Lecture Notes for Quantum Matter
MMathPhys©Professor Steven H. Simon
Oxford University
January 26, 2022

Contents
1 What we will study1

1.1

Bose Superfluids (BECs, Superfluid He, Superconductors) . . . . . . . . . .

1

1.2

Theory of Fermi Liquids . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2

1.3

BCS theory of superconductivity and Majoranas . . . . . . . . . . . . . . .

22 Introduction to Superfluids

3

2.1

Some History and Basics of Superfluid Phenomena . . . . . . . . . . . . . .

3

2.2

Landau and the Two Fluid Model . . . . . . . . . . . . . . . . . . . . . . .

6

2.2.1

More History and a bit of Physics . . . . . . . . . . . . . . . . . . .

6

2.2.2

Landau’s Two Fluid Model . . . . . . . . . . . . . . . . . . . . . . .

7

2.2.3

More Physical Effects and Their Two Fluid Pictures . . . . . . . . .

9

2.2.4

Second Sound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.2.5

Big Questions Remaining . . . . . . . . . . . . . . . . . . . . . . . . 14

2.3

Curl Free Constraint: Introducing the Superfluid Order Parameter . . . . . 14
2.3.1

Vorticity Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.4

Landau Criterion for Superflow . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.5

Superfluid Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.5.1

The Andronikoshvili Experiment . . . . . . . . . . . . . . . . . . . . 21

2.5.2

Landau’s Calculation of Superfluid Density . . . . . . . . . . . . . . 22

3 Charged Superfluid ≈ Superconductor
3.1

25

London Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
3.1.1

Meissner-Ochsenfeld Effect . . . . . . . . . . . . . . . . . . . . . . . 28

3.1.2

Quantum Input and Superfluid Order Parameter . . . . . . . . . . . 29
3

3.1.3

Superconducting Vortices . . . . . . . . . . . . . . . . . . . . . . . . 30

3.1.4

Type I and Type II superconductors . . . . . . . . . . . . . . . . . . 32

3.1.5

How big is Hc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

4 Microscopic Theory of Bosons
4.1

4.2

Mathematical Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
4.1.1

Second quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

4.1.2

Coherent States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

4.1.3

Multiple orbitals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

BECs and the Gross-Pitaevskii Equation . . . . . . . . . . . . . . . . . . . . 41
4.2.1

4.3

4.5

Noninteracting BECs as Coherent States . . . . . . . . . . . . . . . . 42

Interacting Bosons and the Gross-Pitaevskii Equation . . . . . . . . . . . . 42
4.3.1

4.4

37

Rederivation without Coherent States . . . . . . . . . . . . . . . . . 44

Order Parameter and Off-Diagonal Long Ranged Order . . . . . . . . . . . 45
4.4.1

non-interacting bosons at T > 0. . . . . . . . . . . . . . . . . . . . 48

4.4.2

Current and the Order Parameter . . . . . . . . . . . . . . . . . . . 49

Bogoliubov Theory for the Weakly Interacting Bose Gas . . . . . . . . . . . 50
4.5.1

Bogoliubov Transform . . . . . . . . . . . . . . . . . . . . . . . . . . 52

5 Feynman Theory of Helium-4

57

5.1

Ground State and Low Energy Excitations . . . . . . . . . . . . . . . . . . . 57

5.2

Single Mode Excitation Spectrum . . . . . . . . . . . . . . . . . . . . . . . . 64

6 Ginzburg-Landau Theory
6.1

6.2

71

Neutral Superfluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
6.1.1

Spatially Uniform Solution . . . . . . . . . . . . . . . . . . . . . . . 72

6.1.2

Spatial Dependence: Ginzburg Landau Theory . . . . . . . . . . . . 73

Charged Superfluids (i.e., Superconductors) . . . . . . . . . . . . . . . . . . 76
6.2.1

Anderson-Higgs Mechanism . . . . . . . . . . . . . . . . . . . . . . . 76

6.2.2

Equations of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

6.2.3

Energetics from Ginzburg-Landau theory: Type I and Type II superconductors revisited . . . . . . . . . . . . . . . . . . . . . . . . . 80

6.3

Lagrangian Formulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
6.3.1

Nonrelativisitic Neutral Superfluid . . . . . . . . . . . . . . . . . . . 85

6.3.2

Relativisitic Neutral Superfluid . . . . . . . . . . . . . . . . . . . . . 85

6.3.3

Electromagnetism . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

6.3.4

Charged Relativistic Superfluid . . . . . . . . . . . . . . . . . . . . . 86

7 Interacting Fermions

89

7.1

Why study fermions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

7.2

Some Mathematical Preliminaries Regarding Fermions . . . . . . . . . . . . 90

7.3

7.2.1

Second Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

7.2.2

Change of Basis

7.2.3

Writing our Hamiltonian in Second Quantized Form . . . . . . . . . 94

First Order Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . 97
7.3.1

7.4

Hartree-Fock as Optimal Slater Determinant . . . . . . . . . . . . . 103

Application of Hartree-Fock to Translationally Invariant Fermions . . . . . 106
7.5.1

7.6

Hartree and Fock Terms . . . . . . . . . . . . . . . . . . . . . . . . . 98

Hartree-Fock Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . 100
7.4.1

7.5

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

Hartree-Fock Effective Mass . . . . . . . . . . . . . . . . . . . . . . . 111

Appendix: Showing the Log Divergence at The Fermi Surface . . . . . . . . 113

8 Screening and Linear Response

115

8.1

Thomas-Fermi Screening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

8.2

Response More Generally . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

8.3

Lindhard Response Function (Response of Noninteracting Electrons) . . . . 123

8.4

8.5

8.3.1

Interpretations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

8.3.2

High Frequency ω  EF and small k . . . . . . . . . . . . . . . . . . 127

Response of Interacting Electrons: RPA . . . . . . . . . . . . . . . . . . . . 129
8.4.1

Relation to Dielectric Constant . . . . . . . . . . . . . . . . . . . . . 131

8.4.2

High Frequency, Low k response: Plasmons . . . . . . . . . . . . . . 132

Chapter Appendix: Deriving Lindhard Screening expression . . . . . . . . . 135

9 Landau Fermi Liquid Theory

137

9.1

Background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

9.2

Basics Idea of Fermi Liquid Theory . . . . . . . . . . . . . . . . . . . . . . . 138
9.2.1

Landau’s Conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

9.3

Properties of the Dressed Fermi Liquid . . . . . . . . . . . . . . . . . . . . . 141

9.4

Landau Free Energy Functional . . . . . . . . . . . . . . . . . . . . . . . . . 143

9.5

Results from Fermi Liquid Theory . . . . . . . . . . . . . . . . . . . . . . . 147

9.6

9.5.1

Compressibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147

9.5.2

Spin Susceptibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

9.5.3

Mass Renormalization . . . . . . . . . . . . . . . . . . . . . . . . . . 149

9.5.4

More? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150

Further extensions of Fermi Liquid Theory . . . . . . . . . . . . . . . . . . . 150
9.6.1

Local Dynamical Properties and Boltzmann Transport . . . . . . . . 150

9.6.2

Landau-Silin Theory: Long Range Coulomb Interactions . . . . . . . 151

10 Leading up to BCS Theory: Electron Binding

153

10.1 Mechanism of Attraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154
10.1.1 Canonical Transformation . . . . . . . . . . . . . . . . . . . . . . . . 154
10.1.2 RPA Screening and Ion Gas Model . . . . . . . . . . . . . . . . . . . 157
10.2 Bound States and the Cooper Problem . . . . . . . . . . . . . . . . . . . . . 160
10.2.1 Non-Existence of Weakly Bound States in 3D . . . . . . . . . . . . . 160
10.2.2 The Cooper Problem: Bound States in The Presence of a Fermi Sea 161
11 BCS Theory

167

11.1 The BCS wavefunction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
11.1.1 Fixed N wavefunction . . . . . . . . . . . . . . . . . . . . . . . . . . 168
11.1.2 Coherent State Wavefunction . . . . . . . . . . . . . . . . . . . . . . 169
11.2 BCS Hamiltonian, Energy Minimization, and Gap Equation . . . . . . . . . 171
11.3 More Exotic Pairing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
11.3.1 Anisotropic Interactions and Anisotropic Gaps: The Example ofHigh Tc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
11.3.2 Triplet Superconductors: Helium-3 . . . . . . . . . . . . . . . . . . . 179
11.4 BCS Excitation Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 180

11.5 Finite Temperature Gap Equation . . . . . . . . . . . . . . . . . . . . . . . 184
11.6 Inhomogeneous Superconductors . . . . . . . . . . . . . . . . . . . . . . . . 186
11.6.1 Caroli-de Gennes-Matricon States . . . . . . . . . . . . . . . . . . . 186
11.7 Ginzburg-Landau, Number-Phase, and Josephson . . . . . . . . . . . . . . . 187
11.7.1 Josephson Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188
11.8 Appendix: A better argument for the phase associated with the tunnelingmatrix element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191
12 An Introduction to Majoranas

193

12.1 The Kitaev Chain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
12.1.1 Bogoliubov Approach . . . . . . . . . . . . . . . . . . . . . . . . . . 199
12.1.2 Topological Approach . . . . . . . . . . . . . . . . . . . . . . . . . . 200
12.1.3 Influence of the Phase: Operations on the Chain . . . . . . . . . . . 201
12.2 Where else do Majoranas live? . . . . . . . . . . . . . . . . . . . . . . . . . 203

iPrefaceThis course covers interesting quantum states of matter: Superfluids, Superconductors, and Fermi Liquids. This is where condensed matter really starts to get bizarre —and therefore even more fun (at least in my opinion).
I realize that possibly not everyone will have actually attended the official prerequisite courses (or possibly did attend, but didn’t pay attention), so I will try to make thiscourse as self-contained as possible, while still making contact with what has been taught
in prior courses.
Particularly for those who have read the Oxford Solid-State Basics, I will warn you
right from the front that this book is higher level and is likely to be more difficult going.
Whereas that book was written as a third year undergrad course, this book is written ata masters level for students who have a good amount of theoretical background already.
The jokes aren’t as good either.
Lets get started!

ii
At early points in one’s career there is an inevitable tension between learning physics
and learning mathematical formalism. While learning to manipulate commutation relations, and even learning the rules for green’s function expansions can be fun — and indeed,by doing this one gets a feeling that one is actually learning something — the real hardpart is learning to develop intuition for physics. Developing intuition is extremely hard,and it is something that professional physicists struggle with throughout their lives.
This course needs to strike a balance between the two tasks. For some we will notlearn enough formalism. For others we will not learn enough physics. I hope we will beable to satisfy both factions, and I hope everyone will learn at least a bit of both.

Chapter 1

What we will study
1.1

Bose Superfluids (BECs, Superfluid He, Superconductors)

Much of the term will be focused on discussion of superfluids and superconductors.
We will discuss both Bose Einstein Condensates (BECs) and superfluids — the main
example being superfluid Helium 4. The difference here is that a BEC is by definition agas of noninteracting bosons, whereas Helium 4 is strongly interacting. In the modern era
one calls a weakly interacting Bose gas a BEC also – since it can be treated as a weak perturbation of the noninteracting case. Helium 4, however, is very far from noninteracting.
The tension between the physics of the interacting and noninteracting cases will bea theme.
We will then turn to the study of superconductors. First we will view these as simply
being a superfluid of charged bosons.
Of course superconductors are things like aluminum1 are regular metals, where thecharged objects are electrons, which are fermions of charge −e. However, as we will see
later, these charged objects can pair into bosons of charge −2e.
Why are such pairs bosons? Recall that fermions accumulate a minus sign when they
are exchanged. Exchanging a pair of fermions with another pair of fermions accumulates
an even number of minus signs, hence a plus sign, so the pair of fermions is a boson.
This picture of electrons pairing into bosons is not entirely accurate and there aregood reasons to believe that even using this as a cartoon picture is quite dangerous.
However, in some other ways this picture does make sense. In order to figure out how it
makes sense we must first take a detour into a study of Fermions.

1Those using the british spelling and pronounciation ”aluminium” will be ridiculed.

1

2

CHAPTER 1. WHAT WE WILL STUDY

1.2

Theory of Fermi Liquids

Fermi liquids, metals and other systems of interacting fermions have some unusual properties — some of which you have probably studied in your introductory solid state orcondensed matter courses2 . However, in most of these treatments, the interaction between fermions is completely neglected. This is a terrible thing to do being that theinteraction energy scale is usually just as a big (and sometimes even bigger) than the
Fermi energy — and both are immensely bigger than the temperature for usual metals
at room tempertature. We will need to understand some of the properties of these Fermi
systems before we can go on to understand the more exotic physics of electron pairing.

1.3

BCS theory of superconductivity and Majoranas

This is not examinable2

If you have not studied introductory solid state physics, you should do so. I can recommend a goodbook.

Chapter 2

Introduction to Superfluids
Superfluids and superconductors share the unusual property of zero dissipation. For example, we can imagine a toroidal pipe which we can fill with 4 He. If we start 4 He supefluid
flowing around the torus it will continue essentially forever (with caveats that T must be
low enough, the velocity must be small enough, and the system must be big enough).
“Forever” here means an essentially unmeasurable long time.
Similarly with a superconductor. we can take a solid torus (a very thick loop ofwire) and if we start a current flowing around the wire , it will conitnue with no loss
essentially forever (again caveats, the temperature must be low enough, the current mustbe small enough and the the sample must be big enough). Experiments have measured
that the decay time in certain superconducting wires exceeds 105 years. It is probably
much longer, but it becomes very difficult to measure such tiny amounts of decay.
Insert story about crazy Gerrit Flim demonstrating persistant current flow in superconductors.

2.1

Some History and Basics of Superfluid Phenomena

In 1908 in Leiden, Heike Kamerlingh Onnes liquified helium for the first time allowing
access to extremely low temperatures.
On April 8, 1911 Kamerlingh Onnes ”discovered” both supercondctivity and superfluidity.
He noticed that below 4.2K the resistance of mercury (Hg) metal suddenly becomes
unmeasurably small. He spent the next few years convincing himself that this is a realeffect and not some error in the experiment. He also shows that many metals have the same
superconducting behavior — each with its own critical temperature. There was no real
explanation of the observed superconductivity until 1957 (the BCS theory). We will discusssuperconductivity more in later chapters, but this chapter will focus on superfluidity.
3

4

CHAPTER 2. INTRODUCTION TO SUPERFLUIDS

Figure 2.1: The Lambda transition in helium. The superfluid phase is known as He II.
On the very same day as the discovery of superconductivity, he notes in his lab bookthat something different is happening around 2K, but he didn’t follow up on this for many
years.
In 1913 Kamerlingh Onnes wins the Nobel prize.
In 1922-23 Onnes, along with some of his coworkers goes back to re-examine the“something strange” that happens around 2K. Looking at the heat capacity, they findsomething like Fig. 2.1. The discontinuity in the heat capacity is a signal of a thermodynamic phase transition1 The regular phase of helium is known as He I whereas2 thesuperfluid phase is known as He II.
Onnes died in 1926, but the work in his lab continued on. With some exploration,
they uncovered the full phase diagram shown in Fig. 2.2. The line separating the He I andHe II phases is known as the Lambda-line.
Compared to other elements, helium is quite special in that it does not solidify at
low temperature except when under pressure. This special feature is due to both its very
light mass and its weak interactions (no chemical bonding). The light mass is importantsince ∆p∆x > ~/2. If you try to localize x, you have a nonzero p. But since kinetic energy
is p2 /(2m) having a light mass means there is a very high energy cost to localizing theposition of the atoms.
The most remarkable thing about superfluid helium is persistant flow (which Kamerlingh Onnes was never aware of). However, perhaps just as remarkable is the extremely
good transport of heat in superfluids. In fact, going through the supefluid transition, ther1
For those who have studied phase transitions we have Cv ∼ B± + A± |T − Tc |−α where the criticalexponent α ≈ −0.0127 matches that of the XY-universality class and here A± and B± are different
constants depending on whether we approach the critical point from the high or low temperature direction.
When we study the order parameter in section 2.3 it will be clear to those having studied phase transitions
wwhy this is in the XY universality class.
2
This nomenclature is particularly confusing being that we also have 4 He and 3 He, indicating thenumber of nucleons. Everything in this chapter is in reference to phases of regular helium, 4 He, although
we will also discuss 3 He in our chapter below on Fermi liquids.

2.1. SOME HISTORY AND BASICS OF SUPERFLUID PHENOMENA

5

Figure 2.2: The Phase Diagram (P,T) of 4 He. The superfluid phase is known as He II.
The nonsuperfluid phase is known as He I
mal conductance can jump by a factor of 105 or more! Indeed, this was what Kamerlingh
Onnes saw in 1911 when he said “something happens” at 2K). What is happening hereis that the fluid suddenly goes from rapid boiling to no boiling at all in the superfluid
phase. Boiling occurs when a small region of a system is at slightly higher temperaturethan the vaporization transition. This region turns into gas and then boils up to the topof the containier. In a superfluid, the thermal transport is so good that no region in thesample is at higher temperature than any other region. The system still has evaporation,but only directly from the surface.
The discovery of the superfluid properties of He II did not come until the mid 1930s.
The next heros of our story are Peter (Pyotr) Kapitsa and John Allen.
Peter Kaptiza, having survived the flu epidimeic of 1918-19 in Russia which claimed
half of his family, had moved to Cambridge to study with Rutherford. He was made afellow of Trinity College Cambridge in 1925 two years after obtaining his PhD and he went
on to build one of the world’s finest low temperature physics labs at Cambridge. By themid 1930s Kaptiza was earning what was considered a magnificent salary of 800 pounds
per year.
By the early 1930s there were only four labs in the world that were able to liquidify helium and conduct low temperature experiments: Leiden, Oxford3 , Cambridge, andBerlin.
John Allen received his PhD from Toronto in their cryogenics lab, and moved toCambridge to work with Kaptiza. Unfortunately, when he arrived there, he discoveredthat Kapitsa had been “detained” on a visit to the Soviet Union, and was told that he3

Oxford had liquid helium in 1933, one year before Cambridge did. Fredrick Lindemann, the first
Viscount Cherwell, advertised Oxford’s success in the Times and Nature. Nonetheless, the crucial discovery
of superfluid flow was made at Cambridge

6

CHAPTER 2. INTRODUCTION TO SUPERFLUIDS

would not be allowed to return to Cambridge. This was the era of Stalin, and the Soviets
wanted to have the great Kaptiza back in his native land.
Kaptiza was well funded by the Soviets to set up a low temperature physics lab(they purchased much of his equipment from Cambridge).
John Allen, at Cambridge with no supervisor, had to work as an unsupervised
postdoc. Rutherford, the head of the Cavendish lab at Cambridge, made the best of thesituation — he used Kaptiza’s salary to pay both Allen and a young theorist named Rudolf
Peierls.
The cryogenic race was on: it was clear that there was plenty of interesting physics
to be discovered to whoever got there first!
In 1938 both Kaptiza and Allen (with Misener, another student from the Torontolab) simultaneously discovered the effect that gives superfluid helium its name — thatsuperfluid helium flows with apparently no resistance through very thin capillaries.
Another very strange effect, now known as supercreep, was discovered the following
year in Oxford by Sir Francis Simon (no relation that I know of) and Bernard Rollin:
Superfluid helium in an open-topped container will creep up the walls of the containerand eventually find its way to lower gravitational potential by siphoning itself onto thefloor. This effect is often viewed as a combination of wetting and zero viscosity. Manyliquids will form a layer an atom or two thick on the walls of their container all aroundthe container — this is due to the attractive forces between the wall of the container andthe fluid molecules. This phenomenon is known as wetting. Superfluid helium, with noviscosity, can flow readily through a layer which is atomically thin – whereas other fluids
cannot. Thus superfluid helium can siphon itself out of a container onto the floor throughthis very thin layer.
However, the situation is more complicated than simply saying that the fluid has
zero viscocity. Other experiments aiming the measure the viscocity of the liquid come up
with finite results. For example, a vibrating wire, or rotating disk in the helium liquid
will detect a finite viscous damping which only vanishes at zero temperature (Keesom and
MacWood, Leiden Lab, 1938). This connundrum was one of the key points that led to thedevelopment of Landau’s two fluid model.

2.2

Landau and the Two Fluid Model

2.2.1

More History and a bit of Physics

In 1938 just as Kapitsa was discovering superfluidity, the great theorist, Lev Landau wasthrown in prison for publicly comparing Stalin to Hitler4 . He remained imprisoned fora year until Kapitsa put his own head on the line — threatening to quit as a scientist
unless Landau was released, and pleading that only Landau would be able to explain4

He was exceedingly lucky not to have been executed!

2.2. LANDAU AND THE TWO FLUID MODEL

7superfluidity.
While Landau was in prison, the young Hungarian physicist Laszlo Tisza5 working
in Paris with Fritz London tried to explain superfluidity as a result of Bose-Einsteincondensation6 . Fritz and his brother Heinz London, several years earlier working with
Lindemann in Oxford, had already tried to use a similar approach for understanding
superconductors. While they had some success, Landau immediately realized what waswrong with their approach.
 Helium is NOT a noninteracting boson. It has extremely strong short range repulsive
interactions.
 A noninteracting BEC does not have a discontinuity in its specific heat A noninteracting BEC will NOT superflow. Interactions are crucial for obtaining
this effect. (We will explain this in detail below in section 2.4)

At this point Landau gets into the game. He felt that BEC was not the right physics7— perhaps he even felt this more strongly than is actually warranted. Later Feynmann
would clarify the importance of Bose physics for superfluidity (See below in chapter 5).
Landau made two major contributions to the theory of superfluidity.
 He develops the two-fluid model (an expansion on Tisza’s work) He explains the criterion for superflow.

These contributions earned Landau a Nobel prize in 1962.8 . Sadly a few months
before the prize was annouced, Landau was in a car crash and coma. While he came outof the coma, he never fully recovered, he was never scientifically productive again, and hedied in 1968, at age 60, from complications due to these injuries.

2.2.2

Landau’s Two Fluid Model

The general idea of the two fluid model is that we should imagine two interpenetrating
fluids which we call “super” and “normal”9 . We write the total density of the fluid as thesum of the mass densities of the super and normal fractions.
mass/volume = ρtotal = ρS + ρN .
5
Tisza moved to MIT in 1941 and lived in the US the remainder of his life. He passed away in 2009 at
the age of 101.
6
The prediction of BEC by Einstein was much earlier, in 1925.
7
Apparently Landau also really hated Fritz London. It is unclear why this would be true, but Landau
made a point never to cite London.
8
There are many things that Landau could have been awarded a Nobel prize for!
9This is another terrible nomenclature. Above the critical temperature we have only normal fluid. But
below the critical temperature we have both super and normal fluid. However, we also say that He II(below the critical temperature) is superfluid.

8

CHAPTER 2. INTRODUCTION TO SUPERFLUIDS

The normal fraction behaves like we should expect a fluid to behave. The superfluid
fraction, however, has no viscosity, and carries no heat. At zero temperature we expect
all of the fluid to be super, whereas above the critical temperature Tc , i.e., in the He I
phase, we expect all of the fluid to be normal. At nonzero T but still in the He II phase,
we expect to have a mixture of partially super and partially normal fluids.
T =0
Tc > T > 0
T > TcρN = 0ρN (T )
ρN = ρ

ρS = ρρS (T ) = ρtotal − ρN (T )ρS = 0interpolates

One way to think about the super-fraction is to think of it as being the “groundstate” of the system, and the normal fluid being the excitations. For a moment let us
think about the analogy with BEC. In that system we know that only at T = 0 are allof the bosons in the same lowest energy eigenstate. At any finite T there will be somebosons in the lowest energy eigenstate (these we might call “super”) and some will bein the excited eigenstaes (which we might call “normal”). Once we go above the criticaltemperature, there will a microscopic fraction of particles (essentially zero) in the lowest
energy eigenstate, and the entire system will be normal.
The mass current will also be divided into a normal and super component
j = ρS vS + ρN vN
with vN being the normal fluid velocity and vS being the superfluid velocity. Here thesuper part of the fluid will be free from viscosity but the normal part is not.
This two fluid model nicely “explains” the two different measurements of viscosity.
With a vibrating wire or a rotating disk, the object moving through the fluid will haveto push the normal fluid out of the way, and hence will measure the normal viscosity.
However, when trying to push the fluid through a small capillary, no matter how thin thecapillary, the super-fluid component will always flow through – whereas the normal fluid
will stay behind. Since flow occurs through small capillaries one would measure a fluidwith essentially zero viscosity.
From a technical standpoint the two fluid model actually imposes the curl-free constraint∇ × vS = 0(2.1)
We will justify this constraint later; for now let us just treat it as a conjecture. However,
it is worth understanding why this is an appropriate conjecture for a fluid that shows novisosity.
One explanation is to realize that viscosity is a statement transferring force throughshear. However, the curl-free constraint is the statement that the fluid cannot shear atall! (Expand on this! add picture).
Suppose for a moment that ρS is fixed (say we are at zero temperature). Then wehave∇ · vS = 0

2.2. LANDAU AND THE TWO FLUID MODEL

9by current conservation. If we use our conjecture Eq. 2.1 we can use a vector identity toderive∇2 vS = ∇(∇ · vS ) − ∇ × (∇ × vS ) = 0
If we then recall the Navier-Stokes equation we have


∂
1
η
+ (v · ∇) v + ∇P = ∇2 vS∂tρρwhere η is the coefficient of viscocity. By using the curl-free constraint the viscosity termalways vanishes!

2.2.3

More Physical Effects and Their Two Fluid Pictures

Fountain EffectAnother strange effect observed by John Allen10 is known as the “fountain effect”, shownin the left of Fig. 2.3. A tube is inserted in the superfluid and plugged with cotton suchthat only the “super” fraction of the fluid can get through. Heating the cotton, will resultin superfluid being sucked into the tube at a high rate such that it can fountain out the
top. This effect was explained in 1939 by Heinz London11 using the two-fluid model. In
this picture we think of the motion of the fluid as being similar to osmosis. If in someregion (such as in the tube) we increase the temperature from T to T + ∆T , this increases
the concentration of ρN . Using the usual principle of osmosis, this makes ρS flow into thehotter region to try to bring the normal fluid back ot the same concentration as that of
the surrounding area. I.e., the superfluid flows into the hotter region to try to bring thetemperature back down.
This physics may look like a violation of the 2nd law of thermodynamics since fluid
is flowing from cold to hot. However, it is not a violation since the superfluid carries noheat.
The situation can be analyzed in more detail by using the experimental setup shownin the right of Fig. 2.3. Here the tube between the two containers is assumed to be plugged
with cotton, or is such a thin capillary that only superfluid can flow through it (but notnormal fluid). This is sometimes known as a “superleak.”12
We can analyze the situation using simple thermodynamics. In particular, we use
the Gibbs-Duheim relationship for the change in chemical potential which we set to zero10

In fact, the discovery of this effect was a fortuitous error in an experiment!
The London brothers were Jewish refugees from Germany in 1933. Lindeman at Oxford found money
to fund as many German scientists as he could, but his money soon ran out. Fritz moved to Paris andeventually ended up at Duke university in the states. Heinz moved to Bristol, but was declared a potentially”hostile foreigner” and interned for some time on the Isle of Man. He was released to work on the British
nuclear program and was given British citizenship. In the 1950s Heinz invented the dilution refrigerator
which is used heavily in physics research for cooling materials down to as low as a few millikelvin.
12
The superleak is crucial for this experiment since an open tube cannot have a pressure difference across
it!
11

10

CHAPTER 2. INTRODUCTION TO SUPERFLUIDS

Figure 2.3: Left: The fountain effect. A tube is inserted in the superfluid and plugged
with cotton such that only the “super” fraction of the fluid can get through. Heating thecotton, will result in superfluid being sucked into the tube at a high rate such that it canfountain out the top. Right: a similar experiment showing that a temperature difference
implies a pressure difference.

2.2. LANDAU AND THE TWO FLUID MODEL

11

Figure 2.4: Thermomechanical Effect: The propeller turns as if fluid is flowing from hot
to cold, even though there is no net tranport of mass.
in order to have equilibriumdµ = −VS
dT + dP = 0
N
N

Rearranging this we can write
∆P
S
=
∆T
V
and the right hand side can be written in terms of
Z T
S(T ) =0

dT 0

CV (T 0 )T0where CV can be measured experimentally.

Thermomechanical Effect
In this experiment, a propeller is put in a container of helium. One side of the container
is heated as shown in Fig. 2.4. No net current of fluid can flow due to the walls of thecontainer. Nonetheless, the propeller turns as if fluid is flowing from hot to cold.
What is happening here is that the increased temperature (on the left in the figure)increases the density of normal fluid, and the normal fluid then tries to flow to the right.
Similarly, by osmosis the superfluid tries to flow to the left to bring the two sides of thesystem into equilibrium where they have the same density of normal fluid (i.e., have thesame temperature). Thus we have counter flow of normal and super fluid – with no net
flow of mass.
Now the key is that the flow of the normal fluid pushes the propeller while the flowof the superfluid does not. The reason for this is that superfluid flow must not dissipate
mechanical energy — otherwise superflow would not be dissipationless.

12

CHAPTER 2. INTRODUCTION TO SUPERFLUIDS

Figure 2.5: Left: Sound waves are density waves: mass sloshes back and forth. Right:
Second sound waves involves counterflow. There is no mass transport, although heat istransported.

2.2.4

Second Sound

The idea of counter flow of normal and superfluids suggests the phenomenon of second
sound13 .
First let us remind ourselves of how sound waves work. In the left of Fig. 2.5 we see
the density flow in a sound wave — mass sloshes back and forth over the wavelength overthe sound and with the period of the oscillation T . In the right of Fig. 2.5 we see second
sound waves — the supercurrent is always counter-flowing the normal current so there isno net mass transport, jtotal = 0 everywhere. However, heat sloshes back and forth since
only the normal current carries heat.
Such second sound waves provide extremely good thermal conduction — indeed, thebest thermal conductance of any material since heat travels ballistically in second soundrather than diffusively. The heat transport can be over 100 times as efficient as heattransport in, say, copper.
Derivation of second sound
We now turn to do a more formal derivation of second sound. Since we are not interested
in regular sound waves, we will assume that the pressure is constant and the density isconstant (these approximations simplify quite a bit, but are not entirely necessary).
Since we have fixed density and fixed pressure, we will enforce
jtot = 0 = ρN vN + ρS vS
We now assume that there is an osmotic force from temperature imbalance. This isprecisely the force that arises from the fountain effect. Usually we think of this as a force
pushing the superfluid towards the heat, but since the total mass current is fixed to bezero, we can just as well think of this as a force pushing the normal fluid towards lower
13

Second sound was first predicted by Tisza. The correct second sound velocity was predicted by Landau.

2.2. LANDAU AND THE TWO FLUID MODEL

13

temperature regionsv̇N = −α∇T

(2.2)

with α > 0. It is a homework exercise to estimate the value of the coefficient α.
We now write a continuity equation for entropy. Let us write s for the entropy
density. We then have a conservation equationṡ = −∇ · (svN )(2.3)

since the entropy moves with the normal fluid only (the super fraction contains no entropy).
The entropy density is a function of the local temperature, which we can expands(T + δT ) = s(T ) + κδT
where recalling that cV = T (∂s/∂T )|V . we must have
κ = cV (T )/T
Here cV is heat capacity per unit mass.
We can now put this expansion of the entropy into Eq. 2.3 to obtain
κṪ = −s(∇ · vN )Here it may look like there should be another term on the right which would be vN · ∇s.
However, we can throw this out since this is second order in “small” things14 I.e, thegradient of the temperature is assumed small and the velocity is also assumed small.
Differentitating with respect to time we obtains sT̈ = − ∇ · v̇N =α ∇2 Tκκ
where we have again thrown away terms second order in “small”, and we have also used
Eq. 2.2. Thus we obtain the wave equation15
T̈ = γ∇2 Twith γ > 0 which tells us that heat propagates ballistically rather than diffusively as it
usually does.
14
For careful bookkeeping of “small” we start with a system with no wave, for which we have s = s0 ,T = T0 and vN = 0. Then we add perturbations order by order s = s0 + δs1 + 2 δs2 + . . . and vN =
vN 1 + 2 vN 2 + . . . and so forth, then we just keep terms to order .
15
Using the homework assignment where we estimate α as well as some results later about CV and ρN
we can get that the second sound velocity is independent of T at low enough T .

14

CHAPTER 2. INTRODUCTION TO SUPERFLUIDS

2.2.5

Big Questions Remaining

These questions will be addressed in the next sections.
1. What justifies the curl free constaint on the supercurrent ∇ × vS = 0.
2. Why do we get dissipationless superflow? This is NOT only a result of item (1) the
curl free constraint.
3. What do we expect the normal/super fluid density to be as a function of temperatureρN (T ) and ρS (T ) ?

2.3

Curl Free Constraint: Introducing the Superfluid OrderParameter

One of Landau’s other great contributions to physics is the concept of an order parameter.
(If you have studied the Landau theory of phase transitions you will know what this is.)Roughly, the order parameter describes the breaking of a symmetry in a physical system.
As a simpler example, let us reconsider a BEC. In this case, we take the the orderparameter to be the complex wavefunction for the ground state, which we write as ψ,
however, we normalize it so that it goes to zero at the phase transition — i.e., it’s amplitude
is proportional to the number of bosons in the ground state.
What symmetry has been broken when going from the normal state above the criticaltemperature to the superfluid state below the critical temperature? It is actually a gauge
symmetry! When the order parameter ψ becomes nonzero it must choose a complex phase,
which is a spontaneous breaking of a symmetry!
In the case where the bosons are not uniform in space, we generalize the uniform
Landau order parameter to a Landau-Ginzburg order parameter and write it as a functionof position ψ(r).
For interacting bosons we do something quite similar, we define an order parameterasψ(r) = hψ̂(r)i
where ψ̂(r) is a boson annihilation operator. We will be more explicit about this approachin later chapters 4. For now it is acceptable to just think of this order parameter as being
the complex wavefunction of a BEC with the one complication that we want to change
the normalization so thatZ
dr|ψ(r)|2 = N0
rather than normalization to unity. Here N0 is the number of bosons in the condensate16 .
16

It is unfortunately not generally true that N0 /N is the same as the superfluid fraction from the two-2.3. CURL FREE CONSTRAINT: INTRODUCING THE SUPERFLUID ORDER PARAMETER15
We can separate the phase degree of freedom from the magnitude by writing
ψ(r) =√

n0 eiθ(r)(2.4)

where n0 = N0 /N is the condsensate
number density and θ is a phase defined modulo 2π.
√
Here we can think of eiθ / V as being the usual normalization of the wavefunction of aboson in its ground state.
Again thinking of ψ as a wavefunction for superfluid bosons, then we can write thecondensate current in the usual quantum mechanical way
j0 =
=

1 ∗
(ψ p̂ψ − (p̂ψ ∗ )ψ)2−i~ ∗(ψ ∇ψ − (∇ψ ∗ )ψ)
2

(2.5)

where here the momentum operator p̂ gives us a mass current.
Now using Eq. 2.4 we have

∇ψ =


∇n0
+ i∇θ ψ2n0

which gives us a superfluid mass current of
j0 = ~n0 ∇θ

(2.6)

We can write this current as a mass density times a superfluid velocity
j0 = ρ0 vS
which, using ρ0 = mn0 gives us an expression for the superfluid velocity17
vS =

~
∇θ
m

which is a gradient. This then implies the desired curl-free constraint∇ × vS = 0
fluid model in the case of an interacting Bose system. In the two-fluid model ρS is defined in terms of
the amount of mass that flows with no dissipation — which at T = 0 is always the total mass of thefluid. However for interacting bosons the magnitude of the order parameter, which describes the numberof particles in the condensate, will not be the full mass of the fluid even at T = 0. This will be more clear
later when we give more rigorous definitions of the order parameter, and indeeed we calculate N0 for aweakly interacting Bose gas in section 4.5 below.
17
Although as mentioned above in footnote 16 the condensate fraction n0 may not match the superfluid
fraction of the two-fluid model, it is nonetheless true that both pictures give the same expression for the
supefluid velocity.

16

CHAPTER 2. INTRODUCTION TO SUPERFLUIDS

Figure 2.6: Left: A contour in fluid that can be contracted and therefore bounds a disk in
the fluid. In this case the circulation must be zero. Right: A contour in fluid that cannot
be contracted. In this case we can only say that the circulation is quantized in units ofthe elementary circulation quantum h/m.

2.3.1

Vorticity Quantization

Circulation is defined as the line integral of a velocity around a closed contour. Choosing
any given closed contour we haveI
I~
vS · dl =∇θ·dl = 2πp~/m = ph/m
(2.7)
m C
C
where p is an integer. Here we have used the fact that θ must be well defined only modulo
2π so going around a closed path, the value of θ needs to come back to the same value,
but can change by integer multiples of 2π.
Thus we have shown that in a superfluid circulation is quantized in integer multiplesof the circulation quantum h/m.
Suppose now that our contour surrounds only He fluid as shown in the left of Fig. 2.6.
I.e, we have a big sample of He fluid [with nothing else in it but He fluid] and the contour
can be continously contracted to a point. In this case we can use Stoke’s theorem
IZvS · dl =dA · (∇ × vS ) = 0
C=∂D

D

where the area integration is over a disk D bounded by the contour C. Thus we would
conclude that the circulation must be zero. However, this conclusion is only true if superfluid fills the entire disk. If there is a hole within the disk region which has no fluid init (like a bundt cake, or a toroidal pipe) as shown in the right of Fig. 2.6, we cannot use
Stoke’s theorem and therefore we only know that the circulation is quantized in units ofthe elementary circulation quantum h/m.
In fact there is no constraint on the size of the hole — the hole in the fluid can be
extremely small and this still consitutes a hole in the fluid. We only need a single point
where the superfluid density goes to zero such that the phase is no longer defined at thatone point. In this case we can have the phase θ wrap by any integer multiple of 2π as
we go around this single point. This is known as a “vortex”. The existence of vortices in2.3. CURL FREE CONSTRAINT: INTRODUCING THE SUPERFLUID ORDER PARAMETER17

Figure 2.7: If we start a bucket of helium rotating above the critical temperature andthen we cool it below the critical temperature, the circulation is accomodated by the
introduction of many vortex lines.
superfluid helium was predicted by Feynmann in 195518 .
Imagine a bucket of helium above the superfluid transition temperature. We start
the bucket rotating so that the circulation is very large
IvS · dl  0
We now imagine cooling the system through the transition temperature Tc . Since thesuperfluid must be curl free, the only way we can accomodate the circulation is to have
many point vortices, where there is quantized circulation around each such vortex. Lookedat from far away, the velocity flow field is roughly the same in the superfluid phase as it wasin the normal phase. As mentioned above, in the core of the vortex (here one dimensionallines) the superfluid density drops to zero. In actual superfluids He, the vortex core is
only a few angstrom in diameter.
One often hears arguments that quantized circulation explains the persistant flow
phenomenon observed in superfluids. Indeed, one can imagine starting a superfluid flowing
around a toroidal pipe. Since the circulation is quantized, it can only decay in discrete
steps. If there is some impediment, or activation barrier, to making sudden steps incirculation, one might imagine that this prevents the circulation from decaying at all.
While it is true that the circulation can be locked to a single value, this is actually
not sufficient to explain dissipationless flow. The issue is not whether the circulation
decays but whether energy is dissipated. Even without having a decay in the circulation(which is the line integral of the velocity) one can dissipate energy if some sort of friction
heats the fluid. This would slowly reduce the superfluid fraction and increase the normal
fraction, thus reducing the superfluid current, and eventually the system would have nosuperfluid left! So the question boils down to whether or not a particular superfluid will
dissipate energy. In fact in noninteracting BECs, we will see below that, while circulationis quantized, they do dissipate energy to friction and therefore do not have persistent flow.
18

As we will discuss below, as similar prediction was made for superconductors somewhat earlier by
Abrokosov, but he did not publish it until after Feymann convinced the world that the idea was reasonable.

18

CHAPTER 2. INTRODUCTION TO SUPERFLUIDS

Figure 2.8: Left: Flowing a fluid past a wall. Right: In the fluid frame, the wall is moving.

2.4

Landau Criterion for Superflow

Landau developed a brilliant argument to determine whether a fluid can flow without
dissipating energy.
Let us assume that for a superfluid (and normal fluid) at rest there is some spectrum (p) for excitations which we will call “quasiparticles”. In a BEC these are simply
excitations of particles out of the ground state wavefunction. In an interacting superfluid,these are more complex collective excitations. We need only assume that such excitations
exist.
We now imagine flowing the superfluid past a stationary wall at velocity v as in theleft of Fig. 2.8, and we would like to ask whether energy can be dissipated. Another way
to ask this is to ask whether quasiparticles can be excited.
In order to answer this, the argument entirely boils down to figuring out the energy
momentum relation in the rest frame of the superfluid and then asking whether energy andmomentum can be conserved in a process that creates a quasiparticle excitation19 . Thus,
let us switch to the fluid rest frame as in the right of Fig. 2.8. We know the dispersion
f luid (p) = (p) of excitations in this (the fluid) frame, and we would like to determine
the dispersion relation in the lab frame. A classical Galilean transformation gives us
lab (p) = f luid (p) − p · v(2.8)

Note that here the quantum number p describes the momentum of the quasiparticle in
the fluid frame.
To justify this Galilean transformation (Eq. 2.8) we should recall that in quantum
mechanics the phase of the wavefunction oscillates asψ ∼ e−it/~
19
We will often think of superconductors as being superfluids of charged bosons. Using this argumentof switching to the rest frame of fluid is dangerous. With a fluid of Helium, you actually can imagine
making a Galilean tranform and having your rest frame moving along with the fluid. However, with asuperconductor, the charged bosons are flowing past the lattice of ions of the positive ions in the material— i.e., there is a natural rest frame which is the rest frame of the solid material where the electrons live!

2.4. LANDAU CRITERION FOR SUPERFLOW

19

Figure 2.9: A spectrum (p) and the corresponding critical velocity for the superfluid.
So here we are claiming that a wavefunction for a particle with momentum p should
oscillate as
ψ(p) ∼ eip·x/~−it/~
If we transform into a moving frame we then haveψ(p) → eip·(x+vt)/~−it/~=

eip·x/~−i(−p·v)t/~

so we rederive Eq. 2.8.
A more familiar way of understanding this transformation is to think about frequencies ω = /~ instead of energies. We then recognize Eq. 2.8 as being the usual Doppler
shift of a frequency in a moving frame20 .
So given Eq. 2.8 our question is whether in the lab frame, the energy to create
an excitation is ever negative. If so, excitations are created spontaniously and energy is
dissipated. So when does this happen? Obviously  − v · p is minimized when v and p
are parallel. Thus we can get negative values of lab only if
(p)
<v
p
Given a dispersion (p) (in the rest frame), there is thus a critical velocity
vcrit = min
p

(p)
p

(see figure 2.9). Below this critical velocity, there is no way to create a quasiparticle
while conserving energy and momentum. If the fluid flows at velocity greater than thecritical velocity, quasiparticles are spontaneously generated and energy is dissipated fromthe superflow.
20This argument is strictly correct for situations where the dispersion is linear — i.e., we have soundwaves or phonons.

20

CHAPTER 2. INTRODUCTION TO SUPERFLUIDS

Note now that if one considers a BEC, the spectrum of excitations is simply p2 /(2m)i.e., just the spectrum of noninteracting particles. The critical velocity is thenvcrit;BEC = minpp2 /(2m)=0p(!)

We thus conclude (as Landau realized intuitively!) that a noninteracting BEC does notsuperflow!
When we add interacting between the bosons, the spectrum develops an acoustic
wave (we will see this in more detail later!). In that case we have a low energy spectrum
(p) = vsound p + . . .
which then gives us a critical velocity
vcrit = vsoundassuming the correction hidden in the “+ . . .” is positive. If it is a negative correction (asshown in Fig. 2.9) the critical velocity can be somewhat lower.
To be a bit more detailed about interacting Bose systems, we recall what we havelearned in other courses about the Bogoliubov approach to weakly interacting Bose gases.
In fact we will review this technique in section 4.5 below. We derived (and will derive
again later in several ways), that for a weakly interacting Bose gas with interactionUXδ(ri − rj )2
i6=jwith U being weak, the excitation spectrum is given byrp2+ nU )2 − (nU )2
(p) = (2m
where n is the density of bosons. For small p this can be expanded asrpp2
(p) ∼ √
+ 2nU
2m 2m
Thus we obtain a critical velocity
(p)
1
min
= min √
p
p
p
2m

r

p2
+ 2nU =
2mrnU
= vacoustic = vcritmNote: about real experiments
While the Landau theory of critical velocity is quite beautiful and compelling, experiments
do not often agree with the Landau estimate. In fact measured critical flow velocities can2.5. SUPERFLUID DENSITY

21

often be as much as 100 times smaller than the acoustic velocity — and furthermore thecritical velocity often depends on the details of the geometry of the experiment.
In thin tubes of superfluid it is easy to qualitatively explain the physics of the criticalvelocity. This has to do with the physics of vortices and the quantization of circulation. As
we mentioned above if we have a toroidal (but thin) tube of fluid, with a fixed circulation
going around the tube, this circulation is quantized. This circulation can decay if a vortex
is nucleated at one edge of the tube, flows perpendicular to the supefluid flow and thendisintegrates at the other edge of the tube. This process can reduce the circulation indiscrete steps.
Such a process begins to look much less likely in thick tubes of fluid — however,experiments seem to suggest that vortex physics is still involved.

2.5Superfluid Density

Finally we turn to Landau’s prediction of the superfluid density as a function of temperature ρS (T ).

