The Physics of Quantum Mechanics

James Binney
andDavid Skinner

ivThis book is a consequence of the vision and munificence ofWalter of Merton, who in 1264 launched something good

Copyright c 2008–2013 James Binney and David Skinner
Published by Cappella Archive 2008; revised printings 2009, 2010, 2011

Contents
Prefacex

1 Probability and probability amplitudes

1

1.1 The laws of probability• Expectation values 4

1.2 Probability amplitudes• Two-slit interference 6 • Matter waves? 7
1.3 Quantum states• Quantum amplitudes and measurements 7⊲ Complete sets of amplitudes 8 • Dirac notation 9• Vector spaces and their adjoints 9 • The energy representation 12 • Orientation of a spin-half particle 12• Polarisation of photons 14

1.4 Measurement
Problems

3
5
7

15
15

2 Operators, measurement and time evolution

17

2.1 Operators⊲ Functions of operators 20 ⊲ Commutators 20

17

2.2 Evolution in time
• Evolution of expectation values 23

21

2.3 The position representation• Hamiltonian of a particle 26 • Wavefunction for welldefined momentum 27 ⊲ The uncertainty principle 28• Dynamics of a free particle 29 • Back to two-slit interference 31 • Generalisation to three dimensions 31⊲ Probability current 32 ⊲ The virial theorem 33
Problems

24

34

3 Harmonic oscillators and magnetic fields

37

3.1 Stationary states of a harmonic oscillator

37

3.2 Dynamics of oscillators• Anharmonic oscillators 42

41

3.3 Motion in a magnetic field• Gauge transformations 46• Landau Levels 47⊲ Displacement of the gyrocentre 49 • Aharonov-Bohm effect 51
Problems

4 Transformations & Observables
4.1 Transforming kets
• Translating kets 59

45

52
58
58

• Continuous transformations

vi

Contents
and generators 60• The rotation operator 62• Discrete transformations 62 ⊲ (a) The parity operator 62⊲ Mirror operators 63
4.2 Transformations of operators⊲ The parity operator 66 ⊲ Mirror operators 68

64

4.3 Symmetries and conservation laws

68

4.4 The Heisenberg picture70

4.5 What is the essence of quantum mechanics?
Problems

71
73

5 Motion in step potentials

75

5.1 Square potential well
• Limiting cases 78 ⊲ (a) Infinitely deep well 78⊲ (b) Infinitely narrow well 78

75

5.2 A pair of square wells• Ammonia 81 ⊲ The ammonia maser 83

79

5.3 Scattering of free particles⊲ The scattering cross section 86 • Tunnelling through apotential barrier 87 • Scattering by a classically allowed
region 88 • Resonant scattering 89 ⊲ The Breit–Wigner
cross section 92

84

5.4 How applicable are our results?

95

5.5 Summary
Problems

98
99

6 Composite systems

104

6.1 Composite systems• Collapse of the wavefunction 108 • Operators for composite systems 109 • Development of entanglement 110• Einstein–Podolski–Rosen experiment 111⊲ Bell’s inequality 113

105

6.2 Quantum computing

116

6.3 The density operator• Reduced density operators 125
6.4 Thermodynamics

121• Shannon entropy 127

6.5 Measurement
Problems

129
132
135

7 Angular Momentum

139
2

139

7.2 Orbital angular momentum• L as the generator of circular translations 146 • Spectraof L2 and Lz 147 • Orbital angular momentum eigenfunctions 147 • Orbital angular momentum and parity 151• Orbital angular momentum and kinetic energy 151• Legendre polynomials 153

145

7.1 Eigenvalues of Jz and J• Rotation spectra of diatomic molecules 142

7.3 Three-dimensional harmonic oscillator

154

7.4 Spin angular momentum
158• Spin and orientation 159 • Spin-half systems 161 ⊲ The
Stern–Gerlach experiment 161 • Spin-one systems 164• The classical limit 165 • Precession in a magnetic field 168

Contentsvii

7.5 Addition of angular momenta
169• Case of two spin-half systems 173 • Case of spin one andspin half 174 • The classical limit 175
Problems

176

8 Hydrogen

181

8.1 Gross structure of hydrogen• Emission-line spectra 186 • Radial eigenfunctions 186• Shielding 190 • Expectation values for r−k 192

182

8.2 Fine structure and beyond• Spin-orbit coupling 194 • Hyperfine structure 197

193

Problems

9 Perturbation theory

199
203

9.1 Time-independent perturbations• Quadratic Stark effect 205 • Linear Stark effect anddegenerate perturbation theory 206 • Effect of an external magnetic field 208 ⊲ Paschen–Back effect 210⊲ Zeeman effect 210

203

9.2 Variational principle

212

9.3 Time-dependent perturbation theory• Fermi golden rule 214 • Radiative transition rates 215• Selection rules 219

213

Problems

10 Helium and the periodic table

220
226

10.1 Identical particles⊲ Generalisation to the case of N identical particles 227• Pauli exclusion principle 227 • Electron pairs 229

226

10.2 Gross structure of helium• Gross structure from perturbation theory 231• Application of the variational principle to helium 232• Excited states of helium 233• Electronic configurations and spectroscopic terms 236⊲ Spectrum of helium 237

