitly write out this state:+|ψiA0 ΦAB

= (α|0iA0 + β|1iA0 )|00iAB + |11iAB
√
2


.

(6.30)

Distributing terms gives the following equality:1= √ [α|000iA0 AB + β|100iA0 AB + α|011iA0 AB + β|111iA0 AB ] .
2(6.31)We use the relations in Exercise 3.6.6 to rewrite the joint system A0 A in the Bell basis:
1α (|Φ+ iA0 A + |Φ− iA0 A ) |0iB + β (|Ψ+ iA0 A − |Ψ− iA0 A ) |0iB=.
(6.32)
2 +α (|Ψ+ iA0 A + |Ψ− iA0 A ) |1iB + β (|Φ+ iA0 A − |Φ− iA0 A ) |1iB
Simplifying gives the following equality:
1|Φ+ iA0 A (α|0iB + β|1iB ) + |Φ− iA0 A (α|0iB − β|1iB )=.
2 + |Ψ+ iA0 A (α|1iB + β|0iB ) + |Ψ− iA0 A (α|1iB − β|0iB )(6.33)

We can finally rewrite the state as four superposed terms, with a distinct Pauli operator
applied to Bob’s system B for each term in the superposition:
=


1 +
Φ A0 A |ψiB + Φ− A0 A Z|ψiB + Ψ+ A0 A X|ψiB + Ψ− A0 A XZ|ψiB .
2(6.34)

We now outline the three steps of the teleportation protocol (depicted in Figure 6.3):
1. Alice performs a Bell measurement on her systems A0 A. The state collapses to one of
the following four states with uniform probability:
Φ+ A0 A |ψiB ,(6.35)

Φ− A0 A Z|ψiB ,(6.36)+ΨA0 A

X|ψiB ,

(6.37)A0 A

XZ|ψiB .

(6.38)−

Ψ

Notice that the state resulting from the measurement is a product state with respect
to the cut A0 A | B, regardless of the outcome of the measurement. At this point, Aliceknows whether Bob’s state is |ψiB , Z|ψiB , X|ψiB , or XZ|ψiB because she knows theresult of the measurement. On the other hand, Bob does not know anything aboutc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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Figure 6.3: This figure depicts the teleportation protocol. Alice would like to transmit an arbitrary
quantum state |ψiA0 to Bob. Alice and Bob share an ebit before the protocol begins. Alice can “teleport”
her quantum state to Bob by consuming the entanglement and two uses of a noiseless classical bit channel.

the state of his system B—Exercise 4.7.6 states that his local density operator is the
maximally mixed state πB just after Alice performs the measurement. Thus, there
is no teleportation of quantum information at this point because Bob’s local state iscompletely independent of the original state |ψi. In other words, teleportation cannot
be instantaneous.
2. Alice transmits two classical bits to Bob that indicate which of the four measurement
outcomes occurred. After Bob receives the classical information, he is immediately
certain which operation he needs to perform in order to restore his state to Alice’s
original state |ψi. Notice that he does not need to have knowledge of the state in orderto restore it—he only needs knowledge of the restoration operation.
3. Bob performs the restoration operation: one of the identity, a Pauli X operator, aPauli Z operator, or the Pauli operator ZX, depending on the classical informationthat he receives from Alice.
Teleportation is an oblivious protocol because Alice and Bob do not require any knowledgeof the quantum state being teleported in order to perform it. We might also say that thisfeature of teleportation makes it universal—it works independently of the input state.
You might think that the teleportation protocol violates the no-cloning theorem because
a “copy” of the state appears on Bob’s system. But this violation does not occur at any
point in the protocol because the Bell measurement destroys the information about the stateof Alice’s original information qubit while recreating it somewhere else. Also, notice thatthe result of the Bell measurement is independent of the particular probability amplitudes
α and β corresponding to the state Alice wishes to teleport.
The teleportation protocol is not an instantaneous teleportation, as portrayed in the
television episodes of Star Trek. There is no transfer of quantum information instantaneously
after the Bell measurement because Bob’s local description of the B system is the maximally
mixed state π. It is only after he receives the classical bits to “telecorrect” his state that thec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 6. THREE UNIT QUANTUM PROTOCOLS

transfer occurs. It must be this way—otherwise, they would be able to communicate fasterthan the speed of light, and superluminal communication is not allowed by the theory of
relativity.
Finally, we can phrase the teleportation protocol as a resource inequality:[qq] + 2 [c → c] ≥ [q → q] .

(6.39)

Again, we include only non-local resources in the resource count. The above resource inequality is perhaps the most surprising of the three unit protocols we have studied so far. It
combines two resources, noiseless entanglement and noiseless classical communication, that
achieve noiseless quantum communication even though they are both individually weaker
than it. This protocol and super-dense coding are two of the most fundamental protocols inquantum communication theory because they sparked the notion that there are clever waysof combining resources to generate other resources.
In Exercise 6.2.3 below, we discuss a variation of teleportation called remote state preparation, where Alice possesses a classical description of the state that she wishes to teleport.
With this knowledge, it is possible to reduce the amount of classical communication necessary
for teleportation.
Exercise 6.2.3 Remote state preparation is a variation of the teleportation protocol. We
consider a simple example of a remote state preparation √ protocol. Suppose Alice possesses aiφclassical description of a state |ψi ≡ |0i + e |1i / 2 (on the equator of the Bloch sphere)
and she shares an ebit |Φ+ iAB with Bob. Alice would like to prepare the state |ψi on Bob’s
system. Show that Alice can prepare this state on Bob’s system if she measures her system A
in the |ψ ∗ i , ψ ⊥∗ basis, transmits one classical bit, and Bob performs a recovery operation
conditioned on the classical information. (Note that |ψ ∗ i is the conjugate of the vector |ψi).
Exercise 6.2.4 Third-party controlled teleportation is another variation on the teleportation
protocol. Suppose that Alice, Bob, and Charlie possess a GHZ state:|ΦGHZ i ≡|000iABC + |111iABC√.
2(6.40)

Alice would like to teleport an arbitrary qubit to Bob. She performs the usual steps in theteleportation protocol. Give the final steps that Charlie should perform and the informationthat he should transmit to Bob in order to complete the teleportation protocol. (Hint: The
resource inequality for the protocol is as follows:[qqq]ABC + 2 [c → c]A→B + [c → c]C→B ≥ [q → q]A→B ,(6.41)

where [qqq]ABC represents the resource of the GHZ state shared between Alice, Bob, and Charlie, and the other resources are as before with the directionality of communication indicatedby the corresponding subscript.)c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

6.3. OPTIMALITY OF THE THREE UNIT PROTOCOLS

215

Exercise 6.2.5 Gate teleportation is yet another variation of quantum teleportation that isuseful in fault-tolerant quantum computation. Suppose that Alice would like to perform asingle-qubit gate U on a qubit in state |ψi. Suppose that the gate U is difficult to perform,
but that U σi U † , where σi is one of the single-qubit Pauli operators, is much less difficult
to perform. A protocol for gate teleportation is as follows. Alice and Bob first prepare theebit UB |Φ+ iAB . Alice performs a Bell measurement on her qubit |ψiA0 and system A. She
transmits two classical bits to Bob and Bob performs one of the four corrective operations
U σi U † on his qubit. Show that this protocol works, i.e., that Bob’s final state is U |ψi.
Exercise 6.2.6 Show that it is possible to simulate a dephasing qubit channel by the following technique. First, Alice prepares a maximally entangled Bell state |Φ+ i. She sends one
share of it to Bob through a dephasing qubit channel. She and Bob perform the usual teleportation protocol. Show that this procedure gives the same result as sending a qubit througha dephasing channel. (Hint: This result holds because the dephasing channel commutes with
all Pauli operators.)
Exercise 6.2.7 Construct an entanglement swapping protocol from the teleportation protocol. That is, suppose that Charlie and Alice possess a bipartite state |ψiCA . Show that if
Alice teleports her share of the state |ψiCA to Bob, then Charlie and Bob share the state|ψiCB . A special case of this protocol is when the state |ψiCA is an ebit. Then the protocol
is equivalent to an entanglement swapping protocol.

6.3 Optimality of the Three Unit Protocols
We now consider several arguments that may seem somewhat trivial at first, but they are
crucial for having a good theory of quantum communication. We are always thinking about
the optimality of certain protocols—if there is a better, cheaper way to perform a given
protocol, then this would be advantageous. There are several questions that we can ask
about the above protocols:
1. In entanglement distribution, is one ebit per qubit the best that we can do, or is it
possible to generate more than one ebit with a single use of a noiseless qubit channel?
2. In super-dense coding, is it possible to generate two noiseless classical bit channels
with less than one noiseless qubit channel or less than one noiseless ebit? Is it possible
to generate more than two classical bit channels using the given resources?
3. In teleportation, is it possible to teleport more than one qubit using the given resources?
Is it possible to teleport using less than two classical bits or less than one ebit?
In this section, we answer all of these questions in the negative—all the protocols as given
are optimal protocols. Here, we begin to see the beauty of the resource inequality formalism.
It allows us to chain protocols together to make new protocols. We exploit this idea in theforthcoming optimality arguments.
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216

CHAPTER 6. THREE UNIT QUANTUM PROTOCOLSFirst, let us tackle the optimality of entanglement distribution. Is there a protocol that
implements any other resource inequality such as[q → q] ≥ E [qq] ,(6.42)

where the rate E of entanglement generation is greater than one?
We show that such a resource inequality can never occur, i.e., it is optimal for E =1. Suppose such a resource inequality with E > 1 does exist. Under an assumption offree forward classical communication, we can combine the above resource inequality with
teleportation to achieve the following resource inequality:[q → q] ≥ E [q → q] .

(6.43)

We could then simply keep repeating this protocol to achieve an unbounded amount of
quantum communication, which is impossible. Thus, it must be that E = 1.
Next, we consider the optimality of super-dense coding. We again exploit a proof bycontradiction argument. Let us suppose that we have an unlimited amount of entanglement
available. Suppose that there exists some “super-duper”-dense coding protocol that generates an amount of classical communication greater than that which super-dense coding
generates. That is, the classical communication output of super-duper-dense coding is 2C
where C > 1, and its resource inequality is[q → q] + [qq] ≥ 2C [c → c] .

(6.44)

Then this super-duper-dense coding scheme (along with the infinite entanglement) gives thefollowing resource inequality:[q → q] + ∞ [qq] ≥ 2C [c → c] + ∞ [qq] .

(6.45)

An infinite amount of entanglement is still available after executing the super-duper-dense
coding protocol because it consumes only a finite amount of entanglement. We can thenchain the above protocol with teleportation and achieve the following resource inequality:
2C [c → c] + ∞ [qq] ≥ C [q → q] + ∞ [qq] .

(6.46)Overall, we have then shown a scheme that achieves the following resource inequality:[q → q] + ∞ [qq] ≥ C [q → q] + ∞ [qq] .

(6.47)

We can continue with this protocol and perform it k times so that we implement the following
resource inequality:[q → q] + ∞ [qq] ≥ C k [q → q] + ∞ [qq] .
(6.48)The result of this construction is that one noiseless qubit channel and an infinite amountof entanglement can generate an infinite amount of quantum communication. This result
is impossible physically because entanglement does not boost the capacity of a noiselessc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

6.4. EXTENSIONS FOR QUANTUM SHANNON THEORY

217

qubit channel. Also, the scheme is exploiting just one noiseless qubit channel along withthe entanglement to generate an unbounded amount of quantum communication—it mustbe signaling superluminally in order to do so. Thus, the rate of classical communication insuper-dense coding is optimal.
We leave the optimality arguments for teleportation as an exercise because they are similar to those for the super-dense coding protocol. Note that it is possible to prove optimalityof these protocols without assumptions such as free classical communication (for the case ofentanglement distribution), and we do so in Chapter 8.
Exercise 6.3.1 Show that it is impossible for C > 1 in the teleportation protocol where C
is with respect to the following resource inequality:
2 [c → c] + [qq] ≥ C [q → q] .

(6.49)

Exercise 6.3.2 Show that the rates of the consumed resources in the teleportation and superdense coding protocols are optimal.

6.4 Extensions for Quantum Shannon Theory
The previous section sparked some good questions that we might ask as a quantum Shannon
theorist. We might also wonder what types of communication rates are possible if some of
the consumed resources are noisy, rather than being perfect resources. We list some of thesequestions below.
Let us first consider entanglement distribution. Suppose that the consumed noiselessqubit channel in entanglement distribution is instead a noisy quantum channel N . The
communication task is then known as entanglement generation. We can rephrase the communication task as the following resource inequality:
hN i ≥ E [qq] .

(6.50)The meaning of the resource inequality is that we consume the resource of a noisy quantum
channel N in order to generate entanglement between a sender and receiver at some rate E.
We will make the definition of a quantum Shannon-theoretic resource inequality more precise
when we begin our formal study of quantum Shannon theory, but the above definition should
be sufficient for now. The optimal rate of entanglement generation with the noisy quantum
channel N is known as the entanglement generation capacity of N . This task is intimately
related to the quantum communication capacity of N , and we discuss the connection further
in Chapter 24.
Let us now turn to super-dense coding. Suppose that the consumed noiseless qubit
channel in super-dense coding is instead a noisy quantum channel N . The name for thistask is then entanglement-assisted classical communication. The following resource inequality
captures the corresponding communication task:
hN i + E [qq] ≥ C [c → c] .
c 2019(6.51)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

218

CHAPTER 6. THREE UNIT QUANTUM PROTOCOLSThe meaning of the resource inequality is that we consume a noisy quantum channel N
and noiseless entanglement at some rate E to produce noiseless classical communication atsome rate C. We will study this protocol in depth in Chapter 21. We can also consider the
scenario in which the entanglement is no longer noiseless, but it is rather a general bipartite
state ρAB that Alice and Bob share. The task is then known as noisy super-dense coding.1
We study noisy super-dense coding in Chapter 22. The corresponding resource inequality is
as follows (its meaning should be clear at this point):hρAB i + Q [q → q] ≥ C [c → c] .

(6.52)

We can ask the same questions for the teleportation protocol as well. Suppose thatthe entanglement resource is instead a noisy bipartite state ρAB . The task is then noisy
teleportation and has the following resource inequality:hρAB i + C [c → c] ≥ Q [q → q] .

(6.53)The questions presented in this section are some of the fundamental questions in quantum
Shannon theory. We arrived at these questions simply by replacing the noiseless resourcesin the three fundamental noiseless protocols with noisy ones. We will spend a significant
amount of effort building up our knowledge of quantum Shannon-theoretic tools that will be
indispensable for answering these questions.

6.5 Three Unit Qudit Protocols
We end this chapter by studying the qudit versions of the three unit protocols. It is useful
to have these versions of the protocols because we may want to process qudit systems withthem.
The qudit resources are straightforward extensions of the qubit resources. A noiseless
qudit channel is the following map:|iiA → |iiB ,(6.54)
where {|iiA }i∈{0,...,d−1} is some preferred orthonormal basis on Alice’s system and {|iiB }i∈{0,...,d−1}
is some preferred basis on Bob’s system. We can also write the qudit channel map as thefollowing isometry:d−1
X
IA→B ≡|iiB hi|A .
(6.55)
i=0

The map IA→B preserves superposition states so that
d−1
X
i=0

αi |iiA →d−1
Xαi |iiB .

(6.56)i=0

1

The name noisy super-dense coding could just as well apply to the former task of entanglement-assistedclassical communication, but this terminology has “stuck” in the research literature for this specific quantum
information-processing task.
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219

A noiseless classical dit channel or cdit is the following map:
|iihi|A → |iihi|B ,|iihj|A → 0 for i 6= j.

(6.57)(6.58)A noiseless maximally entangled qudit state or an edit is as follows:d−1

1 X|ΦiAB ≡ √|iiA |iiB .
d i=0(6.59)We quantify the “dit” resources with bit measures. For example, a noiseless qudit channel
is the following resource:log d [q → q] ,(6.60)
where the logarithm is base two. Thus, one qudit channel can transmit log d qubits ofquantum information so that the qubit remains our standard unit of quantum information.
We quantify the amount of information transmitted according to the dimension of the spacethat is transmitted. For example, suppose that a quantum system has eight levels. We canthen encode three qubits of quantum information in this eight-level system.
Likewise, a classical dit channel is the following resource:log d [c → c] ,(6.61)

so that a classical dit channel transmits log d classical bits. The parameter d here is thenumber of classical messages that the channel transmits.
Finally, an edit is the following resource:log d [qq] .

(6.62)We quantify the amount of entanglement in a maximally entangled state by its Schmidt rank(see Theorem 3.8.1). We measure entanglement in units of ebits (we return to this issue in
Chapter 19).

6.5.1

Entanglement Distribution

The extension of the entanglement distribution protocol to the qudit case is straightforward.
Alice merely prepares the state |ΦiAA0 in her laboratory and transmits the system A0 througha noiseless qudit channel. She can prepare the state |ΦiAA0 with two gates: the qudit analog
of the Hadamard gate and the CNOT gate. The qudit analog of the Hadamard gate is theFourier gate F introduced in Exercise 3.7.8 whered−1

1 X
F : |li → √
exp
d j=0
c 2019



2πilj
d


|ji,

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CHAPTER 6. THREE UNIT QUANTUM PROTOCOLS

so that
d−1

1 X
F ≡√
exp
d l,j=0



2πilj
d


|jihl|.

(6.64)The qudit analog of the CNOT gate is the following controlled-shift gate:
CNOTd ≡

d−1
X

|jihj| ⊗ X(j),(6.65)

j=0

where X(j) is the shift operator defined in (3.208).
Exercise 6.5.1 Verify that Alice can prepare the maximally entangled qudit state |ΦiAA0
locally by preparing |0iA |0iA0 , applying FA and CNOTd . Show that
|ΦiAA0 = CNOTd ·FA |0iA |0iA0 .

(6.66)The resource inequality for this qudit entanglement distribution protocol is as follows:
log d [q → q] ≥ log d [qq] .

6.5.2(6.67)

Quantum Super-Dense Coding

The qudit version of the super-dense coding protocol proceeds analogously to the qubit case,with some notable exceptions. It still consists of three steps:
1. Alice and Bob begin with a maximally entangled state of the form in (6.59). Aliced−1
to her qudit. The
applies one of d2 unitary operations in the set {X(x)Z(z)}x,z=02
shared state then becomes one of the d maximally entangled qudit states in (3.234).
2. She sends her qudit to Bob with one use of a noiseless qudit channel.
3. Bob performs a measurement in the qudit Bell basis to determine the message Alicesent. The result of Exercise 3.7.11 is that these states are perfectly distinguishable
with a measurement.
This qudit super-dense coding protocol realizes the following resource inequality:
log d [qq] + log d [q → q] ≥ 2 log d [c → c] .

6.5.3

(6.68)

Quantum Teleportation

The operations in the qudit teleportation protocol are again similar to the qubit case. The
protocol proceeds in three steps:
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6.5. THREE UNIT QUDIT PROTOCOLS

221

1. Alice possesses an arbitrary qudit |ψiA0 where|ψiA0 ≡d−1X

αi |iiA0 .

(6.69)

i=0

Alice and Bob share a maximally entangled qudit state |ΦiAB of the form in (6.59).
The joint state of Alice and Bob is then |ψiA0 |ΦiAB . Alice performs a measurementin the basis {|Φi,j iA0 A }i,j .
2. She transmits the measurement result (i, j) to Bob with the use of two classical dit
channels.
3. Bob then applies the unitary transformation ZB (j)XB (i) to his state to “telecorrect”
it to Alice’s original qudit.
We prove that this protocol works by analyzing the probability of the measurement result
and the post-measurement state on Bob’s system. The techniques that we employ here aredifferent from those for the qubit case.
First, let us suppose that Alice would like to teleport the A0 system of a state |ψiRA0 thatshe shares with an inaccessible reference system R. This way, our teleportation protocol
encompasses the most general setting in which Alice would like to teleport a mixed state
on A0 . Also, Alice shares the maximally entangled edit state |ΦiAB with Bob. Alice first
performs a measurement of the systems A0 and A in the basis {|Φi,j iA0 A }i,j where|Φi,j iA0 A = UAij0 |ΦiA0 A ,(6.70)

UAij0 ≡ ZA0 (j)XA0 (i).

(6.71)

andThe measurement operators are thus
|Φi,j ihΦi,j |A0 A .

(6.72)

Then the unnormalized post-measurement state is|Φi,j ihΦi,j |A0 A |ψiRA0 |ΦiAB ,(6.73)

where here and in what follows, we have taken the common practice of omitting tensor
products with identity operators, instead leaving them implicit in order to reduce clutter in
the notation. We can rewrite this state as follows, by exploiting the definition of |Φi,j iA0 A in
(6.70):
†|Φi,j ihΦ|A0 A UAij0 |ψiRA0 |ΦiAB .
(6.74)
Recall the “transpose trick” from Exercise 3.7.12 that holds for any maximally entangled†state |Φi. We can exploit this result to show that the action of the unitary (U ij ) on the A0∗
system is the same as the action of the unitary (U ij ) on the A system:∗|Φi,j ihΦ|A0 A UAij |ψiRA0 |ΦiAB .
(6.75)
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CHAPTER 6. THREE UNIT QUANTUM PROTOCOLS

Then the unitary UAij∗

commutes with the systems R and A0 :|Φi,j ihΦ|A0 A |ψiRA0

UAij

∗

|ΦiAB .

(6.76)

We can again apply the transpose trick from Exercise 3.7.12 to show that the state is equalto†|Φi,j ihΦ|A0 A |ψiRA0 UBij |ΦiAB .
(6.77)
†
Then we can commute the unitary UBij all the way to the left, and we can switch the orderof |ψiRA0 and |ΦiAB without any problem because the system labels are sufficient to track
the states in these systems:
UBij

†

|Φi,j ihΦ|A0 A |ΦiAB |ψiRA0 .

(6.78)

Now let us consider the very special overlap hΦ|A0 A |ΦiAB of the maximally entanglededit state with itself on different systems:d−1hΦ|A0 A |ΦiAB =1 X√
hi|A0 hi|A
d i=0

!

d−1

1 X
√
|jiA |jiB
d j=0

!

d−1
d−1
1X
1X
=
hi|A0 hi|A |jiA |jiB =hi|A0 hi|jiA |jiB
d i,j=0
d i,j=0

=

d−1
d−1
1X
1X
hi|A0 |iiB =|iiB hi|A0
d i=0
d i=0

1
= IA0 →B .
d

(6.79)(6.80)(6.81)(6.82)

The first equality follows by definition. The second equality follows from linearity andrearranging terms in the multiplication and summation. The third and fourth equalities
follow by realizing that hi|A |jiA is an inner product and evaluating it for the orthonormal
basis {|iiA }. The fifth equality follows by rearranging
the bra and the ket. The final equalityP
is our last important realization: the operator d−1|ii
B hi|A0 is the noiseless qudit channel
i=0
IA0 →B that the teleportation protocol creates from the system A0 to B (see the definition ofa noiseless qudit channel in (6.55)). We might refer to this as the “teleportation map.”
We now apply the teleportation map to the state in (6.78):
UBij

c 2019†

1|Φi,j iA0 A IA0 →B |ψiRA0
d

1
ij †
=
U
|Φi,j iA0 A |ψiRBd B
†1= |Φi,j iA0 A UBij |ψiRB .
d|Φi,j ihΦ|A0 A |ΦiAB |ψiRA0 = UBij

†

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6.6. HISTORY AND FURTHER READING223

We can compute the probability of receiving outcome i and j from the measurement when
the input state is |ψiRA0 . It is just equal to the overlap of the above vector with itself:




1
1
ij
ij †
p (i, j|ψ) =hΦi,j |A0 A hψ|RB UB|Φi,j iA0 A UB |ψiRB(6.86)dd†1= 2 hΦi,j |A0 A |Φi,j iA0 A hψ|RB UBij UBij |ψiRB(6.87)d1
1= 2 hΦi,j |A0 A |Φi,j iA0 A hψ|RB |ψiRB = 2 .
(6.88)ddThus, the probability of the outcome (i, j) is completely random and independent of the
input state. We would expect this to be the case for a universal teleportation protocol that
operates independently of the input state. Thus, after normalization, the state on Alice andBob’s system is†|Φi,j iA0 A UBij |ψiRB .
(6.89)At this point, Bob does not know the result of the measurement. We obtain his density
operator by tracing over the systems A0 , A, and R to which he does not have access andtaking the expectation over all the measurement outcomes:
)
( d−1

1 X
†
|Φi,j ihΦi,j |A0 A UBij |ψihψ|RB UBij
TrA0 ARd2 i,j=0
d−1
†
1 X
= 2
UBij ψB UBij = πB . (6.90)
d i,j=0

The first equality follows by evaluating the partial trace and by defining ψB ≡ TrR {|ψihψ|RB } .
The second equality follows because applying a Heisenberg–Weyl operator uniformly at random completely randomizes a quantum state to be the maximally mixed state (see Exercise 4.7.6).
Now suppose that Alice sends the measurement results i and j over two uses of a noiselessclassical dit channel. Bob then knows that the state is†
UBij |ψiRB ,(6.91)and he can apply UBij to make the overall state become |ψiRB . This final step completes
the teleportation process. The resource inequality for the qudit teleportation protocol is asfollows:log d [qq] + 2 log d [c → c] ≥ log d [q → q] .
(6.92)

6.6 History and Further ReadingThis chapter presented the three important protocols that exploit the three unit resources ofclassical communication, quantum communication, and entanglement. We learned, perhaps
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CHAPTER 6. THREE UNIT QUANTUM PROTOCOLS

surprisingly, that it is possible to combine two resources together in interesting ways to simulate a different resource (in both super-dense coding and teleportation). These combinationsof resources turn up quite a bit in quantum Shannon theory, and we see them in their mostbasic form in this chapter.
Bennett and Wiesner (1992) published the super-dense coding protocol, and within ayear, Bennett et al. (1993) realized that Alice and Bob could teleport particles if they swap
their operations with respect to the super-dense coding protocol. These two protocols were
the seeds of much later work in quantum Shannon theory.

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 7

Coherent Protocols
We introduced three protocols in the previous chapter: entanglement distribution, teleportation, and super-dense coding. The last two of these protocols, teleportation and superdense coding, are perhaps more interesting than entanglement distribution because they
demonstrate insightful ways for combining all three unit resources to achieve an informationprocessing task.
It appears that teleportation and super-dense coding might be “inverse” protocols withrespect to each other because teleportation arises from super-dense coding when Alice andBob “swap their equipment.” But there is a fundamental asymmetry between these protocols
when we consider their respective resource inequalities. Recall that the resource inequality
for teleportation is
2 [c → c] + [qq] ≥ [q → q] ,(7.1)
while that for super-dense coding is[q → q] + [qq] ≥ 2 [c → c] .

(7.2)

The asymmetry in these protocols is that they are not dual under resource reversal. Two
protocols are dual under resource reversal if the resources that one consumes are the samethat the other generates and vice versa. Consider that the super-dense coding resource
inequality in (7.2) generates two classical bit channels. Glancing at the left-hand side of theteleportation resource inequality in (7.1), we see that two classical bit channels generated
from super-dense coding are not sufficient to generate the noiseless qubit channel on theright-hand side of (7.1)—the protocol requires the consumption of noiseless entanglement inaddition to the consumption of the two noiseless classical bit channels.
Is there a way for teleportation and super-dense coding to become dual under resource
reversal? One way is if we assume that entanglement is a free resource. This assumption isstrong and we may have difficulty justifying it from a practical standpoint because noiselessentanglement is extremely fragile. It is also a powerful resource, as the teleportation andsuper-dense coding protocols demonstrate. But in the theory of quantum communication,
we often make assumptions such as this one—such assumptions tend to give a dramatic simplification of a problem. Continuing with our development, let us assume that entanglement
225

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CHAPTER 7. COHERENT PROTOCOLS

is a free resource and that we do not have to factor it into the resource count. Under thisassumption, the resource inequality for teleportation becomes
2 [c → c] ≥ [q → q] ,(7.3)

and that for super-dense coding becomes[q → q] ≥ 2 [c → c] .

(7.4)

Teleportation and super-dense coding are then dual under resource reversal under the “freeentanglement” assumption, and we obtain the following resource equality:[q → q] = 2 [c → c] .

(7.5)

Exercise 7.0.1 Suppose that the quantum capacity of a quantum channel assisted by anunlimited amount of entanglement is equal to some number Q. What is the capacity of thatentanglement-assisted channel for transmitting classical information?
Exercise 7.0.2 How can we obtain the following resource equality? (Hint: Assume thatsome resource is free.)[q → q] = [qq] .

(7.6)

Which noiseless protocols did you use to show the above resource equality? The above resource
equality is a powerful statement: entanglement and quantum communication are equivalent
under the assumption that you have found.
Exercise 7.0.3 Suppose that the entanglement generation capacity of a quantum channel isequal to some number E. What is the quantum capacity of that channel when assisted by
free, forward classical communication?
The above assumptions are useful for finding simple ways to make protocols dual underresource reversal, and we will exploit them later in our proofs of various capacity theorems
in quantum Shannon theory. But it turns out that there is a more clever way to maketeleportation and super-dense coding dual under resource reversal. In this chapter, we introduce a new resource—the noiseless coherent bit channel. This resource produces “coherent”
versions of the teleportation and super-dense coding protocols that are dual under resource
reversal. The payoff of this coherent communication technique is that we can exploit it tosimplify the proofs of various coding theorems of quantum Shannon theory. It also leads to a
deeper understanding of the relationship between the teleportation and super-dense coding
protocols from the previous chapter.
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7.1. DEFINITION OF COHERENT COMMUNICATION

227

7.1 Definition of Coherent Communication
We begin by introducing the coherent bit channel as a classical channel that has “quantum
feedback” (in a particular sense). Recall from Exercise 6.1.1 that a classical bit channel isequivalent to a dephasing channel that dephases in the computational basis with dephasing
parameter p = 1/2. The CPTP map corresponding to this completely dephasing channel is
as follows:1(7.7)
N (ρ) = (ρ + ZρZ) .
2
N
An isometric extension UA→BEof the above channel then follows by applying (5.36):
1
N
UA→BE
= √ (IA→B ⊗ |+iE + ZA→B ⊗ |−iE ) ,
2

(7.8)

where we choose the orthonormal basis states of the environment E to be |+i and |−i (recall
that we have unitary freedom in the choice of the basis states for the environment). It isNstraightforward to show that the isometry UA→BE
is as follows by expanding the operators
I and Z and the states |+i and |−i:
N
UA→BE
= |0iB h0|A ⊗ |0iE + |1iB h1|A ⊗ |1iE .

(7.9)

Thus, a classical bit channel is equivalent to the following map, with its action extended bylinearity:|iiA → |iiB |iiE : i ∈ {0, 1} .
(7.10)A coherent bit channel is similar to the above classical bit channel map, with the exceptionthat we assume that Alice somehow regains control of the environment of the channel:|iiA → |iiB |iiA : i ∈ {0, 1} .

(7.11)“Coherence” in this context is also synonymous with linearity—the maintenance and linear
transformation of superposed states. The coherent bit channel is similar to classical copying
because it copies the basis states while maintaining coherent superpositions. We denote theresource of a coherent bit channel as follows:[q → qq] .

(7.12)

Figure 7.1 provides a visual depiction of the coherent bit channel.
Exercise 7.1.1 Show that the following resource inequality holds:[q → qq] ≥ [c → c] .

(7.13)

That is, devise a protocol that generates a noiseless classical bit channel with one use of anoiseless coherent bit channel.
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CHAPTER 7. COHERENT PROTOCOLS

Figure 7.1: This figure depicts the operation of a coherent bit channel. It is the “coherification” of aclassical bit channel in which the sender A has access to the environment’s output.

7.2 Implementations of a Coherent Bit Channel
How might we actually implement a coherent bit channel? The simplest way to do so is withthe aid of a local CNOT gate and a noiseless qubit channel. The protocol proceeds as follows(Figure 7.2 illustrates the protocol):1. Alice possesses an information qubit in the state |ψiA ≡ α|0iA + β|1iA . She preparesan ancilla qubit in the state |0iA0 .
2. Alice performs a local CNOT gate from qubit A to qubit A0 . The resulting state is
α|0iA |0iA0 + β|1iA |1iA0 .

(7.14)

3. Alice transmits qubit A0 to Bob with one use of a noiseless qubit channel idA0 →B . The
resulting state is
α|0iA |0iB + β|1iA |1iB ,(7.15)and it is now clear that Alice and Bob have implemented a noiseless coherent bitchannel as defined in (7.11).
The above protocol realizes the following resource inequality:[q → q] ≥ [q → qq] ,(7.16)

demonstrating that quantum communication generates coherent communication.
Exercise 7.2.1 Show that the following resource inequality holds:[q → qq] ≥ [qq] .

(7.17)

That is, devise a protocol that generates a noiseless ebit with one use of a noiseless coherent
bit channel.
Exercise 7.2.2 Show that the following two resource inequalities cannot hold:[q → qq] ≥ [q → q] ,[qq] ≥ [q → qq] .
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7.2. IMPLEMENTATIONS OF A COHERENT BIT CHANNEL

229

Figure 7.2: A simple protocol to implement a noiseless coherent channel with one use of a noiseless qubit
channel.

We now have the following chain of resource inequalities:[q → q] ≥ [q → qq] ≥ [qq] .

(7.20)

Thus, the power of the coherent bit channel lies in between that of a noiseless qubit channel
and a noiseless ebit.
Exercise 7.2.3 Another way to implement a noiseless coherent bit channel is with a variation of teleportation that we name “coherent communication assisted by entanglement andclassical communication.” Suppose that Alice and Bob share an ebit |Φ+ iAB . Alice can append an ancilla qubit |0iA0 to this state and perform a local CNOT from A to A0 to give thefollowing state:1(7.21)|ΦGHZ iAA0 B = √ (|000iAA0 B + |111iAA0 B ) .
2
Alice prepends an information qubit |ψiA1 ≡ α|0iA1 + β|1iA1 to the above state so that the
global state is as follows:|ψiA1 |ΦGHZ iAA0 B .
(7.22)
Suppose Alice performs the usual teleportation operations on systems A1 , A, and A0 . Give thesteps that Alice and Bob should perform in order to generate the state α|0iA0 |0iB +β|1iA0 |1iB ,thus implementing a noiseless coherent bit channel. Hint: The resource inequality for thisprotocol is as follows:[qq] + [c → c] ≥ [q → qq] .
(7.23)
This should be compared with the teleportation protocol, which corresponds to [qq]+2 [c → c] ≥[q → q].
Exercise 7.2.4 Determine a qudit version of coherent communication assisted by classicalcommunication and entanglement by modifying the steps in the above protocol.
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CHAPTER 7. COHERENT PROTOCOLS

Figure 7.3: This figure depicts the protocol for coherent super-dense coding.

7.3 Coherent Dense Coding
In the previous section, we introduced two protocols that implement a noiseless coherent bitchannel: the simple method in the previous section and coherent communication assisted byclassical communication and entanglement (Exercise 7.2.3). We now introduce a different
method for implementing two coherent bit channels that makes more judicious use of available resources. We name it coherent super-dense coding because it is a coherent version of
the super-dense coding protocol.
The protocol proceeds as follows (Figure 7.3 depicts the protocol):1. Alice and Bob share one ebit in the state |Φ+ iAB before the protocol begins.
2. Alice first prepares two qubits A1 and A2 in the state |a1 iA1 |a2 iA2 and prepends thisstate to the ebit. The global state is as follows:|a1 iA1 |a2 iA2 Φ+ AB ,(7.24)

where a1 and a2 are binary-valued. This preparation step is reminiscent of the superdense coding protocol (recall that, in the super-dense coding protocol, Alice has twoclassical bits she would like to communicate).
3. Alice performs a CNOT gate from register A2 to register A and performs a controlled-Z
gate from register A1 to register A. The resulting state is as follows:|a1 iA1 |a2 iA2 ZAa1 XAa2 Φ+ AB .

(7.25)

4. Alice transmits the qubit in register A to Bob. We rename this register as B1 andBob’s other register B as B2 .
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7.4. COHERENT TELEPORTATION

231

5. Bob performs a CNOT gate from his register B1 to B2 and performs a Hadamard gateon B1 . The final state is as follows:|a1 iA1 |a2 iA2 |a1 iB1 |a2 iB2 .

(7.26)

The above protocol implements two coherent bit channels: one from A1 to B1 and anotherfrom A2 to B2 . You can check that the protocol works for arbitrary superpositions of twoqubit states on A1 and A2 —it is for this reason that this protocol implements two coherent
bit channels. The resource inequality corresponding to coherent super-dense coding is[qq] + [q → q] ≥ 2 [q → qq] .

(7.27)

Exercise 7.3.1 Construct a qudit version of coherent super-dense coding that implements
the following resource inequality:
log d [qq] + log d [q → q] ≥ 2 log d [q → qq] .
(7.28)
P(Hint: The qudit generalization of a controlled-NOT gate is d−1
i=0 |iihi| ⊗ X(i), where X is
Pd−1
defined in (3.208), of a controlled-Z gate is j=0 |jihj|⊗Z(j), where Z is defined in (3.210),
and of the Hadamard gate is the Fourier transform gate.)

7.4 Coherent Teleportation
We now introduce a coherent version of the teleportation protocol that we name coherent
teleportation. Let a Z coherent bit channel ∆Z be one that copies eigenstates of the Z
operator (this is as we defined a coherent bit channel before). Let an X coherent bit channel∆X be one that copies eigenstates of the X operator:∆X : |+iA → |+iA |+iB ,|−iA → |−iA |−iB .

(7.29)(7.30)

It does not really matter which basis we use to define a coherent bit channel—it just mattersthat it copies the orthogonal states of some basis.
Exercise 7.4.1 Show how to simulate an X coherent bit channel using a Z coherent bitchannel and local operations.
The protocol proceeds as follows (Figure 7.4 depicts the protocol):1. Alice possesses an information qubit |ψiA where|ψiA ≡ α|0iA + β|1iA .

(7.31)She sends her qubit through a Z coherent bit channel:|ψiA
c 2019∆Z−→

α|0iA |0iB1 + β|1iA |1iB1 ≡ |ψ̃iAB1 .

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CHAPTER 7. COHERENT PROTOCOLS

Figure 7.4: This figure depicts the protocol for coherent teleportation.

Let us rewrite the above state |ψ̃iAB1 as follows:
|+iA − |−iA|+iA + |−iA√√
|0iB1 + β|1iB1|ψ̃iAB1 = α22
1= √ [|+iA (α|0iB1 + β|1iB1 ) + |−iA (α|0iB1 − β|1iB1 )] .
2

(7.33)(7.34)

2. Alice sends her qubit A through an X coherent bit channel with output systems A
and B2 :|ψ̃iAB1∆X−→

1√ |+iA |+iB2 (α|0iB1 + β|1iB1 )2
1+ √ |−iA |−iB2 (α|0iB1 − β|1iB1 ) . (7.35)
2

3. Bob then performs a CNOT gate from qubit B1 to qubit B2 . Consider that the actionof the CNOT gate with the source qubit in the computational basis and the target
qubit in the +/− basis is as follows:|0i|+i → |0i|+i,|0i|−i → |0i|−i,|1i|+i → |1i|+i,|1i|−i → −|1i|−i,(7.36)(7.37)(7.38)(7.39)

so that the last entry catches a phase of π (eiπ = −1). Then this CNOT gate brings
the overall state to1√ [|+iA |+iB2 (α|0iB1 + β|1iB1 ) + |−iA |−iB2 (α|0iB1 + β|1iB1 )]21 = √ |+iA |+iB2 |ψiB1 + |−iA |−iB2 |ψiB1(7.40)
21= √ [|+iA |+iB2 + |−iA |−iB2 ] |ψiB1(7.41)
2= Φ+ AB2 |ψiB1 .
(7.42)
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7.5. COHERENT COMMUNICATION IDENTITY

233

Thus, Alice teleports her information qubit to Bob, and both Alice and Bob possess
one ebit at the end of the protocol.
The resource inequality for coherent teleportation is as follows:
2[q → qq] ≥ [qq] + [q → q].

(7.43)

Exercise 7.4.2 Show how a cobit channel and an ebit can generate a GHZ state. That is,demonstrate a protocol that realizes the following resource inequality:[qq]AB + [q → qq]BC ≥ [qqq]ABC .

(7.44)

Exercise 7.4.3 Outline the qudit version of the above coherent teleportation protocol. The
protocol should realize the following resource inequality:
2 log d[q → qq] ≥ log d[qq] + log d[q → q].

(7.45)

Exercise 7.4.4 Outline a catalytic version of the coherent teleportation protocol by modifying the original teleportation protocol. Let Alice possess an information qubit |ψiA0 and let
Alice and Bob share an ebit |Φ+ iAB . Replace the Bell measurement with a controlled-NOT
and Hadamard gate, replace the classical bit channels with coherent bit channels, and replace Bob’s conditional unitary operations with controlled unitary operations. The resulting
resource inequality should be of the form:
2 [q → qq] + [qq] ≥ [q → q] + 2 [qq] .

(7.46)

This protocol is catalytic in the sense that it gives the resource inequality in (7.43) when wecancel one ebit from each side.

7.5 Coherent Communication Identity
The fundamental result of this chapter is the coherent communication identity:
2[q → qq] = [qq] + [q → q].

(7.47)

We obtain this identity by combining the resource inequality for coherent super-dense coding in (7.27) and the resource inequality for coherent teleportation in (7.43). The coherent
communication identity demonstrates that coherent super-dense coding and coherent teleportation are dual under resource reversal—the resources that coherent teleportation consumes
are the same as those that coherent super-dense coding generates and vice versa.
The major application of the coherent communication identity is in noisy quantum Shannon theory. We will find later that its application is in the “upgrading” of protocols that
output private classical information. Suppose that a protocol outputs private classical bits.
The super-dense coding protocol is one such example, as the last paragraph of Section 6.2.3
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CHAPTER 7. COHERENT PROTOCOLS

argues. Then it is possible to upgrade the protocol by making it coherent, similar to the way
in which we made super-dense coding coherent by replacing conditional unitary operations
with controlled unitary operations.
We make this idea more precise with an example. The resource inequality for entanglementassisted classical coding (discussed in more detail in Chapter 21) has the following form:
hN i + E [qq] ≥ C [c → c] ,(7.48)

where N is a noisy quantum channel that connects Alice to Bob, E is some rate of entanglement consumption, and C is some rate of classical communication. It is possible to upgrade
the generated classical bits to coherent bits, for reasons that are similar to those that we usedin the upgrading of super-dense coding. The resulting resource inequality has the following
form:
hN i + E [qq] ≥ C [q → qq] .
(7.49)
We can now employ the coherent communication identity in (7.47) and argue that any
protocol that realizes the above resource inequality can realize the following one:hN i + E [qq] ≥CC[q → q] + [qq] ,2
2(7.50)

merely by using the generated coherent bits in a coherent super-dense coding protocol. We
can then make a “catalytic argument” to cancel the ebits on both sides of the resource
inequality. The final resource inequality is as follows:CC[qq] ≥ [q → q] .
(7.51)
hN i + E −2
2
The above resource inequality corresponds to a protocol for entanglement-assisted quantum
communication, and it turns out to be optimal for some channels as this protocol’s converse
theorem shows. This optimality is due to the efficient translation of classical bits to coherent
bits and the application of the coherent communication identity.

7.6 History and Further Reading
Harrow (2004) introduced the idea of coherent communication. Later, the idea of the coherent bit channel was generalized to the continuous-variable case (Wilde et al., 2007). Coherent
communication has many applications in quantum Shannon theory which we will study inlater chapters.

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 8

Unit Resource Capacity Region
In Chapter 6, we presented the three unit protocols of teleportation, super-dense coding,and entanglement distribution. We argued in Section 6.3 that each of these protocols areindividually optimal. For example, recall that the entanglement distribution protocol is
optimal because two parties cannot generate more than one ebit from the use of one noiselessqubit channel.
In this chapter, we show that these three protocols are actually the most important
protocols—we do not need to consider any other protocols when the noiseless resources ofclassical communication, quantum communication, and entanglement are available. Combining these three protocols together is the best that one can do with the unit resources.
In this sense, this chapter gives a good example of a converse proof of a capacity theorem.
We construct a three-dimensional region, known as the unit resource achievable region,that the three unit protocols fill out. The converse proof of this chapter employs physicalarguments to show that the unit resource achievable region is optimal, and we can then refer
to it as the unit resource capacity region. We later exploit the development here when we
get to the study of trade-off capacities (see Chapter 25).

8.1 The Unit Resource Achievable Region
Let us first recall the resource inequalities for the three unit protocols. The resource inequality for teleportation is
2[c → c] + [qq] ≥ [q → q],(8.1)
while that for super-dense coding is[q → q] + [qq] ≥ 2[c → c],(8.2)

and that for entanglement distribution is as follows:[q → q] ≥ [qq].
Each of the resources [q → q], [qq], [c → c] is a unit resource.
235

(8.3)

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CHAPTER 8. UNIT RESOURCE CAPACITY REGION

The above three unit protocols are sufficient to recover all other unit protocols. For
example, we can combine super-dense coding and entanglement distribution to produce thefollowing resource inequality:
2[q → q] + [qq] ≥ 2[c → c] + [qq] .

(8.4)

The above resource inequality is equivalent to the following one:[q → q] ≥ [c → c],(8.5)

after removing the entanglement from both sides and scaling by 1/2 (we can remove theentanglement here because it acts as a catalytic resource).
We can justify this scaling by considering a scenario in which we use the above protocol
N times. For the first run of the protocol, we require one ebit to get it started, but thenevery other run both consumes and generates one ebit, giving
2N [q → q] + [qq] ≥ 2N [c → c] + [qq] .

(8.6)

Dividing by N gives the rate of the task, and as N becomes large, the use of the initial ebit
is negligible. We refer to (8.5) as “classical coding over a noiseless qubit channel.”
We can think of the above resource inequalities in a different way. Let us consider athree-dimensional space with points of the form (C, Q, E), where C corresponds to noiseless classical communication, Q corresponds to noiseless quantum communication, and E
corresponds to noiseless entanglement. Each point in this space corresponds to a protocol
involving the unit resources. A coordinate of a point is negative if the point’s corresponding
resource inequality consumes that coordinate’s corresponding resource, and a coordinate of apoint is positive if the point’s corresponding resource inequality generates that coordinate’s
corresponding resource.
For example, the point corresponding to the teleportation protocol is
xTP ≡ (−2, 1, −1) ,(8.7)

because teleportation consumes two noiseless classical bit channels and one ebit to generate
one noiseless qubit channel. For similar reasons, the respective points corresponding tosuper-dense coding and entanglement distribution are as follows:
xSD ≡ (2, −1, −1) ,xED ≡ (0, −1, 1) .

(8.8)

Figure 8.1 plots these three points in the three-dimensional space of classical communication,
quantum communication, and entanglement.
We can execute any of the three unit protocols just one time, or we can execute any oneof them m times where m is some positive integer. Executing a protocol m times then gives
other points in the three-dimensional space. That is, we can also achieve the points mxTP ,mxSD , and mxED for any positive m. This method allows us to fill up a certain portion of thethree-dimensional space and provides an interpretation for resource inequalities with rationalc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

8.1. THE UNIT RESOURCE ACHIEVABLE REGION

237

Figure 8.1: The three points corresponding to the three respective unit protocols of entanglement distribution (ED), teleportation (TP), and super-dense coding (SD).

coefficients. Since the rational numbers are dense in the real numbers, we can also allow forreal number coefficients. This becomes important later on when we consider combining thethree unit protocols in order to achieve certain rates of transmission (a communication rate
can be any real number). Thus, we can combine the protocols together to achieve any pointof the following form:
αxTP + βxSD + γxED ,(8.9)
where α, β, γ ≥ 0.
Let us further establish some notation. Let L denote a line, Q a quadrant, and O an
octant in the three-dimensional space (it should be clear from the context whether Q refers
to quantum communication or “quadrant”). For example, L−00 denotes a line going in the
direction of negative classical communication:
L−00 ≡ {α (−1, 0, 0) : α ≥ 0} .

(8.10)

Q0+− denotes the quadrant where there is zero classical communication, generation of quantum communication, and consumption of entanglement:
Q0+− ≡ {α (0, 1, 0) + β (0, 0, −1) : α, β ≥ 0} .

(8.11)

O+−+ denotes the octant where there is generation of classical communication, consumptionof quantum communication, and generation of entanglement:α (1, 0, 0) + β (0, −1, 0) + γ (0, 0, 1)
+−+
O
≡
.
(8.12): α, β, γ ≥ 0
It proves useful to have a “set addition” operation between two regions A and B (known
as the Minkowski sum):A + B ≡ {a + b : a ∈ A, b ∈ B}.
(8.13)
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CHAPTER 8. UNIT RESOURCE CAPACITY REGION

The following relations hold
Q0+− = L0+0 + L00− ,O+−+ = L+00 + L0−0 + L00+ ,(8.14)(8.15)

by using the above definition.
The following geometric objects lie in the (C, Q, E) space:
1. The “line of teleportation” LTP is the following set of points:
LTP ≡ {α (−2, 1, −1) : α ≥ 0} .

(8.16)

2. The “line of super-dense coding” LSD is the following set of points:
LSD ≡ {β (2, −1, −1) : β ≥ 0} .

(8.17)

3. The “line of entanglement distribution” LED is the following set of points:
LED ≡ {γ (0, −1, 1) : γ ≥ 0} .

(8.18)eU denote the unit resource achievable region. It consists of all linear
Definition 8.1.1 Let C
combinations of the above protocols:eU ≡ LTP + LSD + LED .
CeU :The following matrix equation gives all achievable triples (C, Q, E) in C
  
 
C
−2 2
0
α
Q =  1 −1 −1 β  ,E−1 −1 1γ
where α, β, γ ≥ 0. We can rewrite the above equation using the matrix inverse:
  
 
α
−1/2 −1/2 −1/2Cβ  =  0−1/2 −1/2Q ,γ−1/2 −10E(8.19)(8.20)(8.21)in order to express the coefficients α, β, and γ as a function of the rate triples (C, Q, E). The
restriction of non-negativity of α, β, and γ gives the following restriction on the achievablerate triples (C, Q, E):
C + Q + E ≤ 0,
Q + E ≤ 0,
C + 2Q ≤ 0.

(8.22)(8.23)(8.24)

eU in (8.19) is equivalent to all rate
The above result implies that the achievable region C
triples satisfying (8.22)–(8.24). Figure 8.2 displays the full unit resource achievable region.
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8.2. THE DIRECT CODING THEOREM

239

eU .
Figure 8.2: This figure depicts the unit resource achievable region C

Definition 8.1.2 The unit resource capacity region CU is the closure of the set of all points(C, Q, E) in the C, Q, E space, satisfying the following resource inequality:
0 ≥ C[c → c] + Q[q → q] + E[qq].

(8.25)The definition states that the unit resource capacity region consists of all those points(C, Q, E) that have corresponding protocols that can implement them. The notation in theabove definition may seem slightly confusing at first glance until we recall that a resourcewith a negative rate implicitly belongs on the left-hand side of the resource inequality.
Theorem 8.1.1 below gives the optimal three-dimensional capacity region for the threeunit resources.
Theorem 8.1.1 The unit resource capacity region CU is equal to the unit resource achievableeU :region CeU .
CU = C(8.26)
Proving the above theorem involves two steps: the direct coding theorem and the converseeU
theorem. For this case, the direct coding theorem establishes that the achievable region C
is in the capacity region CU :eU ⊆ CU .
C(8.27)eU :The converse theorem, on the other hand, establishes optimality of CeU .
CU ⊆ C(8.28)

8.2 The Direct Coding TheoremeU ⊆ CU , is immediate from the definition in
The result of the direct coding theorem, that C
eU , the definition in (8.25) of the unit resource(8.19) of the unit resource achievable region C
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CHAPTER 8. UNIT RESOURCE CAPACITY REGION

capacity region CU , and the theory of resource inequalities. We can achieve points in theunit resource capacity region simply by considering positive linear combinations of the threeunit protocols. The next section shows that the unit resource capacity region consists of all
and only those points in the unit resource achievable region.

8.3 The Converse Theorem
eU in (8.19) and consider the eight octants of the (C, Q, E)
We employ the definition of C
eU ). Let (±, ±, ±)space individually in order to prove the converse theorem (that CU ⊆ Cdenote labels for the eight different octants.
It is possible to demonstrate the optimality of each of these three protocols individuallywith a contradiction argument as we saw in Chapter 6. However, in the converse proofof Theorem 8.1.1, we show that a mixed strategy combining these three unit protocols is
optimal.
We accept the following postulates and exploit them in order to prove the converse:
1. Entanglement alone cannot generate classical communication or quantum communication or both.
2. Classical communication alone cannot generate entanglement or quantum communication or both.
3. Holevo bound: One cannot generate more than one classical bit of communication peruse of a noiseless qubit channel alone.
(+, +, +). This octant of CU is empty because a sender and receiver require some resources to implement classical communication, quantum communication, and entanglement.
(They cannot generate a unit resource from nothing!)(+, +, −). This octant of CU is empty because entanglement alone cannot generate
either classical communication or quantum communication or both.
(+, −, +). The task for this octant is to generate a noiseless classical channel of C bits
and E ebits of entanglement by consuming |Q| qubits of quantum communication. We thus
consider all points of the form (C, Q, E) where C ≥ 0, Q ≤ 0, and E ≥ 0. It suffices to prove
the following inequality:
C + E ≤ |Q| ,(8.29)
because combining (8.29) with C ≥ 0 and E ≥ 0 implies (8.22)–(8.24). The achievabilityof (C, −|Q|, E) implies the achievability of the point (C + 2E, −|Q| − E, 0), because we canconsume all of the entanglement with super-dense coding (8.2):(C + 2E, − |Q| − E, 0) = (C, − |Q| , E) + (2E, −E, −E) .

(8.30)

This new point implies that there is a protocol that consumes |Q|+E noiseless qubit channels
to send C + 2E classical bits. The following bound then applies
C + 2E ≤ |Q| + E,
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8.3. THE CONVERSE THEOREM

241

because the Holevo bound (Exercise 4.2.2 gives a simpler statement of this bound) statesthat we can send only one classical bit per qubit. The bound in (8.29) then follows.
(+, −, −). The task for this octant is to simulate a classical channel of size C bits using|Q| qubits of quantum communication and |E| ebits of entanglement. We consider all pointsof the form (C, Q, E) where C ≥ 0, Q ≤ 0, and E ≤ 0. It suffices to prove the following
inequalities:C ≤ 2|Q|,
C ≤ |Q| + |E|,

(8.32)(8.33)

because combining (8.32)–(8.33) with C ≥ 0 implies (8.22)–(8.24). The achievability of(C, −|Q|, −|E|) implies the achievability of (0, −|Q| + C/2, −|E| − C/2), because we canconsume all of the classical communication with teleportation (8.1):(0, −|Q| + C/2, −|E| − C/2) = (C, −|Q|, −|E|) + (−C, C/2, −C/2) .

(8.34)

The following bound applies (quantum communication cannot be positive):− |Q| + C/2 ≤ 0,

(8.35)

because entanglement alone cannot generate quantum communication. The bound in (8.32)
then follows from the above bound. The achievability of (C, −|Q|, −|E|) implies the achievability of (C, −|Q| − |E|, 0) because we can consume an extra |E| qubit channels with entanglement distribution (8.3):(C, −|Q| − |E|, 0) = (C, −|Q|, −|E|) + (0, − |E| , |E|) .

(8.36)The bound in (8.33) then applies by the same Holevo bound argument as in the previous
octant.
(−, +, +). This octant of CU is empty because classical communication alone cannot
generate either quantum communication or entanglement or both.
(−, +, −). The task for this octant is to simulate a quantum channel of size Q qubits
using |E| ebits of entanglement and |C| bits of classical communication. We consider all
points of the form (C, Q, E) where C ≤ 0, Q ≥ 0, and E ≤ 0. It suffices to prove thefollowing inequalities:
Q ≤ |E| ,2Q ≤ |C| ,(8.37)(8.38)

because combining them with C ≤ 0 implies (8.22)–(8.24). The achievability of the point(−|C|, Q, −|E|) implies the achievability of the point (−|C|, 0, Q − |E|), because we canconsume all of the quantum communication for entanglement distribution (8.3):(−|C|, 0, Q − |E|) = (−|C|, Q, −|E|) + (0, −Q, Q) .
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CHAPTER 8. UNIT RESOURCE CAPACITY REGION

The following bound applies (entanglement cannot be positive):
Q − |E| ≤ 0,

(8.40)

because classical communication alone cannot generate entanglement. The bound in (8.37)
follows from the above bound. The achievability of the point (−|C|, Q, −|E|) implies theachievability of the point (−|C| + 2Q, 0, −Q − |E|), because we can consume all of thequantum communication for super-dense coding (8.2):(−|C| + 2Q, 0, −Q − |E|) = (−|C|, Q, −|E|) + (2Q, −Q, −Q) .

(8.41)

The following bound applies (classical communication cannot be positive):− |C| + 2Q ≤ 0,

(8.42)

because entanglement alone cannot create classical communication. The bound in (8.38)
follows from the above bound.
(−, −, +). The task for this octant is to create E ebits of entanglement using |Q| qubitsof quantum communication and |C| bits of classical communication. We consider all pointsof the form (C, Q, E) where C ≤ 0, Q ≤ 0, and E ≥ 0. It suffices to prove the following
inequality:
E ≤ |Q| ,(8.43)
because combining it with Q ≤ 0 and C ≤ 0 implies (8.22)–(8.24). The achievabilityof (−|C|, −|Q|, E) implies the achievability of (−|C| − 2E, −|Q| + E, 0), because we canconsume all of the entanglement with teleportation (8.1):(−|C| − 2E, −|Q| + E, 0) = (−|C|, −|Q|, E) + (−2E, E, −E) .

(8.44)

The following bound applies (quantum communication cannot be positive):− |Q| + E ≤ 0,

(8.45)

because classical communication alone cannot generate quantum communication. The boundin (8.43) follows from the above bound.
eU completely contains this octant.
(−, −, −). C
We have now proved that the set of inequalities in (8.22)–(8.24) holds for all octants ofthe (C, Q, E) space. The next exercises ask you to consider similar unit resource achievable
regions.
Exercise 8.3.1 Consider the resources of public classical communication:[c → c]pub ,(8.46)[c → c]priv ,(8.47)

private classical communication:c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

8.3. THE CONVERSE THEOREM

243

and shared secret key:[cc]priv .
Public classical communication is equivalent to the following channel:Xρ→hi|ρ|ii |iihi|B ⊗ σEi ,(8.48)(8.49)

iso that an eavesdropper Eve obtains some correlations with the transmitted state ρ. Private
classical communication is equivalent to the following channel:Xρ→hi|ρ|ii |iihi|B ⊗ σE ,(8.50)
iso that Eve’s state is independent of the information that Bob receives. Finally, a secret key
is a state of the following form:!
1XΦAB ⊗ σE ≡|iihi|A ⊗ |iihi|B ⊗ σE ,(8.51)d iso that Alice and Bob share maximal classical correlation and Eve’s state is independent of
it. There are three protocols that relate these three classical resources. Secret key distribution
is a protocol that consumes a noiseless private channel to generate a noiseless secret key. It
has the following resource inequality:[c → c]priv ≥ [cc]priv .

(8.52)The one-time pad protocol exploits a shared secret key and a noiseless public channel togenerate a noiseless private channel (it simply XORs a bit of secret key with the bit that thesender wants to transmit and this protocol is provably unbreakable if the secret key is perfectly
secret). It has the following resource inequality:[c → c]pub + [cc]priv ≥ [c → c]priv .

(8.53)

Finally, private classical communication can simulate public classical communication if we
assume that Bob has a local register where he can place information and he then gives thisto Eve. It has the following resource inequality:[c → c]priv ≥ [c → c]pub .

(8.54)Show that these three protocols fill out an optimal achievable region in the space of publicclassical communication, private classical communication, and secret key. Use the following
three postulates to prove optimality: (1) public classical communication alone cannot generatesecret key or private classical communication, (2) private key alone cannot generate public
or private classical communication, and (3) the net amount of public bit channel uses andsecret key bits generated cannot exceed the number of private bit channel uses consumed.
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CHAPTER 8. UNIT RESOURCE CAPACITY REGION

Exercise 8.3.2 Consider the resource of coherent communication from Chapter 7:[q → qq] .

(8.55)

Recall the coherent communication identity in (7.47):
2 [q → qq] = [q → q] + [qq] .

(8.56)

Recall the other resource inequalities for coherent communication:[q → q] ≥ [q → qq] ≥ [qq] .

(8.57)

Consider a space of points (C, Q, E) where C corresponds to coherent communication, Q
to quantum communication, and E to entanglement. Determine the achievable region one
obtains with the above resource inequalities and another trivial resource inequality:[qq] ≥ 0.

(8.58)

We interpret the above resource inequality as “entanglement consumption,” where Alice simply throws away entanglement.

8.4 History and Further Reading
The unit resource capacity region first appeared in Hsieh and Wilde (2010b) in the context
of trade-off coding. The private unit resource capacity region later appeared in Wilde andHsieh (2012a).

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicensePart IV
Tools of Quantum Shannon Theory

245

CHAPTER 9Distance Measures
We discussed the major noiseless quantum communication protocols such as teleportation,
super-dense coding, their coherent versions, and entanglement distribution in detail in Chapters 6, 7, and 8. Each of these protocols relies on the assumption that noiseless resources
are available. For example, the entanglement distribution protocol assumes that a noiselessqubit channel is available to generate a noiseless ebit. This idealization allowed us to develop
the main principles of the protocols without having to think about more complicated issues,
but in practice, the protocols do not work as expected in the presence of noise.
Given that quantum systems suffer noise in practice, we would like to have a way todetermine how well a protocol is performing. The simplest way to do so is to compare theoutput of an ideal protocol to the output of the actual protocol using a distance measure of
the two respective output quantum states. That is, suppose that a quantum informationprocessing protocol should ideally output some quantum state |ψi, but the actual outputof the protocol is a quantum state with density operator ρ. Then a performance measure
P (ψ, ρ) should indicate how close the ideal output is to the actual output. Figure 9.1 depicts
the comparison of an ideal protocol with another protocol that is noisy.
This chapter introduces two distance measures that allow us to determine how close twoquantum states are to each other. The first distance measure that we discuss is the trace
distance and the second is the fidelity. (However, note that the fidelity is not a distancemeasure in the strict mathematical sense—nevertheless, we exploit it as a “closeness” measure of quantum states because it admits an intuitive operational interpretation.) These two
measures are mostly interchangeable, but we introduce both because it is often times moreconvenient in a given situation to use one or the other.
Distance measures are particularly important in quantum Shannon theory because they
provide a way for us to determine how well a protocol is performing. Recall that Shannon’s
method (outlined in Chapter 2) for both the noiseless and noisy coding theorem is to allow
for a slight error in a protocol, but to show that this error vanishes in the limit of large
block length. In later chapters where we prove quantum coding theorems, we borrow thistechnique of demonstrating asymptotically small error, with either the trace distance or the
fidelity as the measure of performance.
247

248

CHAPTER 9. DISTANCE MEASURES

Figure 9.1: A distance measure quantifies how far the output of a given ideal protocol (depicted on theleft) is from an actual protocol that exploits a noisy resource (depicted as the noisy quantum channel NA→B
on the right).

9.1 Trace Distance
We first introduce the trace distance. Our presentation is somewhat mathematical because
we exploit norms on linear operators in order to define it. Despite this mathematical flavor,
this section offers an intuitive operational interpretation of the trace distance.

9.1.1

Trace NormDefinition 9.1.1 (Trace Norm) The trace norm or Schatten 1-norm kM k1 of an operator
M ∈ L(H, H0 ) is defined askM k1 ≡ Tr {|M |} ,(9.1)√
where |M | ≡ M † M .
Proposition 9.1.1 The trace norm of an operator M ∈ L(H, H0 ) is equal to the sum of its
singular values.
Proof. Recall from Definition 3.3.1 that any function f applied to a Hermitian operator A
is as follows:Xf (A) ≡f (αi )|iihi|,(9.2)
i:αi ∈Dom(f )P

where i αi |iihi| is a spectral decomposition of A. With these two definitions, it is straightforward to show that the trace norm of M is equal to the sum of its singular values. Indeed,
let M = U ΣV be the singular value decomposition of M , where U and V are unitary matrices and Σ is a rectangular matrix with the non-negative singular values along the diagonal.
Then we can writed−1XM=
σi |ui ihvi |,
(9.3)
i=0

where d is the rank of M , {σi } are the strictly positive singular values of M , {|ui i} are
the orthonormal columns of U in correspondence with the set {σi }, and {|vi i} are thec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

9.1. TRACE DISTANCE

249

orthonormal rows of V in correspondence with the set {σi }. ThenM †M ==" d−1
X# " d−1#Xσj |vj ihuj |σi |ui ihvi |

j=0

i=0

d−1
X

(9.4)

σj σi |vj ihuj ||ui ihvi |(9.5)

σi2 |vi ihvi |,

(9.6)

i,j=0

=

d−1
X
i=0

so that√

M †M =d−1 q
X

σi2 |vi ihvi | =

i=0

d−1
X

σi |vi ihvi |,

(9.7)

i=0

finally implying that
Tr {|M |} =

d−1
X

σi .

(9.8)

i=0

This means also that√kM k1 = Tr{ M M † },(9.9)

because the singular values of M M † and M † M are the same (this is the key to Exercise 4.1.4).
One can also easily show that the trace norm of a Hermitian operator is equal to the absolute
sum of its eigenvalues.
The trace norm is indeed a norm because it satisfies the following three properties: nonnegative definiteness, homogeneity, and the triangle inequality.
Property 9.1.1 (Non-Negative Definiteness) The trace norm of an operator M is nonnegative definite:kM k1 ≥ 0.
(9.10)The trace norm is equal to zero if and only if the operator M is the zero operator:
kM k1 = 0⇔

M = 0.

(9.11)

Property 9.1.2 (Homogeneity) For any constant c ∈ C,
kcM k1 = |c| kM k1 .

(9.12)

Property 9.1.3 (Triangle Inequality) For any two operators M, N ∈ L(H, H0 ), the following triangle inequality holds:
kM + N k1 ≤ kM k1 + kN k1 .
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Non-negative definiteness follows because the sum of the singular values of an operator
is non-negative, and the singular values are all equal to zero (and thus the operator is equal
to zero) if and only if the sum of the singular values is equal to zero. Homogeneity follows
directly from the fact that |cM | = |c||M |. We later give a proof of the triangle inequality(however, for a special case only). Exercise 9.1.1 below asks you to prove it for square
operators.
Three other important properties of the trace norm are its invariance under isometries,
convexity, and a variational characterization. Each of the properties below often arise as
useful tools in quantum Shannon theory.
Property 9.1.4 (Isometric Invariance) The trace norm is invariant under multiplication by isometries U and V :U M V † 1 = kM k1 .
(9.14)
Property 9.1.5 (Convexity) For any two operators M, N ∈ L(H, H0 ) and λ ∈ [0, 1], thefollowing inequality holds
kλM + (1 − λ)N k1 ≤ λ kM k1 + (1 − λ) kN k1 .

(9.15)

Isometric invariance holds because M and U M V † have the same singular values. Convexity follows directly from the triangle inequality and homogeneity (thus, any norm is convex
in this sense).
Property 9.1.6 (Variational characterization) For a square operator M ∈ L(H), thefollowing variational characterization of the trace norm holds
kM k1 = max |Tr {M U }| ,U

(9.16)

where the optimization is with respect to all unitary operators.
Proof. The above characterization follows by taking a singular value decomposition of M as
M = W DV , with W and V unitaries and D a diagonal matrix of singular values. Applying
the Cauchy–Schwarz inequality giveson√ √D DV U W(9.17)|Tr {M U }| = |Tr {W DV U }| = Tr
s 
r n

√
† √
√ √ o
≤ Tr
D D
Tr
DV U W
DV U W(9.18)= Tr {D} = kM k1 .

(9.19)The inequality is a consequence of the Cauchy–Schwarz inequality for the Hilbert–Schmidt
inner product:pp
Tr A† B ≤ Tr {A† A} Tr {B † B}.
(9.20)
Equality holds by picking U = V † W † , from which we recover (9.16).
Exercise 9.1.1 Prove that the triangle inequality (Property 9.1.3) holds for square operators
M, N ∈ L(H). (Hint: Use the characterization in Property 9.1.6.)c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

9.1. TRACE DISTANCE

9.1.2

251

Trace Distance from the Trace Norm

The trace norm induces a natural distance measure, called the trace distance.
Definition 9.1.2 (Trace Distance) Given any two operators M, N ∈ L(H, H0 ), the trace
distance between them is as follows:kM − N k1 .

(9.21)The trace distance is especially useful as a measure of the distinguishability of two quantum states with respective density operators ρ and σ. The following bounds apply to thetrace distance between any two density operators ρ and σ:
0 ≤ kρ − σk1 ≤ 2.

(9.22)

Sometimes it is useful to employ the normalized trace distance 21 kρ − σk1 , so that 21 kρ − σk1 ∈[0, 1]. The lower bound in (9.22) applies when two quantum states are equal—quantum statesρ and σ are equal to each other if and only if their trace distance is zero. The physical implication of the trace distance being equal to zero is that no measurement can distinguish ρfrom σ. The upper bound in (9.22) follows from the triangle inequality:kρ − σk1 ≤ kρk1 + kσk1 = 2.

(9.23)The trace distance is maximum when ρ and σ have support on orthogonal subspaces. Later,
we will prove that this is the only case in which this happens, after introducing the fidelity.
The physical implication of maximal trace distance is that there exists a measurement thatcan perfectly distinguish ρ from σ. We discuss these operational interpretations of the trace
distance in more detail in Section 9.1.4.
Exercise 9.1.2 Show that the trace distance between two qubit density operators ρ and σ is−−equal to the Euclidean distance between their respective Bloch vectors →r and →
s , where1−−(I + →
r ·→
σ ),
2
−
−
That is, show that kρ − σk1 = k→r −→s k2 .
ρ=

σ=

1
−
−
(I + →s ·→σ ).
2(9.24)

Exercise 9.1.3 Show that the trace distance obeys a telescoping property:kρ1 ⊗ ρ2 − σ1 ⊗ σ2 k1 ≤ kρ1 − σ1 k1 + kρ2 − σ2 k1 ,(9.25)

for any density operators ρ1 , ρ2 , σ1 , σ2 . (Hint: First prove that kρ ⊗ ω − σ ⊗ ωk1 = kρ − σk1 ,for any density operators ρ, σ, ω.)Exercise 9.1.4 Show that the trace distance is invariant with respect to an isometric quantum channel, in the following sense:kρ − σk1 = U ρU † − U σU † 1 ,(9.26)

where U is an isometry. The physical implication of (9.26) is that an isometric quantum
channel applied to both states does not increase or decrease the distinguishability of the twostates.
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9.1.3

CHAPTER 9. DISTANCE MEASURES

Trace Distance as a Probability Difference

We now state and prove an important lemma that gives an alternative and useful way forcharacterizing the trace distance. This particular characterization finds application in many
proofs of the lemmas that follow concerning trace distance.
Lemma 9.1.1 The normalized trace distance 12 kρ − σk1 between quantum states ρ, σ ∈D(H) is equal to the largest probability difference that two states ρ and σ could give tothe same measurement outcome Λ:1kρ − σk1 = max Tr {Λ (ρ − σ)} .
0≤Λ≤I2(9.27)

The above maximization is with respect to all positive semi-definite operators Λ ∈ L(H) thathave their eigenvalues bounded from above by one.
Proof. Consider that the difference operator ρ − σ is Hermitian and so we can diagonalize
it as follows:Xρ−σ =λi |iihi|,
iwhere {|ii} is an orthonormal basis of eigenvectors and {λi } is a set of real eigenvalues. Let
us defineXXP ≡λi |iihi|,Q≡|λi | |iihi|,(9.28)
i:λi ≥0i:λi <0which implies that P and Q are positive semi-definite and thatρ − σ = P − Q.

(9.29)

Consider also that P Q = 0, and let ΠP and ΠQ denote the projections onto the supports of
P and Q, respectively:
XX
ΠP ≡|iihi|,
ΠQ ≡|iihi|.
(9.30)
i:λi ≥0i:λi <0Then it follows thatΠP P ΠP = P,
ΠP QΠP = 0,

ΠQ QΠQ = Q,
ΠQ P ΠQ = 0.

(9.31)(9.32)

The following property holds as well:|ρ − σ| = |P − Q| = P + Q.

(9.33)

because the supports of P and Q are orthogonal and the absolute value of the operator P −Q
takes the absolute value of its eigenvalues. Therefore,kρ − σk1 = Tr {|ρ − σ|} = Tr {P + Q} = Tr {P } + Tr {Q} .
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9.1. TRACE DISTANCE

253

But
Tr {P } − Tr {Q} = Tr {P − Q} = Tr {ρ − σ}= Tr {ρ} − Tr {σ} = 0.

(9.35)(9.36)

where the last equality follows because both quantum states have unit trace. Therefore,
Tr {P } = Tr {Q} andkρ − σk1 = 2 · Tr {P } .
(9.37)
Consider then thatTr {ΠP (ρ − σ)} = Tr {ΠP (P − Q)} = Tr {ΠP P }1= Tr {P } = kρ − σk1 .
2(9.38)(9.39)

Now we prove that the operator ΠP is the maximizing one. Let Λ be any positive semidefinite operator with spectrum bounded above by one. Then
Tr {Λ (ρ − σ)} = Tr {Λ (P − Q)} ≤ Tr {ΛP }1≤ Tr {P } = kρ − σk1 .
2(9.40)(9.41)

The first inequality follows because Λ and Q are non-negative and so Tr{ΛQ} is non-negative.
The second inequality holds because Λ ≤ I. The final equality follows from (9.37).
Exercise 9.1.5 Let ρ = |0ih0| and σ = |+ih+|. Compute P , Q, ΠP , and ΠQ , as defined in(9.28) and (9.30), for this choice of ρ and σ. Compute the trace distance kρ − σk1 .
Exercise 9.1.6 Show that the trace norm of any Hermitian operator ω is given by the following optimization:kωk1 = max Tr {Λω} .
(9.42)−I≤Λ≤I

9.1.4

Operational Interpretation of the Trace Distance

We now provide an operational interpretation of the trace distance as the distinguishability of
two quantum states. The interpretation results from a hypothesis-testing scenario. Supposethat Bob prepares one of two quantum states ρ0 or ρ1 for Alice to distinguish. Suppose
further that it is equally likely a priori for him to prepare either ρ0 or ρ1 . Let X denote theBernoulli random variable assigned to the prior probabilities so that pX (0) = pX (1) = 1/2.
Alice can perform a binary POVM with elements Λ ≡ {Λ0 , Λ1 } to distinguish the twostates. That is, Alice guesses the state in question is ρ0 if she receives outcome “0” fromthe measurement or she guesses the state in question is ρ1 if she receives outcome “1” fromthe measurement. Let Y denote the Bernoulli random variable assigned to the classicaloutcomes of her measurement. The success probability psucc (Λ) for this hypothesis testingc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 9. DISTANCE MEASURES

scenario is the sum of the probability of detecting “0” when the state is ρ0 and the probabilityof detecting “1” when the state is ρ1 :psucc (Λ) = pY |X (0|0)pX (0) + pY |X (1|1)pX (1)11= Tr {Λ0 ρ0 } + Tr {Λ1 ρ1 } .
2
2(9.43)(9.44)

We can simplify this expression using the completeness relation Λ0 + Λ1 = I:
1
(Tr {Λ0 ρ0 } + Tr {(I − Λ0 ) ρ1 })2
1= (Tr {Λ0 ρ0 } + Tr {ρ1 } − Tr {Λ0 ρ1 })2
1= (Tr {Λ0 ρ0 } + 1 − Tr {Λ0 ρ1 })
2
1= (1 + Tr {Λ0 (ρ0 − ρ1 )}) .
2

psucc (Λ) =(9.45)(9.46)(9.47)(9.48)

Now Alice has freedom in choosing the POVM Λ = {Λ0 , Λ1 } to distinguish the states ρ0 andρ1 , and she would like to choose one that maximizes the success probability psucc (Λ). Thus,
we can define the success probability with respect to all measurements as follows:
psucc ≡ max psucc (Λ) = maxΛΛ1(1 + Tr {Λ0 (ρ0 − ρ1 )}) .
2(9.49)

We can rewrite the above quantity in terms of the trace distance using its characterization in
Lemma 9.1.1 because the expression inside of the maximization involves only the operator Λ0 :11
psucc =
1 + kρ0 − ρ1 k1 .
(9.50)
2
2
Thus, the normalized trace distance has an operational interpretation that it is linearly
related to the maximum success probability in distinguishing two quantum states ρ0 andρ1 in a quantum hypothesis testing experiment. From the above expression for the successprobability, it is clear that the states are indistinguishable when kρ0 − ρ1 k1 is equal to zero.
That is, it is just as good for Alice to guess randomly what the state might be, and in this case,she can do no better than to have 1/2 probability of being correct. On the other hand, thestates are perfectly distinguishable when kρ0 − ρ1 k1 is maximal and the measurement that
distinguishes them consists of two projectors: one projects onto the non-negative eigenspaceof ρ0 − ρ1 and the other projects onto the negative eigenspace of ρ0 − ρ1 . In this sense,
we can say that the normalized trace distance is the bias away from random guessing in ahypothesis testing experiment.
Exercise 9.1.7 Suppose that the prior probabilities in the above hypothesis-testing scenario
are not uniform but are rather equal to p0 and p1 . Show that the success probability is instead
given by1
(9.51)
psucc = (1 + kp0 ρ0 − p1 ρ1 k1 ) .
2
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9.1. TRACE DISTANCE

9.1.5

255

Trace Distance Lemmas

We present several useful corollaries of Lemma 9.1.1 and their corresponding proofs. These
corollaries include the triangle inequality, measurement on close states, and monotonicityof trace distance. Each of these corollaries finds application in many proofs in quantum
Shannon theory.
Lemma 9.1.2 (Triangle Inequality) The trace distance obeys a triangle inequality. For
any three quantum states ρ, σ, τ ∈ D(H), the following inequality holds:kρ − σk1 ≤ kρ − τ k1 + kτ − σk1 .

(9.52)

Proof. Pick Π as the maximizing operator for kρ − σk1 (according to Lemma 9.1.1) so thatkρ − σk1 = 2 · Tr {Π (ρ − σ)}= 2 · Tr {Π (ρ − τ )} + 2 · Tr {Π (τ − σ)}≤ 2 · max [Tr {Π1 (ρ − τ )}] + 2 · max [Tr {Π2 (τ − σ)}]Π1 :0≤Π1 ≤I

Π2 :0≤Π2 ≤I= kρ − τ k1 + kτ − σk1 .

(9.53)(9.54)(9.55)(9.56)The inequality follows because the operator Π maximizing kρ − σk1 in general is not thesame operator that maximizes kρ − τ k1 or kτ − σk1 .
Corollary 9.1.1 (Measurement on Close States) Suppose we have two quantum states
ρ, σ ∈ D(H) and an operator Π ∈ L(H) such that 0 ≤ Π ≤ I. Then1kρ − σk1
2≥ Tr {Πσ} − kρ − σk1 .

Tr {Πρ} ≥ Tr {Πσ} −(9.57)(9.58)

Proof. Consider the following arguments:1kρ − σk1 = max {Tr {Λ (σ − ρ)}}0≤Λ≤I
2
≥ Tr {Π (σ − ρ)}= Tr {Πσ} − Tr {Πρ} .

(9.59)(9.60)(9.61)

The first equality follows from Lemma 9.1.1. The first inequality follows because Λ is the
maximizing operator and can only lead to a probability difference greater than that foranother operator Π such that 0 ≤ Π ≤ I.
The most common way that we employ Corollary 9.1.1 in quantum Shannon theory isin the following scenario. Suppose that a measurement with operator Π succeeds with highprobability on a quantum state σ:
Tr {Πσ} ≥ 1 − ε,c 2019(9.62)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 9. DISTANCE MEASURES

Figure 9.2: The task in this figure is for Bob to distinguish the state ρAB from the state σAB using abinary-valued measurement. Bob could perform an optimal measurement on system A alone if he does
not have access to system B. If he has access to system B as well, then he can perform an optimal joint
measurement on systems A and B. We would expect that he can distinguish the states more reliably if
he performs a joint measurement because there could be more information about the state available in theother system B. Since the trace distance is a measure of distinguishability, we would expect it to obey thefollowing inequality: kρA − σA k1 ≤ kρAB − σAB k1 (the states are less distinguishable if fewer systems areavailable to be part of the distinguishability test).

where ε is some small positive number. Suppose further that another quantum state ρ is
ε-close in trace distance to σ:kρ − σk1 ≤ ε.
(9.63)
Then Corollary 9.1.1 gives the intuitive result that the measurement succeeds with highprobability on the state ρ that is close to σ:
Tr {Πρ} ≥ 1 − 2ε,(9.64)by plugging (9.62) and (9.63) into (9.58).
Exercise 9.1.8 Prove that (9.58) holds for arbitrary Hermitian operators ρ and σ by exploiting the result of Exercise 9.1.6.
We next turn to the monotonicity of trace distance under the discarding of a system. Theinterpretation of this corollary is that discarding of a system does not increase distinguishability of two quantum states. That is, a global measurement on the larger system might beable to distinguish the two states better than a local measurement on an individual subsystem could. In fact, the proof of monotonicity follows this intuition exactly, and Figure 9.2
depicts the intuition behind it.
Corollary 9.1.2 (Monotonicity of Trace Distance) Let ρAB , σAB ∈ D(HA ⊗ HB ). The
trace distance is monotone with respect to discarding of subsystems:kρA − σA k1 ≤ kρAB − σAB k1 .

(9.65)kρA − σA k1 = 2 · Tr {ΛA (ρA − σA )} ,(9.66)

Proof. Consider thatc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

9.1. TRACE DISTANCE

257

for some positive semi-definite operator ΛA ≤ IA . Then
2 · Tr {ΛA (ρA − σA )} = 2 · Tr {(ΛA ⊗ IB ) (ρAB − σAB )}≤ 2 · max Tr {ΛAB (ρAB − σAB )}0≤ΛAB ≤I

= kρAB − σAB k1 .

(9.67)(9.68)(9.69)

The first equality follows because local predictions of the quantum theory should coincidewith its global predictions (as discussed in Section 4.3.3). The inequality follows because thelocal operator ΛA never gives a higher probability difference than a maximization over all
global operators. The last equality follows from the characterization of the trace distance inLemma 9.1.1.
Exercise 9.1.9 (Monotonicity of Trace Distance) Let ρ, σ ∈ D(HA ) and N : L(HA ) →L(HB ) be a quantum channel. Show that the trace distance is monotone with respect to theaction of the channel N :kN (ρ) − N (σ)k1 ≤ kρ − σk1 .
(9.70)(Hint: Use the result of Corollary 9.1.2 and Exercise 9.1.4.)
The result of the previous exercise deserves an interpretation. It states that a quantum
channel N makes two quantum states ρ and σ less distinguishable from each other. That is,a noisy channel tends to “blur” two states to make them appear as if they are more similar
to each other than they are before the quantum channel acts.
Exercise 9.1.10 Prove that a measurement achieves the trace distance, in the following
sense:
Xkρ − σk1 = max|Tr{Λx ρ} − Tr{Λx σ}| ,(9.71){Λx }xwhere ρ, σ ∈ D(H) and the optimization is with respect to all POVMs {Λx }. Hint: Use theresult of Exercise 9.1.9 to show the following bound for any choice of POVM:
Xkρ − σk1 ≥|Tr{Λx ρ} − Tr{Λx σ}| .
(9.72)x

Next, use the developments in the proof of Lemma 9.1.1 to construct an optimal measurementthat saturates this bound. (Further hint: Consider the measurement {ΠP , ΠQ }.)Exercise 9.1.11 Show that the trace distance is strongly convex. That is, for two ensembles{pX1 (x), ρx } and {pX2 (x), σx } such that ρx , σx ∈ D(H) for all x, the following inequality holdsXx

pX1 (x)ρx −X
x

pX2 (x)σx
1

≤

X
x

c 2019|pX1 (x) − pX2 (x)| +X

pX1 (x) kρx − σx k1 . (9.73)

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9.1.6

CHAPTER 9. DISTANCE MEASURES

Channel Distinguishability and the Diamond Norm

Given the operational interpretation of trace distance in terms of the discrimination of quantum states (from Section 9.1.4), a next natural question is to understand how we can distinguish one quantum channel from another. That is, we would like to understand how close
two quantum channels are to each other in an operational sense. For this purpose, there
is a hypothesis testing scenario which extends that from Section 9.1.4. In the protocol forstate discrimination from Section 9.1.4, there were really just two steps: Bob prepares oneof two states at random and sends the state to Alice, who then performs a measurement in
an attempt to figure out which one Bob prepared.
When distinguishing channels, there is an extra degree of freedom: the channel accepts
an input quantum state which then gets transformed to an output quantum state. Thissuggests that we should allow for an extra step in a channel distinguishability scenario, in
which Alice prepares a quantum state. Let N , M : L(HA ) → L(HB ) be quantum channels.
The augmented hypothesis testing scenario consists of the following steps:
1. Alice prepares a state ρA and sends it to Bob.
2. Bob flips a fair coin and based on the outcome, he acts on ρA with either N or M.
Bob sends the output of the channel to Alice.
3. Alice then performs a measurement to figure out which channel Bob applied.
From our development in the previous section, we can immediately conclude that thesuccess probability in distinguishing the channels using such a protocol is equal to1
1
1 + kNA→B (ρA ) − MA→B (ρA )k1 .
(9.74)
2
2
However, it is clear that Alice could potentially increase the success probability by maximizing this quantity with respect to her choice of the input state. This leads to the following
expression for the success probability:


1
1
1+
max kNA→B (ρA ) − MA→B (ρA )k1 .
(9.75)
2
2 ρA ∈D(HA )This suggests that we should consider the quantitymaxρA ∈D(HA )kNA→B (ρA ) − MA→B (ρA )k1(9.76)

to be our measure of distinguishability between channels N and M. However, there is still aproblem because the protocol for distinguishing the channels is not as general as it could be.
That is, it excludes the possibility of Alice preparing an entangled state to distinguish thechannels. The most general protocol for distinguishing the channels consists of the following
steps (depicted in Figure 9.3):
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9.1. TRACE DISTANCE

259

Figure 9.3: Protocol for Alice to distinguish one channel from another (described in main text).

1. Alice prepares a state ρRA on systems R and A and sends system A to Bob. The
reference system R can have an arbitrarily large dimension.
2. Bob flips a fair coin and based on the outcome, he acts on the A system with either
N or M, which produces an output system B. Bob sends the system B to Alice.
3. Alice then performs a measurement on systems R and B to figure out which channel
Bob applied.
In such a protocol, we allow for the possibility of Alice preparing an entangled state. Itturns out that there can sometimes be a huge difference in Alice’s ability to figure out whichchannel was applied, if we allow or do not allow for entangled states to be prepared (thisdepends on the channels).
By the same reasoning as before and allowing for an optimization over all possible input
states that Alice could prepare, Alice’s success probability in distinguishing the channels is
as follows:1
1(9.77)
1 + sup max k(idRn ⊗NA→B ) (ρRn A ) − (idRn ⊗MA→B ) (ρRn A )k1 ,2
2 n ρRn A
where ρRn A ∈ D(HRn ⊗ HA ). In the above formula, n is a positive integer corresponding
to the dimension of the reference system Rn . A priori, we require a supremum over thisdimension size since we have not yet placed a bound on the dimension needed for a reference
system. The channel distance measure appearing in the above formula is known as thediamond-norm distance between the channels:
Definition 9.1.3 (Diamond-Norm Distance) Let N , M : L(HA ) → L(HB ) be quantum
channels. The diamond-norm distance is defined askN − Mk♦ ≡ sup max k(idRn ⊗NA→B ) (ρRn A ) − (idRn ⊗MA→B ) (ρRn A )k1 ,
n

ρRn A

(9.78)

where ρRn A ∈ D(HRn ⊗ HA ).
Given the above definition, a natural question is whether we can place a bound on thedimension of the reference system required. Indeed, this is possible, as the following theorem
states:c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 9. DISTANCE MEASURES

Theorem 9.1.1 Let N , M : L(HA ) → L(HB ) be quantum channels. ThenkN − Mk♦ = max k(idR ⊗NA→B ) (|ψihψ|RA ) − (idR ⊗MA→B ) (|ψihψ|RA )k1 ,|ψiRA(9.79)

where the optimization is with respect to all |ψiRA ∈ HR ⊗ HA such that k|ψiRA k2 = 1, withdim(HR ) = dim(HA ).
Proof. This theorem follows as a consequence of the convexity of the trace norm and theSchmidt
P decomposition. Indeed, let ρRn A be any density operator for systems Rn and A.
Let x pX (x)|ψ x ihψ x |Rn A be a spectral decomposition of ρRn A . From the convexity of the
trace norm, we find thatk(idRn ⊗NA→B ) (ρRn A ) − (idRn ⊗MA→B ) (ρRn A )k1
X
pX (x) [(idRn ⊗NA→B ) (|ψ x ihψ x |Rn A ) − (idRn ⊗MA→B ) (|ψ x ihψ x |Rn A )]
=
x

≤

X

(9.80)
1

x

x

x

x

pX (x) k[(idRn ⊗NA→B ) (|ψ ihψ |Rn A ) − (idRn ⊗MA→B ) (|ψ ihψ |Rn A )]k1

(9.81)x≤ k[(idRn ⊗NA→B ) (|ψ∗x ihψ∗x |Rn A ) − (idRn ⊗MA→B ) (|ψ∗x ihψ∗x |Rn A )]k1 ,(9.82)

where the last inequality follows because the average value is never larger than the maximum
value and we let |ψ∗x iRn A denote the state vector giving the maximum value. From the
Schmidt decomposition theorem (Theorem 3.8.1), the Schmidt rank of |ψ∗x iRn A is no larger
than dim(HA ), implying that |ψ∗x iRn A can be embedded in a tensor-product Hilbert space
HR ⊗ HA such that dim(HR ) = dim(HA ). Since this bound holds for any density operator
ρRn A , the statement of the theorem follows.
As a consequence of the above theorem, we can take the result in (9.79) to be thedefinition of the diamond-norm distance. The main use of the diamond-norm distance isfor comparing quantum channels. For example, when studying the classical capacity of aquantum channel (Chapter 20), we would like to compare how well a given protocol simulatesa noiseless classical channel, and the diamond-norm distance gives a natural way to do so.
The situation is similar with the quantum capacity theorem (Chapter 24): here we would liketo compare how well a protocol simulates a noiseless quantum channel, and we can quantify
the performance using the diamond-norm distance.
Exercise 9.1.12 Suppose that Alice is restricted to use separable states on systems Rn and
A to distinguish two quantum channels N and M. Show that the success probability in doing
so is given by
1
1
1 + max kNA→B (|ψihψ|A ) − MA→B (|ψihψ|A )k1 ,(9.83)
2
2 |ψiA
where |ψihψ|A ∈ D(HA ).
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9.2. FIDELITY

261

Exercise 9.1.13 Suppose that channels N and M are defined as follows:
N (XA ) = Tr{XA }ρB ,M(XA ) = Tr{XA }σB ,(9.84)

where XA ∈ L(HA ) and ρB , σB ∈ D(HB ). Show thatkN − Mk♦ = kρB − σB k1 .

(9.85)

9.2 Fidelity
9.2.1

Pure-State Fidelity

An alternate measure of the closeness of two quantum states is the fidelity. We introduce
its most simple form first. Suppose that we input a particular pure state |ψi to a quantum
information-processing protocol. Ideally, we may want the protocol to output the same statethat is input, but suppose that it instead outputs a pure state |φi. The pure-state fidelity
F (ψ, φ) is a measure of how close the output state is to the input state.
Definition 9.2.1 (Pure-State Fidelity) Let |ψi, |φi ∈ H be pure states. The pure-state
fidelity is the squared overlap of the states |ψi and |φi:F (ψ, φ) ≡ |hψ|φi|2 .

(9.86)The pure-state fidelity has the operational interpretation as the probability that theoutput state |φi would pass a test for being the same as the input state |ψi, conducted bysomeone who knows the input state (see Exercise 9.2.2).
The pure-state fidelity is symmetric F (ψ, φ) = F (φ, ψ), and it obeys the following bounds:
0 ≤ F (ψ, φ) ≤ 1.

(9.87)

It is equal to one if and only if the two states are the same, and it is equal to zero if andonly if the two states are orthogonal to each other. The fidelity measure is not a distancemeasure in the strict mathematical sense because it is equal to one when two states areequal, whereas a distance measure should be equal to zero when two states are equal.
Exercise 9.2.1 Suppose that two pure quantum states |ψi, |φi ∈ H are as follows:
Xp
Xpp(x)|xi,|φi ≡q(x)|xi,|ψi ≡(9.88)

xx

where {|xi} is some orthonormal basis for H. Show that the fidelity F (ψ, φ) between these twostates is equivalent to the Bhattacharyya overlap (classical fidelity) between the distributions
p(x) and q(x):"#2Xp
F (ψ, φ) =p(x)q(x) .
(9.89)x
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9.2.2

CHAPTER 9. DISTANCE MEASURES

Expected Fidelity

Now let us suppose that the output of a given protocol is not a pure state, but it is rather amixed state with density operator ρ. In general, a quantum information-processing protocol
could be noisy and map the pure input state |ψi to a mixed state. We would like a way tocompare these two states.
Definition 9.2.2 (Expected Fidelity) The expected fidelity F (ψ, ρ) between a pure state|ψi ∈ H and a mixed state ρ ∈ D(H) is
F (ψ, ρ) ≡ hψ|ρ|ψi.

(9.90)

We now justify the abovePdefinition of fidelity. Let us decompose ρ according to aspectral decomposition ρ =x pX (x)|φx ihφx |. Recall that we can think of this output
density operator as arising from the ensemble {pX (x), |φx i}. We generalize the pure-state
fidelity from the previous paragraph by defining it as the expected pure-state fidelity, where
the expectation is with respect to states in the ensemble:
F (ψ, ρ) ≡ EX |hψ|φX i|2(9.91)X=
pX (x) |hψ|φx i|2(9.92)x=

X

pX (x) hψ|φx i hφx |ψi(9.93)

x

!
= hψ|

X

pX (x)|φx ihφx | |ψi(9.94)x= hψ|ρ|ψi.

(9.95)

The compact formula F (ψ, ρ) = hψ|ρ|ψi is a good way to characterize the fidelity when the
input state is pure and the output state is mixed. We can see that the above fidelity measure
is a generalization of the pure-state fidelity in (9.86). It obeys the same bounds:
0 ≤ F (ψ, ρ) ≤ 1,

(9.96)

being equal to one if and only if the state ρ is equal to |ψihψ| and equal to zero if and only
if the support of ρ is orthogonal to |ψihψ|.
Exercise 9.2.2 Given a state σ ∈ D(H), we would like to see if it would pass a test forbeing close to a pure state |ϕi ∈ H. We can measure the POVM {|ϕihϕ|, I − |ϕihϕ|} withresult ϕ corresponding to a “pass” and the result I − ϕ corresponding to a “fail.” Show thatthe fidelity is then equal to Pr{“pass”}.
Exercise 9.2.3 Using the result of Corollary 9.1.1, show that the following inequality holds
for a pure state |φi ∈ H and mixed states ρ, σ ∈ D(H):F (φ, ρ) ≤ F (φ, σ) + 12 kρ − σk1 .
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9.2. FIDELITY

9.2.3

263

Uhlmann Fidelity

What is the most general form of the fidelity when both quantum states are mixed? We canborrow the above idea of the pure-state fidelity that exploits the overlap between two purestates. Suppose that we would like to determine the fidelity between two mixed states ρAand σA that represent different states of some quantum system A. Let |φρ iRA and |φσ iRAdenote particular respective purifications of the mixed states to some reference system R(where for now we assume that the reference system has the same dimension as the system
A). We can define the Uhlmann fidelity F (ρA , σA ) between two mixed states ρA and σA as
the maximum overlap between their respective purifications, where the maximization is withrespect to all purifications |φρ iRA and |φσ iRA of the respective states ρA and σA :F (ρA , σA ) ≡max|φρ iRA , |φσ iRA|hφρ |φσ iRA |2 .

(9.98)

We can express the fidelity as a maximization over unitaries instead (recall the result ofTheorem 5.1.1 that all purifications are equivalent up to unitaries on the reference system):


2
ρ †
ρ
σ
σ
F (ρA , σA ) = max
hφ
|
(U
)
⊗
I
(U
⊗
I
)
|φ
i
RA
A
A
R
R
RA
ρ
σ
U ,U

(9.99)

2= max
hφρ |RA (URρ )† URσ ⊗ IA |φσ iRA .
ρ
σ
U ,U

(9.100)

It is unnecessary to maximize over two sets of unitaries because the product (URρ )† URσ represents only a single unitary. The final expression for the fidelity between two mixed statesis then defined as the Uhlmann fidelity.
Definition 9.2.3 (Uhlmann Fidelity) The Uhlmann fidelity F (ρA , σA ) between two mixed
states ρA and σA is the maximum overlap between their respective purifications, where themaximization is with respect to all unitaries U acting on the purification system R:
F (ρA , σA ) = max |hφρ |RA UR ⊗ IA |φσ iRA |2 .
U

(9.101)

We will find that this notion of fidelity generalizes both the pure-state fidelity in (9.86)
and the expected fidelity in (9.95). This holds because the following formula for the fidelityof two mixed states, characterized in terms of the Schatten 1-norm, is equivalent to the above
Uhlmann characterization:√ √2(9.102)F (ρA , σA ) = k ρA σA k1 .
We state this result as Uhlmann’s theorem.
Theorem 9.2.1 (Uhlmann’s Theorem) The following two expressions for fidelity are
equal:√ √
2
F (ρA , σA ) = max |hφρ |RA UR ⊗ IA |φσ iRA |2 = k ρA σA k1 .
(9.103)U

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264

CHAPTER 9. DISTANCE MEASURESProof. Let |φρ iRA denote the canonical purification of ρA (see Exercise 5.1.2):|φρ iRA ≡ (IR ⊗√ρA ) |ΓiRA ,(9.104)

where |ΓiRA is the unnormalized maximally entangled vector:|ΓiRA ≡

X

|iiR |iiA .

(9.105)i

Therefore, the state |φρ iRA is a particular purification of ρ. Let |φσ iRA denote the canonicalpurification of σA :√(9.106)|φσ iRA ≡ (IR ⊗ σA ) |ΓiRA .
Consider that the overlap |hφρ |UR ⊗ IA |φσ i|2 is as follows:√√2ρA ) (IR ⊗ σA ) |ΓiRA |√ √2= |hΓ|RA (UR ⊗ ρA σA ) |ΓiRA |√ √2= hΓ|RA IR ⊗ ρA σA UAT |ΓiRA√ √2
ρA σA UAT.
= Tr

|hφρ |UR ⊗ IA |φσ i|2 = |hΓ|RA (UR ⊗(9.107)(9.108)(9.109)(9.110)The first equality follows by plugging in (9.104) and (9.106). The third equality follows
from Exercise 3.7.12. The last equality follows from Exercise 4.1.3. We can finally invoke
Property 9.1.6 to establish thatmax TrUA√ √ρA σA UAT2

√ √2
= k ρ A σ A k1 ,(9.111)from which (9.103) follows.
√ √2
Exercise 9.2.4 Use the expression
ρA σA 1 for the fidelity and the Cauchy–Schwarz
inequality for the Hilbert–Schmidt inner product (from (9.20)) to prove that the quantum
fidelity between two density operators never exceeds one.
Remark 9.2.1 Note that we can define the fidelity function more generally for any two
positive semi-definite operators, which can sometimes be useful. That is, let P and Q be
positive semi-definite operators acting on the same Hilbert space. Then we define
F (P, Q) ≡

√ p 2
P Q .

(9.112)

1

By applying the Cauchy–Schwarz inequality for the Hilbert–Schmidt inner product (see (9.20))
and the characterization of the trace norm in Property 9.1.6, we find that
F (P, Q) ≤ Tr{P } Tr{Q}.
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9.2. FIDELITY

265

Remark 9.2.2 Note that in the development above, we assumed that the dimension of thereference system R is equal to that of the system A. However, this is the not the most generaldefinition that we could have taken. We could have defined fidelity as
F (ρA , σA ) = sup

max

ρ
σ
dim(HR ) |φ iRA , |φ iRA|hφρ |RA |φσ iRA |2 ,(9.114)in which there is an extra optimization over the dimension of the reference system R in
addition to the optimization over the purifications. However, repeating an analysis similar
to the above one would lead us to the conclusion that√ √2
F (ρA , σA ) = k ρA σA k1(9.115)for this definition as well. Indeed, we have that|hφρ |RA |φσ iRA |2 =

h

i
2
√
√
hΓ|R0 A ρA (VR0 →R )† [UR0 →R σA |ΓiR0 A ] ,(9.116)

where R0 is a reference system with dimension equal to dim(HA ) and UR0 →R and VR0 →R aresome isometries , given that all purifications are related by an isometry acting on the reference
system (Theorem 5.1.1). Carrying through the same analysis along with the characterization√ √ρA σA 1 , so thatof the trace norm in Exercise 9.1.6 then gives that |hφρ |RA |φσ iRA | ≤having an arbitrarily large reference system does not help.

9.2.4

Properties of Fidelity

We discuss some further properties of the fidelity that often prove useful. Some of theseproperties are the counterpart of similar properties of the trace distance. From the characterization of fidelity in (9.103), we observe that it is symmetric in its arguments:
F (ρ, σ) = F (σ, ρ).

(9.117)0 ≤ F (ρ, σ) ≤ 1.

(9.118)It obeys the following bounds:
The lower bound applies if and only if the respective supports of the two states ρ and σ
are orthogonal. √To see this, suppose that the√supports of ρ and σ are orthogonal. This√√implies that ρ σ = 0, so that F (ρ, σ) = k ρ σk21 = 0.√ On the other hand, suppose that√
F (ρ, σ) = 0. Then by definition, this means √that k ρ σk21 = 0, and from non-negative√
definiteness of the trace norm, we find that ρ σ = 0. This then implies that the supportsof ρ and σ are orthogonal. The upper bound in (9.118) applies if and only if the two statesρ and σ are equal to each other.
Exercise 9.2.5 Show that the definition of fidelity in (9.102) reduces to (9.86) when the twostates are pure and to (9.90) when one state is pure and the other is mixed.
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CHAPTER 9. DISTANCE MEASURES

Property 9.2.1 (Multiplicativity) Let ρ1 , σ1 ∈ D(H1 ) and ρ2 , σ2 ∈ D(H2 ). The fidelity
is multiplicative with respect to tensor products:
F (ρ1 ⊗ ρ2 , σ1 ⊗ σ2 ) = F (ρ1 , σ1 )F (ρ2 , σ2 ).

(9.119)This result holds by employing the definition of the fidelity in (9.102).
The following monotonicity lemma is similar to the monotonicity lemma for trace distance(Lemma 9.1.2) and also bears the similar interpretation that quantum states become moresimilar (less distinguishable) under the discarding of subsystems.
Lemma 9.2.1 (Monotonicity) Let ρAB , σAB ∈ D(HA ⊗HB ). The fidelity is non-decreasing
with respect to partial trace:
F (ρAB , σAB ) ≤ F (ρA , σA ),(9.120)whereρA = TrB {ρAB },σA = TrB {σAB }.

(9.121)

Proof. Consider a fixed purification |ψiRAB of ρA and ρAB and a fixed purification |φiRABof σA and σAB . Then|hψ|RAB UR ⊗ IA ⊗ IB |φiRAB |2 ≤ max |hψ|RAB URB ⊗ IA |φiRAB |2
URB= F (ρA , σA ),(9.122)(9.123)where the first inequality follows because the maximization over unitaries URB includes
UR ⊗ IA and the equality is a consequence of Uhlmann’s theorem. Given that the inequality
holds for all unitaries UR , we can conclude that
F (ρAB , σAB ) = max |hψ|RAB UR ⊗ IA ⊗ IB |φiRAB |2 ≤ F (ρA , σA ),UR(9.124)

where the equality is again a consequence of Uhlmann’s theorem.
Property 9.2.2 (Joint Concavity) Let ρx , σx ∈ D(H) for all x and let pX be a probability
distribution. The root fidelity is jointly concave with respect to its input arguments:
!
X
X
X
√
√
F
pX (x)ρx ,pX (x)σx ≥pX (x) F (ρx , σx ).
(9.125)xxx

Proof. We prove joint concavity by exploiting the result of Exercise 5.1.4. Suppose |φρx iRA
and |φσx iRA are respective Uhlmann purifications of ρx and σx (these are purifications thatmaximize the Uhlmann fidelity), so that√√xx
hφρx |φσx iRA = F (φρRA, φσRA) = F (ρx , σx ).
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9.2. FIDELITY

267

Choose some orthonormal basis {|xiX }. Then
Xp
Xp
|φρ i ≡pX (x)|φρx iRA |xiX ,|φσ i ≡pX (x)|φσx iRA |xiX
x(9.127)x

are respective purifications ofby Uhlmann’s theorem:

P

x pX (x)ρx andP

x pX (x)σx . The first inequality below holds

!

√
F

X
x

pX (x)ρx ,X

pX (x)σx≥ |hφρ |φσ i|(9.128)X

pX (x)hφρx |φσx i(9.129)√pX (x) F (ρx , σx ) ,(9.130)

x

=

x

=

X
x

concluding the proof.
Property 9.2.3 (Concavity) Let ρ, σ, τ ∈ D(H) and λ ∈ [0, 1]. The fidelity is concavewith respect to one of its arguments:
F (λρ + (1 − λ) τ, σ) ≥ λF (ρ, σ) + (1 − λ) F (τ, σ).

(9.131)

Proof. Let |ψ σ iRS be a fixed purification of σS . Let |ψ ρ iRS be a purification of ρS such that|hψ σ |ψ ρ i|2 = F (ρ, σ).

(9.132)

Similarly, let |ψ τ iRS be a purification of τS such that|hψ σ |ψ τ i|2 = F (τ, σ).

(9.133)

Then consider that
λF (ρ, σ) + (1 − λ) F (τ, σ)= λ |hψ σ |ψ ρ i|2 + (1 − λ) |hψ σ |ψ τ i|2= λ hψ σ |ψ ρ i hψ ρ |ψ σ i + (1 − λ) hψ σ |ψ τ i hψ τ |ψ σ i= hψ σ |RS (λ|ψ ρ ihψ ρ |RS + (1 − λ) |ψ τ ihψ τ |RS ) |ψ σ iRS= F (|ψ σ ihψ σ |RS , λ|ψ ρ ihψ ρ |RS + (1 − λ) |ψ τ ihψ τ |RS )≤ F (ψSσ , λψSρ + (1 − λ) ψSτ )= F (λρ + (1 − λ) τ, σ).

(9.134)(9.135)(9.136)(9.137)(9.138)(9.139)The first step is a rewriting using (9.132) and (9.133). The fourth equality is a consequenceof Exercise 9.2.5. The inequality follows from monotonicity of the fidelity with respect topartial trace (Lemma 9.2.1).
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268

CHAPTER 9. DISTANCE MEASURES

Exercise 9.2.6 Let ρ, σ ∈ D(H). Show that we can express the root fidelity as
np
o
np
o
√
F (ρ, σ) = Tr
ρ1/2 σρ1/2 = Tr
σ 1/2 ρσ 1/2 ,(9.140)using the definition in (9.102).
Exercise 9.2.7 Let ρ, σ ∈ D(H). Show that the fidelity is invariant with respect to an
isometry U ∈ L(H, H0 ):F (ρ, σ) = F (U ρU † , U σU † ).
(9.141)
Exercise 9.2.8 Let ρ, σ ∈ D(HA ) and let N : L(HA ) → L(HB ) be a quantum channel.
Show that the fidelity is monotone with respect to the channel N :F (ρ, σ) ≤ F (N (ρ), N (σ)).

(9.142)

Exercise 9.2.9 Suppose that Alice uses a noisy quantum channel and a sequence of quantum
channels to generate the following state, shared with Bob and Eve:
1 X√
|miA |miB1 |φm iB2 E ,(9.143)M mwhere {|miA } and {|miB1 } are orthonormal bases and {|φm iB2 E } is a set of states. Alicepossesses the system A, Bob possesses systems B1 and B2 , and Eve possesses the system E.
Let φm
E denote the partial trace of |φm iB2 E over Bob’s system B2 so thatφm
E ≡ TrB2 {|φm ihφm |B2 E } .

(9.144)

Suppose further that F (φmE , θE ) = 1, where θE is some constant density operator (independentof m) for Eve’s system E. Determine a unitary that Bob can perform on his systems B1
and B2 so that he decouples Eve’s system E, in the sense that the state after the decoupling
unitary is as follows:!
1 X√|miA |miB1 ⊗ |φθ iB2 E ,(9.145)M mwhere |φθ iB2 E is a purification of the state θE . The result is that Alice and Bob share maximal
entanglement between the respective systems A and B1 after Bob performs the decoupling
unitary. Figure 9.4 displays the protocol.
Exercise 9.2.10 (Fidelity for Classical–Quantum States) Show that the root fidelity
possesses the following property:
Xp√√
F (ωXB , τXB ) =p(x)q(x) F (ωx , τx ) ,(9.146)
x

where
ωXB ≡

X
x

p(x)|xihx|X ⊗ ωx ,

τXB ≡

X

q(x)|xihx|X ⊗ τx ,(9.147)x

p and q are probability distributions, {|xi} is some orthonormal basis, and ωx , τx ∈ D(H)for all x.

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9.2. FIDELITY

269

Figure 9.4: This figure depicts the protocol relevant to Exercise 9.2.9. Alice transmits one share of anentangled state through a quantum channel with isometric extension channel U. Bob and Eve receive
quantum systems as the output of the isometry. Bob performs some quantum operations so that Alice, Bob,
and Eve share the state in (9.143). Exercise 9.2.9 asks you to determine a decoupling unitary that Bob canperform to decouple his system B1 from Eve.

9.2.5

A Measurement Achieves the Fidelity

There is a classical notion of fidelity for probability distributions, which is sometimes called
the classical fidelity or Bhattacharyya overlap. It is defined as follows:
Definition 9.2.4 (Classical Fidelity) Let p and q be probability distributions defined overa finite alphabet X . The classical fidelity F (p, q) is defined as follows:#2"
Xp
F (p, q) ≡p(x)q(x).

(9.148)x∈X

Exercise 9.2.11 Verify that the classical fidelity is a special case of the quantum fidelity.
That is, let p and q be probability distributions defined over a finite alphabet X , and thenplace the entries of these distributions along the diagonal of commuting matrices ρ and σ,
respectively. Show that F (p, q) = F (ρ, σ).
Now suppose that we have two density operators ρ, σ ∈ D(H), and suppose further that
we perform a POVM {Λx } on these states, leading to the following probability distributions:p(x) = Tr{Λx ρ},

q(x) = Tr{Λx σ}.

(9.149)We can then compute the classical fidelity of the distributions for the measurement outcomesby using the formula in (9.148), and this is a measure of distinguishability of the two quantumc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 9. DISTANCE MEASURES

states, with respect to a particular measurement. From the monotonicity of the quantum
fidelity with respect to quantum channels (Exercise 9.2.8), it follows that the quantum fidelity
F (ρ, σ) never exceeds this classical fidelity"#2Xp
F (ρ, σ) ≤Tr{Λx ρ} Tr{Λx σ} .
(9.150)x∈X

In particular, this bound follows from Exercise
P9.2.8, where the channel here is understood tobe a measurement channel of the form ω → x Tr{Λx ω}|xihx| and then we apply the resultof Exercise 9.2.11. What is perhaps surprising is that there always exists a measurement thatsaturates the bound above, leading to the following alternate characterization of fidelity:
Theorem 9.2.2 (Measurement Achieves Fidelity) Let ρ, σ ∈ D(H). Then#2"
Xp
Tr{Λx ρ} Tr{Λx σ} ,F (ρ, σ) = min{Λx }(9.151)x∈X

where the minimization is with respect to all POVMs.
Proof. As justified before the statement of the theorem, the bound in (9.150) holds for any
POVM. So here we construct a specific POVM (known as the Fuchs-Caves measurement)that saturates the bound. First consider the case in which σ is positive definite (and thus
invertible). Consider the following operator (known as an operator geometric mean of ρ andσ −1 ):

1/2 −1/2
M = σ −1/2 σ 1/2 ρσ 1/2
σ.
(9.152)The operator M is positive semi-definite, and thus has a spectral decomposition:XM=
λy |yihy|,(9.153)ywith {λy } a set of non-negative eigenvalues and {|yi} a corresponding set of eigenvectors.
We will prove that the optimal measurement in (9.151) is {|yihy|}. We begin by notingthat a simple calculation gives
M σM = ρ.
(9.154)
So now consider the classical fidelity of the measurement {|yihy|}:
Xp
Xp
Tr{|yihy|ρ} Tr{|yihy|σ} =hy|ρ|yi hy|σ|yi(9.155)yy

Xp=hy|M σM |yi hy|σ|yi(9.156)yXq=hy|λy σλy |yi hy|σ|yi(9.157)y=X

λy hy|σ|yi.

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9.3. RELATIONS BETWEEN TRACE DISTANCE AND FIDELITY

271

The second equality follows from (9.154). The third equality follows because M |yi = λy |yi.
Continuing, the last line above is equal to()
nX1/2 o √= F (ρ, σ).
Tr
λy |yihy|σ = Tr {M σ} = Tr σ 1/2 ρσ 1/2(9.159)y

The last equality follows from Exercise 9.2.6.
For the case in which σ is not invertible, we repeat the above analysis, replacing ρ withΠσ ρΠσ , where Πσ is the projection onto the support of σ. In this case, the geometric mean
operator M has its support contained in the support of σ, and one can find a spectral
decomposition of M as in (9.153) so thatXp√
F (Πσ ρΠσ , σ) =Tr{|yihy|Πσ ρΠσ } Tr{|yihy|σ}.
(9.160)y

Since the eigenvectors {|yi} do not necessarily span the whole space, we can add additional
orthonormal vectors all orthogonal to those in {|yi}, such that all of them taken together
form a legitimate measurement. Since both Πσ ρΠσ and σ are orthogonal to all of the newvectors, the probabilities for these measurement outcomes are all equal to zero and thus they
do not contribute anything to the sum in (9.160). Finally, we have that
F (Πσ ρΠσ , σ) = F (ρ, σ)because σ 1/2 = Πσ σ 1/2 = σ 1/2 Πσ , so that
np
o
np
o √
√
F (ρ, σ) = Tr
σ 1/2 ρσ 1/2 = Tr
σ 1/2 Πσ ρΠσ σ 1/2 = F (Πσ ρΠσ , σ),(9.161)(9.162)concluding the proof.

9.3 Relations between Trace Distance and Fidelity
In quantum Shannon theory, we are interested in showing that a given quantum informationprocessing protocol approximates an ideal protocol. We might do so by showing that thequantum output of the ideal protocol, say ρ, is close to the quantum output of the actual
protocol, say σ. For example, we may be able to show that the fidelity between ρ and σ ishigh:
F (ρ, σ) ≥ 1 − ε,(9.163)where ε is a small, positive real number that determines how well ρ approximates σ according
to the above fidelity criterion. Typically, in a quantum Shannon-theoretic argument, we willtake a limit to show that it is possible to make ε as small as we would like. As the performanceparameter ε becomes vanishingly small, we expect that ρ and σ are becoming approximately
equal so that they are identically equal when ε vanishes in some limit.
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272

CHAPTER 9. DISTANCE MEASURESWe would naturally think that the trace distance should be small if the fidelity is highbecause the trace distance vanishes when the fidelity is one and vice versa (recall the conditions for saturation of the bounds in (9.22) and (9.118)). The next theorem makes thisintuition precise by establishing several relationships between the trace distance and fidelity.
Theorem 9.3.1 (Relations between Fidelity and Trace Distance) The following boundapplies to the trace distance and the fidelity between two quantum states ρ, σ ∈ D(H):
1−

p
p
1
F (ρ, σ) ≤ kρ − σk1 ≤ 1 − F (ρ, σ).
2(9.164)

Proof. We first show that there is an exact relationship between fidelity and trace distancefor pure states. Let us pick two arbitrary pure states |ψi, |φi ∈ H. We can write the state|φi in terms of the state |ψi and a vector |ψ ⊥ i orthogonal to |ψi:|φi = cos(θ)|ψi + sin(θ)|ψ ⊥ i.

(9.165)First, the fidelity between these two pure states is
F (ψ, φ) = |hφ|ψi|2 = cos2 (θ).

(9.166)

Now let us determine the trace distance. The density operator |φihφ| is as follows:
|φihφ| = cos(θ)|ψi + sin(θ)|ψ ⊥ i cos(θ)hψ| + sin(θ)hψ ⊥ |(9.167)= cos2 (θ)|ψihψ| + sin(θ) cos(θ)|ψ ⊥ ihψ|+ cos(θ) sin(θ)|ψihψ ⊥ | + sin2 (θ)|ψ ⊥ ihψ ⊥ |.

(9.168)The matrix representation of the operator |ψihψ|−|φihφ| with respect to the basis |ψi, |ψ ⊥ iis
1 − cos2 (θ)− sin(θ) cos(θ).
(9.169)− sin(θ) cos(θ)− sin2 (θ)
It is straightforward to show that the eigenvalues of the above matrix are |sin(θ)| and− |sin(θ)| and it then follows that the trace distance between |ψi and |φi is the absolute
sum of the eigenvalues:k|ψihψ| − |φihφ|k1 = 2 |sin(θ)| .
(9.170)Consider the following trigonometric relationship:2
2 |sin(θ)|= 1 − cos2 (θ).
2

(9.171)

Applying it gives the following relation between the fidelity and trace distance for pure states:
2
1
k|ψihψ| − |φihφ|k1 = 1 − F (ψ, φ),(9.172)2
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9.3. RELATIONS BETWEEN TRACE DISTANCE AND FIDELITY

273

by plugging (9.166) into the right-hand side of (9.171) and (9.170) into the left-hand side of(9.171). Thus,p1k|ψihψ| − |φihφ|k1 = 1 − F (ψ, φ).
(9.173)
2
To prove the upper bound for mixed states ρA and σA , choose purifications |φρ iRA and |φσ iRAof respective states ρA and σA such that
F (ρA , σA ) = |hφσ |φρ i|2 = F (φρRA , φσRA ).

(9.174)(Recall that these purifications exist by Uhlmann’s theorem.) Then11kρA − σA k1 ≤ kφρRA − φσRA k122
q

1 − F (φρRA , φσRA )p= 1 − F (ρA , σA ),=(9.175)(9.176)(9.177)

where the first inequality follows by the monotonicity of the trace distance under the discarding of systems (Lemma 9.1.2).
To prove the lower bound for mixed states ρ and σ, recall Exercise 9.1.10 and Theorem 9.2.2. Exercise 9.1.10 states that the trace distance is the maximum classical trace
distance between two probability distributions resulting from a POVM {Λm } acting on thestates ρ and σ:Xkρ − σk1 = max|pm − qm | ,(9.178)
{Λm }

m

where
pm ≡ Tr {Λm ρ} ,qm ≡ Tr {Λm σ} .

(9.179)Furthermore, Theorem 9.2.2 states that the quantum fidelity is the minimum classical fidelity
0
between two probability distributions p0m and qm
resulting from a measurement {Γm } of thestates ρ and σ:
!2
Xp
0
,
(9.180)p0m qmF (ρ, σ) = min
{Γm }

m

where
p0m ≡ Tr {Γm ρ} ,0qm
≡ Tr {Γm σ} .

(9.181)

We return to the proof. Suppose that the POVM {Γm } achieves the minimum classical0
fidelity and results in probability distributions p0m and qm, so that!2
F (ρ, σ) =Xp0p0m qm

.

(9.182)m
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274

CHAPTER 9. DISTANCE MEASURES

Consider thatX pp 2 X 0p0000pm − qm=
pm + qm− 2 p0m qm
m(9.183)mp= 2 − 2 F (ρ, σ).
It also follows thatX pp 2 X p
p
p
p
0
0
0
0
pm − qm
p0m − qm
p0m + qm≤m

(9.184)(9.185)

m

=

X

0
|
|p0m − qm(9.186)|pm − qm |(9.187)m≤

Xm= kρ − σk1 .
(9.188)
√ 0
√ 0
√ 0
√ 0
The first inequality holds because | pm − qm | ≤ | pm + qm |. The second inequality
0minimizing the classical fidelity in general haveholds because the distributions p0m and qm
classical trace distance less than the distributions pm and qm that maximize the classicaltrace distance. Thus, the following inequality resultsp
2 − 2 F (ρ, σ) ≤ kρ − σk1 ,(9.189)and the lower bound in the statement of the theorem follows.
Theorem 9.3.1 allows us to complete our understanding of the extreme values of trace
distance and fidelity. We have already argued that two states ρ, σ ∈ D(H) have trace distanceequal to zero if and only if ρ = σ. Theorem 9.3.1 allows us to conclude that F (ρ, σ) = 1 if andonly if ρ = σ. Similarly, we have argued already that F (ρ, σ) = 0 if and only if the supportof ρ is orthogonal to that of σ. Theorem 9.3.1 allows us to conclude that kρ − σk1 = 2 if
and only if the support of ρ is orthogonal to that of σ.
The following two corollaries are simple consequences of Theorem 9.3.1.
Corollary 9.3.1 Let ρ, σ ∈ D(H) and fix ε ∈ [0, 1]. Suppose that ρ is ε-close to σ in trace
distance:kρ − σk1 ≤ ε.
(9.190)
Then the fidelity between ρ and σ is greater than 1 − ε:F (ρ, σ) ≥ 1 − ε.

(9.191)Corollary 9.3.2 Let ρ, σ ∈ D(H) and fix ε ∈ [0, 1]. Suppose the fidelity between ρ and σ isgreater than 1 − ε:F (ρ, σ) ≥ 1 − ε.
(9.192)√
Then ρ is 2 ε-close to σ in trace distance:√kρ − σk1 ≤ 2 ε.
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275

Exercise 9.3.1 Let ρ, σ ∈ D(H). Prove the following lower bound on the probability of error
pe in a quantum hypothesis test to distinguish ρ from σ:p1
1 − 1 − F (ρ, σ) .
pe ≥2

(9.194)(Hint: Recall the development in Section 9.1.4.)

9.4 Gentle Measurement
The gentle measurement and gentle operator lemmas are particular applications of Theorem 9.3.1, and they concern the disturbance of quantum states. We generally expect inquantum theory that certain measurements might disturb the state which we are measuring.
For example, suppose a qubit is in the state |0i. A measurement along the X direction gives+1 and −1 with equal probability while drastically disturbing the state to become either |+i
or |−i, respectively. On the other hand, we might expect that the measurement does not
disturb the state by very much if one outcome is highly likely. For example, suppose that we
instead measure the qubit along the Z direction. The measurement returns + 1 with unit
probability while causing no disturbance to the qubit. The “gentle measurement lemma”
below quantitatively addresses the disturbance of quantum states by demonstrating that ameasurement with one outcome that is highly likely causes only a little disturbance to thequantum state that we measure (hence, the measurement is “gentle” or “tender”).
Lemma 9.4.1 (Gentle Measurement) Consider a density operator ρ and a measurement
operator Λ where 0 ≤ Λ ≤ I. The measurement operator could be an element of a POVM.
Suppose that the measurement operator Λ has a high probability of detecting state ρ:Tr {Λρ} ≥ 1 − ε,(9.195)where ε ∈ [0, 1] (the probability of detection is high if ε is close to zero). Then the postmeasurement state
√ √
Λρ Λ
0
(9.196)ρ ≡Tr {Λρ}√is 2 ε-close to the original state ρ in trace distance:√kρ − ρ0 k1 ≤ 2 ε.
(9.197)
Thus, the measurement does not disturb the state ρ by much if ε is small.
Proof. Suppose first that ρ is a pure state |ψihψ|. The post-measurement state is then√√Λ|ψihψ| Λ.
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276

CHAPTER 9. DISTANCE MEASURESThe fidelity between the original state |ψi and the post-measurement state above is as follows:
hψ|

2
√
√
√ !
hψ| Λ|ψi|hψ|Λ|ψi|2
Λ|ψihψ| Λ|ψi =≥hψ|Λ|ψihψ|Λ|ψihψ|Λ|ψi= hψ|Λ|ψi ≥ 1 − ε.

(9.199)(9.200)√The first inequality follows because Λ ≥ Λ when Λ ≤ I. The second inequality follows
from the hypothesis of the lemma. Now let us consider when we have mixed states ρA andρ0A . Suppose |ψiRA and |ψ 0 iRA are respective purifications of ρA and ρ0A , where√ΛA |ψiRAI⊗
R|ψ 0 iRA ≡ p.
(9.201)hψ|IR ⊗ ΛA |ψiRA
Then we can apply monotonicity of fidelity (Lemma 9.2.1) and the above result for purestates to show that0
F (ρA , ρ0A ) ≥ F (ψRA , ψRA) ≥ 1 − ε.
(9.202)
We obtain the bound on the trace distance kρA − ρ0A k1 by exploiting Corollary 9.3.2.
The following is a variation on the gentle measurement lemma:
Lemma 9.4.2 (Gentle Operator) Consider a density operator ρ and a measurement operator Λ where 0 ≤ Λ ≤ I. The measurement operator could be an element of a POVM.
Suppose that the measurement operator Λ has a high probability of detecting state ρ:Tr {Λρ} ≥ 1 − ε,where ε ∈ [0, 1] (the probability is high if ε is close to zero). Thenthe original state ρ in trace distance:√ √√ρ − Λρ Λ ≤ 2 ε.

(9.203)√√√Λρ Λ is 2 ε-close to(9.204)

1

Proof. Consider the following chain of inequalities:√ √ρ − Λρ Λ
1

√
√ 
√ √
= I − Λ + Λ ρ − Λρ Λ
1

√ 
√ 
√ 
≤ I− Λ ρ +Λρ I − Λ
1
1


√ 
√ √ √
√ √ √ = Tr I − Λ
ρ · ρ + Tr Λ ρ · ρ I − Λ
s
s 

 


√ 2
√ 2
≤ Tr
I − Λ ρ Tr {ρ} + Tr {Λρ} Tr ρ I − Λpp
Tr {(I − Λ) ρ} + Tr {ρ (I − Λ)}p√= 2 Tr {(I − Λ) ρ} ≤ 2 ε.
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9.4. GENTLE MEASUREMENT

277

The first inequality is a consequence of the triangle inequality. The second equality follows
from the definition of the trace norm and the fact that ρ is a positive semi-definite operator.
The second inequality follows from the Cauchy–Schwarz inequality for the√ Hilbert–Schmidt
2
inner product (see (9.20)). The third inequality follows because (1 − x) ≤ 1 − x for0 ≤ x ≤ 1, Tr {ρ} = 1, and Tr {Λρ} ≤ 1. The final inequality follows from applying (9.203)
and because the square root function is monotone increasing.
Exercise 9.4.1 Show that the gentle operator lemma holds for subnormalized positive semidefinite operators ρ (operators ρ such that Tr {ρ} ≤ 1).
Below is another variation on the gentle measurement lemma that applies to ensemblesof quantum states.
Lemma 9.4.3 (Gentle Measurement
for Ensembles) Let {pX (x), ρx } be an ensemblePwith average density operator ρ ≡ x pX (x)ρx . Given a positive semi-definite operator Λwith Λ ≤ I and Tr {ρΛ} ≥ 1 − ε where ε ∈ [0, 1], thenX√√√pX (x) ρx − Λρx Λ ≤ 2 ε.
(9.211)
1x

Proof. We can apply the same steps in the proof of the gentle operator lemma to get thefollowing inequality, holding for all x:√√ 2
ρx − Λρx Λ ≤ 4 (1 − Tr {Λρx }) .
(9.212)
1

Taking the expectation over both sides produces the following inequality:X√ 2√
pX (x) ρx − Λρx Λ ≤ 4 (1 − Tr {Λρ}) ≤ 4ε.

(9.213)

1x

Taking the square root of the above inequality gives the following one:
s
X
√ 2
√
√
pX (x) ρx − Λρx Λ ≤ 2 ε.
xConcavity of the square root then implies that
r
X
√
√ 2
√
pX (x)
ρx − Λρx Λ ≤ 2 ε,x(9.214)1(9.215)1

concluding the proof.
Exercise 9.4.2
Gentle Measurement) Let
 (Coherentkoperators and ΛA be a POVM such that for all k:Tr ΛkA ρkA ≥ 1 − ε.

 kρAbe a collection of density(9.216)

Let φk RA be a purification of ρkA . Show that there exists a coherent gentle measurement
DA→AK in the sense of Section 5.4 such thatp
DA→AK (φkRA ) − φkRA ⊗ |kihk|K 1 ≤ 2 ε(2 − ε).
(9.217)(Hint: Use the result of Exercise 5.4.1.)
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CHAPTER 9. DISTANCE MEASURES

9.5 Fidelity of a Quantum Channel
It is useful to have measures that determine how well a quantum channel N preserves
quantum information. We developed static distance measures, such as the trace distanceand the fidelity, in the previous sections of this chapter. We would now like to exploit thosemeasures in order to define dynamic measures.
A “first guess” measure of this sort is the minimum fidelity Fmin (N ), where
Fmin (N ) ≡ min F (ψ, N (ψ)).

(9.218)|ψiThis measure seems like it may be a good one because we generally do not know the statethat Alice inputs to a noisy channel before transmitting to Bob.
It may seem somewhat strange that we chose to minimize over pure states in the definitionof the minimum fidelity. Are not mixed states the most general states that occur in thequantum theory? It turns out that joint concavity of the root fidelity (Property 9.2.2) andmonotonicity of the square function implies that we do not have to consider mixed states forthe minimum fidelity. Consider the following sequence of inequalities:
!!
X
X
√
√
F (ρ, N (ρ)) = F
pX (x)|xihx|, N
pX (x)|xihx|(9.219)
x

√
= F

x

≥

x

!
X

pX (x)|xihx|,

X

pX (x)N (|xihx|)(9.220)x√
pX (x) F (|xihx|, N (|xihx|))(9.221)√F (|xmin ihxmin |, N (|xmin ihxmin |)).

(9.222)Xx≥

The first equality follows by expanding the density operator ρ with a spectral decomposition.
The second equality follows from linearity of the quantum operation N . The first inequality
follows from joint concavity of the root fidelity (Property 9.2.2), and the last inequality
follows because there exists some pure state |xmin i (one of the eigenstates of ρ) with fidelity
never larger than the expected fidelity in the previous line.

9.5.1

Expected Fidelity of a Quantum Channel

In general, the minimum fidelity is less useful than other measures of quantum information
preservation over a quantum channel. The difficulty with the minimum fidelity is that it
requires an optimization over the potentially large space of input states. Since it could
be somewhat difficult to manipulate and compute in general, we introduce other ways todetermine the performance of a quantum channel.
We can simplify our notion of fidelity by instead restricting the states that Alice sends
and averaging the fidelity with respect to this set of states. That is, suppose that Alice istransmitting states from an ensemble {pX (x), ρx } and we would like to determine how well ac 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

9.5. FIDELITY OF A QUANTUM CHANNEL

279

Figure 9.5: The entanglement fidelity compares the output of the ideal scenario (depicted on the left) andthe output of the noisy scenario (depicted on the right).

quantum channel N is preserving this source of quantum information. Sending a particularstate ρx through a quantum channel N produces the state N (ρx ). The fidelity between thetransmitted state ρx and the received state N (ρx ) is F (ρx , N (ρx )) as defined before. We
define the expected fidelity of the ensemble as follows:X
F (N ) ≡ EX [F (ρX , N (ρX ))] =pX (x)F (ρx , N (ρx )).
(9.223)x

The expected fidelity indicates how well Alice is able to transmit the ensemble on average
to Bob. It again lies between zero and one, just as the usual fidelity does.
A more general form of the expected fidelity is to consider the expected performancefor any quantum state instead of restricting ourselves to an ensemble. That is, let us fix
some quantum state |ψi and apply a random unitary U to it, where we select the unitaryaccording to the Haar measure (this is the uniform distribution on unitaries). The state
U |ψi represents a random quantum state and we can take the expectation with respect toit in order to define the following more general notion of expected fidelity:
F (N ) ≡ EU F (U |ψihψ|U † , N (U |ψihψ|U † )) .
(9.224)
The above formula for the expected fidelity then becomes the following integral over the
Haar measure:Z
F (N ) = hψ|U † N (U |ψihψ|U † )U |ψi dU.
(9.225)9.5.2

Entanglement Fidelity

We now consider a different measure of the ability of a quantum channel to preserve quantum information. Suppose that Alice would like to transmit a quantum state with density
operator ρA . It admits a purification |ψiRA to a reference system R. Sending the A system
of |ψiRA through the identity channel idA gives back |ψiRA . Sending the A system of |ψiRA
through a quantum channel N : L(HA ) → L(HA ) gives the state σRA ≡ (idR ⊗NA ) (ψRA ).
The entanglement fidelity is defined as follows:
Definition 9.5.1 (Entanglement Fidelity) For ρ, N , σ, and |ψi as defined above, theentanglement fidelity is given by Fe (ρ, N ) ≡ hψ|σ|ψi.
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280

CHAPTER 9. DISTANCE MEASURESIt is a measure of how well the quantum channel N preserves entanglement with anothersystem. Figure 9.5 visually depicts the two states that the entanglement fidelity compares.
The entanglement fidelity is invariant with respect to which purification of the input that
we pick. This follows simply because all purifications are related by an isometry acting on
the purifying system. That is, let |ψiR1 A be one purification of ρA , let |ϕiR2 A be a different
one, and let UR1 →R2 be an isometry that relates them via |ϕiR2 A = UR1 →R2 |ψiR1 A . Thenhϕ|R2 A (idR2 ⊗NA ) (ϕR2 A ) |ϕiR2 Aih= hψ|R1 A UR† 1 →R2 (idR2 ⊗NA ) UR1 →R2 ψR1 A UR† 1 →R2 UR1 →R2 |ψiR1 A(9.226)= hψ|R1 A UR† 1 →R2 UR1 →R2 [(idR1 ⊗NA ) (ψR1 A )] UR† 1 →R2 UR1 →R2 |ψiR1 A(9.227)= hψ|R1 A [(idR1 ⊗NA ) (ψR1 A )] |ψiR1 A ,(9.228)where the second equality follows because the isometry commutes with the identity map idR2
and the last follows because UR1 →R2 is an isometry so that UR† 1 →R2 UR1 →R2 = IR1 .
One of the benefits of considering the task of entanglement preservation is that it implies
the task of quantum communication. That is, if Alice can devise a protocol that preserves
the entanglement with another system, then this same protocol will also be able to preserve
quantum information that she transmits.
The following theorem gives a simple way to represent the entanglement fidelity in termsof the Kraus operators of a given noisy quantum channel.
Theorem 9.5.1 Let ρA ∈ D(HA ) and let N : L(HA ) → L(HA ) be a quantum channel. Suppose that {K m } is a set of Kraus operators for N . Then the entanglement fidelity Fe (ρ, N )is equal to the following expression:X
Fe (ρ, N ) =|Tr {ρA K m }|2 .
(9.229)m

Proof. Given that the entanglement fidelity is invariant with respect to the choice of purification, we can simply use the canonical purification |ψiRA of ρA , i.e.,√|ψiRA = (IR ⊗ ρA ) |ΓiRA ,(9.230)where |ΓiRA is the unnormalized maximally entangled vector from (3.233). We then findthathψ|RA (idR ⊗NA ) (ψRA )|ψiRAX= hψ|RA(IR ⊗ KAm ) |ψihψ|RA IR ⊗ (KAm )† |ψiRA(9.231)

m

=

X



hψ|RA (IR ⊗ KAm ) |ψihψ|RA IR ⊗ (KAm )† |ψiRA(9.232)|hψ|RA (IR ⊗ KAm ) |ψiRA |2 .

(9.233)m=

Xm

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9.5. FIDELITY OF A QUANTUM CHANNEL

281

Then consider thathψ|RA (IR ⊗ KAm ) |ψiRA√√= hΓ|RA (IR ⊗ ρA ) (IR ⊗ KAm ) (IR ⊗ ρA ) |ΓiRA√√= hΓ|RA (IR ⊗ ρA KAm ρA ) |ΓiRA√√= Tr { ρA KAm ρA }= Tr {ρA KAm } ,where we have used Exercise 4.1.3 to establish the third equality. So we find thatX
hψ|RA (idR ⊗NA ) (ψRA )|ψiRA =|Tr {ρA K m }|2 ,(9.234)(9.235)(9.236)(9.237)(9.238)m

concluding the proof.
Exercise 9.5.1 Let ρ1 , ρ2 ∈ D(H) and let N : L(H) → L(H) be a quantum channel. Fix
λ ∈ [0, 1]. Show that the entanglement fidelity is convex in the input state:
Fe (λρ1 + (1 − λ) ρ2 , N ) ≤ λFe (ρ1 , N ) + (1 − λ) Fe (ρ2 , N ).

(9.239)(Hint: The result of Theorem 9.5.1 is useful here.)
Exercise 9.5.2 Prove that the entanglement fidelity does not depend upon the particular
choice of Kraus operators for a given channel. (Hint: Recall that there always exists an
isometry that relates two different Kraus representations of a quantum channel, i.e., for aset {K m } of Kraus operators and another set {Ln }, we have thatX
umn Ln ,(9.240)Km =nwhere umn are the entries of a unitary matrix.)

9.5.3

Expected Fidelity and Entanglement Fidelity

The entanglement fidelity and the expected fidelity provide seemingly different methods forquantifying the ability of a noisy quantum channel to preserve quantum information. Is
there any way that we can show how they are related?
It turns out that they are indeed related. First, consider that the entanglement fidelity
is a lower bound on the channel’s fidelity for preserving the state ρ:
Fe (ρ, N ) ≤ F (ρ, N (ρ)).

(9.241)

The above result follows simply from the monotonicity of fidelity under partial trace (Lemma 9.2.1).
We can show that the entanglement fidelity is always less than the expected fidelity in (9.223)
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282

CHAPTER 9. DISTANCE MEASURES

by combining convexity of entanglement fidelity (Exercise 9.5.1) and the bound in (9.241):!
X
X
Fe
pX (x)ρx , N ≤pX (x)Fe (ρx , N )(9.242)x

x≤

X

pX (x)F (ρx , N (ρx ))(9.243)x= F (N ).

(9.244)Thus, any channel that preserves entanglement with some reference system preserves the
expected fidelity of an ensemble. In most cases, we only consider the entanglement fidelity
as the defining measure of performance of a noisy quantum channel.
The relationship between entanglement fidelity and expected fidelity becomes more exact(and more beautiful) in the case where we select a random quantum state according to theHaar measure. It is possible to show that the expected fidelity in (9.224) relates to theentanglement fidelity as follows:
F (N ) =dFe (π, N ) + 1,
d+1(9.245)where d is the dimension of the input system and π is the maximally mixed state withpurification to the maximally entangled state.
Exercise 9.5.3 Prove that the relation in (9.245) holds for a quantum depolarizing channel.

9.6 The Hilbert–Schmidt Distance Measure
One final distance measure that we develop is the Hilbert–Schmidt distance measure. It
is most similar to the familiar Euclidean distance measure of vectors because an `2 -norm
induces it. This distance measure does not have an appealing operational interpretation
like the trace distance and fidelity do. Furthermore, it is neither generally increasing or
decreasing with respect to the action of quantum channels, and so one should not employ it
as a distinguishability measure of quantum states. Nevertheless, it can sometimes be helpful
in calculations to exploit this distance measure and to relate it to the trace distance via thebound in Exercise 9.6.1 below.
Let us define the Hilbert–Schmidt norm of an operator M ∈ L(H, H0 ) as follows:p(9.246)kM k2 ≡ Tr {M † M }.
It is straightforward to show that the above norm meets the three requirements of a norm:
non-negativity, homogeneity, and the triangle inequality. One can compute this norm simply
by summing the squares of the singular values of the operator M and taking the square root.
The reasoning for this is the same as that in the proof of Proposition 9.1.1.
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9.7. HISTORY AND FURTHER READING

283

The Hilbert–Schmidt norm induces the following Hilbert–Schmidt distance measure:kM − N k2 ,(9.247)

where M, N ∈ L(H, H0 ). We can then apply this distance measure to quantum states ρ andσ simply by plugging ρ and σ into the above formula in place of M and N .
The Hilbert–Schmidt distance measure sometimes finds use in the proofs of coding theorems in quantum Shannon theory because it is often easier to find good bounds on it rather
than on the trace distance. In some cases, we might be taking expectations over ensembles ofdensity operators and this expectation often reduces to computing variances or covariances.
Exercise 9.6.1 Show that the following inequality holds for any operator X
kXk21 ≤ d kXk22 ,(9.248)where d is the rank of X. (Hint: Use the Cauchy–Schwarz inequality for the Hilbert–Schmidt
inner product.)
Exercise 9.6.2 There are explicit counterexamples to the monotonicity of the Hilbert–Schmidt
distance. Let A = |01ih00| + |11ih10| and B = |01ih01| + |11ih11| be Kraus operators for achannel N (ω) = AωA† + BωB † , and consider the states ρ = |0ih0| ⊗ π and σ = |1ih1| ⊗ π,
where π = I/2. First verify that N is a quantum channel and then show by explicit calculation that kρ − σk2 < kN (ρ) − N (σ)k2 for this example. On the other hand, let ρ = |0ih0|,
σ = |1ih1|, and N (ω) = 21 (ω + XωX). Show that kρ − σk2 > kN (ρ) − N (σ)k2 for thisother example. Message: Do not use the Hilbert–Schmidt distance as a measure of distinguishability!

9.7 History and Further Reading
Holevo (1972) introduced and studied a function of two quantum states ρ and σ that is very
closely related to the Uhlmann fidelity from Definition 9.2.3 and Theorem 9.2.1:
√ √ 2(9.249)FH (ρ, σ) ≡ Tr{ ρ σ} .
This function is now known as the Holevo fidelity and was rediscovered as “quantum affinity”
by Luo and Zhang (2004). Holevo (1972) proved that FH (ρ, σ) is related to the normalized
trace distance in the following way:pp
1
1 − FH (ρ, σ) ≤ kρ − σk1 ≤ 1 − FH (ρ, σ),(9.250)2which is very similar to the inequalities presented in Theorem 9.3.1. In fact, by employing
the inequality FH (ρ, σ) ≤ F (ρ, σ), as well as (9.250) and Theorem 9.3.1, we conclude thefollowing generalization of Theorem 9.3.1:
p
p
p
p
1
1 − F (ρ, σ) ≤ 1 − FH (ρ, σ) ≤ kρ − σk1 ≤ 1 − F (ρ, σ) ≤ 1 − FH (ρ, σ) (9.251)2
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284

CHAPTER 9. DISTANCE MEASURESIn spite of these interesting relations and various properties that the Holevo fidelity FH
possesses, we have focused our attention on the Uhlmann fidelity in this chapter due to its
widespread use in quantum information and other fields. For this reason, we refer to the
Uhlmann fidelity simply as “the fidelity.”
Fuchs (1996) and Fuchs and van de Graaf (1998) are a good starting point for learning
more regarding trace distance and fidelity. Other notable sources are Nielsen and Chuang(2000), Yard (2005), and von Kretschmann (2007). Helstrom (1969, 1976) demonstrated theoperational interpretation of the trace distance in the context of quantum hypothesis testing.
Uhlmann (1976) first proved the theorem bearing his name, and Jozsa (1994) later presented
a proof of this theorem for the case of finite-dimensional quantum systems. Theorem 9.2.2
is due to Fuchs and Caves (1995). Schumacher (1996) introduced the entanglement fidelity,
and Barnum et al. (2000) made further observations regarding it. Nielsen (2002) provided asimple proof of the exact relation between entanglement fidelity and expected fidelity.
Winter (1999a,b) originally proved the “gentle measurement” lemma. There, he used it
to obtain a variation of the direct part of the HSW coding theorem. Later, he used it toprove achievable rates for the quantum multiple access √channel (Winter, 2001). Ogawa andNagaoka (2007) subsequently improved this bound to 2 ε.
The counterexample to the monotonicity of the Hilbert–Schmidt distance is due to Ozawa(2000).

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 10

Classical Information and Entropy
All physical systems register bits of information, whether it be an atom, an electrical current,
the location of a billiard ball, or a switch. Information can be classical, quantum, or a hybridof both, depending on the system. For example, an atom or an electron or a superconducting
system can register quantum information because the quantum theory applies to each ofthese systems, but we can safely argue that the location of a billiard ball registers classicalinformation only. These atoms or electrons or superconducting systems can also registerclassical bits because it is always possible for a quantum system to register classical bits.
The term information, in the context of information theory, has a precise meaning thatis somewhat different from our prior “everyday” experience with it. Recall that the notionof the physical bit refers to the physical representation of a bit, and the information bit isa measure of how much we learn from the outcome of a random experiment. Perhaps theword “surprise” better captures the notion of information as it applies in the context of
information theory.
This chapter begins our formal study of classical information. Recall that Chapter 2
reviewed some of the major operational tasks in classical information theory. Here, ourapproach is somewhat different because our aim is to provide an intuitive understandingof information measures, in terms of the parties who have access to the classical systems.
We define precise mathematical formulas that measure the amount of information encoded
in a single physical system or in multiple physical systems. The advantage of developing
this theory is that we can study information in its own right without having to consider thedetails of the physical system that registers it.
We first introduce the entropy in Section 10.1 as the expected surprise of a randomvariable. We extend this basic notion of entropy to develop other measures of informationin Sections 10.2–10.6 that prove useful as intuitive informational measures, but also, andperhaps more importantly, these measures are the answers to operational tasks that one
might wish to perform using noisy resources. While introducing these quantities, we discussand prove several mathematical results concerning them that are important tools for thepracticing information theorist. These tools are useful both for proving results and forincreasing our understanding of the nature of information. Section 10.7 introduces entropy
285

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inequalities that help us to understand the limits on our ability to process information,and Section 10.8 gives several refinements of these entropy inequalities. Section 10.9 ends
the chapter by applying the classical informational measures developed in the forthcoming
sections to the classical information that one can extract from a quantum system.