2.5.1

The Andronikoshvili ExperimentLandau made a prediction for the superfluid density ρS (T ) then convinced a young experimentalist, Elephter Andronikoshvili21 , to make careful measurements of the superfluid
density in 1946, and the results supported his predictions (perhaps unsurprisingly since
Landau was a genius!).
The experiment is shown in Fig. 2.10. A stack of closely spaced disks is hung ina container of helium by a thin wire which acts as a torsion oscillator. The idea is thatwhen the stack of disks rotates, normal fluid, which is viscous will get stuck between thedisks and must rotate with the stack. However superfluid, which has no viscosity slips
through the closely spaced disks and does not rotate. The normal fluid thus contributes
to the total moment of inertia of the stack, and hence changes the oscillation frequency of
the torsion oscillator. By measuring the change in the oscillation frequency as a functionof temperature, one can determine the fraction of helium that is superfluid as a functionof temperature.
Andronikoshvili clearly measured that the normal fluid density is proportional to T 4
at low temperature.
ρN (T ) ∼ T 4
We should compare this result to a simple calculaton for a noninteracting BEC. We
21
In my opinion, this is the coolest name of all physicists. The ending ”vili” is Georgian meaning“descendant of”. Thus, Elephter was part of a royal Georgian family that traces its ancestery back to
Andronikos I of the Eastern Roman Empire in the 1100’s. Andronikos seemed quite violent and fittingly
met a violent end himself.

22

CHAPTER 2. INTRODUCTION TO SUPERFLUIDS

Figure 2.10: The Andronikoshvili experiment. Since normal fluid sticks between the closely
spaced disks (but superfluid does not), the moment of inertia of the stack of disks, andhence frequency of oscillation of the disks tells us the fraction of fluid that is superfluid.
can write the mass of particles that are normal by simply counting the number of particles
that are not in the condensate (i.e., not in p = 0). We thus have
Z
mX
d3 p
1
(2.9)
ρN (T ) =nB (β(p)) = m
V(2π~)3 eβp2 /(2m) − 1
p6=0
where β = 1/(kB T ) as usual, and we have used thatnB (x) =1
ex − 1is the Bose factor and for the noninteracting Bose particles we have the dispersion (p) =p2 /(2m).
We can evaluate Eq. 2.9 by defining q = β 1/2 p giving us
Z
d3 q
1
3/2
ρN (T ) = m(kB T )∼ T 3/2
2 /(2m)3q(2π~) e−1(2.10)

which differs markedly from the experimental result!

2.5.2

Landau’s Calculation of Superfluid Density

Again this calculation relies on thinking about superfluids in both the lab and the moving
superfluid frame as in Fig. 2.8. Here, however, we realize that the normal fluid is draggedby the wall and will have the same velocity as the wall, whereas the superfluid moves
seperately (the fluid velocities drawn in Fig. 2.8 are for the super part of the fluid only).
Thus in the frame where the wall is moving but the superfluid is still, the normal
fluid current (which is the total fluid current) is given byjtotal = jN = ρN vN(2.11)

2.5. SUPERFLUID DENSITY

23

and we would like to calculate ρN .
As before, in the lab frame we havelab−f rame = (p) − v · p
where (p) describes the excitation spectrum when the superfluid is in its rest frame. As
mentioned in the above section where we calculated critical velocity, if lab−f rame becomes
negative, then we get spontaneous generation of excitations and we get dissipation.
However, even if lab−f rame is positive, it can still be excited thermally. We expectthat the density of such particles will be given bynB (β((p) − v · p))
and this gives the occupation probability of the state labeled p in the superfluid rest frame.
So, in the superfluid rest frame we have the momentum (i.e., the mass current) of
the excitations (i.e., of the normal fluid) being given by
Z
d3 p
jN =
p nB (β((p) − v · p))(2π~)3
We can then expand for small v to obtainZ
d3 p 0pn(β(p))
−
pβ(v
·
p)n
(β(p))
jN =B
B(2π~)3
The first term in the brackets vanishes by symmetry. To evaluate the second term, we canassume the velocity v is in the x direction. We then haveZ
d3 p 2  0
jN,x = vx βp−n(β(p))xB(2π~)3
Referring back to Eq. 2.11 and realizing that the wall velocity v is the normal fluid velocity
we obtain the result thatZd3 p 2  0
ρN = βpx −nB (β(p))(2.12)
3(2π~)

Case 1: Noninteracting BEC
Let us try plugging in the dispersion (p) = p2 /(2m) for a noninteracting BEC. We
can scale out the temperature by definingq = β 1/2 p
to obtainρN = β(β −5/2 )Zd3 q 2  0 2
q
−n
(q
/2m)
∼ T 3/2Bx(2π~)3(2.13)

24

CHAPTER 2. INTRODUCTION TO SUPERFLUIDS

which agrees with our prior calculation of the normal fluid density of a BEC in Eq. 2.10.
(With some analytic work it can be shown that the two expressions in fact have the same
prefactor too!).
Case 2: Interacting BEC / Superfluid
For an interacting BEC or superfluid, we expect a low energy acoustic mode so that
(p) = c|p| + . . .
Again using Eq. 2.12 we can here instead define q = βp so thatZ
d3 q 2  0−5qx −nB (c|q|) ∼ T 4
ρN = ββ3(2π~)(2.14)

in agreement with the results of Andronikoshvili! (Score one more for Landau!)

Heat capacity and a caution
It is correct to write the total internal energy of the Bose system at finite temperature as
Z
d3 p
U=
(p)nB (β(p))(2π~)3
Indeed, for a linear (phonon) spectrum this reproduces the usual T 3 Debye heat capacityof phonons at low temperature.
One might wonder whether we could calculate the normal fluid density by justwriting
Z
d3 p
ρ̃N ∼
nB (β(p))(2π~)3
While this redefinition of the normal fluid density worked in Eq. 2.10 for  = p2 /(2m)(giving the same result as Eq. 2.13) it generally is not correct! Indeed, for  = c|p| it fails
to give the same result as Eq. 2.14.
The reason is simply that the quantity ρN is defined only by Eq. 2.11. One could
also define the different quantity ρ̃N but this is in fact not a measurable quantity, whereasρN is.

Chapter 3

Charged Superfluid ≈Superconductor
To a large extent the phenomenology of a superconductor can be understood as just beinga charged superfluid. This approach (roughly) is known as London theory, invented byHeinz and Fritz London when they were in Oxford in 1935. Historically this was before
the discovery of the phenomenon of superfluidity! Indeed, much of the two fluid model
was pioneered first in the context of superconductivity.

3.1

London Theory

As with the picture of superfluid Helium, we postulate both a normal and superfluid partof the charged fluid in a superconductor. We will ignore the normal component for now
and focus on the motion of the super part — which flows with no dissipation.
A dissipationless fluid will display free acceleration when a force F is applied. We
can thus write
∂v
F
−e∗ E
= ∗ =
∂t
m
m∗
This is essentially Drude theory for a fluid with no scattering. We have written here acharge e∗ and a mass m∗ of the charge carrier1 . Note that here only the ratio e∗ /m∗enters. This equation does not know whether the charge carrier is a single electron or apair of electrons or quartets of electrons. However, e∗ /m∗ = e/m in all cases.
The supercurrent is then written as
jsuper = −e∗ n∗S v
In a crystalline solid with band structure we should use the effective mass m∗ of the band at the Fermi
surface. This quantity may also be anisotropic which then gives a directional anisotropy to the London
equations.
1

25

26

CHAPTER 3. CHARGED SUPERFLUID ≈ SUPERCONDUCTOR

where n∗S is the superfluid number density of the charge carriers. We emphasize that
this quantity is the charge density current rather than the mass density current we discussed when we were discussing Helium superfluid above. Note again that the combination−e∗ n∗S = −enS does not know how big the individual particles are. It could be density nSof particles of charge −e or could be density nS /2 of particles of charge 2e and so forth.
The point here is that at level of this London theory, one has no way of knowing that theelectrons might pair up into boson clusters.
Putting the last two equations together we derive the so-called First London Equation(e∗ )2 n∗S∂jsuper=E
1st London Eq.
(3.1)
∂t
m∗
Again, the charge cluster size here does not matter. Note that this equation is reallynothing more than Drude theory with no scattering, and an assumption of a net charge
density −enS .

We can then take the curl of the first London equation and use Faraday’s law toreplace ∇ × E = −∂B/∂t to get(e∗ )2 n∗S∂jsuper−e2 nS ∂B
=
∇
×
E
=
∂t
m∗
m ∂t
R
Then integrating both sides dt starting at some initial time we get∇×∇ × jsuper =−e∗ 2 n∗S
B + C(r)m∗(3.2)

where C(r) is independent of time.
We now need to try to figure out what this integration constant C is. To do so, we
imagine starting with a superconductor in B = 0 in its ground state so that jsuper = 0. In
this situation we can take C(r) = 0 everywhere, at least for this experimental protocol.
Once we have turned on B we want to consider the possible steady-state solutions.
First, we have Ampere’s law∇ × B = µ0 jsuper(3.3)
Here we are ignoring the normal part of any current, since we are only concerned withcurrent that is persistant. Further, since we are concerned with steady state we will ignore
the contribution of ∂E/∂t. Taking the curl of Amperes’ law we get Eq. 3.4. In going from
Eq. 3.4 to 3.5 we use a vector identity on the left and Eq. 3.2 on the right. Finally ingoing to Eq. 3.6 we set C = 0 as suggested above on the right and we use the Maxwell’sequation (no monopoles) on the left.
∇ × (∇ × B) = µ0 (∇ × jsuper ) ∗2 ∗
−e nS
2∇(∇ · B) − ∇ B = µ0
B + C(r)m∗∇2 B =µ0 e∗ 2 n∗S
B(r)
m∗

(3.4)
(3.5)
(3.6)

3.1. LONDON THEORY

27

Figure 3.1: Depiction of magentic field decaying exponentially.

This last equation has solutions of the form (for example)

B(x) = B0 e±x/λ(3.7)

where λ is known as the penetration depth, and is given by


λ=

m∗
µ0 n∗S e∗ 2

1/2
(3.8)

Once again we emphasize that this quantity does not know the size of the charged particle,
but only knows the charge to mass ratio e/m and the total superfluid charge density enS .
For most superconductors the penetration depth at zero temperature is somewhere
between tens and hundreds of nanometers. Note, however, that as T gets very close to thecritical temperature from below the superfluid density n∗S gets very small (and eventuallyvanishes at Tc as the superfluid fraction goes to zero) so that λ diverges at Tc .
The exponential decay we derived in Eqs. 3.6 and 3.7 is depicted in Fig. 3.1. The
magnetic field outside of the superconductor is a constant, but going into the superconductor it decays expontially with a decay length given by λ.
The reason the magnetic field decays going into the superconductor is that the magnetic field is screened by persistant superfluid electric currents. This is similar to Lenz’s
law: We start with a superconductor in no magnetic field, then we turn on a magentic
field and currents flow in the superconductor to prevent the change of the magnetic field
and these currents persist indefinitely since there is no resistance to current flow.
However, as we will see next, it turns out that the screening of magnetic field in asuperconductor is fundamentally more than just Lenz’s law!

28

CHAPTER 3. CHARGED SUPERFLUID ≈ SUPERCONDUCTOR

Figure 3.2: The Meissner Effect

3.1.1

Meissner-Ochsenfeld Effect

The Meissner-Ochsenfeld Effect (or just “Meissner Effect”)2 demonstrates that the screening of magnetic field from superconductors is not just Lenz’s laws.
To demonstrate the effect, see figure 3.2. On the left we have a material in a magnetic
field above its superconducting critical temperture. Then we cool the system down until
it becomes a superconductor. While Lenz’s law would predict that currents would flow
so as to prevent any changes in the magnetic field, what is observed instead is that themagnetic field is completely expelled from the superconductor! So it is not just that thesuperconductor is a perfect conductor: the superconductor is a perfect diamagnet! The
ground state of the total system of superconductor and magnetic field has the magnetic
field expelled from the superconductor. Thus the Meissner effect is thermodynamical not
dynamical (i.e., Lenz’s laws cares about the dynamics of how you get to a situation, thethermodynamic ground state does not).
To explain the Meissner effect the Londons postulated that the integration constantC(r) in Eq. 3.2 is always zero, independent of the initial conditions. Thus we instead write∇ × jsuper =−e∗ 2 n∗S
Bm∗

2nd London Equation(3.9)

which is known as the Second London equation. We should compare this equation to thecurl free condition on superfluid helium, Eq. 2.1. This would be equivalent if we set the2Walther Meissner had established in 1922 one of the world’s only helium liquifiers in Berlin. In 1933,
Meissner’s student, Robert Oschsenfeld discovered that the magnetic field outside fo the superconductorchanged when the superconductor is cooled through the critical tempertaure. Meissner got excited by thisresult and took over the experiment to the large extent.

3.1. LONDON THEORY

29

charge on the superfluid to zero (set e∗ = 0) and fix the superfluid density to a constant. .
London Gauge
Both London equations can be summarized with a single equation in terms of the vector
potential by writing−e∗ 2 n∗S
A
(3.10)
jsuper =m∗if we take the curl of this equation, using ∇ × A = B we immediately recover the second
London equation Eq. 3.9, whereas if we take the time derivative of this equation, usingȦ = −E we obtain the first London equation Eq. 3.1. However we note that in order for
Eq. 3.10 to make sense we have to work in so-called “London gauge” (or Coulomb Gauge)
where
∇·A=0
This guarantees that ∇ · jsuper = 0 which is what we need for a steady state current flow.
Further, we must also choose the gauge
A0 = 0so that we have E = −Ȧ (which we already used) rather than E = −Ȧ − ∇A0 which wewould have otherwise.

3.1.2

Quantum Input and Superfluid Order ParameterAt this point we are going to rederive the London equations using input from quantum
physics. This particular approach is beyond what the London brothers knew. We havea bit of advantage over the Londons’ in that we have most of a century of experience
with quantum physics, whereas in the 1930s when they were first introducing these ideas
quantum mechanics was quite new3 .
Again let us we write an “order parameter” for the superconductor, which is basicallya wavefunction for the superfluid (compare to the discussion of section 2.3). Given such a
wavefunction ψ, we can write the condensate current in the usual way we write a current
in terms of a wavefunctionj=e∗[ψ ∗ (p − e∗ A)ψ − [(p + e∗ A)ψ ∗ ]ψ]2m∗

A few things to note about this equation. First, it is obviously analogous to Eq. 2.5. We
have included a prefactor of e∗ /m∗ because here we are interested in charge current rather
than mass current. (Note, as we have mentioned a number of times, this ratio does notknow how big the charge carrier is, it just knows the charge-to-mass ratio). Secondly we3

We have another advantage over the Londons — we are alive and they are dead.

30

CHAPTER 3. CHARGED SUPERFLUID ≈ SUPERCONDUCTOR

have minimally coupled the momentum to the vector potential in the usual way charged
particles couple to a vector potential.
As in section 2.3 we rewrite the order parameter in terms of a magnitude and aphase4p
ψ(r) = n∗0 eiθ(r)where here n∗S is the density of charge carriers in the condensate, so that the total charge
density associated with the condensate is5ρ0 = n∗0 e∗ .
Using p = −i~∇ and plugging in this form of ψ into our expresion for the superfluid
current we obtaine∗e∗ 2 n∗0j0 = ∗ n∗0 ~∇θ −Am
m∗
The first term we identify as being analogous to Eq. 2.6.
We now insert a bit of physical intuition —in the ground state we expect the phaseθ of the wavefunction will be a constant. We will prove this later, but for now let us justassume it is true. In this case we immediately recover the London equations in the vector
potential form6 as in Eq. 3.10−e∗ 2 n∗0
j0 =A
(3.11)m∗

3.1.3

Superconducting Vortices

In the case of superfluid helium, we derived vorticity quantization in section 2.3.1 above.
Here we would like to do something similar.
Let us consider a thick solid torus of the superconducting substance as shown in
Fig. 3.3. We then consider the superconducting current cirulation around the handle ofthe torus
IIIe∗ n∗0 ~e∗ 2 n∗0
jsuper · dl =∇θ · dl −A · dl(3.12)m∗ C
m∗ C
C
For the first term on the right, just like in Eq. 2.7, the integral must be 2π times aninteger. In the second term on the right, we use Stokes’ theorem to give us
I
Z
Z
A· =
dS · (∇ × A) =dS · B = Φ
C=∂D

D

D

4
As in the case of superfluid Helium we have the subtlety that strictly speaking the condensate density
and the superfluid density do not necessarily match although we think of them as being similar objects.
See footnotefoot:N0rho.
5
In the previous chapter we used the symbol ρ0 as total mass density. I hope this does not cause
confusion!
6
We have gotten something that seems to make sense only in London gauge. If we choose another gauge
we must then not have ∇2 θ 6= 0 in order to preserve current conservation. This then destroys our intuitionof thinking of θ as being similar to the phase in the Helium supefluid where ∇2 θ = 0.

3.1. LONDON THEORY

31

Figure 3.3: A superconducting torus. We are considering the path C as drawn.
where here Φ is the magnetic flux enclosed by the path C. Thus Eq. 3.12 can be rewrittenasIe∗ 2 n∗0e∗ n∗0 ~2πp−Φ
(3.13)
jsuper · dl =m∗m∗Cwhere p is an integer. Rearranging this we haveIm∗h
Φ + ∗2 ∗jsuper · dl = ∗ pee n0 C(3.14)

The combination on the left hand side is often called the fluxoid since it is some modification to the flux. And it is this combination which is quantized. Note also the analogueof this equation to our equation for circulation quantization Eq. 2.7.
If we choose our path C to be deep inside a superconductor, we generally expect thecurrent should be zero (any currents should run mostly on the surface, given the Meissner
effect which expells any magnetic field — which currents would produce) in which case we
obtain flux quantized in units of a fundamental flux quantum h/e∗ .
Φ=

h
p
e∗

(3.15)

Note that here it is now crucial that we know how big the charge carrying cluster e∗ is!
In fact we now know that the “boson” which forms the superconducting superfluid is acharge 2e cluster, so we have the elementary superconducting flux quantum given by7
Φ0 =
7

h
= 2.067 . . . × 10−15 Wb
2e

Quantization of flux through superconducting rings was first observed experimentally in 1961 by Deaver
and Fairbank in Stanford and by Doll and Naubauer in Germany. Critical theory support was provided
by Nina Byers and C. N. Yang. Yang had recently won a Nobel prize (in 1957) for his prediction ofparity breaking in weak nuclear decays. Nina Byers was a fellow of Somerville college Oxford but remainedin superposition between Oxford and California for a decade. She eventually converged in California.
However, late in life she told me personally that she really had always regretted leaving Oxford. She
passed away in 2014.

CHAPTER 3. CHARGED SUPERFLUID ≈ SUPERCONDUCTOR

32

As with a superfluid, we can take the size of the hole down to zero size. This is known
as a superconducting vortex. As with a superfluid, the condensate particle density n0 mustvanish at the center of the vortex and the phase twists around this point8 . However unlike
superfluid helium the rotation of the fluid is screened by the magnetic field. (Compare
the 2nd London Equation, Eq. 3.9 to Eq. 2.1). Thus, the magnetic field penetrates thesuperconductor within a distance λ of the vortex core and screens the rotation of the fluid.
Outside of this region no current flows, and there is no magnetic field.
At whatever distance we draw our path C, we will always have quantized fluxoid.
If the path is far away from the vortex core, then there will be no current, and the flux
enclosed is quantized. If the path is close to the vortex core, there is very little flux
enclosed (let us approximate it as zero) so that the circulation is quantized.

3.1.4

Type I and Type II superconductors

It turns out that there are two types of superconductors, creatively called Type I andType II. In Type I, magnetic field does not penentrate the superconductor at all – it iscompletely expelled. This is known as the Meissner phase. Another way of thinking of thisis that it is a perfect diamagnet9 with susceptibility χ = −1, the induced magnetization
is exactly enough to precisely cancel the applied magentic field.
If enough magnetic field is applied, the superconductivity is destroyed. At temperature T , the field at which the superconductivity is destroyed is known as Hc (T ). Thisquantity goes to zero at the zero-field critical temperature Tc . (See left of Fig. 3.4) Notethat although individual vortices are not stable in type I superconductors, one must still
have a quantized magnetic flux through any hole in the superconductor.
The situation is different for type II superconductors. For small enough magnetic
fields (below Hc1 (T ), see middle of Fig. 3.4), the magnetic field is completely expelled fromthe superconductor giving a Meissner phase, similar to type I superconductors. However,for stronger magnetic fields Hc1 (T ) < H < Hc2 (T ) the magnetic field penetrates thesuperconductor in individual vortex lines. This is known as the vortex phase or Abrikosov
phase. (See right of Fig. 3.4). At strong enough fields, above Hc2 (T ), the superconductivity
is completley destroyed.

8

The prediction of superconducting vortices was first made by Alexei Abrikosov of the Landau School
in 1953. However, Landau didn’t believe the result and prevented Abrikosov from publishing (Landau wasa genius, but he was not always right!). In 1955 Feynman predicted vortices in superfluid Helium andeventually Landau was convinced of the result. Abrikosov’s work was published in 1957. He won a Nobel
prize for this work in 2003.
9
To remind you M = χH defines the susceptibility.

3.1. LONDON THEORY

33

Figure 3.4: The phase diagram of Type I and Type II superconductors. Left: Type I
superconductor. Magnetic field is always completely expelled whenever the material issuperconducting (i.e, for fields lower than Hc (T ). Middle: Type II superconductor. For
fields less than Hc1 (T ) the magnetic field is completely expelled. However for fields between
Hc1 (T ) and Hc2 (T ), the material is superconducting, and the magnetic field penetrates
the material in discrete vortex lines. Right: A depiction of vortices penetrating a sample.
Vortex Pinning
Note that in the vortex phase, if vortices move, then the flux moves with it. This thengenerates a voltage via the Faraday effect
∇×E=−

∂B
∂t .

Thus, in the vortex phase, voltage (and hence resistance) is zero only if the vortices are
not allowed to move.
The way one arranges for vortices to not move is to have them stick to pieces ofdisorder. Indeed, there is good reason that vortices stick to disorder. If some disorderkills superconductivity locally, a vortex that sits at this position costs no energy locally,
because there is no superconductivity to kill. Hence vortices will try to stick to regions
where superconductivity is already killed.
This phenomenon of vortex pinning being necessary in order to have a true zero
resistance state is very similar to our experience with ferromagnets, where domain walls
must be pinned by disorder in order to have a true ferromagnet that retains magnetizationeven in zero field.

3.1.5

How big is Hc

The size of the various critical fields depends on the particular superconductor. Some
superconductors have critical fields in the milli-Tesla range, and others have critical fields

34

CHAPTER 3. CHARGED SUPERFLUID ≈ SUPERCONDUCTOR

that are tens of Telsa. Typically the more “robust” superconductors — meaning thosewith higher critical temperature — also have higher critical fields. Both temperature andmagnetic field are perturbations that destroy superconductivity, and it is typically thecase that a material that is stongly resistant to one is also strongly resistant to the other.
We can understand the critical fields with more detailed thermodynamics. We startby writing a Gibbs free energy, and consider changing the applied magnetic field H, wehave
dG = −µ0 M · V dH
where V is the volume of the system, and M is the magnetization (which is magnetic
moment per unit volume). For simplicity let us assume that both H and M are aligned
along the z-axis. We then integrate to obtain
Z H
G(H, T ) − G(H = 0, T ) = −µ0 V
dH 0 M (H 0 )(3.16)
0

The magnetization is typically given in terms of a magnetic susceptibility
M = χH 0
Let us start by focusing on type-I superconductors. Type-I superconductors completely
expell the magnetic field, meaning χ = −1 as discussed above. Thus we can substituteinto Eq. 3.16 and integrate to obtain
Gsuper (H, T ) − Gsuper (H = 0, T ) = µ0 V H 2 /2

(3.17)

whereas for the normal state of a material, typically χ  1 and we have instead
Gnormal (H, T ) − Gnormal (H = 0, T ) ≈ 0
Now at the transition into the normal state (Hc ) the free energy of the normal state
and the free energy of the superconducting state must be the same (hence the transition),so we have
Gsuper (Hc , T ) = Gnormal (Hc , T ) ≈ Gnormal (0, T )Now substituting into Eq. 3.17 at H = Hc we obtain
Gnormal (H = 0, T ) − Gsuper (H = 0, T ) = µ0 V Hc2 /2(3.18)

This quantity is known as the condensation energy, and it is the amount of energy savedby allowing the material to superconduct rather than remain in the normal state.
On the other hand we can consider the free energy of type-II superconductors
in a magnetic field. Here, the magnetic field penentrates the superconductor, but guided
into flux vortices as shown in the right of Fig. 3.4. We should thus expect that the freeenergy is given by
Gsuper (H) ≈ Gsuper (H = 0) + N L

3.1. LONDON THEORY

35

Figure 3.5: Gibbs free energy for the Meissner phase versus the vortex phase. For H < Hc1
the Meissner phase is the ground state, whereas for H > Hc1 the vortex phase is lower
energy.
where N is the number of vortices, L is the length of the vortex (i.e., the thickness of thesample) and  is some energy per unit length.
Since each vortex accounts for Φ0 worth of flux (one flux quantum), we thus have
N = number of vortices =Area Hµ0
Φ0

So that we can write
Gsuper (H) = Gsuper (H = 0) +V Hµ0Φ0which is linear in magnetic field, compared to Eq. 3.17 for the type I superconductor
where the free energy difference is quadratic in magnetic field. We thus obtain a picture
like that shown in Fig. 3.5. For applied magnetic fields less than Hc1 ≈ 2/Φ0 the complete
expulsion of the magnetic field (i.e., the Meissner phase) is energetically favorable. On theother hand, if H > Hc1 the vortex phase is lower energy.
Now the actual value of Hc1 depends on , the energy per unit length of the vortex
line. If  is small, then vortex lines are not energetically costly and correspondingly Hc1will be small. If on the other hand,  is large, then Hc1 will be large. In some cases Hc1 canbe greater than Hc , meaning that the energy per unit length of a vortex is so large thatat the magnetic field necessary to force a vortex to penetrate the superconductor, the supercondcutivity is already destroyed. This is precisely the case of a type-I superconductor
where there is no vortex phase.

36

CHAPTER 3. CHARGED SUPERFLUID ≈ SUPERCONDUCTOR

Chapter 4

Microscopic Theory of Bosons
We now try to derive some of the above discussed results on both bosonic superfluids andsuperconductors using a microscopic quantum mechanical picture.

4.1

Mathematical Preliminaries

We begin with some mathematical preliminaries on using second quantized operators.
While many people will have seen this before, it is worth repeating because it is easy to
get confused when it comes to some of the more complictated basis transformations andso forth1 .

4.1.1

Second quantization

We start with standard ladder operators for a harmonic oscillator[a, a† ] = 1.
The nth mode of the harmonic oscillator, written in normalized form is(a† )n|ni = √ |0in!
where |0i is the ground state. The states |ni with n > 0 form an orthonormal set.
The magic of second quantization is that it is entirely equivalent to saythat |ni represents n bosons in an orbital!
We can write the number operator, which counts the number of bosons in this orbitalasn̂ = a† a = number operator
1

I always get confused as to where the complex conjugations go.

37

38

CHAPTER 4. MICROSCOPIC THEORY OF BOSONS

4.1.2

Coherent States

We define a coherent state asαa†|αic = e
α2 (a† )2†|0i =
1 + αa ++ . . . |0i
2


α2
=
|0i + α|1i + √ |2i + . . .
2!

(4.1)

A few comments are in order here. First, we have put a subscript c on the ket |αic toindicate that it is a coherent state. If no confusion will result, we may drop this subscript.
Secondly, note that the coherent state |αic is not normalized. In fact it is easy to show
that2
|α|2
6= 1
c hα|αic = e
Similarly the coherent states are not orthogonal0α∗ α0c hα|α ic = e

Finally it√is worth noting that if we examine the prefactors in Eq. 4.1 which are of the√form αn / n! we will discover that this combination peaks at roughly n ≈ |α|.
The point of using coherent states is that it allows us to turn creation and annihilation operators into numbers as we will now show.
Recall thata|ni =√

n|n − 1i

Applying this to Eq. 4.1 we obtain


α3 √
α2 √
2|1i + √3|2i + . . .
a|αic =
0 + α|0i + √2!
3!
= α|αic
Thus the coherent state |αic is an eigenstate of the annihilation operator with eigenvalueα.
Thus, we can take a and replace it by α if the a operator is acting on the coherent
state α. We then get the mappinga→αa|αic = α|αic†

c hα|aa →α∗†= c hα|α∗So that a creation operator a† acting to the left can also be replaced by α∗ .
Let us use this principle to do some simple calculation. For example, let us calculate
the expectation of the number operator n̂ in the coherent state |αic .
2It is a simple excercise to show these! Try it!

4.1. MATHEMATICAL PRELIMINARIES

39

We have (recalling that are working with non-normalized states)hn̂i =†
c hα|a a|αic
c hα|αic

Allowing the a to act to the right and the a† to act to the left we obtainhn̂i =|α|2 c hα|αic= |α|2c hα|αic

Let us next try to calculate hn̂2 i. Similarly we gethn̂2 i =††c hα|a a a a|αic
c hα|αic=† ††c hα|a a a a + a a|αic
c hα|αic= |α|4 + |α|2

where we have used commutations to put the operators into normal order (all creation
operators at the far left).
Using these two results we calcuate the variance of the number of bosons in thecoherent state
q
p
p
p
2 =
VarN = σN
hn̂2 i − hn̂i2 = |α|2 = hn̂i
The important result here is that for large hn̂i the fluctuations in particle number are muchsmaller than the mean – so it won’t matter if we fix the particle number or we allow it tofluctuate. Given that the fluctuations are small, a system with a fixed particle number is
not very different from a coherent state.
A comment for those who have not explored coherent states before: Laser light canbe thought of as a coherent state of photon (which are, appropriately, bosons).
Number-Phase Relation
An interesting feature of coherent states (which will also be true of superfluids) is that
number and phase form a quantum mechanical conjugate pair, like momentum and position — you cannot know both at the same time. To see this, let us write the parameter
α = |α|eiϕ , we then have
|α|2 e2iϕ|α|eiϕ a†iϕ|αic = e|0i = |0i + |α|e |1i + √|2i + . . .
2!
Now let us try differentiating with respect to the phase∂|α|2 e2iϕiϕ−i |αic = 0|0i + 1|α|e |1i + 2 √|2i + . . .
∂ϕ2!
which we realize is exactly the same as applying the number operator to the coherent state(i.e., it multiplies |ni by n). Thus we identify
−i

∂
= n̂
∂ϕ

(4.2)

40

CHAPTER 4. MICROSCOPIC THEORY OF BOSONS

analogous to −i~∂/∂x = p̂. Thus we expect if we are going to have a superfluid with awell defined phase we must also accept that it will have an indefinite number of particles.
We will see this in detail below.

4.1.3

Multiple orbitals

Let us now consider the case of bosons which are allowed to reside in multiple orbitals.
Let us assume we have an orthonormal set of orbitals which we will label i, j, . . .. For
each orbital i we have a corresponding creation operator a†i and annihilation operator ai .
These operators obey the commutations[ai , a†j ] = δij(4.3)[ai , aj ] = [a†i , a†j ] = 0(4.4)

In other words, the creation and annhilation operators in a given orbital are completely
independent from the creation and annihilation operators from another orbital.
We also define a ket |0i to mean the vacuum of all the orbital; i.e., no bosons at all.
Thus a†1 |0i means one boson in orbital 1, whereas a†1
2 and one boson in orbital 1, and so forth.

(a†2 )2
2! |0i means two bosons in orbital

It is crucial that we are able to change basis for our orbitals. Suppose we have twoseperate complete sets of orthonormal bases. Let us call the first set {|φn i} and call thesecond set {ψm }. As an example we might imagine that one set is a set of plane waves,
whereas maybe the other set is a set of harmonic oscillator wavefunctions. The conversion
between the two corresponding sets of creation operators is given byX(4.5)
hψm |φn i a†ψma†φn =m

A very importnat example is the case of particles hopping on a lattice. Here we have anatural tight binding basis where ψr0 (r) = δr,r0 is a wavefunction of a particle localizedat position r and the corresponding second quantized operator that creates a particle inthis orbital is a†r . On the other hand, we also have the plane wave orbitals
1
φk (r) = √ eik·rNwith N the number of sites in the system. We write the corresponding second quantized
operators that create a particle in these plane waves as
1 X ik·r †a†k = √
e ar
N r

(4.6)

Any operator can be written in second quantized notation in any basis — this isdone by representing it in terms of its matrix elements in that basis. For example, a4.2. BECS AND THE GROSS-PITAEVSKII EQUATION

41

one-body operator can be written asÔ =X
hn|Ô|mia†n amn,mSimilarly one can write a two body operator asÔ =X(hn| ⊗ hq|) Ô (|mi ⊗ |pi)n,m,q,pand so forth.

Continuum Limit
Very frequently we take the continuum limit where r lives not on a lattice, but in a continuum. In this case we conventionally replace the creation and annhilation operators a†r andar with the continuum field operator ψ̂ † (r) and ψ̂(r). The corresponding commutation
relations are then given by[ψ̂(r), ψ̂ † (r0 )] = δ(r − r0 )In terms of these field operators the plane wave creation operators are given by1a†k = √

VZ

dr eik·r ψ̂ † (r)with V the volume of the system.
Indeed, any orbital basis {φn } can be written in terms of the continuum field operators asZ
a†φn =

dr φn (r) ψ̂ † (r)The transformation in reverse, assuming we are working with a complete basis, is givenbyψ̂ † (r) =X
hφn |ria†φnn=Xφ∗n (r)a†φn

(4.7)

n

4.2

BECs and the Gross-Pitaevskii Equation

We will now use second quantized operators to examine some of the physics of BECs andinteracting bosons. We begin by considering noninteracting bosons.

42

CHAPTER 4. MICROSCOPIC THEORY OF BOSONS

4.2.1

Noninteracting BECs as Coherent States

The principle of a BEC is that we want to put many bosons in the same oribtal. Let us
call this orbital φ0 , and write a creation operator a†φ0 that creates a boson in this orbital.
Now let us write a coherent state for many bosons in this orbital.
αa†φ|α; φ0 i ≡ e0|0iEnirely analogous to our above calculation we can calculate the expectatoin of the numberof bosons in this orbital
hN0 i =

hα; φ0 |a†φ0 aφ0 |α; φ0 ihα; φ0 |α; φ0 i

= |α|2

meaning that the coherent state puts (on average) |α|2 bosons into the orbital φ0 .
It is very useful to examine the effect of the field operator ψ̂(r) on the coherent
state. To do this we use the decomposition given in Eq. 4.7 to giveψ̂(r)|α; φ0 i =Xφn (r)aφn |α; φ0 i

(4.8)

n

= αφ0 (r)|α; φ0 i

(4.9)

Thus, using a coherent state turns the field operator ψ̂(r) into a scalarψ̂(r) → ψ(r) = αφ0 (r) =pN0 φ0 (r)where we assume the phase of α is real in the last step (although we did not have to makethis assumption). Thus the field operator becomes ψ(r) which is simply the ground state
wavefunction φ0 normalized to have N0 particles.

4.3

Interacting Bosons and the Gross-Pitaevskii Equation

We now turn to consider interacting bosons. We write a first quantized Hamiltonian as3
H=

X  p2ii2m+ V (ri ) +1X
U (ri − rj )2

(4.10)

i6=jHere, the term in bracket is a single paricle Hamiltonian, and we have included possiblya trapping potential V (r). We can also include a chemical potential inside of V (r) if wewish. The final term in the Hamiltonian is some intaction between the bosons.
3

I apologize that I’ve used V (r) both for potential and volume of the system. I hope this does not cause
confusion!

4.3. INTERACTING BOSONS AND THE GROSS-PITAEVSKII EQUATION43

We can now convert the first quantized Hamiltonian to second quantized form. Thisis a straightforward exercise4 yielding
 2 2

Z
−~ ∇
†
H =
dr ψ̂ (r)+ V (r) ψ̂(r)2mZZ1
dr dr0 : ψ̂ † (r)ψ̂(r) U (r − r0 ) ψ̂ † (r0 )ψ̂(r0 ) :+2
In the second line we have the density ρ(r) = ψ̂ † (r)ψ̂(r) so that this term is an interaction
between density at position r and position r0 . We have also included :’s around this term
to indicate that the term should be interpreted as being normal ordered — i.e., all creation
operators moved to the left. It is clear that this is necessary since applying this term to astate with a single boson should give a zero interaction energy (and this will not be thecase unless the term is normal ordered)5 .
From here we will further specialize to a particularly simple delta-function interaction
U (r − r0 ) = U δ(r − r0 )(4.11)

with U a scalar. Such a short range interaction is actually a very good respresentationof the interaction for many physical bosonic systems. For example, for superfluid helium,
the Helium atoms are strongly repulsive only at very short distance, and so this is actuallya fairly good approximation. Similarly for modern cold-atom BECs, it is often the casethat the delta function interaction is actually very representative of the physical system.
This then simplifies our Hamiltonian to the form
 2 2

Z
Z
−~ ∇
U
†
H =
dr ψ̂ (r)+ V (r) ψ̂(r) +dr ψ̂ † (r)ψ̂ † (r)ψ̂(r)ψ̂(r)2m
2(4.12)Let us now consider a coherent state in an orbtial φ(r) as a trial wavefunction.
†|α; φi = eαaφ |0i
As discussed above, application of the√field operator ψ̂(r) to this coherent state generates
the number ψ(r) = αφ(r) with |α| = N .
We now calculate the expectation of the Hamiltonian in this coherent state
hα; φ|H|α; φihα; φ|α; φi

 2 2


Z
−~ ∇
U
=
dr ψ ∗ (r)+ V (r) ψ(r) + |ψ(r)|4
2m
2

hHi =

4
Try it! The trick is to convert terms to a convenient basis — position space or momentum space —as necessary.
5
Note that leaving the term un-normal ordered will only make an order 1 error in an order N term, so
for large numbers of particles it is not a bad mistake!

44

CHAPTER 4. MICROSCOPIC THEORY OF BOSONS

Finally integrating by parts to handle the derivative we get

 2
Z
U
~
2
2
4
|∇ψ(r)| + V (r)|ψ(r)| + |ψ(r)|
hHi =
dr
2m
2
This expression is known as the Gross-Pitaevskii (or Ginzburg-Landau) form. This was
first discussed by Vitali Ginzburg and Landau in 1950 in the context of superconductivity6 .
In 1960 it was rederived by both Gross (in the west) and Pitaevskii (in the USSR) in thecontext of superfluid Helium.
Minimizing the energy by taking a functional derivative and setting it to zero,δhHi=0δψ ∗ (r)we obtain−~2 ∇2+ V (r) + U |ψ(r)|2 ψ(r) = 0(4.13)
2m
which is known as the Gross-Pitaevskii equation, or non-linear-Schroedinger equation. We
will study the solutions to this equation later. This simply the single particle Schroedinger
equation where there is an additional potential U |ψ(r)|2 making the local potential higher
in regions where there are many bosons (the energy eigenvalue can be absorbed into V ).

4.3.1



Rederivation without Coherent States

Is there an easier way to get to the Gross-Pitaevskii equation? Do we need to work in a
“grand-canonical ensemble” where the number of bosons is indefinite? Do we really haveto work with second quantized operators?
The answer to these questions are: Yes, there is an easier way. We don’t need anyof these technical tools. In fact we can work with first quantized operators with a fixed
number of particles.
Recall that a BEC wavefunction puts a macroscopic number of particles in a single
orbital φ(r). So let us write a trial wavefunction for such a BEC as follows:Ψ=NYφ(ri )

i=1

Note that this wavefunction is properly symmetic under exchange of particle positions asa bosonic wavefunction should be.
We then find the expectation of our first quantized Hamiltonian Eq. 4.10 for our
trial state
 2 2

Z
Z
Z
−~ ∇
N (N − 1)∗hΨ|H|Ψi = N dr φ (r)+ V (r) φ(r)+dr dr0 : |φ(r)|2 U (r−r0 ) |φ(r0 )|22m
26

Ginzburg won a Nobel prize for this work in 2003. It is worth reading Ginzburg’s Nobel biography. It
seems his early years, growing up in the Stalinist Soviet Union were somewhat harrowing!

4.4. ORDER PARAMETER AND OFF-DIAGONAL LONG RANGED ORDER

45√
making the substitution ψ = N φ and neglecting the distinction between N and N − 1
we get exactly the same result as when we used coherent states above. Minimizing thetotal energy in either approach will allow us to calculate the best single orbital φ(r), or
equivalently ψ, which we should macroscopically occupy to form a BEC.

4.4

Order Parameter and Off-Diagonal Long Ranged Order

So if it is possible to derive everything with such a simple first quantized approach, whythen did we go through the extra work of working with coherent states and using second
quantization? The reason is that for an interacting BEC (particularly a strongly interacting BEC) it is in fact not correct to put all the bosons in a single orbital. A single orbital
may be multipy occupied, but due to the interactions some bosons will be kicked out of
this single orbital. We will discuss this physics in more detail later. However, despite thiscomplicationhψ̂(r)i(4.14)
will remain a good order parameter and a good description, and a convenient description,of the condensate.
It is worth noting, however, that there are some who object to the use of suchan object as an order paramter. The obvious objection is that it is perfectly possible
to consider a system of bosons with a fixed number, in which case the order parameter
Eq. 4.14 is strictly zero and one needs to think a bit harder7 . It is enlightening to go
through some extra work to try to understand in a bit more detail the physics of thesuperfluid order parameter for the case of fixed particle number.
Let us try to be a bit more precise with our definition of the order paramter fora system with strongly interacting bosons. We can here work with a fixed number ofparticles (a canonical ensemble). Let us write the folowing density operator
ρ1 (r0 , r) = hψ̂ † (r0 )ψ̂(r)i(4.15)

This operator removes a particle from position r and puts a particle back in at position r0 .
While this conserves total particle number, it does not locally conserve particle number if
r and r0 are very different. In terms of the multi-particle wavefunction, this object can be
written asZZZ0ρ1 (r , r) = N dr2 dr3 . . . drN Ψ∗ (r0 , r2 , . . . , rN )Ψ(r, r2 , . . . , rN )(4.16)
We can even write this quantity at finite temperature more generally asZZZ
N X −βEn0ρ1 (r , r) =e
dr2 dr3 . . . drN Ψ∗n (r0 , r2 , . . . , rN )Ψn (r, r2 , . . . , rN )Z n
7

The Nobel Laureate Tony Leggett in particular has emphasized the shortcomings of a number nonconserving order parameter

46

CHAPTER 4. MICROSCOPIC THEORY OF BOSONS

where
here En and Ψn are the nth eigenenergy and eigenvector of the system and Z =P −βEn the partition function.
neNote also that ρ1 (r, r) is just the regular density at position r, and if we wanted to,
we could write even more complicated operators such as
ρ2 (r0 1 , r0 2 ; r1 , r2 )which would be an analogous expression with N − 2 integrals.
For a moment, let us think of positions r0 and r as being discrete. In this case, we
can think of ρ1 (r0 , r) as actually being a matrix (it has two indices, r0 and r). Since thismatrix is hermitian it can be diagonalized and written in terms of its eigenvalues, which
we call Nα and its eigenvectors which we call φα (r). We thus haveX
ρ1 (r0 , r) =Nα φ∗α (r0 )φα (r)(4.17)
αwhere the eigenvectors are normalized
Z

dr|φα (r)|2 = 1

By convention we order the eigenvalues so that N0 is the largest.
Noninteracting BEC at T = 0
For a noninteracting BEC at zero temperature, all of the bosons are in the same
orbital. In this case it is easy to calculate that there is only one nonzero eigenvalue
N0 = N , the number of particles in the system, and φ0 is the single macroscopicallyoccupied orbital.
Let us consider this case in a bit more detail. For a noninteracting BEC without
a trapping√ potential at T = 0, all of the bosons are in the single k = 0 eigenstate,
φ0 = 1/ V . We can then explicitly calculate the order parameter. The easiest way to
calculate this is to go back to second quantized form of Eq. 4.15. We then have
ρ1 (r0 , r) = hψ̂ † (r0 )ψ̂(r)i = N φ∗0 (r0 )φ0 (r) = N/V
the particle density, for all r and r0 . See also the more detailed calculation in section 4.4.1
below.
More generally
However, we expect that even for a BEC for 0 < T < Tc , there should be a single
eigenvalue N0 which will be smaller than N but still extensively large (we will demonstrate
this below). Further, even for an interacting system, we can use N0 being extensive as anindication of the existance of a condensate and φ0 is the condensate wavefunction. Puttingthese two together into an order parameters, we havepψ(r) = N0 φ0 (r)4.4. ORDER PARAMETER AND OFF-DIAGONAL LONG RANGED ORDER

47

Figure 4.1: Measure of Off-Diagonal Long Range Order ρ1 (r1 , r2 ). For r1 = r2 we always
have ρ1 = n = N/V the density. However, for r1 far from r2 a nonzero ρ1 indicates thepresence of a condensate. For a noninteracting BEC at T = 0 (top line) ρ1 = n for all|r1 − r2 |. For a system in the normal phase, T > Tc , we have ρ1 decaying to zero for large|r1 − r2 | as shown in the bottom line. The middle line shows that for a noninteracting
BEC at 0 < T < Tc or for an interacting BEC at any T < Tc , then ρ1 decays to a constantless than n but larger than zero.