230

10.3 The periodic table• From lithium to argon 237
ods 241

237• The fourth and fifth peri-

Problems

242

11 Adiabatic principle

244

11.1 Derivation of the adiabatic principle

245

11.2 Application to kinetic theory246

11.3 Application to thermodynamics

248

11.4 The compressibility of condensed matter

249

11.5 Covalent bonding• A model of a covalent bond 250• Dissociation of molecules 253

250• Molecular dynamics 252

11.6 The WKBJ approximation

253

Problems

255

viii

Contents

12 Scattering Theory257

12.1 The scattering operator• Perturbative treatment of the scattering operator 259

12.2 The S-matrix• The iǫ prescription 261 • Expanding the S-matrix 263• The scattering amplitude 265

12.3 Cross-sections and scattering experiments• The optical theorem 269

257
261

267

12.4 Scattering electrons off hydrogen271

12.5 Partial wave expansions• Scattering at low energy 276

273

12.6 Resonant scattering• Breit–Wigner resonances 280
Problems

278• Radioactive decay 280

282

AppendicesA

The laws of probability

285

B

Cartesian tensors

286

C

Fourier series and transforms

288

D

Operators in classical statistical mechanics

289

E

Lie groups and Lie algebras

291

FThe hidden symmetry of hydrogen

292

G

Lorentz covariant equations

294

H

Thomas precession

297

I

Matrix elements for a dipole-dipole interaction

299

J

Selection rule for j

300

K

Restrictions on scattering potentials

301

Index

303

Preface
This book is the fruit of for many years teaching the introduction to quantum mechanics to second-year students of physics at Oxford University. We
have tried to convey to students that it is the use of probability amplitudes
rather than probabilities that makes quantum mechanics the extraordinary
thing that it is, and to grasp that the theory’s mathematical structure follows
almost inevitably from the concept of a probability amplitude. We have also
tried to explain how classical mechanics emerges from quantum mechanics.
Classical mechanics is about movement and change, while the strong emphasis on stationary states in traditional quantum courses makes the quantum
world seem static and irreconcilably different from the world of every-day
experience and intuition. By stressing that stationary states are merely thetool we use to solve the time-dependent Schrödinger equation, and presenting
plenty of examples of how interference between stationary states gives rise
to familiar dynamics, we have tried to pull the quantum and classical worlds
into alignment, and to help students to extend their physical intuition into
the quantum domain.
Traditional courses use only the position representation. If you stepback from the position representation, it becomes easier to explain that thefamiliar operators have a dual role: on the one hand they are repositories of
information about the physical characteristics of the associated observable,
and on the other hand they are the generators of the fundamental symmetriesof space and time. These symmetries are crucial for, as we show already in
Chapter 4, they dictate the canonical commutation relations, from whichmuch follows.
Another advantage of down-playing the position representation is that it
becomes more natural to solve eigenvalue problems by operator methods thanby invoking Frobenius’ method for solving differential equations in series. A
careful presentation of Frobenius’ method is both time-consuming and rather
dull. The job is routinely bodged to the extent that it is only demonstrated
that in certain circumstances a series solution can be found, whereas inquantum mechanics we need assurance that all solutions can be found by thismethod, which is a priori implausible. We solve all the eigenvalue problems
we encounter by rigorous operator methods and dispense with solution inseries.
By introducing the angular momentum operators outside the position
representation, we give them an existence independent of the orbital angularmomentum operators, and thus reduce the mystery that often surrounds
spin. We have tried hard to be clear and rigorous in our discussions of theconnection between a body’s spin and its orientation, and the implications ofspin for exchange symmetry. We treat hydrogen in fair detail, helium at thelevel of gross structure only, and restrict our treatment of other atoms to an
explanation of how quantum mechanics explains the main trends of atomic
properties as one proceeds down the periodic table. Many-electron atoms
are extremely complex systems that cannot be treated in a first course witha level of rigour with which we are comfortable.
Scattering theory is of enormous practical importance and raises sometricky conceptual questions. Chapter 5 on motion in one-dimensional step
potentials introduces many of the key concepts, such as the connection between phase shifts and the scattering cross section and how and why inresonant scattering sensitive dependence of phases shifts on energy gives riseto sharp peaks in the scattering cross section. In Chapter 12 we discuss fully
three-dimensional scattering in terms of the S-matrix and partial waves.
In most branches of physics it is impossible in a first course to bringstudents to the frontier of human understanding. We are fortunate in being able to do this already in Chapter 6, which introduces entanglement andPrefacexiquantum computing, and closes with a discussion of the still unresolved problem of measurement. Chapter 6 also demonstrates that thermodynamics isa straightforward consequence of quantum mechanics and that we no longer
need to derive the laws of thermodynamics through the traditional, rather
subtle, arguments about heat engines.
We assume familiarity with complex numbers, including de Moivre’s
theorem, and familiarity with first-order linear ordinary differential equations. We assume basic familiarity with vector calculus and matrix algebra.
We introduce the theory of abstract linear algebra to the level we requirefrom scratch. Appendices contain compact introductions to tensor notation,
Fourier series and transforms, and Lorentz covariance.
Every chapter concludes with an extensive list of problems for which
solutions are available. The solutions to problems marked with an asterisk,
which tend to be the harder problems, are available online1 and solutions toother problems are available to colleagues who are teaching a course from thebook. In nearly every problem a student will either prove a useful result or
deepen his/her understanding of quantum mechanics and what it says about
the material world. Even after successfully solving a problem we suspect
students will find it instructive and thought-provoking to study the solution
posted on the web.
We are grateful to several colleagues for comments on the first two editions, particularly Justin Wark for alerting us to the problem with the singlettriplet splitting. Fabian Essler, Andre Lukas, John March-Russell and Laszlo
Solymar made several constructive suggestions. We thank Artur Ekert for
stimulating discussions of material covered in Chapter 6 and for reading thatchapter in draft form.
June 2012

James Binney
David Skinner

1 http://www-thphys.physics.ox.ac.uk/people/JamesBinney/QBhome.htm

1Probability and probability
amplitudes
The future is always uncertain. Will it rain tomorrow? Will Pretty Lady win
the 4.20 race at Sandown Park on Tuesday? Will the Financial Times All
Shares index rise by more than 50 points in the next two months? Nobody
knows the answers to such questions, but in each case we may have information that makes a positive answer more or less appropriate: if we are in
the Great Australian Desert and it’s winter, it is exceedingly unlikely to rain
tomorrow, but if we are in Delhi in the middle of the monsoon, it will almost
certainly rain. If Pretty Lady is getting on in years and hasn’t won a race yet,she’s unlikely to win on Tuesday either, while if she recently won a couple ofmajor races and she’s looking fit, she may well win at Sandown Park. The
performance of the All Shares index is hard to predict, but factors affecting
company profitability and the direction interest rates will move, will makethe index more or less likely to rise. Probability is a concept which enables
us to quantify and manipulate uncertainties. We assign a probability p = 0
to an event if we think it is simply impossible, and we assign p = 1 if wethink the event is certain to happen. Intermediate values for p imply that
we think an event may happen and may not, the value of p increasing withour confidence that it will happen.
Physics is about predicting the future. Will this ladder slip when I
step on it? How many times will this pendulum swing to and fro in anhour? What temperature will the water in this thermos be at when it hascompletely melted this ice cube? Physics often enables us to answer suchquestions with a satisfying degree of certainty: the ladder will not slip provided it is inclined at less than 23.34◦ to the vertical; the pendulum makes
3602 oscillations per hour; the water will reach 6.43◦ C. But if we are pressed
for sufficient accuracy we must admit to uncertainty and resort to probability
because our predictions depend on the data we have, and these are always
subject to measuring error, and idealisations: the ladder’s critical angle depends on the coefficients of friction at the two ends of the ladder, and thesecannot be precisely given because both the wall and the floor are slightly
irregular surfaces; the period of the pendulum depends slightly on the amplitude of its swing, which will vary with temperature and the humidity of
the air; the final temperature of the water will vary with the amount of heattransferred through the walls of the thermos and the speed of evaporation