10.1 Entropy of a Random Variable
Consider a random variable X. Suppose that each realization x of random variable X belongs
to an alphabet X . Let pX (x) denote the probability density function of X so that pX (x)is the probability that realization x occurs. The information content i(x) of a particularrealization x is a measure of the surprise that one has upon learning the outcome of arandom experiment:i(x) ≡ − log (pX (x)) .
(10.1)The logarithm is base two and this choice implies that we measure surprise or informationin units of bits.
Figure 10.1 plots the information content for values in the unit interval. This measure ofsurprise behaves as we would hope—it is higher for lower-probability events that surprise us
more, and it is lower for higher-probability events that do not surprise us as much. Inspectionof the figure reveals that the information content is non-negative for any realization x.
The information content is also additive, due to the choice of the logarithm function.
Given two independent random experiments involving random variable X with respective
realizations x1 and x2 , we have thati(x1 , x2 ) = − log (pX,X (x1 , x2 )) = − log (pX (x1 )pX (x2 )) = i(x1 ) + i(x2 ).

(10.2)

Additivity is a property that we look for in measures of information (so much so that we
dedicate the whole of Chapter 13 to this issue for more general measures of information).
The information content is a useful measure of surprise for particular realizations ofrandom variable X, but it does not capture a general notion of the amount of surprisethat a given random variable X possesses. The entropy H(X) captures this general notionof the surprise of a random variable X—it is the expected information content of randomvariable X:
H(X) ≡ EX {i(X)} .
(10.3)
At a first glance, the above definition may seem strangely self-referential because the argument of the probability density function pX (x) is itself the random variable X, but thisis well-defined mathematically. Evaluating the above formula gives an expression which wetake as the definition for the entropy H(X):Definition 10.1.1 (Entropy) The entropy of a discrete random variable X with probability
distribution pX (x) isX
H(X) ≡ −pX (x) log (pX (x)) .
(10.4)x
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Figure 10.1: The information content or “surprise” in (10.1) as a function of a probability p ranging from0 to 1. An event has a lower surprise if it is more likely to occur and it has a higher surprise if it less likely
to occur.

We adopt the convention that 0 · log(0) = 0 for realizations with zero probability. The
fact that limε→0 ε · log(1/ε) = 0 intuitively justifies this latter convention. (We can interpret
this convention as saying that the fact that the event has probability zero is more important
than or outweighs the fact that you would be infinitely surprised if such an event would
occur.)The entropy admits an intuitive interpretation. Suppose that Alice performs a random
experiment in her lab that selects a realization x according to the density pX (x) of randomvariable X. Suppose further that Bob has not yet learned the outcome of the experiment.
The interpretation of the entropy H(X) is that it quantifies Bob’s uncertainty about X
before learning it—his expected information gain is H(X) bits upon learning the outcome of
the random experiment. Shannon’s noiseless coding theorem, described in Chapter 2, makes
this interpretation precise by proving that Alice needs to send Bob bits at a rate H(X)in order for him to be able to decode a compressed message. Figure 10.2(a) depicts theinterpretation of the entropy H(X), along with a similar interpretation for the conditional
entropy that we introduce in Section 10.2.

10.1.1The Binary Entropy Function

A special case of the entropy occurs when the random variable X is a Bernoulli randomvariable with probability density pX (0) = p and pX (1) = 1 − p. This Bernoulli randomvariable could correspond to the outcome of a random coin flip. The entropy in this case is
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Figure 10.2: (a) The entropy H(X) is the uncertainty that Bob has about random variable X before
learning it. (b) The conditional entropy H(X|Y ) is the uncertainty that Bob has about X when he alreadypossesses Y .

Figure 10.3: The binary entropy function h2 (p) displayed as a function of the parameter p.

Definition 10.1.2 (Binary Entropy) The binary entropy of p ∈ [0, 1] ish2 (p) ≡ −p log p − (1 − p) log (1 − p) .

(10.5)The binary entropy quantifies the number of bits that we learn from the outcome of thecoin flip. If the coin is unbiased (p = 1/2), then we learn a maximum of one bit (h2 (p) = 1).
If the coin is deterministic (p = 0 or p = 1), then we do not learn anything from the outcome(h2 (p) = 0). Figure 10.3 displays a plot of the binary entropy function. The figure reveals
that the binary entropy function h2 (p) is a concave function of the parameter p and has its
peak at p = 1/2.

10.1.2Mathematical Properties of Entropy

We now discuss five important mathematical properties of the entropy H(X).
Property 10.1.1 (Non-Negativity) The entropy H(X) is non-negative for any discreterandom variable X with probability density pX (x):H(X) ≥ 0.
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Proof. Non-negativity follows because entropy is the expected information content of i(X),and the information content itself is non-negative. It is perhaps intuitive that the entropy
should be non-negative because non-negativity implies that we always learn some number ofbits upon learning random variable X (if we already know beforehand what the outcome ofa random experiment will be, then we learn zero bits of information once we perform it). Ina classical sense, we can never learn a negative amount of information!
Property 10.1.2 (Concavity) The entropy H(X) is concave in the probability densitypX (x).
Proof. We justify this result with a heuristic “mixing” argument for now, and provide
a formal proof in Section 10.7.1. Consider two random variables X1 and X2 with two
respective probability density functions pX1 (x) and pX2 (x) whose realizations belong to thesame alphabet. Consider a Bernoulli random variable B with probabilities q ∈ [0, 1] and1 − q corresponding to its two respective realizations b = 1 and b = 2. Suppose that we first
generate a realization b of random variable B and then generate a realization x of randomvariable Xb . Random variable XB then denotes a mixed version of the two random variables
X1 and X2 . The probability density of XB is pXB (x) = qpX1 (x) + (1 − q) pX2 (x). Concavityof entropy is the following inequality:
H(XB ) ≥ qH(X1 ) + (1 − q) H(X2 ).

(10.7)

Our heuristic argument is that this mixing process leads to more uncertainty for the mixed
random variable XB than the expected uncertainty over the two individual random variables.
We can think of this result as a physical situation involving two gases. Two gases each havetheir own entropy, but the entropy increases when we mix the two gases together. We later
give a more formal argument to justify concavity.
Property 10.1.3 (Permutation Invariance) The entropy is invariant with respect to permutations of the realizations of random variable X.
Proof. That is, suppose that we apply some permutation π to realizations x1 , x2 , . . . , x|X |so that they respectively become π(x1 ), π(x2 ), . . . , π(x|X | ). Then the entropy is invariant
under this shuffling because it depends only on the probabilities of the realizations, not thevalues of the realizations.
Property 10.1.4 (Minimum Value) The entropy H(X) vanishes if and only if X is adeterministic variable.
Proof. We would expect that the entropy of a deterministic variable should vanish, given
the interpretation of entropy as the uncertainty of a random experiment. This intuition
holds true and it is the degenerate probability density pX (x) = δx,x0 , where the realization
x0 has all the probability and other realizations have vanishing probability, that gives theminimum value of the entropy: H(X) = 0 when X has a degenerate density. If H(X) = 0,c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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then this means that pX (x) log[1/pX (x)] = 0 for all x ∈ X , which in turn implies that eitherpX (x) = 0 or pX (x) = 1 for all x ∈ X . Since pX is required to be a probability distribution,
we can have that pX (x) = 1 for just one realization and pX (x) = 0 for all others, so that X
is a deterministic random variable.
Sometimes, we may not have any prior information about the possible values of a variable
in a system, and we may decide that it is most appropriate to describe them with a probability
density function. How should we assign this probability density if we do not have any priorinformation about the values? Theorists and experimentalists often resort to a “principle ofmaximum entropy” or a “principle of maximal ignorance”—we should assign the probability
density to be the one that maximizes the entropy.
Property 10.1.5 (Maximum Value) The maximum value of the entropy H(X) for a random variable X taking values in an alphabet X is log |X |:H(X) ≤ log |X | .

(10.8)The inequality is saturated if and only if X is a uniform random variable on X .
Proof. First, note that the result of Exercise 2.1.1 is that log |X | is the entropy of the uniform
random variable on X . Next, we can prove the above inequality with a simple Lagrangian
optimization by solving for the density pX (x) that maximizes the entropy. Lagrangian optimization is well-suited for this task because the entropy is concave in the probability density,and thus any local maximum will be a global maximum. The Lagrangian L is as follows:!
X
L ≡ H(X) + λpX (x) − 1 ,(10.9)
xwhere H(X) is the quantity that we are maximizing, subject to the constraint that theis as follows:
probability density pX (x) sums to one. The partial derivative ∂p∂L
X (x)∂L
= − log (pX (x)) − 1 + λ.
∂pX (x)(10.10)We set the partial derivative ∂p∂Lequal to zero to find the probability density that maxiX (x)mizes L:
0 = − log (pX (x)) − 1 + λ⇒ pX (x) = 2

λ−1.

(10.11)(10.12)

The resulting probability density pX (x) depends only on a constant λ, implying that it must
be uniform pX (x) = 1/ |X |. Thus, the uniform distribution maximizes the entropy H(X)when random variable X is finite.
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10.2 Conditional Entropy
Let us now suppose that Alice possesses random variable X and Bob possesses some otherrandom variable Y . Random variables X and Y share correlations if they are not statisticallyindependent, and Bob then possesses “side information” about X in the form of Y . Let i(x|y)
denote the conditional information content:i(x|y) ≡ − log pX|Y (x|y) .
(10.13)The entropy H(X|Y = y) of random variable X conditioned on a particular realization y of
random variable Y is the expected conditional information content, where the expectation
is with respect to X|Y = y:
H(X|Y = y) ≡ EX|Y =y {i(X|y)}X=−pX|Y (x|y) log pX|Y (x|y) .

(10.14)(10.15)x

The relevant entropy that applies to the scenario where Bob possesses side information isthe conditional entropy H(X|Y ), defined as follows:
Definition 10.2.1 (Conditional Entropy) Let X and Y be discrete random variables
with joint probability distribution pX,Y (x, y). The conditional entropy H(X|Y ) is the expectedconditional information content, where the expectation is with respect to both X and Y :H(X|Y ) ≡ EX,Y {i(X|Y )}X=
pY (y)H(X|Y = y)(10.16)(10.17)

y

=−

X

=−

X

y

pY (y)X

pX|Y (x|y) log(pX|Y (x|y))(10.18)x

pX,Y (x, y) log(pX|Y (x|y)).

(10.19)x,yThe conditional entropy H(X|Y ) as well deserves an interpretation. Suppose that Alicepossesses random variable X and Bob possesses random variable Y . The conditional entropy H(X|Y ) is the amount of uncertainty that Bob has about X given that he alreadypossesses Y . Figure 10.2(b) depicts this interpretation.
The above interpretation of the conditional entropy H(X|Y ) immediately suggests thatit should be less than or equal to the entropy H(X). That is, having access to a side
variable Y should only decrease our uncertainty about another variable. We state this idea
as the following theorem and give a formal proof in Section 10.7.1.
Theorem 10.2.1 (Conditioning Does Not Increase Entropy) The entropy H(X) is
greater than or equal to the conditional entropy H(X|Y ):H(X) ≥ H(X|Y ),c 2019(10.20)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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and equality occurs if and only if X
P and Y are independent random variables. As a consequence of the fact that H(X|Y ) = y pY (y)H(X|Y = y), we see that the entropy is concave.
Non-negativity of conditional entropy follows from non-negativity of entropy because
conditional entropy is the expectation of the entropy H(X|Y = y) with respect to the densitypY (y). It is again intuitive that conditional entropy should be non-negative. Even if we haveaccess to some side information Y , we always learn some number of bits of information uponlearning the outcome of a random experiment involving X. Perhaps strangely, we will seethat quantum conditional entropy can become negative, defying our intuition of informationin the classical sense given here.

10.3 Joint Entropy
What if Bob knows neither X nor Y ? The natural entropic quantity that describes his
uncertainty is the joint entropy H(X, Y ). The joint entropy is merely the entropy of thejoint random variable (X, Y ):Definition 10.3.1 (Joint Entropy) Let X and Y be discrete random variables with joint
probability distribution pX,Y (x, y). The joint entropy H(X, Y ) is defined as
H(X, Y ) ≡ EX,Y {i(X, Y )}X=−
pX,Y (x, y) log(pX,Y (x, y)).

(10.21)(10.22)x,yThe following exercise asks you to explore the relation between joint entropy H(X, Y ),conditional entropy H(Y |X), and marginal entropy H(X). Its proof follows by consideringthat the multiplicative probability relation pX,Y (x, y) = pY |X (y|x)pX (x) of joint probability, conditional probability, and marginal entropy becomes an additive relation under the
logarithms of the entropic definitions.
Exercise 10.3.1 Verify that H(X, Y ) = H(X) + H(Y |X) = H(Y ) + H(X|Y ).
Exercise 10.3.2 Extend the result of Exercise 10.3.1 to prove the following chain rule for
entropy:
H(X1 , . . . , Xn ) = H(X1 ) + H(X2 |X1 ) + · · · + H(Xn |Xn−1 , . . . , X1 ).
Exercise 10.3.3 Prove that entropy is subadditive:nX
H(X1 , . . . , Xn ) ≤H(Xi ),(10.23)(10.24)i=1

by exploiting Theorem 10.2.1 and the entropy chain rule in Exercise 10.3.2.
Exercise 10.3.4 Prove that entropy is additive when the random variables X1 , . . . , Xn are
independent:nX
H(X1 , . . . , Xn ) =H(Xi ).
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10.4 Mutual Information
We now introduce an entropic measure of the common or mutual information that two
parties possess. Suppose that Alice possesses random variable X and Bob possesses randomvariable Y .
Definition 10.4.1 (Mutual Information) Let X and Y be discrete random variables withjoint probability distribution pX,Y (x, y). The mutual information I(X; Y ) is the marginal
entropy H(X) less the conditional entropy H(X|Y ):I(X; Y ) ≡ H(X) − H(X|Y ).

(10.26)The mutual information quantifies the dependence or correlations of the two randomvariables X and Y . It measures how much knowing one random variable reduces the uncertainty about the other random variable. In this sense, it is the common information between
the two random variables. Bob possesses Y and thus has an uncertainty H(X|Y ) about
Alice’s variable X. Knowledge of Y gives an information gain of H(X|Y ) bits about X andthen reduces the overall uncertainty H(X) about X, the uncertainty were he not to haveany side information at all about X.
Exercise 10.4.1 Show that the mutual information is symmetric in its inputs:
I(X; Y ) = I(Y ; X),(10.27)

I(X; Y ) = H(Y ) − H(Y |X).

(10.28)

implying additionally that

We can also express the mutual information I(X; Y ) in terms of the respective joint andmarginal probability density functions pX,Y (x, y) and pX (x) and pY (y):X
pX,Y (x, y).
(10.29)
pX,Y (x, y) log
I(X; Y ) =pX (x)pY (y)x,yThe above expression leads to two insights regarding the mutual information I(X; Y ). Two
random variables X and Y possess zero bits of mutual information if they are statisticallyindependent (recall that the joint density factors as pX,Y (x, y) = pX (x)pY (y) when X and Y
are independent). That is, knowledge of Y does not give any information about X when the
random variables are statistically independent. Later, we show that the converse statement
is true as well. Also, two random variables possess H(X) bits of mutual information if they
are perfectly correlated in the sense that Y = X.
Theorem 10.4.1 below states that the mutual information I(X; Y ) is non-negative forany random variables X and Y —we provide a formal proof in Section 10.7.1. However, thisfollows naturally from the definition of mutual information in (10.26) and “conditioning does
not increase entropy” (Theorem 10.2.1).
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Theorem 10.4.1 The mutual information I(X; Y ) is non-negative for any random variables X
and Y :
I(X; Y ) ≥ 0,
(10.30)and I(X; Y ) = 0 if and only if X and Y are independent random variables (i.e., if pX,Y (x, y) =pX (x)pY (y)).

10.5 Relative Entropy
The relative entropy is another important entropic quantity that quantifies how “far” oneprobability density function p(x) is from another probability density function q(x). It canbe helpful to have a more general definition in which we allow q(x) to be a function taking
non-negative values. Before defining the relative entropy, we need the notion of the support
of a function.
Definition 10.5.1 (Support) Let X denote a finite set. The support of a function f :X → R is equal to the subset of X that takes non-zero values under f :supp(f ) ≡ {x : f (x) 6= 0}.

(10.31)

Definition 10.5.2 (Relative Entropy) Let p be a probability distribution defined on thealphabet X , and let q : X → [0, ∞). The relative entropy D(pkq) is defined as follows: P
x p(x) log (p(x)/q(x)) if supp(p) ⊆ supp(q) .
(10.32)D(pkq) ≡+∞else
According to the above definition, the relative entropy is equal to the following expected
log-likelihood ratio:

 
p(X)
,
(10.33)D(pkq) = EX log
q(X)where X is a random variable distributed according to p.
The above definition implies that the relative entropy is not symmetric under interchangeof p(x) and q(x). Thus, the relative entropy is not a distance measure in the strict mathematical sense because it is not symmetric (nor does it satisfy a triangle inequality).
The relative entropy has an interpretation in source coding, if we let q(x) be a probability
distribution. Suppose that an information source generates a random variable X according
to the density p(x). Suppose further that Alice (the compressor) mistakenly assumes thatthe probability density of the information source is instead q(x) and codes according to thisdensity. Then the relative entropy quantifies the inefficiency that Alice incurs when she
codes according to the mistaken probability density—Alice requires H(X) + D (pkq) bits on
average to code (whereas she would only require H(X) bits on average to code if she used
the true density p(x)).
We might also see now that the mutual information I(X; Y ) is equal to the relative
entropy D(pX,Y kpX ⊗ pY ) by comparing the definition of relative entropy in (10.32) and thec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

10.6. CONDITIONAL MUTUAL INFORMATION

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expression for the mutual information in (10.29) (by pX ⊗ pY we mean the product of themarginal distributions). In this sense, the mutual information quantifies how far the two
random variables X and Y are from being independent because it calculates the “distance”
of the joint density pX,Y to the product pX ⊗ pY of the marginals.
Let pX1 and pX2 be two probability distributions defined over the same alphabet. The
relative entropy D(pX1 kpX2 ) admits a pathological property. It can become infinite if the
distribution pX1 (x1 ) does not have all of its support contained in the support of pX2 (x2 )(i.e., if there is some realization x for which pX1 (x) 6= 0 but pX2 (x) = 0). This can besomewhat bothersome if we like the interpretation of relative entropy as a notion of distance. In an extreme case, we would think that the distance between a deterministic binary
random variable X2 where Pr {X2 = 1} = 1 and one with probabilities Pr {X1 = 0} = ε andPr {X1 = 1} = 1 − ε should be on the order of ε (this is true for the classical trace distance).
However, the relative entropy D(pX1 kpX2 ) in this case is infinite, in spite of our intuition thatthese distributions are close. The interpretation in lossless source coding is that it would
require an infinite number of bits to code a distribution pX1 losslessly if Alice mistakes it
as pX2 . Alice thinks that the symbol X2 = 0 never occurs, and in fact, she thinks that the
typical set consists of just one sequence of all ones and every other sequence is atypical. But
in reality, the typical set is quite a bit larger than this, and it is only in the limit of an
infinite number of bits that we can say her compression is truly lossless.
Exercise 10.5.1 Verify that the definition of relative entropy in Definition 10.5.2 is consistent with the following limit:D(pkq) = lim D(pkq + ε1),
ε&0

(10.34)

where 1 denotes a vector of ones, so that the elements of q + ε1 are q(x) + ε.

10.6 Conditional Mutual Information
What is the common information between two random variables X and Y when we have someside information embodied in a random variable Z? The entropic quantity that quantifies
this common information is the conditional mutual information.
Definition 10.6.1 (Conditional Mutual Information) Let X, Y , and Z be discrete random variables. The conditional mutual information is defined as follows:
I(X; Y |Z) ≡ H(Y |Z) − H(Y |X, Z)= H(X|Z) − H(X|Y, Z)= H(X|Z) + H(Y |Z) − H(X, Y |Z).

(10.35)(10.36)(10.37)

Theorem 10.6.1 (Strong Subadditivity) The conditional mutual information I(X; Y |Z)
is non-negative:
I(X; Y |Z) ≥ 0,
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and the inequality is saturated if and only if X−Z−Y is a Markov chain (i.e., if pX,Y |Z (x, y|z) =pX|Z (x|z)pY |Z (y|z)).
Proof. The proof of the above theorem is a straightforward consequence of the nonnegativity of mutual information (Theorem 10.4.1). Consider the following equality:
X
I(X; Y |Z) =pZ (z)I(X; Y |Z = z),(10.39)z

where I(X; Y |Z = z) is a mutual information with respect to the joint density pX,Y |Z (x, y|z)
and the marginal densities pX|Z (x|z) and pY |Z (y|z). Non-negativity of I(X; Y |Z) then follows
from non-negativity of pZ (z) and I(X; Y |Z = z). The saturation conditions then follow
immediately from those for mutual information given in Theorem 10.4.1 (considering thatthe conditional mutual information is a convex combination of mutual informations).
The proof of the above classical version of strong subadditivity is perhaps trivial inhindsight (it requires only a few arguments). The proof of the quantum version of strong
subaddivity is highly non-trivial on the other hand. We discuss strong subadditivity ofquantum entropy in the next chapter.
Exercise 10.6.1 The expression in (10.38) represents the most compact way to express thestrong subadditivity of entropy. Show that the following inequalities are equivalent ways ofrepresenting strong subadditivity:
H(XY |Z) ≤ H(X|Z) + H(Y |Z),
H(XY Z) + H(Z) ≤ H(XZ) + H(Y Z),
H(X|Y Z) ≤ H(X|Z).

(10.40)(10.41)(10.42)

Exercise 10.6.2 Prove the following chain rule for mutual information:
I(X1 , . . . , Xn ; Y )= I(X1 ; Y ) + I(X2 ; Y |X1 ) + · · · + I(Xn ; Y |X1 , . . . , Xn−1 ). (10.43)
The interpretation of the chain rule is that we can build up the correlations between X1 , . . . , Xn
and Y in n steps: in a first step, we build up the correlations between X1 and Y , and now
that X1 is available (and thus conditioned on), we build up the correlations between X2 andY , etc.

10.7 Entropy Inequalities
The entropic quantities introduced in the previous sections each have bounds associated withthem. These bounds are fundamental limits on our ability to process and store information.
We introduce several bounds in this section: the non-negativity of relative entropy, two dataprocessing inequalities, Fano’s inequality, and a uniform bound for continuity of entropy.
Each of these inequalities plays an important role in information theory, and we describe
these roles in more detail in the forthcoming subsections.
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10.7. ENTROPY INEQUALITIES

10.7.1297

Non-Negativity of Relative Entropy

The relative entropy is always non-negative. This seemingly innocuous result has several
important implications—namely, the maximal value of entropy, conditioning does not increase entropy (Theorem 10.2.1), non-negativity of mutual information (Theorem 10.4.1),
and strong subadditivity (Theorem 10.6.1) are straightforward corollaries of it. A proof ofthis entropy inequality follows from the application of a simple inequality: ln x ≤ x − 1.
Theorem 10.7.1 (Non-Negativity of Relative Entropy) Let p(x) be a probability
disP
tribution over the alphabet X and let q : X → [0, 1] be a function such that x q(x) ≤ 1.
Then the relative entropy D(pkq) is non-negative:D(pkq) ≥ 0,

(10.44)and D(pkq) = 0 if and only if p = q.
Proof. First, suppose that supp(p) 6⊆ supp(q). Then the relative entropy D(pkq) = +∞and the inequality is trivially satisfied.
Now, suppose that supp(p) ⊆ supp(q). A proof relies on the inequality ln x ≤ x − 1 that
holds for all x ≥ 0 and saturates for this range if and only if x = 1. (Brief justification: Letf (x) = x − 1 − ln x. Observe that f (1) = 0, f 0 (1) = 0, f 0 (x) > 0 for x > 1 and f 0 (x) < 0
for x < 1. So f (x) has a minimum at x = 1 and is strictly increasing when x > 1 andstrictly decreasing when x < 1. For the saturation condition: Suppose that f (x) = 0. Thenx = 1 is a solution of the equation. Since the function is strictly decreasing when x < 1 andstrictly increasing when x > 1, x = 1 is the only solution to f (x) = 0.) Figure 10.4 plots thefunctions ln x and x − 1 to compare them.
We first prove the inequality in (10.44). Consider the following chain of inequalities:


X
p(x)
(10.45)p(x) log
D(pkq) =
q(x)
x


q(x)
1 X
=−
p(x) ln(10.46)ln 2 x
p(x)


q(x)
1 X
p(x) 1 −(10.47)≥ln 2 x
p(x)
!
X
X
1
=
p(x) −q(x)(10.48)
ln 2
x
x
≥ 0.

(10.49)The sole inequality follows because − ln x ≥ 1 − x (a simple rearrangementof ln x ≤ x − 1).
P
The last inequality is a consequence of the assumption that x q(x) ≤ 1.
Now suppose that p = q. It is then clear that D(pkq) = 0. Finally, suppose thatD(pkq) = 0. Then we necessarily have supp(p) ⊆ supp(q), and the condition D(pkq) = 0
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Figure 10.4: A plot that compares the functions ln x and x − 1, showing that ln x ≤ x − 1 for all positive x.

implies that both inequalities above are saturated.
So, first we can deduce that q is aP
probability distribution since we assumed that x q(x) ≤ 1 and the last inequality above
is saturated. Next, the inequality in the third line above is saturated, which implies thatln (q(x)/p(x)) = 1 − q(x)/p(x) for all x for which p(x) > 0. But this happens only whenq(x)/p(x) = 1 for all x for which p(x) > 0, which allows us to conclude that p = q.
We can now quickly prove several corollaries of the above theorem.
Proofs of Property 10.1.5, Theorem 10.4.1, Theorem 10.2.1. Recall in Section 10.1.2
that we proved that the entropy H(X) takes the maximal value log |X |, where |X | is thesize of the alphabet of X. The proof method involved Lagrange multipliers. Here, we canprove this result simply by computing the relative entropy D(pX k{1/ |X |}), where pX is theprobability density of X and {1/ |X |} is the uniform density, and applying the non-negativityof relative entropy:
0 ≤ D(pX k{1/ |X |})!
X
pX (x)pX (x) log=
1(10.51)|X |x= −H(X) +(10.50)X

pX (x) log |X |(10.52)x= −H(X) + log |X | .

(10.53)It then follows that H(X) ≤ log |X | by combining the first line with the last. Non-negativityof mutual information (Theorem 10.4.1) follows by recalling that I(X; Y ) = D(pX,Y kpX ⊗pY )and applying the non-negativity of relative entropy. The equality conditions follow from thosefor equality of D(pkq) = 0. Conditioning does not increase entropy (Theorem 10.2.1) follows
by noting that I(X; Y ) = H(X) − H(X|Y ) and applying Theorem 10.4.1.
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10.7. ENTROPY INEQUALITIES

299

Figure 10.5: Two slightly different depictions of the scenario in the data-processing inequality. (a) The
map N1 processes random variable X to produce some random variable Y , and the map N2 processes
the random variable Y to produce the random variable Z. The inequality I(X; Y ) ≥ I(X; Z) applies here
because correlations can only decrease after data processing. (b) This depiction of data processing helps us tobuild intuition for data processing in the quantum world. The protocol begins with two perfectly correlated
random variables X and X 0 —perfect correlation implies that pX,X 0 (x, x0 ) = pX (x)δx,x0 and further that
H(X) = I(X; X 0 ). We process random variable X 0 with a stochastic map N1 to produce a random variable
Y , and then further process Y according to the stochastic map N2 to produce random variable Z. By thedata-processing inequality, the following chain of inequalities holds: I(X; X 0 ) ≥ I(X; Y ) ≥ I(X; Z).

10.7.2

Data-Processing Inequality

Another important inequality in classical information theory is the data-processing inequality.
There are at least two variations of it. The first one states that correlations between randomvariables can only decrease after we process one variable according to some stochastic function that depends only on that variable. The next one states that the relative entropy cannot
increase if a channel is applied to both of its arguments. These data-processing inequalities
find application in the converse proof of a coding theorem (the proof of the optimality of acommunication rate).
Mutual Information Data-Processing Inequality
We detail the scenario that applies for the first data-processing inequality. Suppose that
we initially have two random variables X and Y . We might say that random variable
Y arises from random variable X by processing X according to a stochastic map N1 ≡pY |X (y|x). That is, the two random variables arise by first picking X according to the densitypX (x) and then processing X according to the stochastic map N1 . The mutual information
I(X; Y ) quantifies the correlations between these two random variables. Suppose then that
we process Y according to some other stochastic map N2 ≡ pZ|Y (z|y) to produce a randomvariable Z (note that the map can also be deterministic because the set of stochastic maps
subsumes the set of deterministic maps). Then the first data-processing inequality statesthat the correlations between X and Z must be less than the correlations between X and Y :I(X; Y ) ≥ I(X; Z),(10.54)

because data processing according to any stochastic map N2 cannot increase correlations.
Figure 10.5(a) depicts the scenario described above. Figure 10.5(b) depicts a slightly different
scenario for data processing that helps build intuition for the forthcoming notion of quantumc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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data-processing in Section 11.9.2 of the next chapter. Theorem 10.7.2 below states theclassical data-processing inequality.
The scenario described in the above paragraph contains a major assumption: the stochastic map pZ|Y (z|y) that produces random variable Z depends on random variable Y only—it
has no dependence on X, meaning that
pZ|Y,X (z|y, x) = pZ|Y (z|y).

(10.55)This assumption is called the Markovian assumption and is the crucial assumption in the
proof of the data-processing inequality. We say that the three random variables X, Y , andZ form a Markov chain and use the notation X → Y → Z to indicate this stochastic
relationship.
Theorem 10.7.2 (Data-Processing Inequality) Suppose three random variables X, Y ,and Z form a Markov chain: X → Y → Z. Then the following data-processing inequality
holds
I(X; Y ) ≥ I(X; Z).
(10.56)
Proof. The Markov condition X → Y → Z implies that random variables X and Z areconditionally independent through Y because
pX,Z|Y (x, z|y) = pZ|Y,X (z|y, x)pX|Y (x|y)= pZ|Y (z|y)pX|Y (x|y).

(10.57)(10.58)

We prove the data-processing inequality by manipulating the mutual information I(X; Y Z).
Consider the following equalities:
I(X; Y Z) = I(X; Y ) + I(X; Z|Y ) = I(X; Y ).

(10.59)The first equality follows from the chain rule for mutual information (Exercise 10.6.2). The
second equality follows because the conditional mutual information I(X; Z|Y ) vanishes for a
Markov chain X → Y → Z—i.e., X and Z are conditionally independent through Y (recall
Theorem 10.6.1). We can also expand the mutual information I(X; Y Z) in another way toobtainI(X; Y Z) = I(X; Z) + I(X; Y |Z).
(10.60)
Then the following equality holds for a Markov chain X → Y → Z by exploiting (10.59):
I(X; Y ) = I(X; Z) + I(X; Y |Z).

(10.61)

The inequality in Theorem 10.7.2 follows because I(X; Y |Z) is non-negative for any randomvariables X, Y , and Z (recall Theorem 10.6.1).
By inspecting the above proof, we find the following:
Corollary 10.7.1 The following inequality holds for a Markov chain X → Y → Z:
I(X; Y ) ≥ I(X; Y |Z).
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301

Relative Entropy Data-Processing Inequality
Another kind of data-processing inequality holds for the relative entropy, known as monotonicity of relative entropy. This also is a consequence of the non-negativity of relative
entropy in Theorem 10.7.1.
Corollary 10.7.2 (Monotonicity of Relative Entropy) Let p be a probability distribution on an alphabet X and let q : X → [0, ∞). Let N (y|x) be a conditional probability
distribution (i.e., a classical channel). Then the relative entropy does not increase after thechannel N (y|x) acts on p and q:
D(pkq) ≥ D(N pkN q),(10.63)P
where N p is a probability distribution
Pwith elements (N p)(y) ≡ x N (y|x)p(x) and N q isa vector with elements (N q)(y) =x N (y|x)q(x). Let R be the channel defined by thefollowing set of equations:
R(x|y)(N q)(y) = N (y|x)q(x).
(10.64)The inequality in (10.63) is saturated (i.e., D(pkq) = D(N pkN q))Pif and only if RN p = p,
where RN p is a probability distribution with elements (RN p)(x) = y,x0 R(x|y)N (y|x0 )p(x0 ).
Proof. First, if p and q are such that supp(p) 6⊆ supp(q), then the inequality is triviallytrue because D(pkq) = +∞ in this case. So let us suppose that supp(p) ⊆ supp(q), which
implies that supp(N p) ⊆ supp(N q). Our first step is to rewrite the terms in the inequality.
To this end, consider the following algebraic manipulations:X(N p)(y)(10.65)(N p)(y) log
D(N pkN q) =
(N
q)(y)
y


X
(N p)(y)N (y|x)p(x) log=(10.66)(N q)(y)
y,x
"

#
X
X
(N p)(y)(10.67)=p(x)N (y|x) log
(N
q)(y)
x
y
"
#

X
X
(N p)(y)=p(x) log exp
N (y|x) log.
(10.68)(N q)(y)xyThis implies thatD(pkq) − D(N pkN q) = D(pkr),(10.69)where"r(x) ≡ q(x) exp

Xyc 2019
N (y|x) log(N p)(y)(N q)(y)#.

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Now consider that"Xxr(x) =

X

q(x) expxXN (y|x) logy(N p)(y)(N q)(y)

#

 

(N p)(y)≤q(x)
N (y|x) exp log(N q)(y)
x
y


X
X
(N p)(y)=q(x)
N (y|x)(N q)(y)
x
y
"
#
X X
(N p)(y)=q(x)N (y|x)(N q)(y)yxX=(N p)(y) = 1.
X

X

(10.71)(10.72)(10.73)(10.74)(10.75)y

The inequality in the P
second line follows from convexity of the exponential function. Since r
is a vector such that x r(x) ≤ 1, we can conclude from Theorem 10.7.1 that D(pkr) ≥ 0,
which by (10.69) is the same as (10.63).
We now comment on the saturation conditions. First, suppose that RN p = p. By
monotonicity of relative entropy (what was just proved) under the application of the channel
R, we find thatD(N pkN q) ≥ D(RN pkRN q) = D(pkq),(10.76)
where the equality follows from the assumption that RN p = p, and the fact that RN q = q.
This last statement follows becauseXX
N (y|x)q(x) = q(x).
(10.77)
R(x|y)(N q)(y) =(RN q)(x) =yyThe other implication D(pkq) = D(N pkN q) ⇒ RN p = p is a consequence of a later development, Theorem 10.8.4.

10.7.3Fano’s Inequality

Another entropy inequality that we consider is Fano’s inequality. This inequality also finds
application in the converse proof of a coding theorem.
Fano’s inequality applies to a general classical communication scenario. Suppose Alicepossesses some random variable X that she transmits to Bob over a noisy communication
channel. Let pY |X (y|x) denote the stochastic map corresponding to the noisy communication
channel. Bob receives a random variable Y from the channel and processes it in some way
to produce his best estimate X̂ of the original random variable X. Figure 10.6 depicts this
scenario.
The natural performance metric of this communication scenario is the probability of errorpe ≡ Pr{X̂ 6= X}—a low probability of error corresponds to good performance. On the otherc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

10.7. ENTROPY INEQUALITIES

303

Figure 10.6: The classical communication scenario relevant in Fano’s inequality. Alice transmits a randomvariable X over a noisy channel N , producing a random variable Y . Bob receives Y and processes it according
to some decoding map D to produce his best estimate X̂ of Y .

hand, consider the conditional entropy H(X|Y ). We interpreted it before as the uncertainty
about X from the perspective of someone who already knows Y . If the channel is noiseless(pY |X (y|x) = δy,x ), then there is no uncertainty about X because Y is identical to X:
H(X|Y ) = 0.

(10.78)

As the channel becomes noisier, the conditional entropy H(X|Y ) increases away from zero.
In this sense, the conditional entropy H(X|Y ) quantifies the information about X that islost in the channel noise. We then might naturally expect there to be a relationship between
the probability of error pe and the conditional entropy H(X|Y ): the amount of information
lost in the channel should be low if the probability of error is low. Fano’s inequality provides
a quantitative bound corresponding to this idea.
Theorem 10.7.3 (Fano’s Inequality) Suppose that Alice sends a random variable X througha noisy channel to produce random variable Y and further processing of Y gives an estimate
X̂ of X. Thus, X → Y → X̂ forms a Markov chain. Let pe ≡ Pr{X̂ 6= X} denote the probability of error. Then the following function of the error probability pe bounds the information
lost in the channel noise:
H(X|Y ) ≤ H(X|X̂) ≤ h2 (pe ) + pe log (|X | − 1) ,(10.79)where h2 (pe ) is the binary entropy function. In particular, note that
lim h2 (pe ) + pe log (|X | − 1) = 0.

pe →0(10.80)The bound is sharp in the sense that there is a choice of random variables X and Y thatsaturates it.
Proof. Let E denote an indicator random variable that indicates whether an error occurs:
0 : X = X̂E=.
(10.81)
1 : X 6= X̂
Consider the entropy
H(EX|X̂) = H(X|X̂) + H(E|X X̂).

(10.82)The entropy H(E|X X̂) on the right-hand side vanishes because there is no uncertainty about
the indicator random variable E if we know both X and X̂. Thus,
H(EX|X̂) = H(X|X̂).
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CHAPTER 10. CLASSICAL INFORMATION AND ENTROPY

Also, the data-processing inequality applies to the Markov chain X → Y → X̂:I(X; Y ) ≥ I(X; X̂),(10.84)and implies the following inequality:
H(X|X̂) ≥ H(X|Y ).

(10.85)

Consider the following chain of inequalities:
H(EX|X̂) = H(E|X̂) + H(X|E X̂)≤ H(E) + H(X|E X̂)(10.86)(10.87)= h2 (pe ) + pe H(X|X̂, E = 1)
+ (1 − pe ) H(X|X̂, E = 0)
≤ h2 (pe ) + pe log (|X | − 1) .

(10.88)(10.89)The first equality follows by expanding the entropy H(EX|X̂). The first inequality follows
because conditioning reduces entropy. The second equality follows by explicitly expanding
the conditional entropy H(X|E X̂) in terms of the two possibilities of the error randomvariable E. The last inequality follows from two facts: there is no uncertainty about X
when there is no error (when E = 0) and X̂ is available, and the uncertainty about X when
there is an error (when E = 1) and X̂ is available is less than the uncertainty of a uniform
distribution |X 1|−1 for all of the other possibilities. Fano’s inequality follows from putting
together (10.85), (10.83), and (10.89):
H(X|Y ) ≤ H(X|X̂) = H(EX|X̂) ≤ h2 (pe ) + pe log (|X | − 1) .

(10.90)To establish the statement of saturation, let X be a uniform random variable and let
X be a copy of it (so that the joint distribution is pX,X 0 (x, x0 ) = δx,x0 /|X | where X isthe alphabet). Let pY |X 0 denote a symmetric channel such that the input x0 is transmitted
faithfully with probability 1 − ε and becomes one of the other |X | − 1 letters with probabilityε/(|X | − 1), where ε ∈ [0, 1]. Observe that Y is a uniform random variable, given that pY |X 0
is a symmetric channel. Then Pr{X 6= Y } = ε, and we find that
0

H(X|Y ) = H(Y |X) + H(X) − H(Y ) = H(Y |X) = h2 (ε) + ε log (|X | − 1) ,(10.91)

concluding the proof.

10.7.4

Continuity of Entropy

That the entropy is a continuous function follows from the fact that the function −x log x
is continuous. However, it can be useful in applications to have explicit continuity bounds.
Before we give such bounds, we should establish how we measure distance between probability distributions. A natural way for doing so is to use the classical trace distance, defined
as follows:c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

10.7. ENTROPY INEQUALITIES305

Definition 10.7.1 (Classical Trace Distance) Let p, q : X → R, where X is a finite
alphabet. The classical trace distance between p and q is thenXkp − qk1 ≡|p(x) − q(x)| .
(10.92)x

The classical trace distance is a special case of the trace distance from Definition 9.1.2, in
which we place the entries of p and q along the diagonal of some square matrices. If p and q
are probability distributions, then the operational meaning of the classical trace distance isthe same as it was in the fully quantum case: it is the bias in the probability with which one
could successfully distinguish p and q by means of any binary hypothesis test. The optimal
test is given in the following lemma:
Lemma 10.7.1 Let p and q be probability distributions on a finite alphabet X . Let A ≡{x : p(x) ≥ q(x)}. Then1kp − qk1 = p(A) − q(A),(10.93)
2
PP
where p(A) ≡ x∈A p(x) and q(A) ≡ x∈A q(x).
Proof. A proof for this lemma is very similar to that for Lemma 9.1.1. It is brief and so weprovide it here. Consider thatX
[p(x) − q(x)](10.94)0=x=X[p(x) − q(x)] +X[p(x) − q(x)] ,(10.95)x∈Acx∈Awhich implies thatX

[p(x) − q(x)] =X[q(x) − p(x)] .

(10.96)x∈Ac

x∈A

Now consider thatkp − qk1 =

X

=

X

|p(x) − q(x)|(10.97)x[p(x) − q(x)] +X[q(x) − p(x)](10.98)

x∈Ac

x∈A

=2

X

[p(x) − q(x)] ,(10.99)x∈A

where the last line follows from (10.96).
One can interpret the set A and its complement Ac as a binary hypothesis test that one
could perform after receiving a sample x from either the distribution p or q (without knowing
which distribution was used to generate the sample). If the sample x is in A, then we wouldc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 10. CLASSICAL INFORMATION AND ENTROPY

decide that p generated x, and otherwise, we would decide that q generated x. The successprobability for distinguishing p from q is then
"
#
"
#
X
X
1 X
1
p(x) +
q(x) =
1+[p(x) − q(x)](10.100)
2 x∈A
2
x∈Ac
x∈A


1
1
1 + kp − qk1 .
(10.101)=
2
2
It turns out that this test is the optimal test, which follows from the same reasoning given
in Section 9.1.4. Thus, we can interpret the classical trace distance as an important measureof distinguishability between probability distributions, with an operational interpretation as
given above.
We can now state an important entropy continuity bound.
Theorem 10.7.4 (Zhang–Audenaert) Let X and Y be discrete random variables taking
values in a finite alphabet A. Let pX and pY denote their distributions, respectively. Thenthe following bound holds|H(X) − H(Y )| ≤ T log(|A| − 1) + h2 (T ),(10.102)where T ≡ 21 kpX − pY k1 . Furthermore, this bound is optimal, meaning that there exists apair of random variables saturating the bound for every T ∈ [0, 1] and alphabet size |A|.
In order to prove this theorem, we need to establish the notion of a maximal coupling of
two random variables.
Definition 10.7.2 (Coupling) A coupling of a pair (X, Y ) of two random variables is apair (X̂, Ŷ ) of two other random variables that have the same marginal distributions as thoseof X and Y .
Definition 10.7.3 (Maximal Coupling) A coupling (X̂, Ŷ ) of a pair of two random variables (X, Y ) is maximal if Pr{X̂ = Ŷ } takes on its maximum value, with respect to all
couplings of X and Y .
We can now relate the classical trace distance to a maximal coupling:
Lemma 10.7.2 Let X and Y be discrete random variables taking values in a finite alphabet A. Let pX and pY denote their distributions, respectively. If (X̂, Ŷ ) is a maximal couplingof X and Y , then the following equalities hold
X
Pr{X̂ = Ŷ } =min{pX (u), pY (u)},(10.103)u∈A

Pr{X̂ 6= Ŷ } =c 20191
kpX − pY k1 .
2

(10.104)

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10.7. ENTROPY INEQUALITIES307

Proof. Let B = {u ∈ A : pX (u) < pY (u)}. Let B c = A\B. Then for every coupling (X̂, Ŷ )of (X, Y ), the following holds
Pr{X̂ = Ŷ } = Pr{X̂ = Ŷ ∧ Ŷ ∈ B} + Pr{X̂ = Ŷ ∧ Ŷ ∈ B c }c≤ Pr{X̂ ∈ B} + Pr{Ŷ ∈ B }= Pr{X ∈ B} + Pr{Y ∈ B c }XX=pX (u) +pY (u)
=

u∈B

u∈Bc

X

min{pX (u), pY (u)} +=(10.106)(10.107)(10.108)
min{pX (u), pY (u)}(10.109)

u∈Bc

u∈B

X

X

(10.105)min{pX (u), pY (u)}.

(10.110)u∈A

We now establishP
a construction of a coupling that achieves the bound above, so that it
is maximal. Let q ≡ u∈A min{pX (u), pY (u)}. Let U , V , W , and J be independent discreterandom variables, with pJ (0) = 1 − q and pJ (1) = q. Let U , V , and W have the following
probability distributions:1[min{pX (u), pY (u)}] ,q1pV (v) =[pX (v) − min{pX (v), pY (v)}] ,1−q
1[pY (w) − min{pX (w), pY (w)}] ,pW (w) =1−qpU (u) =(10.111)(10.112)(10.113)

where u, v, w ∈ A. If J = 1, then let X̂ = Ŷ = U , and if J = 0, then let X̂ = V and Ŷ = W .
So we find that for all x, y ∈ ApX̂ (x) = q pX|J (x|1) + (1 − q) pX|J (x|0)= q pU (x) + (1 − q) pV (x)= pX (x).

(10.114)(10.115)(10.116)

By similar reasoning, it follows that pŶ (y) = pY (y), so that we can conclude that theconstructed (X̂, Ŷ ) is a coupling of (X, Y ). It is maximal because
Pr{X̂ = Ŷ } ≥ Pr {J = 1} = q,(10.117)and we can conclude from (10.110) that Pr{X̂ = Ŷ } = q. The equality in (10.104) follows
from (10.103) and the following equality min {a, b} = 21 [a + b − |a − b|], which holds for alla, b ∈ R.
We can finally give a proof of Theorem 10.7.4, by invoking the above results and Fano’s
inequality (Theorem 10.7.3).
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Proof of Theorem 10.7.4. Let (X̂, Ŷ ) be a maximal coupling of (X, Y ). Then
|H(X) − H(Y )| = H(X̂) − H(Ŷ )(10.118)= H(X̂) − H(X̂ Ŷ ) + H(X̂ Ŷ ) − H(Ŷ )(10.119)= H(X̂|Ŷ ) − H(Ŷ |X̂)no≤ max H(X̂|Ŷ ), H(Ŷ |X̂)

(10.120)≤ Pr{X̂ 6= Ŷ } log (|A| − 1) + h2 (Pr{X̂ 6= Ŷ })= T log (|A| − 1) + h2 (T ).

(10.122)(10.123)(10.121)

The first equality follows because (X̂, Ŷ ) is a maximal coupling of (X, Y ). The second
inequality is an application of Fano’s inequality (Theorem 10.7.3). The last equality follows
from (10.104).
The bound given in the statement of the theorem is optimal. An example of a pair ofprobability distributions pX and pY saturating the bound is
pX = 1 0 · · · 0 ,(10.124)
pY = 1 − ε ε/ (|A| − 1) · · · ε/ (|A| − 1) ,(10.125)
where ε ∈ [0, 1]. The normalized classical trace distance between pX and pY is 21 kpX − pY k1 =ε, while an explicit calculation reveals that
|H(X) − H(Y )| = H(Y ) = ε log (|A| − 1) + h2 (ε),(10.126)

concluding the proof.

10.8 Near Saturation of Entropy Inequalities
The entropy inequalities discussed in the previous section are foundational in classical information theory. In fact, one can prove the converse parts of many classical capacity theorems
by making use of these entropy inequalities. Thus, it seems worthwhile to study them inmore detail and to explore the possibility of more refined statements. For example, in eachof these entropy inequalities, there is a condition for when it is saturated: the non-negativityof relative entropy is saturated if and only if p = q and the non-negativity of mutual information is saturated if and only if random variables X and Y are independent. It is thus
a natural question to consider what kind of statement we can make when these entropy
inequalities are nearly saturated.

10.8.1

Pinsker Inequality

One of the main tools for refining entropy inequalities in the classical case is the Pinsker
inequality, which provides a relation between the relative entropy and the classical trace
distance.
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10.8. NEAR SATURATION OF ENTROPY INEQUALITIES

309

Theorem 10.8.1 (Pinsker Inequality) LetP p be a probability distribution on a finite alphabet X and let q : X → [0, 1] be such that x q(x) ≤ 1. Then
D(pkq) ≥1
kp − qk21 .
2 ln 2(10.127)

The utility of the Pinsker inequality is that it relates one measure of distinguishability
to another, and thus allows us to make precise statements about the near saturation of anentropy inequality. For example, Theorem 10.7.1 states that D(pkq) ≥ 0 for p and q as given
above. The Pinsker inequality is a strong refinement of this entropy inequality: it says morethan just D(pkq) ≥ 0, considering that kp − qk1 ≥ 0. Letting q be subnormalized as above
gives us a little more flexibility when applying the Pinsker inequality.
Before proving it, we establish the following lemma:
Lemma 10.8.1 Let a, b ∈ [0, 1]. Then a lnab

+ (1 − a) ln

1−a
1−b



≥ 2 (a − b)2 .

Proof. This bound follows by elementary calculus. First, if b = 0 or b = 1, then the boundholds trivially. So suppose that b ∈ (0, 1). Let us suppose furthermore (for now) that a ≥ b.
Consider the following function:


a
1−a
+ (1 − a) lng(a, b) ≡ a ln
− 2 (a − b)2 ,(10.128)b1−bso that g(a, b) corresponds to the difference of the left-hand side and the right-hand side in
the statement of the lemma. Thena 1−a∂g(a, b)=− +− 4 (b − a)∂bb
1−ba (1 − b) b (1 − a)=−+
− 4 (b − a)b (1 − b)b (1 − b)b−a=− 4 (b − a)b (1 − b)(b − a) (4b2 − 4b + 1)=
b (1 − b)(b − a) (2b − 1)2=
b (1 − b)≤ 0,(10.129)(10.130)(10.131)(10.132)(10.133)(10.134)

with the last step holding from the assumption that a ≥ b and b ∈ (0, 1). Also, observe thatboth ∂g(a, b)/∂b = 0 and g(a, b) = 0 when a = b. Thus, the function g(a, b) is decreasing inb for every a whenever b ≤ a and reaches a minimum when b = a. So g(a, b) ≥ 0 whenevera ≥ b. The lemma holds in general by applying it to a0 ≡ 1 − a and b0 ≡ 1 − b so that b0 ≥ a0 .

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CHAPTER 10. CLASSICAL INFORMATION AND ENTROPY

Proof of Pinsker’s inequality (Theorem 10.8.1). The main idea is to exploit the monotonicity of relative entropy and the lemma above. First let us suppose that q is a probability
distribution. Let A denote a classical “coarse-graining” channel that outputs 1 if x ∈ A and
0 if x ∈ Ac , where A is the set defined in Lemma 10.7.1 (i.e., A ≡ {x : p(x) ≥ q(x)}). Then
D(pkq) ≥ D(A(p)kA(q))




p(A)
1 − p(A)= p(A) log+ (1 − p(A)) logq(A)1 − q(A)2≥(p(A) − q(A))2
ln 2
2

1
2
kp − qk1=
ln 2 2
1
kp − qk21 .
=2 ln 2(10.135)(10.136)(10.137)(10.138)(10.139)

The first inequality follows from monotonicity of the relative entropy (Corollary 10.7.2),
with the quantities p(A) and q(A) defined in Lemma 10.7.1. The second inequality is anapplication of Lemma 10.8.1, where we gain the factor ln 2 by converting from natural
logarithm to binary logarithm. The second equality
P follows from Lemma 10.7.1.
If q is not a probability distribution (so that x q(x) < 1), then we can add an extra
letter to X to make q a probability distribution. That is, we define a probabilityPdistribution
q 0 to have its first |X | entries to be those from q and its last entry to be 1 − x q(x). We
also define an augmented distribution p0 toP
have its first |X | entries to be those from p andits last entry to be 0. Then 0 · log (0/ [1 − x q(x)]) = 0, so thatD(pkq) = D(p0 kq 0 )12≥
kp0 − q 0 k1
2 ln 21=
2 ln 2≥(10.140)(10.141)#!2"
kp − qk1 + 1 −

X

q(x)

(10.142)x1
kp − qk21 ,2 ln 2(10.143)

concluding the proof.

10.8.2

Refinements of Entropy Inequalities

The first refinement of an entropy inequality that we give is regarding the non-negativity ofmutual information and its maximal value:
Theorem 10.8.2 Let X and Y be discrete random variables taking values in alphabets X
and Y, respectively, let pXY denote their joint distribution, and let pX ⊗pY denote the productc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

10.8. NEAR SATURATION OF ENTROPY INEQUALITIES

311

of their marginal distributions. Then
I(X; Y ) ≥

2 2
∆,
ln 2

(10.144)andI(X; Y ) ≤ ∆ log(min{|X | , |Y|} − 1) + h2 (∆),

(10.145)

where ∆ = 21 kpXY − pX ⊗ pY k1 .
Proof. The first inequality is a direct application of the Pinsker inequality (Theorem 10.8.1),
noting that
I(X; Y ) = D(pXY kpX ⊗ pY ).
(10.146)
To prove the second inequality, let X̂ and Ŷ denote a pair of random variables with joint
distribution pX ⊗ pY . Then
I(X; Y ) = I(X; Y ) − I(X̂; Ŷ )(10.147)= H(X|Y ) − H(X̂|Ŷ )(10.148)=XhipY (y) H(X|Y = y) − H(X̂|Ŷ = y) .

(10.149)y

The first equality follows because I(X̂; Ŷ ) = 0. The second equality follows because I(X; Y ) =H(X) − H(X|Y ), I(X̂; Ŷ ) = H(X̂) − H(X̂|Ŷ ), and H(X) = H(X̂). The third equality follows from the definition of conditional entropy. Continuing,X(10.150)≤pY (y) H(X|Y = y) − H(X̂|Ŷ = y)y≤XpY (y) [∆(y) log(|X | − 1) + h2 (∆(y))](10.151)y≤ ∆ log(|X | − 1) + h2 (∆).

(10.152)

The first inequality is a consequence of the triangle inequality. The second inequality follows
from an application of Theorem 10.7.4, and from defining
1X∆(y) =
pX|Y (x|y) − pX (x) .
(10.153)
2 x
The final inequality follows because
1X|pXY (x, y) − pX (x)pY (y)|
∆=
2 x,y
"
#
X
1X
=
pY (y)
pX|Y (x|y) − pX (x)
2
y
x
X
=
pY (y)∆(y),(10.154)(10.155)(10.156)yc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 10. CLASSICAL INFORMATION AND ENTROPY

and from the fact that the binary entropy is concave. We obtain the other boundI(X; Y ) ≤ ∆ log(|Y| − 1) + h2 (∆),

(10.157)

by expanding the mutual information as I(X; Y ) = |I(X; Y ) − I(X̂; Ŷ )| = |H(Y |X) −H(Ŷ |X̂)| and proceeding in a similar fashion.
We can make a similar kind of statement for the conditional mutual information:
Theorem 10.8.3 Let X, Y , and Z be discrete random variables taking values in alphabets
X , Y, and Z, respectively, let pXY Z denote their joint distribution, and let pX|Z pY |Z pZ denote
another distribution which corresponds to the Markov chain X − Z − Y . Then
I(X; Y |Z) ≥

2 2
∆,
ln 2

(10.158)

andI(X; Y |Z) ≤ ∆ log(min{|X | , |Y|} − 1) + h2 (∆),

(10.159)

where ∆ = 21 pXY Z − pX|Z pY |Z pZ 1 .
Proof. A proof
P for the first inequality is similar to the proof of (10.144). One can write
I(X; Y |Z) = z pZ (z)I(X; Y |Z = z), apply the Pinsker inequality (Theorem 10.8.1), andthen convexity of the square function and the trace norm. Alternatively, one can directlycompute that I(X; Y |Z) = D(pXY Z kpX|Z pY |Z pZ ) and then apply the Pinsker inequality(Theorem 10.8.1).
A proof for the second inequality is similar to that for (10.145). Let X̂, Ŷ , and Ẑ denotea triple of random variables with joint distribution pX|Z pY |Z pZ . Then
I(X; Y |Z) = I(X; Y |Z) − I(X̂; Ŷ |Ẑ)

(10.160)hi= H(Y |Z) − H(Y |XZ) − H(Ŷ |Ẑ) − H(Ŷ |X̂ Ẑ)(10.161)= H(Ŷ |X̂ Ẑ) − H(Y |XZ) ,(10.162)

where the first equality follows because I(X̂; Ŷ |Ẑ) = 0, the second by expanding the conditional mutual informations, and the third because H(Y |Z) = H(Ŷ |Ẑ). The rest of the steps
proceed as in the proof of (10.145), and we end up with
I(X; Y |Z) ≤ ∆ log(|Y| − 1) + h2 (∆).

(10.163)

A proof for I(X; Y |Z) ≤ ∆ log(|X | − 1) + h2 (∆) follows by expanding the conditional mutual
information I(X; Y |Z) in the other way, as I(X; Y |Z) = H(X|Z) − H(X|Y Z).
Finally, the following theorem gives a strong refinement of the monotonicity of relative
entropy (Corollary 10.7.2). An important implication is that if the relative entropy decreaseD(pkq) − D(N pkN q) is not too large under the action of a classical channel N , then one
can perform a recovery channel R, satisfying (10.165), such that the recovered distribution
RN p is close to the original distribution p.
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10.8. NEAR SATURATION OF ENTROPY INEQUALITIES

313

Theorem 10.8.4 (Refined Monotonicity of Relative Entropy) Let p be a probability
distribution on a finite alphabet X and let q : X → [0, ∞) be a function such that supp(p) ⊆supp(q). Let N (y|x) be a conditional probability distribution (classical channel). Then thefollowing refinement of the monotonicity of relative entropy holds:
D(pkq) − D(N pkN q) ≥ D(pkRN p)(10.164)
P
where N p is a probability distribution
with elements (N p)(y) ≡ x N (y|x)p(x), N q is a vecP
tor with elements (N q)(y) = x N (y|x)q(x), and the recovery channel R(x|y) (a conditional
probability distribution) is defined by the following set of equations:
R(x|y)(N q)(y) = N (y|x)q(x),(10.165)

which correspond to the Bayes theorem if q is a probability
distribution. Also, RN p is aP
probability distribution with elements (RN p)(x) = y,x0 R(x|y)N (y|x0 )p(x0 ).
Proof. Consider that
(RN p)(x) =X N (y|x)q(x)y= q(x)(N q)(y)[(N p)(y)]X N (y|x)(N p)(y)(N q)(y)y(10.166).

(10.167)

Thus,
D(pkRN p) =X

p(x) log

x

p(x)
q(x)

!
,

P N (y|x)(N p)(y)y(N q)(y)and by definition we have that



p(x)
D(pkq) =p(x) log
,
q(x)
x


X
(N p)(y)(N p)(y) log
D(N pkN q) =.
(N q)(y)yX(10.168)(10.169)

Then
D(pkq) − D(pkRN p) =X

p(x) log

X N (y|x)(N p)(y)!

(N q)(y)XX(N p)(y)≥p(x)
N (y|x) log(N q)(y)
x
y
"
#


X X
(N p)(y)=N (y|x)p(x) log(N q)(y)yxX(N p)(y)=(N p)(y) log(N q)(y)yx= D(N pkN q).
c 2019(10.170)y(10.171)(10.172)(10.173)(10.174)

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CHAPTER 10. CLASSICAL INFORMATION AND ENTROPY

The sole inequality is a consequence of concavity of the logarithm.

10.9 Classical Information from Quantum Systems
We can always process classical information by employing a quantum system as the carrierof information. The inputs and the outputs to a quantum protocol can both be classical.
For example, we can prepare a quantum state according to some random variable X—
the ensemble {pX (x), ρx } captures this idea. We can retrieve classical information from aquantum state in the form of some random variable Y by performing a measurement—
the POVM {Λy } captures this notion (recall that we employ the POVM formalism from
Section 4.2.1 if we do not care about the state after the measurement). Suppose that Alice
prepares a quantum state according to the ensemble {pX (x), ρx } and Bob measures thestate according to the POVM {Λy }. Recall that the following formula gives the conditional
probability pY |X (y|x):
pY |X (y|x) = Tr {Λy ρx } .
(10.175)
Is there any benefit to processing classical information using quantum systems? Later,
in Chapter 20, we see that there indeed is an enhanced performance because we can achieve
higher communication rates in general by processing classical data using quantum resources.
For now, we extend our notions of entropy in a straightforward way to include the above
ideas.

10.9.1

Shannon Entropy of a POVM

The first notion that we can extend is the Shannon entropy, by determining the Shannon
entropy of a POVM. Suppose that Alice prepares a quantum state ρ (there is no classicalindex here). Bob can then perform a particular POVM {Λx } to learn about the quantum
system. Let X denote the random variable corresponding to the classical output of the
POVM. The probability density function pX (x) of random variable X is thenpX (x) = Tr {Λx ρ} .

(10.176)

The Shannon entropy H(X) of the POVM {Λx } is
H(X) = −X=−

X

pX (x) log (pX (x))(10.177)

Tr {Λx ρ} log (Tr {Λx ρ}) .

(10.178)xx

In the next chapter, we prove that the minimum Shannon entropy with respect to all rank-one
POVMs is equal to a quantity known as the quantum entropy of the density operator ρ.
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10.9. CLASSICAL INFORMATION FROM QUANTUM SYSTEMS

10.9.2

315

Accessible Information

Let us consider the scenario introduced at the beginning of this section, in which Alice
prepares an ensemble E ≡ {pX (x), ρx } and Bob performs a POVM {Λy }. Suppose now thatBob is actually trying to retrieve as much information as possible about the random variable
X. The quantity that governs how much information he can learn about random variable
X if he possesses random variable Y is the mutual information I(X; Y ). But here, Bob
can actually choose which measurement he would like to perform, and it would be good for
him to perform the measurement that maximizes his information about X. The resulting
quantity is known as the accessible information Iacc (E) of the ensemble E (because it is theinformation that Bob can access about random variable X):
Iacc (E) ≡ max I(X; Y ),{Λy }(10.179)

where the marginal density pX (x) is that from the ensemble and the conditional density
pY |X (y|x) is given in (10.175). In the next chapter, we show how to obtain a natural boundon this quantity, called the Holevo bound. The bound arises from a quantum generalizationof the data-processing inequality.

10.9.3

Classical Mutual Information of a Bipartite State

A final quantity that we introduce is the classical mutual information Ic (ρAB ) of a bipartite
state ρAB . Suppose that Alice and Bob possess some bipartite state ρAB and would like toextract maximal classical correlation from it. That is, they each retrieve a random variableby performing respective local POVMs {ΛxA } and {ΛyB } on their shares of the bipartite stateρAB . These measurements produce respective random variables X and Y , and they would
like X and Y to be as correlated as possible. A good measure of their resulting classicalcorrelations obtainable from local quantum information processing is as follows:
Ic (ρAB ) ≡max I(X; Y ),{ΛxA },{ΛyB }(10.180)

where the joint distribution
pX,Y (x, y) ≡ Tr {(ΛxA ⊗ ΛyB ) ρAB } .
Suppose that the state ρAB is classical, that is, it has the formXρAB =pX,Y (x, y)|xihx|A ⊗ |yihy|B ,(10.181)(10.182)x,ywhere the states |xiA form an orthonormal basis and so do the states |yiB . Then, the optimalmeasurement in this case is for Alice to perform a complete projective measurement in thebasis |xiA and inform Bob to perform a similar measurement in the basis |yiB . The amountof correlation they extract is then equal to I(X; Y ).
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CHAPTER 10. CLASSICAL INFORMATION AND ENTROPY

Exercise 10.9.1 Prove that it suffices to consider maximizing with respect to rank-one
POVMs when computing (10.180). (Hint: Consider refining the
P POVM {Λx } as the rank-one
POVM {|φx,z ihφx,z |}, where we spectrally decompose Λx as z |φx,z ihφx,z |, and then exploit
the data-processing inequality.)

10.10 History and Further Reading
Cover and Thomas (2006) have given an excellent introduction to entropy and information
theory (some of the material in this chapter is similar to material appearing in that book).
MacKay (2003) also gives a good introduction. E. T. Jaynes was an advocate of the principleof maximum entropy, proclaiming its utility in several sources (Jaynes, 1957a,b, 2003). A
good exposition of Fano’s inequality appears on Scholarpedia (Fano, 2008). Theorem 10.7.4(continuity of entropy bound) is due to Zhang (2007) and Audenaert (2007). The particular
proof that we give in this chapter is Zhang’s (Zhang, 2007), but we used the presentation of
Sason (2013). The Pinsker inequality was first proved by Pinsker (1960), with subsequentenhancements by Csiszar (1967); Kemperman (1969); Kullback (1967). Theorem 10.8.2 wasrealized in private communication with Andreas Winter (August 2015). The refinement of
the monotonicity of relative entropy in Theorem 10.8.4 was established by Li and Winter(2014).

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 11

Quantum Information and Entropy
In this chapter, we discuss several information measures that are important for quantifying
the amount of information and correlations that are present in quantum systems. The first
fundamental measure that we introduce is the von Neumann entropy (or simply quantum
entropy). It is the quantum generalization of the Shannon entropy, but it captures bothclassical and quantum uncertainty in a quantum state.1 The quantum entropy gives meaning
to the notion of an information qubit. This notion is different from that of the physical qubit,
which is the description of a quantum state of an electron or a photon. The information
qubit is the fundamental quantum informational unit of measure, determining how muchquantum information is present in a quantum system.
The initial definitions here are analogous to the classical definitions of entropy, but wesoon discover a radical departure from the intuitive classical notions from the previous chapter: the conditional quantum entropy can be negative for certain quantum states. In theclassical world, this negativity simply does not occur, but it takes on a special meaning inquantum information theory. Pure quantum states that are entangled have stronger-thanclassical correlations and are examples of states that have negative conditional entropy. The
negative of the conditional quantum entropy is so important in quantum information theory
that we even have a special name for it: the coherent information. We discover that the coherent information obeys a quantum data-processing inequality, placing it on a firm footing
as a particular informational measure of quantum correlations.
We then define several other quantum information measures, such as quantum mutual
information, that bear similar definitions as in the classical world, but with Shannon entropies
replaced with quantum entropies. This replacement may seem to make quantum entropy
1

We should point out the irony in the historical development of classical and quantum entropy. The von
Neumann entropy has seen much widespread use in modern quantum information theory, and perhaps thiswould make one think that von Neumann discovered this quantity much after Shannon. But in fact, thereverse is true. Von Neumann first discovered what is now known as the von Neumann entropy and applied
it to questions in statistical physics. Much later, Shannon determined an information-theoretic formula andasked von Neumann what he should call it. Von Neumann told him to call it the entropy for two reasons:
1) it was a special case of the von Neumann entropy and 2) he would always have the advantage in a debate
because von Neumann claimed that no one at the time really understood entropy.

317

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CHAPTER 11. QUANTUM INFORMATION AND ENTROPY

somewhat trivial on the surface, but a simple calculation reveals that a maximally entangledstate on two qubits registers two bits of quantum mutual information (recall that the largest
the mutual information can be in the classical world is one bit for the case of two maximallycorrelated bits). We then discuss several entropy inequalities that play an important rolein quantum information processing: the monotonicity of quantum relative entropy, strong
subadditivity, the quantum data-processing inequalities, and continuity of quantum entropy.

11.1 Quantum Entropy
We might expect a measure of the entropy of a quantum system to be vastly different from theclassical measure of entropy from the previous chapter because a quantum system possesses
not only classical uncertainty but also quantum uncertainty that arises from the uncertainty
principle. But recall that the density operator captures both types of uncertainty and allows
us to determine probabilities for the outcomes of any measurement on a given system. Thus,a quantum measure of uncertainty should be a direct function of the density operator, justas the classical measure of uncertainty is a direct function of a probability density function.
It turns out that this function has a strikingly similar form to the classical entropy, as wesee below.
Definition 11.1.1 (Quantum Entropy) Suppose that Alice prepares some quantum system A in a state ρA ∈ D(HA ). Then the entropy H(A)ρ of the state is defined as follows:
H(A)ρ ≡ − Tr {ρA log ρA } .

(11.1)The entropy of a quantum system is also known as the von Neumann entropy or thequantum entropy but we often simply refer to it as the entropy. We can denote it by H(A)ρ
or H(ρA ) to show the explicit dependence on the density operator ρA . The quantum entropy
has a special relation to the eigenvalues of the density operator, as the following exercise
asks you to verify.
Exercise 11.1.1 Consider a density operator ρA with the following spectral decomposition:XρA =pX (x)|xihx|A .
(11.2)xShow that the quantum entropy H(A)ρ is the same as the Shannon entropy H(X) of a randomvariable X with probability distribution pX (x).
In our definition of quantum entropy, we use the same notation H as in the classical caseto denote the entropy of a quantum system. It should be clear from the context whether weare referring to the entropy of a quantum or classical system.
The quantum entropy admits an intuitive interpretation. Suppose that Alice generates
a quantum state |ψy i in her lab according to some probability density pY (y), corresponding
to a random variable Y . Suppose further that Bob has not yet received the state from Alicec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

11.1. QUANTUM ENTROPY319

and does not know which one she sent. The expected density operator from Bob’s point ofview is thenX
σ = EY {|ψY ihψY |} =pY (y)|ψy ihψy |.
(11.3)y

The interpretation of the entropy H(σ) is that it quantifies Bob’s uncertainty about the state
Alice sent—his expected information gain is H(σ) qubits upon receiving and measuring thestate that Alice sends. Schumacher’s noiseless quantum coding theorem, described in Chapter 18, gives an alternative operational interpretation of the quantum entropy by provingthat Alice needs to send Bob qubits at a rate H(σ) in order for him to be able to decode acompressed quantum state.
The above interpretation of quantum entropy seems qualitatively similar to the interpretation of classical entropy. However, there is a significant quantitative difference that
illuminates the difference between Shannon entropy and quantum entropy. Let us consider
an example. Suppose that Alice generates a sequence |ψ1 i |ψ2 i · · · |ψn i of quantum statesaccording to the following “BB84” ensemble:{{1/4, |0i} , {1/4, |1i} , {1/4, |+i} , {1/4, |−i}} .

(11.4)

Suppose that Alice and Bob share a noiseless classical channel. If she employs Shannon’sclassical noiseless coding protocol, she should transmit classical data to Bob at a rate of twoclassical channel uses per source state |ψi i in order for him to reliably recover the classicaldata needed to reproduce the sequence of states that Alice transmitted (the Shannon entropyof the uniform distribution 1/4 is 2 bits).
Now let us consider computing the quantum entropy of the above ensemble. First, we
determine the expected density operator of Alice’s ensemble:
1(|0ih0| + |1ih1| + |+ih+| + |−ih−|) = π,(11.5)
4
where π is the maximally mixed state. The quantum entropy of the above density operator
is one qubit because the eigenvalues of π are both equal to 1/2. Suppose now that a noiselessquantum channel connects Alice to Bob—this is a channel that can preserve quantum coherence without any interaction with an environment. Then Alice only needs to send qubits at
a rate of one channel use per source symbol if she employs a protocol known as Schumacher
compression (we discuss this protocol in detail in Chapter 18). Bob can then reliably decode
the qubits that Alice sent. The protocol also causes a slight disturbance to the state, which
however vanishes in the limit of many invocations of the source. The above departure fromclassical information theory holds in general—Exercise 11.9.3 of this chapter asks you toprove that the Shannon entropy of any ensemble is never less than the quantum entropy of
its expected density operator.

11.1.1Mathematical Properties of Quantum Entropy

We now discuss several mathematical properties of the quantum entropy: non-negativity, its
minimum value, its maximum value, its invariance with respect to isometries, and concavity.
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The first three of these properties follow from the analogous properties in the classical world
because the quantum entropy of a density operator is the Shannon entropy of its eigenvalues(see Exercise 11.1.1). We state them formally below:
Property 11.1.1 (Non-Negativity) The quantum entropy H(ρ) is non-negative for any
density operator ρ:H(ρ) ≥ 0.
(11.6)
Proof. This follows from non-negativity of Shannon entropy.
Property 11.1.2 (Minimum Value) The minimum value of the quantum entropy is zero,and it occurs when the density operator is a pure state.
Proof. The minimum value occurs when the eigenvalues of a density operator are distributedwith all the probability mass on one eigenvector and zero on the others, so that the density
operator is rank one and corresponds to a pure state.
Why should the entropy of a pure quantum state vanish? It seems that there is quantum
uncertainty inherent in the state itself and that a measure of quantum uncertainty shouldcapture this fact. This last observation only makes sense if we do not know anything about
the state that is prepared. But if we know exactly how it is prepared, we can performa special quantum measurement to verify this, and we do not learn anything from thismeasurement because the outcome of it is always certain. For example, suppose that Alice
prepares the state |φi and Bob knows that she does so. He can then perform the following
measurement {|φihφ|, I − |φihφ|} to verify that she prepared this state. He always receives
the first outcome from the measurement and thus never gains any information from it. Thus,in this sense it is reasonable that the entropy of a pure state vanishes.
Property 11.1.3 (Maximum Value) The maximum value of the quantum entropy is log dwhere d is the dimension of the system, and it occurs for the maximally mixed state.
Proof. A proof of the above property is the same as that for the classical case.
Property 11.1.4 (Concavity) Let ρx ∈ D(H) and let pX (x) be a probability distribution.
The entropy is concave in the density operator:
X
H(ρ) ≥pX (x)H(ρx ),(11.7)
xwhere ρ ≡P

x pX (x)ρx .

The physical interpretation of concavity is as before for classical entropy: entropy cannever decrease under a mixing operation. This inequality is a fundamental property of the entropy, and we prove it after developing some important entropic tools (see Exercise 11.6.10).
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Property 11.1.5 (Isometric Invariance) Let ρ ∈ D(H) and U : H → H0 be an isometry.
The entropy of a density operator is invariant with respect to isometries, in the following
sense:
H(ρ) = H(U ρU † ).
(11.8)
Proof. Isometric invariance of entropy follows by observing that the eigenvalues of a density
operator are invariant with respect to an isometry:X
U ρU † = U
pX (x)|xihx|U †(11.9)x=

X

pX (x)|φx ihφx |,(11.10)x

where {|φx i} is some orthonormal basis such that U |xi = |φx i. The above property follows
because the entropy is a function of the eigenvalues of a density operator.
A unitary or isometric operator is a quantum generalization of a permutation in thiscontext (recall Property 10.1.3 of the classical entropy).

11.1.2

Alternate Characterization of Quantum Entropy

There is an interesting alternate characterization of the quantum entropy of a state ρ as theminimum Shannon entropy when a rank-one POVM is performed on it (we discussed thisbriefly in Section 10.9.1). In this sense, there is some optimal measurement to perform on ρsuch that its entropy is equal to the quantum entropy, and this optimal measurement is the“right question to ask” (as we discussed very early on in Section 1.2.2).
Theorem 11.1.1 Let ρ ∈ D(H). The quantum entropy H(ρ) has the following characterization:#"
X
Tr {Λy ρ} log (Tr {Λy ρ}) ,(11.11)
H(ρ) = min −
{Λy }

y

where the minimum is restricted to be with respect to rank-one
PPOVMs (those with Λy =|φy ihφy | for some vectors |φy i such that Tr {|φy ihφy |} ≤ 1 and y |φy ihφy | = I).
Proof. In order to prove the above result, we should first realize that a complete projective
measurement in the eigenbasis of ρ should achieve the minimum. That is, if ρ =Pp(x)|xihx|,
we should expect that the measurement {|xihx|} achieves the minimum.
x X
In this case, the Shannon entropy of the measurement is equal to the Shannon entropy ofpX (x), as discussed in Exercise 11.1.1.
We now prove that any other rank-one POVM has a higher entropy than that given by thismeasurement. Consider that the distribution of the measurement outcomes for {|φy ihφy |} isequal toX
Tr {|φy ihφy |ρ} =|hφy |xi|2 pX (x),(11.12)x
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so that we can think of |hφy |xi|2 as a conditional probability distribution. Introducing
η(p) ≡ −p log p, which is a concave function, we can write the quantum entropy as
X
H(ρ) =η(pX (x))(11.13)x=X

η(pX (x)) + η(pX (x0 )),(11.14)xwhere x0 is a symbol added to the alphabet of x such thatPpX (x0 ) = 0. Let us denote
thealphabet with the symbols x0 so that H(ρ) = x0 η(pX (x0 )). We know that
P enlarged2
y |hφy |xi| = 1 from the fact that the set {|φy ihφy |} forms a POVM and |xi is a normalizedPstate. We also know that x |hφy |xi|2 ≤ 1 because Tr {|φy ihφy |} ≤ 1 for a rank-one POVM.
Thinking of |hφy |xi|2 as a distribution over x, we can add a symbol x0 with probability
1 − hφy |φy i so that it makes a normalized distribution. Let us call this distribution p(x0 |y).
We then have thatX
η(pX (x))(11.15)
H(ρ) =
x

=

X

=

X

|hφy |xi|2 η(pX (x))(11.16)p(x0 |y)η(pX (x0 ))(11.17)

x,y

x0 ,y

!
=

X X

p(x0 |y)η(pX (x0 ))(11.18)

x0

y

!
≤

X
y

=

X

η

X

p(x0 |y)pX (x0 )(11.19)x0

η (Tr {|φy ihφy |ρ}) .

(11.20)y

The third equality follows from the fact that pX (x0 ) = 0 for the added symbol x0 . The only
inequality follows from concavity of η. The last expression is equal to the Shannon entropyof the POVM {|φy ihφy |} when performed on the state ρ.

11.2 Joint Quantum Entropy
The joint quantum entropy H(AB)ρ of the density operator ρAB ∈ D(HA ⊗ HB ) for a
bipartite system AB follows naturally from the definition of quantum entropy:
H(AB)ρ ≡ − Tr {ρAB log ρAB } .

(11.21)

Now suppose that ρABC is a tripartite state, i.e., in D(HA ⊗ HB ⊗ HC ). Then the entropy
H(AB)ρ in this case is defined as above, where ρAB = TrC {ρABC }. This is a convention
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that we take throughout this book. We introduce a few of the properties of joint quantum
entropy in the subsections below.

11.2.1Marginal Entropies of a Pure Bipartite StateThe five properties of quantum entropy in the previous section may give you the impressionthat the nature of quantum information is not too different from that of classical information.
We proved all these properties for the classical case, and their proofs for the quantum caseseem similar. The first three even resort to the proofs in the classical case!
Theorem 11.2.1 below is where we observe our first radical departure from the classicalworld. It states that the marginal entropies of a pure bipartite state are equal, while theentropy of the overall state is equal to zero. Recall that the joint entropy H(X, Y ) of two
random variables X and Y is never less than either of the marginal entropies H(X) or H(Y ):H(X, Y ) ≥ H(X),

H(X, Y ) ≥ H(Y ).

(11.22)

The above inequalities follow from the non-negativity of classical conditional entropy. But
in the quantum world, these inequalities do not always have to hold, and the following
theorem demonstrates that they do not hold for an arbitrary pure bipartite quantum statewith Schmidt rank greater than one (see Theorem 3.8.1 for a definition of Schmidt rank).
The fact that the joint quantum entropy can be less than the marginal quantum entropy is
one of the most fundamental differences between classical and quantum information.
Theorem 11.2.1 The marginal entropies H(A)φ and H(B)φ of a pure bipartite state |φiAB
are equal:
H(A)φ = H(B)φ ,(11.23)
while the joint entropy H(AB)φ vanishes:
H(AB)φ = 0.

(11.24)

Proof. The crucial ingredient for a proof of this theorem is the Schmidt decomposition(Theorem 3.8.1). Recall that any bipartite state |φiAB admits a Schmidt decomposition of
the following form:
Xp
λi |iiA |iiB ,(11.25)|φiAB =
iP

where λi > 0 for all i, i λi = 1, {|iiA } is some orthonormal set of vectors on system A,and {|iiB } is some orthonormal set on system B. Recall that the Schmidt rank is equal to
the number of non-zero coefficients λi . Then the respective marginal states ρA and ρB on
systems A and B are as follows:XXρA =λi |iihi|A ,ρB =λi |iihi|B .
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Thus, the marginal states admit a spectral decomposition with the same eigenvalues. The
theorem follows because the quantum entropy depends only on the eigenvalues of a given
spectral decomposition.
The theorem applies not only to two systems A and B, but it also applies to any numberof systems if we make a bipartite cut of the systems. For example, if the state is |φiABCDE ,then the following equalities (and others from different combinations) hold by applying Theorem 11.2.1 and Remark 3.8.1:
H(A)φ = H(BCDE)φ ,H(AB)φ = H(CDE)φ ,H(ABC)φ = H(DE)φ ,H(ABCD)φ = H(E)φ .

(11.27)(11.28)(11.29)(11.30)The closest analogy in the classical world to the above property is when we copy a randomvariable X. That is, suppose that X has a distribution pX (x) and X̂ is some copy of it so thatthe distribution of the joint random variable X X̂ is pX (x)δx,x̂ . Then the marginal entropies
H(X) and H(X̂) are both equal. But observe that the joint entropy H(X X̂) is also equalto H(X) and this is where the analogy breaks down. That is, there is not a strong classicalanalogy of the notion of purification.

11.2.2

Additivity

Property 11.2.1 (Additivity) Let ρA ∈ D(HA ) and σB ∈ D(HB ). The quantum entropy
is additive for tensor-product states:
H(ρA ⊗ σB ) = H(ρA ) + H(σB ).

(11.31)

One can verify this property simply by diagonalizing both density operators and resortingto the additivity of the joint Shannon entropies of the eigenvalues.
Additivity is an intuitive property that we would like to hold for any measure of information. For example, suppose that Alice generates a large sequence |ψx1 i |ψx2 i · · · |ψxn i ofquantum states according to the ensemble {pX (x), |ψx i}. She may be aware of the classicalindices x1 x2 · · · xn , but a third party to whom she sends the quantum sequence may not beaware of these values. The description of the state to this third party is then ρ ⊗ · · · ⊗ ρ,
where ρ ≡ EX {|ψX ihψX |}, and the quantum entropy of this n-fold tensor product state isH(ρ ⊗ · · · ⊗ ρ) = nH(ρ), by applying (11.31) inductively.

11.2.3

Joint Quantum Entropy of a Classical–Quantum State

Recall from Definition 4.3.5 that a classical–quantum state is a bipartite state in which aclassical system and a quantum system are classically correlated. An example of such a stateis as follows:XρXB ≡pX (x)|xihx|X ⊗ ρxB .
(11.32)x
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11.2. JOINT QUANTUM ENTROPY325

The joint quantum entropy of this state takes on a special form that appears similar to
entropies in the classical world.
Theorem 11.2.2 The joint entropy H(XB)ρ of a classical–quantum state, as given in(11.32), is as follows:
X
H(XB)ρ = H(X) +pX (x)H(ρxB ),(11.33)xwhere H(X) is the entropy of a random variable X with distribution pX (x).
Proof. Consider that H(XB)ρ = − Tr {ρXB log ρXB } . So we need to evaluate the operator
log ρXB , and we can find a simplified form for it because ρXB is a classical-quantum state:#"X(11.34)log ρXB = log
pX (x)|xihx|X ⊗ ρxB
= log

" x
X

#
|xihx|X ⊗ pX (x)ρxB(11.35)|xihx|X ⊗ log [pX (x)ρxB ] .

(11.36)

x

=

X
x

Then
− Tr {ρXB log ρXB }
("
#"
#)
h
i
X
X
0
= − TrpX (x)|xihx|X ⊗ ρxB|x0 ihx0 |X ⊗ log pX (x0 )ρxB
= − Tr(X(11.37)x0x)pX (x)|xihx|X ⊗ (ρxB log [pX (x)ρxB ])(11.38)x=−

X

pX (x) Tr {ρxB log [pX (x)ρxB ]} .

(11.39)x

Consider thatlog [pX (x)ρxB ] = log (pX (x)) I + log ρxB ,(11.40)which implies that (11.39) is equal to−X

pX (x) [Tr {ρxB log [pX (x)]} + Tr {ρxB log ρxB }](11.41)X(11.42)x=−

pX (x) [log [pX (x)] + Tr {ρxB log ρxB }] .

xThis last line is then equivalent to the statement of the theorem.
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11.3 Potential yet Unsatisfactory Definitions of Conditional Quantum Entropy
The conditional quantum entropy may perhaps seem a bit difficult to define at first because
there is no formal notion of conditional probability in the quantum theory. However, there
are two senses which are perhaps closest to the notion of conditional probability, but both ofthem do not lead to satisfactory definitions of conditional quantum entropy. Nevertheless,
it is instructive for us to explore both of these notions briefly. The first arises in the noisy
quantum theory, and the second arises in the purified quantum theory.
We develop the first notion. Consider an arbitrary bipartite state ρAB . Suppose that
Alice performs a complete projective measurement Π ≡ {|xihx|} of her system, where {|xi}is an orthonormal basis. This procedure leads to an ensemble {pX (x), |xihx|A ⊗ ρxB }, where1
TrA {(|xihx|A ⊗ IB ) ρAB (|xihx|A ⊗ IB )} ,pX (x)pX (x) ≡ Tr {(|xihx|A ⊗ IB ) ρAB } .
ρxB ≡(11.43)(11.44)One could then think of the density operators ρxB as being conditioned on the outcome of
the measurement, and these density operators describe the state of Bob given knowledge ofthe outcome of the measurement.
We could potentially define a conditional entropy as follows:
X
H(B|A)Π ≡pX (x)H(ρxB ),(11.45)xin analogy with the definition of the classical entropy in (10.17). This approach might seem tolead to a useful definition of conditional quantum entropy, but the problem with it is that theentropy depends on the measurement chosen (the notation H(B|A)Π explicitly indicates thisdependence). This problem does not occur in the classical world because the probabilities forthe outcomes of measurements do not themselves depend on the measurement selected, unless
we apply some coarse graining to the outcomes. However, this dependence on measurement
is a fundamental aspect of the quantum theory.
We could then attempt to remove the dependence of the above definition on a particularmeasurement Π by defining the conditional quantum entropy to be the minimization of
H(B|A)Π with respect to all possible measurements. The intuition here is perhaps thatentropy should be the minimal amount of conditional uncertainty in a system after employing
the best possible measurement on the other. However, the removal of one problem leads toanother! This optimized conditional entropy is now difficult to compute as the system grows
larger, whereas in the classical world, the computation of conditional entropy is simple if one
knows the conditional probabilities. The above idea is useful, but we leave it for now because
there is a simpler definition of conditional quantum entropy that plays a fundamental rolein quantum information theory.
The second notion of conditional probability is actually similar to the above notion,
though we present it in the purified viewpoint. Consider a tripartite state |ψiABC and ac 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

11.4. CONDITIONAL QUANTUM ENTROPY

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bipartite cut A|BC of the systems A, B, and C. Theorem 3.8.1 states that every bipartite
state admits a Schmidt decomposition, and the state |ψiABC is no exception. Thus, we canwrite a Schmidt decomposition for it as follows:
Xp|ψiABC =pX (x)|xiA |φx iBC ,(11.46)x

where pX (x) is some probability density, {|xi} is an orthonormal basis for the system A, and{|φx i} is an orthonormal basis for the systems BC. Each state |φx iBC is a pure bipartite
state, so we can again apply a Schmidt decomposition to each of these states:
Xq
pY |X (y|x)|yx iB |yx iC ,(11.47)|φx iBC =ywhere pY |X (y|x) is some conditional probability distribution depending on the value of x,and {|yx iB } and {|yx iC } are both orthonormal bases with dependence on the value x. Thus,
the overall state has the following form:
Xq
pY |X (y|x)pX (x)|xiA |yx iB |yx iC .
(11.48)|ψiABC =x,ySuppose that Alice performs a complete projective measurement in the basis {|xihx|A }.
The state on Bob and Charlie’s systems P
is then |ψx iBC , and each system on B or C has amarginal entropy of H(σx ) where σx ≡ y pY |X (y|x)|yx ihyx |. We could potentially define
the conditional quantum entropy asX
pX (x)H(σx ).
(11.49)x

This quantity does not depend on a measurement as before because we simply choose themeasurement from the Schmidt decomposition. But there are many problems with the above
notion of conditional quantum entropy: it is defined only for pure quantum states, it is notclear how to apply it to a bipartite quantum state, and the conditional entropy of Bob’s
system given Alice’s and that of Charlie’s given Alice’s is the same (which is perhaps thestrangest of all!). Thus this notion of conditional probability is not useful for a definition of
conditional quantum entropy.

11.4 Conditional Quantum Entropy
The definition of conditional quantum entropy that has been most useful in quantum information theory is the following simple one, inspired from the relation between joint entropy
and marginal entropy in Exercise 10.3.1.
Definition 11.4.1 (Conditional Quantum Entropy) Let ρAB ∈ D(HA ⊗HB ). The conditional quantum entropy H(A|B)ρ of ρAB is equal to the difference of the joint quantum
entropy H(AB)ρ and the marginal entropy H(B)ρ :H(A|B)ρ ≡ H(AB)ρ − H(B)ρ .
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The above definition is the most natural one, both because it is straightforward to compute for any bipartite state and because it obeys many relations that the classical conditional
entropy obeys (such as entropy chain rules and conditioning reduces entropy). We explore
many of these relations in the forthcoming sections. For now, we state “conditioning cannot
increase entropy” as the following theorem and tackle its proof later on after developing a
few more tools.
Theorem 11.4.1 (Conditioning Does Not Increase Entropy) Consider a bipartite quantum state ρAB . Then the following inequality applies to the marginal entropy H(A)ρ and theconditional quantum entropy H(A|B)ρ :H(A)ρ ≥ H(A|B)ρ .

(11.51)

We can interpret the above inequality as stating that conditioning cannot increase entropy,
even if the conditioning system is quantum.

11.4.1

Conditional Entropy for Classical–Quantum States

A classical–quantum state is an example of a state for which conditional quantum entropy
behaves as in the classical world. Suppose that two parties share a classical–quantum state
ρXB of the form in (11.32). The system X is classical and the system B is quantum, andthe correlations between these systems are entirely classical, determined by the probability
distribution pX (x). Let us calculate the conditional quantum entropy H(B|X)ρ for thisstate:
H(B|X)ρ = H(XB)ρ − H(X)ρ
X
pX (x)H(ρxB ) − H(X)ρ
= H(X)ρ +(11.52)(11.53)x=

X

pX (x)H(ρxB ).

(11.54)x

The first equality follows from Definition 11.4.1. The second equality follows from Theorem 11.2.2, and the final equality results from algebra.
The above form for conditional entropy is completely analogous with the classical formulain (10.17) and holds whenever the conditioning system is classical.

11.4.2

Negative Conditional Quantum Entropy

One of the properties of the conditional quantum entropy in Definition 11.4.1 that seems
counterintuitive at first sight is that it can be negative. This negativity holds for an ebit
|Φ+ iAB shared between Alice and Bob. The marginal state on Bob’s system is the maximally
mixed state πB . Thus, the marginal entropy H(B) is equal to one, but the joint entropy
vanishes, and so the conditional quantum entropy H(A|B) = −1.
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What do we make of this result? Well, this is one of the fundamental differences between
the classical world and the quantum world, and perhaps is the very essence of the departurefrom an informational standpoint. The informational statement is that we can sometimes
be more certain about the joint state of a quantum system than we can be about any oneof its individual parts, and this is the reason that conditional quantum entropy can be
negative. This is in fact the same observation that Schrödinger made concerning entangledstates (Schrödinger, 1935):“When two systems, of which we know the states by their respective representatives, enter into temporary physical interaction due to known forces between
them, and when after a time of mutual influence the systems separate again, thenthey can no longer be described in the same way as before, viz. by endowing eachof them with a representative of its own. I would not call that one but rather
the characteristic trait of quantum mechanics, the one that enforces its entire
departure from classical lines of thought. By the interaction the two representatives [the quantum states] have become entangled. Another way of expressing
the peculiar situation is: the best possible knowledge of a whole does not necessarily include the best possible knowledge of all its parts, even though they
may be entirely separate and therefore virtually capable of being ‘best possibly
known,’ i.e., of possessing, each of them, a representative of its own. The lack ofknowledge is by no means due to the interaction being insufficiently known —at least not in the way that it could possibly be known more completely — it isdue to the interaction itself.”
These explanations might aid somewhat in understanding a negative conditional entropy,
but the ultimate test for whether we truly understand an information measure is if it is theanswer to some operational task. The task which gives an interpretation of the conditional
quantum entropy is known as state merging. Suppose that Alice and Bob share n copies
of a bipartite state ρAB where n is a large number and A and B are qubit systems. We
also allow them free access to a classical side channel, but we count the number of times
that they use a noiseless qubit channel. Alice would like to send Bob qubits over a noiselessqubit channel so that he receives her share of the state ρAB , i.e., so that he possesses allof the A shares. The naive approach would be for Alice simply to send her shares of thestate over the noiseless qubit channels, i.e., she would use the channel n times to send alln shares. But the state-merging protocol allows her to do much better, depending on thestate ρAB . If the state ρAB has positive conditional quantum entropy, she needs to use
the noiseless qubit channel only ≈ nH(A|B) times (we will prove later that H(A|B) ≤ 1
for any bipartite state on qubit systems). However, if the conditional quantum entropy isnegative, she does not need to use the noiseless qubit channel at all, and at the end of the
protocol, Alice and Bob share ≈ nH(A|B) noiseless ebits! They can then use these ebits for
future communication purposes, such as a teleportation or super-dense coding protocol (seeChapter 6). Thus, a negative conditional quantum entropy implies that Alice and Bob gainc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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the potential for future quantum communication, making clear in an operational sense whata negative conditional quantum entropy means.2 (We will cover this protocol in Chapter 22).
Exercise 11.4.1 Let σABC = ρAB ⊗ τC , where ρAB ∈ D(HA ⊗ HB ) and τC ∈ D(HC ). Show
that H(A|B)ρ = H(A|BC)σ .

11.5 Coherent Information
Negativity of the conditional quantum entropy is so important in quantum information
theory that we even have an information quantity and a special notation to denote thenegative of the conditional quantum entropy:
Definition 11.5.1 (Coherent Information) The coherent information I(AiB)ρ of a bipartite state ρAB ∈ D(HA ⊗ HB ) is as follows:
I(AiB)ρ ≡ H(B)ρ − H(AB)ρ .

(11.55)You should immediately notice that this quantity is the negative of the conditional quantum entropy in Definition 11.4.1, but it is perhaps more useful to think of the coherent
information not merely as the negative of the conditional quantum entropy, but as an information quantity in its own right. This is why we employ a separate notation for it. The“I” is present because the coherent information is an information quantity that measuresquantum correlations, much like the mutual information does in the classical case. For example, we have already seen that the coherent information of an ebit is equal to one. Thus,
it is measuring the extent to which we know less about part of a system than we do about
its whole. Perhaps surprisingly, the coherent information obeys a quantum data-processing
inequality (discussed in Section 11.9.2), which gives further support for it having an “I”
present in its notation. The Dirac symbol “i” is present to indicate that this quantity is aquantum information quantity, having a good meaning really only in the quantum world.
The choice of “i” over “h” also indicates a directionality from Alice to Bob, and this notation will make more sense when we begin to discuss the coherent information of a quantum
channel in Chapter 13.
Exercise 11.5.1 Calculate the coherent information I(AiB)Φ of the maximally entangled
state
d
1 X
|iiA |iiB .
(11.56)|ΦiAB ≡ √d i=1
2

After Horodecki, Oppenheim, and Winter published the state-merging protocol (Horodecki et al., 2005),
the Bristol Evening Post featured a story about Andreas Winter with the amusing title “Scientist Knows
Less Than Nothing,” as a reference to the potential negativity of conditional quantum entropy. Of course,such a title may seem a bit nonsensical to the layman, but it does grasp the idea that we can know lessabout a part of a quantum system than we do about its whole.
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Calculate the coherent information I(AiB)Φ of the maximally correlated state
d

1X
ΦAB ≡
|iihi|A ⊗ |iihi|B .
d i=1(11.57)

Exercise 11.5.2 Let ρAB ∈ D(HA ⊗ HB ). Consider a purification |ψiABE of this state tosome environment system E. Show that
I(AiB)ρ = H(B)ψ − H(E)ψ .

(11.58)Thus, there is a sense in which the coherent information measures the difference in theuncertainty of Bob and the uncertainty of the environment.
Exercise 11.5.3 (Duality of Conditional Entropy) Show that −H(A|B)ρ = I(AiB)ρ =H(A|E)ψ for the purification in the above exercise.
The coherent information can be both negative or positive depending on the bipartite
state for which we evaluate it, but it cannot be arbitrarily large or arbitrarily small. The
following theorem places a useful bound on its absolute value.
Theorem 11.5.1 Let ρAB ∈ D(HA ⊗HB ). The following bound applies to the absolute valueof the conditional entropy H(A|B)ρ :|H(A|B)ρ | ≤ log dim(HA ).

(11.59)The bounds are saturated for ρAB = πA ⊗ σB , where πA is the maximally mixed state andσB ∈ D(HB ), and for ρAB = ΦAB (the maximally entangled state).
Proof. We first prove the inequality H(A|B)ρ ≤ log dim(HA ):H(A|B)ρ ≤ H(A)ρ ≤ log dim(HA ).

(11.60)The first inequality follows because conditioning reduces entropy (Theorem 11.4.1), and thesecond inequality follows because the maximum value of the entropy H(A)ρ is log dim(HA ).
We now prove the inequality H(A|B)ρ ≥ − log dim(HA ). Consider a purification |ψiEAB of
the state ρAB . We then have that
H(A|B)ρ = −H(A|E)ψ
≥ −H(A)ρ
≥ − log dim(HA ).

(11.61)(11.62)(11.63)

The first equality follows from Exercise 11.5.3. The first and second inequalities follow by
the same reasons as the inequalities in the previous paragraph.
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Exercise 11.5.4 (Conditional Coherent Information) Consider a tripartite state ρABC .
Show that
I(AiBC)ρ = I(AiB|C)ρ ,(11.64)
where I(AiB|C)ρ ≡ H(B|C)ρ − H(AB|C)ρ is the conditional coherent information.
Exercise 11.5.5 (Conditional Coherent Information of a Classical–Quantum State)
Suppose we have a classical–quantum state σXAB whereXx,(11.65)σXAB =pX (x)|xihx|X ⊗ σABxx
pX is a probability distribution on a finite alphabet X and σAB∈ D(HA ⊗ HB ) for all x ∈ X .
Show that
X
I(AiBX)σ =pX (x)I(AiB)σx .
(11.66)x

11.6 Quantum Mutual Information
The standard informational measure of correlations in the classical world is the mutual
information, and such a quantity plays a prominent role in measuring classical and quantum
correlations in the quantum world as well.
Definition 11.6.1 (Quantum Mutual Information) The quantum mutual informationof a bipartite state ρAB ∈ D(HA ⊗ HB ) is defined as follows:
I(A; B)ρ ≡ H(A)ρ + H(B)ρ − H(AB)ρ .

(11.67)

The following relations hold for quantum mutual information, in analogy with the classical case:
I(A; B)ρ = H(A)ρ − H(A|B)ρ
= H(B)ρ − H(B|A)ρ .

(11.68)(11.69)These immediately lead to the following relations between quantum mutual information andthe coherent information:
I(A; B)ρ = H(A)ρ + I(AiB)ρ
= H(B)ρ + I(BiA)ρ .

(11.70)(11.71)

The theorem below gives a fundamental lower bound on the quantum mutual information—
we merely state it for now and give a full proof later.
Theorem 11.6.1 (Non-Negativity of Quantum Mutual Information) The quantum
mutual information I(A; B)ρ of any bipartite quantum state ρAB is non-negative:
I(A; B)ρ ≥ 0.
c 2019(11.72)

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Exercise 11.6.1 (Conditioning Does Not Increase Entropy) Show that non-negativityof quantum mutual information implies that conditioning does not increase entropy (Theorem 11.4.1).
Exercise 11.6.2 Calculate the quantum mutual information I(A; B)Φ of the maximally entangled state ΦAB . Calculate the quantum mutual information I(A; B)Φ of the maximallycorrelated state ΦAB .
Exercise 11.6.3 (Bound on Quantum Mutual Information) Let ρAB ∈ D(HA ⊗HB ).
Prove that the following bound applies to the quantum mutual information:
I(A; B)ρ ≤ 2 log [min {dim(HA ), dim(HB )}] .

(11.73)What is an example of a state that saturates the bound?
Exercise 11.6.4 Consider a pure state |ψiRA ∈ HR ⊗ HA . Suppose that an isometry U :HA → HB ⊗ HE acts on the A system of |ψiRA to produce the pure state |φiRBE ∈ HR ⊗HB ⊗ HE . Show that
I(R; B)φ + I(R; E)φ = I(R; A)ψ .
(11.74)
Exercise 11.6.5 Consider a tripartite pure state |ψiSRA ∈ HS ⊗ HR ⊗ HA . Suppose that
an isometry U : HA → HB ⊗ HE acts on the A system of |ψiSRA to produce the state|φiSRBE ∈ HS ⊗ HR ⊗ HB ⊗ HE . Show that
I(R; A)ψ + I(R; S)ψ = I(R; B)φ + I(R; SE)φ .

(11.75)

Exercise 11.6.6 (Entropy, Coherent Information, and Mutual Information) Consider
a pure state |φiABE on systems ABE. Using the Schmidt decomposition with respect to thebipartite cut A | BE, we can write |φiABE as follows:
XppX (x)|xiA ⊗ |φx iBE ,(11.76)|φiABE =xfor some orthonormal states {|xiA }x∈X on system A and some orthonormal states {|φx iBE }on the joint system BE. Prove the following relations:
1
1
I(AiB)φ = I(A; B)φ − I(A; E)φ ,2
2
1
1
H(A)φ = I(A; B)φ + I(A; E)φ .
2
2(11.77)(11.78)Exercise 11.6.7 (Coherent Information and Private Information) We obtain a decohered version φABE of the state in Exercise 11.6.6 by measuring the A system in the basis{|xiA }x∈X . Let us now denote the A system as the X system because it becomes a classicalsystem after the measurement:
X
φXBE =pX (x)|xihx|X ⊗ φxBE .
(11.79)x
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Prove the following relation:
I(AiB)φ = I(X; B)φ − I(X; E)φ .

(11.80)The quantity on the right-hand side is known as the private information, because there isa sense in which it quantifies the classical information in X that is accessible to Bob while
being private from Eve.
Exercise 11.6.8 (Additivity) Let ρA1 B1 ∈ D(HA1 ⊗ HB1 ) and σA2 B2 ∈ D(HA2 ⊗ HB2 ).
Set ωA1 B1 A2 B2 ≡ ρA1 B1 ⊗ σA2 B2 . Prove that I(A1 A2 ; B1 B2 )ω = I(A1 ; B1 )ρ + I(A2 ; B2 )σ .

11.6.1

Holevo Information

Suppose that Alice prepares some classical ensemble E ≡ {pX (x), ρxB } and then hands thisensemble to Bob without telling him the classical index x. The expected density operator ofthis ensemble isXρB ≡ EX ρX=
pX (x)ρxB ,(11.81)Bx

and this density operator ρB characterizes the state from Bob’s perspective because he does
not have knowledge of the classical index x. His task is to determine the classical index x by
performing some measurement on his system B. Recall from Section 10.9.2 that the accessible
information quantifies Bob’s information gain after performing some optimal measurement{Λy } on his system B:
Iacc (E) = max I(X; Y ),(11.82){Λy }where Y is a random variable corresponding to the outcome of the measurement.
What is the accessible information of the ensemble? In general, this quantity is difficult
to compute, but another quantity, called the Holevo information, provides a useful upperbound. The Holevo information χ(E) of the ensemble isXχ(E) ≡ H(ρB ) −pX (x)H(ρxB ).
(11.83)x

Exercise 11.9.2 asks you to prove this upper bound after we develop the quantum dataprocessing inequality for quantum mutual information. The Holevo information characterizes
the correlations between the classical variable X and the quantum system B.
Exercise 11.6.9 (Mutual Information of Classical–Quantum States) Consider the following classical–quantum state representing the ensemble E:XσXB ≡pX (x)|xihx|X ⊗ ρxB .
(11.84)xShow that the Holevo information χ(E) is equal to the mutual information I(X; B)σ :χ(E) = I(X; B)σ .

(11.85)In this sense, the quantum mutual information of a classical–quantum state is most similar
to the classical mutual information of Shannon.
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Exercise 11.6.10 (Concavity of Quantum Entropy) Prove the concavity of entropy (Property 11.1.4) using Theorem 11.6.1 and the result of Exercise 11.6.9.
Exercise 11.6.11 (Dimension Bound) Let σXB ∈ D(HX ⊗ HB ) be a classical–quantum
state of the form:XσXB =pX (x)|xihx|X ⊗ σBx .
(11.86)x

Prove that the following bound applies to the Holevo information:
I(X; B)σ ≤ log [min {dim(HX ), dim(HB )}] .

(11.87)

What is an example of a state that saturates the bound?

11.7 Conditional Quantum Mutual Information
We define the conditional quantum mutual information I(A; B|C)ρ of any tripartite stateρABC ∈ D(HA ⊗ HB ⊗ HC ) similarly to how we did in the classical case:
I(A; B|C)ρ ≡ H(A|C)ρ + H(B|C)ρ − H(AB|C)ρ .

(11.88)In what follows, we sometimes abbreviate “conditional quantum mutual information” as
CQMI.
One can exploit the above definition and the definition of quantum mutual information
to prove a chain rule for quantum mutual information.
Property 11.7.1 (Chain Rule for Quantum Mutual Information) The quantum mutual information obeys a chain rule:
I(A; BC)ρ = I(A; B)ρ + I(A; C|B)ρ .

(11.89)The interpretation of the chain rule is that we can build up the correlations between A andBC in two steps: in a first step, we build up the correlations between A and B, and now that
B is available (and thus conditioned on), we build up the correlations between A and C.
Exercise 11.7.1 Use the chain rule for quantum mutual information to prove that
I(A; BC)ρ = I(AC; B)ρ + I(A; C)ρ − I(B; C)ρ .

11.7.1(11.90)Non-Negativity of CQMI

In the classical world, non-negativity of conditional mutual information follows trivially from
non-negativity of mutual information (recall Theorem 10.6.1). The proof of non-negativityof conditional quantum mutual information is far from trivial in the quantum world, unless the conditioning system is classical (see Exercise 11.7.2). It is a foundational resultc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 11. QUANTUM INFORMATION AND ENTROPY

that non-negativity of this quantity holds because so much of quantum information theory
rests upon this theorem’s shoulders (in fact, we could say that this inequality is one of the“bedrocks” of quantum information theory). The list of its corollaries includes the quantum
data-processing inequality, the answers to some additivity questions in quantum Shannon
theory, the Holevo bound, and others. The proof of Theorem 11.7.1 follows directly frommonotonicity of quantum relative entropy (Theorem 11.8.1), which we prove in Chapter 12.
In fact, it is possible to show that monotonicity of quantum relative entropy follows fromstrong subadditivity as well, so that these two entropy inequalities are essentially equivalentstatements.
Theorem 11.7.1 (Non-Negativity of CQMI) Let ρABC ∈ D(HA ⊗HB ⊗HC ). Then theconditional quantum mutual information is non-negative:
I(A; B|C)ρ ≥ 0.

(11.91)This condition is equivalent to the strong subadditivity inequality in Exercise 11.7.7, so we
also refer to this entropy inequality as strong subadditivity.
Exercise 11.7.2 (CQMI of Classical–Quantum States) Consider a classical–quantum
state σXAB of the form in (11.65). Prove the following relation:X
I(A; B|X)σ =pX (x)I(A; B)σx .
(11.92)x

Conclude that non-negativity of conditional quantum mutual information is trivial in thisspecial case in which the conditioning system is classical, simply by exploiting non-negativityof quantum mutual information (Theorem 11.6.1).
Exercise 11.7.3 (Conditioning Does Not Increase Entropy) Let ρABC ∈ D(HA ⊗HB ⊗HC ). Show that Theorem 11.7.1 is equivalent to the following stronger form of Theorem 11.4.1:
H(B|C)ρ ≥ H(B|AC)ρ .
(11.93)
Exercise 11.7.4 (Conditional Entropy & Recoverability) Show that H(B|C)ρ = H(B|AC)ρ
if there exists a recovery channel RC→AC such that ρABC = RC→AC (ρBC ) for ρABC ∈D(HA ⊗ HB ⊗ HC ). (We will see later that this can be strengthened to “if and only if.”)
Exercise 11.7.5 (Concavity of Conditional Quantum Entropy) Show that strong subadditivity implies that conditional entropy is concave. That is, prove thatX
pX (x)H(A|B)ρx ≤ H(A|B)ρ ,(11.94)xwhere pX is a probability
distribution on a finite alphabet X , ρxAB ∈ D(HA ⊗ HB ) for allPx ∈ X , and ρAB ≡ x pX (x)ρxAB .
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11.8. QUANTUM RELATIVE ENTROPY

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Exercise 11.7.6 (Convexity of Coherent Information) Prove that coherent information is convex:
X
pX (x)I(AiB)ρx ≥ I(AiB)ρ ,(11.95)x

by exploiting the result of the above exercise.
Exercise 11.7.7 (Strong Subadditivity) Theorem 11.7.1 also goes by the name of “strong
subadditivity” because it is an example of a function φ that is strongly subadditive:φ(E) + φ(F ) ≥ φ(E ∩ F ) + φ(E ∪ F ).

(11.96)

Let ρABC ∈ D(HA ⊗ HB ⊗ HC ). Show that non-negativity of conditional quantum mutual
information is equivalent to the following strong subadditivity of quantum entropy:
H(AC)ρ + H(BC)ρ ≥ H(C)ρ + H(ABC)ρ ,(11.97)

where we think of φ in (11.96) as the entropy function H, the argument E in (11.96) as AC,
and the argument F in (11.96) as BC.
Exercise 11.7.8 (Duality of CQMI) Let |ψiABCD ∈ HA ⊗HB ⊗HC ⊗HD be a pure state.
Prove that
I(A; B|C)ψ = I(A; B|D)ψ .
(11.98)Exercise 11.7.9 (Dimension Bound) Let ρABC ∈ D(HA ⊗HB ⊗HC ). Prove the following
dimension bound:
I(A; B|C)ρ ≤ 2 log [min {dim(HA ), dim(HB )}] .
(11.99)Let σXBC ∈ D(HX ⊗ HB ⊗ HC ) be a classical–quantum–quantum state of the formXx
pX (x)|xihx|X ⊗ σBC.
(11.100)x

Prove that
I(X; B|C)σ ≤ log dim(HX ).

(11.101)

Exercise 11.7.10 (Araki–Lieb triangle inequality) Let ρAB ∈ D(HA ⊗HB ). Show that
|H(A)ρ − H(B)ρ | ≤ H(AB)ρ .

(11.102)11.8 Quantum Relative Entropy
The quantum relative entropy is one of the most important entropic quantities in quantum
information theory, mainly because we can reexpress many of the entropies given in theprevious sections in terms of it. This in turn allows us to establish many properties of thesequantities from the properties of relative entropies. Its definition is a natural extension ofthat for the classical relative entropy (see Definition 10.5.2). Before defining it, we need thenotion of the support of an operator:
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Definition 11.8.1 (Kernel and Support) The kernel of an operator A ∈ L(H, H0 ) isdefined asker(A) ≡ {|ψi ∈ H : A|ψi = 0}.
(11.103)
The support of A is the subspace of H orthogonal to its kernel:
supp(A) ≡ ker(A)⊥ ≡ {|ψi ∈ H : hψ|φi = 0 ∀|φi ∈ ker(A)}.
(11.104)
P
If A is Hermitian and thus has a spectral decomposition as A = i:ai 6=0 ai |iihi|, then supp(A) =span{|ii : ai 6= 0}. The projection onto the support of A is denoted byXΠA ≡|iihi|.
(11.105)
i:ai 6=0

Definition 11.8.2 (Quantum Relative Entropy) The quantum relative entropy D(ρkσ)
between a density operator ρ ∈ D(H) and a positive semi-definite operator σ ∈ L(H) isdefined as follows:D(ρkσ) ≡ Tr{ρ [log ρ − log σ]},(11.106)
if the following support condition is satisfied
supp(ρ) ⊆ supp(σ),(11.107)and it is defined to be equal to +∞ otherwise.
This definition is consistent with the classical definition in Definition 10.5.2. However, weshould note that there could be several ways to generalize the classical definition to obtaina quantum definition of relative entropy. For example, one could take
D0 (ρkσ) = Tr ρ log ρ1/2 σ −1 ρ1/2 ,(11.108)
as a definition and it reduces to the classical definition in Definition 10.5.2 as well. In fact,
it is easy to see that there are an infinite number of quantum generalizations of the classicaldefinition of relative entropy. So how do we single out which definition is the right one touse? The definition given in (11.106) is the answer to a meaningful quantum informationprocessing task in the context of quantum hypothesis testing (we do not elaborate on thisfurther here but just mention that it is known as the quantum Stein’s lemma). Furthermore,this definition generalizes the quantum entropic quantities we have given in this chapter,
which all in turn are the answers to meaningful quantum information-processing tasks. For
these reasons, we take the definition given in (11.106) as the quantum relative entropy.
Recall that it was this same line of reasoning that allowed us to single out the entropy andthe mutual information as meaningful measures of information in the classical case.
Similar to the classical case, we can intuitively think of the quantum relative entropy as
a distance measure between quantum states. But it is not strictly a distance measure in themathematical sense because it is not symmetric and it does not obey a triangle inequality.
The following proposition justifies why we take the definition of quantum relative entropy
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11.8. QUANTUM RELATIVE ENTROPY

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Proposition 11.8.1 Let ρ ∈ D(H) and σ ∈ L(H) be positive semi-definite. The quantum
relative entropy is consistent with the following limit:
D(ρkσ) = lim D(ρkσ + εI).
ε&0

(11.109)

Proof. First observe that the operator σ + εI has support equal to H for all ε > 0, sothat D(ρkσ + εI) is finite for all ε > 0. We will see that the limit is finite and consistent
with (11.106) if (11.107) is satisfied, and otherwise the limit blows up to infinity. The idea
in proving this proposition is to represent both ρ and σ with respect to the decomposition
H = supp(σ) ⊕ ker(σ). Let Πσ denote the projection onto supp(σ) and let Π⊥σ denote theprojection onto ker(σ). So we takeρ00 ρ01
σ0 0
ρ=
,
σ=
.
(11.110)ρ10 ρ110 0First suppose that the support condition in (11.107) is satisfied. Then this means that
ρ01 = ρ†10 = 0 and ρ11 = 0. Observe thatD(ρkσ + εI) = Tr {ρ log ρ} − Tr {ρ log (σ + εI)} .

(11.111)

The first term is finite for any ρ, so we should focus on the second term exclusively, since
this is where an issue could arise. Then




ρ00 0
σ0 + εΠσ0
Tr {ρ log (σ + εI)} = Tr
log
0 0
0
εΠ⊥
σ



0 
ρ00 0 log (σ0 + εΠσ )(11.112)= Tr
0 0
0log εΠ⊥σ= Tr {ρ00 log (σ0 + εΠσ )} + Tr 0 · log εΠ⊥(11.113)σ= Tr {ρ00 log (σ0 + εΠσ )} .
(11.114)
Taking the limit ε & 0, we get
lim Tr {ρ00 log (σ0 + εΠσ )} = Tr {ρ00 log σ0 } = Tr {ρ log σ} .

ε&0(11.115)

So we can conclude thatlim D(ρkσ + εI) = Tr {ρ log ρ} − Tr {ρ log σ} ,ε&0(11.116)

in this case.
Now suppose that the support condition in (11.107) is not satisfied. Then ρ11 6= 0, andwe find that



0 
ρ00 ρ01 log (σ0 + εΠσ )Tr {ρ log (σ + εI)} = Tr
ρ10 ρ110log εΠ⊥σ= Tr {ρ00 log (σ0 + εΠσ )} + Tr ρ11 · log εΠ⊥,(11.117)σ
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CHAPTER 11. QUANTUM INFORMATION AND ENTROPY

and thus limε&0 D(ρkσ + εI) = +∞, given that limε&0 [− log ε] = +∞.
One of the most fundamental entropy inequalities in quantum information theory is themonotonicity of quantum relative entropy. When the arguments to the quantum relative
entropy are quantum states, the physical interpretation of this entropy inequality is thatstates become less distinguishable when noise acts on them. We defer a proof of this theorem
until Chapter 12, where we also establish a strengthening of it.
Theorem 11.8.1 (Monotonicity of Quantum Relative Entropy) Let ρ ∈ D(H), σ ∈L(H) be positive semi-definite, and N : L(H) → L(H0 ) be a quantum channel. The quantum
relative entropy can only decrease or stay the same if we apply the same quantum channel
N to ρ and σ:
D(ρkσ) ≥ D(N (ρ)kN (σ)).
(11.118)
Theorem 11.8.1 then implies non-negativity of quantum relative entropy in certain cases.
Theorem 11.8.2 (Non-Negativity) Let ρ ∈ D(H), and let σ ∈ L(H) be positive semidefinite and such that Tr{σ} ≤ 1. Then the quantum relative entropy D(ρkσ) is non-negative:D(ρkσ) ≥ 0,

(11.119)and D(ρkσ) = 0 if and only if ρ = σ.
Proof. The first part of the theorem follows from applying Theorem 11.8.1, taking thequantum channel to be the trace-out map. We then have that


Tr{ρ}
≥ 0.
(11.120)D(ρkσ) ≥ D(Tr{ρ}k Tr{σ}) = Tr{ρ} log
Tr{σ}If ρ = σ, then the support condition in (11.107) is satisfied and plugging into (11.106)
gives that D(ρkσ) = 0. Now suppose that D(ρkσ) = 0. This means that the inequality
above is saturated and thus Tr{σ} = Tr{ρ} = 1, so that σ is a density operator. Let M
be an arbitrary measurement channel. From the monotonicity of quantum relative entropy(Theorem 11.8.1), we can conclude that D(M(ρ)kM(σ)) = 0. The equality condition for
the non-negativity of the classical relative entropy (Theorem 10.7.1) in turn implies that
M(ρ) = M(σ). Now since this equality holds for any possible measurement channel, we
can conclude that ρ = σ. (For example, we could take M to be the optimal measurement
for the trace distance, which would allow us to conclude that maxM kM(ρ) − M(σ)k1 =kρ − σk1 = 0, and hence ρ = σ.)11.8.1

Deriving Other Entropies from Quantum Relative Entropy

There is a sense in which the quantum relative entropy is a “parent quantity” for otherentropies in quantum information theory, such as the quantum entropy, the conditional
quantum entropy, the quantum mutual information, and the conditional quantum mutual
information. The following exercises explore these relations. The main tool needed to solve
some of them is the non-negativity of quantum relative entropy.
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Exercise 11.8.1 (Operator Logarithm) Let PA ∈ L(HA ) and QB ∈ L(HB ) be positive
semi-definite operators. Show that the following identity holds:
log (PA ⊗ QB ) = log (PA ) ⊗ IB + IA ⊗ log (QB ) .

(11.121)

Exercise 11.8.2 (Mutual Information and Relative Entropy) Let ρAB ∈ D(HA ⊗HB ).
Show that the following identities hold:
I(A; B)ρ = D(ρAB kρA ⊗ ρB )= min D(ρAB kρA ⊗ σB )(11.122)(11.123)= min D(ρAB kωA ⊗ ρB )(11.124)= min D(ρAB kωA ⊗ σB ),(11.125)σB

ωA

ωA ,σB

where the optimizations are with respect to ωA ∈ D(HA ) and σB ∈ D(HB ).
Exercise 11.8.3 (Conditional and Relative Entropy) Let ρAB ∈ D(HA ⊗ HB ). Show
that the following identities hold:
I(AiB)ρ = D(ρAB kIA ⊗ ρB )= min D(ρAB kIA ⊗ σB ).
σB ∈D(HB )(11.126)(11.127)

Note that these are equivalent to
H(A|B)ρ = −D(ρAB kIA ⊗ ρB )= − min D(ρAB kIA ⊗ σB ).
σB ∈D(HB )(11.128)(11.129)

Exercise 11.8.4 (CQMI and Relative Entropy) Let ρABC ∈ D(HA ⊗ HB ⊗ HC ). Let
ωABC be the following positive semi-definite operator:
ωABC ≡ 2[log ρAC +log ρBC −log ρC ] ,(11.130)

where identities are implicit if not written (e.g., ρBC is a shorthand for IA ⊗ ρBC ). Showthat
I(A; B|C)ρ = D(ρABC kωABC ).
(11.131)
Exercise 11.8.5 (Dimension Bound) Let ρABC ∈ D(HA ⊗HB ⊗HC ). Prove the following
dimension bound:
I(AiBC)ρ ≤ I(ACiB)ρ + log dim(HC ).
(11.132)(Hint: One way to do this is to use the formula in (11.127). Another way is to use the chain
rule and previous dimension bounds.)
Corollary 11.8.1 (Subadditivity of Quantum Entropy) The quantum entropy is subadditive for a bipartite state ρAB :H(A)ρ + H(B)ρ ≥ H(AB)ρ .

(11.133)

Proof. Subadditivity of entropy is equivalent to non-negativity of quantum mutual information. We can prove non-negativity by exploiting the result of Exercise 11.8.2 andnon-negativity of quantum relative entropy (Theorem 11.8.2).
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11.8.2Mathematical Properties of Quantum Relative Entropy

This section contains several auxiliary mathematical properties of quantum relative entropy,
including its isometric invariance, additivity for tensor-product states, its form for classical–
quantum states (these are left as exercises). There are two other properties given which arecommonly used in relative entropy calculations.
Exercise 11.8.6 (Isometric Invariance) Let ρ ∈ D(H) and σ ∈ L(H) be positive semidefinite. Show that the quantum relative entropy is invariant with respect to an isometry
U : H → H0 :D(ρkσ) = D(U ρU † kU σU † ).
(11.134)
Exercise 11.8.7 (Additivity of Quantum Relative Entropy) Let ρ1 ∈ D(H1 ) and ρ2 ∈D(H1 ) be density operators, and let σ1 ∈ L(H1 ) and σ2 ∈ L(H2 ) be positive semi-definite
operators. Show that the quantum relative entropy is additive in the following sense:D(ρ1 ⊗ ρ2 kσ1 ⊗ σ2 ) = D(ρ1 kσ1 ) + D(ρ2 kσ2 ).

(11.135)

We can apply the above additivity relation inductively to conclude thatD(ρ⊗n kσ ⊗n ) = nD(ρkσ),(11.136)for ρ ∈ D(H) and σ ∈ L(H) positive semi-definite.
Exercise 11.8.8 (Relative Entropy of Classical–Quantum States) Show that the quantum relative entropy between classical–quantum states ρXB and σXB is as follows:X
p(x)D(ρxB kσBx ) + D(pkq), where(11.137)
D(ρXB kσXB ) =
x

ρXB ≡

X
x

p(x)|xihx|X ⊗ ρxB ,

σXB ≡

X

q(x)|xihx|X ⊗ σBx ,(11.138)xwith p and q probability distributions over a finite alphabet X , ρxB ∈ D(HB ) for all x ∈ X ,and σBx ∈ L(HB ) positive semi-definite for all x ∈ X .
Exercise 11.8.9 Let a, b > 0, ρ ∈ D(H), and σ ∈ L(H) be positive semi-definite. Showthat
D(aρkbσ) = a [D(ρkσ) + log (a/b)] .
(11.139)(Note that we only defined quantum relative entropy to have its first argument equal to adensity operator, but one could more generally allow for the first argument to be positive
semi-definite.)
Proposition 11.8.2 Let ρ ∈ D(H) and σ, σ 0 ∈ L(H) be positive semi-definite. Suppose thatσ ≤ σ 0 . Then
D(ρkσ 0 ) ≤ D(ρkσ).
(11.140)
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11.9. QUANTUM ENTROPY INEQUALITIES

343

Proof. The assumption that σ ≤ σ 0 is equivalent to σ 0 − σ being positive semi-definite.
Then the following operator is positive semi-definite: σ ⊗ |0ih0|X + (σ 0 − σ) ⊗ |1ih1|X , andas a consequence
D(ρkσ) = D(ρ ⊗ |0ih0|X k [σ ⊗ |0ih0|X + (σ 0 − σ) ⊗ |1ih1|X ]),(11.141)

which follows by a direct calculation (essentially the same reasoning as that used to solve Exercise 11.8.8). By monotonicity of quantum relative entropy (Theorem 11.8.1), the quantum
relative entropy does not increase after discarding the system X, so thatD(ρ ⊗ |0ih0|X k [σ ⊗ |0ih0|X + (σ 0 − σ) ⊗ |1ih1|X ])≥ D(ρk [σ + (σ 0 − σ)]) = D(ρkσ 0 ), (11.142)
concluding the proof.

11.9 Quantum Entropy Inequalities
Monotonicity of quantum relative entropy has as its corollaries many of the important entropy inequalities in quantum information theory (but keep in mind that some of these also
imply the monotonicity of quantum relative entropy).
Corollary 11.9.1 (Strong Subadditivity) Let ρABC ∈ D(HA ⊗ HB ⊗ HC ). The quantum
entropy is strongly subadditive, in the following sense:
H(AC)ρ + H(BC)ρ ≥ H(ABC)ρ + H(C)ρ .

(11.143)

Proof. Consider from Exercise 11.7.7 that
I(A; B|C)ρ = H(AC)ρ + H(BC)ρ − H(ABC)ρ − H(C)ρ ,(11.144)

I(A; B|C)ρ = H(B|C)ρ − H(B|AC)ρ .

(11.145)so that
From Exercise 11.8.3, we know that
−H(B|AC)ρ = D(ρABC kIB ⊗ ρAC ),H(B|C)ρ = −D(ρBC kIB ⊗ ρC ).

(11.146)(11.147)

D(ρABC kIB ⊗ ρAC ) ≥ D(TrA {ρABC }k TrA {IB ⊗ ρAC })= D(ρBC kIB ⊗ ρC ).

(11.148)(11.149)

Then

The inequality is a consequence of the monotonicity of quantum relative entropy (Theorem 11.8.1), taking ρ = ρABC , σ = IB ⊗ ρAC , and N = TrA . By (11.144)–(11.147), theinequality in (11.148)–(11.149) is equivalent to the inequality in the statement of the corollary.
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CHAPTER 11. QUANTUM INFORMATION AND ENTROPY

Corollary 11.9.2 (Joint Convexity of Quantum Relative Entropy) Let pX be a probability distribution over a finite alphabet X , ρx ∈PD(H) for all x ∈ X , P
and σ x ∈ L(H) bexx
positive semi-definite for all x ∈ X . Set ρ ≡x pX (x)ρ and σ ≡x pX (x)σ . The
quantum relative entropy is jointly convex in its arguments:
X
pX (x)D(ρx kσ x ) ≥ D(ρkσ).
(11.150)x

Proof. Consider classical–quantum states of the following form:XρXB ≡pX (x)|xihx|X ⊗ ρxB ,(11.151)

x

σXB ≡

X

pX (x)|xihx|X ⊗ σBx .

(11.152)x

Observe that ρ = ρB and σ = σB . ThenX
pX (x)D(ρxB kσBx ) = D(ρXB kσXB ) ≥ D(ρB kσB ).

(11.153)x

The equality follows from Exercise 11.8.8, and the inequality follows from monotonicity ofquantum relative entropy (Theorem 11.8.1), where we take the channel to be the partial
trace over the system X.
Corollary 11.9.3 (Unital Channels Increase Entropy) Let ρ ∈ D(H) and let N : L(H) →L(H) be a unital quantum channel (see Definition 4.4.7). Then
H(N (ρ)) ≥ H(ρ).

(11.154)

H(ρ) = −D(ρkI),
H(N (ρ)) = −D(N (ρ)kI) = −D(N (ρ)kN (I)),(11.155)(11.156)

Proof. Consider that

where in the last equality, we have used that N is a unital quantum channel. The inequality in(11.154) is a consequence of the monotonicity of quantum relative entropy (Theorem 11.8.1)
because D(ρkI) ≥ D(N (ρ)kN (I)).
A particular example of a unital channel occurs when we completely dephase a density
operator ρ with respect to some dephasing basis {|yi}. Let ω denote the dephased versionof ρ:Xω ≡ ∆Y (ρ) =|yihy|ρ|yihy|.
(11.157)yThen the entropy H(ω) of the completely dephased state is never smaller than the entropyH(ρ)
of the original state. More P
generally, if we have a set of projectors {Πx } satisfyingPx Πx = I, then the channel ρ →x Πx ρΠx is unital, so that!
XHΠx ρΠx ≥ H(ρ).
(11.158)x
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11.9. QUANTUM ENTROPY INEQUALITIES345

The quantum relative entropy itself is not a distance measure, but it actually gives auseful upper bound on the trace distance between two quantum states. This result is known
as the quantum Pinsker inequality. Thus, in this sense, we can think of the quantum relative
entropy as being comparable to a distance measure when it is small—if the quantum relative
entropy between two quantum states is small, then their trace distance is small as well. We
can view the quantum Pinsker inequality as a refinement of the statement that the quantum
relative entropy is non-negative (Theorem 11.8.2).
Theorem 11.9.1 (Quantum Pinsker Inequality) Let ρ ∈ D(H) and let σ ∈ L(H) be
positive semi-definite such that Tr{σ} ≤ 1. Then
D(ρkσ) ≥1kρ − σk21 .
2 ln 2(11.159)

Proof. This is a direct consequence of the classical Pinsker inequality (Theorem 10.8.1)
and the fact that a measurement achieves the trace distance (see Exercise 9.1.10). To get
the statement for subnormalized σ, we add an extra dimension to the Hilbert space H. Letρ0 ≡ ρ ⊕ [0] and σ 0 ≡ σ ⊕ [1 − Tr{σ}], so that σ 0 is a density operator. Let M denote a
measurement channel that achieves the trace distance for ρ0 and σ 0 (see Exercise 9.1.10).
Then
D(ρkσ) = D(ρ0 kσ 0 )≥ D(M(ρ0 )kM(σ 0 ))1
2
kM(ρ0 ) − M(σ 0 )k1≥
2 ln 212=kρ0 − σ 0 k1
2 ln 21[kρ − σk1 + (1 − Tr{σ})]2=2 ln 21≥kρ − σk21 .
2 ln 2(11.160)(11.161)(11.162)(11.163)(11.164)(11.165)

The first inequality is a consequence of the monotonicity of quantum relative entropy (Theorem 11.8.1). The second inequality follows from the classical Pinsker inequality (Theorem 10.8.1). The second equality follows because we chose the measurement channel M toachieve the trace distance.

11.9.1

Equivalence of Quantum Entropy Inequalities

We have already seen how the monotonicity of quantum relative entropy (Theorem 11.8.1)
implies many of the important entropy inequalities in quantum information theory. However, what is less obvious is that some of these other entropy inequalities also imply themonotonicity of quantum relative entropy. Thus, we can say that together, these entropy
inequalities constitute a “law” of quantum information theory, saying either that information, correlations, or distinguishability decrease under the action of a quantum channel or
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CHAPTER 11. QUANTUM INFORMATION AND ENTROPY

conditional uncertainty increases under the action of a quantum channel on a conditioning
system. We formalize this equivalence in the following theorem:
Theorem 11.9.2 The following statements are equivalent, in the sense that one can prove
the other statements as a consequence of one of them:
1. The quantum relative entropy is monotone with respect to quantum channels: D(ρkσ) ≥D(N (ρ)kN (σ)), where ρ, σ ∈ D(H), and N : L(H) → L(H0 ) is a quantum channel.
2. The quantum relative entropy is monotone with respect to partial trace: D(ρAB kσAB ) ≥D(ρB kσB ), where ρAB , σAB ∈ D(HA ⊗ HB ).
Pxx
3. The quantum relative entropy is jointly convex:
x pX (x)D(ρ kσ ) ≥ D(ρkσ), where
pX isPa probability distribution
over a finite alphabet X , ρx , σ x ∈ D(H) for all x ∈ X ,Pρ ≡ x pX (x)ρx , and σ ≡ x pX (x)σ x .
4. The conditional quantum mutual information is non-negative: I(A; B|C)ρ ≥ 0, whereρABC ∈ D(HA ⊗ HB ⊗ HC ).
P
5. The conditional quantum entropy is concave: H(A|B)ρ ≥ x pX (x)H(A|B)ρx , where
pX is a probability P
distribution on a finite alphabet X , ρxAB ∈ D(HA ⊗ HB ) for allx ∈ X , and ρAB ≡ x pX (x)ρxAB .
Proof. We have already proved 2-4 starting from 1, and 5 from 4 as well (some of these areleft as exercises), so it remains to work our way back up to 1 from the others. Consider that
we can get 1 from 2 by using the Stinespring dilation theorem. That is, let U : H → H0 ⊗ HE
be an isometric extension of the channel N . Consider thatD(ρkσ) = D(U ρU † kU σU † )≥ D(TrE {U ρU † }k TrE {U σU † })= D(N (ρ)kN (σ)).

(11.166)(11.167)(11.168)

The first equality follows from invariance of quantum relative entropy with respect to isometries (Exercise 11.8.6). The inequality follows from monotonicity with respect to partial trace(by assumption), and the last equality follows from the fact that U is an isometric extension
of N .
We can also get 1 from 3 by a related approach. Let d = dim(HE ) and {VEi } be a
Heisenberg–Weyl set of unitaries for the environment system E. Consider that
D(N (ρ)kN (σ)) = D(N (ρ) ⊗ πE kN (σ) ⊗ πE )
!
X
X


1
† 1
†
VEi U ρU † VEi
VEi U σU † VEi
=D
2
2
d i
d i
X
†
†
1
≤ 2
D(VEi U ρU † VEi kVEi U σU † VEi )d i= D(ρkσ).
c 2019(11.169)(11.170)(11.171)(11.172)

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11.9. QUANTUM ENTROPY INEQUALITIES347

The first equality follows from additivity of quantum relative entropy (Exercise 11.8.7) andthe fact that D(πE kπE ) = 0. The second equality follows from the fact that a random
application of a Heisenberg–Weyl unitary is equivalent to a channel that traces out system
E and replaces it with πE (see Exercise 4.7.6). The inequality follows from joint convexity(by assumption), and the last equality follows from invariance of quantum relative entropy
with respect to isometries (Exercise 11.8.6).
We now show that concavity of conditional entropy implies monotonicity of quantum
relative entropy, which has the most involved proof. Before doing so, we need a somewhat
advanced theorem from matrix analysis. Suppose that f is a differentiable function on an
open neighborhood of the spectrum of some self-adjoint operator A. Then its derivative Df
at A is given byX
Df (A) : K →f [1] (λ, η)PA (λ)KPA (η),(11.173)λ,ηP
where A = λ λPA (λ) is the spectral decomposition of A, and f [1] is what is known as thefirst divided difference function. In particular, if x 7−→ A(x) ∈ L(H) (where A(x) is positive
semi-definite) is a differentiable function on an open interval in R, with derivative A0 , thenXdf [1] (λ, η)PA(x) (λ)A0 (x)PA(x) (η),
f (A(x)) =dx
λ,η

(11.174)d
Tr {f (A(x))} = Tr {f 0 (A(x))A0 (x)} .
dx(11.175)

so that

Note how (11.175) appears strikingly similar to the usual chain rule. So if A(x) = A + xB,
thend
Tr {f (A(x))} = Tr {f 0 (A(x))B} ,(11.176)dx
which is the main formula that we need to proceed. In what follows, we will be taking
A(x) = σAB + xρAB , where ρAB , σAB ∈ D(HA ⊗ HB ) and x > 0. If we do not have that0supp(ρAB ) ⊆ supp(σAB ), then we can take σAB≡ (1 − ε)σAB + επAB for ε ∈ (0, 1) andπAB the maximally mixed state. After doing so, we can take a limit as ε → 0 at theend. We also make use of the standard fact that the function f : X → X −1 is everywhere
differentiable on the set of invertible density operators, and at an invertible X, its derivativeis f 0 (X) : Y → −X −1 Y X −1 . Consider that the conditional entropy is homogeneous, in thesense that
H(A|B)xG = xH(A|B)G ,(11.177)

where GAB ∈ L(HA ⊗ HB ) is a positive semi-definite operator. Let
ξY AB ≡
c 2019x1|0ih0|Y ⊗ σAB +|1ih1|Y ⊗ ρAB .
x+1
x+1(11.178)

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CHAPTER 11. QUANTUM INFORMATION AND ENTROPY

Then it follows from homogeneity and concavity of conditional entropy (taking 5 true by
assumption) that
H(A|B)σ+xρ = (x + 1)H(A|B)ξ

x
1
H(A|B)σ +H(A|B)ρ
≥ (x + 1)
x+1x+1= H(A|B)σ + xH(A|B)ρ .

(11.179)(11.180)(11.181)

Manipulating the above inequality then gives
H(A|B)σ+xρ − H(A|B)σ≥ H(A|B)ρ ,x
and taking the limit as x & 0 gives
H(A|B)σ+xρ − H(A|B)σd=
H(A|B)σ+xρ
≥ H(A|B)ρ .
x&0
x
dx
x=0
lim

(11.182)(11.183)

We now evaluate the limit on the left-hand side. So we considerdd
H(A|B)σ+xρ =[− Tr {(σAB + xρAB ) log (σAB + xρAB )}]
dx
dx
d
+
Tr {(σB + xρB ) log (σB + xρB )} . (11.184)dxd[g(y) log g(y)] = [log g(y) + 1] g 0 (y) (up to a scale factor of ln 2
We evaluate this by using dy
from using the binary logarithm) and (11.176) to find thatdTr {(σAB + xρAB ) log (σAB + xρAB )} = Tr {[log (σAB + xρAB ) + IAB ] ρAB } ,dxso thatd
H(A|B)σ+xρ = − Tr {ρAB log (σAB + xρAB )} + Tr {ρB log (σB + xρB )} ,dx
and thusd
H(A|B)σ+xρ= − Tr {ρAB log σAB } + Tr {ρB log σB } .
dx
x=0(11.185)(11.186)(11.187)

Substituting back into the inequality (11.183), we find that
− Tr {ρAB log σAB } + Tr {ρB log σB } ≥ − Tr {ρAB log ρAB } + Tr {ρB log ρB } ,(11.188)

which is equivalent toD(ρAB kσAB ) ≥ D(ρB kσB ).

(11.189)0
If the support condition supp(ρAB ) ⊆ supp(σAB ) does not hold, then we can take σABasmentioned above and all of the above development holds. At the end, we can take the limit
as ε → 0 to find that D(ρAB kσAB ) = +∞, so that the inequality holds trivially in this case.

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11.9. QUANTUM ENTROPY INEQUALITIES

11.9.2

349

Quantum Data Processing

The quantum data-processing inequalities discussed below are similar in spirit to the classical data-processing inequality. Recall that the classical data-processing inequality statesthat processing classical data reduces classical correlations. The quantum data-processing
inequalities state that processing quantum data reduces quantum correlations.
One variant applies to the following scenario. Suppose that Alice and Bob share somebipartite state ρAB . The coherent information I(AiB)ρ is one measure of the quantum correlations present in this state. Bob then processes his system B according to some quantum
channel NB→B 0 to produce some quantum system B 0 and let σAB 0 denote the resulting state.
The quantum data-processing inequality states that this step of quantum data processing
reduces quantum correlations, in the sense that
I(AiB)ρ ≥ I(AiB 0 )σ .

(11.190)

Theorem 11.9.3 (Data Processing for Coherent Information) Let ρAB ∈ D(HA ⊗HB ) and let N : L(HB ) → L(HB 0 ) be a quantum channel. Set σAB 0 ≡ NB→B 0 (ρAB ). Thenthe following quantum data-processing inequality holds
I(AiB)ρ ≥ I(AiB 0 )σ .

(11.191)

Proof. This is a consequence of Exercise 11.8.3 and Theorem 11.8.1. By Exercise 11.8.3,
we know that
I(AiB)ρ = D(ρAB kIA ⊗ ρB ),I(AiB 0 )σ = D(σAB 0 kIA ⊗ σB 0 )= D(NB→B 0 (ρAB )kIA ⊗ NB→B 0 (ρB ))= D(NB→B 0 (ρAB )kNB→B 0 (IA ⊗ ρB )).

(11.192)(11.193)(11.194)(11.195)

The statement then follows from the monotonicity of quantum relative entropy by pickingρ = ρAB , σ = IA ⊗ ρB , and N = idA ⊗NB→B 0 in Theorem 11.8.1.
Theorem 11.9.4 (Data Processing for Mutual Information) Let ρAB ∈ D(HA ⊗HB ),N : L(HA ) → L(HA0 ) be a quantum channel, and M : L(HB ) → L(HB 0 ) be a quantum
channel. Set σA0 B 0 ≡ (NA→A0 ⊗ MB→B 0 )(ρAB ). Then the following quantum data-processing
inequality applies to the quantum mutual information:
I(A; B)ρ ≥ I(A0 ; B 0 )σ .

(11.196)

Proof. From Exercise 11.8.2, we know that
I(A; B)ρ = D(ρAB kρA ⊗ ρB ),I(A0 ; B 0 )σ = D(σA0 B 0 kσA0 ⊗ σB 0 )= D((NA→A0 ⊗ MB→B 0 )(ρAB )kNA→A0 (ρA ) ⊗ MB→B 0 (ρB ))= D((NA→A0 ⊗ MB→B 0 )(ρAB )k(NA→A0 ⊗ MB→B 0 )(ρA ⊗ ρB )).

(11.197)(11.198)(11.199)(11.200)

The statement then follows from the monotonicity of quantum relative entropy by pickingρ = ρAB , σ = ρA ⊗ ρB , and N = NA→A0 ⊗ MB→B 0 in Theorem 11.8.1.
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CHAPTER 11. QUANTUM INFORMATION AND ENTROPY

Exercise 11.9.1 Let ρAB ∈ D(HA ⊗ HB ), N : L(HA ) → L(HA0 ) be a unital quantum channel, and M : L(HB ) → L(HB 0 ) be a quantum channel. Set σA0 B 0 ≡ (NA→A0 ⊗MB→B 0 )(ρAB ).
Prove that
I(AiB)ρ ≥ I(A0 iB 0 )σ .
(11.201)
Exercise 11.9.2 (Holevo Bound) Use the quantum data-processing inequality to showthat the Holevo information χ(E) is an upper bound on the accessible information Iacc (E):
Iacc (E) ≤ χ(E),(11.202)where E is an ensemble of quantum states. (See Section 10.9.2 for a definition of accessible
information.)
Exercise 11.9.3 (Shannon Entropy vs. von Neumann Entropy) Consider an ensemble {pX (x), |ψx i}. The expected density operator of the ensemble isXρ≡pX (x)|ψx ihψx |.
(11.203)x

Use the quantum data-processing inequality to show that the Shannon entropy H(X) is neverless than the quantum entropy of the expected density operator ρ:
H(X) ≥ H(ρ).

(11.204)

P(Hint: Begin with a classical shared randomness state x pX (x)|xihx|X ⊗ |xihx|X 0 and apply
a preparation channel to system X 0 ). Conclude that the Shannon entropy of the ensemble
is strictly greater than the quantum entropy whenever the states in the ensemble are nonorthogonal.
Exercise 11.9.4 Use the idea in the above exercise to show that the conditional entropy
H(X|B)ρ is always non-negative whenever the state ρXB is a classical–quantum state:XpX (x)|xihx|X ⊗ ρxB .
(11.205)
ρXB ≡x

Additionally, show that H(X|B)ρ ≥ 0 is equivalent to H(B)ρ ≤ H(X)ρ + H(B|X)ρ andI(X; B)ρ ≤ H(X)ρ .
Exercise 11.9.5 (Separability and Negativity of Coherent Information) Show thatthe following inequality holds for any separable state ρAB :max {I(AiB)ρ , I(BiA)ρ } ≤ 0.