We can make a plot of ρ1 for severeal different cases as shown in Fig. 4.1.
Recall that ρ1 (r1 , r2 ) is always just the density n for r1 = r2 .
As shown in the figure (and we just discussed) for a noninteracting BEC at T = 0
we have ρ1 (r1 , r2 ) is constant as a function of distance |r1 − r2 |. For a normal system
without a condensate T > Tc , we have ρ1 decaying to zero for large |r1 − r2 |. We will
prove this explicitly below.
Another case shown in the figure, that we will derive below is the case of a noninteracting BEC for 0 < T < Tc . In this case ρ1 decays to a constant larger than zero, butless than n for large |r1 − r2 |.
More difficult to show, however, is the case of an interacting Bose system below its
critical temperature (possibly at T = 0, but possibly at 0 < T < Tc ). The behavior here
is similar to that of a BEC at 0 < T < Tc , that is ρ1 decays to a constant larger than zero,
but less than n for large |r1 − r2 |. Qualitiatively this is again due to the fact that, similar
to temperature, the interactions “kick” some of the bosons out of the single condensed
state. For example, for superfluid Helium at T = 0 at large |r1 − r2 | the value of ρ1 is lessthan 10% of the density.
The fact that the ρ1 correlator becomes a constant for large |r1 −r2 | in the superfluid
phase is known as Off-Diagonal Long Ranged Order8 . It is sometimes abbreviated as
ODLRO. The reason it is called ‘off-diagonal” comes again from thinking about ρ1 asa matrix with two indices, r1 and r2 which are taken different from each other (hence
off-diagonal). The idea of using this quantity as an indicator of superfluid order is due to
8

In comparison, diagonal long range order would be long range order in ρ(r, r) or the density — suchas crystalization.

48

CHAPTER 4. MICROSCOPIC THEORY OF BOSONS

Oliver Penrose9 in 1951 and the Nobel Laureate Lars Onsager in 1956.

4.4.1

non-interacting bosons at T > 0

It is easy enough to calculate ρ1 explicitly for a noninteracting Bose system at finite
temperature (we will again assume no trapping potential so the single particle eigenstates
are plane waves). It is again easiest to work with the second quantized definition of ρ1
given in Eq. 4.15. We can also decompose the field operator into Fourier modes
1 X
ψ̂(r) = √ak eik·r
V k(4.18)We thus have
ρ1 (r0 , r) = hψ̂ † (r0 )ψ̂(r)i =

1 X †
1 X †
0 0
0
hak0 ak ieik·r−ik ·r =hak ak ieik·(r−r )V
V0(4.19)

kk,kWhere we can always writeha†k ak i = nB (k )where k is the energy of the boson in mode k and nB is the Bose factor.
Thus we have explicitlyρ1 (r0 , r) =1 X
0nB (k )eik·(r−r )VkNow for T > Tc we have nB being a smooth function of k. We can then freely
convert the sum into an integral, and get the usual Fourier transform. Since we are
Fourier transforming a smooth function the result ρ1 is smooth and goes to zero for large(r − r0 ).
However, for T < Tc we cannot convert the sum into an integral since a macroscopic
number of particles are in the k = 0 state. In this case we need to separate k = 0 fromthe sum over k, and then the remainder can be considered a smooth integral.
We then have insteadha†0 a0 i = N0 (T )and correspondinglyρ1 (r0 , r) =

N0
1 X
0
+
nB (k )eik·(r−r )
V
V
k6=0

where the sum is again a smooth and decaying function.
9

Brother of Roger Penrose.

4.4. ORDER PARAMETER AND OFF-DIAGONAL LONG RANGED ORDER

4.4.2

49

Current and the Order Parameter

Several times we have treated the order parameter for the condensate as if it is a wavefunction and blithely written things like (See for example, Eq. 2.5)
j(r) =1 ∗(ψ p ψ − (pψ ∗ ) ψ)2

Is this really legitimate? Indeed, we can essentially derive this equation from the definitionof the order parameter as ODLRO.
Let us first write a general expression for the total mass current
Z
Z
Z
1 ∗
(Ψ PΨ − (PΨ∗ ))Ψ)J = dr1 dr2 . . . drN2
whereΨ = Ψ(r1 , . . . rN )is the many-body boson wavefunction, and
P=

N
X

pi

i=1

is the total momentum.
We realize that since the wavefunction is fully symmetric under interchange between
particles we can write
Z
Z
Z
1 ∗
(Ψ p1 Ψ − (p1 Ψ∗ ))Ψ)J = N dr1 dr2 . . . drN
2Further we can write the local mass current density as
Z
Z
Z
1 ∗
j(r) = N dr2 dr3 . . . drN(Ψ p1 Ψ − (p1 Ψ∗ ))Ψ)r1 =r2(4.20)

This expression is entirely general and applies to any bosonic system. Since p1 = −i~∇this can be rewritten as
Z
Z
Z
−i~N
j(r) =
dr2 dr3 . . . drN (Ψ∗ ∇1 Ψ − (∇1 Ψ∗ ))Ψ)r1 =r
2
Then writing[Ψ∗ ∇1 Ψ − (∇1 Ψ∗ ))Ψ]r1 =r= Ψ∗ (r01 , r2 , . . . rN )∇1 Ψ(r1 , r2 , . . . rN ) − (∇10 Ψ∗ (r01 , r2 , . . . rN )))Ψ(r1 , r2 , . . . rN ) r1 =r0 =r1=[∇1 − ∇10 ] Ψ∗ (r01 , r2 , . . . rN )Ψ(r1 , r2 , . . . rN ) r1 =r0 =r1

50

CHAPTER 4. MICROSCOPIC THEORY OF BOSONS

Comparing to the definition of the one-body density matrix in Eq. 4.16 the current density
is thus written as−i~j(r) =(4.21)[∇1 − ∇10 ] ρ1 (r0 1 , r1 ) r1 =r0 =r
1
2
Now recall the decomposition of the density matrix in terms of its eigenvalues and eigenvectors as in Eq. 4.17, where we have
ρ1 (r0 , r) = ψ ∗ (r0 )ψ(r) + . . .

(4.22)√
where ψ = N0 φ0 is the order parameter of the condensate, and the . . . represents many
additional terms where the eigenvalues are small — -terms that do not represent thecondensate.
Using Eq. 4.22 in Eq. 4.21 immediately recovers
j(r) =1 ∗(ψ p ψ − (pψ ∗ ) ψ) + . . .
2o where the . . . are the contributions from the noncondensate part of the system.

4.5

Bogoliubov Theory for the Weakly Interacting Bose Gas

Above we mentioned that we would analyze the effect in more detai. We claimed that theinteractions somehow “kick” bosons out of the condensate leaving ρ1 (r, r0 ) at long distance(large r − r0 ) less than the full physical density even at zero temperate.
The approach we follow here, so-called Bogoliubov Theory10 is a controlled approximation accurate in the limit of weak, but non-zero interaction.
For simplicity let us consider the case where there is no external trapping potential
V (r) = 0 so that the single particle eigenstates are plane waves. We will further, forsimplicity, assume the inter-particle interaction is a delta function as in Eq. 4.11. Neither ofthese assumptions is required for the method to work, although both make the calculationeasier.
The Hamiltonian in second quantized form, as in Eq. 4.12 is given by
 2 2

Z
Z
−~ ∇
U
†
H =
dr ψ̂ (r)+ V (r) ψ̂(r) +dr ψ̂ † (r)ψ̂ † (r)ψ̂(r)ψ̂(r)2m2Since the single particle eigenstaetes are plane waves we can conveniently write thefield operator ψ̂(r) in Fourier modes
1 X
ψ̂(r) = √ak eik·r
V k
10(4.23)

Nikolay Bogoliubov is probably more famous as a mathematician than a physicist. Nonetheless he also
had several important contributions to the physics world, including being one of the B’s in BBGKY fromstatistical physics.

4.5. BOGOLIUBOV THEORY FOR THE WEAKLY INTERACTING BOSE GAS

51

and rewrite the Hamiltonian in the formH=

X ~2 k2 †Uak ak +
2m2VkXa†k1 a†k2 ak3 ak4 δk1 +k2 −k3 −k4(4.24)

k1 ,k2 ,k3 ,k4

Now we expect that even when we turn on the interactions, the k = 0 mode should bemacroscopically occupied at zero temperateure as it is in a BEC. However the commutator[a†0 a0 , Interaction] 6= 0
which means that the interaction must kick some of the particles out of the k = 0 state.
Nonetheless, we still expect macroscopic occupancy of this orbital. Let us thus assume
that the k = 0 orbital is in a coherent state with mean occupancy N0 with 1  N0 < N .
At T = 0 if the interaction were zero, then N0 = N . For sufficiently weak interaction, we
might expect thatN − N0  N
Post-facto, we will verify that this is true for weak interaction.
Since we have a coherent state, as usual this allows us to replace operators a†0 anda0 with numbers via the usual substitutionppa†0 → N0a0 → N0
Returning to our Hamiltonian, we now have a small parameter 1/N0 and we canorganize the parts of the Hamiltonian Eq. 4.24 in terms of which has the most factors of
N0 (i.e., the most factors of a†0 or a0 ). In fact, the kinetic term has no factors of N0 (since
the kinetic energy of the k = 0 state is zero). We thus turn to the interaction term andtry to organize the pieces of the sum in order of which pieces have the most factors of N0 .
Looking at the sum over k1 , k2 , k3 , k4 , the single term wherek1 = k2 = k3 = k4 = 0√
is the largest term, giving us four factors of N0 Thus the value of this term in the sum
is simply the constant
U 2
N(4.25)
2V 0√
At next order We look for a term where there are three factors of N0 meaning
three of the k’s are zero and one is nonzero. However, the delta function in the interaction
term of Eq. 4.24 requires that if three of the k’s are zero than the fourth one is too, so nosuch term exists.
At next order we look for a term where two of the k’s are zero and two are nonzero.
Necessarily the momenta on the remaining two k’s must appropriately sum to zero. We
can choose the two non-zero k’s to be both creation, both annihilation, or one of each

52

CHAPTER 4. MICROSCOPIC THEORY OF BOSONS(which can be chosen in four different ways). Thus the sum of all these terms can be
written asi
U N0 X h †4 ak ak + a†k a†−k + a−k ak(4.26)
2V
k6=0

We can then look for terms where only
√ one of the k’s is zero. There are such terms,
but these will be smaller by a factor of 1/ N0 so we throw these out. (In principle atthe end of the calculation these terms can be considered as a small perturbation on top ofwhat find).
We then assemble the terms we find. First, the kinetic term from the Hamiltonian
Eq. 4.24, secondly, the zeroth order term Eq. 4.25 and the second order term Eq. 4.26.
Finally we use one additional trick, we writeX †N0 = a†0 a0 = N −
ak ak
k6=0

plugged into the N0 in Eq. 4.25, we obtain the resulting HamiltonianX  ~2 k2UρUρ  † ††H=N++ U ρ ak ak +ak a−k + a−k ak
2
2m
2
k6=0

where ρ = N/V . This Hamiltonian is quadratic (and therefore solvable), but it has socalled anomolous terms —those with two creation or two annihilation operators. These
terms allow particles to scatter in or out of the condensate (the state with k = 0). The
scattering terms must conserve total momentum so you can only scatter two-in or two-out
at a time.

4.5.1

Bogoliubov Transform

To solve the quadratic hamiltonian with anomolous terms, we invoke the so-called Bogoliubov transformation (invented 1947, by Bogoliubov). Let us write the following transformation
 



bk
ak
cosh θk sinh θk=(4.27)
sinh θk cosh θk
b†−k
a†−k
It is easy to check that if the a’s satistfy canonical commutations as in Eqs. 4.3 and 4.4,
then the b’s similarly satisfy canonical commutations11[bq , b†p ] = δq,p[bq , bp ] = [b†q , b†p ] = 0
11(Sid’s favorite comment): It may look very strange that the matrix in Eq. 4.27 is not a unitary matrix.
We know we are allowed to make unitary changes of basis in quantum physics, but non-unitary transforms
seem problematic. The reason that this is OK here is because we are making a change of basis on operators.
If we work out what happens to the resulting basis states in the Fock space under this transformation, we
will discover that this actually corresponds to a unitary transformation on the Fock space! The only thing
we need for this to be true is that canonical commutations are preserved.

4.5. BOGOLIUBOV THEORY FOR THE WEAKLY INTERACTING BOSE GAS

53Making this transformation (with a bit of algebra ADD ALGEBRA AS FOOTNOTE
OR APPENDIX?) the Hamiltonian becomes 2 2hi~ k+ U ρ cosh(2θk ) − U ρ sinh(2θk ) b†k bk + b†−k b−k
2m
k6=0

h
 2 2i~ k+ U ρ sinh(2θk ) − U ρ cosh(2θk ) b†k b†−k + b−k bk−
2m

1X
H = const +2

If we then chooseUρtanh(2θk ) = ~2 k2
2m + U ρ(4.28)

we eliminate the anomolous terms and we diagonalize the Hamiltonian, obtaining
H = const +X

Ek b†k bk(4.29)

k6=0

where
s
Ek = +
r
∼

~2 k2
+ Uρ
2m

Uρ
~|k| + . . .
m

2− (U ρ)2for small k

Note that this spectrum of excitations in linear in k at low k and then curves to bequadratic at large k. Because the dispersion is linear it satisfies the Landau criterion forsuperfluidity!
The excitations created by the b†k operators are sometimes known as bogoliubons.
The ground state is obviously given by the state with no bogololiubons present
bk |ground statei = 0

for all k 6= 0

Note however, that since we can invert Eq. 4.27, we can write this condition as(cosh θk ak + sinh θk a†−k )|ground statei = 0
and thus,
ak |ground statei =6 0meaning that in the ground state there is some occupation of bosons in k 6= 0 orbitals. As
we predicted previously, the interaction has pushed some of the bosons out of the k = 0
orbital.

54

CHAPTER 4. MICROSCOPIC THEORY OF BOSONS

Occupancy of k = 0: Depletion of Condensate
Let us try to calculate how many bosons are in the k = 0 orbital. To do this we write as
above
X †
N0 = N −
ak ak
k6=0

so that in the ground state we have
N0
1 X
1 X
=1−
hGS|a†k ak |GSi = 1 −sinh2 θk
N
N
N
k6=0

(4.30)

k6=0

with a bit of nasty algebra (detailed in the appendix below) it can be shown that1−

N0
∼ U 3/2
N

Thus, as we claimed above, for weak interaction, only a few of the bosons are kicked outof the condensate. If U becomes large, then many bosons are kicked out of the condensate
and we cannot use this expansion technique.
ODLRO
It is useful to explicitly calculate the off-diagonal-long-ranged-order (ODLRO). First, we
write the density matrix (see Eq. 4.19
ρ1 (r, r0 ) =
=

1 X †
0
hak ak ieik·(r−r )
V
k
N0
1 X †
0
+
hak ak ieik·(r−r )VV(4.31)

k6=0

Note that if we take r = r0 the second term is precisely the expression we evaluated for the
depletion of the condensate, so that ρ1 (r, r) = N/V as expected. Here the second termis the Fourier transform of a smooth function which goes to zero for r far from r0 . Thus
ρ1 (r, r0 ) for r far from r0 goes to N0 /V .
Appendix: Some nasty algebra
To show the previous result, we first note that from Eq. 4.28 we know tanh 2θ. We thenhavesech2 2θ = 1 − tanh2 2θso1p= cosh 2θ
1 − tanh2 2θ

4.5. BOGOLIUBOV THEORY FOR THE WEAKLY INTERACTING BOSE GASThen we use

55

1(cosh 2θ − 1) = sinh θ2

so that we obtain
N0
1
=1−
N
2N

X
k6=0



1
1 X
−1 =

2
2N
1 − tanh 2θk
k6=0


1
1−



Uρ
U ρ+~2 k2 /2m

2 − 1

Changing the sum to an integral




Z
Z
N0
4πV
1
V
dk 
1



1−
k 2 dk =2 − 1 = 1−
2 − 1 N
2N(2π)3
2N (2π)3 0UρUρ
1 − U ρ+~2 k2 /2m
1 − U ρ+~2 k2 /2m
where in the last step wephave switched to spherical coordinates. Finally, we rescalevariables by defining k = 2mU ρ/~2 z such that we have


Z
p
4πV
1
N0


3/2
= ( 2mU ρ/~2 )3z 2 dz 1−2 − 1 ∼ U
3
N
2N (2π) 01
1 − 1+z 2
as claimed.
For the very brave
The Bogoliubov result is not actually dependent on working with coherent states where
N is indefinite. In fact one can write down a Bogoliubov wavefunction for fixed N . The
form of the wavefunction can be written asZZK̂ =dr r0 ϕ(r, r0 )ψ̂ † (r)ψ̂ † (r0 )(4.32)|Ψi = (K̂)N/2 |0i(4.33)

where K̂ creates two bosons in some wavefunction ϕ. The Bogoliubov result is obtainedby minimizing the energy over all possible shapes of the wavefunction ϕ. While it isalgebraically messy, one can obtain all the same results.

56

CHAPTER 4. MICROSCOPIC THEORY OF BOSONS

Chapter 5

Feynman Theory of Helium-4
For a strongly interacting Bose gas, the Bogoliubov theory fails to be quantitatively accurate, and we need another approach. Feynman came up with a truly ingenious approach1
for understanding strongly interacting Bose system, with the particular application of superfluid Helium-4 in mind. We will follow this method2 in order to elucidate the physics
of superfluidity further. The question we would like to answer is ”what do the low energy
excitations of a strongly interacting Bose gas look like?”.

5.1

Ground State and Low Energy ExcitationsFirst let us consider the (T = 0) ground state for the superfluid at rest in first quantizedwavefunction form. we can write it as
Φ0 (r1 , r2 , . . . , rN )This is some very complicated function of the arguments r1 , . . . , rN since, due to interactions between the bosons, there will be significant correlations between the positions of
the particles. Note however, that the wavefunction must be symmetric in exchange of any
two particle coordinates by bosonic symmetry.
1

Most people agree that Feynman could easily have been given a Nobel prize for his work on superfluidity, had he not won the prize for quantum electrodynamics.
2
The Nobel Laureate Murray Gell-Mann joked that Feynman’s method consisted of three steps:
1. Write down the problem
2. Think very hard
3. Write down the solution
While obviously he was joking, this does characterize much of Feynman’s work. Feynman (similar to
Landau) used physical arguments to simply state what the answer has to be, rather than having toperform lengthy calculations (although apparently in private, Feynmann did plenty of lengthy calculations
that he then hid from the world to maintain the appearance of being a competely intuitive genius). The
negative point of this type of approach is that while it is easy to teach a calculational method, it is hard
to teach someone to come up with brilliant physical insights.

57

58

CHAPTER 5. FEYNMAN THEORY OF HELIUM-4

Figure 5.1: Compression Waves = Longitudinal Phonon
Even if we don’t know the form of the function Φ0 we can still construct from it awavefunction for a superfluid flowing. This will take the form
Ψ(r1 , r2 , . . . , rN ) = Φ0 (r1 , r2 , . . . , rN )NYeiθ(ri )

(5.1)

i=1

where θ(r) is a function of position which must be well defined modulo 2π (and it mustbe single valued if the amplitude of Φ0 nowhere vanishes).
For a uniformly flowing supefluid we can chooseθ(r) = k · r
for some momentum ~k per particle.
We can check that the velocity of the fluid matches our expectation by looking at
the expectation of the current operator p (See Eq. 4.20) Being that Φ0 is defined to be
the fluid at rest, we must havej(r)Φ0 = 0
and thus one easily shows thatj(r)Ψ = ~n∇θ = n~kwith n the density, as expected.
Thus given a ground state wavefunction representing a stationary fluid we can writea wavefunction for a flowing superfluid. However, this is not the type of excitation weare interetsted in. So let us now consider low energy excitations given that there is nosuperflow.
Claim: In absence of superflow, the only low energy excitations of an interacting
Bose gas are (longitudinal) phonons. i.e., compression waves.
A depiction of compression waves is given in Fig.5.1
We will next argue why this claim is true. We constrast this situation with that of asystem of fermions. For fermions (which we will discuss later) the ground state is a Fermi
sea (k states filled up to some Fermi surface). In this case we can have many different
low energy excitation associated with exciting some fermion from slightly below the Fermi
surface to slightly above.

5.1. GROUND STATE AND LOW ENERGY EXCITATIONS

59

Figure 5.2: Left: A wavefunction Φ0 which changes sign as a function of some coordinate
written as x (but is meant to be some combinations of the positions {ri }. Middle: Theabsolute value of Φ0 . Except for (the set of measure zero) precisely at the cusp, this gives
the same argument in Eq. 5.2 as Φ0 itself. Right: Smoothing the wavefunction over a scaleδ will always reduce the energy a little bit.
Point 1.1: . The ground state wavefunction with no superflow, Φ0 (r1 , r2 , . . . , rN ),can be chosen real and positive everywhere.
Firstly, since the Hamiltonian is time-reversal invariant (H = H ∗ ), given that the
ground state is unique, we must be able to take Φ0 = Φ∗0 . Thus we have Φ0 real. We
would now like to show that Φ0 does not change sign.
Since Φ0 is assumed to be the ground state, it must minimize the energy
E = hΦ0 |H|Φ0 i
(5.2)


Z
2
X
X
~
1
=
dr1 . . . drN 
U (ri − rj )|Φ0 (r1 , . . . , rN )|2 |∇i Φ0 (r1 , . . . , rN )|2 +
2m
2
i

i6=j

We now want to show that Φ0 for the ground state can always be taken everywhere
positive. To show this we first note that if Φ0 changes sign, thenE(Φ0 ) = E(|Φ0 |)The reason for this is that, as shown in Fig. 5.2, except for positions {ri } that are a setof measure zero, both |Φ0 |2 and |∇Φ0 |2 are unchanged if you replace Φ0 with |Φ0 |. Thus
the energy of Eq. 5.2 is unchanged.
Given that Φ0 and |Φ0 | have the same energy, let us only consider |Φ0 | for the
moment. Here, we can have a cusp where the wavefunction reaches zero. However, now
we argue that we can always reduce the energy of the wavefunction by smoothing the cusp
a little bit. Consider smoothing the cusp over a distance scale δ as shown in the right ofFig. 5.2. Let us estimate the energy change from doing this smoothing.
Roughly, the slope is reduced from dΦ0 /dx to (say for simplicity) zero, so the kinetic
part of the energy (in 1d) (integrating over x) is changed decreased by roughly δ(dΦ0 /dx)2 .
On the other hand, the magnitude of Φ0 is increased from small (call it zero) to δ(dΦ0 /dx).
Thus the potential part of the energy is increased proportional to δ 3 (dΦ0 /dx)2 . For small

60

CHAPTER 5. FEYNMAN THEORY OF HELIUM-4

Figure 5.3: Particle in a box. Top: Ground state wavefunction is everywhere positive.
Bottom: First excited state — the most positive point is far from the most negative point
to minimize the gradient energy.
energy δ the decrease in kinetic energy always wins, and we conclude that the groundstate Φ0 can be taken everywhere strictly positive.
Point 1.2: Any excited state must change sign so that it is orthogonal to the groundstate. For a low energy excited state, one must keep gradients small, so the most positive
configuration must be “far” from the most negative configuration.
As a demonstration of this, let us consider the simple quantum mechanical problem
of a single particle in a box as shown in Fig. 5.3. The ground state is everywhere postive.
The first excited state must change signs to be orthogonal to the ground state. But tokeep the gradient energy low, the most positive point (marked + in the figure) must be
far from the most negative point (marked - in the figure).
What do these principles mean for a many-particle wavefunction? First, we realizethat the first exctited wavefunction must change signs. But again we want to keep gradients small, so that the postive region should be “far” from the negative region. What
do we mean by “far” in the case of a many-particle wavefunction? This means “far” in
the N d dimensional parameter space {ri } (with N the number of particles and d thespatial dimension). We might naively think that we should move all of the particles as
far as possible. However, this isn’t really correct, because the particles are indistinguishable. Suppose for example, we move all the particles to very far positions as in the left ofFig. 5.4. While this moves every particle “far” from its original position, it is equivalent toa different motion where each particle only moves a small distance, as shown on the rightof Fig. 5.4 (due to the fact that the particles are indistinguishable). In fact any motion
which leaves the local density unchanged cannot have moved all the particles very far fromtheir original positions.
However, if we were to move all the particles so as to change the density locally,
as in Fig. 5.1, this cannot be done simply by moving each particle a small distance –many particles need to be moved a large distance in a way that cannot be reduced byindistinguishibility. Thus two configurations with different density profiles are genuinely

5.1. GROUND STATE AND LOW ENERGY EXCITATIONS

61

Figure 5.4: Left: Moving particles “far” from their initial positions. Right: Since particles
are indistinguishable, one can get to the same configuration by moving each particle only
a little bit.
“far” from each other in the N d dimensional parameter space. (Of course changing thedensity profile too much from uniform will be energetically expensive, and cannot be
part of a low energy excitation). So we should expect that two different density should
correspond to the pieces of the wavefunctions with the most positive and most negativevalues. (I.e., perhaps the wavefunction takes its most positive value when the configuration
looks exactly like Fig. 5.1, and then takes its most negative value when the high and low
density regions are interchanged).
Point 2: A good ansatz for making a low energy excited state from a ground stateis to change a single quantum number (to make sure it is orthogonal to the ground state)
and try to leave everything else unchanged.
Suppose we know a ground state Φ0 . We would like to find the lowest energy excitedstate Ψ which minimizeshΨ|H|ΨiE=hΨ|Ψisubject to the constraint of being orthogonal to the ground statehΨ|Φ0 i = 0(5.3)

Let us try the approach of the approach proposed in Point 2 on the simple case of
the harmonic oscillator.
2

For a harmonic oscillator we know the ground state is Φ0 ∼ e−ax , and the groundstate is even parity (reflects around x = 0). The simplest way we can change the parity
quantum number is just by multiplying the wavefunction by the function x, giving a
trial wavefunction Ψ ∼ xΦ0 . which has odd parity and then necessarily must satisfy theorthogonality condition Eq. 5.3 on account of having different parity from the ground state.
This trial wavefunction is indeed very low energy because qualitatively it is as much likethe ground state as it can possibly be, and yet is orthogonal to the ground state (Indeed, in
this particular case we have actually guessed the exact first excited state wavefunction!).

62

CHAPTER 5. FEYNMAN THEORY OF HELIUM-4

Trial Excited StateWe thus use the same general idea to write a trial wavefunction for the lowest energy
excited state in an interacting Bose gas. We propose the following
1|Ψk i = √ ρk |Φ0 i
N

(5.4)

where Φ0 is the ground state and ρk is the density operator at wavevector k given explicitly
by
X
ρk =
eik·ri = ρ†−ki

= Fourier Transform[ρ(r)]where
ρ(r) =X

δ(r − ri )iis the usual density operator.
Let us now list the reasons why this trial excited state wavefunction |Ψk i is likely
to be an accurate one
1. ρ is symmetric in exchanging particle (switch positions of particles i and j) so Ψkhas the correct bosonic symmetry.
2. ρk changes the momentum quantum number. As described in Point 2 above, thismeans that the proposed excited state is necessarily orthogonal to the ground state
for any k 6= 0.
3. For small k, the operator ρk is fairly smooth. This means that multiplying by ρkdoes not ruin any good correlations that are built into the ground state wavefunction
Φ0 .
4. The operator ρk creates a density wave as in our disucssion in Point 1.2. Taking
P ikxaj
simple case where k is in the x̂ direction, we can check that the prefactor j e
is maximally positive when all x are of the form (2π/k)n for some integer n and it
is maximally negative when all x are of the form (2π/k)(n + 1/2), corresponding to
the two opposite density waves. Of course putting all particles on these particular
positions would result in too high a denisty to be low energy, but nonetheless, thisargument shows where the wavefunction is positive and where it is negative, andwhy the positive and negative regions are “far” from each other.
5. The trial wavefunction does not correspond to a superflow (see Eq. 5.1).
6. The trial wavefunction does not change the local density much, so that it should notchange the interaction energy much either.

5.1. GROUND STATE AND LOW ENERGY EXCITATIONS

63

7. Finally (to be shown below) this trial wavefunction matches up with the low energy
excitations in Bogoliubov theory which is correct at least in the weak interaction
limit.
Let us now confirm item 7, that we would get the same excited state wavefunction
in Bogoliubov theory.
First, let us translate the density operator into second quantized notation. We canwrite
ρ(r) = ψ̂ † (r)ψ̂(r)then Fourier transform using the usual
1 X −iq·r †ψ̂ † (r) = √eaq
V qwhich allows us to write the Fourier transformed density operator
Z
1 X −i(q0 −q)·r †e
aq0 aqρk =dr eik·r
V
0
q,q
X †
=
ak+q aqq

Using the same principles we used in section 4.5 the largest terms are those that include the†q√ = 0 modes which are macroscopically occupied. With the usual replacement a0 , a0 →N0 we obtainp 
ρk ∼ N0 a†k + a−k
Now from Eq. 4.29 we have that the effective hamiltonian can be written as
H = const +X

Ek b†k bk

k6=0

so we can write the low energy excited states as|Ψk i = b†k |Φ0 i
where Φ0 is the ground state, meaning the Bogoliubon vacuum. We can, however, also
use Eq. 4.27 to rewrite b† in terms of a† and a. For small q we have θq in Eq 4.27 being
large which means sinh θ ≈ cosh θ and we haveb†k ∼ a†k + a−kThus for the weakly interacting system, a single bogoliubon excitation is precisely thesame as Ψk where we just apply the density operator to the ground state!

64

CHAPTER 5. FEYNMAN THEORY OF HELIUM-4

5.2

Single Mode Excitation Spectrum

So we have proposed a trial wavefunction Eq. 5.4 for the low energy excitations of ourinteracting Bose gas in terms of the ground state Φ0 . Unfortunately, we do not actuallyknow the ground state Φ0 . Remarkably we can still make progress in calculating the energyof the excited state Ψk . What we would like to calculate is the energy of the excitationcompared to the ground state which we write as∆k = Ek − E0 =hΨk |H − E0 |Ψk if (k)≡hΨk |Ψk i
S(k)

(5.5)

where here we have defined the numerator to be called f and the denominator to be called
S. Further, we have realized that these quantities will be a function only of the magnitude
k = |k|.
Let us attack the denominator first. We haveS(k) = hΨk |Ψk i =1hΦ0 |ρ†k ρk |Φ0 iNthis quantity is known as the structure factor, and it is directly measured by x-ray andneutron scattering3 . We can simplify this result by usingXρk =eik·ri = ρ†−k
(5.6)
i

We then obtain4
E
P −ik·ri P ik·rj
1 D
Φ
Φ0
e
e
0
i
j
N
D
E
P −ik·(ri −rj )
1
=
Φ0
Φ0
i,j e
N
hD
Ei
P
= Fourier Transform Φ0i,j δ(r − (ri − rj )) Φ0

S(k) = hΨk |Ψk i =Experimental neutron diffraction measurements of the structure factor of Helium-4 aregiven in Fig. 5.5 The peak in the structure factor is roughly the analog of a Bragg peak fora crystal — it occurs at a wavevector of roughly 2π/a where a is the typical inter-particle
spacing.
3This is not quite the same structure factor we study in our elementary solid state courses when we
study scattering from crystals, although it is very closely related. See next footnote
4
In elementary solid state physics courses, we define the structure factor to be
X ik·rS̃(k) ∼ he j i,
jand the scattering intensity is ∼ |S̃|2 . However, in a fluid this S̃(k) is zero for any k 6= 0 since all positions
are equally likely. Instead we are now calulating the expectation of the square of this quantity, with theaverage outside of the square.

5.2. SINGLE MODE EXCITATION SPECTRUM

65

Figure 5.5: The structure factor of superfluid Helium. Figure from A. Isihara, Physica
106B (1981) 161-164.
Let us now turn to the numerator of Eq. 5.5. We havef (k) ===

E
1 D
Φ0 |ρ†k (H − E0 )ρk |Φ0
N
Ei
E D
1 hD
†
†
Φ0 |ρk Hρk |Φ0 − Φ0 |ρk ρk H|Φ0N
DE1
Φ0 | ρ†k [H, ρk ]|Φ0
N

(5.7)

where in going to the second line used the fact that the Φ0 is an eigenvalue with eigenenergy
E0 .
On the other hand, we equivalently could have written
f (k) =
=
=
=
=

E
1 D
Φ0 |ρ†k (H − E0 )ρk |Φ0
N
E D
Ei
1 hD
Φ0 |ρ†k Hρk |Φ0 − Φ0 |Hρ†k ρk |Φ0NE−1 D
Φ0 |[H, ρ†k ] ρk |Φ0NE−1 D
Φ0 |[H, ρ†−k ] ρ−k |Φ0NE−1 D
Φ0 |[H, ρk ] ρ†k |Φ0
N

(5.8)

66

CHAPTER 5. FEYNMAN THEORY OF HELIUM-4

where in going to second last line we have used the fact that, due to isotropy of the fluid
we expect f (k) = f (−k) and in going to the last line we use that ρ†k = ρ−k .
Putting together Eq. 5.7 and 5.8 we obtain the double commutator formf (k) =Ehi
1 D
Φ0 ρ†k , [H, ρk ] Φ0
2N

(5.9)

We now must determine the double commutator, which will turn out to be a simple number
rather than an operator! To do this we notice recall the Hamiltonian has three terms, akinetic term K̂, a one body potential term V̂ and a two body interaction term Û (it would
not matter if we had three or four body terms etc). We write
H = K̂ + V̂ + Û
where
X −~2 ∇2
i

K̂ =

2m

i

V̂

X

=

V (ri )iÛ1X
U (ri − rj )
2

=

i,j

And recall thatρk =X

eik·ri = ρ†−k(5.10)i

We now want to calculate [H, ρk ]. Here ρ only contains the operator r (not the conjugate
operator ∇) and this then commutes with both U and V which also only contain r (andnot ∇). However, ρk does not commute with the kinetic term, so we simplify[H, ρk ] = [K̂, ρk ]X −~2 ∇2 Xi,eik·rj 
= 
2m
j

i

=
=

−~2 X h
2m

j
2
X
−~

2m

∇2j , eik·rj

ieik·r −k2 + 2ik · ∇jj

We then want to further calculate the double commutatorhiρ†k , [H, ρk ] =i−~2 X h † ik·rj
ρk , e−k2 + 2ik · ∇j2mj

5.2. SINGLE MODE EXCITATION SPECTRUM

67

The −k2 term is a scalar so it commutes and we haveihi−~2 X h † ik·rj(ik · ∇j )
ρ†k , [H, ρk ] =ρk , emj"#−~2 X X −ik·ri ik·rj=e,e(ik · ∇j )m=j2−~ N hie−ik·r , eik·r (ik · ∇)im
~2 k2 N=
(5.11)mwhich is simply a scalar rather than an operator, as claimed. Thus plugging into Eq. 5.9
we obtain~2 k2
f (k) =
2m
We note that this double commutator calculation is used in several other common contexts,
such as in the derivation of the well-known f -sum rule, and Kohn’s theorem. (Don’t worry
if you don’t know about these! We will run into it again in section 8.3.2 below).
Now plugging back into our original formula for the excitation energy, Eq. 5.5 we
have
∆k =

f (k)
~2 k 2=S(k)2mS(k)(5.12)

This is a rather remarkable result! In this approxiation5 , the low energy excitation spectrum is completely determined by the structure factor!
In Fig. 5.6 the top curve is the prediction of the Feynman theory using Eq. 5.12
with the experimentally measured structure factor S(k) as input. The bottom curve is theexperimentally measured excitation spectrum (using inelastic neutron scattering). The
structure of the two curves is quite similar.
Note that in the long wavelength (small k) limit the Feynman theory matches theexperimental results exactly. At higher k, the Feynman theory gives an excitation energy
which is higher than the real excitation energy. This is to be expected. In fact, theFeynman theory can be taken to be a rigorous upper bound: What we have done is avariational calculation using ρk |Φ0 i as the trial state — even though this isn’t exactly
the eigenstate (except in the long wavelength limit where it becomes exact). In the spaceof states orthogonal to the ground state, the exact state we are looking for is the lowest
energy state, so our trial state is necessarily higher in energy than the exact state.
The minimum in energy at intermediate k is known as the roton minimum6 , and it
5

It is an approximation here because we are using a trial wavefunction for the excited state.
The name “roton” is a historical accident. Landau’s early guess was that the observed low energy
excitations must have something to do with the rotation of the fluid — since all fluids must be able torotate, and the superfluid component has no curl (Landau did not know about vortices at that time). The
excitation was thus named the “roton”. Later it became clear that the roton excitation is just a minimumin the phonon spectrum and is unrelated to rotation. Unfortunately, the name “roton” stuck.
6

68

CHAPTER 5. FEYNMAN THEORY OF HELIUM-4

Figure 5.6: Feynman Theory and Low energy excitations of Helium. Figure from Pines
and Nozieres. Top curve is prediction of Feynman theory. Bottom curve is experimental
measurement of the excitation spectrum. Curves in between are modifications of the
Feynman theory.

5.2. SINGLE MODE EXCITATION SPECTRUM

69

corresponds (approximately) to the peak in the structure factor (See Fig. 5.5) at k roughly
2π/a with a the interparticle spacing.
In fact, Landau predicted this minimum before Feynman’s theory. He based his
prediction on specific heat data. Knowing that the spectrum is linear E(k) = ~vk at
small k, the usual Debye calculation gives a specific heat Cv ∼ T 3 at low T . However,
the experiments showed at higher T an additional contribution that appeared to turn onroughly as ∼ e−Er /(kb T ) for some constant Er . Landau relized that a spectrum of the shape
shown in Fig. 5.6 would give such a term where Er is the energy of the roton minimum.

70

CHAPTER 5. FEYNMAN THEORY OF HELIUM-4

Chapter 6

Ginzburg-Landau Theory
6.1

Neutral Superfluids

Let us begin by recalling the Gross-Pitaevskii energy we derived earlier for weakly interacting (uncharged) Bose gases in Eq. 4.13 which we rewrite here


Z
hHi =

dr

~2
U
|∇ψ(r)|2 + V (r)|ψ(r)|2 + |ψ(r)|4
2m
2


(6.1)

To anyone who has studied the Landau theory of phase transitions (and the associatedGinzburg-Landau1 theory of fluctuations) this form is not surprising. Near a phase transition we typically expand a free energy order by order, writing down all terms that areallowed by symmetry. In the case of a neutral superfluid with a complex order parameter
we generally expect a free energy functional of the formZF [ψ(r)] =dr~2|∇ψ|2 + α2 |ψ|2 + α4 |ψ|4
2m


(6.2)

where |ψ|2 is interpreted as the superfluid density2 . Not only does this agree with GrossPitaevskii, but on symmetry grounds, this must be the form of the free energy near thephase transition.
1

Vitali Ginzburg won a Nobel prize in 2003 for his eary work in the 1950s on what is now called
the Ginzburg-Landau theory of superconductors. (We will come to this later in the current chapter).
Ginzburg’s Nobel biography is well worth a read — as a Jew in Stalin’s USSR, he barely skirted death on
several occassions!
2
Generally we might have thought that the coefficient of the gradient term might have been an arbitrary
fit parameter, but we fix it using the fact that if we write ψ = |ψ|eik·x we want ~k to represent a particle
momentum.

71

72

CHAPTER 6. GINZBURG-LANDAU THEORY

Figure 6.1: Left: The Ginzburg-Landau potential above Tc . Here the minimum is at ψ = 0
indicating no superfluid density. Right: The Ginzburg-Landau potential below Tc . Herethe minimum is at a nonzero value of ψ indicating finite superfluid density.

6.1.1

Spatially Uniform Solution

To begin with, let us consider a uniform system — so that the gradient is zero. As in theusual theory of phase transitions we have
α2 = a(T − Tc )with a > 0 and α4 > 0 as well for stability (i.e, so that for large enough ψ the free energy
always increases.
Often we will think of the non-gradient terms of the free energy as being a potential
V (ψ) = α2 |ψ|2 + α4 |ψ 4 |= a(T − Tc )|ψ|2 + α4 |ψ 4 |with a > 0.
For T > Tc (so α2 > 0) the potential V (ψ) looks like the left of Fig. 6.1, where theminimum is at |ψ| = 0 meaning that the ground state has no supefluid density.
On the other hand, for T < Tc (so α2 < 0) the minimum is at some finite nonzero
value of |ψ| (which we label ψ0 ) meaning there is a finite superfluid density ψ02 in the
ground state.
It is simple to minimize V (|ψ|) to obtain|ψ|2 =|α2 |= ψ02 = ns
2α4

(6.3)

6.1. NEUTRAL SUPERFLUIDS

73

Figure 6.2: Left: The Ginzburg-Landau potential below Tc drawn as a function of thecomplex (2d) field ψ. The two orthogonal directions in the horizontal plane can be taken
to be the real and imaginary parts of ψ. The minimum lies at the bottom of the trough(the bottom of the rim of the hat)which is the density of the superfluid fraction.
Since ψ is actually a complex field, it should really be drawn as the so-called Mexican
hat potential shown in Fig. 6.2. Note that the potential V (ψ) is a function of |ψ| only andis independent of the complex phase of ψ. This means that the ground state is degenerate— the ground state has magnitude ψ0 but all possible complex phases have the sameenergy. In Fig. 6.2 we see this as the symmetry of the picture under rotation around thecentral z-axis. In particular, the minimum of the potential is a (degenerate) circle aroundthe bottom of the rim of the Mexican hat.
Whenever we have multiple global ground states related to each other by a continuous symmetry (the result of spontaneously broken continuous symmetry) we should havea Goldstone boson (or “Nambu-Goldstone” boson3 ). The boson in this case is the low energy excitation associated with the ground state locally reorienting this degree of freedom.
In this case the Goldstone boson is nothing more than phase fluctuations of ψ (withoutchanging the magnitude of ψ). In other words, the phase changes slowly as a function of
position. These excitations are the superfluid phonons, or bogoliubon excitations.

6.1.2

Spatial Dependence: Ginzburg Landau Theory

Let us now return to the free energy Eq. 6.2 and reintroduce the spatial derivative (i.e., nolonger assume ψ has no spatial dependence). Here we will assume we are in the superfluid
phase, i.e., T < Tc . To derive an equation for ψ we take the functional derivative of Eq. 6.2
3

Named for Yoichiro Nambu, a Nobel Laureate, for his work on spontaneous symmetry breaking, andJeffrey Goldstone, who generalized Nambu’s discovery. Nambu joked that he won the Nobel prize mainly
by outliving his enemies. He was 87 when he won the prize.

74

CHAPTER 6. GINZBURG-LANDAU THEORY

to minimize the free energy
δF
δψ ∗
−

= 0

~2 2
∇ ψ + α2 ψ + 2α4 |ψ|2 ψ = 02m

which is a non-linear Schroedinger equation (See Eq. 4.13). Recalling that in a uniform
system below Tc we have |ψ0 |2 = |α2 |/(2α4 ), it is useful to define a new field
f = ψ/|ψ0 |so that in the uniform system we just have f = 1.
Writing ψ = f |ψ0 | and plugging into our nonlinear Schroedinger equation we obtain−~2 2∇ f + α2 f + 2α4 |ψ0 |2 |f |2 f = 0
2m

(6.4)

Dividing through by α2 and using |ψ0 |α4 = |α2 |/2 we obtainξ 2 ∇2 f + f − f |f |2 = 0

(6.5)

where
s
ξ=

~2
2m|α2 |

(6.6)

is known as the Ginzburg-Landau coherence length. This is the natural length scaleassociated with the nonlinear Schroedinger equation. It is effectively a stiffness length fortwising the phase or magnitude of the field ψ.
At low temperatures, the Ginzburg-Landau coherence length for superfluid 4 He is on
the order of ξ ≈ 1 Angstrom. However, note that at the critical temperature α2 = a(T −Tc )changes sign (therefore going through zero) so as we approach Tc we must have ξ divergingasξ ∼ |T − Tc |−1/2
Note that Ginzburg-Landau theory is a mean field theory. A more accurate analysisof the problem will obtain a different exponent4 .
Despite the fact that our equation is nonlinear (and therefore hard to solve) some
exact solutions are in fact possible.
Solution 1. Uniform solution:
f =1
4

everywhere

This is 3d, XY universality class and ν = .671...

6.1. NEUTRAL SUPERFLUIDS

75

Figure 6.3: The healing of the order parameter from a hard wall is a tanh function.
(This one is obvious)
Solution 2. Hard wall boundary condition: Wexf = tanh √
2ξ

(6.7)

This solution, shown roughly in Fig.6.3
f →1

at large x

f →0

at x → 0

The proof that the tanh form in Eq. 6.7 satisfies the nonlinear Schroedinger equation 6.5
is a bit of an exercise in hypergeometric functions. We leave this to the reader.
Solution 3. A Single Vortex
We consider a vortex line going along the z-axis. In cylindrical coordinates, we
imposef = |f (r)|eiθ
Plugging this into our nonlinear Schroedinger Equation 6.5, using

 
1 ∂ ∂
1
∂
2
∇ f=
r
+ 2
f
r ∂r ∂r r∂θ
we obtain


1 ∂ ∂
1
ξ
r
−
|f | + |f | − |f |3 = 0r ∂r ∂r r2
This leaves us with something that is unfortunately not analytically solvable. However,
we can obtain two limits.
2



Small r: First, if we consider small r, we know that f needs to vanish as r → 0.
Thus we can throw out |f |3 term. We then focus on the most singular terms which arethose within the brackets. It is clear these can be made zero by using|f | = αr + . . .

76

CHAPTER 6. GINZBURG-LANDAU THEORY

where the . . . terms are terms higher order in r which one obtains when you reinstate theterms we just threw out. Thus we conclude that |f | ∼ r for small r
Large r: We know far from a vortex, we should get the bulk solution |f | = 1 so welook for a perturbation to this by substituting |f | = 1 − y. Keeping only terms linear iny and looking in the large r limit where the derivatives are suppresed we obtainy = ξ 2 /(2r2 ) + . . .
so|f | = 1 − ξ 2 /(2r2 ) + . . .