2

Chapter 1: Probability and probability amplitudes

from the water’s surface, which depends on draughts in the room as well as
on humidity. If we are asked to make predictions about a ladder that is inclined near its critical angle, or we need to know a quantity like the period of
the pendulum to high accuracy, we cannot make definite statements, we canonly say something like the probability of the ladder slipping is 0.8, or there
is a probability of 0.5 that the period of the pendulum lies between 1.0007 s
and 1.0004 s. We can dispense with probability when slightly vague answers
are permissible, such as that the period is 1.00 s to three significant figures.
The concept of probability enables us to push our science to its limits, andmake the most precise and reliable statements possible.
Probability enters physics in two ways: through uncertain data andthrough the system being subject to random influences. In the first case we
could make a more accurate prediction if a property of the system, such as thelength or temperature of the pendulum, were more precisely characterised.
That is, the value of some number is well defined, it’s just that we don’tknow the value very accurately. The second case is that in which our system
is subject to inherently random influences – for example, to the draughtsthat make us uncertain what will be the final temperature of the water.
To attain greater certainty when the system under study is subject to suchrandom influences, we can either take steps to increase the isolation of oursystem – for example by putting a lid on the thermos – or we can expand thesystem under study so that the formerly random influences become calculable
interactions between one part of the system and another. Such expansionof the system is not a practical proposition in the case of the thermos – theexpanded system would have to encompass the air in the room, and thenwe would worry about fluctuations in the intensity of sunlight through thewindow, draughts under the door and much else. The strategy does workin other cases, however. For example, climate changes over the last ten
million years can be studied as the response of a complex dynamical system– the atmosphere coupled to the oceans – that is subject to random external
stimuli, but a more complete account of climate changes can be made when
the dynamical system is expanded to include the Sun and Moon because
climate is strongly affected by the inclination of the Earth’s spin axis to theplane of the Earth’s orbit and the Sun’s coronal activity.
A low-mass system is less likely to be well isolated from its surroundings
than a massive one. For example, the orbit of the Earth is scarcely affectedby radiation pressure that sunlight exerts on it, while dust grains less than a
few microns in size that are in orbit about the Sun lose angular momentum
through radiation pressure at a rate that causes them to spiral in from near
the Earth to the Sun within a few millennia. Similarly, a rubber duck left
in the bath after the children have got out will stay very still, while tinypollen grains in the water near it execute Brownian motion that carries
them along a jerky path many times their own length each minute. Given
the difficulty of isolating low-mass systems, and the tremendous obstacles
that have to be surmounted if we are to expand the system to the point at
which all influences on the object of interest become causal, it is natural thatthe physics of small systems is invariably probabilistic in nature. Quantum
mechanics describes the dynamics of all systems, great and small. Rather
than making firm predictions, it enables us to calculate probabilities. If thesystem is massive, the probabilities of interest may be so near zero or unitythat we have effective certainty. If the system is small, the probabilistic
aspect of the theory will be more evident.
The scale of atoms is precisely the scale on which the probabilistic aspect
is predominant. Its predominance reflects two facts. First, there is no suchthing as an isolated atom because all atoms are inherently coupled to theelectromagnetic field, and to the fields associated with electrons, neutrinos,
quarks, and various ‘gauge bosons’. Since we have incomplete information
about the states of these fields, we cannot hope to make precise predictions
about the behaviour of an individual atom. Second, we cannot build measuring instruments of arbitrary delicacy. The instruments we use to measure1.1 The laws of probability

3

atoms are usually themselves made of atoms, and employ electrons or photons that carry sufficient energy to change an atom significantly. We rarely
know the exact state that our measuring instrument is in before we bring it
into contact with the system we have measured, so the result of the measurement of the atom would be uncertain even if we knew the precise state thatthe atom was in before we measured it, which of course we do not. Moreover, the act of measurement inevitably disturbs the atom, and leaves it in adifferent state from the one it was in before we made the measurement. On
account of the uncertainty inherent in the measuring process, we cannot be
sure what this final state may be. Quantum mechanics allows us to calculate
probabilities for each possible final state. Perhaps surprisingly, from the theory it emerges that even when we have the most complete information about
the state of a system that is is logically possible to have, the outcomes ofsome measurements remain uncertain. Thus whereas in the classical world
uncertainties can be made as small as we please by sufficiently careful work,in the quantum world uncertainty is woven into the fabric of reality.

1.1 The laws of probability
Events are frequently one-offs: Pretty Lady will run in the 4.20 at Sandown
Park only once this year, and if she enters the race next year, her form andthe field will be different. The probability that we want is for this year’srace. Sometimes events can be repeated, however. For example, there isno obvious difference between one throw of a die and the next throw, so
it makes sense to assume that the probability of throwing a 5 is the sameon each throw. When events can be repeated in this way we seek to assign
probabilities in such a way that when we make a very large number N oftrials, the number nA of trials in which event A occurs (for example 5 comesup) satisfiesnA ≃ pA N.
(1.1)
In any realistic sequence of throws, the ratio nA /N will vary with N , while
the probability pA does not. So the relation (1.1) is rarely an equality. The
idea is that we should choose pA so that nA /N fluctuates in a smaller andsmaller interval around pA as N is increased.
Events can be logically combined to form composite events: if A is theevent that a certain red die falls with 1 up, and B is the event that a white
die falls with 5 up, AB is the event that when both dice are thrown, the reddie shows 1 and the white one shows 5. If the probability of A is pA and theprobability of B is pB , then in a fraction ∼ pA of throws of the two dice thered die will show 1, and in a fraction ∼ pB of these throws, the white die
will have 5 up. Hence the fraction of throws in which the event AB occurs is∼ pA pB so we should take the probability of AB to be pAB = pA pB . In thisexample A and B are independent events because we see no reason why
the number shown by the white die could be influenced by the number that
happens to come up on the red one, and vice versa. The rule for combining
the probabilities of independent events to get the probability of both events
happening, is to multiply them:p(A and B) = p(A)p(B)(independent events).

(1.2)

Since only one number can come up on a die in a given throw, theevent A above excludes the event C that the red die shows 2; A and C are
exclusive events. The probability that either a 1 or a 2 will show is obtainedby adding pA and pC . Thusp(A or C) = p(A) + p(C)(exclusive events).

(1.3)

In the case of reproducible events, this rule is clearly consistent with theprinciple that the fraction of trials in which either A or C occurs should be4

Chapter 1: Probability and probability amplitudes

the sum of the fractions of the trials in which one or the other occurs. If
we throw our die, the number that will come up is certainly one of 1, 2, 3,
4, 5 or 6. So by the rule just given, the sum of the probabilities associatedwith each of these numbers coming up has to be unity. Unless we know thatthe die is loaded, we assume that no number is more likely to come up thananother, so all six probabilities must be equal. Hence, they must all equal
1
6 . Generalising this example we have the rules
With just N mutually exclusive outcomes,NX

pi = 1.

i=1

If all outcomes are equally likely, pi = 1/N.