(11.206)

Exercise 11.9.6 (Quantum Data Processing for CQMI) Let ρABC ∈ D(HA ⊗ HB ⊗HB ), N : L(HA ) → L(HA0 ) be a quantum channel, and M : L(HB ) → L(HB 0 ) be a quantum
channel. Set σA0 B 0 C ≡ (NA→A0 ⊗ MB→B 0 )(ρABC ). Prove that
I(A; B|C)ρ ≥ I(A0 ; B 0 |C)σ .
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11.9. QUANTUM ENTROPY INEQUALITIES

11.9.3351

Entropic Uncertainty Principle

The uncertainty principle reviewed in Section 3.4.2 aims to capture a fundamental feature ofquantum mechanics, namely, that there is an unavoidable uncertainty in the measurement
outcomes of incompatible (non-commuting) observables. This uncertainty principle is aradical departure from classical intuitions, where there it seems as if there should not be any
obstacle to measuring incompatible observables such as position and momentum.
However, the uncertainty principle that we reviewed before (the standard version in most
textbooks) suffers from a few deficiencies. First, the measure of uncertainty used there is
the standard deviation, which is not just a function of the probabilities of measurement
outcomes but also of the values of the outcomes. Thus, the values of the outcomes may skew
the uncertainty measure (however, one could always relabel the values in order to avoid
this difficulty). More importantly however, from an information-theoretic perspective, there
is not a clear operational interpretation for the standard deviation as there is for entropy.
Second, the lower bound in (3.88) depends not only on the observables but also on thestate. In Exercise 3.4.5, we saw how this lower bound can vanish for a state even when the
distributions corresponding to the measurement outcomes in fact do have uncertainty. So,
it would be ideal to separate this lower bound into two terms: one which depends only on
measurement incompatibility and another which depends only on the state.
Additionally, it might seem as if giving two parties access to a maximally entangledstate allows them to defy the uncertainty principle (and this is what confounded Einstein,
Podolsky, and Rosen after quantum mechanics had been established). Indeed, suppose that
Alice and Bob share a Bell state |Φ+ i = 2−1/2 (|00i + |11i) = 2−1/2 (|++i + |−−i). If Alicemeasures the Pauli Z observable on her system, then Bob can guess the outcome of her
measurement with certainty. Also, if Alice were instead to measure the Pauli X observableon her system, then Bob would also be able to guess the outcome of her measurement withcertainty, in spite of the fact that Z and X are incompatible observables. So, a revisionof the uncertainty principle is clearly needed to account for this possibility, in the scenario
where Bob shares a quantum memory correlated with Alice’s system.
The uncertainty principle in the presence of quantum memory is such a revision thatmeets all of the desiderata stated above. It quantifies uncertainty in terms of quantum
entropy rather than with standard deviation, and it also accounts for the scenario in which
an observer has a quantum memory correlated with the system being measured. So, supposethat Alice and Bob share systems A and B, respectively, that are in some state ρAB . If Alice
performs a measurement channel corresponding to a POVM {ΛxA } on her system A, thenthe post-measurement state is as follows:X
σXB ≡|xihx|X ⊗ TrA {(ΛxA ⊗ IB ) ρAB } .
(11.208)x

In the above classical–quantum state, the measurement outcomes x are encoded into orthonormal states {|xi} of the classical register X, and the probability for obtaining outcome
x is Tr {(ΛxA ⊗ IB ) ρAB }. We would like to quantify the uncertainty that Bob has about theoutcome of the measurement, and a natural quantity for doing so is the conditional quanc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

352

CHAPTER 11. QUANTUM INFORMATION AND ENTROPY

tum entropy H(X|B)σ . Similarly, starting from the state ρAB , Alice could choose to perform
some other measurement channel corresponding to POVM {ΓzA } on her system A. In thiscase, the post-measurement state is as follows:
X(11.209)
τZB ≡|zihz|Z ⊗ TrA {(ΓzA ⊗ IB ) ρAB } ,zwith a similar interpretation as before. We can quantify Bob’s uncertainty about the measurement outcome z in terms of the conditional quantum entropy H(Z|B)τ . We define
Bob’s total uncertainty about the measurement outcomes to be the sum of both entropies:
H(X|B)σ + H(Z|B)τ . We will call this the uncertainty sum, in analogy with the uncertainty
product in (3.88).
We stated above that it would be desirable to have a lower bound on the uncertainty
sum consisting of a measurement incompability term and a state-dependent term. One way
to quantify the incompatibility of the POVMs {ΛxA } and {ΓzA } is in terms of the following
quantity:2
p p
ΛxA ΓzA,(11.210)c ≡ maxx,z∞where k·k∞ is the infinity norm of an operator (for the finite-dimensional case, kAk∞ isjust the maximal eigenvalue of |A|). To grasp an intuition for this incompatibility measure,suppose that {ΛxA } and {ΓzA } are actually complete projective measurements with one common element. In this case, it follows that c = 1, so that the measurements are regarded as
maximally compatible. On the other hand, if the measurements are of Pauli observables X
and Z, these are maximally incompatible for a two-dimensional Hilbert space and c = 1/2.
We now state the uncertainty principle in the presence of quantum memory:
Theorem 11.9.5 (Uncertainty Principle with Quantum Memory) Suppose that Alice and Bob share a state ρAB and that Alice performs either of the POVMs {ΛxA } or {ΓzA }on her share of the state (with at least one of {ΛxA } or {ΓzA } being a rank-one POVM). Then
Bob’s total uncertainty about the measurement outcomes has the following lower bound:
H(X|B)σ + H(Z|B)τ ≥ log (1/c) + H(A|B)ρ ,(11.211)

where the states σXB and τZB are defined in (11.208) and (11.209), respectively, and themeasurement incompatibility is defined in (11.210).
Interestingly, the lower bound given in the above theorem consists of both the measurement incompatibility and the state-dependent term H(A|B)ρ . As we know from Exercise 11.9.5, when the conditional quantum entropy H(A|B)ρ becomes negative, this implies
that the state ρAB is entangled (but not necessarily the converse). Thus, a negative conditional entropy implies that the lower bound on the uncertainty sum can become lower thanlog (1/c), and furthermore, that it might be possible to reduce Bob’s total uncertainty about
the measurement outcomes down to zero. Indeed, this is the case for the example we mentioned before with measurements of Pauli X and Z on the maximally entangled Bell state.
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11.9. QUANTUM ENTROPY INEQUALITIES353

One can verify for this case that log(1/c) = 1 and H(A|B) = −1, so that this is consistent
with the fact that H(X|B)σ + H(Z|B)τ = 0 for this example. We now give a path to proving
the above theorem (leaving the final steps as an exercise).
Proof. We actually prove the following uncertainty relation instead:
H(X|B)σ + H(Z|E)ω ≥ log(1/c),(11.212)

where ωZE is a classical–quantum state of the following form:XωZE ≡|zihz|Z ⊗ TrAB {(ΓzA ⊗ IBE ) φρABE } ,(11.213)z

and φρABE is a purification of ρAB . We leave it as an exercise to demonstrate that the above
uncertainty relation implies the one in the statement of the theorem whenever ΓzA is a rankone POVM. Consider the following isometric extensions of the measurement channels for{ΛxA } and {ΓzA }:X
p
UA→XX 0 A ≡|xiX ⊗ |xiX 0 ⊗ ΛxA ,(11.214)x

VA→ZZ 0 A ≡X|ziZ ⊗ |ziZ 0 ⊗

p
ΓzA ,

(11.215)z

where {|xi} and {|zi} are both orthonormal bases. Let ωZZ 0 ABE denote the following state:|ωiZZ 0 ABE ≡ VA→ZZ 0 A |φρ iABE ,(11.216)

so that ωZE = TrZ 0 AB {ωZZ 0 ABE }. Exercise 11.5.3 establishes that
H(Z|E)ω = −H(Z|Z 0 AB)ω ,(11.217)so that (11.212) is equivalent to− H(Z|Z 0 AB)ω ≥ log(1/c) − H(X|B)σ .

(11.218)

Recalling the result of Exercise 11.8.3, we then have that (11.218) is equivalent toD(ωZZ 0 AB kIZ ⊗ ωZ 0 AB ) ≥ log(1/c) + D(σXB kIX ⊗ σB ),(11.219)

where we observe that σB = ωB . So we aim to prove (11.219). Consider the following chainof inequalities:
D(ωZZ 0 AB kIZ ⊗ ωZ 0 AB )≥ D ωZZ 0 AB V V † (IZ ⊗ ωZ 0 AB ) V V= D ρAB V † (IZ ⊗ ωZ 0 AB ) V= D U ρAB U
c 2019††(11.220)
†U V (IZ ⊗ ωZ 0 AB ) V U(11.221)(11.222)†.

(11.223)

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354

CHAPTER 11. QUANTUM INFORMATION AND ENTROPY

The first inequality follows from monotonicity of quantum relative entropy under the channel
ρ → ΠρΠ + (I − Π) ρ (I − Π), where the projector Π ≡ V V † , and also from the fact that(I − Π)ωZZ 0 AB (I − Π) = 0. The first equality follows from invariance of quantum relative
entropy with respect to isometries (Exercise 11.8.6) and the fact that ωZZ 0 AB = V ρAB V † .
The second equality again follows from invariance of quantum relative entropy with respect
to isometries. Let us define σXX 0 ABE as|σiXX 0 ABE ≡ UA→XX 0 A |φρ iABE .

(11.224)

We then have that (11.223) is equal to
D σXX 0 AB U V † (IZ ⊗ ωZ 0 AB ) V U † ,(11.225)and explicitly evaluating U V † (IZ ⊗ ωZ 0 AB ) V U † asU V † (IZ ⊗ ωZ 0 AB ) V U †

q 

X
p 
0
0
0
=U
hz |ziZ hz |Z 0 ⊗ ΓzA ωZ 0 AB |ziZ 0 ⊗ ΓzA U †(11.226)(11.227)z 0 ,z=UX

hz|Z 0 ⊗

p



p 
ΓzA ωZ 0 AB |ziZ 0 ⊗ ΓzA U † ,(11.228)z

gives that (11.225) is equal to
D σXX 0 AB U

X


p 
p 
hz|Z 0 ⊗ ΓzA ωZ 0 AB |ziZ 0 ⊗ ΓzA U †!
.

(11.229)z

We trace out the X 0 A systems and exploit monotonicity of quantum relative entropy andcyclicity of trace to show that the above is not less than
!

o
n
X
p
p
D σXB
|xihx|X ⊗ TrZ 0 A |zihz|Z 0 ⊗ ΓzA ΛxA ΓzA ωZ 0 AB.
(11.230)
z,x

p
p
p p
Using the fact that ΓzA ΛxA ΓzA = | ΓzA ΛxA |2 ≤ cI, it follows that
o
n
h
X
p
p i
|xihx|X ⊗ TrZ 0 A |zihz|Z 0 ⊗ cIA − ΓzA ΛxA ΓzA ωZ 0 AB

(11.231)z,x

is a positive semi-definite operator, or equivalently, that
n
o
X
p
p 
x
z
z
|xihx|X ⊗ TrZ 0 A |zihz|Z 0 ⊗ ΓA ΛA ΓA ωZ 0 AB ≤ c IX ⊗ ωB .

(11.232)z,xProposition 11.8.2 and Exercise 11.8.9 imply that (11.230) is not less thanD(σXB kc IX ⊗ ωB ) = log(1/c) + D(σXB kIX ⊗ ωB )= log(1/c) + D(σXB kIX ⊗ σB ),(11.233)(11.234)

which finally proves the inequality in (11.212). We now leave it as an exercise to prove thestatement of the theorem starting from the inequality in (11.212).
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11.10. CONTINUITY OF QUANTUM ENTROPY

355

Exercise 11.9.7 Prove that (11.212) implies Theorem 11.9.5.
Exercise 11.9.8 Prove that Theorem 11.9.5 implies the following entropic uncertainty relation for a state ρA on a single system:
H(X) + H(Z) ≥ log(1/c) + H(A)ρ ,(11.235)

where H(X) and H(Z) are the Shannon entropies of the measurement outcomes.

11.10 Continuity of Quantum Entropy
Suppose that two density operators ρ and σ are close in trace distance. We might then
expect several properties to hold: the fidelity between them should be close to one and wewould suspect that their entropies should be close. Theorem 9.3.1 states that the fidelity isclose to one if the trace distance is small.
An important theorem below, the Fannes–Audenaert inequality, states that quantum
entropies are close as well. This theorem usually finds application in a proof of a converse
theorem in quantum Shannon theory. Usually, the specification of any good protocol (inthe sense of asymptotically vanishing error) involves placing a bound on the trace distance
between the actual state resulting from a protocol and the ideal state that it should produce.
The Fannes–Audenaert inequality then allows us to translate these statements of error into
informational statements that bound the asymptotic rates of communication in any good
protocol.
Theorem 11.10.1 (Fannes–Audenaert Inequality) Let ρ, σ ∈ D(H) and suppose that1kρ − σk1 ≤ ε ∈ [0, 1]. Then the following inequality holds2
ε log [dim(H) − 1] + h2 (ε) if ε ∈ [0, 1 − 1/ dim(H)]|H(ρ) − H(σ)| ≤.
(11.236)log dim(H)else
Putting these together, a universal bound is|H(ρ) − H(σ)| ≤ ε log dim(H) + h2 (ε).

(11.237)

Proof. A proof of this theorem follows by applying the classical result in Theorem 10.7.4.
We first prove that
H(ρ) − H(σ) ≤ ε log [dim(H) − 1] + h2 (ε)(11.238)
if ε ∈ [0, 1 − 1/ dim(H)]. First note that the function f (ε) ≡ ε log [dim(H) − 1] + h2 (ε) ismonotonedim(H)] because f 0 (ε) = log [dim(H) − 1]+ non-decreasing on the interval [0, 1 − 1/P
1−εlog ε ≥ 0 for ε ∈ [0, 1 − 1/ dim(H)]. Let σ = y p(y)|yihy| be a spectral decompositionof σ. Let ∆σ denote the following completely dephasing channel:X∆σ (ω) =|yihy|ω|yihy|.
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356

CHAPTER 11. QUANTUM INFORMATION AND ENTROPY

Then ∆σ (σ) = σ and Corollary 11.9.3 gives that H(ρ) ≤ H(∆σ (ρ)), so that
H(ρ) − H(σ) ≤ H(∆σ (ρ)) − H(∆σ (σ)).

(11.240)

At the same time, we know from the monotonicity of trace distance with respect to channels(Exercise 9.1.9) that
(11.241)kρ − σk1 ≥ ∆σ (ρ) − ∆σ (σ) 1 .
Putting everything together, we find that
H(ρ) − H(σ) ≤ H(∆σ (ρ)) − H(∆σ (σ))
1≤f
∆σ (ρ) − ∆σ (σ) 1
2


1
≤f
kρ − σk1 ,2(11.242)(11.243)(11.244)

where the second inequality follows from Theorem 10.7.4 and the last inequality follows
because f is monotone non-decreasing on the interval [0, 1 − 1/ dim(H)].
The other inequality H(σ) − H(ρ) ≤ ε log [dim(H) − 1] + h2 (ε) follows by the same
proof method, but instead dephasing with respect to the eigenbasis of ρ. The bound|H(ρ) − H(σ)| ≤ log(dim(H)) follows trivially from the fact that the entropy is non-negativeand cannot exceed log(dim(H)).
There is another variation of this theorem, which we state below. We do not however
give a full proof, but instead just argue for it. The proof sketch makes use of an original
insight by Audenaert (2007).
Theorem 11.10.2 (Fannes–Audenaert Inequality) Let ρ, σ ∈ D(H) and let T ≡ 21 kρ − σk1 .
Then the following inequality holds|H(ρ) − H(σ)| ≤ T log [dim(H) − 1] + h2 (T ).

(11.245)

Furthermore, this bound is optimal because there exists a pair of states that saturates it for
all T ∈ [0, 1] and dimension dim(H).
Proof. The following inequalities are known from the theory of matrix analysis (Bhatia,
1997, Inequality IV.62):
T0 ≡

1
1
Eig↓ (ρ) − Eig↓ (σ) 1 ≤ T = kρ − σk122≤ T1 ≡

1
Eig↓ (ρ) − Eig↑ (σ) 1 , (11.246)
2

where Eig↓ (A) is the list of eigenvalues of Hermitian A in non-increasing order and Eig↑ (A)is the list in non-decreasing order. Then applying the fact that the entropy depends only on
the eigenvalues and is invariant with respect to permutations of them, we find that
|H(ρ) − H(σ)| = H(Eig↓ (ρ)) − H(Eig↓ (σ))≤ T0 log [dim(H) − 1] + h2 (T0 ),c 2019(11.247)(11.248)

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11.10. CONTINUITY OF QUANTUM ENTROPY

357

where the inequality follows from Theorem 10.7.4. Similarly, we find that
|H(ρ) − H(σ)| = H(Eig↓ (ρ)) − H(Eig↑ (σ))≤ T1 log [dim(H) − 1] + h2 (T1 ).

(11.249)(11.250)

We know from (11.246) that T = λT0 + (1 − λ)T1 for some λ ∈ [0, 1]. Applying concavity ofthe binary entropy, we find that
|H(ρ) − H(σ)| ≤ λ [T0 log [dim(H) − 1] + h2 (T0 )]+ (1 − λ) [T1 log [dim(H) − 1] + h2 (T1 )]≤ T log [dim(H) − 1] + h2 (T ). (11.251)
The inequality is optimal because choosing ρ = |0ih0| and σ = (1−ε)|0ih0|+ε/(d−1)|1ih1|+· · · + ε/(d − 1)|d − 1ihd − 1| saturates the bound for all ε ∈ [0, 1] and for all dimensions d.
An important theorem below, the Alicki–Fannes–Winter (AFW) inequality, states thatconditional quantum entropies are close as well. This statement does follow directly fromthe Fannes–Audenaert inequality, but the main advantage of the AFW inequality is thatthe upper bound has a dependence only on the dimension of the first system in the conditional entropy (no dependence on the conditioning system). The AFW inequality also finds
application in a proof of a converse theorem in quantum Shannon theory.
Theorem 11.10.3 (AFW Inequality) Let ρAB , σAB ∈ D(HA ⊗ HB ). Suppose that1kρAB − σAB k1 ≤ ε,2

(11.252)|H(A|B)ρ − H(A|B)σ | ≤ 2ε log dim(HA ) + g2 (ε),(11.253)for ε ∈ [0, 1]. Then

where g2 (ε) ≡ (ε + 1) log2 (ε + 1) − ε log2 ε. If ρXB and σXB are classical–quantum and havethe following form:XρXB =p(x)|xihx|X ⊗ ρxB ,(11.254)
x

σXB =

X

q(x)|xihx|X ⊗ σBx ,(11.255)x

where p and q are probability distributions defined over a finite alphabet X , {|xi} is anorthonormal basis, and ρxB , σBx ∈ D(HB ) for all x ∈ X , then
|H(X|B)ρ − H(X|B)σ | ≤ ε log dim(HX ) + g2 (ε),|H(B|X)ρ − H(B|X)σ | ≤ ε log dim(HB ) + g2 (ε).
c 2019(11.256)(11.257)

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358

CHAPTER 11. QUANTUM INFORMATION AND ENTROPY

Proof. The bounds trivially hold when ε = 0, so henceforth we assume that ε ∈ (0, 1].
All of the upper bounds are monotone non-decreasing with ε, so it suffices to assume that1kρAB − σAB k1 = ε. Let ρAB − σAB = PAB − QAB be a decomposition of ρAB − σAB into its
2positive part PAB ≥ 0 and its negative part QAB ≥ 0 (as in the proof of Lemma 9.1.1). Let∆AB ≡ PAB /ε. Since Tr{PAB } = 12 kρAB − σAB k1 (see the proof of Lemma 9.1.1), it follows
that ∆AB is a density operator. Now consider thatρAB = σAB + (ρAB − σAB )= σAB + PAB − QAB≤ σAB + PAB= σAB + ε∆AB1ε= (1 + ε)
σAB +
∆AB
1+ε
1+ε
= (1 + ε) ωAB ,(11.258)(11.259)(11.260)(11.261)(11.262)(11.263)1εwhere we define ωAB ≡ 1+ε
σAB + 1+ε
∆AB . Now let ∆0AB ≡ 1ε [(1 + ε) ωAB − ρAB ]. It follows
from (11.258)–(11.263) that ∆0AB is positive semi-definite. Furthermore, one can check thatTr{∆0AB } = 1, so that ∆0AB is a density operator. One can also quickly check that

ωAB =

1
ε
1
ε
ρAB +
∆0AB =
σAB +
∆AB .
1+ε
1+ε
1+ε
1+ε

(11.264)

Now consider that
H(A|B)ω = −D(ωAB kIA ⊗ ωB )= H(ωAB ) + Tr{ωAB log ωB }


ε
ε
1
≤ h2
H(ρAB ) +H(∆0AB )
+
1+ε
1+ε
1+ε
1
ε
+
Tr{ρAB log ωB } +Tr{∆0AB log ωB }1+ ε 1+ε1
εD(ρAB kIA ⊗ ωB )
−
= h2
1+ε
1+ε
ε
−
D(∆0AB kIA ⊗ ωB )1+ ε 
ε
1
ε
≤ h2
+
H(A|B)ρ +H(A|B)∆0 .
1+ε
1+ε
1+ε

(11.265)(11.266)(11.267)(11.268)(11.269)

The first equality follows from Exercise 11.8.3, and the second equality follows from thedefinition of quantum relative entropy. The first inequality follows because H(AB) ≤ H(Y )+H(AB|Y ) for a classical–quantum state on systems Y and AB (see Exercise 11.9.4), here
taking the state as1
ε|0ih0|Y ⊗ ρAB +|1ih1|Y ⊗ ∆0AB .
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11.10. CONTINUITY OF QUANTUM ENTROPY359

The third equality follows from algebra and the definition of quantum relative entropy.
The last inequality follows from Exercise 11.8.3. From concavity of the conditional entropy(Exercise 11.7.5), we have that
H(A|B)ω ≥1
εH(A|B)σ +H(A|B)∆ .
1+ε
1+ε(11.271)

Putting together
 the upper and lower bounds on H(A|B)ω , and observing that g2 (ε) =ε(1 + ε) h2 1+ε , we find that
H(A|B)σ − H(A|B)ρ ≤ g2 (ε) + ε [H(A|B)∆0 − H(A|B)∆ ]≤ g2 (ε) + 2ε log dim(HA ),(11.272)(11.273)

where the second inequality follows from a dimension bound for the conditional entropy(Theorem 11.5.1).
The statements for classical–quantum states follow because the density operator ∆ isclassical–quantum in this case and we know that H(X|B)∆ , H(B|X)∆ ≥ 0 (see Exercise 11.9.4).
Exercise 11.10.1 (AFW for Coherent Information) Prove that
|I(AiB)ρ − I(AiB)σ | ≤ 2ε log dim(HA ) + g2 (ε),(11.274)

with 21 kρAB − σAB k1 ≤ ε ∈ [0, 1].
Exercise 11.10.2 (AFW for Quantum Mutual Information) Prove that
|I(A; B)ρ − I(A; B)σ | ≤ 3ε log dim(HA ) + 2g2 (ε),(11.275)

for any ρAB and σAB with 21 kρAB − σAB k1 ≤ ε ∈ [0, 1].
We can also use these results to get a refinement of the non-negativity of mutual information and the dimension upper bounds on mutual information and conditional mutual
information. A refinement of the non-negativity of conditional mutual information (strong
subadditivity) will appear in Chapter 12. The refinements of mutual information are quantified in terms of the trace distance between ρAB and the product of its marginals. The
refinement of conditional mutual information is quantified in terms of the trace distance
between ρABC and a “recovered version” of ρBC , which represents a quantum generalizationof a Markov chain. So these results represent quantum generalizations of the statements in
Theorems 10.8.2 and 10.8.3.
Theorem 11.10.4 Let ρAB ∈ D(HA ⊗ HB ) and let ∆ ≡ 21 kρAB − ρA ⊗ ρB k1 . Then
2 2
∆,
ln 2
I(A; B)ρ ≤ 2∆ log [min {dim(HA ), dim(HB )}] + g2 (∆).
I(A; B) ≥c 2019(11.276)(11.277)

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CHAPTER 11. QUANTUM INFORMATION AND ENTROPY

Proof. The first inequality is a direct application of Exercise 11.8.2 and the quantum Pinsker
inequality (Theorem 11.9.1). Let ωAB ≡ ρA ⊗ ρB . The next inequality follows because
I(A; B)ρ = |I(A; B)ρ − I(A; B)ω |= |H(A)ρ − H(A|B)ρ − [H(A)ω − H(A|B)ω ]|= |H(A|B)ω − H(A|B)ρ |≤ 2∆ log dim(HA ) + g2 (∆),

(11.278)(11.279)(11.280)(11.281)

where in the last line we applied Theorem 11.10.3. The other inequality
I(A; B)ρ ≤ 2∆ log dim(HB ) + g2 (∆)(11.282)

follows by expanding the mutual information in the other way.
Theorem 11.10.5 Let ρABC ∈ D(HA ⊗ HB ⊗ HC ) and let∆≡

1
inf kρABC − RC→AC (ρBC )k1 ,2 RC→AC

(11.283)

where the optimization is with respect to channels R : L(HC ) → L(HA ⊗ HC ). Then
I(A; B|C)ρ , I(A; B|C)σ ≤ 2∆ log dim(HB ) + g2 (∆),

(11.284)

where σABC ≡ R∗C→AC (ρBC ) with R∗C→AC the optimal recovery channel in (11.283).
Proof. Let ∆R ≡ 12 kρABC − RC→AC (ρBC )k1 for some RC→AC and define ωABC ≡ RC→AC (ρBC ).
Consider that
I(A; B|C)ρ = H(B|C)ρ − H(B|AC)ρ
≤ H(B|AC)ω − H(B|AC)ρ
≤ 2∆R log dim(HB ) + g2 (∆R ).

(11.285)(11.286)(11.287)

The first inequality follows from quantum data processing (Theorem 11.9.3) and the second
from Theorem 11.10.3. Since the inequality holds for all recovery channels and the upperbound is monotone non-decreasing in ∆R , we can conclude the inequality in the statementof the theorem. Now consider that
I(A; B|C)σ = H(B|C)σ − H(B|AC)σ≤ H(B|C)σ − H(B|C)ρ
≤ 2∆ log dim(HB ) + g2 (∆).

(11.288)(11.289)(11.290)

The justifications for these inequalities are the same as those for the above ones (we additionally need to use monotonicity of the trace distance with respect to partial trace).
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11.11. HISTORY AND FURTHER READING

361

11.11 History and Further Reading
The quantum entropy and its relatives, such as conditional entropy and mutual information,
are useful information measures and suffice for our studies in this book. However, the quantum entropy is certainly not the only information measure worthy of study. In recent years,
entropic measures such as the min- and max-entropy have emerged (and their smoothedvariants), and they are useful in developing a more general theory that applies beyond thei.i.d. setting that we study in this book. In fact, one could view this theory as more fundamental than the theory presented in this book, since the “one-shot” results often imply thei.i.d. results studied in this book. Rather than developing this theory in full, we point toseveral excellent references on the subject (Renner, 2005; Datta, 2009; Datta and Renner,
2009; Koenig et al., 2009; Tomamichel, 2016).
The fact that the conditional entropy can be negative is discussed in Wehrl (1978);Horodecki and Horodecki (1994); Cerf and Adami (1997). Holevo (1973a) proved the important bound bearing his name. Only later was it understood that this bound is a consequenceof the monotonicity of quantum relative entropy. Lieb and Ruskai (1973a,b) established thestrong subadditivity of quantum entropy and concavity of conditional quantum entropy by
invoking an earlier result of Lieb (1973), and Lieb and Ruskai (1973a) also showed that themonotonicity of quantum relative entropy with respect to partial trace is a consequence ofthe concavity of conditional quantum entropy (the proof of this given in Theorem 11.9.2 isdue to them). Araki and Lieb (1970) proved the inequality in Exercise 11.7.10. Umegaki(1962) established the modern definition of the quantum relative entropy. The fact that it
is non-negative for states is a result known as Klein’s inequality (see Lanford and Robinson(1968) for attribution to Klein). Lindblad (1975) established the monotonicity of quantum relative entropy for separable Hilbert spaces based on the results of Lieb and Ruskai(1973a,b), and Uhlmann (1977) extended Lindblad’s result to more general settings. Coleset al. (2012) proved Proposition 11.8.2. (Ohya and Petz, 1993, Theorem 1.15) proved thequantum Pinsker inequality (stated here as Theorem 11.9.1). The coherent information first
appeared in Schumacher and Nielsen (1996), where they proved that it obeys a quantum
data-processing inequality (this was the first clue that the coherent information would be an
important information quantity for characterizing quantum capacity).
Entropic uncertainty relations have a long and interesting history. We do not review thishistory here but instead point to the survey article (Coles et al., 2015). There has been muchinterest in entropic uncertainty relations, with perhaps the most notable advance being theentropic uncertainty relation in the presence of quantum memory (Berta et al., 2010). The
proof that we give for Theorem 11.9.5 is the same as that in (Coles et al., 2012), which inturn exploits ideas from (Tomamichel and Renner, 2011).
Fannes (1973) proved his eponymous inequality, and Audenaert (2007) gave a significant
improvement of it. Winter (2015b) proved the inequality in Theorem 11.10.3. Earlier, Alicki
and Fannes (2004) proved a weaker version of the inequality in Theorem 11.10.3 (which
however has been extremely useful for many purposes in quantum information theory). Berta
et al. (2015b) proved Theorem 11.10.5 (see also Fawzi and Renner (2015)).

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362

c 2019CHAPTER 11. QUANTUM INFORMATION AND ENTROPYMark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 12

Quantum Entropy Inequalities andRecoverability
The quantum entropy inequalities discussed in the previous chapter lie at the core of quantum Shannon theory and in fact underlie some important principles of physics such as theuncertainty principle (see Section 11.9.3). In fact, we will use these entropy inequalities toprove the converse parts of every coding theorem appearing in the last two parts of this book.
Their prominence in both quantum Shannon theory and other areas of physics motivates us
to study them in more detail. We delved into more depth in Chapter 10 regarding many of
the classical entropy inequalities, and in the process, we established necessary and sufficientconditions for the saturation of the inequalities (Section 10.7.1), while also understanding
the near saturation of the entropy inequalities (Section 10.8). The aim of this chapter is
to carry out a similar program for all of the quantum entropy inequalities presented in theprevious chapter. The outcome will be a proof for the monotonicity of quantum relative
entropy (Theorem 11.8.1), with the added benefit of an understanding of the saturation andnear saturation of this quantum entropy inequality.

12.1 Recoverability Theorem
The main theorem in this chapter can be summarized informally as follows: if the decreasein quantum relative entropy between two quantum states after a quantum channel acts is
relatively small, then it is possible to perform a recovery channel, such that we can perfectly
recover one state while approximately recovering the other. This can be interpreted as
quantifying how well one can reverse the action of a quantum channel. Throughout, we takeρ, σ, and N as given in the following definition:
Definition 12.1.1 Let ρ ∈ D(H) and let σ ∈ L(H) be positive semi-definite, such thatsupp(ρ) ⊆ supp(σ). Let N : L(H) → L(H0 ) be a quantum channel.
The formal statement of the theorem is as follows:363

364 CHAPTER 12. QUANTUM ENTROPY INEQUALITIES AND RECOVERABILITY

Theorem 12.1.1 Given ρ, σ, and N as in Definition 12.1.1, there exists a recovery channel
Rσ,N : L(H0 ) → L(H), depending only on σ and N , such thatD(ρkσ) − D(N (ρ)kN (σ)) ≥ − log F (ρ, (Rσ,N ◦ N )(ρ)),(Rσ,N ◦ N )(σ) = σ.

and(12.1)(12.2)

Given that the quantum fidelity F takes values between zero and one, we can immediately
conclude that− log F (ρ, (Rσ,N ◦ N )(ρ)) ≥ 0,
(12.3)
so that the above theorem implies the monotonicity of quantum relative entropy (Theorem 11.8.1) as a consequence. Furthermore, the recovery channel satisfying (12.1) has theproperty that it perfectly recovers σ from N (σ) (satisfying (12.2)), a fact which we provelater and which makes the inequality in (12.1) non-trivial.
The proof given here for Theorem 12.1.1 relies on the method of complex interpolationand the notion of a Rényi generalization of a relative entropy difference. We review thisbackground first before going through the proof. One of the consequences of Theorem 12.1.1
is to provide physically meaningful improvements to many quantum entropy inequalities
discussed in the previous chapter, such as strong subadditivity, joint convexity of quantum
relative entropy, and concavity of conditional quantum entropy. We explore these consequences in Section 12.6.

12.2 Schatten Norms and Complex Interpolation
The proof of Theorem 12.1.1 given here requires a bit of mathematical background before
we can delve into it. So we first begin by defining the Schatten norms and several of theirproperties. We then review some essential results from complex analysis, that lead to acomplex interpolation theorem known as the Stein–Hirschman interpolation theorem.

12.2.1

Schatten Norms and Duality

An important technical tool in the proof given here is the Schatten p-norm of an operator A,
defined as
1/p
kAkp ≡ [Tr {|A|p }] ,(12.4)√
where A ∈ L(H), |A| ≡ A† A, and p ≥ 1. We have already studied two special cases ofthis norm, which are the trace norm when p = 1 (Section 9.1.1) and the Hilbert–Schmidt
norm when p = 2 (Section 9.6). One can show, along the same lines as the proof for
Proposition 9.1.1, that kAkp is equal to the p-norm of the singular values of A. That is, if
σi (A) is the vector of singular values of A, then
#1/p"
kAkp =X

σi (A)p.

(12.5)

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12.2. SCHATTEN NORMS AND COMPLEX INTERPOLATION365

The convention is for kAk∞ to be defined as the largest singular value of A because kAkp
converges to this in the limit as p → ∞. In the proof of Theorem 12.1.1, we repeatedly use
the fact that kAkp is unitarily invariant. That is, kAkp is invariant with respect to linear
isometries, in the sense that kAkp = U AV † p , where U, V ∈ L(H, H0 ) are linear isometries
satisfying U † U = IH and V † V = IH . Isometric invariance follows from (12.5) and because
these isometries do not change the singular values of A. From these norms, one can define
information measures relating quantum states and channels, with the main one used here
known as a Rényi generalization of a relative entropy difference.
Extending the Cauchy–Schwarz inequality is an important inequality known as the Hölder
inequality:|hA, Bi| = Tr{A† B} ≤ kAkp kBkq ,(12.6)
holding for p, q ∈ [1, ∞] such that p1 + 1q = 1 and A, B ∈ L(H). When p, q ∈ [1, ∞] and1+ 1q = 1, p and q are said to be Hölder conjugates of each other. One can see thatp
Cauchy–Schwarz is a special case by picking p = q = 2. Observe that equality is achieved in(12.6) if A and B are such that A† = a |B|q/p U † for some constant a ≥ 0 and where U is a
unitary such that B = U |B| is a left polar decomposition of B (see Theorem A.0.1). The
Hölder inequality along with the sufficient equality condition is enough for us to conclude
the following variational expression for the p-norm in terms of its Hölder dual q-norm:
kAkp = max Tr{A† B}.
kBkq ≤1(12.7)

This expression can be very useful in calculations.
Exercise 12.2.1 Prove that kABk1 ≤ kAkp kBkq for p, q ∈ [1, ∞] such that p1 + 1q = 1 andA, B ∈ L(H).
Throughout we adopt the convention from Definition 3.3.1 andP define f (A) for a function
f and a positive semi-definite operator A as follows: f (A) ≡ i:λi ∈Dom(f ) f (λi )|iihi|, whereP
A = i λi |iihi| is a spectral decomposition of A. We denote the support of A by supp(A),and we let ΠA denote the projection onto the support of A.

12.2.2Complex Analysis

We now review a few concepts from complex analysis. We will not prove these results in detail, but the purpose instead is to recall them, and the interested reader can follow references
to books on complex analysis for details of proofs. The culmination of the development is
the Stein–Hirschman complex interpolation theorem (Theorem 12.2.3).
The derivative of a complex-valued function f : C → C at a point z0 ∈ C is defined in
the usual way as
df (z)
f (z) − f (z0 )= lim.
(12.8)
dz z=z0 z→z0
z − z0c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License366 CHAPTER 12. QUANTUM ENTROPY INEQUALITIES AND RECOVERABILITY

In order for this limit to exist, it must be the same for all possible directions that one could
take in the complex plane to approach z0 , and this requirement demarcates a substantial
difference between differentiability of real functions and complex ones. Complex differentiability shares several properties with real differentiability: it is linear and obeys the productrule, the quotient rule, and the chain rule. If f is complex differentiable at every point z0 inan open set U , then we say that f is holomorphic on U .
There is a connection between real differentiability and complex differentiability, givenby the Cauchy–Riemann equations. Let f (x + iy) = u(x, y) + iv(x, y) where x, y ∈ R andu, v : R → R. If f is holomorphic, then u and v have first partial derivatives with respect tox and y and satisfy the Cauchy–Riemann equations:
∂v
∂u
=
,
∂x
∂y

∂u
∂v
=− .
∂y
∂x

(12.9)

The converse is not always true. However, if the first partial derivatives of u and v arecontinuous and satisfy the Cauchy–Riemann equations, then f is holomorphic. An important
holomorphic function for our purposes is given in the following exercise:
Exercise 12.2.2 Verify that f (z) = az = e[ln a]z , where a > 0 and z ∈ C, is a holomorphic
function everywhere in the complex plane. (Hint: Use that ez = ex [cos(y) + i sin(y)] forz = x + iy and x, y ∈ R.)
Holomorphic functions have good closure properties. That is, the sums, products, andcompositions of holomorphic functions are holomorphic as well, given that complex differentiation is linear and satisfies the product, quotient, and chain rules. Note that the quotientof two holomorphic functions is holomorphic wherever the denominator is not equal to zero.
The maximum modulus principle is an important principle that holomorphic functions
obey. Formally, it is the following statement: let f : C → C be a function holomorphic onsome connected, bounded open subset U of C. If z0 ∈ U is such that |f (z0 )| ≥ |f (z)| for all
z in a neighborhood of z0 , then the function f is constant on U . A consequence of this isthat if f is not constant on a bounded, connected, open subset U of C, then it achieves its
maximum on the boundary of U .
The maximum modulus principle has an extension to an unbounded strip in C, which wecall the maximum modulus principle on a strip. Let S denote the standard strip in C, S its
closure, and ∂S its boundary:S ≡ {z ∈ C : 0 < Re{z} < 1} ,(12.10)S ≡ {z ∈ C : 0 ≤ Re{z} ≤ 1} ,(12.11)∂S ≡ {z ∈ C : Re{z} = 0 ∨ Re{z} = 1} .

(12.12)

Let f : S → C be bounded on S, holomorphic on S, and continuous on ∂S. Then thesupremum of |f | is attained on ∂S. That is, supz∈S |f (z)| = supz∈∂S |f (z)|.
The maximum modulus principle on a strip implies a result known as the Hadamard
three-lines theorem:
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12.2. SCHATTEN NORMS AND COMPLEX INTERPOLATION

367

Theorem 12.2.1 (Hadamard Three-Lines) Let f : S → C be a function that is bounded
on S, holomorphic on S, and continuous on the boundary ∂S. Let θ ∈ (0, 1) and M (θ) ≡supt∈R |f (θ + it)|. Then ln M (θ) is a convex function on [0, 1], implying thatln M (θ) ≤ (1 − θ) ln M (0) + θ ln M (1).

(12.13)

There is a strengthening of the Hadamard three-lines theorem due to Hirschman, which infact implies the Hadamard three-lines theorem:
Theorem 12.2.2 (Hirschman) Let f (z) : S → C be a function that is bounded on S,
holomorphic on S, and continuous on the boundary ∂S. Then for θ ∈ (0, 1), the following
bound holds
Z ∞

h
i
h
i
ln |f (θ)| ≤dt αθ (t) ln |f (it)|1−θ + βθ (t) ln |f (1 + it)|θ ,(12.14)−∞wheresin(πθ),2(1 − θ) [cosh(πt) − cos(πθ)]
sin(πθ)βθ (t) ≡.
2θ [cosh(πt) + cos(πθ)]αθ (t) ≡For a fixed θ ∈ (0, 1), we have that αθ (t), βθ (t) ≥ 0 for all t ∈ R andZ ∞Z ∞dt αθ (t) =dt βθ (t) = 1 ,−∞(12.15)(12.16)(12.17)−∞(see, e.g., (Grafakos, 2008, Exercise 1.3.8)) so that αθ (t) and βθ (t) can be interpreted asprobability density functions. Furthermore, we have thatlim βθ (t) =θ&0π[cosh(πt) + 1]−1 ≡ β0 (t) ,2(12.18)where β0 is also a probability density function on R. With these observations, we can see that
Hirschman’s theorem implies the Hadamard three-lines theorem, given that an expectation
can never exceed a supremum.

12.2.3Complex Interpolation of Schatten Norms

We can extend much of the development above to operator-valued functions, which is needed
to prove Theorem 12.1.1. Let G : C → L(H) be an operator-valued function. We say that
G(z) is holomorphic if every function mapping z to a matrix entry is holomorphic. For our
purposes in what follows, we are interested in operator-valued functions of the form Az ,where A is a positive semi-definite
In this case, P
we apply the convention from
P operator.
zzDefinition 3.3.1 and take A = i:λi 6=0 λi |iihi|, where A = i λi |iihi| is an eigendecomposition of A with λi ≥ 0 for all i. Given the result of Exercise 12.2.2 combined with thec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License368 CHAPTER 12. QUANTUM ENTROPY INEQUALITIES AND RECOVERABILITY

closure properties of holomorphic functions mentioned above, we can conclude that Az isholomorphic if A is positive semi-definite.
We can now establish a version of the Hirschman theorem which applies to operatorvalued functions and allows for bounding their Schatten norms. This is one of the maintechnical tools that we need to establish Theorem 12.1.1.
Theorem 12.2.3 (Stein–Hirschman) Let G : S → L(H) be an operator-valued functionthat is bounded on S, holomorphic on S, and continuous on the boundary ∂S. Let θ ∈ (0, 1)
and define pθ by
1
1−θ
θ
=
+
,
(12.19)pθ
p0p1
where p0 , p1 ∈ [1, ∞]. Then the following bound holds
Z ∞
i
i
h

h
θ
,
+
β
(t)
ln
kG(1
+
it)k
dt αθ (t) ln kG(it)k1−θ
ln kG(θ)kpθ ≤θ
p1
p0(12.20)−∞where αθ (t) and βθ (t) are defined in (12.15)–(12.16).
Proof. For fixed θ ∈ (0, 1), let qθ be the Hölder conjugate of pθ , defined by
1
1
+
=1.
pθ qθ

(12.21)

Similarly, let q0 and q1 be Hölder conjugates of p0 and p1 , respectively. From the sufficient
equality condition for the Hölder inequality, we can find an operator X such that kXkqθ = 1
and Tr{XG(θ)} = kG(θ)kpθ . We can write the singular value decomposition for X in theform X = U D1/qθ V (implying Tr{D} = 1). For z ∈ S, define
1−zX(z) ≡ U D q0

+ qz

1

V .

(12.22)

As a consequence, X(z) is bounded on S, holomorphic on S, and continuous on the boundary∂S. Also, observe that X(θ) = X. Then the following function satisfies the requirements
needed to apply Theorem 12.2.2:
g(z) ≡ Tr{X(z)G(z)} .

(12.23)

Indeed, we have thatln kG(θ)kpθ = ln |g(θ)|
Z ∞

h
i
h
i
1−θ
θ
≤
dt αθ (t) ln |g(it)|+ βθ (t) ln |g(1 + it)|.

(12.24)(12.25)−∞

Now, from applying Hölder’s inequality and the facts that kX(it)kq0 = 1 = kX(1 + it)kq1 ,we find that|g(it)| = |Tr{X(it)G(it)}| ≤ kX(it)kq0 kG(it)kp0 = kG(it)kp0 ,c 2019(12.26)

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12.3. PETZ RECOVERY MAP

369

and
|g(1 + it)| = |Tr{X(1 + it)G(1 + it)}|≤ kX(1 + it)kq1 kG(1 + it)kp1(12.27)(12.28)= kG(1 + it)kp1 .

(12.29)Bounding (12.25) from above using these inequalities then gives (12.20).
The theorem above is known as a complex interpolation theorem because it allows us
to obtain estimates on the “intermediate” norm in terms of other norms which might beavailable. Furthermore, we are interpolating through the holomorphic family of operators
given by G(z).

12.3 Petz Recovery Map
The channel appearing in the lower bound of Theorem 12.1.1 has an explicit form and isconstructed from a map known as the Petz recovery map, which we define as follows:
Definition 12.3.1 Let σ ∈ L(H) be positive semi-definite, and let N : L(H) → L(H0 ) bea quantum channel. The Petz recovery map Pσ,N : L(H0 ) → L(H) is a completely positive,
trace-non-increasing linear map defined as follows for Q ∈ L(H0 ):


−1/2
−1/2
1/2
†
σ 1/2 .
(12.30)Pσ,N (Q) ≡ σ N [N (σ)]
Q [N (σ)]
The Petz recovery map Pσ,N is linear, and it is completely positive because it is equal
to a serial concatenation of three completely positive maps: Q → [N (σ)]−1/2 Q [N (σ)]−1/2 ,Q → N † (Q), and M → σ 1/2 M σ 1/2 for M ∈ L(H). It is trace-non-increasing because thefollowing holds for positive semi-definite Q:no
Tr{Pσ,N (Q)} = Tr σN † [N (σ)]−1/2 Q [N (σ)]−1/2(12.31)no= Tr N (σ) [N (σ)]−1/2 Q [N (σ)]−1/2(12.32)= Tr{ΠN (σ) Q} ≤ Tr{Q}.

(12.33)

An important special case of the Petz recovery map occurs when σ and N are effectively
classical.
That is, suppose that N is a classical-to-classical channel with Kraus operatorsp{ N (y|x)|yihx|} (see Section
P 4.6.4), where N (y|x) is a conditional probability distribution.
Suppose further that σ =x q(x)|xihx|, with q(x) ≥ 0 for all x. In this case, one cancheckp that the Petz recovery map is a classical-to-classical channel with Kraus operators{ R(x|y)|xihy|}, where R(x|y) is a conditional probability distribution given by the Bayes
theorem, satisfying
R(x|y)(N q)(y) = N (y|x)q(x),(12.34)c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

370 CHAPTER 12. QUANTUM ENTROPY INEQUALITIES AND RECOVERABILITYP
for all x and y, where (N q)(y) = x N (y|x)q(x). We leave the details of this calculation as
an exercise for the reader and point out that this recovery channel appears in the refinementof the monotonicity of classical relative entropy from Theorem 10.8.4.
We can also define a partial isometric map Uσ,t in the following way:Uσ,t (M ) ≡ σ it M σ −it .

(12.35)Since σ it σ −it = Πσ , we can conclude thatUσ,t (σ) = σ,(12.36)so that this isometric map does not have any effect when σ is input. In the case that σ ispositive definite, Uσ,t is a unitary channel. We can then define a rotated or “swiveled” Petzmap, which plays an important role in the construction of a recovery channel satisfying thelower bound in Theorem 12.1.1.
Definition 12.3.2 (Rotated Petz Map) Let σ ∈ L(H) be positive semi-definite, and let
N : L(H) → L(H0 ) be a quantum channel. A rotated Petz map is defined as follows for
Q ∈ L(H0 ):Rtσ,N (Q) ≡ (Uσ,−t ◦ Pσ,N ◦ UN (σ),t )(Q).
(12.37)
Proposition 12.3.1 (Perfect Recovery) Let σ ∈ L(H) be positive semi-definite, and let
N : L(H) → L(H0 ) be a quantum channel. A rotated Petz map Rtσ,N perfectly recovers σfrom N (σ):
Rtσ,N (N (σ)) = σ.
(12.38)Proof. Consider that
Pσ,N (N (σ)) = σ 1/2 N † [N (σ)]−1/2 N (σ) [N (σ)]−1/2 σ 1/2= σ 1/2 N † (ΠN (σ) )σ 1/2
≤σ

1/2

†

N (I)σ1/2=σ(12.39)(12.40)

1/2

Iσ

1/2

= σ.

(12.41)

The inequality follows because ΠN (σ) ≤ I and N † is a completely positive map. The second
to last equality follows because the adjoint is unital.
Now we prove the other operator inequality Pσ,N (N (σ)) ≥ σ, which will allow us toconclude that Pσ,N (N (σ)) = σ. Let U : H → H0 ⊗ HE be an isometric extension of thechannel N . From Lemma A.0.4, we know that supp U σU † ⊆ supp (N (σ) ⊗ IE ), which
implies that ΠU σU † ≤ ΠN (σ)⊗IE = ΠN (σ) ⊗ IE . Then for any vector |ψi ∈ H, we have thathψ|Πσ |ψi = hψ|U † ΠU σU † U |ψi(12.42)≤ hψ|U † ΠN (σ) ⊗ IE U |ψi(12.43)= Tr{U |ψihψ|U † ΠN (σ) ⊗ IE }= Tr{N (|ψihψ|)ΠN (σ) }(12.44)(12.45)= Tr{|ψihψ|N † (ΠN (σ) )}(12.46)†= hψ|N (ΠN (σ) )|ψi.
c 2019(12.47)

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12.4. RÉNYI INFORMATION MEASURE

371

Since |ψi is arbitrary, this establishes that Πσ ≤ N † (ΠN (σ) ). We can then use this operator
inequality to see that
Pσ,N (N (σ)) = σ 1/2 N † (ΠN (σ) )σ 1/2(12.48)≥ σ 1/2 Πσ σ 1/2 = σ.

(12.49)Finally, we can conclude that
Rtσ,N (N (σ)) = (Uσ,−t ◦ Pσ,N ◦ UN (σ),t )(N (σ))(12.50)= (Uσ,−t ◦ Pσ,N )(N (σ))= Uσ,−t (σ) = σ,(12.51)(12.52)where the second and last equalities follow from (12.36).
Exercise 12.3.1 Verify that a rotated Petz map satisfies the following:
Tr{Rtσ,N (Q)} = Tr{ΠN (σ) Q} ≤ Tr{Q},(12.53)for positive semi-definite Q ∈ L(H0 ), which implies that it is trace non-increasing.
Exercise 12.3.2 Let ω be positive semi-definite and let hA, Biω ≡ Tr{A† ω 1/2 Bω 1/2 }. The
adjoint of the Petz recovery map (with respect to the Hilbert–Schmidt inner product) is equal
to

†
Pσ,N
(M ) = [N (σ)]−1/2 N σ 1/2 M σ 1/2 [N (σ)]−1/2 ,(12.54)†
where M ∈ L(H). Show that Pσ,Nis the unique linear map with domain supp(σ) and range
supp(N (σ)), which satisfies the following for all M ∈ L(H) and Q ∈ L(H0 ):D
E††
M, N (Q) σ = Pσ,N (M ), Q.
(12.55)N (σ)(Hint: Consider picking M = |iihj| and Q = |kihl|.)

12.4 Rényi Information Measuree α known
Given ρ, σ, and N as in Definition 12.1.1, we define a Rényi information measure ∆as a Rényi generalization of a relative entropy difference:1eα (ρ, σ, N ),
ln Q
α−1


2α
1−α
α−1
1−α
e
2α
2α
Qα (ρ, σ, N ) ≡ [N (ρ)][N (σ)]⊗ IE U σ 2α ρ1/2,

e α (ρ, σ, N ) ≡∆(12.56)(12.57)2αwhere α ∈ (0, 1) ∪ (1, ∞) and U : H → H0 ⊗ HE is an isometric extension of the channel
N . That is, U is a linear isometry satisfying TrE {U (·)U † } = N (·) and U † U = IH . Recall
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372 CHAPTER 12. QUANTUM ENTROPY INEQUALITIES AND RECOVERABILITY

that all isometric extensions of a channel are related by an isometry acting on the environment system E, so that the definition in (12.56) is invariant under any such choice. Recall
from Proposition 5.2.1 that the adjoint N † of a channel is given in terms of an isometricextension U as N † (·) = U † ((·) ⊗ IE ) U .
e α (ρ, σ, N ) is a Rényi
The following lemma is one of the main reasons that we say that ∆generalization of a relative entropy difference.
Lemma 12.4.1 The following limit holds for ρ, σ, and N as given in Definition 12.1.1:
1e α (ρ, σ, N ) = D(ρkσ) − D(N (ρ)kN (σ)).
lim ∆ln 2 α→1(12.58)Proof. Let Πω denote the projection onto the support of ω. From the condition supp(ρ) ⊆supp(σ), it follows that supp(N (ρ)) ⊆ supp (N (σ)) (see Lemma A.0.5). We can then conclude thatΠσ Πρ = Πρ ,
ΠN (ρ) ΠN (σ) = ΠN (ρ) .
(12.59)We also know that supp(U ρU † ) ⊆ supp (N (ρ) ⊗ IE ) (see Lemma A.0.4), so thatΠN (ρ) ⊗ IE ΠU ρU † = ΠU ρU † .

(12.60)

When α = 1, we find from the above facts that
2
ΠN (ρ) ΠN (σ) ⊗ IE U Πσ ρ1/2 2
2
ΠN (ρ) ⊗ IE U Πρ ρ1/2 2
2
ΠN (ρ) ⊗ IE ΠU ρU † U ρ1/2 2

e1 (ρ, σ, N ) =
Q
=
=

2

(12.61)(12.62)(12.63)2= ΠU ρU † U ρ1/2 2 = ρ1/2 2 = 1.

(12.64)

So from the definition of the derivative, this means thateα (ρ, σ, N ) − ln Q
e1 (ρ, σ, N )
ln Q
α→1
α−1
h
i
d
eα (ρ, σ, N )
=
ln Q
dα
α=1
i
1
d he
=
Qα (ρ, σ, N )e1 (ρ, σ, N ) dα
Q
α=1
h
i
d e
=
Qα (ρ, σ, N )
.
dα
α=1

e α (ρ, σ, N ) = lim
lim ∆

α→1

(12.65)(12.66)(12.67)(12.68)Let α0 ≡ α−1. Now consider thatαeα (ρ, σ, N ) = TrQ
c 2019nhρ

1/2 −α0 /2

σ

N

†



α0 /2

N (σ)

−α0

N (ρ)

α0 /2

N (σ)



σ

−α0 /2 1/2

ρ

iα o

.

(12.69)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

12.4. RÉNYI INFORMATION MEASURE

373

Define the function

eα,β (ρ, σ, N ) ≡ TrQh1/2 −α0 /2

ρ

σ

N

†



N (σ)

α0 /2

N (ρ)

−α0

N (σ)

α0 /2



σ

−α0 /2 1/2

ρ

iβ 

,

(12.70)and consider that
i
d he
d e
Qα (ρ, σ, N )Qα,α (ρ, σ, N )
=
dα
dα
α=1
α=1
d e
d e
=
Qα,1 (ρ, σ, N )+Q1,β (ρ, σ, N )
.
dα
dβ
α=1
β=1

(12.71)(12.72)e1,β (ρ, σ, N ) as follows:
We first compute Qnβ oρ1/2 Πσ N † (ΠN (σ) ΠN (ρ) ΠN (σ) )Πσ ρ1/2nβ o= Tr ρ1/2 N † (ΠN (ρ) )ρ1/2
n

β o
= Tr ρ1/2 U † ΠN (ρ) ⊗ IE U ρ1/2
n

β o
= Tr
ΠN (ρ) ⊗ IE U ρU † ΠN (ρ) ⊗ IE
n

 o
† β
= Tr U ρU= Tr ρβ .

e1,β (ρ, σ, N ) = TrQ

(12.73)(12.74)(12.75)(12.76)(12.77)So thendd e
Q1,β (ρ, σ, N )
=
Tr ρβ
dβ
dβ
β=1


= Tr ρβ ln ρβ=1(12.78)β=1= Tr {ρ ln ρ} .

(12.79)d e
Now we turn to the other term dα
Qα,1 (ρ, σ, N ). First consider that





d
d 1−α
d 1
1
0
(−α ) ==
− 1 = − 2,
dα
dα
α
dα α
α
n


o
0
0
0
eα,1 (ρ, σ, N ) = Tr ρσ −α /2 N † N (σ)α /2 N (ρ)−α N (σ)α0 /2 σ −α0 /2 .
Q
c 2019(12.80)(12.81)

Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License374 CHAPTER 12. QUANTUM ENTROPY INEQUALITIES AND RECOVERABILITYd e
Qα,1 (ρ, σ, N ) is equal to
Now we show that dα

n


o
d
0
0
0
0
0
Tr ρσ −α /2 N † N (σ)α /2 N (ρ)−α N (σ)α /2 σ −α /2
dα


 


d −α0 /2
†
α0 /2
−α0
α0 /2
−α0 /2
N N (σ) N (ρ) N (σ)σσ= Tr ρ
dα





d
−α0 /2†−α0 /2
−α0
α0 /2
α0 /2
+ Tr ρσNσ
N (ρ) N (σ)
N (σ)
dα






d
−α0
α0 /2
−α0 /2α0 /2−α0 /2†+ Tr ρσN N (σ)
N (ρ)
N (σ)
σ
dα





d
α0 /2
−α0
−α0 /2−α0 /2†α0 /2σ+ Tr ρσN N (σ) N (ρ)
N (σ)
dα



 d
−α0 /2
†
α0 /2
−α0
α0 /2
−α0 /2+ Tr ρσN N (σ) N (ρ) N (σ)(12.82)
σ
dα

"


o
1
1 n
0
0
0
0
0
= 2 − Tr ρ [ln σ] σ −α /2 N † N (σ)α /2 N (ρ)−α N (σ)α /2 σ −α /2
α
2


o
1 n
0
0
0
0
0
+ Tr ρσ −α /2 N † [ln N (σ)] N (σ)α /2 N (ρ)−α N (σ)α /2 σ −α /2
2 n


o
0
0
0
0
0
− Tr ρσ −α /2 N † N (σ)α /2 [ln N (ρ)] N (ρ)−α N (σ)α /2 σ −α /2


o
1 n
0
0
0
0
0
+ Tr ρσ −α /2 N † N (σ)α /2 N (ρ)−α N (σ)α /2 [ln N (σ)] σ −α /22#o1 n −α0 /2 † 0000
N N (σ)α /2 N (ρ)−α N (σ)α /2 σ −α /2 [ln σ] . (12.83)− Tr ρσ2
Taking the limit as α → 1 gives

1 
d e
Qα,1 (ρ, σ, N )= − Tr ρ [ln σ] Πσ N † ΠN (σ) ΠN (ρ) ΠN (σ) Πσ
dα
2
α=1

1 
+ Tr ρΠσ N † [ln N (σ)] ΠN (σ) ΠN (ρ) ΠN (σ) Πσ
2 − Tr ρΠσ N † ΠN (σ) [ln N (ρ)] ΠN (ρ) ΠN (σ) Πσ1 + Tr ρΠσ N † ΠN (σ) ΠN (ρ) ΠN (σ) [ln N (σ)] Πσ21 − Tr ρΠσ N † ΠN (σ) ΠN (ρ) ΠN (σ) [ln σ] Πσ . (12.84)
2
We now simplify the first three terms and note that the last two are Hermitian conjugatesc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

12.4. RÉNYI INFORMATION MEASURE

375

of the first two:Tr ρ [ln σ] Πσ N † (ΠN (σ) ΠN (ρ) ΠN (σ) )Πσ= Tr ρ [ln σ] N † (ΠN (ρ) )= Tr N (ρ [ln σ]) ΠN (ρ)
= Tr U ρ [ln σ] U † ΠN (ρ) ⊗ IE
= Tr ΠU ρU † U ρU † U [ln σ] U † ΠN (ρ) ⊗ IE

= Tr U ρU † U [ln σ] U †= Tr {ρ [ln σ]} ,(12.85)(12.86)(12.87)(12.88)(12.89)(12.90)Tr ρΠσ N † ([ln N (σ)] ΠN (σ) ΠN (ρ) ΠN (σ) )Πσ= Tr ρN † ([ln N (σ)] ΠN (ρ) )= Tr N (ρ) [ln N (σ)] ΠN (ρ)= Tr {N (ρ) [ln N (σ)]} ,(12.91)(12.92)(12.93)Tr ρΠσ N † (ΠN (σ) [ln N (ρ)] ΠN (ρ) ΠN (σ) )Πσ= Tr ρN † ([ln N (ρ)] ΠN (ρ) )= Tr N (ρ) [ln N (ρ)] ΠN (ρ)= Tr {N (ρ) [ln N (ρ)]} .

(12.94)(12.95)(12.96)This then implies that the following equality holds
d eQα,1 (ρ, σ, N )= − Tr {N (ρ [ln σ])}dα
α=1
+ Tr {N (ρ) [ln N (σ)]} − Tr {N (ρ) [ln N (ρ)]} . (12.97)
Putting together (12.68), (12.72), (12.79), and (12.97), we can then conclude the statementof the theorem.
For α = 1/2, observe that2
e 1/2 (ρ, σ, N ) = − ln [N (ρ)]1/2 [N (σ)]−1/2 ⊗ IE U σ 1/2 ρ1/2∆
(12.98)1= − ln F (ρ, Pσ,N (N (ρ))).

(12.99)√ √ 2e α (ρ, σ, N ) were monotone
where F (ρ, σ) ≡
ρ σ 1 is the quantum fidelity. Thus, if ∆non-decreasing with respect to α, we could combine these observations to conclude that
1 e
∆1 (ρ, σ, N )
ln 2
?
1 e
≥
∆1/2 (ρ, σ, N )ln 2= − log F (ρ, Pσ,N (N (ρ))) .

D(ρkσ) − D(N (ρ)kN (σ)) =c 2019(12.100)(12.101)(12.102)

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If this were true, then we could conclude that Theorem 12.1.1 would be true with therecovery channel taken to be the Petz recovery map. However, it is not known whether thisis true, and we will instead invoke the Stein–Hirschman theorem to conclude that a convex
combination of rotated Petz maps satisfies the bound stated in Theorem 12.1.1.

12.5 Proof of the Recoverability Theorem
This section presents the proof of Theorem 12.1.1. In fact, we prove a stronger statement,
which implies Theorem 12.1.1 for a particular recovery channel that we discuss below.
Theorem 12.5.1 Let ρ, σ, and N be as given in Definition 12.1.1. Then the following
inequality holds
Z ∞
h 
i
t/2
(12.103)
dt β0 (t) log F ρ, (Rσ,N ◦ N )(ρ) ,D(ρkσ) − D(N (ρ)kN (σ)) ≥ −−∞where β0 (t) = π2 [cosh(πt) + 1]−1 is a probability density function for t ∈ R and Rσ,N is arotated Petz recovery map from Definition 12.3.2.
t/2

Proof. We can prove this result by employing Theorem 12.2.3. We first establish theinequality in (12.103). Let U : H → H0 ⊗ HE be an isometric extension of the channel N .
Pick
G(z) ≡ [N (ρ)]z/2 [N (σ)]−z/2 ⊗ IE U σ z/2 ρ1/2 ,(12.104)
2. The operator valued-function
for z ∈ S, p0 = 2, p1 = 1, and θ ∈ (0, 1), which fixes pθ = 1+θ
G(z) satisfies the conditions needed to apply Theorem 12.2.3. For the choices above, we find



θ/2

−θ/2



,
⊗ IE U σ θ/2 ρ1/2
2/(1+θ)kG (it)k2 = [N (ρ)]it/2 [N (σ)]−it/2 ⊗ IE U σ it ρ1/2

kG(θ)k2/(1+θ) =[N (ρ)][N (σ)](12.105)2≤ ρ1/2 2= 1,(12.106)kG (1 + it)k1 = [N (ρ)](1+it)/2 [N (σ)]−(1+it)/2 ⊗ IE U σ (1+it)/2 ρ1/2
1


1
it
1
it
1
it 1
−
−
= [N (ρ)] 2 [N (ρ)] 2 [N (σ)] 2 [N (σ)] 2 ⊗ IE U σ 2 σ 2 ρ 2
1


1/2
−it/2
−1/2
1/2 it/2 1/2= [N (ρ)] [N (σ)][N (σ)]⊗ IE U σ σ ρ1√= F ρ, Uσ,−t/2 ◦ Pσ,N ◦ UN (σ),t/2 (N (ρ))√
t/2
= F (ρ, (Rσ,N ◦ N )(ρ)).
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12.5. PROOF OF THE RECOVERABILITY THEOREM

377

Then we can apply Theorem 12.2.3 to conclude that
ln [N (ρ)]θ/2 [N (σ)]−θ/2 ⊗ IE U σ θ/2 ρ1/2
2/(1+θ) Z ∞θ/2 
t/2
dt βθ (t) ln F ρ, (Rσ,N ◦ N )(ρ). (12.108)≤−∞

This implies that


2
θ/2
−θ/2
⊗ IE U σ θ/2 ρ1/2
− ln [N (ρ)] [N (σ)]θ
2/(1+θ)
Z ∞
h 
i
t/2
dt βθ (t) ln F ρ, (Rσ,N ◦ N )(ρ) . (12.109)≥−−∞

Letting θ = (1 − α) /α, we see that this is the same as
Z ∞
h 
i
t/2
e
∆α (ρ, σ, N ) ≥ −dt β(1−α)/α (t) ln F ρ, (Rσ,N ◦ N )(ρ) .

(12.110)−∞Since the inequality in (12.109) holds for all θ ∈ (0, 1) and thus (12.110) holds for allα ∈ (1/2, 1), we can take the limit as α % 1 and apply (12.58) and the dominated convergence
theorem to conclude that (12.103) holds.
With the theorem above in hand, Theorem 12.1.1 follows as a consequence by taking
Rσ,N to be the following recovery channel:
Z ∞t/2
dt β0 (t) Rσ,N (Q) + Tr{(I − ΠN (σ) )Q}ω,(12.111)Rσ,N (Q) ≡−∞0where Q ∈ L(H ) and ω ∈ D(H). This is because
Z ∞
h 
i
t/2
−
dt β0 (t) log F ρ, (Rσ,N ◦ N )(ρ)−∞
   Z ∞t/2
dt β0 (t)Rσ,N ◦ N (ρ)≥ − log F ρ,−∞

≥ − log [F (ρ, (Rσ,N ◦ N )(ρ))] , (12.112)
where the first inequality is due to the concavity of both the logarithm and the fidelity, andthe second inequality follows from the assumption that supp(ρ) ⊆ supp(σ) and a reasoning
similar to that in the proof of Proposition 11.8.2 (one can also argue this last step from theoperator monotonicity of the square root function).
The extra term Tr{(I − ΠN (σ) )Q}ω is needed to ensure that Rσ,N is trace-preserving in
addition to being completely positive. Trace preservation of Rσ,N follows because
Z ∞t/2
Tr{Rσ,N (Q)} =dt β0 (t) Tr{Rσ,N (Q)} + Tr{(I − ΠN (σ) )Q}(12.113)−∞Z ∞=dt β0 (t) Tr{ΠN (σ) Q} + Tr{(I − ΠN (σ) )Q}(12.114)−∞= Tr{ΠN (σ) Q} + Tr{(I − ΠN (σ) )Q} = Tr{Q},c 2019(12.115)

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where the second equality follows from Exercise 12.3.1. Observe that this recovery channel
has the “perfect recovery of σ” property mentioned in (12.2), which follows from Proposition 12.3.1 and the particular form in (12.111).
As a corollary of Theorem 12.5.1, we obtain equality conditions for the monotonicity ofquantum relative entropy:
Corollary 12.5.1 (Equality Conditions) Let ρ, σ, and N be as given in Definition 12.1.1.
Then
D(ρkσ) = D(N (ρ)kN (σ))(12.116)
if and only if all rotated Petz recovery maps perfectly recover ρ from N (ρ):∀t ∈ R : (Rtσ,N ◦ N )(ρ) = ρ.

(12.117)

Proof. Recall from Proposition 12.3.1 that, independent of the conditions in the statementof the corollary, we always have that (Rtσ,N ◦ N )(σ) = σ for all t ∈ R.
We start by proving the “only if” part. Suppose that ∀t ∈ R : (Rtσ,N ◦ N )(ρ) = ρ. Thenfor a particular t ∈ R, the monotonicity of quantum relative entropy implies thatD(ρkσ) ≥ D(N (ρ)kN (σ))
D(N (ρ)kN (σ)) ≥ D((Rtσ,N ◦ N )(ρ)k(Rtσ,N ◦ N )(σ))= D(ρkσ),(12.118)(12.119)(12.120)

which in turn imply that D(ρkσ) = D(N (ρ)kN (σ)).
We now prove the “if part” of the theorem. Suppose that D(ρkσ) = D(N (ρ)kN (σ)). By
Theorem 12.5.1, we can conclude that
Z ∞
h
h 
ii
t/2
dt β0 (t) − log F ρ, (Rσ,N ◦ N )(ρ)= 0.
(12.121)−∞t/2

Since β0 (t) is a positive definite function for all t ∈ R, − log F ≥ 0, the recovery maps Rσ,N
are continuous in t and so is the fidelity, we can conclude that
h 
i
t/2
− log F ρ, (Rσ,N ◦ N )(ρ) = 0(12.122)t/2

for all t ∈ R, which is the same as F (ρ, (Rσ,N ◦ N )(ρ)) = 1 for all t ∈ R. We can thenconclude that (12.117) holds because the fidelity between two states is equal to one if andonly if the states are the same.

12.6 Refinements of Quantum Entropy Inequalities
Theorem 12.1.1 leads to a strengthening of many quantum entropy inequalities, including
strong subadditivity of quantum entropy, concavity of conditional entropy, joint convexityof relative entropy, non-negativity of quantum discord, and the Holevo bound. We list theseas corollaries and give brief proofs for them in the following subsections.
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12.6. REFINEMENTS OF QUANTUM ENTROPY INEQUALITIES

12.6.1

379

Strong Subadditivity

Recall the conditional quantum mutual information of a tripartite state ρABC :I(A; B|C)ρ ≡ H(AC)ρ + H(BC)ρ − H(C)ρ − H(ABC)ρ .

(12.123)

Strong subadditivity is the statement that I(A; B|C)ρ ≥ 0 for all tripartite states ρABC .
Corollary 12.6.1 below gives an improvement of strong subadditivity. It is a direct consequence of Theorem 12.1.1 after choosingρ = ρABC ,σ = ρAC ⊗ IB ,N = TrA ,(12.124)

N † (·) = (·) ⊗ IA ,(12.125)so thatN (ρ) = ρBC ,N (σ) = ρC ⊗ IB ,andD(ρkσ) − D(N (ρ)kN (σ)) = D(ρABC kρAC ⊗ IB ) − D(ρBC kρC ⊗ IB )= I(A; B|C)ρ ,Pσ,N (·) = σ 1/2 N † [N (σ)]−1/2 (·) [N (σ)]−1/2 σ 1/2
h
i
1/2
−1/2
−1/2
1/2
= ρAC ρC (·)ρC ⊗ IA ρAC .

(12.126)(12.127)

Corollary 12.6.1 Let ρABC ∈ D(HA ⊗ HB ⊗ HC ). Then the following inequality holds
I(A; B|C)ρ ≥ − log [F (ρABC , RC→AC (ρBC ))] ,(12.128)R∞
t/2
where the recovery channel RC→AC = −∞ dt β0 (t) RρAC ,TrA perfectly recovers ρAC from ρC .

12.6.2

Concavity of Conditional Quantum Entropy

Let E ≡ {pX (x), ρxAB } be an ensemble of bipartite quantum states with expectation ρAB ≡P
xx pX (x)ρAB . Concavity of conditional entropy is the statement that
X
H(A|B)ρ ≥pX (x)H(A|B)ρx .
(12.129)x

Let ωXAB denote the following classical-quantum state in which we have encoded theensemble E:X(12.130)
ωXAB ≡pX (x)|xihx|X ⊗ ρxAB .
x

We can rewrite
H(A|B)ρ −X

pX (x)H(A|B)ρx = H(A|B)ω − H(A|BX)ω

(12.131)x= I(A; X|B)ω
= H(X|B)ω − H (X|AB)ω
= D(ωXAB kIX ⊗ ωAB ) − D(ωXB kIX ⊗ ωB ).
c 2019(12.132)(12.133)(12.134)

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380 CHAPTER 12. QUANTUM ENTROPY INEQUALITIES AND RECOVERABILITY

We can see the last line above as a relative entropy difference (as defined in the right-handside of (12.58)) by picking ρ = ωXAB , σ = IX ⊗ ωAB , and N = TrA .
Applying Theorem 12.1.1 and Exercise 9.2.10, we find the following improvement of
concavity of conditional entropy:
Corollary 12.6.2 Let an ensemble E be as given above. Then the following inequality holds
XX√(12.135)
H(A|B)ρ −pX (x)H(A|B)ρx ≥ −2 log
pX (x) F (ρxAB , RB→AB (ρxB )),xxwhere the recovery map RB→AB ≡12.6.3R∞−∞t/2dt β0 (t) RρAB ,TrA perfectly recovers ρAB from ρB .

Joint Convexity of Quantum Relative Entropy

Let {pX (x), ρx } be an ensemble of density operators and {pX (x), σx } be an ensemble of positive P
semi-definite operatorsPsuch that supp (ρx ) ⊆ supp(σx ) for all x and with expectations
ρ ≡ x pX (x)ρx and σ ≡ x pX (x)σx . Joint convexity of quantum relative entropy is thestatement that distinguishability of these ensembles does not increase under the loss of theclassical label:X(12.136)
pX (x)D(ρx kσx ) ≥ D(ρkσ).
x

By pickingρ = ρXB ≡X

pX (x)|xihx|X ⊗ ρx ,(12.137)

pX (x)|xihx|X ⊗ σx ,(12.138)xσ = σXB ≡Xx

N = TrX ,(12.139)

and applying Theorem 12.1.1, we arrive at the following improvement of joint convexity ofquantum relative entropy:
Corollary 12.6.3 Let ensembles be as given above. Then the following inequality holds
X
pX (x)D(ρx kσx ) − D(ρkσ) ≥ − log F (ρXB , RσXB ,TrX (ρ)),(12.140)xwhere the recovery map RσXB ,TrX =12.6.4R∞−∞t/2

dt β0 (t) RσXB ,TrX perfectly recovers σXB from σB .

Non-Negativity of Quantum DiscordLet ρAB be a bipartite density operator and let {|ϕx ihϕ
P x |A } be a rank-one quantum measurement on system A (i.e., the vectors |ϕx iA satisfy x |ϕx ihϕx |A = IA ). It suffices for us
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12.6. REFINEMENTS OF QUANTUM ENTROPY INEQUALITIES

381

to consider rank-one measurements for our discussion here because every quantum measurement can be refined to have a rank-one form, such that it delivers more classical information
to the experimentalist observing the apparatus. Then the (unoptimized) quantum discord
is defined to be the difference between the following mutual informations:
I(A; B)ρ − I(X; B)ω , whereωXB ≡ MA→X (ρAB ),X
MA→X (·) ≡hϕx |A (·)|ϕx iA |xihx|X .

(12.141)(12.142)(12.143)x

The quantum channel MA→X is a measurement channel, so that the state ωXB is theclassical-quantum state resulting from the measurement. The set {|xiX } is an orthonormal basis so that X is a classical system. The quantum discord is non-negative, and by
applying Theorem 12.1.1, we find the following improvement of this entropy inequality:
Corollary 12.6.4 Let ρAB and MA→X be as given above. Then the following inequality
holds
I(A; B)ρ − I(X; B)ω ≥ − log F (ρAB , EA (ρAB )),(12.144)
where

Z ∞
EA ≡

dt β0 (t) (UρA ,t ◦ PρA ,MA→X ◦ MA→X )(12.145)−∞

is an entanglement-breaking map, PρA ,MA→X ◦ MA→X is an entanglement-breaking map:
(PρA ,MA→X ◦ MA→X )(·) =

1/2
1/2
X
ρ |ϕx ihϕx |A ρA,
hϕx |A (·)|ϕx iA A
hϕ
|
ρ
|ϕ
i
x
A
A
x
A
x

and the partial isometric map UρA ,t is defined from (12.35). The recovery map
PρA ,MA→X ) perfectly recovers ρA from MA→X (ρA ).

(12.146)R∞−∞

dt β0 (t) (UρA ,t ◦Proof. We start with the rewriting
I(A; B)ρ − I(X; B)ω = D (ρAB kρA ⊗ IB ) − D (ωXB kωX ⊗ IB ) ,(12.147)and follow by picking ρ = ρAB , σ = ρA ⊗IB , and N = MA→X , and12.1.1.
R ∞applying Theorem
t/2This then shows the corollary with a recovery map of the form −∞ dt β0 (t) RρA ,MA→X .
The channel RtρA ,MA→X ◦ MA→X is entanglement-breaking because it consists of a measurement channel MA→X followed by a preparation. We
P now work out the form for therecovery map in (12.144). Consider that MA→X (ρA ) = x hϕx |A ρA |ϕx iA |xihx|X , so that
UMA→X (ρA ),−t (·) =
# "
#
"
X
X
[hϕx |A ρA |ϕx iA ]−it |xihx|X (·)[hϕx0 |A ρA |ϕx0 iA ]it |x0 i hx0 |X . (12.148)x

c 2019x0Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

382 CHAPTER 12. QUANTUM ENTROPY INEQUALITIES AND RECOVERABILITY

Thus, when composing MA→X with UMA→X (ρA ),−t , the phases cancel out to give the following
relation: UMA→X (ρA ),−t (MA→X (·)) = MA→X (·). One can then work out that
(PρA ,MA→X ◦ MA→X )(·)


1/2
1/2
= ρA M† [MA→X (ρA )]−1/2 MA→X (·) [MA→X (ρA )]−1/2 ρA
=

1/2
1/2
X
ρ |ϕx ihϕx |A ρA,
hϕx |A (·)|ϕx iA A
hϕ
|
ρ
|ϕ
i
x
A
A
x
A
x

(12.149)(12.150)

concluding the proof.

12.6.5

Holevo Bound

The Holevo bound is a special case of the non-negativity of quantum discord in which ρAB
is a quantum-classical state, which we write explicitly asX
pY (y)ρyA ⊗ |yihy|Y ,(12.151)
ρAB =yP
where each ρyA is a density operator, so that ρA = y pY (y)ρyA . The Holevo bound statesthat the mutual information of the state ρAB in (12.151) is never smaller than the mutual
information after system A is measured. By applying Corollary 12.6.4 and (9.146), we findthe following improvement:
Corollary 12.6.5 (Holevo Bound) Let ρAB be as in (12.151), and let MA→X and ωXB
be as in Section 12.6.4, respectively. Then the following inequality holds
X√
I(A; B)ρ − I(X; B)ω ≥ −2 log
pY (y) F (ρyA , EA (ρyA )) ,(12.152)ywhere EA is an entanglement-breaking map of the form in (12.146) and the partial isometricmap UρA ,t is defined from (12.35).

12.7 History and Further Reading
Section 11.11 reviews the history of many quantum entropy inequalities in quantum information. Here, we detail the history of the refinements. Petz (1986, 1988) considered theequality conditions for monotonicity of quantum relative entropy, defined what is now known
as the Petz recovery map (there called “transpose channel”), and gave many such equality
conditions, one of which is that D(ρkσ) = D(N (ρ)kN (σ)) if and only if the Petz recovery
map perfectly recovers ρ from N (ρ): (Pσ,N ◦N )(ρ) = ρ. Later, Hayden et al. (2004a) invoked
Petz’s result to elucidate the structure of tripartite states that saturate the strong subadditivity of quantum entropy with equality. Mosonyi and Petz (2004) then invoked Petz’s resultc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

12.7. HISTORY AND FURTHER READING383

to establish the structure of triples (ρ, σ, N ) for which the monotonicity of quantum relative
entropy is saturated with equality (see also (Mosonyi, 2005)). The transpose channel was
independently discovered many years after Petz’s work by Barnum and Knill (2002) in thecontext of approximate quantum error correction.
The topic of the near saturation of quantum entropy inequalities is a more recent development. Brandao et al. (2011) established a lower bound on conditional mutual information
related to how entangled the two unconditioned systems are with respect to the 1-LOCC
norm. Interest then turned to obtaining lower bounds in terms of the trace norm. Much of
the very recent work was inspired by the posting of Winter and Li (2012) and a presentationof Kim (2013). The main conjecture of Winter and Li (2012) (hitherto unproven) is thatD(ρkσ) − D(N (ρ)kN (σ)) ≥ D(ρk(Rσ,N ◦ N )(ρ)),(12.153)

for some recovery channel Rσ,N , depending only on σ and N and such that (Rσ,N ◦N )(σ) = σ.
Carlen and Lieb (2014) then established an interesting lower bound on the monotonicity ofquantum relative entropy with respect to partial trace, which directly leads to a related
lower bound on conditional quantum mutual information, as pointed out by Zhang (2014)
by making use of the well known relation I(A; B|C)ρ = D(ρABC kIB ⊗ρAC )−D(ρBC kIB ⊗ρC ).
Motivated by these developments, Berta et al. (2015b) defined a Rényi generalization of theconditional mutual information. This notion and known properties of Rényi entropies ledthem to conjecture that I(A; B|C)ρ ≥ − log F (ρABC , PρAC ,TrA (ρBC )), which hitherto remains
unproven. Shortly thereafter, Seshadreesan et al. (2015a) put forward the notion of a Rényi
generalization of a relative entropy difference, which is one of the main tools used in the
proof of Theorem 12.1.1.
Fawzi and Renner (2015) established the following:
I(A; B|C)ρ ≥ − log F (ρABC , RC→AC (ρBC )),(12.154)

where RC→AC is a recovery channel consisting of some unitary acting on C, followed by thePetz recovery map PρAC ,TrA , and then some unitary acting on AC. Since their argument
made use of the probabilistic method, they could not give further information about the
aforementioned unitaries. Concurrently, Seshadreesan and Wilde (2015) defined the “fidelityof recovery”
F (A; B|C)ρ ≡ sup F (ρABC , RC→AC (ρBC ))(12.155)
RC→AC

as an information measure analogous to conditional mutual information and they proved
many of its properties. Li and Winter (2014), based on earlier arguments in Winter andLi (2012), used the result of Fawzi and Renner (2015) to establish a lower bound on an
entanglement measure known as squashed entanglement and further elaborated on the conjecture in (12.153). Brandao et al. (2014) proved the bound I(A; B|C)ρ ≥ − log F (A; B|C)ρ
by invoking quantum state redistribution, but their method gave less information about thestructure of the recovery channel. Wilde (2014) showed how to apply the bound in (12.154)
to give lower bounds on multipartite information measures, which had an interpretation inc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License384 CHAPTER 12. QUANTUM ENTROPY INEQUALITIES AND RECOVERABILITY

terms of local recoverability. Berta et al. (2015a) used the techniques of Fawzi and Renner (2015) to establish a non-trivial lower bound on a relative entropy difference. Dattae α (ρ, σ, N ) ≥ 0 for all α ∈ (1/2, 1) ∪ (1, ∞) in addition toand Wilde (2015) proved that ∆other related statements. Berta and Tomamichel (2016) proved that the fidelity of recovery
is multiplicative with respect to tensor-product states, which then simplified the proof ofthe bound I(A; B|C)ρ ≥ − log F (A; B|C)ρ . Sutter et al. (2016a) showed that the recovery
map in (12.154) possesses a universal property in the sense that RC→AC could be taken to
depend only on the marginal state ρAC . Their argument also led to the conclusion that theunitaries could be taken to commute with ρC and ρAC .
Wilde (2015) invoked the Hadamard three-lines theorem and the notion of a Rényi generalization of a relative entropy difference to establish the following bound:t(12.156)D(ρkσ) − D(N (ρ)kN (σ)) ≥ − log sup F (ρ, (Rσ,N ◦ N )(ρ)) ,t∈R

where (Rtσ,N ◦ N )(σ) = σ. Notice that the recovery map above is not universal, because the optimal t could depend on ρ. Wilde (2015) also established an upper boundon D(ρkσ) − D(N (ρ)kN (σ)), which in some cases has an interpretation in terms of recoverability. Dupuis and Wilde (2016) subsequently defined “swiveled Rényi entropies” in an
attempt to address some of the open questions raised in (Berta et al., 2015b; Seshadreesan
et al., 2015a). An upshot of this work was to establish further lower and upper bounds onD(ρkσ) − D(N (ρ)kN (σ)). Sutter et al. (2016b) then showed the following lower bound by
a different method of proof, known as “the pinched Petz recovery” approach:D(ρkσ) − D(N (ρ)kN (σ)) ≥ DM (ρk(Rρ,σ,N ◦ N )(ρ)),(12.157)

where (Rρ,σ,N ◦ N )(σ) = σ and Rρ,σ,N is some convex combination of rotated Petz maps
and DM is the “measured relative entropy”, which satisfies DM ≥ − log F . Junge et al.
(2015) invoked Hirschman’s improvement (Hirschman, 1952) of the Hadamard three-lines
theorem and the method of proof from (Wilde, 2015) to establish Theorem 12.1.1, with theconsequence of having an explicit recovery channel Rσ,N which depends only on σ and N
and satisfies (Rσ,N ◦ N )(σ) = σ. This approach also directly recovers the equality conditions
from Petz (1986, 1988).
More information about the theory of complex interpolation is available in Bergh andLöfström (1976); Reed and Simon (1975). Grafakos (2008) provides an excellent review
of Hirschman’s theorem. Stein (1956) proved the Stein–Hirschman interpolation theorem,
which is essential to the proof of Theorem 12.1.1. Zurek (2000) and Ollivier and Zurek (2001)
defined the quantum discord. Jencova (2012) proved Proposition 12.3.1 for t = 0.

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 13

The Information of Quantum
Channels
We introduced several classical and quantum entropic quantities in Chapters 10 and 11:
entropy, conditional entropy, joint entropy, mutual information, relative entropy, and conditional mutual information. Each of these entropic quantities is static, in the sense thateach is evaluated with respect to random variables or quantum systems that certain parties
possess.
In this chapter, we introduce several dynamic entropic quantities for channels, whether
they be classical or quantum. We derive these measures by exploiting the static measuresfrom the two previous chapters. They come about by sending one share of a state througha channel, computing a static measure with respect to the input–output state, and thenmaximizing the static measure with respect to all possible states that we can transmit
through the channel. This process then gives rise to a dynamic measure that quantifies theability of a channel to preserve correlations. For example, we could send one share of a pureentangled state |φiAA0 through a quantum channel NA0 →B —this transmission gives rise tosome bipartite state NA0 →B (φAA0 ). We would then evaluate the mutual information of theresulting state and maximize the mutual information with respect to all such pure input
states:max I(A; B)ω ,(13.1)
φAA0

where ωAB ≡ NA0 →B (φAA0 ). The above quantity is a dynamic information measure of thechannel’s ability to preserve correlations—Section 13.4 introduces this quantity as the mutual
information of the channel N .
For now, we simply think of the quantities in this chapter as measures of a channel’s ability to preserve correlations. Later, we show that these quantities have explicit operational
interpretations in terms of a channel’s ability to perform a certain task, such as the transmission of classical or quantum information.1 Such an operational interpretation gives meaning
1

Giving operational interpretations to information measures is in fact one of the main goals of this book!

385

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CHAPTER 13. THE INFORMATION OF QUANTUM CHANNELS

to an entropic measure—otherwise, it is difficult to understand a measure in an informationtheoretic sense without having a specific operational task to which it corresponds.
Recall that the entropy obeys an additivity property for any two independent randomvariables X1 and X2 :H(X1 , X2 ) = H(X1 ) + H(X2 ).
(13.2)
The above additivity property extends to a large sequence X1 , . . . , Xn of independent andidentically distributed random variables. That is, applying (13.2) inductively shows thatnH(X) is equal to the entropy of the sequence:
H(X1 , . . . , Xn ) =

n
X
i=1

H(Xi ) =n
X

H(X) = nH(X),(13.3)

i=1

where random variable X has the same distribution as each of X1 , . . . , Xn . Similarly, quantum entropy is additive for any two quantum systems in a product state ρ ⊗ σ:H(ρ ⊗ σ) = H(ρ) + H(σ),(13.4)

and applying (13.4) inductively to a sequence of quantum states gives the following similar simple formula: H(ρ⊗n ) = nH(ρ). Additivity is a desirable property and a natural
expectation that we have for any measure of information evaluated on independent systems.
In analogy with the static measures, we would like additivity to hold for the dynamic
information measures. Without additivity holding, we cannot really make sense of a given
measure because we would have to evaluate the measure on a potentially infinite number of
independent channel uses. This evaluation on so many channel uses is generally an impossible
optimization problem. Additionally, the requirement to maximize over so many uses of thechannel does not identify a given measure as a unique measure of a channel’s ability toperform a certain task. There could generally be other measures that are equal to theoriginal one when we take the limit of many channel uses. Thus, a measure does not havemuch substantive meaning if additivity does not hold.
We devote this chapter to the discussion of several dynamic measures. Additivity holds
in the general case for only three of the dynamic measures presented here: the mutual information of a classical channel, the private information of a classical wiretap channel, andthe mutual information of a quantum channel. For all other measures, there are known
counterexamples of channels for which additivity does not hold. In this chapter, we do notdiscuss the counterexamples, but instead focus only on classes of channels for which additivity does hold, in an effort to understand it in a technical sense. The proof techniques for
additivity exploit many of the ideas introduced in the two previous chapters and give us achance to practice with what we have learned there on one of the most important problems
in quantum Shannon theory.

13.1 Mutual Information of a Classical Channel
Suppose that we would like to determine how much information we can transmit through aclassical channel N . Recall our simplified model of a classical channel N from Chapter 2,
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13.1. MUTUAL INFORMATION OF A CLASSICAL CHANNEL387

in which some conditional probability density pY |X (y|x) models the effects of noise. That is,
we obtain some random variable Y if we input a random variable X to the channel.
What is a good measure of the information throughput of this channel? The mutual
information is perhaps the best starting point. Suppose that random variables X and Y are
Bernoulli. If the classical channel is noiseless and X is completely random, the input andoutput random variables X and Y are perfectly correlated, the mutual information I(X; Y )is equal to one bit, and the sender can transmit one bit per transmission as we would expect.
If the classical channel is completely noisy (in the sense that it prepares an output that has
a constant probability distribution irrespective of the input), the input and output randomvariables are independent and the mutual information is equal to zero bits. This observation
matches our intuition that the sender should not be able to transmit any information throughthis completely noisy channel.
In the above model for a classical channel, the conditional probability density pY |X (y|x)
remains fixed, but we can “play around” with the input random variable X by modifying its
probability density pX (x).2 Thus, we still “have room” for optimizing the mutual informationof the channel N by modifying this input density. This gives us the following definition:
Definition 13.1.1 (Mutual Information of a Classical Channel) The mutual information I(N ) of a classical channel N ≡ pY |X is defined as follows:
I(N ) ≡ max I(X; Y ).

(13.5)pX (x)

13.1.1

Regularized Mutual Information of a Classical Channel

We now consider whether exploiting multiple uses of a classical channel N and allowing forcorrelations between its inputs can increase its mutual information. That is, suppose that
we have two independent uses of a classical channel N available. Let X1 and X2 denote therandom variables input to the respective first and second copies of the channel, and let Y1 andY2 denote the output random variables. Each of the two uses of the channel is equivalent to
the mapping pY |X (y|x) so that the channel uses are independent and identically distributed.
Let N ⊗ N denote the joint channel that corresponds to the mapping
pY1 ,Y2 |X1 ,X2 (y1 , y2 |x1 , x2 ) = pY1 |X1 (y1 |x1 )pY2 |X2 (y2 |x2 ),(13.6)

where both pY1 |X1 (y1 |x1 ) and pY2 |X2 (y2 |x2 ) are equal to the mapping pY |X (y|x). The mutual
information of a classical joint channel is as follows:I(N ⊗ N ) ≡max
pX1 ,X2 (x1 ,x2 )I(X1 , X2 ; Y1 , Y2 ).

(13.7)We might think that we could increase the mutual information of this classical channel by
allowing for correlations between the inputs to the channels through a correlated distribution
2

Recall the idea from Section 2.2.4 where Alice and Bob actually choose a code for the channel randomly
according to the density pX (x).
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CHAPTER 13. THE INFORMATION OF QUANTUM CHANNELS

Figure 13.1: This figure displays the scenario for determining whether the mutual information of twoclassical channels N1 and N2 is additive. The question of additivity is equivalent to the possibility ofclassical correlations being able to enhance the mutual information of two classical channels. The result
proved in Theorem 13.1.1 is that the mutual information is additive for any two classical channels, so thatclassical correlations cannot enhance it.

pX1 ,X2 (x1 , x2 ). That is, there could be some superadditive effect if the mutual information of
the classical joint channel N ⊗N is strictly greater than two individual mutual informations:?

I(N ⊗ N ) > 2I(N ).

(13.8)

Figure 13.1 displays the scenario corresponding to the above question.
In fact, we can take the above argument to its extreme, by defining the regularized mutual
information Ireg (N ) of a classical channel as follows:1
I(N ⊗n ).
n→∞ n

Ireg (N ) ≡ lim(13.9)In the above definition, the quantity I(N ⊗n ) is defined as follows:I(N ⊗n ) ≡ maxn I(X n ; Y n ),(13.10)pX n (x )and N ⊗n denotes n channels corresponding to the following conditional probability distribution:nYn n
pY n |X n (y |x ) =pYi |Xi (yi |xi ),(13.11)i=1where X n ≡ X1 , X2 , . . . , Xn , xn ≡ x1 , x2 , . . . , xn , and Y n ≡ Y1 , Y2 , . . . , Yn . The potential
superadditive effect would have the following form after bootstrapping the inequality in(13.8) to the regularization:?

Ireg (N ) > I(N ).

(13.12)

Exercise 13.1.1 Determine a finite upper bound on Ireg (N ). Thus, this quantity is always
finite.
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13.1. MUTUAL INFORMATION OF A CLASSICAL CHANNEL389

The next section shows that the above strict inequalities do not hold for a classicalchannel, implying that no such superadditive effect occurs for its mutual information. In fact,
the mutual information of a classical channel obeys an additivity property that represents
one of the cornerstones of our understanding of classical information theory. This additivity
property is the statement that
I(N ⊗ M) = I(N ) + I(M)(13.13)for any two classical channels N and M, and thus
Ireg (N ) = I(N ),(13.14)by an inductive argument. Thus, classical correlations between inputs do not increase themutual information of a classical channel.
We are stressing the importance of additivity in classical information theory because
recent research has demonstrated that superadditive effects can occur in quantum Shannon
theory (see Section 20.5, for example). These quantum results imply that our understandingof quantum Shannon theory is not complete, but they also demonstrate the fascinating
possibility that quantum correlations can increase the information throughput of a quantum
channel.

13.1.2

Additivity

The mutual information of classical channels satisfies the important and natural property ofadditivity. We prove the strongest form of additivity that occurs for the mutual informationof two different classical channels. Let N1 and N2 denote two different classical channels
corresponding to the respective mappings pY1 |X1 (y1 |x1 ) and pY2 |X2 (y2 |x2 ), and let N1 ⊗ N2
denote the joint channel that corresponds to the mapping
pY1 ,Y2 |X1 ,X2 (y1 , y2 |x1 , x2 ) = pY1 |X1 (y1 |x1 )pY2 |X2 (y2 |x2 ).

(13.15)The mutual information of the joint channel is then as follows:
I(N1 ⊗ N2 ) ≡max
pX1 ,X2 (x1 ,x2 )I(X1 , X2 ; Y1 , Y2 ).

(13.16)

The following theorem states the additivity property.
Theorem 13.1.1 (Additivity of Mutual Information of Classical Channels) The mutual information of the classical joint channel N1 ⊗ N2 is the sum of their individual mutual
informations:
I(N1 ⊗ N2 ) = I(N1 ) + I(N2 ).
(13.17)
Proof. We first prove the inequality I(N1 ⊗ N2 ) ≥ I(N1 ) + I(N2 ). This inequality is moretrivial to prove than the other direction. Let p∗X1 (x1 ) and p∗X2 (x2 ) denote distributions thatc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 13. THE INFORMATION OF QUANTUM CHANNELS

achieve the respective maximums of I(N1 ) and I(N2 ). The joint probability distribution for
all input and output random variables is then as follows:
pX1 ,X2 ,Y1 ,Y2 (x1 , x2 , y1 , y2 ) = p∗X1 (x1 )p∗X2 (x2 )pY1 |X1 (y1 |x1 )pY2 |X2 (y2 |x2 ).

(13.18)Observe that X1 and Y1 are independent of X2 and Y2 . Then the following chain of inequalities holds:
I(N1 ) + I(N2 ) = I(X1 ; Y1 ) + I(X2 ; Y2 )= I(X1 , X2 ; Y1 , Y2 )≤ I(N1 ⊗ N2 ).

(13.19)(13.20)(13.21)

The first equality follows by evaluating the mutual informations I(N1 ) and I(N2 ) withrespect to the maximizing distributions p∗X1 (x1 ) and p∗X2 (x2 ). The second equality follows
because the mutual information is additive with respect to the independent joint randomvariables (X1 , Y1 ) and (X2 , Y2 ). The final inequality follows because the input distributionp∗X1 (x1 )p∗X2 (x2 ) is a particular input distribution of the more general form pX1 ,X2 (x1 , x2 )needed in the maximization of the mutual information of the joint channel N1 ⊗ N2 .
We now prove the non-trivial inequality I(N1 ⊗ N2 ) ≤ I(N1 ) + I(N2 ). Let p∗X1 ,X2 (x1 , x2 )denote a distribution that maximizes I(N1 ⊗ N2 ), and let
qX1 |X2 (x1 |x2 ) and qX2 (x2 )(13.22)

p∗X1 ,X2 (x1 , x2 ) = qX1 |X2 (x1 |x2 )qX2 (x2 ).

(13.23)be distributions such that

Recall that the conditional probability distribution for the joint channel N1 ⊗N2 is as follows:
pY1 ,Y2 |X1 ,X2 (y1 , y2 |x1 , x2 ) = pY1 |X1 (y1 |x1 )pY2 |X2 (y2 |x2 ).

(13.24)

By summing over y2 , we observe that Y1 and X2 are independent when conditioned on X1
becausepY1 |X1 ,X2 (y1 |x1 , x2 ) = pY1 |X1 (y1 |x1 ).
(13.25)
Also, the joint distribution pX1 ,Y1 ,Y2 |X2 (x1 , y1 , y2 |x2 ) has the form
pX1 ,Y1 ,Y2 |X2 (x1 , y1 , y2 |x2 ) = pY1 |X1 (y1 |x1 )qX1 |X2 (x1 |x2 )pY2 |X2 (y2 |x2 ).

(13.26)

Then Y2 is conditionally independent of X1 and Y1 when conditioning on X2 . Consider thefollowing chain of inequalities:
I(N1 ⊗ N2 ) = I(X1 , X2 ; Y1 , Y2 )= H(Y1 , Y2 ) − H(Y1 , Y2 |X1 , X2 )= H(Y1 , Y2 ) − H(Y1 |X1 , X2 ) − H(Y2 |Y1 , X1 , X2 )= H(Y1 , Y2 ) − H(Y1 |X1 ) − H(Y2 |X2 )≤ H(Y1 ) + H(Y2 ) − H(Y1 |X1 ) − H(Y2 |X2 )= I(X1 ; Y1 ) + I(X2 ; Y2 )≤ I(N1 ) + I(N2 ).
c 2019(13.27)(13.28)(13.29)(13.30)(13.31)(13.32)(13.33)

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13.1. MUTUAL INFORMATION OF A CLASSICAL CHANNEL

391

The first equality follows from the definition of I(N1 ⊗ N2 ) in (13.16) and by evaluating
the mutual information with respect to the distributions p∗X1 ,X2 (x1 , x2 ), pY1 |X1 (y1 |x1 ), and
pY2 |X2 (y2 |x2 ). The second equality follows by expanding the mutual information I(X1 , X2 ; Y1 , Y2 ).
The third equality follows from the entropy chain rule. The fourth equality follows because
H(Y1 |X1 , X2 ) = H(Y1 |X1 ) given that Y1 is independent of X2 when conditioned on X1 as
pointed out in (13.24). Also, the equality follows because H(Y2 |Y1 , X1 , X2 ) = H(Y2 |X2 )given that Y2 is conditionally independent of X1 and Y1 as pointed out in (13.26). The
first inequality follows from subadditivity of entropy (Exercise 10.3.3). The last equality
follows from the definition of mutual information, and the final inequality follows because
the marginal distributions for X1 and X2 can only achieve a mutual information less thanor equal to the respective maximizing marginal distributions for I(N1 ) and I(N2 ).
A simple corollary of Theorem 13.1.1 is that correlations between input random variables cannot increase the mutual information of a classical channel. The proof follows by
a straightforward induction argument. Thus, the single-letter expression in (13.5) for themutual information of a classical channel suffices for understanding the ability of a classicalchannel to maintain correlations between its input and output.
Corollary 13.1.1 The regularized mutual information of a classical channel is equal to its
mutual information:
Ireg (N ) = I(N ).
(13.34)
Proof. We prove the result using induction on n, by showing that I(N ⊗n ) = nI(N ) for
all n, implying that the limit in (13.9) is not necessary. The base case for n = 1 is trivial.
Suppose the result holds for n: I(N ⊗n ) = nI(N ). The following chain of equalities thenproves the inductive step:I(N ⊗n+1 ) = I(N ⊗ N ⊗n )= I(N ) + I(N ⊗n )= I(N ) + nI(N ).

(13.35)(13.36)(13.37)

The first equality follows because the channel N ⊗n+1 is equivalent to the parallel concatenation of N and N ⊗n . The second critical equality follows from the application of Theorem 13.1.1 because the distributions of N and N ⊗n factor as in (13.24). The final equality
follows from the induction hypothesis.

13.1.3

The Problem with Regularization

The main problem with a regularized information quantity is that it does not uniquely
characterize the capacity of a channel for a given information-processing task, whether it bea classical or quantum one. This motivates the task of proving additivity of an information
quantity.
We illustrate this point now with an example. Let Ia (N ) denote the following functionof a classical channel N :Ia (N ) = max [H(X) − aH(X|Y )] ,(13.38)pX (x)

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CHAPTER 13. THE INFORMATION OF QUANTUM CHANNELS

where a > 1. From the definitions we can conclude that
I(N ) ≥ Ia (N ),(13.39)and in fact that

1
1
I(N ⊗n ) ≥ Ia (N n ),(13.40)nn
for any fixed positive integer n. However, something interesting happens when we regularize
the quantity Ia (N ), i.e., when we consider the following limit:
1
1
Ia (N ⊗n ) = limmax [H(X n ) − aH(X n |Y n )] .
n→∞ n
n→∞ n pX n (xn )lim(13.41)

Let n be a large fixed integer (large enough so that the law of large numbers comes intoplay). That is, we can fix constants ε ∈ (0, 1) and δ > 0. From Shannon’s channel coding
theorem, we know that there exists a code {xn (m)}m∈M of rate n1 log |M| = I(N ) − δ andlength n for the channel N , such that the probability of error when decoding is no larger
than ε. So we can pick pX n (xn ) to be the uniform distribution over the codewords for thiscode, and for this choice, we get
H(X n ) = log |M| = n [I(N ) − δ] .

(13.42)Furthermore, we can apply Fano’s inequality (Theorem 10.7.3) to conclude thatH(X n |Y n ) ≤ h2 (ε) + ε log [|M| − 1]
≤ h2 (ε) + εn [I(N ) − δ] .

(13.43)(13.44)

Putting these inequalities together implies that11
Ia (N ⊗n ) =max [H(X n ) − aH(X n |Y n )]n
n pX n (xn )1≥ (n [I(N ) − δ] − a [h2 (ε) + εn [I(N ) − δ]])n1
= (1 − aε) I(N ) − aδε − [δ − ah2 (ε)] .
n(13.45)(13.46)(13.47)

Taking the limit n → ∞ then gives1
Ia (N ⊗n ) ≥ (1 − aε) I(N ) − aδε.
n→∞ n
lim

(13.48)

However, as n gets larger, we can make both ε and δ go to zero (in principle we could write
ε and δ as explicit functions of n that go to zero as n → ∞). This establishes that1
Ia (N ⊗n ) ≥ I(N ).
n→∞ n
lim

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13.1. MUTUAL INFORMATION OF A CLASSICAL CHANNEL

393

However, we have for every n that
1
1
[H(X n ) − aH(X n |Y n )] ≤ [H(X n ) − H(X n |Y n )]nn1
≤ lim I(N ⊗n )n→∞ n
= I(N ),

(13.50)(13.51)(13.52)where the last equality follows from Corollary 13.1.1. These two bounds lead us to theconclusion that1
lim Ia (N ⊗n ) = I(N ).
(13.53)n→∞ nThus, even though for a given channel N we can have a strict inequality between Ia (N ) andI(N ) on the single-copy level (and even on the multi-copy level for finite n), this difference
gets washed away or “blurred” in the limit n → ∞ for any finite a > 1. For this reason, we
should always be wary of a regularized characterization of capacity, knowing that it is incomplete. It is only when we have additivity of an information quantity that we can conclude
that it fully characterizes capacity for a given information-processing task. Unfortunately,
regularization is a problem that plagues much of quantum Shannon theory (except for a few
tasks, such as entanglement-assisted classical communication, discussed in Chapter 21).

13.1.4

Optimizing the Mutual Information of a Classical Channel

The definition in (13.5) seems like a suitable definition for the mutual information of a classical channel, but how difficult is the maximization problem that it sets out? Theorem 13.1.2
below states an important property of the mutual information I(X; Y ) that allows us toanswer this question. Suppose that we fix the conditional density pY |X (y|x), but can vary
the input density pX (x). Theorem 13.1.2 below states that the mutual information I(X; Y )is a concave function of the density pX (x). In particular, this result implies that the channel
mutual information I(N ) has a global maximum, and the optimization problem is therefore
a straightforward computation that can exploit convex optimization methods.
Theorem 13.1.2 Suppose that we fix the conditional probability density pY |X (y|x). Thenthe mutual information I(X; Y ) is concave in the marginal density pX (x):λI(X1 ; Y ) + (1 − λ) I(X2 ; Y ) ≤ I(Z; Y ),(13.54)

where random variable X1 has density pX1 (x), X2 has density pX2 (x), and Z has density
λpX1 (x) + (1 − λ) pX2 (x) for λ ∈ [0, 1].
Proof. Let us fix
Pthe density pY |X (y|x). The density pY (y) is a linear function of pX (x)because pY (y) = x pY |X (y|x)p
P X (x). Thus H(Y ) is concave in pX (x). Recall that the conditional entropy H(Y |X) = x pX (x)H(Y |X = x). The entropy H(Y |X = x) is fixed when
the conditional probability density pY |X (y|x) is fixed. Thus, H(Y |X) is a linear functionof pX (x). These two results imply that the mutual information I(X; Y ) is concave in themarginal density pX (x) when the conditional density pY |X (y|x) is fixed.
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CHAPTER 13. THE INFORMATION OF QUANTUM CHANNELS

Figure 13.2: The setting for the classical wiretap channel.

13.2 Private Information of a Wiretap Channel
Suppose now that we extend the above two-user classical communication scenario to a threeuser communication scenario, where the parties involved are Alice, Bob, and Eve. Supposethat Alice would like to communicate to Bob while keeping her messages private from Eve.
The channel N corresponding to this setting is called the wiretap channel, which has thefollowing conditional probability density:
pY,Z|X (y, z|x).

(13.55)

Alice has access to the input random variable X, Bob receives output random variable Y ,and Eve receives the random variable Z. Figure 13.2 depicts this setting.
We would like to establish a measure of information throughput for this scenario. It
might seem intuitive that it should be the amount of correlations that Alice can establish
with Bob, less the correlations that Eve receives: I(X; Y ) − I(X; Z). However, a moregeneral procedure would allow Alice to pick an auxiliary random variable U and then to pick
X given the value of U , leading to the following information quantity:
I(U ; Y ) − I(U ; Z).

(13.56)

But Alice can maximize over all such input distributions pU,X (u, x) on her end. This leads
us to the following definition:
Definition 13.2.1 (Private Information of a Wiretap Channel) The private information P (N ) of a classical wiretap channel N ≡ pY,Z|X is defined as follows:
P (N ) ≡ max [I(U ; Y ) − I(U ; Z)].

(13.57)

pU,X (u,x)
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13.2. PRIVATE INFORMATION OF A WIRETAP CHANNEL395

It is possible to provide an operational interpretation of the private information, but we
do not do that here. We instead focus on the additivity properties of the private information
P (N ).
One may wonder if the above quantity is non-negative, given that it is equal to the difference of two mutual informations. Non-negativity does hold, and a simple proof demonstrates
this fact.
Property 13.2.1 The private information P (N ) of a wiretap channel is non-negative:
P (N ) ≥ 0.

(13.58)Proof. We can choose the joint density pU,X (u, x) in the maximization of P (N ) to be thedegenerate distribution pU,X (u, x) = δu,u0 δx,x0 for some values u0 and x0 . Then both mutual
informations I(U ; Y ) and I(U ; Z) vanish, and their difference vanishes as well. The private
information P (N ) can then only be greater than or equal to zero because the above choice
is a particular choice of the density pU,X (u, x) and P (N ) requires a maximization over all
such distributions.
Exercise 13.2.1 Show that adding an auxiliary random variable cannot increase the mutual information of a classical channel. That is, let N ≡ pY |X and show that I(N ) =maxpU,X I(U ; Y ). Explain why it is possible for an auxiliary random variable to increase theprivate information of a classical wiretap channel.

13.2.1

Additivity of Private Information

The private information of general classical wiretap channels is additive. This result follows
essentially from an application of the chain rule for mutual information. Figure 13.3 displays the scenario corresponding to the analysis involved in determining whether the private
information is additive.
Theorem 13.2.1 (Additivity of Private Information) Let Ni be a classical wiretap channel with distribution pYi ,Zi |Xi , for i ∈ {1, 2}. The private information of the classical joint
channel N1 ⊗ N2 is the sum of the individual private informations of the channels:
P (N1 ⊗ N2 ) = P (N1 ) + P (N2 ).

(13.59)Proof. The inequality P (N1 ⊗ N2 ) ≥ P (N1 ) + P (N2 ) is trivial and we leave it as an exercise
for the reader to complete.
We thus prove the following non-trivial inequality: P (N1 ⊗ N2 ) ≤ P (N1 ) + P (N2 ). Let
pU,X1 ,X2 (u, x1 , x2 ) be any probability distribution that we could consider for the quantity
P (N1 ⊗ N2 ), where U is an auxiliary random variable. The joint channel has the following
probability distribution:
pY1 ,Z1 |X1 (y1 , z1 |x1 )pY2 ,Z2 |X2 (y2 , z2 |x2 ).
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CHAPTER 13. THE INFORMATION OF QUANTUM CHANNELS

Figure 13.3: This figure displays the scenario for determining whether the private information of twoclassical wiretap channels N1 and N2 is additive. The question of additivity is equivalent to the possibilityof classical correlations being able to enhance the private information of two classical wiretap channels. The
result stated in Theorem 13.2.1 is that the private information is additive for two wiretap channels, so thatclassical correlations cannot enhance the private information.

Consider the following chain of equalities:
I(U ; Y1 Y2 ) − I(U ; Z1 Z2 )= I(U ; Y1 ) + I(U ; Y2 |Y1 ) − I(U ; Z2 ) − I(U ; Z1 |Z2 )= I(U ; Y1 |Z2 ) + I(U ; Y2 |Y1 ) − I(U ; Z2 |Y1 ) − I(U ; Z1 |Z2 )= I(U ; Y1 |Z2 ) − I(U ; Z1 |Z2 ) + I(U ; Y2 |Y1 ) − I(U ; Z2 |Y1 ).

(13.61)(13.62)(13.63)

The first equality follows from an application of the chain rule for mutual information. The
second equality follows from the identity
I(U ; Y1 ) − I(U ; Z2 ) = I(U ; Y1 |Z2 ) − I(U ; Z2 |Y1 ),(13.64)

which can be verified using definitions. The third equality is a simple rearrangement. We
now focus on the term I(U ; Y1 |Z2 ) − I(U ; Z1 |Z2 ). Note that the probability distribution forthese random variables can be written as
pY1 ,Z1 |X1 (y1 , z1 |x1 )pX1 |U (x1 |u)pU |Z2 (u|z2 )pZ2 (z2 ),(13.65)P
where pU |Z2 (u|z2 ) = x2 pU,X2 |Z2 (u, x2 |z2 ). (The fact that the distribution factors in thisway is where the assumption of independent wiretap channels comes into play.) Consider
that
I(U ; Y1 |Z2 ) − I(U ; Z1 |Z2 )X=
pZ2 (z2 ) [I(U ; Y1 |Z2 = z2 ) − I(U ; Z1 |Z2 = z2 )](13.66)z2≤ max [I(U ; Y1 |Z2 = z2 ) − I(U ; Z1 |Z2 = z2 )](13.67)≤ P (N1 ).

(13.68)z2

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13.2. PRIVATE INFORMATION OF A WIRETAP CHANNEL397

The first equality follows because the conditional mutual informations can be expanded as aconvex combination of mutual informations. The last inequality follows because pU |Z2 (u|z2 )is a particular distribution for an auxiliary random variable U |Z2 = z2 . By the same kind of
reasoning, we find that
I(U ; Y2 |Y1 ) − I(U ; Z2 |Y1 ) ≤ P (N2 ),(13.69)and can then conclude that
I(U ; Y1 Y2 ) − I(U ; Z1 Z2 ) ≤ P (N1 ) + P (N2 ).

(13.70)Since this inequality holds for any auxiliary random variable U , we can finally conclude that
P (N1 ⊗ N2 ) ≤ P (N1 ) + P (N2 ).
Exercise 13.2.2 Show that the sum of the individual private informations can never begreater than the private information of the classical joint channel:
P (N1 ⊗ N2 ) ≥ P (N1 ) + P (N2 ).

(13.71)

Exercise 13.2.3 Show that the regularized private information of a wiretap channel N isequal to its private information: limn→∞ n1 P (N ⊗n ) = P (N ).

13.2.2Degraded Wiretap Channels

The formula for the private information simplifies for a particular type of wiretap channel,
called a physically degraded wiretap channel. A wiretap channel is physically degraded if X,
Y , and Z form the following Markov chain: X → Y → Z. That is, the channel distribution
factors as pY,Z|X (y, z|x) = pZ|Y (z|y)pY |X (y|x), so that there is some channel pZ|Y (z|y) thatBob can apply to his output to simulate the channel pZ|X (z|x) to Eve:pZ|X (z|x) =X

pZ|Y (z|y)pY |X (y|x).

(13.72)yThis condition allows us to apply the data-processing inequality to give the following simplified formula for the private information of a degraded wiretap channel, in which there isno need for an auxiliary random variable:
Proposition 13.2.1 The private information P (N ) of a degraded classical wiretap channel
N ≡ pY,Z|X simplifies as follows:
P (N ) ≡ max [I(X; Y ) − I(X; Z)].

(13.73)pX (x)

Proof. Consider that we always havemax [I(U ; Y ) − I(U ; Z)] ≥ max [I(X; Y ) − I(X; Z)],
pU,X (u,x)
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CHAPTER 13. THE INFORMATION OF QUANTUM CHANNELS

by picking the distribution on the left-hand side as pU,X (u, x) = pX (x)δx,u . That is, we set U
to be a random variable with the same alphabet size as the channel input and then just sendU directly into the channel. To prove the other inequality, we need to use the assumptionof degradability, which implies that U − X − Y − Z is a Markov chain. Consider that
I(U ; Y ) = I(X; Y ) + I(U ; Y |X) − I(X; Y |U )= I(X; Y ) − I(X; Y |U ).

(13.75)(13.76)

The first equality follows from two applications of the chain rule for mutual information.
The second equality follows because U − X − Y is a Markov chain, so that I(U ; Y |X) = 0.
By the same reasoning, we have that I(U ; Z) = I(X; Z) − I(X; Z|U ). Then
I(U ; Y ) − I(U ; Z) = I(X; Y ) − I(X; Y |U ) − [I(X; Z) − I(X; Z|U )]= I(X; Y ) − I(X; Z) − [I(X; Y |U ) − I(X; Z|U )]≤ I(X; Y ) − I(X; Z)≤ max [I(X; Y ) − I(X; Z)].

(13.77)(13.78)(13.79)(13.80)pX (x)The inequality follows from the assumption that the wiretap channel is degraded, so that we
can apply the data-processing inequality and conclude that I(X; Y |U ) ≥ I(X; Z|U ). Since
the inequality holds for any auxiliary random variable U , we can conclude thatmax [I(U ; Y ) − I(U ; Z)] ≤ max [I(X; Y ) − I(X; Z)],
pU,X (u,x)(13.81)pX (x)

which completes the proof.
An analogous notion of degradability exists in the quantum setting, and Section 13.5
demonstrates that degradable quantum channels have additive coherent information. The
coherent information of a quantum channel is a measure of how much quantum informationa sender can transmit through that channel to a receiver and thus is an important quantity
to consider for quantum data transmission.
Exercise 13.2.4 Show that the private information of a degraded wiretap channel can also
be written as maxpX (x) I(X; Y |Z).

13.3 Holevo Information of a Quantum Channel
We now turn our attention to the case of dynamic information measures for quantum channels, and we begin with a measure of classical correlations. Suppose that Alice would like toestablish classical correlations with Bob, by means of a quantum channel. Alice can prepare
an ensemble {pX (x), ρx } in her laboratory, where the states ρx are acceptable inputs to thequantum channel. She keeps a copy of the classical index x in some classical register X. The
expected density operator of this ensemble is the following classical–quantum state:XρXA ≡pX (x)|xihx|X ⊗ ρxA .
(13.82)x
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13.3. HOLEVO INFORMATION OF A QUANTUM CHANNEL

399

Such a preparation is the most general way that Alice can correlate classical data with aquantum state to input to the channel. Let ρXB be the state that arises from sending the A
system through the quantum channel NA→B :X(13.83)ρXB ≡pX (x)|xihx|X ⊗ NA→B (ρxA ).
x

We would like to determine a measure of the ability of the quantum channel to preserve
classical correlations. We can appeal to ideas from the classical case in Section 13.1, while
incorporating the static quantum measures from Chapter 11. A good measure of the input–
output classical correlations is the Holevo information of the above classical–quantum state:
I(X; B)ρ . This measure corresponds to a particular preparation that Alice chooses, butobserve that she can prepare the input ensemble in such a way as to achieve the highest
possible correlations. Maximizing the Holevo information over all possible preparations
gives a dynamic measure called the Holevo information of the channel.
Definition 13.3.1 (Holevo Information of a Quantum Channel) The Holevo information χ(N ) of a channel N is a measure of the classical correlations that Alice can establish
with Bob:
χ(N ) ≡ max I(X; B)ρ ,(13.84)
ρXA

where the maximization is with respect to all input classical–quantum states of the form in(13.82) and I(X; B)ρ is evaluated with respect to the state in (13.83).

13.3.1

Additivity of the Holevo Information for Specific Channels

The Holevo information of a quantum channel is generally not additive (by no means is thisobvious!). The question of additivity for this case is not whether classical correlations canenhance the Holevo information, but it is rather whether quantum correlations can enhance
it. That is, Alice can choose an ensemble of the form {pX (x), ρxA1 A2 } for input to two uses of
the quantum channel. The conditional density operators ρxA1 A2 can be entangled and thesequantum correlations can potentially increase the Holevo information.
The question of additivity of the Holevo information of a quantum channel was a longstanding open conjecture in quantum information theory—many researchers thought thatquantum correlations would not enhance it and that additivity would hold. But recent
research has demonstrated a counterexample to the additivity conjecture, and perhaps unsurprisingly in hindsight, this counterexample exploits maximally entangled states to demonstrate superadditivity (see Section 20.5). Figure 13.4 displays the scenario corresponding to
the question of additivity of the Holevo information.
Additivity of Holevo information may not hold for all quantum channels, but it is possible
to prove its additivity for certain classes of quantum channels. One such class for which
additivity holds is the class of entanglement-breaking channels, and the proof of additivity
is perhaps the simplest for this case.
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CHAPTER 13. THE INFORMATION OF QUANTUM CHANNELS

Figure 13.4: The scenario for determining whether the Holevo information of two quantum channels N1
and N2 is additive. The question of additivity is equivalent to the possibility of quantum correlations being
able to enhance the Holevo information of two quantum channels. The result stated in Theorem 13.3.1 isthat the Holevo information is additive for the tensor product of an entanglement-breaking channel and anyother quantum channel, so that quantum correlations cannot enhance the Holevo information in this case.
This is perhaps intuitive because an entanglement-breaking channel destroys quantum correlations in theform of quantum entanglement.

Theorem 13.3.1 (Additivity for Entanglement-Breaking Channels) Suppose that aquantum channel N EB is entanglement breaking and another channel M is arbitrary. Thenthe Holevo information χ(N EB ⊗ M) of the tensor-product channel N EB ⊗ M is the sum of
the individual Holevo informations χ(N EB ) and χ(M):χ(N EB ⊗ M) = χ(N EB ) + χ(M).

(13.85)

Proof. The trivial inequality χ(N EB ⊗ M) ≥ χ(N EB ) + χ(M) holds for any two quantum
channels N EB and M because we can choose the input ensemble on the left-hand side to bea tensor product of the ones that individually maximize the terms on the right-hand side.
We now prove the non-trivial inequality χ(N EB ⊗M) ≤ χ(N EB )+χ(M) that holds when
N EB is entanglement breaking. Let ρXB1 B2 be a state that maximizes the Holevo information
χ(N EB ⊗ M), whereρXB1 B2 ≡ (NAEB→B ⊗ M)(ρXA1 A2 ),X1 1
ρXA1 A2 ≡pX (x)|xihx|X ⊗ ρxA1 A2 .

(13.86)(13.87)x

Let ρXB1 A2 be the state after only the entanglement-breaking channel NAEB
acts. We can
1 →B1write this state as follows:XρXB1 A2 ≡ NAEB(ρ)=
pX (x)|xihx|X ⊗ NAEB
(ρxA1 A2 )(13.88)
XA
A
1
2
→B
1
1
1 →B1
x

=
=

X

pX (x)|xihx|X ⊗X

x,ypY |X (y|x) σBx,y1 ⊗ θA
2

x

y

X

x,y
pY |X (y|x)pX (x)|xihx|X ⊗ σBx,y1 ⊗ θA.
2

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13.3. HOLEVO INFORMATION OF A QUANTUM CHANNEL401

The third equality follows because the P
channel N EB breaks any entanglement in the statex,yρxA1 A2 , leaving behind a separable state y pY |X (y|x) σBx,y1 ⊗ θA
. Then the state ρXB1 B2 has
2the form
Xx,y).
(13.91)
ρXB1 B2 =pY |X (y|x)pX (x)|xihx|X ⊗ σBx,y1 ⊗ M(θA2x,yLet ωXY B1 B2 be an extension of ρXB1 B2 whereX
x,yωXY B1 B2 ≡pY |X (y|x)pX (x)|xihx|X ⊗ |yihy|Y ⊗ σBx,y1 ⊗ M(θA),
2

(13.92)x,yand TrY {ωXY B1 B2 } = ρXB1 B2 . Then the following chain of inequalities holds
χ(N EB ⊗ M) = I(X; B1 B2 )ρ
= I(X; B1 )ρ + I(X; B2 |B1 )ρ
≤ χ(N EB ) + I(X; B2 |B1 )ρ

(13.93)(13.94)(13.95)

The first equality follows because we took ρXB1 B2 to be a state that maximizes the Holevo
information χ(N EB ⊗ M) of the tensor-product channel N EB ⊗ M. The second equality is
an application of the chain rule for conditional mutual information (Property 11.7.1). The
inequality follows because the Holevo information I(X; B1 )ρ is with respect to the following
state:XρXB1 ≡pX (x)|xihx|X ⊗ NAEB
(ρxA1 ),
(13.96)
1 →B1
x

whereas the Holevo information of the channel NAEB
is defined to be the maximal Holevo
1 →B1
information with respect to all input ensembles. Now let us focus on the term I(X; B2 |B1 )ρ .
Consider that
I(X; B2 |B1 )ρ = I(X; B2 |B1 )ω
≤ I(XB1 ; B2 )ω
≤ I(XY B1 ; B2 )ω
= I(XY ; B2 )ω + I(B1 ; B2 |XY )ω
= I(XY ; B2 )ω
≤ χ(M).

(13.97)(13.98)(13.99)(13.100)(13.101)(13.102)

The first equality follows because the reduced state of ωXY B1 B2 on systems X, B1 , andB2 is equal to ρXB1 B2 . The first inequality follows from the chain rule: I(X; B2 |B1 ) =I(XB1 ; B2 ) − I(B1 ; B2 ) ≤ I(XB1 ; B2 ). The second inequality follows from the quantum
data-processing inequality. The second equality is again from the chain rule for conditional
mutual information. The third equality is the crucial one that exploits the entanglementbreaking property. It follows by examining (13.92) and observing that the state ωXY B1 B2 on
systems B1 and B2 is product when conditioned on classical variables X and Y , so that theconditional mutual information between systems B1 and B2 given both X and Y is equal tozero. The final inequality follows because ωXY B2 is a particular state of the form needed in
the maximization of χ(M).
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CHAPTER 13. THE INFORMATION OF QUANTUM CHANNELS

Corollary 13.3.1 The regularized Holevo information of an entanglement-breaking quantum channel N EB is equal to its Holevo information:
χreg (N EB ) = χ(N EB ).

(13.103)

Proof. A proof of this property uses the same induction argument as in Corollary 13.1.1
and exploits the additivity property in Theorem 13.3.1 above.

13.3.2

Optimizing the Holevo Information

Pure States are Sufficient
The following theorem allows us to simplify the optimization problem that (13.84) sets out—
we show that it is sufficient to consider ensembles of pure states at the input.
Theorem 13.3.2 It is sufficient to maximize the Holevo information with respect to purestates:χ(N ) = max I(X; B)ρ = max I(X; B)τ ,(13.104)
ρXA

τXA

where
τXA ≡

X

pX (x)|xihx|X ⊗ |φx ihφx |A ,(13.105)x

and ρXB and τXB are the states that result from sending the A system of ρXA and τXAthrough the quantum channel NA→B , respectively.
Proof. Suppose that ρXA is any state of the form in (13.82). Consider a spectral decomposition of the states ρxA :X
pY |X (y|x)ψAx,y ,ρxA =(13.106)ywhere the states ψAx,y are pure. Then let σXY A denote the following state:X
σXY A ≡
pY |X (y|x)pX (x)|xihx|X ⊗ |yihy|Y ⊗ ψAx,y ,(13.107)x

so that TrY {σXY A } = ρXA . Also, observe that σXY A is a state of the form τXA with XY
as the classical system. Let σXY B denote the state that results from sending the A systemthrough the quantum channel NA→B . Then the following relations hold:
I(X; B)ρ = I(X; B)σ ≤ I(XY ; B)σ .

(13.108)

The equality follows because TrY {σXY B } = ρXB and the inequality follows from the quantum
data-processing inequality. It then suffices to consider ensembles with only pure states
because the state σXY B is a state of the form τXB with the combined system XY acting as
the classical system.
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Concavity in the Distribution and Convexity in the Signal States
We now show that the Holevo information is concave as a function of the input distribution
when the signal states are fixed.
Theorem 13.3.3 The Holevo information I(X; B) is concave in the input distribution when
the signal states are fixed, in the sense thatλI(X; B)σ + (1 − λ) I(X; B)τ ≤ I(X; B)ω ,(13.109)

where σXB and τXB are of the form
σXB ≡

X

pX (x)|xihx|X ⊗ N (ρx ),(13.110)qX (x)|xihx|X ⊗ N (ρx ),(13.111)

x

τXB ≡

X
x

and ωXB is a mixture of the states σXB and τXB of the formX
[λpX (x) + (1 − λ) qX (x)] |xihx|X ⊗ N (ρx ),ωXB ≡(13.112)x

where 0 ≤ λ ≤ 1.
Proof. Let ωXU B be the stateXωXU B ≡[pX (x)|xihx|X ⊗ λ|0ih0|U + qX (x)|xihx|X ⊗ (1 − λ) |1ih1|U ] ⊗ N (ρx ). (13.113)
x

Observe that TrU {ωXU B } = ωXB . Then the statement of concavity is equivalent to I(X; B|U )ω ≤I(X; B)ω . We can rewrite this as H(B|U )ω − H(B|U X)ω ≤ H(B)ω − H(B|X)ω . Observethat H(B|U X)ω = H(B|X)ω , i.e., one can calculate that both of these conditional entropies
are equal toX[λpX (x) + (1 − λ) qX (x)] H(N (ρx )).
(13.114)x

The statement of concavity then becomes H(B|U )ω ≤ H(B)ω , which follows from concavityof quantum entropy.
The Holevo information is convex as a function of the signal states when the input
distribution is fixed.
Theorem 13.3.4 The Holevo information I(X; B) is convex in the signal states when the
input distribution is fixed, in the sense thatλI(X; B)σ + (1 − λ) I(X; B)τ ≥ I(X; B)ω ,c 2019(13.115)

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CHAPTER 13. THE INFORMATION OF QUANTUM CHANNELS

where σXB and τXB are of the form
σXB ≡

X

τXB ≡

X

pX (x)|xihx|X ⊗ N (σ x ),(13.116)pX (x)|xihx|X ⊗ N (τ x ),(13.117)xx

and ωXB is a mixture of the states σXB and τXB of the formXωXB ≡pX (x)|xihx|X ⊗ N (λσ x + (1 − λ) τ x ),(13.118)x

where 0 ≤ λ ≤ 1.
Proof. Let ωXU B be the stateXωXU B ≡pX (x)|xihx|X ⊗ [λ|0ih0|U ⊗ N (σ x ) + (1 − λ) |1ih1|U ⊗ N (τ x )] .

(13.119)x

Observe that TrU {ωXU B } = ωXB . Then convexity in the input states is equivalent to thestatement I(X; B|U )ω ≥ I(X; B)ω . Consider that I(X; B|U )ω = I(X; BU )ω − I(X; U )ω , by
the chain rule for the quantum mutual information. Since the input distribution pX (x) is
fixed, there are no correlations between X and the convexity variable U , so that I(X; U )ω =0. Thus, the above inequality is equivalent to I(X; BU )ω ≥ I(X; B)ω , which follows fromthe quantum data-processing inequality.
In the above two theorems, we have shown that the Holevo information is either concave or convex depending on whether the signal states or the input distribution are fixed,
respectively. Thus, the computation of the Holevo information of a general quantum channel
becomes difficult as the input dimension of the channel grows larger, since a local maximumof the Holevo information is not necessarily a global maximum. However, if the channel
has a classical input and a quantum output, the computation of the Holevo information isstraightforward because the only input parameter is the input distribution, and we provedthat the Holevo information is a concave function of the input distribution.

13.4 Mutual Information of a Quantum Channel
We now consider a measure of the ability of a quantum channel to preserve quantum correlations. The way that we arrive at this measure is similar to what we have seen before.
Alice prepares some pure quantum state φAA0 in her laboratory, and inputs the A0 system toa quantum channel NA0 →B —this transmission gives rise to the following bipartite state:ρAB = NA0 →B (φAA0 ).

(13.120)

The quantum mutual information I(A; B)ρ is a static measure of correlations present in
the state ρAB . To maximize the correlations that the quantum channel can establish, Alicec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

13.4. MUTUAL INFORMATION OF A QUANTUM CHANNEL

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should maximize the quantum mutual information I(A; B)ρ with respect to all possible purestates that she can input to the channel NA0 →B . This procedure leads to the definition of
the mutual information I(N ) of a quantum channel:
I(N ) ≡ max I(A; B)ρ .
φAA0(13.121)

The mutual information of a quantum channel corresponds to an important operational
task that is not particularly obvious from the above discussion. Suppose that Alice andBob share unlimited bipartite entanglement in whatever form they wish, and suppose they
have access to a large number of independent uses of the channel NA0 →B . Then the mutual
information of the channel corresponds to the maximal amount of classical information thatthey can transmit in such a setting. This setting is the noisy analog of the super-dense coding
protocol from Chapter 6 (recall the discussion in Section 6.4). By teleportation, the maximal
amount of quantum information that they can transmit is half of the mutual information of
the channel. We discuss how to prove these statements rigorously in Chapter 21.

13.4.1

Additivity

There might be little reason to expect that the quantum mutual information of a quantum
channel is additive, given that the Holevo information is not. But perhaps surprisingly,
additivity does hold for the mutual information of a quantum channel! This result meansthat we completely understand this measure of information throughput, and it also meansthat we understand the operational task to which it corresponds (entanglement-assistedclassical coding discussed in the previous section).
We might intuitively attempt to explain this phenomenon in terms of this operationaltask—Alice and Bob already share unlimited entanglement between their terminals and soentangled correlations at the input of the channel do not lead to any superadditive effectas it does for the Holevo information. This explanation is somewhat rough, but perhaps
the additivity proof explains best why additivity holds. The crucial tool in the proof is thechain rule for mutual information and one application of subadditivity of entropy (Corollary 11.8.1). Figure 13.5 illustrates the setting corresponding to the analysis for additivityof the mutual information of a quantum channel.
Theorem 13.4.1 (Additivity of Q. Mutual Information of Q. Channels) Let N and
M be any quantum channels. Then the mutual information of the tensor-product channel
N ⊗ M is the sum of their individual mutual informations:I(N ⊗ M) = I(N ) + I(M).

(13.122)

Proof. We first prove the trivial inequality I(N ⊗ M) ≥ I(N ) + I(M). Let φA1 A01 andψA2 A02 be states that maximize the respective mutual informations I(N ) and I(M). Let
ρA1 A2 B1 B2 ≡ (NA01 →B1 ⊗ MA02 →B2 )(φA1 A01 ⊗ ψA2 A02 ).
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Figure 13.5: This figure displays the scenario for determining whether the mutual information of two
quantum channels N1 and N2 is additive. The question of additivity is equivalent to the possibility ofquantum correlations between channel inputs being able to enhance the mutual information of two quantum
channels. The result stated in Theorem 13.4.1 is that the mutual information is additive for any two quantum
channels, so that quantum correlations cannot enhance it.

Observe that the state ρA1 A2 B1 B2 is a particular state of the form required in the maximizationof I(N ⊗ M), by taking A ≡ A1 A2 . Then the following holds
I(N ) + I(M) = I(A1 ; B1 )N (φ) + I(A2 ; B2 )M(ψ)= I(A1 A2 ; B1 B2 )ρ
≤ I(N ⊗ M).

(13.124)(13.125)(13.126)

The first equality follows by evaluating the mutual informations I(N ) and I(M) with respect
to the maximizing states φA1 A01 and ψA2 A02 . The second equality follows from the fact thatmutual information is additive with respect to tensor-product states (Exercise 11.6.8). The
final inequality follows because the input state φA1 A01 ⊗ψA2 A02 is a particular input state of themore general form φAA01 A02 needed in the maximization of the quantum mutual informationof the tensor-product channel N ⊗ M.
We now prove the non-trivial inequality I(N ⊗ M) ≤ I(N ) + I(M). Let φAA01 A02 be astate that maximizes the mutual information I(N ⊗ M) and let
ρAB1 B2 ≡ (NA01 →B1 ⊗ MA02 →B2 )(φAA01 A02 ),(13.127)

Consider the following chain of inequalities:
I(N ⊗ M) = I(A; B1 B2 )ρ
= I(A; B1 )ρ + I(AB1 ; B2 )ρ − I(B1 ; B2 )ρ
≤ I(A; B1 )ρ + I(AB1 ; B2 )ρ
≤ I(N ) + I(M).
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13.4. MUTUAL INFORMATION OF A QUANTUM CHANNEL407

The first equality follows from the definition of I(N ⊗ M) in (13.121) and evaluating
I(A; B1 B2 ) with respect to the maximizing state φ. The second equality follows by expanding the quantum mutual information, using Exercise 11.7.1. The first inequality follows because I(B1 ; B2 )ρ ≥ 0. The last inequality follows because I(A; B1 )ρ ≤ I(N ) andI(AB1 ; B2 )ρ ≤ I(M), where we have applied the result of Exercise 13.4.4 and the fact that
the A system extends the A1 system which is input to the channel N and the fact that theAB1 systems extend the A2 system which is input to the channel M.
Corollary 13.4.1 The regularized mutual information of a quantum channel N is equal toits mutual information:
Ireg (N ) = I(N ).
(13.132)
Proof. A proof of this property uses the same induction argument as in Corollary 13.1.1
and exploits the additivity property in Theorem 13.4.1 above.
Exercise 13.4.1 (Alternate Mutual Information of a Quantum Channel) Let ρXAB
denote a state of the following form:XpX (x)|xihx|X ⊗ NA0 →B (φxAA0 ).
(13.133)
ρXAB ≡
x

Consider the following alternate definition of the mutual information of a quantum channel:
Ialt (N ) ≡ max I(AX; B),
ρXAB(13.134)

where the maximization is with respect to states of the form ρXAB . Show that
Ialt (N ) = I(N ).

(13.135)

Exercise 13.4.2 Compute the mutual information of a dephasing channel with dephasing
parameter p.
Exercise 13.4.3 Compute the mutual information of an erasure channel with erasure parameter ε.
Exercise 13.4.4 (Pure States are Sufficient) Let NA0 →B be a quantum channel. Show
that it is sufficient to consider pure states φAA0 for determining the mutual informationof a quantum channel. That is, one does not need to consider mixed states ρAA0 in the
optimization task because
max I(A; B)N (φ) = max I(A; B)N (ρ) .
φAA0

ρAA0

(13.136)(Hint: Consider purifying and using the quantum data-processing inequality.)c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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13.4.2

Optimizing the Mutual Information of a Quantum Channel

We now show that the mutual information of a quantum channel is concave as a functionof the input state. This result allows us to compute this quantity with standard convex
optimization techniques.
Theorem 13.4.2 The mutual information I(A; B) is concave in the input state, in the sensethatX
pX (x)I(A; B)ρx ≤ I(A; B)σ ,(13.137)x

where ρxAB ≡ NA0 →B (φxAA0 ), σA0≡xx pX (x)ρA0 , φAA0

P

is a purification of σA0 , and σAB ≡NA0 →B (φAA0 ).
Proof. Let ρXABE be the following classical–quantum state:X
ρXABE ≡pX (x)|xihx|X ⊗ UAN0 →BE (φxAA0 ),(13.138)x

where UAN0 →BE is an isometric extension of the channel. Consider the following chain of
inequalities:
X
pX (x)I(A; B)ρx = I(A; B|X)ρ
(13.139)x= H(A|X)ρ + H(B|X)ρ − H(AB|X)ρ
= H(BE|X)ρ + H(B|X)ρ − H(E|X)ρ
= H(B|EX)ρ + H(B|X)ρ
≤ H(B|E)ρ + H(B)ρ
= H(B|E)σ + H(B)σ= I(A; B)σ .

(13.140)(13.141)(13.142)(13.143)(13.144)(13.145)

The first equality follows because the conditioning system X in I(A; B|X)ρ is classical. The
second equality follows by expanding the quantum mutual information. The third equality
follows because the state on ABE is pure when conditioned on X. The fourth equality
follows from the definition of conditional quantum entropy. The inequality follows fromstrong subadditivity and concavity of quantum entropy. The equality follows by inspecting
the definition of the state σ, and the final equality follows because the state is pure on
systems ABE.

13.5 Coherent Information of a Quantum Channel
This section presents an alternative, important measure of the ability of a quantum channel
to preserve quantum correlations: the coherent information of the channel. The way wearrive at this measure is similar to how we did for the mutual information of a quantumc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

13.5. COHERENT INFORMATION OF A QUANTUM CHANNEL

409

channel. Alice prepares a pure state φAA0 and inputs the A0 system to a quantum channel
NA0 →B . This transmission leads to a bipartite state ρAB whereρAB = NA0 →B (φAA0 ).

(13.146)

The coherent information of the state that arises from the channel is as follows: I(AiB)ρ =H(B)ρ − H(AB)ρ , leading to our next definition.
Definition 13.5.1 (Coherent Information of a Quantum Channel) The coherent information Q(N ) of a quantum channel is the maximum of the coherent information withrespect to all input pure states:
Q(N ) ≡ max I(AiB)ρ .
φAA0(13.147)

The coherent information of a quantum channel corresponds to an important operational
task (perhaps the most important for quantum information). It is a good lower bound on
the capacity for Alice to transmit quantum information to Bob, but it is actually equal tosuch a quantum communication capacity of a quantum channel in some special cases. We
prove these results in Chapter 24.
Exercise 13.5.1 Let Ic (ρ, N ) denote the coherent information of a channel N when stateρ is its input:
Ic (ρ, N ) ≡ H(N (ρ)) − H(N c (ρ)),(13.148)
where N c is a channel complementary to the original channel N . Show that
Q(N ) = max Ic (ρ, N ).
ρ

(13.149)

An equivalent way of writing the above expression on the right-hand side ismax [H(B)ψ − H(E)ψ ] ,φAA0(13.150)

where |ψiABE ≡ UAN0 →BE |φiAA0 and UAN0 →BE is an isometric extension of the channel N .
The following property points out that the coherent information of a channel is always
non-negative, even though the coherent information of any given state can sometimes be
negative.
Property 13.5.1 (Non-Negativity of Channel Coherent Information) The coherent
information Q(N ) of a quantum channel N is non-negative:
Q(N ) ≥ 0.
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CHAPTER 13. THE INFORMATION OF QUANTUM CHANNELS

Proof. We can choose the input state φAA0 to be a product state of the form ψA ⊗ ϕA0 . The
coherent information of this state vanishes:
I(AiB)ψ⊗N (ϕ) = H(B)N (ϕ) − H(AB)ψ⊗N (ϕ)= H(B)N (ϕ) − H(A)ψ − H(B)N (ϕ)= 0.

(13.152)(13.153)(13.154)

The first equality follows by evaluating the coherent information for the product state. The
second equality follows because the state on AB is product. The last equality follows because the state on A is pure. Non-negativity then holds because the coherent informationof a channel can only be greater than or equal to this amount, given that it involves amaximization over all input states and the above state is a particular input state.

13.5.1

Additivity of Coherent Information for Some Channels

The coherent information of a quantum channel is generally not additive for arbitrary quantum channels. One might potentially view this situation as unfortunate, but it implies thatquantum Shannon theory is a richer theory than its classical counterpart. Attempts to understand why and how this quantity is not additive have led to many breakthroughs (seeSection 24.8).
Degradable quantum channels form a special class of channels for which the coherent
information is additive. These channels have a property that is analogous to a property ofthe degraded wiretap channels from Section 13.2. To understand this property, recall thatany quantum channel NA0 →B has a complementary channel NAc 0 →E , realized by considering
an isometric extension of the channel and tracing over Bob’s system.
Definition 13.5.2 (Degradable Quantum Channel) A quantum channel NA0 →B is degradable if there exists a degrading channel DB→E such that
NAc 0 →E (ρA0 ) = DB→E (NA0 →B (ρA0 )),(13.155)

for any input state ρA0 and where NAc 0 →E is a channel complementary to NA0 →B .
The intuition behind a degradable quantum channel is that the channel from Alice to
Eve is noisier than the channel from Alice to Bob, in the sense that Bob can simulate thechannel to Eve by applying a degrading channel to his system. The picture to consider forthe analysis of additivity is the same as that in Figure 13.5.
There are also antidegradable channels, defined in the opposite way:
Definition 13.5.3 (Antidegradable Quantum Channel) A quantum channel NA0 →B isantidegradable if there exists a degrading channel DE→B such that
DE→B (NAc 0 →E (ρA0 )) = NA0 →B (ρA0 ),(13.156)

for any input state ρA0 and where NAc 0 →E is a channel complementary to NA0 →B .
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13.5. COHERENT INFORMATION OF A QUANTUM CHANNEL

411

Theorem 13.5.1 (Additivity for Degradable Quantum Channels) Let N and M be
any quantum channels that are degradable. Then the coherent information of the tensorproduct channel N ⊗ M is the sum of their individual coherent informations:
Q(N ⊗ M) = Q(N ) + Q(M).

(13.157)

Proof. We leave the proof of the inequality Q(N ⊗ M) ≥ Q(N ) + Q(M) as Exercise 13.5.3
below, and we prove the non-trivial inequality Q(N ⊗ M) ≤ Q(N ) + Q(M) that holds whenquantum channels N and M are degradable. Consider a pure state φAA01 A02 that serves as
the input to the two quantum channels. Let UAN0 →B1 E1 denote an isometric extension of the
1
first channel and let UAM0 →B2 E2 denote an isometric extension of the second channel. Let
2

σAB1 E1 A02 ≡ U N φ U N
θAA01 B2 E2 ≡ U M φ U

†

,

M †

(13.158)
,

 
†
† 
ρAB1 E1 B2 E2 ≡ U N ⊗ U M φ U N ⊗ U M.

(13.159)(13.160)

We need to show that Q(N ⊗ M) = Q(N ) + Q(M) when both channels are degradable.
Furthermore, let ρAB1 E1 B2 E2 be a state that maximizes Q(N ⊗ M). Consider the following
chain of inequalities:
Q(N ⊗ M) = I(AiB1 B2 )ρ
= H(B1 B2 )ρ − H(AB1 B2 )ρ
= H(B1 B2 )ρ − H(E1 E2 )ρ
= H(B1 )ρ − H(E1 )ρ + H(B2 )ρ − H(E2 )ρ
− [I(B1 ; B2 )ρ − I(E1 ; E2 )ρ ]≤ H(B1 )ρ − H(E1 )ρ + H(B2 )ρ − H(E2 )ρ
= H(B1 )σ − H(AA02 B1 )σ + H(B2 )θ − H(AA01 B2 )θ
= I(AA02 iB1 )σ + I(AA01 iB2 )θ
≤ Q(N ) + Q(M).

(13.161)(13.162)(13.163)(13.164)(13.165)(13.166)(13.167)(13.168)

The first equality follows from the definition of Q(N ⊗ M) and because we set ρ to be astate that maximizes the tensor-product channel coherent information. The second equality
follows from the definition of coherent information, and the third equality follows because
the state ρ is pure on systems AB1 E1 B2 E2 . The fourth equality follows by expanding theentropies in the previous line. The first inequality follows because there is a degrading
channel from both B1 to E1 and B2 to E2 , allowing us to apply the quantum data-processing
inequality to get I(B1 ; B2 )ρ ≥ I(E1 ; E2 )ρ . The fifth equality follows because the entropiesof ρ, σ, and θ on the given reduced systems are equal and because the state σ on systems
AA02 B1 E1 is pure and the state θ on systems AA01 B2 E2 is pure. The last equality follows fromthe definition of coherent information, and the final inequality follows because the coherent
informations are less than their respective maximizations over all possible input states.
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Corollary 13.5.1 The regularized coherent information of a degradable quantum channel isequal to its coherent information: Qreg (N ) = Q(N ).
Proof. A proof of this property uses the same induction argument as in Corollary 13.1.1
and exploits the additivity property in Theorem 13.5.1 above.
Exercise 13.5.2 Consider the quantum erasure channel where the erasure parameter ε ∈[0, 1/2]. Find the channel that degrades this one, reproducing the channel from input toenvironment.
Exercise 13.5.3 (Superadditivity of Coherent Information) Show that the coherent
information of the tensor-product channel N ⊗ M is never less than the sum of their individual coherent informations: Q(N ⊗ M) ≥ Q(N ) + Q(M).
Exercise 13.5.4 Prove using monotonicity of relative entropy that the coherent information
is subadditive for a degradable channel: Q(N ) + Q(M) ≥ Q(N ⊗ M).
Exercise 13.5.5 Consider a quantity known as the reverse coherent information:
Qrev (N ) ≡ max I(BiA)ω ,φAA0(13.169)

where ωAB ≡ NA0 →B (φAA0 ). Show that the reverse coherent information is additive withrespect to any quantum channels N and M: Qrev (N ⊗ M) = Qrev (N ) + Qrev (M).
Exercise 13.5.6 Prove that the coherent information of an antidegradable channel is equal
to zero. (Hint: Consider using the identity from Exercise 11.6.6.)
Exercise 13.5.7 Prove that an entanglement-breaking channel is antidegradable.

13.5.2

Optimizing the Coherent Information

We would like to determine how difficult it is to maximize the coherent information of aquantum channel. For general channels, this problem is difficult, but it turns out to bestraightforward for the class of degradable quantum channels. Theorem 13.5.2 below states
an important property of the coherent information Q(N ) of a degradable quantum channel
N that allows us to answer this question. The theorem states that the coherent information
Q(N ) of a degradable quantum channel is a concave function of the input density operator
ρA0 over which we maximize it. In particular, this result implies that a local maximum of
the coherent information Q(N ) is a global maximum since the set of density operators isconvex, and the optimization problem is therefore a straightforward computation that can
exploit convex optimization methods. The theorem below exploits the characterization of
the channel coherent information from Exercise 13.5.1.
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13.6. PRIVATE INFORMATION OF A QUANTUM CHANNEL

413

Theorem 13.5.2 Suppose that a quantum channel N is degradable. Then the coherent
information Ic (ρ, N ) is concave in the input density operator:!
XX
pX (x)Ic (ρx , N ) ≤ Ic
pX (x)ρx , N ,(13.170)xxwhere pX (x) is a probability density function and each ρx is a density operator.
Proof. Consider the following states:XσXB ≡pX (x)|xihx|X ⊗ N (ρx ),(13.171)

x

θXE ≡

X

pX (x)|xihx|X ⊗ (T ◦ N ) (ρx ) ,(13.172)x

where T is a degrading channel for the channel N , so that T ◦ N = N c . Then the following
statements hold:
I(X; B)σ ≥ I(X; E)θ
∴ H(B)σ − H(B|X)σ ≥ H(E)θ − H(E|X)θ
∴ H(B)σ − H(E)θ ≥ H(B|X)σ − H(E|X)θ
!!
∴

H

N

X

pX (x)ρx(13.173)(13.174)(13.175)

!!
−H

N

c

X

x

pX (x)ρx

x≥

X

pX (x) [H (N (ρx )) − H (N c (ρx ))] (13.176)

x

!
∴

Ic

X
x

pX (x)ρx , N≥

X

pX (x)Ic (ρx , N ) .