6.2

Charged Superfluids (i.e., Superconductors)

The same approach can be applied to charged superfluids with the additional complicationthat we need to keep to keep track of the electromagnetic field. We can write a free energyfunctional now in the form


Z
1
∗
2
2
4
F [ψ, E, B] =dr|(−i~∇ − e A)ψ| + α2 |ψ| + α4 |ψ|2m∗

Z 0  2
B
E2
dr
+
+ 0
(6.8)
2µ0
2
and as usual α4 > 0 and α2 = a(T − Tc ) with a > 0. Here we have minimally coupled
the momentum operator to the vector potential, and we have added the usual energy ofan electromagnetic field (the second line). Note that the integral on the second line has aprime on it, this is to point out that integral of the electromagnetic field extends over allof space – the field energy still needs to be considered in regions outside of the physicalsample, which may be finite.
As in the case of the neutral superfluid, below Tc we haves
|α2 | p ∗|ψ| = ψ0 == ns
2α4with n∗s the number of particles in the superfluid (superconducting) fraction (where here
we are assuming a uniform solution with no applied vector potential to make more precise
analogy with the neutral superfluid).

6.2.1

Anderson-Higgs Mechanism

As in the case of the neutral superfluid, one might expect that there would be a Goldstone
mode associated with changing the phase of the order parameter — i.e., moving the fieldaround the rim of the Mexican hat as a function of position. But in fact, there is no6.2. CHARGED SUPERFLUIDS (I.E., SUPERCONDUCTORS)

77

such low energy mode of the superfluid in a superconductor! This is due to the so-called“Anderson-Higgs” mechanism5 which we will now discuss.
It is perhaps more natural to discuss the Higgs mechanisms in terms of Lagrangians
rather than free energies — since that makes it easier to keep track of dynamics. However,so far we have been working in terms of free energies, and we will therefore continue using
this approach since it is easier to compare to what we have done earlier. However, one
really must be careful here since one can draw incorrect conclusions about the dynamics.67 .
For those who are worried about honesty we do a brief Lagrangian derivation in the chapter
appendix.
First we note that the free energy is invariant under the gauge transform:ψ → eiα ψe∗ A → e∗ A + ∇α/~with α real. This gauge transform can be a function of both time and position α(t, r).
Recalling that for T < Tc we haves|ψ| = ψ0 =

|α2 |
2α4

which minimizes the potential V (ψ). Here ψ0 is the so-called Vacuum Expectation Valueof the field, or VEV in high-energy language. Let us now consider fluctuations aroundthis minimum.
We writeψ = (ψ0 + h)eiχ
5

(6.9)

Perhaps the history of the Higgs mechanism starts with Nambu, who wrote a key paper in 1960
starting to understand the ideas of spontaneous breaking of a gauge symmetry. Then in 1962, Anderson
fully described what we now call the Higgs mechanism in the context of superconductors. In 1964, in Higgs’
first paper (in the first paragraph) he credits Anderson saying (a) This is nothing more than extending
what Anderson did to the electro-weak interactions and (b) Anderson himself suggested that a similar
mechanism might exist in high energy. More detailed papers appeared the same year expanding further on
the idea by Brout and Englert; and by Guralnik, Hagen, and Kibble. Anderson probably was entitled topart of the Nobel shared by Higgs and Englert (Brout had passed away before the prize was awarded), butAnderson already had a prize for what is now called “Anderson Localization”, which is totally unrelated(except for the fact that both require someone very smart to figure them out).
6
The right way to do this is to consider F to be the Hamiltonian for ψ and write a Lagrangian density
as L = |∂t ψ|2 − H with H the Hamiltonian density. And here lies a crucial subtlety. The dynamical termis |∂t ψ|2 which looks relativistic rather than iψ ∗ ∂t ψ which looks more like the non-relativistic lagrangian.
The reason for this form is that superconductors have a natural particle-hole symmetry (i.e., they aredescribed by a “relativistic” 2-spinor) as we will see in chapter 11. For a chargeless unpaired superfluid,
however, we should use the nonrelativistic version! . This issue is discussed further in the chapter appendix
7
In quantum mechanics, the dynamics of the Hamiltonian are not defined unless you specify the commutation relations of the operators (or in classical mechanics, specify the canonical coordinates). For example,
if we write H = p̂2 /(2m) + (κ/2)x̂2 this looks like a Harmonic oscillator, but it only has that dynamics
once we specify that p and x are canonical coordinates so dp/dt = −∂H/∂x and dx/dt = ∂H/∂p. Or inquantum mechanics we specify commutations between p̂ and x̂. However, if we start with a Lagrangian,we can derive the canonical commutations!

78

CHAPTER 6. GINZBURG-LANDAU THEORY

where h and χ are both real. Here, h is the fluctuation of the magnitude of ψ, and χ isfluctuation of the angle around the “rim” of the Mexican hat. Near the bottom of the rim
the energy should be quadratic in h describing deviation up the hill and away from thebottom of the rim. Inedeed,
α2 |ψ|2 + α4 |ψ|4 = α2 (ψ0 + h)2 + α4 (ψ0 + h)4 = const + 2|α2 |h2 + . . .
Now we would like to plug Eq. 6.9 into the free energy Eq. 6.8. However, before doing
this, let us choose a gauge that absorbes the phase χ in Eq. 6.9. I.e., we gauge transformto make ψ real
ψ → ψ 0 = e−iχ ψ = ψ0 + h∗∗

0(6.10)∗

e A → e A = e A − ~∇χWith this transformation we have


Z
1
∗ 0
2
2
F =
dr
|(−i~∇ − e A )(ψ0 + h)| + 2|α2 |h + . . .
2m∗
Z 0  2
B
E2
+
dr
+ 0
2µ0
2

(6.11)

Note that both the electric and magnetic fields corresponding to A0 are the same as thosecorresponding to A (the fields are gauge invariant).
This free energy can be rewritten as
 2


Z
~
ψ02 (e∗ )2 0 2
2
2
F =
dr
+
+
.
.
.
|∇h|
+
2|α
|h
A
2
2m∗
2m∗

Z 0  2
B
E2
+
dr
+ 0
2µ0
2

(6.12)

One might naively expect there would have been terms i(∇h) · A0 but such terms have tocancel on account of being imaginary (they do!).
We will now examine the “dynamics” implied by this free energy. As mentioned
above, one has to be very careful about deducing dynamics from a free energy and for lesshand-waving arguments, see the chapter appendix.
Let us first consider the terms involving h. This field is massive, meaning that there
are no low energy excitation, the mass here being given by the α2 term8 .
The term with A0 2 represents a mass for the electromagnetic photon. Like the h2term, this term gives a finite energy to any nonzero value of the electromagentic field, andhence is a mass. We should compare this to the case of conventional electromagnetism
where we have only the B2 and E2 terms.
B2 = (∇ × A)2E2 = (−∇A0 − ∂t A)28

To see this, we see that any nonzero value of h gives a positive energy. The lowest energy we can get
for a normalized function h(r) is 2|α2 | if h is uniform in space.

6.2. CHARGED SUPERFLUIDS (I.E., SUPERCONDUCTORS)
T > TcOscillations of ψ in real and complex direciton
2 polarizations of massless photon
4 total degrees of freedom=

79

T < TcOscillation of h in real direction only
3 polarizations of massive photon
4 total degrees of freedom

Table 6.1: Conservation of total number of degrees of freedom in the Higgs mechanism.
Since both of these have derivatives, the energy of the electromagnetic field goes to zero atlong wavelength (we should know this since the energy of a photon is 2π~c/λ). However,
now with the A0 term added, the energy of a photon is finite even at very long wavelenth.
What does it mean that the photon is massive? What is means is that even along wavelength photon costs energy when it is inside a superconductor. If we imagine
a photon outside a superconductor, when it tries to go into the superconductor it will
experience an energetic barrier and will be repelled — this is precisely the Meissner effect:
electromagnetism is expelled from superconductors!
What has happened here? The so-called Higgs mechanism is this process by which
the Goldstone mode associated with the phase of the order parameter disappears (we got
rid of it by absorbing the phase in Eq. 6.10), but in the process the photon becomes
massive9 . Note also that where a massless photon has only two polarizations (due to thefact that it moves at the speed of light), a massive photon which moves slower than the
speed of light (like a phonon) has three possible polarizations — inlcuding a longitudinalmode. Thus, although we lose a degree of freedom from the oscillation of the field ψ, we
gain a degree of freedom in a new oscillation mode of the photon. As shown in Table 6.1
the total number of degrees of freedom is the same above and below Tc .
A comment I feel compelled to add before leaving the discussion of Higgs mechanism:
Particularly in the media, one often hears people say that the Higgs boson “gives” mass
to particles10 . This is not true. The Anderson-Higgs boson in this case is the field h.
However, it is the vacuum expectation value (VEV) ψ0 that gives mass to photon. The
Higgs boson represents the oscillations of the ψ field that remain once it has aquired a
VEV (And further these oscillations become massive since the massless goldstone mode
disappeared when we fixed the phase).
One final comment to remove some possible confusion is regarding the non-existenceof a Higgs mode for superfluid Helium. In Helium we have a mexican hat potential(Fig. 6.2), and we concluded that there is a low energy (Goldstone) mode associated withlong wavelength phase twists around the bottom rim of the hat. Why then do we nothave a high energy “Higgs” mode associated with oscillations that go “up the Hill” of thepotential? Indeed we do not have such a mode. We know this because we calculated the9
The easy to remember catch phrase is that the gauge boson has become massive because it ate thegoldstone boson... with the obvious reference being that one becomes massive when one eats. There may
be other mechanisms by which the gauge boson gets some exercise and loses weight (that is a joke).
10
It is also often said that the Higgs mechanism gives mass to all the particles in the universe. This isalso not true. Hadrons get their mass mainly from gluon energy!

80

CHAPTER 6. GINZBURG-LANDAU THEORY

spectrum of excitations for example, in section 4.5 or for homework (question 5), and we
did not find such a gapped mode! The reason for this absence is hidden in footnote 6:
For the superconductor, due to particle-hole symmetry (discussed later in chapter 11) the
dynamical terms of the action takes a different form from the dynamical terms for thesuperfluid and so the analogy of a superconductor being simply a charged superfluid is
not really perfect here!11

6.2.2

Equations of Motion

As in our Gross-Pitaevskii approach, one can differentiate the free energy to obtain equations of motion.
δF
=0
δψ ∗
yielding the nonlinear Schroedinger equation coupled to a vector potential


1
∗
2
2
(−i~∇ − e A) + α2 + 2α4 |ψ(r)| ψ(r) = 0
2m∗
Similarly we can differentiate with respect to the vector potential in our fixed gauge
δF
=0
δAµ
and after some exercize in functional differentiation we obtain
∂E
e∗
∇×B
− 0
=
[ψ ∗ {(p − e∗ A)ψ} − {(p + e∗ A)ψ ∗ } ψ]
µ0
∂t
2m∗
and from Maxwell’s equation on the left we can identify that the right hand side must
be the electrical current density j as we had previously claimed. If the ground state ψ isuniform, we thus recover the London equationj=(e∗ )2 2(e∗ )2 ∗
|ψ|
A
=
n A
m∗
m∗ s

where we have now identified |ψ|2 to be the superfluid density.

6.2.3

Energetics from Ginzburg-Landau theory: Type I and Type IIsuperconductors revisited

From the Ginzburg-Landau potential
V (ψ) = α2 |ψ|2 + α4 |ψ|4
11

A fun exercise is to use the method of homework questoin 5 to calculate the spectrum of a relativistic
uncharged superfluid, and see that we do indeed develop an additional excitation mode of finite energy!

6.2. CHARGED SUPERFLUIDS (I.E., SUPERCONDUCTORS)

81

recall from Eq. 6.3 that we derived that the minimum free energy occurs at|ψ|2 = ψ02 =

|α2 |
= n∗s
2α4

where we have identified this quantity as the superfluid density (we have inserted the ∗as previously to again indicate that we do not know how many electrons might be boundtogether to form one boson).
Recall also from Eq. 3.8 the London penetration depth is given byssm∗
2α4 m∗λ==(e∗ )2 n∗s µ0(e∗ )2 |α2 |µ0(6.13)

which gives us the decay length of a magnetic field when it penetrates into a superconductor.
On the other hand, we also derived in Eq. 6.6 a phase stiffness length (or coherence
length)sξ=~22m∗ |α2 |(6.14)

Note that both of these lengths scale as1ξ and λ ∼ p∼ |T − Tc |−1/2|α2 |(6.15)

when we are close to Tc .
Note that this exponent of −1/2 is a so-called mean-field exponent. A more careful
renormalization group calculation will give a different exponent. However, it turns outthat mean field theory tends to be extremely good for superconductors, except when one
gets exceedingly close to the critical temperature12 .
The free energy density is given bymin(F )/V = F [ψ0 ]/V= α2 |ψ0 |2 + α4 |ψ0 |4
−α22
=
4α4

(6.16)

This energy is the energy that is gained by allowing ψ0 to become nonzero — i.e., it is
the condensation energy per unit volume. Back in Eq. 3.18 we already calculated the12
The reason that mean field theory works so well is that that the coherence length ξ in a superconductor
is extremely large compared to micscopic length scales. The typical fluctations are of energy kb T over asize ξ D with D the dimensionality of space giving an energy density for the fluctuations of kB T /ξ D ∼|T − Tc |D/2 . We should compare this mean fluctuation energy to the mean field ground state energy
density which is α22 /(4α4 ) ∼ |T − Tc |2 (See Eq. 6.16). Close enough to Tc the fluctuation always becomeslarger than the mean field energy (for D < 4) so mean field theory always fails. However, if the prefactor
y in ξ = y|T − Tc |−1/2 is extremely large, as it is in superconductors, then one must go very close to Tc
before this failure happens.

82

CHAPTER 6. GINZBURG-LANDAU THEORY

Figure 6.4: The critical field scales linearly with T − Tc close to Tc , at least within a mean
field theory calculation.
condensation energy per unit volume and using thermodynamics we related it to thecritical field Hc ascondensation energy per volume =α2µ0 Hc2= 2
2
4α4Thus in terms of the Ginzburg-Landau parameters we obtainr
1∼ (T − Tc )Hc = |α2 |2µ0 α4(6.17)as shown in Fig. 6.4√
In Eq. 6.17 we have the combination |α2 |/ α4 and it is useful to try to constructthis combination out of the known quantities of the coherence length ξ in Eq. 6.14 andthe penentration depth λ in Eq. 6.13. If we take the combination 
1|α2 | e∗ √
=√
µ0 ,
ξλ
α4 ~
we note that a factor of 2m∗ has cancelled here. Recalling that the superconducting flux
quantum is φ0 = 2π~/(e∗ ) we can rearrange the prior two equations to get
Hc =

1
φ0
√
ξλ 2πµ0 2(6.18)

What this means is that if we draw a box of area λ on one side and ξ on the other side,
Hc , the magnetic field necessary to completely destroy superconductivity, is the magnetic
field where we have roughly one flux quantum penetrating this area as shown in the leftof Fig. 6.5.
Now recall from the end of section 3.1.5 that the distinction between Type I andType II superconductors depends on the energy per unit length , or “tension” of a vortex
line. To remind you we have the lower critical field (where magnetic field first can penetrate
a superconductor)

Hc1 =
2φ0

6.2. CHARGED SUPERFLUIDS (I.E., SUPERCONDUCTORS)

Hc

83

Hc1

Figure 6.5: Pictoral Descriptions of Critical Fields. Left: Hc can be described as a fieldsuch that one flux quantum penetrates a box of size λ by ξ. Right: Hc1 can be described
as a field such that one flux quantum penetrates a box of size λ by λ.
and if this quantity is lower than Hc then we have a Type II superconductor with a vortex(Abrikosov) phase for Hc1 < H < Hc2 , but if Hc1 > Hc then no vortex phase exists. It
is thus useful to try to make an estimate of the vortex line tension , or energy per unit
length of the vortex.
Roughly the energy of a vortex is due to the fact that the superconductivity is
destroyed in the vortex core. Since the healing length of the order parameter is thecoherence length, this means the core should be roughly size ξ. Thus the energy per unit
length of the core is the condensation energy denisty Hc2 /(2µ0 ) times the area of the vortex
core, πξ 2 . We thus have
EnergyH2H2=  = c πξ 2 ≈ c ξ 2
Length2µ0µ0where here we are dropping order unity factors (since we are not going to get them rightanyway!). We thus haveµ0 Hc2 ξ 2
φ0 1

≈
≈ 2
(6.19)
Hc1 =
2φ0
φ0
λ µ0
where we are again dropping order one factors and in the last step we have used theexpression Eq. 6.18 for Hc . The meaning of this result is that Hc1 , the lowest field wherea vortex can penetrate a superconductor, is given by the magnetic field necessary suchthat a single flux quantum penetrates a square area of size λ by λ, as shown in the rightof Fig. 6.5.
Comparing Eq. 6.18 to 6.19 we see that in order for Hc1 to be less than Hc we must
have
λ
&1
ξ
which should be the necessary condition for a Type II superconductor. (We have dropped
constants of order unity, so we should not expect to get the precise value correct.)Conventionally one defines the dimensionless ratio known as the Ginzburg-Landau

84

CHAPTER 6. GINZBURG-LANDAU THEORY

Figure 6.6: The order parameter heals over a short length scale ξ whereas the magnetic
field is screened over a longer scale λ.
parameterrλ(T )α4 m ∗κ==2ξ(T )µ0 ~e∗

which is temperature independent (at least whenever Ginzburg-Landau theory is valid).
A more precise calculation can be done to give the condition for a Type II superconductor is that κ > √12 .
In order to have a vortex we must have a type II superconductor, meaning, κ & 1.
This means the London penetration depth, the screening length for magnetic fields, islonger than the coherence length, the healing length for the order parameter. Thus arounda vortex core, the order parameter heals quickly whereas the magnetic field is screened
slowly. This is roughly depicted in Fig. 6.6.
In fact a more precise calculation will show that the energy of an isolated vortex
will have a log κ correction that we ignored. By calculating the energy of the vortex core,
we found that the energy per unit length is (See Eq. 6.19)
 ≈ 2µ0 Hc2 ξ 2
However, the actual result includes a term of the form
µ0 Hc2 ξ 2 log(λ/ξ)
To understand where this term comes from, we roughly say that at radius less than λwe can neglect the screening due to the electromagnetism and the order parameter aroundthe vortex core will be similar to what it is for a neutral superfluid. In this case, theorder parameter is ψ ∼ eiθ so there is a velocity v ∼ ∇θ ∼ 1/r in the θ̂ direction. The
Ginzburg-Landau kinetic energy density will give us something of the form
KE =

mv 2
1
∼ 2
2
r

6.3. LAGRANGIAN FORMULATIONS

85

we then have a total energy
ZE∼1
d r 2 ∼
r
2

Z λ
dr
ξ

1
∼ log(λ/ξ)
rwhere we have cut off the integral at ξ because below this, the order parameter drops
and we also cut off at λ because above this radius the screening makes the current drop
exponentially.

6.3

Lagrangian Formulations

6.3.1

Nonrelativisitic Neutral Superfluid

Let us start by writing an appropriate Lagrangian density for a neutral (nonrelativistic)
bose superfluid. This will look like the Ginzburg Landau free energy (Eq. 6.2), except(1) there is a dynamical term with a time derivative and (2) the sign of the free energy is
flipped. We have
 2~ |∇ψ|2∗
2
4
L = i~ψ ∂t ψ −+ α2 |ψ| + α4 |ψ |2mWriting the Euler-Lagrange equations for this gives us
i~∂t ψ = −~2 ∇2
ψ + α2 ψ + 2α4 |ψ|2 ψ
2mwhich is the time-dependent Ginzburg-Landau equation. It is a homework assignment toderive the excitation spectrum of this equation. We find a single gapless acoustic mode.
You might try deriving the same result by plugging
Ψ = (|ψ0 | + h)eiχ

(6.20)

into thisp equation (with h and θ both real) and expanding for small v and small χ, whereψ0 = −α2 /(2α4 ) is the magnitude of the order parameter at the bottom of the mexican
hat.

6.3.2

Relativisitic Neutral Superfluid

Now let us consider the relativistic version of a (neutral) bose superfluid. We will write
this in relativistic notation for simplicity (and drop factors of ~ etc). We start with our
Lagrangian density
L = (∂ µ ψ ∗ )(∂µ ψ) − α2 |ψ|2 − α4 |ψ|4
The Euler-Lagrange equations give us∂µ ∂ µ ψ + α2 ψ + 2α4 |ψ|2 ψ = 0

86

CHAPTER 6. GINZBURG-LANDAU THEORY

If we plug in the form of Eq. 6.20 and expand for small h and small χ we obtain two
equations (one for the real and one for the imaginary part of the Euler-Lagrange equation)
which are given by(∂µ ∂ µ + 2|α2 |)h = (∂t2 − ∇2 + 2|α2 |)h = 0(∂µ ∂ µ )χ = (∂t2 − ∇2 )χ = 0
These correspondingly have spectraω =pq 2 + 2|α2 |ω = |q|The latter is a gapless (massless) mode, corresponding to the “Goldstone” mode (ie motion
aroundthe rim of the mexican hat), the former is gapped (massive) with minimum energyp2|α2 |.

6.3.3

Electromagnetism

Before looking at the case of a charged superfluid, let us consider pure electromagnetism.
We can the field strength tensor in terms of the vector potential as Fµν = ∂µ Aν − ∂ν Aµ .
The Lagrangian density can then be written as
1
L = − Fµν F µν
4
The Euler lagrange equations ∂µF µν = 0 then generate the free space maxwell equations∂t E = ∇ × B and ∇ · E = 0. (The remaining maxwell equations are guaranteed by thefact that we are writing the fields E and B in terms of a vector potential).

6.3.4

Charged Relativistic Superfluid

We couple the relativisitic superfluid to electromagnetism by using the covraiant derivativeDµ = (∂µ − ieAµ )So that our Lagrangian can now be written as
1
L = (Dµ ψ ∗ )(Dµ ψ) − α2 |ψ|2 − α4 |ψ|4 − Fµν F µν
4where we have added the Lagrangian for electromagneism at the end. Here as in Eq. 6.9
and 6.10 we choose a gauge to absorb the phase χ given by
eA0µ = eAµ + i∂µ χ
Then expanding around the minimum (i.e., for small h) we obtain a new Lagrangian12
L = (∂µ h)(∂ µ h) + 2|α2 |h2 + |ψ0 |2 A0 − Fµν F µν + . . .
4

6.3. LAGRANGIAN FORMULATIONS

87

Where F is now the field strength for A0 which is the same as the field strenght of A bygauge invariance. The first two terms in this Lagrangaian are the massive higgs mode.
Applying Euler-Lagrange for h we obtain(∂µ ∂ µ − 2|α2 |)h = (∂t2 − ∇2 − 2|α2 |)h = 0
which gives us the massive field with spectumpω = q 2 + 2|α2 |The last two terms in this Lagrangian, are known as the “Proca” lagrangian for electromagnetism, and they have massive dynamics. It is easy enough to derive the euler lagrange
equations∂µ F µν = |ψ0 |2 Aνor∂µ (∂ µ Aν − ∂ ν Aµ ) = |ψ0 |2 Aν
µTo check that this is massive, lets attempt a plane wave solution Aν = B ν eikµ x , then lets
go to the long wavelength limit where the spatial part of k is zero, and k0 = ω so we getω 2 = |ψ0 |2
as the lowest frequency solution we can have. I..e, there is a mass gap.

88

CHAPTER 6. GINZBURG-LANDAU THEORYChapter 7

Interacting Fermions
7.1

Why study fermions

Having spent the entire course so far discussing bosons, we now turn our attention tofermions, and interacting fermions in particular.
Why should we study interacting fermions? There are several good reasons for
thisElectrons (which are fermions) are the world1 . The physics of every metal,
every semiconductor, every insulator, is completely controlled by the behavior of its electrons. In fact essentially all of chemistry is nothing more than the study of the behaviorof electrons in different environments.
Usually when a condensed matter physicist says they are studying fermions, theymean they are studying electrons. There is one notable exception2 to this: the fermion 3 He.
This fermionic isotope3 of Helium is a favorite of condensed matter. As with the bosonicversion, the light mass of Helium prevents it from forming a solid at low temperatures, andinstead we get a quantum fluid — or Fermi liquid. However, as compared to electrons, it
has short range interactions rather than long range Coulomb interactions, and this makes
its physics somewhat simpler in many ways4 . An enormous amount has been learned from1
You can imagine all the electrons getting together and holding hands and singing “We are the world...!”.
At this point, anyone reading this footnote probably thinks I have gone crazy, but I’ve been typing thesenotes for a lot of hours in a row, and yes, maybe I’m going a bit batty.
2
In the modern era of cold atomic gases, there are now experiments on fermionic cold atoms too.
3
Helium-3 (3 He) was discovered experimentally in 1934 by Mark Oliphant at Cambridge. A mere few
milligrams of the material became available for experiments as an offshoot of the post-war nuclear program.
Now it is much more plentiful due to the fact that it is a waste product of nuclear missiles (Tritium, afuel of hydrogen bombs, has a half-life of 12 years, and decays into 3 He). The price of Helium 3 fluctuates
between 100 and 10000 dollars per litre, depending on the global supply at the time.
4
One interesting feature we will not discuss in depth is that at extremely low temperaturures (meaning
sub milli Kelvin) 3 He actually becomes a superfluid. The experimental discovery of this in 1972 earned
Lee, Osheroff and Richardson a Nobel prize in 1996. Theoretical work on the subject earned a Nobel prize

89

90

CHAPTER 7. INTERACTING FERMIONS

the study of this fermion. We will return to discussion of this below in section 9.1.
We need to understand strongly interacting fermions. In all of our introductory courses on solid state physics, we learn about metals, semiconductors, and insulatorscompletely ignoring the interactions between electrons. But in fact in typical materials,
the interaction energy is on the order of the kinetic energy, or even larger. It seems crazyto have ignored it! So why on earth is it OK for us to ignore interactions altogether inthese simple pictures. At some level the interactions must have some effect — what arethese effects?
Eventually we want to study superconductivity!. We have so far treatedsuperconductivity crudely as simply thinking about bosons with charge e∗ = 2e, but thisis completely wrong for a number of reasons: Electrons repel each other very strongly. It seems crazy to think of them forming
pairs6 .
 Even if you could manufacture some weak attraction between fermions, to make atrue boson that you could condense, you would need a binding radius smaller than
the inter-electron distance. This would require a binding energy on the order of the
Fermi energy, which is huge!7

7.2

Some Mathematical Preliminaries Regarding Fermions

Although it is possible to do almost every calculation in first quantized notation, it becomes
rapidly extremely messy to do so. It is really better to just bite the proverbial bullett
and introduce second quantization and then talk in that language. The preliminaries
introduced here are the fermionic analogs of the preliminaries introduced in section 4.1 forbosons.
for Tony Leggett5 in 2003. Similar to electons pairing at low temperature to superconduct, Helium 3 forms
pairs at low tempertaure in order to form a superfluid.
5
See also comment about Leggett’s opinion of number non-conserving order parameters in section 4.4.
Tony Leggett was an undergraduate at Oxford, in Classics!. (More properly the degree was called “greats”
or something like that at the time). After his undergrad degree he managed to convince a tutor to let him
study physics, which he says he struggled with at first. In his Nobel biography he tells an entertaining
story about rowing in his first year at Oxford (since everyone was doing it) and feeling like he was actually
doing well, until they swapped him for the coxswain.
6
Landau famously said “You cannot repeal Coulomb’s law!”
7
This very strong binding is precisely what happens to create the boson 4 He out of two neutrons, two
protons, and two electrons. However, with electrons interacting with other electrons, there is no analogous
strong attraction mechanism.

7.2. SOME MATHEMATICAL PRELIMINARIES REGARDING FERMIONS

7.2.1

91

Second Quantization

Orthonormal Single Particle Basis
Suppose we have a set of orthonormal orbitals which we will label i, j, k, . . .. What we meanby this is wavefunctions ϕi (r, σ), ϕj (r, σ), ϕk (r, σ), . . .. Here note that the wavefunction isa function of both position and particle spin. The orthonormality condition is8
XZ
dr ϕ∗i (r, σ)ϕj (r, σ)hi|ji =
σwe will frequently abbreviate this kind of integral asZ
hi|ji = d(1) ϕ∗i (1)ϕj (1)where here (1) is meant to indicate positions and spins (if we have particles with spin).

Many-Fermion Slater Determinant
A many-body fermionic wavefunction must be fully antisymmetric. We can write down
a convenient basis of fully antisymmetric states using so-called Slater determinants9 . Wewrite
ϕ1 (1̃) ϕ1 (2̃) . . . ϕ1 (M̃ )ϕ2 (1̃) ϕ2 (2̃) . . . ϕ2 (M̃ )1Ψ(1̃, 2̃, . . . , M̃ ) = √
..
..
..
M!
.
.
.
ϕM (1̃) ϕM (2̃) . . . ϕM (M̃ )Here for clarity, the numbers with tildes over them 1̃, 2̃, . . . are the position and spin
variables, whereas the subscripts 1 , 2 , . . . are orbital indices. If the constituent orbitals are
orthonormal then the Slater determinants are properly normalizedZhΨ|Ψi = d(1̃)d(2̃) . . . d(M̃ )Ψ∗ (1̃, 2̃, . . . , M̃ ) Ψ(1̃, 2̃, . . . , M̃ ) = 1(7.1)
where again the integrals d(1̃) means both to integrate over space and sum over spin.
Further, two Slater determinants made from different sets of orthonormal orbitals will beorthogonal.
8

The observant reader will realize this is nothing more than a completeness relationshipXhi|ji =
hi|xihx|jixwhere x is the position and spin basis.
9
Slater was a professor at MIT for many years.

92

CHAPTER 7. INTERACTING FERMIONS

Fermionic Creation Operators
We define a set of Fermionic creation operators c†i corresponding to the orbitals ϕi . These
operators are defined by the following statement: Applying creation operators to thevacuum gives a Slater determinant of the corresponding orbitals. In particularc†1 c†2 . . . c†M |0i = √1
M!

ϕ1 (1̃)ϕ2 (1̃)..
.

ϕ1 (2̃)ϕ2 (2̃)
..
.

...
...

ϕ1 (M̃ )ϕ2 (M̃ )..
.

ϕM (1̃) ϕM (2̃) . . . ϕM (M̃ )Let us look at the normalization condition Eq. 7.1. In terms of these operators we have
1 = hΨ|Ψi = h0|cM . . . c1 c†1 . . . c†M |0i(7.2)

We should thus think of c†i as an operator that adds a fermion to orbital i, and correspondingly ci is an operator that removes a fermion from orbital i. So in Eq. 7.2 we starton the far right with the vacuum, we add a fermion in orbital M and then in M − 1 and soforth until we add a fermion in orbital 1, then continuing towards the left we remove thefermions in the opposite order until we are back to the vacuum. Alternately we can thinkof the ci operators as just being creation operators that have been hermitian conjugated
so that they act on the dual space to the left. I.e., h0|ci is the dual (hermitian conjugate)
to the wavefunction c†i |0i, and both represent one particle created in an orbital.
Anticommutations
Note that the fermionic creation operators must anticommutec†i c†j = −c†j c†iThis is because if we switch the order of creation of two orbitals, this corresponds toswitching two rows of the Slater determinant — which incurs a minus sign. Similarly, we
can conclude
ci cj = −cj ci
Further, we havec†i c†i = 0
since you cannot put two fermions in the same orbital (or equivalently, if two rows of a
Slater determinant are the same, the determinant vanishes). Similarly we have the dual
statement
ci ci = 0
meaning that you can only annihilate a particle, at most, once.
Finally we note thatc†i ci + ci c†i = 1(7.3)

7.2. SOME MATHEMATICAL PRELIMINARIES REGARDING FERMIONS

93

To see this is true we apply this expression to an arbitrary |statei. We can decompose the|statei, into two parts, one in which orbital i is filled and one in which orbital i is empty(there are no other possibilities for fermions). If the orbital is filled thenc†i ci |statei = |statei
since the fermion in orbital i is removed and then put back. On the other handci c†i |statei = 0
since it is trying to create another fermion in orbital i which is already filled.
On the other hand, if orbital i is empty we haveci c†i |statei = |statei
since we can add a particle to orbital i then remove it again. And also we havec†i ci |statei = 0
since we are trying to annihilate a particle from an orbital that is already empty. Puttingthese statements together we find that Eq. 7.3 is satisfied independent of whether it acts
on a filled or empty orbital.
We can summarize many of these results by defining anticommutator brackets. For
any operators let us define{A, B} = AB + BA
All of the previous results in this subsection are summarized with the following statements10 :

7.2.2{ci , cj } = {c†i , c†j } = 0(7.4){ci , c†j } = δij(7.5)

Change of Basis

As in the bosonic case we can consider changing the basis for our orbitals. Let us suppose
we have two sets of orthonormal basis orbitals. One set we call {|αi} and the other set
10

The observant reader will notice that in fact we have not yet shown Eq. 7.5 for i 6= j. This is requiredby consistency. To see this let us try writing (with i 6= j)c†j |0i = c†j ci c†i |0i
But alsoc†j |0i = ci c†i c†j |0i = −ci c†j c†i |0i
Since these two lines must equal each other, we must have c†j ci = −ci c†j94

CHAPTER 7. INTERACTING FERMIONS

we call {|ii}. We can derive creation operators in a new basis in terms of the creation
operators in the old basis (this is entirely analogous to Eq. 4.5 for bosons)Xc†α =hi|αic†ii

A useful example of such a basis change is from tight binding orbitals to plane waves
on a lattice (where we leave the spin unchanged in this particular case).
XX
c†r;σ =
hk, σ|r, σic†k;σ =e−ik·r c†k;σkkOften we would like to use continuum fields r rather than lattice points, in which casewe write ψ̂σ† (r) instead of c†σ (r). In this case the fields ψ̂ and ψ̂ † obey the anticommutations{ψ̂σ† (r), ψ̂σ0 (r0 )} = δ(r − r0 )δσ,σ0
If we have an orbital basis ϕi (r, σ) we can write the creation operators in this basis interms of the ψ̂ operators as
XZ
†
ci =
dr ϕi (r, σ)ψ̂σ (r)Zσ=d(1)ϕi (1)ψ̂ † (1)Important Example: Fermi Sea As an example, let us consider a sytem of kstates filling a Fermi sea. We can write
Y Y †
(7.6)
|Fermi Seai =ck,σ |0i|k|<kf σ=↑,↓We note that there is some presumed fiducial11 ordering of operators in these products.
Without an ordering we cannot establish the sign of the wavefunction. Fortunately, we
may never need to actually specify the ordering, but we should be very careful to keep
track of any changes to the ordering, since any exchange incurs a minus sign.

7.2.3

Writing our Hamiltonian in Second Quantized Form

We will typically be concerned with a Hamiltonian of the form
H = H0 + Hint
where12H0 =X  p2
i

i
11

2m

 X
ĥi
+ U (ri ) =iFiducial means essentially “accepted as a fixed reference”.
I apologize that here I’ve used U for single particle potential and V for interaction, whereas in priorsections it was the other way around. I hope this does not create too much confusion!
127.2. SOME MATHEMATICAL PRELIMINARIES REGARDING FERMIONS

95

is the single particle Hamiltonian of particles in a trapping potential U (r) and
Hint =1X
V (ri − rj )2
i6=j

is the interaction between particles. This might be, for example, a Coulomb interaction if
we are considering electrons.
We now want to rewrite these in second quantized form.
To do this we can write the particle density,Xρ(r) =
ψ̂σ† (r)ψ̂σ (r)σso that we can write the interaction asZZ
1d(1) d(2) : ρ(1) V (1 − 2) ρ(2) :Hint =
2ZZ
1d(1) d(2) : ψ̂ † (1)ψ̂(1) V (1 − 2) ψ̂ † (2)ψ̂(2) :=2As in section 4.3 we have added colons : to indicate that the expression should be normal
ordered which implies we should move all the creation operators to the left (and remember
to add a minus sign for each exchange). To justify this we consider a system with a single
fermion — without the normal ordering colons, this single fermion would have a nonzero
self-interaction.
We can thus rewrite the interaction term as
ZZ
1d(1) d(2)ψ̂ † (1)ψ̂ † (2) V (1 − 2) ψ̂(2)ψ̂(1)Hint =
2
The single particle term, we can write as
 2 2

Z
−~ ∇
H0 = d(1) ψ̂ † (1)+ U (r) ψ̂(1)
2m

(7.7)

(7.8)

To see that this is correct, we rewrite ψ † in terms of some single particle basis |αi.
Xψ̂ † (1) =ϕ∗α (1)c†ααso that we have
H0 =X

hαβ c†α cβ

α,β

where
Z
hα,β = hα|ĥ|βi =d(1) ϕ∗α (1)−~2 ∇2+ U (r) ϕβ (1)
2m

(7.9)

96

CHAPTER 7. INTERACTING FERMIONSIt is most convenient at this point (although certainly not necessary) to choose abasis |ai that are eigenvectors of the operator ĥ with eigenvalues 0a , so we haveĥ|ai = 0a |ai
Thus we can write
H0 =X

0a c†a ca(7.10)a

The meaning of this expression is quite clear. The operators n̂a = c†a ca counts the numberof fermions (0 or 1) in orbital a, and if there is a fermion in the orbital, it is assigned anenergy 0a .
It is then useful to write the interaction part of the Hamiltonian Eq. 7.8 in an orbital
basis to obtain
1 X
Hint =(7.11)
vαβγδ c†α c†β cδ cγ
2α,β,γ,δ

where the matrix element vαβγδ is given byZ
vα,β,γ,δ = hαβ|V |γδi =Zd(1)d(2) ϕ∗α (1)ϕ∗β (2) V (1 − 2) ϕγ (1)ϕδ (2)(7.12)

Note carefully the ordering of the indices in Eq. 7.11 (the last two indices in the two setsof subscript are switched). To see where this comes from compare Eq. 7.7 to Eq. 7.12.
We will continue to use this convention, although other conventions are possible as well.
Obviously the matrix element hαβ|V |γδi in Eq. 7.12 uses a two-particle basis|γδi → ϕγ (1)ϕδ (2)which is not properly symmetrized for fermions. The symmetrization is automatic though
once we put these matrix elements into Eq. 7.11 using the fermionic creation operators.
Note also that the matrix element has a number of obvious symmetries including
vαβγδ = vβαδγ(7.13)The Hamiltonian H = H0 + Hint with Hint written in the form of Eq. 7.11 and H0
written in the form of Eq. 7.9 or Eq. 7.14 will be our object of study for the rest of thischapter. This form is extremely general and can be used for study of any interacting Fermi
system, such as electrons in a metal, electrons in an atom, Helium 3, or even neutrons ina neutron star.
We will use several different “first pass” approximations (first order perturbation
theory, Hartree approximation, Hartree-Fock, etc). Sometimes these approximations will
give the same results, but sometimes they will not.

7.3. FIRST ORDER PERTURBATION THEORY

97

Figure 7.1: (Ignore the lined paper!). This depicts filling all orbitals up to the chemical
potential.

7.3

First Order Perturbation Theory

Let us work in a basis such that the single particle Hamiltonian H0 is diagonal. We thus
have
X
H0 =
0a c†a ca(7.14)
a

Ignoring the interaction (“zeroth” order in interaction) the ground state is just to fill up
all orbitals below the chemical potential µ.
"
#
Y
0
†
|Noninteracting Ground Statei = |GS i =ca |0i(7.15)
a occupied

where here an orbital a is occupied if
0a < µas depicted schematically in Fig.7.1. As in Eq. 7.6, there is some presumed ordering oforbitals in Eq. 7.15 which we will not specify.
Next we want to include the interaction term. We write this term as above in theform (Same as Eq. 7.11)
1 X
Hint =(7.16)
vαβγδ c†α c†β cδ cγ
2α,β,γ,δ

At first order in perturbation theory the correction to the energy isδE = hGS0 |Hint |GS0 i
We now have to evaluate this expectation. We could do this by commuting a bunch ofoperators around, but there is a simple “shortcut” way of thinking:YXY †δE = h0|ce | 21vabcd
c†a c†b
cd cc|
cf |0i|{z}|{z}e occupiedabcd
f occupied
return same tworemove two
|
{z
}
|
{z
}
same filled state

1filled state

Here we note that the initial ket and final bra are exactly the same state. This means
whichever particles are removed by cd cc have to be returned by c†a c†b . There are exactly
two ways in which this can happen: Either a matches c and b matches d,c†a c†b cd cc

98

CHAPTER 7. INTERACTING FERMIONS

or a matches d and b matches cc†a c†b cd cc
We thus obtain
δE = hHint i =1
2X(vabab − vabba )(7.17)

a,b occupied

It is clear that the two terms should have opposite signs since one commutation will makethem look the same. To be more precise about the signs, note that in the first case (werea matches c) we can write the contribution asY †Y
ce | c†a c†b cb ca |cf |0i = hstate|stateih0|e occupiedf occupied

so we end up with a positive sign. On the other hand, to get the second case into thisform we need to anticommute two of the operators and thus pick up an minus sign.
This approach, can at least in principle, be extended systematically to higher order
perturbation theory.

7.3.1

Hartree and Fock Terms

The two terms in Eq. 7.17 have different meanings. Let us study them one at a time.
Hartree, or Direct, Term
Using Eq. 7.12 the first term in Eq. 7.17 is given byZZ
vabab = d(1) d(2)|ϕa (1)|2 |ϕb (2)|2 V (1 − 2)

(7.18)

This is known as the Hartree13 or Direct interaction term. This is simply the interaction of
the density in orbital a with the density in orbital b. If we keep only this term in Eq. 7.17
we have what is known as the Hartree approximation, which is somewhat simpler than
the full first order perturbation calculation.
Self-Consistent Hartree We can also consider doing a self-consistent version ofthis calculation. This is a common approach known as self-consistent Hartree. It amounts
to solving the Schroenger equation in a potential which includes the interaction from all
the other electrons in the system14 . In other words we writeZX
ef fU (r) = U (r) + dr0|ϕn (r0 )|2 V (r − r0 )n occupiedZ= U (r) +dr0 hρ(r0 )i V (r − r0 )13
Douglas Hartree was a British Numerical expert. He has a unit, the Hartree, named after him (whichis e2 /(4π0 a0 ) = 2Ry ≈ 27.2eV).
14
Self-consistent Hartree for bosons is precisely the Gross-Pitaevskii equation! See discussion near
Eq.4.13.

7.3. FIRST ORDER PERTURBATION THEORY99

The procedure for implementing self-consistent Hartree is as follows:
1. Solve the Schroedinger equation in the presence of the potential U ef f (r)2. Fill the lowest N orbitals, and calculate hρ(r)i
3. Recalculate U ef f
4. Go back to step 1, and repeat until converged.
Fock, or Exchange, Term
Using Eq. 7.12 the second term in Eq. 7.17 is given byZZ
vabba = d(1) d(2) ϕ∗a (1)ϕ∗b (2) V (1 − 2) ϕb (1)ϕa (2)(7.19)

Note in particular that this term is zero unless the orbitals a and b overlap15 . Note inparticular that this means that orbital a and b need to overlap both in space and in spin
space16 . Eq. 7.19 is known as the Fock17 or exchange term. This term cannot be describedas simply some sort of effective potential that an electron moves in, as the Hartree term
was. Note also that if the potential V (r) is repulsive, the Fock term (which enters in
Eq. 7.17 with a minus sign) is attractive.
Further Comments on First Order Perturbation Theory
It should be noted that, despite the fact that we have a Hartree term and a Fock term, what
we have calculated so far (Eq. 7.17) is NOT the Hartree-Fock approximation (admittedly
the nominclature is confusing).
We should also note that first order perturbation theory can also be used to (approximately) calculate the energy shift, due to interactions, of excited states. Again, if we
15
If the support of ϕa and ϕb (meaning the region where they are nonzero) is disjoint then there is novalue of the variable 1 for which the integrand is nonzero
16
For example, sinceZZX ∗d(1)ϕ∗a (1)ϕb (1) = dr
ϕa (r, σ)ϕb (r, σ)σ

If we work with orbitals having a definite spin, the two orbitals would have to have the same spin for thisto be nonzero.
17
Vladimir Fock was the creator of the idea of second quantization in 1932 (although some of the ideas
were presented by Dirac as early as 1927). Fock is immortalized through the idea of “Fock space”. Muchof Fock’s later career was devoted to philosophy of science. He became a defender of relativity in a
hostile Marxist world. Unfortunately, some of the Marxist philosophers thought that modern physics was“Bourgeois” and therefore dangerous to the state. Fock was forced to become a philosopher, speak theirlanguage, and clarify what is physics and what is philosophy. It is unclear how well Soviet physics would
have survived were it not for Fock.

100

CHAPTER 7. INTERACTING FERMIONS

begin with the noninteracting system, we can write an excited state as|Excitedi = c†j ci |GS0 i
where i is one of the orbitals which is filled in the noninteracting ground state, and j is oneof the orbitals which is empty in the noninteracting ground state. Without interactions
the energy of this excited state is00
Eexcited= 0j − 0i + Egroundstate

Again we can calculate the energy shift due to interactions as
1 X
δE = hHint i =(vabab − vabba )2
a,b occupied

with the expectation given in the excited state, meaning that the sum over occupied states
includes orbital j but not orbital i.