(1.4)

1.1.1 Expectation values
A random variable x is a quantity that we can measure and the value that
we get is subject to uncertainty. Suppose for simplicity that only discretevalues xi can be measured. In the case of a die, for example, x could be thenumber that comes up, so x has six possible values, x1 = 1 to x6 = 6. If pi
is the probability that we shall measure xi , then the expectation value ofx isXpi xi .
(1.5)
hxi ≡
i

If the event is reproducible, it is easy to show that the average of the values
that we measure on N trials tends to hxi as N becomes very large. Consequently, hxi is often referred to as the average of x.
Suppose we have two random variables, x and y. Let pij be the probability that our measurement returns xi for the value of x and yj for the valueof y. Then the expectation of the sum x + y ishx + yi =Xijpij (xi + yj ) =Xij

pij xi +X

pij yj

(1.6)

ij

P
xi regardless of what weBut
j pij is the probability that we measurePmeasure for y, so it must equal pi . Similarly i pij = pj , the probability ofmeasuring yj irrespective of what we get for x. Inserting these expressions
in to (1.6) we findhx + yi = hxi + hyi .
(1.7)
That is, the expectation value of the sum of two random variables is thesum of the variables’ individual expectation values, regardless of whether
the variables are independent or not.
A useful measure of the amount by which the value of a random variablefluctuates from trial to trial is the variance of x:
E
D
2
(1.8)
(x − hxi)2 = x2 − 2 hx hxii + hxi ,where we have made use of equation (1.7). The expectation hxi is not a2
random
E variable, but has a definite value. Consequently hx hxii = hxi andD
hxi2 = hxi2 , so the variance of x is related to the expectations of x and
x2 by

2

∆2x ≡ (x − hxi)2 = x2 − hxi .

(1.9)

1.2 Probability amplitudes

5

Figure 1.1 The two-slit interference experiment.

1.2 Probability amplitudes
Many branches of the social, physical and medical sciences make extensive
use of probabilities, but quantum mechanics stands alone in the way that it
calculates probabilities, for it always evaluates a probability p as the modsquare of a certain complex number A:
p = |A|2 .

(1.10)

The complex number A is called the probability amplitude for p.
Quantum mechanics is the only branch of knowledge in which probability amplitudes appear, and nobody understands why they arise. They
give rise to phenomena that have no analogues in classical physics throughthe following fundamental principle. Suppose something can happen by two(mutually exclusive) routes, S or T , and let the probability amplitude for it
to happen by route S be A(S) and the probability amplitude for it to happen
by route T be A(T ). Then the probability amplitude for it to happen by one
route or the other isA(S or T ) = A(S) + A(T ).

(1.11)

This rule takes the place of the sum rule for probabilities, equation (1.3).
However, it is incompatible with equation (1.3), because it implies that theprobability that the event happens regardless of route isp(S or T ) = |A(S or T )|2 = |A(S) + A(T )|2= |A(S)|2 + A(S)A∗ (T ) + A∗ (S)A(T ) + |A(T )|2= p(S) + p(T ) + 2ℜe(A(S)A∗ (T )).

(1.12)

That is, the probability that an event will happen is not merely the sumof the probabilities that it will happen by each of the two possible routes:
there is an additional term 2ℜe(A(S)A∗ (T )). This term has no counterpartin standard probability theory, and violates the fundamental rule (1.3) ofprobability theory. It depends on the phases of the probability amplitudes
for the individual routes, which do not contribute to the probabilities p(S) =|A(S)|2 of the routes.
Whenever the probability of an event differs from the sum of the probabilities associated with the various mutually exclusive routes by which it
can happen, we say we have a manifestation of quantum interference.
The term 2ℜe(A(S)A∗ (T )) in equation (1.12) is what generates quantum
interference mathematically. We shall see that in certain circumstances theviolations of equation (1.3) that are caused by quantum interference are notdetectable, so standard probability theory appears to be valid.
How do we know that the principle (1.11), which has these extraordinaryconsequences, is true? The soundest answer is that it is a fundamental
postulate of quantum mechanics, and that every time you look at a digital
watch, or touch a computer keyboard, or listen to a CD player, or interact
with any other electronic device that has been engineered with the helpof quantum mechanics, you are testing and vindicating this theory. Our
civilisation now quite simply depends on the validity of equation (1.11).

6

Chapter 1: Probability and probability amplitudes

Figure 1.2 The probability distributions of passing through each of thetwo closely spaced slits overlap.

1.2.1 Two-slit interference
An imaginary experiment will clarify the physical implications of the principle and suggest how it might be tested experimentally. The apparatus
consists of an electron gun, G, a screen with two narrow slits S1 and S2 ,and a photographic plate P, which darkens when hit by an electron (seeFigure 1.1).
When an electron is emitted by G, it has an amplitude to pass throughslit S1 and then hit the screen at the point x. This amplitude will clearly
depend on the point x, so we label it A1 (x). Similarly, there is an amplitude
A2 (x) that the electron passed through S2 before reaching the screen at x.
Hence the probability that the electron arrives at x is
P (x) = |A1 (x) + A2 (x)|2 = |A1 (x)|2 + |A2 (x)|2 + 2ℜe(A1 (x)A∗2 (x)). (1.13)|A1 (x)|2 is simply the probability that the electron reaches the plate after
passing through S1 . We expect this to be a roughly Gaussian distribution
p1 (x) that is centred on the value x1 of x at which a straight line from G
through the middle of S1 hits the plate. |A2 (x)|2 should similarly be a roughly
Gaussian function p2 (x) centred on the intersection at x2 of the screen andthe straight line from G through the middle of S2 . It is convenient to write√
Ai = |Ai |eiφi = pi eiφi , where φi is the phase of the complex number Ai .
Then equation (1.13) can be writtenp(x) = p1 (x) + p2 (x) + I(x),where the interference term I isp
I(x) = 2 p1 (x)p2 (x) cos(φ1 (x) − φ2 (x)).

(1.14a)(1.14b)Consider the behaviour of I(x) near the point that is equidistant from theslits. Then (see Figure 1.2) p1 ≃ p2 and the interference term is comparable
in magnitude to p1 + p2 , and, by equations (1.14), the probability of anelectron arriving at x will oscillate between ∼ 2p1 and 0 depending on thevalue of the phase difference φ1 (x) − φ2 (x). In §2.3.4 we shall show that thephases φi (x) are approximately linear functions of x, so after many electrons
have been fired from G to P in succession, the blackening of P at x, which
will be roughly proportional to the number of electrons that have arrived atx, will show a sinusoidal pattern.
Let’s replace the electrons by machine-gun bullets. Then everyday experience tells us that classical physics applies, and it predicts that the probability p(x) of a bullet arriving at x is just the sum p1 (x) + p2 (x) of theprobabilities of a bullet coming through S1 or S2 . Hence classical physics
does not predict a sinusoidal pattern in p(x). How do we reconcile the very
different predictions of classical and quantum mechanics? Firearms manufacturers have for centuries used classical mechanics with deadly success, so is1.3 Quantum states7