(13.177)x

The first statement is the crucial one and follows from the quantum data-processing inequality and the fact that the map T degrades Bob’s state to Eve’s state. The second andthird statements follow from the definition of quantum mutual information and rearranging
entropies. The fourth statement follows by plugging in the density operators into the entropies in the previous statement. The final statement follows from the alternate definitionof coherent information in Exercise 13.5.1.

13.6 Private Information of a Quantum Channel
The private information of a quantum channel is the last information measure that we
consider in this chapter. Alice would like to establish classical correlations with Bob, but
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CHAPTER 13. THE INFORMATION OF QUANTUM CHANNELS

does not want the environment of the channel to have access to these classical correlations.
The ensemble that she prepares is similar to the one we considered for the Holevo information.
The expected density operator of the ensemble she prepares is a classical–quantum state of
the form
XρXA0 ≡pX (x)|xihx|X ⊗ ρxA0 .
(13.178)x

Sending the A0 system through an isometric extension UAN0 →BE of a quantum channel N leads
to a state ρXBE . A good measure of the private classical correlations that she can establish
with Bob is the difference of the classical correlations she can establish with Bob, less theclassical correlations that Eve can obtain: I(X; B)ρ −I(X; E)ρ , leading to our next definition(Chapter 23 discusses the operational task corresponding to this information quantity).
Definition 13.6.1 (Private Information of a Quantum Channel) The private information P (N ) of a quantum channel N is defined as follows:
P (N ) ≡ max I(X; B)ρ − I(X; E)ρ ,ρXA0(13.179)

where the maximization is with respect to all states of the form in (13.178) and the entropic
quantities are evaluated with respect to the state UAN0 →BE (ρXA0 ).
Property 13.6.1 The private information P (N ) of a quantum channel N is non-negative:
P (N ) ≥ 0.

(13.180)

Proof. We can choose the input state ρXA0 to be a state of the form |0ih0|X ⊗ ψA0 , whereψA0 is pure. The private information of the output state vanishes
I(X; B)|0ih0|⊗N (ψ) − I(X; E)|0ih0|⊗N c (ψ) = 0.

(13.181)The equality follows just by evaluating both mutual informations for the above state. The
above property then holds because the private information of a channel can only be greaterthan or equal to this amount, given that it involves a maximization over all input states andthe above state is a particular input state.
The regularized private information is as follows:1
P (N ⊗n ).
n→∞ nPreg (N ) = lim

13.6.1(13.182)

Private Information and Coherent Information

The private information of a quantum channel bears a special relationship to that channel’scoherent information. It is always at least as great as the coherent information of the channel
and is equal to it for certain channels. The following theorem states the former inequality,
and the next theorem states the equivalence for degradable quantum channels.
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13.6. PRIVATE INFORMATION OF A QUANTUM CHANNEL

415

Theorem 13.6.1 The private information P (N ) of any quantum channel N is never smaller
than its coherent information Q(N ):Q(N ) ≤ P (N ).

(13.183)

Proof. We can see this relation through a few steps. Consider a pure state φAA0 thatmaximizes the coherent information Q(N ), and let φABE denote the state that arises fromsending the A0 system through an isometric extension UAN0 →BE of the channel N . Let φA0
denote the reduction of this state to the A0 system. Suppose that it admits the following
spectral decomposition:X
φA0 =pX (x)|φx ihφx |A0 .
(13.184)x

We can create an augmented classical–quantum state that correlates a classical variable with
the index x:
X
pX (x)|xihx|X ⊗ |φx ihφx |A0 .
(13.185)
σXA0 ≡x

Let σXBE denote the state that results from sending the A0 system through an isometricextension UAN0 →BE of the channel N . Then the following chain of inequalities holds:
Q(N ) = I(AiB)φ= H(B)φ − H(E)φ= H(B)σ − H(E)σ= H(B)σ − H(B|X)σ − H(E)σ + H(B|X)σ= I(X; B)σ − H(E)σ + H(E|X)σ= I(X; B)σ − I(X; E)σ≤ P (N ).

(13.186)(13.187)(13.188)(13.189)(13.190)(13.191)(13.192)

The first equality follows from evaluating the coherent information of the state φABE thatmaximizes the coherent information of the channel. The second equality follows because thestate φABE is pure. The third equality follows from the definition of σXBE in (13.185) and its
relation to φABE . The fourth equality follows by adding and subtracting H(B|X)σ , and thenext one follows from the definition of the mutual information I(X; B)σ and the fact thatthe state of σXBE on systems B and E is pure when conditioned on X. The last equality
follows from the definition of the mutual information I(X; E)σ . The final inequality follows
because the state σXBE is a particular state of the form in (13.178), and P (N ) involves amaximization over all states of that form.
Theorem 13.6.2 Suppose that a quantum channel N is degradable. Then its private information P (N ) is equal to its coherent information Q(N ):P (N ) = Q(N ).
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Proof. We prove the inequality P (N ) ≤ Q(N ) for degradable quantum channels because
we have already proven that Q(N ) ≤ P (N ) for any quantum channel N . Consider aclassical–quantum state ρXBE that arises from transmitting the A0 system of the state in(13.178) through an isometric extension UAN0 →BE of the channel. Suppose further that thisstate maximizes P (N ). We can take a spectral decomposition of each ρxA0 in the ensemble
to be as follows:
X
pY |X (y|x)ψAx,y0 ,(13.194)
ρxA0 =ywhere each state ψAx,y0 is pure. We can construct the following extension of the state ρXBE :σXY BE ≡X

pY |X (y|x)pX (x)|xihx|X ⊗ |yihy|Y ⊗ UAN0 →BE (ψAx,y0 ).

(13.195)x,yThen the following chain of inequalities holds:
P (N ) = I(X; B)ρ − I(X; E)ρ
= I(X; B)σ − I(X; E)σ= I(XY ; B)σ − I(Y ; B|X)σ − [I(XY ; E)σ − I(Y ; E|X)σ ]= I(XY ; B)σ − I(XY ; E)σ − [I(Y ; B|X)σ − I(Y ; E|X)σ ] .

(13.196)(13.197)(13.198)(13.199)

The first equality follows from the definition of P (N ) and because we set ρ to be the statethat maximizes it. The second equality follows because ρXBE = TrY {σXY BE }. The third
equality follows from the chain rule for quantum mutual information. The fourth equality
follows from a rearrangement of entropies. Continuing,≤ I(XY ; B)σ − I(XY ; E)σ= H(B)σ − H(B|XY )σ − H(E)σ + H(E|XY )σ= H(B)σ − H(B|XY )σ − H(E)σ + H(B|XY )σ= H(B)σ − H(E)σ≤ Q(N ).

(13.200)(13.201)(13.202)(13.203)(13.204)

The first inequality (the crucial one) follows because there is a degrading channel from B toE, so that the quantum data-processing inequality implies that I(Y ; B|X)σ ≥ I(Y ; E|X)σ .
The second equality is a rewriting of entropies, the third follows because the state of σ on
systems B and E is pure when conditioned on classical systems X and Y , and the fourth
follows by canceling entropies. The last inequality follows because the entropy difference
H(B)σ − H(E)σ is less than the maximum of that difference over all possible input states.

13.6.2

Additivity of Private Information of Degradable Channels

The private information of general quantum channels is not additive, but it is so in thecase of degradable quantum channels. The method of proof is somewhat similar to that in
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13.6. PRIVATE INFORMATION OF A QUANTUM CHANNEL

417

Figure 13.6: This figure displays the scenario for determining whether the private information of two
quantum channels N1 and N2 is additive. The question of additivity is equivalent to the possibility ofquantum correlations between channel inputs being able to enhance the private information of two quantum
channels. The result stated in Theorem 13.6.3 is that the private information is additive for any two
degradable quantum channels, so that quantum correlations cannot enhance it in this case.

the proof of Theorem 13.5.1, essentially exploiting the degradability property. Figure 13.6
illustrates the setting to consider for additivity of the private information.
Theorem 13.6.3 (Additivity for Degradable Quantum Channels) Let N and M be
any quantum channels that are degradable. Then the private information of the tensorproduct channel N ⊗ M is the sum of their individual private informations:
P (N ⊗ M) = P (N ) + P (M).

(13.205)

P (N ⊗ M) = Q(N ⊗ M) = Q(N ) + Q(M).

(13.206)

Furthermore, it holds that

Proof. We first prove the more trivial inequality P (N ⊗ M) ≥ P (N ) + P (M). Let ρX1 A01
and σX2 A02 be states of the form in (13.178) that maximize the respective private informations
P (N ) and P (M). Let θX1 X2 A01 A02 be the tensor product of these two states: θ = ρ ⊗ σ. Let
ρX1 B1 E1 and σX2 B2 E2 be the states that arise from sending ρX1 A01 and σX2 A02 through therespective isometric extensions UAN0 →B1 E1 and UAM0 →B2 E2 . Let θX1 X2 B1 B2 E1 E2 be the state that12arises from sending θX1 X2 A01 A02 through the tensor-product channel UAN0 →B1 E1 ⊗ UAM0 →B2 E2 .
12Then
P (N ) + P (M)= I(X1 ; B1 )ρ − I(X1 ; E1 )ρ + I(X2 ; B2 )σ − I(X2 ; E2 )σ= I(X1 X2 ; B1 B2 )θ − I(X1 X2 ; E1 E2 )θ
≤ P (N ⊗ M).
c 2019(13.207)(13.208)(13.209)

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CHAPTER 13. THE INFORMATION OF QUANTUM CHANNELS

The first equality follows from the definition of the private informations P (N ) and P (M)and by evaluating them on the respective states ρX1 A01 and σX2 A02 that maximize them.
The second equality follows because the mutual information is additive on tensor-productstates (see Exercise 11.6.8). The final inequality follows because the state θX1 X2 B1 B2 E1 E2 is
a particular state of the form needed in the maximization of the private information of thetensor-product channel N ⊗ M.
We now prove the inequality P (N ⊗ M) ≤ P (N ) + P (M). Let ρXA01 A02 be a state thatmaximizes P (N ⊗ M) where
XρXA01 A02 ≡pX (x)|xihx|X ⊗ ρxA01 A02 ,(13.210)x

and let ρXB1 B2 E1 E2 be the state that arises from sending ρXA01 A02 through the tensor-productchannel UAN0 →B1 E1 ⊗ UAM0 →B2 E2 . Consider a spectral decomposition of each state ρxA0 A0 :
1
2
1 2
X
(13.211)
ρxA01 A02 =pY |X (y|x)ψAx,y0 A0 ,12ywhere each state ψAx,y0 A0 is pure. Let σXY A01 A02 be an extension of ρXA01 A02 where
1 2
X
pY |X (y|x)pX (x)|xihx|X ⊗ |yihy|Y ⊗ ψAx,y0 A0 ,σXY A01 A02 ≡1

2(13.212)x,yand let σXY B1 E1 B2 E2 be the state that arises from sending σXY A01 A02 through the tensorproduct channel UAN0 →B1 E1 ⊗ UAM0 →B2 E2 . Consider the following chain of inequalities:
1

2

P (N ⊗ M)= I(X; B1 B2 )ρ − I(X; E1 E2 )ρ
= I(X; B1 B2 )σ − I(X; E1 E2 )σ= I(XY ; B1 B2 )σ − I(XY ; E1 E2 )σ− [I(Y ; B1 B2 |X)σ − I(Y ; E1 E2 |X)σ ]≤ I(XY ; B1 B2 )σ − I(XY ; E1 E2 )σ= H(B1 B2 )σ − H(B1 B2 |XY )σ − H(E1 E2 )σ + H(E1 E2 |XY )σ= H(B1 B2 )σ − H(B1 B2 |XY )σ − H(E1 E2 )σ + H(B1 B2 |XY )σ= H(B1 B2 )σ − H(E1 E2 )σ= H(B1 )σ − H(E1 )σ + H(B2 )σ − H(E2 )σ− [I(B1 ; B2 )σ − I(E1 ; E2 )σ ]≤ H(B1 )σ − H(E1 )σ + H(B2 )σ − H(E2 )σ≤ Q(N ) + Q(M)= P (N ) + P (M).

(13.213)(13.214)(13.215)(13.216)(13.217)(13.218)(13.219)(13.220)(13.221)(13.222)(13.223)(13.224)

The first equality follows from the definition of P (N ⊗ M) and from evaluating it on thestate ρ that maximizes it. The second equality follows because the state σXY B1 E1 B2 E2 is
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13.7. SUMMARY

419

equal to the state ρXB1 E1 B2 E2 after tracing out the system Y . The third equality follows
from the chain rule for mutual information: I(XY ; B1 B2 ) = I(Y ; B1 B2 |X) + I(X; B1 B2 ).
It holds that I(Y ; B1 B2 |X)σ ≥ I(Y ; E1 E2 |X)σ because there is a degrading channel from
B1 to E1 and from B2 to E2 . Then the first inequality follows because I(Y ; B1 B2 |X)σ −I(Y ; E1 E2 |X)σ ≥ 0. The fourth equality follows by expanding the mutual informations, andthe fifth equality follows because the state σ on systems B1 B2 E1 E2 is pure when conditioning
on the classical systems X and Y . The sixth equality follows from algebra, and the seventh
follows by rewriting the entropies. It holds that I(B1 ; B2 )σ ≥ I(E1 ; E2 )σ because there
is a degrading channel from B1 to E1 and from B2 to E2 . Then the inequality follows
because I(B1 ; B2 )σ − I(E1 ; E2 )σ ≥ 0. The third inequality follows because the entropy
difference H(Bi ) − H(Ei ) is always less than the coherent information of the channel, andthe final equality follows because the coherent information of a channel is equal to its private
information when the channel is degradable (Theorem 13.6.2).
Corollary 13.6.1 Suppose that a quantum channel N is degradable. Then the regularized
private information Preg (N ) of the channel is equal to its private information P (N ):Preg (N ) = P (N ).

(13.225)

Proof. A proof follows by the same induction argument as in Corollary 13.1.1 and by
exploiting the result of Theorem 13.6.3 and the fact that the tensor power channel N ⊗n isdegradable if the original channel N is.

13.7 Summary
We conclude this chapter with a table that summarizes the main results regarding the mutual information of a classical channel I(pY |X ), the private information of a classical wiretap
channel P (pY,Z|X ), the Holevo information of a quantum channel χ(N ), the mutual information of a quantum channel I(N ), the coherent information of a quantum channel Q(N ),and the private information of a quantum channel P (N ). The table exploits the following
definitions:XρXA0 ≡pX (x)|xihx|X ⊗ φxA0 ,(13.226)
X
σXA0 ≡pX (x)|xihx|X ⊗ ρxA0 .
(13.227)
Quantity
I(pY |X )P (pY,Z|X )
χ(N )
I(N )
Q(N )
P (N )
c 2019

Input
pX
pX
ρXA0
φAA0
φAA0
σXA0

OutputpX pY |X
pX pY,Z|X
NA0 →B (ρXA0 )NA0 →B (φAA0 )NA0 →B (φAA0 )UAN0 →BE (σXA0 )FormulamaxpX I(X; Y )maxpU,X I(U ; Y ) − I(U ; Z)maxρ I(X; B)
maxφ I(A; B)
maxφ I(AiB)
maxσ I(X; B) − I(X; E)

Single-letter
all channels
all channels
some channels
all channels
degradable
degradable

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CHAPTER 13. THE INFORMATION OF QUANTUM CHANNELS

13.8 History and Further Reading
Boyd and Vandenberghe (2004) is a good reference for the theory and practice of convex
optimization, which is helpful for computing capacity formulas. Wyner (1975) introduced
the classical wiretap channel and proved that the private information is additive for degradedwiretap channels. Csiszár and Körner (1978) proved that the private information is additivefor general wiretap channels. Holevo (1998) and Schumacher and Westmoreland (1997) provided an operational interpretation of the Holevo information of a quantum channel. Shor
(2002a) showed the additivity of the Holevo information for entanglement-breaking channels.
Adami and Cerf (1997) introduced the mutual information of a quantum channel, and they
proved several of its important properties that appear in this chapter: non-negativity, additivity, and concavity. Bennett et al. (1999, 2002) later gave an operational interpretation
for this information quantity as the entanglement-assisted classical capacity of a quantum
channel. Lloyd (1997), Shor (2002b), and Devetak (2005) gave increasingly rigorous proofs
that the coherent information of a quantum channel is an achievable rate for quantum communication. Devetak and Shor (2005) showed that the coherent information of a quantum
channel is additive for degradable channels. Yard et al. (2008) proved that the coherent information of a quantum channel is a concave function of the input state whenever the channel
is degradable. Devetak et al. (2006); Garcı́a-Patrón et al. (2009) discussed the reverse coherent information of a quantum channel and showed that it is additive for all quantum
channels. Devetak (2005) and Cai et al. (2004) independently introduced the private classical capacity of a quantum channel, and both papers proved that it is an achievable rate
for private classical communication over a quantum channel. Smith (2008) showed that theprivate classical information is additive and equal to the coherent information for degradable
quantum channels.

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 14

Classical Typicality
This chapter begins our first technical foray into the asymptotic theory of information. We
start with the classical setting in an effort to build up our intuition of asymptotic behavior
before delving into the asymptotic theory of quantum information.
The central concept of this chapter is the asymptotic equipartition property. The nameof this property may sound somewhat technical at first, but it is merely an application of
the law of large numbers to a sequence drawn independently and identically from a distribution pX (x) for some random variable X. The asymptotic equipartition property reveals that
we can divide sequences into two classes when their length becomes large: those that are
overwhelmingly likely to occur and those that are overwhelmingly likely not to occur. The
sequences that are likely to occur are the typical sequences, and the ones that are not likely
to occur are the atypical sequences. Additionally, the size of the set of typical sequences
is exponentially smaller than the size of the set of all sequences whenever the random variable generating the sequences is not uniform. These properties are an example of a moregeneral mathematical phenomenon known as “measure concentration,” in which a smoothfunction over a high-dimensional space or over a large number of random variables tends toconcentrate around a constant value with high probability.
The asymptotic equipartition property immediately leads to the intuition behind Shannon’s scheme for compressing classical information. The scheme first generates a realizationof a random sequence and asks the question: Is the produced sequence typical or atypical? If
it is typical, compress it. Otherwise, throw it away. The error probability of this compression
scheme is non-zero for any fixed length of a sequence, but it vanishes in the asymptotic limit
because the probability of the sequence being in the typical set converges to one, while theprobability that it is in the atypical set converges to zero. This compression scheme has astraightforward generalization to the quantum setting, in which we wish to compress qubits
instead of classical bits.
The bulk of this chapter is here to present the many technical details needed to makerigorous statements in the asymptotic theory of information. We begin with an example,
follow with the formal definition of a typical sequence and a typical set, and prove thethree important properties of a typical set. We then discuss other forms of typicality such
421

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CHAPTER 14. CLASSICAL TYPICALITY

Figure 14.1: This figure depicts the sample entropy of a realization of a random binary sequence as afunction of its length. The source is a binary random variable with distribution ( 34 , 14 ). For the realizations
generated, the sample entropy of the sequences is converging to the true entropy of the source.

as joint typicality and conditional typicality. These other notions turn out to be useful for
proving Shannon’s classical capacity theorem as well (recall that Shannon’s theorem gives theultimate rate at which a sender can transmit classical information over a classical channel
to a receiver). We also introduce the method of types, which is a powerful technique inclassical information theory, and apply this method in order to develop a stronger notionof typicality. The chapter then features a development of the strong notions of joint andconditional typicality and ends with a concise proof of Shannon’s channel capacity theorem.

14.1 An Example of Typicality
Suppose that Alice possesses a binary random variable X that takes the value zero withprobability 34 and the value one with probability 14 . Such a random source might produce
the following sequence:
0110001101,(14.1)
if we generate ten independent realizations. The probability that such a sequence occurs is 5  5
1
3,(14.2)
4
4
determined simply by counting the number of ones and zeros in the above sequence and by
applying the assumption that the source is i.i.d.
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14.2. WEAK TYPICALITY

423

The information content of the above sequence is the negative logarithm of its probability
divided by its length: 5  5 !
 
 
1
3
5
1
5
3
1
log
= − log−log≈ 1.207.
(14.3)
−
10
4
4
10
4
10
4
We also refer to this quantity as the sample entropy. The true entropy of the source is
 
 
1
1
3
3
− log
− log
≈ 0.8113.
(14.4)
44
4
4
We would expect that the sample entropy of a random sequence tends to approach thetrue entropy as its length increases because the number of zeros should be approximatelyn (3/4) and the number of ones should be approximately n (1/4) according to the law oflarge numbers.
Another sequence of length 100 might be as follows:
00000000100010001000000000000110011010000000100000
00000110101001000000010000001000000010000100010000,
featuring 81 zeros and 19 ones. Its sample entropy is 19  81 !
 
 
1
1
3
19
1
81
3
−
log
=−
log
−
log
≈ 0.7162.
100
4
4
100
4
100
4

(14.5)(14.6)

The above sample entropy is closer to the true entropy in (14.4) than the sample entropy ofthe previous sequence, but it still deviates significantly from it.
Figure 14.1 continues this game by generating random sequences according to the distribution 34 , 14 , and the result is that a concentration around the true entropy begins tooccur around n ≈ 106 . That is, it becomes highly likely that the sample entropy of a random
sequence is close to the true entropy if we increase the length of the sequence, and this holds
for the realizations generated in Figure 14.1.

14.2 Weak Typicality
This first section generalizes the example from the introduction to an arbitrary discrete,
finite-cardinality random variable. Our first notion of typicality is the same as discussed in
the example—we define a sequence to be typical if its sample entropy is close to the true
entropy of the random variable that generates it. This notion of typicality is known as weaktypicality. Section 14.7 introduces another notion of typicality that implies weak typicality,
but the implication does not hold in the other direction. For this reason, we distinguish thetwo different notions of typicality as weak typicality and strong typicality.
Suppose that a random variable X takes values in an alphabet X with cardinality |X |.
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CHAPTER 14. CLASSICAL TYPICALITY

samples independently from the distribution of random variable X and emits n realizations
x1 , . . . , xn . Let X n ≡ X1 · · · Xn denote the n random variables that describe the information
source, and let xn ≡ x1 · · · xn denote an emitted realization of X n . The probability pX n (xn )of a particular string xn is as follows:pX n (xn ) ≡ pX1 ,...,Xn (x1 , . . . , xn ),(14.7)and pX n (xn ) factors as follows because the source is i.i.d.:pX n (xn ) = pX1 (x1 ) · · · pXn (xn ) = pX (x1 ) · · · pX (xn ) =nQ

pX (xi ).

(14.8)

i=1

Roughly speaking, we expect a long string xn to contain about npX (a1 ) occurrences ofsymbol a1 , npX (a2 ) occurrences of symbol a2 , etc., when n is large, due to the law of large
numbers. If this is occurring, the probability that the source emits a particular string xn isapproximately
pX n (xn ) = pX (x1 ) · · · pX (xn ) ≈ pX (a1 )npX (a1 ) · · · pX (a|X | )npX (a|X | ) ,(14.9)and the information content of a given string is thus roughly
|X |
X
1
n
pX (ai ) log(pX (ai )) = H(X).
− log(pX n (x )) ≈ −ni=1(14.10)

The above intuitive argument shows that the information content divided by the length of
the sequence is roughly equal to the entropy in the limit of large n. It then makes sense tothink of this quantity as the sample entropy of the sequence xn .
Definition 14.2.1 (Sample Entropy) The sample entropy H(xn ) of a sequence xn withrespect to a probability distribution pX (x) is defined as follows:1
H(xn ) ≡ − log(pX n (xn )),nwhere pX n (xn ) =(14.11)Qn

i=1 pX (xi ).

This definition of sample entropy leads us to our first important definitions in asymptotic
information theory.
Definition 14.2.2 (Typical Sequence) A sequence xn is δ-typical if its sample entropyH(xn ) is δ-close to the entropy H(X) of random variable X, where this random variable isthe source of the sequence.
n

Definition 14.2.3 (Typical Set) The δ-typical set TδX is the set of all δ-typical sequences xn :nTδX ≡ xn : H(xn ) − H(X) ≤ δ .
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14.3. PROPERTIES OF THE TYPICAL SET

425

Figure 14.2: This figure depicts the idea that the typical set is exponentially smaller than the set ofnall sequences because |X | = 2n log|X | > 2nH(X) whenever X is not a uniform random variable. Yet, thisexponentially smaller set contains nearly all of the probability.

14.3 Properties of the Typical Set
The set of typical sequences enjoys three useful and beautifully surprising properties that
occur when we step into the “land of large numbers.” We can summarize these properties
as follows: the typical set contains almost all the probability, yet it is exponentially smaller
than the set of all sequences, and each typical sequence has almost uniform probability.
Figure 14.2 attempts to depict the main idea of the typical set.
Property 14.3.1 (Unit Probability) The typical set asymptotically has probability one.
So as n becomes large, it is highly likely that a source emits a typical sequence. We formallystate this property as follows:Xn
pX n (xn ) ≥ 1 − ε,(14.13)
Pr X n ∈ TδX =xn ∈TδXn

for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
nProperty 14.3.2 (Exponentially Smaller Cardinality) The number TδX of δ-typicalsequences is exponentially smaller than the total number |X |n of sequences for every randomvariable X besides the uniform random variable. We formally state this property as follows:TδXn≤ 2n(H(X)+δ) .

(14.14)

We can also bound the size of the δ-typical set from below:
TδXn≥ (1 − ε) 2n(H(X)−δ) ,(14.15)for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
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Property 14.3.3 (Equipartition) The probability of a particular δ-typical sequence xn isapproximately uniform:
2−n(H(X)+δ) ≤ pX n (xn ) ≤ 2−n(H(X)−δ) .

(14.16)This last property represents the “equipartition” in “asymptotic equipartition property” because all typical sequences occur with nearly the same probability when n is large.
The size TδX of the δ-typical set is approximately equal to the total number |X |n ofsequences only when random variable X is uniform because H(X) = log |X | in such a caseand thusnTδX ≤ 2n(H(X)+δ) = 2n(log|X |+δ) = |X |n · 2nδ ≈ |X |n .
(14.17)n

14.3.1

Proofs of Typical Set Properties

Proof of the Unit Probability Property (Property 14.3.1). The weak law of large
numbers states that the sample mean converges in probability to the expectation. More precisely, consider a sequence of i.i.d. random variables Z1 , . . . , Zn that each have expectation µ.
The sample average of this sequence is as follows:
n

1X
Z=
Zi .
n i=1

(14.18)The formal statement of the law of large numbers is that ∀ε ∈ (0, 1), δ > 0 ∃n0 : ∀n > n0
(14.19)Pr Z − µ ≤ δ ≥ 1 − ε.
We can now consider the sequence of random variables − log(pX (X1 )), . . . , − log(pX (Xn )).
The sample average of this sequence is equal to the sample entropy of X n :
n

1
1X
−
log(pX (Xi )) = − log(pX n (X n ))n i=1n= H(X n ).

(14.20)(14.21)

Recall from (10.3) that the expectation of the random variable − log(pX (X)) is equal to the
Shannon entropy:
EX {− log(pX (X))} = H(X).
(14.22)
Then we can apply the law of large numbers and find that ∀ε ∈ (0, 1), δ > 0 ∃n0 : ∀n > n0
such that
Pr H(X n ) − H(X) ≤ δ ≥ 1 − ε.
(14.23)The event H (X n ) − H(X) ≤ δ is precisely the condition for a random sequence X n tonbe in the typical set TδX , and the probability of this event goes to one as n becomes large.

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14.4. APPLICATION: DATA COMPRESSION

427

Proof of the Exponentially Smaller Cardinality Property (Property 14.3.2). Consider the following chain of inequalities:
1=X
xn ∈X npX n (xn ) ≥X
xn ∈TδX

pX n (xn )nX≥2−n(H(X)+δ) = 2−n(H(X)+δ) TδXn. (14.24)

nxn ∈TδX

The first inequality uses the fact that the probability of the typical set is smaller thanthe probability of the set of all sequences. The second inequality uses the equipartition
property of typical sets (proved below). After rearranging the leftmost side of (14.24) with
its rightmost side, we find thatnTδX ≤ 2n(H(X)+δ) .
(14.25)The second part of the property follows because the “unit probability” property holds for
sufficiently large n. Then the following chain of inequalities holds:Xn
1 − ε ≤ Pr X n ∈ TδX =pX n (xn )xn ∈TδXn≤X

2−n(H(X)−δ) = 2−n(H(X)−δ) TδXn. (14.26)nxn ∈TδX

We can then bound the size of the typical set as follows:TδXn≥ 2n(H(X)−δ) (1 − ε) ,(14.27)for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
Proof of the Equipartition Property (Property 14.3.3). The property follows immediately by manipulating the definition of a typical set.

14.4 Application: Data Compression
The above three properties of typical sequences immediately give our first application inasymptotic information theory. It is Shannon’s compression protocol, which is a scheme forcompressing the output of an i.i.d. information source.
We begin by defining the information-processing task and a corresponding (n, R, ε) sourcecode. It is helpful to recall the picture in Figure 2.1. An information source outputs asequence xn drawn independently according to the distribution of some random variable X.
A sender Alice encodes this sequence according to some encoding map E where
E : X n → {0, 1}nR .
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CHAPTER 14. CLASSICAL TYPICALITY

The encoding takes elements from the set X n of all sequences to a set {0, 1}nR of size 2nR .
She then transmits the codewords over nR uses of a noiseless classical bit channel. Bob
decodes according to some decoding map D : {0, 1}nR → X n . The probability of error foran (n, R, ε) source code isp(e) ≡ Pr {(D ◦ E) (X n ) 6= X n } ≤ ε,(14.29)where ε ∈ (0, 1). The rate of the source code is equal to the number of channel uses divided
by the length of the sequence, and it is equal to R for the above scheme. A particularcompression rate R is achievable for X if there exists an (n, R + δ, ε) source code for allε ∈ (0, 1), δ > 0, and sufficiently large n. We can now state Shannon’s lossless compression
theorem.
Theorem 14.4.1 (Shannon Compression) The entropy of an information source specified by a discrete random variable X is the smallest achievable rate for compression:
inf {R : R is achievable for X} = H(X).

(14.30)

A proof of this theorem consists of two parts, traditionally called the direct coding theorem and the converse theorem. The direct coding theorem is the direction LHS ≤ RHS—theproof exhibits a coding scheme with an achievable rate and demonstrates that its rate converges to the entropy in the asymptotic limit. The converse theorem is the direction LHS ≥RHS and is a statement of optimality—it establishes that any coding scheme with rate below
the entropy is not achievable. The proofs of each part are usually completely different. We
employ typical sequences and their properties for proving a direct coding theorem, while theconverse part resorts to entropy inequalities from Chapter 10.1 For now, we prove the directcoding theorem and hold off on the converse part until we reach Schumacher compression for
quantum information in Chapter 18. Our main goal here is to illustrate a simple applicationof typical sequences, and we can wait on the converse part because Shannon compression isin some sense a special case of Schumacher compression.
The idea behind the proof of the direct coding theorem is simple: just keep the typicalsequences and throw away the rest. This coding strategy succeeds with asymptoticallyvanishing probability of error because the typical set asymptotically has all of the probability.
Since we are only concerned with error probabilities in communication protocols, it makes
sense that we should only be keeping track of a set where all of the probability concentrates.
We can formally state a proof as follows. Pick an ε ∈ (0, 1), a δ > 0, and a sufficiently
large n such that Property 14.3.1 holds. Consider that Property 14.3.2 then holds so thatthe size of the typical set is no larger than 2n[H(X)+δ] . We choose the encoding to be a
one-to-one function f that maps a typical sequence to a binary sequence in {0, 1}nR , where
R = H(X) + δ. We define f to map any atypical sequence to 0n . This scheme gives up on
encoding the atypical sequences because they have vanishingly small probability. We define
1

The direct part of a quantum coding theorem can employ the properties of typical subspaces (discussed
in Chapter 15), and the proof of a converse theorem for quantum information usually employs the quantum
entropy inequalities from Chapter 11.
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14.5. WEAK JOINT TYPICALITY

429

Figure 14.3: Shannon’s scheme for the compression of classical data. The encoder f is a map from thetypical set to a set of binary sequences of size ≈ 2nH(X) where H(X) is the entropy of the information source.
The map f is invertible on the typical set but maps an atypical sequence to a constant. Alice then transmits
the compressed data over ≈ nH(X) uses of a noiseless classical channel. The inverse map f −1 (the decoder)is the inverse of f on the typical set and decodes to some error sequence otherwise.

the decoding operation to be the inverse of f . This scheme has probability of error less thanε, by considering Property 14.3.1. Figure 14.3 depicts this coding scheme.
Shannon’s scheme for compression suffers from a problem that plagues all results inclassical and quantum information theory. The proof guarantees that there exists a schemethat can compress at the rate of entropy in the asymptotic limit. But the complexityof encoding and decoding is far from practical—without any further specification of theencoding, it could require resources that are prohibitively exponential in the length of thesequence.
The above scheme certainly gives an achievable rate for compression of classical information, but how can we know that it is optimal? The converse theorem addresses this point(recall that a converse theorem gives a sense of optimality for a particular protocol) andcompletes the operational interpretation of the entropy as the fundamental limit on thecompressibility of classical information. For now, we do not prove a converse theorem andinstead choose to wait until we cover Schumacher compression because its converse proof
applies to Shannon compression as well.

14.5 Weak Joint Typicality
Joint typicality is a concept similar to typicality, but the difference is that it applies to any
two random variables X and Y . That is, there are analogous notions of typicality for thejoint random variable (X, Y ).
Definition 14.5.1 (Joint Sample Entropy) Consider n independent realizations xn =x1 · · · xn and y n = y1 · · · yn of respective random variables X and Y . The sample jointc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 14. CLASSICAL TYPICALITY

entropy H(xn , y n ) of these two sequences is1
H(xn , y n ) ≡ − log (pX n ,Y n (xn , y n )) ,n

(14.31)

where we assume that the joint distribution pX n ,Y n (xn , y n ) has the i.i.d. property:
pX n ,Y n (xn , y n ) ≡ pX,Y (x1 , y1 ) · · · pX,Y (xn , yn ).

(14.32)This notion of joint sample entropy immediately leads to the following definition of joint
typicality. Figure 14.4 attempts to depict the notion of joint typicality.
Definition 14.5.2 (Jointly Typical Sequence) Two sequences xn , y n are δ-jointly typical if their sample joint entropy H(xn , y n ) is δ-close to the joint entropy H(X, Y ) of randomvariables X and Y and if both xn and y n are marginally typical.
nDefinition 14.5.3 (Jointly Typical Set) The δ-jointly typical set TδX Y
δ-jointly typical sequences:nconsists of all


n
n
n n
TδX Y ≡ (xn , y n ) : H(xn , y n ) − H(X, Y ) ≤ δ, xn ∈ TδX , y n ∈ TδY .

(14.33)The extra conditions on the marginal sample entropies are necessary to have a sensible
definition of joint typicality. That is, it does not necessarily follow that the marginal sample
entropies are close to the marginal true entropies if the joint ones are close, but it intuitively
makes sense that this condition should hold. Thus, we add these extra conditions to thedefinition of jointly typical sequences. Later, we find in Section 14.7 that the intuitive
implication holds (it is not necessary to include the marginals) when we introduce a different
definition of typicality.

14.5.1

Properties of the Jointly Typical Setnn

The set TδX Y of jointly typical sequences enjoys three properties similar to what we haveseen in Section 14.2, and the proofs of these properties are nearly identical to those in
Section 14.2.
Property 14.5.1 (Unit Probability) The jointly typical set asymptotically has probability
one. So as n becomes large, it is highly likely that a source emits a jointly typical sequence.
We formally state this property as follows:n nPr (X n , Y n ) ∈ TδX Y ≥ 1 − ε,(14.34)for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
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14.5. WEAK JOINT TYPICALITY

431

Figure 14.4: A depiction of the jointly typical set. Some sequence pairs (xn , y n ) are such that xn is typicalor such that y n is typical, but fewer are such that the pair is jointly typical. The jointly typical set has
size roughly equal to 2nH(X,Y ) , which is smaller than the Cartesian product of the marginally typical sets if
random variables X and Y are not independent.

nnProperty 14.5.2 (Exponentially Smaller Cardinality) The number TδX Y of δ-jointly
typical sequences is exponentially smaller than the total number (|X | |Y|)n of sequences forany joint random variable (X, Y ) that is not uniform. We formally state this property as
follows:n n
TδX Y ≤ 2n(H(X,Y )+δ) .
(14.35)
We can also bound the size of the δ-jointly typical set from below:
n

TδX Y

n

≥ (1 − ε) 2n(H(X,Y )−δ) ,(14.36)

for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
Property 14.5.3 (Equipartition) The probability of a particular δ-jointly typical sequence
xn y n is approximately uniform:
2−n(H(X,Y )+δ) ≤ pX n ,Y n (xn , y n ) ≤ 2−n(H(X,Y )−δ) .

(14.37)

Exercise 14.5.1 Prove the above three properties of the jointly typical set.
The above three properties may be similar to what we have seen before, but there isanother interesting property of jointly typical sequences that we give below. It states that
two sequences drawn independently according to the marginal distributions pX (x) and pY (y)are jointly typical according to the joint distribution pX,Y (x, y) with probability ≈ 2−nI(X;Y ) .
This property gives a simple interpretation of the mutual information that is related to its
most important operational interpretation as the classical channel capacity discussed briefly
in Section 2.2.
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CHAPTER 14. CLASSICAL TYPICALITY

Property 14.5.4 (Probability of Joint Typicality) Consider two independent randomvariables X̃ n and Ỹ n whose respective probability density functions pX̃ n (xn ) and pỸ n (y n ) areequal to the marginal densities of the joint density pX n ,Y n (xn , y n ):(X̃ n , Ỹ n ) ∼ pX n (xn )pY n (y n ).

(14.38)nThen we can bound the probability that two random sequences X̃ and Ỹn ntypical set TδX Y :non nPr (X̃ n , Ỹ n ) ∈ TδX Y≤ 2−n(I(X;Y )−3δ) .

nare in the jointly

Exercise 14.5.2 Prove Property 14.5.4. (Hint: Consider that
n
o
X
n n
Pr (X̃ n , Ỹ n ) ∈ TδX Y=
pX n (xn )pY n (y n ),(14.39)(14.40)n n
xn ,y n ∈TδX Y

and use the properties of typical and jointly typical sets to bound this probability.)

14.6 Weak Conditional Typicality
Conditional typicality is a property that we expect to hold for any two random sequences—itis also a useful tool in the proofs of coding theorems. Suppose two random variables X andY have respective alphabets X and Y and a joint distribution pX,Y (x, y). We can factor thejoint distribution pX,Y (x, y) as the product of a marginal distribution pX (x) and a conditional
distribution pY |X (y|x), and this factoring leads to a particular way that we can think about
generating realizations of the joint random variable. We can consider random variable Y to
be a noisy version of X, where we first generate a realization x of the random variable X
according to the distribution pX (x) and follow by generating a realization y of the randomvariable Y according to the conditional distribution pY |X (y|x).
Suppose that we generate n independent realizations of random variable X to obtain thesequence xn = x1 · · · xn . We then record these values and use the conditional distribution
pY |X (y|x) n times to generate n independent realizations of random variable Y . Let y n =y1 · · · yn denote the resulting sequence.
Definition 14.6.1 (Conditional Sample Entropy) The conditional sample entropy H(y n |xn )of two sequences xn and y n with respect to pX,Y (x, y) = pX (x)pY |X (y|x) is1
H(y n |xn ) = − log pY n |X n (y n |xn ),(14.41)n
where
pY n |X n (y n |xn ) ≡ pY |X (y1 |x1 ) · · · pY |X (yn |xn ).
(14.42)
Definition 14.6.2 (Conditionally Typical Set) Let xn ∈ X n . The δ-conditionally typY n |xn
ical set Tδ
consists of all sequences whose conditional sample entropy is δ-close to thetrue conditional entropy:Y n |xnTδ≡ y n : H(y n |xn ) − H(Y |X) ≤ δ .
(14.43)Figure 14.5 provides an intuitive picture of the notion of conditional typicality.
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14.6. WEAK CONDITIONAL TYPICALITY

433n

Figure 14.5: The notion of the conditionally typical set. A typical sequence xn in TδX maps stochasticallythrough many instantiations of a conditional distribution pY |X (y|x) to some sequence y n . It is overwhelmY n |xn

ingly likely that y n is in a conditionally typical set Tδ
when n becomes large. This conditionally typicalset has size around 2nH(Y |X) . It contains nearly all of the probability but is exponentially smaller than theset of all sequences Y n .

14.6.1

Properties of the Conditionally Typical Set
Y n |xn

The set Tδ
of conditionally typical sequences enjoys properties similar to what we haveseen before, and we list them for completeness.
Y n |xn

Property 14.6.1 (Unit Probability) The set Tδ
asymptotically has probability onenon average with respect to a random sequence X . So as n becomes large, it is highly likelythat random sequences Y n and X n are such that Y n is a conditionally typical sequence (onaverage with respect to X n ). We formally state this property as follows:on
Y n |X nn
EX n
Pr n Y ∈ Tδ
≥ 1 − ε,(14.44)n
Y |X

for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
Y n |xn

Property 14.6.2 (Exponentially Smaller Cardinality) The number Tδ

of δ-conditionallyntypical sequences is exponentially smaller than the total number |Y| of sequences for any
conditional random variable Y |X that is not uniform. We formally state this property as
follows:
Y n |xnTδ≤ 2n(H(Y |X)+δ) .
(14.45)
We can also bound the expected size of the δ-conditionally typical set from below when xn isa random sequence:no
Y n |X n
EX n Tδ≥ (1 − ε) 2n(H(Y |X)−δ) ,(14.46)
for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
Property 14.6.3 (Equipartition) The probability of a given δ-conditionally typical sequence y n (corresponding to the sequence xn ) is approximately uniform:
2−n(H(Y |X)+δ) ≤ pY n |X n (y n |xn ) ≤ 2−n(H(Y |X)−δ) .
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CHAPTER 14. CLASSICAL TYPICALITY

In summary,
averaged over realizations of the random variable X n , the conditionallynn
Y |Xtypical set Tδ
has almost all the probability, and its size is exponentially smaller than
the size of the set of all sequences. For each realization of X n , each δ-conditionally typicalsequence has an approximate uniform probability of occurring.
Our last note on the weak conditionally typical set is that there is a subtlety in thestatement of Property 14.6.1 that allows for a relatively straightforward proof. This subtlety
is that we average over the sequence X n as well, and this allows one to exploit the extra
randomness in order to establish a proof. We do not impose such a constraint later on in
Section 14.9 when we introduce the notion of a strong conditionally typical sequence. We
instead impose the constraint that the sequence xn is a strongly typical sequence, and thisproperty is sufficient to prove that similar properties hold for a strong conditionally typicalset.
We now prove the first property. This is just again another application of the law of large
numbers. Consider thatno
Y n |X nn
EX n
Pr Y ∈ Tδ
Y n |X n
n
n
oo
n
n
n
= EX n EY n |X n IT Y |X (Y )(14.48)δno= EX n ,Y n IT Y n |X n (Y n )(14.49)δX=
pX n ,Y n (xn , y n )IT Y n |xn (y n ),(14.50)δ

xn ∈X n ,y n ∈Y n

where I denotes an indicator function. Given random variables X and Y , let us define therandom variable g(X, Y ) = − log pY |X (Y |X). The expectation of this random variable isEX,Y {g(X, Y )} = EX,Y { − log pY |X (Y |X) }(14.51)XpX,Y (x, y) − log pY |X (y|x)(14.52)=x,y= H(Y |X).

(14.53)

Consider that the conditional sample entropy H(Y n |X n ) for the random sequences X n andY n factors as follows:1
H(Y n |X n ) = − log pY n |X n (Y n |X n )
n
n

1 X
=
− log pY |X (Yi |Xi )n i=1(14.54)(14.55)n

1X=
g(Xi , Yi ).
n i=1(14.56)This is the sample average of the random variables g(Xi , Yi ) for all i ∈ {1, . . . , n}. Given
all of the above and the definition of weak conditional typicality, we can rewrite (14.50) as
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14.7. STRONG TYPICALITY

follows:

(
Pr

XnY n

435

n

1X
g (Xi , Yi ) − EX,Y {g(X, Y )} ≤ δ
n i=1

)
.

(14.57)

By applying the law of large numbers, this is ≥ 1 − ε for all ε ∈ (0, 1), δ > 0, and sufficiently
large n.
Exercise 14.6.1 Prove Properties 14.6.2 and 14.6.3 for the weak conditionally typical set.

14.7 Strong Typicality
In the development in the previous sections, we showed how the law of large numbers is theunderpinning method to prove many of the interesting results regarding typical sequences.
These results are satisfactory and provide an intuitive notion of typicality through the ideaof the sample entropy approaching the true entropy for sufficiently long sequences.
It is possible to develop a stronger notion of typicality with a different definition. Insteadof requiring that the sample entropy of a random sequence is close to the true entropy ofa distribution for sufficiently long sequences, strong typicality requires that the empirical
distribution or relative frequency of symbols of a random sequence has a small deviationfrom the true probability distribution for sufficiently large sequence length.
We begin with a simple example to help illustrate this stronger notion of typicality.
Suppose that we generate a binary sequence i.i.d. according to the distribution p(0) = 1/4
and p(1) = 3/4. Such a random generation process could lead to the following sequence:
0110111010.

(14.58)Rather than computing the sample entropy of this sequence and comparing it with the true
entropy, we can count the number of zeros or ones that appear in the sequence and compare
their normalizations with the true distribution of the information source. For the above
example, the number of zeros is equal to 4, and the number of ones (the Hamming weightof the sequence) is equal to 6:
N (0 | 0110111010) = 4,

N (1 | 0110111010) = 6.

(14.59)We can compute the empirical distribution of this sequence by normalizing the above numbers by the length of the sequence:21
N (0 | 0110111010) = ,
10
5

1
3
N (1 | 0110111010) = .
10
5(14.60)This empirical distribution deviates from the true distribution by the following amount:3
1 2 3 3
max
− , −
= ,
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CHAPTER 14. CLASSICAL TYPICALITY

which is a fairly significant deviation. However, suppose that the length of the sequence
grows large enough so that the law of large numbers comes into play. We would then expect
it to be highly likely that the empirical distribution of a random sequence does not deviate
much from the true distribution, and the law of large numbers again gives a theoreticalunderpinning for this intuition. This example gives the essence of strong typicality.
We wish to highlight another important aspect of the above example. The particularsequence in (14.58) has a Hamming weight of six, but this sequence is not the only one
with this Hamming weight. By a simple counting argument, there are 10− 1 = 209 other6
sequences with the same length and Hamming weight. That is, all these other sequences havethe same empirical distribution and thus have the same deviation from the true distribution
as the original sequence in (14.58). We say that all these sequences are in the same “typeclass,” which simply means that they have the same empirical distribution. The type classis thus an equivalence class on sequences where the equivalence relation is the empirical
distribution of the sequence.
We mention a few interesting properties of the type class before giving more formal
definitions. We can partition the set of all possible sequences according to type classes.
10Consider that the set of all binary sequences of length ten has
 size 2 . There is one sequence
10
10
with all zeros, 1 sequences with Hamming weight one, 2 sequences with Hamming weight
two, etc. The binomial theorem guarantees that the total number of sequences is equal to
the number of sequences in all of the type classes:
10  
X
10
10
2 =
.
(14.62)ii=0Suppose now that we generate ten i.i.d. realizations of the Bernoulli distribution p(0) =1/4 and p(1) = 3/4. Without knowing anything else, our best description of the distributionof the random sequence isp(x1 , . . . , x10 ) = p(x1 ) · · · p(x10 ),(14.63)where x1 , . . . , x10 are different realizations of the binary random variable. But suppose thata third party tells us the Hamming weight w0 of the generated sequence. This information
allows us to update our knowledge of the distribution of the sequence, and we can say that any
sequence with Hamming weight not equal to w0 has zero probability. All the sequences withthe same Hamming weight have the same distribution because we generated the sequence inan i.i.d. way, and each sequence with Hamming weight w0 has a uniform distribution after
renormalizing. Thus, conditioned on the Hamming weight w0 , our best description of the
distribution of the random sequence is(0: w (x1 , . . . , x10 ) 6= w0
p(x1 , . . . , x10 |w0 ) =,(14.64)10 −1: w (x1 , . . . , x10 ) = w0
w0
where w is a function that gives the Hamming weight of a binary sequence. This property
has important consequences for asymptotic information processing because it gives us away to extract uniform randomness from an i.i.d. distribution, and we later see that it hasapplications in several quantum information-processing protocols as well.
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14.7. STRONG TYPICALITY

14.7.1

437

Types and Strong Typicality

We now formally develop the notion of a type and strong typicality. Let xn denote a sequencex1 x2 . . . xn , where each xi belongs to the alphabet X . Let |X | be the cardinality of X . Let
N (x|xn ) be the number of occurrences of the symbol x ∈ X in the sequence xn .
Definition 14.7.1 (Type) The type or empirical distribution txn of a sequence xn is a
probability mass function whose elements are txn (x) wheretxn (x) ≡1
N (x|xn ).
n(14.65)nDefinition 14.7.2 (Strongly Typical Set) The δ-strongly typical set TδX is the set of all
sequences with an empirical distribution n1 N (x|xn ) that has maximum deviation δ from thetrue distribution pX (x). Furthermore, the empirical distribution n1 N (x|xn ) of any sequencenin TδX vanishes for any letter x for which pX (x) = 0:
n
TδX ≡




1
1
n
n
N (x|x ) − pX (x) ≤ δ if pX (x) > 0, else N (x|x ) = 0 .
x : ∀x ∈ X ,nn(14.66)n

The extra condition where n1 N (x|xn ) = 0 when pX (x) = 0 is a somewhat technicalcondition, nevertheless intuitive, that is necessary to prove the three desired properties fornthe strongly typical set. Also, we are using the same notation TδX to indicate both theweakly and strongly typical set, but which one is appropriate should be clear from thecontext, or we will explicitly indicate which one we are using.
The notion of type class becomes useful for us in our later developments—it is simply
a way for grouping together all the sequences with the same empirical distribution. Its
most important use is as a way for obtaining a uniform distribution from an arbitrary
i.i.d. distribution (recall that we can do this by conditioning on a particular type).
nDefinition 14.7.3 (Type Class) Let TtX denote the type class of a particular type t. Thentype class TtX is the set of all sequences with length n and type t:nTtX ≡ {xn ∈ X n : txn = t}.

(14.67)

Property 14.7.1 (Bound on the Number of Types) The number of types for a given
sequence of length n containing symbols from an alphabet X is exactly equal ton + |X | − 1.
(14.68)|X | − 1
A good upper bound on the number of types is (n + 1)|X | .
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CHAPTER 14. CLASSICAL TYPICALITY

Proof. The number of types is equal to the number of ways that the symbols in a sequenceof length n can form |X | distinct groups. Consider the following visual aid:• • • • • | • • • • • •| • • • • • • • • • | • • • • .

(14.69)We can think of the number of types as the number of different ways of arranging |X | − 1
vertical bars to group the n dots into |X | distinct groups, which gives (14.68). The upperbound follows from a simple argument. The number of types is the number of different waysthat |X | positive numbers can sum to n. Overestimating the count, we can choose the firstnumber in n + 1 different ways (it can be any number from 0 to n), and we can choose the|X | − 1 other numbers in n + 1 different ways. Multiplying all of these possibilities together
gives an upper bound (n + 1)|X | on the number of types. This bound illustrates that the
number of types is only polynomial in the length n of the sequence (compare with the total
number |X |n of sequences of length n being exponential in the length of the sequence).
Definition 14.7.4 (Typical Type) Let pX (x) denote the true probability distribution ofsymbols x in the alphabet X . For δ > 0, let τδ denote the set of all typical types that havemaximum deviation δ from the true distribution pX (x):τδ ≡ {t : ∀x ∈ X , |t(x) − pX (x)| ≤ δ if pX (x) > 0 else t(x) = 0}.

(14.70)We can then equivalently define the set of strongly δ-typical sequences of length n as aunion over all the type classes of the typical types in τδ :
[ n
n
TδX =
TtX .
(14.71)t∈τδ14.7.2

Properties of the Strongly Typical Set

The strongly typical set enjoys many useful properties (similar to the weakly typical set).
Property 14.7.2 (Unit Probability) The strongly typical set asymptotically has probability one. So as n becomes large, it is highly likely that a source emits a strongly typicalsequence. We formally state this property as follows:n
Pr X n ∈ TδX ≥ 1 − ε,(14.72)for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
nProperty 14.7.3 (Exponentially Smaller Cardinality) The number TδX of δ-typicalsequences is exponentially smaller than the total number |X |n of sequences for most randomvariables X. We formally state this property as follows:TδXn≤ 2n(H(X)+cδ) ,(14.73)where c is some positive constant. We can also bound the size of the δ-typical set from below:
TδXn≥ (1 − ε) 2n(H(X)−cδ) ,(14.74)for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
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14.7. STRONG TYPICALITY

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Property 14.7.4 (Equipartition) The probability of a given δ-typical sequence xn occurring is approximately uniform:
2−n(H(X)+cδ) ≤ pX n (xn ) ≤ 2−n(H(X)−cδ) .

(14.75)This last property of strong typicality demonstrates that it implies weak typicality up toa constant c.

14.7.3

Proofs of the Properties of the Strongly Typical Set

Proof of the Unit Probability Property (Property 14.7.2). A proof proceeds similarly
to the proof of the unit probability property for the weakly typical set. The law of large
numbers states that the sample mean of a random sequence converges in probability to the
expectation of the random variable from which we generate the sequence. So consider asequence of i.i.d. random variables Z1 , . . . , Zn where each random variable in the sequence
has expectation µ. The sample average of this sequence is as follows:
n

1X
Z≡
Zi .
n i=1

(14.76)

The precise statement of the weak law of large numbers is that ∀ε ∈ (0, 1), δ > 0 ∃n0 : ∀n >n0 such that
Pr Z − µ > δ < ε.
(14.77)
We can now consider the indicator random variables I(X1 = a), . . . , I(Xn = a). The sample
mean of a random sequence of indicator variables is equal to the empirical distribution
N (a|X n )/n:n
1
1X
I(Xi = a) = N (a|X n ),(14.78)n i=1nand the expectation of the indicator random variable I(X = a) is equal to the probabilityof the symbol a:
EX {I(X = a)} = pX (a).
(14.79)Also, any random sequence X n has probability zero if one of its symbols xi is such thatpX (xi ) = 0. Thus, the probability that n1 N (a|X n ) = 0 is equal to one whenever pX (a) = 0:

Pr


1
n
N (a|X ) = 0 : pX (a) = 0 = 1,n(14.80)and we can consider the cases when pX (a) > 0. We apply the law of large numbers to find
that


ε
1
n
∀ε ∈ (0, 1), δ > 0 ∃n0,a : ∀n > n0,aPr
N (a|X ) − pX (a) > δ <.
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CHAPTER 14. CLASSICAL TYPICALITY

Choosing n0 = maxa∈X {n0,a }, the following condition holds by the union bound of probability theory:∀ε ∈ (0, 1), δ > 0 ∃n0 : ∀n > n0)(
[ 1
N (a|X n ) − pX (a) > δ
Pr
n
a∈X

X  1
n
N (a|X ) − pX (a) > δ < ε. (14.82)
≤
Pr
n
a∈X
Thus the complement of the above event on the left has high probability. That is, ∀ε ∈(0, 1), δ > 0 ∃n0 : ∀n > n0 such that1nPr ∀a ∈ X ,N (a|X ) − pX (a) ≤ δ ≥ 1 − ε.
(14.83)nThe event ∀a ∈ X , n1 N (a|X n ) − pX (a) ≤ δ is the condition for a random sequence X nnto be in the strongly typical set TδX , and the probability of this event goes to one as n
becomes sufficiently large.
Proof of the Exponentially Smaller Cardinality Property (Property 14.7.3). By
the proof of Property 14.7.4 (proved below), we know that the following relation holds forany sequence xn in the strongly typical set:2−n(H(X)+cδ) ≤ pX n (xn ) ≤ 2−n(H(X)−cδ) ,(14.84)

where c is some positive constant that we define when we prove Property 14.7.4. Summingover all sequences in the typical set, we get the following inequalities:XXn
2−n(H(X)+cδ) ≤ Pr X n ∈ TδX ≤2−n(H(X)−cδ) ,(14.85)xn ∈TδXnxn ∈TδX⇒ 2−n(H(X)+cδ) TδXn≤ Pr X n ∈ TδXnn≤ 2−n(H(X)−cδ) TδXn.

(14.86)

By the unit probability property of the strongly typical set, we know that the following
relation holds for sufficiently large n:n
1 ≥ Pr X n ∈ TδX ≥ 1 − ε.

(14.87)

Then the following inequalities result by combining the above inequalities:
2n(H(X)−cδ) (1 − ε) ≤ TδXn≤ 2n(H(X)+cδ) ,(14.88)concluding the proof.
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14.7. STRONG TYPICALITY

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Proof of the Equipartition Property (Property 14.7.4). The following relation holds
from the i.i.d. property of the distribution pX n (xn ) and because the sequence xn is strongly
typical:Yn
pX n (xn ) =pX (x)N (x|x )(14.89)x∈X ++where X denotes all the letters x in X with pX (x) > 0. (The fact that the sequence xn isstrongly typical according to Definition 14.7.2 allows us to employ this modified alphabet.)
Take the logarithm of the above expression:Xlog (pX n (xn )) =N (x|xn ) log (pX (x)) .
(14.90)x∈X +Multiply both sides by − n1 :−X 1
1log (pX n (xn )) = −N (x|xn ) log (pX (x)) .
nn+(14.91)x∈X

The following relation holds because the sequence xn is strongly typical:∀x ∈ X + :1
N (x|xn ) − pX (x) ≤ δ,n

(14.92)and it implies that⇒ ∀x ∈ X + : −δ + pX (x) ≤1
N (x|xn ) ≤ δ + pX (x).
n(14.93)

Now multiply (14.93) by − log (pX (x)) > 0, sum over all letters in the alphabet X + , andapply the substitution in (14.91). This procedure gives the following set of inequalities:−1(−δ + pX (x)) log (pX (x)) ≤ − log (pX n (xn ))
n
x∈X +
X
(δ + pX (x)) log (pX (x)) , (14.94)
≤−
X

x∈X +

1
⇒ −cδ + H(X) ≤ − log (pX n (xn )) ≤ cδ + H(X),n−n(H(X)+cδ)⇒2≤ pX n (xn ) ≤ 2−n(H(X)−cδ) ,(14.95)(14.96)

where
c≡−

X

log (pX (x)) ≥ 0.

(14.97)x∈X +It now becomes apparent why we require the technical condition in the definition of strong
typicality (Definition 14.7.2). Were it not there, then the constant c would not be finite, andwe would not be able to obtain a reasonable bound on the probability of a strongly typicalsequence.
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CHAPTER 14. CLASSICAL TYPICALITY

14.7.4

Cardinality of a Typical Type Class

Recall that a typical type class is defined to be the set of all sequences with the same empirical
distribution, and the empirical distribution happens to have maximum deviation δ from thenntrue distribution. It might seem that the size TtX of a typical type class TtX should besmaller than the size of the strongly typical set. But the following property overrides thisnintuition and shows that a given typical type class TtX in some sense has almost as manynsequences in it as the strongly typical set TδX for sufficiently large n.
Property 14.7.5 (Minimal Cardinality of a Typical Type Class) Let X be a randomnvariable with alphabet X . Fix δ ∈ (0, 2/|X |]. For t ∈ τδ , the size TtX of the typical typenclass TtX is bounded from below as follows:
TtX

n

≥

1
(n + 1)1|X |2n[H(X)−η(|X |δ)] = 2n[H(X)−η(|X |δ)−|X | n log(n+1)] ,(14.98)where η(δ) is some function such that η(δ) → 0 as δ → 0. Thus, a typical type class is ofsize roughly 2nH(X) when n → ∞ and δ → 0 (it is about as large as the typical set when nbecomes large).
Proof. We first show that if X1 , . . . , Xn are random variables drawn i.i.d. from a distribution
q(x), then the probability q n (xn ) of a particular sequence xn depends only on its type:q n (xn ) = 2−n(H(txn )+D(txn kq)) ,(14.99)where D(txn kq) is the relative entropy between txn and q. To see this, consider the following
chain of equalities:nnq (x ) ==nYq(xi ) =Ynq(x)N (x|x ) =

i=1

x∈X

Y

2ntxn (x) log q(x) = 2nY

q(x)ntxn (x)(14.100)x∈X
P

x∈X txn (x) log q(x)x∈XPn x∈X txn (x) log q(x)−txn (x) log txn (x)+txn (x) log txn (x)(14.101)=2(14.102)= 2−n(D(txn kq)+H(txn )) .

(14.103)

It then follows that the probability of the sequence xn is 2−nH(txn ) if the distribution q(x) =ntxn (x). Now consider that each type class TtX has sizen,(14.104)
ntxn (x1 ), . . . , ntxn (x|X | )where the distribution t = txn (x1 ), . . . , txn (x|X | ) and the letters of X are x1 , . . . , x|X | .
This result follows because the size of a type class is just the number of ways of arranging
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443

ntxn (x1 ), . . . , ntxn (x|X | ) in a sequence of length n. We now prove that the type class TtX
has the highest probability among all type classes when the probability distribution is t:nntn (TtX ) ≥ tn (TtX0 ) for all t0 ∈ Pn ,n(14.105)

where tn is the i.i.d. distribution induced by the type t and Pn is the set of all types. Consider
the following equalities:QnTtX
txn (x)ntxn (x)nXnt (Tt )x∈X
= Xn Q(14.106)0Xnnt (Tt0 )|Tt0 |
txn (x)ntxn (x)
x∈X
 Q
n
txn (x)ntxn (x)ntxn (x1 ),...,ntxn (x|X | ) x∈X=(14.107)Q
0ntxn (x)ntxn (x)nt0 n (x1 ),...,nt0 n (x|X | )x=x

Y nt0 n (x)!
xntxn (x)!
x∈Xx∈Xtxn (x)n(txn (x)−txn (x)) .
0(14.108)

Now apply the bound m!
≥ nm−n (that holds for any positive integers m and n) to getn!
nYtn (TtX )0n(t0xn (x)−txn (x))
n (x)][nt
txn (x)n(txn (x)−txn (x))
≥
x
n
X
n
t (Tt0 ) x∈XY0nn(txn (x)−txn (x))=

x∈XPn x∈X t0xn (x)−txn (x)=n(14.109)(14.110)(14.111)n(1−1)=n= 1.

(14.112)(14.113)nnThus, it holds that tn (TtX ) ≥ tn (TtX0 ) for all t0 . Now we are close to obtaining the desired
bound in Property 14.7.5. Consider the following chain of inequalities:
X
X
X
n
n
n
Xn
1=
tn (TtX0 ) ≤
max
t
(T
)
=
tn (TtX )(14.114)0t0
t0 ∈Pntt0 ∈Pnt0 ∈PnXn≤ (n + 1)|X | tn (Tt ) = (n + 1)|X |X

tn (xn )(14.115)nxn ∈TtX= (n + 1)|X |X

2−nH(t)(14.116)nxn ∈TtX= (n + 1)|X | 2−nH(t) TtXn.

(14.117)

Recall that t is a typical type, implying that |t(x) − p(x)| ≤ δ for all x. This then implies
that the variational distance between the distributions is small:X|t(x) − p(x)| ≤ |X | δ.
(14.118)x
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CHAPTER 14. CLASSICAL TYPICALITY

We can apply the continuity of entropy (Theorem 11.10.1) to get a bound on the differenceof entropies:1(14.119)|H(t) − H(X)| ≤ |X | δ log |X | + h2 (|X | δ/2).
2
The desired bound then follows with η(|X | δ) ≡ |X | δ/2 log |X | + h2 (|X | δ/2).
Exercise 14.7.1 Prove that 2nH(t) is an upper bound on the number of sequences xn oftype t:n
TtX ≤ 2nH(t) .
(14.120)
Use this bound and (14.103) to prove the following upper bound on the probability of a typeclass where each sequence is generated i.i.d. according to a distribution q(x): nPr TtX ≤ 2−nD(tkq) .
(14.121)14.8 Strong Joint Typicality
It is possible to extend the above notions of strong typicality to jointly typical sequences. In
a marked difference with the weakly typical case, we can show that strong joint typicality
implies marginal typicality. Thus there is no need to impose this constraint in the definition.
Let N (x, y|xn , y n ) be the number of occurrences of the symbol x ∈ X , y ∈ Y in therespective sequences xn and y n . The type or empirical distribution txn yn of sequences xn andy n is a probability mass function whose elements are txn yn (x, y) where
txn yn (x, y) ≡1
N (x, y|xn , y n ).
n(14.122)

Definition 14.8.1 (Strong Jointly Typical Sequence) Two sequences xn , y n are δ-strongly
jointly typical if their empirical distribution has maximum deviation δ from the true distribution and vanishes for any two symbols x and y for which pX,Y (x, y) = 0.
nDefinition 14.8.2 (Strong Jointly Typical Set) The δ-jointly typical set TδX Y
set of all δ-jointly typical sequences:nnis thenTδX Y ≡
xn , y n : ∀(x, y) ∈ X × Y

1
N (x, y|xn , y n ) − pX,Y (x, y)n1
N (x, y|xn , y n ) = 0n≤ δ if pX,Y (x, y) > 0.
otherwise(14.123)

It follows from the above definitions that strong joint typicality implies marginal typicality
for both sequences xn and y n . We leave justifying this statement as the following exercise.
Exercise 14.8.1 Prove that strong joint typicality implies marginal typicality for either thesequence xn or y n .
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14.8. STRONG JOINT TYPICALITY

14.8.1445

Properties of the Strong Jointly Typical Setnn

The set TδX Y of strong jointly typical sequences enjoys properties similar to what we haveseen before.
nnProperty 14.8.1 (Unit Probability) The strong jointly typical set TδX Y asymptoticallyhas probability one. So as n becomes large, it is highly likely that a source emits a strong
jointly typical sequence. We formally state this property as follows:n n(14.124)
Pr X n Y n ∈ TδX Y ≥ 1 − ε,for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
nnProperty 14.8.2 (Exponentially Smaller Cardinality) The number TδX Y of δ-jointly
typical sequences is exponentially smaller than the total number (|X | |Y|)n of sequences forany joint random variable (X, Y ) that is not uniform. We formally state this property as
follows:
n n(14.125)
TδX Y ≤ 2n(H(X,Y )+cδ) ,where c is a positive constant. We can also bound the size of the δ-jointly typical set frombelow:n n
TδX Y ≥ (1 − ε) 2n(H(X,Y )−cδ) ,(14.126)
for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
Property 14.8.3 (Equipartition) The probability of a given δ-jointly typical sequence
xn y n occurring is approximately uniform:
2−n(H(X,Y )+cδ) ≤ pX n ,Y n (xn , y n ) ≤ 2−n(H(X,Y )−cδ) ,(14.127)

where c is a positive constant.
Property 14.8.4 (Probability of Strong Joint Typicality) Consider two independent
random variables X̃ n and Ỹ n whose respective probability density functions pX̃ n (xn ) andpỸ n (y n ) are equal to the marginal densities of the joint density pX n ,Y n (xn , y n ):(X̃ n , Ỹ n ) ∼ pX n (xn )pY n (y n ).

(14.128)

Then we can bound the probability that two random sequences X̃ n and Ỹ n are in the jointly
n ntypical set TδX Y :non

Pr (X̃ n , Ỹ n ) ∈ TδX Yn≤ 2−n(I(X;Y )−3cδ) .

(14.129)

Proofs of the first three properties are the same as in the previous section, and the proofof the last property is the same as that for the weakly typical case.
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CHAPTER 14. CLASSICAL TYPICALITY

14.9 Strong Conditional Typicality
Strong conditional typicality bears some similarities to strong typicality, but it is sufficiently
different for us to provide a discussion of it. We first introduce it with a simple example.
Suppose that we draw a sequence from an alphabet {0, 1, 2} according to the distribution:11
1
pX (0) = ,pX (1) = ,pX (2) = .
442
A particular realization sequence could be as follows:
2010201020120212122220202222.

(14.130)(14.131)

We count up the occurrences of each symbol and find them to be
N (0 | 2010201020120212122220202222) = 8,
N (1 | 2010201020120212122220202222) = 5,
N (2 | 2010201020120212122220202222) = 15.

(14.132)(14.133)(14.134)

The maximum deviation of the sequence’s empirical distribution from the true distributionof the source is as follows:




1
8
5
1
1 15
1 2 1
1
max
−
, −
, −
, ,
(14.135)= max= .
4 28 4 28 2 28
28 28 28
14
We now consider generating a different sequence from an alphabet {a, b, c}. However, we
generate it according to the following conditional probability distribution:

pY |X (a|0) = 51 pY |X (a|1) = 16 pY |X (a|2) = 24
 pY |X (b|0) = 2 pY |X (b|1) = 3 pY |X (b|2) = 1  .
(14.136)5
6
4
pY |X (c|0) = 52 pY |X (c|1) = 26 pY |X (c|2) = 41
The second generated sequence should thus have correlations with the original sequence. A
possible realization of the second sequence could be as follows:
abbcbccabcabcabcabcbcbabacba.

(14.137)

We would now like to analyze how close the empirical conditional distribution is to the true
conditional distribution for all input and output sequences. A useful conceptual first step isto apply a permutation to the first sequence so that all of its symbols appear in lexicographic
order, and we then apply the same permutation to the second sequence:
2 0 1 0 2 0 1 0 2 0 1 2 0 2 1 2 1 2 2 2 2 0 2 0 2 2 2 2a b b c b c c a b c a b c a b c a b c b c b a b a c b a
permute−−−−−→
0 0 0 0 0 0 0 0 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2.
b c c a c c b b b c a b a a b b b a c b c b c a a c b a(14.138)c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

14.9. STRONG CONDITIONAL TYPICALITY

447

This rearrangement makes it easy to count up the empirical conditional distribution of thesecond sequence. We first place the joint occurrences of the symbols into the following
matrix:N (0, a) = 1 N (1, a) = 2 N (2, a) = 5
 N (0, b) = 3 N (1, b) = 2 N (2, b) = 6  ,(14.139)
N (0, c) = 4 N (1, c) = 1 N (2, c) = 4
and we obtain the empirical conditional distribution matrix by dividing these entries by themarginal distribution of the first sequence: N (0,a)N (1,a)N (2,a)
1
2
5
=
=
=
N (0)
8
N (1)
5
N (2)
15
 N (0,b)
N (1,b)N (2,b)3
2
6 (14.140)
 N (0) = 8 N (1) = 5 N (2) = 15  .
N (0,c)N (1,c)N (2,c)41
4= 8 N (1) = 5 N (2) = 15
N (0)We then compare the maximal deviation of the elements in this matrix with the elements inthe stochastic matrix in (14.136):


1
1
5
6
4
1 1 2 3 2 4 1 2 3 2 2 1 2− , − , − , − , − , − , −
max
, −
, −
5 8 5 8 5 8 6 5 6 5 6 5 4 15 4 15 4 15

3 1 1 7 1 2 1 3 1
7= max, , , , , , , ,= . (14.141)
40 40 10 30 10 15 6 20 60
30
The above analysis applies to a finite realization to illustrate the notion of conditional typicality, and there is a large deviation from the true distribution in this case. We would
again expect this deviation to vanish for a random sequence in the limit as the length of thesequence becomes large.

14.9.1

Definition of Strong Conditional Typicality

We now give a formal definition of strong conditional typicality.
Definition 14.9.1 (Conditional Empirical Distribution) The conditional empirical distribution tyn |xn (y|x) is as follows:
tyn |xn (y|x) =txn yn (x, y).
txn (x)(14.142)

Definition 14.9.2 (Strong Conditional Typicality) Suppose that a sequence xn is an
Y n |xn
strongly typical sequence in TδX . Then the δ-strong conditionally typical set Tδ
corresponding to the sequence xn consists of all sequences whose joint empirical distribution n1 N (x, y|xn , y n )is δ-close to the product of the true conditional distribution pY |X (y|x) with the marginal empirical distribution n1 N (x|xn ):Y n |xnTδ≡|N (x, y|xn , y n ) − p(y|x)N (x|xn )| ≤ nδ if p(y|x) > 0
y : ∀(x, y) ∈ X × Y
N (x, y|xn , y n ) = 0
otherwisen, (14.143)

where we abbreviate pY |X (y|x) as p(y|x).
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CHAPTER 14. CLASSICAL TYPICALITY

The above definition of strong conditional typicality implies that the conditional empirical
distribution is close to the true conditional distribution, in the sense that1
txn yn (x, y)− pY |X (y|x) ≤δ.
txn (x)txn (x)(14.144)

Of course, such a relation only makes sense if the marginal empirical distribution txn (x) is
non-zero.
The extra technical condition (N (x, y|xn , y n ) = 0 if pY |X (y|x) = 0) in Definition 14.9.2
is present again for a reason that we found in the proof of the Equipartition Property for
Strong Typicality (Property 14.7.4).

14.9.2

Properties of the Strong Conditionally Typical Set
Y n |xn

The set Tδ
of conditionally typical sequences enjoys a few useful properties that are
similar to what we have for the weak conditionally typical set, but the initial sequence xn
can be deterministic. However, we do impose the constraint that it has to be strongly typicalso that we can prove useful properties for the corresponding strong conditionally typical set.
n
So first suppose that a given sequence xn ∈ TδX0 for some δ 0 > 0.
Y n |xn

Property 14.9.1 (Unit Probability) The set Tδ
asymptotically has probability one.
So as n becomes large, it is highly likely that a random sequence Y n corresponding to a given
typical sequence xn is a conditionally typical sequence. We formally state this property as
follows:no
Y n |xn
Pr Y n ∈ Tδ
≥ 1 − ε,(14.145)
for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
Y n |xn

Property 14.9.2 (Exponentially Smaller Cardinality) The number Tδ

of δ-conditionallyntypical sequences is exponentially smaller than the total number |Y| of sequences for any
conditional random variable Y that is not uniform. We formally state this property as follows:
0
Y n |xnTδ≤ 2n(H(Y |X)+c(δ+δ )) .
(14.146)
We can also bound the size of the δ-conditionally typical set from below:
Y n |xnTδ0≥ (1 − ε) 2n(H(Y |X)−c(δ+δ )) ,(14.147)

for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
Property 14.9.3 (Equipartition) The probability of a particular δ-conditionally typicalsequence y n is approximately uniform:
002−n(H(Y |X)+c(δ+δ )) ≤ pY n |X n (y n |xn ) ≤ 2−n(H(Y |X)−c(δ+δ )) .
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14.9. STRONG CONDITIONAL TYPICALITY

449

In summary,given a realization xn of the random variable X n , the conditionally typicalnn
Y |x
set Tδ
has almost all the probability, its size is exponentially smaller than the size of theset of all sequences, and each δ-conditionally typical sequence has an approximately uniform
probability of occurring.

14.9.3

Proofs of the Properties of the Strong Conditionally Typical Set

Proof of the Unit Probability Property (Property 14.9.1). A proof of this property
is somewhat more complicated for strong conditional typicality. Since we are dealing with ani.i.d. distribution, we can assume that the sequence xn is lexicographically ordered with anorder on the alphabet X . We write the elements of X as x1 , . . . , x|X | . Then the lexicographic
ordering means that we can write the sequence xn as follows:
xn = x1 · · · x1 x2 · · · x2 · · · x|X | · · · x|X | .
| {z } | {z }| {z }
N (x1 |xn )N (x2 |xn )(14.149)

N (x|X | |xn )It follows that N (x|xn ) ≥ n (pX (x) − δ 0 ) from the typicality of xn , and the law of large
numbers comes into play for each block xi · · · xi with length N (xi |xn ) when this length is large
enough. Let pY |X=x (y) be the distribution for the conditional random variable Y | (X = x).
Then the following is an equivalent way to write the notion of conditional typicality:
o
n
n o
^n
n
(Y |(X=x))N (x|x )Y n |xn,y N (x|x ) ∈ Tδ⇔
y n ∈ Tδ

(14.150)x∈X

where the symbol ∧ denotes concatenation (note that the lexicographic ordering of xn appliesn(Y |(X=x))N (x|x )to the ordering of the sequence y n as well). Also, Tδ
is the typical set for asequence of conditional random variables Y | (X = x) with length N (x|xn ):n(Y |(X=x))N (x|x )
Tδ
≡


y

N (x|xn ): ∀y ∈ Y,n
N (y|y N (x|x ) )− pY |X=x (y) ≤ δ .
N (x|xn )(14.151)n(Y |(X=x))N (x|x )We can apply the law of large numbers to each of these typical sets Tδ
the length N (x|xn ) becomes large. It then follows that

n
o Y n
n o
Y n |xn(Y |(X=x))N (x|x )nN (x|xn )Pr Y∈ TδPr Y ∈ Tδ
=

where(14.152)x∈X≥ (1 − ε)|X |≥ 1 − |X | ε,(14.153)(14.154)

concluding the proof.
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CHAPTER 14. CLASSICAL TYPICALITY

Proof of the Equipartition Property (Property 14.9.3). The following relation holds
from the i.i.d. property of the conditional distribution pY n |X n (y n |xn ) and because the sequence y n is strong conditionally typical according to Definition 14.9.2:
Yn n
pY n |X n (y n |xn ) =pY |X (y|x)N (x,y|x ,y )(14.155)(X ,Y)+

where (X , Y)+ denotes all the letters x, y in X , Y with pY |X (y|x) > 0. Take the logarithmof the above expression:Xlog pY n |X n (y n |xn ) =N (x, y|xn , y n ) log pY |X (y|x) .
(14.156)x,y∈(X ,Y)+

Multiply both sides by − n1 :−

1log pY n |X n (y n |xn ) = −nXx,y∈(X ,Y)1
N (x, y|xn , y n ) log pY |X (y|x) .
n+(14.157)

The following relations hold because the sequence xn is strongly typical and y n is strong
conditionally typical:
1
N (x|xn ) − pX (x) ≤ δ 0 ,n1⇒ ∀x ∈ X + : −δ 0 + pX (x) ≤ N (x|xn ) ≤ δ 0 + pX (x),n11∀x, y ∈ (X , Y)+ : N (x, y|xn , y n ) − pY |X (y|x) N (x|xn ) ≤ δnn∀x ∈ X + :(14.158)(14.159)(14.160)

1
1⇒ ∀x, y ∈ (X , Y)+ : −δ + pY |X (y|x) N (x|xn ) ≤ N (x, y|xn , y n )nn1≤ δ + pY |X (y|x) N (x|xn ). (14.161)n
Now multiply (14.161) by − log pY |X (y|x) > 0, sum over all letters in the alphabet (X , Y)+ ,and apply the substitution in (14.157). This procedure gives the following set of inequalities:
X

−




1
n
−δ + pY |X (y|x) N (x|x ) log pY |X (y|x)
n+x,y∈(X ,Y)≤−1≤ − log pY n |X n (y n |xn )
n

X 

1
n
δ + pY |X (y|x) N (x|x ) log pY |X (y|x) . (14.162)n+x,y∈(X ,Y)

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14.9. STRONG CONDITIONAL TYPICALITY

451

Now apply the inequalities in (14.159) (assuming that pX (x) ≥ δ 0 for x ∈ X + ) to get thatX⇒−−δ + pY |X (y|x) (−δ 0 + pX (x)) log pY |X (y|x)(14.163)x,y∈(X ,Y)+≤−1
≤ − log pY n |X n (y n |xn )nδ + pY |X (y|x) (δ 0 + pX (x)) log pY |X (y|x)

X(14.164)(14.165)+x,y∈(X ,Y)1⇒ −c(δ + δ 0 ) + H(Y |X) ≤ − log pY n |X n (y n |xn ) ≤ c(δ + δ 0 ) + H(Y |X),n0−n(H(Y |X)+c(δ+δ 0 ))⇒2
≤ pY n |X n (y n |xn ) ≤ 2−n(H(Y |X)−c(δ+δ )) ,(14.166)(14.167)

where
c≡−

X


log pY |X (y|x) ≥ 0.

(14.168)+x,y∈(X ,Y)

It again becomes apparent why we require the technical condition in the definition of strong
conditional typicality (Definition 14.9.2). Were it not there, then the constant c would not
be finite, and we would not be able to obtain a reasonable bound on the probability of astrong conditionally typical sequence.
We close this section with a lemma that relates strong conditional, marginal, and joint
typicality.
Y n |xn

Lemma 14.9.1 Suppose that y n is a conditionally typical sequence in Tδ
and its condinXnnntioning sequence x is a typical sequence in Tδ0 . Then x and y are jointly typical in the
XnY nYn
set Tδ+δ
, and y n is a typical sequence in T|X
0|(δ+δ 0 ) .
Proof. It follows from the above that ∀x ∈ X , y ∈ Y :1
N (x|xn ) ≤ δ 0 + pX (x),n1
1
1
pY |X (y|x) N (x|xn ) − δ ≤ N (x, y|xn , y n ) ≤ δ + pY |X (y|x) N (x|xn ).
nnnpX (x) − δ 0 ≤(14.169)(14.170)

Substituting the upper bound on n1 N (x|xn ) gives1
N (x, y|xn , y n ) ≤ δ + pY |X (y|x) (δ 0 + pX (x))
n= δ + pY |X (y|x)δ 0 + pX (x)pY |X (y|x)≤ δ + δ 0 + pX,Y (x, y).

(14.171)(14.172)(14.173)

Similarly, substituting the lower bound on n1 N (x|xn ) gives1
N (x, y|xn , y n ) ≥ pX,Y (x, y) − δ − δ 0 .
n
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CHAPTER 14. CLASSICAL TYPICALITY

Putting both of the above bounds together, we get the following bound:1
N (x, y|xn , y n ) − pX,Y (x, y) ≤ δ + δ 0 .
n(14.175)nn

X Y
This then implies that the sequences xn and y n lie in the strong jointly typical set Tδ+δ.
0nYn
It follows from the result of Exercise 14.8.1 that y ∈ T|X |(δ+δ0 ) .

14.10 Application: Channel Capacity Theorem
We close the technical content of this chapter with a remarkable application of conditional
typicality: Shannon’s channel capacity theorem. As discussed in Section 2.2.3, this theorem
is one of the central results of classical information theory, appearing in Shannon’s seminal
paper. The theorem establishes that the highest achievable rate for communication overmany independent uses of a classical channel is equal to a simple function of the channel.
We begin by defining the information-processing task and a corresponding (n, R, ε) channel code. It is helpful to recall Figure 2.4 depicting a general protocol for communication
over a classical channel N ≡ pY |X (y|x). Before communication begins, the sender Alice andreceiver Bob have already established a codebook {xn (m)}m∈M , where each codeword xn (m)corresponds to a message m that Alice might wish to send to Bob. If Alice wishes to send
message m, she inputs the codeword xn (m) to the i.i.d. channel N n ≡ pY n |X n (y n |xn ). Moreformally, her encoding is some map E n : M → X n . She then exploits n uses of the channel tosend xn (m). Bob receives some sequence y n from the output of the channel, and he performs
a decoding Dn : Y n → M in order to recover the message m that Alice transmits. The rate
R of the code is equal to [log |M|] /n, measured in bits per channel use. The probability oferror pe for an (n, R, ε) channel code is bounded from above as
pe ≡ max Pr {Dn (N n (E n (m))) 6= m} ≤ ε.
m

(14.176)

A communication rate R is achievable for N if there exists an (n, R − δ, ε) channel code
for all ε ∈ (0, 1), δ > 0 and sufficiently large n. The channel capacity C(N ) of N is thesupremum of all achievable rates for N . We can now state Shannon’s channel capacity
theorem:
Theorem 14.10.1 (Shannon Channel Capacity) The maximum mutual information I(N )is equal to the capacity C(N ) of a channel N ≡ pY |X (y|x):
C(N ) = I(N ) ≡ max I(X; Y ).

(14.177)

pX (x)

Proof. A proof consists of two parts. The first part, known as the direct coding theorem,
demonstrates that the RHS ≤ LHS. That is, there is a sequence of channel codes with rate
I(N ), demonstrating that this rate is achievable. The second part, known as the converse
part, demonstrates that the LHS ≤ RHS. That is, it demonstrates that the rate on the RHS
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14.10. APPLICATION: CHANNEL CAPACITY THEOREM

453

is optimal, and it is impossible to have achievable rates exceeding it. Here, we prove thedirect coding theorem and hold off on proving the converse part until we reach the HSW
theorem in Chapter 20 because the converse theorem there suffices as the converse part forthis classical theorem. We have already outlined the proof of the direct coding theorem in
Section 2.2.4, and it might be helpful at this point to review this section. In particular, theproof breaks down into three parts: random coding to establish the encoding, the decoding
algorithm for the receiver, and the error analysis. We now give all of the technical details of
the proof because this chapter has established all the tools that we need.
Code Construction. Before communication begins, Alice and Bob agree upon a code by
the following random selection procedure. For every message m ∈ M, generate a codeword
xn (m) i.i.d. according to the product distribution pX n (xn ), where pX (x) is a distributionthat maximizes I(N ). Importantly, this random construction is such that every codeword
is generated independently of the other codewords.
Encoding. If Alice wishes to send message m, she inputs the codeword xn (m) to thechannels.
Decoding Algorithm. After receiving the sequence y n from the channel outputs,nYnBob
P tests whether y is in the typical set Tδ corresponding to the distribution pY (y) ≡x pY |X (y|x)pX (x). If it is not, then he reports an error. He then tests if there is someY n |xn (m)message m such that the sequence y n is in the conditionally typical set Tδ. If m is
Y n |xn (m)nthe unique message such that y ∈ Tδ
, then he declares m to be the transmitted
Y n |xn (m)message. If there is no message m such that y n ∈ Tδor multiple messages m0 suchn |xn (m0 )Ythat y n ∈ Tδ
, then he reports an error. Observe that the decoder is a function of thechannel, so that we might say that we construct channel codes “from the channel.”
Error Analysis. As discussed in the above decoding algorithm, there are three kinds oferrors that can occur in this communication scheme when Alice sends the codeword xn (m)over the channels:n

E0 (m): The event that the channel output y n is not in the typical set TδY .
n

E1 (m): The event that the channel output y n is in TδY but not in the conditionally typicalY n |xn (m)set Tδ
.
n

E2 (m): The event that the channel output y n is in TδY but it is in the conditionally typicalset for some other message: nny ∈ TδYno
Y n |xn (m0 )0nand ∃m 6= m : y ∈ Tδ.

(14.178)

Recall from Section 2.2.4 that it is helpful to analyze the expectation of the average
error probability, where the expectation is with respect to the random selection of thecode and the average is with respect to a uniformly random choice of the message m. Let
C ≡ {X n (1), X n (2), . . . , X n (|M|)} denote the random variable corresponding to the randomc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 14. CLASSICAL TYPICALITY

selection of a code. The expectation of the average error probability of a randomly selectedcode is as follows:)(
1 X
Pr {E0 (m) ∪ E1 (m) ∪ E2 (m)} .
(14.179)
EC|M| m
Our first “move” is to exchange the expectation and the sum, following from linearity of the
expectation:
1 X
EC {Pr {E0 (m) ∪ E1 (m) ∪ E2 (m)}} .
(14.180)|M| m
Since all codewords are selected in the same way (randomly and independently of the message
m and according to the same distribution pX n (xn )), the following equality holds for all
m, m0 ∈ M:
EC {Pr {E0 (m) ∪ E1 (m) ∪ E2 (m)}} = EC {Pr {E0 (m0 ) ∪ E1 (m0 ) ∪ E2 (m0 )}} ,(14.181)

implying that it suffices to analyze EC {Pr {E0 (m) ∪ E1 (m) ∪ E2 (m)}} for just a single message
m. Without loss of generality, we can pick m = 1 (the first message). Using the above, wefind that the expectation of the average error probability simplifies as follows:
1 X
EC {Pr {E0 (m) ∪ E1 (m) ∪ E2 (m)}} = EC {Pr {E0 (1) ∪ E1 (1) ∪ E2 (1)}} .
|M| m(14.182)

So we can then apply the union bound:
EC {Pr {E0 (1) ∪ E1 (1) ∪ E2 (1)}}≤ EC {Pr {E0 (1)}} + EC {Pr {E1 (1)}} + EC {Pr {E2 (1)}} . (14.183)
We now analyze each error individually. For each of the above events, we can exploit
indicator functions in order to simplify the error analysis (we are also doing this to help
build a bridge between this classical proof and the packing lemma approach for the quantum
case in Chapter 16—projectors in some sense replace indicator functions later on). Recall
that an indicator function IA (x) is equal to one if x ∈ A and equal to zero otherwise. So
the following three functions being equal to one or larger then corresponds to error events
E0 (1), E1 (1), and E2 (1), respectively:
1 − ITδY n (y n ),ITδY n (y n ) 1 − IT Y n |xn (1) (y n ) ,δXnITδY n (y )IT Y n |xn (m0 ) (y n ).
m0 6=1(14.184)(14.185)(14.186)

δ(The last sum of indicators is a consequence of applying the union bound again to the error
E2 (1), which itself is a union of events.)
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14.10. APPLICATION: CHANNEL CAPACITY THEOREM

455

By exploiting the indicator function from (14.184), we have thatEC {Pr {E0 (1)}}nnoo= EX n (1) EY n |X n (1) 1 − ITδY n (Y n )nonn= 1 − EX n (1),Y n ITδY (Y )no
= 1 − EY n ITδY n (Y n )n= Pr Y n ∈/ TδY ≤ ε,(14.187)(14.188)(14.189)(14.190)

where the first line follows because Y n is generated according to the conditional distribution
pY n |X n and from X n (1) (since the first message was transmitted) and all other codewords
have no role in the test, so that we marginalize over them. In the last line we have exploitednthe high probability property of the typical set TδY . In the above, we are also exploiting thefact that E {IA } = Pr {A}. By exploiting the indicator function from (14.185), we have thatEC {Pr {E1 (1)}}
n
n

oo
n
n
n
n
n
= EX n (1) EY n |X n (1) ITδY (Y ) 1 − IT Y |X (1) (Y )
δ
n
n
oo
n
≤ EX n (1) EY n |X n (1) 1 − IT Y n |X n (1) (Y )nn δoo
= 1 − EX n (1) EY n |X n (1) IT Y n |X n (1) (Y n )δon
Y n |X n (1)n= EX n (1)
Prn
Y ∈
/ Tδ
≤ ε,
n
Y |X (1)(14.191)(14.192)(14.193)(14.194)

where in the lastline we have exploited the high probability property of the conditionallyY n |X n (1)typical set Tδ. We finally consider the probability of the last kind of error by exploiting
the indicator function in (14.186):
EC {Pr {E2 (1)}})(
X
ITδY n (y n )IT Y n |X n (m0 ) (y n )≤ EC
m0 6=1=Xno
EC ITδY n (y n )IT Y n |X n (m0 ) (y n )(14.196)no
EX n (1),X n (m0 ),Y n ITδY n (y n )IT Y n |X n (m0 ) (y n )(14.197)δ

m0 6=1=Xδ

m0 6=1=X(14.195)

δX

pX n (xn (1))pX n (xn (m0 ))×

m0 6=1 xn (1),xn (m0 ),y n

pY n |X n (y n |xn (1))ITδY n (y n )IT Y n |xn (m0 ) (y n )
δ
X X
n
0
n
=
pX n (x (m ))pY n (y )ITδY n (y n )IT Y n |xn (m0 ) (y n ).
m0 6=1 xn (m0 ),y n

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CHAPTER 14. CLASSICAL TYPICALITY

The first inequality is from the union bound, and the first equality follows from the waythat we select the random code: for every message m, the codewords are selected independently and randomly according to pX n so that the distribution for the joint random variable
X n (1)X n (m0 )Y n ispX n (xn (1)) pX n (xn (m0 )) pY n |X n (y n |xn (1)).
(14.200)
The second equality follows from marginalizing over X n (1). Continuing, we have
X X
≤ 2−n[H(Y )−δ]pX n (xn (m0 ))IT Y n |xn (m0 ) (y n )= 2−n[H(Y )−δ]X X

pX n (xn (m0 ))≤ 2−n[H(Y )−δ] 2n[H(Y |X)+δ]X
yn

m0 6=1 xn (m0 )X X(14.201)δ

m0 6=1 xn (m0 ),y n

IT Y n |xn (m0 ) (y n )(14.202)

δ

pX n (xn (m0 ))(14.203)m0 6=1 xn (m0 )≤ |M| 2−n[I(X;Y )−2δ] .

(14.204)

The first inequality follows from the bound pY n (y n )ITδY n (y n ) ≤ 2−n[H(Y )−δ] that holds for
Y n |xn (m0 )typical sequences. The second inequality follows from the cardinality bound Tδ

≤

2n[H(Y |X)+δ] on the conditionally typical set. The last inequality follows becauseX
pX n (xn (m0 )) = 1,

(14.205)xn (m0 )P|M| is an upper bound on m0 6=1 1 = |M| − 1, and by the identity I(X; Y ) = H(Y ) −H(Y |X). Thus, we can make this error arbitrarily small by choosing the message set size
|M| = 2n[I(X;Y )−3δ] . Putting everything together, we have the following bound on (14.179):
ε0 ≡ 2ε + 2−nδ ,(14.206)

as long as we choose the message set size as given above. It follows that there exists a
particular code with the same error bound on its average error probability. We can thenexploit an expurgation argument as discussed in Section 2.2.4 to convert an average error
bound into a maximal one (the expurgation step throws away the worse half of the codewords,
guaranteeing a bound of 2ε0 on the maximum error probability). Thus, we have shown theachievability of an (n, C(N ) − δ 0 , 2ε0 ) channel code for all δ 0 > 0, ε0 ∈ (0, 1/2) and sufficiently
large n (where δ 0 = 3δ + 1/n). Finally, as a simple observation, our proof above does notrely on whether the definition of conditional typicality employed is weak or strong.

14.11 Concluding Remarks
This chapter deals with many different definitions and flavors of typicality in the classicalworld, but the essential theme is Shannon’s central insight—the application of the law ofc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

14.12. HISTORY AND FURTHER READING

457

large numbers in information theory. Our main goal in information theory is to analyze theprobability of error in the transmission or compression of information. Thus, we deal withprobabilities and we do not care much what happens for all sequences, but we instead only
care what happens for the likely sequences. This frame of mind immediately leads to thedefinition of a typical sequence and to a simple scheme for the compression of information—
keep only the typical sequences and performance is optimal in the asymptotic limit. Despite
the seemingly different nature of quantum information when compared to its classical counterpart, the intuition developed in this chapter carries over to the quantum world in the next
chapter where we define several different notions of quantum typicality.

14.12 History and Further Reading
Cover and Thomas (2006) contains a great presentation of typicality in the classical case.
The proof of Property 14.7.5 is directly from the Cover and Thomas book. Berger (1977)
introduced strong typicality, and Csiszár and Körner (2011) systematically developed it.
Other helpful books on information theory are those of Berger (1971) and Yeung (2002).
There are other notions of typicality which are useful, including those presented in Gamal
and Kim (2012) and Wolfowitz (1978). Our proof of Shannon’s channel capacity theorem issimilar to that in Savov (2012).

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c 2019CHAPTER 14. CLASSICAL TYPICALITY

Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 15

Quantum Typicality
This chapter marks the beginning of our study of the asymptotic theory of quantum information, where we develop the technical tools underpinning this theory. The intuition for it
is similar to the intuition we developed in the previous chapter on typical sequences, but wewill find some important differences between the classical and quantum cases.
So far, there is not a single known information-processing task in quantum Shannon
theory where the tools from this chapter are not helpful in proving the achievability partof a coding theorem. For the most part, we can straightforwardly import many of theideas from the previous chapter about typical sequences for use in the asymptotic theoryof quantum information. However, one might initially think that there are some obstacles
to doing so. For example, what is the analogy of a quantum information source? Once wehave established this notion, how would we determine if a state emitted from a quantum
information source is a typical state? In the classical case, a simple way of determining
typicality is to inspect all of the bits in the sequence. But there is a problem with thisapproach in the quantum domain—“looking at quantum bits” is equivalent to performing ameasurement and doing so destroys delicate superpositions that we would want to preservein any subsequent quantum information-processing task.
So how can we get around the aforementioned problem and construct a useful notion ofquantum typicality? Well, we should not be so destructive in determining the answer to aquestion when it has only two possible answers. After all, we are only asking “Is the state
typical or not?”, and we can be a bit more delicate in the way that we ask this question. Asan analogy, suppose Bob is curious to determine whether Alice could join him for dinner at anice restaurant on the beach. He would likely just phone her and politely ask, “Sweet Alice,
are you available for a lovely oceanside dinner?”, as opposed to barging into her apartment,
probing through all of her belongings in search of her calendar, and demanding that she join
him if she is available. This latter infraction would likely disturb her so much that she would
never speak to him again (and what would become of quantum Shannon theory without
these two communicating!). It is the same with quantum information—we must be gentle
when handling quantum states. Otherwise, we will disturb the state so much that it will not
be useful in any future quantum information-processing task.
459

460

CHAPTER 15. QUANTUM TYPICALITY

We can gently ask a binary question of a quantum system by constructing an incomplete
measurement with only two outcomes. If one outcome has a high probability of occurring,then we do not learn much about the state after learning this outcome, and thus we would
expect that this inquiry does not disturb the state very much. For the case above, we canformulate the question, “Is the state typical or not?” as a binary measurement that returns
only the answer to this question and no more information. Since it is highly likely that thestate is indeed a typical state, we would expect this inquiry not to disturb the state very
much, and we could use it for further quantum information-processing tasks. This is theessential content of this chapter, and there are several technicalities necessary to provide arigorous underpinning.
We structure this chapter as follows. We first discuss the notion of a typical subspace (thequantum analogy of the typical set). We can employ weak or strong notions of typicality
in the definition of quantum typicality. Section 15.2 then discusses conditional quantum
typicality, a form of quantum typicality that applies to quantum states chosen randomly
according to a classical sequence. We end this chapter with a brief discussion of the methodof types for quantum systems. All of these developments are important for understanding
the asymptotic nature of quantum information and for determining the ultimate limits ofstorage and transmission with quantum media.

15.1 The Typical Subspace
Our first task is to establish the notion of a quantum information source. It is analogous to
the notion of a classical information source, in the sense that the source randomly outputs
a quantum state according to some probability distribution, but the states that it outputs
do not necessarily have to be distinguishable as in the classical case.
Definition 15.1.1 (Quantum Information Source) A quantum information source issome device that randomly emits pure qudit states in a Hilbert space HA of finite dimension.
We use the symbol A to denote the quantum system for the quantum information source.
Suppose that the source outputs states |ψy i randomly according to some probability distribution pY (y). Note that the states |ψy i do not necessarily have to form an orthonormal set.
Then the density operator ρA of the source is the expected state emitted:
XρA ≡ EY {|ψY ihψY |A } =pY (y)|ψy ihψy |A .
(15.1)y

There are many decompositions of a density operator as a convex sum of rank-one projectors(and the above decomposition is one such example), but perhaps the most important suchdecomposition is a spectral decomposition of the density operator ρ:XρA =pX (x)|xihx|A .
(15.2)
x∈X
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15.1. THE TYPICAL SUBSPACE

461

The above states |xiA are eigenvectors of ρA and form a complete orthonormal basis for
Hilbert space HA , and the non-negative, convex real numbers pX (x) are the eigenvaluesof ρA .
We have written the states |xiA and the eigenvalues pX (x) in a suggestive notation
because it is actually possible to think of our quantum source as a classical information
source—the emitted states {|xiA }x∈X are orthonormal and each corresponding eigenvaluepX (x) acts as a probability for choosing |xiA . We can say that our source is classical because
it is emitting the orthogonal, and thus distinguishable, states |xiA with probability pX (x).
This description is equivalent to the ensemble {pY (y), |ψy i}y because the two ensembles leadto the same density operator (recall that two ensembles that have the same density operator
are essentially equivalent because they lead to the same probabilities for outcomes of any
measurement performed on the system). Our quantum information source then corresponds
to the pure-state ensemble:{pX (x), |xiA }x∈X .
(15.3)
Recall that the quantum entropy H(A)ρ of the density operator ρA is as follows (Definition 11.1.1):
H(A)ρ ≡ − Tr {ρA log ρA } .
(15.4)
It is straightforward to show that the quantum entropy H(A)ρ is equal to the Shannon
entropy H(X) of a random variable X with distribution pX (x) because the basis states |xiA
are orthonormal.
Suppose now that the quantum information source emits a large number n of random
quantum states so that the density operator describing the emitted state is as follows:ρ A1 ⊗ · · · ⊗ ρ An{z
} = (ρA )⊗n .
ρAn ≡ |
n(15.5)The labels A1 , . . . , An denote the Hilbert spaces corresponding to the different quantum
systems, but the density operator is the same for each quantum system A1 , . . . , An and isequal to ρA . The above description of a quantum source is within the i.i.d. setting for thequantum domain. A spectral decomposition of the state in (15.5) is as follows:
X
X
ρAn =
pX (x1 )|x1 ihx1 |A1 ⊗ · · · ⊗pX (xn )|xn ihxn |An(15.6)
x1 ∈X

=

xn ∈X

X

pX (x1 ) · · · pX (xn ) (|x1 i · · · |xn i) (hx1 | · · · hxn |)A1 ,...,An(15.7)

x1 ,··· ,xn ∈X=X

pX n (xn )|xn ihxn |An ,(15.8)

xn ∈X nwhere we employ the shorthand:pX n (xn ) ≡ pX (x1 ) · · · pX (xn ),|xn iAn ≡ |x1 iA1 · · · |xn iAn .

(15.9)

The above quantum description of the density operator is essentially equivalent to the classical picture of n realizations of random variable X with each eigenvalue pX1 (x1 ) · · · pXn (xn )c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 15. QUANTUM TYPICALITY

acting as a probability because the set of states {|x1 i · · · |xn iA1 ,...,An }x1 ,··· ,xn ∈X is an orthonormal set.
We can now “quantize” or extend the notion of typicality to the quantum information
source. The definitions follow directly from the classical definitions in Chapter 14. The
quantum definition of typicality can employ either the weak notion as in Definition 14.2.3
or the strong notion as in Definition 14.7.2. We do not distinguish the notation for a typicalsubspace and a typical set because it should be clear from the context which kind of typicality
we are employing.
Definition 15.1.2 (Typical Subspace) The δ-typical subspace TAδ n is a subspace of thefull Hilbert space HAn = HA1 ⊗ · · · ⊗ HAn , associated with many copies of a density operator,
such as the one in (15.2). It is spanned by states |xn iAn whose corresponding classicalsequences xn are δ-typical:
nTAδ n ≡ span |xn iAn : xn ∈ TδX ,(15.10)where it is implicit that the typical subspace TAδ n on the left-hand side is with respect tona density operator ρ and the typical set TδX on the right-hand side is with respect to the
distribution pX (x) from the spectral decomposition of ρ in (15.2). We could also denote thetypical subspace as TAρ,δn if we would like to make the dependence of the space on ρ more
explicit.

15.1.1The Typical Subspace Measurement

The definition of the typical subspace (Definition 15.1.2) gives a way to divide up the Hilbert
space of n qudits into two subspaces: the typical subspace and the atypical subspace. The
properties of the typical subspace are similar to what we found for typical sequences. Thatis, the typical subspace is exponentially smaller than the full Hilbert space of n qudits, yet it
contains nearly all of the probability (in a sense that we show below). The intuition for theseproperties of the typical subspace is the same as it is classically, as depicted in Figure 14.2,
once we have a spectral decomposition of a density operator.
The typical projector is a projector onto the typical subspace, and the complementary
projector projects onto the atypical subspace. These projectors play an important operational role in quantum Shannon theory because we can construct a quantum measurementfrom them. That is, this measurement is the best way of asking the question, “Is the state
typical or not?” because it minimally disturbs the state while still retrieving this one bit of
information.
Definition 15.1.3 (Typical Projector) Let ΠδAn denote the typical projector for the typical subspace of a density operator ρA with spectral decomposition in (15.2). It is a projector
onto the typical subspace:XΠδAn ≡|xn ihxn |An ,(15.11)xn ∈TδX
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15.1. THE TYPICAL SUBSPACE

463

where it is implicit that the xn below the summation is a classical sequence in the typical setnTδX , and the state |xn i is a quantum state given in (15.9) and associated with the classicalsequence xn via the spectral decomposition of ρ in (15.2). We can also denote the typicalprojector as Πρ,δAn if we would like to make its dependence on ρ explicit.
The action of multiplying the density operator ρAn by the typical projector ΠδAn is toselect out all the basis states of ρAn that are in the typical subspace and form a “sliced”
operator ρ̃An that is close to the original density operator ρAn :X
pX n (xn )|xn ihxn |An .
(15.12)ρ̃An ≡ ΠδAn ρAn ΠδAn =xn ∈TδXnThat is, the effect of projecting a state onto the typical subspace TAδ n is to “slice” out any
component of the state ρAn that does not lie in the typical subspace TAδ n .
Exercise 15.1.1 Show that the typical projector ΠδAn commutes with the density operator ρAn :ρAn ΠδAn = ΠδAn ρAn .

(15.13)

The typical projector allows us to formulate an operational method for delicately asking
the question: “Is the state typical or not?” We can construct a quantum measurement thatconsists of two outcomes: the outcome “1” reveals that the state is in the typical subspace,and “0” reveals that it is not. This typical subspace measurement is often one of the firstimportant steps in most protocols in quantum Shannon theory.
Definition 15.1.4 (Typical Subspace Measurement) The following map is a quantum
instrument (see Section 4.6.8) that realizes the typical subspace measurement:σ → I − ΠδAn σ I − ΠδAn ⊗ |0ih0| + ΠδAn σΠδAn ⊗ |1ih1|,(15.14)
where σ is some density operator acting on the Hilbert space HAn . It associates a classicalregister with the outcome of the measurement—the value of the classical register is |0i forthe support of the state σ that is not in the typical subspace, and it is equal to |1i for the
support of the state σ that is in the typical subspace.
The implementation of a typical subspace measurement is currently far from the realityof what is experimentally accessible if we would like to have the measure concentration
effects necessary for proving many of the results in quantum Shannon theory. Recall from
Figure 14.1 that we required a sequence of about a million bits in order to have the needed
measure concentration effects. We would need a similar number of qubits emitted froma quantum information source, and furthermore, we would require the ability to performnoiseless coherent operations over about a million or more qubits in order to implement
the typical subspace measurement. Such a daunting requirement firmly places quantum
Shannon theory as a “highly theoretical theory,” rather than being a theory that can makeclose connection to current experimental practice.1
1

We should note that this was certainly the case as well for information theory when Claude Shannon
developed it in 1948, but in the many years since then, there has been much progress in the development ofpractical classical codes for achieving the classical capacity of a classical channel.
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CHAPTER 15. QUANTUM TYPICALITY

15.1.2

The Difference between the Typical Set and the Typical
Subspace

We now offer a simple example to discuss the difference between the classical viewpoint associated with the typical set and the quantum viewpoint associated with the typical subspace.
Suppose that a quantum information source emits the state |+i with probability 1/2 and itemits the state |0i with probability 1/2. For the moment, let us ignore the fact that the twostates |+i and |0i are not perfectly distinguishable and instead suppose that they are. Thenit would turn out that nearly every sequence emitted from this source is a typical sequence
because the distribution of the source is uniform. Recall that the typical set has size roughly
equal to 2nH(X) , and in this case, the entropy of the distribution 12 , 12 is equal to one bit.
Thus the size of the typical set is roughly the same as the size of the set of all sequences forthis distribution because 2nH(X) = 2n .
Now let us take into account the fact that the states |+i and |0i are not perfectly
distinguishable and use the prescription given in Definition 15.1.2 for the typical subspace.
The density operator of the above ensemble is as follows:11|+ih+| + |0ih0| =
2
2

3
4
1
4

1
4
1
4


,

(15.15)where its matrix representation is with respect to the computational basis. The spectral
decomposition of the density operator iscos2 (π/8)|ψ0 ihψ0 | + sin2 (π/8)|ψ1 ihψ1 |,(15.16)

where the states |ψ0 i and |ψ1 i are orthogonal, and thus distinguishable from one another.
The quantum information source that outputs |0i and |+i with equal probability is thus
equivalent to a source that outputs |ψ0 i with probability cos2 (π/8) and |ψ1 i with probability
sin2 (π/8).
We construct the projector onto the typical subspace by taking sums of typical stringsof the states |ψ0 ihψ0 | and |ψ1 ihψ1 | rather than the states |0ih0|
and |+ih+|, where typicality22
is with respect to the distribution cos (π/8), sin (π/8) . The dimension of the typicalsubspace corresponding to the quantum information source is far different from the size ofthe aforementioned typical set corresponding to the distribution (1/2, 1/2).
It is roughly

2
0.6n
2
equal to 2
because the entropy of the distribution cos (π/8), sin (π/8) is about 0.6 bits.
This stark contrast in the sizes has to do with the non-orthogonality of the states from theoriginal description of the ensemble. That is, non-orthogonality of states in an ensemble
implies that the size of the typical subspace can potentially be dramatically smaller than thesize of the typical set corresponding to the distribution of the states in the ensemble. This
result has implications for the compressibility of quantum information, and we will discussthese ideas in more detail in Chapter 18. For now, we continue with the technical details oftypical subspaces.
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15.1. THE TYPICAL SUBSPACE

15.1.3465

Properties of the Typical Subspace

The typical subspace TAδ n enjoys several useful properties that are “quantized” versions of
the typical sequence properties:
Property 15.1.1 (Unit Probability) Suppose that we perform a typical subspace measurement of a state ρAn . Then the probability that the quantum state ρAn is in the typicalsubspace TAδ n approaches one as n becomes large. That is,(15.17)Tr ΠδAn ρAn ≥ 1 − ε,for all ε ∈ (0, 1), δ > 0, and sufficiently large n, where ΠδAn is the typical subspace projector
from Definition 15.1.3.
Property 15.1.2 (Exponentially Smaller Dimension) The dimension dim(TAδ n ) of theδ-typical subspace is exponentially smaller than the dimension |A|n of the entire space ofquantum states when the output of the quantum information source is not maximally mixed.
We formally state this property as follows:Tr ΠδAn ≤ 2n(H(A)+cδ) ,(15.18)where c is some positive constant that depends on whether we employ the weak or strong
notion of typicality. We can also bound the dimension dim(TAδ n ) of the δ-typical subspacefrom below:Tr ΠδAn ≥ (1 − ε) 2n(H(A)−cδ) ,(15.19)for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
Property 15.1.3 (Equipartition) The operator ΠδAn ρAn ΠδAn corresponds to a “slicing” ofthe density operator ρAn where we slice out and keep only the part with support in the typicalsubspace. We can then bound all of the eigenvalues of the sliced operator ΠδAn ρAn ΠδAn as
follows:2−n(H(A)+cδ) ΠδAn ≤ ΠδAn ρAn ΠδAn ≤ 2−n(H(A)−cδ) ΠδAn .
(15.20)
The above inequality is an operator inequality. It is a statement about the eigenvalues ofthe operators ΠδAn ρAn ΠδAn and ΠδAn , and these operators have the same eigenvectors because
they commute. Therefore, the above inequality is equivalent to the following inequality thatapplies in the classical case:n∀xn ∈ TδX : 2−n(H(A)+cδ) ≤ pX n (xn ) ≤ 2−n(H(A)−cδ) .

(15.21)This equivalence holds because each probability pX n (xn ) is an eigenvalue of ΠδAn ρAn ΠδAn .
The dimension dim(TAδ n ) of the δ-typical subspace is approximately equal to the dimension |X |n of the entire space only when the density operator of the quantum information
source is maximally mixed becauseTr ΠδAn ≤ |A|n · 2nδ ' |A|n .
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CHAPTER 15. QUANTUM TYPICALITY

Proofs of the above properties are essentially identical to those from the classical casein Sections 14.7.3 and 14.9.3, regardless of whether we employ a weak or strong notion ofquantum typicality. We leave the proofs as the three exercises below.
Exercise 15.1.2 Prove the unit probability property of the δ-typical subspace (Property 15.1.1).
First show that the probability that many copies of a density operator is in the δ-typical subspace is equal to the probability that a random sequence is δ-typical:
n(15.23)Tr ΠδAn ρAn = Pr X n ∈ TδX .
Exercise 15.1.3 Prove the exponentially smaller dimension property of the δ-typical subspace (Property 15.1.2). First show that the trace of the typical projector ΠδAn is equal to thedimension of the typical subspace TAδ n :
dim(TAδ n ) = Tr ΠδAn .
(15.24)
Then prove the property.
Exercise 15.1.4 Prove the equipartition property of the δ-typical subspace (Property 15.1.3).
First show thatXΠδAn ρAn ΠδAn =pX n (xn )|xn ihxn |An ,(15.25)xn ∈TδXn

and then argue the proof.
The result of the following exercise shows that the sliced operator ρ̃An ≡ ΠδAn ρAn ΠδAn isa good approximation to the original state ρAn in the limit of many copies of the states, andit effectively gives a scheme for quantum data compression (more on this in Chapter 18).
√
Exercise 15.1.5 Use the gentle operator lemma (Lemma 9.4.2) to show that ρAn is 2 εclose to the sliced operator ρ̃An when n is large:√kρAn − ρ̃An k1 ≤ 2 ε.
(15.26)
Use the gentle measurement lemma (Lemma 9.4.1) to show that the sliced state
  δ
−1
Tr ΠAn ρAnρ̃An(15.27)√is 2 ε-close in trace distance to ρ̃An .

Exercise 15.1.6 Show that the purity Tr (ρ̃An )2 of the sliced state ρ̃An satisfies the following bound for sufficiently large n and any ε ∈ (0, 1) (use weak quantum typicality):(1 − ε) 2−n(H(A)+δ) ≤ Tr (ρ̃An )2 ≤ 2−n(H(A)−δ) .
(15.28)Exercise 15.1.7 Show that the following bounds hold for the rank and the ∞-norm of thesliced state ρ̃An for any ε ∈ (0, 1) and sufficiently large n:(1 − ε) 2n(H(A)−δ) ≤ rank(ρ̃An ) ≤ 2n(H(A)+δ) ,(15.29)2−n(H(A)+δ) ≤ kρ̃An k∞ ≤ 2−n(H(A)−δ) .

(15.30)(Recall that the rank of an operator is equal to the size of its support and that the infinity
norm is equal to its maximum eigenvalue. Again use weak quantum typicality.)c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

15.1. THE TYPICAL SUBSPACE

15.1.4467

The Typical Subspace for Bipartite or Multipartite States

Recall from Section 14.5 that two classical sequences xn and y n are weak jointly typical if
the joint sample entropy of xn y n is close to the joint entropy H(X, Y ) and if the sample
entropies of the individual sequences are close to their respective marginal entropies H(X)and H(Y ) (where the entropies are with respect to some joint distribution pX,Y (x, y)). How
would we then actually check that these conditions hold? The most obvious way is simply
to look at the sequence xn y n , compute its joint sample entropy, compare this quantity to
the true joint entropy, determine if the difference is under the threshold δ, and do the samefor the marginal sequences. These two operations both commute in the sense that we candetermine first if the marginals are typical and then if the joint sequence is typical or viceversa without any difference in which one we do first.
But such a commutation does not necessarily hold in the quantum world. The waythat we determine whether a quantum state is typical is by performing a typical subspacemeasurement. If we perform a typical subspace measurement of the whole system followedby such a measurement on the marginals, the resulting state is not necessarily the same asif we performed the marginal measurements followed by the joint measurements. For thisreason, the notion of weak joint typicality as given in Definition 14.5.3 does not really exist in
general for the quantum case. Nevertheless, we still briefly overview how one would handle
such a case and later give an example of a restricted class of states for which weak joint
typicality holds.
Suppose that we have a quantum system in the mixed state ρAB shared between two
parties A and B. We can decompose the mixed state with the spectral theorem:XpZ (z)|ψz ihψz |AB ,(15.31)ρAB =z∈Z

where the states {|ψz iAB }z∈Z form an orthonormal basis for the joint quantum system AB
and each of the states |ψz iAB can be entangled in general.
We can consider the nth extension ρAn B n of the above state and abbreviate its spectral
decomposition as follows:XρAn B n ≡ (ρAB )⊗n =pZ n (z n )|ψzn ihψzn |An B n ,(15.32)z n ∈Z n

where
pZ n (z n ) ≡ pZ (z1 ) · · · pZ (zn ),|ψzn iAn B n ≡ |ψz1 iA1 B1 · · · |ψzn iAn Bn .

(15.33)(15.34)This development immediately leads to the definition of the typical subspace for a bipartite
state.
Definition 15.1.5 (Typical Subspace of a Bipartite State) The δ-typical subspace TAδ n B nof ρAB is the space spanned by states |ψzn iAn B n whose corresponding classical sequence z n is
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468

CHAPTER 15. QUANTUM TYPICALITYnin the typical set TδZ :n
TAδ n B n ≡ span |ψzn iAn B n : z n ∈ TδZ .

(15.35)The states |ψzn iAn B n are from a spectral decomposition of ρAB , and the distribution to consider for typicality of the classical sequence z n is pZ (z) from the spectral decomposition.
Definition 15.1.6 (Typical Projector of a Bipartite State) Let ΠδAn B n denote the projector onto the typical subspace of ρAB :X
ΠδAn B n ≡|ψzn ihψzn |An B n .
(15.36)z n ∈TδZn

Thus, there is ultimately no difference between the typical subspace for a bipartite state
and the typical subspace for a single-party state because the spectral decomposition gives away for determining the typical subspace and the typical projector in both cases. Perhaps
the only difference is a cosmetic one because AB denotes the bipartite system while Z
indicates a random variable with a distribution given from a spectral decomposition. Finally,
Properties 15.1.1–15.1.3 hold for quantum typicality of a bipartite state.

15.1.5The Jointly Typical Subspace for Classical States

The notion of weak joint typicality may not hold in the general case, but it does hold for a
special class of states that are completely classical. Suppose now that the mixed state ρABshared between two parties A and B has the following special form:XXρAB =pX,Y (x, y) (|xi ⊗ |yi) (hx| ⊗ hy|)AB(15.37)x∈X y∈Y=XX

pX,Y (x, y)|xihx|A ⊗ |yihy|B ,(15.38)x∈X y∈Y

where the states {|xiA }x∈X and {|yiB }y∈Y form an orthonormal basis for the respective
systems X and Y. This state has only classical correlations because Alice and Bob can
prepare it simply by local operations and classical communication. That is, Alice can sample
from the distribution pX,Y (x, y) in her laboratory and send Bob the variable y. Furthermore,
the states on A and B locally form a distinguishable set.
We can consider the nth extension ρAn B n of the above state:
ρAn B n ≡ (ρAB )⊗n
X=(15.39)pX n ,Y n (xn , y n ) (|xn i ⊗ |y n i) (hxn | ⊗ hy n |)An B n(15.40)pX n ,Y n (xn , y n )|xn ihxn |An ⊗ |y n ihy n |B n .

(15.41)xn ∈X n ,y n ∈Y n=X
xn ∈X n ,y n ∈Y n

This development immediately leads to the definition of the weak jointly typical subspace
for this special case.
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15.2. CONDITIONAL QUANTUM TYPICALITY

469

Definition 15.1.7 (Jointly Typical Subspace) The weak δ-jointly typical subspace TAδ n B n
is the space spanned by states |xn i|y n iAn B n whose corresponding classical sequence xn y n isin the jointly typical set:n n
TAδ n B n ≡ span |xn iAn |y n iB n : xn y n ∈ TδX Y .
(15.42)
Definition 15.1.8 (Jointly Typical Projector) Let ΠδAn B n denote the jointly typical projector. It is the projector onto the jointly typical subspace:XΠδAn B n ≡|xn ihxn |An ⊗ |y n ihy n |B n .
(15.43)n

xn ,y n ∈TδX Y

nProperties of the Jointly Typical Projector for Classical States
Properties 15.1.1–15.1.3 apply to the jointly typical subspace TAδ n B n because it is a typicalsubspace. The following property, analogous to Property 14.5.4 for classical joint typicality,
holds whenever the state ρAB has the special form in (15.37):
Property 15.1.4 (Probability of Joint Typicality) Let ρAn B n be a classical state asgiven in (15.39). Consider the following marginal density operators:ρAn ≡ TrB n {ρAn B n } ,ρB n ≡ TrAn {ρAn B n } .

(15.44)

Let us define ρÃn B̃ n as the following density operator:
ρÃn B̃ n ≡ ρAn ⊗ ρB n 6= ρAn B n .

(15.45)The marginal density operators of ρÃn B̃ n are therefore equivalent to the marginal density
operators of ρAn B n . Then we can bound the probability that the state ρÃn B̃ n lies in the typicalsubspace TAδ n B n :Tr ΠδAn B n ρÃn B̃ n ≤ 2−n(I(A;B)−3δ) .
(15.46)
Exercise 15.1.8 Prove the bound in Property 15.1.4:
Tr ΠδAn B n ρÃn B̃ n ≤ 2−n(I(A;B)−3δ) .

(15.47)15.2 Conditional Quantum Typicality
The notion of conditional quantum typicality is somewhat similar to the notion of conditional
typicality in the classical domain, but we again quickly notice some departures because
different quantum states do not have to be perfectly distinguishable. The technical tools
for conditional quantum typicality developed in this section are important for developing
schemes that send public or private classical information over a quantum channel (topics
discussed in Chapters 20 and 23).
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We first develop the notion of a conditional quantum information source. Consider arandom variable X with probability distribution pX (x). Let X be the alphabet of the randomvariable, and let |X | denote its cardinality. We also associate a quantum system X with therandom variable X and use an orthonormal set {|xi}x∈X to represent its realizations. We
again label the elements of the alphabet X as {x}x∈X .
Suppose we generate a realization x of random variable X according to its distributionpX (x), and we follow by generating a random quantum state according to some conditional
distribution. This procedure then gives us a set of |X | quantum information sources (eachof them are as in Definition 15.1.1). We index them by the classical index x, and thequantum information source has expected density operator ρxB if the emitted classical index
is x. Furthermore, we impose the constraint that each ρxB has the same dimension (onecould achieve this by embedding the lower dimensional states into a larger Hilbert space).
This quantum information source is therefore a “conditional quantum information source.”
Let HB and B denote the respective Hilbert space and system label corresponding to thequantum output of the conditional quantum information source. Let us call the resulting
ensemble the “classical–quantum ensemble” and say that a “classical–quantum information
source” generates it. The classical–quantum ensemble is as follows:{pX (x), |xihx|X ⊗ ρxB }x∈X ,(15.48)where we correlate the classical state |xiX with the density operator ρxB of the conditional
quantum information source. The expected density operator of the above classical–quantum
ensemble is the following classical–quantum state (discussed in Section 4.3.4):
ρXB ≡

X

pX (x)|xihx|X ⊗ ρxB .

(15.49)x∈X

The conditional quantum entropy H(B|X)ρ of the classical–quantum state ρXB is asfollows:
X
pX (x)H(ρxB ).
H(B|X)ρ =(15.50)x∈X

We can write a spectral decomposition of each conditional density operator ρxB as follows:
X

pY |X (y|x)|yx ihyx |B ,(15.51)y∈Ywhere the elements of the set {y}y∈Y label the elements of an alphabet Y, the orthonorxmal set {|yx iB }y∈Y is the set of eigenvectors of ρB , and the corresponding eigenvalues are
pY |X (y|x) y∈Y . We need the x label for the orthonormal set {|yx iB }y∈Y because the decomposition may be different for different density operators ρxB . The above notation is again
suggestive because the eigenvalues pY |X (y|x) correspond to conditional probabilities, and theset {|yx iB } of eigenvectors corresponds to an orthonormal set of quantum states conditionedon label x. With this respresentation, the conditional entropy H(B|X) reduces to a formulac 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

15.2. CONDITIONAL QUANTUM TYPICALITY

471

that looks like that for the classical conditional entropy:
X
H(B|X) =pX (x)H(ρxB )(15.52)x∈X=

X

pX (x)pY |X (y|x) log

x∈X ,y∈Y

1.
pY |X (y|x)(15.53)

We now consider when the classical–quantum information source emits a large number
n of states. The density operator for the output state ρX n B n is as follows:ρX n B n≡ (ρXB )⊗n!
X=pX (x1 )|x1 ihx1 |X1 ⊗ ρxB11 ⊗ · · · ⊗=XpX (xn )|xn ihxn |Xn ⊗ ρxBnn(15.55)

xn ∈X

x1 ∈X

X

(15.54)!

pX (x1 ) · · · pX (xn )|x1 i · · · |xn ihx1 | · · · hxn |X n ⊗ ρxB11 ⊗ · · · ⊗ ρxBnn .

(15.56)

x1 ,...,xn ∈X

We can abbreviate the above state asXn
pX n (xn )|xn ihxn |X n ⊗ ρxB n ,(15.57)xn ∈X n

where
pX n (xn ) ≡ pX (x1 ) · · · pX (xn ),|xn iX n ≡ |x1 iX1 · · · |xn iXn ,(15.58)ρxn
Bn≡ ρxB11 ⊗ · · · ⊗ ρxBnn ,(15.59)n

and a spectral decomposition for the state ρxB n isXn
ρxB n =
pY n |X n (y n |xn )|yxnn ihyxnn |B n ,(15.60)y n ∈Y n

where
pY n |X n (y n |xn ) ≡ pY1 |X1 (y1 |x1 ) · · · pYn |Xn (yn |xn ),|yxnn iB n ≡ | y1 x1 iB1 · · · | yn xn iBn .

(15.61)(15.62)

The above developments are a step along the way for formulating the definitions of weak
and strong conditional quantum typicality.

15.2.1

Weak Conditional Quantum Typicality

We can “quantize” the notion of weak classical conditional typicality so that it applies to aclassical–quantum information source.
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CHAPTER 15. QUANTUM TYPICALITY

Definition 15.2.1 (Weak Conditionally Typical Subspace) The conditionally typicalsubspace TBδ n |xn corresponds to a particular sequence xn and an ensemble {pX (x), ρxB }. It is
the subspace spanned by the states |yxnn iB n whose conditional sample entropy is δ-close to thetrue conditional quantum entropy:TBδ n |xn ≡ span |yxnn iB n : H(y n |xn ) − H(B|X) ≤ δ ,(15.63)where the states |yxnn iB n are formed from the eigenstates of the density operators ρxB (they areof the form in (15.62)) and the sample entropy is with respect to the distribution pY |X (y|x)
from (15.51).
Definition 15.2.2 (Weak Conditionally Typical Projector) The projector ΠδB n |xn onto
the conditionally typical subspace TBδ n |xn is as follows:ΠδB n |xn ≡X|yxnn ihyxnn |B n .

(15.64)

Y n |xn
yxnn ∈Tδ

15.2.2

Properties of the Weak Conditionally Typical Subspace

The weak conditionally typical subspace TBδ n |xn enjoys several useful properties that are“quantized” versions of the properties for weak conditionally typical sequences discussed in
Section 14.6. We should point out that we cannot really say much for several of the properties
for a particular sequence xn , but we can do so on average for a random sequence X n . Thus,
several of the properties give expected behavior for a random sequence X n . This convention
for quantum weak conditional typicality is the same as we had for classical weak conditional
typicality in Section 14.6.
Property 15.2.1 (Unit Probability) The expectation of the probability that we measurenδa random quantum state ρX
B n to be in the conditionally typical subspace TB n |X n approaches
one as n becomes large: n
EX n Tr ΠδB n |X n ρX
≥ 1 − ε,(15.65)Bn
for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
Property 15.2.2 (Exponentially Smaller Dimension) The dimension dim(TBδ n |xn ) of
the δ-conditionally typical subspace is exponentially smaller than the dimension |Y|n of theentire space of quantum states for most classical–quantum sources. We formally state thisproperty as follows:Tr ΠδB n |xn ≤ 2n(H(B|X)+δ) .
(15.66)
We can also bound the dimension dim(TBδ n |xn ) of the δ-conditionally typical subspace frombelow:
 
EX n Tr ΠδB n |X n
≥ (1 − ε) 2n(H(B|X)−δ) ,(15.67)for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
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15.2. CONDITIONAL QUANTUM TYPICALITY

473

n

Property 15.2.3 (Equipartition) The density operator ρxB n looks approximately maximally mixed when projected to the conditionally typical subspace:n2−n(H(B|X)+δ) ΠδB n |xn ≤ ΠδB n |xn ρxB n ΠδB n |xn ≤ 2−n(H(B|X)−δ) ΠδB n |xn .

(15.68)Exercise 15.2.1 Prove all three of the above properties for weak conditional quantum typicality.

15.2.3

Strong Conditional Quantum Typicality

We now develop the notion of strong conditional quantum typicality. This notion again
applies to an ensemble or to a classical–quantum state such as that given in (15.49). However,it differs from weak conditional quantum typicality because we can prove stronger statements
about the asymptotic behavior of conditional quantum systems (just as we could for theclassical case in Section 14.9). We begin this section with an example to build up ourintuition. We then follow with the formal definition of strong conditional quantum typicality,
and we end by proving some properties of the strong conditionally typical subspace.
Recall the example from Section 14.7. In a similar way to this example, we can draw asequence from an alphabet {0, 1, 2} according to the following distribution:1
pX (0) = ,41
pX (1) = ,41
pX (2) = .
2(15.69)One potential realization sequence is as follows:
201020102212.

(15.70)

The above sequence has four “zeros,” three “ones,” and five “twos,” so that the empirical
distribution of this sequence is (1/3, 1/4, 5/12) and has maximum deviation 1/12 from thetrue distribution in (15.69).
For each symbol in the above sequence, we could then draw from one of three quantum
information sources based on whether the classical index is 0, 1, or 2. Suppose that the
expected density operator of the first quantum information source is ρ0 , that of the second
is ρ1 , and that of the third is ρ2 . Then the density operator for the resulting sequence ofquantum states is as follows:
ρ2B1 ⊗ ρ0B2 ⊗ ρ1B3 ⊗ ρ0B4 ⊗ ρ2B5 ⊗ ρ0B6 ⊗ ρ1B7 ⊗ ρ0B8 ⊗ ρ2B9 ⊗ ρ2B10 ⊗ ρ1B11 ⊗ ρ2B12 ,(15.71)

where the subscripts label the systems as usual. So, the state of systems B1 , B5 , B9 , B10 ,and B12 is equal to five copies of ρ2 , the state of systems B2 , B4 , B6 , and B8 is equal to four
copies of ρ0 , and the state of systems B3 , B7 , and B11 is equal to three copies of ρ1 . Let Ix
be an indicator set for each x ∈ {0, 1, 2}, so that Ix consists of all the indices in the sequence
for which a symbol is equal to x. For the above example,
I0 = {2, 4, 6, 8} ,c 2019I1 = {3, 7, 11} ,I2 = {1, 5, 9, 10, 12} .

(15.72)

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CHAPTER 15. QUANTUM TYPICALITY

These sets serve as a way of grouping all of the density operators that are the same because
they correspond to the same classical symbol, and it is important to do so if we would like toconsider concentration of measure effects when we go to the asymptotic setting. As a visual
aid, we could permute the sequence of density operators in (15.71) if we would like to see
systems with the same density operator grouped together:
ρ0B2 ⊗ ρ0B4 ⊗ ρ0B6 ⊗ ρ0B8 ⊗ ρ1B3 ⊗ ρ1B7 ⊗ ρ1B11 ⊗ ρ2B1 ⊗ ρ2B5 ⊗ ρ2B9 ⊗ ρ2B10 ⊗ ρ2B12 .

(15.73)

There is then a typical projector for the first four systems with density operator ρ0 , a different
typical projector for the next three systems with density operator ρ1 , and an even different
typical projector for the last five systems with density operator ρ2 (however, the length of theabove quantum sequence is certainly not large enough to observe any measure concentration
effects!). Thus, the indicator sets Ix serve to identify which systems have the same density
operator so that we can know upon which systems a particular typical projector should act.
This example helps build our intuition of strong conditional quantum typicality, and we
can now begin to state what we would expect in the asymptotic setting. Suppose that theoriginal classical sequence is large and strongly typical, so that it has roughly n/4 occurrencesof “zero,” n/4 occurrences of “one,” and n/2 occurrences of “two.” We would then expect
the law of large numbers to come into play for n/4 and n/2 when n is large enough. Thus,
we can use the classical sequence to identify which quantum systems have the same density
operator, and apply a typical projector to each of these subsets of quantum systems. Then
all of the useful asymptotic properties of typical subspaces apply whenever n is large enough.
We can now state the definition of the strong conditionally typical subspace and thestrong conditionally typical projector, and we prove some of their asymptotic properties by
exploiting the properties of typical subspaces.
Definition 15.2.3 (Strong Conditionally Typical Subspace) The strong conditionallytypical subspace corresponds to a sequence xn and an ensemble {pX (x), ρxB }. Let a spectral
decomposition of each state ρxB be as in (15.51) with distribution pY |X (y|x) and corresponding
eigenstates |yx i. The strong conditionally typical subspace TBδ n |xn is then as follows:
)
(
O
|Ix |
(Y |x)
TBδ n |xn ≡ spany Ix ∈ Tδ
,(15.74)|yxIx iB Ix : ∀x,x∈X

where Ix ≡ {i : xi = x} is an indicator set that selects the indices i in the sequence xn forwhich the ith symbol xi is equal to x ∈ X , B Ix selects the systems from B n where the classicalsequence xn is equal to the symbol x, |yxIx i is some string of states from the set {|yx i}, y Ix
is a classical string corresponding to this string of states, Y |x is a random variable with
distribution pY |X (y|x), and |Ix | is the cardinality of the indicator set Ix .
Definition 15.2.4 (Strong Conditionally Typical Projector) The strong conditionallytypical projector again corresponds to a sequence xn and an ensemble {pX (x), ρxB }. It is atensor product of typical projectors for each state ρxB in the ensemble:
O ρ ,δ
ΠδB n |xn ≡ΠBxIx ,
(15.75)x∈X
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15.2. CONDITIONAL QUANTUM TYPICALITY

475

where Ix is defined in Definition 15.2.3, and B Ix indicates the systems onto which a particulartypical projector for ρx projects.2

15.2.4

Properties of the Strong Conditionally Typical Subspace

The strong conditionally typical subspace admits several useful asymptotic properties similar
to what we have seen before, and the proof strategy for proving all of them is similar to theway that we proved the analogous properties for the strong conditionally typical set in
Section 14.9.3. Suppose that we draw a sequence xn from a probability distribution pX (x),and we are able to draw as many samples as we wish so that it is very likely that the sequencexn is strongly typical and the occurrences N (x|xn ) of each symbol x are as large as we wish.
Then the following properties hold for xn strongly typical and each N (x|xn ) large.
Property 15.2.4 (Unit Probability) The probability that we measure a quantum staten
ρxB n to be in the conditionally typical subspace TBδ n |xn has the following lower bound:nTr ΠδB n |xn ρxB n ≥ 1 − ε,(15.76)

for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
Property 15.2.5 (Exponentially Smaller Dimension) The dimension dim(TBδ n |xn ) of
the δ-conditionally typical subspace is exponentially smaller than the dimension |B|n of theentire space of quantum states for all classical–quantum information sources besides ones
where all their density operators are maximally mixed. We formally state this property as
follows:00
Tr ΠδB n |xn ≤ 2n(H(B|X)+δ ) ,(15.77)
Y n |xn

where δ 00 is given in (15.92). We can also bound the dimension dim(Tδ
conditionally typical subspace from below:00Tr ΠδB n |xn ≥ (1 − ε) 2n(H(B|X)−δ ) ,) of the δ-(15.78)for all ε ∈ (0, 1), δ > 0, and sufficiently large n.
nProperty 15.2.6 (Equipartition) The state ρxB n is approximately maximally mixed when
projected onto the strong conditionally typical subspace:
00

n

00

2−n(H(B|X)+δ ) ΠδB n |xn ≤ ΠδB n |xn ρxB n ΠδB n |xn ≤ 2−n(H(B|X)−δ ) ΠδB n |xn ,(15.79)where δ 00 is given in (15.92).
2

Having the conditional density operators in the subscript breaks somewhat from our convention throughout this chapter, but it is useful here to indicate explicitly which density operator corresponds to a typicalprojector.
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CHAPTER 15. QUANTUM TYPICALITY

15.2.5

Proofs of the Properties of the Strong Conditionally Typical Subspace

Proof of the Unit Probability Property (Property 15.2.4). A proof of this property
is similar to the proof of Property 14.9.1 for the strong conditionally typical set. Since weare dealing with an i.i.d. distribution, we can assume without loss of generality that thesequence xn is lexicographically ordered with an order on the alphabet X . We write theelements of X as a1 , . . . , a|X | . Then the lexicographic ordering means that we can write thesequence of quantum states ρxn as follows:ρxn = ρa1 ⊗ · · · ⊗ ρa1 ⊗ ρa2 ⊗ · · · ⊗ ρa2 ⊗ · · · ⊗ ρa|X | ⊗ · · · ⊗ ρa|X | .
{z
} |
{z
}
|
{z
}
|
N (a1 |xn )N (a2 |xn )(15.80)N (a|X | |xn )It follows that N (ai |xn ) ≥ n (pX (ai ) − δ 0 ) from the typicality of xn , and thus the law oflarge numbers comes into play for each block ai · · · ai with length N (ai |xn ). The strong
conditionally typical projector ΠδB n |xn for this system is as follows:ΠδB n |xn ≡O

ΠρBxN,δ(x|xn ) ,(15.81)x∈X

because we assumed the lexicographic ordering of the symbols in the sequence xn . Eachprojector ΠρBxN,δ(x|xn ) in the above tensor product is a typical projector for the density operator
ρx when N (x|xn ) ≈ npX (x) becomes very large. Then we can apply the unit probability
property (Property 15.1.1) for each of these typical projectors, and it follows that()
O ρ ,δ δnnTr ΠB n |xn ρxB n = TrΠBxN (x|xn ) ρx⊗N (x|x )(15.82)
x∈X

=

Y

n
o
(x|xn )Tr ΠρBxN,δ(x|xn ) ρ⊗N
x(15.83)x∈X≥ (1 − ε)|X |≥ 1 − |X | ε,(15.84)(15.85)concluding the proof.
Proof of the Equipartition Property (Property 15.2.6). We first assume without lossnof generality that we can write the state ρxB n in lexicographic order as in (15.80). Then thestrong conditionally typical projector is again as in (15.81). It follows thatΠδB n |xn ρxn ΠδB n |xn =OnΠρBxN,δ(x|xn ) ρx⊗N (x|x ) ΠρBxN,δ(x|xn ) .

(15.86)x∈X

We can apply the equipartition property of the typical subspace for each typical projector
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15.2. CONDITIONAL QUANTUM TYPICALITY

477

ΠρBxN,δ(x|xn ) (Property 15.1.3):
OnO

ΠρBxN,δ(x|xn ) 2−N (x|x )(H(ρx )+cδ) ≤x∈Xn(x|x ) ρx ,δΠρBxN,δ(x|xn ) ρ⊗N
ΠB N (x|xn )
x

x∈X

≤

O

n

ΠρBxN,δ(x|xn ) 2−N (x|x )(H(ρx )−cδ) . (15.87)

x∈X

The following inequalities hold because the sequence xn is strongly typical as defined in
Definition 14.7.2:
O ρ ,δ
0
ΠBxN (x|xn ) 2−n(pX (x)+δ )(H(ρx )+cδ) ≤ ΠδB n |xn ρxn ΠδB n |xn
x∈X

≤

O

0

ΠρBxN,δ(x|xn ) 2−n(pX (x)−δ )(H(ρx )−cδ) . (15.88)

x∈X
0

We can factor out each term 2−n(pX (x)+δ )(H(ρx )+cδ) from the tensor products:
Y02−n(pX (x)+δ )(H(ρx )+cδ)

x∈X

O

ρx ,δ
δ
δ
ΠB
N (x|xn ) ≤ ΠB n |xn ρxn ΠB n |xn

x∈X

≤

Y

0

2−n(pX (x)−δ )(H(ρx )−cδ)x∈XOΠρBxN,δ(x|xn ) . (15.89)

x∈X

0

We then multiply out the |X | terms 2−n(pX (x)+δ )(H(ρx )+cδ) :2−n(H(B|X)+P) Πδ n n ≤ Πδ n n ρ n Πδ n n00x (H(ρx )δ +cpX (x)δ+cδδ )

B |x

B |x

x

−n(H(B|X)+

B |x

0
0
x cδδ −H(ρx )δ −cpX (x)δ ) Πδ≤2
B n |xn . (15.90)
PP
The final step below follows because x pX (x) = 1 and because the boundH(ρx ) ≤|X | log d applies where d is the dimension of the density operator ρx :

P

00

00

2−n(H(B|X)+δ ) ΠδB n |xn ≤ ΠδB n |xn ρxn ΠδB n |xn ≤ 2−n(H(B|X)−δ ) ΠδB n |xn ,(15.91)whereδ 00 ≡ δ 0 |X | log d + cδ + |X | cδδ 0 .

(15.92)This concludes the proof.
Exercise 15.2.2 Prove Property 15.2.5.

15.2.6Strong Conditional and Marginal Quantum Typicality

We end this section on strong conditional quantum typicality by proving a final property
that applies to a state drawn from an ensemble and the typical subspace of the expected
density operator of the ensemble.
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CHAPTER 15. QUANTUM TYPICALITY

Property 15.2.7
Consider an ensemble of the form {pX (x), ρx } with expected density opP
erator ρ ≡ x pX (x)ρx . Suppose that xn is a strongly typical sequence with respect to the
distribution pX (x) and leads to a conditional density operator ρxn . Then the probability ofmeasuring ρxn in the strongly typical subspace of ρ is high:Tr Πnρ,δ ρxn ≥ 1 − ε,(15.93)for all ε ∈ (0, 1), δ > 0, and sufficiently large n, where the typical projector Πnρ,δ is withrespect to the density operator ρ.
Proof. Let the expected density operator have the following spectral decomposition:Xρ=pZ (z)|zihz|.
(15.94)z

We define the “pinching” operation as a dephasing with respect to the basis {|zi}:X|zihz|σ|zihz|.
(15.95)σ → ∆(σ) ≡zLet ρx denote the pinched version of the conditional density operators ρx :X
X
pZ|X (z|x)|zihz|,|zihz|ρx |zihz| =ρx ≡ ∆(ρx ) =(15.96)zz

where pZ|X (z|x) ≡ hz|ρx |zi. This pinching is the crucial insight for the proof because all of
the pinched density operators ρx have a common eigenbasis and the analysis reduces from aquantum one to a classical one that exploits the properties of strong marginal, conditional,
and joint typicality. The following chain of inequalities then holds by exploiting the above
definitions:


 X

 n
n
n
Tr Πρ,δ ρxn = Tr|z ihz |ρxn
(15.97)
 n Zn

z ∈Tδ


 X

= Tr
|z n i hz n |z n i hz n |ρxn
(15.98)
 n Zn

z ∈Tδ


 X

|z n ihz n |ρxn |z n ihz n |(15.99)= Tr
 n Zn

z ∈Tδ


 X

= Tr
pZ n |X n (z n |xn )|z n ihz n |(15.100)
 n Znz ∈TδX=pZ n |X n (z n |xn ).
(15.101)z n ∈TδZ

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15.3. THE METHOD OF TYPES FOR QUANTUM SYSTEMS

479

The first equality follows from the definition of the typical projector Πnρ,δ . The second equality
follows because |z n ihz n | is a projector, and the third follows from linearity and cyclicity ofthe trace. The fourth equality follows becausennhz |ρxn |z i =nYhzi | ρxi |zi i =

i=1

n
Y

pZ|X (zi |xi ) ≡ pZ n |X n (z n |xn ).

(15.102)i=1

Now consider this final expression zn ∈T Z n pZ n |X n (z n |xn ). It is equal to the probability that
δa random conditional sequence Z n |xn is in the typical set for pZ (z):nPr Z n |xn ∈ TδZ .
(15.103)P

By taking n large enough, the law of large numbers guarantees that it is highly likely (withprobability greater than 1 − ε for any ε > 0) that this random conditional sequence Z n |xn
Z n |xn
is in the conditionally typical set Tδ0for some δ 0 . It then follows that this conditionalnsequence has a high probability of being in the unconditionally typical set TδZ because we
assumed that the sequence xn is strongly typical and Lemma 14.9.1 states that a sequencez n is unconditionally typical if xn is strongly typical and z n is strong conditionally typical.

15.3 The Method of Types for Quantum Systems
Our final development in this chapter is to establish the method of types in the quantum
domain, and the classical tools from Section 14.7 have a straightforward generalization.
We can partition the Hilbert space of n qudits into different type class subspaces, just as
we can partition the set of all sequences into different type classes. For example, consider
the Hilbert space of three qubits. The computational basis is an orthonormal basis for theentire Hilbert space of three qubits:{|000i, |001i, |010i, |011i, |100i, |101i, |110i, |111i} .

(15.104)

Then the computational basis states with the same Hamming weight form a basis for eachtype class subspace. So, for the above example, the type class subspaces are as follows:
T0 ≡ {|000i} ,T1 ≡ {|001i, |010i, |100i} ,T2 ≡ {|011i, |101i, |110i} ,T3 ≡ {|111i} ,(15.105)(15.106)(15.107)(15.108)

and the projectors onto the different type class subspaces are as follows:Π0 ≡ |000ih000|,Π1 ≡ |001ih001| + |010ih010| + |100ih100|,
Π2 ≡ |011ih011| + |101ih101| + |110ih110|,Π3 ≡ |111ih111|.
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CHAPTER 15. QUANTUM TYPICALITY

We can generalize the above example to an n-fold tensor product of qudit systems using
the method of types.
Definition 15.3.1 (Type Class Subspace) The type class subspace is the subspace spannedby all states with the same type:n
TAt n ≡ span |xn iAn : xn ∈ TtX ,(15.113)
where the notation TAt n on the left-hand side indicates the type class subspace, and the notantion TtX on the right-hand side indicates the type class of the classical sequence xn .
Definition 15.3.2 (Type Class Projector) Let ΠtAn denote the type class subspace projector:
XΠtAn ≡|xn ihxn |An .
(15.114)xn ∈TtXn

Property 15.3.1 (Resolution of the Identity with Type Class Projectors) The sumof all type class projectors forms a resolution of the identity on the full Hilbert space HAn ofn qudits:X
IAn =ΠtAn ,(15.115)twhere IAn is the identity operator on HAn .
Definition 15.3.3 (Maximally Mixed Type Class State) The maximally mixed density operator proportional to the type class subspace projector isπAt n ≡ Dt −1 ΠtAn ,(15.116)

where Dt is the dimension of the type class:
Dt ≡ Tr ΠtAn .