7.4

Hartree-Fock Approximation

The so-called Hartree-Fock approximation is a self-consistent mean-field theory approach.
It is often the first approach a theorists tries, and it is used extensively in condensed
matter physics, but also in nuclear physics, atomic physics, molecular chemistry18 , and soforth.
Let us briefly recall the (Weiss) mean field method for understanding ferromagnets.
Here we have a Hamiltonian like the Heisenberg HamiltonianJX
H=Si · Sj2hi,jiwhere, for example, the sum is over neighboring spins i, j. To handle this interaction
we average one of the spins and leave the other one un-averaged,. So we have instead a
Hamiltonian for a single spin given byX
hj = J
Si · hSj i
j neighbors iwith hSj i = hSi is the average of the spin on any site. We can then solve the single site
problem, and we need to “self-consist” meaning that the solution of the single-site problem
should give us hSi.
We hope to use the same mean-field idea to handle the interaction term of ourHamiltonian. We thus take
1 X
vαβγδ c†α c†β cδ cγ(7.20)
Hint =
2
α,β,γ,δ

18

A Nobel Prize in chemistry was awarded to John Pople in 1998 for developing numerical Hartree Fockmethods for simulating molecules.

7.4. HARTREE-FOCK APPROXIMATION

101

and we replace it with
ef f
Hint
=

h
i
1 X
vαβγδ hc†α cγ ic†β cδ + hc†β cδ ic†α cγ − hc†α cδ ic†β cγ − hc†β cγ ic†α cδ
2(7.21)

α,β,γ,δ

Here we have taken all possible (nonzero) averages of two operators and left two otheroperators unaveraged19 Note also that due to the symmetry Eq. 7.13 the first two terms inthe square brackets are identitical to each other and the last two terms are also identicalto each other.
We can thus write a more abbreviated form
X
ef f
Hint
=
fαδ c†α cδ(7.22)α,δ

wherehc†β cγ i [vαβδγ − vαβγδ ](7.23)

Including the single-body terms hαβ (See Eq. 7.9) we haveX
H ef f =[hαδ + fαδ ] c†α cδ(7.24)

fαδ =

X
β,γ

αδ

The two equations Eq. 7.23 and 7.24 constitute the Hartree-Fock approximation. Our
procudure for solving these two equations at the same time is iterative (similar to theabove described, self-consistent Hartree approach). While in principle one can produce
a solution in any basis, it is convenient to successively change basis as we iterate toconvergence. We thus follow the following procedure
1. Let µ be the eigenvalues of hαδ + fαδ with corresponding eigenvectors |µi.
2. Change to this eigenbasis by definingc̃†µ =

X

hα|µic†ααIn this basis, the Hamiltonian is of the formX
H ef f =µ c̃†µ c̃µµ3. Fill the lowest N states with our fermions (Assuming we have an N fermion system)Y|Ψic̃†µ |0ilowest N

When we get to superconductivity, we will also consider the possibility of needing terms like hc† c† i andhcci. However if we insist on a system with a definite number of particles, such expectations are strictly
zero.
19

102

CHAPTER 7. INTERACTING FERMIONS

4. Recalculate fαδ in this new ground state using Eq. 7.23. Note that this recalculation
is why we wanted to work in the diagonal basis. In this basis we havehc†µ cν i = δµν ×

1 if µ is filled
0 if µ is empty

We thus have20
fαδ =

X

[vαµδµ − vαµµδ ](7.25)µ occupied

5. Go back to step 1 and repeat until convergence. When it converges, the old basis
and the new basis match. This means, that in this basis, h + f is already diagonal.
Thus we haveX
µ = hµµ + fµµ = hµµ +[vµνµν − vµννµ ]ν occupied

The first term in the square brackets being from the Hartree term and the second
term being from the Fock term.
Note that if we were to follow exactly the same procedure, but using only the first(Hartree) of the two v terms, we would have the self-consistent Hartree approximation, as
described above.
The energies µ can be thought of as the Hartree-Fock ionization energies. I.e.,
this is the energy (in Hartree-Fock approximation) required to remove the electron inthe µ orbital. This results is known as Koopmans’ theorem21 . The total energy of the
Hartree-Fock system is given byX 
1Etotal =(7.26)
hµµ + fµµ2
µ filled
The factor of 1/2 here is similar to what occurs in Weiss mean field theory. In writing
down the total energy of the system, one does not want to overcount the energy of spin
i interacting with spin j and also spin j interacting with spin i. It is quite similar here.
Simple addition of all of the µ Hartree Fock ionization energies would count interactionof each particle µ with all the other particles ν but then would again count the interactionof each ν with µ. Thus we need to introduce the factor of 1/2 in the interaction.
20
Note here that the value of fαδ depends on what basis we have chosen for the indices α, δ, µ. On each
iteration of Hartree-Fock, the basis changes, and f correspondingly changes until convergence.
21
Tjalling Koopmans is a Nobel laureate in the field of economics (never mind that economics was not
one of the original subjects that Nobel had established prizes for — so it is strictly speaking a Nobel
memorial medal, not a Nobel prize). He did his work on Hartree-Fock in 1934 when he was a 24 year old
grad student, but shortly thereafter switched fields to economics. His prize in 1975 was awared for thetheory of optimal allocation of resources. Apparently his own resource was more optimally allocated in
the field of economics. The word “Koopman” in dutch means “Trader” or “Merchant”, possibly justifying
his attraction to the field of economics.

7.4. HARTREE-FOCK APPROXIMATION

7.4.1

103

Hartree-Fock as Optimal Slater Determinant

We claim that the solution of the Hartree-Fock equations provides the best (lowest energy) variational wavefunction in the form of a single Slater determinant. This optimal
wavefunction isY|HFi =c†µ |0iµ filled

for the Hartree-Fock orbitals µ. This brings us to an equivalent definition of the HartreeFock approximation:
Hartree-Fock is an approximation by which one finds the optimal ground state
wavefunction in the form of a single Slater determinant.
Proof that this statement is equivalent to the Hartree-Fock equations derived as amean field theory above (Eq. 7.23 and 7.24) is nontrivial. We will present it here.
Proof:
Consider a basis of orbitals µ and a single Slater determinant (or “filled Fermi sea”)
made of these orbitals.
Y|GSi =c†µ |0iµ filled

We would like to find a condition such that |GSi is a variational minimum. To find thiscondition we want to try varying the basis µ and seeing if the energy decreases. So let us
imagine switching to a new basis
XXc̃†α =
hµ|αic†µ =Uαµ c†µ(7.27)µµ

where U is a unitary matrix. Note that we are only concerned with a change of basis
which mixes filled and empty orbitals. If we change basis in a way that that mixes two
filled orbitals with each other, or in a way that mixes two empty orbitals with each other,
the resulting Slater determinant remains unchanged22,23 . At the same time we want the22

Let us see an example of this. Suppose we make the basis transform√c̃†a = (c†1 + c†2 )/ 2√c̃†b = (c†1 − c†2 )/ 2

It is easy to check that if both orbitals are filled, the basis change does not change the resulting Slater
determinant:c̃†a c̃†b |0i = c†1 c†2 |0i
However if we only fill one of these two orbitalsc̃†a |0i 6= c†1 |0i 6= c†2 |0i
Thus to change the Slater determinant, we need to mix a filled state with an empty state.
23
We should be cautious that rewriting a determinant of orbitals in terms of an equivalent set of orbitals
can sometimes incur an overall sign to the wavefunction if we somehow reorder the orbitals in the process.
We will not care about this sign as it does not change the energy of the wavefunction.

104

CHAPTER 7. INTERACTING FERMIONS

variation to be very small, so we can write
U = eiM
with M = M † a Hermitian matrix24 . With  small, we haveU = 1 + iM + . . .
which we plug into Eq. 7.27 to givec̃†µ = c†µ + iXMµν c†ν + . . .

νHere, as mentioned above, to have any change in the resulting Slater determinant we are
only interested in M that mixes filled and empty orbitals. Thus we can assume that Mµν isnonzero only if either µ is filled and ν is empty or ν is filled and µ is empty. (in particularµ 6= ν).
Thus we consider trial states of the formY|Ψi =c̃†µ |0iµ filled

!
=

Y

X

c†µ + iµ filled

Mµν c†ν|0i(7.28)

ν emptywhere the sum over ν must be only over unoccupied states, such that we have µ occupiedand ν unoccupied (or equivalently M is simply zero for all other cases). We now want toexpand this product order by order in . We get
"
#
"
#
Y
X
Y †
†
†
|Ψi =
cµ |0i + iMµν cν cµcλ |0i + . . .
µ filled

ν empty; µ filled

λ filled

Let us walk through this result in some more detail. The first term in brackets, the productof the c†µ is the product of all of the first terms in Eq. 7.28. This just gives the unperturbed
ground state |GSi (ie., the  = 0 wavefunction). The order 1 term is more complicated.
To get this term from the product of Eq. 7.28 one chooses only a single µ in the productwhere instead of taking the first term c†µ one instead chooses the Mµν term25 . We write
this term in a creative way, by starting with |GSi (the term in square brackets on the
far right) and removing the orbital µ with cµ and then putting in the sum over Mµν c†νinstead. As a shorthand we can write|Ψi = |GSi + |δΨi + . . .
24

Let us check that exponentiating a hermitian matrix gives a unitary matrix. For a matrix to be unitary
we must have U U † = 1. So let us check (eiM )(eiM )† = (eiM )(e−iM ) = 1.
25
If a sign occurs in reordering creation operators this can be absorbed into M .

7.4. HARTREE-FOCK APPROXIMATION

105

where
|δΨi = iXMµν c†ν cµ |GSi

(7.29)

ν empty; µ filled

At this linear order in  it is easy to see thathδΨ|GSi = 0
since δΨ always has some orbitals ν filled which are not filled in the ground state, GS.
This then implies that2 hΨ|Ψi = 1 + O( )Finally we turn to the variational condition. We would like our wavefunction to beat an extremum of the energy. Thus we want∂E=0∂ =0which we write as∂ hΨ|H|Ψi=0∂ =0  hΨ|Ψi

where this condition should now be true for any hermitian matrix M that mixes filled andunfilled orbitals. Note that the denominator of this fraction we just found to be 1 up tocorrections of order 2 so we can throw this out. Expanding our condition and taking the derivative gives us the condition
0 = hδΨ|H|GSi + hGS|H|δΨi = RehGS|H|δΨi
And indeed since this needs to be true for any matrix M in Eq. 7.29, we can equivalently
state the condition
RehGS|Hc†ν cµ |GSi = 0
where µ is filled in the ground state GS and ν is empty in the ground state GS. Writing
H out explicitly we get


X
X 1
RehGS| hαβ c†α cβ +vαβγδ c†α c†β cδ cγ  c†ν cµ |GSi = 0(7.30)2αβαβγδ

Let us attack this term by term. In the first term, we havehGS|c†α cβ c†ν cµ |GSi
Since µ is occupied (filled) and ν is unoccupied (empty) in |GSi, this expression can only
be nonzero if µ = α and ν = β.
hGS|c†α cβ c†ν cµ |GSi = δνβ δαµ106

CHAPTER 7. INTERACTING FERMIONS

Thus the first term in the correlator Eq.7.30 is simply hµν .
We next examine the second, more complicated, term in Eq.

7.30. This has theform
hGS| c†α c†β cδ cγ c†ν cµ |GSi
Now there are four different ways in which this can be nonzero, which we now list here.
For all of these we must have µ initially filled and ν initially empty.
hGS| c†α c†β cδ cγ c†ν cµ |GSi → +δαµ δβδ δγνδ initially filled

hGS| c†α c†β cδ cγ c†ν cµ |GSi → +δαγ δβµ δδνγ initially filled

hGS| c†α c†β cδ cγ c†ν cµ |GSi → −δαµ δβγ δδν

γ initially filled

hGS| c†α c†β cδ cγ c†ν cµ |GSi → −δαδ δβµ δγν

δ initially filled

Putting these results together we obtain the condition
hµν +

1 X
[vµxνx + vxµxν − vµxxν − vxµxν ] = 0
2 x occupied

Now using the symmetry of v given in Eq. 7.13 we find that the first two v terms are thesame and the last two v terms are the same thus giving usXhµν +[vµxνx − vµxxν ] = 0
x occupied

for all µ filled and ν empty. We recognize the sum over x as being precisely fµν as given
in Eq. 7.25. So our condition ishµν + fµν = 0
for µ initially filled and ν initially unfilled. Thus if we started with a Hartree-Fock basis,
where h + f is diagonal, this condition in necessarily satisfied (since it will only be nonzero
for µ = ν). Thus we conclude that the solution to the Hartree-Fock equations indeed
produces a single Slater determinant with its energy extremized! QED. . Yay. etc.

7.5

Application of Hartree-Fock to Translationally Invariant Fermions

We now would like to apply Hartree-Fock theory to a very simple, but also very importantcase — the translationally invariant interacting Fermi gas.

7.5. APPLICATION OF HARTREE-FOCK TO TRANSLATIONALLY INVARIANT FERMIONS107
This situation is supposed to model electrons in metals, but we have thrown away
the crystal lattice for simplicity and just kept the electrons. (So, for example, we will havea perfectly spherical Fermi sea, as compared to what we get in real metals). Note that if
we completely throw away the positive ions in the crystal lattice, the Coulomb energy of
all the electrons interacting with each other becomes infinite. To avoid this problem, one
usually smears out a neutralizing positive charge which we keep along with the electrons,
so that the entire system is charge neutral (but still completely translationally invariant).
This model is known as the “jellium” model26 (with the idea that the positive charge islike a smeared out jelly that has been spread evenly around the system).
Another way to handle the problem coming from the infinite Coulomb interaction is
to just consider fermions with shorter range interaction V (r). While this does not apply
to real electrons, it does apply to Fermi systems such as 3 He.
Whether we use a jellium background or not, our Hamiltonian will always be of theform
H = H0 + Hint
where
H0 =

X ~2 k2 †c c
2m kσ kσk,σis the kinetic (single particle, noninteracting part) of the Hamiltonian, andHint =

1
2

X

v(k1 σ1 )(k2 σ2 )(k3 σ3 )(k4 σ4 ) c†k1 σ1 c†k2 σ2 ck4 σ4 ck3 σ3(k1 σ1 )(k2 σ2 )(k3 σ3 )(k4 σ4 )Note when comparing to the form written in Eq. 7.11 the indices we used as a, b, c, d are
now replaced by the combined indices (k1 σ1 ), (k2 σ2 ), (k3 σ3 ), (k4 σ4 ). The matrix element
is given byv(k1 σ1 )(k2 σ2 )(k3 σ3 )(k4 σ4 ) = hk1 σ1 ; k2 σ2 |V (r − r0 )|k3 σ3 ; k4 σ4 i
Here the kets are normalized (but unsymmetrized) plane wave state of two particles
hrσ3 ; r0 σ4 |k3 σ3 ; k4 σ4 i =1 ik3 ·r+ik4 ·r0eVwith V the volume of the system. Thus we have
Z
Z
1
0
0
v(k1 σ1 )(k2 σ2 )(k3 σ3 )(k4 σ4 ) = δσ1 σ3 δσ2 σ4 dr dr0 2 e−ik1 ·r−ik2 ·r eik3 ·r+ik4 ·r V (r − r0 )V1
= δσ1 σ3 δσ2 σ4 δk1 +k2 ,k3 +k4 Ṽ (k4 − k2 )V
26
The Jellium model (including the name) was constructed by Conyers Herring. Herring was an extremely important condensed matter physicist in the mid 1900s — and his influence was often in the
background of other great works. Herring was the first head of the famous Bell Labs theory department.
Many years later, when Bell Labs was closing down its fundamental research effort, I was the last head of
the Bell Labs theory department – a rather dubious distinction I’m afraid.

108

CHAPTER 7. INTERACTING FERMIONS

Figure 7.2: Left: The interaction in Eq. 7.31 drawn as a diagram. Middle: connecting up
the legs of the diagram to make the Hartree contribution. Right: Connecting the legs up
to make the Fock contribution.
whereZṼ (k) =dre−ik·r V (r)We can thus rewrite
Hint =

1
2V

c†k1 σ1 c†(k4 −q)σ4 ck4 σ4 c(k1 −q)σ1 Ṽ (q)X(7.31)(k1 ,σ1 ),(k4 ,σ4 ),q

Note that there is another, potentially more physical way to write this expression
Hint =X: ρ(q)Ṽ (q)ρ(−q) :qwhere ρ is the density written in Fourier spaceρ(q) =X †c(k+q)σ ckσ = FourierTransformq [ψ̂ † (r)ψ̂(r)](7.32)

k,σ

Although we don’t intend to express our perturbation theory in terms of Feynman
diagrams, it is occassionally useful to write these diagrams anyway — not the least reason
being that one will often see these diagrams in the literature, and it is useful to be able
to convert the diagram into physical expressions. The interaction 7.31 can be written
diagrammatically as shown in the left of Fig. 7.2. When we make the Hartree-Fock approximation, we want to average one creation and one annihilation operator to give hc†a cb i,and this can be done in two ways, corresponding to the direct and exchange (Hartree andFock) terms.
The first possibility is the direct or Hartree termc†k1 σ1 c†(k4 −q)σ4 ck4 σ4 c(k1 −q)σ1

7.5. APPLICATION OF HARTREE-FOCK TO TRANSLATIONALLY INVARIANT FERMIONS109
and the second is the exchange or Fock termc†k1 σ1 c†(k4 −q)σ4 ck4 σ4 c(k1 −q)σ1
We will examine these one at a time.
Note that we are going to assume that the ground state is in the form of a Fermiseac†kσ |0i

Y|FSi =

σ,k<kFand we will justify this assumption post-facto.

Direct (Hartree) Term
We haveX †1 X X †HHartree =hck1 σ1 c(k1 −q)σ1 i
c(k4 −q)σ4 ck4 σ4 Ṽ (q)V q
k1 ,σ1

k4 ,σ4

Here we examine the expectation. If we are taking the expection in a Fermi sea state, wehavehc†k1 σ1 c(k1 −q)σ1 i = δq,0 × nF (k)where
nF (k) =1 |k| < kF
0 otherwise(7.33)

is the Fermi occupation factor at zero temperature for the Fermi sea where kF is the Fermi
wavevector. Thus the expression in the square brackets is



X



hc†k1 σ1 c(k1 −q)σ1 i = N δq,0

k1 ,σ1

with N the total particle number. We then get
HHartree =X †NṼ (0)
ck4 σ4 ck4 σ4V
k4 ,σ4This is nothing more than a shift of the chemical potential and is therefore trivial. Indeed,
if one considers a jellium model where the positive charge is neutralized by a uniformbackground charge, Ṽ (0) = 0, and this term vanishes exactly. Diagramatically, this isexpressed by the diagram in the middle of Fig. 7.2.

110

CHAPTER 7. INTERACTING FERMIONS

Exchange (Fock) Term
Here we have
HF ock =

−1 X
V




X


k1 ,σ1hc†(k4 −q)σ4 c(k1 −q)σ1 iṼ (q) c†k1 σ1 ck4 σ4

k4 ,σ4 ,q

We again examine the expectation in the Fermi sea ground state giving us
hc†((k4 −q)σ4 c(k1 −q)σ1 i = δσ1 ,σ4 δk1 ,k4 nF (k4 − q)so the Fermi occupation factor here is unity only if |k4 − q| < kF and is otherwise zero.
We can then write the Fock term asX
HF ock =ΣF ock (k) c†kσ ckσk,σwhere ΣF ock is known as the self-energy and is given byΣF ock (k) = −1XṼ (q)nF (|k − q|)
V q

Diagrammatically this result is shown in the far right of Fig. 7.2.
Putting the Hartree-Fock pieces together
Adding together the Hartree and Fock contributions we get the effective Hartree-Fock
HamiltonianX(7.34)HHF =HF (k) c†k,σ ckσ
k,σ

where

~2 k2 N
+ Ṽ (0) + ΣF ock (k)(7.35)
2m
V
What is crucial to note here is that Eq. 7.34 is “diagonal” in the variables k and σ. Thismeans we have the correct Hartree-Fock eigenstates. Recall that we were supposed toiterate Hartree-Fock until we converge to diagonal. Here we simply guessed the rightstarting wavefunction, and we found it was diagonal immediately. This may seem surprising, but actually it is not. Before we turned on the interaction, we had a translationallyinvariant system with a ground state which was a simple Fermi sea. When the interaction
was turned on, we did not ruin translational invariance, and this greatly constrains whatkind of wavefunctions we can write down. In particular, each Hartree-Fock eigenstatemust be an eigenstate of momentum. As a result, the Hartree-Fock best single Slater
determinant is simply the filled non-interacting Fermi sea. And in this case, Hartree-Fockjust gives a modification to (k). This results is the same as what we would get for firstHF (k) =7.5. APPLICATION OF HARTREE-FOCK TO TRANSLATIONALLY INVARIANT FERMIONS111
order perturbation theory in this case, although that is not always true for Hartree-Fockin non-translationally invariant systems.
It is worth looking at the three terms in Eq. 7.35. The first term is simply thenoninteracting kinetic energy. When we sum this up over the entire Fermi sea in Eq. 7.34
we obtain (this is an easy exercise to show!)3
E0 = N EF0
5(7.36)

where EF0 is the noninteracting Fermi energy. The total Hartree energy summed over allof the particles is
1 N2Ṽ (0)(7.37)
EHartree =2 Vwith V the volume of the system as usual; and the total Fock energy summed over all theparticles is−1 X
EF ock =nF (q)nF (q0 )Ṽ (q − q0 )(7.38)
2V
0
q,q ,σNote that this rather intuitive expresion gives only interaction between aligned spins, as
is required for the Fock term. The factors of 1/2 in Eq.7.37 and 7.38 are included to avoid
overcounting as described in Eq. 7.26.

7.5.1

Hartree-Fock Effective Mass

The interesting physics of a metal is often the low energy physics compared to EF , since
EF for a metal is often huge. We are thus interested in the low energy excitations near
the Fermi surface.
For a noninteracting Fermi gas we can expand around the Fermi surface and write~kF~(|k| − kF )m= EF + vF ~(|k| − kF )

0 (k) = EF +where we have defined the Fermi velocity to bevF = ~kF /mNote that we are dropping terms which are (|k| − kF )2 and higher powers being that weare assuming we are close to the Fermi surface.
For an interacting Fermi system we can write instead~kF~(|k| − kF )m∗where we have defined an effective mass m∗ by
(k) = EF∗ +~2∂2=
HF (k)
m∗
∂k 2
k=kF

112

CHAPTER 7. INTERACTING FERMIONSFor example, for 3 He, the effective mass ranges from 2-6 times the bare mass of the Helium
atom, depending on the pressure applied to the system.
Effective Mass for Coulomb Interaction
We now return to the Coulomb interaction, which is slightly troublesome due to its longrange nature. We start withe2
V (r) =
4π0 |r|which we Fourier transform to get27
ZṼ (k) = dre−ik·re2e2=
4π0 |r|0 |k|2(7.39)

Here we note that Ṽ (0) (which we need for the Hartree energy) is divergent for the Coulombinteraction. This is a result of having not included the neutralizing positive (jellium)
background. If we put this in, it will fully cancel Ṽ (0) and we don’t have to worry about
this divergence.
Let us now calculate the Fock self-energy
1X
ΣF ock (k) = −Ṽ (q)nF (|k − q|)
V q
=

e2 1
0 V

1
e2
=
0
2
|k − k |0X|k0 |<kF

Z |k0 |=kF
0

dk0
1
3
(2π) |k − k0 |2(7.40)

which is sketched in Fig. 7.3. For k = 0 it is easy to calculate that Σ = e2 kf /(0 2π 2 ). And
it is also easy to see that Σ must decay for very large k. However, what is not as obvious
is that the self energy has a log-divergent slope at kF . The calculation that shows this isgiven just below.
Given the log divergent slope, m∗ then goes to zero at the Fermi surface. The reason
for the log divergence is the long-ranged Coulomb interaction. Over long distances, otherelectrons rearrange themselves to cancel the Coulomb potential, and this is not properly
accounted for in simple Hartee-Fock. This rearrangement of charges to remove the longrange Coulomb interaction is known as screening and will be treated in some detail in thenext chapter.
27

On dimensional grounds, and by rotational invariance we know that it has to be proportional to |k|1 2 ,α
so let us call the result |k|2 . To get the prefactor right it is useful to look at the inverse Fourier transformZdk ik·r α
V (r) =
e
(2π)3
|k|2
Then take the laplacian∇2 V (r) = αZdk ik·re= αδ(r)(2π)3and by Gauss’ law we can then fix the prefactor α = e2 /0 .

7.6. APPENDIX: SHOWING THE LOG DIVERGENCE AT THE FERMI SURFACE113

Figure 7.3: A plot of the self-energy in the case of the Coulomb interaction

Figure 7.4: A plot of the self-energy in the case of the Coulomb interaction

7.6

Appendix: Showing the Log Divergence at The Fermi
Surface

We want to calculate the self energy Eq. 7.40 as k approaches the Fermi surface. Let
us consider k (without loss of generality) in the x direction and approaching the Fermi
surface kF , as shown in Fig. 7.4. We divide the integral over k0 into an integral in the x
direction and an inegral in the directions perpendicular to x, which we call k0⊥ .
Z
Z
10ΣHF (k) =dk ⊥ dkx0|k − k0 |2√
Z kF
Z k2 −(kx0 )2
F
1
0 0
=
dkx0 2πdk⊥k⊥
0 |2 + k 2
|k
−
k
−kF
0
x
⊥
Z kF
≈ 2π
dkx0 log(|k − kx0 |2 /cutoff)−kF∼ (k − kF ) log(|k − kF |)Thus we have

and

∂HF
∼ log(|k − kF |)∂k1∂ 2 HF
1
∼
∼
m∗
∂k 2
|k − kF |114

CHAPTER 7. INTERACTING FERMIONSChapter 8

Screening and Linear Response
In the previous chapter we had some problems with the long-rangedness of the Coulombinteraction. While we will not completely solve this problem here, we understand thatthe solution comes from allowing other particles to respond to the potential created, thus
screening the long range interaction. Indeed, we know that if we try to put a postive
charge in a metal at one point and a negative charge at another point (so as to keep thetotal charge neutral) the other charges in the metal should rearrange so as to make themetal look overall neutral everywhere — at least when looked at from far away.
More generally we are interested in how a system responds to any arbitrary perturbation. We will study this at various levels of sophisitication.

8.1

Thomas-Fermi Screening

Let us consider a uniform electron gas and we imagine there is some potential U (r) thatthe electrons feel. We thus have a Hamiltonian which is
H=

X p2
i

i

2m

+ U (ri )Let us further assume that U (r) varies slowly in space and hU i = 0 for simplicity. For
example, we might assume that U (r) = U0 cos(q · r) with small |q|.
To determine what happens in this situation we start by thinking locally: Locally
we have a Fermi sea which is filled up to the chemical potential µ. If the potential U
increases, this pushes some of the electrons above the chemical potential and the density
drops byδn(r) = −D(EF )U (r)where D(EF ) is the density of states per unit volume. This is shown in Fig. 8.1.
115

(8.1)

116

CHAPTER 8. SCREENING AND LINEAR RESPONSE

Figure 8.1: Left: States are filled up to the chemical potential µ. Right: If an additional
potential U is added to the system, some states are pushed above the chemical potential,
and the local density drops by D(EF )U where D is the density of states per unit volume.
In this situation, we sometimes call the quantity −D(EF ) in Eq. 8.1 a response
function1 .
At this point, we are still ignoring interactions between the electrons, we are only
keeping the interaction between the electrons and the potential U . With more careful
thought, we realize that the interaction beween electrons should matter. When the electrons move around in response to the potential U , they then create a new electrostatic
potential that the other electrons then should respond to. Thus we need to consider some
sort of self-consistent approach. First, however, we should convert our response equation
into an equation that explains how charge moves around, rather than how density moves
around.2δρ(r) = −e δn(r)The potential felt by an electron moving in an electrostatic potential φ(r) is
U (r) = −eφ(r)Thus, our above response equation Eq. 8.1 can be rewritten as a charge response to anelectrostatic potentialδρ(r) = −e2 D(EF )φ(r)(8.2)
Now if there is a change in density, there is a resulting change in the electrostatic
potential. The resulting electrostatic potential is given by Gauss’ law∇2 φ =
=
1

−δρ
0
2
e
D(EF )φ0(8.3)

Later we will call this quantity the response function χ0 (up to some factors of e). The superscript 0
indicating that it does not yet include interaction between electrons, but only interaction of electrons withthe potential U .
2This is going to be one of those times when we really wish we could go back in time and tell BenjaminFranklin to switch the definition of positive charge so that the electron is positive! The minus signs here
are a common source of error!

8.1. THOMAS-FERMI SCREENING

117

where we have used Eq. 8.2 in going to the second line. This resulting equation for φis precisely the self-consistent Hartree equation: Electrons respond to a density which iscreated by all the other electrons.
It is useful now to define the so-called Thomas-Fermi3 wavevector4
kT2 F =e2
D(EF )
0(8.4)

So that our self consistent screening equation 8.3 is just∇2 φ = kT2 F φSuch an equation has a characteristic decay of the formφ ∼ e−kT F |r|
The potential decays exponentially because it is screened by other electrons. To be more
precise, let us consider the field near a point charge of strength Q inserted into our systemof electrons. The resulting electrostatic potential is given by the Yukawa5 form
φ=Qe−kT F |r|4π0 |r|(8.5)

This potential solves the equation∇2 φ = kT2 F φ +

Q
δ(r)
0

where the second term on the right represents the source charge that is inserted.
Let us now estimate the Thomas-Fermi wavevector. The density of states in anoninteracting Fermi liquid is given by6
D(EF ) =

3 n
mkF
= 2 2
2 EF
~ π

3
Llewelyn Thomas invented this in 1927, frighteningly soon after Fermi-Dirac statistics was invented(1925).
4
It is worth checking that kT F does indeed have the dimensions of a wavevector. First, note that
D(EF ) is is a density of states per unit volume, so it has dimension 1/(Energy − Volume). Next note thate2 /(0 × length) is an energy. The rest is easy.
5
Yukawa won a Nobel prize in 1949 for his work on nuclear forces.
6
To derive this E ∼ kF2 ∼ n2/3 , so dE/dn = (2/3)E/n, and dn/dE = (2/3)E/n. We also need
EF = ~2 kF2 /(2m) and an expression for the density in terms of kF which is given by

n=2

1 4 3
πkF
(2π)3 3where the prefactor of 2 is for two spin states, the 2π factors are the usual factor that comes with k space
integration, and the remaining factor is volume of sphere with radius kF .

118

CHAPTER 8. SCREENING AND LINEAR RESPONSE

We thus haveskT F =

D(EF )e2=
0s

mkF e2=~2 π 2 0

r r
4 kF
π a0

where a0 is the Bohr radius. Now in a typical metal we have a screening length 1/kT F ≈a0 ≈ 1 Angstrom which is very short. However, in a semiconductor, this screening length
can be 10-100 Angstroms or more due to a small kF and a large a0 (the small kF comingfrom a low density of electrons, and the large a0 coming from a low mass and a large
effective dielectric constant).
The Thomas-Fermi approximtion is a good first approach for understanding screening. In particular it removes the long range singularity of the Coulomb interactions.
However, there are some things that are obviously wrong with the Thomas-Fermi approach. The main problem is that it is an “instantaneous” approximation, whereas if we
add a charge to a system it actually takes some time for the electrons to rearrange so as
to screen the charge. This time delay can turn out to be very important. This brings us
to the topic of dynamical screening and dynamical response.

8.2

Response More Generally

We will now consider response functions more generally — including time dependent
response. Let us generally write our Hamiltonian
H = Ĥ0 + δ Ĥ(t)where δ Ĥ(t) is the small perturbation applied to the system (and Ĥ0 is the Hamiltoninan
in the absence of the perturbation). Although we will here be interested in the response of
an interacting Fermi liquid to a perturbation, the same principles we use here are far moregeneral and can be applied to many other systems. It is always conventient to assume
that the perturbation goes to zero at time t = −∞ and turns on slowly.
An important example of such a perturbation (for our electron system) would be aweak time dependent externally applied electrostatic potential, in which case we haveZ
δ Ĥ(t) = dr (−eφext (r, t)n̂(r))(8.6)
However, we could perturb the system with many other things, such as a magnetic field,
or pressure, etc, and much of the calculation of response will be similar.
We now want to ask how the system responds to this perturbation. To probe this
response we measure some observable operator B̂. So we are interested inhB̂it = hψ(t)|B̂|ψ(t)i
where |ψ(t)i is the time dependent many-particle wavefunction.

8.2. RESPONSE MORE GENERALLY

119

An example here is that we might be interested in the density operator
B̂ = δn(r).
So, for example, if we perturb (in Eq. 8.6) with the potential φ(r, t) we might ask what is
the resulting change in density at some later time.
One key priniciple to keep in mind is that of causality: If a perturbation occurs attime t we should only have a response at times t0 > t.
Time Dependent Perturbation Theory
So how are we going to find the general response hB̂it ? The strategy here is to assume
the perturbation δ Ĥ(t) is small and use time dependent perturbation theory.
Let us start with the time dependent Schroedinger equation.
hi∂i~ |ψ(t)iS = H0 + δ Ĥ(t) |ψ(t)iS∂t
The subscript S here indicates that the kets are in the Schroedinger representation. Also
note that inside the square brackets is the full Hamiltonian — which is a combination of
the bare part H0 and the small perturbation.
We now switch to so-called “interaction” representation.
|ψ(t)iI = eiH0 t/~ |ψ(t)iS
In this representation, operators have an additiona time dependence
B̂→ B̂I (t)= eiH0 t/~ B̂S e−iH0 t/~δ Ĥ → δ ĤI (t) = eiH0 t/~ δ ĤS e−iH0 t/~
In this representation, the Schroedinger equation becomesi~∂|ψ(t)iI = δ Ĥ(t) |ψ(t)iI
∂t

(8.7)

Note that in this representation, if δH = 0 then |ψiI is actually time independent.
We can now write out a formal solution of the Schroedinger equation in the form 2 Z t
Z
Z t0
i t 0
i
0
0
dt
dt00 δ ĤI (t0 )δ ĤI (t00 )|ψiI0 + . . .
|ψ(t)iI = |ψiI0 −dt δ ĤI (t )|ψiI0 +
~ −∞
~
−∞
−∞
(8.8)
where |ψiI0 is the wavefunction in the absence of the perturbation, in the interaction
rep (and it is therefore a time independent ket). We can confirm that Eq. 8.8 solves
the Schroedinger equation Eq. 8.7 by plugging the expansion into both sides of the
Schroedinger equation.

120

CHAPTER 8. SCREENING AND LINEAR RESPONSE

We will now make use of the fact that δ Ĥ is small and we will truncate the expansion
at first order, throwing away terms of order (δ Ĥ)2 and higher. We can then use the timedependent wavefunction in Eq. 8.8 to calculate the time dependence of some operator tobe observedhB̂it = hψI (t)|B̂I (t)|ψI (t)i




Z
Z
i t 0
i t 0
0
0
= I0 hψ| +dt I0 hψ|δ ĤI (t ) B̂I (t) |ψiI0 −dt δ ĤI (t )|ψiI0
~ −∞
~ −∞
Z
i t 0
= I0 hψI |B̂I (t)|ψI iI0 −dt I0 hψ| [B̂I (t), δ ĤI (t0 )]|ψiI0~ −∞The first term in the last line is just the value of hBit in the absence of the perturbation.
Thus we have a change in the value of B given byZi t 0hδ B̂it = −dt h[B̂I (t), δ ĤI (t0 )]i~ −∞where the expectation is taken in the unperturbed state |ψiI0 as above. This formula is a
general linear response equation, otherwise known as the Kubo formula7 . The importanceof this type of formula can hardly be overstated, as it is used universally to calculate theresponse of a system to a perturbation.
It is crucial to note that the response here is linear in the perturbation (hence thename “linear response”). Linearity gives us the important principle that response of asum of perturbations is the sum of the responses of each individual perturbation. This
will allow us to decompose a general perturbation into a convenient basis of elementary
simple perturbations, which will simplify the calculation greatly.
Example of Externally Applied Potential
Let us go back to the main example we want to study: the density response to an externallyapplied potential. The perturbation is generally of the form
Z0
δ ĤS (t ) = dr −eφext (r, t0 )n̂S (r)(8.9)
with the operators being expressed here in the Schroedinger representation. The operator
we would like to measure is again the density operator
B̂ = n̂S (r)It is often easier to work in Fourier space. By using Parseval’s theorem8 we have−e X
δ ĤS (t) =φext (k, t0 ) n̂S (−k)
V

(8.10)(8.11)

k

7

Named for the Japanese physicist Ryogo Kubo who constructed this general formula in 1957.
Named for Marc-Antoine Parseval des Chênes who published it in 1799 without proof, stating that it
was self-evident.
88.2. RESPONSE MORE GENERALLY

121

with V the volume as usual. Since we are considering linear response, we can imagine
applying a single wavevector perturbation k — and if we have an initial perturubation
which is a sum of different wavevectors we can just add up the responses to each wavevector
at the end of the calculation.
We also write the quantity we want to measure (the density) in k-spaceδB = δn̂S (k)again in the Schroedinger representation.
We can then write our Kubo formula asZ
ie t
dt0 h[n̂I (k, t), n̂I (−k, t0 )]iφext (k, t0 )hδn(k)it =
~V −∞
Z ∞
=
dt0 χ̃(k, t − t0 )φext (k, t0 )(8.12)(8.13)

−∞

where

ie
h[n̂I (k, t), n̂I (−k, t0 )]i
~V

1 x>0
θ(x) =0 x<0χ̃(k, t − t0 ) = θ(t − t0 )andis the usual step function. The insertion of the step function in Eq.8.13 implements thecorrect boundary conditions given explicitily in Eq. 8.12. The function χ̃ is known as thelinear response function (or the susceptibility or the linear response kernel) and note that
it is translationally invariant in time. I.e., if we perturb a system at time 0 and measure
at time a, this is the same as perturbing at time b and measuring at time a + b.
Note that here on the left hand side of Eq. 8.12, we have indicated the response at
wavevector k which was the same wavevector which we perturbed with on the right. In
other words we are assuming that the response is diagonal in wavevector. It turns outthat this is necessarily true in any tranlationally invariant system — a system that isperturbed at some wavevector k, in linear response theory, will only only respond at thesame wavector. We will prove this statement below.
Our next step is to realize that the integral in Eq. 8.13 is a convolution in the tvariables. As is usual with a convolution, it is then conventient to Fourier transform.
Further, we remember that since we are thinking about linear response, we can always
decompose the perturbation into a sum of pieces and think about the response to eachpiece separately. Thus let us just think about a perturbation at a fixed frequency
φext (k, t) = e−iωt φext (k, ω)
If we have a more complicated time dependence we can always sum over many frequencies.
Note that we had previously decided that to make sure things are well-behaved, wewant to make sure the perturbation is turned off at time t = −∞. Thus we write insteade−i(ω+iε)t(8.14)

122

CHAPTER 8. SCREENING AND LINEAR RESPONSE

where ε is an infinitesimle positive number. This will then properly vanish at t = −∞.
We will often not write the +iε but we should remember that it is there.
The integral Eq. 8.13 in Fourier space instead reads
δn(k, ω) = χ(k, ω)φext (k, ω)
where the response χ is given byZ ∞χ(k, ω) =dt eiωt χ̃(k, t)Zie ∞dt eiωt h[n̂I (k, t), n̂I (−k, 0)]i
~V 0
−∞

=

(8.15)

Note that the +iε we added to ω in Eq. 8.14 regularizes this integral at t = +∞.
Eq. 8.15 has two terms in the commutator. We will consider one of these terms
explicitly (and we assume the reader can fill in the rest of the calculation for the second
term!). We will assume we are at temperature T = 0 and that the system is initiallyin the ground state, which we write as |GSi. The first term in the commutator is (withsubscripts I and being interaction representation, and unsubscripted being Schroedinger
representation)
I hGS|n̂I (k, t)n̂I (−k, 0)|GSiI= hGS|n̂(k)e−iH0 t/~ n̂(−k)eiH0 t/~ |GSi
X=
hGS|n̂(k)e−iH0 t/~ |mi hm|n̂(−k)eiH0 t/~ |GSimwhere we have inserted a complete set of eigenstates |mi in the last line (their corresponding energies will be taken to be Em with EGS being the ground state energy). Here, andhereafter, we have returned to the Schroedinger representation but for simplicity we nolonger put subscripts S on kets and operators.
We now make use of the fact that n̂(−k) = n̂† (k) to writeX
hGS|n̂(k, t)n̂(−k, 0)GSi =ei(EGS −Em )t hGS|n̂(k)|mi hm|n̂† (k)|GSim

At this point we think back to the claim made after Eq. 8.13: That a perturbation at
wavevetor k will only elicit a response at the same wavevector k. We can now see why
this is the case. Reading this equation from right to left: We start with the grounds state|GSi which we assume has momentum zero. The operator n̂† (k) adds momentum −~kto the ground state and generates an excited state |mi with momentum −~k. Then theoperator n̂(k) removes this momentum and returns us to the ground state. If we hadmeasured a response at a different wavevector k0 6= k, the operator n̂(k0 ) would not beable to return us to the ground state and we would end up getting zero, as claimed.
We thus have the resultZoie ∞ iωt X n i(EGS −Em )t/~χ(k, ω) =ee|hGS|n̂(k)|mi|2 − e−i(EGS −Em )t/~ |hGS|n̂(−k)|mi|2~V 0m

8.3. LINDHARD RESPONSE FUNCTION (RESPONSE OF NONINTERACTING ELECTRONS)123
where the second term in the brackets is from the other term of the commutator. Taking
the integral now gives|hGS|n̂(k)|mi|2−e X|hGS|n̂(−k)|mi|2(8.16)
χ(k, ω) =−~V m ω + iε − (Em − EGS )/~ ω + iε − (EGS − Em )/~We now are left with the problem of calculating the matrix elements such as hGS|n̂(k)|mi.
This depends entriely on what particular Hamiltonian we are working with. So far, what
we have written is entirely general, and it is still exact at linear response level.
The linear response calculation we have done can be generalized to give the responseof any measured operator to any perturbation at linear order. For example, we cangeneralize the perturbation of Eq. 8.9 to couple to any operator Â
Z
δH(t) = drφext (r, t)Â(r)where φext is not necessarily an electrostatic potential, and we can measure the respnonse
in any operator B̂(r). If we again Fourier transoform to frequency and wavevector we canvery generally write
δB(k, ω) = χBA (k, ω)φext (k, ω)and essentially the same calculation will give the response function()
1 X hGS|B̂(k)|mihm|Â(−k)|GSi hGS|Â(−k)|mihm|B̂(k)|GSiχAB (k, ω) =−~V m
ω + iε − (Em − EGS )/~ω + iε − (EGS − Em )/~8.3

Lindhard Response Function (Response of Noninteracting Electrons)

Let us return to the density-response function Eq. 8.16. While this is a very generalequation it is usually very hard to know what the eigenstates |mi and eigenenergies Em
are for a complicated system. So, let us consider a very system to start with: free electrons.
We take
X p2
i
H0 =
2m
i

only. (In principle, one might want to include interactions in the Hamiltonian — which
we will do later, at least in some approximation. However, writing down a complete setof eigenstates is extremely hard when we have interactions, whereas it is fairly easy forthe noninteracting case). In this noninteracting case, the ground state |GSi is simply thefilled Fermi sea. We can write the density operator again as (See Eq. 7.32, with apologies
that we changed notation for ρ to n̂)X †n̂(k) =c(k+q)σ cqσ = FourierTransformk [ψ̂ † (r)ψ̂(r)](8.17)q,σ124

CHAPTER 8. SCREENING AND LINEAR RESPONSE

Figure 8.2: The relevant excitations for this calculation are excitations of a single electron
out of the Fermi sea.
We thus see that n̂(k) can excite a single electron out of the Fermi sea. Thus, the intermediate states |mi are all of the form
|mi = c†(k+q)σ cqσ |GSi
with |q| < kF and |k + q| > kF as shown in Fig. 8.2. The corresponding energy differenceis given by~2 |k + q|2 ~2 |q|2
Em − EGS =−= 0k+q − 0q2m2mwhere 0q = ~2 |q|2 /(2m) as usual, the superscript 0 indicating that this is the energy ofnoninteracting electrons. Plugging these results into Eq. 8.16, and using χ(k) = χ(−k)
we obtain (See chapter appendix for the relevant algebra!)
χ0 (k, ω) =−e X nF (q) − nF (q + k)
V q,σ ~(ω + iε) − (0q+k − 0q )(8.18)

with nF the Fermi function, which is simply a step function at the Fermi surface at zerotemperature (See Eq. 7.33). The superscript 0 here on χ0 indicates that we are thinking
about noninteracting electrons. This rather important result is known as the Lindhard
response function9 .

Aside: Green’s function language
For those who like to think in Feynman diagrams and Green’s functions it is sometimes
useful to rephrase our result in this language. For those who are not familiar with thistechnique, it is OK to mostly ignore it (it is not examinable in this course) and skip down
to section 8.3.1. It is simply useful to be able to speak different languages sometimes.
First, the “source” for the response function is the density n̂(k), which we think ofin the form of Eq. 8.17 which therefore includes terms of the form c†(k+q)σ cqσ . We write
this diagrammatically as the vertex shown in the left of Fig.8.3 We then also write the9

Named for Jens Lindhard who wrote it down for the first time in 1954 — rather late for such animportant result to be first discovered, although perhaps this is not surprising being that the Kubo formula
had not yet been discovered at this time.

8.3. LINDHARD RESPONSE FUNCTION (RESPONSE OF NONINTERACTING ELECTRONS)125

Figure 8.3: Left: Vertex representing a density operator. Right: Bubble representing
density-density response
Greens’ function for the electron as
G(k, Ω) =1Ω − (0k − µ) + iεsgn(0k − µ)which we write as a line with an arrow. The response function χ0 is then given by the
diagram in the right of Fig. 8.3. This diagram corresponds to the integral over the Green’s
functionsXZ
χ0 (k, ω) =dΩ G(q, Ω)G(k + q, ω + Ω)q

Carrying out the integrals here will generate precisely the same result as Eq 8.18 (up to aprefactor which I may have dropped).
For complicated diagrams, meaning high order perturbation theory calculations,these Green’s functions and diagrammatics are a very useful bookkeeping tool. However,at this level it is easier to just do the calculation directly without bothering with diagrams.
Nonetheless, it is useful to draw the diagram, because one often sees it in the literature,and it is good to know what it means!