the resolution that bullets do not obey quantum mechanics? We believe they
do, and the probability distribution for the arrival of bullets should show asinusoidal pattern. However, in §2.3.4 we shall find that quantum mechanics
predicts that the distance ∆ between the peaks and troughs of this pattern
becomes smaller and smaller as we increase the mass of the particles we arefiring through the slits, and by the time the particles are as massive as abullet, ∆ is fantastically small ∼ 10−29 m. Consequently, it is not experimentally feasible to test whether p(x) becomes small at regular intervals.
Any feasible experiment will probe the value of p(x) averaged over manypeaks and troughs of the sinusoidal pattern. This averaged value of p(x)agrees with the probability distribution we derive from classical mechanics
because the average value of I(x) in equation (1.14) vanishes.
1.2.2 Matter waves?
The sinusoidal pattern of blackening on P that quantum mechanics predicts
proves to be identical to the interference pattern that is observed in Young’s
double-slit experiment. This experiment established that light is a wave phenomenon because the wave theory could readily explain the existence of theinterference pattern. It is natural to infer from the existence of the sinusoidal
pattern in the quantum-mechanical case, that particles are manifestations ofwaves in some medium. There is much truth in this inference, and at anadvanced level this idea is embodied in quantum field theory. However, inthe present context of non-relativistic quantum mechanics, the concept ofmatter waves is unhelpful. Particles are particles, not waves, and they passthrough one slit or the other. The sinusoidal pattern arises because probability amplitudes are complex numbers, which add in the same way as wave
amplitudes. Moreover, the energy density (intensity) associated with a wave
is proportional to the mod square of the wave amplitude, just as the probability density of finding a particle is proportional to the mod square of theprobability amplitude. Hence, on a mathematical level, there is a one-to-one
correspondence between what happens when particles are fired towards apair of slits and when light diffracts through similar slits. But we cannot
consistently infer from this correspondence that particles are manifestationsof waves because quantum interference occurs in quantum systems that aremuch more complex than a single particle, and indeed in contexts where
motion through space plays no role. In such contexts we cannot ascribe theinterference phenomenon to interference between real physical waves, so it isinconsistent to take this step in the case of single-particle mechanics.

1.3 Quantum states
1.3.1 Quantum amplitudes and measurements
Physics is about the quantitative description of natural phenomena. A quantitative description of a system inevitably starts by defining ways in which
it can be measured. If the system is a single particle, quantities that we canmeasure are its x, y and z coordinates with respect to some choice of axes,
and the components of its momentum parallel to these axes. We can alsomeasure its energy, and its angular momentum. The more complex a systemis, the more ways there will be in which we can measure it.
Associated with every measurement, there will be a set of possible numerical values for the measurement – the spectrum of the measurement.
For example, the spectrum of the x coordinate of a particle in empty spaceis the interval (−∞, ∞), while the spectrum of its kinetic energy is (0, ∞).
We shall encounter cases in which the spectrum of a measurement consists of discrete values. For example, in Chapter 7 we shall show thatthe angular momentum of a particle parallel to any given axis has spectrum (. . . , (k − 1)h̄, kh̄, (k + 1)h̄, . . .), where h̄ is Planck’s constant h =8Chapter 1: Probability and probability amplitudes

6.63 × 10−34 J s divided by 2π, and k is either 0 or 21 . When the spectrum isa set of discrete numbers, we say that those numbers are the allowed valuesof the measurement.
With every value in the spectrum of a given measurement there will bea quantum amplitude that we will find this value if we make the relevantmeasurement. Quantum mechanics is the science of how to calculate such
amplitudes given the results of a sufficient number of prior measurements.
Imagine that you’re investigating some physical system: some particles
in an ion trap, a drop of liquid helium, the electromagnetic field in a resonantcavity. What do you know about the state of this system? You have two typesof knowledge: (1) a specification of the physical nature of the system (e.g.,
size & shape of the resonant cavity), and (2) information about the current
dynamical state of the system. In quantum mechanics information of type(1) is used to define an object called the Hamiltonian H of the system thatis defined by equation (2.5) below. Information of type (2) is more subtle.
It must consist of predictions for the outcomes of measurements you could
make on the system. Since these outcomes are inherently uncertain, yourinformation must relate to the probabilities of different outcomes, and in thesimplest case consists of values for the relevant probability amplitudes. For
example, your knowledge might consist of amplitudes for the various possible
outcomes of a measurement of energy, or of a measurement of momentum.
In quantum mechanics, then, knowledge about the current dynamical
state of a system is embodied in a set of quantum amplitudes. In classicalphysics, by contrast, we can state with certainty which value we will measure,and we characterise the system’s current dynamical state by simply giving
this value. Such values are often called ‘coordinates’ of the system. Thusin quantum mechanics a whole set of quantum amplitudes replaces a single
number.
Complete sets of amplitudes Given the amplitudes for a certain set of
events, it is often possible to calculate amplitudes for other events. The phenomenon of particle spin provides the neatest illustration of this statement.
Electrons, protons, neutrinos, quarks, and many other elementary particles turn out to be tiny gyroscopes: they spin. The rate at which theyspin and therefore thep
the magnitude of their spin angular momentum neverchanges; it is always 3/4h̄. Particles with this amount of spin are called
spin-half particles for reasons that will emerge shortly. Although the spinof a spin-half particle is fixed in magnitude, its direction can change. Consequently, the value of the spin angular momentum parallel to any given axis
can take different values. In §7.4.2 we shall show that parallel to any given
axis, the spin angular momentum of a spin-half particle can be either ± 12 h̄.
Consequently, the spin parallel to the z axis is denoted sz h̄, where sz = ± 21
is an observable with the spectrum {− 21 , 12 }.
In §7.4.2 we shall show that if we know both the amplitude a+ that sz
will be measured to be + 21 and the amplitude a− that a measurement will
yield sz = − 12 , then we can calculate from these two complex numbers theamplitudes b+ and b− for the two possible outcomes of the measurement of
the spin along any direction. If we know only a+ (or only a− ), then we cancalculate neither b+ nor b− for any other direction.
Generalising from this example, we have the concept of a complete
set of amplitudes: the set contains enough information to enable one
to calculate amplitudes for the outcome of any measurement whatsoever.
Hence, such a set gives a complete specification of the physical state of thesystem. A complete set of amplitudes is generally understood to be a minimal
set in the sense that none of the amplitudes can be calculated from the others.
The set {a− , a+ } constitutes a complete set of amplitudes for the spin of anelectron.