(15.117)

Recall from Definition 14.7.4 that a δ-typical type is one for which the empirical distribution has maximum deviation δ from the true distribution, and τδ is the set of all δ-typicaltypes. For the quantum case, we determine the maximum deviation δ of a type from thetrue distribution pX (x) (this is the distribution from a spectral decomposition of a density
operator ρ). This definition allows us to write the strongly δ-typical subspace projector ΠδAnof ρ as a sum over all of the δ-typical type class projectors ΠtAn :XΠδAn =
ΠtAn .
(15.118)t∈τδSome protocols in quantum Shannon theory such as entanglement concentration in Chapter 19 employ the above decomposition of the typical subspace projector into types. The
way that such a protocol works is first to perform a typical subspace measurement on manyc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

15.4. CONCLUDING REMARKS

481

copies of a state, and this measurement succeeds with high probability. One party involvedin the protocol then performs a type class measurement {ΠtAn }t . We perform this lattermeasurement in a protocol if we would like the state to have a uniform distribution overstates in the type class. One might initially think that the dimension of the remaining state
would not be particularly large, but it actually holds that the dimension is large because
we can obtain the following useful lower bound on the dimension of any typical type classprojector.
Property 15.3.2 (Minimal Dimension of a Typical Type Class Projector) Supposethat pX (x) is the distribution from a spectral decomposition of a density operator ρ, and τδcollects all the type class subspaces with maximum deviation δ from the distribution pX (x).
Then for any type t ∈ τδ and for sufficiently large n, we can bound the dimension of the typeclass projector ΠtAn from below as follows:1
Tr ΠtAn ≥ 2n[H(ρ)−η(dδ)−d n log(n+1)] ,(15.119)

where d is the dimension of the Hilbert space on which ρ acts and the function η(dδ) → 0 as
δ → 0.
Proof. A proof follows directly by exploiting Property 14.7.5 from the previous chapter.

15.4 Concluding Remarks
This chapter is about the asymptotic nature of quantum information in the i.i.d. setting.
The main technical development is the notion of the typical subspace, and our approachhere is simply to “quantize” the definition of the typical set from the previous chapter. The
typical subspace enjoys properties similar to those of the typical set—the probability thatmany copies of a density operator lie in the typical subspace approaches one as the numberof copies approaches infinity, the dimension of the typical subspace is exponentially smaller
than the dimension of the full Hilbert space, and many copies of a density operator lookapproximately maximally mixed on the typical subspace. The rest of the content in thischapter involves an extension of these ideas to conditional quantum typicality.
The content in this chapter is here to provide a rigorous underpinning that we can quickly
cite later on, and after having mastered the results in this chapter along with the tools in
the next two chapters, we will be ready to prove many of the important results in quantum
Shannon theory.

15.5 History and Further Reading
Ohya and Petz (1993) devised the notion of a typical subspace, and later Schumacher (1995)
independently devised it when he proved the quantum data-compression theorem bearing his
name. Holevo (1998) and Schumacher and Westmoreland (1997) introduced the conditionallyc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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typical subspace in order to prove the HSW coding theorem. Winter’s thesis is a good sourcefor proofs of several properties of quantum typicality (Winter, 1999b). Nielsen and Chuang(2000) use weak conditional quantum typicality to prove the HSW theorem. Bennett et al.
(2002) and Holevo (2002b) introduced frequency typical (or strongly typical) subspaces toquantum information theory in order to prove the entanglement-assisted classical capacity
theorem. Devetak used strong typicality to prove the HSW coding theorem in Appendix Bof (Devetak, 2005).

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 16

The Packing Lemma
The packing lemma is a general method for one party to “pack” or encode classical messages
into a Hilbert space so that another party can distinguish the encoded messages. The firstparty can prepare an ensemble of quantum states, and the other party has access to a setof projectors using which he can form a quantum measurement. If the ensemble and the
projectors satisfy the conditions of the packing lemma, then it guarantees the existence of ascheme by which the second party can distinguish the classical messages that the first party
prepares.
The statement of the packing lemma is quite general, and this approach has a great
advantage because we can use it as a primitive for many coding theorems. Examples ofcoding theorems that we can prove using the packing lemma are the Holevo–Schumacher–Westmoreland (HSW) theorem for the transmission of classical information over a quantum
channel and the entanglement-assisted classical capacity theorem for the transmission ofclassical information over an entanglement-assisted quantum channel (furthermore, Chapter 22 shows that these two protocols are sufficient to generate most known protocols inquantum Shannon theory). Combined with the covering lemma of the next chapter, thepacking lemma gives a method for transmitting private classical information over a quantum
channel, and this technique in turn gives a way to communicate quantum information over aquantum channel. As long as we can determine an ensemble and a set of projectors satisfying
the conditions of the packing lemma, we can apply it in a straightforward way. For example, we prove the HSW coding theorem in Chapter 20 largely by relying on the propertiesof typical and conditionally typical subspaces that we proved in the previous chapter, andsome of these properties are equivalent to the conditions of the packing lemma.
The packing lemma is a “one-shot” lemma because it applies to a general scenario thatis not limited only to i.i.d. uses of a quantum channel. This “one-shot” approach is part of
the reason that we can apply it to a variety of situations. The technique of proving a “oneshot” result and applying it to the i.i.d. scenario is a common method of attack in quantum
Shannon theory (Chapter 17 does this also by establishing a covering lemma, which helps indetermining a method for sending private classical information over a quantum channel).
We begin in the next section with a simple example that illustrates the main ideas of
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the packing lemma. We then generalize this setting and give the statement of the packing
lemma. We dissect its proof in several sections that explain the random selection of a code,
the construction of a quantum measurement (called the “square-root measurement”), andthe error analysis. We then show how to derandomize the packing lemma so that there exists
some scheme for packing classical messages into Hilbert space with negligible probability oferror for determining each classical message. Finally, we show how a different quantum
measurement, called a sequential decoder, can also be used by a receiver to decode themessages transmitted.

16.1 Introductory Example
Suppose that Alice would like to communicate classical information to Bob, and suppose
further that she can prepare a message for Bob using the following BB84 ensemble:{|0i, |1i, |+i, |−i} ,(16.1)

where each state occurs with equal probability. Let us label each of the above states by
the classical indices a, b, c, and d so that a labels |0i, b labels |1i, etc. She cannot use
all of the states for transmitting classical information because, for example, |0i and |+i are
non-orthogonal states and there is no measurement that can distinguish them with highprobability.
How can Alice communicate to Bob using this ensemble? She can choose a subset of thestates in the BB84 ensemble for transmitting classical information. She can choose the states|0i and |1i for encoding one classical bit of information. Bob can then perform a complete
projective measurement in the basis {|0i, |1i} to determine the message that Alice encodes.
Alternatively, Alice and Bob can use the states |+i and |−i in a similar fashion for encoding
one classical bit of information.
In the above example, Alice can send two messages by using the labels a and b only. We
say that the labels a and b constitute the code. The states |0i and |1i are the codewords, theprojectors |0ih0| and |1ih1| are each a codeword projector, and the projector |0ih0| + |1ih1| isthe code projector (in this case, the code projector projects onto the whole Hilbert space).
The construction in the above example gives a way to use a certain ensemble for “packing”
classical information into Hilbert space, but there is only so much room for packing. For
example, it is impossible to encode more than one bit of classical information into a qubit
such that someone else can access this classical information reliably—this is the statementof the Holevo bound (Exercise 11.9.2).

16.2 The Setting of the Packing Lemma
We generalize the above example to show how Alice can effectively pack classical information
into a Hilbert space such that Bob can retrieve it with high probability. Suppose that Alice’s
resource for communication is an ensemble {pX (x), σx }x∈X of quantum states that she canc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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prepare for Bob, where the states σx are not necessarily perfectly distinguishable. We define
the ensemble as follows:
Definition 16.2.1 (Ensemble) Suppose X is a set of size |X | with elements x, and suppose
X is a random variable with probability density function pX (x). Suppose we have an ensemble{pX (x), σx }x∈X of quantum states where we encode each realization x into a quantum stateσx ∈ D(H). The expected density operator of the ensemble isXσ≡pX (x)σx .
(16.2)x∈X

How can Alice transmit classical information reliably to Bob by making use of this ensemble? As suggested in the example from the previous section, Alice can select a subsetof messages from the set X , and Bob’s task is to distinguish this subset of states as best
he can. We equip him with certain tools: a code subspace projector Π and a set of codeword subspace projectors {Πx }x∈X with certain desirable properties (we explain these terms
in more detail below). As a rough description, he can use these projectors to construct aquantum measurement that determines the message Alice sends. He would like to be almost
certain that the received state lies in the subspace onto which the code subspace projector Π
projects. He would also like to use the codeword subspace projectors {Πx }x∈X to determine
the classical message that Alice sends. If the ensemble and the projectors satisfy certain
conditions, the four conditions of the packing lemma, then it is possible for Bob to build up
a measurement such that Alice can communicate reliably with him.
Suppose that Alice chooses some subset C of X for encoding classical information. The
subset C that Alice chooses constitutes a code. Let us index the code C by a message set
M with elements labeled by m. The set M contains messages that Alice would like totransmit to Bob, and we assume that she chooses each message m with equal probability.
The subensemble that Alice uses for transmitting classical information is thus as follows:


1
, σc
,
(16.3)|M| mwhere each cm is a codeword that depends on the message m and takes a value in X .
Bob needs a way to determine the classical message that Alice transmits. The mostgeneral way that quantum mechanics offers for retrieving classical information is a POVM.
Thus, Bob performs some measurement described by a POVM {Λm }m∈M . Bob constructs
this POVM by using the codeword subspace projectors {Πx }x∈X and the code subspace
projector Π (we give an explicit construction in the proof of the packing lemma, called the“square-root” measurement—later we give a different construction called sequential decoding). If Alice transmits a message m, the probability that Bob correctly retrieves the message
m is as follows:
Tr {Λm σcm } .
(16.4)
Thus, the probability of error for a given message m while using the code C is as follows:pe (m, C) ≡ 1 − Tr {Λm σcm }= Tr {(I − Λm ) σcm } .
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We are interested in the performance of the code C that Alice and Bob choose, and we
consider three different measures of performance.
1. The first and strongest measure of performance is the maximal probability of error of
the code C. A code C has maximum probability of error ε if
ε = max pe (m, C).
m∈M(16.7)

2. A weaker measure of performance is the average probability of error p̄e (C) of the code
C, where|M|

1 X
pe (m, C).
p̄e (C) ≡|M| m=1(16.8)

3. The third measure of performance is even weaker than the previous two but turns out
to be the most useful in the mathematical proofs. It uses a conceptually different notionof code called a random code. Suppose that Alice and Bob choose a code C randomly
from the set of all possible codes according to some probability density pC (the code C
itself therefore becomes a random variable!). The third measure of performance is the
expectation of the average probability of error of a random code C where the expectation
is with respect to the set of all possible codes, with each code chosen according to somedensity pC :


|M|

 1 X
pe (m, C)
EC {p̄e (C)} ≡ EC

 |M|
m=1


|M|
X
X
1
pe (m, C) .
=
pC 
|M|
m=1
C

(16.9)(16.10)We will see that considering this performance criterion simplifies the mathematics
in the proof of the packing lemma. Then we will employ a series of arguments tostrengthen the result from this weakest performance criterion to the first and strongest
performance criterion.

16.3 Statement of the Packing Lemma
Lemma 16.3.1 (Packing Lemma) Suppose that we have an ensemble as in Definition 16.2.1.
Suppose that a code subspace projector Π and codeword subspace projectors {Πx }x∈X exist,
they project onto subspaces of H, and these projectors and the ensemble satisfy the followingc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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conditions:
Tr {Πσx } ≥ 1 − ε,Tr {Πx σx } ≥ 1 − ε,Tr {Πx } ≤ d,1
ΠσΠ ≤ Π,D

(16.11)(16.12)(16.13)(16.14)

where ε ∈ (0, 1), D > 0, and d ∈ (0, D). Suppose that M is a set of size |M| with elements
m. We generate a set C = {Cm }m∈M of random variables Cm independently at random
according to pX (x), so that each random variable Cm takes a value in X and corresponds tothe message m, but its distribution is independent of the particular message m. The set C
constitutes a random code. Then there exists a corresponding POVM {Λm }m∈M that reliably
distinguishes the states {σCm }m∈M , in the sense that the expectation of the average probabilityof detecting the correct state is high:
)
(
√ 
d
1 X
(16.15)Tr {Λm σCm } ≥ 1 − 2 ε + 2 ε − 4 |M| ,EC|M| m∈M
D
when D/d is large, |M|  D/d, and ε is small.
Condition (16.11) states that the code subspace with projector Π contains each message
σx with high probability. Condition (16.12) states that each codeword subspace projector
Πx contains its corresponding state σx with high probability. Condition (16.13) states thatthe dimension of each codeword subspace projector Πx is less than some positive numberd ∈ (0, D). Condition (16.14) states that the expected density operator σ of the ensemble
is approximately maximally mixed when projecting it onto the subspace with projector Π.
Conditions (16.11) and (16.14) imply thatTr {Π} ≥ D (1 − ε) ,(16.16)so that the dimension of the code subspace projector Π is approximately D if ε is small.
We show how to construct a code with messages that Alice wants to send. These four
conditions are crucial for constructing a decoding POVM with the desirable property thatit can distinguish the encoded messages with high probability.
The main idea of the packing lemma is that we can pack |M| classical messages into asubspace with corresponding projector Π. There is then a small probability of error whentrying to detect the classical messages using codeword subspace projectors Πx . The intuitionis the same as that depicted in Figure 2.6. We are trying to pack as many subspaces of sized into a larger space of size D. In the proof of the HSW coding theorem in Chapter 20, D
will be of size ≈ 2nH(B) and d will be of size ≈ 2nH(B|X) , suggesting that we can pack in≈ 2n[H(B)−H(B|X)] = 2nI(X;B) messages while still being able to distinguish them reliably.
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16.4 Proof of the Packing Lemma
The proof technique employs a Shannon-like argument in which we generate a code at random. We first show how to construct a POVM, the “pretty good” or “square-root” measurement, that can decode a classical message with high probability. We then prove that the
expectation of the average error probability is small (where the expectation is with respect
to all random codes). In a corollary in the next section, we finally use standard Shannon-likearguments to show that a code exists whose maximal probability of error for all messages issmall.

16.4.1

Code Construction

We present a Shannon-like random coding argument to simplify the mathematics that follow.
We construct a code C at random by independently generating |M| codewords according to
the distribution pX (x). Let C ≡ {cm }m∈M be a collection of the realizations cm of |M|
independent random variables Cm . Each Cm takes a value cm in X with probability pX (cm )and represents a classical codeword in the random code C. The probability p(C) of choosing
a particular code C is equal to the following:p(C) =|M|Y

pX (cm ).

(16.17)m=1

There is a great advantage to choosing the code in this way. The expectation of any productf (Cm )g(Cm0 ) of two functions f and g of two different random codewords Cm and Cm0 , where
the expectation is with respect to the random choice of code, factors as follows:X
EC {f (Cm )g(Cm0 )} =p(c)f (cm )g(cm0 )(16.18)
c

=

X
c1 ∈X

=

X

pX (c1 ) · · ·XpX (c|M| )f (cm )g(cm0 )(16.19)X(16.20)c|M| ∈X

pX (cm )f (cm )cm ∈XpX (cm0 )g(cm0 )cm0 ∈X= EX {f (X)} EX {g(X)} .

(16.21)This factoring happens because of the random way in which we choose the code, and we
exploit this fact in the proof of the packing lemma. We employ the following events insequence:
1. We choose a random code as described above.
2. We reveal the code to the sender and receiver (i.e., they are allowed to meet beforehand
to agree on a strategy before communication begins).
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3. The sender chooses a message m at random (with uniform probability according tosome random variable M ) from M and encodes it in the codeword cm . The quantum
state that the sender transmits is then equal to σcm .
4. The receiver performs a POVM {Λm }m∈M to determine the message that the sendertransmits, and each POVM element Λm corresponds to a message m in the code. The
receiver obtains a classical result from the measurement, and we model it with therandom variable M 0 . The conditional probability Pr {M 0 = m|M = m} of obtaining
the correct result from the measurement is equal to
Pr {M 0 = m|M = m} = Tr {Λm σcm } .

(16.22)

5. The receiver decodes correctly if M 0 = M and decodes incorrectly if M 0 6= M .

16.4.2

POVM Construction

We cannot directly use the projectors Πx in a POVM because they
P do not satisfy the conditions for being a POVM. Namely, it is not necessarily true that x∈X Πx = I. Furthermore,
the codeword subspace projectors Πx may have support outside that of the code subspace
projector Π. It is necessary for us to have the code subspace projector involved in the analysis
because we will need to invoke condition (16.14) of the packing lemma.
To remedy these issues, first consider the following set of operators:∀x Υx ≡ ΠΠx Π.

(16.23)The operator Υx is a positive semi-definite operator, and the effect of “coating” the codeword
subspace projector Πx with the code subspace projector Π is to slice out any part of the
support of Πx that is not in the support of Π. From the conditions (16.11)–(16.12) of thepacking lemma, there should be little probability for our states of interest to lie in thepart of the support of Πx outside the support of Π. The operators Υx have the desirable
property that they only have support inside the subspace corresponding to the code subspace
projector Π. So we have remedied the second issue stated above. Exercise 16.5.4 explores
an alternative way of resolving this issue.
We now remedy the first problem stated above by constructing a set {Λm }m∈M with thefollowing elements:

−1/2

−1/2
|M|
|M|
X
X
Λm ≡ 
Υcm0 
Υcm 
Υcm0 
.
(16.24)
m0 =1

m0 =1

The elements of the set {Λm }m∈M constitute the “pretty good” or “square-root” measureP
ment. These elements generally have the property that |M|I, and we can “comm=1 Λm ≤P
plete” the set to be a full POVM by inserting the element Λ0 ≡ I − |M|
m=1 Λm into the set{Λm }m∈M . The extra element Λ0 corresponds to a failed decoding. Note that the inversec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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square root A−1/2 of a positive semi-definite operator A is defined as the inverse square root
operation only on the support of A, per the usual convention from Definition 3.3.1. The
idea of the pretty good measurement is that the POVM elements {Λm }|M|
m=1 correspond tothe messages sent and the element Λ0 corresponds to an error result (an inability to identify
any of the messages).

16.4.3

Error Analysis

Before proceeding with the error analysis, we need the following operator inequality, which
will be helpful in analyzing the error probability:
Lemma 16.4.1 (Hayashi–Nagaoka) Let S, T ∈ L(H) be positive semi-definite operatorssuch that I − S is positive semi-definite also. Then for a strictly positive constant c, thefollowing operator inequality holds
(16.25)I − (S + T )−1/2 S (S + T )−1/2 ≤ (1 + c) (I − S) + 2 + c + c−1 T.
Proof. For any two operators A, B ∈ L(H), the following operator inequality holds(A − cB)† (A − cB) ≥ 0,

(16.26)

which is equivalent to
Now pick A =that√

c−1 A† A + cB † B ≥ A† B + B † A.
(16.27)√
T R and B = T (I − R), where R ∈ L(H), and plug into the above to findc−1 R† T R + c (I − R)† T (I − R) ≥ R† T (I − R) + (I − R)† T R.

(16.28)Thus, we find that
T = R† T R + R† T (I − R) + (I − R)† T R + (I − R)† T (I − R)
≤ 1 + c−1 R† T R + (1 + c) (I − R)† T (I − R) .

(16.29)(16.30)Setting R = (S + T )1/2 , we find thatT ≤ 1 + c−1 (S + T )1/2 T (S + T )1/2 + (1 + c) I − (S + T )1/2 T I − (S + T )1/2 . (16.31)
Since T ≤ S + T , we can conclude thatT ≤ 1 + c−1 (S + T )1/2 T (S + T )1/2
+ (1 + c) I − (S + T )1/2 (S + T ) I − (S + T )1/2
h

i

1/2
1/2
−1
= (S + T )1+c
T + (1 + c) I + S + T − 2 (S + T )(S + T )1/2hi= (S + T )1/2 2 + c + c−1 T + (1 + c) I + S − 2 (S + T )1/2 (S + T )1/2
≤ (S + T )1/2 2 + c + c−1 T + (1 + c) (I + S − 2S) (S + T )1/2
= (S + T )1/2 2 + c + c−1 T + (1 + c) (I − S) (S + T )1/2 .
c 2019(16.32)(16.33)(16.34)(16.35)(16.36)

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The last inequality follows because S ≤ S 1/2 ≤ (S + T )1/2 . This makes use of the assumptionthat S ≤ I and that fact that the square root function is operator monotone, meaning that
if X ≤ Y for positive semi-definite X and Y , then X 1/2 ≤ Y 1/2 . Multiplying both sides ofthis operator inequality by (S + T )−1/2 , we find that
(16.37)(S + T )−1/2 T (S + T )−1/2 ≤ 2 + c + c−1 T + (1 + c) (ΠS+T − S) ,where ΠS+T denotes the projector onto the support of S + T , so that ΠS+T T ΠS+T = T andΠS+T SΠS+T = S. Then consider thatI − (S + T )−1/2 S (S + T )−1/2= I − (S + T )−1/2 (S + T ) (S + T )−1/2 + (S + T )−1/2 T (S + T )−1/2
−1/2

−1/2

= I − ΠS+T + (S + T )T (S + T )≤ (1 + c) (I − ΠS+T ) + 2 + c + c−1 T + (1 + c) (ΠS+T − S)
= 2 + c + c−1 T + (1 + c) (I − S) .

(16.38)(16.39)(16.40)(16.41)This concludes the proof.
Suppose we have chosen a particular code C. Let pe (m, C) be the probability of decoding
incorrectly given that message m was sent while using the code C:pe (m, C) ≡ Tr {(I − Λm ) σcm } .

(16.42)

In Lemma 16.4.1, we can set
T =

|M|
X

Υcm0 ,

S = Υcm ,(16.43)m0 6=mand pick c = 1,1 so that the bound in (16.25) becomesI − Λm ≤ 2 (I − Υcm ) + 4|M|XΥcm0 .

(16.44)m0 6=mThen using (16.44) and linearity of trace, we obtain the following upper bound on the error
probability:|M|Xpe (m, C) ≤ 2 Tr {(I − Υcm ) σcm } + 4
Tr Υcm0 σcm .
(16.45)m0 6=mThe above bound on the message error probability for code C has a similar interpretation
as that in classical Shannon-like proofs. We bound the error probability by the probability
1

For our purposes here, it suffices to pick the parameter c = 1, but for more fine-tuned error analyses, it
is essential to choose the parameter c to vary with other code parameters.
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CHAPTER 16. THE PACKING LEMMA

of incorrectly decoding the message m with the message operator Υcm (the first term in(16.45)) summed with the probability of confusing the transmitted message with a message
cm0 different from the transmitted one (the second term in (16.45)).
Consider the first term Tr {(I − Υcm ) σcm } on the right-hand side of (16.45). We canbound it from above by a small number, simply by applying (16.11)–(16.12) and the gentle
operator lemma (Lemma 9.4.2). Consider the following chain of inequalities:
Tr {Υcm σcm } = Tr {ΠΠcm Πσcm }= Tr {Πcm Πσcm Π}≥ Tr {Πcm σcm } − kΠσcm Π − σcm k1√≥ 1 − ε − 2 ε.

(16.46)(16.47)(16.48)(16.49)The first equality follows from the definition of Υcm in (16.23). The second equality follows
from cyclicity of the trace. The first inequality follows from applying Exercise 9.1.8. The lastinequality follows from applying (16.11) to Tr {Πcm σcm } and from applying (16.12) and thegentle operator lemma (Lemma 9.4.2) to kΠσcm Π − σcm k1 . The above bound then implies
the following one:√Tr {(I − Υcm ) σcm } = 1 − Tr {Υcm σcm } ≤ ε + 2 ε,(16.50)and by substituting into (16.45), we get the following bound on the probability of error:
|M|

X

√ 
pe (m, C) ≤ 2 ε + 2 ε + 4
Tr Υcm0 σcm .

(16.51)

m0 6=mThe average error probability p̄e (C) over all transmitted messages for code C is|M|

1 X
pe (m, C),p̄e (C) =|M| m=1(16.52)

if we assume that Alice chooses the message m according to the uniform distribution. From
(16.51), we see that the average error probability p̄e (C) obeys the following bound:
|M|

|M|


√ 
4 X X
Tr Υcm0 σcm .
p̄e (C) ≤ 2 ε + 2 ε +|M| m=1 m0 6=m(16.53)At this point, bounding the average error probability further is a bit difficult, given the
sheer number of combinations of terms Tr{Υcm0 σcm } that we would have to consider to doso. Thus, we now invoke the classic Shannon argument in order to simplify the mathematics.
Instead of considering the average probability of error, we consider the expectation of theaverage error probability EC {p̄e (C)} with respect to all possible random codes C. Considering
this error quantity significantly simplifies the mathematics because of the way in whichc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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we constructed the code. We can use the probability distribution pX (x) to compute theexpectation EC because we constructed our code according to this distribution. The boundabove becomes as follows, now denoting each codeword as a random variable Cm :
|M| |M|


X
X

√ 
4
Tr ΥCm0 σCm(16.54)EC {p̄e (C)} ≤ EC 2 ε + 2 ε +|M| m=1 0
m 6=m

|M|

|M|

 
√ 
4 X X
=2 ε+2 ε +EC Tr ΥCm0 σCm|M| m=1 m0 6=m,

where the equality follows from the linearity of expectation.
We now calculate the expectation of Tr{ΥCm0 σCm } over all random codes C:  EC Tr ΥCm0 σCm= EC Tr ΠΠCm0 ΠσCm 
= EC Tr ΠCm0 ΠσCm Π = ECm ,Cm0 Tr ΠCm0 ΠσCm Π .

(16.55)(16.56)(16.57)(16.58)The first equality follows from the definition in (16.23), and the second equality follows fromcyclicity of trace. The third equality follows because only the codewords for m and m0 are
involved. Independence of random variables Cm and Cm0 (from the code construction) givesthat the above expression equals(16.59)Tr ECm0 {ΠCm0 }ΠECm {σCm }Π = Tr ECm0 {ΠCm0 }ΠσΠ1
≤ Tr ECm0 {ΠCm0 } Π(16.60)D
1(16.61)= Tr ECm0 {ΠCm0 }Π ,D
P
where the first equality uses the fact that ECm {σCm } = x∈X p(x)σx = σ and Π is a constantwith respect to the expectation. The first inequality uses the fourth condition (16.14) of thepacking lemma, the fact that ΠσΠ, Π, and ΠCm0 are all positive semi-definite operators, and
Tr {CA} ≥ Tr {CB} for C ≥ 0 and A ≥ B. Continuing, we have11
Tr ECm0 {ΠCm0 }Π ≤ Tr ECm0 {ΠCm0 }(16.62)
DD
1= ECm0 Tr{ΠCm0 }(16.63)Dd≤ .
(16.64)D
The first inequality follows from the fact that Π ≤ I and ΠCm0 is a positive semi-definite
operator. The last inequality follows from (16.13). The following inequality then holds byconsidering the development from (16.56) to (16.64):
 
EC Tr ΥCm0 σCm
c 2019≤d.
D(16.65)

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494

CHAPTER 16. THE PACKING LEMMA

We substitute into (16.55) to show that the expectation EC {p̄e (C)} of the average error
probability p̄e (C) over all codes obeys the bound stated in the packing lemma:
|M|

|M|

 
√ 
4 X X
EC Tr σcm Υcm0
EC {p̄e (C)} ≤ 2 ε + 2 ε +|M| m=1 m0 6=m|M|(16.66)|M|√ 
4 X X d≤2 ε+2 ε +|M| m=1 m0 6=m D√ d≤ 2 ε + 2 ε + 4 (|M| − 1)D√ d≤ 2 ε + 2 ε + 4 |M| .
D(16.67)(16.68)(16.69)16.5 Derandomization and Expurgation
The above version of the packing lemma is a randomized version that shows how the expectation of the average probability of error is small. We now prove a derandomized version thatguarantees the existence of a code with small maximal error probability for each message.
The last two arguments are traditionally called derandomization and expurgation.
Corollary 16.5.1 Suppose we have the ensemble as in Definition 16.2.1. Suppose that acode subspace projector Π and codeword subspace projectors {Πx }x∈X exist, they project onto
subspaces of H, and these projectors and the ensemble have the following properties:
Tr {Πσx } ≥ 1 − ε,Tr {Πx σx } ≥ 1 − ε,Tr {Πx } ≤ d,1
ΠσΠ ≤ Π,D

(16.70)(16.71)(16.72)(16.73)where ε ∈ (0, 1) and d ∈ (0, D). Suppose that M is a set of size |M| with elements m. Then
there exists a code C0 = {cm }m∈M with codewords cm depending on the message m and taking
values in X , and there exists a corresponding POVM {Λm }m∈M that reliably distinguishes
the states {σcm }m∈M in the sense that the probability of detecting the correct state is high:∀m ∈ M√ dTr {Λm σcm } ≥ 1 − 4 ε + 2 ε − 16 |M| ,D(16.74)

if ε is small and |M|  D/d. We can then use the code C0 and the POVM {Λm }m∈M to
encode and decode, respectively, |M| classical messages with high success probability.
Proof. Generate a random code according to the construction in the previous lemma. The
expectation of the average error probability then satisfies the bound in the statement of thec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

16.5. DERANDOMIZATION AND EXPURGATION

495

packing lemma. We now make a few standard Shannon-like arguments to strengthen the
result of the previous lemma.
Derandomization. The expectation of the average error probability EC {p̄e (C)} satisfies
the following bound:√ d(16.75)EC {p̄e (C)} ≤ 2 ε + 2 ε + 4 |M| .
D
It then follows that the average error probability of at least one code C0 = {cm }m∈M satisfies
the same bound:√ dp̄e (C0 ) ≤ 2 ε + 2 ε + 4 |M| .
(16.76)D
Choose this code C0 as the code, and it is possible to find this code C0 in practice by exhaustive
search. This process is known as derandomization.
0
Exercise
Markov’s inequality to prove the following strong√ statement: if ε ≡√ 16.5.1 Used
2 (ε + 2 ε) + 4 |M| D is small, then the overwhelming fraction 1 − ε0 of codes contructed√
randomly have average error probability less than ε0 .

Expurgation. We now go from average error probability to maximal error probability
insteadby√employing an expurgation argument. We know from Exercise 2.2.1 that pe (m) ≤
2 2 (ε + 2 ε) + 4 |M| Dd for at least half of the indices (if it were not true, then these indiceswould contribute more than ε0 to the average error probability p̄e ). Throw out the half of thecodewords with the worst decoding probability and redefine the code according to the new0
set of indices. If we redefine the message set
size |M0 | = |M| /2,
 M such√ that the0 message
then the error bound becomes pe (m) ≤ 2 2 (ε + 2 ε) + 8 |M | Dd . We could then use thedecoding POVM from the original code and be guaranteed that every codeword has an error
probability no larger than this amount (alternatively, we could also use a modified squareroot decoding POVM that is built from the codeword subspace projectors remaining after
expurgation). These steps have a negligible effect on the parameters of the code when welater consider a large number of uses of a noisy quantum channel.
Exercise 16.5.2 Use Markov’s inequality to prove an even stronger expurgation argument(following
on the result of Exercise√16.5.1). Prove that we can retain a large fraction 1 −√4 0
ε of the codewords√(expurgating 4 ε0 of them) so that each remaining codeword has error
probability less than 4 ε0 .
Exercise 16.5.3 Prove that the packing lemma and its corollary hold for the same ensemble
and a set of projectors for which the following conditions hold:XpX (x) Tr {σx Π} ≥ 1 − ε,(16.77)x∈X

X

pX (x) Tr {σx Πx } ≥ 1 − ε,(16.78)Tr {Πx } ≤ d,1
ΠσΠ ≤ Π.
D

(16.79)x∈X

c 2019(16.80)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

496

CHAPTER 16. THE PACKING LEMMA

Exercise 16.5.4 Prove that a variation of the packing lemma holds in which the POVM isof the following form:Λm ≡ 

|M|
X

−1/2
Πcm0 

m0 =1


Πcm 

|M|
X

−1/2
Πcm0 

.

(16.81)m0 =1That is, it is not actually necessary to “coat” each operator in the square-root measurementwith the overall message subspace projector.

16.6 Sequential Decoding
We now prove a variation of the packing lemma by making use of a completely different
decoding scheme, called sequential decoding. This scheme has the receiver perform sequential
tests using the codeword subspace projectors to “ask” sequentially, “Was the first codeword
sent?”, “The second?”, “The third?”, etc., until the outcome of one of these measurements
is “yes.”
It is perhaps unintuitive that such a measurement strategy should work. After all, we
are well aware by this point that measurements can disturb the state of a quantum system.
However, what the packing lemma demonstrates is that if good codeword projectors areavailable and we do not try to pack in too many messages, then reliable decoding is possible.
This also means that the codewords selected are approximately orthogonal. In this sense, it isperhaps less surprising that such a sequential decoding strategy should work if we know thatthe states to be distinguished are approximately orthogonal. The main tool for analyzing theperformance of a sequential decoder is the “non-commutative union bound” (Lemma 16.6.1)
which is a generalization of the union bound from probability theory.
Here we divide up the proof into several parts, and some of the analysis is similar to that
presented in the previous section. The parts of the proof consist of codebook generation,
POVM construction, and the error analysis. We begin with all of the objects and premisesof the Packing Lemma (in Lemma 16.3.1).
Codebook Generation. This part is exactly the same as before, so we instead point
to the discussion of this part in Section 16.4.1.
POVM Construction. The method for Bob to decode the state that Alice transmits is
as follows: Bob should first ask “Is the received state in the code subspace?” He can do thisoperationally by performing the measurement {Π, I − Π}. Next, he asks in sequential order,“Is the received codeword in the ith codeword subspace?” This is in some sense equivalent
to the question, “Is the received codeword the ith transmitted codeword?” He can ask thesequestions operationally by performing the codeword subspace measurements {Πci , I − Πci }.
Why should this sequential decoding scheme work well? Supposing that Alice transmits
message m, one reason is that the encoded state σcm lies in the code subspace with highprobability, as indicated by (16.11). Also, the projector Πcm is a “good detector” for theencoded state σcm , due to (16.12).
c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

16.6. SEQUENTIAL DECODING

497

Error Analysis. The probability of detecting the mth codeword correctly under oursequential decoding scheme is equal tonoTr Πcm Π̂cm−1 · · · Π̂c1 Πσcm ΠΠ̂c1 · · · Π̂cm−1 Πcm ,(16.82)where we make the abbreviation Π̂cm ≡ I − Πcm . That is, the receiver should get an initial“yes” for projecting into the code subspace, m − 1 “no’s,” and finally a “yes” for the mth
test. Thus, the probability of an incorrect detection for the mth codeword is given byno
1 − Tr Πcm Π̂cm−1 · · · Π̂c1 Πσcm ΠΠ̂c1 · · · Π̂cm−1 Πcm ,(16.83)and the average error probability of this scheme is equal too
1 X n
Tr Πcm Π̂cm−1 · · · Π̂c1 Πσcm ΠΠ̂c1 · · · Π̂cm−1 Πcm .
1−
|M| m

(16.84)Instead of analyzing the average error probability, we analyze the expectation of the average
error probability, where the expectation is with respect to the random choice of code:
(
)
o
1 X n
1 − ECTr ΠCm Π̂Cm−1 · · · Π̂C1 ΠσCm ΠΠ̂C1 · · · Π̂Cm−1 ΠCm.
(16.85)|M| m
We rewrite the above expression just slightly, by observing that)(
1 X
Tr {σCm }1 = EC
|M| m
(
)
o
n
1 X
= EC
Tr {ΠσCm } + Tr Π̂σCm
|M| m
(
)
1 X
≤ EC
Tr {ΠσCm Π} + ε,
|M| m

(16.86)(16.87)(16.88)where we have used (16.11) in the last line. Substituting into (16.85) (and forgetting about
the small ε term for now) gives an upper bound of
(
)
n
o
1 X
EC
Tr {ΠσCm Π} − Tr ΠCm Π̂Cm−1 · · · Π̂C1 ΠσCm ΠΠ̂C1 · · · Π̂Cm−1 ΠCm. (16.89)|M| m
We now need a tool for analyzing this error probability, which is known as the noncommutative union bound:
Lemma 16.6.1 (Non-Commutative Union Bound) Let ω be such that ω ≥ 0 and Tr{ω} ≤1. Let P1 , . . . , PL be Hermitian projectors. Thenv
u LuX
Tr{ω} − Tr{PL · · · P1 ωP1 · · · PL } ≤ 2t
Tr{(I − Pi )ω}.
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498

CHAPTER 16. THE PACKING LEMMA

Proof. It suffices to prove the following bound for a vector |ψi such that k|ψik22 ≤ 1:
v
u L
uX
2
2
k|ψik2 − kPL · · · P1 |ψik2 ≤ 2tk(I − Pi ) |ψik22 .
(16.91)i=1This is becausek|ψik22 = Tr{|ψihψ|},(16.92)

kPL · · · P1 |ψik22 = Tr{PL · · · P1 |ψihψ|P1 · · · PL },(16.93)k(I − Pi ) |ψik22 = Tr{(I − Pi )|ψihψ|},(16.94)andP any ω satisfying the conditions given can be written as a convex combination2 ω =z p(z)|ψz ihψz | where p(z) is a probability distribution and each |ψz i satisfies k|ψz ik2 ≤ 1.
Then (16.90) follows from (16.91) by concavity of the square root function. So we now focus
on proving (16.91). We begin by showing thatk|ψi − PL · · · P1 |ψik22 ≤LX

k(I − Pi ) |ψik22 .

(16.95)i=1

To see this, consider thatk|ψi − PL · · · P1 |ψik22 = k(I − PL ) |ψik22 + kPL (|ψi − PL−1 · · · P1 |ψi)k22≤ k(I − PL ) |ψik22 + k|ψi − PL−1 · · · P1 |ψik22≤

L
X

k(I − Pi ) |ψik22 .

(16.96)(16.97)(16.98)i=1

The first equality follows from the Pythagorean theorem. The first inequality follows because
a projection cannot increase the norm of a vector. The last inequality is by induction. Now
we take the square root of (16.95):v
u L
uX(16.99)k|ψi − PL · · · P1 |ψik2 ≤ tk(I − Pi ) |ψik22 ,i=1

from which we can conclude the following by the triangle inequality:v
u LuXk|ψik2 − kPL · · · P1 |ψik2 ≤ tk(I − Pi ) |ψik22 .

(16.100)

i=1

Then rearrange this as follows:v
u LuXk|ψik − tk(I − P ) |ψik2 ≤ kP · · · P |ψik
i

2

2

L

1

2

(16.101)i=1
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16.6. SEQUENTIAL DECODING

499

and square both sides to get
v
2
u L
uX
k|ψik − tk(I − P ) |ψik2 


i

2

2

i=1

v
u L
L
uX
X
2
2
t
= k|ψik − 2 k|ψikk(I − P ) |ψik +k(I − P ) |ψik2
2

i

2

i

2

i=1

2

(16.102)

i=1≤ kPL · · · P1 |ψik22 .

(16.103)

This then implies (16.91) by dropping the non-negative term

PL2
i=1 k(I − Pi ) |ψik2 .

Remark 16.6.1 We can think of the bound in (16.90) as a “non-commutative union bound”
because it is analogous to the following union bound from probability theory:
Pr {(A1 ∩ · · · ∩ AN )c } = Pr {Ac1 ∪ · · · ∪ AcN } ≤NXPr {Aci } ,(16.104)i=1

where A1 , . . . , AN are events. The analogous bound for projector logic would be
Tr {(I − P1 · · · PN · · · P1 ) ρ} ≤N
X

Tr {(I − Pi ) ρ} ,(16.105)

i=1

if we think of P1 · · · PN as a projector onto the intersection of subspaces. However, theabove bound only holds if the projectors P1 , . . . , PN are commuting (choosing P1 = |+ih+|,P2 = |0ih0|, and ρ = |0ih0| gives a counterexample). If the projectors are non-commuting,then the non-commutative union bound seems to be the next best thing and suffices for our
purposes here.
Continuing, we then apply the non-commutative union bound to the expression in (16.89)
with ω = ΠσCm Π and the sequential projectors as ΠCm , Π̂Cm−1 , . . . , Π̂C1 . This gives thefollowing upper bound on (16.89):
"#1/2 
m−1
 1 X

X
EC
2 Tr {(I − ΠCm ) ΠσCm Π} +Tr {ΠCi ΠσCm Π}
 |M|

m
i=1
" (
)#1/2
X
1 X
≤ 2 EC
Tr {(I − ΠCm ) ΠσCm Π} +Tr {ΠCi ΠσCm Π}, (16.106)
|M| m
i6=m
where the inequality follows from concavity of the square root function and by summingover all of the codewords not equal to the mth codeword (these terms can only increase thec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

500

CHAPTER 16. THE PACKING LEMMA

expression). At this point, we have two error terms to analyze that are essentially the sameas those in (16.45). Thus, we can invoke the analysis from before to conclude that√
Tr {(I − Πcm ) Πσcm Π} ≤ ε + 2 ε,(16.107)()
1 XX
d(16.108)EC
Tr {ΠCi ΠσCm Π} ≤ |M| ,|M| m i6=mD
which leads to a final bound of
(
)
o
1 X n
1 − ECTr ΠCm Π̂Cm−1 · · · Π̂C1 ΠσCm ΠΠ̂C1 · · · Π̂Cm−1 ΠCm
|M| m
r
√
d
≤ 2 ε + 2 ε + |M| + ε. (16.109)
D
We can then derandomize and expurgate codewords as before to establish the existence√of acode and a sequential decoder with maximum error probability no larger than 4(ε + 2 ε +2 |M| d/D)1/2 + 2ε.
Exercise 16.6.1 Following Exercise 16.5.4, show that a variation of the sequential decoder,in which there is no initial code subspace projection, works well for decoding the messages
transmitted.
Exercise 16.6.2 Show that the sequential decoding strategy works when assuming only theconditions in Exercise 16.5.3.

16.7 History and Further Reading
Holevo (1998) and Schumacher and Westmoreland (1997) did not prove the classical coding
theorem using the packing lemma, but they instead used other arguments to bound theprobability of error. The operator inequality in (16.25) is at the heart of the packing lemma.
Hayashi and Nagaoka (2003) proved this operator inequality in order to develop the moregeneral setting of the quantum information spectrum method, where there is no i.i.d. constraint and essentially no structure to a channel. Hsieh et al. (2008a) later exploited thisoperator inequality in the context of entanglement-assisted classical coding and followed theapproach in Hayashi and Nagaoka (2003) to prove the packing lemma.
Giovannetti et al. (2012); Lloyd et al. (2011) introduced the method of sequential decoding
to quantum information theory. Sen (2011) proved the non-commutative union bound andapplied it to various problems in quantum information theory. Wilde (2013) extended thenon-commutative union bound to non-projective tests (ones of the form {Λ, I − Λ} forΛ ≥ 0) and established a “one-shot” characterization of sequential decoding for classicalcommunication.

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 17

The Covering Lemma
The goal of the covering lemma is perhaps opposite to that of the packing lemma because
it applies to a setting in which one party wishes to make messages indistinguishable toanother party (instead of trying to make them distinguishable, as in the packing lemma of
the previous chapter). That is, the covering lemma is helpful when one party is trying tosimulate a noisy channel to another party, rather than trying to simulate a noiseless channel.
One party can accomplish this task by randomly “covering” the Hilbert space of the otherparty (this viewpoint gives the covering lemma its name).
One can certainly simulate noise by choosing a quantum state uniformly at random froma large set of quantum states and passing along the chosen quantum state to a third party
without indicating which state was chosen. But the problem with this approach is that it
could potentially be expensive if the set from which we choose a random state is large, andwe would really like to use as few resources as possible in order to simulate noise. That is, wewould like the set from which we choose a quantum state uniformly at random to be as small
as possible when simulating noise. The covering lemma is similar to the packing lemma inthe sense that its conditions for application are general (involving bounds on projectors andan ensemble), and it gives an effective scheme for simulating noise when we apply it in an
i.i.d. setting.
One application of the covering lemma in quantum Shannon theory is in the constructionof a code for transmitting private classical information over a quantum channel (discussed
in Chapter 23). The method of proof for private classical transmission involves a clevercombination of packing messages so that Bob can distinguish them, while covering Eve’s
space in such a way that Eve cannot distinguish the messages intended for Bob. A few otherapplications of the covering lemma are in secret key distillation, determining the amount ofnoise needed to destroy correlations in a bipartite state, and compressing the outcomes ofan i.i.d. measurement on an i.i.d. quantum state.
We begin this chapter with a simple example to explain the main idea behind the coveringlemma. Section 17.2 then discusses its general setting and gives its statement. We dissect its
proof into several different parts: the construction of a “Chernoff ensemble,” the constructionof a “Chernoff code,” the application of the operator Chernoff bound, and the error analysis.
501

502

CHAPTER 17. THE COVERING LEMMA

The main tool that we use to prove the covering lemma is the operator Chernoff bound.
This bound is a generalization of the standard Chernoff bound from probability theory, whichstates that the sample mean of a sequence of i.i.d. random variables converges exponentiallyfast to its true mean. A proof of the operator version of the Chernoff bound appears in
Section 17.3. The exponential convergence in the Chernoff bound is much stronger thanthe polynomial convergence from Chebyshev’s inequality and is helpful for establishing theexistence of good private classical codes in Chapter 23.

17.1 Introductory Example
Suppose that Alice is trying to communicate with Bob as before, but now there is an eavesdropper Eve listening in on their communication. Alice wants the messages that she issending to Bob to be private so that Eve does not gain any information about the message
that she is sending.
How can Alice make the information that she is sending private? The strongest criterion
for security is to ensure that whatever Eve receives is independent of what Alice is sending.
Alice may have to sacrifice the amount of information she can communicate to Bob in orderto have privacy, but this sacrifice is worth it to her because she really does not want Eve toknow anything about the intended message for Bob.
We first give an example to motivate a general method that Alice can use to make her
information private. Suppose Alice can transmit one of four messages {a, b, c, d} to Bob,and suppose he receives them perfectly as distinguishable quantum states. She chooses fromthese messages with equal probability. Suppose further that Alice and Eve know that Eve
receives one of the following four states corresponding to each of Alice’s messages:
a → |0i,b → |1i,c → |+i,d → |−i.

(17.1)

Observe that each of Eve’s states lies in the two-dimensional Hilbert space of a qubit. We
refer to the quantum states in the above ensemble as “Eve’s ensemble.”
We are not so much concerned for what Bob receives for the purposes of this example,
but we just make the assumption that he can distinguish the four messages that Alice sends.
Without loss of generality, let us just assume that he receives the messages unaltered insome preferred orthonormal basis such as {|ai , |bi , |ci , |di} so that he can distinguish thefour messages, and let us call this ensemble “Bob’s ensemble.”
Both Alice and Eve then know that the expected density operator of Eve’s ensemble isthe maximally mixed state if Eve does not know which message Alice chooses:
111
1|0ih0| + |1ih1| + |+ih+| + |−i h−| = π,4444(17.2)

where π ≡ I/2 is the maximally mixed state of a qubit. How can Alice ensure that Eve’s
information is independent of the message Alice is sending? Alice can choose subsets or
subensembles of the states in Eve’s ensemble to simulate the expected density operator of
Eve’s ensemble. Let us call these new simulating ensembles the “fake ensembles.” Alicec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

17.1. INTRODUCTORY EXAMPLE

503

chooses the member states of the fake ensembles according to the uniform distribution inorder to randomize Eve’s knowledge. The density operator for each new fake ensemble is its“fake expected density operator.”
Which states work well for being members of the fake ensembles? An equiprobablemixture of the states |0i and |1i suffices to simulate the expected density operator of
Eve’s ensemble because the fake expected density operator of this new ensemble is as follows: [|0ih0| + |1ih1|] /2 = π. An equiprobable mixture of the states |+i and |−i also
works because the fake expected density operator of this other fake ensemble is as follows:[|+ih+| + |−ih−|] /2 = π.
So it is possible for Alice to encode a private bit in this way. She first generates arandom bit that selects a particular message within each fake ensemble. So she selects aor b according to the random bit if she wants to transmit a “0” privately to Bob, and she
selects c or d according to the random bit if she wants to transmit a “1” privately to Bob.
In each of these cases, Eve’s resulting expected density operator is the maximally mixed
state. Thus, there is no measurement that Eve can perform to distinguish the original
message that Alice transmits, in the sense that she cannot do better than a random guessing
strategy. Bob, on the other hand, can perform a measurement in the basis {|ai , |bi , |ci , |di}to determine Alice’s private bit. In the case in which one private bit is being transmitted,
Eve can guess its value correctly with probability 1/2, but Alice and Bob can make thisprobability exponentially small if Alice sends more private bits using this technique (theguessing probability is 2−n if n private bits are transmitted in this way).
We can explicitly calculate Eve’s information about the private bit. Consider Eve’s
description of the state if she does not know which message Alice transmits—it is an equal
mixture of the following states: {|0i, |1i, |+i, |−i} (equal to the maximally mixed state π).
Eve’s description of the state “improves” to an equal mixture of the states {|0i, |1i} or{|+i, |−i}, each having density operator π, if she does know which message Alice transmits.
The following classical–quantum state describes this setting:11
ρM KE ≡ |0ih0|M ⊗ |0ih0|K ⊗ |0ih0|E + |0ih0|M ⊗ |1ih1|K ⊗ |1ih1|E4
411+ |1ih1|M ⊗ |0ih0|K ⊗ |+ih+|E + |1ih1|M ⊗ |1ih1|K ⊗ |−ih−|E , (17.3)
44
where we suppose that Eve never has access to the K register. Tracing over the register K
gives the reduced state ρM E :11
ρM E = |0ih0|M ⊗ |0ih0|E + |0ih0|M ⊗ |1ih1|E4
411+ |1ih1|M ⊗ |+ih+|E + |1ih1|M ⊗ |−ih−|E(17.4)
4
4
1
1
1
1= |0ih0|M ⊗ [|0ih0|E + |1ih1|E ] + |1ih1|M ⊗ [|+ih+|E + |−ih−|E ](17.5)
22
2
2
1
1= |0ih0|M ⊗ πE + |1ih1|M ⊗ πE
(17.6)
2
2= πM ⊗ πE .
(17.7)
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CHAPTER 17. THE COVERING LEMMA

Then Eve’s register is completely independent of the private bit in M and her information
about the private bit in M is given by evaluating the mutual information of the reduced
state ρM E : I(M ; E)ρ = 0, because the state is product. Thus, using this scheme, Eve has
no information about the private bit as we argued before.
We are interested in making this scheme use as little noise as possible because Alicewould like to transmit as much information as she can to Bob while still retaining privacy.
Therefore, Alice should try to make the fake ensembles use as little randomness as possible.
In the above example, Alice cannot make the fake ensembles any smaller because a smaller
size would leak information to Eve.

17.2 Setting and Statement of the Covering Lemma
The setting of the covering lemma is a generalization of the setting in the above example.
It essentially uses the same strategy for making information private, but the mathematicalanalysis becomes more involved in the more general setting. In general, we cannot haveperfect privacy as in the above example, but instead we ask only for approximate privacy.
Approximate privacy then becomes perfect in the asymptotic limit in the i.i.d. setting.
We first define the relevant ensemble for the covering lemma. We call it the “true ensemble” in order to distinguish it from the “fake ensemble.”
Definition 17.2.1 (True Ensemble) Suppose X is a set of size |X | with elements x. Suppose we have an ensemble {pX (x), σx }x∈X of quantum states where each value x occurs
with probability pX (x) according to some random variable X, and suppose we encode eachvaluePx into a quantum state σx ∈ D(H). The expected density operator of the ensemble isσ ≡ x∈X pX (x)σx .
The definition for a fake ensemble is similar to the way that we constructed the fake ensembles
in the example. It is merely a subset of the states in the true ensemble chosen according toa uniform distribution.
Definition 17.2.2 (Fake Ensemble) Let S be a set such that S ⊆ X . The fake ensemble
is defined as follows:{1/ |S| , σs }s∈S .
(17.8)Let σ denote the “fake expected density operator” of the fake ensemble:
σ(S) ≡1 X
σs .
|S| s∈S(17.9)In the example, Alice was able to obtain perfect privacy from Eve. We need a goodmeasure of privacy because it is not possible in general to obtain perfect privacy, but Alicecan instead obtain only approximate privacy. We call this measure the “obfuscation error”
because it determines how well Alice can obfuscate the state that Eve receives.
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17.3. OPERATOR CHERNOFF BOUND

505

Definition 17.2.3 (Obfuscation Error) The obfuscation error oe (S) of set S is a measure of how close the fake expected density operator σ(S) is to the actual expected density
operator:(17.10)oe (S) = kσ(S) − σk1 .
The goal for Alice is to make the size of her fake ensembles as small as possible whilestill having privacy from Eve. The covering lemma quantifies this trade-off by determining
how small each fake ensemble can be in order to obtain a certain obfuscation error.
The hypotheses of the covering lemma are somewhat similar to those of the packing
lemma. But as stated in the introduction to this chapter, the goal of the covering lemma ismuch different.
Lemma 17.2.1 (Covering Lemma) Let {pX (x), σx }x∈X be an ensemble as in Definition 17.2.1.
Suppose a total subspace projector Π and codeword subspace projectors {Πx }x∈X are given,
they project onto subspaces of H, and these projectors and each state σx satisfy the following
conditions:
Tr {σx Π} ≥ 1 − ε,Tr {σx Πx } ≥ 1 − ε,Tr {Π} ≤ D,1
Πx σx Πx ≤ Πx ,d(17.11)(17.12)(17.13)(17.14)

where ε ∈ (0, 1), D > 0, and d ∈ (0, D). Suppose that M is a set of size |M| with elements
m. Let a random covering code C ≡ {Cm }m∈M consist of random codewords Cm where thecodewords Cm are chosen independently according to the distribution pX (x) and give rise toa fake ensemble {1/ |M| , σCm }m∈M . Then there is a high probability that the obfuscation
error oe (C) of the random covering code C is small:
 3


√
√
ε |M| d4,
(17.15)
Pr oe (C) ≤ ε + 4 ε + 24 ε ≥ 1 − 2D exp −C
4 D
when ε is small and |M|  ε3 d/D. Thus, it is highly likely that a given fake ensemble {1/ |M| , σcm }m∈M has its expected density operator indistinguishable from the expected
density operator of the original ensemble {pX (x), σx }x∈X . It is in this sense that the fakeensemble {1/ |M| , σcm }m∈M “covers” the original ensemble {pX (x), σx }x∈X .

17.3 Operator Chernoff Bound
Before giving the proof of the covering lemma, we first state and prove the operator Chernoff
bound, which is the most critical tool for establishing the covering lemma. The operator
Chernoff bound is a theorem from the theory of large deviations and essentially states thatthe sample average of a large number of i.i.d. operator-valued random variables is close to
their expectation (with some constraints on the operator-valued random variables).
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CHAPTER 17. THE COVERING LEMMA

Lemma 17.3.1 (Operator Chernoff Bound) Let ξ1 , . . . , ξK ∈ L(H) be K i.i.d. positive
semi-definite operator-valued random variables. Suppose that each ξk has all of its eigenvalues
between zero and one:∀k ∈ [K] : 0 ≤ ξk ≤ I.
(17.16)
Let ξ denote the sample average of the K operator-valued random variables:
K

1 X
ξ=
ξk .
K k=1

(17.17)

Suppose that the expectation Eξ {ξk } ≡ µ of each operator ξk is positive definite, so that µ
exceeds the identity operator scaled by a number a ∈ (0, 1): µ ≥ aI. Then for every η where0 < η < 1/2 and (1 + η) a ≤ 1, we can bound the probability that the sample average ξ lies
inside the operator interval [(1 ± η) µ]:Kη 2 aPr (1 − η) µ ≤ ξ ≤ (1 + η) µ ≥ 1 − 2 dim(H) exp −.
(17.18)ξ
4
Thus it is highly likely that the sample average operator ξ becomes close to the true expected
operator µ as K becomes large.
We prove the above lemma by making a progression through the operator Markov inequality all the way to the proof of the operator Chernoff bound. Recall that we writeA ≥ B if A − B is a positive semi-definite operator,
P and we write A  B otherwise. In
what follows, we take the convention exp{A}=i exp(ai )|iihi| for a Hermitian operaP
tor A withP
spectral decomposition A = i ai |iihi| (this differs from our usual convention
exp{A} = i:ai 6=0 exp(ai )|iihi|).
Lemma 17.3.2 (Operator Markov Inequality) Let X ∈ L(H) be a positive semi-definite
operator-valued random variable. Let E {X} denote its expectation. Let A be a fixed positive
definite operator in L(H). Then
Pr {X  A} ≤ Tr E {X} A−1 .
(17.19)Proof. Observe that if X  A then A−1/2 XA−1/2  I. This then implies that the largest
eigenvalue of A−1/2 XA−1/2 exceeds one: A−1/2 XA−1/2 ∞ > 1. Let IXA denote an indicatorfunction for the event X  A. We then have that
(17.20)
IXA ≤ Tr A−1/2 XA−1/2 .
The above inequality follows because the right-hand side is non-negative if the indicatoris zero. If the indicator is one, then the right-hand side exceeds one because its largest
eigenvalue is greater than one and the trace exceeds the largest eigenvalue for a positive
semi-definite operator. We then have the following inequalities: 
Pr {X  A} = E IXA ≤ E Tr A−1/2 XA−1/2(17.21) 
−1−1= E Tr XA= Tr E {X} A,(17.22)where the inequality follows from (17.20) and the second equality from cyclicity of trace.
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17.3. OPERATOR CHERNOFF BOUND

507

Lemma 17.3.3 (Bernstein Trick) Let X, X1 , . . . , XK ∈ L(H) be i.i.d. Hermitian operatorvalued random variables, and let A be a fixed Hermitian operator. Then for any invertible
operator T , the following bound holds
(K
)
X


K
Pr
Xk  KA ≤ dim(H) E exp T (X − A) T †.
(17.23)∞
k=1

Proof. The proof of this lemma relies on the Golden–Thompson trace inequality fromstatistical mechanics, which holds for any two Hermitian operators A and B (we state it
without proof):
Tr {exp {A + B}} ≤ Tr {exp {A} exp {B}} .
(17.24)
Consider the following chain of inequalities:
(K
)
(K
)
X
X
Pr
Xk  KA = Pr
(Xk − A)  0
k=1

= Pr

k=1
(K
X

(17.25))

T (Xk − A) T †  0(17.26)k=1(= Pr exp(KX)
T (Xk − A) T †)I(17.27)k=1( (≤ Tr E exp

(K
X

)))
T (Xk − A) T†.

(17.28)k=1

The first two equalities are straightforward and the third follows because A ≤ B is equivalent
to exp {A} ≤ exp {B} for commuting operators A and B. The inequality follows by applying
the operator Markov inequality (Lemma 17.3.2). Continuing, we have
( (
(K
)))
X
= E Tr exp
T (Xk − A) T †(17.29)k=1((≤ E Tr exp

(K−1
X

)
T (Xk − A) T†))

exp T (XK − A) T †(17.30)

k=1(= EX1 ,...,XK−1

Tr exp(= EX1 ,...,XK−1((

Tr exp

(K−1
X
k=1
(K−1
X

)
T (Xk − A) T †))EXK exp T (XK − A) T †)
T (Xk − A) T †(17.31)))

EX exp T (X − A) T †.

(17.32)

k=1

The first equality follows from exchanging the expectation and the trace. The inequality follows from applying the Golden–Thompson trace inequality. The second and third equalities
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CHAPTER 17. THE COVERING LEMMA

follow from the i.i.d. assumption. Continuing,
( (
(K−1
)))
X


≤ EX1 ,...,XK−1 Tr exp
T (Xk − A) T †EX exp T (X − A) T †∞(17.33)k=1

≤ Tr {I} EX exp T (X − A) T †= dim(H) EX exp T (X − A) T †

K
∞
K
∞

(17.34).

(17.35)

The first inequality follows from Tr {AB} ≤ Tr {A} kBk∞ for A positive semi-definite. The
second inequality follows from a repeated application of the same steps. The final equality
follows because the trace of the identity operator is equal to the dimension of the Hilbert
space. This proves the “Bernstein trick” lemma.
Proof of the Operator Chernoff Bound (Lemma 17.3.1). We first prove that thefollowing inequality holds for i.i.d. Hermitian operator-valued random variables X, X1 , . . . ,XK such that E {X} ≤ mI, A ≥ aI, and 1 ≥ a > m ≥ 0:
(K
)
X
Pr
Xk  KA ≤ dim(H) exp {−KD(akm)} ,(17.36)k=1

where D(akm) is the binary relative entropy:
D(akm) = a ln a − a ln m + (1 − a) ln (1 − a) − (1 − a) ln (1 − m) .
√We first apply the Bernstein Trick (Lemma 17.3.3) with T = tI, for t > 0:
(K
)
(K
)
X
X
Pr
Xk  KA ≤ Pr
Xk  KaI
k=1(17.37)(17.38)k=1≤ dim(H) kE {exp {tX} exp {−ta}}kK∞.

(17.39)So it is clear that it is best to optimize t in such a way thatkE {exp {tX} exp {−ta}}k∞ < 1,

(17.40)so that we have exponential decay with increasing K. Now consider the following inequality:
exp {tX} − I ≤ X (exp {t} − 1) ,(17.41)

which holds because a similar one holds for all real x ∈ [0, 1]:(exp {tx} − 1) ≤ x(exp {t} − 1).

(17.42)E {exp {tX}} ≤ E {X} (exp {t} − 1) + I≤ mI (exp {t} − 1) + I= (m exp {t} + 1 − m) I.

(17.43)(17.44)(17.45)

Applying this inequality givesc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

17.3. OPERATOR CHERNOFF BOUND

509

which in turn implies
kE {exp {tX} exp {−ta}}k∞ ≤ (m exp {t} + 1 − m) exp {−ta} .

(17.46)

Choosing
t = ln

a 1−m
·
m 1−a


> 0,

(17.47)

which follows from the assumption that a > m, gives(m exp {t} + 1 − m) exp {−ta}
 




 
a 1−m
a 1−m
= m
·
+ 1 − m exp − ln
·
a
m 1−a
m 1−a





a
1−m
1−m
= a·
+ 1 − m exp −a ln− a ln
1−a
m
1−a







1−m
a
1−m
=
exp −a ln− a ln
1−a
m
1−a



a
1−a
− (1 − a) ln= exp −a lnm
1−m= exp {−D(akm)} ,(17.48)(17.49)(17.50)(17.51)(17.52)

proving the desired bound in (17.36).
By substituting Yk = I − Xk and B = I − A into (17.36) and having the opposite
conditions E {X} ≥ mI, A ≤ aI, and 0 ≤ a < m ≤ 1, we can show that the following
inequality holds for i.i.d. operators X, X1 , . . . , XK :
(K
)
X
Pr
Xk  KA ≤ dim(H) exp {−KD(akm)} .
(17.53)k=1

To finish off the proof of the operator Chernoff bound, consider the variables Zk =Lµ−1/2 Xk µ−1/2 with µ ≡ E {X} ≥ LI. Then E {Zk } = LI and 0 ≤ Zi ≤ I. The following
events are thus equivalent
K
K
1 X
1 X
Xk ≤ (1 + η) µ ⇐⇒ (1 − η) LI ≤Zk ≤ (1 + η) LI,(1 − η) µ ≤
K k=1
K k=1

(17.54)and we can apply (17.36), (17.53), and the union bound to obtain
(
!
!)
K
K
[ 1 X
1 X
Pr
(1 − η) µ 
Xk
Xk  (1 + η) µ
K k=1
K k=1
≤ dim(H) exp {−KD((1 − η) LkL)} + dim(H) exp {−KD((1 + η) LkL)}
η2L≤ 2 dim(H) exp −K,
4c 2019(17.55)(17.56)

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CHAPTER 17. THE COVERING LEMMA

where the last line exploits the following inequality valid for −1/2 ≤ η ≤ 1/2 and (1 + η) L ≤1:
η2L
.
(17.57)
D((1 + η) LkL) ≥4
This last inequality follows from a calculus argument. To see it, define f (η) ≡ D((1 + η) LkL),
and note that f 0 (η) = L ln(1+η)−L ln(1−ηL/(1−L)) and f 00 (η) = L/[(1+η)(1−(1+η)L)] ≥2
L/(1 + η). By Taylor’s theorem, it follows that f (η) = f (0) + ηf 0 (0) + η2 f 00 (ξ), for some
2
η2 L
η2 L
ξ ∈ [0, η]. Since f (0) = f 0 (0) = 0, it follows that f (η) = η2 f 00 (ξ) ≥ 2(1+ξ)≥ 2(1+η). Since22

η L
η < 1, then 1/(1 + η) ≥ 1/2, implying that 2(1+η)≥ η 4L and finally gives (17.57). Thisconcludes the proof of Lemma 17.3.1.

17.4 Proof of the Covering Lemma
The first step of the proof of the covering lemma is to construct an alternate ensemble thatis close to the original ensemble yet satisfies the conditions of the operator Chernoff bound(Lemma 17.3.1). We call this alternate ensemble the “Chernoff ensemble.” We then generate
a random code, a set of M i.i.d. random variables, using the Chernoff ensemble. Call thisrandom code the “Chernoff code.” We apply the operator Chernoff bound to the Chernoff
code to obtain a good bound on the obfuscation error of the Chernoff code. We finally showthat the bound holds for a covering code generated by the original ensemble because theoriginal ensemble is close to the Chernoff ensemble in trace distance.

17.4.1

Construction of the Chernoff Ensemble

We first establish a few definitions to construct intermediary ensembles. We then use theseintermediary ensembles to construct the Chernoff ensemble. We construct the first “primed”
ensemble {pX (x), σx0 } by using the projection operators Πx to slice out some of the supportof the states σx :∀x σx0 ≡ Πx σx Πx .
(17.58)The above “slicing” operation cuts outs any part of the support of σx that is not in the
support of Πx . The expected operator σ 0 for the first primed ensemble is as follows:X
σ0 ≡pX (x)σx0 .
(17.59)x∈X

We then continue slicing with the projector Π and form the second primed ensemble {pX (x), σx00 }as follows:∀x σx00 ≡ Πσx0 Π.
(17.60)The expected operator for the second primed ensemble is as follows:X
σ 00 ≡pX (x)σx00 .

(17.61)x∈X
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17.4. PROOF OF THE COVERING LEMMA

511

Let Π̂ be the projector onto the subspace spanned by the eigenvectors of σ 00 whose corresponding eigenvalues are greater than ε/D. We would expect that this extra slicing does notchange the state very much when D is large and ε is small. We construct states ωx in the
Chernoff ensemble by using the projector Π̂ to slice out some more elements of the supportof the original ensemble:∀x ωx ≡ Π̂σx00 Π̂.
(17.62)The expected operator ω for the Chernoff ensemble is then as follows:Xω≡pX (x)ωx .

(17.63)x∈X

The Chernoff ensemble satisfies the conditions necessary to apply the operator Chernoff
bound. We wait to apply the operator Chernoff bound and for now show how to constructa random covering code.

17.4.2Chernoff Code Construction

We present a Shannon-like random coding argument. We construct a covering code C at
random by independently generating |M| codewords according to the distribution pX (x).
Let C = {cm }m∈M be a collection of the realizations cm of |M| independent random variables
Cm . Each Cm takes a value cm in X with probability pX (cm ) and represents a codeword in
the random code C. This process generates the Chernoff code C consisting of |M| quantum
states {ωcm }m∈M . The fake expected operator ω(C) of the states in the Chernoff code is asfollows:|M|
1 X
ω(C) ≡
ωc ,(17.64)|M| m=1 m
because we assume that Alice randomizes codewords in the Chernoff code according to a
uniform distribution (notice that there is a difference in the distribution that we use tochoose the code and the distribution that Alice uses to randomize the selected codewords).
The expectation EC {ωCm } of each operator ωCm is equal to the expected operator ω becauseof the way that we constructed the coveringcode. We can also define codes with respect to theprimed ensembles as follows: {σcm }m∈M , σc0 m m∈M , σc00m m∈M . These codes respectively
have fake expected operators of the following form:|M|

1 X
σ(C) ≡σc ,|M| m=1 m(17.65)|M|

1 X 0
σ (C) ≡σ ,|M| m=1 cm
0(17.66)|M|

1 X 00σ (C) ≡σ .
|M| m=1 cm
00

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CHAPTER 17. THE COVERING LEMMA

Applying the Operator Chernoff Bound. We make one final modification before
applying the operator Chernoff bound. The operators ωcm are in the operator interval
between the zero operator 0 and d1 Π̂:
1∀m ∈ M : 0 ≤ ωcm ≤ Π̂.
d(17.68)The above statement holds because the operators σx0 satisfy σx0 = Πx σx Πx ≤ d1 Πx (the fourth
condition of the covering lemma) and this condition implies the following inequalities:1
σx0 = Πx σx Πx ≤ Πx
d1
1
⇒ Πσx0 Π = σx00 ≤ ΠΠx Π ≤ Πdd11⇒ ωx = Π̂σx00 Π̂ ≤ Π̂ΠΠ̂ ≤ Π̂.
dd(17.69)(17.70)(17.71)

Therefore, we consider another set of operators (not necessarily density operators) where we
scale each ωcm by d so that(17.72)∀m ∈ M : 0 ≤ dωcm ≤ Π̂.
This code satisfies the conditions of the operator Chernoff bound with a = εd/D and with Π̂
acting as the identity on the subspace onto which it projects. We can now apply theoperator−1 P
Chernoff bound to bound the probability that the sample average ω ≡ |M|m∈M ωcm
falls in the operator interval [(1 ± ε) ω]:Pr {(1 − ε) ω ≤ ω ≤ (1 + ε) ω} = Pr {d (1 − ε) ω ≤ dω ≤ d (1 + ε) ω}n o|M| ε2 (εd/D)≥ 1 − 2 Tr Π̂ exp −4 3
ε |M| d.
≥ 1 − 2D exp −4 D

17.4.3(17.73)(17.74)(17.75)

Obfuscation Error of the Covering CodeThe random covering code is a set of |M| quantum states {σCm }m∈M where the quantum
states arise from the original ensemble. Recall that our goal is to show that the obfuscation
error of the random covering code C,oe (C) = kσ(C) − σk1 ,(17.76)has a high probability of being small.
We now show that the obfuscation error of this random covering code is highly likely tobe small, by relating it to the Chernoff ensemble. Our method of proof is simply to exploitc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

17.4. PROOF OF THE COVERING LEMMA

513

the triangle inequality, the gentle operator lemma (Lemma 9.4.2), and Exercise 9.1.8 several
times. The triangle inequality gives the following bound for the obfuscation error:oe (C) = kσ(C) − σk1= kσ(C) − σ 00 (C) − (ω(C) − σ 00 (C)) + (ω(C) − ω) + (ω − σ 00 ) − (σ − σ 00 )k1≤ kσ(C) − σ 00 (C)k1 + kω(C) − σ 00 (C)k1+ kω(C) − ωk1 + kω − σ 00 k1 + kσ − σ 00 k1 .

(17.77)(17.78)We show how to obtain a good bound for each of the above five terms.
First consider the rightmost term kσ − σ 00 k1 in (17.78). Consider that the projected state0
σx = Πx σx Πx is close to the original state σx by applying (17.12) and the gentle operator
lemma:√(17.79)kσx − σx0 k1 ≤ 2 ε.
Consider thatqkσx0 − σx00 k1 ≤ 2

√
ε+2 ε

(17.80)because σx00 = Πσx0 Π and from applying the gentle operator lemma to
Tr {Πσx0 } ≥ Tr {Πσx } − kσx − σx0 k1√≥ 1 − ε − 2 ε,(17.81)(17.82)where the first inequality follows from Exercise 9.1.8 and the second from (17.11) and (17.79).
Then the state σx00 is close to the original state σx for all x because
kσx − σx00 k1 ≤ kσx − σx0 k1 + kσx0 − σx00 k1q√√≤ 2 ε + 2 ε + 2 ε,(17.83)(17.84)

where we first applied the triangle inequality and the bounds from (17.79) and (17.80).
Convexity of the trace distance then gives a bound on kσ − σ 00 k1 :kσ − σ 00 k1 =X

pX (x)σx −

x∈X

=

X

≤

X

X

pX (x)σx00

x∈X(17.85)1pX (x) (σx − σx00 )x∈X(17.86)1pX (x) kσx − σx00 k1(17.87)q√√pX (x) 2 ε + 2 ε + 2 ε(17.88)

x∈X

≤

X
x∈X

√
=2 ε+2c 2019q√
ε + 2 ε.

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CHAPTER 17. THE COVERING LEMMA

We now consider the second rightmost term kω − σ 00 k1 in (17.78). The support of σ 00 has
dimension less than D by (17.13), the third condition in the covering lemma.PTherefore,00
eigenvalues smaller than ε/D contribute at most ε to Tr {σP }. In particular, if i λi |iihi| is00
a spectral decomposition of σ , with λi ≥ 0 for all i and i λi ≤ 1, we have thatTr {σ 00 } − Tr {ω} =

X

λi −

i

λi

(17.90)ε· D = ε.
D

(17.91)i:λi ≥ε/D

X

=

X

λi ≤

i:λi <ε/D

We can use this to bound the trace of ω as follows:
Tr {ω} ≥ Tr {σ 00 } − εX=pX (x) Tr {σx00 } − ε(17.92)(17.93)

x∈X

!
≥

X

pX (x)

x∈X=1−2 ε+√ 
1−ε−2 ε −ε

√ 
ε ,

(17.94)(17.95)where the first inequality applies the above “eigenvalue-bounding” argument and the second
inequality employs the bound in (17.82). This argument shows that the average operator of
the Chernoff ensemble has trace√ almost equal to one. We can then apply the gentle operator
lemma to Tr {ω} ≥ 1 − 2 (ε + ε) to giveq√ kω − σ 00 k1 ≤ 2 2 ε + ε .
(17.96)
We now consider the middle term kω − ωk1 in (17.78). The Chernoff bound gives us aprobabilistic estimate and not a deterministic estimate like the other two bounds we haveshown above. So we suppose for now that the fake operator ω of the Chernoff code is close
to the average operator ω of the Chernoff ensemble:
ω(C) ≡1 X
ωc ∈ [(1 ± ε) ω] .
|M| m∈M m(17.97)kω(C) − ωk1 ≤ ε,(17.98)With this assumption, it holds thatby employing Lemma A.0.2 from Appendix A and Tr {ω} ≤ 1.
We consider the second leftmost term kω(C) − σ 00 (C)k1 in (17.78). The following inequality holds:√
Tr {ω(C)} ≥ 1 − 3ε − 2 ε,(17.99)c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

17.4. PROOF OF THE COVERING LEMMA

because in (17.95) we showed that Tr {ω} ≥ 1 − 2 (ε +implies that515√ε), and the triangle inequality

Tr {ω(C)} = kω(C)k1= kω − (ω − ω(C))k1≥ kωk1 − kω − ω(C)k1= Tr {ω} − kω − ω(C)k1√ ≥ 1−2 ε+ ε −ε√
= 1 − 3ε − 2 ε.

(17.100)(17.101)(17.102)(17.103)(17.104)(17.105)√
Applying the gentle operator lemma to Tr {ω(C)} ≥ 1 − 3ε − 2 ε gives thatq√kω(C) − σ (C)k1 ≤ 2 3ε + 2 ε.
00

(17.106)

We finally bound the leftmost term kσ(C) − σ 00 (C)k1 in (17.78). We can use convexity oftrace distance and (17.84) to obtain the following bounds:
1 X
σCm − σC00 m 1|M| m∈Mq√√≤ 2 ε + 2 ε + 2 ε.

kσ(C) − σ 00 (C)k1 ≤(17.107)(17.108)

We now combine all of the above bounds with the triangle inequality in order to boundthe obfuscation error of the covering code C:oe (C)= kσ(C) − σk1= kσ(C) − σ 00 (C) − (ω(C) − σ 00 (C)) + (ω(C) − ω) + (ω − σ 00 ) − (σ − σ 00 )k1≤ kσ(C) − σ 00 (C)k1 + kω(C) − σ 00 (C)k1+ kω(C) − ωk1 + kω − σ 00 k1 + kσ − σ 00 k1

  q

q
√
√
√
≤ 2 ε + 2 ε + 2 ε + 2 3ε + 2 ε + ε
 

 q
q
√ 
√
√
+ 2 2 ε+ ε + 2 ε+2 ε+2 ε
q
q
q
√
√
√
√ 
= ε + 4 ε + 4 ε + 2 ε + 2 3ε + 2 ε + 2 2 ε + ε√√≤ ε + 4 ε + 24 4 ε.

(17.109)(17.110)(17.111)(17.112)(17.113)(17.114)

Observe from the above that the event that the quantity ε bounds the obfuscation√
error oe (C)√of the Chernoff code with states ωCm implies the event when the quantity ε + 4 ε + 24 4 ε
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CHAPTER 17. THE COVERING LEMMA

bounds the obfuscation error oe (C) of the original code with states σCm . Thus, we can boundthe probability of obfuscation error of the covering code by applying the Chernoff bound:√√Pr oe (C, {σCm }) ≤ ε + 4 ε + 24 4 ε ≥ Pr {oe (C, {ωCm }) ≤ ε}(17.115) 3
ε |M| d.
(17.116)≥ 1 − 2D exp −4 DThis argument shows that it is highly likely that a random covering code is good in the sensethat it has a low obfuscation error.
Exercise 17.4.1 Prove that the covering lemma holds for the same ensemble and a set ofprojectors for which the following conditions hold:XpX (x) Tr {σx Π} ≥ 1 − ε,(17.117)x∈X

X

pX (x) Tr {σx Πx } ≥ 1 − ε,(17.118)

Tr {Π} ≤ D,1
Πx σx Πx ≤ Πx .
d(17.119)x∈X(17.120)

Exercise 17.4.2 Show that there exists a particular covering code with the property that theobfuscation error is small.

17.5 History and Further Reading
Ahlswede and Winter (2002) introduced the operator Chernoff bound in the context of quantum identification. Winter and Massar (2001); Winter (2004) later applied it to quantum
measurement compression. Devetak and Winter (2003) applied the covering lemma to classical compression with quantum side information and to distilling secret key from quantum
states (Devetak and Winter, 2005). Devetak (2005) and Cai et al. (2004) applied it to private
classical communication over a quantum channel, and Groisman et al. (2005) applied it to
study the destruction of correlations in a bipartite state.

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicensePart V
Noiseless Quantum Shannon Theory

517

CHAPTER 18

Schumacher Compression
One of the fundamental tasks in classical information theory is the compression of information. Given access to many uses of a noiseless classical channel, what is the best that asender and receiver can make of this resource for compressed data transmission? Shannon’scompression theorem demonstrates that the Shannon entropy is the fundamental limit forthe compression rate in the i.i.d. setting (recall the development in Section 14.4). Thatis, if one compresses at a rate above the Shannon entropy, then it is possible to recover
the compressed data perfectly in the asymptotic limit, and otherwise, it is not possible to
do so.1 This theorem establishes the prominent role of the entropy in Shannon’s theory of
information.
In the quantum world, it very well could be that one day a sender and a receiver would
have many uses of a noiseless quantum channel available,2 and the sender could use thisresource to transmit compressed quantum information. But what exactly does this meanin the quantum setting? A simple model of a quantum information source is an ensembleof quantum states {pX (x), |ψx i}; i.e., the source outputs the state |ψx i with probabilitypX (x), and the states {|ψx i} do not necessarily have to form an orthonormal basis. Let
us suppose for the moment that the classical data x is available as well, even though thismight not necessarily be the case in practice. A naive strategy for compressing this quantum
information source would be to ignore the quantum states coming out, handle the classicaldata instead, and exploit Shannon’s compression protocol from Section 14.4. That is, thesender compresses the sequence xn emitted from the quantum information source at a rateequal to the Shannon entropy H(X), sends the compressed classical bits over the noiselessquantum channels, the receiver reproduces the classical sequence xn at his end, and finallyreconstructs the sequence |ψxn i of quantum states corresponding to the classical sequence xn .
The above strategy will certainly work, but it makes no use of the fact that the noiseless quantum channels are quantum! It is clear that noiseless quantum channels will be
1

Technically, we did not prove the converse part of Shannon’s data-compression theorem, but the converseof this chapter suffices for Shannon’s classical theorem as well.
2
How we hope so! If working, coherent fault-tolerant quantum computers come along one day, they standto benefit from quantum compression protocols.

519

520

CHAPTER 18. SCHUMACHER COMPRESSION

expensive in practice, and the above strategy is wasteful in this sense because it could havemerely exploited classical channels (channels that cannot preserve superpositions) to achieve
the same goals. Schumacher compression is a strategy that makes effective use of noiselessquantum channels to compress a quantum information source down to a rate equal to thequantum entropy. This has a great benefit from a practical standpoint—recall from Exercise 11.9.3 that the quantum entropy of a quantum information source is strictly lower than
the source’s Shannon entropy if the states in the ensemble are non-orthogonal. In order
to execute
P the protocol, the sender and receiver simply need to know the density operator
ρ ≡x pX (x)|ψx ihψx | of the source. Furthermore, Schumacher compression is provably
optimal in the sense that any protocol that compresses a quantum information source of
the above form at a rate below the quantum entropy cannot have a vanishing error in theasymptotic limit.
Schumacher compression thus gives an operational interpretation of the quantum entropy
as the fundamental limit on the rate of quantum data compression. Also, it sets the term
“qubit” on a firm foundation in an information-theoretic sense as a measure of the amountof quantum information “contained” in a quantum information source.
We begin this chapter by giving the details of the general information-processing taskcorresponding to quantum data compression. We then prove that the quantum entropy
is an achievable rate of compression and follow by showing that it is optimal (these two
respective parts are the direct coding theorem and the converse theorem for quantum datacompression). We illustrate how much savings one can gain in quantum data compression
by detailing a specific example. The final section of the chapter closes with a presentationof more general forms of Schumacher compression.

18.1 The Information-Processing Task
We first discuss the general task that any quantum compression protocol attempts to accomplish. Three parameters n, R, and ε, corresponding to the length of the original quantum data sequence, the rate, and the error, respectively, characterize any such protocol.
An (n, R, ε) quantum compression code consists of four steps: state preparation, encoding,transmission, and decoding. Figure 18.1 depicts a general protocol for quantum compression.
State Preparation. The quantum information source outputs a sequence |ψxn iAn ofquantum states according to the ensemble {pX (x), |ψx i} where|ψxn iAn ≡ |ψx1 iA1 ⊗ · · · ⊗ |ψxn iAn .

(18.1)The density operator, from the perspective of someone ignorant of the classical sequence xn ,is equal to the tensor power state ρ⊗n whereXρ≡pX (x)|ψx ihψx |.
(18.2)
x

Also, we can think about the purification of the above density operator. That is, a related
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18.1. THE INFORMATION-PROCESSING TASK

521

Figure 18.1: The most general protocol for quantum compression. Alice begins with the output of somequantum information source whose density operator is ρ⊗n on some system An . The inaccessible reference
system holds the purification of this density operator. She performs some CPTP encoding map E, sends thecompressed qubits through nR uses of a noiseless qubit channel, and Bob performs some CPTP decoding
map D to decompress the qubits. The scheme is successful if the initial state and the final state areindistinguishable in the asymptotic limit n → ∞.

picture is to imagine that the quantum information source produces states of the formXp
|ϕρ iRA ≡pX (x)|xiR |ψx iA ,(18.3)
xwhere R is the label for an inaccessible reference system (not to be confused with the rateR!). The resulting i.i.d. state produced is (|ϕρ iRA )⊗n .
Encoding. Alice encodes the systems An according to some compression channel EAn →W
where W is a quantum system of size 2nR . Recall that R is the rate of compression:R=1
log dim(HW ).
n(18.4)

Transmission. Alice transmits the system W to Bob using nR noiseless qubit channels.
Decoding. Bob sends the system W through a decompression channel DW →Ân .
The protocol has ε ∈ [0, 1] error if the compressed and decompressed state is ε-close innormalized trace distance to the original state (|ϕρ iRA )⊗n :1(ϕρRA )⊗n − (DW →Ân ◦ EAn →W )((ϕρRA )⊗n ) 1 ≤ ε.
2

(18.5)

We say that a quantum compression rate R is achievable if there exists an (n, R + δ, ε)quantum compression code for all δ > 0, ε ∈ (0, 1), and sufficiently large n. The quantum
data compression limit of ρ is equal to the infimum of all achievable quantum compression
rates.
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CHAPTER 18. SCHUMACHER COMPRESSION

Exercise 18.1.1 An alternate figure of merit for the performance of Schumacher compression is the average ensemble trace distance, given by
1X
pX n (xn ) kψxn − (DW →Ân ◦ EAn →W )(ψxn )k1 .
2 xn(18.6)

Prove that a quantum compression protocol satisfying (18.5) also has an average ensemble
trace distance no larger than ε. (Hint: Consider using the monotonicity of trace distance withrespect to quantum channels and acting on the reference systems with a particular channel.)

18.2 The Quantum Data-Compression Theorem
Schumacher’s compression theorem establishes the quantum entropy as the fundamental
limit on quantum data compression.
Theorem 18.2.1 (Quantum Data Compression) Suppose that ρA is the density operator corresponding to a quantum information source. Then the quantum entropy H(A)ρ is
equal to the quantum data compression limit of ρ.

18.2.1

The Direct Coding Theorem

Schumacher’s compression protocol demonstrates that the quantum entropy H(A)ρ is anachievable rate for quantum data compression. It is remarkably similar to Shannon’s compression protocol from Section 14.4, but it has some subtle differences that are necessary
for the quantum setting. The basic steps of the encoding are to perform a typical subspacemeasurement and an isometry that compresses the typical subspace. The decoder then performs the inverse of the isometry to decompress the state. The protocol is successful if thetypical subspace measurement successfully projects onto the typical subspace, and it fails
otherwise. Just like in the classical case, the law of large numbers guarantees that the protocol is successful in the asymptotic limit as n → ∞. Figure 18.2 provides an illustration of
the protocol, and we now provide a rigorous argument.
Alice begins with n copies of the state (ϕρRA )⊗n . Suppose that a spectral decompositionof ρ is as follows:Xρ=pZ (z)|zihz|,(18.7)zwhere pZ (z) is some probability distribution, and {|zi} is some orthonormal basis. Her firststep is to perform a typical subspace measurement onto the typical subspace of An , where
the typical projector is with respect to the density operator ρ. Recall from the Shannon
compression protocol in Section 14.4 that we exploited a one-to-one function f that mappedfrom the set of typical sequences to a set of binary sequences {0, 1}n[H(ρ)+δ] . Now, we canconstruct a linear map Uf that is a coherent version of this classical function f . It simply
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18.2. THE QUANTUM DATA-COMPRESSION THEOREM

523

Figure 18.2: Schumacher’s compression protocol. Alice begins with many copies of the output of thequantum information source. She performs a measurement onto the typical subspace corresponding to thestate ρ and then performs a compression isometry of the typical subspace to a space of dimension 2n[H(ρ)+δ] ,corresponding to n [H(ρ) + δ] qubits. She transmits these compressed qubits over n [H(ρ) + δ] uses of anoiseless qubit channel. Bob performs the inverse of the isometry to uncompress the qubits. The protocol issuccessful in the asymptotic limit due to the properties of typical subspaces.

maps the orthonormal basis {|z n iAn } to the basis {|f (z n )iW }:Uf ≡X|f (z n )iW hz n |An ,(18.8)nz n ∈TδZnwhere Z is a random variable corresponding to the distribution pZ (z), so that TδZ is its
typicalinverse of the above operator is an isometry because the input space nset. Thennnspan |z iA : z ∈ TδZ is a subspace of size at most 2n[H(ρ)+δ] (recall Property 15.1.2) embedded in a larger space of size [dim(HA )]n and the output space is of size at most 2n[H(ρ)+δ] .
So her next step is to perform the compression conditioned on the typical subspace measurement being successful. We can thus write the encoding as a single quantum channel asfollows:
EAn →W (XAn ) ≡ Uf ΠδAn XAn ΠδAn Uf† + Tr{(IAn − ΠδAn )XAn }σW ,(18.9)
where XAn ∈ L(HAn ) and σW is any density operator with support in span{|f (z n )iW : z n ∈nTδZ }. Alice then sends the W system of EAn →W ((ϕρRA )⊗n ) (the compressed qubits) overn [H(ρ) + δ] uses of the noiseless qubit channel.
Bob’s decoding DW →An essentially performs the inverse of the linear map Uf , implemented as the following quantum channel:
DW →An (YW ) ≡ Uf† YW Uf + Tr{(I − Uf Uf† )YW }τAn ,(18.10)where YW ∈ L(HW ) and τAn ∈ D(HAn ).
We now can analyze how this protocol performs with respect to our performance criterionc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

524

CHAPTER 18. SCHUMACHER COMPRESSIONin (18.5). Consider that
(DW →An ◦ EAn →W ) ((ϕρRA )⊗n )= DW →An Uf ΠδAn (ϕρRA )⊗n ΠδAn Uf† + TrAn {(IAn − ΠδAn ) (ϕρRA )⊗n } ⊗ σW


⊗n δ
†
ρ
δ
= DW →An Uf ΠAn (ϕRA ) ΠAn Uf
+ DW →An TrAn {(IAn − ΠδAn ) (ϕρRA )⊗n } ⊗ σW

(18.12)= ΠδAn (ϕρRA )⊗n ΠδAn + TrAn {(IAn − ΠδAn ) (ϕρRA )⊗n } ⊗ DW →An (σW ).

(18.13)(18.11)Then(ϕρRA )⊗n − (DW →An ◦ EAn →W ) ((ϕρRA )⊗n ) 1≤ (ϕρRA )⊗n − ΠδAn (ϕρRA )⊗n ΠδAn 1+ TrAn {(IAn − ΠδAn ) (ϕρRA )⊗n } ⊗ DW →An (σW ) 1√≤ 2 ε + ε.

(18.14)(18.15)

The first inequality follows from the triangle inequality. The second inequality follows fromthe first property of typical subspaces:Tr ΠδAn (ϕρRA )⊗n = Tr ΠδAn ρ⊗n ≥ 1 − ε,(18.16)
the gentle operator lemma (Lemma 9.4.2), and the fact that
TrAn {(IAn − ΠδAn ) (ϕρRA )⊗n } ⊗ DW →An (σW ) 1= TrAn {(IAn − ΠδAn ) (ϕρRA )⊗n } 1 kDW →An (σW )k1(18.17)≤ Tr{(IAn − ΠδAn ) (ϕρRA )⊗n } ≤ ε.

(18.18)We remark that it is important for the typical subspace measurement in (18.9) to be
implemented (as indicated) as a non-destructive quantum measurement. That is, the only
information that this measurement should learn is whether the state is typical or not. Otherwise, there would be too much disturbance to the quantum information, and the protocol
would fail at the desired task of compression. Such precise control on so many qubits ispossible in principle, but it is rather daunting to implement in practice!

18.2.2The Converse Theorem

We now prove the converse theorem for quantum data compression by considering the mostgeneral compression protocol that meets the success criterion in (18.5) and demonstratingthat a sequence of such protocols with error approaching zero should have their rate ofcompression above the quantum entropy of the source. Alice would like to compress astate ρ⊗n that acts on a Hilbert space An . A purification φRn An ≡ (ϕρRA )⊗n of this state
represents the state of the joint systems An and Rn where Rn is the purifying system (again,c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

18.2. THE QUANTUM DATA-COMPRESSION THEOREM

525

we should not confuse reference system Rn with rate R). If she can compress any system
on An and recover it faithfully, then she should be able to do so for the purification of thestate. An (n, R + δ, ε) compression code has the property that it can compress at a rateR + δ ≡ [log dim(HW )] /n with only error ε. The quantum data processing is
An

DW →Ân Ân(18.19)−−−−−→and the following inequality holds for a quantum compression protocol with error ε:
EAn →W W−−−−→

1
ωRn Ân − (ϕρRA )⊗n 1 ≤ ε,2

(18.20)ωRn Ân ≡ DW →Ân (EAn →W ((ϕρRA )⊗n )).

(18.21)

where
Let τRn W ≡ EAn →W ((ϕρRA )⊗n ). Consider the following chain of inequalities:
2 log dim(HW ) ≥ I(W ; Rn )τ
nn≥ I(Â ; R )ω≥ I(An ; Rn )ϕ⊗n − f (n, ε)= nI(A; R)ϕ − f (n, ε)= 2nH(A)ϕ − f (n, ε).

(18.22)(18.23)(18.24)(18.25)(18.26)

The first inequality is a consequence of a dimension bound for the quantum mutual information I(E; F ) ≤ 2 log (min {|E| , |F |}) (see Exercise 11.6.3). The second inequality follows
from the quantum data-processing inequality (Bob processes W using the decoder to get
Ân ). The third inequality follows because H(Rn )ω = H(Rn )ϕ⊗n , so that
I(Ân ; Rn )ω − I(An ; Rn )ϕ⊗n
= H(Rn )ω − H(Rn |Ân )ω − [H(An )ϕ⊗n − H(Rn |An )ϕ⊗n ](18.27)= H(Rn |An )ϕ⊗n − H(Rn |Ân )ω

(18.28)≤ f (n, ε) ≡ 2εn log dim(HR ) + g2 (ε).

(18.29)The inequality directly above follows from the AFW inequality (Theorem 11.10.3) applied tothe success criterion in (18.20). The function f (n, ε) has the property that limε→0 limn→∞ n1 f (n, ε) =0. The equality in (18.25) follows because the quantum mutual information is additive fortensor-product states. The equality in (18.26) follows because the quantum mutual information of a pure, bipartite state is equal to twice the marginal entropy. Putting everything
together, we find that11
log dim(HW ) ≥ H(A)ϕρ − f (n, ε).
(18.30)n2nTaking the limit as n → ∞ and ε, δ → 0 allows us to conclude that an achievable rate R ofquantum data compression necessarily satisfies R ≥ H(A)ϕρ .
R+δ =c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 18. SCHUMACHER COMPRESSION

Exercise 18.2.1 We proved the converse theorem for Schumacher compression with respect
to the error criterion in (18.5). However, it might be the case that if we use the less stringent
error criterion in (18.6) that we could achieve a smaller rate of quantum data compression.
Show that this is not the case, by establishing that the converse theorem holds with this lessstringent error criterion. (Hint: The development is essentially the same as in (18.22)–(18.26), except you should have the reference system be classical, containing a classical label
for which state in the i.i.d. ensemble was chosen.)

18.3 Quantum Compression Example
We now highlight a particular example where Schumacher compression gives a big savingsin compression rates if noiseless qubit channels are available. Suppose that the ensemble isof the following form:

 

1
1
, |0i ,
, |+i.
(18.31)
2
2This ensemble is known as the Bennett-92 ensemble because it is useful in Bennett’s protocol
for quantum key distribution. The naive strategy would be for Alice and Bob to exploit
Shannon’s compression protocol. That is, Alice would ignore the quantum nature of thestates, and supposing that the classical label for them were available, she would encode theclassical label. However, the entropy of the uniform distribution on two states is equal to
one bit, and she would have to transmit classical messages at a rate of one bit per channeluse.
A different strategy is to employ Schumacher compression. The density operator of theabove ensemble is1
1|0ih0| + |+ih+|,(18.32)2
2
which has the following spectral decomposition:cos2 (π/8) |+0 i h+0 | + sin2 (π/8) |−0 i h−0 | ,(18.33)|+0 i ≡ cos(π/8)|0i + sin(π/8)|1i,|−0 i ≡ sin(π/8)|0i − cos(π/8)|1i.

(18.34)(18.35)whereThe binary entropy h2 (cos2 (π/8)) of the distribution cos2 (π/8), sin2 (π/8) is approximately
equal to0.6009 qubits,(18.36)and thus they can save a significant amount in terms of compression rate by employing
Schumacher compression. This type of savings will always occur whenever the ensemble
includes non-orthogonal quantum states.
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18.4. VARIATIONS ON THE SCHUMACHER THEME

527

Exercise 18.3.1 In the above example, suppose that Alice associates a classical label withthe states, so that the ensemble instead is
 


1
1
, |0ih0| ⊗ |0ih0| ,, |1ih1| ⊗ |+ih+|.
(18.37)
2
2
Does this help in reducing the amount of qubits she has to transmit to Bob?

18.4 Variations on the Schumacher Theme
We can propose several variations on the Schumacher compression theme. For example,
suppose that the quantum information source corresponds to the following ensemble instead:{pX (x), ρxA } ,(18.38)where each ρx is a mixed state. Then the situation is not as “clear-cut” as in the simpler
model for a quantum information source, and the entropy of the source does not necessarily
serve as a lower bound on the ultimate compressibility rate. This depends on the figure of
merit that we choose for mixed-state compression, and there are at least three interesting
figures of merit that we could consider. To see these different ones, observe that the following
state serves as a purification of the mixed-state source in (18.38):
Xpx
pX (x)|xiX |xiX 0 φρ RA ,(18.39)|φiXX 0 RA ≡xρx

where |φ iRA is a purification of ρxA , so that the purifying system of A is the joint system
XX 0 R. The three figures of merit to consider for any encoding-decoding pair (EAn →W , DW →Ân )are as follows:
1 ⊗n
φXX 0 RA − (DW →Ân ◦ EAn →W )(φ⊗n(18.40)
XX 0 RA ) 1 ,
2
xn
xn
1X
pX n (xn ) φρRn An − (DW →Ân ◦ EAn →W )(φρRn An ) ,(18.41)
2 xn
1
1X
n
n
pX n (xn ) ρxAn − (DW →Ân ◦ EAn →W )(ρxAn ) 1 .
(18.42)
2 xn
Consider that satisfying the first error criterion up to some ε ∈ [0, 1] implies that second
criterion is satisfied as well, which in turn implies that the third is satisfied (this follows froma reasoning similar to that in the hint for Exercise 18.1.1).
How should we handle the mixed source case in general? Let us consider the direct coding
theorem and the converse theorem. The direct coding theorem for this case is essentiallyequivalent to Schumacher’s protocol for quantum compression—there does not appear to bea better approach in the general case. The density operator of the source is equal toXρA =pX (x)ρxA .
(18.43)x
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CHAPTER 18. SCHUMACHER COMPRESSION

A compression rate R ≥ H(A)ρ is achievable if we form the typical subspace measurementfrom the typical subspace projector ΠδAn onto the state (ρA )⊗n , and the error analysis frombefore shows that this holds for the error criterion in (18.40), which implies that it holds forthe other two error criteria mentioned above.
Although the direct coding theorem stays the same, the converse theorem changes somewhat. If we demand that the converse hold for the error criterion in (18.40), then the methodof proof in (18.22)–(18.26) demonstrates that the entropy H(A)ρ serves as a converse bound.
Thus, in this case, we have a statement of optimality. However, if we demand that the converse hold for the error criterion in (18.41), then the same method of proof gives a converse
bound of 21 I(XR; A)φ . If we demand that the converse hold for the error criterion in (18.42),
then the method of proof from Exercise 18.2.1 gives a converse bound of I(X; A)φ . In general, these latter two lower bounds are incomparable, but we can deduce that a sequence ofcompression schemes each meeting the error criterion in (18.41) should have a compression
rate larger than or equal to max{ 12 I(XR; A)φ , I(X; A)φ }, given that the error criterion in(18.41) is more stringent than that in (18.42).
Let us consider a special example of the above situation, which allows for comparing thetwo different error criteria in (18.40) and (18.42) and the P
optimal rates. Suppose that themixed states ρx act on orthogonal subspaces, and let ρA = x pX (x)ρx denote the expected
density operator of the ensemble. The states in {ρx } are then perfectly distinguishable
by a measurement whose projectors project onto the different orthogonal subspaces. As aconsequence, Alice could perform this measurement and associate classical labels with eachof the states, leading to the following classical–quantum state:XσXA ≡pX (x)|xihx|X ⊗ ρxA .
(18.44)x

Furthermore, she can do this in principle without disturbing the state in any way, andtherefore the entropy of the state σXA is equivalent to the original entropy of the state ρA :H(A)ρ = H(XA)σ .

(18.45)

Applying Schumacher compression to such a source meets the error criterion in (18.40)
at an optimal rate equal to H(A)ρ . This compression rate is equal to
H(XA)σ = H(X)σ + H(A|X)σ ,(18.46)and in this case, H(A|X)σ ≥ 0 because the conditioning system is classical. Furthermore,
if at least one ρx is truly mixed then we have a strict inequality H(A|X)σ > 0. However, if
we are only interested in a scheme which meets the error criterion in (18.42), then a muchbetter strategy than Schumacher compression is for Alice to measure the classical variable
X directly, compress it with Shannon compression, and transmit to Bob so that he canreconstruct the quantum states at his end of the channel. The rate of compression hereis equal to the Shannon entropy H(X)σ , which is provably lower than H(XA)ρ for thisexample. Since I(X; A)σ = H(X)σ (see exercise below), this example has an optimal rate
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18.5. CONCLUDING REMARKS

529

for the error criterion in (18.42) and we see that there can be a strict difference in optimal
rates if we consider different error criteria in mixed-state compression.
The next exercise asks you to verify that ensembles of mixed states on orthogonal subspaces saturate the lower bound of I(X; A)φ .
Exercise 18.4.1 Show that the Holevo information of an ensemble of mixed states on orthogonal subspaces has its Shannon information equal to its Holevo information. Thus, thisis an example of a class of ensembles that meet the lower bound I(X; A)φ on compressibility.

18.5 Concluding Remarks
Schumacher compression was the first quantum Shannon-theoretic result discovered andis the simplest one that we encounter in this book. The proof is remarkably similar tothe proof of Shannon’s noiseless coding theorem, with the main difference being that weshould be more careful in the quantum case not to be learning any more information thannecessary when performing measurements. The intuition that we gain for future quantum
protocols is that it often suffices to consider only what happens to a high probability subspacerather than the whole space itself if our primary goal is to have a small probability oferror in a communication task. In fact, this intuition is the same needed for understanding
information-processing tasks such as entanglement concentration, classical communication,
private classical communication, and quantum communication.
The problem of characterizing the lower and upper bounds for the quantum compression
rate of a mixed state quantum information source still remains open, despite considerable
efforts in this direction. It is only in special cases, such as the example mentioned in Section 18.4, that we know of a matching lower and upper bound as in Schumacher’s original
theorem.

18.6 History and Further Reading
Ohya and Petz (1993) devised the notion of a typical subspace, and Schumacher (1995)
independently introduced typical subspaces and additionally proved the quantum datacompression theorem. Jozsa and Schumacher (1994) later generalized this proof, and Lo(1995) further generalized the theorem to mixed state sources. There are other generalizations in Horodecki (1998); Barnum et al. (2001a). Several schemes for universal quantum
data compression exist (Jozsa et al., 1998; Jozsa and Presnell, 2003; Bennett et al., 2006), in
which the sender does not need to have a description of the quantum information source inorder to compress its output. There are also practical schemes for quantum data compression
discussed in work about quantum Huffman codes (Braunstein et al., 2000).
Going beyond the settings considered here, researchers have considered error exponents,
strong converses, and second-order characterizations for quantum data compression (we explain what these terms mean in Section 20.7). Winter (1999b) established a strong conversec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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theorem, Hayashi (2002) derived error exponents, and Datta and Leditzky (2015) establishedsecond-order characterizations.

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 19

Entanglement Manipulation
Entanglement is one of the most useful resources in quantum information. If Alice andBob share noiseless entanglement in the form of maximally entangled states, then they canteleport quantum bits between each other with the help of classical communication, or they
can double the capacity of a noiseless qubit channel for transmitting classical information. We
will see further applications in Chapter 21 in which they can exploit noiseless entanglement
to assist in the transmission of classical or quantum data over a quantum channel.
Given the utility of maximal entanglement, a reasonable question to ask is what two
spatially separated parties can accomplish if they share pure entangled states that are notmaximally entangled. In the quantum Shannon-theoretic setting, we make the further assumption that the two parties share many copies of the same pure entangled state. We findout in this chapter that they can “concentrate” these non-maximally entangled states tomaximally entangled ebits by performing local operations on their systems, and the optimal rate at which they can do so is equal to the “entropy of entanglement” (the quantum
entropy of the marginal density operator of the original state). Entanglement concentration is thus another fundamental task in noiseless quantum Shannon theory, and it gives adifferent operational interpretation to the quantum entropy. Entanglement concentration isperhaps complementary to Schumacher compression in the sense that it gives a firm quantum
information-theoretic interpretation of the term “ebit” (just as Schumacher compression didfor the term “qubit”), and it plays a part in demonstrating how the entropy of entanglement
is the unique measure of entanglement for pure bipartite states.
More generally, Alice and Bob could try to convert a large number of copies of a pure state|ψiAB into as many copies as possible of another bipartite pure state |φiAB , by performing
only local operations and exchanging classical messages (called “LOCC,” an abbreviation
for “local operations and classical communication”). It is important to place a constraint
on their allowed operations—if they could perform arbitrary global operations on their systems, then the task becomes trivial. We call such a task “entanglement manipulation” as a
generalization of the aforementioned entanglement concentration task.
The main result in this chapter is that the optimal rate of conversion for such an entanglement manipulation task is equal to H(A)ψ /H(A)φ . That is, if the goal is to convert
531

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CHAPTER 19. ENTANGLEMENT MANIPULATION

⊗nE
|ψi⊗n
AB by LOCC to a state that has very high fidelity with |φiAB , then this transformationis possible for large n if and only if E ≤ H(A)ψ /H(A)φ . This conversion rate H(A)ψ /H(A)φis known as the “entanglement manipulation limit.” The achievability part of this theorem
follows by breaking the task into two parts. In the first part, Alice and Bob perform entanglement concentration to convert |ψi⊗n
AB to ≈ nH(A)ψ ebits. The next part makes use of a
protocol called “entanglement dilution,” which converts nH(A)φ ebits to ≈ n copies of |φiABby means of LOCC. Scaling the conversion rate appropriately, it follows that nH(A)ψ ebits
can be converted to ≈ n [H(A)ψ /H(A)φ ] copies of |φiAB , concluding the direct part. The
converse part of this theorem follows by exploiting the properties of an information quantity
called the relative entropy of entanglement (the main properties that we need are that it
does not increase under the action of an LOCC channel, it is never smaller than the coherent
information, and it is equal to the entropy of entanglement for pure states).
The entanglement manipulation theorem mentioned above is one of the most importantresults in the resource theory of quantum entanglement. It demonstrates that the conversionof pure-state entanglement from one form to another is essentially reversible in the limit ofmany copies. That is, when n is very large, Alice and Bob could first concentrate |ψi⊗n
AB to≈ nH(A)ψ ebits. Then they could execute an entanglement dilution protocol to transform⊗n(1−δ)these ≈ nH(A)ψ ebits back to the original state |ψiAB, where δ ∈ (0, 1) is a small
number that can be made to go to zero as n becomes large. Alternatively, they could takethe original state to some other “in-between” state besides the ebit at a rate given by theratio of the entropies of entanglement, but the main advantage of the ebit is that its entropyof entanglement is equal to one, so that we can think of it as a unit resource.
The technique for proving that the quantum entropy is an achievable rate for entanglement concentration exploits the method of types outlined in Sections 14.7 and 15.3 for classical and quantum typicality, respectively (the most important property is Property 14.7.5
which states that the exponentiated entropy is a lower bound on the size of a typical typeclass). In hindsight, it is perhaps surprising that a typical type class is exponentially large
in the large n limit (on the same order as the typical set itself), and we soon discover thequantum Shannon-theoretic consequences of this result. The protocol for entanglement dilution is in some sense just that for entanglement concentration “run backwards,” additionallymaking use of quantum teleportation.
We begin this chapter by discussing a simple example of entanglement concentration forthree copies of a state, and then we sketch out how more general entanglement concentrationand dilution protocols operate. Section 19.2 gives a formal definition of LOCC and introduces the relative entropy of entanglement (an LOCC monotone). Section 19.3 then details
the information-processing task for entanglement manipulation, and Section 19.4 states theentanglement manipulation theorem and proves both the direct coding and converse parts.

19.1 Sketch of Entanglement Manipulation
This section sketches the main ideas underlying entanglement concentration and dilution.
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19.1. SKETCH OF ENTANGLEMENT MANIPULATION

19.1.1

533

Three-Copy Entanglement Concentration Example

A simple example illustrates the main idea underlying the concentration of entanglement.
Consider the following partially entangled state:|Φθ iAB ≡ cos(θ)|00iAB + sin(θ)|11iAB ,(19.1)

where θ is some parameter such that 0 < θ < π/2. The Schmidt decomposition (Theorem 3.8.1) guarantees that the above state is the most general form to consider for a pure
bipartite entangled state on qubits. Now suppose that Alice and Bob share three copies of
the above state. We can rewrite the three copies of the above state using simple algebra:|Φθ iA1 B1 |Φθ iA2 B2 |Φθ iA3 B3 = cos3 (θ)|000iA |000iB + sin3 (θ)|111iA |111iB√
1+ 3 cos(θ) sin2 (θ) √ (|110iA |110iB + |101iA |101iB + |011iA |011iB )3√
1+ 3 cos2 (θ) sin(θ) √ (|100iA |100iB + |010iA |010iB + |001iA |001iB ) , (19.2)
3
where we have relabeled all of the systems on Alice and Bob’s respective sides as A ≡ A1 A2 A3and B ≡ B1 B2 B3 . Observe that the subspace with coefficient cos3 (θ) whose states havezero “ones” is one-dimensional. The subspace whose states have three “ones” is also onedimensional. But the subspace with coefficient cos(θ) sin2 (θ) whose states have two “ones”
is three-dimensional, and the same holds for the subspace whose states each have one “one.”
A protocol for entanglement concentration in this scenario is then straightforward. Alice
performs a projective measurement consisting of the operators Π0 , Π1 , Π2 , Π3 whereΠ0 ≡ |000ih000|A ,Π1 ≡ |001ih001|A + |010ih010|A + |100ih100|A ,Π2 ≡ |110ih110|A + |101ih101|A + |011ih011|A ,Π3 ≡ |111ih111|A .

(19.3)(19.4)(19.5)(19.6)The subscript i of the projection operator Πi corresponds to the Hamming weight of thebasis states in the corresponding subspace. Bob can perform the same “Hamming weight”
measurement on his side. With probability cos6 (θ) + sin6 (θ), the procedure fails because it
results in |000iA |000iB or |111iA |111iB which are both product states with no entanglement
at all. But with probability 3 cos2 (θ) sin4 (θ), the state is in the subspace with Hamming
weight two, and it has the following form:1√ (|110iA |110iB + |101iA |101iB + |011iA |011iB ) ,3(19.7)

and with probability 3 cos4 (θ) sin2 (θ), the state is in the subspace with Hamming weight one,and it has the following form:1√ (|100iA |100iB + |010iA |010iB + |001iA |001iB ) .
3
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CHAPTER 19. ENTANGLEMENT MANIPULATION

Alice and Bob can then perform local isometric operations on their respective systems torotate either of these states to a maximally entangled state with Schmidt rank three:1√ (|0iA |0iB + |1iA |1iB + |2iA |2iB ) .
3

19.1.2(19.9)

Sketch of Entanglement Concentration

The simple protocol outlined above is the basis for the entanglement concentration protocol,
but it unfortunately fails with a non-negligible probability in this case. On the other hand, if
we allow Alice and Bob to have a large number of copies of a pure bipartite entangled state,
the probability of failing becomes negligible in the asymptotic limit due to the properties oftypicality, and each type class subspace contains an exponentially large maximally entangledstate. The proof of the direct coding theorem in Section 19.4.2 makes this intuition precise.
Generalizing the procedure outlined above to an arbitrary number of copies is straightforward. Suppose Alice and Bob share n copies of the partially entangled state |Φθ i. We
can then write the state as follows:nXX|Φθ iAn B n =cosn−k (θ) sink (θ)|xiAn |xiB n(19.10)k=0

x:w(x)=k


s 
n
X
n
1
cosn−k (θ) sink (θ)  q 
=
n
k
k=0
k


X

|xiAn |xiB n  ,(19.11)x:w(x)=k

where w(x) is the Hamming weight of the binary vector x. Alice performs a “Hamming
weight” measurement whose projective operators are as follows:XΠk =|xihx|An ,(19.12)x:w(x)=k
and the Schmidt rank of the maximally entangled state that they then share is nk .
A rough analysis of the protocol’s performance for large n follows by exploiting√ Stirling’s
approximation and typicality. Recalling that Stirling’s approximation is n! ≈ 2πn (n/e)nand thinking of k/n as a constant (as is the case for typical sequences), this gives
√
 
n
n!
2πn (n/e)np=≈√(19.13)k!n − k!
k
2πk (k/e)k 2π (n − k) ((n − k) /e)n−krnnn=(19.14)2πk (n − k) (n − k)n−k k ks−(n−k)  −k

k
1
n−k


=
(19.15)nn
2πn nk 1 − nk

c 2019= [poly(n)]−1 · 2n[−((n−k)/n) log((n−k)/n)−(k/n) log(k/n)](19.16)= [poly(n)]−1 · 2nh2 (k/n) ,(19.17)

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19.1. SKETCH OF ENTANGLEMENT MANIPULATION

535

where h2 is the binary entropy function in (1.1) and poly(n) indicates a term at most polynomial in n. When n is large, the exponential term 2nh2 (k/n) dominates the polynomial−1/2, so that the polynomial term begins to behave merely as a constant.
2πn nk 1 − nk
So, the protocol is for Alice to perform a strongly typical subspace measurement with respect
to the distribution cos2 (θ), sin2 (θ) , and the state then reduces to the following one withhigh probability:


s 
n
X
X
n
1
1
√
cosn−k (θ) sink (θ)  q |xiAn |xiB n  ,(19.18)nkN
k=0 :k x:w(x)=k|k/n−sin2 (θ)|≤δwhere N ≥ 1−ε is an appropriate normalization constant. Alice and Bob then both performa Hamming weight measurement and the state reduces to a state of the form
X
1
p|xiAn |xiB n ,(19.19)[poly(n)]−1 · 2nh2 (k/n) x:w(x)=kdepending on the outcome k of the measurement. In the above, we have employed theapproximation in (19.13)–(19.17), being valid since k/n is approximately constant k/n ≈sin2 (θ), due to typicality. The above state is a maximally entangled state with Schmidt rank
[poly(n)]−1 · 2nh2 (k/n) , and it follows thath2 (k/n) ≥ h2 (sin2 (θ)) − δ = h2 (cos2 (θ)) − δ,(19.20)from the assumption that the state first projects into the typical subspace. Alice and Bob
can then perform local operations to rotate this state to approximately nh2 (cos2 (θ)) ebits.
Thus, this procedure concentrates the original non-maximally entangled state to ebits at a
rate equal to the entropy of entanglement of the state |Φθ iAB in (19.1). The above proof isa bit rough, and it applies only to entangled qubit systems in a pure state. The proof ofthe direct coding theorem in Section 19.4.2 generalizes this proof to pure entangled stateson d-dimensional systems.

19.1.3

Sketch of Entanglement Dilution

Entanglement dilution is easier to sketch out if we are not concerned about the classicalcommunication cost of the protocol. Suppose that the goal is to create n copies of |φiAB
from ≈ nH(A)φ ebits. Then Alice can prepare n copies of |φiAB in her laboratory. She
performs Schumacher compression on the B systems, which compresses these systems to≈ nH(A)φ qubits while causing only a small disturbance to the state (when n is large). If
she and Bob share ≈ nH(A)φ ebits, then she can teleport the compressed qubits to Bob(this costs ≈ n2H(A)φ classical bits). Bob receives the compressed qubits and decompresses
them, which is the end of the protocol. So this version of the entanglement dilution protocol
is rather straightforward. Later on, we see how the classical communication cost can bemuch smaller—in fact a modified protocol requires a number of classical bits required which
is sublinear in n, so that the rate of classical communication needed vanishes in the large nlimit.
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CHAPTER 19. ENTANGLEMENT MANIPULATION

19.2 LOCC and Relative Entropy of Entanglement
Before describing the information-processing task for entanglement manipulation, we should
formally define what we mean by LOCC, the allowed set of operations. An LOCC channel
consists of a finite number of compositions of the following:
1. Alice performs a quantum instrument, which has both a quantum and classical output. She forwards the classical output to Bob, who then performs a quantum channel
conditioned on the classical data received. This sequence of actions corresponds to achannel of the following form:X(19.21)
FAx ⊗ GBx ,x
Px
where {FA } is a collection of completely positive maps such that x FAx is a quantum
channel and {GBx } is a collection of quantum channels.

2. The situation is reversed, with Bob performing the initial instrument, who forwards
the classical data to Alice, who then performs a quantum channel conditioned on theclassical data. This sequence of actions corresponds to a channel of the form in (19.21),
with the A and B labels switched.
An information measure is an LOCC monotone if it is non-increasing with respect toan LOCC channel. One such information measure is the relative entropy of entanglement,
defined as follows:
Definition 19.2.1 (Relative Entropy of Entanglement) Let ρAB ∈ D(HA ⊗ HB ). The
relative entropy of entanglement of ρAB is equal to the “relative entropy distance” betweenρAB and the closest separable state:
ER (A; B)ρ ≡min

σAB ∈SEP(A:B)

D(ρAB kσAB ).

(19.22)That ER (A; B)ρ is an LOCC monotone follows from the monotonicity of relative entropy
with respect to channels (Theorem 11.8.1) and the fact that LOCC channels take separable
states to separable states. That is, from the definition of LOCC channels given above, we
can see that such operations have Kraus operators of thePform {FAz ⊗ GzB }, so that an LOCC
channel ΛAB acts as follows on a separable state σAB = y p(y)ωAy ⊗ τBy :"#XX
ΛAB (σAB ) =(FAz ⊗ GzB )(19.23)p(y)ωAy ⊗ τBy (FAz ⊗ GzB )†
z

=

X

y

p(y)FAz ωAy (FAz )† ⊗ GzB τBy (GzB )† ,(19.24)z,ywhich is clearly a separable state. So this means that for any separable state σAB and LOCC
channel ΛAB→A0 B 0 , we find that
D(ρAB kσAB ) ≥ D(ΛAB→A0 B 0 (ρAB )kΛAB→A0 B 0 (σAB )) ≥ ER (A0 ; B 0 )Λ(ρ) ,c 2019(19.25)

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19.2. LOCC AND RELATIVE ENTROPY OF ENTANGLEMENT

537

where the first inequality follows from the monotonicity of relative entropy (Theorem 11.8.1).
Since the inequality holds for all σAB ∈ SEP(A : B), we can conclude that
ER (A; B)ρ ≥ ER (A0 ; B 0 )Λ(ρ) ,(19.26)

which is equivalent to the following:
Theorem 19.2.1 The relative entropy of entanglement is an LOCC monotone.
We need two other properties of the relative entropy of entanglement:
Proposition 19.2.1 Let ρAB ∈ D(HA ⊗ HB ). Then the relative entropy of entanglement isnever smaller than the coherent information:
ER (A; B)ρ ≥ max{I(AiB)ρ , I(BiA)ρ }.

(19.27)

Proof. Let σAB ∈ SEP(A : B). Then
X
X
p(y)IA ⊗ τBy = IA ⊗ σB .
p(y)ωAy ⊗ τBy ≤σAB =(19.28)yyThe operator inequality follows because ωAy ≤ IA , which implies thatis positive semi-definite. We can then conclude thatPyyy p(y) (IA − ωA ) ⊗ τB

D(ρAB kσAB ) ≥ D(ρAB kIA ⊗ σB )≥ min D(ρAB kIA ⊗ σB )(19.29)(19.30)= I(AiB)ρ .

(19.31)σB

The first inequality follows from Proposition 11.8.2. The equality follows from Exercise 11.8.3.
Since the inequality holds for all σAB ∈ SEP(A : B), we can conclude that ER (A; B)ρ ≥I(AiB)ρ . The other inequality follows by a symmetric proof.
Proposition 19.2.2 Let |ψiAB ∈ HA ⊗ HB be a pure bipartite state. Then the relative
entropy of entanglement is equal to the entropy of entanglement:
H(A)ψ = ER (A; B)ψ .

(19.32)

Proof. From Proposition 19.2.1, we know that ER (A; B)ψ ≥ I(AiB)ψ = H(A)ψ . So it
remains to prove the other inequality. Suppose that |ψiAB has a Schmidt decomposition as
follows:
Xp|ψiAB =
p(x)|xiA ⊗ |xiB .
(19.33)x

Let ∆ denote the following channel:∆(X) ≡ P XP + (I − P )X(I − P ),c 2019(19.34)

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CHAPTER 19. ENTANGLEMENT MANIPULATION

where P ≡Px |xihx|A ⊗ |xihx|B . Let

ψ AB ≡

X

p(x)|xihx|A ⊗ |xihx|B .

(19.35)x

Note that ψ AB is a separable state. Consider that
H(A)ψ = I(AiB)ψ
= D(ψAB kIA ⊗ ψB )≥ D(∆(ψAB )k∆(IA ⊗ ψB ))(19.36)(19.37)(19.38)= D(ψAB kψ AB )≥min
D(ψAB kσAB )(19.39)(19.40)= ER (A; B)ψ .

(19.41)σAB ∈SEP(A:B)

The second equality follows from Exercise 11.8.3. The first inequality follows from themonotonicity of quantum relative entropy (Theorem 11.8.1). The third equality follows
because ∆(ψAB ) = ψAB and∆(IA ⊗ ψB ) = P (IA ⊗ ψB )P + (I − P )(IA ⊗ ψB )(I − P ).

(19.42)= ψ AB + (I − P )(IA ⊗ ψB )(I − P ).

(19.43)Since ψAB does not have support outside of the subspace onto which P projects, it follows
that(19.44)
D(ψAB kψ AB + (I − P )(IA ⊗ ψB )(I − P )) = D(ψAB kψ AB ).
This concludes the proof.

19.3 Entanglement Manipulation Task
We can now define an (n, m/n, ε) entanglement manipulation protocol. Alice and Bob begin
with n copies of a pure bipartite, entangled state |ψiAB . They then perform an LOCC
(n)channel ΛAn B n →Am B m in an attempt to convert the original state (|ψiAB )⊗n to m copies ofanother bipartite pure state |φiAB . Let ωAm B m denote the state after the LOCC channel:(n)⊗nωAm B m ≡ ΛAn B n →Am B m (ψAB).

(19.45)The protocol has ε error if the final state ωAm B m is ε-close to φ⊗m
AB :1
ωAm B m − φ⊗m(19.46)
AB 1 ≤ ε,2where ε ∈ [0, 1] and the rate of entanglement conversion is m/n.
We say that a particular rate E of entanglement manipulation is achievable if there exists
an (n, E − δ, ε) entanglement manipulation protocol for all ε ∈ (0, 1), δ > 0, and sufficiently
large n. The entanglement manipulation limit E(ψ → φ) for the conversion |ψiAB → |φiAB
is equal to the supremum of all achievable rates.
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19.4. THE ENTANGLEMENT MANIPULATION THEOREM

539

19.4 The Entanglement Manipulation Theorem
We first state the entanglement manipulation theorem and then prove it below in two parts(the converse theorem and the direct coding theorem).
Theorem 19.4.1 (Entanglement Manipulation) Let |ψiAB , |φiAB ∈ HA ⊗ HB be pure
bipartite states. The entanglement manipulation limit for the conversion |ψiAB → |φiAB isequal to the ratio H(A)ψ /H(A)φ :E(ψ → φ) =

H(A)ψ
.
H(A)φ

(19.47)

Remark 19.4.1 Theorem 19.4.1 implies that the entanglement concentration and dilution
protocols are individually optimal. That is, if the goal is to convert n copies of |ψiAB to as
many ebits as possible, then the maximal rate of ebit generation is equal to H(A)ψ . Furthermore, if the goal is to convert nR ebits to n copies of |ψiAB , then the minimal rate R of ebit
consumption is equal to H(A)ψ .

19.4.1The Converse Theorem

We now prove the converse theorem for entanglement manipulation, i.e., that the entanglement manipulation limit for the conversion |ψiAB → |φiAB does
not exceed H(A)ψ /H(A)φ .
 (n)Suppose that there is a sequence of LOCC transformations Λ, each of which takes ncopies of a pure state |ψiAB to mn approximate copies of a pure state |φiAB . That is, Λ(n)is such that1⊗nn
Λ(n) (ψAB≤ ε,(19.48)) − φ⊗mAB
1
2⊗n
where ε ∈ (0, 1). Let ωAmn B mn ≡ Λ(n) (ψAB). To bound the rate mn /n, we use the relative
entropy of entanglement. Consider thatnH(A)ψ = H(An )ψ⊗n
= ER (An ; B n )ψ⊗n
≥ ER (Amn ; B mn )ω
≥ I(Amn iB mn )ω
≥ I(Amn iB mn )φ⊗mn − f (mn , ε)
= H(Amn )φ⊗mn − f (mn , ε)= mn H(A)φ − f (mn , ε).

(19.49)(19.50)(19.51)(19.52)(19.53)(19.54)(19.55)

The first equality follows because the entropy is additive for a tensor-product state. The
second equality follows from Proposition 19.2.2. The first inequality follows because therelative entropy of entanglement is an LOCC monotone (Theorem 19.2.1). The second
inequality follows from Proposition 19.2.1. The third inequality is a consequence of continuity
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CHAPTER 19. ENTANGLEMENT MANIPULATION

of conditional entropy (the AFW inequality), with f (mn , ε) ≡ 2εmn log dim(HA ) + g2 (ε).
The third equality follows because the coherent information of a pure state is equal to themarginal entropy. The last equality follows because the entropy of a tensor-product state isadditive. Putting everything together, we find thatH(A)ψg2 (ε)
mn(1 − 2ε log dim(HA )/H(A)φ ) ≤+.
nH(A)φ nH(A)φ(19.56)

Thus, if we are considering a sequence of (n, mn /n, ε) entanglement manipulation protocols
with rate E − δn = mn /n, such that limn→∞ δn = 0, then the above bound becomes(E − δn ) (1 − 2ε log dim(HA )/H(A)φ ) ≤H(A)ψ
g2 (ε)+.
H(A)φ nH(A)φ(19.57)

Taking the limit as n → ∞ and ε → 0, we find that any achievable rate E of entanglement
manipulation for ψ → φ necessarily satisfies the following bound:E≤

19.4.2

H(A)ψ
.
H(A)φ

(19.58)

The Direct Coding Theorem

As discussed in the introduction of this chapter, we break the direct coding theorem into
two parts: entanglement concentration and entanglement dilution. We begin by discussing
entanglement concentration. To do so, it is helpful to discuss a related, exclusively classicaltask known as randomness concentration (also known as randomness extraction).
Randomness Concentration
In a randomness concentration protocol, the goal is to extract as many approximately uniformly random bits as possible from a given distribution. Since we are operating in a quantum
Shannon theoretic regime, let us suppose that a sequence xn is generated according to an
i.i.d. distributionnYnpX n (x ) ≡pX (xi ).
(19.59)i=1Recall the method of types from Section 14.7.1. Suppose that Alice performs a mapping
xn → (t(xn ), ft (xn )),(19.60)where t(xn ) is the type (or empirical distribution) of the sequence xn and the function ft (xn )is an index keeping track of the ordering of the symbols in the sequence xn for a given typenclass TtX . Note that this mapping is reversible (given an output, one can determine theinput uniquely).
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19.4. THE ENTANGLEMENT MANIPULATION THEOREM

541

For example, all three-bit sequences map in this way as follows:
000 → (0, 0) ,011 → (2, 0) ,110 → (2, 2) ,001 → (1, 0) ,100 → (1, 2) ,111 → (3, 0) .

010 → (1, 1) ,101 → (2, 1) ,(19.61)(19.62)(19.63)For binary sequences, the type is the Hamming weight. Thus, 000 has type 0 and is the only
sequence in this type class, thus receiving an index 0. Also, 011 has type 2 and we label itas the first sequence in this type class, indexed with 0. The sequence 101 has type 2 and itis the second sequence in this type class, indexed with 1, etc.
What is the use of performing this mapping? First consider that the joint probability ofobserving a particular type t and index f is equal topt(X n ),ft (X n ) (t, f ) = pX n (xn ),(19.64)nwhere xn ∈ TtX . This is because the mapping xn → (t(xn ), ft (xn )) is reversible. Considernfurthermore that for xn ∈ TtX , we have thatYYnnpX (x)N (x|xt ) = pX n (xnt ),(19.65)pX (x)N (x|x ) =pX n (xn ) =
x∈X

x∈X

n

where xnt is a representative sequence of the type class TtX (having type t as well). Thenthe probability for observing a particular type t is equal to
X
X
Y
n
n
pt(X n ) (t) =pX n (xn ) =pX n (xnt ) = TtX
pX (x)N (x|xt ) .
(19.66)xn ∈TtXn

xn ∈TtXnx∈X
nThat is, pt(X n ) (t) just depends on the size of the type class TtX and the empirical distribution of the type t. These considerations then imply that the conditional probability
distribution pft (X n )|t(X n ) is uniform because
pft (X n )|t(X n ) (f |t) =pt(X n ),ft (X n ) (t, f )1= Xn .
pt(X n ) (t)|Tt |(19.67)

That is, conditioned on observing a particular type t, all of the sequences in the type classn
TtX are uniformly distributed. Thus, the mapping xn → (t(xn ), ft (xn )) “reshapes” thedistribution of xn in the above way.
This “distribution reshaping” leads to a first idea for a randomness concentration protocol. Given a sequence xn , send it through the reversible mapping xn → (t(xn ), ft (xn )).
nGiven the type t, the value f = ft (xn ) is uniformly random on a set of size TtX , so thatn
we recover log TtX uniformly random bits.
There are two main problems with the above method. First, some type classes are verysmall and lead to little randomness or none at all. Second, the amount of randomness thatthe above procedure yields is not consistent: it varies from type to type. What we wouldc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 19. ENTANGLEMENT MANIPULATION

prefer is to have a method which takes a random sequence X n and maps it to exactly nR
bits, such that the distribution of these nR bits is nearly indistinguishable from a uniform
distribution.
How can we accomplish this? To start, we should first have a preprocessing step in which
we only proceed with the above method if xn is a strongly typical sequence and otherwise
declare failure. Equivalently, we only proceed if t(xn ) is a strongly typical type, such thatthe empirical distribution N (x|xn ) satisfies maxx |N (x|xn ) − pX (x)| ≤ δ for some δ > 0. For
sufficiently large n, we are guaranteed that this preprocessing step fails with probability nolarger than an arbitrarily small constant ε ∈ (0, 1), due to the law of large numbers (orthe “high probability” property of typicality). By Property 14.7.5, this preprocessing stepguarantees that every strongly typical type class has size bounded as follows:
TtX

n

1

≥ 2n[H(X)−η(|X |δ)−|X | n log(n+1)] .

(19.68)Note that we also have the following upper bound:TtXn≤ 2n[H(X)+cδ] ,(19.69)because the size of a typical type class cannot exceed the size of the strongly typical set,for some constant c > 0. Thus, the preprocessing
step solves the first problem
 because
it guarantees that we will have at least n H(X) − η(|X | δ) − |X | n1 log(n + 1) uniformly
random bits if it is successful.
From here, we can then solve the second problem by performing a hashing function, in1
order to hash down every typical type class to a set of size 2n[H(X)−η(|X |δ)−|X | n log(n+1)] 2−nδ ,so that we decrease the size of the set by a factor of 2nδ . Even though the size drops byan amount that is exponential in n, we only lose δ on the rate of randomness concentration, which is the main parameter of interest in the large n limit. That is, at the end of
the protocol, we will be left with n H(X) − η(|X | δ) − |X | n1 log(n + 1) − δ bits that arenearly indistinguishable from uniformly random bits. As n → ∞, the rate of randomness
concentration is equal to H(X) uniformly random bits per source symbol.
To have a complete proof, we need the following hashing lemma, which gives a way forhashing down a uniform random variable on a larger set to one on a much smaller set:Lemma 19.4.1 (Hashing) Let k and l be positive integers such that k ≥ l (in fact thinkof k as being much larger than l). Let Wk be uniformly distributed on the set {1, . . . , k},Wl be uniformly distributed on the set {1, . . . , l}, and Wr be uniformly distributed on the set{1, . . . , r = dk/le}. Then there exists a one-to-one function g : {1, . . . , k} → {1, · · · , l} ×
{1, . . . , r} such that1lpWl × pWr − pg(Wk ) 1 ≤ .
(19.70)2k
Proof. Divide the set {1, . . . , k} into l bins, each of which has size no more than r = dk/le.
Then take g(wk ) to be the one-to-one mapping which outputs the bin index and the location
inside of that bin. The distribution of g(Wk ) is uniformly random, equal to 1/k for k of thec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

19.4. THE ENTANGLEMENT MANIPULATION THEOREM

543

possible l · dk/le output values and zero for the others (the unfilled locations if k/l is not an
integer). The distribution of the joint random variable (Wl , Wr ) is uniformly random also,
equal to 1/ (l · dk/le) for each of the possible l · dk/le values. The distributions then overlap
on exactly k of the possible output values, implying that the trace distance between the two
distributions is as follows:111+ (l · dk/le − k).
(19.71)pg(Wk ) − pWl × pWr 1 = k −k l · dk/lel · dk/leConsider that
k

1
dk/le
1
k/l
−
=k
−
k l · dk/lek · dk/le k · dk/le1
1l=|dk/le − k/l| ≤
≤ .
dk/le
dk/le
k

(19.72)(19.73)Furthermore,(l · dk/le − k)11
= (dk/le − k/l)l · dk/le
dk/le
1
l
≤
≤ .
dk/le
k

(19.74)(19.75)This concludes the proof.
We can now specify the complete protocol for randomness concentration. It begins witha sequence xn being generated at random according to pX n (xn ). Alice computes whether it
e n beis a strongly typical sequence, declaring failure of the protocol in case it is not. Let X
a random variable with the following distribution:Pn
pX n (xn )/ xn ∈T X n pX n (xn ) if xn ∈ TδXnδ
pXe n (x ) ≡,(19.76)0elsenwhere TδX is the strongly typical set. This random variable represents the distribution of
the sequence conditioned on it being strongly typical. Set the rate1log(n + 1) − δ.
(19.77)n
Let τδ denote the set of all strongly typical types for the distribution pX , for sequences oflength n and tolerance δ. Note that |τδ | ≤ (n + 1)|X | (see Property 14.7.1), so that only
|X | log(n + 1) bits are required to record the type. If xn is strongly typical, then Aliceapplies the one-to-one mapping
R = H(X) − η(|X | δ) − |X |xn → (t(xn ), gt (ft (xn ))),

(19.78)wherent : TδX → τδ ,(19.79)1

gt : TtX → {0, 1}nR × {0, 1}n[(1+c)δ+η(|X |δ)+|X | n log(n+1)] ,nc 2019(19.80)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

544

CHAPTER 19. ENTANGLEMENT MANIPULATIONnwith gt the one-to-one function guaranteed by Lemma 19.4.1, with k = TtX and l = 2nR .
e n ) = t deLet Wout denote a uniform random variable over the set {0, 1}nR , and let Wrem |t(X1note a conditional random variable that is uniform over a subset of {0, 1}n[(1+c)δ+η(|X |δ)+|X | n log(n+1)]
of size dk/le. Consider from our discussion around (19.67) that the conditional random varie n )|t(Xe n ) = t, for t ∈ τδ , is a uniform random variable. Thus, we can apply
able ft (XnLemma 19.4.1, taking k = TtX and l = 2nR , and find that12nR
pgt (ft (Xe n ))|t(Xe n )=t − pWout × pWrem |t(Xe n )=t ≤ X n
2
|Tt |
1
1
2n[H(X)−η(|X |δ)−|X | n log(n+1)−δ]≤= 2−nδ . (19.81)
1n[H(X)−η(|X |δ)−|X | nlog(n+1)]
2
This bound is then sufficient for us to conclude that1p en
≤ 2−nδ ,e n − pt(Xe n ),Wrem × pWout
2 t(X ),gt (ft (X ))1(19.82)

becausept(Xe n ),gt (ft (Xe n )) − pt(Xe n ),Wrem × pWout
1X=
pt(Xe n ) (t) pgt (ft (Xe n ))|t(Xe n )=t − pWout × pWrem |t(Xe n )=tt∈τδ. (19.83)
1P
For all ε ∈ (0, 1) and sufficiently large n, we know that P ≡ xn ∈T X n pX n (xn ) ≥ 1 − ε.
δThen we can conclude that the distributions pXe n and pX n are nearly indistinguishable because
pX n − pXe n 1 =XX|1 − P |n
xn ∈TδX

pX n (xn )(19.84)

n
xn ∈T
/ δX

n
xn ∈TδX=

X|pX n (xn ) − pX n (xn )/P | +X
pX n (xn )+pX n (xn )P
Xnn(19.85)

x ∈T
/ δ

≤ 2ε.

(19.86)

We can then finally conclude that1pt(X n ),gt (ft (X n )) − pt(Xe n ),Wrem × pWout ≤ ε + 2−nδ ,2
1(19.87)

by applying the triangle inequality to (19.82) and (19.86), and using the fact thatpr(Z) − pr(Y ) 1 = kpZ − pY k1(19.88)for random variables Z and Y and a one-to-one mapping r.
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19.4. THE ENTANGLEMENT MANIPULATION THEOREM

545

The final step of the randomness concentration protocol consists of discarding the typeregister containing t and the Wrem register. Monotonicity of the trace distance with respect
to discardings implies that
1
TrWrem {pgt (ft (X n )) } − pWout 1 ≤ ε + 2−nδ .
2(19.89)Even though the final step of randomness concentration consists of this discarding, we havedeveloped the protocol using one-to-one functions because it is essential for our development of the entanglement dilution protocol, which follows just after we discuss entanglement
concentration next.
Entanglement Concentration
We have actually done the bulk of the “hard work” in the previous section, when developing
the method for randomness concentration. We now just need to apply the one-to-one mapping in (19.78) in a coherent way in order to have a method for entanglement concentration.
In the setting of entanglement concentration, Alice and Bob begin with n copies of the state|ψiAB . Suppose that |ψiAB has a Schmidt decomposition of the following form:
XppX (x)|xiA |xiB .
(19.90)|ψiAB =
x∈X

Then the state |ψi⊗n
AB has the following form:
X p
|ψi⊗n
=
pX n (xn )|xn iAn |xn iB n .
AB(19.91)xn ∈X nSo we need to figure out local quantum channels that Alice and Bob can each performin order to convert this state to as many ebits as possible. Before doing so, we state thefollowing lemma, which will allow us to quantify the performance of a coherent version of aclassical protocol:
Lemma 19.4.2 Let pX and qX be probability distributions such thatkpX − qX k1 ≤ ε.
(19.92)P p
P p
Then the quantum states |ψ p iAB ≡ x pX (x)|xiA |xiB and |ψ q iAB ≡ x qX (x)|xiA |xiB
are such that√qp
k1 ≤ 2 ε.
kψAB
− ψAB
(19.93)
Proof. Consider that the quantum fidelity between |ψ p iAB and |ψ q iAB is equal to theclassical fidelity of the distributions pX and qX :Xp√pq
F (ψAB, ψAB) = |hψ q |ψ p iAB | =qX (x)pX (x).
(19.94)x
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546

CHAPTER 19. ENTANGLEMENT MANIPULATIONpq
From Corollary 9.3.1, we can conclude that F (ψAB, ψAB
) ≥ 1 − ε. Corollary 9.3.2 in turnimplies (19.93).
Let UAn →T Wout Wrem denote the following isometric implementation of the one-to-one mapping in (19.78):
X
UAn →T Wout Wrem ≡|t(xn ), gt (ft (xn ))iT Wout Wrem hxn |An .
(19.95)xn ∈TδXn

Then we set Alice’s encoding to be the following quantum channel:
EAn →T Wout Wrem (YAn ) ≡ U ΠδAn YAn ΠδAn U † + Tr{(IAn − ΠδAn )YAn }σT Wout Wrem ,(19.96)P
where ΠδAn is the strongly typical projector for the density operator x∈X pX (x)|xihx|A andσT Wout Wrem is some state of the registers T Wout Wrem . This channel isometrically embeds
the typical subspace for Alice’s system into the registers T Wout Wrem . Let EB n →(T Wout Wrem )B
denote the same encoding for Bob, and let EAn and EB n be a shorthand for Alice and Bob’s
encodings, respectively.
Let |ψen iAn B n denote the following state:
X q(19.97)
pXe n (xn )|xn iAn |xn iB n ,|ψen iAn B n ≡xn ∈X nwhere the distribution pXe n is defined in (19.76). By invoking (19.86) and Lemma 19.4.2, wefind that√⊗nψAB− ψeAn n B n ≤ 2 2ε.
(19.98)
1

The monotonicity of trace distance then implies that⊗n(EAn ⊗ EB n )(ψAB) − (EAn ⊗ EB n )(ψeAn n B n )√≤ 2 2ε.

(19.99)1

Consider that the coherent version of the distribution pt(Xe n ),Wrem × pWout is the following
state:⊗nR
ΥTA WA0 TB WB0 ⊗ Φ+,
(19.100)
AB
where ΥTA WA0 TB WB0 is a coherent version of the distribution pt(Xe n ),Wrem , defined asX qpt(Xe n ),Wrem (t, w)|t, wiTA WA0 |t, wiTB WB0 ,(19.101)|ΥiTA WA0 TB WB0 ≡t∈τδ ,w√
and |Φ+ iAB ≡ [|00iAB + |11iAB ] / 2. By invoking (19.82) and Lemma 19.4.2, we find that√⊗nR
(EAn ⊗ EB n )(ψeAn n B n ) − ΥTA WA0 TB WB0 ⊗ Φ+
≤ 2 2 · 2−nδ .
(19.102)
AB
1

Applying the triangle inequality to (19.99) and (19.102), we find that
h√
i
√
⊗nR
⊗n
−nδ .
(EAn ⊗ EB n )(ψAB) − ΥTA WA0 TB WB0 ⊗ Φ+
2ε
+
2
·
2
≤
2
AB

(19.103)1

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19.4. THE ENTANGLEMENT MANIPULATION THEOREM

547

The final step is for Alice and Bob to discard the registers TA TB WA0 WB 0 , which implies
that
i
h√
√
⊗nR
⊗n
−nδ .
2ε
+
2
·
2
((TrTA WA0 ◦EAn ) ⊗ (TrTB WB0 ◦EB n ))(ψAB) − Φ+≤
2
AB1(19.104)
The rate of ebit generation is equal to R = H(A)ψ −η(|X | δ)−|X | n1 log(n+1)−δ. No classicalcommunication is required. This concludes the proof for entanglement concentration.
Entanglement Dilution
We have already outlined an entanglement dilution protocol in Section 19.1.3. However, theprotocol sketched there uses far more classical communication than is necessary. Here, weshow that entanglement dilution requires a rate of classical communication that vanishes asn becomes large. This result demonstrates that the resource theory of entanglement for purestates is truly a reversible theory, in the sense that the only resource we need to count is therate of entanglement conversion, given that the classical communication rate is negligible.
We now discuss such an entanglement dilution protocol. The main idea is really just to
take the entanglement concentration protocol from the previous section and “run it backwards.” So we keep the system labels as they were in the previous section. Let Alice andBob share the following maximally entangled state at the beginning:
ΦTA WA0 TB WB0 ⊗ Φ+
AB

⊗nR

,

(19.105)

Consider thatlog dim(HTA ) = |X | log n,log dim(HWA0 ) = n(1 + c)δ + nη(|X | δ) + |X | log n,(19.106)(19.107)

implying that the total number of ebits in the state ΦTA WA0 TB WB0 is equal to n(1 + c)δ +nη(|X | δ) + 2 |X | log n, and the total number of ebits that they share overall is equal tonH(A)ψ + ncδ + |X | log n. Alice prepares the state ΥTA WA0 TB WB0 locally in her lab and uses
the state ΦTA WA0 TB WB0 to teleport the TB WB 0 systems of Υ to Bob. This requires
2 [n(1 + c)δ + nη(|X | δ) + 2 |X | log n](19.108)

bits of classical communication. At this point, Alice and Bob share the following state:
ΥTA WA0 TB WB0 ⊗ Φ+
AB

⊗nR

.

(19.109)

They each then perform the following quantum channel, which is essentially the “inverse”
of the encoding in (19.96):
E (−1) (Z) = U † ZU + Tr{(I − U U † )Z}ω,c 2019(19.110)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 19. ENTANGLEMENT MANIPULATION

where U is the isometry defined in (19.95) and ω is any state having support in span{|xn i :nnxn ∈ TδX }. In fact, since ψeAn n B n exclusively has support in span{|xn i : xn ∈ TδX }, it follows
that
(−1)
(−1)
(19.111)(EAn ⊗ EB n )(EAn ⊗ EB n )(ψeAn n B n ) = ψeAn n B n ,which along with the monotonicity of trace distance applied to (19.102), implies that√⊗nR 
(−1)(−1)
ψeAn n B n − (EAn ⊗ EB n ) ΥTA WA0 TB WB0 ⊗ Φ+
2 · 2−nδ .
(19.112)≤2
AB
1

Applying the triangle inequality to (19.98) and (19.112), we find thath√i√⊗nR 
(−1)
(−1)
⊗n
−nδ ,
ψAB
− (EAn ⊗ EB n ) ΥTA WA0 TB WB0 ⊗ Φ+
≤
2
ε
+
2
·
2
AB

(19.113)1which concludes the error analysis.
|X |⊗n
The rate of ebits needed to form ψAB
is equal to H(A) ψ + cδ + n log n ebits per 1copy ofψAB , and the rate of classical communication needed is 2 (1 + c)δ +√ η(|X | δ) + 2 |X | n log ncbits per copy of ψAB . For large n, we can take δ to be order n and the central limit
theorem implies that we can achieve any constant error ε ∈ (0, 1) (this is a modified versionof typicality in which δ changes with n). At the same time, the ebit rate converges to H(A)ψ
and the classical communication rate vanishes.
Entanglement Manipulation
We can now put together entanglement concentration and dilution to give a general achievable strategy for an entanglement manipulation protocol. The goal here is to convert ncopies of ψAB to as many copies of φAB as possible. In order to do so, Alice and Bob first
conduct an entanglement concentration protocol, which takes n copies of ψAB to ≈ nH(A)ψ
approximate ebits. Then, using entanglement dilution and a negligible rate of classical communication, they can convert these ≈ nH(A)ψ approximate ebits to ≈ n [H(A)ψ /H(A)ψ ]approximate copies of φAB . The accuracy of the protocol becomes arbitrarily small and therate of entanglement conversion approaches H(A)ψ /H(A)ψ in the limit as n becomes large.