8.3.1Interpretations

We now would like to interpret the result we have derived Eq. 8.18. We will look at someproperties of χ0 , partially to understand it, and partially as a sanity check that we havegotten a reasonable answer!
Imaginary Dissipative Part
First, let us examine the imaginary part of χ. What does it mean that our response
function has an imaginary part rather than all real? This simply means that the response is dissipative – energy is absorbed by they system (this is precisely analogous to a
conductivity being real rather than imaginary).
To find out when χ0 has an imaginary part recall that
lim

ε→0+

1
1
= −iπδ(x) + P( )x + iεx

126

CHAPTER 8. SCREENING AND LINEAR RESPONSE

Figure 8.4: The region of q,ω space where there is an imaginary part of χ. This is precisely
the region where Eq. 8.19 can be satisfied.
with P here the principle part, thus giving the imaginary and real part of the expression.
In Eq. 8.18, it is implied that we are taking a limit of ε → 0+ . Thus the response has an
imaginary part when
 |q + k| > kF and |q| < kF
0
0
OR
(8.19)~ω = q+k − q where|q + k| < kF and |q| > kF
In Fig. 8.4 we sketch the region of k, ω space where this condition is satisfied. This is theregion of k, ω space where we can add a wavevector k to the system and get an excitationof energy ω.
Let us make a few quick calculations to determine the rough shape of this region.
First we notice that that the longest vector across the Fermi surface is the diameter whichis k = 2kF . It is impossible to make any zero energy excitations (ie., from the Fermi
surface to the Fermi surface) for wavevectors larger than this. This is why the shaded
region in Fig. 8.4 has a boundary at ω = 0 and k = 2kF . Secondly it is useful to look atwhat happens for small k. Here we can have the condition~ω = 0q+k − 0q =~2 |k + q|2 ~2 |q|2−2m
2m

but for small k we need to have q ≈ kF to satisfy the conditions in Eq. 8.19. Thus we can
write
~kF
~ω ≈
k cos θmwhere θ is the angle between k and q. The largest ω is then given by cos θ = 1 or
ω = vF kwith vF = ~kF /m the Fermi velocity, which is shown in Fig. 8.4.

8.3. LINDHARD RESPONSE FUNCTION (RESPONSE OF NONINTERACTING ELECTRONS)127
Low Frequency, Small k
Let us look at the limit ω → 0 with small k. We then have00
e X nF (q ) − nF (q+k )χ0 (k, ω) =V q,σ(0q+k − 0q )where nF is now a step function in energy at the chemical potential µ. If k is small, thenthe difference in the denominator is small, as is the difference of the arguments of nF in
the numerator, and we can view the fraction on the right as being a derivative. Since thederivative of a step function is a delta function, we then haveeXχ0 (k, ω) =δ(q − µ)V q,σZ= e d D()δ( − µ) = eD(µ)
where in going to the second line, we have replaced a sum over q as an integral over energy
times a density of states D(). The final result here is just the density of states, which isthe static compressibility10 we derived previously (See Eq. 8.2).

8.3.2

High Frequency ω  EF and small k

Another limit we can consider is that of high frequency.
Explicit Limit
Here we expand the denominator of Eq. 8.18 for large ω obtainingX−enF (q) − nF (q + k) 1χ0 (k, ω) =(0q+k −0q )
V q,σ
~ω
1−

(8.20)~ωThe leading term in this expansion replaces the square brackets with unity, but then thesum over q vanishes. For the next term it is useful to write for small k,(0q+k − 0q ) ≈ k · ∇q 0q
and similarlynF (q + k) − nF (q) ≈ k · ∇q nF (q)We thus have
χ0 (k, ω) =

=
10

X
e
(k · ∇q nF (q))(k · ∇q 0q )2
V(~ω) q,σ
"
#

X
−e
2 0
nF (q) (k · ∇q ) q )V(~ω)2 q,σSee section 9.5.1 footnote 9 for why we call this compressibility.

(8.21)(8.22)

128

CHAPTER 8. SCREENING AND LINEAR RESPONSE

In obtaining the second line, we have integrated by parts (convert to integral, integrate byparts and convert back to a sum). The quantity in the second square bracket is ~2 k 2 /(m)and is q independent. The quantity in the first square bracket can thus be summed alone
to just give N . We thus have in this limit of large ωχ0 (k, ω) =−ek 2 nmω 2

(8.23)

where n is the electron density. In fact this result is extremely general — independent ofmany details of the system, as we will see next.

More general f -sum rule approach
The limit can be obtained by invoking the so-called f -sum rule. We return to the primitive
integral form of our response function (See Eq. 8.15)
χ(k, ω) =

ie
~V

Z ∞

dt ei(ω+iε)t h[n̂(k, t), n̂(−k, 0)]i

0

Where
n̂(k, t) = eiHt/~ n̂(k)e−iHt/~
and here we would like to be completely general, so we consider any Hamiltonian H atthis point (which may include electron-electron interactions, for example).
Since we are interested in high frequency, we equivalently want to look at small
times t. Thus we expand for small t to writen̂(k, t) ≈ n̂(k) +it[H, n̂(k)] + . . .
~so that we have11ieχ(k, ω) =
~V

Z ∞
0

dt ei(ω+iε)tith[[H, n̂(k, t)], n̂(−k, 0)]i~

This double commutator is familiar from our discussion of the Feynman theory of superfluids. In Eq. 5.11 we discovered that this double commutator is independent of interactions
between particles and yields the constant ~2 k2 N/m. Thus we can perform the integral toobtain the same result−ek 2 nχ=
(8.24)mω 2
11

The term h[n(k), n(−k)]i has to give zero for a rotationally invariant ground state.

8.4. RESPONSE OF INTERACTING ELECTRONS: RPA

129

Drude Theory Approach
In fact one can get the same result out of simple Drude theory for a translationally invariant
system. Translational invariance here means there should be no scatterers. Let us startwith the current conservation equation (with the overdot being time derivative)˙
δn(r,t) + ∇ · j(r, t) = 0
where here δn is the local change in particle density and j is the local particle current.
Let us rewrite this in terms of its frequency and wavevector Fourier modes−iω δn(k, ω) + i k · j(k, ω) = 0
which we can write as
jk =

ω δn
k

(8.25)

We can then use the equations of motion when an electric field is appliedmv̇ = −eE
where the velocity is related to the current density by
j = n̄v
with n̄ the average density12 . We thus havemj̇ = −en̄E = en̄∇φagain moving to Fourier space we have−iωmjk (k, ω) = ien̄k φ(k, ω)
Plugging in Eq. 8.25 we obtainδn(k, ω) =8.4

−k 2 n̄eφ(k, ω)mω 2

Response of Interacting Electrons: RPA

In the last section we have been considering non-interacting electrons. We have the charge
density response to an applied potentialδn(q, ω) = χ0 (q, ω)φext (q, ω)
12

We would also have a term δnv but this is order (small)2 , since both δn and v are assumed small.

130

CHAPTER 8. SCREENING AND LINEAR RESPONSE

where the superscript 0 indicates noninteracting. We would like to put the interactions
back into the system in the simplest way possible. What do the interactions do? Well, if
a nonuniform density δn occurs, this will induce a Coulomb interactionZ
φinduced (r, t) = dr0 V (r − r0 )δn(r0 , t)
(8.26)
where we will assume here that the interaction V is of Coulomb form
V (r) =−e4π0 ralthough our approach will work for any form of V .
Equation 8.26 is a convolution so we can write it more simply in Fourier space
φinduced (q, ω) = Ṽ (q)δn(q, ω)(8.27)

where Ṽ is the Fourier transform of V (r).
Now the system must respond to both the externally applied φext and the induced
φinduced . A simple approximation is to assume the system responds like a noninteracting
system — and let the entire effect of the interaction be included in the fact that we let
the system respond to both the external and the induced potentials. Thus we writeδn(q, ω) = χ0 (q, ω) [φext (q, ω) + φinduced (q, ω))]hi
= χ0 (q, ω) φext (q, ω) + Ṽ (q)δn(q, ω))We can then solve this to obtainδn(q, ω) =

χ0 (q, ω)φext (q, ω)
1 − χ0 (q, ω)Ṽ (q)or equivalently
δn(q, ω) = χRP A (q, ω)φext (q, ω)
whereχRP A (q, ω) =

χ0 (q, ω)
1 − χ0 (q, ω)Ṽ (q)(8.28)

This approximation of the response function of an interacting system is known as the
RPA approximation13 or time-dependent self-consistent Hartree approximation, since weare simply treating the effect of the interaction as being an effective potential (Eq. 8.26)
that the electrons respond to, analogous to the Hartree approximation we used previously
in section 7.3.1.
13
RPA stands for “Random Phase Approximation”. However these words mean almost nothing now.
The approximation was first used by Bohm and Pines in 1952, and it was derived in a very roundabout
way which is no longer used. In that language it was related to random phases, but this is only of historicalinterest.

8.4. RESPONSE OF INTERACTING ELECTRONS: RPA

131

Figure 8.5: The RPA diagrams or bubble sum
One way of understanding the RPA approximation is to expand the denominatorand writeχRP A = χ0 + χ0 Ṽ χ0 + χ0 Ṽ χ0 Ṽ χ0 + . . .
which is shown in diagrammatic language in Fig. 8.5. The physical interpretation of thisseries is as follows: In the first term, χ0 , the electrons directly respond to the external
perturbation. In the second term, χ0 Ṽ χ0 the electrons responsd to the perturbation
and build up a density which creates a potential and then the electrons respond to thispotential. And so forth.

8.4.1

Relation to Dielectric Constant

The response function χ is closely related to the relative dielectric constant. Recall that
1
D
r

E=
meaning that14
φtotal =

1
φext
r

But we haveφtotal = φext + φinducedhi= 1 + Ṽ χ φext
This equation is exact, so long as one uses the exact response function χ in the equation.
So we have the exact result1r =1 + Ṽ χ
Note that r in principle depends on frequency and wavevector.
Using the RPA approximation for χ we substitute in to obtain
r =
14

1
0

χ
1 + Ṽ 1−V
χ0= 1 − V χ0(8.29)

Recall that D is the field applied normal externaly to a material, whereas E is the physical field.

132

CHAPTER 8. SCREENING AND LINEAR RESPONSE

Static Limit
It is useful to look a the static ω = 0 limit. Recall that in this limit (Eq. 8.2)
χ0 = eD(EF )and we have the Coulomb interaction in Fourier space (Eq. 7.39)
Ṽ (k) =−e
0 k 2We thus plug into Eq. 8.29 to getr (k) = 1 +

kT2 F
e2
D(E
)
=
1
+
F
0 k 2k2

where kT F is the Thomas-Fermi wavevector, Eq. 8.4.
We can then try to find out what happens if we insert a test charge of charge Q into
our system. We then haveQ
φext (k) =0 k 2
which gives us
φtotal (k) =1Q/0
φext = 2
r
kT F + k 2

Fourier transforming this immediately gives
φtotal (r) =Qe−kT F |r|4π0 |r|(8.30)

matching the Yukawa form we derived in Eq. 8.5.

8.4.2

High Frequency, Low k response: Plasmons

Let us consider interactions in the small k high frequency limit. As we derived above (inseveral ways!) we have the noninteracting response (Eq. 8.23) given byχ0 =−en̄k 2mω 2

If we plug this into the RPA equation (Eq. 8.28) we discover that the denominator 1− Ṽ χ0diverges whene en̄k 2
1 = Ṽ χ0 =
0 k 2 mω 2
or
s
e2 n
ω=
0 m

8.4. RESPONSE OF INTERACTING ELECTRONS: RPA

133

Figure 8.6: Displacing all of the mobile charges by a distance x builds up a surface chargeρsurf ace = n̄xe.
The divergence in χ indicates an excitation mode of the system — in other words, thesystem responds at finite amplitude to even an infinitessimly small amplitude perturbation
if you perturb it in resonance with the excitation energy. This particular excitation mode
is known as plasma oscillation or plasmon (when we think of it as a quantized object likea phonon or photon).
There is actually a very simple physical explanation for the plasmon. We consider
a big cube of our material — with mobile negative charges and a stationary positive
background. Now let us imagine moving all of our negative charge a distance x in the x
direction. This builds up a charge on the two x faces of our cube as shown in Fig. 8.6.
The surface charge density that is built up is n̄xe. We thus have a capacitor formed by
the two charged faces. We may recall that from Gauss’ law ∇ · E = ρ/0 we can obtain
the electric field from a capacitor
E = ρ/0 = n̄xe/0
The electric field then applies a force
F = −eρ/
to all of the electrons. Newtons law then gives us
mẍ = −eρ/0 =

−n̄e2
x
0

thus resulting in harmonic motion with frequency
s
e2 n
ω=
0 m .
It is important to note that this result holds with the mass m not being renormalized
by any interactions (We discussed mass renormalization in the previous chapter, but that
does not matter here). The reason the mass remains unrenormalized is because the plasma
frequency describes the motion of the center of mass of all of the electrons in the system,and this is completely independent from the details of interaction between electrons. Thisis quite related to the argument leading up to Eq. 8.24, that the high frequency response
is unchanged by interactions between particles.

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CHAPTER 8. SCREENING AND LINEAR RESPONSE

Figure 8.7: The excitation spectrum of an electron gas in three dimensions, with Coulombinteraction between particles. This is qualitiatively similar to Fig. 8.4 except that the
Coulomb interaction gives an isolated plasma mode at small k and high frequency. In
RPA calculation, the upper edge of the particle-hole excitation band is at ω = vF q withvF the unrenormalized Fermi velocity. However, in more detailed calculations this will be
modified to ω = vF∗ q with a renormalized effective mass. However, the frequency of theplasma mode remains unrenormalized.

In Fig. 8.7 we sketch the excitation spectrum of a three dimensional electron gas with
Coulomb interactions. The spectrum is qualitatively similar to the noninteracting fermion
spectrum shown in Fig. 8.4 except for the addition of the plasma mode at ωplasmon at
small wavevector k. In the RPA calculation, the upper edge of the particle-hole excitation
band is at ω = vF q with vF the unrenormalized Fermi velocity. In a more detailed
calculation this will be modified to ω = vF∗ q with a renormalized effective mass. However,the frequency of the plasma mode remains unrenormalized.

8.5. CHAPTER APPENDIX: DERIVING LINDHARD SCREENING EXPRESSION135

8.5

Chapter Appendix: Deriving Lindhard Screening expression

We start with Eq. 8.16. Making use of inversion symmetry of the system we have χ(k, ω) =χ(−k, ω) so we can write|hGS|n̂(−k)|mi|2−e X|hGS|n̂(k)|mi|2
χ(k, ω) =−~V m ω + iε − (Em − EGS )/~ ω + iε − (EGS − Em )/~−e X|hm|n̂(k)|GSi|2|hm|n̂(−k)|GSi|2=(8.31)−~V m ω + iε − (Em − EGS )/~ ω + iε − (EGS − Em )/~Conisder the first term in the brackets of Eq. 8.31. Writing
X †n̂(k) =
cq+k,σ cq,σq,σthe intermediate state is clearly |mi = c†q+k,σ cq,σ |GSi. We can thus write the first termof the sum as−e X Θ(|q + k| > kF ) Θ(|q| < kF )(8.32)~V q,σω + i − (k+q − q )/~with Θ the step function.
Similarly consider the second term in the brackets of Eq. 8.31. The intermediate
state is |mi = c†q−k,σ cq,σ |GSi. So we can write the second term in the sum ase X Θ(|q − k| > kF ) Θ(|q| < kF )~V q,σω + i − (q − q−k )/~In this second term we now shift summation variables q → q + k. So the term becomese X Θ(|q| > kF ) Θ(|q + k| < kF )~V q,σω + i − (q+k − q )/~Summing this with Eq. 8.32 yields
χ(k, ω) =−e X Θ(|q + k| > kF ) Θ(|q| < kF ) − Θ(|q| > kF ) Θ(|q + k| < kF )~V q,σω + i − (k+q − q )/~This form is equivalent to that of Eq. 8.18 except that the numerators look differnt. By
looking at all four cases where |q| is greater or less than kf and |q + k| is greater or lessthan kF it is easy to see that these numerators are in fact identical, thus completing the
calculation.

136

CHAPTER 8. SCREENING AND LINEAR RESPONSE

Chapter 9

Landau Fermi Liquid Theory
9.1

Background

What we have done so far (first order perturbation theory, Hartree-Fock, RPA) is fairly
accurate for weakly interacting Fermi systems. However, real physical fermion systems
usually have strong interactions1 . Whether we are talking about metals, 3 He, neutronstars, or semiconductors2 , we typically have the interaction larger than the Fermi energy
Einteraction > EF
As a result the calculations we have done so far cannot be used quantitatively3 , although
they can be used for qualitative inspiration. Despite this strong warning, we almost
always get away with ignoring interactions completely. Indeed, almost all of semiconductor
physics (including the whole semiconductor industry) is built on the idea of mostly noninteracting electrons. Why does this make any sense at all? I’m glad you asked!
Understanding of strongly interacting fermions really began with experiments on
This system is much simpler than systems of electrons because of the absence oflong ranged Coulomb interactions. In 1953-1954, William Fairbank4 manged to cool 3 He
to below about 1K which is less than the Fermi energy EF ≈ 5K. At low temeperature he
3 He.

1

In recent years people have been able to make cold atomic Fermi gases where one can tune theinteraction strength, and almost turn it off if desired. A huge amount of the pioneering research on cold
fermionic atomic systems was done by Debbie Jin, who very sadly died of cancer at the young age of 47 in2016.
2
One might think that the low density of semiconductors might help. However, this turns out not tobe true. The interaction is on the scale of e2 /(a) where a is the distance between fermions, whereas the
kinetic energy is on the scale ~2 /(2ma2 ), so as we make the density lower, in fact the ratio of interaction
to Fermi energy actually increases!
3
There is an exception: At small enough q and high enough ω, RPA becomes exact even for stronglyinteracting systems.
4
Fairbank was at Duke university at the time (where he recruited Fritz London in 1952). A few
milligrams of 3 He was supplied by Oak Ridge National Lab, where much of the research for the American
nuclear project was being done. This small amount was plenty for conducting the key experiments.

137

138

CHAPTER 9. LANDAU FERMI LIQUID THEORY

was able to see evidence of the Fermi-Dirac distribution. In particular, he measured
Cv ∼ Tχ ∼ indep of T
where here χ means the pauli paramagnetic susceptibility (and as usual Cv is heat capacity). Both of these relations are indicative of fermions. However, in both cases, theproportionality constants did not match the predictions of the noninteracting fermion
model. Now in our discussion of Hartree-Fock approximation above, we already have seenthat the particle mass can be renormalized by interactions. However, this alone does not
fix the problem in a consistent manner5 . .

9.2

Basics Idea of Fermi Liquid Theory

In 1956, Landau, in one of his classic papers6 , explained how we should understand interactions of fermions. This work, based on brilliant intuition, set the language for our
understanding of Fermions, and was fully justified many years later using both perturbation theory and renormalization group methods.

9.2.1

Landau’s Conjecture

Landau conjectured that interacting fermions should be just like noninteracting fermions,
but “dressed” by interactions. What he meant by this is that
 The ground state shold be a “dressed” Fermi sea
 Low energy excitations, can be described in terms of quasiparticle excitations of theground state. These are “dressed” versions of the noninteracting excitations of the
Fermi sea.

The key question here is what we mean by “dressed”. Here we mean that it needs toinclude a cloud of interactions with its surroundings. This very vague statement we will
make more definite by the following argument:
Landau Adiabaticity Argument
Let us imagine starting with the noninteracting case adiabatically turning on the interactions. Since this is a procedure that can be implimented using a Hamiltonian that we5
If one were to try to just replace the mass with an effective mass, one would require two different
effective masses to explain the two different experiments!
6
Titled, “The Theory of Fermi Liquids”. This is a rather remarkable paper. It is only 6 pages long andhas only a single citation (to the experiment by Fairbank).

9.2. BASICS IDEA OF FERMI LIQUID THEORY

139

Figure 9.1: Top: turning on interaction adiabatically turns a noninteracting Fermi sea into
the interacting “dressed” ground state. Bottom: If we start with a Fermi sea plus a single
particle outside the Fermi sea, when we turn on the interaction, we obtain a quasiparticle.

Figure 9.2: Fluid backflow around a moving object.
modify in time, it must be representable using a unitary (time evolution) operator. The
result of this procedure gives the dressed Fermi sea. This is depicted as a cartoon in the
top of Fig. 9.1. Similarly, we can start with a single fermion added to the noninteracting
Fermi sea and adiabatically turn on the interaction. Here we get a dressed Fermi sea witha dressed additional particle outside of the Fermi sea as shown in the bottom of Fig. 9.1.
We call this dressed additional particle a quasiparticle.
Here it is key to realize that in turning on the interaction, the quantum numbersof the system do not change — in particular, the momentum and spin are unchanged.
Thus the dressed quasiparticles are in one-to-one correspondence with the noninteracting
particles. However, in contrast to the noninteracting particles, the quasiparticles carrywith them a cloud of interactions with their environment. This is very much like the ideaof backflow around a moving object in a fluid as shown in Fig. 9.2.
Why does this picture of dressed quasiparticles make sense? To answer this weshould ask the opposite question: how could this picture fail? (Of course one could always

140

CHAPTER 9. LANDAU FERMI LIQUID THEORY

Figure 9.3: Two different depictions of quasiparticle decay of a single particle added above
a Fermi surface.
have a first order transition when the interaction turns on and the new ground state could
be something completely different, like a crystal, but let us assume that such a first ordertransition does not occur.) The only way our picture of dressed particles can fail is if theintroduction of interactions lets the quasiparticle decay into multiple other quasiparticles— then the idea that we have a “single-particle” excitation will fail.
Let us think about what would have to be the case if such a decay were to happen. Of
course we need to conserve both energy and momentum. If we have an initial momentumof our additional particle is k1 above the Fermi surface (|k1 | > kF ), we can imagine that it
decays into two quasipartcles with momenta k2 , k4 above the Fermi surface and a quasiholewith momentum k3 below the Fermi surface. Such a process is shown in Fig. 9.3.
The key claim here is that due to phase space restrictions associated with energy
conservation, the scattering lifetime for such a process isτ ∼ |k1 − kF |−2(9.1)

so if we are considering quasielectrons or quasiholes very close to the Fermi surface, they
are very long lived, i.e., they do not fall apart. If a particle is “close enough” to the Fermi
surface it lives essentially forever, meaning it is in an eigenstate.
Here is a quasiproof of the statement Eq. 9.1. In the process shown in Fig. 9.3, all of
the particles have energy vF |k − kF | with their given k, independent of whether they are
quasiparticles or quasiholes (a quasihole below the Fermi surface costs positive energy).
We then enforce that the intitial energy must be the same as the final energy, so that in
calculating the scattering rate using Fermi’s golden rule we will have integrals of the form
Z ∞
Z ∞
Z ∞
1
∼
d2
d3
d4 δ(1 − (2 + 3 + 4 ))τ
0009.3. PROPERTIES OF THE DRESSED FERMI LIQUID

141

Figure 9.4: The region that conserves energy is the area of triangular piece of a plane a
distance 1 out along each axis.
where 1 is the initial energy and 2 , 3 , 4 are the three final energies of the three final
particles. As shown in Fig. 9.4 the value of this integral is equal to the area of a triangularregion in (2 , 3 , 4 ) space which intersects each axis at a value 1 . Thus we obtain
Area ∼ 21 ∼ |k1 − kF |2
Thus the scattering rate decreases as we approach the Fermi surface and we have welldefined quasiparticles7 .

9.3

Properties of the Dressed Fermi Liquid

Let us then ask what the dressing of the Fermi liquid looks like in more detail. As written
figuratively in Fig. 9.1, the dressing is given by some unitary transformation U . So thatthe interacting ground state can be written in terms of the noninteracting ground state|GSi = U |GS0 i
Similarly we can write creation operators for the dressed quasiparticles c̃†k,σ by dressing
the noninteracting quasiparticles c†k,σc̃†k,σ = U c†k,σ U †Now notice that the occupation of these dressed quasiparticles in the ground state is simply
a step function, since
nqp (k, σ) = hn̂qp (k, σ)i = hGS|c̃†k,σ c̃k,σ |GSi
= hGS0 |U † U c̃†k,σ U † U c̃k,σ U † U |GS0 i= hGS0 |c†k,σ ck,σ |GS0 i = nF (k)It is much more complicated to find the occupancy of the noninteracting plane wave
orbitals in the interacting ground state. Generally, this will be a hard task. However, at
7

There has recently been a lot of study of so-called non-Fermi liquid cases where the quasiparticle is
not well defined because 1/τ > . Certain models of high temperature superconductors have this property,
for example.

142

CHAPTER 9. LANDAU FERMI LIQUID THEORY

Figure 9.5: An interacting Fermi liquid still has a step in the occupation at kF , although
the step is not as big as for a noninteracting Fermi system.
least if the interaction is weak, we can apply perturbation theory to write the interacting
ground state in terms of the noninteracting one:
X |nihn|V |GS0 i+ . . .(9.2)
|GSi = η |GS0 i +En − EGS
n6=GS


X
vq1 ,q2 ,q3 ,q4= η |GS0 i +c†q1 c†q2 cq4 cq3 |GS0 i+ . . .Eq1 + Eq2 − Eq3 − Eq4
q1 ,q2 >kF ;q3 ,q4 <kF

where η is a factor included to keep the wavefunction properly normalized (and we have
dropped spin indices for simplicity).
Now when we calculate the occupation of the original orbitals we haven(k, σ) = hn̂(k, σ)i = hGS|c†k,σ ck,σ |GSi
= hGS0 |U † c†k,σ ck,σ U |GS0 i
and the unitary transform is precisely the series shown in Eq. 9.2. The key thing to notehere is that, at least perturbatively, there must remain a step in the occupation n(k)coming from the first term in the series. It is a less pronounced step than for a purely
noninteracting system, but it is a step nonetheless as shown in Fig. 9.5.
This brings us to an important fact about interacting Fermi systems. (We can evengeneralize our consideration now to arbitrary band structures in crystals, not just simple
spherically symmetric systems).
Let us define a Fermi surface to be the locus of points in k-space where there is a
discontinuity in n̂(k).
Luttinger’s Theorem:8 The volume of the Fermi sea in k-space is unchanged byinteractions.
8

Joaquin (Quin) Mazdak Luttinger discovered this (partially with John Ward) in 1960. Those whoknow field theory know Ward from the famous Ward identities, which are key to the derivation of thistheorem.

9.4. LANDAU FREE ENERGY FUNCTIONAL

143

In a rotationally invariant system, this means kF is unchanged by interactions.

9.4

Landau Free Energy Functional

Given this quasiparticle concept, how are we going to describe the physics of a Fermi liquid.
The general idea is to always describe a system by the occupation of the quasiparticle statesnp,σ . (Above we called this nqp (k, σ) to distinguish it from the noninteracting particle
occupation. Here we will drop the qp and also write this as a function of p rather than k
as is more conventional in the field.) Remember that we are describing the quasiparticle
occupancy here!
As mentioned above, at T = 0 we expect to have the usual step function occupancy

1 |p| < pF0
T =0
np,σ =
0 |p| > pF
The superscript 0 means this is the occupancy in equilibrium (i.e., unperturbed). At finite
temperature we would have a Fermi function instead of a step function.
Let us now consider small perturbations to this distribution such that any excitedquasiparticle or quasihole stays near the Fermi surface. We can then Taylor expand
δnp,σ = np,σ − n0p,σWe the expect the change of free energy to be given by
X
F − F0 ≈(p − µ)δnp,σ + . . .
p,σSince we are near the Fermi surface we havep − µ = vF∗ (|p| − pF )where we have linearized around the Fermi surface. Note in particular that we are using
the effective mass herevF∗ = pF /m∗which tells us the energy of a single quasiparticle above (or below) the Fermi surface. Here
our intuition is from Hartree-Fock, where we learned that the effective mass of a particle
can be renormalized by interactions.
A key realization by Landau is that the energy difference F − F0 is second order insmall quantities: Both |p| − pF is small and δn is also small. As a result, to be consistent,
we must consider other terms of roughly the same size — i.e., the next terms in the Taylor
expansion giving
F − F0 =X1(p − µ)δnp,σ +2p,σXp,p0 ,σ,σ 0

fσ,σ0 (p, p0 )δnp,σ δnp0 ,σ0 + . . .

(9.3)

144

CHAPTER 9. LANDAU FERMI LIQUID THEORY

The quantity fσ,σ0 (p, p0 ) is some arbitrary interaction between a quasiparticle with momentum p with spin σ and another quasiparticle with momentum p0 with spin σ 0 . Whatis crucial here is that this allows for a shift in energy, but it does not allow for scattering— i.e, changing momentum! (The reason for this being the idea that the quasiparticle isvery close to being an eigenstate when it is near the Fermi surface).
Aside: We’ve been thinking in terms of spins σ which can be ↑ or ↓. However, moregenerally we would have to consider cases when the spin is polarized in the x-direction. As
such we would write a density matrix δnα,β (p) rather than just a density. So for example,
the spin of the quasiparticle would be−−h→s i = Tr[→σ β,α δnα,β ]and the interaction function would be a more complicated fα,β;γ,δ (p, p0 ). Fortunately, we
will never need to deal with this more complicated situation. end of aside
Differentiating Eq. 9.3 we can obtain the energy of a quasiparticle ˜ when it is in
the presence of other quasiparticlesXδF= ˜q,σ − µ = p − µ +fσ,σ0 (p, p0 )δnp0 ,σ0
δnp
0 0
p ,σ

The term including f here gives the energy shift of the quasiparticle due to the presenceof other quasiparticles!
We now implement a number of simplifications.
1. We assume p and p0 are both near the Fermi surface, so that
fσ,σ0 (p, p0 ) = fσ,σ0 (p̂, p̂0 )with p̂ being the direction of p.
2. Further assuming rotational invariance we havefσ,σ0 (p, p0 ) = fσ,σ0 (θp̂,p̂0 )with θ being the angle between p̂ and p̂0 .
We can then further decomposte f into harmonics
Xfσ,σ0 (θ) =fl,σ,σ0 Pl (cos θ)l>0fl,σ,σ0=2l + 1
2

Z 1
Pl (cos θ)fσ,σ0 (θ) dcos θ−1where Pl is the Legendre polynomial.
In particular, this implies
1f0,σ,σ0 =4πZdΩ̂ fσ,σ0 (θΩ̂ )sphere9.4. LANDAU FREE ENERGY FUNCTIONAL

145

3. We further assume our system is spin rotationally invariant this means thatfl,↑↑ = fl,↓↓fl,↑↓ = fl,↓↑So that it is convenient to define the symmetric and antisymmetric combinations
fls =
fla =

1
(fl,↑↑ + fl,↑↓ )2
1(fl,↑↑ − fl,↑↓ )2

Finally we note that f has dimensions of energy (everything else in the least termof Eq. 9.3 is dimensionless). It is then convenient to define a dimensionless quantity. Let
us define the density of statesVm∗ pFD∗ (F ) = 3 2~ πNote that this quantity has dimensionsl of 1/energy, and represents the density of statesfor quasiparticles (having a factor of m∗ in it rather than m). We can then define thedimensionless quantities
Fls,a = D∗ (F )fls,aknown as Fermi liquid coefficients. These parameters along with m∗ and kF completely
describe our Fermi liquid. Note that not all these parameters are completely independent:
in particular we will see that m∗ is related to F1s .
These Fl parameters describe the energy renormalizations associated with making
a deformation of the Fermi surface in the shape of the lth spherical harmonic The s
superscript means both spin Fermi surfaces are deformed in the same way whereas the asuperscript means the two Fermi surfaces are deformed in opposite ways.
Aside: Although the following may look complicated, I think it clarifies a number
of issues (Ref: See book by Baym and Pethick). If we write the shape of the deformed
surface asδpF (p̂, σ) = pσ Pl (cos θ)with θ measured from some given point on the sphere, we get an energy of deformation
given by




Fls
Fla
2
2
F − F0 ∼ ps 1 ++ pa 1 +2l + 12l + 1
where ps = p↑ + p↓ and pa = p↑ − p↓ . Note the stability condition that Fl > −(2l + 1). If
this condiditon is violated then the Fermi sea deforms spontaneously in the ground state.
end aside

146

CHAPTER 9. LANDAU FERMI LIQUID THEORY

Figure 9.6: The two processes contributing to the interaction energy between p and p0
included at first order in interaction strength.

Example: f at first order
It is useful to try calculating the function f at first order in the interactions. Recall thatf (p, p0 ) is the interaction between quasiparticles at p and p0 but does not allow scattering
to another momentum p00 . (Again this is the well-defined quasiparticle assumption).
There are two processes in this interaction which we can label direct and exchange,
analogous to the interactions we discussed in prior chapters. These two processes areshown in Fig. 9.6. The direct process (as in prior chapters) simply contributes Ṽ (q = 0)(the interaction bewtween two uniform densities). This is independent of the spin stateof the two quasiparticles. The more interesting contribution is the exchange term which
gives −Ṽ (p − p0 ), and this contribution requires both spin states to be aligned. Thus wehavef↑,↓ (p, p0 ) = Ṽ (0)f↑,↑ (p, p0 ) = Ṽ (0) − Ṽ (p − p0 ) = Ṽ (0) − Ṽ (2pF cos θp,p0 )Example: 3 He
Examples of the Fermi liquid parameters for 3 He are given in the following table (BookHelium 3 by Dobbs)
1 atm
33.5 atm

m∗ /m
2.93
5.81

F0s
11.5
87.1

F0a
-0.71
-0.75

F1s
5.8
14.4

F1a
-0.61
-1.01

F2s
-0.18
0.83

F2a
-0.8
0.0

Note that the F0a parameter is close to the stability boundary −1 where it deforms
spontaneously in the ground state (in this case meaning that it becomes ferromagnetic,see section 9.5.2 below).

9.5. RESULTS FROM FERMI LIQUID THEORY

147

Figure 9.7: To find compressibility we imagine an expansion of the Fermi surface by dpF

9.5

Results from Fermi Liquid Theory

We are now in a position to use Landau’s Fermi liquid theory to calculate some physicalresponses to small external perturbations.

9.5.1

Compressibility

Let us try to find9 the compressibility ∂N/∂µ. Changing the number density involves
an expansion or compression of the Fermi surface as shown in Fig. 9.7 Let us imagine
expanding the Fermi surface by dpF . We then want to calculate the new chemical potentialµnew = Energy of qp on the new Fermi surface= ˜pF +dpF
= pF +dpF +X

fσ,σ0 (p, p0 )δnp0 ,σ0p0 ,σHere we havepF +dpF = µold + vF∗ dpF
with vF∗ = pF /m∗ , and in the interaction term we should use

1 pF < |p| < pF + dpF
δnp,σ =
0 otherwise
representing the change in the Fermi surface. We thus haveX
µnew = µold + vF∗ dpF +fσσ0 (p, p0 )
pF <|p0 |<pF +dpF ,σ 0

I.e., the renormalization of the energy at the Fermi surface contains the usual dispersionof energy (vF∗ dpF ) plus the change in the interaction term. We can rewrite this as
#

" Z
X
1
numberofpstatesfσ,σ0 (Ω̂)
dΩ̂µnew = µold + vF∗ dpF +between pF and pF + dpF
4π
0
σ

9

It may not be obvious that this quantity is related to (but not quite exacly the same as) the usual∂V∂P
definition of compressibility κT = −1
. To see this first using n = N/V write κ−1. Then
T = n ∂V
V ∂P N,T
T,V∂µ
use Gibbs Duhiem N dµ = V dP − SdT to derive ∂n= n1 ∂P. Thus deriving κT = n12 ∂n∂n T,V
∂µ
T,V

T,V

.

148

CHAPTER 9. LANDAU FERMI LIQUID THEORY

The quantity in the square brackets we recognize as 2f0s and the quantity in the roundbrackets we recognize asnumber of p states
between pF and pF + dpF

1
1= D(EF )vF dpF = D∗ (EF )vF∗ dpF
2
2

Putting these together and recalling the definition of F0s we obtain∂µ= vF∗ (1 + F0s )∂pF
Noninteracting Case: For noninteracting fermions we have
∂µ
= vF
∂pF
So the compressibility is∂N∂N ∂pF=∂µ∂pF ∂µand here it is easy to show ∂N/∂pF = D(EF )vF , so we have∂N1
= [D(EF )vF ]= D(EF )∂µvF
which is what we usually expect.
Interacting case: Here we use instead∂µ= vF∗ (1 + F0s )∂pF
So that∂N∂µ=∂N ∂pF∂pF ∂µ1= [D(EF )vF ] ∗vF (1 + F0s )But using the fact that
D(EF )vF = D∗ (EF )vF∗
we get the final result∂N
D∗ (EF )=∂µ
1 + F0swhich shows that the compressibility is renormalized both by the expected mass renormalization (resulting in the ∗ on D∗ (EF )) but also by the Fermi liquid coefficient F0s .

9.5. RESULTS FROM FERMI LIQUID THEORY

149

Figure 9.8: To determine spin susceptibility, we increase the size of the Fermi surface forspin up and reduce the size of the Fermi surface for spin down

9.5.2

Spin Susceptibility

Here we want to consider a similar expansion/compression of the Fermi surface. However,in this case we let the up spins expand and the down spins compress, corresponding to a netchange in magentization, as shown in Fig. 9.8. This is an spin-antisymmetric deformationof the Fermi surface, and hence will couple to F0a as compared to the spin-symmetricdeformation we considered in calculating the compressibility. For noninteracting fermions,this gives the usual Pauli spin susceptibility calculation. Yielding the result1
χP auli = ( gµB )2 µ0 D(EF )2with µB being the Bohr magneton and g the g-factor. However, with a very similar calculation as we did for compressibility above, we determine the susceptibility for interacting
fermions is
1
D∗ (EF )χ = ( gµB )2 µ0
2
1 + F0aso that the spin susceptibility is again renormalized both by the mass renormalization andthe Fermi liquid parameter, F0a in this case.
Note that if F0a < −1 we have χ < 0. This means that it is energetically favorable10
for the ground state to deform by making a spin-antisymmetric deformation of the Fermi
surface — i.e., the Fermi surface spontaneously magnetizes and we have a ferromagnet!

9.5.3

Mass Renormalization

Since this is a homework assignment, I’m not going to solve it here, but I’ll give hints.
The renormalization of mass is related to the F1s Fermi liquid parameter via
m∗
Fs
=1+ 1
m
3
10

(9.4)

Recall that for any magnetic system we can write the energy as a function of the magnetization asE(M ) =

M 2 µ0− MB + . . .
2χ

150

CHAPTER 9. LANDAU FERMI LIQUID THEORY

Figure 9.9: Shifting the Fermi surface is an F1 deformation. It is also a Galilean boost.
Why is this related to F1s ? The l = 1 Fourier spherical mode of the Fermi surface, shownin Fig. 9.9 corresponds to a simple displacement of the Fermi surface — which is thesame as a Galilean boost. One could calculate the energy of this Galilean boost either bystarting with the Fermi liquid expression Eq. 9.3, or by realizing that a boost must haveenergy N 21 mv 2 , with m the bare mass of the particles! Equating these two gives Eq. 9.4.

9.5.4

More?

It may seem at this point that we have simply introduced a new fudge factor F for eachexperiment we want to do. However, once we fix these Fermi liquid parameters, it turns
out that many different experiments can be explained. Further almost any experiment we
can think of will probe only the first few of the Fermi liquid coefficients.

9.6

Further extensions of Fermi Liquid Theory

9.6.1

Local Dynamical Properties and Boltzmann Transport

Similar to the idea of going from Landau theory of phase transitions to Landau-Ginzburg
theory of phase transitions, our next bold step will be to promote the occupations fromglobal to local quantities
np,σ → np,σ (r)Admittedly this appears to violate the uncertainty relation, since we can’t specify p andr at the same time. However, we can indulge in some amount of semiclassical thinking.
As long as we don’t try to specify precise values of either, it will still be acceptable. As
such, n(p, σ, r) will become some sort of momentum and position phase space density. We
can then construct a Boltzmann equation or dynamical theory. We will not derive theBoltzmann equation here, but ...
.. it has solutions that are oscillations of the shape of the Fermi surface, without
oscillations in the local density. This phenomenon is known as zero sound. It was predicted
by Landau and then observed in the 1960s in 3 He. Note that as the temperature goes tozero, regular sound vanishes. The reason for this is the lack of quasiparticle scattering
when the quasiparticles are at low energy. Regular sound is derived by assuming local
thermodynamic quantities, such as pressure, which is calculated from a local equation ofstate (pressure as a function of density). However, if the quasiparticles are not scattering,
one never gets thermodynamic equilibrium and resular sound then no longer makes sense.

9.6. FURTHER EXTENSIONS OF FERMI LIQUID THEORY

151

In contrast, zero sound persists down to zero temeprature, since it is derived without
assuming any scattering of quasiparticles (the assumption of Fermi liquid theory!).

9.6.2

Landau-Silin Theory: Long Range Coulomb Interactions

Finally we return to Fermi liquids in metals. These are omnipresent in our world andso are of great interest. The long range Coulomb interaction presents some problems
for Fermi liquid theory. However, Landau and Silin (1956) developed a good method toaddress this, which can be thought of as a combination of Landau theory of Fermi liquids
and the RPA or self-consistent Hartree approximation which we apply in two steps:(1) First, we imagine a Fermi system with short range interaction only. Since theFock interaction is short range (it requires wavefunction overlap), we can treat this part ofthe interaction between particles, but we throw out the long range hartree part. With thismodified short range interaction, we have something we can treat properly with conventional Fermi liquid theory. We then calculate the properties of this fictitious interactiontruncated Fermi system. For example, we might have
χ̃0 =

∂n
D∗ (EF )=∂µ
1 + F0s(2) Once we have calculated all of these responses, we then add back in the longrange Hartree part of the interaction at RPA level. For example,χ=χ̃0
1 − Ṽ χ̃0(9.5)

So this approach is just like RPA, except instead of using the noninteracting response
9.5, we use a response χ̃0 which includes Fermi liquid corrections due to theshort range part of the interaction. So we see that the compressibility is modified by
the mass renormalization and the F0s Fermi liquid correction. For example, that in thisapproach, the Thomas-Fermi wavevector becomes modified to getχ0 in Eq.

kT2 F =

e2 D∗ (EF )e= χ̃0 (q = 0, ω = 0)s
0 1 + F00152

CHAPTER 9. LANDAU FERMI LIQUID THEORYChapter 10

Leading up to BCS Theory:
Electron Binding
As mentioned way back in chapter 2, the era of superconductivity started in 1911. Yet
there was no microscopic understanding of the effect until 1957 when Bardeen1 , Cooper,
and Schrieffer (BCS) produced what still remains the theory of superconductors.
We already have discussed the general idea that a superconductor can roughly bedescribed as a superfluid (roughly a Bose-Einstein condensate) of charge 2e bosons. While
this had been discussed prior to BCS, no one took it seriously for the two reasons mentioned
at the start of chapter 7: electrons repel and even if you can get electrons to attract eachother, in order to get them to bind into a small boson you can treat individually(i.e.,binding radius smaller than interboson distance) you need a binding energy on order of
the Fermi energy, which is entirely crazy. What will turn out to be crucial is that evena very weak attraction2 will be sufficient to bind electrons together, at least in a sense.
Landau’s quip that “you cannot repeal Coulomb’s law” was somewhat incorrect3 . The
screening of the Coulomb interaction on very short scales makes the Coulomb repulsion
much less effective than you might otherwise fear, and even a weak attraction mechanism
can then be important.

1

John Bardeen is one of only four people to win two Nobel prizes. In the late 1940’s Bardeen was oneof the co-inventors of the transistor — the fundamental element of circuitry that powers the informationage. However, his boss at Bell labs, William Shockley (who won the prize with Bardeen), was also apathological personality and made it essentially impossible for Bardeen to continue doing any work ontransistors. Bardeen left Bell labs for Urbana where he began work on superconductivity.
2
Indeed often a weak attraction even in the presence of lots of repuslion!
3
You might not be able to repeal it, but you can screen it!

153

154

10.1

CHAPTER 10. LEADING UP TO BCS THEORY: ELECTRON BINDING

Mechanism of Attraction

First we need to figure out why electrons attract each other. In “conventional” superconductors, the mechanism involves the interaction between the electrons and the vibrationsof the crystal lattice – i.e, phonons. This idea was put forth as early as 1950 by people
such as Feynman and Frölich.
In more exotic superconductors, including the high Tc perovskite superconductors
discovered in the 1980s, the pnictide superconductors discovered in the 2000s’ and 3 He
superfluid (3 He pairs at low temperature like a superconductor) it is believed that themechanism of attraction does not involve phonons. These are much more complicatedsystems and they remain popular topics of research. In fact, there have been (no exaggeration) over 106 publications aimed to understand why the high Tc superconductors
superconduct, and still there is no agreement!
For conventional superconductors (which includes the vast majority of supercondutors known), fortunately, there is quite a bit of good evidence that it is the coupling to thephonons that is crucial. A very strong piece of evidence comes from the so-called “isotope
effect”. One considers different isotopes of the same material — i.e., adding and subtracting a few neutrons from the nucleus of atoms. The neutrons do nothing to the electronic
properties of the material, or the band structure. The only thing they do is to change
the frequency of vibration of the atoms by changing their masses. In these conventional
superconductors it is usually found that the critical temperature for superconductivity
scales as
Tc ∼ M −α(10.1)
where M is the mass of the atomic nucleus, where α ≈ .5. The fact that the nuclear
mass plays any role in the superconductivity is a sure sign that phonons are crucial to themechanism.
Since this point is fairly important we will give two different calculations which
describe how the coupling to phonons generates an attraction between electrons.