1.3 Quantum states9

1.3.2 Dirac notation
Dirac introduced the symbol |ψi, pronounced ‘ket psi’, to denote a complete
set of amplitudes for the system. If the system consists of a particle1 trappedin a potential well, |ψi could consist of the amplitudes an that the energy
is En , where (E1 , E2 , . . .) is the spectrum of possible energies, or it might
consist of the amplitudes ψ(x) that the particle is found at x, or it might
consist of the amplitudes a(p) that the momentum is measured to be p.
Using the abstract symbol |ψi enables us to think about the system withoutcommitting ourselves to what complete set of amplitudes we are going touse, in the same way that the position vector x enables us to think about
a geometrical point independently of the coordinates (x, y, z), (r, θ, φ) or
whatever by which we locate it. That is, |ψi is a container for a complete setof amplitudes in the same way that a vector x is a container for a complete
set of coordinates.
The ket |ψi encapsulates the crucial concept of a quantum state, whichis independent of the particular set of amplitudes that we choose to quantify
it, and is fundamental to several branches of physics.
We saw in the last section that amplitudes must sometimes be added: if
an outcome can be achieved by two different routes and we do not monitor
the route by which it is achieved, we add the amplitudes associated with eachroute to get the overall amplitude for the outcome. In view of this additivity,
we write|ψ3 i = |ψ1 i + |ψ2 i(1.15)
to mean that every amplitude in the complete set |ψ3 i is the sum of thecorresponding amplitudes in the complete sets |ψ1 i and |ψ2 i. This rule isexactly analogous to the rule for adding vectors because b3 = b1 +b2 implies
that each component of b3 is the sum of the corresponding components ofb1 and b2 .
Since amplitudes are complex numbers, for any complex number α we
can define|ψ ′ i = α|ψi
(1.16)
to mean that every amplitude in the set |ψ ′ i is α times the corresponding
amplitude in |ψi. Again there is an obvious parallel in the case of vectors:
3b is the vector that has x component 3bx , etc.

1.3.3 Vector spaces and their adjoints
The analogy between kets and vectors proves extremely fruitful and is worth
developing. For a mathematician, objects, like kets, that you can add andmultiply by arbitrary complex numbers inhabit a vector space. Since welive in a (three-dimensional) vector space, we have a strong intuitive feel forthe structures that arise in general vector spaces, and this intuition helps
us to understand problems that arise with kets. Unfortunately our everyday experience does not prepare us for an important property of a generalvector space, namely the existence of an associated ‘adjoint’ space, because
the space adjoint to real three-dimensional space is indistinguishable fromreal space. In quantum mechanics and in relativity the two spaces are distinguishable. We now take a moment to develop the mathematical theoryof general vector spaces in the context of kets in order to explain the relationship between a general vector space and its adjoint space. When weare merely using kets as examples of vectors, we shall call them “vectors”.
Appendix G explains how these ideas are relevant to relativity.
1 Most elementary particles have intrinsic angular momentum or ‘spin’ (§7.4). A complete set of amplitudes for a particle such as electron or proton that has spin, includes
information about the orientation of the spin. In the interests of simplicity, in our discussions particles are assumed to have no spin unless the contrary is explicitly stated, eventhough spinless particles are rather rare.

10

Chapter 1: Probability and probability amplitudes

For any vector space V it is natural to choose a set of basis vectors,
that is, a set of vectors |ii that is large enough for it to be possible toexpress any given vector |ψi as a linear combination of the set’s members.
Specifically, for any ket |ψi there are complex numbers ai such that
|ψi =

X
i

ai |ii.

(1.17)The set should be minimal in the sense that none of its members can be
expressed as a linear combination of the remaining ones. In the case of ordinary three-dimensional space, basis vectors are provided by the unit vectors
i, j and k along the three coordinate axes, and any vector b can be expressedas the sum b = a1 i + a2 j + a3 k, which is the analogue of equation (1.17).
In quantum mechanics an important role is played by complex-valued
linear functions on the vector space V because these functions extract theamplitude for something to happen given that the system is in the state |ψi.
Let hf | (pronounced ‘bra f’) be such a function. We denote by hf |ψi theresult of evaluating this function on the ket |ψi. Hence, hf |ψi is a complex
number (a probability amplitude) that in the ordinary notation of functions
would be written f (|ψi). The linearity of the function hf | implies that for
any complex numbers α, β and kets |ψi, |φi, it is true that
hf | α|ψi + β|φi = αhf |ψi + βhf |φi.

(1.18)

Notice that the right side of this equation is a sum of two products of complex
numbers, so it is well defined.
To define a function on V we have only to give a rule that enables us
to evaluate the function on any vector in V . Hence we can define the sum
hh| ≡ hf | + hg| of two bras hf | and hg| by the rulehh|ψi = hf |ψi + hg|ψi(1.19)

Similarly, we define the bra hp| ≡ αhf | to be result of multiplying hf | bysome complex number α through the rulehp|ψi = αhf |ψi.

(1.20)

Since we now know what it means to add these functions and multiply themby complex numbers, they form a vector space V ′ , called the adjoint spaceof V .
The dimension of a vector space is the number of vectors required tomake up a basis for the space. We now show that V and V ′ have the same
dimension. Let2 {|ii} for i = 1, N be a basis for V . Then a linear function
hf | on V is fully defined once we have given the N numbers hf |ii. To seethat this is true, we use (1.17)
P and the linearity of hf | to calculate hf |ψi for
an arbitrary vector |ψi = i ai |ii:
hf |ψi =NX
i=1

ai hf |ii.

(1.21)

This result implies that we can define N functions hj| (j = 1, N ) throughthe equations
hj|ii = δij ,(1.22)
where δij is 1 if i = j and zero otherwise, because these equations specify thevalue that each bra hj| takes on every basis vector |ii and therefore through2 Throughout this book the notation {x } means ‘the set of objects x ’.
ii

1.3 Quantum states

11(1.21) the value that hj| takes on any vector |ψi. Now consider the following
linear combination of these bras:hF | ≡

NX
j=1

hf |jihj|.

(1.23)

It is trivial to check that for any i we have hF |ii = hf |ii, and from this
it follows that hF | = hf | because we have already agreed that a bra is fully
specified by the values it takes on the basis vectors. Since we have now shownthat any bra can be expressed as a linear combination of the N bras specifiedby (1.22), and the latter are manifestly linearly independent, it follows thatthe dimensionality of V ′ is N , the dimensionality of V .
In summary, we have established that every N -dimensional vector space
V comes with an N -dimensional space V ′ of linear functions on V , called the
adjoint space. Moreover, we have shown that once we have chosen a basis{|ii} for V , there is an associated basis {hi|} for V ′ . Equation (1.22) showsthat there is an intimate relation between the ket |ii and the bra hi|: hi|ii = 1
while hj|ii = 0 for j 6= i. We acknowledge this relationship by saying that hi|
is the adjoint of |ii. We extend this definition of an adjoint to an arbitrary
ket |ψi as follows: if
|ψi =

X
i

ai |ii then

hψ| ≡

X
i

a∗i hi|.

(1.24)

With this choice, when we evaluate the function hψ| on the ket |ψi we findhψ|ψi =

Xia∗i hi|X
j X|ai |2 ≥ 0.
aj |ji =(1.25)

i

Thus for any state the number hψ|ψi is real and non-negative, and it canvanish only if |ψi = 0 because every ai vanishes. We call this number thelength of |ψi.
The components of an ordinary three-dimensional vector b = bx i +by j + bz k are real. Consequently, we evaluate the length-square of b as
simply (bx i + by j + bz k) · (bx i + by j + bz k) = b2x + b2y + b2z . The vector on theextreme left of this expression is strictly speaking the adjoint of b but it isindistinguishable from it because we have not modified the components inany way. In the quantum mechanical case eq. 1.25, the components of theadjoint vector are complex conjugates of the components of the vector, so
the difference between a vector and its adjoint is manifest.
PP
If |φi = i bi |ii and |ψi = i ai |ii are any two states, a calculation
analogous to that in equation (1.25) shows thathφ|ψi =
Similarly, we can show that hψ|φi =XPb∗i ai .