19.5 Concluding Remarks
Entanglement concentration was one of the earliest discovered protocols in quantum Shannon
theory. The protocol exploits one of the fundamental tools of classical information theory(the method of types), but it applies the method in a coherent fashion so that a type classmeasurement learns only the type and nothing more. The protocol is similar to Schumacher
compression in this regard (in that it learns only the necessary information required to
execute the protocol and preserves coherent superpositions), and we will continue to see thisidea of applying classical techniques in a coherent way in future quantum Shannon-theoretic
protocols. For example, the protocol for quantum communication over a quantum channel
is a coherent version of a protocol to transmit private classical information over a quantum
channel.
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19.6. HISTORY AND FURTHER READING

549

19.6 History and Further Reading
Elias (1972) constructed a protocol for randomness concentration in an early paper. Bennett et al. (1996a) offered two different protocols for entanglement concentration (one ofwhich we developed in this chapter). Nielsen (1999) later connected entanglement concentration protocols to the theory of majorization. Lo and Popescu (2001, 1999) studied
entanglement concentration and the classical communication cost of the inverse protocol(entanglement dilution). Hayden and Winter (2003) characterized the classical communication cost of entanglement dilution, as did Harrow and Lo (2004). Kaye and Mosca (2001)
developed practical networks for entanglement concentration, and recently, Blume-Kohout
et al. (2014) took this line of research a step further by considering streaming protocols forentanglement concentration. Hayashi and Matsumoto (2001) also developed protocols foruniversal entanglement concentration.
Vedral and Plenio (1998) introduced the relative entropy of entanglement as one of thefirst LOCC monotones in quantum information theory. Proposition 19.2.1 is due to Plenio
et al. (2000).
Going beyond the settings considered here, researchers have considered error exponents,
strong converses, and second-order characterizations for entanglement manipulation tasks(we explain what these terms mean in Section 20.7). Lo and Popescu (2001) establisheda strong converse theorem for entanglement concentration. Hayashi et al. (2003) derived
error exponents and an exact strong converse for entanglement concentration. Kumagai and
Hayashi (2013) established exact second-order characterizations of entanglement concentration and dilution.

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c 2019CHAPTER 19. ENTANGLEMENT MANIPULATIONMark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicensePart VI
Noisy Quantum Shannon Theory

551

553

Before quantum information theory became an established discipline, John R. Pierce issued
the following quip at the end of his 1973 retrospective article on the history of information
theory (Pierce, 1973):“I think that I have never met a physicist who understood information theory. I wish
that physicists would stop talking about reformulating information theory and would give us
a general expression for the capacity of a channel with quantum effects taken into accountrather than a number of special cases.”
Since the publication of Pierce’s article, we have learned much more about quantum
mechanics and information theory than he might have imagined at the time, but we havealso realized that there is much more to discover. In spite of all that we have learned, we
still unfortunately have not been able to address Pierce’s concern in the above quote in fullgenerality.
The most basic question that we could ask in quantum Shannon theory (and the one
with which Pierce was concerned) is how much classical information can a sender transmit
to a receiver by exploiting a quantum channel. We have determined many special cases ofquantum channels for which we do know their classical capacities, but we also now knowthat this most basic question is still wide open in the general case.
What Pierce might not have imagined at the time is that a quantum channel has amuch larger variety of capacities than does a classical channel. For example, we might wish
to determine the classical capacity of a quantum channel assisted by entanglement shared
between the sender and receiver. We have seen that in the simplest of cases, such as thenoiseless qubit channel, shared entanglement boosts the classical capacity up to two bits,
and we now refer to this phenomenon as the super-dense coding effect (see Chapter 6).
Interestingly, the entanglement-assisted capacity of a quantum channel is one of the few
scenarios where we can claim to have a complete understanding of the channel’s transmission
capabilities. From the results regarding the entanglement-assisted capacity, we have learned
that shared entanglement is often a “friend” because it tends to simplify results in bothquantum Shannon theory and other subfields of quantum information science.
Additionally, we might consider the capacity of a quantum channel for transmitting quantum information. In 1973, it was not even clear what was meant by “quantum information,”
but we have since been able to formulate what it means, and we have been able to characterize the quantum capacity of a quantum channel. The task of transmitting quantum
information over a quantum channel bears some similarities with the task of transmitting
private classical information over that channel, where we are concerned with keeping the classical information private from the environment of the channel. This connection has given
insight for achieving good rates of quantum communication over a noisy quantum channel,
and there is even a certain class of channels for which we already have a good expression
for the quantum capacity (the expression being the coherent information of the channel).
However, the problem of determining a good expression for the quantum capacity in thegeneral case is still wide open.
The remaining chapters of the book are an attempt to summarize many items the quantum information community has learned in the past few decades, all of which are an attempt
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554

to address Pierce’s concern in various ways. The most important open problem in quantum
Shannon theory is to find better expressions for these capacities so that we can actually
compute them for an arbitrary quantum channel.

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 20

Classical Communication
This chapter begins our exploration of “dynamic” information-processing tasks in quantum
Shannon theory, where the term “dynamic” indicates that a quantum channel connects
a sender to a receiver and their goal is to exploit this resource for communication. We
specifically consider the scenario in which a sender Alice would like to communicate classicalinformation to a receiver Bob, and the capacity theorem that we prove here is one particular
generalization of Shannon’s noisy channel coding theorem from classical information theory(reviewed in Section 2.2). In later chapters, we will see other generalizations of Shannon’s
theorem, depending on what resources are available to assist their communication or whether
they are trying to communicate classical or quantum information. For this reason and others,
quantum Shannon theory is quite a bit richer than classical information theory.
The naive approach to communicate classical information over a quantum channel isfor Alice and Bob simply to mimic the approach used in Shannon’s noisy channel coding
theorem. That is, they randomly select a classical code according to some distribution pX (x),and Bob performs individual measurements of the outputs of a quantum channel according
to some POVM. The POVM at the output induces some conditional probability distribution
pY |X (y|x), which we can in turn think of as an induced classical channel. The classicalmutual information I(X; Y ) of this channel is an achievable rate for communication, andthe best strategy for Alice and Bob is to optimize the mutual information over all of Alice’s
inputs to the channel and over all measurements that Bob could perform at the output. The
resulting quantity is equal to Bob’s optimized accessible information, which we previously
discussed in Section 10.9.2.
If the aforementioned coding strategy were optimal, then there would not be anythingmuch interesting to say for the information-processing task of classical communication (infact, there would not be any need for all of the tools we developed in Chapters 15 and 16!).
This is perhaps one first clue that the above strategy is not necessarily optimal. Furthermore,
we know from Chapter 11 that the Holevo information is an upper bound on the accessible
information, and this bound might prompt us to wonder if it is also an achievable rate forclassical communication, given that the accessible information is achievable.
The main theorem of this chapter is the classical capacity theorem (also known as the
555

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Holevo–Schumacher–Westmoreland theorem), and it states that the Holevo information of aquantum channel is an achievable rate for classical communication. The Holevo information
is easier to manipulate mathematically than is the accessible information. The proof of its
achievability demonstrates that the aforementioned strategy is not generally optimal, andthe proof also shows how performing a collective measurement on all of the channel outputs
allows the sender and receiver to achieve the Holevo information as a rate for classical communication. Thus, this strategy fundamentally makes use of quantum-mechanical effects atthe decoder and suggests that such an approach is necessary to achieve the Holevo information. Although this strategy exploits a collective measurement at the decoder, it does notmake use of entangled states at the encoder. That is, the sender could input quantum statesthat are entangled across all of the channel inputs, and this encoder entanglement might
potentially increase classical communication rates.
One major drawback of the classical capacity theorem (also the case for many otherresults in quantum Shannon theory) is that it only demonstrates that the Holevo information
is an achievable rate for classical communication—the converse theorem is a “multi-letter”
converse, meaning that it might be necessary in the general case to evaluate the Holevo
information over a potentially infinite number of uses of the channel. The multi-letter natureof the capacity theorem implies that the optimization task for general channels is intractable
and thus further implies that we know very little about the actual classical capacity of generalquantum channels. Now, there are many natural quantum channels such as the depolarizing
channel and the dephasing channel for which the classical capacity is known (the Holevo
information becomes “single-letter” for these channels), and these results imply that wehave a complete understanding of the classical information-transmission capabilities of thesechannels. All of these results have to do with the additivity of the Holevo information of aquantum channel, which we studied previously in Chapter 13.
We mentioned that the Holevo–Schumacher–Westmoreland coding strategy does notmake use of entangled inputs at the encoder. But a natural question is to wonder whether
entanglement at the encoder could boost classical information-transmission rates, given that
it is a resource for many quantum protocols. This question was known as the additivity conjecture and went unsolved for many years, but Hastings (2009) offered a proof that entangled
inputs can increase communication rates for certain channels. Thus, for these channels, thesingle-letter Holevo information is not the proper characterization of classical capacity (however, this is not to say that there could be some alternate characterization of the classicalcapacity other than the Holevo information which would be single-letter). These results
demonstrate that we still know little about classical communication in the general case andfurthermore that quantum Shannon theory is an active area of research.
We structure this chapter as follows. We first discuss the aforementioned naive strategy
in detail, so that we can understand the difference between it and the Holevo–Schumacher–Westmoreland strategy. Section 20.2 describes the steps needed in any protocol for classicalcommunication over a quantum channel. Section 20.3 provides a statement of the classicalcapacity theorem, and its two subsections prove the corresponding direct coding theorem
and the converse theorem. The direct coding theorem exploits two tools: quantum typicalityc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

20.1. NAIVE APPROACH: PRODUCT MEASUREMENTS

557

Figure 20.1: The most naive strategy for Alice and Bob to communicate classical information over many
independent uses of a quantum channel. Alice wishes to send some message M and selects some tensor
product state to input to the channel, conditioned on the message M . She transmits the codeword overthe channel, and Bob then receives a noisy version of it. He performs individual measurements of hisquantum systems and produces some estimate M 0 of the original message M . This scheme is effectively aclassical scheme because it makes no use of quantum-mechanical features such as entanglement or a collective
measurement.

from Chapter 15 and the packing lemma from Chapter 16. The converse theorem exploits
two tools from Chapter 11: continuity of entropies (the AFW inequality in Theorem 11.10.3)
and the quantum data-processing inequality (Theorem 11.9.4). We then detail how to calculate the classical capacity of several exemplary channels such as entanglement-breaking
channels, quantum Hadamard channels, erasure channels, and depolarizing channels—these
are channels for which we have a complete understanding of their classical capacity. Finally,
we end with a discussion of the proof that the Holevo information can be superadditive(that is, entangled inputs at the encoder can enhance classical communication rates forsome channels).

20.1 Naive Approach: Product Measurements
We begin by discussing in more detail the most naive strategy that a sender and receiver can
exploit for the transmission of classical information over many uses of a quantum channel.
Figure 20.1 depicts this naive approach. This first approach mimics certain features of
Shannon’s classical approach without making any use of quantum-mechanical effects. Aliceand Bob agree on a codebook beforehand, such that each classical codeword xn (m) in thecodebook corresponds to some message m that Alice wishes to transmit. Alice can exploit
some set {ρx } of density operators to act as input to the quantum channel. That is, thequantum codewords are of the formρxn (m) ≡ ρx1 (m) ⊗ ρx2 (m) ⊗ · · · ⊗ ρxn (m) .

(20.1)

Bob then performs individual measurements of the outputs of the quantum channel by
exploiting some POVM {Λy }. This scheme induces the following conditional probability
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CHAPTER 20. CLASSICAL COMMUNICATION

Figure 20.2: A coding strategy that can outperform the previous naive strategy, simply by making use ofentanglement at the encoder and decoder.

distribution:
pY1 ···Yn |X1 ···Xn (y1 · · · yn |x1 (m) · · · xn (m))
= Tr Λy1 ⊗ · · · ⊗ Λyn (N ⊗ · · · ⊗ N ) ρx1 (m) ⊗ · · · ⊗ ρxn (m)= Tr (Λy1 ⊗ · · · ⊗ Λyn ) N (ρx1 (m) ) ⊗ · · · ⊗ N (ρxn (m) )nY=Tr Λyi N (ρxi (m) ) ,(20.2)(20.3)(20.4)

i=1

which we immediately realize is equivalent to many i.i.d. instances of the following classicalchannel:pY |X (y|x) ≡ Tr {Λy N (ρx )} .
(20.5)
Thus, if they exploit this scheme, the optimal rate at which they can communicate is equal
to the following expression:
Iacc (N ) ≡max{pX (x),ρx ,Λ}I(X; Y ),(20.6)

where the maximization of the classical mutual information is with respect to all input
distributions, all input density operators, and all POVMs that Bob could perform at theoutput of the channel. This information quantity is known as the accessible information of
the channel.
The above strategy is not necessarily an optimal strategy if the channel is truly a quantum
channel—it does not make use of any quantum effects such as entanglement or collective
measurements (an example of a collective measurement is a Bell measurement, as in quantum
teleportation). A first simple modification of the protocol to allow for such effects would be
to consider coding for the tensor-product channel N ⊗ N rather than the original channel.
The input states would be entangled across two channel uses, and the output measurements
would be over two channel outputs at a time. In this way, they would be exploiting entangled
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20.2. THE INFORMATION-PROCESSING TASK

559

states at the encoder and collective measurements at the decoder. Figure 20.2 illustrates themodified protocol, and the rate of classical communication that they can achieve with such a
strategy is 12 Iacc (N ⊗N ). This quantity is always at least as large as Iacc (N ) because a special
case of the strategy for the tensor-product channel N ⊗N is to choose the distribution pX (x),the states ρx , and the POVM Λ to be tensor products of the ones that maximize Iacc (N ).
We can then extend this construction inductively by forming codes for the tensor-productchannel N ⊗k (where k is a positive integer), and this extended strategy achieves the classicalcommunication rate of k1 Iacc (N ⊗k ) for any finite k. These results then suggest that theultimate classical capacity of the channel is the regularization of the accessible informationof the channel:
1(20.7)
Ireg (N ) ≡ lim Iacc (N ⊗k ).
k→∞ k
The regularization of the accessible information is intractable for general quantum channels, but the optimization task could simplify immensely if the accessible information isadditive (additive in the sense discussed in Chapter 13). In this case, the regularized accessible information Ireg (N ) would be equal to the accessible information Iacc (N ). However, even
if the quantity is additive, the optimization could still be difficult to perform in practice.
A simple upper bound on the accessible information is the Holevo information χ(N ) of thechannel, defined as
χ(N ) ≡ max I(X; B),(20.8)
ρ

where the maximization is with respect to classical–quantum states ρXB of the followingform:XpX (x)|xihx|X ⊗ NA0 →B (ψAx 0 ).
(20.9)ρXB ≡x

The Holevo information is a more desirable quantity to characterize classical communication
over a quantum channel because it is always an upper bound on the accessible informationand it does not involve an optimization over measurements.
Thus, a natural question to ask is whether Alice and Bob can achieve the Holevo information rate, and the main theorem of this chapter states that it is possible to do so. The
resulting coding scheme bears some similarities with the techniques in Shannon’s channel
coding theorem, but the main difference is that the decoding POVM is a collective measurement over all of the channel outputs.

20.2 The Information-Processing Task
20.2.1

Classical Communication

We now discuss the most general form of the information-processing task and give the criterion for a classical communication rate C to be achievable—i.e., we define an (n, C, ε) codefor classical communication over a quantum channel. Alice begins by selecting some classical message m that she would like to transmit to Bob—she selects from a set of messagesc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 20. CLASSICAL COMMUNICATION

Figure 20.3: The most general protocol for classical communication over a quantum channel. Alice selects
some message M and encodes it as a quantum codeword for input to many independent uses of the noisy
quantum channel. Bob performs some POVM over all of the channel outputs to determine the message that
Alice transmits.

{1, . . . , |M|}. Let M denote the random variable corresponding to Alice’s choice of message,and let |M| denote its cardinality. She then prepares some state ρm
A0n as input to the many
independent uses of the channel—the input systems are n copies of the channel input system
A0 . She transmits this state over n independent uses of the channel N , and the state at
Bob’s receiving end isN ⊗n (ρm(20.10)A0n ).
Bob has some decoding POVM {Λm } that he can exploit to determine which message Alicetransmits. Figure 20.3 depicts such a general protocol for classical communication over aquantum channel.
Let M 0 denote the random variable for Bob’s estimate of the message. The probability
that he determines the correct message m is as follows:
Pr {M 0 = m|M = m} = Tr Λm N ⊗n (ρm(20.11)A0n ) ,and thus the probability of error for a particular message m ispe (m) ≡ 1 − Pr {M 0 = m|M = m}
= Tr (I − Λm ) N ⊗n (ρmA0n ) .

(20.12)(20.13)The maximal probability of error for any coding scheme is thenp∗e ≡ max pe (m).

(20.14)1
log |M| ,n(20.15)m∈M

The rate C of communication isC≡

and the code has ε ∈ [0, 1] error if p∗e ≤ ε.
A rate C of classical communication is achievable for a channel N if there exists an(n, C − δ, ε) code for all δ > 0, ε ∈ (0, 1), and sufficiently large n. The classical capacity
C(N ) of N is equal to the supremum of all achievable rates for classical communication.
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20.2. THE INFORMATION-PROCESSING TASK

20.2.2561

Randomness Distribution

A sender and receiver can exploit a quantum channel for the alternate but related task of
randomness distribution. Here, Alice prepares a local classical system in a uniformly random
state, she makes a copy of it, and the goal is for Bob to have the copy, such that Alice andBob share a state of the following form at the end of the protocol:
X 1|mihm|M ⊗ |mihm|M 0 .
(20.16)
ΦM M 0 ≡|M|m∈MSuch shared randomness is not particularly useful as a resource, but this viewpoint is helpful
for proving the converse theorem of this chapter and later on when we encounter otherinformation-processing tasks in quantum Shannon theory. The main point to note is thata noiseless classical bit channel can always generate one bit of noiseless shared randomness.
Thus, if a quantum channel has a particular capacity for classical communication, it canalways achieve the same capacity for randomness distribution. In fact, the capacity for
randomness distribution can only be larger than that for classical communication because
shared randomness is a weaker resource than classical communication. This relationshipgives a simple way to bound the capacity for classical communication from above by thecapacity for randomness distribution.
The most general protocol for randomness distribution is as follows. Alice begins bylocally preparing a state of the form in (20.16). She then performs an encoding channel thattransforms this state to the following one:
X 1
|mihm|M ⊗ ρm
(20.17)
A0n ,
|M|
m∈M
and she transmits the A0n systems over n independent uses of the quantum channel N ,producing the following state:
X 1|mihm|M ⊗ N ⊗n (ρmωM B n ≡(20.18)A0n ).
|M|m∈M
Bob then performs a quantum instrument on the received systems (exploiting some POVM
{Λm }), and the resulting state isXpp1|mihm|M ⊗ Λm0 N ⊗n (ρm)Λm0 ⊗ |m0 ihm0 |M 0 .
(20.19)0nA|M|0m,m ∈M
The state
ωM M 0 =
1
00
Tr Λm0 N ⊗n (ρmA0n ) |mihm|M ⊗ |m ihm |M 0
|M|m,m0 ∈M
X(20.20)should then be ε-close in trace distance to the original state in (20.16) for an (n, C, ε) protocol
for randomness distribution:1
ΦM M 0 − ωM M 0 1 ≤ ε.
(20.21)
2
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CHAPTER 20. CLASSICAL COMMUNICATION

A rate C for randomness distribution is achievable if there exists an (n, C − δ, ε) randomness distribution code for all δ > 0, ε ∈ (0, 1), and sufficiently large n. The capacity for
randomness distribution is equal to the supremum of all achievable rates. Clearly, from thedefinitions, it follows that the classical capacity of a channel can never exceed the capacity
for randomness distribution.

20.3 The Classical Capacity Theorem
We now state the main theorem of this chapter, the classical capacity theorem.
Theorem 20.3.1 (Holevo–Schumacher–Westmoreland) The classical capacity of a quantum channel is equal to the regularization of the Holevo information of the channel:
C(N ) = χreg (N ),(20.22)where1
χ(N ⊗k ),k→∞ kχreg (N ) ≡ lim(20.23)and the Holevo information χ(N ) of a channel N is defined in (20.8).
The regularization in the above characterization is a reflection of our ignorance of a better
formula for the classical capacity of a quantum channel. The proof of the above theorem in
the next two sections demonstrates that the above quantity is indeed equal to the classicalcapacity, but the regularization implies that the above characterization is intractable forgeneral quantum channels. However, if the Holevo information of a particular channel isadditive (in the sense discussed in Chapter 13), then χreg (N ) = χ(N ), the classical capacity
formula simplifies for such a channel, and we can claim to have a complete understanding of the channel’s classical transmission capabilities. This “all-or-nothing” situation withcapacities is quite common in quantum Shannon theory, and it implies that we still havemuch remaining to understand regarding classical information transmission over quantum
channels.
The next two sections prove the above capacity theorem in two parts: the direct coding
theorem and the converse theorem. The proof of the direct coding theorem demonstrates theinequality LHS ≥ RHS in (20.22). That is, it shows that the regularized Holevo information
is an achievable rate for classical communication, and it exploits typical and conditionallytypical subspaces and the packing lemma to do so. The proof of the converse theorem shows
the inequality LHS ≤ RHS in (20.22). That is, it shows that any sequence of protocols
with achievable rate C (with vanishing error in the large n limit) should have this rate
below the regularized Holevo information. The proof of the converse theorem exploits theaforementioned idea of randomness distribution, continuity of entropy, and the quantum
data-processing inequality.
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20.3. THE CLASSICAL CAPACITY THEOREM

20.3.1

563

The Direct Coding Theorem

We first prove the direct coding theorem. Suppose that a quantum channel N connects Aliceto Bob, and they are allowed access to many independent uses of this quantum channel. Alicecan choose some ensemble {pX (x), ρx } of states which she can exploit to make a random codefor this channel. She selects |M| codewords {xn (m)}m∈{1,...,|M|} independently according to
the following distribution:
( hP
i−1
n
n
n pX n (x )pX n (xn ) : xn ∈ TδX
X
n
n
0
x
∈T
,
(20.24)pX 0n (x ) =δn0: xn ∈/ TδX
where X 0n is a random variable with distribution p0X 0n (xn ), pX n (xn ) = pX (x1 ) · · · pX (xn ),nand TδX denotes the set of strongly typical sequences for the distribution pX n (xn ) (seeSection 14.7). This “pruned” distribution is approximately close to the i.i.d. distribution
pX n (xn ) because the probability mass of the typical set is nearly one. In fact, from (19.86),n
we have that the following holds if Pr X n ∈ TδX ≥ 1 − ε:X
|p0X 0n (xn ) − pX n (xn )| ≤ 2ε.
(20.25)xn ∈X nnIndeed, we know from typicality that Pr X n ∈ TδX ≥ 1−ε for all ε ∈ (0, 1) and sufficiently
large n.
These classical codewords {xn (m)}m∈{1,...,|M|} lead to quantum codewords of the followingform:nρx (m) ≡ ρx1 (m) ⊗ · · · ⊗ ρxn (m) ,(20.26)by exploiting the quantum states in the ensemble {pX (x), ρx }. Alice then transmits thesecodewords through the channel, leading to the following tensor-product density operators:nσ x (m) ≡ σ x1 (m) ⊗ · · · ⊗ σ xn (m)≡ N (ρx1 (m) ) ⊗ · · · ⊗ N (ρxn (m) ).

(20.27)(20.28)

Bob then detects which codeword Alice transmits by exploiting some detection POVM {Λm }that acts on all of the channel outputs.
At this point, we would like to exploit the packing lemma (Lemma 16.3.1 from Chapter 16). Recall that four objects are needed to apply the packing lemma, and they should
satisfy four inequalities. The first object needed is an ensemble from which we can selectna code randomly, and the ensemble in our case is p0X 0n (xn ), σ x . The next object is the
expected density operator of this ensemble:
n 0n oXn
EX 0n σ X=
p0X 0n (xn )σ x .
(20.29)xn ∈X nFinally, we need a message subspace projector and a total subspace projector, and we letnthese respectively be the conditionally typical projector ΠδB n |xn for the state σ x and thec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 20. CLASSICAL COMMUNICATION

P
typical projector ΠδB n for the tensor product state σ ⊗n where σ ≡ x pX (x)σ x . Intuitively,
0n
the tensor product state σ ⊗n should be close to the expected state EX 0n σ X , and the next
exercise asks you to verify this statement.
 0n
Exercise 20.3.1 Prove that the trace distance between the expected state EX 0n σ X
andthe tensor product state σ ⊗n is small for all sufficiently large n:
n 0n oEX 0n σ X
− σ ⊗n ≤ 2ε,(20.30)1nwhere ε is an arbitrarily small positive number such that Pr X n ∈ TδX ≥ 1 − ε.
If the four conditions of the packing lemma are satisfied (see (16.11)–(16.14)), then there
exists a coding scheme with a detection POVM that has an arbitrarily low maximal probability of error as long as the number of messages in the code is not too high. We now show
how to satisfy these four conditions by exploiting the properties of typical and conditionally typical projectors. The following three conditions follow from the properties of typicalsubspaces:nTr ΠδB n σBx n ≥ 1 − ε,(20.31) δxn
Tr ΠB n |xn σB n ≥ 1 − ε,(20.32) δ
Tr ΠB n |xn ≤ 2n(H(B|X)+cδ) ,(20.33)where c is a strictly positive constant. The first inequality follows from Property 15.2.7.
The second inequality follows from Property 15.2.4, and the third from Property 15.2.5. Weleave the proof of the fourth inequality for the packing lemma as an exercise.
Exercise 20.3.2 Prove that the following inequality holds:
n 0n o0ΠδB n EX 0n σBXn ΠδB n ≤ [1 − ε]−1 2−n(H(B)−c δ) ΠδB n , 0n
where c0 is a strictly positive constant. (Hint: First show that EX 0n σBXn
and then apply the third property of typical subspaces—Property 15.1.3.)(20.34)≤ [1 − ε]−1 σB nWith these four conditions holding, it follows from Corollary 16.5.1 (the derandomized
version of the packing lemma) that there exists a deterministic code and a POVM {Λm }that can detect the transmitted states with arbitrarily low maximal probability of error as
long as the size |M| of the message set is small enough:np∗e ≡ max Tr (I − Λm ) N ⊗n (ρx (m) )(20.35)m√ 
0(20.36)≤ 4 ε + 2 ε + 16 [1 − ε]−1 2−n(H(B)−H(B|X)−(c+c )δ) |M|√ −1 −n(I(X;B)−(c+c0 )δ)= 4 ε + 2 ε + 16 [1 − ε] 2
|M| .
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20.3. THE CLASSICAL CAPACITY THEOREM

565

0

Thus, we can choose the size of the message set to be |M| = 2n(I(X;B)−(c+c +1)δ) so that therate of communication is the Holevo information I(X; B):1
log |M| = I(X; B) − (c + c0 + 1)δ,n
and the bound on the maximal probability of error becomes√ p∗e ≤ 4 ε + 2 ε + 16 [1 − ε]−1 2−nδ .

(20.38)(20.39)Let ε0 ∈√(0, 1) and δ 0 > 0. By picking n large enough, it is clear that we can have both4 (ε + 2 ε) + 16 [1 − ε]−1 2−nδ ≤ ε0 and (c + c0 + 1)δ ≤ δ 0 . Thus, the Holevo information
I(X; B)ρ , with respect to the following classical–quantum state:XρXB ≡pX (x)|xihx|X ⊗ N (ρx ),(20.40)x∈X

is an achievable rate for the transmission of classical information over N .
Alice and Bob can achieve the Holevo information χ(N ) of the channel N simply by
selecting a random code according to the ensemble {pX (x), ρx } that maximizes I(X; B)ρ .
Lastly, they can achieve the rate k1 χ(N ⊗k ) by coding instead for the tensor-product channel
N ⊗k , and this last result implies that they can achieve the regularization χreg (N ) by making
the blocks for which they are coding be arbitrarily large. This concludes the proof of thedirect part of the coding theorem.
We comment more on the role of entanglement at the encoder before moving on to theproof of the converse theorem. First, the above coding scheme for the channel N does notnmake use of entangled inputs at the encoder because the codeword states ρx (m) are separable
across the channel inputs. It is only when we code for the tensor-product channel N ⊗k thatentanglement comes into play. Here, the codeword states are of the formx (m)ρxn (m) ≡ ρA10k

x (m)⊗ · · · ⊗ ρAn0k.

(20.41)x (m)That is, the states ρAi0k act on the tensor-product Hilbert space of k channel inputs and canbe entangled across these k systems. Whether entanglement at the encoder could increaseclassical communication rates over general quantum channels was the subject of much intense
work over the past few years, but it is now known that there exists a channel for which
exploiting entanglement at the encoder is strictly better than not exploiting entanglement(see Section 20.5).
It is worth re-examining the proof of the packing lemma (Lemma 16.3.1) in order tounderstand better the decoding POVM at the receiving end. The particular decoding POVM
elements employed in the packing lemma have the following form:
!− 12
!− 12
X
X
Γm
,
(20.42)
Γm0
Λm ≡
Γm0
m0 ∈M
δ
Γm ≡ ΠB n ΠδB n |xn (m) ΠδB n .

c 2019m0 ∈M(20.43)

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566

CHAPTER 20. CLASSICAL COMMUNICATION(Simply substitute the conditionally typical projector ΠδB n |xn (m) and the typical projector ΠδB ninto (16.24).) A POVM with the above elements is known as a “square-root” measurement
because of its particular form. We employ such a measurement at the decoder because it
has nice analytic properties that allow us to obtain a good bound on the expectation of
the average error probability (in particular, we can exploit the operator inequality from
Lemma 16.4.1). This measurement is a collective measurement because the conditionallytypical projector and the typical projector are both acting on all of the channel outputs, andwe construct the square-root measurement from these projectors. Such a decoding POVM is
far more exotic than the naive strategy overviewed in Section 20.1 where Bob measures thechannel outputs individually—it is for the construction of this decoding POVM and the proof
that it is asymptotically good that Holevo, Schumacher, and Westmoreland were given muchpraise for their work. However, there is no known way to implement this decoding POVM
efficiently, and the original efficiency problems with the decoder in the proof of Shannon’s
noisy classical channel coding theorem plague the decoders in the quantum world as well.
Alternatively, Bob’s decoding operation could take the form of a sequential decoder, as
discussed in Section 16.6. That is, Bob would perform the measurements {ΠδB n |xn (m) , IB n −ΠδB n |xn (m) } one after another in an attempt to learn which message was transmitted. This
decoding strategy is also a collective measurement strategy because each of the above measurements is a collective measurement acting on all of the channel outputs. This scheme isalso inefficient: even if there is an efficient implementation for each of the individual tests,
there are an exponential number of them to perform in the worst case since there are an
exponential number of messages in a codebook (exponential in n).
Exercise 20.3.3 Show that a measurement with POVM elements of the following form issufficient to achieve the Holevo information of a quantum channel:
!−1/2
Λm ≡

X

ΠδB n |xn (m0 )

m0 ∈M

!−1/2
ΠδB n |xn (m)XΠδB n |xn (m0 ).

(20.44)m0 ∈M

Alternate Proofs of the Direct Part
There are at least two alternate proofs of the direct part of the HSW theorem, which can be
useful as building blocks for other scenarios. The first is based on weak typicality and thesecond is called the constant-composition coding approach. We discuss these briefly here.
Let {pX (x), ρxA0 } be an ensemble of states for the input of the channel NA0 →B ,P
which leadsxxx0
to an ensemble {pX (x), σB } at the output, where σB ≡ NA →B (ρA0 ). Let σB ≡ x pX (x)σBx
be the expected density operator of the output ensemble. In the weak typicality approach,the codewords {xn (m)} are chosen independently at random according to the product disnQ
tribution pX n (xn ) =pX (xi ). One then sets the total subspace projector to be the weakly
i=1

typical projection ΠδB n for σB⊗n and each message subspace projection (for the codeword
xn (m)) to be the weak conditionally typical projection ΠδB n |xn (m) . From the properties ofc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

20.3. THE CLASSICAL CAPACITY THEOREM

567

weak typicality (see Chapter 15), the following conditions for the average version of thepacking lemma (see Exercise 16.5.3) are satisfied:n
pX n (xn ) Tr ΠδB n σBx n ≥ 1 − ε,(20.45)n
pX n (xn ) Tr ΠδB n |xn σBx n ≥ 1 − ε,(20.46)X
xn ∈X nX
xn ∈X nTr ΠδB n |xn ≤ 2n(H(B|X)+δ) ,(20.47)

ΠδB n σB⊗n ΠδB n ≤ 2−n(H(B)−δ) ΠδB n .

(20.48)

We can then conclude that the rate I(X; B) is achievable for classical communication overNA0 →B , by making use of this slightly different scheme. (The Holevo information is withrespect to the output ensemble {pX (x), σBx }.)The next scheme that we mention is the constant-composition
ConsiderPcoding scheme.
xx
again the output ensemble {pX (x), σB } with expectation σB ≡ x pX (x)σB . Now select a
typical type class Tt , as discussed in Definition 14.7.4 and Section 14.7.4. This is a set of
all the sequences xn with empirical distribution t(x) that deviates from the true distributionpX (x) by no more than δ > 0. All the sequences in the same type class are related to one
another by a permutation, and all of them are strongly typical. The idea for selecting a codeat random is now to pick all of the codewords independently and uniformly at random fromthe typical type class t. Thus, the ensemble from which we are selecting codewords is nown{1/ |Tt | , σBx n }xn ∈Tt and we would like to show that in doing so, we can still achieve a rateequal to I(X; B). LetσeB n ≡ |Tt |−1X
xn ∈Ttn

σBx n ,σB ≡X

t(x)σBx .

(20.49)x∈X

Observe that 21 kσ B − σB k1 ≤ |X | δ/2, and thus |H(B)σ − H(B)σ | ≤ η(|X | δ), with η(·) defined just after (14.119) and such that limδ→0 η(|X | δ) = 0. We can take the total subspace
projector to be the strongly typical projection ΠδB n for σ ⊗n
B , and we take the message subn
space projection (for the codeword x (m)) to be the strong conditionally typical projectorΠδB n |xn (m) . We then need to verify that the conditions of the packing lemma (Corollary 16.5.1)
hold. Consider thatnTr ΠδB n σBx n ≥ 1 − ε,nTr ΠδB n |xn σBx n ≥ 1 − ε,Tr ΠδB n |xn ≤ 2n(H(B|X)+cδ) .

(20.50)(20.51)(20.52)

The first inequality follows because each xn ∈ Tt is strongly typical with respect to t(x),so that we can apply Property 15.2.7. The second two inequalities are properties of strong
conditional typicality. So we need to figure out something about the last condition. Let
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CHAPTER 20. CLASSICAL COMMUNICATION

tn (xn ) ≡nQt(xi ), i.e., the product distribution realized by t(x). Consider thati=1

X I(xn ∈ Tt ) n
1 X xnσB n =σBx n|Tt | xn ∈T|T|t
xn ∈X ntXn≤(n + 1)|X | tn (xn )σBx n = (n + 1)|X | σ ⊗n
B .

σeB n =(20.53)(20.54)xn ∈X nThe inequality follows from the development in (14.114)–(14.117). Using this, we find that1ΠδB n σeB n ΠδB n ≤ 2−n(H(B)σ − n |X | log(n+1)) ΠδB n ,(20.55)which is the last condition we need for the packing lemma (Corollary 16.5.1). We canthen conclude that the rate H(B)σ − H(B|X) is achievable for classical communication
using constant-composition codes. However, since |H(B)σ − H(B)σ | ≤ η(|X | δ), we can also
conclude that the rate I(X; B) is achievable for classical communication using constantcomposition codes, where the Holevo information is with respect to the original output
ensemble {pX (x), σBx }.

20.3.2The Converse Theorem

The second part of the classical capacity theorem is the converse theorem, and we provide
a simple proof of it in this section. Suppose that Alice and Bob are trying to accomplish
randomness distribution rather than classical communication—the capacity for such a task
can only be larger than that for classical communication as we argued before in Section 20.2.
Recall that in such a task, Alice first prepares a maximally correlated state ΦM M 0 so thatthe rate of randomness distribution is equal to n1 log |M|. Alice and Bob share a state of
the form in (20.19) after encoding, channel transmission, and decoding. We now show thatthe regularized Holevo information in (20.23) bounds the capacity of randomness distribution. As a result, the regularized Holevo information also bounds the capacity for classicalcommunication from above. Consider the following chain of inequalities:
log |M| = I(M ; M 0 )Φ≤ I(M ; M 0 )ω + f (|M| , ε)
≤ I(M ; B n )ω + f (|M| , ε)≤ χ(N ⊗n ) + f (|M| , ε).

(20.56)(20.57)(20.58)(20.59)The first equality follows because the mutual information of the shared randomness stateΦM M 0 is equal to log |M| bits. The first inequality follows from the error criterion in (20.21)
and by applying the AFW inequality (Theorem 11.10.3). That is, since H(M )Φ = H(M )ω ,c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

20.4. EXAMPLES OF CHANNELS

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we know that
|I(M ; M 0 )Φ − I(M ; M 0 )ω |= |H(M )Φ − H(M |M 0 )Φ − [H(M )ω − H(M |M 0 )ω ]|= |H(M |M 0 )ω − H(M |M 0 )Φ |≤ f (|M| , ε) ≡ ε log |M| + g2 (ε).

(20.60)(20.61)(20.62)(20.63)The second inequality results from the quantum data-processing inequality for quantum mutual information (Theorem 11.9.4)—recall that Bob processes the B n system with a quantum
instrument to get the classical system M 0 . Also, the quantum mutual information is evaluated on a classical–quantum state of the form in (20.18). The final inequality follows
because the classical–quantum state in (20.18) has a particular distribution and choice ofstates, and this choice always leads to a value of the quantum mutual information thatcannot be greater than the Holevo information of the tensor-product channel N ⊗n . Putting
everything together, we find that1
1
1
log |M| (1 − ε) ≤ χ(N ⊗n ) + g2 (ε).
nnn(20.64)Thus, if we are considering a sequence of (n, [log |M|] /n, εn ) classical communication protocols with rate C − δn = n1 log |M|, such that limn→∞ εn = limn→∞ δn = 0, then the above
bound becomes
1
1(20.65)(C − δn ) (1 − εn ) ≤ χ(N ⊗n ) + g2 (εn ).
nnTaking the limit as n → ∞ then establishes that an achievable rate C necessarily satisfies
C ≤ χreg (N ), where χreg (N ) is the regularized Holevo formula given in (20.23).

20.4 Examples of Channels
Observe that the final upper bound in (20.59) on the rate C is the multi-letter Holevo
information of the channel. It would be more desirable to have χ(N ) as the upper bound on
C rather than n1 χ(N ⊗n ) because the former is simpler, and the optimization problem set out
in the latter quantity is simply impossible to compute in general using finite computational
resources (for large n). However, the upper bound in (20.59) is the best known upper bound
if we do not know anything else about the structure of the channel, and for this reason, thebest known characterization of the classical capacity is the one given in (20.22).
If we know that the Holevo information of the tensor product of a certain channel withan arbitrary number of copies of itself is additive, then there is no need for the regularization
χreg (N ), and the characterization in Theorem 20.3.1 reduces to a very good one: the Holevo
information χ(N ). There are many examples of channels for which the classical capacity
reduces to the Holevo information of the channel, and we detail a few such classes of examples in this section: entanglement-breaking channels, quantum Hadamard channels, erasurechannels, and quantum depolarizing channels. The proof that demonstrates additivity of thec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 20. CLASSICAL COMMUNICATION

Holevo information for each of these channels depends explicitly on structural properties ofeach one, and there is unfortunately not much to learn from these proofs in order to say anything about additivity of the Holevo information of general quantum channels. Nevertheless,
it is good to have some natural channels for which we can compute their classical capacity,and it is instructive to examine these proofs in detail to understand what it is about eachchannel that makes their Holevo information additive.

20.4.1

Classical Capacity of Entanglement-Breaking Channels

We have already seen in Section 13.3.1 that the Holevo information of an entanglementbreaking channel is additive. As a consequence, we can conclude that the capacity of anentanglement-breaking channel N is given by χ(N ).
We now focus our discussion on cq channels. Recall from Section 4.6.7 that a quantum
channel is a particular kind of entanglement-breaking channel (cq channel) if the action of thechannel is equivalent to performing first a complete projective measurement of the input andthen preparing a quantum state conditioned on the value of the classical variable resulting
from the measurement. Additionally, Theorem 13.3.3 states that the Holevo information isa concave function of the input distribution over which we are optimizing for such channels.
Thus, computing the classical capacity of cq channels can be performed by optimization
techniques because the Holevo information is additive for them.
The Relation to General Channels
We can always exploit the above result regarding cq entanglement-breaking channels to get areasonable lower bound on the classical capacity of any quantum channel N . The sender Alicecan simulate an entanglement-breaking channel by modifying the processing at the input ofan arbitrary quantum channel. She can first measure the input to her simulated channel
in the basis {|xihx|}, prepare a state ρx conditioned on the outcome of the measurement,
and subsequently feed this state into the channel N . These actions are equivalent to thefollowing channel:
Xhx|σ|xiN (ρx ),(20.66)σ→xand the capacity of this simulated channel is equal toI(X; B)ρ ,(20.67)pX (x)|xihx|X ⊗ N (ρx ),(20.68)

where
ρXB ≡

X
x

pX (x) ≡ hx|σ|xi.

(20.69)

Of course, Alice has the freedom to prepare whichever state σ she would like to be input to thesimulated channel, and she also has the ability to prepare whichever states ρx she would likec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

20.4. EXAMPLES OF CHANNELS

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to be conditioned on the outcomes of the first measurement, so we should let her maximize theHolevo information over all these inputs. Thus, the capacity of the entanglement-breaking
channel composed with the actual channel is equal to the Holevo information of the original
channel:max I(X; B)ρ .
(20.70)pX (x),ρxThis capacity is also known as the product-state capacity of the channel because it is thecapacity achieved by inputting unentangled, separable states at the encoder (Alice can in fact
just input product states), and it can be a good lower bound on the true classical capacityof a quantum channel, even if it does not allow for entanglement at the encoder.

20.4.2

Classical Capacity of Quantum Hadamard Channels

Recall from Section 5.2.3 that quantum Hadamard channels are those with a complementary channel that is entanglement breaking, and this property allows us to prove that theHolevo information of the original channel is additive. Several important natural channels
are quantum Hadamard channels. A trivial example is the noiseless qubit channel because
Bob could perform a projective measurement of his system and send a constant state to Eve.
A less trivial example of a quantum Hadamard channel is a generalized dephasing channel(see Section 5.2.3), but this channel trivially has a maximal classical capacity of log d bits perchannel use because this channel transmits a preferred orthonormal basis without error. A
quantum Hadamard channel with a more interesting classical capacity is known as a cloning
channel, the channel induced by a universal cloning machine (however we will not discuss
this channel in any detail).
Theorem 20.4.1 The Holevo information of a quantum Hadamard channel NH and anyother channel N is additive:χ(NH ⊗ N ) = χ(NH ) + χ(N ).

(20.71)

Proof. First, recall from Theorem 13.3.2 that it is sufficient to consider ensembles of purestates at the input of the channel when maximizing its Holevo information. That is, we only
need to consider classical–quantum states of the following form:X
σXA0 ≡pX (x)|xihx|X ⊗ |φx ihφx |A0 ,(20.72)xwhere A0 is the input to some channel NA0 →B . Let ωXBE ≡ UAN0 →BE (σXA0 ) where UAN0 →BE is
an isometric extension of the channel and UAN0 →BE denotes the corresponding channel. Thus,
the Holevo information of NA0 →B is equal to a different expression:χ(N ) ≡ max I(X; B)ω
σ= max [H(B)ω − H(B|X)ω ](20.74)= max [H(B)ω − H(E|X)ω ] ,(20.75)σσc 2019(20.73)

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CHAPTER 20. CLASSICAL COMMUNICATION

where the second equality follows from the definition of the quantum mutual information,and the third equality follows because, conditioned on X, the input to the channel is pureand the entropies H(B|X)ω and H(E|X)ω are equal.
Exercise 20.4.1 Prove that it is sufficient to consider pure-state inputs when maximizing
the following entropy difference over classical–quantum states:max [H(B)ω − H(E|X)ω ] .
σ(20.76)

Suppose now that σ is a state that maximizes the Holevo information of the joint channel
NH ⊗ N , and suppose it has the following form:X
σXA01 A02 ≡pX (x)|xihx|X ⊗ |φx ihφx |A01 A02 .
(20.77)x

Let
ωXB1 B2 E1 E2 ≡ (UAN0H→B1 E1 ⊗ UAN02 →B2 E2 )(σXA01 A02 ).
1(20.78)The Hadamard channel is degradable, and the degrading channel from Bob to Eve takes
a particular form: it is a measurement that produces a classical variable Y , followed by1
the preparation of a state conditioned on the outcome of the measurement. Let DB
1 →Y
be the first part of the degrading channel that produces the classical variable Y , and let
1(ωXB1 B2 E1 E2 ). Let DY2 →E 0 be the second part of the degrading channel
θXY E1 B2 E2 ≡ DB
1 →Y
1that produces the state of E1 conditioned on the classical variable Y , and let τXE10 E1 B2 E2 ≡
DY2 →E 0 (θXY E1 B2 E2 ). Figure 20.4 summarizes these structural relationships. Consider the1
following chain of inequalities:
I(X; B1 B2 )ω = H(B1 B2 )ω − H(B1 B2 |X)ω= H(B1 B2 )ω − H(E1 E2 |X)ω≤ H(B1 )ω + H(B2 )ω − H(E1 |X)ω − H(E2 |E1 X)ω
= H(B1 )ω − H(E1 |X)ω + H(B2 )ω − H(E2 |E10 X)τ
≤ H(B1 )ω − H(E1 |X)ω + H(B2 )θ − H(E2 |Y X)θ
≤ χ(NH ) + χ(N ).

(20.79)(20.80)(20.81)(20.82)(20.83)(20.84)

The first equality follows from the definition of the quantum mutual information. The
second equality follows because H(B1 B2 |X)ω = H(E1 E2 |X)ω when the conditional inputs|φx iA01 A02 to the channel are pure states. The next inequality follows from subadditivity ofentropy H(B1 B2 )ω ≤ H(B1 )ω +H(B2 )ω and from the chain rule for entropy: H(E1 E2 |X)ω =H(E1 |X)ω + H(E2 |E1 X)ω . The third equality follows from a rearrangement of terms andrealizing that the state of τ on systems E10 E2 X is equal to the state of ω on the same systems.
The second inequality follows from the quantum data-processing inequality I(E2 ; E10 |X)τ ≤I(E2 ; Y |X)θ . The final inequality follows because the state ω is a state of the form in(20.75), because the entropy difference is never greater than the Holevo information of thefirst channel, and from the result of Exercise 20.4.1. The same reasoning follows for the otherentropy difference and by noting that the classical system is the composite system XY .
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20.4. EXAMPLES OF CHANNELS

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Figure 20.4: A summary of the structural relationships for the additivity question if one channel is aquantum Hadamard channel. Alice first prepares a state of the form in (20.77). She transmits one system
A01 through the quantum Hadamard channel and the other A02 through the other channel. The first Bob
B1 at the output of the Hadamard channel can simulate the channel to the first Eve E1 because the firstchannel is a quantum Hadamard channel. He performs a complete projective measurement of his system,
leading to a classical variable Y , followed by the preparation of some state conditioned on the value of theclassical variable Y . The bottom of the figure labels the state of the systems at each step.

20.4.3

Classical Capacity of the Quantum Erasure Channel

The quantum erasure channel is one of the simplest channels for which we can compute theclassical capacity. Recall from Section 4.7.6 that the qudit erasure channel is defined asfollows:
ρ → (1 − ε)ρ + ε|eihe|,(20.85)where ρ is a d-dimensional qudit input state, ε ∈ [0, 1], and |ei is an erasure symbol orthogonal to the input space of the channel.
Theorem 20.4.2 (Classical Capacity of the Erasure Channel) The classical capacityof the d-dimensional quantum erasure channel is equal to (1 − ε) log d.
Proof. The rate (1 − ε) log d is achievable by picking the input ensemble to be {1/d, |ii}
and evaluating the Holevo information. So we need to show that it is not possible to achievea rate higher than this. Let MA1 →B1 be some quantum channel and let NAε 2 →B2 denote thequantum erasure channel. Let ρXA1 A2 be the following classical–quantum stateXρXA1 A2 ≡pX (x)|xihx|X ⊗ φxA1 A2 ,(20.86)xand suppose that it achieves the Holevo information χ(M ⊗ N ε ). Consider that
(20.87)
ωXB1 B2 ≡ (MA1 →B1 ⊗ NAε 2 →B2 )(ρXA1 A2 )X
=pX (x)|xihx|X ⊗ (1 − ε)MA1 →B1 (φxA1 A2 ) + εMA1 →B1 (φxA1 ) ⊗ |eihe|B2 .
x(20.88)c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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Consider that the isometry [|0ih0|B2 + . . . |d − 1ihd − 1|B2 ] ⊗ |0iY + |eihe|B2 ⊗ |1iY takes theabove state to the following one:XωXB1 B2 Y ≡pX (x)|xihx|X ⊗ MA1 →B1 (φxA1 A2 ) ⊗ (1 − ε)|0ih0|Yx+X

pX (x)|xihx|X ⊗ MA1 →B1 (φxA1 ) ⊗ |eihe|B2 ⊗ ε|1ih1|Y , (20.89)

xso that the Y register is a flag indicating whether an erasure occurred. Thenχ(M ⊗ N ε ) = I(X; B1 B2 )ω
= I(X; B1 B2 Y )ω= I(X; B1 B2 |Y )ω + I(X; Y )ω= I(X; B1 B2 |Y )ω
= (1 − ε) I(X; B1 A2 )M(φx ) + εI(X; B1 )M(φx )≤ (1 − ε) χ(M ⊗ id) + εχ(M)= χ(M) + (1 − ε) χ(id)= χ(M) + (1 − ε) log d.

(20.90)(20.91)(20.92)(20.93)(20.94)(20.95)(20.96)(20.97)The second equality follows because the aforementioned isometry takes B1 B2 to B1 B2 Y . The
third equality follows from the chain rule for mutual information, and the fourth because
I(X; Y )ω = 0. The fifth equality follows because Y is a classical system, so that we can
expand the mutual information as a convex combination of individual mutual informations.
The inequality follows by maximizing the Holevo informations with respect to all input
ensembles. The second-to-last equality follows because the identity channel is a Hadamard
channel, so that Theorem 20.4.1 implies that χ(M ⊗ id) = χ(M) + χ(id).
By setting M = (N ε )⊗[n−1] and iterating the above, we see that χ((N ε )⊗n ) ≤ n (1 − ε) log d,so that the regularized Holevo information cannot exceed (1 − ε) log d. Since this rate is alsoachievable, this concludes the proof.

20.4.4

Classical Capacity of the Depolarizing Channel

The qudit depolarizing channel is another example of a channel for which we can computeits classical capacity. Additionally, we will see that achieving the classical capacity of thischannel requires a strategy which is very “classical”—it is sufficient to prepare classical states{|xihx|} at the input of the channel and to measure each channel output in the same basis(see Exercise 20.4.3). However, we will later see in Chapter 24 that the depolarizing channel
has some rather bizarre, uniquely quantum features when considering its quantum capacity,even though the features of its classical capacity are rather classical.
Recall from Section 4.7.4 that the depolarizing channel is the following map:
ND (ρ) = (1 − p)ρ + pπ,(20.98)where π is the maximally mixed state.
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Theorem 20.4.3 (Classical Capacity of the Depolarizing Channel) The classical capacity of the qudit depolarizing channel ND is as follows:

p

p
p
p
log 1 − p ++ (d − 1) log.
(20.99)χ(ND ) = log d + 1 − p +ddddProof. The first part of the proof of this theorem relies on a somewhat technical result,namely, that the Holevo information of the tensor-product channel ND ⊗N is additive (where
the first channel is the depolarizing channel and the other is arbitrary):χ(ND ⊗ N ) = χ(ND ) + χ(N ).

(20.100)

This result is due to King (2003), and it exploits a few properties of the depolarizing channel.
The result implies that the classical capacity of the depolarizing channel is equal to its Holevo
information. We now show how to compute the Holevo information of the depolarizing
channel. To do so, we first determine the minimum output entropy of the channel.
Definition 20.4.1 (Minimum Output Entropy) The minimum output entropy H min (N )of a channel N is the minimum of the entropy at the output of the channel:
H min (N ) ≡ min H(N (ρ)),ρ(20.101)

where the minimization is over all states input to the channel.
Exercise 20.4.2 Prove that it is sufficient to minimize over only pure-state inputs to thechannel when computing the minimum output entropy. That is,
H min (N ) = min H(N (|ψihψ|)).
|ψi(20.102)

The depolarizing channel is a highly symmetric channel. For example, if we input a purestate |ψi to the channel, the output is as follows:p(1 − p)ψ + pπ = (1 − p)ψ + I
dp= (1 − p)ψ + (ψ + I − ψ)dpp
= 1−p+
ψ + (I − ψ).
dd(20.103)(20.104)(20.105)

Observe that the eigenvalues of the output state are the same for any pure state and areequal to 1 − p + dp with multiplicity one and dp with multiplicity d − 1. Thus, the minimum
output entropy of the depolarizing channel is just


p
p
p
p
H min (ND ) = − 1 − p +log 1 − p +− (d − 1) log.
(20.106)ddddc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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We now compute the Holevo information of the depolarizing channel. Recall from Theorem 13.3.2 that it is sufficient to consider optimizing the Holevo information over a classical–
quantum state with conditional states that are pure (a state σXA0 of the form in (20.72)).
Also, the Holevo information has the following form:
max I(X; B)ω = max [H(B)ω − H(B|X)ω ] ,σσ(20.107)

where ωXB is the output state. Consider the following augmented input ensemble:
ρXIJA0 ≡d−1

1 XX
pX (x)|xihx|X ⊗ |iihi|I ⊗ |jihj|J ⊗ X(i)Z(j)ψAx 0 Z † (j)X † (i), (20.108)d2 x i,j=0
where X(i) and Z(j) are the generalized Pauli operators from Section 3.7.2. Suppose that
we trace over the IJ system. Then the state ρXA0 is as follows:X
pX (x)|xihx|X ⊗ πA0 ,(20.109)
ρXA0 =
x

by recalling the result of Exercise 4.7.6. Also, note that inputting the maximally mixed state
to the depolarizing channel results in the maximally mixed state at its output. Consider thefollowing chain of inequalities:
I(X; B)ω = H(B)ω − H(B|X)ω
≤ H(B)ρ − H(B|X)ω
= log d − H(B|XIJ)ρ
X= log d −pX (x)H(B)ND (ψx )(20.110)(20.111)(20.112)(20.113)x≤ log d − min H(B)ND (ψx )(20.114)≤ log d − H(20.115)xmin(ND ).

The first equality follows by expanding the quantum mutual information. The first inequality
follows from concavity of entropy. The second equality follows because the state of ρ on
system B is the maximally mixed state π and from the following chain of equalities:
d−1

1 XX
H(B|XIJ)ρ = 2
pX (x)H(B)ND (X(i)Z(j)ψx Z † (j)X † (i))
d x i,j=0(20.116)d−1

1 XX
pX (x)H(B)X(i)Z(j)ND (ψx )Z † (j)X † (i)d2 x i,j=0
X=
pX (x)H(B)ND (ψx )=(20.117)(20.118)x= H(B|X)ω .
c 2019(20.119)

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20.4. EXAMPLES OF CHANNELS

577

Figure 20.5: The classical capacity of the quantum depolarizing channel as a function of the dimension dof the channel and the depolarizing parameter p. The classical capacity vanishes when p = 1 because thechannel replaces the input with the maximally mixed state. The classical capacity is maximal at log d whenp = 0 because there is no noise. In between these two extremes, the classical capacity is a smooth functionof p and d given by the expression in (20.99).

The third equality in (20.113) follows from the above chain of equalities. The second inequality in (20.114) follows because the expectation can never be smaller than the minimum(this step is not strictly necessary for the depolarizing channel). The last inequality follows
because minx H(B)ND (ψx ) ≥ H min (ND ) (though it is actually an equality for the depolarizing
channel).
An ensemble of the following form suffices to achieve the classical capacity of the depolarizing channel:
d−11X|iihi|I ⊗ |iihi|A0 ,(20.120)d i=0because we only require that the reduced state on A0 be equal to the maximally mixed
state. The final expression for the classical capacity of the depolarizing channel is as statedin Theorem 20.4.3, which we plot in Figure 20.5 as a function of the dimension d and thedepolarizing parameter p.
Exercise 20.4.3 (Achieving the Classical Capacity of the Depolarizing Channel)
We actually know that even more is true regarding the method for achieving the classical capacity of the depolarizing channel. Prove that it is possible to achieve
the classical capacity

of the depolarizing channel by choosing states from an ensemble d1 , |xihx| and performing
a complete projective measurement in the same basis at the output of each channel. Thatis, the naive scheme outlined in Section 20.1 is sufficient to attain the classical capacity ofthe depolarizing channel. (Hint: First show that the classical channel pY |X (y|x) induced by
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CHAPTER 20. CLASSICAL COMMUNICATION

inputting a state |xi to the depolarizing channel and measuring |yi at the output is as follows:p
pY |X (y|x) = (1 − p)δx,y + .
d(20.121)

Then show that the distribution pY (y) is uniform if pX (x) is uniform. Finally, show that

p
p

p
p
log 1 − p +− (d − 1)log.
(20.122)
H(Y |X) = − 1 − p +ddddConclude that the classical capacity of the induced channel pY |X (y|x) is the same as that for
the quantum depolarizing channel.)
Exercise 20.4.4 A covariant channel NC is one for which the state resulting from a unitary
U acting on the input state before the channel occurs is equivalent to one where there is a
unitary representation RU of the unitary U acting on the output of the channel:
NC (U ρU † ) = RU NC (ρ)RU† .

(20.123)Show that the Holevo information χ(NC ) of a covariant channel is equal toχ(NC ) = log d − H(NC (ψ)),(20.124)

where ψ is an arbitrary pure state.

20.5 Superadditivity of the Holevo Information
Many researchers thought for some time that the Holevo information would be additive for all
quantum channels, implying that it would be a good characterization of the classical capacity
in the general case—this conjecture was known as the additivity conjecture. Researchers
thought that this conjecture would hold because they discovered a few channels for which it
did hold, but without any common theme occurring in the proofs for the different channels,
they soon began looking in the other direction for a counterexample to disprove it. After sometime, Hastings (2009) found the existence of a counterexample to the additivity conjecture,demonstrating that it cannot hold in the general case. This result demonstrates that evenone of the most basic questions in quantum Shannon theory still remains wide open and thatentanglement at the encoder can help increase classical communication rates over a quantum
channel.
We first review a relation between the Holevo information and the minimum output
entropy of a tensor-product channel. Suppose that we have two channels N and M. The
Holevo information of the tensor-product channel is additive if
χ(N ⊗ M) = χ(N ) + χ(M).

(20.125)

Since the Holevo information is always superadditive for any two channels:χ(N ⊗ M) ≥ χ(N ) + χ(M)c 2019(20.126)

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20.5. SUPERADDITIVITY OF THE HOLEVO INFORMATION

579(recall the statement at the beginning of the proof of Theorem 13.3.1), we say that it isnon-additive if it is strictly superadditive:χ(N ⊗ M) > χ(N ) + χ(M).

(20.127)

The minimum output entropy H min (N ⊗ M) of the tensor-product channel is a quantity
related to the Holevo information (see Definition 20.4.1). It is additive if
H min (N ⊗ M) = H min (N ) + H min (M).

(20.128)

Since the minimum output entropy is always subadditive:
H min (N ⊗ M) ≤ H min (N ) + H min (M),(20.129)

we say that it is non-additive if it is strictly subadditive:
H min (N ⊗ M) < H min (N ) + H min (M).

(20.130)

Additivity of these two quantities is in fact related—it is possible to show that additivity of
the Holevo information implies additivity of the minimum output entropy and vice versa (weleave one of these implications as an exercise). Thus, researchers focused on additivity ofminimum output entropy rather than additivity of Holevo information because it is a simpler
quantity to manipulate.
Exercise 20.5.1 Prove that non-additivity of the minimum output entropy implies nonadditivity of the Holevo information:
H min (N1 ⊗ N2 ) < H min (N1 ) + H min (N2 )⇒ χ(N1 ⊗ N2 ) > χ(N1 ) + χ(N2 ). (20.131)(Hint: Consider an augmented version Ni0 of each channel Ni , that has its first input be thesame as the input to Ni and its second input be a control input, and the action of the channel
is equivalent to measuring the auxiliary input σ and applying a generalized Pauli operator:
X
X(k)Z(l)Ni (ρ)Z † (l)X † (k) hk|hl|σ|ki |li .
(20.132)
Ni0 (ρ ⊗ σ) ≡k,lWhat is the Holevo information of the augmented channel Ni0 ? What is the Holevo information of the tensor product of the augmented channels N10 ⊗ N20 ?) After proving the above
statement, we can also conclude that additivity of the Holevo information implies additivityof the minimum output entropy.
We briefly overview the main ideas behind the construction of a channel for which theHolevo information is not additive. Consider a random-unitary channel of the following
form:
D
X
E(ρ) ≡
pi Ui ρUi† ,(20.133)i=1
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CHAPTER 20. CLASSICAL COMMUNICATION

where the dimension of the input state is N and the number of random unitaries is D. This
channel is “random unitary” because it applies a particular unitary Ui with probability pi tothe state ρ. The cleverness behind the construction is not actually to provide a deterministic
instance of this channel, but rather, to provide a random instance of the channel where
both the distribution and the unitaries are chosen at random, and the dimension N and thenumber D of chosen unitaries satisfy the following relationships:
1  D  N.

(20.134)

The other channel to consider to disprove additivity is the conjugated channelE(ρ) ≡DX†

pi U i ρU i ,(20.135)i=1

where pi and Ui are the same respective probability distribution and unitaries from thechannel E, and here U i denotes the complex conjugate of Ui . The goal is then to show that
there is a non-zero probability over all channels of these forms that the minimum output
entropy is non-additive:
H min (E ⊗ E) < H min (E) + H min (E).

(20.136)

A good candidate for a state that could saturate the minimum output entropy H min (E ⊗E)of the tensor-product channel is the maximally entangled state |Φi, where
N −1

1 X
|Φi ≡ √
|ii|ii.
N i=0(20.137)

Consider the effect of the tensor-product channel E ⊗ E on the maximally entangled state Φ:(E ⊗ E)(Φ)
=

D
X

†

pi pj (Ui ⊗ U j )Φ(Ui† ⊗ U j )(20.138)

i,j=1

=

X

†

p2i (Ui ⊗ U i )Φ(Ui† ⊗ U i ) +

i=j

=

D
X
i=1

X

†

pi pj (Ui ⊗ U j )Φ(Ui† ⊗ U i )(20.139)

i6=j

!
p2i

Φ+

X

†

pi pj (Ui ⊗ U j )Φ(Ui† ⊗ U i ),(20.140)i6=jwhere the last line uses the fact that (M ⊗ I) |Φi = I ⊗ M T |Φi for any operator M (this
implies that U ⊗ U |Φi = |Φi). When comparing the above state to one resulting from
inputting a product state to the channel, there is a sense in which the above state is lessnoisy than the product state because D of the combinations of the random unitaries (theones which have the same index) have no effect on the maximally entangled state. Usingc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

20.6. CONCLUDING REMARKS

581

techniques from Hastings (2009), we can make this intuition precise and obtain the following
upper bound on the minimum output entropy:
H min (E ⊗ E) ≤ H((E ⊗ E)(Φ))
ln D≤ 2 ln D −,D(20.141)(20.142)for N and D large enough. However, using techniques in the same paper, we can also showthat
H min (E) ≥ ln D − δS max ,(20.143)
wherec+ poly(D)OδS max ≡
D

r

ln N
N

!
,

(20.144)c is a constant, and poly(D) indicates a term polynomial in D. Thus, for large enough D
and N , it follows thatln D,
(20.145)
2δS max <D
and we get the existence of a channel for which a violation of additivity occurs, becauseln D
D< 2 ln D − 2δS max

H min (E ⊗ E) ≤ 2 ln D −≤ H min (E) + H min (E).

(20.146)(20.147)(20.148)

20.6 Concluding Remarks
The HSW theorem offers a good characterization of the classical capacity of certain classesof channels, but at the same time, it also demonstrates our lack of understanding of classicaltransmission over general quantum channels. To be more precise, the Holevo information is a
useful characterization of the classical capacity of a quantum channel whenever it is additive,
but the regularized Holevo information is not particularly useful as a characterization of it
because we cannot even compute this quantity. This suggests that there could be some otherformula that better characterizes the classical capacity (if such a formula were additive). Asof the writing of this book, such a formula is unknown.
Despite the drawbacks of the HSW theorem, it is still interesting because it at least
offers a step beyond the most naive characterization of the classical capacity of a quantum
channel in terms of the regularized accessible information. The major insight of HSW was
the construction of an explicit POVM (corresponding to a random choice of code) thatallows the sender and receiver to communicate at a rate equal to the Holevo informationof the channel. This theorem is also useful for determining achievable rates in different
communication scenarios: for example, when two senders are trying to communicate over ac 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 20. CLASSICAL COMMUNICATION

noisy medium to a single receiver or when a single sender is trying to transmit both classicaland quantum information to a receiver.
The depolarizing channel is an example of a quantum channel for which there is a simple
expression for its classical capacity. Furthermore, the expression reveals that the scheme
needed to achieve the capacity of the channel is rather classical—it is only necessary for thesender to select codewords uniformly at random from some orthonormal basis, and it is only
necessary for the receiver to perform measurements of the individual channel outputs in thesame orthonormal basis. Thus, the coding scheme is classical because entanglement plays norole at the encoder and the decoding measurements act on the individual channel outputs.
Finally, we discussed Hastings’ construction of a quantum channel for which the heralded
additivity conjecture does not hold. That is, there exists a channel for which entanglement
at the encoder can improve communication rates. This superadditive effect is a uniquelyquantum phenomenon (recall that Theorem 13.1.1 states that the classical mutual information of a classical channel is additive, and thus correlations at the input cannot increase
capacity). This result implies that our best known characterization of the classical capacity of a quantum channel in terms of the channel’s Holevo information is far from being asatisfactory characterization of the true capacity, and we still have much more to discoverhere.

20.7 History and Further Reading
Holevo (1973a) was the first to prove the bound bearing his name, regarding the transmissionof classical information using a quantum channel, and Holevo (1998) and Schumacher andWestmoreland (1997) many years later proved that the Holevo information is an achievablerate for classical data transmission. Just prior to these works, Hausladen et al. (1996)
proved achievability of the Holevo information for the special case of a channel that accepts
a classical input and outputs a pure state conditioned on the input. They also published apreliminary article (Hausladen et al., 1995) in which they answered the catchy question (forthe special case of pure states), “How many bits can you fit into a quantum-mechanical it?”
Shor (2002a) established the additivity of the Holevo information for entanglementbreaking channels. Bowen and Nagarajan (2005) proved that the classical capacity of anentanglement-breaking channel is not enhanced by a classical feedback channel from receiverto sender. King (2002) first proved additivity of the Holevo information for unital qubit
channels and later showed it for the depolarizing channel (King, 2003). Shor (2004a) later
showed the equivalence of several additivity conjectures (that they are either all true or all
false). Hayden (2007), Winter (2007), and a joint paper between them (Hayden and Winter,2008) proved some results leading up to the work of Hastings (2009), who demonstrated a
counterexample to the additivity conjecture. Thus, by Shor’s aforementioned paper, all of
the additivity conjectures are false in general. There has been much follow-up work in an
attempt to understand Hastings’ result (Brandao and Horodecki, 2010; Fukuda and King,
2010; Fukuda et al., 2010; Aubrun et al., 2011).
In a landmark result, Giovannetti et al. (2015) established the classical capacity of all
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20.7. HISTORY AND FURTHER READING

583

phase-insensitive quantum Gaussian channels, building on a long series of works startingwith that of Holevo and Werner (2001).
Some other papers have tried to understand the HSW coding theorem from the perspective of hypothesis testing. Hayashi and Nagaoka (2003) began much of this work, andHayashi (2006) covers quite a bit of quantum hypothesis testing in his book. Mosonyi andDatta (2009) followed up with some work along these lines, as did Wang and Renner (2012)
and Wilde (2013).
Researchers have invested quite a bit of effort in refining the HSW theorem, with regard
to error exponents, strong converse exponents and second-order characterizations. From theproof of the HSW theorem given in this chapter, we see that if we pick the rate R of classicalcommunication to be smaller than the capacity by an additive constant, then it is possible
to make the decoding error decay exponentially fast to zero with an increasing number n ofchannel uses. The optimal exponential decay rate of the error for a given communication rateis known as the error exponent. Burnashev and Holevo (1998) derived a lower bound on the
optimal error exponent for pure-state classical–quantum channels, and Holevo (2000) thenderived a lower bound on the error exponent for mixed-state classical–quantum channels.
Hayashi (2007) later improved upon this result for mixed-state classical–quantum channels.
Dalai (2013) derived an upper bound on the optimal error exponent of classical–quantum
channels, called the sphere-packing bound.
One can also ask about the behavior of the error probability when the communication rate
exceeds the capacity by an additive constant. This regime is known as the strong converse
regime, and a channel obeys the strong converse property if the error probability tends to one
when the communication rate exceeds the capacity. Winter (1999a) and Ogawa and Nagaoka(1999) proved the strong converse for the classical capacity of classical–quantum channels,
Koenig and Wehner (2009) for channels with certain symmetry, Wilde et al. (2014) for
entanglement-breaking and Hadamard channels, Bardhan et al. (2015) for phase-insensitive
quantum Gaussian channels, and Ding and Wilde (2015) for entanglement-breaking channels
assisted by a noiseless classical feedback channel.
A second-order characterization asks how fast the communication rate can converge tocapacity if the error probability is fixed to be a constant. One of the main tools here is the
Berry–Esseen refinement of the central limit theorem (see, e.g., Feller (1971); Tyurin (2010)).
Tomamichel and Tan (2015) derived an optimal second-order characterization for classical–
quantum channels (even going beyond and establishing it for “image-additive” channels).
Wilde et al. (2016) characterized an achievable second-order strategy for the pure-loss bosonicchannel (with a converse part remaining open).

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584

c 2019CHAPTER 20. CLASSICAL COMMUNICATIONMark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 21

Entanglement-Assisted ClassicalCommunication
We have learned that shared entanglement is often helpful in quantum communication. Thisis certainly true for the case of a noiseless qubit channel. Without shared entanglement, themost classical information that a sender can reliably transmit over a noiseless qubit channel
is just one classical bit (recall Exercise 4.2.2 and the Holevo bound in Exercise 11.9.2).
With shared entanglement, they can achieve the super-dense coding resource inequality from
Chapter 7:[q → q] + [qq] ≥ 2 [c → c] .
(21.1)
That is, using one noiseless qubit channel and one shared noiseless ebit, the sender canreliably transmit two classical bits.
A natural question then for us to consider is whether shared entanglement could be helpfulin transmitting classical information over a noisy quantum channel N . As a first simplifying
assumption, we let Alice and Bob have access to an infinite supply of entanglement, inwhatever form they wish, and we would like to know how much classical information Alicecan reliably transmit to Bob over such an entanglement-assisted quantum channel. That is,
we would like to determine the highest achievable rate C of classical communication in thefollowing resource inequality:
hN i + ∞ [qq] ≥ C [c → c] .

(21.2)

The answer to this question is one of the strongest known results in quantum Shannon theory, and it is given by the entanglement-assisted classical capacity theorem. This
theorem states that the mutual information I(N ) of a quantum channel N is equal to its
entanglement-assisted classical capacity, whereI(N ) ≡ max I(A; B)ρ ,ϕAA0(21.3)

ρAB ≡ NA0 →B (ϕAA0 ), and the maximization is with respect to all pure bipartite states ofthe form ϕAA0 . We should stress that there is no need to regularize this formula in order
585

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CHAPTER 21. ENTANGLEMENT-ASSISTED CLASSICAL COMMUNICATION

to characterize the capacity (as done in the previous chapter and as is so often neededin quantum Shannon theory). The value of this formula is equal to the capacity. Also, theoptimization task that the formula in (21.3) sets out is a straightforward convex optimization
program. Any local maximum is a global maximum because the quantum mutual informationis concave with respect to the input state ϕA0 (recall Theorem 13.4.2 from Chapter 13) andthe set of density operators is convex.
From the perspective of an information theorist, we should only say that a capacity
theorem has been solved if there is a tractable formula equal to the optimal rate for achievinga particular operational task. The formula should apply to an arbitrary quantum channel,
and it should be a function of that channel. Otherwise, the capacity theorem is still unsolved.
There are several operative words in the above sentences that we should explain in moredetail. The formula should be tractable, meaning that it sets out an optimization task whichis efficient to solve in the dimension of the channel’s input system. The formula should givethe optimal achievable rate for the given information-processing task, meaning that if a rate
exceeds the capacity of the channel, then the probability of error for any such protocol should
be bounded away from zero as the number of channel uses grows large.1 Finally, perhaps themost stringent (though related) criterion is that the formula itself (and not its regularization)
should give the capacity of an arbitrary quantum channel. Despite the success of the HSW
coding theorem in demonstrating that the Holevo information of a channel is an achievablerate for classical communication, the classical capacity of a quantum channel is still unsolved
because there is an example of a channel for which the Holevo information is not equal tothat channel’s capacity (see Section 20.5). Thus, it is rather impressive that the formula in(21.3) is equal to the entanglement-assisted classical capacity of an arbitrary channel, given
the stringent requirements that we have set out for a formula to give the capacity. In thissense, shared entanglement simplifies quantum Shannon theory.
This chapter presents a comprehensive study of the entanglement-assisted classical capacity theorem. We begin by defining the information-processing task, consisting of all thesteps in a general protocol for classical communication over an entanglement-assisted quantum channel. We then present a simple example of a strategy for entanglement-assistedclassical coding that is inspired by super-dense coding, and in turn, that inspires a strategy
for the general case. Section 21.3 states the entanglement-assisted classical capacity theorem.
Section 21.4 gives a proof of the direct coding theorem, making use of strong quantum typicality from Chapter 15 and the packing lemma from Chapter 16. It demonstrates that therate in (21.3) is an achievable rate for entanglement-assisted classical communication. Aftertaking a step back from the protocol, we can realize that it is merely a glorified super-dense
coding applied to noisy quantum channels. Section 21.5 gives a proof of the converse of theentanglement-assisted classical capacity theorem. It exploits familiar tools such as the AFW
inequality, the quantum data-processing inequality, and the chain rule for quantum mutual
information (all from Chapter 11), and the last part of it exploits additivity of the mutual1

We could strengthen this requirement even more by demanding that the probability of error increases
exponentially to one in the asymptotic limit. Fulfilling such a demand would constitute a proof of a strong
converse theorem.
c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License21.1. THE INFORMATION-PROCESSING TASK

587

information of a quantum channel (from Chapter 13). The converse theorem establishes
that the rate in (21.3) is optimal. With the proof of the capacity theorem complete, we thenshow the interesting result that the classical capacity of a quantum channel assisted by aquantum feedback channel is equal to the entanglement-assisted classical capacity of thatchannel. We close the chapter by computing the entanglement-assisted classical capacityof both a quantum erasure channel and an amplitude damping channel, and we leave thecomputation of the entanglement-assisted capacities of two other channels as exercises.

21.1 The Information-Processing Task
We begin by explicitly defining the information-processing task of entanglement-assistedclassical communication. That is, we define an (n, C, ε) entanglement-assisted classical code
and what it means for a rate C to be achievable. Prior to the start of the protocol, we
assume that Alice and Bob share pure-state entanglement in whatever form they wish. Let
ΨTA TB denote this state. Alice selects some message m from a set M of messages. Let M
denote the random variable corresponding to Alice’s choice of message, and let |M| be thecardinality of the set M. She applies some encoding channel ETmA →A0n to her share of theentangled state ΨTA TB depending on her choice of message m. The global state then becomes
ETmA →A0n (ΨTA TB ).

(21.4)

Alice transmits the systems A0n over n independent uses of a noisy channel NA0 →B , leading
to the following state:
NA0n →B n (ETmA →A0n (ΨTA TB )),(21.5)
where NA0n →B n ≡ (NA0 →B )⊗n . Bob receives the systems B n , combines them with his share TBnof the entanglement, and performs a POVM {Λm
B n TB } on the channel outputs B and his share
TB of the entanglement in order to detect the message m that Alice transmits. Figure 21.1
depicts such a general protocol for entanglement-assisted classical communication.
Let M 0 denote the random variable corresponding to the output of Bob’s decoding POVM
(this represents Bob’s estimate of the message). The probability of Bob correctly decoding
Alice’s message ismPr{M 0 = m|M = m} = Tr{Λm
B n TB NA0n →B n (ETA →A0n (ΨTA TB ))},and thus the probability of error pe (m) for message m ismpe (m) ≡ Tr{ I − Λm
B n TB NA0n →B n (ETA →A0n (ΨTA TB ))}.

(21.6)(21.7)The maximal probability of error p∗e for the coding scheme isp∗e ≡ max pe (m).

(21.8)

1
log |M| ,n(21.9)m∈M

The rate C of communication isC≡
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CHAPTER 21. ENTANGLEMENT-ASSISTED CLASSICAL COMMUNICATION

Figure 21.1: The most general protocol for entanglement-assisted classical communication. Alice applies
an encoder to her classical message M and her share TA of the entanglement, and she inputs the encoded
systems A0n to many uses of the channel. Bob receives the outputs of the channels, combines them with hisshare of the entanglement, and performs some decoding operation to estimate Alice’s transmitted message.

and the code has ε error if p∗e ≤ ε.
A rate C of entanglement-assisted classical communication is achievable if there exists
an (n, C − δ, ε) entanglement-assisted classical code for all ε ∈ (0, 1), δ > 0, and sufficiently
large n. The entanglement-assisted classical capacity CEA (N ) of a quantum channel N isequal to the supremum of all achievable rates.

21.2 A Preliminary Example
Let us first recall
P a few items about qudits. The maximally entangled qudit state is defined as|ΦiAB ≡ d−1/2 d−1
i=0 |iiA |iiB . Recall from Section 3.7.2 that the Heisenberg–Weyl operators
X(x) and Z(z) are an extension of the Pauli matrices to d dimensions:X(x) ≡

d−1
X

0

0

|x + x ihx |,Z(z) ≡

x0 =0

d−1
X

0

e2πizz /d |z 0 ihz 0 |.

(21.10)z 0 =0Let |Φx,z iAB denote the state that results when Alice applies the operator X(x)Z(z) to hershare of the maximally entangled state |ΦiAB :|Φx,z iAB ≡ (XA (x)ZA (z) ⊗ IB ) |ΦiAB .

(21.11)

Recall from Exercise 3.7.11 that the set of states {|Φx,z iAB }d−1x,z=0 forms a complete orthonormal basis:hΦx1 ,z1 |Φx2 ,z2 iAB = δx1 ,x2 δz1 ,z2 ,d−1X|Φx,z ihΦx,z |AB = IAB .

(21.12)x,z=0
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21.2. A PRELIMINARY EXAMPLE

589

Figure 21.2: A simple scheme, inspired by super-dense coding, for Alice and Bob to exploit shared entanglement and a noisy channel in order to establish an ensemble at Bob’s receiving end.

Let πAB denote the maximally mixed state on Alice and Bob’s system: πAB ≡ IAB /d2 , andlet πA and πB denote the respective maximally mixed states on Alice and Bob’s systems:
πA ≡ IA /d and πB ≡ IB /d. Observe that πAB = πA ⊗ πB .
We now consider a simple strategy, inspired by super-dense coding and the HSW coding
scheme from Theorem 20.3.1, that Alice and Bob can employ for entanglement-assistedclassical communication. That is, we show how a strategy similar to super-dense coding
induces a particular ensemble at Bob’s receiving end, to which we can then apply the HSW
coding theorem in order to establish the existence of a good code for entanglement-assistedclassical communication. Suppose that Alice and Bob possess a maximally entangled qudit
state |ΦiAB . Alice chooses two symbols x and z uniformly at random, each in {0, . . . , d − 1}.
She applies the operators X(x)Z(z) to her share of the maximally entangled state |ΦiAB ,and the resulting state is |Φx,z iAB . She then sends her system A over the noisy channel
NA→B 0 , and Bob receives the output B 0 from the channel. The noisy channel on the wholesystem is NA→B 0 ⊗ idB , and the ensemble that Bob receives is as follows: −2d , (NA→B 0 ⊗ idB ) (Φx,zAB ) .

(21.13)This constitutes an ensemble that they can prepare with one use of the channel and one
shared entangled state (Figure 21.2 depicts all of these steps). But, in general, we allow
them to exploit many uses of the channel and however much entanglement that they need.
Bob can then perform a collective measurement on both his share of the entanglement andthe channel outputs in order to determine a message that Alice is transmitting.
Consider that the above scenario is similar to HSW coding. Theorem 20.3.1 from theprevious chapter states that the Holevo information of the above ensemble is an achievablerate for classical communication over this entanglement-assisted quantum channel. Thus, we
can already state and prove the following corollary of Theorem 20.3.1, simply by calculating
the Holevo information of the ensemble in (21.13).
Corollary 21.2.1 The quantum mutual information I(A; B)σ of the state σAB ≡ NA0 →B (ΦAA0 )is an achievable rate for entanglement-assisted classical communication over a quantum channel NA0 →B .
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Proof. Observe that the ensemble in (21.13) maps to the following classical–quantum state:
ρXZB 0 B ≡

d−1
X
1
x,z=0

d2

|xihx|X ⊗ |zihz|Z ⊗ (NA→B 0 ⊗ idB ) (Φx,zAB ).

(21.14)

The Holevo information of this classical–quantum state is I(XZ; B 0 B)ρ = H(B 0 B)ρ −H(B 0 B|XZ)ρ , and it is an achievable rate for entanglement-assisted classical communication
over the channel NA0 →B by the direct part of Theorem 20.3.1. We now proceed to calculate
it. First, we determine the entropy H(B 0 B)ρ by tracing over the classical registers XZ:
TrXZ {ρXZB 0 B } =

d−1
X
1

d2
x,z=0

(NA→B 0 ⊗ idB ) (Φx,zAB )= (NA→B 0 ⊗ idB )

d−1
X
1

d2
x,z=0

(21.15)!
Φx,zAB= (NA→B 0 ⊗ idB ) (πAB )= NA→B 0 (πA ) ⊗ πB ,(21.16)(21.17)(21.18)where the third equality follows from (21.12). Thus, the entropy H(B 0 B) is as follows:H(B 0 B) = H(NA→B 0 (πA )) + H(πB ).

(21.19)We now determine the conditional quantum entropy H(B 0 B|XZ)ρ :H(B 0 B|XZ)ρ
=

d−1
X
1

d2
x,z=0

H ((NA→B 0 ⊗ idB ) (Φx,z
AB ))

d−1
h

i
1 X 
= 2
H NA→B 0 (XA (x)ZA (z)) (ΦAB ) ZA† (z)XA† (x)d x,z=0(21.20)(21.21)

d−1

 T
1 X
T
∗
∗
0 Z (z)X (x)(ΦAB )X (x)Z (z)
H
N
A→B
B
B
B
B
d2 x,z=0(21.22)

d−1

1 X
= 2
H ZBT (z)XBT (x) [(NA→B 0 ) (ΦAB )] (XB∗ (x)ZB∗ (z))
d x,z=0(21.23)= H(NA→B 0 (ΦAB )).

(21.24)=

The first equality follows because the system XZ is classical (recall the result in Section 11.4.1). The second equality follows from the definition of the state Φx,z
AB . The third
equality follows by exploiting the “transpose trick” in Exercise 3.7.12. The fourth equality
follows because the transposed unitaries acting on Bob’s system commute with the action
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21.3. ENTANGLEMENT-ASSISTED CAPACITY THEOREM

591

of the channel. Finally, the entropy of a state is invariant with respect to unitaries. So theHolevo information I(XZ; B 0 B)ρ of the state ρXZB 0 B in (21.14) is equal toI(XZ; B 0 B)ρ = H(N (πA )) + H(πB ) − H((NA→B 0 ⊗ idB ) (ΦAB )).

(21.25)Equivalently, we can write it as the quantum mutual information I(A; B)σ , evaluated withrespect to the state σAB ≡ NA0 →B (ΦAA0 ).
For some channels, the quantum mutual information in Corollary 21.2.1 is equal to thatchannel’s entanglement-assisted classical capacity. This occurs for the depolarizing channel,
a dephasing channel, and an erasure channel to name a few. But there are examples ofchannels, such as the amplitude damping channel, where the quantum mutual informationin Corollary 21.2.1 is not equal to the entanglement-assisted capacity. In the general case,it might perhaps be intuitive that the quantum mutual information of the channel in (21.3)
is equal to the entanglement-assisted capacity of the channel, and it is the goal of the next
sections to prove this result.
Exercise 21.2.1 Consider the following strategy for transmitting and detecting classical information over an entanglement-assisted depolarizing channel. Alice selects a state |Φx1 ,z1 iABD
uniformly at random and sends the A system over the quantum depolarizing channel NA→B
0,
where
D
NA→B
(21.26)0 (ρ) ≡ (1 − p)ρ + pπ.
Bob receives the output B 0 of the channel and combines it with his share B of the entanglement. He then performs a measurement of these systems in the Bell basis {|Φx2 ,z2 ihΦx2 ,z2 |B 0 B }.
Determine a simplified expression for the induced classical channel pZ2 X2 |Z1 X1 (z2 , x2 |z1 , x1 )
where

D
x1 ,z1pZ2 X2 |Z1 X1 (z2 , x2 |z1 , x1 ) ≡ hΦx2 ,z2 | NA→B
ihΦx1 ,z1 |AB ) |Φx2 ,z2 i .
(21.27)0 ⊗ idB (|ΦShow that the classical capacity of the channel pZ2 X2 |Z1 X1 (z2 , x2 |z1 , x1 ) is equal to the entanglementassisted classical capacity of the depolarizing channel (you can take it for granted that theentanglement-assisted classical capacity of the depolarizing channel is given by Corollary 21.2.1).
Thus, there is no need for the receiver to perform a collective measurement on many channel
outputs in order to achieve capacity—it suffices to perform single-channel Bell measurements
at the receiving end.

21.3 Entanglement-Assisted Capacity Theorem
We now state the entanglement-assisted classical capacity theorem. Section 21.4 proves the
direct part of this theorem, and Section 21.5 proves its converse part.
Theorem 21.3.1 (Bennett–Shor–Smolin–Thapliyal) The entanglement-assisted classical capacity of a quantum channel is equal to the channel’s mutual information:
CEA (N ) = I(N ),(21.28)where the mutual information I(N ) of a channel N is defined as I(N ) ≡ maxϕAA0 I(A; B)ρ ,ρAB ≡ NA0 →B (ϕAA0 ), and ϕAA0 is a pure bipartite state.
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CHAPTER 21. ENTANGLEMENT-ASSISTED CLASSICAL COMMUNICATION

21.4 The Direct Coding Theorem
The direct coding theorem is a statement of achievability:
Theorem 21.4.1 (Direct Coding) The following resource inequality corresponds to an
achievable protocol for entanglement-assisted classical communication over a quantum channel NA0 →B :hN i + H(A)ρ [qq] ≥ I(A; B)ρ [c → c] ,(21.29)where ρAB ≡ NA0 →B (ϕAA0 ).
Proof. We suppose that Alice and Bob share n copies of an arbitrary pure, bipartite
entangled state |ϕiAB . This is allowed in the setting of entanglement-assisted communication as discussed in Section 21.1. Alternatively, they could convert nH(A) shared ebits to≈ n copies of |ϕiAB by making use of the entanglement dilution protocol discussed in Section 19.4.2 (they would need a sublinear amount of classical communication to do so, which
has negligible rate). We would like to apply a similar coding technique as outlined in Section 21.2. For example, it would be useful to exploit the transpose trick from Exercise 3.7.12,
but we cannot do so directly because this trick only applies to maximally entangled states.
However, we can instead exploit the fact that Alice and Bob share many copies of the state|ϕiAB that decompose into a direct sum of maximally entangled states. First, recall thatevery pure, bipartite state has a Schmidt decomposition (see Theorem 3.8.1):|ϕiAB ≡Xp
pX (x)|xiA |xiB ,(21.30)xPwhere pX (x) > 0 for all x, x pX (x) = 1, and {|xiA } and {|xiB } are orthonormal bases for
Alice and Bob’s respective systems. Let us take n copies of the above state, giving a stateof the following form:
Xp|ϕiAn B n ≡pX n (xn )|xn iAn |xn iB n ,(21.31)xn

where xn ≡ x1 · · · xn , pX n (xn ) ≡ pX (x1 ) · · · pX (xn ), and |xn i ≡ |x1 i · · · |xn i. We can write
the above state in terms of its type decomposition (see Section 14.7.1):|ϕiAn B n =X X p
pX n (xn )|xn iAn |xn iB n
t

=

Xp
t

=

Xp
t

=

(21.32)xn ∈TtXp

pX n (xnt )X|xn iAn |xn iB n(21.33)xn ∈Tt

1 X n
pX n (xnt )dt √|x iAn |xn iB n
dt xn ∈T

(21.34)p(t)|Φt iAn B n ,(21.35)tt

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21.4. THE DIRECT CODING THEOREM

593

with the following definitions:p(t) ≡ pX n (xnt )dt ,1 X n|Φt iAn B n ≡ √|x iAn |xn iB n .
dt xn ∈T

(21.36)(21.37)t

The first equality in (21.32) follows by decomposing the state into its different type classsubspaces. The next equality follows because pX n (xn ) is the same for all sequences xn in thesame type class and because the distribution is i.i.d. (let xnt be some representative sequenceof all sequences in the type class Tt ). The third equality follows by introducing dt as thedimension of a type class subspace Tt , and the final equality in (21.35) follows from thedefinitions. Observe that the state |Φt iAn B n is maximally entangled.
Each state |Φt iAn B n is maximally entangled with Schmidt rank dt , and we can thus apply
the transpose trick for operators acting on the type class subspaces. Inspired by the densecoding-like strategy from Section 21.2, we allow Alice to choose unitary operators from theHeisenberg–Weyl set of d2t operators that act on the An share of |Φt iAn B n . We denote oneof these operators as V (xt , zt ) ≡ X(xt )Z(zt ) where xt , zt ∈ {0, . . . , dt − 1}. If she does thisfor every type class subspace and applies a phase (−1)bt in each subspace, then the resulting
unitary operator U (s) acting on all of her An systems is a direct sum of all of these unitaries:
M
U (s) ≡(−1)bt V (xt , zt ),(21.38)twhere s is a vector containing all of the indices needed to specify the unitary U (s):s ≡ ((xt , zt , bt ))t .

(21.39)Let S denote the set of all possible vectors s. The transpose trick holds for these particular
unitary operators:
(UAn (s) ⊗ IB n ) |ϕiAn B n = IAn ⊗ UBT n (s) |ϕiAn B n(21.40)because it applies in each type class subspace:(UAn (s) ⊗ IB n ) |ϕiAn B n
!
M
Xp
bt
p(t)|Φt iAn B n=(−1) VAn (xt , zt )t=

(21.41)t

Xpp(t) (−1)bt VAn (xt , zt )|Φt iAn B n(21.42)t=Xpp(t) (−1)bt VBTn (xt , zt )|Φt iAn B n(21.43)t!
=M(−1)bt VBTn (xt , zt )t
= IAn ⊗ UBT n (s) |ϕiAn B n .
c 2019Xp
p(t)|Φt iAn B n(21.44)t(21.45)

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CHAPTER 21. ENTANGLEMENT-ASSISTED CLASSICAL COMMUNICATION

Now we need to establish a means by which Alice can select a random code. For every
message m ∈ M that Alice would like to transmit, she chooses the elements of the vectors ∈ S uniformly at random, leading to a particular unitary operator U (s). We can writes(m) instead of just s to denote the explicit association of the vector s with the message m—
we can think of each chosen vector s(m) as a classical codeword, with the codebook being{s(m)}m∈{1,...,|M|} . This random selection procedure leads to entanglement-assisted quantum
codewords of the following form:|ϕm iAn B n ≡ (UAn (s(m)) ⊗ IB n ) |ϕiAn B n .

(21.46)

Alice then transmits her systems An through many uses of the quantum channel NA→B 0 ,leading to the following state that is entirely in Bob’s control:
NAn →B 0n (|ϕm ihϕm |An B n ).

(21.47)

Interestingly, the above state is equal to the state in (21.50) below, by exploiting the transpose trick from (21.40):
NAn →B 0n (|ϕm ihϕm |An B n )= NAn →B 0n (UAn (s(m))|ϕihϕ|An B n UA† n (s(m)))(21.48)= NAn →B 0n (UBT n (s(m))|ϕihϕ|An B n UB∗ n (s(m)))(21.49)= UBT n (s(m))NAn →B 0n (|ϕihϕ|An B n )UB∗ n (s(m)).

(21.50)Observe that the transpose trick allows us to commute the action of the channel with Alice’s
encoding unitary U (s(m)), just as we did in the example in the proof of Corollary 21.2.1.
LetρB 0n B n ≡ NAn →B 0n (|ϕihϕ|An B n )(21.51)so thatNAn →B 0n (|ϕm ihϕm |An B n ) = UBT n (s(m))ρB 0n B n UB∗ n (s(m)).

(21.52)

Remark 21.4.1 (Tensor-Power Channel Output States) When using the coding scheme
given above, the reduced state on the channel output (obtained by ignoring Bob’s share of theentanglement in B n ) is the same tensor-power state, independent of the unitary that Aliceapplies at the channel input:
TrB n {NAn →B 0n (|ϕm ihϕm |An B n )} = ρB 0n= NAn →B 0n (ϕAn ),(21.53)(21.54)

where ϕAn = (TrB {ϕAB })⊗n . This follows directly from (21.52) and taking the partial traceover B n . We exploit this feature in the next chapter, where we construct codes for transmitting both classical and quantum information with the help of shared entanglement.
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21.4. THE DIRECT CODING THEOREM

595

⊗n

Figure 21.3: (a) Alice shares many copies of a pure, bipartite state |ϕiwith Bob. She encodes amessage m according to some unitary of the form in (21.38). She transmits her share of the entanglementassisted quantum codeword over many uses of the quantum channel, and it is Bob’s task to determine which
message she transmits. (b) Alice acting locally with the unitary U (s(m)) on her share An of the entanglement⊗n|ϕiis the same as her acting non-locally with U T (s(m)) on Bob’s share B n of the entanglement. This
follows because of the particular structure of the unitaries in (21.38).

After Alice has transmitted her entanglement-assisted quantum codewords over the channel, it becomes Bob’s task to determine which message m Alice transmitted, and he should
do so with some POVM {Λm } that depends on the random choice of code. Figure 21.3
depicts the protocol.
At this point, we would like to exploit the packing lemma from Chapter 16 in order toestablish the existence of a reliable decoding POVM for Bob. Recall that the packing lemma
requires four objects, and these four objects should satisfy the four inequalities in (16.11)–(16.14). The first object required is an ensemble from which Alice and Bob can select a code
randomly, and in our case, the ensemble is


1
T
∗
, UB n (s)ρB 0n B n UB n (s).
(21.55)|S|s∈S
The next object required is the expected density operator of this ensemble:
ρB 0n B n ≡ ES U T (S)B n ρB 0n B n U ∗ (S)B n
1 X T
U n (s)ρB 0n B n UB∗ n (s).
=|S| s∈S B(21.56)(21.57)

We later prove that this expected density operator has the following simpler form:X(21.58)
ρB 0n B n =p(t) NAn →B 0n (πAt n ) ⊗ πBt n ,twhere p(t) is the distribution from (21.36) and πAt n is the maximally mixed state on a typeclass subspace: πAt n ≡ It /dt . The final two objects that we require for the packing lemma
are the message subspace projectors and the total subspace projector. We assign theserespectively asc 2019∗UBT n (s)Πρ,δ
B 0n B n UB n (s),(21.59)ρ,δ
Πρ,δ
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596

CHAPTER 21. ENTANGLEMENT-ASSISTED CLASSICAL COMMUNICATION

ρ,δρ,δ
where Πρ,δB 0n B n , ΠB 0n , and ΠB n are the typical projectors for many copies of the states ρB 0 B ≡NA→B 0 (ϕAB ), ρB 0 = TrB {ρB 0 B }, and ρB = TrB 0 {ρB 0 B }, respectively. Observe that the0
size of each message subspace projector is ≈ 2nH(B B) , and the size of the total subspace0
projector is ≈ 2n[H(B )+H(B)] . By dimension counting, this is suggesting that we can pack in000≈ 2n[H(B )+H(B)] /2nH(B B) = 2nI(B ;B) messages using this coding technique.
If the four conditions of the packing lemma are satisfied (see (16.11)–(16.14)), then there
exists a detection POVM that can reliably decode Alice’s transmitted messages as long as
the number of messages in the code is not too high. The four conditions in (16.11)–(16.14)
translate to the following four conditions for our case:
n

o
ρ,δ
∗
T
0n
n
(s)
≥ 1 − ε,(21.61)(s)ρ
UTr Πρ,δ
⊗
Π
U
B B
Bn
Bn
Bn
B 0n
n

o

∗
T
∗
0n B n U n (s)Tr UBT n (s)Πρ,δ
U
(s)
U
(s)ρ
≥ 1 − ε,(21.62)
n
n
0n
n
B
B
B
B
B B
o
n
n[H(B 0 B)ρ +cδ]∗,
(21.63)
Tr UBT n (s)Πρ,δ
B 0n B n UB n (s) ≤ 2






ρ,δ
ρ,δ
ρ,δ
Πρ,δ
⊗
Π
Π
ρ
⊗
Π
0n
n
n
n
0n
0n
B B
B
B
B
B


0
ρ,δ
≤ 2−n[H(B )ρ +H(B)ρ −η(n,δ)−cδ] Πρ,δ⊗ΠB n , (21.64)
B 0nwhere c is some positive constant and η(n, δ) is a function that approaches zero as n → ∞and δ → 0.
We now prove the four inequalities in (21.61)–(21.64), attacking them in the order ofincreasing difficulty. The condition in (21.62) holds because
n

o
ρ,δ
T
∗
T
∗
Tr UB n (s)ΠB 0n B n UB n (s) UB n (s)ρB 0n B n UB n (s)no0nn= Tr Πρ,δρ(21.65)
0n
n
B B
B B
≥ 1 − ε.

(21.66)The equality holds by cyclicity of the trace and because U ∗ U T = I. The inequality holds by
exploiting the unit probability property of typical projectors (Property 15.1.1). From thisinequality, observe that we choose each message subspace projector so that it is exactly the
one that should identify the entanglement-assisted quantum codeword UBT n (s)ρB 0n B n UB∗ n (s)with high probability.
We next consider the condition in (21.63):
o
n
o
n
ρ,δ
∗
U
(s)
=
Tr
Π
(21.67)
Tr UBT n (s)Πρ,δ
n
B 0n B n
B 0n B n B0≤ 2n[H(B B)ρ +cδ] .

(21.68)The equality holds again by cyclicity of trace, and the inequality follows from the “exponentially smaller cardinality” property of the typical subspace (Property 15.1.2).
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21.4. THE DIRECT CODING THEOREM

597

Consider the condition in (21.61). First, define P̂ = I − P . Then
ρ,δ
ρ,δ
ρ,δ
Πρ,δ
B 0n ⊗ ΠB n = (I − Π̂B 0n ) ⊗ (I − Π̂B n )(21.69)= (IB 0n ⊗ IB n ) − (Π̂ρ,δ
B 0n ⊗ IB n )ρ,δρ,δ
− (IB 0n ⊗ Π̂B n ) + (Π̂ρ,δ
B 0n ⊗ Π̂B n )ρ,δ
≥ (IB 0n ⊗ IB n ) − (Π̂ρ,δ
B 0n ⊗ IB n ) − (IB 0n ⊗ Π̂B n ).

(21.70)(21.71)

Consider the following chain of inequalities:
n
o
ρ,δ
ρ,δ
T
∗
Tr (ΠB 0n ⊗ ΠB n ) UB n (s)ρB 0n B n UB n (s)≥ Tr UBT n (s)ρB 0n B n UB∗ n (s)noT∗n ) U n (s)ρB 0n B n U n (s)− Tr (Π̂ρ,δ
⊗
I
0n
B
B
B
B
n
o
T
∗
0n
n
− Tr (IB 0n ⊗ Π̂ρ,δ
)
U
(s)ρ
U
(s)
n
n
n
B B
B
B
B
n
o
n
o
ρ,δ
= 1 − Tr Π̂ρ,δ
B 0n ρB 0n − Tr Π̂B n ρB n

(21.73)≥ 1 − 2ε.

(21.74)(21.72)

The first inequality follows from the development in (21.69)–(21.71). The first equality
follows because Tr UBT n (s)ρB 0n B n UB∗ n (s) = 1 and from performing a partial trace on B nand B 0n , respectively (while noting that we can apply the transpose trick for the second one).
The final inequality follows from the unit probability property of the typical projectors Πρ,δB 0nand Πρ,δ
(Property
15.1.1).
n
B
The last inequality in (21.64) requires the most effort to prove. We first need to provethat the expected density operator ρB 0n B n takes the form given in (21.58). To simplify thedevelopment, we evaluate the expectation without the channel applied, and we then apply
the channel to the state at the end of the development. Consider thatρAn B n =1 X T
UB n (s)|ϕihϕ|An B n UB∗ n (s)|S| s∈SXp
1 X T=
UB n (s)p(t)|Φt iAn B n|S| s∈St(21.75)!
X!
phΦt0 |An B n p(t0 ) UB∗ n (s)
1 X Xp=p(t) (−1)bt (s) VBTn ((zt , xt )(s)) |Φt iAn B n|S| s∈St
X(21.76)t0!

!
phΦt0 |An B n (−1)bt0 (s) (VB∗n ((zt0 , xt0 )(s))) p(t0 ) .

(21.77)t0

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598

CHAPTER 21. ENTANGLEMENT-ASSISTED CLASSICAL COMMUNICATIONLet us first consider the case when t = t0 . Then the expression in (21.77) becomes
1 XX
p(t) VBTn ((zt , xt )(s)) |Φt ihΦt |An B n (VB∗n ((zt , xt )(s)))|S| s∈S t
"
#
X
1 X T
=
p(t) 2
VB n (zt , xt )|Φt ihΦt |An B n VB∗n (zt , xt )dt xt ,zttX=p(t)πAt n ⊗ πBt n .

(21.78)(21.79)(21.80)t

These equalities hold because the sum over all the elements in S implies that we are uniformly mixing the maximally entangled states |Φt iAn B n on the type class subspaces andExercise 4.7.6 gives us that the resulting state on each type class subspace is equal to
TrB n {[Φt ]An B n } ⊗ πBt n = πAt n ⊗ πBt n . Let us now consider the case when t 6= t0 . Thenthe expression in (21.77) becomes
1 X X pp(t)p(t0 ) (−1)bt (s)+bt0 (s) ×|S| s∈S t0 ,t6=t0

VBTn ((zt , xt )(s)) |Φt ihΦt0 |An B n (VB∗n ((zt0 , xt0 )(s)))
X
X 1
p
p(t)p(t0 ) (−1)bt +bt0 ×=
2 2
d d 0 4 b ,b ,x ,z ,x ,zt0 ,t6=t0 t t
t0

t

=

t

t

t0

t0

VBTn (zt , xt )|Φt ihΦt0 |An B n VB∗n (zt0 , xt0 )bt +bt0XX
1

d2 d20t0 ,t6=t0 t t bt ,bt0(−1)
4

(21.81)

×




X



p
p(t)p(t0 )VBTn (zt , xt )|Φt ihΦt0 |An B n VB∗n (zt0 , xt0 )

(21.82)

xt ,zt ,xt0 ,zt0

= 0.

(21.83)It then follows thatX
1 X T
p(t)πAt n ⊗ πBt n ,UB n (s)|ϕihϕ|An B n UB∗ n (s) =|S| s∈St(21.84)and by linearity, that
1 X T
U n (s)NAn →B 0n (|ϕihϕ|An B n )UB∗ n (s)|S| s∈S B=

X

p(t)NAn →B 0n (πAt n ) ⊗ πBt n . (21.85)t

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21.4. THE DIRECT CODING THEOREM

599

We now prove the final condition in (21.64) for the packing lemma. Consider the following
chain of inequalities:





ρ,δ
ρ,δ
ρ,δ
ρ
Π
⊗
Π
Πρ,δ
⊗
Π
0n
n
n
n
0n
0n
B B
B
B
B
B


 X
ρ,δ
= Πρ,δ
⊗Πp(t)NAn →B 0n (πAt n ) ⊗ πBt n
n
0n
B
B

!



ρ,δ
Πρ,δ
B 0n ⊗ ΠB n

(21.86)

t

=

X

=

X

≤

X

p(t)



ρ,δ
ρ,δ t
ρ,δ
t
Πρ,δ
B 0n NAn →B 0n (πAn )ΠB 0n ⊗ ΠB n πB n ΠB n

(21.87)

t


ρ,δ
ρ,δ
t
p(t) Πρ,δ
B 0n NAn →B 0n (πAn )ΠB 0n ⊗ ΠB n

t

p(t)



ΠtB n
Πρ,δn
Tr {ΠtB n } Bρ,δ
−n[H(B)ρ −η(n,δ)] ρ,δ
t
ΠB n
Πρ,δ
B 0n NAn →B 0n (πAn )ΠB 0n ⊗ 2(21.88).

(21.89)t

The first equality follows from (21.85). The second equality follows by a simple manipulation. The third equality follows because the maximally mixed state πBt n is equal to thenormalized type class projection operator ΠtB n . The inequality follows from Property 15.3.2
ρ,δ
ρ,δ
t
and Πρ,δB n ΠB n ΠB n ≤ ΠB n (the support of a typical type projector is always in the support of
the typical projector and the intersection of the support of an atypical type with the typicalprojector is null). Continuing, by linearity, the last line above is equal to!
Πρ,δ
B 0n NAn →B 0nXp(t)πAt n−n[H(B)ρ −η(n,δ)] ρ,δ
Πρ,δ
ΠB n
B 0n ⊗ 2tρ,δ
−n[H(B)ρ −η(n,δ)] ρ,δ= ΠB 0n NAn →B 0n (ϕAn )Πρ,δ
ΠB n
B 0n ⊗ 20ρ,δρ,δ
≤ 2−n[H(B )ρ −cδ] ΠB 0n ⊗ 2−n[H(B)ρ −η(n,δ)] ΠB n0ρ,δ= 2−n[H(B )ρ +H(B)ρ −η(n,δ)−cδ] Πρ,δB 0n ⊗ ΠB n .

(21.90)(21.91)(21.92)P
The first equality follows because ϕAn = t p(t)πAt n . The inequality follows from the equipartition property of typical projectors (Property 15.1.3). The final equality follows by rearranging terms.
With the four conditions in (21.61)–(21.64) holding, it follows from Corollary 16.5.1(the derandomized version of the packing lemma) that there exists a deterministic code
and a POVM {Λm
B 0n B n } that can detect the transmitted states with arbitrarily low maximal
probability of error as long as the size |M| of the message set is small enough:T∗p∗e ≡ max Tr (I − Λm
B 0n B n ) UB n (s(m))ρB 0n B n UB n (s(m))
m

√ 
0
0
≤ 4 2ε + 2 2ε + 16 · 2−n[H(B )ρ +H(B)ρ −η(n,δ)−cδ] 2n[H(B B)ρ +cδ] |M|√ 0= 4 2ε + 2 2ε + 16 · 2−n[I(B ;B)ρ −η(n,δ)−2cδ] |M| .
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CHAPTER 21. ENTANGLEMENT-ASSISTED CLASSICAL COMMUNICATION0

We can choose the size of the message set to be |M| = 2n[I(B ;B)−η(n,δ)−3cδ] so that the rateof communication is1
log |M| = I(B 0 ; B) − η(n, δ) − 3cδ,(21.96)n
and the bound on the maximal probability of error becomes√ ∗pe ≤ 4 2ε + 2 2ε + 16 · 2−ncδ .
(21.97)Let ε0 ∈ (0, 1) and δ 0 > 0. It is thenthat by picking n large enough and δ small√ clear
enough, we can have both 4 2ε + 2 2ε + 16 · 2−ncδ ≤ ε0 and η(n, δ) + 3cδ ≤ δ 0 . Thus,
the quantum mutual information I(B 0 ; B)ρ , with respect to the state ρB 0 B ≡ NA→B 0 (ϕAB )is an achievable rate for the entanglement-assisted transmission of classical information overthe channel N . To obtain the precise statement in Theorem 21.3.1, we can simply rewrite
the quantum mutual information as I(A; B)ρ with respect to the state ρAB ≡ NA0 →B (ϕAA0 ).
Alice and Bob can achieve the maximum rate of communication simply by determining astate ϕAA0 that maximizes the quantum mutual information I(A; B)ρ and by generating
entanglement-assisted classical codes from the state ρAB .

21.5 The Converse Theorem
This section contains a proof of the converse part of the entanglement-assisted classicalcapacity theorem. Let us begin by supposing that Alice and Bob are trying to use theentanglement-assisted channel many times to accomplish the task of randomness distribution(recall that we took this approach for the converse of the classical capacity theorem in
Section 20.3.2). An upper bound on the rate at which Alice can distribute randomness to
Bob also serves as an upper bound on the rate at which they can communicate because anoiseless classical channel can distribute randomness. In such a task, Alice and Bob shareentanglement in some state ΨTA TB . Alice first prepares the maximally correlated state ΦM M 0 ,as defined in (20.16), and the rate of randomness in this state is n1 log |M|. Alice then applies
some encoding channel EM 0 TA →An to the classical system M 0 and her share TA of ΨTA TB . The
resulting state isθM An TB ≡ EM 0 TA →An (ΦM M 0 ⊗ ΨTA TB ).
(21.98)She sends her An systems through many uses NAn →B n of the channel NA→B , and Bob receives
the systems B n , producing the state
ωM TB B n ≡ NAn →B n (EM 0 TA →An (ΦM M 0 ⊗ ΨTA TB )).

(21.99)Finally, Bob performs some decoding channel DB n TB →M̂ on the above state to give0ωM
≡ DB n TB →M̂ (ωM TB B n ).
M̂(21.100)

0
An (n, [log |M|] /n, ε) protocol for randomness distribution is such that the actual state ωMM̂resulting from the protocol is ε-close in trace distance to the ideal shared randomness state:

1 0ω− ΦM M̂ 1 ≤ ε.
2 M M̂
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21.5. THE CONVERSE THEOREM

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We now show that the quantum mutual information of the channel serves as an upperbound on the rate of any reliable protocol for entanglement-assisted randomness distribution(a protocol meeting the error criterion in (21.101)). Consider the following:log |M| = I(M ; M̂ )Φ
≤ I(M ; M̂ )ω0 + f (|M| , ε)
≤ I(M ; B n TB )ω + f (|M| , ε)
= I(TB M ; B n )ω + I(M ; TB )ω − I(B n ; TB )ω + f (|M| , ε)
= I(TB M ; B n )ω − I(B n ; TB )ω + f (|M| , ε)
≤ I(TB M ; B n )ω + f (|M| , ε)
≤ I(N ⊗n ) + f (|M| , ε)= nI(N ) + f (|M| , ε).

(21.102)(21.103)(21.104)(21.105)(21.106)(21.107)(21.108)(21.109)

The first equality follows by evaluating the quantum mutual information of the shared randomness state ΦM M̂ . The first inequality follows for the exact same reasons as (20.57) does(the first inequality of the converse for the HSW theorem), with f (|M| , ε) ≡ ε log |M| +(1 + ε) h2 (ε/ [1 + ε]). The second inequality follows from quantum data processing (Theorem 11.9.4)—Bob processes the state ωM TB B n with the decoder DB n TB →M̂ to get the state0. The second equality follows from the chain rule for quantum mutual information (seeωMM̂
Exercise 11.7.1). The third equality follows because the systems M and TB are in a productstate, so that I(M ; TB )ω = 0. The third inequality follows because I(B n ; TB )ω ≥ 0. Observethat the state ωM TB B n = NAn →B n (θM An TB ), so that the systems M and TB extend the An
system that is input to N ⊗n . Thus, the mutual information between M TB and B n cannever exceed the maximum mutual information of the channel, where we need to apply the
result of Exercise 13.4.4. Finally, the mutual information of a quantum channel is additive(Theorem 13.4.1), and Corollary 13.4.1 implies that I(N ⊗n ) = nI(N ). We can then rewrite
the above bound as follows:
1+ε1
log |M| (1 − ε) ≤ I(N ) +h2 (ε/ [1 + ε]).
(21.110)nnThus, if we are considering a sequence of (n, [log |M|] /n, εn ) entanglement-assisted classical communication protocols with rate C−δn = n1 log |M|, such that limn→∞ εn = limn→∞ δn =0, then the above bound becomes
1 + εn(C − δn ) (1 − εn ) ≤ I(N ) +h2 (εn / [1 + εn ]).
(21.111)n
Taking the limit as n → ∞ then establishes that an achievable rate C necessarily satisfies
C ≤ I(N ). This demonstrates a single-letter upper bound on the entanglement-assistedclassical capacity of a quantum channel and completes the proof of Theorem 21.3.1.

21.5.1

Feedback Does Not Increase Capacity

The entanglement-assisted classical capacity formula is the closest formal analogy to Shannon’s capacity formula for a classical channel. The mutual information I(N ) of a quantumc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 21. ENTANGLEMENT-ASSISTED CLASSICAL COMMUNICATION

channel N is the optimum of the quantum mutual information over all bipartite input states:
I(N ) = max I(A; B),
φAA0(21.112)and it is equal to the channel’s entanglement-assisted classical capacity by Theorem 21.4.1.
The mutual information I(pY |X ) of a classical channel pY |X is the optimum of the classicalmutual information over all correlated inputs to the channel:
I(pY |X ) = max0 I(X; Y ),XX

(21.113)

where XX 0 are correlated random variables with the distribution pX,X 0 (x, x0 ) = pX (x)δx,x0 .
The formula is equal to the classical capacity of a classical channel by Shannon’s noisy coding
theorem. Both formulas not only appear similar in form, but they also have the important
property of being “single-letter,” meaning that the above formulas are equal to the capacity(this was not generally the case for the Holevo information from the previous chapter).
We now consider another way in which the entanglement-assisted classical capacity isa good candidate for being the fully quantum generalization of Shannon’s formula to thequantum world. Though it might be surprising, it is well known that free access to a classicalfeedback channel from receiver to sender does not increase the capacity of a classical channel.
We state this result as the following theorem.
Theorem 21.5.1 (Feedback Does Not Increase Classical Capacity) The feedback capacity of a classical channel pY |X (y|x) is equal to the mutual information of that channel:
sup C : C is achievable for pY |X with feedback = I(pY |X ),(21.114)
where I(pY |X ) is defined in (21.113).
Proof. We first define an (n, C − δ, ε) classical feedback code as one in which every symbolxi (m, Y i−1 ) of a codeword xn (m) is a function of the message m ∈ M and all of the previously
received values Y1 , . . . , Yi−1 from the receiver. The decoder consists of the decoding function
g : Y n → {1, 2, . . . , |M|} such thatPr {M 0 6= M } ≤ ε,(21.115)

where M 0 ≡ g(Y n ). The lower bound LHS ≥ RHS follows because we can always avoid theuse of the feedback channel and achieve the mutual information of the classical channel by
employing Shannon’s noisy coding theorem. The upper bound LHS ≤ RHS is less obvious,
but it follows from the memoryless structure of the channel and the structure of a feedback
code. Consider the following chain of inequalities:
log |M| = H(M ) = I(M ; M 0 ) + H(M |M 0 )≤ I(M ; M 0 ) + 1 + ε log |M|≤ I(M ; Y n ) + 1 + ε log |M| .
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21.5. THE CONVERSE THEOREM

603

The first equality follows because we assume that the message M is uniformly distributed.
The first inequality follows from Fano’s inequality (see Theorem 10.7.3) and the assumptionin (21.115) that the protocol is good up to error ε. The last inequality follows from classicaldata processing. Continuing, we can bound I(M ; Y n ) from above:I(M ; Y n ) = H(Y n ) − H(Y n |M ) = H(Y n ) −nX

H(Yk |Y k−1 M )(21.119)k=1n= H(Y ) −n
X

H(Yk |Yk−1

n

M Xk ) = H(Y ) −

k=1

≤

n
X

H(Yk ) − H(Yk |Xk ) =

k=1

n
X

H(Yk |Xk )(21.120)

k=1
n
X

I(Xk ; Yk )(21.121)k=1≤ n max0 I(X; Y ).

(21.122)

XX

The first equality follows from the definition of mutual information. The second equality follows from the chain rule for entropy (see Exercise 10.3.2). The third equality follows because
Xk is a function of Y k−1 and M . The fourth equality follows because Yk is conditionallyindependent of Y k−1 and M through Xk (Y k−1 M → Xk → Yk forms a Markov chain). The
first inequality follows from subadditivity of entropy. The fifth equality follows by definition,and the final inequality follows because the individual mutual informations in the sum cannever exceed the maximum over all inputs. Putting everything together, our final bound on
the feedback-assisted capacity of a classical channel is
C − δ ≤ I(pY |X ) +ε1+ log |M| ,n n(21.123)

which becomes C ≤ I(pY |X ) as n → ∞ and ε, δ → 0.
Given the above result, we might wonder if a similar result could hold for the entanglementassisted classical capacity. Such a result would more firmly place the entanglement-assistedclassical capacity as a good generalization of Shannon’s coding theorem. Indeed, the following theorem states that this result holds.
Theorem 21.5.2 (Quantum Feedback Does Not Increase EAC Capacity) The classical capacity of a quantum channel assisted by a quantum feedback channel is equal to thatchannel’s entanglement-assisted classical capacity:
sup {C | C is achievable for N with quantum feedback} = I(N ),(21.124)

where I(N ) is defined in (21.112).
Proof. We define free access to a quantum feedback channel to mean that there is a noiselessquantum channel of arbitrarily large dimension connecting the receiver Bob to the sender
Alice. The bound LHS ≥ RHS follows because Bob can use the quantum feedback channel
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CHAPTER 21. ENTANGLEMENT-ASSISTED CLASSICAL COMMUNICATION

Figure 21.4: Three rounds of the most general protocol for classical communication with a quantum
feedback channel.

to establish an arbitrarily large amount of entanglement with Alice. They then just executethe protocol from Section 21.4 to achieve the entanglement-assisted classical capacity.
The bound LHS ≤ RHS is much less obvious, and it requires a proof that is different
from the proof of Theorem 21.5.1. We first need to determine the most general protocol
for classical communication with the assistance of a quantum feedback channel. Figure 21.4
depicts such a protocol. The protocol begins with Alice preparing a classical register M
with a uniformly random message to be sent, which is correlated with some system A00 .
Bob uses the quantum feedback channel to send a quantum system X0 to Alice, which iscorrelated with some quantum system B10 . Alice performs an encoding EA1 0 X0 →A0 A1 . Alice0
1
sends system A1 through the first use of the channel N . Bob now applies the decoding
1channel DB00 . The next encoder of Alice occurs, and the procedure repeats. The
1 B1 →X1 B2last decoding channel of Bob outputs a classical system M 0 which contains Bob’s estimateof the message that Alice transmitted. The state of registers M Bn Bn0 after the nth channel
NAn →Bn has been applied has the following form:
(n)

(n)

ωM Bn Bn0 ≡ NAn →Bn (ρM Bn0 An ),(21.125)(n)where ρM Bn0 An is the state of registers M Bn0 An after the nth encoding channel has been
(n)

(n)

applied. Let ψR(n) M B 0 An be a purification of ρM Bn0 An , and let
n

(n)

(n)

ωR(n) M Bn B 0 ≡ NAn →Bn (ψR(n) M B 0 An ).
n(21.126)nThis protocol is the most general for classical communication with quantum feedback.
We can now proceed with proving the upper bound LHS ≤ RHS. To do so, we assume thatthe random variable M modeling Alice’s message selection is a uniform random variable, andBob obtains a random variable M 0 by measuring all of his systems Bn and Bn0 at the end of the
protocol. For any ε-good protocol for classical communication, the bound Pr {M 0 6= M } ≤ εapplies. Consider the following chain of inequalities (these steps are essentially the same as
those in (21.116)–(21.118)):
log |M| = H(M ) = I(M ; M 0 ) + H(M |M 0 )≤ I(M ; M 0 ) + 1 + ε log |M|≤ I(M ; Bn Bn0 )ω(n) + 1 + ε log |M| ,c 2019(21.127)(21.128)(21.129)

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21.5. THE CONVERSE THEOREM

605

where the last mutual information is with respect to the state in (21.125). This chain of
inequalities follows for the same reason as those in (21.116)–(21.118), with the last step
following from quantum data processing. Continuing, we haveI(M ; Bn Bn0 )ω(n) = I(M ; Bn |Bn0 )ω(n) + I(M ; Bn0 )ω(n)≤ I(M Bn0 ; Bn )ω(n) + I(M ; Bn0 )ω(n)≤ I(R(n) M Bn0 ; Bn )ω(n) + I(M ; Bn0 )ω(n) .

(21.130)(21.131)(21.132)

The first equality is the chain rule for mutual information. The first inequality follows because
I(M ; Bn |Bn0 ) = I(M Bn0 ; Bn ) − I(Bn0 ; Bn ) ≤ I(M Bn0 ; Bn ). The second inequality follows fromquantum data processing. Now, given that the mutual information I(R(n) M Bn0 ; Bn ) is withrespect to the state in (21.125) and this state has the following form
NAn →Bn (φRAn ),(21.133)

where φRAn is some pure state and R is some system not going into the channel (hereidentified with R(n) M Bn0 ), we can optimize over all such inputs to find that
I(R(n) M Bn0 ; Bn )ω(n) ≤ I(N ),(21.134)

where I(N ) is the quantum mutual information of the channel. So this means that
I(M ; Bn Bn0 )ω(n) ≤ I(N ) + I(M ; Bn0 )ω(n)0≤ I(N ) + I(M ; Bn−1 Bn−1
)ω(n−1) .

(21.135)(21.136)

where the last inequality follows from quantum data processing (the system Bn0 results from0). At this point, we realize that theapplying the n − 1 decoder to the systems Bn−1 Bn−10)ω(n−1) , so we iterabove chain of steps (21.130)–(21.136) can be applied to I(M ; Bn−1 Bn−1
ate this sequence until we go all the way back to the beginning of the protocol. Putting
everything together, we get the following upper bound on any achievable rate C for classicalcommunication with quantum feedback:C − δ ≤ I(N ) +ε1+ log |M| ,n n(21.137)

which becomes C ≤ I(N ) as n → ∞ and ε, δ → 0.
Corollary 21.5.1 The capacity of a quantum channel with unlimited entanglement and classical feedback is equal to the entanglement-assisted classical capacity of N .
Proof. This result follows because I(N ) is a lower bound on this capacity (simply by
avoiding use of the classical feedback channel). Also, I(N ) is an upper bound on this capacity
because the entanglement and classical feedback channel can simulate an arbitrarily large
quantum feedback channel via teleportation, and the above theorem gives an upper boundof I(N ) for this setting.
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CHAPTER 21. ENTANGLEMENT-ASSISTED CLASSICAL COMMUNICATION

21.6 Examples of Channels
This section shows how to compute the entanglement-assisted classical capacity of both thequantum erasure channel and the amplitude damping channel, while leaving the capacity ofthe quantum depolarizing channel and the dephasing channel as exercises. For three of thesechannels (erasure, depolarizing, and dephasing), a super-dense-coding-like strategy suffices
to achieve capacity. This strategy involves Alice locally rotating an ebit shared with Bob,sending one share of it through the noisy channel, and Bob performing measurements in
the Bell basis to determine what Alice sent. This process induces a classical channel from
Alice to Bob, for which its capacity is equal to the entanglement-assisted capacity of theoriginal quantum channel (in the case of depolarizing, dephasing, and erasure channels).
For the amplitude damping channel, this super-dense-coding-like strategy does not achieve
capacity—in general, it is necessary for Bob to perform a large, collective measurement on
all of the channel outputs in order for him to determine Alice’s message.
Figure 21.5 plots the entanglement-assisted capacities of these four channels as a functionof their noise parameters. As expected, the depolarizing channel has the worst performance
because it is a “worst-case scenario” channel—it either sends the state through or replaces
it with a completely random state. The erasure channel’s capacity is just a line of constantslope down to zero—this is because the receiver can easily determine the fraction of the timethat he receives something from the channel. The dephasing channel eventually becomes acompletely classical channel, for which entanglement cannot increase capacity beyond one
bit per channel use. Finally, perhaps the most interesting curve is for the amplitude damping
channel. This channel’s capacity is convex when its noise parameter is less than 1/2 andconcave when it is greater than 1/2.

21.6.1The Quantum Erasure Channel

Recall that the quantum erasure channel acts as follows on an input density operator ρA0 :ρA0 → (1 − ε) ρB + ε|eihe|B ,(21.138)

where ε ∈ [0, 1] is the erasure probability and |eiB is an erasure state that is orthogonal to
the support of the input state ρ.
Proposition 21.6.1 The entanglement-assisted classical capacity of a quantum erasure channel with erasure probability ε is equal to 2 (1 − ε) log dA , where dA is the dimension of the
input system.
Proof. To determine the entanglement-assisted classical capacity of this channel, we needto compute its mutual information. So, consider that sending one share of a bipartite state
φAA0 through the channel produces the output
σAB ≡ (1 − ε) φAB + εφA ⊗ |eihe|B .
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Figure 21.5: The entanglement-assisted classical capacity of the amplitude damping channel, the erasurechannel, the depolarizing channel, and the dephasing channel as a function of each channel’s noise parameter.

We could now attempt to calculate and optimize the quantum mutual information I(A; B)σ .
However, observe that Bob can apply the following isometry UB→BX to his state:
UB→BX ≡ ΠB ⊗ |0iX + |eihe|B ⊗ |1iX ,(21.140)

where ΠB is a projector onto the support of the input state (for qubits, it would be just|0ih0| + |1ih1|). Applying this isometry leads to a state σABX where†
σABX ≡ UB→BX σAB UB→BX= (1 − ε) φAB ⊗ |0ih0|X + εφA ⊗ |eihe|B ⊗ |1ih1|X .

(21.141)(21.142)The quantum mutual information I(A; BX)σ is equal to I(A; B)σ because entropies do notchange under the isometry UB→BX . We now calculate I(A; BX)σ :I(A; BX)σ = H(A)σ + H(BX)σ − H(ABX)σ= H(A)φ + H(B|X)σ − H(AB|X)σ= H(A)φ + (1 − ε) [H(B)φ − H(AB)φ ]+ ε H(B)|eihe| − H(AB)φA ⊗|eihe|= H(A)φ + (1 − ε) H(B)φ − ε H(A)φ + H(B)|ei= (1 − ε) [H(A)φ + H(B)φ ]= 2 (1 − ε) H(A)φ≤ 2 (1 − ε) log dA .
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The first equality follows by the definition of quantum mutual information. The second
equality follows from φA = TrBX {σABX }, from the chain rule of entropy, and by canceling
H(X) on both sides. The third equality follows because the X register is a classical register,
indicating whether the erasure occurs. The fourth equality follows because H(AB)φ = 0,
H(B)|eihe| = 0, and H(AB)φA ⊗|eihe| = H(A)φ + H(B)|eihe| . The fifth equality follows again
because H(B)|eihe| = 0 and by collecting terms. The final equality follows because H(A)φ =H(B)φ (φAB is a pure bipartite state). The final inequality follows because the entropyof a state on system A is never greater than the logarithm of the dimension of A. We
can conclude that the maximally entangled state ΦAA0 achieves the entanglement-assistedclassical capacity of the quantum erasure channel because H(A)Φ = log dA .
The strategy for achieving the entanglement-assisted classical capacity of the quantum
erasure channel is straightforward. Alice and Bob simply employ a super-dense coding
strategy on all of the channel uses (this means that Bob performs measurements on eachchannel output with his share of the entanglement—there is no need for a large, collective
measurement on all of the channel outputs). For a good fraction 1 − ε of the time, thisstrategy works and Alice can communicate 2 log dA bits to Bob. For the other fraction ε, allis lost to the environment. In order for this to work, Alice and Bob need to make use of a
feedback channel from Bob to Alice so that Bob can report which messages come throughand which do not, but Corollary 21.5.1 states that this feedback cannot improve the capacity.
Thus, the rate of communication they can achieve is equal to the capacity 2 (1 − ε) log dA .

21.6.2The Amplitude Damping Channel

We now compute the entanglement-assisted classical capacity of the amplitude damping
channel NAD . Recall that this channel acts as follows on an input qubit in state ρ:
NAD (ρ) = A0 ρA†0 + A1 ρA†1 ,(21.150)

where, for γ ∈ [0, 1],
A0 ≡ |0ih0| +p

1 − γ|1ih1|,

A1 ≡√γ|0ih1|.

(21.151)

Proposition 21.6.2 The entanglement-assisted classical capacity of an amplitude damping
channel with damping parameter γ ∈ [0, 1] is equal toI(NAD ) = max h2 (p) + h2 ((1 − γ) p) − h2 (γp),(21.152)p∈[0,1]

where h2 (x) ≡ −x log x − (1 − x) log(1 − x) is the binary entropy function.
Proof. Suppose that a matrix representation of the input qubit density operator ρ in thecomputational basis is
1 − p η∗ρ=.
(21.153)
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21.6. EXAMPLES OF CHANNELS

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One can readily verify that the density operator for Bob has the following matrix representation:


√
∗
1
−
γη
1−
(1
−
γ)
p
√
,
(21.154)
NAD (ρ) =1 − γη
(1 − γ) p
and by calculating the elements Tr{Ai ρA†j }|iihj|, we can obtain a matrix representation for
Eve’s density operator:√ ∗
1 − γpγη
c,(21.155)
NAD (ρ) = √γηγpcwhere NADis the complementary channel to Eve. By comparing (21.154) and (21.155), we
can see that the channel to Eve is an amplitude damping channel with damping parameter
1−γ. The entanglement-assisted classical capacity of NAD is equal to its mutual information:

I(NAD ) = max I(A; B)σ ,φAA0(21.156)

where φAA0 is some pure bipartite input state and σAB = NAD (φAA0 ). We need to determine
the input density operator that maximizes the above formula as a function of γ. As it stands
now, the optimization depends on three parameters: p, Re {η}, and Im {η}. We can showthat it is sufficient to consider an optimization over only p with η = 0. The formula in(21.156) also has the following form:c
I(NAD ) = max [H(ρ) + H(NAD (ρ)) − H(NAD(ρ))] ,

(21.157)

I(A; B)σ = H(A)φ + H(B)σ − H(AB)σ= H(A0 )φ + H(NAD (ρ)) − H(E)σc= H(ρ) + H(NAD (ρ)) − H(NAD(ρ))≡ Imut (ρ, NAD ).

(21.158)(21.159)(21.160)(21.161)ρ

because

The three entropies in (21.157) depend only on the eigenvalues of the three density operatorsin (21.153)–(21.155), respectively, which are as follows:


q
1
2
2
1 ± (1 − 2p) + 4 |η| ,(21.162)
2


q
1
2
2
1 ± (1 − 2 (1 − γ) p) + 4 |η| (1 − γ) ,(21.163)2q1
1 ± (1 − 2γp)2 + 4 |η|2 γ .
(21.164)2
The above eigenvalues are in the order of Alice, Bob, and Eve. All of the above eigenvalues
have a similar form, and their dependence on η is only through its magnitude. Thus, itsuffices to consider η ∈ R (this eliminates one parameter). Next, the eigenvalues do not
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CHAPTER 21. ENTANGLEMENT-ASSISTED CLASSICAL COMMUNICATION

change if we flip the sign of η (this is equivalent to rotating the original state ρ by Z, toZρZ), and thus, the mutual information does not change as well:
Imut (ρ, NAD ) = Imut (ZρZ, NAD ).

(21.165)

By the above relation and concavity of quantum mutual information in the input density
operator (Theorem 13.4.2), the following inequality holds:
1[Imut (ρ, NAD ) + Imut (ZρZ, NAD )]2≤ Imut ((ρ + ZρZ) /2, NAD )(21.167)= Imut (∆(ρ), NAD ),(21.168)

Imut (ρ, NAD ) =(21.166)

where ∆ is a completely dephasing channel in the computational basis. This demonstrates
that it is sufficient to consider diagonal density operators ρ when optimizing the quantum
mutual information. Thus, the eigenvalues in (21.162)–(21.164) respectively become{p, 1 − p} ,{(1 − γ) p, 1 − (1 − γ) p} ,
{γp, 1 − γp} ,(21.169)(21.170)(21.171)

giving our final expression in the statement of the proposition.
Exercise 21.6.1 Consider the qubit depolarizing channel: ρ → (1 − p)ρ + pπ. Prove that its
entanglement-assisted classical capacity is equal to
2 + (1 − 3p/4) log(1 − 3p/4) + (3p/4) log(p/4).

(21.172)

Exercise 21.6.2 Consider the dephasing channel: ρ → (1 − p/2) ρ + (p/2)ZρZ. Prove thatits entanglement-assisted classical capacity is equal to 2 − h2 (p/2), where p is the dephasing
parameter.

21.7 Concluding Remarks
Shared entanglement has the desirable property of simplifying quantum Shannon theory. The
entanglement-assisted capacity theorem is one of the strongest known results in quantum
Shannon theory because it states that the quantum mutual information of a channel isequal to its entanglement-assisted capacity. This function of the channel is concave in the
input state and the set of input states is convex, implying that finding a local maximum
is equivalent to finding a global one. The converse theorem and additivity of the channel
mutual information demonstrates that there is no need to take the regularization of theformula. Furthermore, quantum feedback does not improve this capacity, just as it does
not for the classical case of Shannon’s setting. In these senses, the entanglement-assisted
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21.8. HISTORY AND FURTHER READING

611

classical capacity is the most natural generalization of Shannon’s capacity formula to thequantum setting.
The direct coding part of the capacity theorem exploits a strategy similar to super-dense
coding—effectively the technique is to perform super-dense coding in the type class subspacesof many copies of a shared entangled state. This strategy is equivalent to super-dense coding
if the initial shared state is a maximally entangled state. The particular protocol that
we outlined in this chapter has the appealing feature that we can easily make it coherent,
similar to the way that coherent dense coding is a coherent version of the super-dense coding
protocol. We take this approach in the next chapter and show that we can produce a whole
host of other protocols using this technique, eventually leading to a proof of the direct coding
part of the quantum capacity theorem.
This chapter features the calculation of the entanglement-assisted classical capacity ofcertain channels of practical interest: the depolarizing channel, the dephasing channel, theamplitude damping channel, and the erasure channel. Each one of these channels has asingle parameter that governs its noisiness, and the capacity in each case is a straightforward
function of this parameter. One could carry out a similar type of analysis to determine theentanglement-assisted capacity of any channel, although it generally will be necessary toemploy techniques from convex optimization.
Unfortunately, quantum Shannon theory only gets more complicated from here onward.2
For the other capacity theorems that we will study, such as the private classical capacityor the quantum capacity, the best expressions that we have for them are good only up
to regularization of the formulas. In certain cases, these formulas completely characterize
the capabilities of the channel for these particular operational tasks, but these formulas
are not particularly useful in the general case. One important goal for future research inquantum Shannon theory would be to improve upon these formulas, in the hopes that we
could further our understanding of the best strategy for achieving the information-processing
tasks corresponding to these other capacity questions.

21.8 History and Further Reading
Adami and Cerf (1997) figured that the mutual information of a quantum channel would
play an important role in quantum Shannon theory, and they proved several of its mostimportant properties. Bennett et al. (1999, 2002) later demonstrated that the quantum mutual information of a channel has an operational interpretation as its entanglement-assistedclassical capacity. Our proof of the direct part of the entanglement-assisted classical capacity
theorem is the same as that in Hsieh et al. (2008a). We exploit this approach because itleads to all of the results in the next chapter, implying that this protocol is sufficient to generate all of the known protocols in quantum Shannon theory (with the exception of private
classical communication). Giovannetti and Fazio (2005) determined several capacities of the2

We could also view this “unfortunate” situation as being fortunate for conducting open-ended researchin quantum Shannon theory.
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CHAPTER 21. ENTANGLEMENT-ASSISTED CLASSICAL COMMUNICATION

amplitude damping channel, and Wolf and Pérez-Garcı́a (2007) made some further observations regarding it. Bowen (2004) proved that the classical capacity of a channel assisted byunbounded quantum feedback is equal to its entanglement-assisted classical capacity.
There has also been work on the strong converse and second-order characterization for
entanglement-assisted capacity (see Section 20.7 for a discussion of these terms). Bennett
et al. (2014) proved the quantum reverse Shannon theorem, which quantifies the rate ofclassical communication needed to simulate a quantum channel in the presence of unlimited
entanglement shared between sender and receiver. Berta et al. (2011) provided an alternate
proof of the quantum reverse Shannon theorem. By a simulation argument, the quantum
reverse Shannon theorem implies a strong converse for entanglement-assisted capacity (Bennett et al., 2014). Gupta and Wilde (2015) gave a direct proof for the strong converse bymaking use of Rényi entropies. Cooney et al. (2016) later showed that the same strong
converse bound still holds in the presence of a quantum feedback channel, strengthening theresult of Bowen (2004). Datta et al. (2016) established a second-order achievability result
for entanglement-assisted classical communication and proved that this characterization istight for some channels, by making use of prior results of Matthews and Wehner (2014).

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 22

Coherent Communication with Noisy
Resources
This chapter demonstrates the power of both coherent communication from Chapter 7 andthe particular protocol for entanglement-assisted classical coding from the previous chapter.
Recall that coherent dense coding is a version of the dense coding protocol in which thesender and receiver perform all of its steps coherently.1 Since our protocol for entanglementassisted classical coding from the previous chapter is really just a glorified dense coding
protocol, the sender and receiver can perform each of its steps coherently, generating a
protocol for entanglement-assisted coherent coding. Then, by exploiting the fact that two
coherent bits are equivalent to a qubit and an ebit, we obtain a protocol for entanglementassisted quantum coding that consumes far less entanglement than a naive strategy would inorder to accomplish this task. We next combine this entanglement-assisted quantum coding
protocol with entanglement distribution (Section 6.2.1) and obtain a protocol for quantum
communication at a rate equal to the channel’s coherent information (Section 13.5). Thissequence of steps demonstrates an alternate proof of the direct part of the quantum capacity
theorem stated in Chapter 24.
Entanglement-assisted classical communication is one generalization of super-dense coding, in which the noiseless qubit channel becomes an arbitrary noisy quantum channel while
the noiseless ebits remain noiseless. Another generalization of super-dense coding is a protocol named noisy super-dense coding, in which the shared entanglement becomes a shared
noisy state ρAB and the noiseless qubit channels remain noiseless. Interestingly, the protocol that we employ in this chapter for noisy super-dense coding is essentially equivalent to
the protocol from the previous chapter for entanglement-assisted classical communication,with some slight modifications to account for the different setting. We can also construct a
coherent version of noisy super-dense coding, leading to a protocol that we name coherent
state transfer. Coherent state transfer accomplishes not only the task of generating coherent
communication between Alice and Bob, but it also allows Alice to transfer her share of the1

Performing a protocol coherently means that we replace conditional unitaries with controlled unitaries
and measurements with controlled gates (e.g., compare Figures 6.2 and 7.3).

613

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CHAPTER 22. COHERENT COMMUNICATION WITH NOISY RESOURCES

state ρAB to Bob. By combining coherent state transfer with both the coherent communication identity and teleportation, we obtain protocols for quantum-assisted state transfer andclassical-assisted state transfer, respectively. The latter protocol gives an operational interpretation to the conditional quantum entropy H(A|B)ρ —if it is positive, then the protocol
consumes entanglement at the rate H(A|B)ρ , and if it is negative, the protocol generates
entanglement at the rate |H(A|B)ρ |.
The final part of this chapter shows that our particular protocol for entanglement-assistedclassical communication is even more powerful than suggested in the first paragraph. It
allows for a sender to communicate both coherent bits and incoherent classical bits to areceiver, and they can trade off these two resources against one another. The structure of
the entanglement-assisted protocol allows for this possibility, by taking advantage of Remark 21.4.1 and by combining it with the HSW classical communication protocol from
Chapter 20. Then, by exploiting the coherent communication identity, we obtain a protocol
for entanglement-assisted communication of classical and quantum information. Chapter 25
demonstrates that this protocol, teleportation, super-dense coding, and entanglement distribution are sufficient to accomplish any task in dynamic quantum Shannon theory involving
the three unit resources of classical bits, qubits, and ebits. These four protocols give athree-dimensional achievable rate region that is the best known characterization for any
information-processing task that a sender and receiver would like to accomplish using aquantum channel and the three unit resources. Chapter 25 discusses this triple trade-off
scenario in full detail.

22.1 Entanglement-Assisted Quantum Communication
The entanglement-assisted classical capacity theorem states that the quantum mutual information of a channel is equal to its capacity for transmitting classical information with thehelp of shared entanglement, and the direct coding theorem from Section 21.4 provides a
protocol that achieves the capacity. We were not much concerned with the rate at which thisprotocol consumes entanglement, but a direct calculation reveals that it consumes H(A)ϕebits per channel use, where |ϕiAB is the bipartite state that they share before the protocol begins. This result follows because they can concentrate n copies of the state |ϕiAB to≈ nH(A)ϕ ebits, as we learned in Chapter 19. Also, they can “dilute” nH(A)ϕ ebits to ≈ ncopies of |ϕiAB with the help of a sublinear amount of classical communication that does
not factor into the resource count, as also discussed in Chapter 19.
Suppose now that Alice is interested in exploiting the channel and shared entanglement
in order to transmit quantum information to Bob. There is a simple (and as we will see,naive) way that we can convert the protocol in Section 21.4 to one that transmits quantum
information: they can just combine it with teleportation. This naive strategy requires consuming ebits at an additional rate of 12 I(A; B)ρ in order to have enough entanglement to
combine with teleportation, where ρAB ≡ NA0 →B (ϕAA0 ). To see this, consider the followingc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License22.1. ENTANGLEMENT-ASSISTED QUANTUM COMMUNICATION

resource inequalities:1
1
hN i + H(A)ρ + I(A; B)ρ [qq] ≥ I(A; B)ρ [c → c] + I(A; B)ρ [qq]22
1
≥ I(A; B)ρ [q → q] .
2

615

(22.1)(22.2)

The first inequality follows by having them exploit the channel and the nH(A)ρ ebits togenerate classical communication at a rate I(A; B)ρ (while doing nothing with the extra
n 21 I(A; B)ρ ebits). Alice then exploits the ebits and the classical communication in a teleportation protocol to send n 12 I(A; B)ρ qubits to Bob. This rate of quantum communication
is provably optimal—were it not so, it would be possible to combine the protocol in (22.1)–(22.2) with super-dense coding and beat the optimal rate for classical communication givenby the entanglement-assisted classical capacity theorem.
Although the above protocol achieves the entanglement-assisted quantum capacity, we
are left thinking that the entanglement consumption rate of H(A)ρ + 21 I(A; B)ρ ebits perchannel use might be a bit more than necessary because teleportation and super-dense coding
are not dual under resource reversal. That is, if we combine the protocol with super-dense
coding and teleportation ad infinitum, then it consumes an infinite amount of entanglement.
In practice, this “back and forth” with teleportation and super-dense coding would be a poor
way to consume the precious resource of entanglement.
How might we make more judicious use of shared entanglement? Recall that coherent
communication from Chapter 7 was helpful for doing so, at least in the noiseless case. A
sender and receiver can combine coherent teleportation and coherent dense coding ad infinitum without any net loss in entanglement, essentially because these two protocols are dual
under resource reversal. The following theorem shows how we can upgrade the protocol in
Section 21.4 to one that generates coherent communication instead of just classical communication. The resulting protocol is one way to have a version of coherent dense coding inwhich one noiseless resource is replaced by a noisy one.
Theorem 22.1.1 (Entanglement-Assisted Coherent Communication) The following
resource inequality corresponds to an achievable protocol for entanglement-assisted coherent
communication over a quantum channel NA0 →B :hN i + H(A)ρ [qq] ≥ I(A; B)ρ [q → qq] ,(22.3)

where ρAB ≡ NA0 →B (ϕAA0 ).
Proof. Suppose that Alice and Bob share many copies of some pure, bipartite entangledstate |ϕiAB . Consider the code from the direct coding theorem in Section 21.4. We cansay that it is a set of D2 ≈ 2nI(A;B)ρ unitaries U (s(m)), from which Alice can select, andshe applies a particular unitary U (s(m)) to her share An of the entanglement in order toencode message m. Also, Bob has a detection POVM {Λm
B 0n B n } acting on his share of theentanglement and the channel outputs that he can exploit to detect message m. Just as wec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 22. COHERENT COMMUNICATION WITH NOISY RESOURCES

were able to construct a coherent super-dense coding protocol in Chapter 7 by performing
all the steps in dense coding coherently, we can do so for the entanglement-assisted classicalcoding protocol in Section 21.4. We track the steps in such a protocol. Suppose Alice sharesa state with a reference system R to which she does not have access:2|ψiRA1 ≡D
X

αl,m |liR |miA1 ,(22.4)

l,m=1

where {|li} and {|mi} are some orthonormal bases for R and A1 , respectively. We saythat Alice and Bob have implemented a coherent channel if they execute the map |miA1 →|miA1 |miB1 , which transforms the above state to2DX

αl,m |liR |miA1 |miB1 .

(22.5)

l,m=1

We say that they have implemented a coherent channel approximately if the state resulting
from the protocol is ε-close in trace distance to the above state. If we can show that ε ∈ (0, 1)
approaches zero in the limit of many channel uses, then the simulation of an approximate
coherent channel becomes perfect in the asymptotic limit. Alice’s first step is to append hershares of the entangled state |ϕiAn B n to |ψiRA1 and apply the following controlled unitaryfrom her system A1 to her system An :X
|mihm|A1 ⊗ UAn (s(m)).
(22.6)m

The resulting global state is as follows:X
αl,m |liR |miA1 UAn (s(m))|ϕiAn B n .