10.1.1

Canonical Transformation

Let us write a simple model of electrons and phonons
H = H0 + H1
where H0 is the Hamiltonian for uncoupled phonons and electrons and H1 includes thesmall coupling between the two. We write explicitlyXX
H0 =k c†k ck + ~ωa†q aqkqHere the first term is the electrons (and for simplicity of notation, we suppress the spin
indices) and the second term is Einstein phonons (with polarization indices suppressed).

10.1. MECHANISM OF ATTRACTION

155

Figure 10.1: Electron-Phonon coupling
The use of Einstein phonons is simply the statement that all of our phonons have thesame frequency ω independent of q. We can switch to a real phonon dispersion without
too much added complication, but this simplified model will be enough to get the idea.
The coupling between electrons and phonons is given by
H1 =

i Xh
i
Xh
M c†k+q ck a†q + h.c. =M ρ̂q a†q + h.c.
q,k(10.2)q

where M is an interaction matrix element that usually would be Mq but we will take it
as a constant for simplicity. The physics here is that any charge density that builds up isable to couple to phonons4 . Diagrammatically this coupling is shown in Fig. 10.1
The Hamiltonian H = H0 + H1 is a complicated interacting Hamiltonian, and isgenerally difficult to solve. However, we can take advantage of the fact that the coupling
constant M is small. This enables us to work perturbatively in this parameter. Our idea
is to remove the phonons from consideration and determine what the effective interaction
is between electrons. This is known as “integrating out” the phonons5 . To do this we
make a canonical transformation on our Hamiltoninan
1
H = e−S HeS = H + [H, S] + [[H, S], S] + . . .
2
The idea is to choose S so that the electron part of the Hamiltonian becomes completely
decoupled from the phonons (at the price of introducing an effective interaction between
the electrons). We can make this decoupling order by order in the small parameter M ,where H1 is order M and we assume that S is also order M (since if M is zero, we aredecoupled with S = 0). Let us rewrite the series, order by order in the small parameter
  
 

1
H = H0 + H1 + [H0 , S] + [H1 , S] + [[H0 , S], S] + O(small3 )2where each set of terms in parenthesis () are of the same order. Let us choose S so as toeliminate the first order term
H1 + [H0 , S] = 0(10.3)
4

There are two main mechanisms by which phonons couple to electrons. The first is piezoelectric
coupling whenever a crystal is non-centro-symmetric, where compressing the crystal actually creates anelectric field that the electrons then respond to. The second mechanism is so-called deformation potential
coupling – where compressing the crystal changes the hopping strength between atomic orbitals and changes
the energies of the bands, and hence changes the energies of the electrons.
5
The nominclature is from path integral language where it really is an integral.

156

CHAPTER 10. LEADING UP TO BCS THEORY: ELECTRON BINDING

Plugging this back into the expansion we get
1
H = H0 + [H1 , S] + O(small3 )2
We have thus eliminated the coupling between electrons and phonons at first order, leaving
only the second order term, which we call
1
Hint = [H1 , S]
2
So that
H = H0 + Hint + O(small3 )Let us now back up to Eq. 10.3 and try to figure out what operator S satisfies thisequation. To determine this, we take the matrix element between two kets
hn|H1 |mi = −hn|[H0 , S]|mi= −(En − Em )hn|S|misohn|S|mi =hn|H1 |miEm − En(10.4)

We now want to figure out what Hint does to make an effective interaction between
electrons. Let us consider two kets |ai and |bi having the same number of phonons excited
and we are interested in finding the effective interaction matrix element
ha|Hint |bi
This can be evaluated as
ha|Hint |bi ===1
ha|[H1 , S]|bi2
1X[ha|H1 |cihc|S|bi − ha|S|cihc|H1 |bi]
2 c


1X
1
1
+
ha|H1 |cihc|H1 |bi
2 c
Eb − Ec Ea − Ec

where in going to the last line we used Eq. 10.4.
Now let us take as given that |ai and |bi both have no excited phonons. Then thestate |ci must have one excited phonon, considering the form of the coupling H1 given in
Eq. 10.2. Note that if the intermediate state |ci is of higher energy than either |ai or |bithen the term in the brackets is negative and we have an attractive interaction.
To see this in more detail, consider consider a case where we have for example|bi = c†k c†p |F Si
|ai =
|ci =

c†k−q c†p+q |F Sic†k−q c†p a†q |F Si(10.5)(10.6)(10.7)

10.1. MECHANISM OF ATTRACTION

157

Figure 10.2: Electron-Phonon scattering makes an effective electron-electron interaction.
Thus the process looks like Fig. 10.2 The effective interaction Hamiltonian can then be
rewritten as


1X †
1
1
ha|Hint |bi =cp+q cp c†k+q ck |M |2+2
k − k−q − ~ω p+q − p − ~ωp,k,q

Now if the energy differences between the fermion states are small compared to ~ωthen the interaction is attractive. Further if we can drop the energies of the fermions thissimplifies even further to the very simple attractive form
Hef f = −|M |2 Xρq ρ−q~ω q

Fourier transforming, an effective interaction which is constnat in q, one can see that thiscorresponds to a short ranged attractive interaction between electrons.

10.1.2RPA Screening and Ion Gas Model

Because the above calculation may seem a bit mysterious, we are going to do a second,almost classical, calculation to show where the attraction comes from. This calculation isbased on self-consistent calculations very similar to the RPA that we discussed above insection 8.4.
In this approach we are going to crudely model the background ions as a structureless
fluid of moving positive charge. We throw away the lattice and consider the ions as some
sort of liquid.
To impliment this approach, let us first remind ourselves of high frequency behaviorof charged particles. We derived (in several ways!) in sections 8.3.2 this response given byδρ(q, ω) = χ0 (q, ω)φ(q, ω)
where here δρ is the induced charge density6 andq2 n̄e2χ0 (q, ω) = 2= q 2 0ω m
6

Ω2e,pl!

ω2Note that above we calculated the induced number density, so this differs by a factor of −e.

158

CHAPTER 10. LEADING UP TO BCS THEORY: ELECTRON BINDING

where n̄ is the averege density and Ωe,pl is the plasma frequencysΩe,pl =n̄e2
m0Recall that this is nothing more than the harmonic oscillations of charged particles in aneutralizing background (See section 8.4.2). For most metals this plasma frequency is very
high — up in the several eV range (> 10,000’s of K)7 .
We now apply the same idea to the background ions which we view as simply a free
fluid of charged particles too — only here the particles have positive charge, and theirmass is much much greater than that of an electron. Here the mass is that of the entire
positive ion (and recall that a single nucleon is some 1800 times heavier than an electron,and further each ion has many nucleons!). Thus, even for a light metal like aluminum,
we have the mass of the nucleon being almost 50,000 times that of the electron. Thus wehave a response for the background ion fluid being given by!
2Ω
ion,plχ0ion (q, ω) = q 2 0
ω2
where
s
Ωion,pl =

n̄e2
M 0

(10.8)

with n̄ the density of ions e their charge, and M their (very large) mass. This descriptionof the motion of ions as a fluid is a very crude approximation of phonons in a solid. We
haveΩion,pl ≈ ωDebye(10.9)
the characteristic scale of phonons, the which is typically on the order of 100 Kelvin.
At these very low frequencies (compared to the electron plasma frequency) the electron response is approximately the same as if ω were zero. (I.e., ion are extremely slow
compared to electrons!). The low frequency response of noninteracting electrons is givenby the compressibility
χ0e (ω ≈ 0) = −e2 D(EF )with D(EF ) being the density of states at the Fermi surface.
Let us now consider an RPA calculation (See section 8.4) of the the response of thecoupled system of electrons and ions together. We write the total electrostatic potential
as the sum of an external potential and an induced potential
φtot (q, ω) = φex (q, ω) + φind (q, ω)
7

To see this, consider a case where the density is 1/a20 with a0 the Bohr radius. In this case it is easy
to see that the plasma frequency is essentially the Rydberg.

10.1. MECHANISM OF ATTRACTION

159

where the induced potential is due to the Coulomb potential from both the electrons andthe ions, which in turn is created by their own response to the total electrostatic potential
φtot (q, ω) = φex (q, ω) + Ṽ (q) δρe (q, ω) + δρion (q, ω)= φex (q, ω) + Ṽ (q) χ0e φtot (q, ω) + χ0ion (q, ω)φtot (q, ω)Solving, we obtainφtot (q, ω) =φext (q, ω)
1 − Ṽ (q) χ0e + χ0ion (q, ω)which gives the general screening of an externally applied potential. Similarly any potential
between any two charged particles will be screened, and we have an effective interaction
V ef f (q, ω) ==Ṽ (q)

1 − Ṽ (q) χ0e + χ0ion (q, ω)Ṽ (q)
k2

1 + qT2F −Ω2ion,plω2where kT F is the Thomas-Fermi screening wavevector of section 8.1.
There are several key points to note about this result.
 Without phonons this just recovers regular Thomas-Fermi screening. (We can Fouriertransform to obtain the same Yukawa interaction as in Eq. 8.5 and 8.30).
 For high frequency ω  Ωion,pl ≈ ωDebye the ion term vanishes and the phonons
don’t matter.
 In the limit of ω → 0 we obtain perfect screening as the ions fluid can always perfectly
screen a fully static potential (the denominator explodes and gives a zero result).
 Crucially, for small ω the ion term in the denominator dominates and we get an
attractive interaction!8

The physical picture we should have here is that when an electron moves throughthe ions, it polarizes the ions, but the ions move slowly. The ions respond and build up
a screening charge that is maintained for a long time after the electron is gone. Anotherelectron can be attracted to this region for some time after the electron is gone. This
constitutes the low frequency attractive interaction between electrons.
8

It turns out that this oversimplified model where we treat the ions as being a fluid and perform RPAlike calculations greatly over-estimates the strength of the resulting attractive interaction. Nonetheless, it
gives an idea of where the attraction comes from.

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CHAPTER 10. LEADING UP TO BCS THEORY: ELECTRON BINDING

Figure 10.3: A square well potential extending to distance R and a wavefunction extended
to a distance L.

10.2

Bound States and the Cooper Problem

10.2.1

Non-Existence of Weakly Bound States in 3D

We have thus developed a way to justify that a (weak) attractive interaction may be
developed between electrons. We can now ask a fundamental question:
Question:
Given a very weak attractive potential in D dimensions, does there exist a bound state
Answers:
In D < 2 the answer is yes.
In D > 2 the answer is no.
D = 2 is complicated, but the answer is usually yes.
Since the result of this statement is important and (except for D = 2) the result iseasy to prove, it is worth actually proving it. (In fact our proof will not be fully rigorous,
but it will give the right ideas).
Note that since we are thinking about superconductivity we are really concernedwith forming a bound state between two particles, but we can go to relative coordinateswith a reduced mass and think about this as forming a bound state of a single particle ina potential.
Let us consider a potential in general dimension D although we will draw the potential in 1d for obvious reasons as shown in Fig. 10.3. The detailed shape of the potential
doesn’t matter much, so we will choose it to be a square well which extends to some radius R. What is crucial is that at infinite distance from the origin, the potential must go
strictly to zero.
If the potential is infinitessimly small (but attractive), we would like to know whether
there a bound state. In other words, is there an eigenstate with energy below zero? If so,then by WKB approximation we can show that the state is bound near position zero andthe wavefunction must decay as e−|x|/L with L very long if the bound state is very close
to zero energy.
If the eigenstate is very weakly bound, we must assume that the binding radius L10.2. BOUND STATES AND THE COOPER PROBLEM

161

is large. However, we must also have normalizationZ
dD r |ψ|2 = 1
so that we must assume the wavefunction magnitude is roughly|ψ| ∼ L−D/2
Now let us try to estimate the potential and kinetic energy of the system. With the boxlike potential in D dimensions we simply want the integral of the squared wavefunction
times the potential inside the box
Z
2
D
Potential Energy = −V0dD
r |ψ| ∼ −V0 (R/L)|r|<R

On the other hand, the kinetic energy should be
Kinetic Energy =

~2
2mL2

Thus the total energy is~2
KE+PE =
− V0
2mL2

 D
R
L

For D < 2 this always becomes negative for large enough L, for arbitrarily small V0 . Thus
there is always a bound state in D < 2. For D > 2 there is generally no bound state forvery small V0 (although a stronger V0 can certainly form a bound state). Since we aremainly concerned with three dimensional materials this seems like a rather discouraging
result — no bound states will form if an attraction is weak9 .
*** STOPPED EDITS HERE ***10.2.2The Cooper Problem: Bound States in The Presence of a Fermi
Sea

The crucial discovery by Leon Cooper (1956) is that the presence of a Fermi sea completelychanges the situation. We will see that in any dimension, in the presence of a Fermi sea
an arbitrarily weak attractive potential will form a bound state! The key here is that thePauli exclusion principle facilitates electron binding. Let us see how this works.
Suppose we start with a Fermi sea and we add two electrons. These two electrons
can only be added above the Fermi wavevector kF .
Let us write an appropriate trial bound state for the two added electronsψ(r1 , σ1 ; r2 , σ2 ) = χ(σ1 , σ2 )φ(r1 , r2 )9(10.10)

Clearly the D = 2 case will be complicated since the kinetic and potential energies scale the same way.

162

CHAPTER 10. LEADING UP TO BCS THEORY: ELECTRON BINDING

where, as usual, we have now separated the spatial and spin part of the wavefunction.
We can choose to have the electrons in either a spin singlet or a spin triplet – either ispossible, and in fact the argument doesn’t change much. Since electron pairing in mostsuperconductors have the spins in singlet pairs, we will choose this case to study. So wewrite10 the singlet form for the spin wavefunction1|χi = √ ( | ↑1 ↓2 i − | ↓1 ↑2 i )2(10.11)Since the spin part of the wavefunction is antisymmetric we must have the spatial part of
the wavefunction be symmetric, so we propose the following:
1 X ik·(r1 −r2 )φ(r1 , r2 ) =e
gk(10.12)
V|k|>kF

where we have added a factor of the volume of the system V for convenient normalization.
In order to have the wavefunction be symmetric between r1 and r2 we have to choose
gk = g−k . In fact for simplicity, if we are considering a rotationally invariant system wecan choose gk to be a function of |k| only so we write g|k| or sometimes gk ..
There are several key things to note about this trial wavefunction.
 Note that in the sum over wavevectors we have only included wavevectors k > kF .
This is because we are assuming there is already a Fermi sea and all states below kF
are already filled.
 The trial wavefunction has no dependence on r1 +r2 . This means that we are putting
the center of mass of the pair into a K = 0 momentum state. This is equivalent tosaying that we always occupy k with −k at the same time in a pair.
 It is perhaps simpler to write the wavefunction 10.10-10.12 in second quantized form
where we have
1 X
|Ψi =
gk c†k↑ c†−k,↓ |F Si
V|k|>kF

where |F Si is the filled Fermi sea. The anti-commutations of the creation operator
ensure we obtain a singlet state.
Let us now assume there is a weak attractive interaction U (r1 − r2 ) between the two
added electrons. Let us try to solve the Schroedinger equation for the two electrons to see
if we find a bound state. We have the Schroedinger equaiton (in shorthand form)
ˆ + PEˆ − E |ψi = 0
KE
Letting k be the kinetic energy of an electron with wavevector k. We then haveX
eik·(r1 −r2 ) gk [2k + U (r1 − r2 ) − E] = 0|k|>kF
10√
In another language the singlet is χ(↑, ↓) = −χ(↓, ↑) = 1/ 2 and χ(↑, ↑) = χ(↓, ↓) = 0.

10.2. BOUND STATES AND THE COOPER PROBLEM

163

To solve this, we Fourier transform applyingZdD (r1 − r2 ) e−iq·(r1 −r2 )where we define the Fourier tranformed potential via
U (r) =

1 X
Ũq eiq·r
V q

The Schroedinger equation becomes(2q − E) gq +1 X
Ũq−k gk = 0
V(10.13)|k|>kF

This is still a rather difficult equation to solve. The trick here is to choose a particularly
simple interaction U that roughly represents the type of interaction we might actually
have in a real system, but is still simple enough to work with. We recall that when we
derived a phonon-mediated attraction, it was attractive for low frequencies only. We thus
propose the following very simplified interaction
−U0
if k and q are bothwithin ~ωDebye of the Fermi surface
Ũq,k =0otherwise
Note that this is a bit of a strange interaction. We write this as a function of two wavevectors k and q which are the Fourier transforms of r1 and r2 . However, the resulting
interaction is not a function of k − q only, meaning that in real space this is not actuallytranslationally invariant! Nonetheless, this interaction has nice analytic properties, so wewill use it.
Replacing Ũq−k with Ũq,k in Eq. 10.13 we obtain(2q − E) gq −

U0
V

X

gk = 0(10.14)|k|>kFk <EF +~ωDebye

which is now simple enough to solve.
Next we use the fact that we wanted to take gk to be a function of |k| only, andfurther for a rotationally invariant system, we can say that k is a function of the kinetic
energy . (Usually we write it the other way, (k) but here we will write k()). We canthen replace the second term in Eq. 10.14 withZ EF +~ωDebye
X
U0
gk = U0 D(EF )d0 gk(0 ) ≡ C(10.15)VEF|k|>kFk <EF +~ωDebye

164

CHAPTER 10. LEADING UP TO BCS THEORY: ELECTRON BINDING

where D(EF ) is the density of states per unit volume at the Fermi surface (which we willassume is just a constant over the small energy range of interest). We have defined thevalue of this integral to be called C.
We can then invert Eq. 10.14 to givegk() ==

R E +~ωDebye 0
U0 D(EF ) EFFd gk(0 )2 − EC
2 − E

We can then multiply both sides of this equation by U0 D(EF ) and integrate both sides ofthis equation as in Eq. 10.15 in order to obtain C on the left
Z EF +~ωDebye
C = U0 D(EF )
EF

d0

C
20 − E

Cancelling C we then obtain
1=
D(EF )U0

Z EF +~ωDebyeEF
2(EF + ~ωDebye ) − E
1
1
= ln
d 0
2 − E2
2EF − E0which can then be exponentiated to give2~ωDebye2
exp=1+
U0 D(EF )2EF − E

(10.16)

Now for small U0 , the left is very large, which means that the right needs to be large
too, meaning that 2EF − E is very small but positive. This means that there is aneigenstate with energy E < 2EF . This must be a bound state!! (Since the energy of two
noninteracting electrons above the Fermi sea can be no lower than 2EF .)We can simplify a bit more here. For small U0 both left and right of Eq. 10.16 arelarge, which means we can drop the 1 on the right hand side. Thud giving
2~ωDebye2
exp=(10.17)
U0 D(EF )2EF − Eor equivalently the binding energy is given by
Binding Energy = 2EF − E = 2 ~ωDebye exp−2
U0 D(EF )
(10.18)There are two crucial things to point out about this result:First, without the Fermi sea, there is no bound state. To see this, let the densityof states at the Fermi surface D(EF ) go to zero (as it does in 3d as the Fermi sea getssmaller and smaller). Then the binding energy goes to zero.

10.2. BOUND STATES AND THE COOPER PROBLEM

165

Second: Note that the expression for the binding energy is non-analytic in thepotential strength U0 . No order in perturbation theory would ever find this result. To see
this consider expanding a function like e−1/x around x = 0 in a taylor series. All derivatives
at x = 0 vanish — so this function, while smooth, has zero radius of convergence aroundx = 0. So too does the expression for the binding energy if we try to expand it for small
U0 . This nonanalyticity is one reason why discovering the binding of electrons on top ofa Fermi sea was so difficult!

166

CHAPTER 10. LEADING UP TO BCS THEORY: ELECTRON BINDING

Chapter 11

BCS Theory
Given that two electrons just above the Fermi surface form a bound state even for weakly
attractive interactions, we can see that the whole Fermi surface must become unstable:
First two electrons jump above the Fermi sea to form a bound state and lower theirenergies, then two more, and so forth, until all of the electrons near the Fermi surface arepaired up. The question is how do we describe such a state of matter now?
In 1956, many of the top physics minds in the world, both in the West and in theSoviet Union, were working on superconductivity – people like Feynman and Landau1 .
After seeing Cooper’s calculation of the binding of electrons above a Fermi sea, the world
was hot on the chase to figure out the theory of superconductivity. Bardeen had hired
Cooper to come to Urbana and work with him. Unfortunately (or perhaps fortunately)in the fall of 1956, Bardeen won the Nobel prize2 for the invention of the transisitor, andhe had to pause his research to go to Sweden to collect his prize3 . While he was gone,
his graduate student, Robert Schrieffer, was left to puzzle over the problem with Cooper.
The two of them went to some conferences in New York City, and while Schrieffer was on
the subway he cracked the problem. Within a few months they had essentially workedout most of the details of the modern theory of superconductivity4 . The general idea is
to make something that looks a bit like a BEC of pairs of electrons. We will see that thetrial wavefunction has some properties that makes it look like a BEC and some properties
1

This was an era when many top scientists thought there was not much interesting in high energy
physics for a while.
2
Along with William Shockley and Walter Brattain3
Bardeen only brought one of his three children with him to Stockholm for the prize ceremony since theother two were studying at Harvard and he didn’t want to interrupt their studies (Although he was very
happy to win the prize he fundamentally did not think prizes should be that important!). King Gustav
chided him for not bringing all his children, and Bardeen promised that he would bring the whole family
for his next Nobel prize. Sixteen years later he kept that promise.
4
Schrieffer’s PhD thesis is one of the most beautiful pieces of PhD research one could possibly imagine.
It is essentially identical to his later book on superconductivity which is considered a classic of the field.
Like many geniuses Schrieffer struggled with his mental health. Much later in life Schrieffer was sentenced
to two years in prison for vehicular manslaughter having killed two people in a crash while he was drivingwith a suspended license.

167

168

CHAPTER 11. BCS THEORY

that make it distinct from a BEC.

11.1

The BCS wavefunction

11.1.1

Fixed N wavefunction

Recall how we construct a BEC from bosons. We can write a creation operator for bosons
in some orbital χZ†aχ = dr χ(r)ψ̂ † (r)Then a BEC with N particles can be written as
BEC|ψNi = (a†χ )N |0i

Can we do something similar with fermion pairs? Let us define a pair creation operator(assuming the pair is a singlet)
†

ã

Z

Z

dr1 dr2 g(r1 − r2 ) ψ̂↑† (r1 )ψ↓† (r2 )X=
gk c†k,↑ c†−k,↓=kwhich creates a pair of electrons in the orbital1g(r1 , σ1 ; r2 , σ2 ) = √ ( | ↑1 ↓2 i − | ↓1 ↑2 i )g(r1 − r2 )2
In the simple case where the orbital g(r) is radially symmetric, we can take gk to be a realfunction of |k| only which we might write as gk .
It is sometimes useful to define a pair creation operator for creating singlet pairs in
opposite momentum eigenstates.
Pk† = c†k,↑ c†−k,↓(11.1)
such that the creation operator for a pair in orbital g(r1 − r2 ) can be written as
ã† =

X

gk Pk†k

We might wonder if the operator ã† and ã are proper canonical boson creation andannihilation operators. For this we would like to have[ã, ã† ] = 1

11.1. THE BCS WAVEFUNCTION

169

or, indeed, any constant on the right hand side would work since we could rescale theoperators to make it into unity. However, direct calculation instead gives ushiX[ã, ã† ] =gk∗ 0 gk c−k0 ,↓ ck0 ,↑ , c†k,↑ c†−k,↓
k,k0

=

X

|gk |2 (1 − nk,↑ − nk,↓ )kSo we see that these operators are not quite bosons. However, for most k’s not near the
Fermi surface the operators nk on the right hand side are going to be fixed anyway (farbelow the Fermi surface they will be filled, far above they will be empty) so we can thinkof the right hand side as being “mostly” a constant, and see how far we get with these“pseudo”-bosons.
Let us now consider trying to make a BEC-like wavefunction out of these operators.
We try to multiply occupy this orbital by writingBCS|ψN
i = (ã† )N |0i

This wavefunction can be explicitly evaluated to give the following result
BCS
ψN= Pf(g) ≡ A [g(1, 2)g(3, 4)g(5, 6) . . . g(2N − 1, 2N )](11.2)

here Pf stands for “Pfaffian”5 which is a mathematical shorthand for what is written on
the far write. We have used the notation g(1, 2) to mean g(r1 , σ1 ; r2 , σ2 ) and A means weshould antisymmetrize over all possible pairings of electrons with each other. For example
A [g(1, 2)g(3, 4)] = g(1, 2)g(3, 4) − g(1, 3)g(2, 4) + g(1, 4)g(2, 3)This expression Eq. 11.2 is the fixed N version of the BCS wavefunction. While it is niceto write it out, and see how this wavefunction generates a sum of all possible pairings ofelectrons with each other, this particular form of the wavefunction is close to useless fordoing any calculations!

11.1.2

Coherent State Wavefunction

Recall we can write a coherent state expression for a BEC†|ψBEC i = eαa |0i
So let us try something similar for our paired fermions†|ψBCS i = eαã |0iP††= eα k gk ck,↑ c−k,↓ |0i"#
Y αg c† c†=
e k k,↑ −k,↓ |0ik
5

Named for Johann Friedrich Pfaff, 1765-1825.

170

CHAPTER 11. BCS THEORY

However here we note that since the pairs are made up of fermions we cannot create twoof them in the same k states, i.e.,c†k,↑ c†−k,↓

2=0

So when we expand the exponential we have††eαgk ck,↑ c−k,↓ = 1 + αgk c†k,↑ c†−k,↓with no quadratic or higher order term. Thus our wavefunction can be written asY|ψBCS i =1 + αgk c†k,↑ c†−k,↓ |0i(11.3)k

As in the case of the coherent state BEC, this wavefuntion has an indefinite number ofparticles. Furthermore, it is not properly normalized, that ishψBCS |ψBCS i =6 1
So our first task will be to calculate the normalization so that we can fix this problem.
YYhψBCS |ψBCS i = h0|
1 + α∗ gk∗ 0 c−k0 ,↓ ck0 ,↑1 + αgk c†k,↑ c†−k,↓ |0ik0k

We can simplify this calculation by considering each k orbital separately where |0k i means
the vacuum for that orbital (i.e., the unfilled orbital for both k, ↑ and −k, ↓). We thus
have



Y
hψBCS |ψBCS i =h0k | 1 + α∗ gk∗ c−k,↓ ck,↑ 1 + αgk c†k,↑ c†−k,↓ |0k ik=Y

1 + |αgk |2

(11.4)kThis non-standard normalization is inconvenient so we make the following definitions
uk =
vk =

1
p
1 + |αgk |2
αgkp
1 + |αgk |2(11.5)(11.6)

such that
αgk = vk /uk(11.7)

and we have the normalization condition|uk| |2 + |vk |2 = 1(11.8)

11.2. BCS HAMILTONIAN, ENERGY MINIMIZATION, AND GAP EQUATION 171
If we now take our BCS wavefunction, Eq. 11.3 and divide through by the squareroot of the normalization constant Eq. 11.4 we obtain the normalized form of the BCS
wavefunction
Y
normalized|ψBCSi=uk + vk c†k,↑ c†−k,↓(11.9)kwhich properly satisfies
normalized normalized
hψBCS
|ψBCS
i=1

This is the form of the BCS wavefunction which is most commonly used. As such we will
drop the superscript normalized henceforth.
At this point we have a form for a trial wavefunction, but we still don’t know what
to use as the parameters uk and vk , or equivalently we don’t know αgk , or equivalently we
don’t know the shape of the orbital g(r) of the bound state of two electrons. To find this
we must choose a Hamiltonian and our plan will be to treat the shape of the orbital g(r)(or equivalently uk and vk ) as a variational parameter, and we will minimize the energyof the Hamiltonian with respect to variations in this quantity.

11.2

BCS Hamiltonian, Energy Minimization, and Gap Equation

The type of Hamiltonian for interacting electrons that we have been working with so far
has been roughly of the formH=Xξk,σ c†k,σ ck,σ +X
k1 ,k2 ,k3 ,k4
σ1 ,σ2 ,σ3 ,σ4k,σvk1 ,k2 ,k3 ,k4 c†k1 ,σ1 c†k2 ,σ2 ck4 ,σ4 ck3 ,σ3
σ1 ,σ2 ,σ3 ,σ4where we have made the conventional substitution
ξk,σ = k,σ − µwhich is just the energy measured from the Fermi surface6 .
While even this is an approximation (our interaction, recall, actually is frequency
dependent strictly speaking!), it is still too complicated to work with.
Instead let us follow the ideas of Cooper and grossly simplify the Hamiltonian, So
let us write the very approximate interaction
HBCS =

Xk,σ

6

ξk,σ c†k,σ ck,σ +X

Vk,k0 c†k,↑ c†−k,↓ c−k0 ,↓ ck0 ,↑(11.10)k,k0Strictly speaking here our Hamiltonian with this shift of µ is actually giving us E − µN . It is quite
crucial to keep track of the chemical potential here since our system has an indefinite particle number and
we must count a free energy µ each time a particle enters the system.

172

CHAPTER 11. BCS THEORY

Figure 11.1: this is the caption
where we assume Vk,k0 = Vk0 ,k so that the Hamiltonian is hermitian. For simplicity wewill usually assume V is real and attractive (although this is not entirely necessary).
Let us examine this Hamiltonian in a bit of depth. First of all, it is conceptually
useful to use the pair creation operator from Eq. 11.1. The BCS Hamiltonian can then be
rewritten more succinctly asXX(11.11)
Vk,k0 Pk† Pk0
HBCS =
ξk,σ c†k,σ ck,σ +
k,σ

k,k0

This simply scatters a pair of electrons (k0 , ↑; −k0 , ↓) to new momenta (k, ↑; −k, ↓) withthe assumption that both k and k0 are sufficiently close to the Fermi surface. This process
is schematically shown in Fig. 11.1.
The interaction term of the BCS Hamiltonian may seem like it is a tremendously
crude simplifiction. However, it turns out to give the right physics of superconductivity
so we will use it.7
So now our plan is to calculate the energy of the BCS Hamiltonian
E = hψBCS |HBCS |ψBCS i(11.12)and then minimize over the variational parameters uk , vk subject to the normalization
constraint Eq. 11.8.
To evaluate the expectation of the energy we will need to evaluate a number ofsimple expectations, likehψBCS |c†k,↑ ck,↑ |ψBCS i = h0k | u∗k + vk∗ c−k,↓ ck,↑; |c†k,↑ ck,↑ | uk + vk c†k,↑ c†−k,↓ |0k i= |vk |2and similarly for the occupation of down spin electrons. We will also want to knowhψBCS |c†k,↑ c†−k,↓ |ψBCS i = h0k | u∗k + vk∗ c−k,↓ ck,↑; |c†k,↑ c†−k,↓ | uk + vk c†k,↑ c†−k,↓ |0k i= vk∗ uk
7

(11.13)Many years after BCS theory was first developed, renormalization group calculations were used toshow that the only “relevant” terms in any interaction close to a Fermi surface are precisely those that are
kept in the BCS Hamiltonian.

11.2. BCS HAMILTONIAN, ENERGY MINIMIZATION, AND GAP EQUATION 173
and similarlyhψBCS |c−k,↓ ck,↑ |ψBCS i = u∗k vk(11.14)Using these relations we evaluate Eq. 11.12 with the Hamiltonian 11.10 to obtainXX
E=
2ξk |vk |2 +Vk,k0 vk∗ uk u∗k0 vk0k,k0k

where the factor of 2 in front of the first term is the sum over two spin states.
Now recalling that we are interested in a pairing wavefunction g(r1 − r2 ) which issymmetric, this means we can take g(r) = g(−r) and hence gk real. This further means
we can take uk and vk both real. With the constraint 11.8 it is then convenient to simply
definevk = cos θk(11.15)uk = sin θk(11.16)

in which case the variational energy is given byX
XE=2ξk cos2 θk +Vk,k0 sin θk cos θk sin θk0 cos θk0k,k0k

We now use some convenient trig identities cos2 θ = (1 + cos 2θ)/2 and sin θ cos θ =(sin 2θ)/2 to rewrite the energy asE=Xξk (1 + cos 2θk ) +k1X
Vk,k0 sin 2θk sin 2θk0
4 0k,k

We minimize this energy with respect to the parameters θq by setting ∂E/∂θq = 0
to obtainX
0 = −2ξq sin 2θq + cos 2θq
Vq,k sin 2θk(11.17)kwhere the second term is twice as big as one might expect because we differentiate bothsign terms and we have used the expected symmetry Vk,k0 = Vk0 ,k .
It is now convenient to define the following object known as the “BCS gap function”(for reasons we will discover soon this will be closely related the superconducting energy
gap for low energy excitations):∆q = −

1X
Vq,k sin 2θk2(11.18)k

I realize we are in the midst of doing a lot of algebra, but this object is of great physicalsignificance. In order to better understand it, let us unpack the definition — recalling that

174

CHAPTER 11. BCS THEORY

sin 2θ = 2 sin θ cos θ we have∆q = −X= −X

Vq,k sin θk cos θkVq,k u∗k vk(11.19)k

Even though we had assumed above that u and v were real, we write them here as generally
complex. The reason for this is that in the more general case where we consider more
complicated pairing with complex u and v, we will still want to use Eq. 11.19 to define∆q . Unpacking this one step further by using Eq. 11.14 we obtain the more physicaldefinition of the gap function in terms of the so-called “anomalous correlator”8X∆q = −
Vq,k hc−k,↓ ck,↑ i(11.20)k

Returning now to Eq. 11.17 and substituting in the gap function we obtain−∆q
1 1X
Vq,k sin 2θk =ξq 2
ξqtan 2θq =(11.21)k

From this expression we hope to determine sin θ and cos θ (and hence u and v and hence
the actual wavefunction!). To do this we need a bit more trigonometry. Using 1 + tan2 θ =sec2 θ gives us
cos 2θq =

−ξq
±1
q
=q
ξq2 + |∆q |2
1 + tan2 2θq(11.22)sin 2θq =± tan 2θq∆qq=q
ξq2 + |∆q |2
1 + tan2 2θq(11.23)The unknown signs are assigned on the far right so that Eq. 11.21 comes out correctly(We could alternately multiply both terms by -1 and everything would still work). Also
note that we have used |∆q | in the denominator here which turns out to be the correct
expression when the quantity is complex.
Since we have now obtained sin 2θ we can plug Eq. 11.23 into Eq. 11.18 to obtain∆q = −

1X∆k
Vq,k q2
ξ 2 + |∆ |2kk(11.24)kThis important equation is known as the “BCS gap equation” and it holds extremely
generally for any singlet superconductor.
8
As in the case of using hψ † i as an order parameter for a Bose condensate, we have chosen an orderparameter which involves having an indefinite number of particles in the system, and is strictly zero
if the number of particles is fixed. As in the Bose case with greater work it is possible to formulate
superconductivity for systems of fixed particle number, although it is substantially more complicated.

11.2. BCS HAMILTONIAN, ENERGY MINIMIZATION, AND GAP EQUATION 175
At this point, although the gap equation is in principle solvable for any Vq,k it is
only analytically solvable for extremely simple forms — we will choose one such form.
Recall that we expect the interaction to be attractive only if the energy of the electron
is within ~ωDebye of the Fermi surface. Thus a very crude representaion of this physics
would be
−V0 if|ξk |, |ξq | < ~ωDebye
Vq,k =(11.25)0otherwisewith V0 positive corresponding to an attractive interaction. We then claim that with thisform of V , the gap equation is solved by the gap function∆ if|ξq | < ~ωDebye∆q =(11.26)0otherwise
Plugging in this this form of the interaction (Eq. 11.25) and gap function (Eq. 11.26) into
the gap equation (Eq. 11.24), cancelling ∆ on both sides, we obtain a solution for thescalar gap function ∆ given by the solution of the equation (with ∆ assumed real here)
1=

V0
2

X
k
ξk <~ωDebye1qξk2 + ∆2

(11.27)

We will see below that ∆ will represent the energy range around the Fermi surface where
there are no excitations — i.e., a gap (hence the name).
This equation can now be simplified a bit further by replacing the sum over k withan integral with a density of statesZZX
1 µ+~ωDebye
1 µ+~ωDebye→d D() ≈ D(EF )d2 µ−~ωDebye
2 µ−~ωDebyekξk <~ωDebye

where D(EF ) is the density of states at the Fermi surface, and the factor of 1/2 is included
because the density of states includes a sum over spins and the sum over k does not. Using
this, the gap equation (Eq. 11.27) is then rewritten as
Z ~ωDebye~ωDebye1
1 = V0 D(EF )= V0 D(EF )arcsinhdξ p∆ξ 2 + ∆2−~ωDebyewhich can then be inverted to give∆=~ωDebye1sinh V0 D(EF)

Assuming the attractive interaction V0 is very weak, this can be approximated as−1∆ = 2~ωDebye exp(11.28)
V0 D(EF )176

CHAPTER 11. BCS THEORY

which is extremely similar to the solution to the Cooper problem Eq. 10.18. In particular
it is also non-perturbative (nonanalytic) in the strength of the interaction V0 .
Exercise: An exercise worth trying is to calculate the condensation energy
associated with superconductivity. This is given by
Econdensation = hψBCS |HBCS |ψBCS i − hF S|HBCS |F Si
with |F Si the usual Fermi sea. If you do this correctly you will get
1
Econdensation /V = − D(EF )∆2
2(11.29)and we recall from Eq. 3.18 that this same quantity should also be given by1
Econdensation /V = µ0 Hc2
2(11.30)

It is worth looking at some of the details of the BCS wavefunction. In particular wewould like to know what the functions uk and vk actually look like. Using Eqs. 11.22 and11.23 we have


ξq
2θk
1
√
(11.31)|uk |2 = sin2 θk = 1−cos
=
1
+
2 +∆2
2
2
ξq
q


ξq
2
1
2
2
|vk | = cos θk = 1 − sin θk = 2 1 − √ 2 2(11.32)ξq +∆q

These functions are sketched in Fig. 11.2. Roughly vk looks like a Fermi function — unity
sufficiently far below the Fermi surface, and zero sufficiently far above. The functions only
differ appreciably from 0 and 1 in a region roughly ∆ around the Fermi surface. Notethat in more detail this function is not like a Fermi function because the Fermi function
converges exponentially to 1 and 0 whereas far from the Fermi surface whereas theseconverge algebraically.
How Big is a Cooper Pair?
Even if the interaction V between the electrons is very short ranged, the size of a cooper
pair can be large. Let us attempt to make an estimate of this size. The important
realization is that uk and vk are constant except for range of energy of roughly ∆ aroundthe Fermi surface. In terms of wavevector this means that these parameters are changing
over a range of wavevectorsδk ∼ ∆/vF. Since the Fourier transform of the pair wavefunction is gk ∼ vk /uk we may conclude
that the pair wavefunction has size of approximately
ξpair =

1
~vF
=
δk
∆

11.2. BCS HAMILTONIAN, ENERGY MINIMIZATION, AND GAP EQUATION 177

Figure 11.2: The u and v functions sketched as a function of energy (ie we plot uk(E) andvk(E)Crucially this distance can be much much larger than the distance between electrons inmost metals. Thus, as expected, we do not have tightly bound pairs that can be thoughtof as individual bosons — as many pairs must spatially overlap with each other.
This parameter ξpair at low temperature is essentially the superconducting coherence length which we discussed in Ginzburg-Landau theory. Let us try to elucidate thisconnection.
To see this connection, let us equate the two expressions for the condensation energy
prt unit volume given in Eq. 11.29 and 11.30
1
1
D(EF )∆2 = µ0 Hc2
2
2(11.33)

Now recall from our discussion of Ginzburg-Landau theory that we can write Hc in termsof the London length λ and the coherence length ξ (See Eq. 6.18)
Hc ∼

1 φ0
ξλ µ0

where we are dropping order one factors like π and 2 for simplicity. The right hand sideof Eq. 11.33 is thus
1
1 1 φ21 (e∗ )2 ns 2
1 ~2 n∗sµ0 Hc2 ∼ 2 2 0 ∼ 2
φ=
02
ξ λ µ0ξm∗
ξ 2 m∗(11.34)

where we have used (from London theory) λ2 = m∗ /((e∗ )2 n∗s µ0 ) with n∗s the superfluid
density of bosons and φ0 = 2π~/e∗ ) where we are still dropping order unity factors. On theother hand, on the left of Eq. 11.33 we have D(EF ) ∼ (n/EF ) and we set ∆ = ~vF /ξpair
thus giving us the condensation energy (dropping order unity factors)~2 v 2 n1
1 ~2 nD(EF )∆2 ∼ 2 F∼ 22
ξpair EF
ξpair m(11.35)

178

CHAPTER 11. BCS THEORY

Figure 11.3: Square Latice Antiferromagnet
where we have used EF = 12 mvF2 . Thus up to factors of order unity (factors of 2 etc),
comparing Eq. 11.34 to 11.35, and setting n∗s = n (or n/2) and m∗ = m (or 2m) we
conclude thatξpair = ξthe size of a Cooper pair is roughly the coherence length. However, we should be a bit
cautious here that the equivalence is true at zero temperature but not near the criticaltemperature. We must keep in mind that in Eq. 11.34 the factor of n∗s is the superfluid
density — which is the total density at zero temperature, but goes to zero at T = Tc
implying that the ratio ξ/ξpair divereges as T approaches Tc .

11.3

More Exotic Pairing

The simple BCS story we have just explained can be extended in many ways. Here wewill consider two interesting such extensions.

11.3.1

Anisotropic Interactions and Anisotropic Gaps: The Example ofHigh Tc

We have assumed that the pairing interaction Eq. 11.25 and the gap function Eq. 11.26
are isotropic – i.e, they depend on the magnitude of k only and not on the direction of k.
This need not be the case.
A very prominent case where interactions are expected to be anisotropic is that of theperovskite high-temperature superconductors. While there is no agreement on the detailed
mechanism for superconductivity in these materials9 there is rough general agreement onsome general things. To begin with, the so-called parent compound for these material
is a two-dimensional square lattice antiferromagnet (which is stabilized by strong on-site
repulsive Hubbard interactions). This means that the spins are arranged in an up-down
checkerboard pattern as shown in Fig. 11.3
When a small density of additional holes are added to the parent, these mobile
holes run around and destroy (or “melt”) the antiferromagnetic order. Nonetheless, there
remains a tendency of sites neighboring each other (in vertical or horizontal direction)
to have spins pointing in opposite directions — whereas spins along the diagonal tend topoint in the same direction. If there is a large on-site repulsive interaction, we cannot have
9

Despite over 105 publications on the topic.

11.3. MORE EXOTIC PAIRING

179

on-site pairing, and we must think about pairing between neigboring sites. If we want tomake singlet pairs of electrons we should then want to make pairs in the horizontal andvertical directions (where spins point opposite), but not along the diagonal (where spinsare aligned), thus implying anisotropic pairing.
In the more conventional superconductor case we discussed above, we choose out
pairing function g(r), and hence gk to be independent of the direction of the argument.
However, it is perfectly consistent to have a pairing function which depends on the directionin space. In particular, for the high Tc compounds a natural choice is to have (withr = (x, y)g(x, y) = (x2 − y 2 )g̃(r)which then results analogously in
gk = (kx2 − ky2 )g̃|k|
We call this type of pairing “d-wave”. This name comes from the name of atomic orbitals.
Recall atomic orbitals are named by their angular momentum s, p, d, f, . . .. This particularshape of orbital has angular momentum L = 2, hence is d − wave. As with an atom, an
orbital which is isotropic in space is called s-wave. Note that we must consider only evenangular momenta by fermionic (Pauli) symmetry — odd angular momenta for the spatial
orbital would give us a minus sign when we exchange the two particles, and hence would
require the spins to be in a triplet state (symmetric) rather than a singlet.
A model interaction which would give a d-wave pairing in 2-dimensional BCS theoryis given by−V0 cos 2φq cos 2φk if|ξk |, |ξq | < ~ωDebye(11.36)
Vq,k =
0otherwise
which can be solved by the gap function∆ cos 2φq
∆q =
0

if
|ξq | < ~ωDebye
otherwise(11.37)

where φq is the polar angle of q in the plane. Note that the factor of 2 in cos 2φ showsthat we have angular momentum 2.

11.3.2

Triplet Superconductors: Helium-3

So far we have only discussed condensing pairs of fermions in a singlet state. However,
one can also pair fermions into a triplet state and condense those. While there are a few
materials which are believed to be triplet superconductors, the most compelling case10 oftriplet pairing is that of the fermion Helium 3, which turns into a superfluid below a few
hundred microkelvin in temperature11 .
10
Of the many thousands of materials that superconduct, perhaps 20 or so are thought to be triplets
superconductors. Needless to say, these materials are pretty interesting to study.
11
A Nobel prize was awarded in 1996 to Osheroff, Richardson, and Lee for their discovery of Helium 3
superfluidity. A Nobel prize was awareded in 2003 to Tony Leggett for theoretical understanding of thisstate of matter.