(1.26)

i∗
i ai bi , and from this it follows that∗hψ|φi = hφ|ψi .

(1.27)We shall make frequent use of this equation.
Equation (1.26) shows that there is a close connection between extracting the complex number hφ|ψi from hφ| and |ψi and the operation of taking
the dot product between two vectors b and a.

12

Chapter 1: Probability and probability amplitudes

1.3.4 The energy representation
Suppose our system is a particle that is trapped in some potential well. Thenthe spectrum of allowed energies will be a set of discrete numbers E0 , E1 , . . .
and a complete set of amplitudes are the amplitudes ai whose mod squares
give the probabilities pi of measuring the energy to be Ei . Let {|ii} be a setof basis kets for the space V of the system’s quantum states. Then we use
the set of amplitudes ai to associate them with a ket |ψi through
|ψi =

X
i

ai |ii.

(1.28)

This equation relates a complete set of amplitudes {ai } to a certain ket|ψi. We discover the physical meaning of a particular basis ket, say |ki, by
examining the values that the expansion coefficients ai take when we apply
equation (1.28) in the case |ki = |ψi. We clearly then have that ai = 0 for
i 6= k and ak = 1. Consequently, the quantum state |ki is that in which
we are certain to measure the value Ek for the energy. We say that |ki isa state of well defined energy. It will help us remember this important
identification if we relabel the basis kets, writing |Ei i instead of just |ii, sothat (1.28) becomesX
ai |Ei i.
(1.29)|ψi =i

Suppose we multiply this equation through by hEk |. Then by the linearity of this operation and the orthogonality relation (1.22) (which in our
new notation reads hEk |Ei i = δik ) we findak = hEk |ψi.

(1.30)

This is an enormously important result because it tells us how to extract froman arbitrary quantum state |ψi the amplitude for finding that the energy is
Ek .
Equation (1.25) yields
hψ|ψi =

Xi|ai |2 =X

pi = 1,(1.31)iwhere the last equality follows because if we measure the energy, we mustfind some value, so the probabilities pi must sum to unity. Thus kets that
describe real quantum states must have unit length: we call kets with unit
length properly normalised. During calculations we frequently encounterkets that are not properly normalised, and it is important to remember thatthe key rule (1.30) can be used to extract P
predictions only from properly
normalised kets. Fortunately, any ket |φi = i bi |ii is readily normalised: itis straightforward to check that
|ψi ≡

X
i

bi
p
|ii
hφ|φi

(1.32)

is properly normalised regardless of the values of the bi .
1.3.5 Orientation of a spin-half particle
Formulae for the components of the spin angular momentum of a spin-halfparticle that we shall derive in §7.4.2 provide a nice illustration of how theabstract machinery just introduced enables us to predict the results of experiments.
If you measure one component, say sz , of the spin s of an electron, youwill obtain one of two results, either sz = 21 or sz = − 21 . Moreover the state1.3 Quantum states13|+i in which a measurement of sz is certain to yield 21 and the state |−i inwhich the measurement is certain to yield − 21 form a complete set of states
for the electron’s spin. That is, any state of spin can be expressed as a linear
combination of |+i and |−i:|ψi = a− |−i + a+ |+i.

(1.33)

Let n be the unit vector in the direction with polar coordinates (θ, φ).
Then the state |+, ni in which a measurement of the component of s alongn is certain to return 21 turns out to be (Problem 7.6)|+, ni = sin(θ/2) eiφ/2 |−i + cos(θ/2) e−iφ/2 |+i.

(1.34a)Similarly the state |−, ni in which a measurement of the component of salong n is certain to return − 12 is|−, ni = cos(θ/2) eiφ/2 |−i − sin(θ/2) e−iφ/2 |+i.

(1.34b)By equation (1.24) the adjoints of these kets are the brash+, n| = sin(θ/2) e−iφ/2 h−| + cos(θ/2) eiφ/2 h+|h−, n| = cos(θ/2) e−iφ/2 h−| − sin(θ/2) eiφ/2 h+|.

(1.35)

From these expressions it is easy to check that the kets |±, ni are properly
normalised and orthogonal to one another.
Suppose we have just measured sz and found the value to be 12 and wewant the amplitude A− (n) to find − 12 when we measure n · s. Then the stateof the system is |ψi = |+i and the required amplitude isA− (n) = h−, n|ψi = h−, n|+i = − sin(θ/2)eiφ/2 ,(1.36)

so the probability of this outcome isP− (n) = |A− (n)|2 = sin2 (θ/2).

(1.37)

This vanishes when θ = 0 as it should since then n = (0, 0, 1) so n · s = sz ,and we are guaranteed to find sz = 12 rather than − 21 . P− (n) rises to 12 when
θ = π/2 and n lies somewhere in the x, y plane. In particular, if sz = 21 , ameasurement of sx is equally likely to return either of the two possible values± 21 .
Putting θ = π/2, φ = 0 into equations (1.34) we obtain expressions forthe states in which the result of a measurement of sx is certain
1|+, xi = √ (|−i + |+i)2;

1|−, xi = √ (|−i − |+i) .
2(1.38)

Similarly, inserting θ = π/2, φ = π/2 we obtain the states in which the resultof measuring sy is certain
eiπ/4|+, yi = √ (|−i − i|+i)
2

;

eiπ/4
|−, yi = √ (|−i + i|+i) .
2(1.39)

Notice that |+, xi and |+, yi are both states in which the probability ofmeasuring sz to be 12 is 21 . What makes them physically distinct states isthat the ratio of the amplitudes to measure ± 12 for sz is unity in one caseand i in the other.