(22.7)

l,m

By the structure of the unitaries U (s(m)) (see (21.38) and (21.40)), the above state isequivalent to the following one:X
αl,m |liR |miA1 UBT n (s(m))|ϕiAn B n .
(22.8)
l,mInterestingly, observe that Alice applying
P the controlledT gate in (22.6) is the same as her
applying the non-local controlled gate m |mihm|A1 ⊗ UB n (s(m)), due to the non-local (andperhaps spooky!) properties of the entangled state |ϕiAn B n . Alice then sends her systems
N
An through many uses of the quantum channel NA→B 0 , whose isometric extension is UA→B
0E .
N
Let |ϕiB 0n E n B n denote the state resulting from n instances of the isometric extension UA→B 0 E
of the channel acting on the state |ϕiAn B n :|ϕiB 0n E n B n ≡ UANn →B 0n E n |ϕiAn B n ,c 2019(22.9)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License22.1. ENTANGLEMENT-ASSISTED QUANTUM COMMUNICATION

617

N
⊗n
where UANn →B 0n E n ≡ (UA→B. After Alice transmits through the isometric extension, the0E )state becomesX
αl,m |liR |miA1 UBT n (s(m))|ϕiB 0n E n B n ,(22.10)l,mwhere Bob now holds his shares B n of the entanglement and the channel outputs B 0n .
(Observe that the action of the controlled unitary in (22.6) commutes with the action of thechannel.) Rather than perform an incoherent measurement with the POVM {Λm
B 0n B n }, Bobapplies a coherent gentle measurement (see Section 5.4), an isometry of the following form:
XpΛm(22.11)
B 0n B n ⊗ |miB1 .
m√
Using the result of Exercise 5.4.1, we can readily check that the resulting state is 2 2ε-closein trace distance to the following state:X
αl,m |liR |miA1 UBT n (s(m))|ϕiB 0n E n B n |miB1 .
(22.12)l,mThus, for the rest of the protocol, we pretend that they are acting on the above state. Aliceand Bob would like to coherently remove the coupling of their index m to the environment,so Bob performs the following controlled unitary:X|mihm|B1 ⊗ UB∗ n (s(m)),(22.13)m

and the final state is2DX

αl,m |liR |miA1 |ϕiB 0n E n B n |miB1

l,m=1
2

=

D
X

!
αl,m |liR |miA1 |miB1⊗ |ϕiB 0n E n B n . (22.14)l,m=1

Thus, this protocol implements a D2 -dimensional coherent channel up to an arbitrarily small
error, which implies that the resource inequality in the statement of the theorem holds.
Figure 22.1 depicts the entanglement-assisted coherent coding protocol.
It is now a straightforward task to convert the protocol from Theorem 22.1.1 into one forentanglement-assisted quantum communication, by exploiting the coherent communication
identity from Section 7.5.
Corollary 22.1.1 (Entanglement-Assisted Quantum Communication) The following
resource inequality corresponds to an achievable protocol for entanglement-assisted quantum
communication over a quantum channel NA0 →B :1
1
hN i + I(A; E)ϕ [qq] ≥ I(A; B)ϕ [q → q] ,2
2

(22.15)where |ϕiABE ≡ UAN0 →BE |ϕiAA0 and UAN0 →BE is an isometric extension of the channel NA0 →B .
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CHAPTER 22. COHERENT COMMUNICATION WITH NOISY RESOURCES

Figure 22.1: The protocol for entanglement-assisted coherent communication. Observe that it is thecoherent version of the protocol for entanglement-assisted classical communication, just as coherent dense
coding is the coherent version of super-dense coding (compare this figure and Figure 21.3 with Figures 6.2
and 7.3). Instead of applying conditional unitaries, Alice applies a controlled unitary from her system A1 toher share of the entanglement and sends the encoded state through many uses of the noisy channel. Rather
than performing a POVM, Bob performs a coherent gentle measurement from his systems B 0n and B n toan ancilla B1 . Finally, he applies a similar controlled unitary in order to decouple the environment from thestate of his ancilla B1 .

Proof. Consider the coherent communication identity from Section 7.5. This identity statesthat a D2 -dimensional coherent channel can perfectly simulate a D-dimensional quantum
channel and a maximally entangled state |ΦiAB with Schmidt rank D. In terms of cobits,
qubits, and ebits, the coherent communication identity is the following resource equality for
D-dimensional systems:
2 log D [q → qq] = log D [q → q] + log D [qq] .

(22.16)

Consider the following chain of resource inequalities:
hN i + H(A)ϕ [qq] ≥ I(A; B)ϕ [q → qq](22.17)1
1(22.18)≥ I(A; B)ϕ [q → q] + I(A; B)ϕ [qq] .
2
2
The first resource inequality is the statement of Theorem 22.1.1, and the second resource
inequality follows from an application of coherent teleportation. If we then allow for catalytic protocols, in which we allow for some use of a resource with the demand that it be
returned at the end of the protocol, we have a protocol for entanglement-assisted quantum
communication:
1
1
hN i + I(A; E)ϕ [qq] ≥ I(A; B)ϕ [q → q] ,(22.19)2
2
because H(A)ϕ − 21 I(A; B)ϕ = 21 I(A; E)ϕ (see Exercise 11.6.6).
When comparing the entanglement consumption rate of the naive protocol in (22.1)–(22.2) with that of the protocol in Corollary 22.1.1, we see that the former requires anadditional I(A; B)ρ ebits per channel use. Also, Corollary 22.1.1 leads to a simple proof of
the achievability part of the quantum capacity theorem, as we see in the next section.
c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License22.2. QUANTUM COMMUNICATION

619

Exercise 22.1.1 Suppose that Alice can obtain the environment E of the channel UAN0 →BE .
Such a channel is known as a coherent feedback isometry. Show how they can achieve thefollowing resource inequality with the coherent feedback isometry UAN0 →BE :1
1
hUAN0 →BE i ≥ I(A; B)ϕ [q → q] + I(E; B)ϕ [qq] ,2
2(22.20)where |ϕiABE = UAN0 →BE |ϕiAA0 and ρA0 = TrA {ϕAA0 }. This protocol is a generalization ofcoherent teleportation from Section 7.4 because it reduces to coherent teleportation in thecase that UAN0 →BE is equivalent to two coherent channels.

22.2 Quantum Communication
We can obtain a protocol for quantum communication simply by combining the protocol from
Theorem 22.1.1 further with entanglement distribution. The resulting protocol again makes
catalytic use of entanglement, in the sense that it exploits some amount of entanglement
shared between Alice and Bob at the beginning of the protocol, but it generates the sameamount of entanglement at the end, so that the net entanglement consumption rate of the
protocol is zero. The resulting rate of quantum communication turns out to be the same as
we find for the quantum channel coding theorem in Chapter 24 (though the protocol given
there does not make catalytic use of shared entanglement).
Corollary 22.2.1 (Quantum Communication) The coherent information Q(N ) is anachievable rate for quantum communication over a quantum channel N . That is, the following resource inequality holds:
hN i ≥ Q(N ) [q → q] ,(22.21)
where Q(N ) ≡ maxϕ I(AiB)ρ and ρAB ≡ NA0 →B (ϕAA0 ).
Proof. If we further combine the entanglement-assisted quantum communication protocol
from Theorem 22.1.1 with entanglement distribution at a rate 21 I(A; E)ρ , we obtain thefollowing resource inequalities:1
hN i + I(A; E)ρ [qq]2
1
1
≥ [I(A; B)ρ − I(A; E)ρ ] [q → q] + I(A; E)ρ [q → q]2
2
1
1
≥ [I(A; B)ρ − I(A; E)ρ ] [q → q] + I(A; E)ρ [qq] ,2
2(22.22)(22.23)

which after resource cancelation, becomeshN i ≥ I(AiB)ρ [q → q] ,(22.24)

because I(AiB)ρ = 21 [I(A; B)ρ − I(A; E)ρ ] (see Exercise 11.6.6). They can achieve thecoherent information of the channel simply by generating codes from the state ϕAA0 thatmaximizes the channel’s coherent information.
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CHAPTER 22. COHERENT COMMUNICATION WITH NOISY RESOURCES

22.3 Noisy Super-Dense Coding
Recall that the resource inequality for super-dense coding is[q → q] + [qq] ≥ 2 [c → c] .

(22.25)The entanglement-assisted classical communication protocol from the previous chapter is
one way to generalize this protocol to a noisy setting, simply by replacing the noiseless qubit
channels in (22.25) with many uses of a noisy quantum channel. This replacement leads
to the setting of entanglement-assisted classical communication presented in the previous
chapter.
Another way to generalize super-dense coding is to let the entanglement be noisy while
keeping the quantum channels noiseless. We allow Alice and Bob access to many copies ofsome shared noisy state ρAB and to many uses of a noiseless qubit channel with the goalof generating noiseless classical communication. One might expect the resulting protocol tobe similar to that for entanglement-assisted classical communication, and this is indeed thecase. The resulting protocol is known as noisy super-dense coding:
Theorem 22.3.1 (Noisy Super-Dense Coding) The following resource inequality corresponds to an achievable protocol for quantum-assisted classical communication with a shared
quantum state:hρAB i + H(A)ρ [q → q] ≥ I(A; B)ρ [c → c] ,(22.26)
where ρAB is some bipartite state that Alice and Bob share at the beginning of the protocol.
Proof. The proof of the existence of a protocol proceeds similarly to the proof of Theorem 21.4.1, with a few modifications to account for our different setting here. We simply
need to establish a way for Alice and Bob to select a code randomly, and then we can invoke
the packing lemma (Lemma 16.3.1) to establish the existence of a detection POVM thatBob can employ to detect Alice’s messages. The method by which they select a random
code is exactly the same as they do in the proof of Theorem 21.4.1, and for this reason, weonly highlight the key aspects of the proof. First consider the state ρAB , and suppose that
|ϕiABR is a purification of this state, with R a reference system to which Alice and Bob do
not have access. We can say that the state |ϕiABR arises from some isometry UAN0 →BR acting
on system A0 of a pure state |ϕiAA0 , so that |ϕiAA0 is defined by |ϕiABR = UAN0 →BR |ϕiAA0 . We
can also then think that the state ρAB arises from sending the state |ϕiAA0 through a channel
NA0 →B , obtained by tracing out the environment R of UAN0 →BR . Our setting here is becoming
closer to the setting in the proof of Theorem 21.4.1, and we now show how it becomes nearly
identical. Observe that the state |ϕi⊗nAA0 admits a type decomposition, similar to the typedecomposition in (21.32)–(21.35):
|ϕi⊗n
AA0 =

Xp
p(t)|Φt iAn A0n .

(22.27)tc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

22.3. NOISY SUPER-DENSE CODING

621

Similarly, we can write |ϕi⊗nABR as
|ϕi⊗n
ABR =

Xp
p(t)|Φt iAn |B n Rn ,(22.28)twhere the vertical line in An |B n Rn indicates the bipartite cut between systems An and
B n Rn . Alice can select a unitary UAn (s) of the form in (21.38) uniformly at random, andthe expected density operator with respect to this random choice of unitary isnoρAn B n ≡ ES UAn (S)ρAn B n UA† n (S)(22.29)X(22.30)=p(t)πAt n ⊗ NA0n →B n (πAt 0n ),tby exploiting the development in (21.75)–(21.85). For each message m that Alice would liketo send, she selects a vector s of the form in (21.39) uniformly at random, and we can writes(m) to denote the explicit association of the vector s with the message m after Alice makes
the assignment. This leads to quantum-assisted codewords2 of the following form:
UAn (s(m))ρAn B n UA† n (s(m)).

(22.31)

We would now like to exploit the packing lemma (Lemma 16.3.1), and we require message
subspace projectors and a total subspace projector in order to do so. We choose them
respectively as†
UAn (s)Πρ,δ
An B n UAn (s),(22.32)ρ,δΠρ,δ
An ⊗ Π B n ,(22.33)ρ,δ
ρ,δ
where Πρ,δAn B n , ΠAn , and ΠB n are typical projectors for ρAn B n , ρAn , and ρB n , respectively.
The following four conditions for the packing lemma hold, for the same reasons that theyhold in (21.61)–(21.64):
n

o
ρ,δ
†
n (s)ρAn B n U n (s)Tr Πρ,δ
⊗ΠU≥ 1 − ε,(22.34)
n
n
A
A
B
A
n

o
†
Tr UAn (s)Πρ,δ
UAn (s)ρAn B n UA† n (s)≥ 1 − ε,(22.35)
An B n UAn (s)on†n[H(AB)ρ +cδ]
Tr UAn (s)Πρ,δ
,(22.36)
An B n UAn (s) ≤ 2






ρ,δ
ρ,δ
ρ,δ
ρ
Π
⊗
Π
⊗
Π
Πρ,δ
n
n
n
n
n
n
A B
A
B
B
A


ρ,δ
≤ 2−n[H(A)ρ +H(B)ρ −η(n,δ)−cδ] Πρ,δ⊗Π
, (22.37)
n
n
A
B

2

We say that the codewords are “quantum-assisted” because we will allow the assistance of quantum
communication in transmitting them to Bob.
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CHAPTER 22. COHERENT COMMUNICATION WITH NOISY RESOURCES

Figure 22.2: The protocol for noisy super-dense coding that corresponds to the resource inequality in
Theorem 22.3.1. Alice first projects her share into its typical subspace (not depicted). She then applies
a unitary encoding U (s(m)), based on her message m, to her share of the state ρAn B n . She compresses
her state to approximately nH(A)ρ qubits and transmits these qubits over noiseless qubit channels. Bob
decompresses the state and performs a decoding POVM that gives Alice’s message m with high probability.

where c is some positive constant and η(n, δ) is a function that approaches zero as n → ∞and δ → 0. Let us assume for the moment that Alice simply sends her An systems to
Bob with many uses of a noiseless qubit channel. It then follows from Corollary 16.5.1(the derandomized version of the packing lemma) that there exists a code and a POVM
{Λm
An B n } that can detect the transmitted codewords of the form in (22.31) with arbitrarily
low maximal probability of error, as long as the size |M| of Alice’s message set is smallenough:∗p∗e ≡ max Tr {(I − Λm
An B n ) U (s(m))B n ρAn B n UB n (s(m))}
m√ ≤ 4 ε + 2 ε + 16 · 2−n[H(A)ρ +H(B)ρ −η(n,δ)−cδ] 2n[H(AB)ρ +cδ] |M|√ = 4 ε + 2 ε + 16 · 2−n[I(A;B)ρ −η(n,δ)−2cδ] |M| .

(22.38)(22.39)(22.40)Thus, we can choose the size of the message set to be |M| = 2n[I(A;B)−η(n,δ)−3cδ] so that therate of classical communication is1
log |M| = I(A; B)ρ − η(n, δ) − 3cδ,nand the bound on the maximal probability of error becomes√ p∗e ≤ 4 ε + 2 ε + 16 · 2−ncδ .

(22.41)(22.42)Let ε0 ∈√(0, 1) and δ 0 > 0. By picking n large enough and δ small enough, we can make
4 (ε + 2 ε) + 16 · 2−ncδ ≤ ε0 and η(n, δ) + 3cδ ≤ δ 0 . Thus, the quantum mutual information
I(A; B)ρ , with respect to the state ρAB is an achievable rate for noisy super-dense codingwith ρ.
We now summarize the protocol (with a final modification). Alice and Bob begin withthe state ρAn B n . Alice first performs a typical subspace measurement of her system An .
This measurement succeeds with high probability and reduces the size of her system An to a
subspace with size approximately equal to nH(A)ρ qubits. If Alice wishes to send message m,
she applies the unitary UAn (s(m)) to her share of the state. She then performs a compression
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22.4. STATE TRANSFER

623

isometry from her subspace of An to nH(A)ρ qubits. She transmits her qubits over nH(A)ρ
noiseless qubit channels, and Bob receives them. Bob performs the decompression isometry
from the space of nH(A)ρ noiseless qubits to a space isomorphic to Alice’s original systems
An . He then performs the decoding POVM {Λm
An B n } and determines Alice’s message m withvanishingly small error probability. Note: The only modification to the protocol is the typicalsubspace measurement at the beginning, and one can readily check that this measurement
does not affect any of the conditions in (22.34)–(22.37). Figure 22.2 depicts the protocol.

22.4 State Transfer
We can also construct a coherent version of the noisy super-dense coding protocol, in amanner similar to the way in which the proof of Theorem 22.1.1 constructs a coherent
version of entanglement-assisted classical communication. However, the coherent version of
noisy super-dense coding achieves an additional task: the transfer of Alice’s share of thestate ρ⊗n
AB to Bob. The resulting protocol is known as coherent state transfer, and from thisprotocol, we can derive a protocol for quantum-communication-assisted state transfer, or
quantum-assisted state transfer3 for short.
Theorem 22.4.1 (Coherent State Transfer) The following resource inequality corresponds
to an achievable protocol for coherent state transfer using a state ρAB :hWS→AB : ρS i + H(A)ρ [q → q] ≥ I(A; B)ρ [q → qq] + hidS→B̂B : ρS i,(22.43)where ρAB is a bipartite state that Alice and Bob share at the beginning of the protocol.
The resource inequality in (22.43) features some notation that we have not seen yet. The
expression hWS→AB : ρS i means that a source party S distributes many copies of the stateρS to Alice and Bob, by applying some isometry WS→AB to the state ρS . This resource iseffectively equivalent to Alice and Bob sharing many copies of the state ρAB , a resource we
expressed in Theorem 22.3.1 as hρAB i. The expression hidS→B̂B : ρS i means that a source
party applies the identity map to ρS and gives the full state to Bob. We can now state themeaning of the resource inequality in (22.43): Using n copies of the state ρAB and nH(A)ρ
noiseless qubit channels, Alice can simulate nI(A; B)ρ noiseless coherent channels to Bob
while at the same time transferring her share of the state ρ⊗n
AB to him.
Proof. A proof proceeds similarly to the proof of Theorem 22.1.1. Let |ϕiABR be a purification of ρAB . Alice begins with a state that she shares with a reference system R1 , on which
she would like to simulate coherent channels:2|ψiR1 A1 ≡D
X

αl,m |liR1 |miA1 ,(22.44)l,m=1
3This protocol goes by several other names in the quantum Shannon theory literature: state transfer,fully quantum Slepian–Wolf, state merging, and the merging mother.
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CHAPTER 22. COHERENT COMMUNICATION WITH NOISY RESOURCES

where D2 ≈ 2nI(A;B)ρ . She appends |ψiR1 A1 to |ϕiAn B n Rn ≡ |ϕi⊗nABR and applies a typicalnsubspace measurement to her system A . (In what follows, we use the same notation for the
typical projected state because the states are the same up to a vanishingly small error.) She
applies the following controlled unitary to her systems A1 An :X
|mihm|A1 ⊗ UAn (s(m)),(22.45)mresulting in the overall stateX

αl,m |liR1 |miA1 UAn (s(m))|ϕiAn B n Rn .

(22.46)l,m

Alice compresses her An systems, sends them over nH(A)ρ noiseless qubit channels, and Bob
receives them. He decompresses them and places them in systems B̂ n isomorphic to An . The
resulting state is the same as |ϕiAn B n Rn , with the systems An replaced by B̂ n . Bob performs
a coherent gentle measurement of the following form:XqΛm⊗ |miB1 ,(22.47)B̂ n B nmresulting in a state that is close in trace distance toX
αl,m |liR1 |miA1 |miB1 UB̂ n (s(m))|ϕiB̂ n B n Rn .

(22.48)l,m

He finally performs the controlled unitaryX|mihm|B1 ⊗ UB̂† n (s(m)),(22.49)m

resulting in the state!
X

αl,m |liR1 |miA1 |miB1⊗ |ϕiB̂ n B n Rn .

(22.50)l,mThus, Alice has simulated nI(A; B)ρ coherent channels to Bob with arbitrarily small error,
while also transferring her share of the state |ϕiAn B n Rn to him. Figure 22.3 depicts the
protocol.
We obtain the following resource inequality for quantum-assisted state transfer, by combining the above protocol with the coherent communication identity:
Corollary 22.4.1 (Quantum-Assisted State Transfer) The following resource inequality corresponds to an achievable protocol for quantum-assisted state transfer using a sharedstate ρAB :1
1
hWS→AB : ρS i + I(A; R)ϕ [q → q] ≥ I(A; B)ϕ [qq] + hidS→B̂B : ρS i,(22.51)2
2where ρAB is a bipartite state that Alice and Bob share at the beginning of the protocol, and|ϕiABR is a purification of it.
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22.4. STATE TRANSFER

625

Figure 22.3: The protocol for coherent state transfer, a coherent version of the noisy super-dense coding
protocol that accomplishes the task of state transfer in addition to coherent communication.

Proof. Consider the following chain of resource inequalities:
hWS→AB : ρS i + H(A)ϕ [q → q]
≥ I(A; B)ϕ [q → qq] + hidS→B̂B : ρS i
1
1
≥ I(A; B)ϕ [q → q] + I(A; B)ϕ [qq] + hidS→B̂B : ρS i,
2
2(22.52)(22.53)where the first follows from coherent state transfer and the second follows from the coherent
communication identity. By resource cancelation, we obtain the resource inequality in thestatement of the theorem because 21 I(A; R)ϕ = H(A)ρ − 21 I(A; B)ρ .
Corollary 22.4.2 (Classical-Assisted State Transfer) The following resource inequality corresponds to an achievable protocol for classical-assisted state transfer using a sharedstate ρAB :hWS→AB : ρS i + I(A; R)ϕ [c → c] ≥ I(AiB)ϕ [qq] + hidS→B̂B : ρS i,(22.54)

where ρAB is a bipartite state that Alice and Bob share at the beginning of the protocol, and|ϕiABR is a purification of it.
Proof. We simply combine the protocol above with teleportation:
1hWS→AB : ρS i + I(A; R)ϕ [c → c] + I(A; R)ϕ [qq]21
≥ hWS→AB : ρS i + I(A; R)ϕ [q → q]
2
1
≥ I(A; B)ϕ [qq] + hidS→B̂B : ρS i.
2
c 2019(22.55)(22.56)

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CHAPTER 22. COHERENT COMMUNICATION WITH NOISY RESOURCES

Using the fact that 12 I(A; B)ϕ − 21 I(A; R)ϕ = I(AiB)ϕ , we obtain the resource inequality inthe statement of the corollary.
The above protocol gives a wonderful operational interpretation to the coherent information (or negative conditional entropy −H(A|B)ρ ). When the coherent information ispositive, Alice and Bob share that rate of entanglement at the end of the protocol (andthus the ability to teleport if extra classical communication is available). When the coherent
information is negative, they need to consume entanglement at a rate of H(A|B)ρ ebits per
copy in order for the state transfer process to complete.
Exercise 22.4.1 Suppose that Alice actually possesses the reference R in the above protocols.
Show that Alice and Bob can achieve the following resource inequality:11hψABR i + I(A; R)ψ [q → q] ≥ (H(A)ψ + H(B)ψ + H(R)ψ ) [qq] ,2
2(22.57)

where |ψiABR is some pure state.

22.4.1

The Dual Roles of Quantum Mutual Information

The resource inequality for entanglement-assisted quantum communication in (22.15) andthat for quantum-assisted state transfer in (22.51) appear to be strikingly similar. Bothcontain a noisy resource and both consume a noiseless quantum resource in order to generate
another noiseless quantum resource. We say that these two protocols are related by source–channel duality because we obtain one protocol from another by changing channels to statesand vice versa.
Also, both protocols require the consumed rate of the noiseless quantum resource to beequal to half the quantum mutual information between the system A for which we are trying
to preserve quantum coherence and the environment to which we do not have access. In bothcases, our goal is to break the correlations between the system A and the environment, andthe quantum mutual information is quantifying how much quantum coherence is required
to break these correlations. Both protocols in (22.15) and (22.51) have their rates for thegenerated noiseless quantum resource equal to half the quantum mutual information between
the system A and the system B. Thus, the quantum mutual information is also quantifying
how much quantum correlations we can establish between two systems—it plays the dualrole of quantifying both the destruction and creation of correlations.

22.5 Trade-off Coding
Suppose that you are a communication engineer working at a quantum communication company. Suppose further that your company has made quite a profit from entanglement-assistedclassical communication, beating out the communication rates that other companies canachieve simply because your company has been able to generate high-quality noiseless entanglement between several nodes in its network, while the competitors have not been able
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22.5. TRADE-OFF CODING

627

to do so. But now suppose that your customer base has become so large that there is notenough entanglement to support protocols that achieve the rates given in the entanglementassisted classical capacity theorem (Theorem 21.3.1). Your boss would like you to make thebest of this situation, by determining the optimal rates of classical communication for a fixedentanglement budget. He is hoping that you will be able to design a protocol such that there
will only be a slight decrease in communication rates. You tell him that you will do yourbest.
What should you do in this situation? Your first thought might be that we have already determined unassisted classical codes with a communication rate equal to the channel
Holevo information χ(N ) and we have also determined entanglement-assisted codes with acommunication rate equal to the channel mutual information I(N ). It might seem that areasonable strategy is to mix these two strategies, using some fraction λ of the channel uses
for the unassisted classical code and the other fraction 1 − λ of the channel uses for theentanglement-assisted code. This strategy achieves a rate ofλχ(N ) + (1 − λ) I(N ),(22.58)and it has an error no larger than the sum of the errors of the individual codes (thus,
this error vanishes asymptotically). Meanwhile, it consumes entanglement at a lower rate of(1 − λ) E ebits per channel use, if E is the amount of entanglement that the original protocol
for entanglement-assisted classical communication consumes. This simple mixing strategy
is known as time sharing. You figure this strategy might perform well, and you suggest it
to your boss. After your boss reviews your proposal, he sends it back to you, telling youthat he already thought of this solution and suggests that you are going to have to be a bit
more clever—otherwise, he suspects that the existing customer base will notice the drop in
communication rates.
Another strategy for communication is known as trade-off coding. We explore this strategy in the forthcoming section and in a broader context in Chapter 25. Trade-off coding
beats time sharing for many channels of interest, but for other channels, it just reduces totime sharing. It is not clear a priori how to determine which channels benefit from tradeoff coding, but it certainly depends on the channel for which Alice and Bob are coding.
Chapter 25 follows up on the development here by demonstrating that this trade-off coding
strategy is provably optimal for certain channels, and for general channels, it is optimal in
the sense of regularized formulas. Trade-off coding is our best known way to deal with theabove situation with a fixed entanglement budget, and your boss should be pleased withthese results. Furthermore, we can upgrade the protocol outlined below to one that achieves
entanglement-assisted communication of both classical and quantum information.

22.5.1

Classical Communication with Limited Entanglement

We first show that the resource inequality given in the following theorem is achievable, and wefollow up with an interpretation of it in the context of trade-off coding. We name the protocol
CE trade-off coding because it captures the trade-off between classical communication andentanglement consumption.
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CHAPTER 22. COHERENT COMMUNICATION WITH NOISY RESOURCES

Theorem 22.5.1 (CE Trade-off Coding) The following resource inequality corresponds
to an achievable protocol for entanglement-assisted classical communication over a quantum
channel NA0 →B :hN i + H(A|X)ρ [qq] ≥ I(AX; B)ρ [c → c] ,(22.59)where ρXAB is a state of the following form:
XρXAB ≡pX (x)|xihx|X ⊗ NA0 →B (ϕxAA0 ),(22.60)x

and the states ϕxAA0 are pure.
Proof. The proof of the above trade-off coding theorem exploits the direct parts of boththe HSW coding theorem (Theorem 20.3.1) and the entanglement-assisted classical capacity
theorem (Theorem 21.4.1). In particular, we exploit the constant-composition coding variant of the HSW theorem, described in Section 20.3.1, and that the entanglement-assisted
quantum codewords from Theorem 21.4.1 are tensor-power states after tracing over Bob’s
shares of the entanglement (this is the observation mentioned in Remark 21.4.1). Supposethat Alice and Bob exploit a constant-composition
HSW code for the channel NA0 →B . Such xn (m)nI(X;B)ρ
with ≈ 2
quantum codewords. The Holevo
a code consists of a codebook ρminformation I(X; B)ρ is with respect to some classical–quantum state ρXB whereXρXB ≡pX (x)|xihx|X ⊗ NA0 →B (ρxA0 ),(22.61)xnand each codeword ρx (m) is a tensor-product state of the formρxn (m) = ρx1 (m) ⊗ ρx2 (m) ⊗ · · · ⊗ ρxn (m) .

(22.62)

Corresponding to the codebook is some decoding POVM {Λm
B n }, which Bob can employ to
decode each codeword transmitted through the channel with arbitrarily high probability forall ε ∈ (0, 1):noxn (m)
m∀m Tr ΛB n NA0n →B n (ρA0n ) ≥ 1 − ε.
(22.63)
Recall from the constant-composition HSW coding variant described in Section 20.3.1 that weselect each codeword xn (m) from a typical type class, typical with respect to the distributionpX (x). Let t(x) denote the empirical distribution for the typical type class, and it is suchthat maxx |t(x) − pX (x)| ≤ δ for some δ > 0. This implies that each classical codeword
xn (m) has approximately npX (a1 ) occurrences of the symbol a1 ∈ X , npX (a2 ) occurrences ofthe symbol a2 ∈ X , and so on, for all letters in the alphabet X . However, for a typical typeclass, all sequences have exactly the same empirical distribution, so that there exists somepermutation πm that rearranges each sequence xn (m) in lexicographical order according to
the alphabet X . That is, this permutation πm arranges the sequence xn (m) into |X | blocks,
each of length nt(a1 ), . . . , nt(a|X | ):πm (xn (m)) = a1 · · · a1 a2 · · · a2 · · · a|X | · · · a|X | .
| {z } | {z }| {z }
nt(a1 )

c 2019nt(a2 )

(22.64)nt(a|X | )Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

22.5. TRADE-OFF CODING

629

The same holds true for the corresponding permutation operator πm applied to a quantumnstate ρx (m) corresponding to the sequence xn (m):nπm (ρx (m) ) = ρa1 ⊗ · · · ⊗ ρa1 ⊗ ρa2 ⊗ · · · ⊗ ρa2 ⊗ · · · ⊗ ρa|X | ⊗ · · · ⊗ ρa|X | .
|
{z
} |
{z
}
|
{z
}
nt(a1 )

nt(a2 )

(22.65)nt(a|X | )Now, we assume that n is quite large, so large that each of nt(a1 ), . . . , nt(a|X | ) are large
enough for the law of large numbers to come into play for each block in the permuted sequencenπm (xn (m)) and tensor-product state πm (ρx (m) ). Let ϕxAA0 be a purification of each ρxA0 in
the ensemble {pX (x), ρxA0 }, where we assume that Alice has access to system A0 and Bob
xn (m)
has access to A. Then, for every HSW quantum codeword ρA0n , there is some purificationxn (m)ϕAn A0n , wherexn (m)
x (m)
x (m)
x (m)
(22.66)
ϕAn A0n ≡ ϕA11 A0 ⊗ ϕA22 A0 ⊗ · · · ⊗ ϕAnn A0n ,
1

0n

2

≡ A01 · · · A0n , and Bob has access to An
xn≡ A1 · · · An .

πm (ϕx (m) ) = ϕa1 ⊗ · · · ⊗ ϕa1 ⊗ ϕa2 ⊗ · · · ⊗ ϕa2 ⊗ · · · ⊗ ϕa|X | ⊗ · · · ⊗ ϕa|X | ,
|
{z
} |
{z
}
{z
}
|

(22.67)

Alice has access to the systems A
Applying the permutation πm to any purified tensor-product state ϕ

gives

n

nt(a1 )

nt(a2 )

nt(a|X | )where we have assumed that the permutation applies on both the purification systems An
and the systems A0n .
We can now formulate a strategy for trade-off coding. Alice begins with a standard
classical sequence x̂n that is in lexicographical order, of the form in (22.64). Accordi
ing to this sequence, she arranges the states {ϕaAA0 } to be in |X | blocks, each of lengthni ≡ nt(ai ) ≈ npX (ai )—the resulting state is of the same form as in (22.67). Since nt(ai )is large enough for the law of large numbers to come into play, for each block, there exists
an entanglement-assisted classical code with ≈ 2ni I(A;B)N (ϕai ) entanglement-assisted quantum codewords, where the quantum mutual information I(A; B)N (ϕai ) is with respect to theistate NA0 →B (ϕaAA0 ). Then each of these |X | entanglement-assisted classical codes consumesa≈ ni H(A)ϕAi ebits (i.e., each state (ϕai )⊗ni is produced from ≈ ni H(A)ϕaAi ebits via entanglement dilution and a negligible rate of classical communication). The entanglement-assisted
quantum codewords for each block are of the form
UAni (s(li ))(ϕaAini A0ni )UA† ni (s(li )),(22.68)where li is a message in the message set of size ≈ 2ni I(A;B)ϕai , the state ϕaAini A0ni = ϕaAi1 A0 ⊗1· · · ⊗ ϕaAin A0n , and the unitaries UAni (s(li )) are of the form in (21.38). Observe that theiicodewords in (22.68) are all equal to ρaAi0ni after tracing over Bob’s systems Ani , regardless ofthe particular unitary that Alice applies (this is the content of Remark 21.4.1). Alice then−1
determines the permutation πmneeded to permute the standard sequence x̂n to a codeword−1
sequence xn (m), and she applies the permutation operator πmto her systems A0n so thatxn (m)
her channel input density operator is the HSW quantum codeword ρA0n (we are tracingc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 22. COHERENT COMMUNICATION WITH NOISY RESOURCES

over Bob’s systems An and applying Remark 21.4.1 to obtain this result). She transmits
her systems A0n over the channel to Bob. If Bob ignores his shareof the entanglement in
xn (m)n
A , the state that he receives from the channel is NA0n →B n (ρA0n ). He then applies hisn
HSW measurement {ΛmB n } to the systems B received from the channel, and he determines
the sequence xn (m), and hence the message m, with nearly unit probability. Also, thismeasurement
has negligible disturbance on the state, so that the post-measurement state√is 2 ε-close in trace distance to the state that Alice transmitted through the channel (inwhat follows, we assume that the measurement does not change the state, and we collecterror terms at the end of the proof). Now that he knows m, he applies the permutation
operator πm to his systems B n , and we are assuming that he already has his share An of theentanglement arranged in lexicographical order according to the standard sequence x̂n . His
state is then as follows:
|X |
O


UAni (s(li )) ϕaAini A0ni UA† ni (s(li )).

(22.69)i=1

At this point, he can decode the message li in the ith block by performing a collective measurement on the systems Ani A0ni . He does this for each of the |X | entanglement-assistedclassical codes, and this completes the protocol for trade-off coding. The total error accumulated√ is no larger than the entanglement dilution error, the sum of ε for the first measurement,
2 ε for the disturbance of the state, and |X | ε for the error from the final measurement of
the |X | blocks. Figure 22.4 depicts this protocol for an example.
We now show how the total rate of classical communication adds up to I(AX; B)ρ where
ρXAB is a state of the form in (22.60). First, we can apply the chain rule for quantum mutual
information to observe that the total rate I(AX; B)ρ is the sum of a Holevo information
I(X; B)ρ and a conditional quantum mutual information I(A; B|X)ρ :I(AX; B)ρ = I(X; B)ρ + I(A; B|X)ρ .

(22.70)

They achieve the rate I(X; B)ρ because Bob first decodes the HSW quantum codeword,of which there can be ≈ 2nI(X;B) . His next step is to permute and decode the |X | blocks,
each consisting of an entanglement-assisted classical code on ≈ npX (x) channel uses. Eachentanglement-assisted classical code communicates ≈ npX (x)I(A; B)N (ϕx ) bits and consumes≈ npX (x)H(A)ϕx ebits. Thus, the total rate of classical communication for this last part is# of bits generated≈# of channel uses

P

=

X

x n pX (x)I(A; B)N (ϕx )P

x n pX (x)pX (x)I(A; B)N (ϕx )(22.71)(22.72)x= I(A; B|X)ρ ,c 2019(22.73)

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631

Figure 22.4: A simple protocol for trade-off coding between assisted and unassisted classical communication. Alice wishes to send the classical message m while also sending the messages l1 , l2 , and l3 . Her HSW
codebook has the message m map to the sequence 1231213, which in turn gives the HSW quantum codeword
ρ1 ⊗ ρ2 ⊗ ρ3 ⊗ ρ1 ⊗ ρ2 ⊗ ρ1 ⊗ ρ3 . A purification of these states is the following tensor product of pure states:
ϕ1 ⊗ ϕ2 ⊗ ϕ3 ⊗ ϕ1 ⊗ ϕ2 ⊗ ϕ1 ⊗ ϕ3 , where Bob possesses the purification of each state in the tensor product.
She begins with these states arranged in lexicographic order in three blocks (there are three letters in thisalphabet). For each block i, she encodes the message li with the local unitaries for an entanglement-assistedclassical code. She then permutes her shares of the entangled states according to the permutation associatedwith the message m. She inputs her systems to many uses of the channel, and Bob receives the outputs. His
first action is to ignore his shares of the entanglement and perform a collective HSW measurement on allof the channel outputs. With high probability, he can determine the message m while causing a negligible
disturbance to the state of the channel outputs. Based on the message m, he performs the inverse of the
permutation that Alice used at the encoder. He combines his shares of the entanglement with the permuted
channel outputs. His final three measurements are those given by the three entanglement-assisted codes
Alice used at the encoder, and they detect the messages l1 , l2 , and l3 with high probability.

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CHAPTER 22. COHERENT COMMUNICATION WITH NOISY RESOURCES

and similarly, the total rate of entanglement consumption isP
n pX (x)H(A)ϕx# of ebit consumed≈ xP# of channel uses
n pX (x)X x=
pX (x)H(A)ϕx(22.74)(22.75)x= H(A|X)ρ .

(22.76)This gives the resource inequality in the statement of the theorem.

22.5.2

Trade-off Coding Subsumes Time Sharing

Before proceeding to other trade-off coding settings, we show how time sharing emerges as a
special case of a trade-off coding strategy. Recall from (22.58) that time sharing can achieve
the rate λχ(N ) + (1 − λ) I(N ) for any λ such that 0 ≤ λ ≤ 1. Suppose that φAA0 is the purestate that maximizes the channel mutual information I(N ), and suppose that {pX (x), ψAx 0 }is an ensemble of pure states that maximizes the channel Holevo information χ(N ) (recall
from Theorem 13.3.2 that it is sufficient to consider pure states for maximizing the Holevo
information of a channel). Time sharing simply mixes between these two strategies, and we
can construct a classical–quantum state of the form in (22.60), for which time sharing turns
out to be the strategy executed by the constructed trade-off code:σU XAB ≡ (1 − λ) |0ih0|U ⊗ |0ih0|X ⊗ NA0 →B (φAA0 )X
pX (x)|xihx|X ⊗ |0ih0|A ⊗ NA0 →B (ψAx 0 ). (22.77)+ λ|1ih1|U ⊗x

In the above, the register U is acting as a classical binary flag to indicate whether the codeshould be an entanglement-assisted classical capacity-achieving code or a code that achieves
the channel’s Holevo information. The amount of classical bits that Alice can communicate
to Bob using a trade-off code is I(AU X; B)σ , where we have assumed that U and X together
form the classical register. We can then evaluate this mutual information by applying thechain rule:
I(AU X; B)σ = I(A; B|XU )σ + I(X; B|U )σ + I(U ; B)σ"#X= (1 − λ) I(A; B)N (φ) + λpX (x)I(A; B)|0ih0|⊗N (ψx ) +(22.78)x(1 − λ) I(X; B)|0ih0|⊗N (φ) + λI(X; B){p(x),ψx } + I(U ; B)σ≥ (1 − λ) I(N ) + λχ(N ).

(22.79)(22.80)The second equality follows by evaluating the first two conditional mutual informations. The
inequality follows from the assumptions that I(N ) = I(A; B)N (φ) and χ(N ) = I(X; B){p(x),ψx } ,the fact that quantum mutual information vanishes on product states, and I(U ; B)σ ≥ 0.
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22.5. TRADE-OFF CODING

633

Thus, in certain cases, thisPstrategyx might do slightly better than time sharing, but forchannels for which φA0 = x p(x)ψA0 , this strategy is equivalent to time sharing because
I(U ; B)σ = 0 in this latter case.
Thus, time sharing emerges as a special case of trade-off coding. In general, we can try
to see if trade-off coding beats time sharing for certain channels by optimizing the rates in
Theorem 22.5.1 over all possible choices of states of the form in (22.60).

22.5.3

Trading between Coherent and Classical Communication

We obtain the following corollary of Theorem 22.5.1, simply by upgrading the |X | entanglementassisted classical codes to entanglement-assisted coherent codes. The upgrading is along thesame lines as that in the proof of Theorem 22.1.1, and for this reason, we omit the proof.
Corollary 22.5.1 The following resource inequality corresponds to an achievable protocol
for entanglement-assisted coherent communication over a quantum channel N :hN i + H(A|X)ρ [qq] ≥ I(A; B|X)ρ [q → qq] + I(X; B)ρ [c → c] ,where ρXAB is a state of the following form:XpX (x)|xihx|X ⊗ NA0 →B (ϕxAA0 ),ρXAB ≡(22.81)(22.82)x

and the states ϕxAA0 are pure.

22.5.4

Trading between Classical Communication and EntanglementAssisted Quantum Communication

We end this section with a protocol that achieves entanglement-assisted communication ofboth classical and quantum information. It is essential to the trade-off between a quantum
channel and the three resources of noiseless classical communication, noiseless quantum communication, and noiseless entanglement. We study this trade-off in full detail in Chapter 25,
where we show that combining this protocol with teleportation, super-dense coding, andentanglement distribution is sufficient to achieve any task in dynamic quantum Shannon
theory involving the three unit resources.
Corollary 22.5.2 (CQE Trade-off Coding) The following resource inequality corresponds
to an achievable protocol for entanglement-assisted communication of classical and quantum
information over a quantum channel NA0 →B :1
1
(22.83)hN i + I(A; E|X)ρ [qq] ≥ I(A; B|X)ρ [q → q] + I(X; B)ρ [c → c] ,22where ρXABE is a state of the following form:XρXABE ≡pX (x)|xihx|X ⊗ UAN0 →BE (ϕxAA0 ),(22.84)x

the states ϕxAA0 are pure, and UAN0 →BE is an isometric extension of the channel NA0 →B .
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CHAPTER 22. COHERENT COMMUNICATION WITH NOISY RESOURCES

Proof. Consider the following chain of resource inequalities:
hN i + H(A|X)ρ [qq]≥ I(A; B|X)ρ [q → qq] + I(X; B)ρ [c → c]
11
≥ I(A; B|X)ρ [qq] + I(A; B|X)ρ [q → q] + I(X; B)ρ [c → c] .
22(22.85)(22.86)

The first inequality is the statement in Corollary 22.5.1, and the second inequality follows from the coherent communication identity. After resource cancelation and noting that
H(A|X)ρ − 12 I(A; B|X)ρ = 12 I(A; E|X)ρ , the resulting resource inequality is equivalent to
the one in (22.83).

22.5.5

Trading between Classical and Quantum Communication

Our final trade-off coding protocol that we consider is that between classical and quantum
communication. The proof of the resource inequality below follows by combining the protocolin Corollary 22.5.2 with entanglement distribution, in much the same way as we did in
Corollary 22.2.1. Thus, we omit the proof.
Corollary 22.5.3 (CQ Trade-off Coding) The following resource inequality corresponds
to an achievable protocol for simultaneous classical and quantum communication over a quantum channel NA0 →B :hN i ≥ I(AiBX)ρ [q → q] + I(X; B)ρ [c → c] ,where ρXAB is a state of the following form:
XρXAB ≡pX (x)|xihx|X ⊗ NA0 →B (ϕxAA0 ),(22.87)(22.88)x

and the states ϕxAA0 are pure.

22.6 Concluding Remarks
The maintainence of quantum coherence is the theme of this chapter. Alice and Bob can
execute powerful protocols if they perform encoding and decoding in superposition. In bothentanglement-assisted coherent communication and coherent state transfer, Alice performs
controlled gates instead of conditional gates and Bob performs coherent measurements thatplace measurement outcomes in an ancilla register without destroying superpositions. Also,
Bob’s final action in both of these protocols is to perform a controlled decoupling unitary,
ensuring that the state of the environment is independent of Alice and Bob’s final state.
Thus, the same protocol accomplishes the different tasks of entanglement-assisted coherent communication and coherent state transfer, and these in turn can generate a wholec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

22.7. HISTORY AND FURTHER READING

635

host of other protocols by combining them with entanglement distribution and the coherent and incoherent versions of teleportation and super-dense coding. Among these othergenerated protocols are entanglement-assisted quantum communication, quantum communication, quantum-assisted state transfer, and classical-assisted state transfer. The exercisesin this chapter explore further possibilities if Alice has access to the environments of thedifferent protocols—the most general version of coherent teleportation arises in such a case.
Trade-off coding is the theme of the last part of this chapter. Here, we are addressing
the question: Given a fixed amount of a certain resource, how much of another resource canAlice and Bob generate? Noisy quantum channels are the most fundamental description ofa medium over which information can propagate, and it is thus important to understand thebest ways to make effective use of such a resource for a variety of purposes. We determineda protocol that achieves the task of entanglement-assisted communication of classical andquantum information, simply by combining the protocols we have already found for classicalcommunication and entanglement-assisted coherent communication. Chapter 25 continues
this theme of trade-off coding in a much broader context and demonstrates that the protocol given here, when combined with teleportation, super-dense coding, and entanglement
distribution, is optimal for some channels of interest and essentially optimal in the generalcase.

22.7 History and Further Reading
Devetak et al. (2004, 2008) showed that it is possible to make the protocols for entanglementassisted classical communication and noisy super-dense coding coherent, leading to Theorems 22.1.1 and 22.4.1. They called these protocols the “father” and “mother,” respectively,
because they generated many other protocols in quantum Shannon theory by combining
them with entanglement distribution, teleportation, and super-dense coding. Horodecki et al.
(2001) formulated a protocol for noisy super-dense coding, but our protocol here makes useof the coding technique in Hsieh et al. (2008a). Shor (2004b) first proved a coding theorem
for trading between assisted and unassisted classical communication, and Devetak and Shor
(2005) followed up on this result by finding a scheme for trade-off coding between classicaland quantum communication. Some time later, Hsieh and Wilde (2010a) generalized thesetwo coding schemes to produce the result of Theorem 22.5.2. The proofs given in this chapter
for these trade-off results are different from those which appeared in (Shor, 2004b; Devetak
and Shor, 2005; Hsieh and Wilde, 2010a).

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c 2019CHAPTER 22. COHERENT COMMUNICATION WITH NOISY RESOURCES

Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 23

Private Classical Communication
We have now seen in Chapters 20–22 how Alice can communicate classical or quantum information to Bob, perhaps even with the help of shared entanglement. One might argue thatthese communication tasks are the most fundamental tasks in quantum Shannon theory,
given that they have furthered our understanding of the nature of information transmission
over quantum channels. However, when discussing the communication of classical information, we made no stipulation as to whether this classical information should be public, sothat any third party might have partial or full access to it, or private, so that no third party
has access.
This chapter establishes the private classical capacity theorem, which gives the maximum
rate at which Alice can communicate classical information privately to Bob without anyone
else in the universe knowing what she sent to him. A variation of the information-processing
task corresponding to this theorem was one of the earliest studied in quantum information
theory, with the Bennett–Brassard-84 quantum key distribution protocol being the first
proposed protocol for exploiting quantum mechanics to establish a shared secret key between
two parties. The private classical capacity theorem is important for quantum key distribution
because it establishes the maximum rate at which two parties can generate a shared secretkey.
Another equally important, but less obvious utility of private classical communication
is in establishing a protocol for quantum communication at the coherent information rate.
Section 22.2 demonstrated a somewhat roundabout way of arriving at the conclusion that it
is possible to communicate quantum information reliably at the coherent information rate—
recall that we “coherified” the entanglement-assisted classical capacity theorem and thenexploited the coherent communication identity and catalytic use of entanglement. Establishing achievability of the coherent information rate via private classical coding is anotherway of arriving at the same result, with the added benefit that the resulting protocol does
not require the catalytic use of entanglement.
The intuition for quantum communication via privacy arises from the no-cloning theorem.
Suppose that Alice is able to communicate private classical messages to Bob, so that thechannel’s environment (Eve) is not able to distinguish which message Alice is transmitting
637

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CHAPTER 23. PRIVATE CLASSICAL COMMUNICATION

to Bob. That is, Eve’s state is completely independent of Alice’s message if the transmitted
message is private. Then we might expect it to be possible to make a coherent version ofthis private classical code by exploiting superpositions of the private classical codewords.
Since Eve’s states are independent of the quantum message that Alice is sending throughthe channel, she is not able to “steal” any of the coherence in Alice’s superposed states.
Given that the overall evolution of the channel to Bob and Eve is unitary and the fact that
Eve does not receive any quantum information with this scheme, we should expect that thequantum information appears at the receiving end of the channel so that Bob can decode
it. Were Eve able to obtain any information about the private classical messages, then Bob
would not be able to decode all of the quantum information when they construct a coherent
version of this private classical code. Otherwise, they would violate the no-cloning theorem.
We discuss this important application of private classical communication in the next chapter.
This chapter follows a similar structure as previous chapters. We first detail the informationprocessing task for private classical communication. Section 23.2 then states the private
classical capacity theorem, with the following two sections proving the achievability partand the converse part. We end with a general discussion of the private classical capacity anda brief overview of the secret-key-assisted private classical capacity.

23.1 The Information-Processing Task
We begin by describing the information-processing task for private classical communication(we define an (n, P, ε) private classical code). Alice selects a message m from a set M of messages. Alice prepares some state ρm
A0n as input to many uses of the quantum channel NA0 →B
and transmits it, producing the following state at Bob’s receiving end:
NA0n →B n (ρmA0n ),(23.1)

where NA0n →B n ≡ (NA0 →B )⊗n .
Bob employs a decoding POVM {Λm } in order to detect Alice’s transmitted message m.
The probability of error for a particular message m is as follows:pe (m) = Tr {(I − Λm ) NA0n →B n (ρmA0n )} ,(23.2)

so that the maximal probability of error isp∗e ≡ max pe (m),m∈M

(23.3)

where p∗e ≤ ε ∈ [0, 1] for an (n, P, ε) code. The rate P of the code isP ≡1
log |M| .
n(23.4)

So far, the above specification of a private classical code is nearly identical to that for thetransmission of classical information outlined in Section 20.2. What distinguishes a private
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23.1. THE INFORMATION-PROCESSING TASK

639

Figure 23.1: The information-processing task for private classical communication. Alice encodes some
private message m into a quantum codeword ρm
A0n and transmits it over many uses of a quantum channel.
The goal of such a protocol is for Bob to be able to reliably distinguish the message, while the channel’senvironment Eve should not be able to learn anything about it.

classical code from a public one is the following extra condition for privacy. Let UAN0 →BE be
bA0 →E toan isometric extension of the channel NA0 →B , so that the complementary channel N
the environment Eve is as follows:
bA0 →E (σ) ≡ TrB {UAN0 →BE (σ)}.
N(23.5)

If Alice transmits a message m, then the state for Eve is as follows:
bA0n →E n (ρm0n ).
ωEmn ≡ N
A(23.6)

Our condition for ε-privacy is that Eve’s state is always close to a constant state σE n ,regardless of which message m Alice transmits through the channel:∀m ∈ M :1 m
kω n − σE n k1 ≤ ε.
2 E

(23.7)

This definition is the strongest definition of privacy because it implies that Eve cannot learn
anything about the message m that Alice transmits through the channel. Figure 23.1 depicts
the information-processing task for private classical communication.
A rate P of private classical communication is achievable for NA0 →B if there exists an(n, P − δ, ε) private classical code for all ε ∈ (0, 1), δ > 0, and sufficiently large n, where εcharacterizes both the reliability and the privacy of the code. The private classical capacity
CP (N ) of a channel NA0 →B is equal to the supremum of all achievable rates for private
classical communication.
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23.1.1

Mutual Information of Eve

We comment briefly how the condition in (23.7) implies that Eve has little mutual information about the transmitted message. It follows from (23.7) that
ε≥

1
1 X 1
kωEmn − σE n k1 = kωM E n − πM ⊗ σE n k1 ,2 m∈M |M|2

where
ωM E n ≡1|mihm|M ⊗ ωEmn .
|M|
m∈M
X

(23.8)(23.9)

The criterion in (23.8) implies that Eve’s Holevo information with M is small:I(M ; E n )ω = H(M )ω − H(M |E n )ω= H(M |E n )π⊗σ − H(M |E n )ω≤ ε log |M| + g2 (ε).

(23.10)(23.11)(23.12)The inequality follows from applying the AFW inequality (Theorem 11.10.3) to both entropies. Thus, if ε is exponentially small in n (which will be the case for our codes), thenit is possible to make Eve’s information about the message become arbitrarily small in theasymptotic limit.

23.2 The Private Classical Capacity Theorem
We now state the main theorem of this chapter, the private classical capacity theorem.
Theorem 23.2.1 (Devetak–Cai–Winter–Yeung) The private classical capacity CP (N )of a quantum channel NA0 →B is equal to the regularized private information of the channel:
CP (N ) = Preg (N ),(23.13)

where

1
P (N ⊗k ).
k→∞ k
The private information P (N ) is defined as
Preg (N ) ≡ lim

P (N ) ≡ max [I(X; B)σ − I(X; E)σ ] ,ρwhere ρXA0 is a classical–quantum state of the following form:
XρXA0 ≡pX (x)|xihx|X ⊗ ρxA0 ,(23.14)(23.15)(23.16)x

and σXBE ≡ UAN0 →BE (ρXA0 ), with UAN0 →BE an isometric extension of the channel NA0 →B .
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23.3. THE DIRECT CODING THEOREM

641

We first prove the achievability part of the coding theorem and follow with the converse
proof. Recall that the private information is additive whenever the channel is degradable(Theorem 13.6.3). Thus, for this class of channels, the regularization in (23.14) is not necessary and the private information of the channel is equal to the private classical capacity(in fact, the results from Theorem 13.6.2 and the next chapter demonstrate that the private
information of a degradable channel is also equal to its quantum capacity). The regularization of the private information seems to be necessary in general in order to characterize theprivate capacity because there is an example of a channel for which the private information
is superadditive.

23.3 The Direct Coding Theorem
This section gives a proof that the private information in (23.15) is an achievable rate forprivate classical communication over a quantum channel NA0 →B . We first give the intuition
behind the protocol. Alice’s goal is to build a doubly indexed codebook {xn (m, k)}m∈M,k∈K
that satisfies two properties:
1. Bob should be able to detect the message m and the “junk” variable k with highprobability. From the classical coding theorem of Chapter 20, our intuition is that heshould be able to do so as long as |M| |K| ≈ 2nI(X;B) .
2. Randomizing over the “junk” variable k should approximately cover the typical subspace of Eve’s system, so that every state of Eve depending on the message m looks
like a constant, independent of the message m Alice sends (we would like the code tosatisfy (23.7)). Our intuition from the covering lemma (Chapter 17) is that the sizeof the “junk” variable set K needs to be at least |K| ≈ 2nI(X;E) in order for Alice toapproximately cover Eve’s typical subspace.
Our method for generating a code is again random because we can invoke the typicality
properties that hold in the asymptotic limit of many channel uses. Thus, if Alice chooses
a code that satisfies the above criteria, she can send approximately |M| ≈ 2n[I(X;B)−I(X;E)]
distinguishable signals to Bob such that they are indistinguishable to Eve. We devote theremainder of this section to proving that the above intuition is correct. Figure 23.2 displays
the anatomy of a private classical code.

23.3.1

Dimensionality Arguments

Before giving the proof of achievability, we confirm the above intuition with some dimensionality arguments and show how to satisfy the conditions of both the packing and coveringlemmas. Suppose that Alice has some ensemble {pX (x), ρxA0 } from which she can generaterandom codes. Let UAN0 →BE denote an isometric extension of the channel NA0 →B , and let ρxBE
denote the joint state of Bob and Eve after Alice inputs ρxA0 :ρxBE ≡ UAN0 →BE (ρxA0 ).
c 2019(23.17)

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CHAPTER 23. PRIVATE CLASSICAL COMMUNICATION

Figure 23.2: The anatomy of a code for private classical communication. In this illustrative example,
Alice has eight codewords, with each depicted as a • and indexed by m ∈ {1, 2} and k ∈ {1, 2, 3, 4}. Thus,
she is interested in sending one of two messages and has the “junk” variable k available for randomizing
Eve’s state. Each classical codeword xn (m, k) maps to a distinguishable subspace on Bob’s typical subspace(we show two of the mappings in the figure, while displaying eight distinguishable subspaces). From thepacking lemma, our intuition is that Alice can reliably send about 2nI(X;B) distinguishable signals. The
codewords {xn (1, k)}k∈{1,2,3,4} and {xn (2, k)}k∈{1,2,3,4} are each grouped in a box to indicate that they
form a privacy amplification set. When randomizing k, the codewords {xn (1, k)}k∈{1,2,3,4} uniformly coverEve’s typical subspace (and so does the set {xn (2, k)}k∈{1,2,3,4} ), so that it becomes nearly impossible in
the asymptotic limit for Eve to distinguish whether Alice is sending a codeword in {xn (1, k)}k∈{1,2,3,4} or{xn (2, k)}k∈{1,2,3,4} . In this way, Eve cannot determine which message Alice is transmitting. The minimum
size for each privacy amplification set in the asymptotic limit is ≈ 2nI(X;E) .

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23.3. THE DIRECT CODING THEOREM

Party
Alice
Bob
Bob conditioned on xn
Eve
Eve conditioned on xn

Quantity
X
ρB n
n
ρxB n
ρE n
n
ρxE n

643

Typical Set/Subspace
n
TδX
TBδ n
TBδ n |xnTEδ n
TEδ n |xn

Projector
N/AΠδB n
ΠδB n |xn
ΠδE n
ΠδE n |xn

Table 23.1: This table lists several mathematical quantities involved in the construction of a random
private code. The first column lists the party to whom the quantities belong. The second column lists therandom classical or quantum states. The third column gives the appropriate typical set or subspace. The
final column lists the appropriate projector onto the typical subspace for the quantum states.

The local respective density operators for Bob and Eve given a letter x are as follows:ρxB ≡ TrE {ρxBE } ,ρxE ≡ TrB {ρxBE } .

The expected respective density operators for Bob and Eve are as follows:
XX
pX (x)ρxE .
pX (x)ρxB ,
ρE =
ρB =

(23.18)(23.19)x

x

Given a particular input sequence xn , we define the nth extensions of the above states asfollows: nn
ρxB n ≡ TrE n ρxB n E n ,(23.20)nn
ρxE n ≡ TrB n ρxB n E n ,(23.21)Xn
ρB n =
pX n (xn )ρxB n ,(23.22)xn ∈X n

ρE n =Xn

pX n (xn )ρxE n .

(23.23)xn ∈X nTable 23.1 organizes these various density operators and their corresponding typical subspaces and projectors.
following four conditions corresponding to the packing lemma hold for Bob’s states xThen
ρB n , Bob’s average density operator ρB n , Bob’s typical subspace TBδ n , and Bob’s conditionally typical subspace TBδ n |xn :nTr ΠδB n ρxB n ≥ 1 − ε,nTr ΠδB n |xn ρxB n ≥ 1 − ε,Tr ΠδB n |xn ≤ 2n(H(B|X)+cδ) ,ΠδB n ρB n ΠδB n ≤ 2−n(H(B)−cδ) ΠδB n ,(23.24)(23.25)(23.26)(23.27)

where c is some positive constant (see Properties 15.2.7, 15.1.1, 15.1.2, and 15.1.3).
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CHAPTER 23. PRIVATE CLASSICAL COMMUNICATION

following four conditions corresponding to the covering lemma hold for Eve’s states xThen
ρE n , Eve’s typical subspace TEδ n , and Eve’s conditionally typical subspace TEδ n |xn :nTr ΠδE n ρxE n ≥ 1 − ε,nTr ΠδE n |xn ρxE n ≥ 1 − ε,Tr ΠδE n ≤ 2n(H(E)+cδ) ,ΠδE n |xn ρxnEnΠδE n |xn ≤ 2−n(H(E|X)−cδ) ΠδE n |xn .

(23.28)(23.29)(23.30)(23.31)

The above properties suggest that we can use the methods of both the packing lemma
and the covering lemma for constructing a private code. Consider two sets M and K withthe following respective sizes:|M| = 2n[I(X;B)−I(X;E)−6cδ] ,(23.32)|K| = 2n[I(X;E)+3cδ] ,(23.33)so that the product set M × K indexed by the ordered pairs (m, k) is of size
|M × K| = |M| |K| = 2n[I(X;B)−3cδ] .

(23.34)The sizes of these sets suggest that we can use the product set M × K for sending classicalinformation, but we can use |M| “privacy amplification” sets each of size |K| for reducing
Eve’s knowledge of the message m (see Figure 23.2).

23.3.2

Random Code Construction

We now argue for the existence of a good private classical code with rate P ≈ I(X; B) −I(X; E) if Alice selects it randomly according to the ensemblen{p0X 0n (xn ), ρxA0n },(23.35)where p0X 0n (xn ) is the pruned distribution (see Section 20.3.1—recall that this distribution isclose to the i.i.d. distribution). Let us choose |M| |K| random variables X n (m, k) according
to the distribution p0X 0n (xn ) where the realizations of the random variables X n (m, k) takevalues in X n . After selecting these codewords randomly, the code C = {xn (m, k)}m∈M,k∈K isthen a fixed set of codewords xn (m, k) depending on the message m and the randomization
variable k.
We first consider how well Bob can distinguish the pair (m, k) and argue that the random
code is a good code in the sense that the expectation of the average error probability overall codes is low. The packing lemma is the basis of our argument. By applying the packing lemma (Lemma 16.3.1) to (23.24)–(23.27), there exists a POVM {Λm,k }(m,k)∈M×K correX n (m,k)
sponding to the random choice of code that reliably distinguishes the states {ρB n}m∈M,k∈K
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23.3. THE DIRECT CODING THEOREM

645

in the following sense:
(

)
o
X X n
1
X n (m,k)
EC {p̄e (C)} = 1 − ECTr Λm,k ρB n
|M| |K| m∈M k∈K n(H(B|X)+cδ)√ 
2|M × K|≤2 ε+2 ε +42n(H(B)−cδ) n(H(B|X)+cδ) n[I(X;B)−3cδ] √ 
22=2 ε+2 ε +4n(H(B)−cδ)
2
√ 
−ncδ
=2 ε+2 ε +4·2
≡ ε0 ,(23.36)(23.37)(23.38)(23.39)where the first equality follows by definition, the first inequality follows by application of thepacking lemma to the conditions in (23.24)–(23.27), the second equality follows by substitution of (23.34), and the last equality follows by a straightforward calculation. We can make
ε0 arbitrarily small by choosing n large enough.
X n (m,k)
Let us now consider the corresponding density operators ρE n
for Eve. Consider
dividing the random code C into |M| privacy amplification sets, each of size |K|. Each
X n (m,k)privacy amplification set Cm ≡ {ρE n}k∈K of density operators forms a good covering
code according to the covering lemma (Lemma 17.2.1). The fake density operator of eachprivacy amplification set Cm is as follows:
1 X X n (m,k)ρ n,(23.40)ρ̂m≡nE|K| k∈K E
because Alice chooses the randomizing variable k uniformly at random. The obfuscation
error oe (Cm ) of each privacy amplification set Cm is as follows:oe (Cm ) ≡ kρ̂m
E n − ρ E n k1 ,(23.41)

where ρE n is defined in (23.23). The covering lemma (Lemma 17.2.1) states that the obfuscation error for each random privacy amplification set Cm has a high probability of being
small if n is sufficiently large and |K| is chosen as in (23.33):
√√
Pr oe (Cm ) ≤ ε + 4 ε + 24 4 ε 3−ε |K| 2n(H(E|X)−cδ)n≥ 1 − 2dE exp(23.42)42n(H(E)+cδ) 3 n[I(X;E)+3cδ] n(H(E|X)−cδ) −ε 22n= 1 − 2dE exp(23.43)42n(H(E)+cδ) 3
−ε ncδ
n= 1 − 2dE exp
2.
(23.44)4
In particular, let us choose n large enough so that the following bound holds:√√εPr oe (Cm ) ≤ ε + 4 ε + 24 4 ε ≥ 1 −.
|M|
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CHAPTER 23. PRIVATE CLASSICAL COMMUNICATION
That we can do so follows from the important fact that exp −ε3 2ncδ / (4) is doubly exponentially decreasing in n. (We also see here why it is necessary to have the “wiggle room”
given by an arbitrarily small, yet strictly positive δ.)This random construction already has some of the desirable features that we are looking
for in a private code just by choosing n to be sufficiently large. The expectation of Bob’s
average error probability for detecting the pair m, k is small, and the obfuscation error ofeach privacy amplification set has a high probability of being small. Our hope is that there
exists some code for which Bob can retrieve the message m with the guarantee that Eve’s
state is independent of this message m. We argue in the next two sections that such a good
private code exists.

23.3.3

Derandomization

We now apply a derandomization argument similar to the one that is needed in the proofof the HSW coding theorem. The argument in this case is more subtle because we would
like to find a code that has good classical communication with the guarantee that it also has
good privacy. We need to determine the probability over all codes that there exists a good
private code. If this probability is non-zero, then we are sure that a good private code exists.
As we have said at the beginning of this section, a good private code has two qualities:
the code is ε-good for classical communication and it is ε-private as well. Let E0 denote theevent that the random code C is ε-good for classical communication:
E0 = {p̄e (C) ≤ ε} ,(23.46)

where we restrict the performance criterion to the average probability of error for now. Let
Em denote the event that the mth message in the random code is ε-private:
Em = {oe (Cm ) ≤ ε} .

(23.47)

We would like all of the above events to be true, or, equivalently, we would like the intersectionof the above events to occur:\
Epriv ≡ E0 ∩Em .
(23.48)m∈M

If there is a positive probability over all codes that the above event is true, then there exists
a particular code that satisfies the above conditions. Let us instead consider the complementof the above event (the event that a good private code does not exist):[cc.
(23.49)Epriv= E0c ∪Emm∈M

We can then exploit the union bound from probability theory to bound the probability ofcthe complementary event Epriv
as follows:
(
)
[
X
c
c
Pr E0c ∪Em≤ Pr {E0c } +Pr {Em}.
(23.50)m∈M
c 2019m∈M

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23.3. THE DIRECT CODING THEOREM

647

csmall, then the probability of the event
So if we can make the probability of the event Epriv
Epriv , that there exists a good private code, is high.
Let us first bound the probability of the event E0c . Markov’s inequality states that thefollowing holds for a non-negative random variable Y and strictly positive α:
E {Y }Pr {Y ≥ α} ≤.
(23.51)αWe can apply Markov’s inequality because the random average error probability p̄e (C) isalways non-negative:no E {p̄ (C)}
√
ε0
4 0
C
e
c
0 3/4
Pr {E0 } = Pr p̄e (C) ≥ (ε )≤≤=ε.
(23.52)
3/4
3/4
0
0
(ε )
(ε )
So we now have a good bound on the probability of the complementary event E0c .
c. The bounds in the previous section
Let us now bound the probability of the events Em
already give us what we need:√√c
Pr {Em} = Pr oe (Cm ) > ε + 4 ε + 24 4 ε(23.53)ε<,(23.54)|M|
implying thatXεc
Pr {Em} < |M|= ε.
(23.55)|M|m∈M

So it now follows that the probability of the complementary event is small:√ c
4
Pr Epriv≤ ε0 + ε,
and there is a high probability that there is a good code:√
4 0
ε +ε .
Pr {Epriv } ≥ 1 −(23.56)(23.57)

Thus, there exists a particular code C such that its average probability of error is small
for decoding the classical information:p̄e (C) ≤ (ε0 )3/4,and the obfuscation error of each privacy amplification set is small:√√∀m : oe (Cm ) ≤ ε + 4 ε + 24 4 ε.

(23.58)(23.59)The derandomized code C is as follows:C ≡ {xn (m, k)}m∈M,k∈K ,(23.60)so that each codeword xn (m, k) is a deterministic variable. Each privacy amplification set
for the derandomized code is as follows:Cm ≡ {xn (m, k)}k∈K .

(23.61)

The result in (23.57) is perhaps astonishing in hindsight. By choosing a private code ina random way and choosing the block length n of the private code to be sufficiently large,
the overwhelming majority of codes constructed in this fashion are good private codes!
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CHAPTER 23. PRIVATE CLASSICAL COMMUNICATION

23.3.4

Expurgation

We would like to strengthen the above result even more, so that the code has a low maximal
probability of error, not just a low average error probability. We expurgate codewords fromthe code as before, but we have to be careful with the expurgation argument because weneed to make sure that the code still has good privacy after expurgation.
We can apply Markov’s inequality for the expurgation in a way similar as in Exercise 2.2.1.
It is possible to apply Markov’s inequality√ to the bound on the average error probability in(23.52)to show that at most a fraction ε0 of the√codewords have error probability greater√
4 0
than ε . We could merely expurgate the worst ε0 codewords from the private code. But
expurgating in this fashion does not guarantee that each privacy amplification
set has the√0same number of codewords. Therefore, we expurgate the worst fraction ε√of the codewords
the privacy
in each privacy amplification set. We then expurgate the worst fraction ε0 of √00amplification sets. The expurgated sets M and K both become a fraction 1 − ε0 of their
original size. We denote the expurgated code as follows:C 0 ≡ {xn (m, k)}m∈M0 ,k∈K0 ,(23.62)and the expurgated code has the following privacy amplification sets:0Cm≡ {xn (m, k)}k∈K0 .

(23.63)The expurgation has a negligible impact on the rate of the private code when n is large.
Does each privacy amplification set still have good privacy properties after performing
the above expurgation? The fake density operator for each expurgated privacy amplification
set is as follows:
1 X xn (m,k)
ρ̂m0
≡
ρE n
.
(23.64)
n
E
0 |
|Cm
k∈K0
√
It is possible to show that the fake density operators in the derandomized code are 2 ε0 -closein trace distance to the fake density operators in the expurgated code:√m∀m ∈ M0
kρ̂m0
−
ρ̂
k
≤
2
ε0 ,
(23.65)nnEE
1because these operators only lose a small fraction of their mass after expurgation.
We now drop the primed notation to denote the expurgated code. It follows that the
expurgated code C has good privacy:
√
√
√
4
0
∀m ∈ Mkρ̂m(23.66)E n − ρE n k1 ≤ ε + 4 ε + 24 ε + 2 ε ,and reliable communication:∀m ∈ M, k ∈ K

pe (C, m, k) ≤

√
4 0
ε.

(23.67)

The first expression follows by application of the triangle inequality to (23.59) and (23.65).
c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

23.4. THE CONVERSE THEOREM

649

We end the proof by summarizing the operation of the private code. Alice chooses a
message m from the message set M and the randomization variable k uniformlyn at randomx (m,k)from K. She encodes these as xn (m, k) and inputs the quantum codeword ρA0n
to thexn (m,k)channel. Bob receives the state ρB n
and performs a POVM
{Λm,k }(m,k)∈M×K that deter√
4 0mines the pair m and k correctly with probability 1 − ε . The code guarantees that Eve
has almost no knowledge about the message m. The private communication rate P of theprivate code is equal to the following expression:1
log |M| = I(X; B) − I(X; E) − 6cδ.
(23.68)nThis concludes the proof of the direct coding theorem.
We remark that the above proof applies even in the scenario in which Eve does not get thefull purification of the channel. That is, suppose that the channel has one input A0 for Aliceand two outputs B and E for Bob and Eve, respectively. Then the channel has an isometricextension to some environment F . In this scenario, the private information I(X; B)−I(X; E)
is still achievable for some classical–quantum state input such that the Holevo informationdifference is non-negative. However, one could always give both outputs E and F to aneavesdropper (this is the setting that we proved in the above theorem). Giving the full
purification of the channel to the environment ensures that the transmitted information isprivate from the “rest of the universe” (anyone other than the intended receiver), and it thus
yields the highest standard of security in any protocol for private information transmission.
P ≡23.4 The Converse Theorem
We now prove the converse part of the private classical capacity theorem, which demonstrates
that the regularization of the private information is an upper bound on the private classicalcapacity. We suppose instead that Alice and Bob are trying to accomplish the task of secretkey generation. As we have argued in other converse proofs (see Sections 20.3.2 and 21.5), thecapacity for generating this static resource can only be larger than the capacity for private
classical communication because Alice and Bob can always use a noiseless private channel
to establish a shared secret key. In such a task, Alice first prepares a maximally correlated
state ΦM M 0 and encodes the M 0 variable as a codeword ρm
A0n . This encoding leads to a stateof the following form, after Alice transmits her systems A0n over many independent uses of
the channel:
1 X(23.69)
ωM B n E n ≡|mihm|M ⊗ UAN0n →B n E n (ρmA0n ).
|M| m∈M
Bob finally applies a decoding channel DB n →M 0 to recover his share of the secret key:
ωM M 0 E n ≡ DB n →M 0 (ωM B n E n ).

(23.70)The following condition holds for an (n, [log |M|] /n, ε) protocol for secret key generation:1
ωM M 0 E n − ΦM M 0 ⊗ σE n 1 ≤ ε,2
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CHAPTER 23. PRIVATE CLASSICAL COMMUNICATION

so that Eve’s state σE n is a constant state independent of the secret key ΦM M 0 . In particular,
the above condition implies that Eve’s information about M is small:I(M ; E n )ω ≤ f (|M| , ε),(23.72)where we apply the reasoning in Section 23.1.1, with f (|M| , ε) ≡ ε log |M| + g2 (ε). The rateof secret key generation is equal to n1 log |M|. Consider the following chain of inequalities:
log |M| = I(M ; M 0 )Φ≤ I(M ; M 0 )ω + f (|M| , ε)
≤ I(M ; B n )ω + f (|M| , ε)
≤ I(M ; B n )ω − I(M ; E n )ω + 2f (|M| , ε)
≤ P (N ⊗n ) + 2f (|M| , ε).

(23.73)(23.74)(23.75)(23.76)(23.77)

The first equality follows because the mutual information of the shared randomness stateΦM M 0 is equal to log |M|. The first inequality follows from applying the AFW inequality to(23.71). The second inequality follows from quantum data processing. The third inequality
follows from (23.72), and the final inequality follows because the classical–quantum state in(23.69) has a particular distribution and choice of states, and this choice always leads to avalue of the private information that cannot be larger than the private information of thetensor-product channel N ⊗n . Putting everything together, we find that1
2
1
log |M| (1 − 2ε) ≤ P (N ⊗n ) + g2 (ε).
nnn(23.78)Thus, if we are considering a sequence of (n, [log |M|] /n, εn ) private classical communication
protocols with rate P − δn = n1 log |M|, such that limn→∞ εn = limn→∞ δn = 0, then theabove bound becomes1
2(P − δn ) (1 − 2εn ) ≤ P (N ⊗n ) + g2 (εn ).
(23.79)nn
Taking the limit as n → ∞ then establishes that an achievable rate P necessarily satisfies
P ≤ Preg (N ), where Preg (N ) is the regularized private information given in (23.14).
Exercise 23.4.1 Prove that free access to a forward public classical channel from Alice to
Bob cannot improve the private classical capacity of a quantum channel.

23.5 Discussion of Private Classical Capacity
This last section discusses some important aspects of the private classical capacity. Twoof these results have to do with the fact that Theorem 23.2.1 only provides a regularized
characterization of the private classical capacity, and the last asks what rates of private
classical communication are achievable if the sender and receiver share a secret key before
communication begins. For full details, we refer the reader to the original papers in thequantum Shannon theory literature.
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23.5. DISCUSSION OF PRIVATE CLASSICAL CAPACITY

23.5.1

651

Superadditivity of the Private Information

Theorem 23.2.1 states that the private classical capacity of a quantum channel is equal to
the regularized private information of the channel. As we have said before (at the beginningof Chapter 21), a regularized formula is not particularly useful from a practical perspective
because it is impossible to perform the optimization task that it sets out, and it is notdesirable from an information-theoretical perspective because such a regularization does not
identify a formula as a unique measure of capacity.
In light of the unsatisfactory nature of a regularized formula, is it really necessary tohave the regularization in Theorem 23.2.1 for arbitrary quantum channels? Interestingly,
the answer seems to be “yes” in the general case (however, we know it is not necessary if
the channel is degradable). The reason is that there exists an example of a channel N for
which the private information is strictly superadditive:mP (N ) < P N ⊗m ,(23.80)for some positive integer m. Specifically, Smith et al. showed that the private information ofa particular Pauli channel exhibits this superadditivity (Smith et al., 2008). To do so, theycalculated the private information P (N ) for such a channel. Next, they consider performing
an m-qubit “repetition code” before transmitting qubits into the channel. A repetition code
is a quantum code that performs the following encoding:α|0i + β|1i → α|0i⊗m + β|1i⊗m .

(23.81)

Evaluating the private information when sending a particular state through the repetitioncode and then through m instances of the channel leads to a higher value than mP (N ),implying the strict inequality in (23.80). Thus, additivity of the private information formula P (N ) cannot hold in the general case.
The implications of this result are that we really do not understand the best way oftransmitting information privately over a quantum channel that is not degradable, and it isthus the subject of ongoing research.

23.5.2

Superadditivity of Private Classical Capacity

The private information of a particular channel can be superadditive (as discussed in theprevious section), and so the regularized private information is our best characterization of
the capacity for this information-processing task. In spite of this, we might hope that someeventual formula for the private classical capacity would be additive (some formula otherthan the private information P (N )). Interestingly, this is also not the case.
To clarify this point, suppose that P ? (N ) is some formula for the private classical capacity.
If it were an additive formula, then it should be additive as a function of channels:
P ? (N ⊗ M) = P ? (N ) + P ? (M).

(23.82)Li et al. have shown that this cannot be the case for any proposed private capacity formula,by making a clever argument with a construction of channels (Li et al., 2009). Specifically,c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 23. PRIVATE CLASSICAL COMMUNICATION

they constructed a particular channel N which has a single-letter classical capacity. The
fact that the channel’s classical capacity is sharply upper bounded implies that its private
classical capacity is as well. Let D be the upper bound so that P ? (N ) ≤ D. Also, they
considered a 50% erasure channel, one which gives the input state to Bob and an erasuresymbol to Eve with probability 1/2 and gives the input state to Eve and an erasure symbol
to Bob with probability 1/2. Such a channel has zero capacity for sending private classicalinformation essentially because Eve is getting the same amount of information as Bob doeson average. Thus, P ? (M) = 0. In spite of this, Li et al. show that the tensor-productchannel N ⊗ M has a private classical capacity that exceeds D. We can then make theconclusion that these two channels allow for superadditivity of private classical capacity:
P ? (N ⊗ M) > P ? (N ) + P ? (M),(23.83)and that (23.82) cannot hold in the general case. More profoundly, their results demonstratethat the private classical capacity itself is non-additive, even if a characterization of it isfound that is more desirable than that with the formula in Theorem 23.2.1. Thus, it will
likely be difficult to obtain a desirable characterization of the private classical capacity for
general quantum channels.

23.5.3

Secret-key Assisted Private Classical Communication

The direct coding part of Theorem 23.2.1 demonstrates how to send private classical information over a quantum channel N at the private information rate P (N ). A natural extension
to consider is the scenario in which Alice and Bob share a secret key before communication
begins. A secret key shared between Alice and Bob and secure from Eve is a tripartite stateof the following form:ΦAB ⊗ σE ,(23.84)
where ΦAB is the maximally correlated state and σE is a state on Eve’s system that is independent of the key shared between Alice and Bob. Like the entanglement-assisted capacity
theorem, we assume that they obtain this secret key from some third party, and the third
party ensures that the key remains secure.
The resulting capacity theorem is known as the secret-key-assisted private classical capacity theorem, and it characterizes the trade-off between secret key consumption and private
classical communication. The main idea for this setting is to show the existence of a protocolthat transmits private classical information at a rate of I(X; B) private bits per channel use
while consuming secret key at a rate of I(X; E) secret key bits per channel use, where theinformation quantities are with respect to the state in Theorem 23.2.1. The protocol forachieving these rates is almost identical to the one we gave in the proof of the direct coding
theorem, though with one difference. Instead of sacrificing classical bits at a rate of I(X; E)
in order to randomize Eve’s knowledge of the message (recall that our randomization variable
had to be chosen uniformly at random from a set of size ≈ 2nI(X;E) ), the sender exploits
the secret key to do so. The converse proof shows that this strategy is optimal (with amulti-letter characterization). Thus, we have the following capacity theorem.
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23.6. HISTORY AND FURTHER READING

653

Theorem 23.5.1 (Secret-key-assisted capacity theorem) The secret-key-assisted private classical capacity region CSKA (N ) of a quantum channel N is given by
CSKA (N ) =

∞
[
1

k
k=1

e(1) (N ⊗k ),C
SKA(23.85)e(1) (N ) is the set of all P, S ≥ 0 such that
where the overbar indicates the closure of a set. C
SKA
P ≤ I(X; B)σ − I(X; E)σ + S,
P ≤ I(X; B)σ .

(23.86)(23.87)

where P is the rate of private classical communication, S is the rate of secret key consumption, the state σXBE is of the following form:X
σXBE ≡pX (x)|xihx|X ⊗ UAN0 →BE (ρxA0 ) ,(23.88)x

and UAN0 →BE is an isometric extension of the channel.
Showing that the above inequalities are achievable follows by time sharing between the
protocol from the direct coding part of Theorem 23.2.1 and the aforementioned protocol for
secret-key-assisted private classical communication.

23.6 History and Further Reading
Bennett and Brassard (1984) devised the first protocol for sending private classical data overa quantum channel. The protocol given there became known as quantum key distribution,
which has now become a thriving field in its own right (Scarani et al., 2009). Devetak(2005) and Cai et al. (2004) proved the characterization of the private classical capacity
given in this chapter (both using the techniques that we reviewed in this chapter). Hsieh
et al. (2008b) proved achievability of the secret-key-assisted protocol given in Section 23.5.3,
and Wilde (2011) proved the converse and stated the secret-key-assisted capacity theorem.
Later work characterized the full trade-off between public classical communication, private
classical communication, and secret key (Hsieh and Wilde, 2009; Wilde and Hsieh, 2012a).
Smith et al. (2008) showed that the private information can exhibit superadditivity, and Li
et al. (2009) showed that the private classical capacity is generally non-additive. Elkouss
and Strelchuk (2015) demonstrated a striking superadditivity effect, which suggests that aregularized expression is necessary to determine the private capacity of an arbitrary channel.
Smith (2008) later showed that the symmetric-side-channel-assisted private classical capacity
is additive. Datta and Hsieh (2010) demonstrated universal private codes for quantum
channels.
A different, weaker notion of security is based on the eavesdropper not being able to learnmuch from any measurement performed on her system. Although this notion might seemc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 23. PRIVATE CLASSICAL COMMUNICATION

similar to the notion of security that we discussed in this chapter, it is in fact much different(König et al., 2007) and is the basis for the information locking effect (DiVincenzo et al.,2004). Impressive information locking schemes exist (Hayden et al., 2004b; Dupuis et al.,2013; Fawzi et al., 2013), and recently, the locking capacity of a channel was introduced
and bounded (Guha et al., 2014), calculated for particular channels (Winter, 2015a), anddeveloped further (Lupo and Lloyd, 2014, 2015).

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 24

Quantum Communication
The quantum capacity theorem is one of the most important theorems in quantum Shannon
theory. It is a fundamentally “quantum” theorem in that it demonstrates that a fundamentally quantum information quantity, the coherent information, is an achievable rate forquantum communication over a quantum channel. The fact that the coherent information
does not have a strong analog in classical Shannon theory truly separates the quantum andclassical theories of information.
The no-cloning theorem (Section 3.5.4) provides the intuition behind the quantum capacity theorem. The goal of any quantum communication protocol is for Alice to establish
quantum correlations with the receiver Bob. We know well now that every quantum channel
has an isometric extension, so that we can think of another receiver, the environment Eve,
who is at a second output port of a larger unitary evolution. Were Eve able to learn anything
about the quantum information that Alice is attempting to transmit to Bob, then Bob could
not be retrieving this information—otherwise, they would violate the no-cloning theorem.
Thus, Alice should figure out some subspace of the channel input where she can place herquantum information such that only Bob has access to it, while Eve does not. That thedimensionality of this subspace is exponential in the coherent information is perhaps thenunsurprising in light of the above no-cloning reasoning. The coherent information is an entropy difference H(B) − H(E)—a measure of the amount of quantum correlations that Alicecan establish with Bob less the amount that Eve can gain.1
We proved achievability of the coherent information for quantum data transmission in
Corollary 22.2.1, but the roundabout path that we followed to prove achievability there
perhaps does not give much insight into the structure of a quantum code that achieves thecoherent information. Our approach in this chapter is different and should shed more lighton this structure. Specifically, we show how to make coherent versions of the private classicalcodes from the previous chapter. By exploiting the privacy properties of these codes, we canform subspaces where Alice can store her quantum information such that Eve does not haveaccess to it. Thus, this approach follows the above “no-cloning intuition” more closely.
1

Recall from Exercise 11.6.6 that we can also write the coherent information as half the difference of
Bob’s mutual information with Alice less Eve’s: I(AiB) = 1/2 [I(A; B) − I(A; E)].

655

656

CHAPTER 24. QUANTUM COMMUNICATIONThe best characterization that we have for the quantum capacity of a general quantum
channel is the regularized coherent information. It turns out that the regularization is notnecessary for the class of degradable channels, implying that we have a complete understanding of the quantum data transmission capabilities of these channels. However, if a channel
is not degradable, there can be some startling consequences, and these results imply that
we have an incomplete understanding of quantum data transmission in the general case.
First, the coherent information can be strictly superadditive for the depolarizing channel.
This means that the best strategy for achieving the quantum capacity is not necessarily thefamiliar one where we generate random quantum codes from a single instance of a channel.
This result is also in marked contrast with the “classical” strategies that achieve the unassisted and entanglement-assisted classical capacities of the depolarizing channel. Second,
perhaps the most surprising result in quantum Shannon theory is that it is possible to “superactivate” the quantum capacity. That is, suppose that two channels on their own havezero capacity for transmitting quantum information (for the phenomenon to occur, thesechannels are specific channels). Then it is possible for the joint channel (the tensor productof the individual channels) to have a non-zero quantum capacity, in spite of them being
individually useless for quantum data transmission. This latter result implies that we arerather distant from having a complete quantum theory of information, in spite of the manysuccesses reviewed in this book.
We structure this chapter as follows. We first overview the information-processing task
relevant for quantum communication. Next, we discuss the no-cloning intuition for quantum capacity in some more detail, presenting the specific example of a quantum erasurechannel. Section 24.3 states the quantum capacity theorem, and the following two sections
prove the direct coding and converse theorems corresponding to it. Section 24.7 computes
the quantum capacity of two degradable channels: the quantum erasure channel and theamplitude damping channel. We then discuss superadditivity of coherent information andsuperactivation of quantum capacity in Section 24.8. Finally, we prove the existence of anentanglement distillation protocol, whose proof bears some similarities to the proof of thedirect coding part of the quantum capacity theorem.

24.1 The Information-Processing Task
We begin the technical development in this chapter by describing the information-processing
task for quantum communication (we define an (n, Q, ε) quantum communication code).
First, there are several different tasks that we can consider as quantum communication, but
the strongest definition of quantum capacity corresponds to a task known as entanglementtransmission. Suppose that Alice shares entanglement with a reference system to which she
does not have access. Then their goal is to devise a quantum coding scheme such that Alicecan transfer this entanglement to Bob. To this end, suppose that Alice and the reference
share an arbitrary state |ϕiRA1 , where the systems R and A1 have the same dimension. Alicethen performs some encoder on system A1 to prepare it for input to many instances of aquantum channel NA0 →B . The resulting state is as follows: EA1 →A0n (ϕRA1 ). Alice transmits
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24.1. THE INFORMATION-PROCESSING TASK

657

Figure 24.1: The information-processing task for entanglement transmission. Alice is trying to preserve theentanglement with some inaccessible reference system by encoding her system and transmitting the encoded
quantum data over many independent uses of a noisy quantum channel. Bob performs a decoding of thesystems he receives, and the state at the end of the protocol is close to the original state shared between
Alice and the reference if the protocol is any good for entanglement transmission.

the systems A0n through many independent uses of the channel NA0 →B , resulting in thefollowing state: NA0n →B n (EA1 →A0n (ϕRA1 )), where NA0n →B n ≡ (NA0 →B )⊗n . After Bob receives
the systems B n from the channel outputs, he performs some decoding channel DB n →B1 ,where B1 is some system of the same dimension as A1 . The final state after Bob decodes is
as follows:ωRB1 ≡ DB n →B1 (NA0n →B n (EA1 →A0n (ϕRA1 ))).
(24.1)
Figure 24.1 depicts all of the above steps. For an (n, Q, ε) protocol, the following condition
should hold for all states |ϕiRA1 :1
kϕRA1 − ωRB1 k1 ≤ ε.
2

(24.2)The rate Q of this scheme is equal to the number of qubits transmitted per channel use:Q≡1
log dim(HA1 ).
n(24.3)

A rate Q is achievable for N if there exists an (n, Q − δ, ε) quantum communication code
for all ε ∈ (0, 1), δ > 0, and sufficiently large n. The quantum capacity CQ (N ) is defined as
the supremum of all achievable rates for N .
The above notion of quantum communication encompasses other quantum informationprocessing tasks such as mixed-state transmission, pure-state transmission, and entanglement
generation. Alice can transmit any mixed or pure state if she can preserve the entanglementwith a reference system. Also, she can generate entanglement with Bob if she can preserve
entanglement with a reference system—she just needs to create an entangled state locallyand apply the above protocol to one system of the entangled state.
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CHAPTER 24. QUANTUM COMMUNICATION

Exercise 24.1.1 Show that an (n, Q, ε) protocol for quantum communication satisfies thefollowing:1
idA1 →B1 −DB n →B1 ◦ NA⊗n≤ ε,(24.4)
0 →B ◦ EA1 →A0n♦2where k·k♦ denotes the diamond norm, defined in Section 9.1.6.

24.2 No-Cloning and Quantum Communication
We first discuss quantum communication over a quantum erasure channel before stating andproving the quantum capacity theorem. Consider the quantum erasure channel that gives
Alice’s input state to Bob with probability 1−ε and an erasure flag to Bob with probability ε:
ρ → (1 − ε) ρ + ε|eihe|,(24.5)

where he|ρ|ei = 0 for all inputs ρ. Recall that an isometric extension of this channel is asfollows (see Exercise 5.2.6):√√(24.6)|ψiRA → 1 − ε|ψiRB |eiE + ε|ψiRE |eiB ,so that the channel now has the other interpretation that Eve gets the state with probability εwhile giving her the erasure flag with probability 1 − ε.
Now suppose that the erasure parameter is set to 1/2. In such a scenario, the channel to
Eve is the same as the channel to Bob, namely, both have the channel ρ → 1/2 (ρ + |eihe|).
We can argue that the quantum capacity of such a channel should be equal to zero, by
invoking the no-cloning theorem. More specifically, suppose there is a scheme (an encoderand decoder as given in Figure 24.1) for Alice and Bob to communicate quantum information
reliably at a non-zero rate over such a channel. If so, Eve could simply use the same decoderthat Bob does, and she should also be able to obtain the quantum information that Aliceis sending. But the ability for both Bob and Eve to decode the quantum information that
Alice is transmitting violates the no-cloning theorem. Thus, the quantum capacity of sucha channel should vanish.
Exercise 24.2.1 Argue that the quantum capacity of an amplitude damping channel vanishes if its damping parameter is equal to 1/2.
The no-cloning theorem plays a more general role in the analysis of quantum communication over quantum channels. In the construction of a quantum code, we are trying to finda “no-cloning” subspace of the input Hilbert space that is protected from Eve. If Eve is able
to obtain any of the quantum information in this subspace, then this information cannot begoing to Bob by the same no-cloning argument featured in the previous paragraph. Thus,
we might then suspect that the codes from the previous chapter for private classical communication might play a role for quantum communication because we constructed them insuch a way that Eve would not be able to obtain any information about the private message
that Alice is transmitting to Eve. The main insight needed is to make a coherent versionof these private classical codes, so that Alice and Bob conduct every step in superposition(much like we did in Chapter 22).
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24.3. THE QUANTUM CAPACITY THEOREM

659

24.3 The Quantum Capacity Theorem
The main theorem of this chapter is the following quantum capacity theorem.
Theorem 24.3.1 (Quantum Capacity) The quantum capacity CQ (N ) of a quantum channel NA0 →B is equal to the regularized coherent information of the channel:
CQ (N ) = Qreg (N ),where
Qreg (N ) ≡ lim

1

k→∞ k

(24.7)

Q(N ⊗k ).

(24.8)

The channel coherent information Q(N ) is defined as
Q(N ) ≡ max I(AiB)σ ,(24.9)

φ

where the optimization is with respect to all pure, bipartite states φAA0 and σAB ≡ NA0 →B (φAA0 ).
We prove this theorem in two parts: the direct coding theorem and the converse theorem.
The proof of the direct coding theorem proceeds by exploiting the private classical codesfrom the previous chapter. The proof of the converse theorem is similar to approaches fromprevious chapters—we exploit the AFW inequality and quantum data processing in orderto obtain an upper bound on the quantum capacity. In general, the regularized coherent
information is our best characterization of the quantum capacity, but the regularization is notnecessary for the class of degradable channels. Since many channels of interest are degradable(including dephasing, amplitude damping, and erasure channels), we can calculate theirquantum capacities.

24.4 The Direct Coding Theorem
The proof of the direct coding part of the quantum capacity theorem follows by taking advantage of the properties of the private classical codes constructed in the previous chapter(see Section 23.3). We briefly recall this construction. Suppose that a classical–quantum–
quantum channel connects Alice to Bob and Eve. Specifically, if Alice inputs a classicalletter x to the channel, then Bob receives a density operator ρxB and Eve receives a density
operator ωEx . The direct coding part of Theorem 23.2.1 establishes the existence of a codebook {xn (m, k)}m∈M,k∈K selected from a distribution pX (x) and a corresponding decoding
POVM {Λm,k
B n } such that Bob can detect Alice’s message m and randomizing variable k withhigh probability:noxn (m,k)∀m ∈ M, k ∈ K : Tr Λm,k
ρ
≥ 1 − ε,(24.10)nnBBwhile Eve obtains a negligible amount of information about Alice’s message m:∀m ∈ M :c 2019

1 X xn (m,k)ω n− ω ⊗n|K| k∈K E≤ ε,(24.11)

1

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CHAPTER 24. QUANTUM COMMUNICATION

where ω is Eve’s expected density operator:Xω≡pX (x)ωEx .

(24.12)x

The above statements hold true for all ε ∈ (0, 1) and sufficiently large n as long as|M| ≈ 2n[I(X;B)−I(X;E)] ,(24.13)|K| ≈ 2nI(X;E) .

(24.14)

We can now construct a coherent version of the above code that is good for quantum datatransmission. First, suppose that there is some density operator with the following spectral
decomposition:
Xρ A0 ≡pX (x)|ψx ihψx |A0 .
(24.15)x

Now suppose that the channel NA0 →B has an isometric extension UAN0 →BE , so that inputting|ψx iA0 leads to the following state shared between Bob and Eve:|ψx iBE ≡ UAN0 →BE |ψx iA0 .

(24.16)

From the direct coding part of Theorem 23.2.1, we know that there exists a private classicalcode {xn (m, k)}m∈M,k∈K with the properties in (24.10)–(24.11) and with rate
I(X; B)σ − I(X; E)σ ,where σXBE is a classical–quantum state of the following form:XpX (x)|xihx|X ⊗ |ψx ihψx |BE .
σXBE ≡(24.17)(24.18)x

The following identity demonstrates that the private information in (24.17) is equal to thecoherent information for the particular state σXBE above:
I(X; B)σ − I(X; E)σ = H(B)σ − H(B|X)σ − H(E)σ + H(E|X)σ= H(B)σ − H(B|X)σ − H(E)σ + H(B|X)σ= H(B)σ − H(E)σ .

(24.19)(24.20)(24.21)

The first equality follows from the identity I(C; D) = H(D) − H(D|C). The second equality
follows because the state on systems BE is pure when conditioned on the classical variable X.
Observe that the last expression is a function solely of the density operator ρ in (24.15), and
it is also equal to the coherent information of the channel for the particular input state ρ(see Exercise 11.5.2).
Now we show how to construct a quantum code achieving the coherent information rate
H(B)σ − H(E)σ by making a coherent version of the above private classical code. Supposethat Alice shares a state |ϕiRA1 with a reference system R, whereX|ϕiRA1 ≡αl,m |liR |miA1 ,(24.22)l,m∈M
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24.4. THE DIRECT CODING THEOREM

661

and {|liR } is some orthonormal basis for R while {|miA1 } is some orthonormal basis for A1 .
Also, we set |M| ≈ 2n[H(B)σ −H(E)σ ] . We would like for Alice and Bob to execute a quantum
communication protocol such that Bob can reconstruct Alice’s share of the above state on
his system with Alice no longer entangled with the reference (we would like for the final
state to be close to |ϕiRA1 where Bob is holding the A1 system). To this end, Alice createsa quantum codebook {|φm iA0n }m∈M with quantum codewords:
1 X iγm,k xn (m,k)|ψ
iA0n ,e|φm iA0n ≡ p|K| k∈K(24.23)nwhere the states |ψ x (m,k) iA0n are the nth extensions of the states arising from the spectral
decomposition in (24.15), the classical sequences xn (m, k) are from the codebook for private
classical communication, and we specify how to choose the phases γm,k later. All the statesn|ψ x (m,k) iA0n are orthonormal because they are chosen from the spectral decomposition in(24.15) and the expurgation from Section 23.3.4 guarantees that they are distinct (otherwise,nthey would not be good codewords!). The fact that the states |ψ x (m,k) iA0n are orthonormal
implies that the quantum codewords |φm iA0n are also orthonormal.
Alice’s first action is to coherently copy the value of m in the A1 register to anotherregister A2 , so that the state in (24.22) becomesX
αl,m |liR |miA1 |miA2 .
(24.24)l,mAlice then performs some isometric encoding from A2 to A0n that takes the above unencodedstate to the following encoded state:X
αl,m |liR |miA1 |φm iA0n ,(24.25)l,mwhere each |φm iA0n is a quantum codeword of the form in (24.23). Alice transmits the systems
A0n through many uses of the quantum channel, leading to the following state shared between
the reference, Alice, Bob, and Eve:X
(24.26)αl,m |liR |miA1 |φm iB n E n ,l,mwhere |φm iB n E n is defined from (24.16) and (24.23). Recall from (24.10) that Bob can detect
the message m and the variable k in the private classical code with high probability:onxn (m,k)∀m, k : Tr Λm,k
ψ
≥ 1 − ε.
(24.27)nn
B
B
So Bob instead constructs a coherent version of this POVM:
X q m,kΛB n ⊗ |miB1 |kiB2 .

(24.28)m∈M,k∈K
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662

CHAPTER 24. QUANTUM COMMUNICATIONHe then performs this coherent POVM, resulting in the stateq
X XX 1
0 ,k 0niγk,mp αl,m |liR |miA1 Λm|ψ x (m,k) iB n E n |m0 , k 0 iB1 B2 .
Bn e|K|
m0 ∈M, l,m k∈K(24.29)k0 ∈KWe would like for the above state to be close in trace distance to the following state:
XX 1np αl,m |liR |miA1 eiδm,k |ψ x (m,k) iB n E n |miB1 |kiB2 ,(24.30)|K|
l,m k∈K
where δm,k are some phases that we will specify shortly. To this end, consider that the sets{|χm,k iB n E n B1 B2 }m,k and {|ϕm,k iB n E n B1 B2 }m,k form orthonormal bases, wheren|χm,k iB n E n B1 B2 ≡ |ψ x (m,k) iB n E n |miB1 |kiB2 ,X q m0 ,k0 nnn|ϕm,k iB E B1 B2 ≡
ΛB n |ψ x (m,k) iB n E n |m0 iB1 |k 0 iB2 .

(24.31)(24.32)m0 ∈M,k0 ∈K

Also, consider that the overlap between corresponding states in the different bases is high:hχm,k |ϕm,k i= hψxn (m,k)X|B n E n hm|B1 hk|B2q
0 ,k 0xn (m,k)Λm
iB n E n |m0 iB1 |k 0 iB2
B n |ψ

(24.33)m0 ∈M,k0 ∈K

X

=

hψ

xn (m,k)q
0 ,k 0xn (m,k)iB n E n hm|m0 iB1 hk|k 0 iB2|B n E n Λm
B n |ψ

(24.34)m0 ∈M,k0 ∈K= hψ

xn (m,k)xn (m,k)qxn (m,k)iB n E n|B n E n Λm,k
B n |ψ

|B n E n Λm,k
B n |ψ
nonm,k x (m,k)= Tr ΛB n ψB n≥ hψxn (m,k)iB n E n(24.35)(24.36)(24.37)≥ 1 − ε,(24.38)qm,km,kwhere the first inequality follows from the fact that Λm,k
B n ≥ ΛB n for ΛB n ≤ I and thesecond inequality follows from (24.27). By applying Lemma A.0.3 from Appendix A, weknow that there exist phases γm,k and δm,k such thathχm |ϕm i ≥ 1 − ε,(24.39)where
1 X iδm,k|χm iB n E n B1 B2 ≡ pe|χm,k iB n E n B1 B2 ,|K| k
1 X iγm,k|ϕm iB n E n B1 B2 ≡ pe|ϕm,k iB n E n B1 B2 .
|K| kc 2019(24.40)(24.41)

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24.4. THE DIRECT CODING THEOREM

663

So we choose the phases in a way such that the above inequality holds. We can then apply
the above result to show that the state in (24.29) has high fidelity with the state in (24.30):
!
!
X
X
∗
αl0 ,m0 |l0 iR |m0 iA1 |ϕm0 iB n E n B1 B2
αl,mhl|R hm|A1 hχm |B n E n B1 B2
l0 ,m0

l,m

=

X

∗
αl,m
αl0 ,m0 hl|l0 iR hm|m0 iA1 hχm |ϕm0 iB n E n B1 B2(24.42)l,m,l0 ,m0=X|αl,m |2 hχm |ϕm iB n E n B1 B2(24.43)l,m≥ 1 − ε.

(24.44)Thus, the state resulting after Bob performs the coherent POVM is close in trace distanceto the following state:
1 X iδm,k xn (m,k)αl,m |liR |miA1 pe|ψ
iB n E n |miB1 |kiB2
|K|
l,m∈M
k∈K
X
αl,m |liR |miA1 |φem iB n E n B2 |miB1 , (24.45)
=
X

l,m∈M

where

1 X iδm,k xn (m,k)
e|ψiB n E n |kiB2 .
|φem iB n E n B2 ≡ p|K| k∈K

Consider the state of Eve for a particular value of m:
h i
n
o
φem
= TrB n B2 |φem ihφem |B n E n B2
En
)
(
X 1
n
n
0
ei(δm,k0 −δm,k ) |ψ x (m,k) ihψ x (m,k ) |B n E n ⊗ |kihk 0 |B2= TrB n B2
|K|
k,k0 ∈K
1 X xn (m,k)=ψ n.
|K| k∈K E(24.46)(24.47)(24.48)(24.49)We are now in a position to apply the second property of the private classical code. Recall
from the privacy condition in (24.11) that Eve’s state is guaranteed to be ε-close in trace
distance to the tensor-power state [NAc 0 →E (ρ)]⊗n , where NAc 0 →E is the complementary channeland ρ is the density operator in (24.15). Let |θN c (ρ) iE n B3 be some purification of this tensorpower state. By Uhlmann’s theorem and the relation between trace distance and fidelitym(see Definition 9.2.3 and Theorem 9.3.1),√ there is some isometry UB n B2 →B3 for each value ofm such that the following states are 2 ε-close in trace distance (see Exercise 9.2.9):
UBmn B2 →B3 |φem iB n E n B2
c 2019√2 ε≈|θN c (ρ) iE n B3 .

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CHAPTER 24. QUANTUM COMMUNICATION

Bob then performs the following controlled isometry on his systems B n , B1 , and B2 :X
|mihm|B1 ⊗ UBmn B2 →B3 ,(24.51)m

leading to a state that is close in trace distance to the following state:!
X
αl,m |liR |miA1 |miB1 ⊗ |θN c (ρ) iE n B3 .

(24.52)l,m∈M

At this point, the key observation is that the state on E n B3 is effectively decoupled from thestate on systems R, A1 , and B1 , so that Bob can just throw away his system B3 . Thus, they
have successfully implemented an approximate coherent channel from system A1 to A1 B1 .
We now allow for Alice to communicate classical information to Bob in order for them to
implement a quantum communication channel rather than just a mere coherent channel (ina moment we argue that this free forward classical communication is not necessary). Alice
performs a Fourier transform on the register A1 , leading to the following state:X1pαl,m exp {2πimj/dA1 } |liR |jiA1 |miB1 ,(24.53)dA1 l,m,j∈M
where dA1 ≡ dim(HA1 ). She then measures register A1 in the computational basis, leading
to some outcome j and the following post-measurement state:!
X
αl,m exp {2πimj/dA1 } |liR |miB1 ⊗ |jiA1 .
(24.54)l,m∈M

She sends Bob the outcome j of her measurement over a classical channel, and the protocol
ends with Bob performing the following unitary:Z † (j)|miB1 = exp {−2πimj/dA1 } |miB1 ,leaving the desired state on the reference and Bob’s system B1 :X
αl,m |liR |miB1 .

(24.55)(24.56)l,m∈M

All of the errors accumulated in the above protocol are some finite sum of ε terms, andapplying the triangle inequality several times implies that the actual state is close to thedesired state in the asymptotic limit of large block length. Figure 24.2 depicts all of thesteps in this protocol for quantum communication.
We now argue that the classical communication is not necessary—there exists a schemethat does not require the use of this forward classical channel. After reviewing the above
protocol and glancing at Figure 24.2, we realize that Alice’s encoder is a quantum instrumentof the following form:X
E(ρ) ≡
Ej (ρ) ⊗ |jihj|.
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24.4. THE DIRECT CODING THEOREM

665

Figure 24.2: All of the steps in the protocol for quantum communication. Alice and Bob’s goal is tocommunicate as much quantum information as they can while making sure that Eve’s state is independent ofwhat Alice is trying to communicate to Bob. The figure depicts the series of controlled unitaries that Alice andBob perform and the final measurement and classical communication that enables quantum communicationfrom Alice to Bob at the coherent information rate.

Each map Ej is trace-non-increasing and has the following action on a pure-state input|ϕiRA1 :!
!
X
X
0
hj|A1 FA1
|φm0 ihm |A2|mihm|A1 ⊗ |miA2 |ϕiRA1 ,(24.58)m0mP0
where m |mihm|A1 ⊗ |miA2 is Alice’s coherent copier in (24.24),
m0 |φm0 ihm |A2 is her
quantum encoder in (24.25), FA1 is the Fourier transform, and hj|A1 represents the projection
onto a particular measurement outcome j. We can simplify the above expression as follows:!
XX= hj|A1|m
e 0 ihm0 |A1|mihm|A1 ⊗ |φm iA2 |ϕiRA1(24.59)P

m0= hj|A1

Xm|mihm|e
A1 ⊗ |φm iA2 |ϕiRA1(24.60)

m

=

1
p

!
X

|M| m

ei2πmj/|M| |φm iA2 hm|A1|ϕiRA1 .

(24.61)

It follows that the trace of each Ej is uniform and independent of the input state |ϕiRA1 :Tr {Ej (ϕRA1 )} =

1
.
|M|

(24.62)pObserve that multiplying the map in (24.61) by |M| gives a proper isometry that could
suffice as an encoding. Let Ej0 denote the rescaled isometry. Corresponding to each encoder
is a decoding map Dj consisting of Bob’s coherent measurement in (24.28), his decoupler in(24.51), and his phase shifter in (24.55). We can thus represent the state output from ourc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

666

CHAPTER 24. QUANTUM COMMUNICATION

classically coordinated protocol as follows:X
Dj N ⊗n (Ej (ϕRA1 )) .

(24.63)j

From the analysis in the preceding paragraphs, we know that the trace distance between theideal state and the actual state is small for the classically coordinated scheme:X≤ ε0 ,


Dj N ⊗n (Ej (ϕRA1 )) − ϕRA1

j(24.64)1

where ε0 is some arbitrarily small positive number. Thus, the fidelity between these twostates is high:!
X(24.65)F
Dj N ⊗n (Ej (ϕRA1 )) , ϕRA1 ≥ 1 − ε0 .
j

But we can rewrite the fidelity as follows:
!
F

X


Dj N ⊗n (Ej (ϕRA1 )) , ϕRA1

jXDj N ⊗n (Ej (ϕRA1 )) |ϕiRA1(24.66)


hϕ|RA1 Dj N ⊗n (Ej (ϕRA1 )) |ϕiRA1(24.67)= hϕ|RA1

j

=

X
j

=

X 1 


hϕ|RA1 Dj N ⊗n Ej0 (ϕRA1 ) |ϕiRA1|M|j≥ 1 − ε0 ,(24.68)(24.69)

implying that at least one of the encoder–decoder pairs (Ej0 , Dj ) has arbitrarily high fidelity.
Thus, Alice and Bob simply agree beforehand to use a scheme (Ej0 , Dj ) with high fidelity,
obviating the need for the forward classical communication channel.
The protocol given here achieves communication at the coherent information rate. In
order to achieve the regularized coherent information rate in the statement of the theorem, Alice and Bob apply the same protocol to the superchannel (NA0 →B )⊗k instead of thechannel NA0 →B .

24.5 Converse Theorem
This section proves the converse part of the quantum capacity theorem, demonstrating thatthe regularized coherent information is an upper bound on the quantum capacity of any
quantum channel. For the class of degradable channels, the coherent information itself is an
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24.5. CONVERSE THEOREM

667

upper bound on the quantum capacity—this demonstrates that we completely understandthe quantum data transmission capabilities of these channels.
For this converse proof, we assume that Alice is trying to generate entanglement with
Bob. The capacity for this task is an upper bound on the capacity for quantum data transmission because we can always use a noiseless quantum channel to establish entanglement.
We also allow Alice free forward classical communication to Bob, and we demonstrate that
this resource cannot increase the quantum capacity (essentially because the coherent information is convex). In a protocol for entanglement generation, Alice begins by preparing the
maximally entangled state ΦAA1 of Schmidt rank |A| in her local laboratory, where n1 log |A|is the rate of this entangled state. She performs some encoding operation EA1 →A0n M that
outputs many systems A0n and a classical register M . She then inputs the systems A0n to
many independent uses of a quantum channel NA0 →B , resulting in the state
ωAM B n ≡ NA0n →B n (EA1 →A0n M (ΦAA1 )),(24.70)where NA0n →B n ≡ (NA0 →B )⊗n . Bob takes the outputs B n of the channels and the classicalregister M and performs some decoding operation DB n M →B1 , resulting in the state0ωAB≡ DB n M →B1 (ωAM B n ).
1

(24.71)

The following condition holds for an (n, [log |A|] /n, ε) protocol for entanglement generation:
1 0
ω− ΦAB1 1 ≤ ε.
2 AB1
The converse proof then proceeds in the following steps:
log |A| = I(AiB1 )Φ
≤ I(AiB1 )ω0 + f (|A| , ε)
≤ I(AiB n M )ω + f (|A| , ε).

(24.72)(24.73)(24.74)(24.75)

The first equality follows because the coherent information of a maximally entangled stateis equal to the logarithm of the dimension of one of its systems. The first inequality follows
from an application of the AFW inequality to the condition in (24.72), with f (|A| , ε) ≡2ε log |A|+g2 (ε). The second inequality follows from quantum data processing. Now consider
that the state ωM AB n is a classical–quantum state of the following form:X(24.76)
ωM AB n ≡pM (m)|mihm|M ⊗ NA0n →B n (ρmAA0n ).
m

We can then perform a spectral decomposition of each state ρm as follows:Xρm=pL|M (l|m)|φl,m ihφl,m |AA0n ,0nAA(24.77)l

and augment the above state as follows:XωM LAB n ≡pM (m)pL|M (l|m)|mihm|M ⊗ |lihl|L ⊗ NA0n →B n (φl,mAA0n ),(24.78)m,lc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 24. QUANTUM COMMUNICATION

so that ωM AB n = TrL {ωM LAB n }. We continue with bounding the relevant term:
I(AiB n M )ω ≤ I(AiB n M L)ω
X=
pM (m)pL|M (l|m)I(AiB n )N (φl,m )(24.79)(24.80)m,l≤ I(AiB n )N (φ∗l,m )(24.81)≤ Q(N ⊗n ).

(24.82)

The first inequality follows from the quantum data-processing inequality. The first equality
follows because the registers M and L are both classical, and we can apply the result of
Exercise 11.5.5. The second inequality follows because the expectation is always less thanthe maximal value (where we define φ∗l,m to be the state that achieves this maximum).
The final inequality follows from the definition of the channel coherent information as the
maximum of the coherent information over all pure, bipartite inputs. Putting everything
together, we find that1g2 (ε)
1
log |A| (1 − 2ε) ≤ Q(N ⊗n ) +.
nnnThus, if we are considering a sequence of (n, [log |A|] /n, εn ) quantum communication protocols with rate Q − δn = n1 log |A|, such that limn→∞ εn = limn→∞ δn = 0, then the above
bound becomesg( εn )
1.
(24.83)(Q − δn ) (1 − 2εn ) ≤ Q(N ⊗n ) +nnTaking the limit as n → ∞ then establishes that an achievable rate Q necessarily satisfies
Q ≤ Qreg (N ), where Qreg (N ) is the regularized coherent information given in (24.8). Thisconcludes the proof of the converse part of the quantum capacity theorem.
There are a few comments we should make regarding the converse theorem. First, we
see that classical communication cannot improve quantum capacity because the coherent
information is convex. We could obtain the same upper bound on quantum capacity even if
there were no classical communication. Second, it is sufficient to consider isometric encoders
for quantum communication—that is, it is not necessary to exploit general noisy CPTP maps
at the encoder. This makes sense intuitively because it would seem odd if noisy encodings
could help in the noiseless transmission of quantum data. Our augmented state in (24.78) andthe subsequent development reveals that this is so (again because the coherent informationis convex).
We can significantly strengthen the statement of the quantum capacity theorem for theclass of degradable quantum channels because the following equality holds for them:Q(N ⊗n ) = nQ(N ).

(24.84)This inequality follows from the additivity of coherent information for degradable channels(Theorem 13.5.1). Also, the task of optimizing the coherent information for these channels
is straightforward because it is a concave function of the input density operator (Theorem 13.5.2) and the set of density operators is convex.
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24.6. AN INTERLUDE WITH QUANTUM STABILIZER CODES

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24.6 An Interlude with Quantum Stabilizer Codes
We now describe a well-known class of quantum error-correcting codes known as the stabilizer
codes, and we prove that a randomly chosen stabilizer code achieves a quantum communication rate known as the hashing bound of a Pauli channel (the hashing bound is equal to thecoherent information of a Pauli channel when sending one share of a Bell state through it).
The proof of this theorem is different from our proof above that the coherent information
rate is achievable, and we consider it instructive to see this other approach for the special
case of stabilizer codes used for protecting quantum information sent over many independent instances of a Pauli channel. Before delving into the proof, we first briefly introduce
the simple repetition code and the more general stabilizer quantum codes.

24.6.1The Qubit Repetition CodeThe simplest quantum error-correction code is the repetition code, which encodes one qubit|ψi ≡ α|0i + β|1i into three physical qubits as follows:α|0i + β|1i → α|000i + β|111i.

(24.85)

A simple way to perform this encoding is to attach two ancilla qubits in the state |0i to theoriginal qubit and perform a CNOT gate from the first qubit to the second and from the firstto the last. This encoding illustrates one of the fundamental principles of quantum error
correction: the quantum information is spread across the correlations between the threephysical qubits after the encoding takes place. (Of course, this was also the case for thecodes we constructed in the direct part of the quantum capacity theorem.)
The above encoding will protect the encoded qubit against an artificial noise where either
the first, second, or third qubit is subjected to a bit flip (and no other errors occur). For
example, if a bit flip occurs on the second qubit, the encoded state changes as follows:
X2 (α|000i + β|111i) = α|010i + β|101i,(24.86)

where the notation X2 indicates that a Pauli operator X acts on the second qubit. The
procedure for the receiver to recover from such an error is to perform collective measurementson all three qubits that learn only about the error and nothing about the encoded quantum
data. In this case, the receiver can perform a measurement of the operators Z1 Z2 and Z2 Z3
to learn only about the error, so that the coherent superposition is preserved. One can easily
verify that Z1 Z2 and Z2 Z3 are as follows:
Z1 Z2 ≡ Z ⊗ Z ⊗ I = [(|00ih00| + |11ih11|) − (|01ih01| + |10ih10|)] ⊗ I,
Z2 Z3 ≡ I ⊗ Z ⊗ Z = I ⊗ [(|00ih00| + |11ih11|) − (|01ih01| + |10ih10|)] ,(24.87)(24.88)revealing that these measurements return a +1 if the parity of the basis states is even and−1 if the parity is odd. So, for our example error in (24.86), the syndrome measurements
will return −1 for Z1 Z2 and −1 for Z2 Z3 , which the receiver can use to identify the error
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CHAPTER 24. QUANTUM COMMUNICATION

that occurs. He can then perform the bit-flip operator X2 to invert the action of the error.
One can verify that the following syndrome table identifies which type of error occurs:
Measurement result+1, +1+1, −1−1, +1−1, −1

Error
I
X3
X1
X2

(24.89)Thus, if the only errors that occur are either no error or a single-qubit bit-flip error, then it
is possible to perfectly correct these. If errors besides these ones occur, then it is not possible
to correct them using this code.

24.6.2

Stabilizer Codes

We can generalize the main idea behind the above qubit repetition code to formulate theclass of quantum stabilizer codes. These stabilizer codes then generalize the classical theoryof linear error correction to the quantum case.
In the repetition code, observe that the encoded state in (24.85) is a +1-eigenstate of theoperators Z1 Z2 and Z2 Z3 , i.e., it holds that
Z1 Z2 (α|000i + β|111i) = α|000i + β|111i = Z2 Z3 (α|000i + β|111i) .

(24.90)We say that the operators Z1 Z2 and Z2 Z3 stabilize the encoded state. The stabilizing
operators form a group under multiplication because we obtain another stabilizing operator
if we multiply two of them: one can check that the operator Z1 Z3 stabilizes the encoded state
and that Z1 Z3 = (Z1 Z2 )(Z2 Z3 ). Also, the two operators Z1 Z2 and Z2 Z3 commute, implyingthat the encoded state is in the simultaneous eigenspace of these operators, and that it ispossible to measure the operators Z1 Z2 and Z2 Z3 in any order, in order to learn about errorsthat occur.
We now describe the theory of quantum stabilizer codes. Recall that the Pauli matrices
for one qubit are I, X, Y , and Z, whose action on the computational basis is as follows:I|0i = |0i,
X|0i = |1i,
Y |0i = i|1i,
Z|0i = |0i,

I|1i = |1i,
X|1i = |0i,
Y |1i = −i|0i,
Z|1i = −|1i.

(24.91)(24.92)(24.93)(24.94)

The X operator is known as the “bit-flip” operator, Z as the “phase-flip” operator, and Y as
the “bit and phase flip” operator. The Pauli group Gn acting on n qubits consists of n-foldtensor products of these operators along with the phase factors ±1 and ±i:
Gn ≡ {±1, ±i} ⊗ {I, X, Y, Z}⊗n .
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24.6. AN INTERLUDE WITH QUANTUM STABILIZER CODES

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The inclusion of the phase factors, along with the relations Y = iXZ, Z = iY X, and
X = iZY and the fact that any one of X, Y , and Z anticommutes with the other two
ensures that the set Gn is closed under multiplication. It is useful in the theory of quantum
error correction to consider the Pauli group quotiented out by its center: Gn / {±1, ±i},
essentially because global phases are not physically observable. This reduced version of thePauli group has 4n elements.
Let S be an abelian subgroup of the Pauli group Gn . Any such subgroup S has size 2n−k
for some integer k such that 0 ≤ k ≤ n. This subgroup S can be generated by a set of sizen − k, so that S = hS1 , . . . , Sn−k i. A state |ψi is stabilized by the subgroup S if
S|ψi = |ψi∀S ∈ S.

(24.96)The 2k -dimensional subspace of the full 2n -dimensional space for the n qubits that is stabilized by S is known as the codespace, or equivalently, an [n, k] stabilizer code that encodes
k logical qubits into n physical qubits. The decoding operation that the receiver performs is
analogous to that for the repetition code—he just measures the n − k operators constituting
some generating set of S and performs a recovery operation based on the results of thesemeasurements.
We can define logical operations on the quantum information encoded inside an [n, k]stabilizer code. These are operations that manipulate the quantum information inside thecodespace without taking the encoded information outside the codespace. These logicaloperations are part of the normalizer of S, defined as
N (S) ≡ U ∈ U(2n ) : U SU † = S ,(24.97)
where U(2n ) denotes the unitary group for n qubits. We can easily see that any U ∈ N (S)does not take a state |ψi in the codespace outside it. First, for all U ∈ N (S), it follows thatU † ∈ N (S), so that for all S ∈ S, we have
SU |ψi = U U † SU |ψi = U SU |ψi = U |ψi,

(24.98)where SU = U † SU and SU ∈ S from the definition of the normalizer. From the above,
we conclude that the state U |ψi is in the codespace since it is stabilized by all S ∈ S:
SU |ψi = U |ψi. It also follows that S ⊆ N (S) because S is abelian, implying thatS1 S2 S1† = S2 S1 S1† = S2

∀S1 , S2 ∈ S.

(24.99)In quantum error correction, we are concerned with correcting a fixed set of errors E ⊆ Gn
such that each element of E acts on the n physical qubits. In doing so, we might not be able
to correct all of the errors in a set E if there exists a pair E1 , E2 ∈ E such that
E1† E2 ∈ N (S).

(24.100)

E1† E2 S = (−1)g(S,E1 )+g(S,E2 ) SE1† E2 ,(24.101)

Consider that for all S ∈ S, we havec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

672

CHAPTER 24. QUANTUM COMMUNICATION

where we define g(P, Q) by P Q = (−1)g(P,Q) QP for all P, Q ∈ Gn . The above relation thenimplies the following one for all S ∈ S:
E1† E2 S(E1† E2 )† = (−1)g(S,E1 )+g(S,E2 ) S.

(24.102)

Since we assumed that E1† E2 ∈ N (S), the only way that the above relation can be true for
all S ∈ S is if g(S, E1 ) = g(S, E2 ). Thus, during the error correction procedure, Bob willmeasure a set {Sj } of generators, and since the outcome of a measurement of Sj on E|ψi isg(S, E), the errors E1 and E2 will be assigned the same syndrome. Since they have the same
syndrome, the receiver will have to reverse these errors with the same recovery operation,and this is only possible if E1 |ψi = E2 |ψi for all states |ψi in the codespace. This lattercondition is only true if E1† E2 ∈ S, leading us to the error-correcting conditions for quantum
stabilizer codes:
Theorem 24.6.1 It is possible to correct a set of errors E with a quantum stabilizer code if
every pair E1 , E2 ∈ E satisfies
E1† E2 ∈/ N (S)\S.
(24.103)
A simple way to satisfy the error-correcting conditions is just to demand that every pairof errors in E be such that E1† E2 ∈/ N (S). In such a case, each error is assigned a uniquesyndrome, and codes along with an error set satisfying this property are known as nondegenerate codes. Codes with a corresponding error set not satisfying this are known asdegenerate codes.

24.6.3

The Hashing Bound

We now provide a proof that the hashing bound for a Pauli channel (coherent information
when sending one share of a Bell state through a Pauli channel) is an achievable rate forquantum communication. Our proof of the direct part of the quantum capacity theorem
already suffices as a proof of this statement, but we think it is instructive to provide a proofof this statement using the theory of stabilizer codes. The main idea of the proof is to choosea stabilizer code randomly from the set of all stabilizer codes and show that such a code cancorrect the typical errors issued by a tensor-product Pauli channel.
Theorem 24.6.2 (Hashing Bound) There exists a stabilizer quantum error-correcting codethat achieves the hashing bound R = 1 − H(p) for a Pauli channel of the following form:ρ → pI ρ + pX XρX + pY Y ρY + pZ ZρZ,

(24.104)

where p = (pI , pX , pY , pZ ) and H(p) is the entropy of this probability vector.
Proof. We consider a decoder that corrects only the typical errors. That is, consider defining
the typical error set as follows:


1
pn
n
(24.105)Tδ ≡ a : − log (Pr {Ean }) − H(p) ≤ δ ,nc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

24.6. AN INTERLUDE WITH QUANTUM STABILIZER CODES

673

where an is some sequence consisting of letters corresponding to the Pauli operators {I, X, Y, Z}
and Pr {Ean } is the probability that an i.i.d. Pauli channel issues some tensor-product error
Ean ≡ Ea1 ⊗ · · · ⊗ Ean . This typical set consists of the likely errors in the sense thatX

Pr {Ean } ≥ 1 − ε,(24.106)nan ∈Tδpfor all ε ∈ (0, 1) and sufficiently large
n. The error-correcting conditions for a stabilizer codepnnin this case are that {Ean : a ∈ Tδ } is a correctable set of errors ifEa†n Ebn ∈/ N (S)\S,(24.107)n

for all error pairs Ean and Ebn such that an , bn ∈ Tδp . Also, we consider the expectation of
the error probability under a random choice of a stabilizer code. We proceed as follows:

ES {pe } = ES(
X)
Pr {Ean } I(Ean is uncorrectable using S)(24.108)

an

≤ ES


 X


=



Pr {Ean } I(Ean is uncorrectable using S) + εpn(24.109)an ∈Tδ

X

Pr {Ean } ES {I(Ean is uncorrectable using S)} + ε(24.110)Pr {Ean } Pr {Ean is uncorrectable using S} + ε.

(24.111)nan ∈Tδp=Xnan ∈Tδp

S

The first equality follows by definition—I is an indicator function equal to one if Ean is uncorrectable using S and equal to zero otherwise. The first inequality follows from (24.106)—wecorrect only the typical errors because the atypical error set has negligible probability mass.
The second equality follows by exchanging the expectation and the sum. The third equality
follows because the expectation of an indicator function is the probability that the event it
selects occurs. Continuing, we now bound the probability PrS {Ean is uncorrectable usingc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

674

CHAPTER 24. QUANTUM COMMUNICATIONnS} when an ∈ Tδp :Pr {Ean is uncorrectable using S}
Sonn= Pr ∃Ebn : bn ∈ Tδp , bn 6= an , Ea†n Ebn ∈ N (S)\S
S
o
n
†
pn
n
n
n
≤ Pr ∃Ebn : b ∈ Tδ , b 6= a , Ean Ebn ∈ N (S)
S




[
Ea†n Ebn ∈ N (S)
= Pr
S 

n
bn ∈Tδp , bn 6=an
o
n
X
†
≤
Pr Ean Ebn ∈ N (S)bn ∈Tδp≤nS(24.112)(24.113)(24.114)(24.115), bn 6=an

X

2−(n−k)(24.116)nbn ∈Tδp , bn 6=an

≤ 2n[H(p)+δ] 2−(n−k)(24.117)= 2−n[1−H(p)−k/n−δ] .

(24.118)

The first equality follows from the error-correcting conditions for a quantum stabilizer code,
where N (S) is the normalizer of S. The first inequality follows by ignoring any potential
degeneracy in the code—we consider an error uncorrectable if it lies in the normalizer N (S)and the probability can only be larger because N (S)\S ⊆ N (S). The second equality follows
by realizing that the probabilities for the existence criterion and the union of events are equal.
The second inequality follows by applying the union bound. The third inequality follows fromthe fact that the probability for a fixed operator Ea†n Ebn not equal to the identity commuting
with the stabilizer operators of a random stabilizer can be upper bounded as follows:no 2n+k − 1
Pr Ea†n Ebn ∈ N (S) = 2n
≤ 2−(n−k) .
S
2 −1(24.119)

The random choice of a stabilizer code is equivalent to fixing operators Z1 , . . . , Zn−k andperforming a uniformly random Clifford unitary U . The probability that a fixed operator
commutes with U Z1 U † , . . . , U Zn−k U † is then just the number of non-identity operators in
the normalizer (2n+k − 1) divided by the total number of non-identity operators(22n − 1).
pn
After applying the above bound, we then exploit the typicality bound |Tδ | ≤ 2n[H(p)+δ] .
Plugging back into (24.111), we find thatES {pe } ≤ 2−n[1−H(p)−k/n−δ] + ε.

(24.120)

We conclude that as long as the rate k/n = 1 − H(p) − 2δ, the expectation of the error
probability becomes arbitrarily small, so that there exists at least one choice of a stabilizer
code with the same bound on the error probability.
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24.7. EXAMPLE CHANNELS

675

24.7 Example Channels
We now show how to calculate the quantum capacity for two exemplary channels: thequantum erasure channel and the amplitude damping channel. Both of these channels aredegradable for particular channel parameters, simplifying the calculation of their quantum
capacities.

24.7.1The Quantum Erasure Channel

Recall that the quantum erasure channel acts as follows on an input density operator ρA0 :ρA0 → (1 − ε) ρB + ε|eihe|B ,(24.121)

where ε ∈ [0, 1] is the erasure probability and |eiB is an erasure state that is orthogonal to
the support of any input state ρ.
Proposition 24.7.1 Let dA be the dimension of the input system for the quantum erasurechannel. The quantum capacity of a quantum erasure channel with erasure probability ε isequal to (1 − 2ε) log dA when ε ∈ [0, 1/2] and it is equal to zero otherwise.
Proof. The quantum erasure channel is antidegradable for ε ∈ [1/2, 1]. This follows from
Exercise 5.2.6 and the fact that the erasure channel is degradable for ε ∈ [0, 1/2] (seeExercise 13.5.2). From Exercise 13.5.6 and the fact that N ⊗n is antidegradable if N is, we
can then conclude that the quantum capacity is equal to zero for ε ∈ [1/2, 1].
To determine the quantum capacity for ε ∈ [0, 1/2], we know that it is degradable forthis range (see Exercise 13.5.2), so it suffices to compute its coherent information. We cando so in a similar way as we did in Proposition 21.6.1. Consider that sending one share of apure, bipartite state φAA0 through the channel produces the output
σAB ≡ (1 − ε) φAB + εφA ⊗ |eihe|B .

(24.122)

Recall that Bob can apply the following isometry UB→BX to his state:
UB→BX ≡ ΠB ⊗ |0iX + |eihe|B ⊗ |1iX ,(24.123)

where ΠB is a projector onto the support of the input state (for qubits, it would be just|0ih0| + |1ih1|). Applying this isometry leads to a state σABX where†
σABX ≡ UB→BX σAB UB→BX= (1 − ε) φAB ⊗ |0ih0|X + εφA ⊗ |eihe|B ⊗ |1ih1|X .

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676

CHAPTER 24. QUANTUM COMMUNICATIONThe coherent information I(AiBX)σ is equal to I(AiB)σ because entropies do not change
under the isometry UB→BX . We now calculate I(AiBX)σ :I(AiBX)σ = H(BX)σ − H(ABX)σ= H(B|X)σ − H(AB|X)σ= (1 − ε) [H(B)φ − H(AB)φ ]+ ε H(B)|eihe| − H(AB)φA ⊗|eihe|= (1 − ε) H(B)φ − ε H(A)φ + H(B)|eihe|= (1 − 2ε) H(A)φ≤ (1 − 2ε) log dA .

(24.126)(24.127)(24.128)(24.129)(24.130)(24.131)

The first equality follows from the definition of coherent information. The second equality
follows from φA = TrBX {σABX }, from the chain rule of entropy, and by canceling H(X)on both sides. The third equality follows because the X register is a classical register,
indicating whether the erasure occurs. The fourth equality follows because H(AB)φ = 0,
H(B)|eihe| = 0, and H(AB)φA ⊗|eihe| = H(A)φ + H(B)|eihe| . The fifth equality follows again
because H(B)|eihe| = 0, by collecting terms, and because H(A)φ = H(B)φ (φAB is a pure
bipartite state). The final inequality follows because the entropy of a state on system A isnever greater than the logarithm of the dimension of A. We can conclude that the maximallyentangled state ΦAA0 achieves the quantum capacity of the quantum erasure channel forε ∈ [0, 1/2] because H(A)Φ = log dA .

24.7.2The Amplitude Damping Channel

We now compute the quantum capacity of the amplitude damping channel NAD . Recall that
this channel acts as follows on an input qubit in state ρ:
NAD (ρ) = A0 ρA†0 + A1 ρA†1 ,(24.132)

where, for γ ∈ [0, 1],
A0 ≡ |0ih0| +p

1 − γ|1ih1|,

A1 ≡√γ|0ih1|.

(24.133)

The development here is similar to the development in the proof of Proposition 21.6.2.
Proposition 24.7.2 The quantum capacity of an amplitude damping channel with damping
parameter γ ∈ [0, 1] is equal to the following
max h2 ((1 − γ)p) − h2 (γp),(24.134)p∈[0,1]

whenever γ ∈ [0, 1/2] (recall that h2 (x) is the binary entropy function). For γ ∈ [1/2, 1], thequantum capacity is equal to zero.
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24.7. EXAMPLE CHANNELS

677

Proof. Suppose that a matrix representation of the input qubit density operator ρ in thecomputational basis is
1 − p η∗.
(24.135)ρ=ηp
One can readily verify that the density operator for Bob has the following matrix representation:√
1−(1 − γ) p
1 − γη ∗√,(24.136)
NAD (ρ) =1 − γη
(1 − γ) p
and by calculating the elements Tr{Ai ρA†j }|iihj|, we can obtain a matrix representation for
Eve’s density operator:√ ∗
1 − γpγη
c
NAD (ρ) = √,(24.137)γη
γpcwhere NADis the complementary channel to Eve. By comparing (24.136) and (24.137), we
can see that the channel to Eve is an amplitude damping channel with damping parameter1 − γ. One can verify that the channel is antidegradable for γ ∈ [1/2, 1] and degradable forγ ∈ [0, 1/2] (Exercise: find the degrading channel). By the same reasoning in the previous
proposition, the quantum capacity is equal to zero for γ ∈ [1/2, 1] and it is equal to the
optimized coherent information for γ ∈ [0, 1/2]. So we now focus on this latter case. The
quantum capacity of NAD is equal to its coherent information:

Q(NAD ) = max I(AiB)σ ,φAA0(24.138)

where φAA0 is some pure bipartite input state and σAB = NAD (φAA0 ). We need to determine
the input density operator that maximizes the above formula as a function of γ. So far, theoptimization depends on three parameters: p, Re {η}, and Im {η}. We can show that it issufficient to consider an optimization over only p with η = 0. The formula in (24.138) also
has the following form:c
Q(NAD ) = max [H(NAD (ρ)) − H(NAD(ρ))] ,

(24.139)

I(AiB)σ = H(B)σ − H(AB)σ= H(NAD (ρ)) − H(E)σc= H(NAD (ρ)) − H(NAD(ρ))≡ Icoh (ρ, NAD ).

(24.140)(24.141)(24.142)(24.143)ρ

because

The two entropies in (24.139) depend only on the eigenvalues of the two density operatorsin (24.136)–(24.137), respectively, which are as follows:


q
1
2
2
1 ± (1 − 2 (1 − γ) p) + 4 |η| (1 − γ) ,(24.144)
2


q
1
2
2
1 ± (1 − 2γp) + 4 |η| γ .
(24.145)2
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CHAPTER 24. QUANTUM COMMUNICATION

The above eigenvalues are in the order of Bob and Eve. All of the above eigenvalues have asimilar form, and their dependence on η is only through its magnitude. Thus, it suffices toconsider η ∈ R (this eliminates one parameter). Next, the eigenvalues do not change if we
flip the sign of η (this is equivalent to rotating the original state ρ by Z, to ZρZ), and thus,
the coherent information does not change as well:
Icoh (ρ, NAD ) = Icoh (ZρZ, NAD ).

(24.146)

By the above relation and concavity of coherent information in the input density operator
for degradable channels (Theorem 13.5.2), the following inequality holds:
1[Icoh (ρ, NAD ) + Icoh (ZρZ, NAD )]2≤ Icoh ((ρ + ZρZ) /2, NAD )Icoh (ρ, NAD ) == Icoh (∆(ρ), NAD ),(24.147)(24.148)(24.149)

where ∆ is a completely dephasing channel in the computational basis. This demonstrates
that it is sufficient to consider diagonal density operators ρ when optimizing the coherent
information. Thus, the eigenvalues in (24.144)–(24.145) respectively become{(1 − γ) p, 1 − (1 − γ) p} ,
{γp, 1 − γp} ,(24.150)(24.151)

giving our final expression in the statement of the proposition.
Exercise 24.7.1 Consider the dephasing channel: ρ → (1 − p/2) ρ + (p/2)ZρZ. Prove thatits quantum capacity is equal to 1 − h2 (p/2), where p is the dephasing parameter.

24.8 Discussion of Quantum Capacity
The quantum capacity is particularly well-behaved and understood for the class of degradable
channels. Thus, we should not expect any surprises for this class of channels. If a channel is
not degradable, we currently cannot say much about the exact value of its quantum capacity,
but the study of non-degradable channels has led to many surprises in quantum Shannon
theory and this section discusses two of these surprises. The first is the superadditivity ofcoherent information for the depolarizing channel, and the second is a striking phenomenon
known as superactivation of quantum capacity, where two channels that individually havezero quantum capacity can combine to make a channel with non-zero quantum capacity.

24.8.1Superadditivity of Coherent Information

Recall that the depolarizing channel N D transmits its input with probability 1 − p andreplaces it with the maximally mixed state π with probability p ∈ [0, 1]:
N D (ρ) = (1 − p)ρ + pπ.
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1
Optimizing State
Maximum Coherent Information

0.8
0.6
0.4
0.2
0

0

0.2

0.4
0.6
0.8
Channel parameter p

1

Figure 24.3: This figure plots the maximum coherent information of the depolarizing channel (dashed line)
versus the depolarizing parameter p ∈ [0, 1]. It also plots the value µ ∈ [0, 1] for the state which optimizes
the value of the coherent√ information (solid line) versus the depolarizing parameter p, considering states of√the form µ|00iAA0 + 1 − µ|11iAA0 . It demonstrates that the maximum value is achieved for µ = 1/2 (themaximally entangled state) for all values for which the coherent information is larger than zero and for µ = 0
otherwise.

We focus on the case in which the input and output of this channel is a qubit. The depolarizing channel is an example of a quantum channel that is not degradable.2 As such,
we might expect it to exhibit some strange behavior with respect to its quantum capacity.
Indeed, it is known that its coherent information is strictly superadditive when the channel
becomes very noisy:
5Q(N D ) < Q((N D )⊗5 ).
(24.153)
How can we show that this result is true? First, we can calculate the coherent informationof this channel with respect to one channel use. It is possible to show that the maximallyentangled state ΦAA0 maximizes the channel coherent information Q(N D ) for all values of thecoherent information for which it is non-negative. To see this, consider that the depolarizing
channel is unitarily covariant, so that N D (U ρU † ) = U N D (ρ)U † for any unitary U andD
any qubit input density operator ρ. Thus, for optimizing√ the coherent information of N ,√it suffices to consider states of the form µ|00iAA0 + 1 − µ|11iAA0 where µ ∈ [0, 1]. A
numerical optimization over all such states gives the plot in Figure 24.3, which demonstrates
that the maximally entangled state (µ = 1/2) is optimal for all values of the depolarizing
parameter p for which the coherent information is non-negative, and for all other values, aproduct state with µ = 0 is optimal. Thus, we can calculate the coherent information as
follows:
Q(N D ) = H(B)Φ − H(AB)N D (Φ) = 1 − H(AB)N D (Φ) ,(24.154)
2

Smith and Smolin (2007) have given an explicit condition for whether a channel is degradable.

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680

CHAPTER 24. QUANTUM COMMUNICATION

where H(B)Φ = 1 follows because the output state on Bob’s system is the maximally mixed
state whenever the input to the channel is one share of a maximally entangled state. In
order to calculate H(AB)N D (Φ) , observe that the state on AB isp(1 − p)ΦAB + pπA ⊗ πB = (1 − p)ΦAB + IAB
4p= (1 − p)ΦAB + ([IAB − ΦAB ] + ΦAB )
4


3p
p
= 1−
ΦAB + (IAB − ΦAB ) .
4
4(24.155)(24.156)(24.157)

Since ΦAB and IAB − ΦAB are orthogonal, the eigenvalues of this state are 1 − 3p/4 withmultiplicity one and p/4 with multiplicity three. Thus, the entropy H(AB)N D (Φ) is




p
3p
3p
3p
log 1 −−log,
H(AB)N D (φ) = − 1 −4444
and our final expression for the single-copy coherent information is




p
3p
3p
3p
D
Q(N ) = 1 + 1 −log 1 −
+
log
.
4
4
4
4

(24.158)(24.159)

Another strategy for transmitting quantum data is to encode one share of the maximallyentangled state with a five-qubit repetition code:1√ (|00iAA1 + |11iAA1 )2
1→ √ (|000000iAA1 A2 A3 A4 A5 + |111111iAA1 A2 A3 A4 A5 ) , (24.160)
2and calculate the following coherent information with respect to the state resulting fromsending the systems A1 · · · A5 through the channel:
1
I(AiB1 B2 B3 B4 B5 ).
5(24.161)(We normalize the above coherent information by five in order to make a fair comparison
between a code achieving this rate and one achieving the rate in (24.159).) We know thatthe rate in (24.161) is achievable by applying the direct part of the quantum capacity theorem to the channel (N D )⊗5 , and operationally, this strategy amounts to concatenating arandom quantum code with a five-qubit repetition code. The remarkable result is that thisconcatenation strategy can beat the single-copy coherent information when the channel becomes very noisy. Figure 24.4 demonstrates that the concatenation strategy has positive
coherent information even when the single-copy coherent information in (24.159) vanishes.
This demonstrates superadditivity of coherent information.
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Figure 24.4: The figures plot the coherent information in (24.159) (bottom curves in both figures) and thatin (24.161) (top curves in both figures) versus the depolarizing noise parameter p. The figure on the left ison a linear scale, and the one on the right is on a logarithmic scale. The notable features of the figure on theleft are that the quantum data rate of the top curve is equal to one and the quantum data rate of the bottom
curve is 1/5 when the channel is noiseless (the latter rate is to be expected for a five-qubit repetition code).
Both data rates become small when p is near 0.25, but the figure on the right reveals that the repetitioncode concatenation strategy still gets positive coherent information even when the rate of the random coding
strategy vanishes. This is an example of a channel for which the coherent information can be superadditive.

Why does this phenomenon occur? The simplest (though perhaps not completely satisfying) explanation is that it results from a phenomenon known as degeneracy. Consider aqubit α|0i + β|1i encoded in a repetition code:
α|00000i + β|11111i.

(24.162)

If the “error” Z1 ⊗ Z2 occurs, then it actually has no effect on this state. The same holds
for other two-qubit combinations of Z errors. When the channel noise is low, degeneracy ofthe code with respect to these errors does not help very much because these two-qubit error
combinations are less likely to occur. However, when the channel becomes really noisy, theseerrors are more likely to occur, and the help from degeneracy of the repetition code offsets
the loss in rate.
It is perhaps strange that the coherent information of a depolarizing channel behaves inthis way. The channel seems simple enough, and we could say that the strategies for achieving
the unassisted and entanglement-assisted classical capacity of this channel are very “classical”
strategies. Recall that the best strategy for achieving the unassisted classical capacity is togenerate random codes by picking states uniformly at random from some orthonormal basis,and the receiver measures each channel output in this same orthonormal basis. For achieving
the entanglement-assisted classical capacity, we choose a random code by picking Bell states
uniformly at random and the receiver measures each channel output and his share of eachentangled state in the Bell basis. Both of these results follow from the additivity of therespective capacities. In spite of these other results, the best strategy for achieving thequantum capacity of the depolarizing channel remains very poorly understood.
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24.8.2

Superactivation of Quantum Capacity

Perhaps the most startling result in quantum communication is a phenomenon known assuperactivation. Suppose that Alice is connected to Bob by a quantum channel N1 with zero
capacity for transmitting quantum data. Also, suppose that there is some other zero quantum
capacity channel N2 connecting them. Intuitively, we would expect that Alice should not beable to transmit quantum data reliably over the tensor-product channel N1 ⊗ N2 . That is,using these channels in parallel seems like it should not give any advantage over using theindividual channels alone if they are both individually useless for quantum data transmission(this is the intuition that we have whenever a capacity formula is additive). But two examplesof zero-capacity channels are known that can superactivate each other, such that the joint
channel has a non-zero quantum capacity. How is this possible?
First, consider a 50% quantum erasure channel N1 that transmits its input state withprobability 1/2 and replaces it with an orthogonal erasure state with probability 1/2. As wehave argued before with the no-cloning theorem, such a channel has zero capacity for sending
quantum data reliably (see also Proposition 24.7.1). Now consider some other channel N2 .
The following theorem states that the coherent information of the joint channel N1 ⊗ N2 is
at least equal to half the private information of N2 alone.
Theorem 24.8.1 Let {pX (x), ρxA2 } be an ensemble of inputs for the channel N2 , and let
N1 be a 50% erasure channel. Then there exists a pure state ϕRA1 A2 such that the coherent
information H(B1 B2 ) − H(E1 E2 ) of the joint channel N1 ⊗ N2 is equal to half the private
information I(X; B2 ) − I(X; E2 ) of the second channel:
H(B1 B2 )ω − H(E1 E2 )ω =1
[I(X; B2 )ρ − I(X; E2 )ρ ] ,2(24.163)

whereωRB1 B2 E1 E2 ≡ (UAN11→B1 E1 ⊗ UAN22→B2 E2 ) (ϕRA1 A2 ) ,X
ρXB2 E2 ≡pX (x)|xihx|X ⊗ UAN22→B2 E2 (ρxA2 ),(24.164)(24.165)x

and UAN11→B1 E1 and UAN22→B2 E2 are respective isometric extensions of N1 and N2 . This impliesthat
Q(N1 ⊗ N2 ) ≥ P (N2 )/2.
(24.166)
Proof. Consider the following classical–quantum state corresponding to the ensemble {pX (x), ρxA2 }:X
ρXA2 ≡pX (x)|xihx|X ⊗ ρxA2 .
(24.167)x

Let ρXB2 E2 ≡ UAN22→B2 E2 (ρXA2 ). A purification of this state is
XppX (x)|xiX |xiA01 |φx iA001 A2 ,|ϕiXA1 A2 ≡(24.168)x
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where we identify A1 ≡ A01 A001 and each |φx iA001 A2 is a purification of ρxA2 , so that the state|ϕiXA1 A2 is a purification of ρXA2 . Let |ϕiXB1 E1 B2 E2 be the state resulting from sending A1
and A2 through the tensor-product channel UAN11→B1 E1 ⊗ UAN22→B2 E2 . Identifying the system
B1 ≡ B10 B100 and E1 ≡ E10 E100 , we can write this state as follows by recalling the isometricextension of the erasure channel in (24.6):
1 Xp|ϕiXB1 E1 B2 E2 ≡ √pX (x)|xiX |xiB10 |φx iB100 B2 E2 |eiE1
2 x
1 Xp
+√
pX (x)|xiX |xiE10 |φx iE100 B2 E2 |eiB1 . (24.169)
2 x
Recall that Bob can perform an isometry on B1 of the form in (24.123) that identifies
whether he receives the state or the erasure symbol, and let ZB be a flag register indicating
the outcome. Eve can do the same, and let ZE indicate her flag register. The resulting stateis as follows:1|ψiXB1 E1 B2 E2 ZB ZE ≡ √ |ψ 0 iXB1 E1 B2 E2 |0iZB |1iZE
2
1+ √ |ψ 1 iXB1 E1 B2 E2 |1iZB |0iZE , (24.170)2
where|ψ 0 iXB1 E1 B2 E2 ≡Xp

pX (x)|xiX |xiB10 |φx iB100 B2 E2 |eiE1 ,(24.171)

pX (x)|xiX |xiE10 |φx iE100 B2 E2 |eiB1 .