180

CHAPTER 11. BCS THEORY

A simple example of a triplet superconductor is given by considering supercondcutivity in a completely spin-polarized system12 . In this case we can write a trial wavefunctionas
Y
triplet|ψBCSi=uk + vk c†k,↑ c†−k,↑(11.38)kwhere the pair is created with aligned spins. With fermionic symmetry we havehc†k,↑ c†−k,↑ i = −hc†−k,↑ c†k,↑ iwhich then means that∆q = −

X

Vq,k hc−k,↑ ck,↑ i = −∆−q(11.39)kwhere we have only assumed Vq,k = V−q,−k . This then implies that the spatial pairing
must be antisymmetric to accomodate the symmetric pairing of the spins. An example ofthis is p-wave pairing−V0 cos φq cos φk if|ξk |, |ξq | < ~ωDebye(11.40)Vq,k =
0otherwise
wher e φq is the angle of q on the Fermi surface. This type of pairing results in a gap
function∆q ∼ cos φq
Another type of p-wave pairing is the so-called “chiral” p-wave pairing, where the (complex) gap function instead takes the form∆q ∼ eiφq

11.4

BCS Excitation Spectrum

Having determined the ground state of the BCS superconducting system, we now turnto ask whether we can figure out the excited states as well. Our approach is to use
what is known as BCS mean-field theory, although it is essentially a small extension toHartree-Fock.
Let us recall in Hartree-Fock theory (See Eq. 7.21) following mean field theory, we
have approximated a four-fermion by taking an expectation of two of these Fermi operatorsand leaving the other two as operators:c†α c†β cδ cγ −→hc†α cγ ic†β cδ + hc†β cδ ic†α cγ − hc†α cδ ic†β cγ − hc†β cγ ic†α cδ(11.41)In BCS theory these terms still exist, but they are not very interesting or important.
They will do the same things we discussed in prior chapters: renormalize masses and so
12

A version of this occurs in the ν = 5/2 fractional quantum Hall state11.4. BCS EXCITATION SPECTRUM

181

forth. Here instead we will consider some additional mean field terms that we have notpreviously consideredhc†α c†β icδ cγ + c†α c†β hcδ cγ i(11.42)
We ignored such terms previously when we studied Hartree-Fock, as the correlators mustbe zero for any wavefunction with a fixed number of particles. However, now that weare using the BCS trial wavefunction, where the number of particles is indefinite, these
correlators are nonzero. Thus, throwing out the usual Hartree-fock terms of Eq. 11.41 but
keeping the terms in Eq. 11.42, our BCS Hamiltonian Eq. 11.10 is approximated as themean-field formXX
HM F =
ξk,σ c†k,σ ck,σ +Vk,k0 hc†k,↑ c†−k,↓ ic−k0 ,↓ ck0 ,↑ + c†k,↑ c†−k,↓ hc−k0 ,↓ ck0 ,↑ i
k,k0

k,σ

Using the expression for the gap functionX∆q = −
Vq,k hc−k,↓ ck,↑ ikwe then obtain
HM F =

Xk,σ

ξk,σ c†k,σ ck,σ −X ∆∗k c−k0 ,↓ ck0 ,↑ + ∆k c†k,↑ c†−k,↓(11.43)k,k0

Since this Hamiltonian is quadratic in fermion operators it is completely solvable by Bogoliubov transformation13 — extremely similar to what we did for bosons in section 4.5.1
above. Let us define fermionic Bogoliubon operatorsγk,↑ = u∗k ck,↑ − vk c†−k,↓†= vk∗ ck,↑ + uk c†−k,↓γ−k,↓(11.44)

It is easy to check that this set of operators satisfies fermionic canonical commutations
relations{γk,σ , γk† 0 ,σ0 } = δk,k0 δσ,σ0{γk,σ , γk0 ,σ0 } =†{γk,σ, γk† 0 ,σ0 } = 0(11.45)(11.46)

With some algebra, once can check that the mean-field Hamiltonian then can be rewrittenas
X†
HM F = Ground State Energy +Ek γk,σγk,σ(11.47)k,σwhere the (positive) Bogoliubon energy isq
Ek = + ξk2 + |∆k |2
13(11.48)In the context of superconductivity this transformation was first used independently by Bogoliubov
in the Soviet Union and by Valatin who was in Birmingham UK. It was published in the same journal,
Nuovo Cimento, on the same day in 1958.

182

CHAPTER 11. BCS THEORY
Exercise: The algebra involved in showing that Eq. 11.47 is the same as
Eq. 11.43 is tedious but straightfoward — one simply substitutes the expressions for γ and γ † in terms of c and c† as given in Eq. 11.44 into Eq. 11.47. It
is worth working through this as an exercise. To get you started, we will try
deriving a particular term on the right of Eq. 11.43, the kinetic term withspin ↑. We will make the desired substitution and only keep track of terms
where there is exactly one ck,↑ and one c†k,↑ and throw out all other terms.
††
Ek γk,↑
γk,↑ + γ−k,↓
γ−k,↓


= Ek (uk c†k,↑ − . . .)(u∗k − . . .) + (vk∗ ck,↑ + . . .)(vk c†k,↑ + . . .)
= Ek |uk |2 − |vk |2 c†k,↑ ck,↑ + const
The factor (|uk |2 − |vk |2 ) = sin2 θk − cos2 θk = − cos 2θk from Eq. 11.16.
Then using Eq. 11.22 (and finally Eq. 11.48) our expression becomes= Ek |uk |2 − |vk |2 c†k,↑ ck,↑ + const = Ek cos 2θk c†k,↑ ck,↑


q
ξ
k
 c† ck,↑ = ξk c† ck,↑
ξk2 + |∆k |2  q
=
k,↑
k,↑
2
2
ξk + |∆k |which is exactly the term we expect in Eq. 11.43.

Having derived the mean field Hamitlonian Eq. 11.47 and the spectrum Eq. 11.48
we note that the BCS ground state is the state is simply the Bogoliubov vacuum
γk,σ |ψBCS i = 0†.
and excitations of (always nonnegative) energy Ek are created by the operator γk,σLet us think a bit more carefully about the structure of these quasiparticles . First,
the quasiparticle in Eqs. 11.44 is a superposition of an electron and a hole. As ξk gets
much larger than zero (looking at the structure of u and v as shown in Eq. 11.2) the
quasiparticle operator γ † becomes mostly an electron creation operator (and only a tiny
bit of a hole creation). Whereas when ξ becomes much less than zero, the quasiparticle
operator is mostly a hole creation operator and only a little bit of an electron creation
operator.
Let us consider the simple isotropic (s-wave) case we have considered in Eq. 11.26
where ∆ is just a constant indpendent of direction on the Fermi surface. In this case, theminimum energy excitation (Eq. 11.47) is of energy ∆, hence the name “gap” assigned tothe paramter ∆.
The existence of the energy gap has a number of profound ramifications. For example, due to the lack of low energy excitations, the electronic heat capacity goes tozero exponentially at low temperature roughly as e−∆/T . Similarly electronic thermal

11.4. BCS EXCITATION SPECTRUM

183

transport, and electronic ultrasonic attenuation14 drop exponentially at low temperature.
Another classic experiment is known as Knight shift, which measures the spin susceptibility (how easy it is to flip a spin). For singlet superconductors15 , at low temperature when
there are no quasiparticles present, the Knight shift drops to zero, since to flip a spin one
needs to break a pair.
Finally we note that the absence of low energy excitations assures that a superconductor satisfies the Landau criterion for superfluidity!
Density of States
It is useful to try to figure out the density of states in a superconductor (again assuming
the simple s-wave form). The easiest way to do this is to always go back to the density ofstates in k space which is completely fixed by the volume of the system. As usual, let thenonsuperconducting kinetic energy of an electron be called ξ = ~2 k 2 /(2m) − µ ( we thenconvert k-space integrals to energy integrals as
Z
Z
d3 k
2V
. . . = dξDN (ξ) . . .
(2π)3with DN (ξ) the normal-state density of states. Near the Fermi surface, DN (ξ) can beapproximated as a constant density of states, which we usually call D(EF ).
Similarly, however, we could write this as an integral over the energy of the superconducting quasiparticles
Z=

dEDS (E) . . .

where DS (E) is the density of states of superconducting quasiparticles. Thus we have therelation
DS (E)dE = DN (ξ)dξ ≈ D(EF )dξor equivalently
DS (E) = D(EF )dξdEUsing Eq. 11.48 we haveξ=p
E 2 − ∆2

where we have assumed the isotropic s-wave case where ∆ is just a constant. We thenhave
E
DS (E) = D(EF ) √E>∆
E 2 − ∆2
as sketched in Fig. 11.4. We see that the states that had energy below ∆ in the normal
state are pushed up above ∆ to form a peak in the superconducting density of states. This
14
In ultrasonic attenuation, sound is sent through a metal – the energy of the sound wave can be absorbedby the electrons in the system (if there is no energy gap) thus attenuating the sound.
15
For triplet superconductors one can smoothly rotate the spin of the pair without breaking the pairand the Knight shift does not drop to zero at low temperature.

184

CHAPTER 11. BCS THEORY

Figure 11.4: The density of states of a superconductor.
density of states is easily measured in tunneling experiments where one tunnels unpaired
electrons into the superconductor.

11.5

Finite Temperature Gap Equation

At finite temperature the quasiparticles can be thermally excited. We should expect that†hγk,σγk,σ i = nF (Ek )where nF is the Fermi functionnF (x) =1
eβx + 1

We can then think of the system at finite temperature as having the γ states thermallyoccupied. However, the occupation of these thermal states will then have a feedback effecton the gap function itself.
To see how this happens, we re-do the calculation of the gap function of Eq. 11.20
but now at finite temperature. To do this we need to invert Eq. 11.44 to write c and c† interms of γ as follows†
c−k,↓ = −vk γk↑+ u∗k γ−k↓†c†k,↑ = u∗k γk↑ + vk γ−k↓Then plugging into Eq. 11.20 we get∆q = −X= −X

Vq,k hc−k,↓ ck,↑ i(11.49)††Vq,k u∗k vk h1 − γ−k↑γ−k↑ − γk↓γk↓ i(11.50)kk†where we have dropped terms like γk↑γk↓ since they will give zero in a sector with a fixed11.5. FINITE TEMPERATURE GAP EQUATION

185

Figure 11.5: Gap as a function of T
occupation of quasiparticles. We then have∆q = −

X

Vq,k u∗k vk (1 − 2nF (Ek ))(11.51)∆k(1 − 2nF (Ek ))2Ek(11.52)

k

= −

X

Vq,k

k

where we have used that 2u∗k vk = sin 2θq along with Eq. 11.23.
With the same assumption of isotropic interaction, and isotropic gap function (Eqs. 11.25
and 11.26) we obtain a modified version of the BCS gap equation that includes a thermal
occupation factor (here using that 1 − 2nF (x) = tanh(x/2)) given by

1 =

=

V0
2

X

tanh



1
2β

q

ξk2 + ∆2q
ξk2 + ∆2
 p
Z ωDebye tanh 1 β ξ + ∆2
2
V0 D(EF )pdξ2ξ 2 + ∆2
−ωDebyekξk <~ωDebye(11.53)where we have converted the sum over k to an integral over energy analogous to thediscussion after Eq. 11.24. This equation now implicitly defines the gap ∆ as a functionof temperature (β = 1/kb T ). While the integral is not possible to do analytically, it canbe done numerically to give a dependency roughly as that shown in Fig. 11.5. At sometemperature, which we now define as Tc , the gap ∆ goes to zero and above this temperature
there is no solution. Further, the ratio of kB Tc to the gap at zero temperature is a universal
constant∆(T = 0) = 1.76 kB Tc
a relationship that holds true reasonably well in most conventional superconductors.

186

11.6

CHAPTER 11. BCS THEORY

Inhomogeneous Superconductors

So far we have only studied superconductors in clean systems where k is a good quantum
number. However, this is not necessary. Let us introduce some amount of disorder U (r)and write a single-particle Hamiltonian asĥ =−~2 ∇2+ U (r) − µ2m(11.54)

We can then write a more general effective Schroedinger equation in real-space as





ĥ
∆(r)
u(r)
u(r)
=
E
v(r)
v(r)
∆∗ (r) −ĥ∗
which is known as the Bogoliubov-de Gennes equation. This is supplemented by thereal-space definition of the gap functionX∆(r) = −V0
un (r)vn (r)nIn the translationally invariant case these equations are equivalent to the above discussion:
The eigenstates are plane waves with momentum k, and the u and v are the same uk andvk factors we discussed above. However, the Bogoliubov-de Gennes equations also allow
for disordered eigenfunctions as well. We can similarly define quasiparticle operators whichthen take the real-space form such asZγn↑ = dr ψ̂ † (r)↑ u∗n (r) − vn (r)ψ↓ (r)analogous to the momentum space version in Eq. 11.44.
An important theorem due to Anderson (creatively called “Anderson’s theorem”)
states that weak (nonmagnetic) disorder added in Eq. 11.54 does not reduce the size ofthe gap or reduce the critical temperature. This is the reason that superconductivity cansurvive disorder. The essence of this theorem is that (at least in non-magnetic systems–i.e., systems that obey time reversal invariance) it is always possible to pair an electronwith its time reversed partner. In the absence of disorder we pair k ↑ with −k ↓. When
we break translational invariance we may not have k states any more, but nonetheless wecan pair an eigenstate with its time-reversed partner.

11.6.1Caroli-de Gennes-Matricon States

An interesting application of the Bogoliubov-de Gennes equation is is the calculation of
the quasiparticle spectrum near the core of a vortex in a superconductor. Recall thatthe gap ∆(r) drops to zero at the core of the vortex, but then rises (over a length scaleroughly the coherence length ξ) to its bulk value far away from the vortex core. Thismeans that at very low energy, quasiparticles cannot exist far from the vortex core, but

11.7. GINZBURG-LANDAU, NUMBER-PHASE, AND JOSEPHSON

187

Figure 11.6: The density of states near the vortex core includes discrete low energy eigenstates below the bulk gap.
they can exist close to the vortex core. In this way we can view teh Bogoliubov-de Gennes
equation as providing a “particle in a box” problem for the quasiparticles — with a box
being a region of radius roughly ξ around the vortex core. As a result there are discrete
low energy eigenstates (below the bulk gap) for quasiparticles living in the vortex core
— these are known as Caroli-de Gennes-Matricon states. The density of states near thevortex core thus looks roughly like shown in Fig. 11.6.

11.7

Ginzburg-Landau, Number-Phase, and Josephson

Since the gap parameter vanishes at the critical temperature, it is natural to ask what its
relationship is to the Ginzburg-Landau order parameter introduced in section 6.2. Indeed,
these can be roughly considered to be the same thing (once we include an appropriate
normalization, they become equivalent). We can go a step further and ask about thephase of the order parameter.
The superconductor can be given an arbitrary complex phase in the same way wewould introduce a phase for a neutral superfluid or a coherent state (See for example,
section 4.1.2). For a neutral superfluid, we would write
e|α|e

iϕ a†

|0i

Applying the same philosophy here, we obtain a BCS wavefunction of the form
|ψϕ i =

Y

|uk | + eiϕ |vk |c†k↑ c†−k↓(11.55)k

As with the coherent state in section 4.1.2, since each pair comes with a factor of eiϕ (justlike with a coherent state) the action of −i∂/∂ϕ and the action of the number operator

188

CHAPTER 11. BCS THEORY

N̂ which counts pairs are equivalent. Thus we have16
∂
= N̂
∂ϕ

(11.56)∂=ϕ∂N(11.57)−iand conversely
ifrom which we also have∂e−iϕ = e ∂N = N̂ +(11.58)which is a shift operator which increase the value of N by one17 . For example, consider
the pair creation operator in Eq. 11.13. We see that this has value v ∗ u, which in Eq. 11.55
is indeed proportional to e−iϕ .
We can check that this additional phase included in Eq. 11.55 is reflected in the gap
function by using Eq. 11.19
X
X∆q = −
Vq,k u∗k vk = −Vq,k |uk ||vk |eiϕ
=

k
0 iϕ
∆q e

k

We can check that no energies depend on the overall phase, since this is just a gauge
degree of freedom as expected. However relative phases are physical quantities.

11.7.1

Josephson EffectLet us now consider two superconductors connected by a voltage meter as shown in
Fig. 11.7 The BCS wavefunction of each superconductor has an indefinite number of particles, but has a definite phase. Recalling that the Hamiltonian is actually a free energy(since energy is measured with respect to the chemical potential µ, see footnote in section
11.2). This means that adding a pair into the condensate does not generally increase or
decrease the free energy. However, once we add a voltage between two superconductors,
then transferring a pair from one superconductor to the other does cost energy. Thus we
can write an effective Hamiltonian of the form
We can represent their effective Hamiltonians as
Hef f = −eV (N̂L − N̂R )16

Compare these to the conjugate pair p = −id/dx and x = id/dp.
To see this note that ed/dx is an operator that when applied to a function f (x) returns f (x + 1). Thisis demonstrated by expanding the exponential and recognizing it as a Taylor series


d
1 d2
ed/dx f (x) = 1 +
+
+
.
.
.
f (x) = f (x + 1)
dx
2 dx2
17

11.7. GINZBURG-LANDAU, NUMBER-PHASE, AND JOSEPHSON

189

Figure 11.7: this is the caption
where N̂L,R is the number of pairs on the left, or right respectively (note that transferring
a pair from left to right changes the energy by 2eV as it should). Using the Heisenberg
equations of motion to determine the evolution of the difference in the phases of the two
superconductors18 we havei~∂(ϕL − ϕR )= [(ϕL − ϕR ), H] = [(ϕL − ϕR ), −eV (N̂L − N̂R )]∂t

Now using Eq. 11.56 we obtain∂(ϕL − ϕR )∂∂i~
= −eV (ϕL − ϕR ) , (−i
+i
)
∂t
∂ϕL
∂ϕR
or

∂(ϕL − ϕR )−2eV=
(11.59)∂t~This conversion of voltage into frequency is used metrologically to fix the definition of thevolt.
If we now set the voltage between the two sides to zero, and let current flow between
the two sides, let us ask how the current depends on the phases of the two sides.19 We
can write Heisenberg equations of motioni~(ṄL − ṄR ) = [(NL − NR ), H]

But then since number and phase are conjugate∂NL − NR = −i∂(φL − φR )18

The absolute phase is gauge dependent, but the difference in phases is well defined.
In the lecture I instead use Hamilton’s equations of motion, which is OK for large N where thedynamics is pretty classical. The argument is as follows. Number and phase are conjugate variables
analogous to momentum and position. Let us recall Hamilton’s equations for momentum and position∂H= ∂xand ∂H= − ∂pNow replace the conjugate variables x and p with ϕL − ϕR and NL − NR .
∂p
∂t
∂x
∂t
We then have ∂H/∂(ϕL − ϕR ) = −∂(NL − NR )/∂t and ∂H/∂(NL − NR ) = ∂(ϕL − ϕR )/∂t The latterof these two equations we immediately identify as giving us exactly Eq. 11.59. The former of these two
equations, instead gives us Eq. 11.60.
19

190

CHAPTER 11. BCS THEORY

So we have i~(ṄL − ṄR ) = −i∂∂(φL − φR )







, H = −i∂H∂(φL − φR )
(11.60)where we have used [d/dx, f (x)]g(x) = d(f g)/dx − f dg/dx = (df /dx)g or just [d/dx, f ] =(df /dx). We thus obtain∂hHi
hJi = 2e(11.61)∂(ϕL − ϕR )with J the current between the two subsystems. We thus must determine how the energyof the two superconductors depend on their phases. Since within each subsystem, thephase is a gauge degree of freedom, the overall dependence on the relative phase can only
come from the coupling between the two subsystems which we assume is weak. Let us
focus on this coupling. We write a tunneling Hamiltonain beween the two systems asfollows
Htun = t̂ + t̂†
where t̂ tunnels a single electron from right to left. More explicitly we can writeXt̂ =Tk,q c†Lk,σ cR,q,σk,q,σWe now treat the coupling order by order in perturbation theory. At first order in perturbation theory we have
δE1 = hHtun i = 0
The reason for this is that the ground state has an even number of electrons on both side.
Applying Htun once will put an odd number on both sides thus being orthogonal to theground state.
At second order in perturbation theory, we have the following (written in a bit ofshorthand)
δE2 =X hGS|t̂† |mihm|t̂|GSim+

E0 − EmX hGS|t̂|mihm|t̂|GSim

E0 − Em+X hGS|t̂|mihm|t̂† |GSim+

E0 − EmX hGS|t̂† |mihm|t̂† |GSim

E0 − Em

where |GSi indicates the BCS ground state on both sides, and |mi is some excitation
where a single electron is moved from one side to the other, leaving a broken pair on bothsides. The terms on the first line move a particle from one side to the other and thenmove it back from where it came in order to get back into the ground state — implying
no net motion of electrons from one side to the other. On the other hand, the terms on
the second line transfer a pair of electrons either left to right or right to left. By Eq. 11.58
these terms must be proportional to ei(ϕR −ϕL ) or ei(ϕL −ϕR ) respectively. A more explicit
argument for this is given in the appendix 11.8 below. Thus we haveδE2 = Const + Xei(ϕR −ϕL ) + X ∗ ei(ϕL −ϕR )11.8. APPENDIX: A BETTER ARGUMENT FOR THE PHASE ASSOCIATED WITH THE TUNNELINGwith X some complex constant. We can argue that for a system with no time reversal
breaking, all matrix elements should be real (in the absence of a complex superconducting
phase). This allows us to conclude thta X should be real and we obtain
δE2 = Const + 2X cos(ϕR − ϕL )Hence using Eq. 11.61 we have
hJi = Ic sin(ϕR − ϕL )with Ic = 4eX known as the critical current. This relationship between tunnelling current
and superconducting phase difference is known as the DC-Josephson effect20 . Roughly
one can think of this as the discrete analog of supercurrent being given as the gradient of
the phase.

11.8

Appendix: A better argument for the phase associatedwith the tunneling matrix element

We are interested in computing a matrix element such as
hGS|t̂|mihm|t̂|GSi(11.62)and we would like to know how it depends on the relative phases of the two sides of thesystem where m is an eigenstate that you can arrive at by applying t̂ = c†L cR to |GSi.
Let us first rewrite the c operators in terms of the quasiparticle operators, invertingEq. 11.44 to obtain expressions such as†ck↑ = u∗k γk↑ + vk γk↓†c−k↓ = −vk∗ γk↑ + uk γk↓Within t̂ we have terms like (here we choose σ =↑ there are similar terms with ↓)



†
†
∗
t̂1 = c†↑Lk c↑Rq = uLk γLk↑
+ vLk
γLk↓ u∗Rq γRq↑ + vRq γRq↓
and also terms like



†
†
∗
t̂2 = c†↓L−k c↓R−q = −vLk
γLk↑ + uLk γLk↓
−vRk γRk↑+ u∗Rk γRk↓
Since the ground state is the vacuum of the γ operators we then have∗hGS|t̂2 |mihm|t̂1 |GSi = uLk vLkvRq u∗Rq
20
Predicted by Josephson when he was 22 years old in 1962. He won a Nobel prize eleven years later.
Most of his life after the Nobel prize has been devoted to studying what other scientists term “fringe”
theories — including topics such as meditation, telepathy, higher states of consciousness, parapsychology
and so forth.

192

CHAPTER 11. BCS THEORY

Recalling that u∗ v is the order parameter including phase (See Eq. 11.19 or note in
Eq. 11.55 that a relative phase is included between u and v to represent the phase ofteh order parameter. We thus find the matrix element will be proportional to
hGS|t̂|mihm|t̂|GSi ∼ ei(ϕR −ϕL )as claimed. It does not matter how many different scattering terms are added together
here, the overall phase prefactor of each term will be the same.
Another way to understand this is to just look at a matrix element which transfers
two electrons in a cooper pair from right to left
hGS|(c†L↑ cR↑ )(c†L↓ cR↓ |GSi = hBCSL |c†L↑ c†L↓ |BCSL ihBCSR |c†R↑ c†R↓ |BCSR i= uL vL∗ vR u∗R ∼ ei(ϕR −ϕL )where we have used Eq.11.14.

Chapter 12

An Introduction to Majoranas
The hunt for Majoranas1 has been an obsession of the condensed matter community forover a decade. While several recent experiments, in a number of different experimental
systems, have been extremely suggestive that the elusive Majorana has been found, forthe sceptics definitive proof is still awaited.
This chapter introduces the concept of Majoranas, how they may appear, and why
they are so interesting.
Let us begin by recalling our definition of fermionic operators from Eq.12.2:{ci , cj } = {c†i , c†j } = 0(12.1){ci , c†j } = δij(12.2)

We will now be more specific and call these fermionic operators Dirac fermions. The
Hilbert space associated with a given Fermion orbital j is two dimensional, it can beeither empty, which we notate as |0j i or filled which we notate as |1j i. The filled andempty states can be transformed into each other by applying the Dirac fermion operators.
c†j |0j i = |1j i

cj |1j i = |0j i.

Ettore Majorana2 found a way to to constructed fermionic operators that are theirown Hermitian conjugates. We can construct such operators from the Dirac fermion
operators. For a set of Fermion orbitals j = 1, 2, . . . N , let us define so-called Majorana
1“Majorana” is pronounced roughly my-or-ah-nah. I’ve heard Italians argue over how you should or
should not emphasize the syllables.
2
Majorana was a young genius who dissapeared off the face of the earth in 1938 at the young age of
32. There are many theories as to what happened to him. Some say he committed suicide by jumping off
a ship. Others say he ran afoul of the Mafia and is at the bottom of the ocean with cement shoes. Stillothers think he escaped to South America to start his life over under another name. Due to the extensive
recent interest in Majorana, the Rome attorney’s office released a statement in 2015 saying that the believe
he had moved to Venezuela. They then declared the case closed.

193

194

CHAPTER 12. AN INTRODUCTION TO MAJORANAS

operators
γ1 = (c†1 + c1 )γ2 = i(c†1 − c1 )γ3 = (c†2 + c2 )γ4 = i(c†2 − c2 )..
.

γ2N −1 = (c†N + cN )(12.3)
γ2N = i(c†N − cN )(12.4)

Notice that these Majorana operators are their own Hermitian conjugatesγj† = γjThis essentially means that the Majorana describes a particle that is its own anti-particle.
It is easy to check that the anti-commutation relations of the Majorana operators are{γi , γj } = 2δij
and in particular this means that
γi2 = 1
Note that the γ operators change the particle number in an indefinite way — they area superposition of creating and annihilating a fermion. In both cases, the parity of thenumber of femions is changed. Given our discussion of quasiparticles in superconductors,
this should now not look so unusual.
Note that Majoranas always come in pairs3 . Since the number of fermion orbitals
in a system is an integer, if we convert to Majorans we will always have an even numberof Majoranas. .
The tranformation from Dirac fermions to Majoranas can of course be inverted. A
system of 2N Majoranas can be converted into Dirac fermion via
1c†1 = (γ1 − iγ2 )21c†2 = (γ3 − iγ4 )21c1 = (γ1 + iγ2 )21c2 = (γ3 + iγ4 )
2
..
.

1
c†N = (γ2N −1 − iγ2N )21cN = (γ2N −1 + iγ2N )2(12.5)

In partiuclar note that the number operator for one of these orbitals (say the first orbital)
is given by1n̂1 = c†1 c1 = (1 + iγ1 γ2 )2meaning that the operator iγ1 γ2 has eigenvalues ±1 corresponding to filled or empty
orbitals respectively
3

Like the Sith.

12.1. THE KITAEV CHAIN

195

Figure 12.1: The KitaevWire. Individual sites arranged in a wire on the surface of a
spinless superconductor

It is important to note that we cannot say that a single Majorana is filled or empty
the same way we can say this for a regular Dirac fermion orbital. In order to say somethingis filled or empty we must put two Majorana operators together in pairs in order to make
Dirac fermions, whose orbitals can be filled or empty. (As we will discuss further below, wehave choices in how we choose to pair these Majorana operators to make Dirac fermions.)
Since we can tranform from N Dirac fermions to 2N Majoranas, any Hamiltonianthat we can write in terms of Dirac fermions (such as the interacting fermion Hamiltonians
from chapter 7. See section 7.2.3 for example) can be written equally well in terms ofMajoranas. So far we haven’t done anything except re-express the same Hamiltonian
in a different language. However, soon we will give interesting physical meaning to theindividual Majoranas!

12.1

The Kitaev Chain

The explosion of interest in Majoranas was perhaps spurred by a pivotal paper by Alexei
Kitaev in 20004 . This chapter will summarize much of that work.
Let us consider a bulk (fully gapped) superconductor of spinless fermions (or more
realistically, spin polarized fermions, so we can forget about the spin degree of freedom).
As discussed in section 11.3.2 for spinless electrons, we need to have triplet pairing suchas chiral p-wave (we need chiral so that the superconductor is gapped in all directions on
the Fermi surface). This is not much more complicated than singlet pairing. Now, on
the surface of this superconductor, let us put a number N of sites where these spinlessfermions may sit — and we arrange these sites into a one-dimensional wire as shown in
Fig. 12.1.
4

See arXiv:cond-mat/0010440

196

CHAPTER 12. AN INTRODUCTION TO MAJORANASThe Hamiltonian for this wire is given as

H =

N
−1
X

−t(c†j+1 cj + c†j cj+1 )

j=1

+

N
X
j=1

+

1
−µ(c†j cj − )
2

N
−1
X

∆j c†j c†j+1 + ∆∗j cj cj+1

(12.6)

j=1

The first term is the usual hopping term whereby a ferrmion hops from one site to theneighboring site. The second term is just a chemical potential term — with a shift inenergy so that at µ = 0 is particle-hole symmetric. Finally the last term is the interesting
one —which looks like a pairing term in a Bogoliubov-de Gennes Hamiltonian. In thiscontext, this term creates or annihilates fermions on the chain in pairs. In fact what thisis representing is that fermions can jump on and off of the chain in pairs into the bulk
superconductor. They cannot jump one at a time because that would give an odd numberof fermions in the superconductor which would require breaking a pair and would cost theenergy gap in the superconductor (which we assume is large). Note that this term creates
or annihilates fermions at neighboring sites of the chain: We cannot put two fermions on
the same site of the chain because of the Pauli principle.
For simplicity let us assume that the gap function ∆ is homogenious in space∆j = ∆ = |∆|eiθ
although being independent of position is not required.
Now this Hamiltonian is quadratic in fermion operators, so it can be solved by Bogoliubov transform. (Maybe add an appendix doing this?). However, it is more instructive
to look at a few special cases to understand what can happen. Before doing this, it isuseful to transform this Hamiltonian to its Majorana representation:
To do this we transform
γ2j−1 = e−iθ/2 a†j + eiθ/2 cj
γ2j= ie−iθ/2 a†j − ieiθ/2 cjThis is exaclty the same as the tranform of Eq. 12.4 except that we have absorbed thephase of the gap function θ to simply the result. With this transform we obtain the
Hamiltonian12.1. THE KITAEV CHAIN

197

N

−µ X
iγ2j−1 γ2 j
2

H =

j=1

N −1

(|∆| + t) X
iγ2j γ2j+1
2

+

j=1

N
−1
X

(|∆| − t)2+iγ2j−1 γ2j+2(12.7)

j=1

Again this is quadratic in fermion operators and is therefore solvable analytically. However,instead let us look at some simple cases.
Case 1: t = ∆ = 0
In this case the Hamiltonian is simply
H=

N

N

j=1

j=1

X
−µ X
1
iγ2j−1 γ2 j =−µ(c†j cj − )22

The ground state is simply to fill all the orbitals with a fermion, or leave them all empty,
depending on the sign of µ.
Case 2: µ = 0 and |∆| = t
This is a more interesting case. Here the Hamiltonian is just
H=t

N
−1
X

iγ2j γ2j+1

j=1

To solve this Hamiltonian, we need to assemble the Majorana operators γ into Diracfermions. But these fermions are not the original Dirac fermions cj and c†j but instead
different linear combinations. Let us define1c̃1 = (γ2 + iγ3 )21c̃2 = (γ4 + iγ5 )21c̃†1 = (γ2 − iγ3 )21c̃†2 = (γ4 − iγ5 )2..
.
1c̃†N −1 = (γ2N −2 − iγ2N −1 )21c̃N −1 = (γ2N −2 + iγ2N −1 )2

In terms of these new Dirac fermions we have
H = 2t

N
−1
X
j=1

1
(c̃†j c̃j − )2(12.8)

198

CHAPTER 12. AN INTRODUCTION TO MAJORANAS

Figure 12.2: Here how the Kitaev chain takes a system of dirac fermions (top), splits theminto majoranas (middle) and then regroups them into new dirac fermions (bottom) leaving
one majorana zero mode at each end of the chain.
The ground state of this is to either fill all of the new tilde Dirac fermion orbitals or leavethem all empty depending on the sign of t.
What we have done here is depicted in Fig. 12.2. Originally the Majoranas γ1 andγ2 come from the first Dirac fermion site, but then we re-assemble new Dirac fermions outof γ2 and γ3 .
What is interesting here is that there are two remaining Majoranas that are left overat the end of this transformation and are not reassembled into the tilde Dirac fermions.
These two Majoranas, γ1 and γ2N do not enter into the Hamiltonian. We can assemble
these two remaining Majoranas into a single Dirac fermion via
1f † = (γ1 − iγ2N )21
f = (γ1 + iγ2N )2where, since it does not enter the Hamiltonian, the energy of the system is indepdendentof whether this Dirac fermion is in the occupied or unoccupied state. Since this mode iszero energy, the two Majorana γ1 and γ2N are known as Majorana zero modes.
The two states of the system, corresponding to having the f orbital filled or empty
differ in fermion parity but not in energy. It is interesting, and important, that this twostate system is delocalized between the two ends of the system – which can be arbitrarily
far apart.
The excitement over Majoranas is partially due to the their potential use for quantum computation and quantum information processing — a feature first pointed out by

12.1. THE KITAEV CHAIN

199

Kitaev. The general idea is to use this type of two-state system as a quantum bit (orqubit). We will see that, although this might look like an inconvenient two-state system,
there are significant advantages to storing information this way.
The obvious question to ask is whether we obtained zero modes simply because wefine-tuned our Hamiltonian, setting µ = 0 exactly and t = |∆| exactly (and dropping any
term that is not of the form of .Eq. 12.6 such as four fermi terms). I claim that for a wide
range of parameters (not fine-tuned), we will still have Majorana zero modes, although
the solution of the Hamiltonian is not as simple. We will give two approaches to analyzing
the system

12.1.1

Bogoliubov Approach

As mentioned above, since the Hamiltonian Eq. 12.6 is quadratic in Fermi operators it isalways solvable by Bogoliubov transformation. reults:
Bulk spectrum: For an infinitely long chain we have a spectrum of excitationsqE(q) = ± (2t cos q + µ)2 + 4|∆|2 sin2 q
This calculation is left as a challenge for the reader (or maybe I add an appendix?). This
spectrum has gapless excitations only when ∆ = 0 or |2t| = |µ|, otherwise the spectrum
is completely gapped.
Edge spectrum For chain with two ends, but an infinite distance between the twoends (N arbitarily large), one finds extra Majorana zero modes at the edge only in thecase where |2t| > |µ|. One can define a Majorana zero mode operator on each end5
X
X
γ̃1 =
CjL γjγ̃2N =CjR γjjjwhere the coefficients CjL decay exponentially as j increases (meaning that the mode islocalized near the left of the system, and similarly CjR decays exponentially as j decreases(meaning the mode is localized near the right of the system). The exponential decay length
depends on the parameters of the Hamiltonian, but diverges near the phase transition
where the system becomes gapless.
As in the “case 2” discussed above, one can assemble these two Majorana zero modes
into a Dirac fermion
1f † = (γ̃1 − iγ̃2N )21
f = (γ̃1 + iγ2N )2and this Dirac fermion does not enter the Hamiltonian at all — meaning that each eigenstate of the system is 2-fold degenerate corresponding to the two possibly occupancies ofthe f fermion orbital. (This degeneracy occurring exactly in the cases where |2t| > |µ|).
5

Proving this is not trivial, but it is not all that hard either. We are looking for functions Cj such that[γ̃, H] = 0.

200

CHAPTER 12. AN INTRODUCTION TO MAJORANAS

Thus the exisitence of these Majorana zero modes and the resulting 2-fold degeneraties are thus not specific ot the fine-tuned Hamiltonian we studied in “case 2” above(although the Hamiltonian studied in case 2 is special in that the Majorana γ̃1 is just γ1
meaning the zero mode is isolated on a single site).
On a system of finite size (finite N ), the two Majoranas zero modes γ̃1 and γ̃2N
while decaying in the bulk do overlap a very small amount in the middle of the system.
This allows the Hamiltonian to include a term
1
2iJ γ̃1 γ̃2N = J(f † f − )2where the overlap J is exponentially small in the length of the system. Such a term added
to the Hamiltonian splits the putative degeneracy an exponentially small amount.

12.1.2Topological Approach

While we can solve the above discussed Hamiltonian by Bogoliubov transformation wemight worry about whether other terms that might be added to the Hamiltonian (maybe
multi-fermion interaction terms, for example) could ruin our zero modes. Such a more
complicatd Hamiltonian would not be exactly solveable. Nonetheless, I claim that no
sufficiently small local terms added to the Hamiltonian can make the zero modes go away.
To make this argument, let us start with our exactly solvable Hamiltonian H, with
its zero modes which are localized near the ends of the system. For simplicity (although
this is not necessary) let us assume that these modes are localized exactly at the ends ofthe system (as in case 2 above) so we call them γ1 and γ2N .
Now let us add an additional local term δH to the Hamiltonian (like some fermionfermion interaction, for example). Now let us ask whether a small δH can break thedegeneracy of, say, the 2-fold degenerate ground state. Let us treat this term in perturbation theory. What happens to the energies of the two ground states Eα ? (with |αi being
the two different possible ground states). We can write a Brillouin-Wigner perturbation
series for each stateδEα = hα|δH|αi + hα|δH G δH|αi + hα|δH G δH G δH|αi + . . .
where G is a Greens’ function of the form
G=

X |nihn|n

E − Enwhich includes an energy denominator in each term which is at least the size of the gap.
This perturbation series is exaclty the same for the two degenerate ground states up to avery high order in peturbation theory. Why is this? In order for the energies to be different,
one of the operators in the series (such as δHGδH) must somehow make measurement of
which state the system is in, i.e., it must measure 2iγ1 γ2N = (f † f − 1/2). Do to this the12.1. THE KITAEV CHAIN

201

operator must connect γ1 to γ2N . However, these two operators are spatially separated
and δH is local (and G is diagonal so it doesn’t connect any state to any other state). So,
let us say that δH only connects Majorana operators a distance p apart from each other.
If the system is length N , then the two perturbation series given identical results up toorder N/p. As such the splitting of the putatively degenerate ground states can be nobigger than

N/p
|δH|
Splitting ∼Energy Gap
which, for small δH is exponentially small for a large system. This argument is very
general and applies for any local perturbation δH.
Analogously if some noise is in the environment (imagine the noise being included inδH) the qubit will be exponentially well protected from the noise! This is the reason for
the excitement over Majoranas in condensed matter. By splitting a qubit into two pieces,and spatially separating these two pieces, the qubit becomes extremely well protected fromany noise in the environment.

12.1.3

Influence of the Phase: Operations on the Chain

Let us now return to our definition of the Majorana operators
γ2j−1 = e−iθ/2 a†j + eiθ/2 cj
γ2j= ie−iθ/2 a†j − ieiθ/2 cj

and take particular note of the phases we have included. The phase θ is a superconducting
phase and can be manipulated in experiment by using Josephson circuitry for example.
What happens if we allow the phasSe to wrap by 2π. Interestingly this results in each
Majorana picking up a minus sign
Phase wrapping by 2π :γj → −γj

Since the Dirac fermions are linear in the Majorana operators this has the effect thata†j → −a†jand also in the tilde basisã†j → −ã†jand also
f † → −f †Thus when we wrap the phase by 2π a state with an odd number of fermions will pick up
a minus sign whereas a state with an even number of fermions does not pick up a minus
sign (or equivalently picks up an even number of minus signs). Since the two degenerate
ground states have different fermion parity, we can express this operation as|0f i → |0f i|1f i → −|1f i

202

CHAPTER 12. AN INTRODUCTION TO MAJORANAS

Figure 12.3: Dragging a vortex around the wire rotates the superconducting phase by 2π,
and as a result implements a Z gate on the qubit.

Figure 12.4: Dragging a vortex aound one end of the wire the wire rotates the phase by
2π at one end of the wire, but does not rotate the phase at the other end of the wire. This
implements an X gate on the qubit.
or in the usual language of a qubit, this is a Z operation (or σz in the language of spins).
We can implement such a phase rotation in an experimental system by dragging a
superconducting vortex all the way around our wire as shown in Fig. 12.3. Since goingaround a vortex one experiences a phase rotation of 2π, equivalently dragging a vortex
around the wire makes the phase on the wire rotate by 2π.
We can analogously consider the motion shown in Fig. 12.4. Here the vortex passes
through the middle of the wire, thereby encircling the Majorana zero mode on one end,
but not the Majorana zero mode at the other end. Accordingly this flips the sign of the
Majorana at one end of the wire, but not the otherγ1 → γ1
γ2N → γ2N

This then results in1(f † f − ) = 2iγ1 γ2N
2=⇒

1− 2iγ1 γ2N = −(f † f − )2meaning that the fermion parity, and hence the state of the qubit, is flipped.
|0f i → |1f i|1f i → |0f i
which we call an X gate in quantum information langugae (or σx in spin language).

12.2. WHERE ELSE DO MAJORANAS LIVE?

203

Thus we have constructed a (topologically protected) and we can peform operations
on this qubit! If this can be achieved we would be well on our way to building a quantum
computer!

12.2

Where else do Majoranas live?

Majorana zero modes in condensed matter are not unique to the Kitaev chain. It turns
out that there are a number of other contexts where Majorana zero modes appear. These
contexts all seem to be somehow related to triplet superconductors.
Let us examine the paradigmatic case of a spinless two-dimensional chiral p-wave
superconductor (see the discussion in section 11.3.2). A generic Bogoliubov quasiparticle
can be written as
Z


γ=

dr ψ̂(r)u(r) + ψ † (r)v(r)The quasiparticle wil be a Majorana zero mode if and only if γ = γ † , or if
u(r) = v ∗ (r).

(12.9)

Examining the Bogoliubov-de Gennes equations


ˆ
ĥ
∆
∗
ˆ
∆ −ĥ∗



u
v




=E

u
v



it is clear that that the condition 12.9 is satisfied exactly when E = 0. Since in bulk
superconductors the quasiparticle energy is always E > ∆0 with ∆0 the gap, it seems thatnone of our quasiparticles are Majorana zero modes6
However, recall the fact (see section 11.6.1) that there can exist low energy quasiparticle excitations in a vortex core. If it turns out that we happen to have a zero energy
bound state in the vortex core, then we have our sought-after Majorana zero mode. Theinteresting fact is that in chiral p-wave spinless superconductors a zero energy bound state(and hence a Majorana zero mode) always exists in the core of a vortex!7 While the proofof this statement is not too difficult, we will not give it here (See Read and Green 2000
for a fairly easily followed proof.).
Thus if we have a spinless chiral p-wave superconductor we have Majorana zeromodes associated with each vortex. If we have a spinful chiral superconductor, then thesituation is more complicated. We can think about such a superconductor as having an
6

For certain order parameters we have discussed, such as achiral p-wave, ∆(k) = cos(φk ) the orderparameter drops to zero in certain directions on the fermi surface. In this case there can also be zeroenergy states, but these states are uninteresting in the sense that they are effectively plane wave states offermions as if there were no superconducting gap at all!
7
This fact was first realized in the context of 3-dimensional Helium-3 superfluid in 1991 by Kopnin and
Salomaa and then in 2d by Read and Green in 2000 mainly being concerned with the quantum Hall analog.

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Figure 12.5: In a chiral p-wave superconductor, each vortex harbors a single Majorana
zero mode.
order parameter ∆↑ for the up-up pairs, and another order parameter ∆↓ for the downdown pairs. If there is a vortex in either one of these two order parameters (i.e., if theorder parameter twists by 2π around some point), then there is a resulting Majorana zeromode in the vortex core. However, if there is a vortex in both order parameters at thesame point, then one gets two Majoranas at the same point in space (call them γ↑ andγ↓ ) and generically there is nothing to prevent these two Majoranas from talking to eachother to generate a perturbing term in the Hamiltonian δH = iγ↑ γ↓ which then splits theputative zero mode into finite energy states.
Returning to the spinless (or spin-polarized) chiral p-wave superconductor, we canimagine a two dimensional system with many vortices in it, where each vortex core harbors
a single Majorana zero mode as shown in Fig. 12.5. We may conceptually draw a circle
around any two such Majoranas and declare that to be a qubit. In the figure we havedrawn a circle around γ2 and γ3 , and we would define1f † = (γ2 − iγ3 )2The situation is then identical to what we found for the Kitaev wire: If we wrap a vortex(say the vortex associated with γ1 ) fully around our qubit we will implement a minus
sign only if the parity of fermions enclosed is odd — hence this is a Z-gate for our qubit.
Similarly if we wrap our vortex associated with γ1 around γ3 but not around γ2 then we
will flip the parity of the qubit made from γ2 and γ3 , thus implementing an X-gate.
There are a number of experimental systems that might have Majorana zero modes.
All of these systems (including the ν = 5/2 fractional quantum Hall state) are effectively
equivalent to either Kitaev wire and the chiral p-wave superconductor. Tunneling experiments, for various of these expermental systems, have found zero energy peaks which arevery suggestive of Majorana zero modes. Unfortunately, there are multiple other things
that can sometimes give zero energy tunneling peaks in experiment so this signature is
not definitive.
Nonetheless hope in the community springs eternal. Microsoft research, in particular, has invested vast sums of money (over 108 $) into pursuing Majorana zero modes in12.2. WHERE ELSE DO MAJORANAS LIVE?

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condensed matter. While it has been a bit frustrating for the community that Majorana
zero modes have been so elusive, we have also learned an enormous amount about physics
during the pursuit.

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