(24.172)x|ψ 1 iXB1 E1 B2 E2 ≡Xp
x

Then we can evaluate the coherent information of the state resulting from sending system
A1 through the erasure channel and A2 through the other channel N2 :H(B1 B2 )ϕ − H(E1 E2 )ϕ= H(B1 ZB B2 )ψ − H(E1 ZE E2 )ψ
= H(B1 B2 |ZB )ψ + H(ZB )ψ − H(E1 E2 |ZE )ψ − H(ZE )ψ
= H(B1 B2 |ZB )ψ − H(E1 E2 |ZE )ψ
1
1= [H(B1 B2 )ψ0 + H(B2 )ψ1 ] − [H(E2 )ψ0 + H(E1 E2 )ψ1 ]2
2
1
1= [H(XE2 )ψ0 + H(B2 )ψ1 ] − [H(E2 )ψ0 + H(XB2 )ψ1 ]2
2
1
1= [H(XE2 )ρ + H(B2 )ρ ] − [H(E2 )ρ + H(XB2 )ρ ]2
2
1
= [I(X; B2 )ρ − I(X; E2 )ρ ] .
2
c 2019(24.173)(24.174)(24.175)(24.176)(24.177)(24.178)(24.179)

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CHAPTER 24. QUANTUM COMMUNICATION

The first equality follows because Bob and Eve can perform the isometries that identify
whether they receive the state or the erasure flag. The second equality follows from thechaining rule for entropy, and the third follows because the entropies of the flag registers
ZB and ZE are equal for a 50% erasure channel. The fourth equality follows because theregisters ZB and ZE are classical when tracing over the other Z register, so that we canevaluate the conditional entropies as a uniform convex sum of different possibilities: Bob
obtaining the state transmitted or not, and Eve obtaining the state transmitted or not. The
fifth equality follows because the states ψ 0 and ψ 1 are pure. The sixth equality follows
01
because ψXE= ρXE2 , ψB1 2 = ρB2 , ψE0 2 = ρE2 , and ψXB= ρXB2 . The final equality follows
2
2
from the definition of quantum mutual information.
Armed with the above theorem, we need to find an example of a quantum channel thathas zero quantum capacity, but for which there exists an ensemble that registers a nonzero private information. If such a channel were to exist, we could combine it with a 50%
erasure channel in order to achieve a non-zero coherent information (and thus a non-zero
quantum capacity) for the joint channel. Indeed, such a channel exists, and it is known as anentanglement-binding channel. It has the ability to generate private classical communication
but no ability to transmit quantum information (we point the reader to Horodecki et al.
(1996); Horodecki (1997) for further details on these channels). Thus, the 50% erasurechannel and the entanglement-binding channel can superactivate each other.
The startling phenomenon of superactivation has important implications for quantum
data transmission. First, it implies that a quantum channel’s ability to transmit quantum
information depends on the context in which it is used. For example, if other seemingly
useless channels are available, it could be possible to transmit more quantum information
than would be possible were the channels used alone. Next, and more importantly forquantum Shannon theory, it implies that whatever formula might eventually be found tocharacterize quantum capacity (some characterization other than the regularized coherent
information in Theorem 24.3.1), it should be strongly non-additive in some cases (strongly
non-additive in the sense of superactivation). That is, suppose that Q? (N ) is some unknown
formula for the quantum capacity of N and Q? (M) is the same formula characterizing thequantum capacity of M. Then this formula in general should be strongly non-additive insome cases:
Q? (N ⊗ M) > Q? (N ) + Q? (M).
(24.180)
The discovery of superactivation has led us to realize that at present we are much farther
than we might have thought from understanding reliable quantum communication rates overquantum channels.

24.9 Entanglement Distillation
We close out this chaper with a final application of the techniques in the direct coding partof Theorem 24.3.1 to the task of entanglement distillation. Entanglement distillation is a
protocol where Alice and Bob begin with many copies of some bipartite state ρAB . They
attempt to distill ebits from it at some positive rate by employing local operations andc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

24.9. ENTANGLEMENT DISTILLATION

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forward classical communication from Alice to Bob. If the state is pure, then Alice and Bob
should simply perform the entanglement concentration protocol from Chapter 19, and there
is no need for forward classical communication in this case. Otherwise, they can perform theprotocol given in the proof of the following theorem.
Theorem 24.9.1 (Devetak–Winter) Suppose that Alice and Bob share the state ρ⊗nAB
where n is an arbitrarily large positive integer. Then it is possible for them to distill ebits at
the rate I(AiB)ρ if they are allowed forward classical communication from Alice to Bob.
We should mention that we have already proved the statement in the above theorem
with the protocol given in Corollary 22.4.2. Nevertheless, it is still instructive to exploit
the techniques from this chapter in proving the existence of an entanglement distillation
protocol.
Proof. Suppose that Alice and Bob begin with a general bipartite state ρAB with purification
ψABE . We can write the purification in Schmidt form as follows:|ψiABE ≡Xp
pX (x)|xiA ⊗ |ψx iBE .

(24.181)x∈X

The nth extension of the above state is
X p
pX n (xn )|xn iAn ⊗ |ψxn iB n E n .
|ψiAn B n E n ≡(24.182)xn ∈X nThe protocol begins with Alice performing a type class measurement given by the typeprojectors (recall from (15.118) that the typical projector decomposes into a sum of the typeclass projectors):XΠnt ≡|xn ihxn |.
(24.183)xn ∈TtXn

If the type resulting from the measurement is not a typical type, then Alice aborts the
protocol (this result happens with arbitrarily small probability). If it is a typical type, they
can then consider a code over a particular type class t with the following structure:
LM K ≈ |Tt | ≈ 2nH(X) ,(24.184)

K ≈ 2nI(X;E) ,(24.185)

M K ≈ 2nI(X;B) ,(24.186)

where t is the type class and the entropies are with respect to the following dephased state:XpX (x)|xihx|X ⊗ |ψx ihψx |BE .

(24.187)x∈X
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CHAPTER 24. QUANTUM COMMUNICATION

It follows that M ≈ 2n(I(X;B)−I(X;E)) = 2n[H(B)−H(E)] and L ≈ 2nH(X|B) . We label thecodewords as xn (l, m, k) where xn (l, m, k) ∈ Tt . Thus, we instead operate on the following
state |ψ̃t iAn B n E n resulting from the type class measurement:X1|xn iAn ⊗ |ψxn iB n E n .
(24.188)|ψ̃t iAn B n E n ≡ p|Tt | xn ∈Tt
The protocol proceeds as follows. Alice first performs an incomplete measurement of thesystem An , with the following measurement operators:
(
)
X
Γl ≡
|m, kihxn (l, m, k)|An .
(24.189)m,klThis measurement collapses the above state as follows:
1 X√|m, kiAn ⊗ ψxn (l,m,k) B n E n .
M K m,k(24.190)

Alice transmits the classical information in l to Bob, using nH(X|B) bits of classical information. Bob needs to know l so that he can know in which code they are operating.
Bob then constructs the following isometry, a coherent POVM similar to that in (24.28)(constructed from the POVM for a private classical communication code):X q m,kΛB n ⊗ |m, kiB .
(24.191)m,kAfter performing the above coherent POVM, the state is close to the following one:
1 X√
(24.192)|m, kiAn ⊗ |m, kiB ψxn (l,m,k) B n E n .
M K m,kAlice then performs a measurement of the k register in the Fourier-transformed basis:()
1 X i2πks/K|ŝi ≡ √e|ki.
(24.193)K ks∈{1,...,K}

Alice performs this particular measurement because she would like Bob and Eve to maintaintheir entanglement in the k variable. The state resulting from this measurement is
1 X√|miAn ⊗ ei2πks/K |m, kiB ψxn (l,m,k) B n E n .
(24.194)
M K m,kAlice then uses nI(X; E) bits to communicate the s variable to Bob. Bob then applies thephase transformation Z † (s), whereX
Z † (s) =e−i2πsk/K |kihk|,(24.195)k
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to his k variable in register B. The resulting state is√

1 X|miAn ⊗ |m, kiB ψxn (l,m,k) B n E n .
M K m,k(24.196)

They then proceed as in the final steps (24.45)–(24.52) of the protocol from the direct coding
part of Theorem 24.3.1, and they extract a state close to a maximally entangled state of thefollowing form:
1 X√|miAn ⊗ |miB ,(24.197)
M mwith rate equal to (log M ) /n = H(B) − H(E).
Exercise 24.9.1 Argue that the above protocol cannot perform the task of state transfer as
can the protocol in Corollary 22.4.2.

24.10 History and Further Reading
The quantum capacity theorem has a long history that led to many important discoveries inquantum information theory. Shor (1995) first stated the problem of finding the quantum
capacity of a quantum channel in his seminal paper on quantum error correction. DiVincenzo
et al. (1998) demonstrated that the coherent information of the depolarizing channel is superadditive by concatenating a random code with a repetition code (this result in hindsight was
remarkable given that the coherent information was not even known at the time). Smith andSmolin (2007) later extended this result to show that the coherent information is strongly
super-additive for several examples of Pauli channels. Schumacher and Nielsen (1996) demonstrated that the coherent information obeys a quantum data-processing inequality, much
like the classical data-processing inequality for mutual information. Schumacher and Westmoreland (1998) started making connections between private communication and quantum
communication. Bennett et al. (1996c) and Barnum et al. (2000) demonstrated that forward
classical communication cannot increase the quantum capacity. In the same paper, Bennett
et al. (1996c) introduced the idea of entanglement distillation, which has connections with
the quantum capacity.
Schumacher (1996); Schumacher and Nielsen (1996); Barnum et al. (1998, 2000) madeimportant progress on the quantum capacity theorem in a series of papers that establishedthe coherent information upper bound on the quantum capacity. Lloyd (1997), Shor (2002b),
and Devetak (2005) are generally credited with proving the coherent information lower boundon the quantum capacity, though an inspection of Lloyd’s proof reveals that it is perhaps
not as rigorous as the latter two proofs. Shor (2002b) delivered his proof of the lower boundin a lecture, though he never published this proof in a journal. Later, Hayden et al. (2008b)
published a paper detailing a proof of the quantum capacity theorem that they considered to
be close in spirit to the proof in Shor (2002b). After Shor’s proof, Devetak (2005) provided
a detailed proof of the lower bound on the quantum capacity, by analyzing superpositions
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CHAPTER 24. QUANTUM COMMUNICATIONof the codewords from private classical codes. This is the approach we have taken in thischapter. We should also mention that Hamada (2005) showed how to achieve the coherent
information for certain input states by using random stabilizer codes, and Harrington and
Preskill (2001) showed how to achieve the coherent information rate for a very specific classof channels.
Gottesman (1997) established the stabilizer formalism for quantum error correction. Our
discussion of stabilizer codes in this chapter follows closely the development in the PhD
thesis of Smith (2006).
Another approach to proving the quantum capacity theorem is known as the decoupling
approach (Hayden et al., 2008a). This approach exploits a fundamental concept introducedin Schumacher and Westmoreland (2002). Suppose that the reference, Bob, and Eve sharea tripartite pure entangled state |ψiRBE after Alice transmits her share of the entanglementwith the reference through a noisy channel. Then if the reduced state ψRE on the reference
system and Eve’s system is approximately decoupled, meaning thatkψRE − ψR ⊗ σE k1 ≤ ε,(24.198)

where σE is some arbitrary state, this implies that Bob can decode the quantum informationthat Alice intended to send to him. Why is this so? Let us suppose that the state is exactly
decoupled. Then one purification of the state ψRE is the state |ψiRBE that they share after
the channel acts. Another purification of ψRE = ψR ⊗ σE is|ψiRB1 ⊗ |σiB2 E ,(24.199)

where |ψiRB1 is the original state that Alice sent through the channel and |σiB2 E is someother state that purifies the state σE of the environment. Since all purifications are related
by isometries and since Bob possesses the purification of R and E, there exists some unitary
UB→B1 B2 such that
UB→B1 B2 |ψiRBE = |ψiRB1 ⊗ |σiB2 E .
(24.200)
This unitary is then Bob’s decoder! Thus, the decoupling condition implies the existenceof a decoder for Bob, so that it is only necessary to show the existence of an encoderthat decouples the reference from the environment. Simply put, the structure of quantum
mechanics allows for this way of proving the quantum capacity theorem.
Many researchers have now exploited the decoupling approach in a variety of contexts.
This approach is implicit in Devetak’s proof of the quantum capacity theorem (Devetak,2005). Horodecki et al. (2005, 2007) exploited it to prove the existence of a state-merging
protocol. Yard and Devetak (2009) and Ye et al. (2008) used it in their proofs of the state
redistribution protocol. Dupuis et al. (2010) proved the best known characterization of
the entanglement-assisted quantum capacity of the broadcast channel using this approach.
The thesis of Dupuis and subsequent work generalized this decoupling approach to settings
beyond the traditional i.i.d. setting (Dupuis, 2010; Dupuis et al., 2014). Datta and coworkers
have also applied this approach in a variety of contexts (Buscemi and Datta, 2010; Datta
and Hsieh, 2011, 2013), and Wilde and Hsieh (2010) used the approach to study quantum
communication using a noisy channel and a noisy state.
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Bennett et al. (1997) found the quantum capacity of the erasure channel, and Giovannetti
and Fazio (2005) computed the quantum capacity of the amplitude damping channel. Smith
and Yard (2008) showed superactivation and later showed superactivation for channels thatcan be realized more easily in the laboratory (Smith et al., 2011). Devetak and Winter (2005)
established that the coherent information is achievable for entanglement distillation. Cubitt
et al. (2015) and Elkouss and Strelchuk (2015) demonstrated a striking superadditivity effect,
which suggests that a regularized expression is necessary to determine the quantum capacityof an arbitrary channel.
There have also been some results on error exponents, the strong converse, and secondorder characterizations of the quantum capacity (see Section 20.7 for a discussion of themeaning of these terms). Berta et al. (2013) proved that a quantity called the entanglement
cost of a quantum channel is a strong converse rate for quantum communication. Morgan
and Winter (2014) established what they called a “pretty strong converse” for the quantum
capacity of degradable channels, meaning that there is a sharp transition in the fidelity from
one to 1/2, when the rate of communication goes from below to above the quantum capacity(this is in the limit of many channel uses). Wilde and Winter (2014) demonstrated that
randomly selected codes with a communication rate exceeding the quantum capacity of thequantum erasure channel lead to a fidelity that decreases exponentially fast as the numberof channel uses increases (a strong converse would however demonstrate that this behavior
occurs for all codes). Tomamichel et al. (2017) proved that a quantity known as the Rainsbound (defined in Rains (2001)—see also the later work of Audenaert et al. (2002)) is a strong
converse rate for quantum communication over any channel, which in turn establishes thestrong converse for any dephasing channel.
Beigi et al. (2015) and Tomamichel et al. (2015) established second-order achievability
characterizations for quantum capacity. Beigi et al. (2015) did so by making use of a “Petzrecovery map” decoder and Tomamichel et al. (2015) with a version of the decoupling theorem
from Morgan and Winter (2014). Tomamichel et al. (2015) also gave a second-order conversefor quantum communication by making use of the Rains bound, and they obtained an exact
second-order characterization of quantum communication for dephasing channels.

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690

c 2019CHAPTER 24. QUANTUM COMMUNICATIONMark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 25

Trading Resources for Communication
This chapter unifies all of the channel coding theorems that we have studied in this book.
One of the most general information-processing tasks that a sender and receiver can accomplish is to transmit classical and quantum information and generate entanglement with many
independent uses of a quantum channel and with the assistance of classical communication,
quantum communication, and shared entanglement.1 The resulting rates for communication are net rates that give the generation rate of a resource less its consumption rate.
Since we have three resources, all achievable rates are rate triples (C, Q, E) that lie in athree-dimensional capacity region, where C is the net rate of classical communication, Q isthe net rate of quantum communication, and E is the net rate of entanglement consumption/generation. The capacity theorem for this general scenario is known as the quantum
dynamic capacity theorem, and it is the main theorem that we prove in this chapter. All of
the rates given in the channel coding theorems of previous chapters are special points in thisthree-dimensional capacity region.
The proof of the quantum dynamic capacity theorem comes in two parts: the direct coding theorem and the converse theorem. The direct coding theorem demonstrates that thestrategy for achieving any point in the three-dimensional capacity region is remarkably simple: we just combine the protocol from Corollary 22.5.2 for entanglement-assisted classicaland quantum communication with the three unit protocols of teleportation, super-dense coding, and entanglement distribution. The interpretation of the achievable rate region is that
it is the unit resource capacity region from Chapter 8 translated along the points achievablewith the protocol from Corollary 22.5.2. In the proof of the converse theorem, we analyze
the most general protocol that can consume and generate classical communication, quantum
communication, and entanglement along with the consumption of many independent usesof a quantum channel, and we show that the net rates for such a protocol are bounded by
the regularization of the achievable rate region. In the general case, our characterization ismulti-letter, meaning that the computation of the capacity region requires an optimization
over a potentially infinite number of channel uses and is thus intractable. However, both
1

Recall that Chapter 8 addressed a special case of this task that applies to the scenario in which thesender and receiver do not have access to many independent uses of a quantum channel.

691

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CHAPTER 25. TRADING RESOURCES FOR COMMUNICATION

the quantum Hadamard channels from Section 5.2.3 and the quantum erasure channel are
special classes of channels for which the regularization is not necessary, and we can compute
their capacity regions with respect to a single instance of the channel. Another importantclass of channels for which the capacity region is known is the class of pure-loss bosonicchannels (though the optimality proof is only up to a long-standing conjecture which many
researchers believe to be true). These pure-loss bosonic channels model free-space communication or loss in a fiber optic cable and thus have an elevated impetus for study becauseof their importance in practical applications.
One of the most important questions for communication in this three-dimensional setting is whether it is really necessary to exploit the trade-off coding strategy given in Corollary 22.5.2. That is, would it be best simply to use a classical communication code fora fraction of the channel uses, a quantum communication code for another fraction, an
entanglement-assisted code for another fraction, etc.? Such a strategy is known as timesharing and allows the sender and receiver to achieve convex combinations of any rate triples
in the capacity region. The answer to this question depends on the channel. For example,
time sharing is optimal for the quantum erasure channel, but it is not for a dephasing channel or a pure-loss bosonic channel. In fact, trade-off coding for a pure-loss bosonic channel
can give tremendous performance gains over time sharing. How can we know which one will
perform better in the general case? It is hard to say, but at the very least, we know thattime sharing is a special case of trade-off coding as we argued in Section 22.5.2. Thus, fromthis perspective, it might make sense simply to always use a trade-off strategy.
We organize this chapter as follows. We first review the information-processing task corresponding to the quantum dynamic capacity region. Section 25.2 states the quantum dynamic
capacity theorem and shows how many of the capacity theorems we studied previously arise
as special cases of it. The next two sections prove the direct coding theorem and the converse theorem. Section 25.4.2 introduces the quantum dynamic capacity formula, which isimportant for analyzing whether the quantum dynamic capacity region is single-letter. In
the final section of this chapter, we compute and plot the quantum dynamic capacity region
for the dephasing channels, erasure channels, and the pure-loss bosonic channels.

25.1 The Information-Processing Task
Figure 25.1 depicts the most general protocol for generating classical communication, quantum communication, and entanglement with the consumption of a noisy quantum channel
NA0 →B and the same respective resources. Alice possesses two classical registers (each labeled by M and of dimension 2nCout ), a quantum register A1 of dimension 2nQout entangledwith a reference system R, and another quantum register TA of dimension 2nEin that contains
her share of the state ΦTA TB maximally entangled with Bob:ωM M RA1 TA TB ≡ ΦM M ⊗ ΨRA1 ⊗ ΦTA TB .

(25.1)She passes one of the classical registers and the registers A1 and TA into an encoding channel
EM A1 TA →A0n SA LA2 that outputs a quantum register SA of dimension 2nEout and a quantumc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

25.1. THE INFORMATION-PROCESSING TASK

693

Figure 25.1: The most general protocol for generating classical communication, quantum communication,and entanglement with the help of the same respective resources and many uses of a quantum channel.
Alice begins with her classical register M , her quantum register A1 , and her share of the entanglement inregister TA . She encodes according to some encoding channel E that outputs a quantum register SA , many
registers A0n , a quantum register A2 , and a classical register L. She inputs A0n to many uses of the quantum
channel N and transmits A2 over a noiseless quantum channel and L over a noiseless classical channel.
Bob receives the channel outputs B n , the quantum register A2 , and the classical register L and performs
a decoding D that recovers the quantum information and classical message. The decoding also generates
entanglement with system SA . Many protocols are a special case of the above one. For example, the protocol
is entanglement-assisted communication of classical and quantum information if the registers L, SA , SB , andA2 are empty.

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CHAPTER 25. TRADING RESOURCES FOR COMMUNICATION

register A2 of dimension 2nQin , a classical register L of dimension 2nCin , and many quantum
systems A0n . The register SA is for creating entanglement with Bob. The state after theencoding E is as follows:ωM A0n SA LA2 RTB ≡ EM A1 TA →A0n SA LA2 (ωM M RA1 TA TB ).

(25.2)She sends the systems A0n through many uses NA0n →B n ≡ (NA0 →B )⊗n of the quantum channel
NA0 →B , transmits L over a noiseless classical channel, and transmits A2 over a noiselessquantum channel, producing the following state:
ωM B n SA LA2 RTB ≡ NA0n →B n (ωM A0n SA LA2 RTB ).
The above state has the following form:XpX (x)|xihx|X ⊗ NA0n →B n (ρxAA0n ),(25.3)(25.4)

xwith A ≡ RTB A2 SA and X ≡ M L. Bob then applies a decoding channel DB n A2 TB L→B1 SB M̂
that outputs a quantum system B1 , a quantum system SB , and a classical register M̂ . Letω 0 denote the final state. The following condition holds for a protocol with error ε ∈ (0, 1):10ΦM M̂ ⊗ ΨRB1 ⊗ ΦSA SB − ωM
≤ ε,B1 SB M̂ SA R 1
2(25.5)

implying that Alice and Bob establish maximal classical correlations in M and M̂ andmaximal entanglement between SA and SB . The above condition also implies that thecoding scheme preserves the entanglement with the reference system R. The net rate triple
for the protocol is as follows: (Cout − Cin , Qout − Qin , Eout − Ein ). The protocol generates aresource if its corresponding rate is positive, and it consumes a resource if its corresponding
rate is negative. A protocol of the above form is an (n, Cout − Cin , Qout − Qin , Eout − Ein , ε)protocol.
We say that a rate triple (C, Q, E) is achievable for N if there exists a sequence of(n, C − δ, Q − δ, E − δ, ε) protocols for all δ > 0, ε ∈ (0, 1) and sufficiently large n. The
quantum dynamic capacity region CCQE (N ) is equal to the union of all achievable rates.

25.2 The Quantum Dynamic Capacity Theorem
The dynamic capacity theorem gives bounds on the reliable communication rates of a noisy
quantum channel when combined with the noiseless resources of classical communication,
quantum communication, and shared entanglement. The theorem applies regardless ofwhether a protocol consumes the noiseless resources or generates them.
Theorem 25.2.1 (Quantum Dynamic Capacity) The dynamic capacity region CCQE (N )of a quantum channel N is equal to the following expression:
CCQE (N ) =

∞
[
1
k=1

c 2019k(1)

CCQE (N ⊗k ),(25.6)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

25.2. THE QUANTUM DYNAMIC CAPACITY THEOREM

695

(1)

where the overbar indicates the closure of a set. The region CCQE (N ) is equal to the union(1)of the state-dependent regions CCQE,σ (N ):
[ (1)
(1)
CCQE (N ) ≡CCQE,σ (N ).
(25.7)σ(1)

The state-dependent region CCQE,σ (N ) is the set of all rates C, Q, and E, such thatC + 2Q ≤ I(AX; B)σ ,Q + E ≤ I(AiBX)σ ,C + Q + E ≤ I(X; B)σ + I(AiBX)σ .

(25.8)(25.9)(25.10)

The above entropic quantities are with respect to a classical–quantum state σXAB , whereX(25.11)σXAB ≡pX (x)|xihx|X ⊗ NA0 →B (φxAA0 ),xand the states φxAA0 are pure. It is implicit that one should consider states on A0k instead of0
A when taking the regularization in (25.6).

The above theorem is a “multi-letter” capacity theorem due to the regularization in(25.6). However, we show in Section 25.5.1 that the regularization is not necessary for the
Hadamard class of channels. We prove the above theorem in two parts:
1. The direct coding theorem in Section 25.3 shows that combining the protocol from
Corollary 22.5.2 with teleportation, super-dense coding, and entanglement distribution
achieves the above region.
2. The converse theorem in Section 25.4 demonstrates that any coding scheme cannot dobetter than the regularization in (25.6), in the sense that a sequence of protocols withdecreasing error should have the communication rates below the above amounts.
Exercise 25.2.1 Show that it suffices to evaluate the following four entropies in order todetermine the state-dependent region in Theorem 25.2.1:
X
pX (x)H(A)φx ,(25.12)
H(A|X)σ =x!
H(B)σ = HX

pX (x)NA0 →B (φxA0 ) ,(25.13)x

H(B|X)σ =X

H(E|X)σ =X

pX (x)H(NA0 →B (φxA0 )),(25.14)pX (x)H(NAc 0 →E (φxA0 )),(25.15)xxwhere the state σXABE extends the state in (25.11) and is of the formX
σXABE ≡pX (x)|xihx|X ⊗ UAN0 →BE (φxAA0 ),(25.16)x

where UAN0 →BE is an isometric extension of the channel NA0 →BE .
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CHAPTER 25. TRADING RESOURCES FOR COMMUNICATION

25.2.1

Special Cases of the Quantum Dynamic Capacity Theorem

We first consider five special cases of the above capacity theorem that arise when Q andE both vanish, C and E both vanish, or one of C, Q, or E vanishes. The first two cases
correspond respectively to the classical capacity theorem from Chapter 20 and the quantum capacity theorem from Chapter 24. Each of the other special cases traces out a twodimensional achievable rate region in the three-dimensional capacity region. The five coding
scenarios are as follows:
1. Classical communication (C) when there is no entanglement assistance or quantum
communication. The achievable rate region lies on the (C, 0, 0) ray extending from theorigin.
2. Quantum communication (Q) when there is no entanglement assistance or classicalcommunication. The achievable rate region lies on the (0, Q, 0) ray extending from theorigin.
3. Entanglement-assisted quantum communication (QE) when there is no classical communication. The achievable rate region lies in the (0, Q, −E) quarter-plane of thethree-dimensional region in Theorem 25.2.1.
4. Classically enhanced quantum communication (CQ) when there is no entanglement
assistance. The achievable rate region lies in the (C, Q, 0) quarter-plane of the threedimensional region in Theorem 25.2.1.
5. Entanglement-assisted classical communication (CE) when there is no quantum communication. The achievable rate region lies in the (C, 0, −E) quarter-plane of thethree-dimensional region in Theorem 25.2.1.
Classical Capacity
The following theorem gives the one-dimensional capacity region CC (N ) of a quantum channel N for classical communication.
Theorem 25.2.2 (Holevo–Schumacher–Westmoreland) The classical capacity region
CC (N ) is given by∞[
1 (1) ⊗k
CC (N ).
(25.17)
CC (N ) =
k
k=1
(1)

(1)

(1)

The region CC (N ) is the union of the state-dependent regions CC,σ (N ), where CC,σ (N ) is
the set of all C ≥ 0, such thatC ≤ I(X; B)σ + I(AiBX)σ .

(25.18)The entropic quantity is with respect to the state σXAB in (25.11).
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25.2. THE QUANTUM DYNAMIC CAPACITY THEOREM

697

The bound in (25.18) is never larger than the bound in (25.10) with Q = 0 and E = 0,
given that I(X; B)σ + I(AiBX)σ ≤ I(AX; B)σ . The above characterization of the classical capacity region may seem slightly different from the characterization in Chapter 20,
until we make a few observations. First, we rewrite the coherent information I(AiBX)σas H(B|X)σ − H(E|X)σ . Then I(X; B)σ + I(AiBX)σ = H(B)σ − H(E|X)σ . Next, purestates of the form |ϕx iA0 are sufficient to attain the classical capacity of a quantum channel (see Theorem 13.3.2). Then H(E|X)σ = H(B|X)σ so that I(X; B)σ + I(AiBX)σ =H(B)σ − H(B|X)σ = I(X; B)σ for states of this form. Thus, the expression in (25.18) cannever exceed the classical capacity and finds its maximum exactly at the Holevo information.
Quantum Capacity
The following theorem gives the one-dimensional quantum capacity region CQ (N ) of a quantum channel N .
Theorem 25.2.3 (Quantum Capacity) The quantum capacity region CQ (N ) is given by
CQ (N ) =

∞
[
1

(1)

CQ (N ⊗k ).
k
k=1

(1)

(25.19)(1)(1)

The region CQ (N ) is the union of the state-dependent regions CQ,σ (N ), where CQ,σ (N ) is
the set of all Q ≥ 0, such thatQ ≤ I(AiBX)σ .
(25.20)The entropic quantity is with respect to the state σXAB in (25.11) with the restriction thatthe density pX (x) is degenerate.
The bound in (25.20) is a special case of the bound in (25.9) with E = 0. The other bounds
in Theorem 25.2.1 are looser than the bound in (25.9) when C, E = 0.
Entanglement-Assisted Quantum Capacity
The following theorem gives the two-dimensional entanglement-assisted quantum capacity
region CQE (N ) of a quantum channel N .
Theorem 25.2.4 (Devetak–Harrow–Winter) The entanglement-assisted quantum capacity region CQE (N ) is given by
CQE (N ) =

∞
[
1

(1)

CQE (N ⊗k ).
k
k=1

(1)

(25.21)(1)(1)

The region CQE (N ) is the union of the state-dependent regions CQE,σ (N ), where CQE,σ (N ) is
the set of all Q, E ≥ 0, such that
2Q ≤ I(AX; B)σ ,Q ≤ I(AiBX)σ + |E| .
c 2019(25.22)(25.23)

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698

CHAPTER 25. TRADING RESOURCES FOR COMMUNICATIONThe entropic quantities are with respect to the state σXAB in (25.11) with the restriction thatthe density pX (x) is degenerate.
The bounds in (25.22) and (25.23) are a special case of the respective bounds in (25.8) and(25.9) with C = 0. The other bounds in Theorem 25.2.1 are looser than the bounds in (25.8)
and (25.9) when C = 0. Observe that the region is a union of general pentagons (see theQE-plane in Figure 25.2 for an example of one of these general pentagons in the union).
Classically-Enhanced Quantum Capacity
The following theorem gives the two-dimensional capacity region CCQ (N ) for classically enhanced quantum communication over a quantum channel N .
Theorem 25.2.5 (Devetak–Shor) The classically enhanced quantum capacity region CCQ (N )is given by∞[
1 (1) ⊗k
CCQ (N ).
(25.24)
CCQ (N ) =
k
k=1
(1)

(1)

(1)

The region CCQ (N ) is the union of the state-dependent regions CCQ,σ (N ), where CCQ,σ (N ) is
the set of all C, Q ≥ 0, such thatC + Q ≤ I(X; B)σ + I(AiBX)σ ,Q ≤ I(AiBX)σ .

(25.25)(25.26)The entropic quantities are with respect to the state σXAB in (25.11).
The bounds in (25.25) and (25.26) are a special case of the respective bounds in (25.9) and(25.10) with E = 0. The first inequality in (25.8) is redundant because Q ≤ I(AiBX)σ =H(A|EX)σ ≤ H(A|X)σ and combining this with the inequality in (25.25) gives (25.8).
Observe that the region is a union of trapezoids (see the CQ-plane in Figure 25.2 for an
example of one of these rectangles in the union).
Entanglement-Assisted Classical Capacity with Limited Entanglement
Theorem 25.2.6 (Shor) The entanglement-assisted classical capacity region CCE (N ) of aquantum channel N is∞[
1 (1) ⊗k
CCE (N ).
(25.27)
CCE (N ) =
k
k=1
(1)

(1)

(1)

The region CCE (N ) is the union of the state-dependent regions CCE,σ (N ), where CCE,σ (N ) is
the set of all C, E ≥ 0, such thatC ≤ I(AX; B)σ ,C ≤ I(X; B)σ + I(AiBX)σ + |E| ,(25.28)(25.29)where the entropic quantities are with respect to the state σXAB in (25.11).
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25.3. THE DIRECT CODING THEOREM

699

The bounds in (25.28) and (25.29) are a special case of the respective bounds in (25.8)
and (25.10) with Q = 0. Observe that the region is a union of general polyhedra (see theCE-plane in Figure 25.2 for an example of one of these general polyhedra in the union).

25.3 The Direct Coding Theorem
The unit resource achievable region is what Alice and Bob can achieve using the protocols
entanglement distribution, teleportation, and super-dense coding (see Chapter 8). It is thecone of the rate triples corresponding to these protocols:{α (0, −1, 1) + β (2, −1, −1) + γ (−2, 1, −1) : α, β, γ ≥ 0} .

(25.30)We can also write any rate triple (C, Q, E) in the unit resource capacity region with a matrix
equation:
  
 
C
0
2 −2 αQ = −1 −1 1  β  .
(25.31)E
1 −1 −1γ
The inverse of the above matrix is as follows: 1− 2 −1 0 0 −1 −1 ,2
2− 12 − 12 − 21

(25.32)and gives the following set of inequalities for the unit resource achievable region:C + 2Q ≤ 0,
Q + E ≤ 0,
C + Q + E ≤ 0,

(25.33)(25.34)(25.35)by inverting the matrix equation in (25.31) and applying the constraints α, β, γ ≥ 0.
Now, let us include the protocol from Corollary 22.5.2 for entanglement-assisted communication of classical and quantum information. Corollary 22.5.2 states that we can achieve
the following rate triple by channel coding for NA0 →B :11
I(X; B)σ , I(A; B|X)σ , − I(A; E|X)σ ,(25.36)2
2
for any state σXABE of the formσXABE ≡X

pX (x)|xihx|X ⊗ UAN0 →BE (φxAA0 ),(25.37)x

where UAN0 →BE is an isometric extension of the quantum channel NA0 →B . Specifically, we
showed in Corollary 22.5.2 that one can achieve the above rates with vanishing error in thec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 25. TRADING RESOURCES FOR COMMUNICATION

limit of large blocklength. Thus the achievable rate region is the following translation of theunit resource achievable region in (25.31):    
C0
2 −2 αI(X; B)σQ = −1 −1 1  β  +  1 I(A; B|X)σ  .
(25.38)2
1− 2 I(A; E|X)σE
1 −1 −1γ
We can now determine bounds on an achievable rate region that employs the above coding
strategy. We apply the inverse of the matrix in (25.31) to the left-hand side and right-handside, giving
 1   1  − 2 −1 0C
I(X; B)σα− 2 −1 0 0 − 1 − 1  Q −  0 − 1 − 1   1 I(A; B|X)σ  = β  .
(25.39)222
2
2− 12 − 12 − 12
− 21 − 12 − 21
− 12 I(A; E|X)σEγ
Then using the following identities:
I(X; B)σ + I(A; B|X)σ = I(AX; B)σ ,1
1
I(A; B|X)σ − I(A; E|X)σ = I(AiBX)σ ,2
2(25.40)(25.41)and the constraints α, β, γ ≥ 0, we obtain the inequalities in (25.8)–(25.10), corresponding
exactly to the state-dependent region in Theorem 25.2.1. Taking the union over all possible
states σ in (25.11) and taking the regularization gives the full dynamic achievable rate region.
Figure 25.2 illustrates an example of the general polyhedron specified by (25.8)–(25.10),
where the channel is the qubit dephasing channel ρ → (1 − p)ρ + pZρZ with dephasing
parameter p = 0.2, and the input state is1
σXAA0 ≡ (|0ih0|X ⊗ φ0AA0 + |1ih1|X ⊗ φ1AA0 ),2(25.42)wherepp
1/4|00iAA0 + 3/4|11iAA0 ,p
p
φ1 AA0 ≡ 3/4|00iAA0 + 1/4|11iAA0 .
φ0 AA0 ≡(25.43)(25.44)The state σXABE resulting from the channel is UAN0 →BE (σXAA0 ) where UAN0 →BE is an isometricextension of the qubit dephasing channel. The figure caption provides a detailed explanation(1)of the state-dependent region CCQE,σ (note that Figure 25.2 displays the state-dependent
region and does not display the full capacity region).

25.4 The Converse Theorem
We provide a catalytic, information-theoretic converse proof of the dynamic capacity region,
showing that (25.6) gives a multi-letter characterization of it. The catalytic approach meansc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

25.4. THE CONVERSE THEOREM

701

(1)

Figure 25.2: An example of the state-dependent achievable region CCQE σ (N ) corresponding to a stateσXABE that arises from a qubit dephasing channel with dephasing parameter p = 0.2. The figure depicts
the octant corresponding to the consumption of entanglement and the generation of classical and quantum
communication. The state input to the channel N is σXAA0 , defined in (25.42). The plot features sevenachievable corner points of the state-dependent region. We can achieve the convex hull of these seven points
by time sharing any two different coding strategies. We can also achieve any point above an achievable pointby consuming more entanglement than necessary. The seven achievable points correspond to entanglementassisted quantum communication (EAQ), the protocol from Corollary 22.5.3 for classically enhanced quantum
communication (CEQ), the protocol from Theorem 22.5.1 for entanglement-assisted classical communication
with limited entanglement (EAC), quantum communication (LSD), combining CEF with entanglement distribution and super-dense coding (CEF-SD-ED), the protocol from Corollary 22.5.2 for entanglement-assisted
communication of classical and quantum information (CEF), and combining CEF with teleportation (CEFTP). Observe that we can obtain EAC by combining CEF with super-dense coding, so that the points CEQ,
CEF, EAC, and CEF-SD-ED all lie in plane III. Observe that we can obtain CEQ from CEF by entanglement distribution and we can obtain LSD from EAQ and EAQ from CEF-TP, both by entanglement
distribution. Thus, the points CEF, CEQ, LSD, EAQ, and CEF-TP all lie in plane II. Finally, observe that
we can obtain all corner points by combining CEF with the unit protocols of teleportation, super-dense
coding, and entanglement distribution. The bounds in (25.8)–(25.10) uniquely specify the respective planes
I-III. We obtain the full achievable region by taking the union over all states σ of the state-dependent regions(1)
Cσ (N ) and taking the regularization, as outlined in Theorem 25.2.1. The above region is a translation of
the unit resource capacity region from Chapter 8 to the protocol for entanglement-assisted communicationof classical and quantum information.

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CHAPTER 25. TRADING RESOURCES FOR COMMUNICATION

that we are considering the most general protocol that consumes and generates classicalcommunication, quantum communication, and entanglement in addition to the uses of thenoisy quantum channel. This approach has the advantage that we can prove the converse
theorem in “one fell swoop.” We employ the AFW inequality, the chain rule for quantum
mutual information, elementary properties of quantum entropy, and the quantum dataprocessing inequality to prove the converse.
We show that the bounds in (25.8)–(25.10) hold for randomness distribution instead ofclassical communication because a capacity for generating shared randomness can only bebetter than that for generating classical communication (classical communication can generate shared randomness). We also consider a protocol that preserves maximal entanglementwith a reference system instead of one that generates quantum communication.
We prove that the converse theorem holds for a state of the following form:Xx
σXAB n ≡p(x)|xihx|X ⊗ NA⊗n(25.45)0 →B (ρAA0n ),xwhere the states ρxAA0n are mixed. We identify this state with ωXAB n defined in (25.3), setting
A ≡ RSA A2 TB and X ≡ LM . We do this rather than proving it for a state of the form in(25.11). Then we show in Section 25.4.1 that it is not necessary to consider an ensembleof mixed states—i.e., we can do just as well with an ensemble of pure states, giving thestatement of Theorem 25.2.1.
We first prove the bound in (25.8). Consider the following chain of inequalities:n(Cout + 2Qout )= I(M ; M̂ )Φ + I(R; B1 )Φ

(25.46)= I(RM ; B1 M̂ )Φ⊗Φ

(25.47)≤ I(RM ; B1 M̂ )ω0 + nδ 0
≤ I(RM ; B n A2 LTB )ω + nδ 0
= I(RA2 TB LM ; B n )ω + I(RM ; A2 TB L)ω − I(B n ; A2 TB L)ω + nδ 0
≤ I(RSA A2 TB LM ; B n )ω + I(RM ; A2 TB L)ω + nδ 0
= I(AX; B n )ω + I(RM ; TB )ω + I(RM ; L|TB )ω + I(RM ; A2 |TB L)ω + nδ 0
≤ I(AX; B n )ω + n(Cin + 2Qin ) + nδ 0 .

(25.48)(25.49)(25.50)(25.51)(25.52)(25.53)

The first equality holds by evaluating the quantum mutual informations on the respective
states ΦM M̂ and ΦRB1 . The second equality follows because the mutual information is additive with respect to tensor-product states (see Exercise 11.6.8). The first inequality follows
from the condition in (25.5) and an application of the AFW inequality where δ 0 is a parameter that vanishes as n → ∞ and ε → 0. The second inequality follows from quantum dataprocessing. The third equality is a consequence of an identity from Exercise 11.7.1. The third
inequality follows from quantum data processing and the fact that I(B n ; A2 TB L)ω ≥ 0. The
fourth equality follows by identifying A ≡ RSA A2 TB and X ≡ LM , and also from the chain
rule, which implies that I(RM ; A2 TB L)ω = I(RM ; TB )ω +I(RM ; L|TB )ω +I(RM ; A2 |TB L)ω .
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25.4. THE CONVERSE THEOREM

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The final inequality follows because I(RM ; TB )ω = 0 and from the dimension bounds
I(RM ; L|TB )ω ≤ log dim(HL ) = nCin and I(RM ; A2 |TB L)ω ≤ 2 log dim(HA2 ) = n2Qin(see Exercise 11.7.9). Thus, (25.8) holds for the net rates.
We now prove the second bound in (25.9). Consider the following chain of inequalities:n(Qout + Eout ) = I(RiB1 )Φ + I(SA iSB )Φ
= I(RSA iB1 SB )Φ⊗Φ
≤ I(RSA iB1 SB )ω0 + nδ 0
≤ I(RSA iB n A2 TB LM )ω + nδ 0
≤ I(RSA A2 TB iB n LM )ω + log dim(HA2 ⊗ HTB ) + nδ 0
= I(AiB n X)ω + n(Qin + Ein ) + nδ 0 .

(25.54)(25.55)(25.56)(25.57)(25.58)(25.59)The first equality follows by evaluating the coherent informations of the respective statesΦRB1 and ΦSA SB . The second equality follows because ΦRB1 ⊗ ΦSA SB is a product state andcoherent information is additive with respect to tensor-product states. The first inequality
follows from the condition in (25.5) and an application of the AFW inequality with parameter δ 0 vanishing when n → ∞ and ε → 0. The second inequality follows from quantum
data processing. The third inequality follows from the dimension bound in Exercise 11.8.5.
The final equality follows by identifying A ≡ RSA A2 TB and X ≡ LM , and noting thatlog dim(HA2 ) = nQin and log dim(HTB ) = nEin . Thus, (25.9) holds for the net rates.
We prove the last bound in (25.10). Consider the following chain of inequalities:n(Cout + Qout + Eout )= I(M ; M̂ )Φ + I(RSA iB1 SB )Φ⊗Φ

(25.60)≤ I(M ; M̂ )ω0 + I(RSA iB1 SB )ω0 + nδ 0
≤ I(M ; B n A2 TB L)ω + I(RSA iB n A2 TB LM )ω + nδ 0
= I(M L; B n )ω + I(RSA A2 TB iB n LM )ω + I(M ; L)ω
+ H(A2 TB |B n )ω − I(A2 B n TB ; L)ω + nδ 0
≤ I(X; B n )ω + I(AiB n X)ω + I(M ; L)ω + H(A2 TB |B n )ω + nδ 0
≤ I(X; B n )ω + I(AiB n X)ω + n(Cin + Qin + Ein ) + nδ 0 .

(25.61)(25.62)(25.63)(25.64)(25.65)

The first equality follows from evaluating the mutual information of the state ΦM M̂ and thecoherent information of the product state ΦRB1 ⊗ ΦSA SB . The first inequality follows fromthe condition in (25.5) and an application of the AFW inequality with δ 0 vanishing whenn → ∞ and ε → 0. The second inequality follows from quantum data processing. The
second equality is an identity, verified by expanding all quantities as unconditional entropies
and seeing that both terms are equal to H(M )ω + H(B n A2 TB L)ω − H(RSA B n A2 TB LM )ω .
The third inequality follows by identifying A ≡ RTB A2 SA and X ≡ M L, and also because
I(A2 B n TB ; L)ω ≥ 0. The last inequality follows from the dimension bounds I(M ; L)ω ≤log dim(HL ) = nCin and H(A2 TB |B n )ω ≤ log dim(HA2 ⊗ HTB ) = n(Qin + Ein ). Thus, (25.10)
applies to the net rates. This concludes the proof of the converse theorem.
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704

CHAPTER 25. TRADING RESOURCES FOR COMMUNICATION

25.4.1

Pure-state Ensembles are Sufficient

We prove that it is sufficient to consider an ensemble of pure states as in the statement ofTheorem 25.2.1 rather than an ensemble of mixed states as in (25.45) in the proof of ourconverse theorem. We first determine a spectral decomposition of the mixed-state ensemble,
model the index of the pure states in the decomposition as a classical variable Y , and thenplace this classical variable Y in a classical register. It follows that the communication rates
can only improve, and it is sufficient to consider an ensemble of pure states.
Consider that each mixed state in the ensemble in (25.45) admits a spectral decompositionof the following form:
Xx,y
ρxAA0 =
p(y|x)ψAA(25.66)0.
y

We can thus represent the ensemble as follows:X
x,yρXAB ≡
p(x)p(y|x)|xihx|X ⊗ NA0 →B (ψAA
0 ).

(25.67)x,yThe inequalities in (25.8)–(25.10) for the dynamic capacity region involve the mutual information I(AX; B)ρ , the Holevo information I(X; B)ρ , and the coherent information I(AiBX)ρ .
As we show below, each of these entropic quantities can only improve in each case if we makethe variable y be part of the classical variable. This improvement then implies that it is only
necessary to consider pure states in the dynamic capacity theorem.
Let θXY AB denote an augmented state of the following form:
Xx,yθXY AB ≡p(x)p(y|x)|xihx|X ⊗ |yihy|Y ⊗ NA0 →B (ψAA
(25.68)0 ).
xThis state is actually a state of the form in (25.11) if we subsume the classical variables
X and Y into one classical variable. The following three inequalities each follow from anapplication of the quantum data-processing inequality:
I(X; B)ρ = I(X; B)θ ≤ I(XY ; B)θ ,I(AX; B)ρ = I(AX; B)θ ≤ I(AXY ; B)θ
I(AiBX)ρ = I(AiBX)θ ≤ I(AiBXY )θ .

(25.69)(25.70)(25.71)Each of these inequalities proves the desired result for the respective Holevo information,
mutual information, and coherent information, and it suffices to consider an ensemble ofpure states in Theorem 25.2.1. The same argument holds with NA0 →B replaced by NA⊗n0 →B .

25.4.2The Quantum Dynamic Capacity Formula

Here we introduce the quantum dynamic capacity formula and show how its additivity
implies that the computation of the Pareto optimal trade-off surface of the capacity region
requires an optimization over a single copy of the channel, rather than a potentially infinite
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25.4. THE CONVERSE THEOREM

705

number of them. The Pareto optimal trade-off surface consists of all points in the capacity
region that are Pareto optimal, in the sense that it is not possible to make improvements in
one resource without offsetting another resource (these are essentially the boundary pointsof the region in our case). We then show how several important capacity formulas discussed
previously in this book are special cases of the quantum dynamic capacity formula.
Definition 25.4.1 (Quantum Dynamic Capacity Formula) The quantum dynamic capacity formula of a quantum channel N is defined as follows:D~λ (N ) ≡ max λ1 I(AX; B)σ + λ2 I(AiBX)σ + λ3 [I(X; B)σ + I(AiBX)σ ] ,σ(25.72)where σ is a state of the form in (25.11) and ~λ ≡ (λ1 , λ2 , λ3 ) is a vector of Lagrange multipliers such that λ1 , λ2 , λ3 ≥ 0.
Definition 25.4.2 The regularized quantum dynamic capacity formula is defined as follows:
1
D~ (N ⊗k ).
k→∞ k λD~λreg (N ) ≡ lim(25.73)

Lemma 25.4.1 Suppose the quantum dynamic capacity formula is additive for a channel N
and any other arbitrary channel M:
D~λ (N ⊗ M) = D~λ (N ) + D~λ (M).

(25.74)

Then the regularized quantum dynamic capacity formula for N is equal to the quantum
dynamic capacity formula:D~λreg (N ) = D~λ (N ).
(25.75)In this sense, the regularized formula “single-letterizes.”
Proof. We prove the result using induction on n. The base case for n = 1 is trivial.
Suppose the result holds for n: D~λ (N ⊗n ) = nD~λ (N ). Then the following chain of equalities
establishes the inductive step:D~λ (N ⊗n+1 ) = D~λ (N ⊗ N ⊗n ) = D~λ (N ) + D~λ (N ⊗n ) = D~λ (N ) + nD~λ (N ).

(25.76)

The first equality follows by expanding the tensor product. The second critical equality
follows from the assumption in (25.74), setting M = N ⊗n . The final equality follows fromthe induction hypothesis.
Theorem 25.4.1 Single-letterization of the quantum dynamic capacity formula implies thatthe computation of the Pareto optimal trade-off surface of the dynamic capacity region requires an optimization over a single copy of the channel.
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706

CHAPTER 25. TRADING RESOURCES FOR COMMUNICATION

Proof. We employ ideas from optimization theory for the proof (see Boyd and Vandenberghe(2004)). We would like to characterize all the points in the capacity region that are Pareto
optimal. Such a task is a standard vector optimization in the theory of Pareto trade-off
analysis (see Section 4.7 of Boyd and Vandenberghe (2004)).
~ ≡ (C, Q, E) a rate vector, andLet w~ ≡ (wC , wQ , wE ) ∈ R3 be a weight vector, RxE ≡ {pX (x), φAA0 } an ensemble. Our main goal here is to show that the computational
problem reduces to computing the quantum dynamic capacity formula in (25.72) for all
values of ~λ such that λ1 , λ2 , λ3 ≥ 0. As such, we can focus for now on a single copy of thechannel and deduce the statement of the theorem once we are done. We can phrase the taskof computing the boundary of the single-copy capacity region as the following optimization
problem:~~ ·R
P ∗ (w)~ ≡ sup w(25.77)~
R,E

C + 2Q ≤ I(AX; B)σ ,Q + E ≤ I(AiBX)σ ,C + Q + E ≤ I(X; B)σ + I(AiBX)σ ,subject to(25.78)(25.79)(25.80)~ = (C, Q, E) and ensembles
where the optimization is with respect to all rate vectors R
E = {pX (x), φxAA0 }, with σXAB a state of the form in (25.11).
The geometric interpretation of the optimization task is that we are trying to find a
supporting plane of the dynamic capacity region such that the weight vector w~ is the normal~
vector of the plane and the value of its inner product with R characterizes the offset of theplane, so that P ∗ (w)~ for all w~ ∈ R3 characterizes the boundary of the region. Note that the
optimal value P ∗ (w)~ can sometimes be infinite. For example, if w
~ = (−1, −1, −1), then the~
optimal rates are R = (−∞, −∞, −∞), which corresponds to simply consuming all resources(the dynamic capacity theorem does not give any constraint on all resources being consumed
at the same time, but rather on the generation of some resources while others are consumed).
~ subject to the constraints
For now, let us fix an ensemble E and let P ∗ (w,~ E) ≡ supR~ w~ · R,in (25.78)–(25.80). The optimization problem then becomes what is known as a linear~ and the constraints are linear
program, given that the objective function is linear in R
~ We can then define the following Lagrangian, which introduces
inequalities involving R.
Lagrange multipliers λ1 , λ2 , λ3 for the respective constraints in (25.78)–(25.80):~ E, ~λ)L(w,~ R,≡ wC C + wQ Q + wE E + λ1 [I(AX; B)σ − (C + 2Q)]
+ λ2 [I(AiBX)σ − (Q + E)]+ λ3 [I(X; B)σ + I(AiBX)σ − (C + Q + E)]= (wC − λ1 − λ3 ) C + (wQ − 2λ1 − λ2 − λ3 ) Q + (wE − λ2 − λ3 ) E+ λ1 I(AX; B)σ + λ2 I(AiBX)σ + λ3 [I(X; B)σ + I(AiBX)σ ] .
c 2019(25.81)(25.82)

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25.4. THE CONVERSE THEOREM

707

The Lagrange dual function is defined as~ E, ~λ).
~ R,g(w,~ E, ~λ) ≡ sup L(w,(25.83)~R

By inspection, g(w,~ E, ~λ) is infinite unless wC = λ1 + λ3 , wQ = 2λ1 + λ2 + λ3 , wE = λ2 + λ3 ,or equivalently (by inverting these equations), g(w,~ E, ~λ) is infinite unless
1(wQ − wE ) ,21
λ2 = (−2wC + wQ + wE ) ,21
λ3 = (2wC − wQ + wE ) .
2
λ1 =(25.84)(25.85)(25.86)

So we see thatg(w,~ E, ~λ) = λ1 I(AX; B)σ + λ2 I(AiBX)σ + λ3 [I(X; B)σ + I(AiBX)σ ](25.87)

if (25.84)–(25.86) hold and g(w,~ E, ~λ) = +∞ otherwise. Observe that (25.87) contains theexpressions in the quantum dynamic capacity formula in (25.72).
When λ1 , λ2 , λ3 ≥ 0, the Lagrange dual function gives upper bounds on P ∗ (w,~ E). To see~
this, let R be a rate vector satisfying (25.78)–(25.80) for fixed E. Then
~ ≤ L(w,~ E, ~λ) ≤ g(w,
w
~ ·R
~ R,
~ E, ~λ).

(25.88)

We can then obtain the tightest upper bound on P ∗ (w,~ E) by taking an infimum over all ~λsatisfying λ1 , λ2 , λ3 ≥ 0:
P ∗ (w,~ E) ≤ D∗ (w,~ E) ≡

inf

λ1 ,λ2 ,λ3 ≥0g(w,~ E, ~λ).

(25.89)The optimization problem on the right-hand side is known as the dual optimization problem
and the inequality P ∗ (w,~ E) ≤ D∗ (w,~ E) is known as weak duality, which always holds. Let
1(wQ − wE ) ,21
λ∗2 = (−2wC + wQ + wE ) ,21∗
λ3 = (2wC − wQ + wE ) .
2
λ∗1 =(25.90)(25.91)(25.92)

Then by inspection, observe thatD∗ (w,~ E) = λ∗1 I(AX; B)σ + λ∗2 I(AiBX)σ + λ∗3 [I(X; B)σ + I(AiBX)σ ](25.93)if λ∗1 , λ∗2 , λ∗3 ≥ 0 and it is equal to +∞ otherwise.
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708

CHAPTER 25. TRADING RESOURCES FOR COMMUNICATION~ E, ~λ)In computing the dual optimal value D∗ (w,~ E), we first maximized the Lagrangian L(w,~ R,
~ and then we minimized with respect to the Lagrange mulwith respect to the rate vector R~
tipliers λ, i.e.,~ E, ~λ).
~ R,(25.94)D∗ (w,~ E) = inf sup L(w,λ1 ,λ2 ,λ3 ≥0~RWhat happens if we conduct the optimizations in the opposite order? By inspection, considerthat~ if (25.78)–(25.80) hold
w
~ ·R
~
~
inf L(w,~ R, E, λ) =.
(25.95)λ1 ,λ2 ,λ3 ≥0−∞else
As a consequence, we find that
sup
~
R

inf

λ1 ,λ2 ,λ3 ≥0~ E, ~λ) = P ∗ (w,
L(w,
~ R,
~ E).

(25.96)The statement of strong duality is that P ∗ (w,~ E) = D∗ (w,~ E), which we see from the above
amounts to an exchange of a minimum and a maximum. Strong duality holds for any linear
program (Boyd and Vandenberghe, 2004, Exercise 5.23), provided that at least one of the
primal or dual problems is feasible. For our primal problem, this means that there should~ meeting the constraints in (25.78)–(25.80). Since this is always the case, we
exist a value R
can conclude that strong duality holds:
P ∗ (w,~ E) = D∗ (w,~ E),(25.97)

which in turn implies that the optimal value P ∗ (w)~ in (25.77) can be rewritten as
P ∗ (w)~ = sup P ∗ (w,~ E) = sup D∗ (w,~ E).
E(25.98)E

By combining with (25.93), we see that the optimal primal value is given byP ∗ (w)~ = sup D∗ (w,~ E)(25.99)E= sup λ∗1 I(AX; B)σ + λ∗2 I(AiBX)σ + λ∗3 [I(X; B)σ + I(AiBX)σ ] ,(25.100)E

if λ∗1 , λ∗2 , λ∗3 ≥ 0 and it is equal to +∞ otherwise, with λ∗1 , λ∗2 , λ∗3 defined in (25.90)–(25.92).
Thus, it is now clear that the quantum dynamic capacity formula in (25.72) plays an essential role in computing the quantum dynamic capacity region. If it is additive, then thecomputation of the Pareto optimal trade-off surface requires an optimization with respect
to a single copy of the channel.
Special Cases of the Quantum Dynamic Capacity Formula
We now show how several capacity formulas of a quantum channel, including the entanglementassisted classical capacity (Theorem 21.3.1), the quantum capacity formula (Theorem 24.3.1),
and the classical capacity formula (Theorem 20.3.1) are special cases of the quantum dynamic
capacity formula.
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25.4. THE CONVERSE THEOREM

709

We first give a geometric interpretation of these special cases before proceeding to theproofs. Recall that the dynamic capacity region has the simple interpretation as a translationof the three-faced unit resource capacity region along the trade-off curve for entanglementassisted classical and quantum communication (see Figure 25.4 for an example of the region
for the dephasing channel). Any particular weight vector (wC , wQ , wE ) in (25.77) gives a set ofparallel planes that slice through the (C, Q, E) space, and the goal of the scalar optimization
task is to find one of these planes that is a supporting plane, intersecting a point (or a setof points) on the trade-off surface of the dynamic capacity region. We consider three special
planes:1. The first corresponds to the plane containing the vectors of super-dense coding andteleportation. The normal vector of this plane is (1, 2, 0), and suppose that we set theweight vector in (25.77) to be this vector. Then the optimization program finds theset of points on the trade-off surface such that a plane with this normal vector is a
supporting plane for the region. The optimization program singles out the constraintin (25.78), and by inspecting (25.90)–(25.92), this is equivalent to setting ~λ∗ = (1, 0, 0)
in the quantum dynamic capacity formula in (25.100). We show below that the optimization program becomes equivalent to finding the entanglement-assisted capacity(Theorem 21.3.1), in the sense that the quantum dynamic capacity formula becomesthe entanglement-assisted capacity formula.
2. The next plane contains the vectors of teleportation and entanglement distribution.
The normal vector of this plane is (0, 1, 1). Setting the weight vector in (25.77) to be
this vector makes the optimization program single out the constraint in (25.79), andand by inspecting (25.90)–(25.92), this is equivalent to setting ~λ∗ = (0, 1, 0) in thequantum dynamic capacity formula in (25.100). We show below that the optimization
program becomes equivalent to finding the quantum capacity (Theorem 24.3.1), in thesense that the quantum dynamic capacity formula becomes the LSD formula for thequantum capacity.
3. A third plane contains the vectors of super-dense coding and entanglement distribution.
The normal vector of this plane is (1, 1, 1). Setting the weight vector in (25.77) tobe this vector makes the optimization program single out the constraint in (25.80),
and by inspecting (25.90)–(25.92), this is equivalent to setting ~λ∗ = (0, 0, 1) in thequantum dynamic capacity formula in (25.100). We show below that the optimization
becomes equivalent to finding the classical capacity (Theorem 20.3.1), in the sense thatthe quantum dynamic capacity formula becomes the HSW formula for the classicalcapacity.
Corollary 25.4.1 The quantum dynamic capacity formula is equal to the entanglementassisted classical capacity formula when λ1 = 1, λ2 = 0, and λ3 = 0:
max I(AX; B)σ = max I(A; B)ρ ,σXAA0

where σXAA0 ≡c 2019φAA0(25.101)xx
x pX (x)|xihx|X ⊗|φ ihφ |AA0 , σXAB ≡ NA0 →B (σXAA0 ), and ρAB ≡ NA0 →B (φAA0 ).

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710

CHAPTER 25. TRADING RESOURCES FOR COMMUNICATION

Proof. The inequality maxσ I(AX; B)σ ≥ maxφAA0 I(A; B)ρ follows because the state σXAA0
0
is of the form in (25.11) and we can always choose pX (x) = δx,x0 and φxAA
0 to be the statethat maximizes I(A; B).
We now show the other inequality maxσ I(AX; B)σ ≤ maxφAA0 I(A; B)ρ . Consider thatthe following state purifies σXAA0 :Xp
pX (x)|xiX |xiR |φx iAA0 .
(25.102)|ϕiXRAA0 ≡x

Then by quantum data processing, we find that
I(AX; B)σ ≤ I(RAX; B)N (ϕ) ≤ max I(A; B)ρ ,φAA0(25.103)

where the second inequality follows because systems RAX of ϕXRAA0 purify A0 and theformula on the right involves an optimization over all such purifications.
Corollary 25.4.2 The quantum dynamic capacity formula is equal to the LSD quantum
capacity formula when λ1 = 0, λ2 = 1, and λ3 = 0:
max I(AiBX) = max I(AiB).
σ(25.104)

φAA0

Proof. The inequality maxσ I(AiBX) ≥ maxφAA0 I(AiB) follows because the state σ is0of the form in (25.11) and we can always choose pX (x) = δx,x0 and φxAA
0 to be the statethat maximizes
P I(AiB). The inequality maxσ I(AiBX) ≤ maxφAA0 I(AiB) follows because
I(AiBX) = x pX (x)I(AiB)N (φx ) and a maximum is never smaller than the average.
Corollary 25.4.3 The quantum dynamic capacity formula is equal to the HSW classicalcapacity formula when λ1 = 0, λ2 = 0, and λ3 = 1:
max [I(AiBX)σ + I(X; B)σ ] =σmax{pX (x),ψx }I(X; B).

(25.105)

Proof. The inequality maxσ I(AiBX)σ +I(X; B)σ ≥ max{pX (x),ψx } I(X; B) follows by choosing σ to be the pure-state ensemble that maximizes I(X; B) and noting that I(AiBX)σ vanishes for a pure-state ensemble. We now prove the inequality maxσ I(AiBX)σ + I(X; B)σ ≤max{pX (x),ψx } I(X; B). Consider a state ωXY BE obtained by performing a complete projective
measurement on the A system of the state σXABE . Then
I(AiBX)σ + I(X; B)σ = H(B)σ − H(E|X)σ= H(B)ω − H(E|X)ω
≤ H(B)ω − H(E|XY )ω
= H(B)ω − H(B|XY )ω
= I(XY ; B)ω
≤ max I(X; B).
{pX (x),ψx }c 2019(25.106)(25.107)(25.108)(25.109)(25.110)(25.111)

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25.5. EXAMPLES OF CHANNELS

711

The first equality follows by expanding the conditional coherent information and the Holevo
information. The second equality follows because the measured A system is not involved in
the entropies. The first inequality follows because conditioning does not increase entropy.
The third equality follows because the state ω is pure when conditioned on X and Y . The
fourth equality follows by definition, and the last inequality follows for clear reasons.

25.5 Examples of Channels
In this final section, we prove that a broad class of channels, known as the Hadamard channels(see Section 5.2.3), have a single-letter dynamic capacity region. We prove this result by
analyzing the quantum dynamic capacity formula for this class of channels. A dephasing
channel is a special case of a Hadamard channel, and so we can compute its dynamic capacity
region. We also show that the quantum erasure channel has a single-letter dynamic capacity
region, and in the process, we see that time-sharing is an optimal coding strategy for thischannel. We finally overview the dynamic capacity region of a pure-loss bosonic channel,
which is a good model for free-space communication or loss in an optical fiber. However,
we only state the main results and do not get into too many details of this channel (doing
so requires the theory of quantum optics and infinite-dimensional Hilbert spaces which isbeyond the scope of this book). The upshot for this channel is that trade-off coding can give
remarkable gains over time sharing.

25.5.1

Quantum Hadamard Channels

Below we show that the regularization in (25.6) is not necessary if the quantum channel is a
Hadamard channel. This result holds because a Hadamard channel has a special structure(see Section 5.2.3).
Theorem 25.5.1 The dynamic capacity region CCQE (NH ) of a quantum Hadamard channel(1)
NH is equal to its single-letter region CCQE (NH ).
The proof of the above theorem follows in two parts: 1) the lemma below states that thequantum dynamic capacity formula is additive when one of the channels is Hadamard and2) the induction argument in Lemma 25.4.1 establishes single-letterization.
Lemma 25.5.1 The following additivity relation holds for a Hadamard channel NH , anyother channel N , and for all ~λ such that λ1 , λ2 , λ3 ≥ 0:
D~λ (NH ⊗ N ) = D~λ (NH ) + D~λ (N ).

(25.112)

Proof. We first note that the inequality D~λ (NH ⊗ N ) ≥ D~λ (NH ) + D~λ (N ) holds for any
two channels simply by selecting the state σ in the maximization to be a tensor product of
the ones that individually maximize D~λ (NH ) and D~λ (N ).
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CHAPTER 25. TRADING RESOURCES FOR COMMUNICATION

So we prove that the non-trivial inequality D~λ (NH ⊗ N ) ≤ D~λ (NH ) + D~λ (N ) holds when
the first channel is a Hadamard channel. Since the first channel is Hadamard, it is degradable1
and its degrading channel has a particular structure: there are channels DB
and DY2 →E1
1 →Y
2
1where Y is a classical register and such that the degrading channel is DY →E1 ◦ DB.
1 →Y
Suppose the state we are considering to input to the tensor-product channel is
X
ρXAA01 A02 ≡pX (x)|xihx|X ⊗ φxAA01 A02 ,(25.113)x

and this state is the one that maximizes D~λ (NH ⊗ N ). Suppose that the output of the firstchannel isθXAB1 E1 A02 ≡ UAN0H→B1 E1 (ρXAA01 A02 ),(25.114)1

and the output of the second channel isωXAB1 E1 B2 E2 ≡ UAN02 →B2 E2 (θXAB1 E1 A02 ).

(25.115)

Finally, we define the following state as the result of applying the first part of the degrading
channel (a complete projective measurement) to ω:
1
σXY AE1 B2 E2 ≡ DB
(ωXAB1 E1 B2 E2 ).
1 →Y(25.116)

In particular, the state σ on systems AE1 B2 E2 is pure when conditioned on X and Y . Thenthe following chain of inequalities holds:
D~λ (NH ⊗ N )= λ1 I(AX; B1 B2 )ω + λ2 I(AiB1 B2 X)ω + λ3 [I(X; B1 B2 )ω + I(AiB1 B2 X)ω ]= λ1 H(B1 B2 E1 E2 |X)ω + λ2 H(B1 B2 |X)ω + (λ1 + λ3 ) H(B1 B2 )ω
− (λ1 + λ2 + λ3 ) H(E1 E2 |X)ω= λ1 H(B1 E1 |X)ω + λ2 H(B1 |X)ω + (λ1 + λ3 ) H(B1 )ω
− (λ1 + λ2 + λ3 ) H(E1 |X)ω + λ1 H(B2 E2 |B1 E1 X)ω + λ2 H(B2 |B1 X)ω
+ (λ1 + λ3 ) H(B2 |B1 )ω − (λ1 + λ2 + λ3 ) H(E2 |E1 X)ω
≤ λ1 H(B1 E1 |X)θ + λ2 H(B1 |X)θ + (λ1 + λ3 ) H(B1 )θ
− (λ1 + λ2 + λ3 ) H(E1 |X)θ + λ1 H(B2 E2 |Y X)σ + λ2 H(B2 |Y X)σ+ (λ1 + λ3 ) H(B2 )σ − (λ1 + λ2 + λ3 ) H(E2 |Y X)σ= λ1 I(AA02 X; B1 )θ + λ2 I(AA02 iB1 X)θ + λ3 [I(X; B1 )θ + I(AA02 iB1 X)θ ]+ λ1 I(AE1 Y X; B2 )σ + λ2 I(AE1 iB2 Y X)σ+ λ3 [I(Y X; B2 )σ + I(AE1 iB2 Y X)σ ]≤ D~λ (NH ) + D~λ (N ).

(25.117)(25.118)(25.119)(25.120)(25.121)(25.122)

The first equality follows by evaluating the quantum dynamic capacity formula D~λ (NH ⊗ N )on the state ρ. The next two equalities follow by rearranging entropies and because the statec 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

25.5. EXAMPLES OF CHANNELS713

ω on systems AB1 E1 B2 E2 is pure when conditioned on X. The inequality in the middle is the
crucial one and follows from the Hadamard structure of the channel: we exploit monotonicity
of conditional entropy with respect to quantum channels so that
H(B2 |B1 X)ω ≤ H(B2 |Y X)σ ,H(B2 E2 |B1 E1 X)ω ≤ H(B2 E2 |Y X)σ ,H(E2 |Y X)σ ≤ H(E2 |E1 X)ω ,H(B2 |B1 )ω ≤ H(B2 )ω .

(25.123)(25.124)(25.125)(25.126)

The next equality follows by rearranging entropies and the final inequality follows because θ
is a state of the form (25.11) for the first channel while σ is a state of the form (25.11) for
the second channel.

25.5.2

The Dephasing Channel

The theorem below states that the full dynamic capacity region admits a particularly simple
form when the noisy quantum channel is a qubit dephasing channel ∆p where∆p (ρ) ≡ (1 − p)ρ + p∆(ρ),(25.127)∆(ρ) ≡ h0|ρ|0i|0ih0| + h1|ρ|1i|1ih1|,(25.128)

and p ∈ [0, 1]. Figure 25.3 plots this region for the case of a dephasing channel with dephasing
parameter p = 0.2. Figure 25.4 plots special two-dimensional cases of the full region forvarious values of the dephasing parameter p. The figure demonstrates that trade-off coding
just barely improves upon time sharing.
Theorem 25.5.2 The dynamic capacity region CCQE (∆p ) of a dephasing channel with dephasing parameter p ∈ [0, 1] is the set of all C, Q, and E such thatC + 2Q ≤ 1 + h2 (ν) − h2 (γ(ν, p)),Q + E ≤ h2 (ν) − h2 (γ(ν, p)),C + Q + E ≤ 1 − h2 (γ(ν, p)),(25.129)(25.130)(25.131)

where ν ∈ [0, 1/2], h2 is the binary entropy function, and
1 1γ(ν, p) ≡ +2 2

r
1 − 16 ·pp
1−ν(1 − ν).
2
2

(25.132)

Proof. The dephasing channel is a Hadamard channel, so that the region single-letterizes
and it suffices to simplify the quantum dynamic capacity formula evaluated on a single copyof the channel. We then notice that it suffices to consider an ensemble of pure states whose
reductions to A0 are diagonal in the dephasing basis (see the following exercise).
c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 25. TRADING RESOURCES FOR COMMUNICATION

Figure 25.3: A plot of the dynamic capacity region for a qubit dephasing channel with dephasing parameterp = 0.2. The plot shows that the CEF trade-off curve (the protocol from Corollary 22.5.2) lies along theboundary of the dynamic capacity region. The rest of the region is simply the combination of the CEF points
with the unit protocols teleportation (TP), super-dense coding (SD), and entanglement distribution (ED).

Figure 25.4: Plot of (a) the CQ trade-off curve and (b) the CE trade-off curve for a p-dephasing qubit
channel for p = 0, 0.1, 0.2, . . . , 0.9, 1. The trade-off curves for p = 0 correspond to those of a noiseless qubit
channel and are the rightmost trade-off curves in each plot. The trade-off curves for p = 1 correspond to
those for a classical channel and are the leftmost trade-off curves in each plot. Each trade-off curve between
these two extremes beats a time-sharing strategy, but these two extremes do not beat time sharing.

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

25.5. EXAMPLES OF CHANNELS

715

Exercise 25.5.1 Prove that the following properties hold for a generalized dephasing channel
ND , its complement NDc , the completely dephasing channel ∆, and all input states ρ:
ND (∆(ρ)) = ∆(ND (ρ)),(25.133)

NDc (∆(ρ)) = NDc (ρ).

(25.134)

Conclude that
H(ρ) ≤ H(∆(ρ)),(25.135)

H(ND (ρ)) ≤ H(∆(ND (ρ))) = H(ND (∆(ρ))),(25.136)

H(NDc (ρ)) = H(NDc (∆(ρ))),(25.137)

so that it suffices to consider diagonal input states for the dephasing channel.
Next we prove below that it is sufficient to consider an ensemble of the following form tocharacterize the boundary points of the region:1
10
1|0ih0|X ⊗ ψAA|1ih1|X ⊗ ψAA
0 +
0,
2
2

(25.138)10where ψAA
0 and ψAA0 are pure states, defined as follows for ν ∈ [0, 1/2]:

 0
TrA ψAA= ν|0ih0|A0 + (1 − ν) |1ih1|A0 ,0 1
TrA ψAA0 = (1 − ν) |0ih0|A0 + ν|1ih1|A0 .

(25.139)(25.140)

We now prove the above claim. We assume without loss of generality that the dephasingbasis is the computational basis. Consider a classical–quantum state with a finite number
N of conditional density operators φxAA0 whose reduction to A0 is diagonal:
ρXAA0 ≡N
−1X

pX (x)|xihx|X ⊗ φxAA0 .

(25.141)x=0

We can form a new classical–quantum state with double the number of conditional density
operators by “bit-flipping” the original conditional density operators:
N −1

1X
pX (x) (|xihx|X ⊗ φxAA0 + |x + N ihx + N |X ⊗ XA0 φxAA0 XA0 ) ,σXAA0 ≡
2 x=0

(25.142)

where X is the σX “bit-flip” Pauli operator. Consider the following chain of inequalities thatc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

716

CHAPTER 25. TRADING RESOURCES FOR COMMUNICATION

holds for all λ1 , λ2 , λ3 ≥ 0:
λ1 I(AX; B)ρ + λ2 I(AiBX)ρ + λ3 [I(X; B)ρ + I(AiBX)ρ ]= λ1 H(A|X)ρ + (λ1 + λ3 ) H(B)ρ + λ2 H(B|X)ρ − (λ1 + λ2 + λ3 ) H(E|X)ρ
≤ (λ1 + λ3 ) H(B)σ + λ1 H(A|X)σ + λ2 H(B|X)σ − (λ1 + λ2 + λ3 ) H(E|X)σ= (λ1 + λ3 ) + λ1 H(A|X)σ + λ2 H(B|X)σ − (λ1 + λ2 + λ3 ) H(E|X)σX= (λ1 + λ3 ) +pX (x) [λ1 H(A)φx + λ2 H(B)φx − (λ1 + λ2 + λ3 ) H(E)φx ](25.143)(25.144)(25.145)(25.146)x≤ (λ1 + λ3 ) + max [λ1 H(A)φx + λ2 H(B)φx − (λ1 + λ2 + λ3 ) H(E)φx ](25.147)= (λ1 + λ3 ) + λ1 H(A)φx∗ + λ2 H(B)φx∗ − (λ1 + λ2 + λ3 ) H(E)φx∗ .

(25.148)x

The first equality follows by standard entropic manipulations. The second equality follows
because the conditional entropy H(B|X) is invariant under a bit-flipping unitary on the
input state that commutes with the channel: H(B)XρxB X = H(B)ρxB . Furthermore, a bit flipon the input state does not change the eigenvalues for the output of the dephasing channel’s
complementary channel:
H(E)N c (XρxA0 X) = H(E)N c (ρxA0 ) .
(25.149)
The first inequality follows because entropy is concave, i.e., the local state σB is a mixedversion of ρB . The third equality follows because
!
!
X1
X
1
pX (x)(ρxB + XρxB X) = H
pX (x)I = 1.
(25.150)
H(B)σB = H
22xx
The fourth equality follows because the system X is classical. The second inequality follows because the maximum value of a realization of a random variable is not less than its
expectation. The final equality simply follows by defining φx∗ to be the conditional density
operator on systems A, B, and E that arises from sending through the channel a state whose
reduction to A0 is of the form ν|0ih0|A0 + (1 − ν) |1ih1|A0 . Thus, an ensemble of the kind in(25.138) is sufficient to attain a point on the boundary of the region. Evaluating the entropic
quantities in Theorem 25.2.1 on a state of the above form then gives the expression for theregion in Theorem 25.5.2.

25.5.3The Quantum Erasure Channel

We can also obtain a simple characterization of the quantum dynamic capacity region fora quantum erasure channel for ε ∈ [0, 1/2]. Recall that the erasure channel is defined asfollows:
N ε (ρ) = (1 − ε) ρ + ε|eihe|,(25.151)
where ρ is a d-dimensional input state, |ei is an erasure flag state orthogonal to all inputs(so that the output space has dimension d + 1), and ε ∈ [0, 1] is the erasure probability.
Theorem 25.5.3 characterizes the quantum dynamic capacity region for such a channel andc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

25.5. EXAMPLES OF CHANNELS

717

Figure 25.5: The quantum dynamic capacity region for the (qubit) quantum erasure channel with ε = 1/4.
The plot demonstrates that time-sharing is optimal.

establishes that time-sharing is an optimal strategy. Figure 25.5 plots an example of theregion for a qubit erasure channel with ε = 1/4.
Theorem 25.5.3 Let N ε be a quantum erasure channel with ε ∈ [0, 1/2]. Then the quantum
dynamic capacity region CCQE (N ε ) is equal to the union of the following regions, obtained byvarying λ ∈ [0, 1]:
C + 2Q ≤ (1 − ε) (1 + λ) log d,
Q + E ≤ (1 − 2ε) λ log d,
C + Q + E ≤ (1 − ε − ελ) log d.
The region is achievable for ε ∈ (1/2, 1].
Proof. We need to show that the above region is achievable and that the regularized
quantum dynamic capacity formula simplifies to it as well. For the achievability part, wechoose a particularP p ensemble and show that the above rate region is achievable using it. Let|φiAA0 ≡pX (x)|xiA |xiA0 where {|xiA } and {|xiA0 } are the standard computationalxbases for systems A and A0 , respectively. Observe that H(pX ) = H(A)φ . We take the input
ensemble as {1/d, X(x)A0 |φiAA0 }, where X(x)A0 are the Heisenberg–Weyl shift operators.
This ensemble has the property that the expected density operator on system A0 is the maximally mixed state πA0 . A classical–quantum state for evaluating the rates in Theorem 25.2.1
is as follows:
X1
ρXAB ≡|xihx|X ⊗ NAε 0 →B [X(x)A0 |φihφ|AA0 X † (x)A0 ](25.152)dx= (1 − ε) ρ0XAA0 + ερ1XAB ,c 2019(25.153)

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718

CHAPTER 25. TRADING RESOURCES FOR COMMUNICATION

where
ρ0XAA0 ≡X1x

ρ1XAB ≡

d

X1
x

d

|xihx|X ⊗ X(x)A0 |φihφ|AA0 X † (x)A0 ,(25.154)|xihx|X ⊗ φA ⊗ |eihe|B .

(25.155)

Then, from Theorem 25.2.1, we see that it suffices to compute the following three information
quantities: I(AX; B)ρ , I(AiBX)ρ , and I(X; B)ρ . Consider that we can perform the following
isometry on the B system without affecting any of the information quantities:[|0ih0| + . . . + |d − 1ihd − 1|] ⊗ |0iY + |eihe| ⊗ |1iY .

(25.156)

Let ωXABY denote the resulting state:
ωXABY ≡ (1 − ε) ρ0XAA0 ⊗ |0ih0|Y + ερ1XAB ⊗ |1ih1|Y .

(25.157)

Then, observing that ωXAY = πX ⊗ φA ⊗ [(1 − ε) |0ih0|Y + ε|1ih1|Y ], we find that
I(AX; B)ρ = I(AX; BY )ω = I(AX; B|Y )ω ,I(AiBX)ρ = I(AiBXY )ω = I(AiBX|Y )ω ,I(X; B)ρ = I(X; BY )ω = I(X; B|Y )ω .

(25.158)(25.159)(25.160)

We now evaluate each of these in turn:I(AX; B|Y )ω = (1 − ε) I(AX; A0 )ρ0 + εI(AX; B)ρ1
= (1 − ε) (log d + H(pX )) ,I(AiBX|Y )ω = (1 − ε) I(AiA0 X)ρ0 + εI(AiBX)ρ1= (1 − ε) H(pX ) + ε [−H(pX )]= (1 − 2ε) H(pX ),I(X; B|Y )ω = (1 − ε) I(X; A0 )ρ0 + εI(X; B)ρ1
= (1 − ε) (log d − H(pX )) .

(25.161)(25.162)(25.163)

Then, by Theorem 25.2.1, we can achieve the rate region in the statement of the theorem.
That is, we can pick distributions pX such that H(pX ) = λ log d for all λ ∈ [0, 1].
We now bound the quantum dynamic capacity formula for (N ε )⊗n from above for all
positive integers n, and as a consequence, establish that the region given in the statementof the theorem is optimal for ε ∈ [0, 1/2]. To begin with, let us consider the quantum dynamic capacity formula for an erasure channel NAε 1 →B1 and an arbitrary degradable channel
MA2 →B2 . We use the same idea as above, applying the isometry in (25.156) to the output
B1 of the erasure channel. The quantum dynamic capacity formula for NAε 1 →B1 ⊗ MA2 →B2
becomesλ1 I(AX; B1 Y B2 )σ + λ2 I(AiB1 Y B2 X)σ+ λ3 [I(X; B1 Y B2 )σ + I(AiB1 Y B2 X)σ ] , (25.164)
c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

25.5. EXAMPLES OF CHANNELS

719

where
0
1
σXAB1 B2 Y ≡ σXAA
⊗ (1 − ε) |0ih0|Y + σXAB
⊗ ε|1ih1|Y ,
1 B2
1 B2
X
0
σXAA
≡
pX (x)|xihx|X ⊗ MA2 →B2 (φxAA1 A2 ),1 B2(25.165)(25.166)

x
1
σXAB
≡
1 B2

X

pX (x)|xihx|X ⊗ MA2 →B2 (φxAA2 ) ⊗ |eihe|B1 .

(25.167)x
01Observe that σXAB
= σXAB= σXAB2 , so that2
2

σXAB2 Y = σXAB2 ⊗ [(1 − ε) |0ih0|Y + ε|1ih1|Y ] .

(25.168)

First we handle the quantum mutual information term:
I(AX; B1 Y B2 )σ = I(AX; B1 B2 |Y )σ= (1 − ε) I(AX; A1 B2 )σ0 + εI(AX; B2 )σ1= (1 − ε) [I(AX; B2 )σ0 + I(AX; A1 |B2 )σ0 ] + εI(AX; B2 )σ1= (1 − ε) [I(AX; B2 )σ + I(AX; A1 |B2 )σ0 ] + εI(AX; B2 )σ= (1 − ε) I(AX; A1 |B2 )σ0 + I(AX; B2 )σ .

(25.169)(25.170)(25.171)(25.172)(25.173)

We can bound the term I(AX; A1 |B2 )σ0 from above as follows:
I(AX; A1 |B2 )σ0 = H(A1 |B2 )σ0 − H(A1 |B2 AX)σ0= H(A1 |B2 )σ0 + H(A1 |E2 X)σ0
≤ log d + H(A1 |E2 X)σ0 ,(25.174)(25.175)(25.176)

where σ 0 is defined below in (25.184). This then gives the following bound:
I(AX; B1 Y B2 )σ ≤ (1 − ε) [log d + H(A1 |E2 X)σ0 ] + I(AX; B2 )σ .

(25.177)

Now we handle the Holevo information term, and the development is similar:
I(X; B1 Y B2 )σ = I(X; B1 B2 |Y )σ= (1 − ε) I(X; A1 B2 )σ0 + εI(X; B2 )σ1= (1 − ε) [I(X; B2 )σ0 + I(X; A1 |B2 )σ0 ] + εI(X; B2 )σ1= (1 − ε) [I(X; B2 )σ + I(X; A1 |B2 )σ0 ] + εI(X; B2 )σ= (1 − ε) I(X; A1 |B2 )σ0 + I(X; B2 )σ .

(25.178)(25.179)(25.180)(25.181)(25.182)

We finally handle the coherent information term. To√ this end, note that√ an isometric extension of the erasure channel is as follows: |ψiA → 1 − ε|ψiB |eiE + ε|eiB |ψiE . Both thereceiver and the environment can perform the isometry in (25.156) on their systems, withoutc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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CHAPTER 25. TRADING RESOURCES FOR COMMUNICATION

affecting the information quantities. Let UAM2 →B2 E2 be an isometric extension of MA2 →B2 anddefine√|σiXX 0 AB1 B2 E1 E2 Y Z ≡ |σ 0 iXX 0 AA1 B2 E1 E2 ⊗ 1 − ε|0iY |1iZ√
+ |σ 1 iXX 0 AB1 B2 A1 E2 ⊗ ε|1iY |0iZ , (25.183)
where|σ 0 iXX 0 AA1 B2 E1 E2 ≡Xp

pX (x)|xiX |xiX 0 ⊗ UAM2 →B2 E2 |φx iAA1 A2 |eiE1 ,(25.184)

pX (x)|xiX |xiX 0 ⊗ UAM2 →B2 E2 |φx iAA1 A2 |eiB1 .

(25.185)x1|σ iXX 0 AB1 B2 A1 E2 ≡

Xp
x

Then
I(AiB1 Y B2 X)σ = I(AiB1 B2 |XY )σ= H(B1 B2 |Y X)σ − H(E1 E2 |Y X)σ= [(1 − ε) H(A1 B2 |X)σ0 + εH(B2 |X)σ1 ]− [(1 − ε) H(E2 |X)σ0 + εH(A1 E2 |X)σ1 ]= [(1 − ε) (H(B2 |X)σ0 + H(A1 |B2 X)σ0 ) + εH(B2 |X)σ1 ]− [(1 − ε) H(E2 |X)σ0 + ε (H(E2 |X)σ1 + H(A1 |E2 X)σ1 )]= [(1 − ε) (H(B2 |X)σ + H(A1 |B2 X)σ0 ) + εH(B2 |X)σ ]− [(1 − ε) H(E2 |X)σ + ε (H(E2 |X)σ + H(A1 |E2 X)σ1 )]= H(B2 |X)σ − H(E2 |X)σ + (1 − ε) H(A1 |B2 X)σ0 − εH(A1 |E2 X)σ1
= H(B2 |X)σ − H(E2 |X)σ + (1 − ε) H(A1 |B2 X)σ0 − εH(A1 |E2 X)σ0 .

(25.186)(25.187)(25.188)(25.189)(25.190)(25.191)

Applying the assumption that the channel M is degradable (so that H(A1 |B2 X)σ0 ≤H(A1 |E2 X)σ0 ), we find the upper bound≤ H(B2 |X)σ − H(E2 |X)σ + (1 − ε) H(A1 |E2 X)σ0 − εH(A1 |E2 X)σ0
= H(B2 |X)σ − H(E2 |X)σ + (1 − 2ε) H(A1 |E2 X)σ0 .

(25.192)(25.193)

It also follows from (25.182) and (25.191) that
I(X; B1 Y B2 )σ + I(AiB1 Y B2 X)σ= (1 − ε) I(X; A1 |B2 )σ0 + I(X; B2 )σ + H(B2 |X)σ − H(E2 |X)σ+ (1 − ε) H(A1 |B2 X)σ0 − εH(A1 |E2 X)σ0
= I(X; B2 )σ + H(B2 |X)σ − H(E2 |X)σ+ (1 − ε) H(A1 |B2 )σ0 − (1 − ε) H(A1 |B2 X)σ0
+ (1 − ε) H(A1 |B2 X)σ0 − εH(A1 |E2 X)σ0
= I(X; B2 )σ + H(B2 |X)σ − H(E2 |X)σ + (1 − ε) H(A1 |B2 )σ0 − εH(A1 |E2 X)σ0
≤ I(X; B2 )σ + H(B2 |X)σ − H(E2 |X)σ + (1 − ε) log d − εH(A1 |E2 X)σ0
c 2019(25.194)(25.195)(25.196)(25.197)

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25.5. EXAMPLES OF CHANNELS

721

Now putting together (25.177), (25.193), and (25.197), we find the following upper boundon the quantum dynamic capacity formula in (25.164):
λ1 I(AX; B1 Y B2 )σ + λ2 I(AiB1 Y B2 X)σ+ λ3 [I(X; B1 Y B2 )σ + I(AiB1 Y B2 X)σ ](25.198)≤ λ1 (1 − ε) [log d + H(A1 |E2 X)σ0 ] + λ2 (1 − 2ε) H(A1 |E2 X)σ0
+ λ3 [(1 − ε) log d − εH(A1 |E2 X)σ0 ]+ λ1 I(AX; B2 )σ + λ2 [H(B2 |X)σ − H(E2 |X)σ ]+ λ3 [I(X; B2 )σ + H(B2 |X)σ − H(E2 |X)σ ](25.199)= (λ1 + λ3 ) (1 − ε) log d + [λ1 (1 − ε) + λ2 (1 − 2ε) − λ3 ε] H(A1 |E2 X)σ0
+ λ1 I(AX; B2 )σ + λ2 [H(B2 |X)σ − H(E2 |X)σ ]+ λ3 [I(X; B2 )σ + H(B2 |X)σ − H(E2 |X)σ ] .
(25.200)
In the case that λ1 (1 − ε)+λ2 (1 − 2ε)−λ3 ε ≥ 0, we can apply data processing (H(A1 |E2 X)σ0 ≤H(A1 |X)σ0 ) to find that the last line is never larger than(λ1 + λ3 ) (1 − ε) log d + [λ1 (1 − ε) + λ2 (1 − 2ε) − λ3 ε] H(A1 |X)σ0+ λ1 I(AX; B2 )σ + λ2 [H(B2 |X)σ − H(E2 |X)σ ]+ λ3 [I(X; B2 )σ + H(B2 |X)σ − H(E2 |X)σ ] . (25.201)
If we now take MA2 →B2 = (N ε )⊗n−1 and iterate this development n − 1 more times, we findthat
1
D~ ((N ε )⊗n ) ≤ (λ1 + λ3 ) (1 − ε) log d
n λ
"

n

1X
+ [λ1 (1 − ε) + λ2 (1 − 2ε) − λ3 ε]H(Ai |X)n i=1≤ (λ1 + λ3 ) (1 − ε) log d+ [λ1 (1 − ε) + λ2 (1 − 2ε) − λ3 ε] max H(Ai )φx .
i,x#(25.202)(25.203)

This establishes the optimality of the region whenever λ1 (1 − ε) + λ2 (1 − 2ε) − λ3 ε ≥ 0.
For the case when λ1 (1 − ε) + λ2 (1 − 2ε) − λ3 ε < 0, we start from (25.200) and use dataprocessing for the channel TrB2 {UAM2 →B2 E2 (·)}, giving H(A1 |E2 X)σ0 ≥ H(A1 |A2 X)σ0 , to findthat
(λ1 + λ3 ) (1 − ε) log d + [λ1 (1 − ε) + λ2 (1 − 2ε) − λ3 ε] H(A1 |E2 X)σ0 + λ1 I(AX; B2 )σ+ λ2 [H(B2 |X)σ − H(E2 |X)σ ] + λ3 [I(X; B2 )σ + H(B2 |X)σ − H(E2 |X)σ ]≤ (λ1 + λ3 ) (1 − ε) log d + [λ1 (1 − ε) + λ2 (1 − 2ε) − λ3 ε] H(A1 |A2 X)σ0
+ λ1 I(AX; B2 )σ + λ2 [H(B2 |X)σ − H(E2 |X)σ ]+ λ3 [I(X; B2 )σ + H(B2 |X)σ − H(E2 |X)σ ] . (25.204)
c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

722

CHAPTER 25. TRADING RESOURCES FOR COMMUNICATION

If we now take MA2 →B2 = (N ε )⊗n−1 and iterate this development n − 1 more times, we findthat
1
D~ ((N ε )⊗n ) ≤ (λ1 + λ3 ) (1 − ε) log d
n λ
"

#
n
1X
i−1
+ [λ1 (1 − ε) + λ2 (1 − 2ε) − λ3 ε]H(Ai |A X)
n i=1(25.205)= (λ1 + λ3 ) (1 − ε) log d1n+ [λ1 (1 − ε) + λ2 (1 − 2ε) − λ3 ε] H(A |X)n≤ (λ1 + λ3 ) (1 − ε) log d.
(25.206)(25.207)

The last line follows because entropy is non-negative. This inequality shows that the inequality λ1 (1 − ε) + λ2 (1 − 2ε) − λ3 ε < 0 holding implies that it is optimal to pick λ = 0 in thestatement of the theorem (which just corresponds to the case in which we are maximizing
the unassisted classical capacity). This completes the proof.

25.5.4

The Pure-Loss Bosonic Channel

One of the most important practical channels in quantum communication is known as thepure-loss bosonic channel. This channel can model the communication of photons throughfree space or over a fiber optic cable because the main source of noise in these settings isjust the loss of photons. The pure-loss bosonic channel has one parameter η ∈ [0, 1] thatcharacterizes the fraction of photons that make it through the channel to the receiver on
average. The environment Eve is able to collect all of the photons that do not make it to
the receiver—this fraction is 1 − η. Usually, we also restrict the mean number of photons
that the sender is allowed to send through the channel (if we do not do so, then there could
be an infinite amount of energy available, which is unphysical from a practical perspective,
and furthermore, some of the capacities become infinite, which is less interesting from an
information-theoretical perspective). So, we let NS ∈ [0, ∞) be the mean number of photons
available at the transmitter. Capacities of this channel are then a function of these twoparameters η and NS .
Exercise 25.5.2 Prove that the quantum capacity of a pure-loss bosonic channel vanishes
when η = 1/2.
In this section, we show how trade-off coding for this channel can give a remarkable gain
over time sharing. Trade-off coding for this channel amounts to a power-sharing strategy,
in which the sender dedicates a fraction λ ∈ [0, 1] of the available photons to the quantum
part of the code and the other fraction 1 − λ to the classical part of the code. This powersharing strategy is provably optimal (up to a long-standing conjecture) and can beat timesharing by significant margins (much more so than the dephasing channel does, for example).
c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

25.5. EXAMPLES OF CHANNELS

723

Specifically, recall that a trade-off coding strategy has the sender and receiver generaterandom codes from an ensemble of the following form:{pX (x), |φx iAA0 } ,(25.208)

where pX (x) is some distribution and the states |φx iAA0 are correlated with this distribution,with Alice feeding system A0 into the channel. For the pure-loss bosonic channel, it turns
out that the best ensemble to choose is of the following form:p(1−λ)NS (α), DA0 (α)|ψTMS (λ)iAA0 ,(25.209)
where α is a complex variable. The distribution p(1−λ)NS (α) is an isotropic Gaussian distribution with variance (1 − λ) NS :
1exp − |α|2 / [(1 − λ) NS ] ,(25.210)p(1−λ)NS (α) ≡π (1 − λ) NS
where λ ∈ [0, 1] is the power-sharing or photon-number-sharing parameter, indicating how
many photons to dedicate to the quantum part of the code, while 1 − λ indicates how
many photons to dedicate to the classical part. In (25.209), DA0 (α) is a “displacement”
unitary operator acting on system A0 (more on this below), and |ψTMS (λ)iAA0 is a “twomode squeezed” (TMS) state of the following form:s∞
X[λNS ]n|niA |niA0 ,(25.211)|ψTMS (λ)iAA0 ≡[λNS + 1]n+1
n=0
where |ni is a state of definite photon number n. Let θ(λ) denote the state resulting fromtracing over the mode A:
θ(λ) ≡ TrA {|ψTMS (λ)ihψTMS (λ)|AA0 }∞X[λNS ]n0=
n+1 |nihn|A .
[λN
+
1]
S
n=0

(25.212)(25.213)

The reduced state θ(λ) is known as a thermal state with mean photon number λNS . We canreadily check that its mean photon number is λNS simply by computing the expectation of
the photon number n with respect to the geometric distribution [λNS ]n / [λNS + 1]n+1 :
∞
X
n=0

n

[λNS ]n= λNS .
[λNS + 1]n+1(25.214)

The most important property of the displacement operators DA0 (α) for our purposes is that
averaging over a random choice of them according to the Gaussian distribution p(1−λ)NS (α),
where each operator acts on the state θ, gives a thermal state with mean photon number NS :Zθ ≡ dα p(1−λ)NS (α) D(α)θ(λ)D† (α)(25.215)=

c 2019∞X[NS ]n0
n+1 |nihn|A .
[N
+
1]
S
n=0

(25.216)

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724

CHAPTER 25. TRADING RESOURCES FOR COMMUNICATION

Thus, the choice of ensemble in (25.209) meets the constraint that the average number ofphotons input to the channel be equal to NS .
In order to calculate the quantum dynamic capacity region for this pure-loss bosonicchannel, it is helpful to observe that the entropy of a thermal state with mean number ofphotons NS is equal tog(NS ) ≡ (NS + 1) log (NS + 1) − NS log(NS ),(25.217)

because we will evaluate all of the relevant entropies on thermal states. From Exercise 25.2.1,
we know that we should evaluate just the following four entropies:
Z
H(A|X)σ = dα p(1−λ)NS (α) H(D(α)θ(λ)D† (α)),(25.218)
H(B)σ = H(N (θ)),
Z
H(B|X)σ = dα p(1−λ)NS (α) H(N (D(α)θ(λ)D† (α))),Z
H(E|X)σ = dα p(1−λ)NS (α) H(N c (D(α)θ(λ)D† (α))),(25.219)(25.220)(25.221)

where N is the pure-loss bosonic channel that transmits η of the input photons to thereceiver and N c is the complementary channel that transmits 1 − η of the input photons to
the environment Eve. We proceed with calculating the above four entropies:
ZZ†
dα p(1−λ)NS (α) H(D(α)θ(λ)D (α)) = dα p(1−λ)NS (α) H(θ(λ))(25.222)= H(θ(λ)) = g (λNS ) .

(25.223)

The first equality follows because D(α) is a unitary operator, and the third equality follows
because θ is a thermal state with mean photon number NS . Continuing, we haveH(N (θ)) = g(ηNS ),(25.224)

because θ is a thermal state with mean photon number NS , but the channel only lets a
fraction η of the input photons through on average. The third entropy in (25.220) equalsZdα p(1−λ)NS (α) H(N (D(α)θ(λ)D† (α)))
Z√√= dα p(1−λ)NS (α) H(D( ηα)N (θ(λ))D† ( ηα))(25.225)Z= dα p(1−λ)NS (α) H(N (θ(λ)))(25.226)= H(N (θ(λ))) = g(ληNS ).

(25.227)

The first equality follows because a displacement operator is covariant with respect to thepure-loss channel (we do not justify this rigorously here). The second equality follows because
c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

25.5. EXAMPLES OF CHANNELS

725

D(α) is a unitary operator. The final equality follows because θ(λ) is a thermal state withmean photon number λNS , but the channel only lets a fraction η of the input photons
through on average. By the same line of reasoning (except that the complementary channel
lets through only a fraction 1 − η of the input photons), the fourth entropy in (25.221) isequal toZdα p(1−λ)NS (α) H(N c (D(α)θ(λ)D† (α)))
Zpp= dα p(1−λ)NS (α) H(D( 1 − ηα)N c (θ(λ))D† ( 1 − ηα))Z
= dα p(1−λ)NS (α) H(N c (θ(λ)))(25.228)= H(N c (θ(λ))) = g(λ(1 − η)NS ).

(25.230)(25.229)

Then, by the result of Exercise 25.2.1 and a matching converse that holds whenever η ≥ 1/2,2
we have the following characterization of the quantum dynamic capacity region of the pureloss bosonic channel.
Theorem 25.5.4 Provided that Strong Conjecture 2 of (Guha, 2008) is true, the quantum
dynamic capacity region for a pure-loss bosonic channel with transmissivity η ≥ 1/2 is theunion of regions of the form:
C + 2Q ≤ g(λNS ) + g(ηNS ) − g((1 − η) λNS ),Q + E ≤ g(ηλNS ) − g((1 − η) λNS ),C + Q + E ≤ g(ηNS ) − g((1 − η) λNS ),(25.231)(25.232)(25.233)

where λ ∈ [0, 1] is a photon-number-sharing parameter and g(N ) is the entropy defined in(25.217). The region is achievable for all η ∈ [0, 1].
Figure 25.6 depicts two important special cases of the region in the above theorem: (a) the
trade-off between classical and quantum communication without entanglement assistanceand (b) the trade-off between entanglement-assisted and unassisted classical communication.
The figure indicates the remarkable improvement over time sharing that trade-off coding
gives.
Other special cases of the above capacity region are the unassisted classical capacityg(ηNS ) when λ1 , λ2 , Q, E = 0 and λ3 = 1, the quantum capacity g(ηNS ) − g((1 − η)NS )when λ2 = 1, λ1 , λ3 , C, E = 0, and the entanglement-assisted classical capacity g(NS ) +g(ηNS ) − g((1 − η)NS ) when λ1 = 1, λ2 , λ3 , Q = 0, and E = −∞.
2

We should clarify that the converse holds only if a long-standing minimum-output entropy conjecture istrue (researchers have collected much evidence that it should be true).
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726

CHAPTER 25. TRADING RESOURCES FOR COMMUNICATION

Figure 25.6: (a) Suppose a channel transmits on average 3/4 of the photons to the receiver, while losing
the other 1/4 en route. Such a channel can reliably transmit a maximum of log (3/4) − log (1/4) ≈ 1.58
qubits per channel use, and a mean photon budget of about 200 photons per channel use at the transmitteris sufficient to nearly achieve this quantum capacity. A trade-off coding strategy which lowers the quantum
data rate to about 1.4 qubits per channel use while retaining the same mean photon budget allows for asender to reliably transmit an additional 4.5 classical bits per channel use, while time-sharing would only
allow for an additional 1 classical bit per channel use with this photon budget. The 6.5 dB increase in theclassical data rate that trade-off coding gives over time-sharing for this example is strong enough to demandthat quantum communication engineers employ trade-off coding strategies in order to take advantage of suchtheoretical performance gains. (b) The sender and the receiver share entanglement, and the sender would liketo transmit classical information while minimizing the consumption of entanglement. With a mean photon
budget of 200 photons per channel use over a channel that propagates only 3/4 of the photons input to it,
the sender can reliably transmit a maximum of about 10.7 classical bits per channel use while consuming
entanglement at a rate of about 9.1 entangled bits per channel use. With trade-off coding, the sender cansignificantly reduce the entanglement consumption rate to about 5 entangled bits per channel use whilestill transmitting about 10.5 classical bits per channel use, only a 0.08 dB decrease in the rate of classicalcommunication for a 2.6 dB decrease in the entanglement consumption rate. The savings in entanglement
consumption could be useful for them if they would like to have the extra entanglement for future rounds ofassisted communication.

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25.6. HISTORY AND FURTHER READING

727

25.6 History and Further Reading
Shor (2004b) considered the classical capacity of a channel assisted by rate-limited entanglement. He calculated a trade-off curve that determines how a sender can optimally trade
the consumption of noiseless entanglement with the generation of noiseless classical communication. This trade-off curve also bounds a rate region consisting of rates of entanglement
consumption and generated classical communication. Shor’s result then inspired Devetak
and Shor (2005) to consider a scenario in which a sender exploits a quantum channel tosimultaneously transmit both noiseless classical and quantum information, a scenario later
dubbed “classically enhanced quantum coding” in (Hsieh and Wilde, 2010a,b) after schemes
formulated in the theory of quantum error correction (Kremsky et al., 2008; Wilde and Brun,2008). Devetak and Shor (2005) provided a multi-letter characterization of the classically enhanced quantum capacity region for general channels, but they were able to show that bothgeneralized dephasing channels and erasure channels admit single-letter capacity regions.
The above scenarios are a part of the dynamic, double-resource quantum Shannon theory,in which a sender can exploit a quantum channel to generate two noiseless resources, or asender can exploit a quantum channel in addition to a noiseless resource to generate anothernoiseless resource. This theory culminated with the work of Devetak et al. (2004, 2008), thatprovided a multi-letter characterization for virtually every combination of two resources anda quantum channel which one can consider. Other researchers concurrently considered how
noiseless resources might trade off against each other in tasks outside of the dynamic, doubleresource quantum Shannon theory, such as quantum compression (Koashi and Imoto, 2001;
Barnum et al., 2001b; Hayden et al., 2002), remote state preparation (Abeyesinghe andHayden, 2003; Bennett et al., 2005), and hybrid quantum memories (Kuperberg, 2003).
Hsieh and Wilde (2010a,b); Wilde and Hsieh (2012b) considered the dynamic, tripleresource quantum Shannon theory by providing a multi-letter characterization of an entanglementassisted quantum channel’s ability to transmit both classical and quantum information.
Hsieh and Wilde (2010a) also constructed a new protocol, dubbed the “classically enhanced
father protocol,” that outperforms a time-sharing strategy for transmitting both classicaland quantum information over an entanglement-assisted quantum channel. Brádler et al.
(2010) showed that the quantum Hadamard channels have a single-letter capacity region.
Later studies continued these efforts of exploring information trade-offs (Jochym-O’Connor
et al., 2011; Wilde and Hsieh, 2012a).
Wilde et al. (2012c,b) found the quantum dynamic capacity region of the pure-loss bosonicchannel (up to a long-standing minimum-output entropy conjecture). The results there build
on a tremendous body of literature for bosonic channels. Giovannetti et al. (2004b) found theclassical capacity of the pure-loss bosonic channel. Others found the entanglement-assistedclassical and quantum capacities of the pure-loss bosonic channel (Bennett et al., 2002;
Holevo and Werner, 2001; Giovannetti et al., 2003a,b) and its quantum capacity (Wolf et al.,2007; Guha et al., 2008). The long-standing minimum-output entropy conjecture (different
from the one proved by Giovannetti et al. (2015)) is detailed in the papers of Giovannetti
et al. (2004a); Guha et al. (2008); Giovannetti et al. (2010).

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728

c 2019CHAPTER 25. TRADING RESOURCES FOR COMMUNICATIONMark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseCHAPTER 26

Summary and Outlook
This brief final chapter serves as a compact summary of many results presented in thisbook, it highlights information-processing tasks that we did not cover, and it discusses new
directions. We exploit the resource inequality formalism in our summary.
A resource inequality is a statement of achievability:
X
X
αk ≥
βj ,
(26.1)
jkmeaning that the resources {αk } on the left-hand side can simulate the resources {βj } on
the right-hand side. The simulation can be exact and finite or asymptotically perfect. We
can classify resources as follows:
1. Unit, noiseless, or noisy.
2. Dynamic or static. Moreover, dynamic resources can be relative (see below).
3. Classical, quantum, or hybrid.
The unit resources are as follows: [c → c] represents one noiseless classical bit channel,[q → q] represents one noiseless qubit channel, [qq] represents one noiseless ebit, and [q → qq]
represents one noiseless coherent bit channel. We also have [c → c]priv representing a noiselessprivate classical bit channel and [cc]priv representing a noiseless bit of secret key. An example
of a noiseless resource is a pure bipartite state |φiAB shared between Alice and Bob or an
identity channel idA→B from Alice to Bob. An example of a noisy resource could be a mixed
bipartite state ρAB or a noisy channel NA0 →B . Unit resources are a special case of noiselessresources, which are in turn a special case of noisy resources.
A shared state ρAB is an example of a noisy static resource, and a channel N is an example
of a noisy dynamic resource. We indicate these by hρi or hN i in a resource inequality. We
can be more precise if necessary and write hN i as a dynamic, relative resource hNA0 →B : σA0 i,
meaning that the protocol only works as it should if the state input to the channel is σA0 .
729

730

CHAPTER 26. SUMMARY AND OUTLOOKIt is obvious when a resource is classical or when it is quantum, and an example of a
hybrid resource is a classical–quantum stateX(26.2)ρXA =pX (x)|xihx|X ⊗ ρxA .
x

26.1 Unit Protocols
Chapter 6 discussed entanglement distribution[q → q] ≥ [qq] ,(26.3)

2 [c → c] + [qq] ≥ [q → q] ,(26.4)[q → q] + [qq] ≥ 2 [c → c] .

(26.5)

teleportationand super-dense coding
Chapter 7 introduced coherent dense coding[q → q] + [qq] ≥ 2 [q → qq] ,(26.6)

2 [q → qq] ≥ [q → q] + [qq] .

(26.7)and coherent teleportation
The fact that these two resource inequalities are dual under resource reversal implies thecoherent communication identity:
2 [q → qq] = [q → q] + [qq] .

(26.8)We also have the following resource inequalities:[q → q] ≥ [q → qq] ≥ [qq] .

(26.9)

Other unit protocols not covered in this book are the one-time pad[c → c]pub + [cc]priv ≥ [c → c]priv ,(26.10)[c → c]priv ≥ [cc]priv ,(26.11)[c → c]priv ≥ [c → c]pub .

(26.12)secret key distribution
and private-to-public transmission

The last protocol assumes a model where the receiver can locally copy information and place
it in a register to which Eve has access.
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26.2. NOISELESS QUANTUM SHANNON THEORY

731

26.2 Noiseless Quantum Shannon Theory
Noiseless quantum Shannon theory consists of resource inequalities involving unit resourcesand one non-unit, noiseless resource, such as an identity channel or a pure bipartite state.
Schumacher compression from Chapter 18 gives a way to simulate an identity channel
idA→B acting on a mixed state ρA by exploiting noiseless qubit channels at a rate equal tothe entropy H(A)ρ :H(A)ρ [q → q] ≥ hidA→B : ρA i .
(26.13)We also know that if n uses of an identity channel are available, then achievability of thecoherent information for quantum communication (Chapter 24) implies that we can sendquantum data down this channel at a rate equal to H(B) − H(E), where the entropies arewith respect to some input density operator ρA . But H(E) = 0 because the channel is theidentity channel (the environment gets no information) and H(B) = H(A)ρ because Alice’s
input goes directly to Bob. This gives us the following resource inequality:
hidA→B : ρA i ≥ H(A)ρ [q → q] ,(26.14)and combining (26.13) and (26.14) gives the following resource equality:
hidA→B : ρA i = H(A)ρ [q → q] .

(26.15)

Entanglement concentration from Chapter 19 converts many copies of a pure, bipartite
state |φiAB into ebits at a rate equal to the entropy of entanglement:
hφAB i ≥ H(A)φ [qq] .

(26.16)

Entanglement dilution exploits a sublinear amount of classical communication to dilute ebits
into n copies of a pure, bipartite state |φiAB . Ignoring the sublinear rate of classical communication gives the following resource inequality:
H(A)φ [qq] ≥ hφAB i .

(26.17)

Combining entanglement concentration and entanglement dilution gives the following resource equality:hφAB i = H(A)φ [qq] .
(26.18)The noiseless quantum Shannon theory is satisfactory in the sense that we can obtain resource equalities, illustrating the interconvertibility of noiseless qubit channels with a relative
identity channel and pure, bipartite states with ebits.

26.3 Noisy Quantum Shannon Theory
Noisy quantum Shannon theory has resource inequalities with one noisy resource, such as anoisy channel or a noisy state, interacting with other unit resources. We can further classify
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732

CHAPTER 26. SUMMARY AND OUTLOOK

a resource inequality as dynamic or static, depending on whether the noisy resource involved
is dynamic or static.
We first review the dynamic resource inequalities presented in this book. These protocols
involve a noisy channel interacting with the other unit resources. Many of the protocols innoisy quantum Shannon theory generate random codes from a state of the following form:XρXABE ≡pX (x)|xihx|X ⊗ UAN0 →BE (φxAA0 ),(26.19)xwhere φxAA0 is a pure, bipartite state and UAN0 →BE is an isometric extension of a channel
NA0 →B . Also important is a special case of the above form:
σABE ≡ UAN0 →BE (φAA0 ),(26.20)where φAA0 is a pure, bipartite state. Holevo–Schumacher–Westmoreland coding for classicalcommunication over a quantum channel (Chapter 20) is the following resource inequality:
hN i ≥ I(X; B)ρ [c → c] .

(26.21)

Devetak–Cai–Winter–Yeung coding for private classical communication over a quantum
channel (Chapter 23) is as follows:hN i ≥ (I(X; B)ρ − I(X; E)ρ ) [c → c]priv .

(26.22)Upgrading the private classical code to one that operates coherently gives Devetak’s method
for coherent communication over a quantum channel (Chapter 24):hN i ≥ I(AiB)σ [q → qq] ,(26.23)

which we showed can be converted asymptotically into a protocol for quantum communication:hN i ≥ I(AiB)σ [q → q] .
(26.24)Bennett–Shor–Smolin–Thapliyal coding for entanglement-assisted classical communication
over a quantum channel (Chapter 21) is the following resource inequality:
hN i + H(A)σ [qq] ≥ I(A; B)σ [c → c] .

(26.25)We showed how to upgrade this protocol to one for entanglement-assisted coherent communication (Chapter 22):hN i + H(A)σ [qq] ≥ I(A; B)σ [q → qq] ,(26.26)and combining with the coherent communication identity gives the following protocol for
entanglement-assisted quantum communication:
1
1
hN i + I(A; E)σ [qq] ≥ I(A; B)σ [q → q] .
22c 2019(26.27)

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26.3. NOISY QUANTUM SHANNON THEORY

733

Further combining with entanglement distribution gives the resource inequality in (26.24)
for quantum communication. By combining the HSW and BSST protocols together (thisneeds to be done at the level of coding and not at the level of resource inequalities—see
Chapter 22), we recover a protocol for entanglement-assisted communication of classical andquantum information:
1
1
hN i + I(A; E|X)σ [qq] ≥ I(A; B|X)σ [q → q] + I(X; B)σ [c → c] .
22(26.28)This protocol recovers any protocol in dynamic quantum Shannon theory that involves anoisy channel and the three unit resources after combining it with the three unit protocolsin (26.3)–(26.5). Important special cases are entanglement-assisted classical communication
with limited entanglement:
hN i + H(A|X)σ [qq] ≥ I(AX; B)σ [c → c] ,(26.29)and simultaneous classical and quantum communication:hN i ≥ I(X; B)σ [c → c] + I(AiBX)σ [q → q] .

(26.30)Chapter 22 touched on some important protocols in static quantum Shannon theory.
These protocols involve some noisy state ρAB interacting with the unit resources. The
protocol for coherent-assisted state transfer is the static couterpart to the protocol in (26.26):
hWS→AB : ρS i + H(A)ρ [q → q] ≥ I(A; B)ρ [q → qq] + hidS→B̂B : ρS i,(26.31)

where W is some isometry that distributes the state from a source S to two parties A and
B and idS→B̂B is the identity. Ignoring the source and state transfer in the above protocol
gives a protocol for quantum-assisted coherent communication:hρi + H(A)ρ [q → q] ≥ I(A; B)ρ [q → qq] .

(26.32)

We can also combine (26.31) with the unit protocols to obtain quantum-assisted state transfer:
1
1hWS→AB : ρS i + I(A; R)ϕ [q → q] ≥ I(A; B)ϕ [qq] + hidS→B̂B : ρS i,(26.33)2
2
and classical-assisted state transfer:hWS→AB : ρS i + I(A; R)ϕ [c → c] ≥ I(AiB)ϕ [qq] + hidS→B̂B : ρS i,(26.34)

where |ϕiABR is a purification of ρAB . We also have noisy super-dense coding
hρi + H(A)ρ [q → q] ≥ I(A; B)ρ [c → c] ,(26.35)hρi + I(A; B)ρ [c → c] ≥ I(AiB)ρ [q → q] ,(26.36)and noisy teleportationby combining (26.32) with the coherent communication identity and the unit protocols.
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734

CHAPTER 26. SUMMARY AND OUTLOOK

26.4 Protocols Not Covered In This Book
There are many important protocols that we have not covered in this book because our focus
here was mostly on communication over quantum channels. One such example is quantum
state redistribution. Suppose that Alice and Bob share many copies of a tripartite state ρACB
where Alice has the shares AC and Bob has the share B. The goal of state redistribution isfor Alice to transfer the C part of the state to Bob using the minimal resources needed to
do so. It is useful to identify a pure state ϕRACB as a purification of ρACB , where R is the
purifying system. Devetak and Yard (2008); Yard and Devetak (2009) showed the existenceof the following state redistribution protocol:
11hWS→AC|B : ρS i + I(C; RB)ϕ [q → q] + I(C; A)ϕ [qq] ≥2
2
1
1hWS→A|CB : ρS i + I(C; B)ϕ [q → q] + I(C; B)ϕ [qq] , (26.37)
22
where WS→AC|B is some isometry that distributes the system S as AC for Alice and B for Bob
and WS→A|CB is defined similarly. They also demonstrated that the above resource inequality
gives an optimal cost pair for the quantum communication rate Q and the entanglement
consumption rate E, with1
(26.38)Q = I(C; R|B)ϕ ,2
1
E = [I(C; A)ϕ − I(C; B)ϕ ] .
(26.39)2
Thus, their protocol gives a direct operational interpretation to the conditional quantum
mutual information 12 I(C; R|B)ϕ as the net rate of quantum communication required inquantum state redistribution.
A simple version of the quantum reverse Shannon theorem gives a way to simulate theaction of a channel NA0 →B on some input state ρA0 by exploiting classical communicationand entanglement (Bennett et al., 2002, 2014; Berta et al., 2011):
H(B)σ [qq] + I(R; B)σ [c → c] ≥ hNA0 →B : ρA0 i,(26.40)where σRB ≡ NA0 →B (ϕRA0 ), with ϕRA0 a purification of ρA0 . One utility of the quantum
reverse Shannon theorem is that it gives an indication of how one channel might simulate
another in the presence of shared entanglement. In the simulation of the channel NA0 →B ,the environment is also simulated and ends up in Alice’s possession. Thus, they end up
simulating the quantum feedback channel UAN0 →AB , and we can restate (26.40) as follows:
H(B)σ [qq] + I(R; B)σ [c → c] ≥ hUAN0 →AB : ρA0 i.

(26.41)

It is possible to upgrade the classical communication to coherent communication (Devetak, 2006), leading to the following coherent, fully-quantum version of the quantum reverse
Shannon theorem (Abeyesinghe et al., 2009):
1
1
I(A; B)σ [qq] + I(R; B)σ [q → q] ≥ hUAN0 →AB : ρA0 i.
22
c 2019(26.42)

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26.5. NETWORK QUANTUM SHANNON THEORY

735

Combining this resource inequality with the following one from Exercise 22.1.1
11hUAN0 →AB : ρA0 i ≥ I(A; B)σ [qq] + I(R; B)σ [q → q]2
2

(26.43)gives the following satisfying resource equality:11
hUAN0 →AB : ρA0 i = I(A; B)σ [qq] + I(R; B)σ [q → q] .
2
2

(26.44)

The above resource equality is a generalization of the coherent communication identity.
A more general version of the quantum reverse Shannon theorem quantifies the resources
needed to simulate many independent instances of a quantum channel on an arbitrary input
state, and the proof in this case is more involved (Bennett et al., 2014; Berta et al., 2011).
Other protocols that we did not cover are remote state preparation (Bennett et al., 2001,
2005; Abeyesinghe and Hayden, 2003), classical compression with quantum side information (Devetak and Winter, 2003), trade-offs between public and private resources and channels (Wilde and Hsieh, 2012a), trade-offs in compression (Hayden et al., 2002), a trade-off fora noisy state with the three unit resources (Hsieh and Wilde, 2010b), measurement compression (Winter, 2004), measurement compression with quantum side information (Wilde et al.,2012a), and measurement channel simulation (Berta et al., 2014). The resource inequality
formalism is helpful for devising new protocols in quantum Shannon theory by imaginingsome resources to be unit and others to be noisy.

26.5 Network Quantum Shannon Theory
The field of network quantum Shannon theory has arisen in recent years, motivated by theidea that one day we will be dealing with a quantum Internet in which channels of increasing
complexity can connect a number of senders to a number of receivers. A quantum multiple
access channel has multiple senders and one receiver. Various authors have considered classical communication over a multiple access channel (Winter, 2001; Fawzi et al., 2012; Wildeand Savov, 2012; Wilde and Guha, 2012; Boche and Notzel, 2014), quantum communication
over multiple access channels (Horodecki et al., 2005; Yard et al., 2008), entanglementassisted protocols (Hsieh et al., 2008a), and nonadditivity effects (Czekaj and Horodecki,
2009; Grudka and Horodecki, 2010). A quantum broadcast channel has one sender and multiple receivers. Various authors have addressed similar scenarios in this setting (Yard et al.,2011; Dupuis et al., 2010; Guha and Shapiro, 2007; Savov and Wilde, 2015; Radhakrishnan
et al., 2014; Hirche and Morgan, 2015; Seshadreesan et al., 2015b). A quantum interference
channel has multiple senders and multiple receivers in which certain sender-receiver pairs
are interested in communicating. Recent progress in this direction is in (Fawzi et al., 2012;
Sen, 2011; Hirche et al., 2016). One could also consider distributed compression tasks, andvarious authors have contributed to this direction (Ahn et al., 2006; Abeyesinghe et al.,2009; Savov, 2008). We could imagine a future textbook containing several chapters thatsummarize all of the progress in network quantum Shannon theory and the novel techniquesc 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

736

CHAPTER 26. SUMMARY AND OUTLOOK

needed to handle coding over such channels. Savov (2012) highlights much of this direction
in his PhD thesis (at least for classical communication).

26.6 Future Directions
Quantum Shannon theory has evolved from the first and simplest result regarding Schumacher compression to a whole host of protocols that indicate how much data we can transmit over noisy quantum channels or how much we can compress information of varying
types—the central question in any task is, “How many unit resources can we extract from agiven non-unit resource, perhaps with the help of other non-unit resources?” This book may
give the impression that so much has been solved in the area of quantum Shannon theory
that little remains for the future, but this is actually far from the truth! There remains muchto do to improve our understanding, and this final section briefly outlines just a few of theseimportant questions.
Find a better formula for the classical capacity other than the HSW formula. Our best
characterization of the classical capacity is with a regularized version of the HSW formula,and this is unsatisfying in several ways that we have mentioned before. In a similar vein, finda better formula for the private classical capacity, the quantum capacity, and even for the
trade-off capacities. All of these formulas are unsatisfying because their regularizations seemto be necessary in the general case. It could be the case that an entropic expression evaluatedon some finite tensor power of the channels would be sufficient to characterize the capacity for
different tasks, but this is a difficult question to answer. Interestingly, recent work suggests
pursuing to find out whether this question is algorithmically undecidable (see Wolf et al.
(2011)). Effects such as superactivation of quantum capacity (see Section 24.8.2) and nonadditivity of private capacity (see Section 23.5.2) have highlighted how little we actuallyknow about the corresponding information-processing tasks in the general case. Also, it isimportant to understand these effects more fully and to see if there is any way of exploiting
them in a practical communication scheme. Finally, a different direction is to expand thenumber of channels that have additive capacities. For example, finding the quantum capacityof a non-degradable quantum channel would be a great result. Many questions remain open
regarding second-order characterizations, error exponents, and strong converses. Results inthese directions give much finer characterizations of communication tasks. We have alreadyhighlighted progress in these directions at the end of relevant chapters.
Continue to explore network quantum Shannon theory. The single-sender, single-receiverchannel setting is a useful model for study and applies to many practical scenarios, but
eventually, we will be dealing with channels connecting many inputs to many outputs. Havingsuch an understanding for information transmission in these scenarios could help guide thedesign of practical communication schemes and might even shed light on the open problems
in the preceding paragraph.

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseAPPENDIX A

Supplementary Results
This section collects various useful definitions and lemmas that we use throughout the proofsof certain theorems in this book.
Lemma A.0.1 Suppose that M and N are positive semi-definite operators. Then the operators M + N , M N M , and N M N are positive semi-definite.
Lemma A.0.2 Suppose that the operators ω̂ and ω have trace less than or equal to one.
Suppose ω̂ lies in the operator interval [(1 − ε) ω, (1 + ε) ω]. Thenkω̂ − ωk1 ≤ ε.

(A.1)

Proof. The statement “ω̂ lies in the operator interval [(1 − ε) ω, (1 + ε) ω]” is equivalent to
the following two conditions:(1 + ε) ω − ω̂ = εω − (ω̂ − ω) ≥ 0,
ω̂ − (1 − ε) ω = (ω̂ − ω) + εω ≥ 0.

(A.2)
(A.3)

Let α ≡ ω̂ − ω. Let us rewrite α in terms of the positive semi-definite operators P and Q:
α = P − Q,(A.4)

as we did in the proof of Lemma 9.1.1. The above conditions become as follows:εω − α ≥ 0,
α + εω ≥ 0.

(A.5)
(A.6)

Let the positive projectors ΠP and ΠQ project onto the respective supports of P and Q. We
then apply the projector ΠP to the first condition:
ΠP (εω − α) ΠP ≥ 0,
⇒ εΠP ωΠP − ΠP αΠP ≥ 0,
⇒ εΠP ωΠP − P ≥ 0,
737

(A.7)
(A.8)
(A.9)

738

APPENDIX A. SUPPLEMENTARY RESULTS

where the first inequality follows from Lemma A.0.1. We apply the projector ΠQ to thesecond condition:
ΠQ (α + εω) ΠQ ≥ 0,
⇒ ΠQ αΠQ + εΠQ ωΠQ ≥ 0,
⇒ −Q + εΠQ ωΠQ ≥ 0,

(A.10)(A.11)(A.12)

where the first inequality again follows from Lemma A.0.1. Adding the two positive semidefinite operators together gives another positive semi-definite operator by Lemma A.0.1:
εΠP ωΠP − P − Q + εΠQ ωΠQ ≥ 0,
⇒ εΠP ωΠP − |ω̂ − ω| + εΠQ ωΠQ ≥ 0.

(A.13)(A.14)

Apply the trace operation to get the following inequality:
ε Tr {ω} ≥ Tr {|ω̂ − ω|} = kω̂ − ωk1 .

(A.15)

Using the hypothesis that Tr {ω} ≤ 1 gives the desired result.
Theorem A.0.1(Polar Decomposition) Any operator√√ A admits a left polar decomposition A = U A† A, and a right polar decomposition A = AA† V .
Proof. We give a simple proof for just the right polar decomposition by appealing to
the singular value decomposition. Any operator A admits a singular value decomposition
A = U1 ΣU2 , where U1 and U2 are unitary operators and Σ is an
positive√ operator with
†
†
†
†
2 †
†
singular values. Then AA = U1 ΣU2 U2 ΣU1 = U1 Σ U1 , and thus √AA = U1 ΣU1 . We cantake V = U1 U2 and we obtain the right polar decomposition of A as AA† V = U1 ΣU1† U1 U2 =U1 ΣU2 = A.
Lemma A.0.3 Consider two collections of orthonormal states {|χj i}j∈[N ] and {|ζj i}j∈[N ]such that hχj |ζj i ≥ 1 − ε for all j. There exist phases γj and δj such thathχ̂|ζ̂i ≥ 1 − ε,(A.16)whereNN

1 X iγj|χ̂i = √e |χj i,
N j=1

1 X iδj|ζ̂i = √e |ζj i.
N j=1

(A.17)

Proof. Define the Fourier transformed statesN

1 X 2πijs/N|χ̂s i ≡ √e|χj i,
N j=1
c 2019

(A.18)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

739

and similarly define |ζ̂s i. By Parseval’s relation, it follows that
N
N
1 X
1 X
hχ̂s |ζ̂s i =hχj |ζj i ≥ 1 − ε.
N s=1
N j=1

(A.19)

Thus, at least one value of s obeys the following inequality: eiθs hχ̂s |ζ̂s i ≥ 1 − ε, for somephase θs . Setting γj = 2πjs/N and δj = γj + θs satisfies the statement of the lemma.
The following “support lemmas” are taken directly from (Renner, 2005, Appendix B).
Lemma A.0.4 Let XAB ∈ L(HA ⊗ HB ) be positive semi-definite, and let XA ≡ TrB {XAB }and XB ≡ TrA {XAB }. Then supp(XAB ) ⊆ supp(XA ) ⊗ supp(XB ).
Proof. First suppose that XAB is rank one, so that XAB = |ΨihΨ|AB for some vector|ΨiAB ∈ HA ⊗ HB . Due to the Schmidt decomposition theorem (Theorem 3.8.1), we havethat
X
|ΨiAB =
γz |θz iA ⊗ |ξz iB ,(A.20)
z∈Z

where |Z| ≤ min{dim(HA ), dim(HB )}, {γz } is a set of strictly positive numbers, and {|θz iA }and {|ξz iB } are orthonormal bases. Thensupp(XAB ) = span{|ΨiAB }⊆ span{|θz iA : z ∈ Z} ⊗ span{|ξz iB : z ∈ Z}.

(A.21)(A.22)The statement then follows for this case because supp(XA ) = span{|θz iA : z ∈ Z} andsupp(XB ) = span{|ξz iB : z ∈ Z}.
Now suppose that XAB is not rank one. It admits a decomposition into rank-one vectorsof the following form:X
XAB =|Ψx ihΨx |AB ,(A.23)
x∈X

where |Ψx iAB ∈ HA ⊗ HB for all x ∈ X . Set ΨxAB = |Ψx ihΨx |AB , and let ΨxA ≡ TrB {ΨxAB }and ΨxB ≡ TrA {ΨxAB }. Thensupp(XAB ) = span{|Ψx iAB : x ∈ X }"#[⊆ span[supp(ΨxA ) ⊗ supp(ΨxB )](A.24)(A.25)x∈X"⊆ span#[supp(ΨxA ) ⊗ spanx∈X= supp(XA ) ⊗ supp(XB ),"#[supp(ΨxB )(A.26)x∈X(A.27)

concluding the proof.
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740

APPENDIX A. SUPPLEMENTARY RESULTS

Lemma A.0.5 Let XAB , YAB ∈ L(HA ⊗ HB ) be positive semi-definite, and suppose thatsupp(XAB ) ⊆ supp(YAB ). Then supp(XA ) ⊆ supp(YA ), where XA ≡ TrB {XAB } and YA ≡TrB {YAB }.
Proof. First suppose that XAB is rank one, as in the first part of the proof of the previous
lemma, and let us use the same notation as given there. Applying the same lemma givesthat
supp(XAB ) ⊆ supp(YAB ) ⊆ supp(YA ) ⊗ supp(YB ),(A.28)
which in turn implies that supp(XAB ) = span{|ΨiAB } ⊆ supp(YA ) ⊗ supp(YB ). This implies
that |θz iA ∈ supp(YA ) for all z ∈ Z, and thus that span{|θz iA } ∈ supp(YA ). We can thenconclude the statement in this case because span{|θz iA } = supp(XA ).
Now suppose that XAB is not rank one. Then it admits a decomposition as given in the
proof of the previous lemma. Using the same notation, we have that supp(ΨxAB ) ⊆ supp(YAB )holds for all x ∈ X . Since we have proven the lemma for rank-one operators, we can conclude
that supp(ΨxA ) ⊆ supp(YA ) holds for all x ∈ X . As a consequence, we find that"#[supp(XA ) = spansupp(ΨxA ) ⊆ supp(YA ),(A.29)
x∈X

concluding the proof.

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseAPPENDIX B

Unique Linear Extension of a
Quantum Physical Evolution
Recall in Section 4.4.1 that we argued on physical grounds how any quantum physical evolution N should be convex linear when acting on the space D(HA ) of density operators:
N (λρA + (1 − λ)σA ) = λN (ρA ) + (1 − λ)N (σA ),(B.1)

where ρA , σA ∈ D(HA ) and λ ∈ [0, 1]. Here we show how to construct a unique linear
e of N , whose action is well defined on the space of all operators XA ∈ L(HA ).
extension N
The development follows the approach given in Proposition 2.30 of Heinosaari and Ziman(2012).
e (0) ≡ 0, where the inputs and outputs are understood to be the zero
We first define N
operator. We next extend the action of N to all positive semi-definite operators PA 6= 0 as
follows:e (PA ) ≡ Tr{PA }N ([Tr{PA }]−1 PA ),N(B.2)
where it is clear that this is well defined from N because [Tr{PA }]−1 PA is a density operator.
Now consider for a constant s > 0 that we have scale invariance:e (sPA ) = Tr{sPA }N ([Tr{sPA }]−1 sPA )N−1= s Tr{PA }N ([Tr{PA }]e (PA ).
= sN

PA )

(B.3)
(B.4)
(B.5)

Furthermore, for two non-zero positive semi-definite operators PA and QA , we have thefollowing additivity relation:e (PA + QA ) = N
e (PA ) + N
e (QA ),
N
741

(B.6)

742

APPENDIX B. UNIQUE LINEAR EXTENSION OF A QUANTUM PHYSICALEVOLUTION

which follows becausee (PA + QA )N
= Tr{PA + QA }N ([Tr{PA + QA }]−1 (PA + QA ))


1
1
PA +
QA
= Tr{PA + QA }N
Tr{PA + QA }Tr{PA + QA }
Tr{PA }PA
Tr{QA }QA
= Tr{PA + QA }N+
Tr{PA + QA } Tr{PA } Tr{PA + QA } Tr{QA }




PA
QA
= Tr{PA }N+ Tr{QA }N
Tr{PA }
Tr{QA }e (PA ) + N
e (QA ),
=N

(B.7)
(B.8)
(B.9)
(B.10)(B.11)

where in the fourth equality, we exploited convex linearity of the quantum physical evolution N .
For the next step, recall that any Hermitian operator TA can be written as a linear
combination of a positive part and a negative part: TA = TA+ − TA− , where both TA+ and TA−
e on any Hermitian operator
are positive semi-definite operators. So we define the action of N
TA as follows:e (TA ) ≡ Ne (T + ) − Ne (T − ).
N(B.12)AA
To see that the following additivity relation holds for all Hermitian SA and TAe (SA + TA ) = Ne (SA ) + N
e (TA ),
N

(B.13)SA + TA = (SA + TA )+ − (SA + TA )− ,(B.14)SA + TA = SA+ + TA+ − SA− − TA− .

(B.15)

consider thatwhile also
Equating both sides, we find that(SA + TA )+ + SA− + TA− = (SA + TA )− + SA+ + TA+ .

(B.16)

Now we exploit this equality, (B.6), and the definition in (B.12) to establish (B.13).
e to all operators XA ∈ L(HA ). Here, we recall
The final step is to extend the action of N
that any linear operator can be written in terms of a real and imaginary part as follows:
XAR ≡


1
XA + XA† ,
2

XAI ≡


1 
XA − XA† ,2i(B.17)

where by inspection, XAR and XAI are Hermitian operators. So we definee (XA ) ≡ N
e (XAR ) + iNe (XAI ).
N
c 2019(B.18)Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

743

e of the quantum physicalThis completes the development of a well defined linear extension N
evolution N .
To show that it is unique, recall that any operator XA can be expanded as a linear
combination of density operators from the basis {ρx,yA }, defined in (4.185), as follows:X
XA =αx,y ρx,y(B.19)A ,x,ywhere αx,y ∈ C for all x and y. It is straightforward to show from the above developmentthat
X
e (XA ) =N
αx,y N (ρx,y(B.20)A ).
x,y

Now suppose that N 0 is some other linear map for which N 0 (ρA ) = N (ρA ) for all ρA ∈ D(HA ).
Then the following equality holds for all XA ∈ L(HA ):X
X
eαx,y N (ρx,y(B.21)
αx,y N 0 (ρx,y)=
N 0 (XA ) =A ) = N (XA ).
A
x,y

x,y

e , given that they have the same action on every operator XA ∈ L(HA ).
As a result, N 0 = N

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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APPENDIX B. UNIQUE LINEAR EXTENSION OF A QUANTUM PHYSICALEVOLUTION

c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

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c 2019Mark M. Wilde—Licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported LicenseIndexaccessible information, 315, 334
of a channel, 558
adjoint, 155
Alicki-Fannes-Winter inequality, 357
amplitude damping channel, 174, 608, 676
anticommutator, 79
antidegradable channel, 410, 412, 675, 677
Araki–Lieb triangle inequality, 337
asymptotic equipartition theorem, 49, 421
Bell inequality, 26, 99
Bell measurement, 210
Bell states, 106, 112, 208, 210
Bell’s theorem, 99
binary entropy, 27binary symmetric channel, 51
Bloch sphere, 72, 124
Born rule, 71, 83, 96, 118bosonic channel, 722
Cauchy–Riemann equations, 366
channel distinguishability, 257
Choi operator, 149
Choi rank, 149
Choi–Kraus theorem, 149
CHSH game, 99, 137classical capacity theorem, 562
classical-quantum state, 144, 176classical-to-classical channel, 161classical-to-quantum channel, 162
coherent bit channel, 227
coherent communication, 613
coherent communication identity, 233, 618, 733
coherent dense coding, 613
coherent information, 317, 330, 655
conditional, 332
convexity, 337
of a channel, 408
additivity for a degradable channel, 410
positivity, 409
superadditivity, 678

coherent state transfer, 623
commutator, 79
complementary channel, 188, 192
completely positive map, 148
completely positive trace-preserving map, 148
composite quantum systems, 88
conditional quantum channel, 176conditional typicality, 60
constant-composition coding, 567
controlled gate, 92
covering lemma, 501
data compression, 41, 427
theorem, 44
data processing inequality, 299
coherent information, 349
conditional quantum mutual information, 350
mutual information, 299
quantum mutual information, 349
relative entropy, 301
decoupling approach, 688
degradable channel, 194, 410–417, 419, 420, 572, 641,
651, 656, 659, 666, 668, 675, 678, 712
density operator, 118, 120
local, 141dephasing channel, 171, 610, 713
generalized, 193
depolarizing channel, 173, 574, 610, 678
diamond-norm distance, 257
displacement operator, 723
ebit, 208
elementary coding, 209
entangled state, 136
entanglement concentration, 531, 533, 534, 545, 731
entanglement dilution, 532, 547, 731
entanglement distillation, 684
entanglement distribution, 203, 206, 219
entanglement manipulation, 531
entanglement swapping, 215
entanglement-assisted

769

770

INDEX

classical communication, 31, 217, 585, 732
with feedback, 601
coherent communication, 615
quantum communication, 617, 697, 732
entanglement-assisted capacity theorem, 591
converse, 600
direct part, 592
entanglement-breaking channel, 164–167, 179, 193–
195, 381, 382, 399–402, 412, 420, 557, 569,
570, 582, 583
erasure channel, 175, 606, 658, 675
expurgation, 59, 495, 648
extension, 185
Fannes-Audenaert Inequality, 355, 356
Fano’s inequality, 302
fidelity, 30, 261
achieving measurement, 269
concavity, 267
entanglement fidelity, 279
expected, 262
joint concavity, 266
monotonicity, 266
multiplicativity, 265
Uhlmann, 262
Fourier transform, 109
gentle measurement, 275
coherent, 277
for ensembles, 277
Hölder inequality, 365
Hadamard channel, 194, 571, 692, 711
Hadamard gate, 80
Hadamard three-lines theorem, 366
hashing bound, 672
Hayashi–Nagaoka operator inequality, 490
Heisenberg-Weyl operators, 108, 593
Hilbert–Schmidt distance, 282
Hilbert–Schmidt inner product, 155
Hirschman’s theorem, 367
Holevo bound, 34, 315, 350, 382, 484, 585
Holevo information, 334
dimension bound, 335
of a channel, 398
concavity, 403
superadditivity, 578
holomorphic function, 366
operator-valued, 367
HSW theorem, 31, 483, 562, 696, 732
converse, 568
direct part, 563
c 2019indeterminism, 24
indicator function, 128
information content, 43, 286
information locking, 654
isometric channel, 160
isometric extension, 186, 188
isometry, 160
kernel, 338
Kraus representation, 118
law of large numbers, 47, 421
local operations and classical communication, 536
LOCC, 531, 536
LOCC monotone, 536
locking capacity, 654
LSD theorem, 32, 732
Markov’s inequality, 56, 495
matrix representation, 78
maximally mixed state, 123
maximum modulus principle, 366
measurement channel, 164
minimum output entropy, 575
mutual information, 62, 293
conditional, 295
of a classical channel, 386
additivity, 389
network quantum Shannon theory, 735
no-cloning theorem, 34, 94, 637, 655, 658
no-deletion theorem, 95
noisy channel coding theorem, 54
noisy super-dense coding, 613
non-commutative union bound, 497
observable, 83
Operator Chernoff Bound, 506
packing lemma, 483, 564, 595, 621
partial trace, 139
Pauli channel, 172
Pauli matrices, 80
Petz recovery map, 369
rotated, 370
phase-flip operator, 79
Pinsker inequality, 309
quantum, 345
polar decomposition, 738
positive map, 147
positive operator-valued measure, 132
positive semi-definite, 122
private classical capacity

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INDEX

superadditivity, 651
private classical capacity theorem, 640
converse, 649
direct part, 641
private classical communication, 637, 732
secret-key assisted, 652
private informationof a quantum channel, 413
additivity for a degradable channel, 416
of a wiretap channel, 394
additivity for degraded channels, 395
positivity, 395
suparadditivity, 651
probability amplitudes, 71, 92
product state, 134
purification, 182
equivalence, 183
interpretation, 183
purity, 123
quantum capacity theorem, 20, 32, 217, 655, 659, 697
converse, 666
direct part, 659
quantum data processing inequality, 349
quantum discord, 381
quantum dynamic capacity formula, 705
quantum dynamic capacity theorem, 691, 694
converse, 700
direct part, 699
quantum erasure channel, 573
quantum hypothesis testing, 253
quantum information source, 460, 519
quantum instrument, 169
coherent, 196
quantum interference, 24, 25
quantum key distribution, 35, 637
quantum mutual information, 332
chain rule, 335
conditional
dimension bound, 337
duality, 337
non-negativity, 336
dimension bound, 333
non-negativity, 332
of a channel, 404
additivity, 405
quantum relative entropy, 338
additivity, 342
for classical–quantum states, 342
isometric invariance, 342
joint convexity, 344, 380
monotonicity, 340
c 2019771

non-negativity, 340
quantum reverse Shannon theorem, 612, 734
quantum teleportation, 26, 31, 36, 203, 211, 220, 236
coherent, 231
noisy, 218, 733
quantum typicality, 459
conditional, 469
strong, 473
weak, 471
quantum-to-classical channel, 164
qudit, 108, 218Rényi generalization of a relative entropy difference,
371
randomness concentration, 540
relative entropy, 294
non-negativity, 297
positivity, 297
relative entropy of entanglement, 536
remote state preparation, 214, 735
repetition code, 51
quantum, 669
rotation operators, 82
sample entropy, 47, 423, 424
conditional, 432
joint, 430
Schatten norm, 364
Schmidt decomposition, 113, 592
Schumacher compression, 30, 319, 519, 731
separable states, 118, 136
sequential decoding, 496, 566
Shannon compression, 44, 427
Shannon entropy, 44, 286
binary, 287
concavity, 289
conditional, 291
continuity, 304
joint, 292
positivity, 289
square-root measurement, 489, 566
stabilizer codes, 669
state merging, 38, 330
state redistribution, 734
state transfer
classical-assisted, 625, 733
coherent, 623
quantum-assisted, 624, 733
Stein–Hirschman interpolation theorem, 368
Stinespring dilation, 188
strong subadditivity, 295, 336, 337, 343, 379
super-dense coding, 203, 210, 220

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772

INDEX

coherent, 230
noisy, 620, 733
superactivation, 38, 656, 682
superposition, 24, 25, 71, 84
supportof a function, 294
of an operator, 338
surprisal, 43
tensor product, 89
thermal state, 723
time-sharing, 627
trace, 119
cyclicity, 120
trace distance, 30, 248, 251
monotonicity, 256
operational interpretation, 253
triangle inequality, 255
trace norm, 248
trace preserving map, 148trade-off coding, 626, 733
Tsirelson’s bound, 105
twirling, 173
two-mode squeezed state, 723
type, 437
bound on number of types, 437
type class, 437
projector, 480
subspace, 480, 592
typical, 442
typical projector, 462
conditionallystrong, 474
typical sequence, 48, 421, 424
jointly, 430
typical set, 424, 425
conditionally, 432
jointly, 430
strongly, 437
typical subspace, 30, 462
conditionallystrong, 474
weak, 472
measurement, 463
typicality
strong, 435
conditional, 446
joint, 444
weak, 423
conditional, 432
joint, 429
c 2019uncertainty principle, 87, 351
unit resource capacity region, 235, 691
unital map, 155
von Neumann entropy, 317, 318
additivity, 324
concavity, 320, 335
conditional, 327
concavity, 336, 379
duality, 331
continuity, 355
joint, 322
operational interpretation, 520, 532
positivity, 320
subadditivity, 341
von Neumann measurement, 86

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