14

Chapter 1: Probability and probability amplitudes

1.3.6 Polarisation of photons
A discussion of the possible polarisations of a beam of light displays aninteresting connection between quantum amplitudes and classical physics.
At any instant in a polarised beam of light, the electric vector E is in one
particular direction perpendicular to the beam. In a plane-polarised beam,
the direction of E stays the same, while in a circularly polarised beam it
rotates. A sheet of Polaroid transmits the component of E in one direction
and blocks the perpendicular component. Consequently, in the transmitted
beam |E| is smaller than in the incident beam by a factor cos θ, where θ isthe angle between the incident field and the direction in the Polaroid thattransmits the field. Since the beam’s energy flux is proportional to |E|2 , afraction cos2 θ of the beam’s energy is transmitted by the Polaroid.
Individual photons either pass through the Polaroid intact or are absorbed by it depending on which quantum state they are found to be inwhen they are ‘measured’ by the Polaroid. Let |→i be the state in which thephoton will be transmitted and |↑i that in which it will be blocked. Thenthe photons of the incoming plane-polarised beam are in the state|ψi = cos θ|→i + sin θ|↑i,(1.40)

so each photon has an amplitude a→ = cos θ for a measurement by the
Polaroid to find it in the state |→i and be transmitted, and an amplitudea↑ = sin θ to be found to be in the state |↑i and be blocked. The fractionof the beam’s photons that are transmitted is the probability get throughP→ = |a→ |2 = cos2 θ. Consequently a fraction cos2 θ of the incident energy
is transmitted, in agreement with classical physics.
The states |→i and |↑i form a complete set of states for photons thatmove in the direction of the beam. An alternative complete set of states is
the set {|+i, |−i} formed by the state |+i of a right-hand circularly polarised
photon and the state |−i of a left-hand circularly polarised photon. In thelaboratory a circularly polarised beam is often formed by passing a plane
polarised beam through a birefringent material such as calcite that has its
axes aligned at 45◦ to the incoming plane of polarisation. The incoming
beam is resolved into its components parallel to the calcite’s axes, and one
component is shifted in phase by π/2 with respect to the other. In terms ofunit vectors êx and êy parallel to the calcite’s axes, the incoming field isE 
E = √ ℜ (êx + êy )e−iωt2(1.41)and the outgoing field of a left-hand polarised beam isE 
E− = √ ℜ (êx + iêy )e−iωt ,2(1.42a)while the field of a right-hand polarised beam would beE 
E+ = √ ℜ (êx − iêy )e−iωt .
2(1.42b)The last two equations express the electric field of a circularly polarised
beam as a linear combination of plane polarised beams that differ in phase.
Conversely, by adding (1.42b) to equation (1.42a), we can express the electricfield of a beam polarised along the x axis as a linear combination of the fieldsof two circularly-polarised beams.
Similarly, the quantum state of a circularly polarised photon is a linear
superposition of linearly-polarised quantum states:1|±i = √ (|→i ∓ i|↑i) ,2(1.43)

Problems

15

and conversely, a state of linear polarisation is a linear superposition of statesof circular polarisation:1|→i = √ (|+i + |−i) .
2(1.44)

Whereas in classical physics complex numbers are just a convenient way ofrepresenting the real function cos(ωt + φ) for arbitrary phase φ, quantum
amplitudes are inherently complex and the operator ℜ is not used. Whereas
in classical physics a beam may be linearly polarised in a particular direction,or circularly polarised in a given sense, in quantum mechanics an individual
photon has an amplitude to be linearly polarised in a any chosen direction
and an amplitude to be circularly polarised in a given sense. The amplitude
to be linearly polarised may vanish in one particular direction, or it may
vanish for one sense of circular polarisation. In the general case the photon
will have a non-vanishing amplitude to be polarised in any direction and any
sense. After it has been transmitted by an analyser such as Polaroid, it willcertainly be in whatever state the analyser transmits.

1.4 Measurement
Equation (1.28) expresses the quantum state of a system |ψi as a sum overstates in which a particular measurement, such as energy, is certain to yield aspecified value. The coefficients in this expansion yield as their mod-squares
the probabilities with which the possible results of the measurement will beobtained. Hence so long as there is more than one term in the sum, the resultof the measurement is in doubt. This uncertainty does not reflect shortcomings in the measuring apparatus, but is inherent in the physical situation –any defects in the measuring apparatus will increase the uncertainty above
the irreducible minimum implied by the expansion coefficients, and in §6.3
the theory will be adapted to include such additional uncertainty.
Here we are dealing with ideal measurements, and such measurements
are reproducible. Therefore, if a second measurement is made immediately
after the first, the same result will be obtained. From this observation it
follows that the quantum
P state of the system is changed by the first measurement from |ψi = i ai |ii to |ψi = |Ii, where |Ii is the state in which
the measurement is guaranteed to yield the value that was obtainedPby the
first measurement. The abrupt change in the quantum state from i ai |ii
to |Ii that accompanies a measurement is referred to as the collapse of the
wavefunction.
What happens when the “wavefunction collapses”? It is tempting tosuppose that this event is not a physical one but merely an updating ofour knowledge of the system: that the system was already in the state |Ii
before the measurement, but we only became aware of this fact when themeasurement was made. It turns out that this interpretation is untenable,
and that wavefunction collapse is associated with a real physical disturbanceof the system. This topic is explored further in §6.5.

Problems
1.1 What physical phenomenon requires us to work with probability amplitudes rather than just with probabilities, as in other fields of endeavour?
1.2 What properties cause complete sets of amplitudes to constitute theelements of a vector space?
1.3 V ′ is the dual space of the vector space V . For a mathematician, what
objects comprise V ′ ?

16Problems

1.4 In quantum mechanics, what objects are the members of the vector
space V ? Give an example for the case of quantum mechanics of a memberof the dual space V ′ and explain how members of V ′ enable us to predict
the outcomes of experiments.
1.5 Given that |ψi = eiπ/5 |ai + eiπ/4 |bi, express hψ| as a linear combinationof ha| and hb|.
1.6 What properties characterise the bra ha| that is associated with the ket|ai?
1.7 An electron can be in one of two potential wells that are so close thatit can “tunnel” from one to the other (see §5.2 for a description of quantummechanical tunnelling). Its state vector can be written|ψi = a|Ai + b|Bi,(1.45)

where |Ai is the state of being in the first well and |Bi is the state of being inthe second well and all kets are correctly normalised. What is the probabilityof finding the√particle in the first well given that: (a) a = i/2; (b) b = eiπ ;(c) b = 13 + i/ 2?
1.8 An electron can “tunnel” between potential wells that form a chain, so
its state vector can be written
|ψi =

∞
X
−∞

an |ni,(1.46a)where |ni is the state of being in the nth well, where n increases from left toright. Let |n|/21−ieinπ .
(1.46b)an = √
2 3
a. What is the probability of finding the electron in the nth well?
b. What is the probability of finding the electron in well 0 or anywhere to
the right of it?

2
Operators, measurement and time
evolution
In the last chapter we saw that each quantum state of a system is represented
by a point or ‘ket’ |ψi that lies in an abstract vector space. We saw thatstates for which there is no uncertainty in the value that will be measured
for a quantity such as energy, form a set of basis states for this space –these basis states are analogous to the unit vectors i, j and k of ordinary
vector geometry. In this chapter we develop these ideas further by showing
how every measurable quantity such as position, momentum or energy is
associated with an operator on state space. We shall see that the energy
operator plays a special role in that it determines how a system’s ket |ψimoves through state space over time. Using these operators we are ableat the end of the chapter to study the dynamics of a free particle, and tounderstand how the uncertainties in the position and momentum of a particle
are intimately connected with one another, and how they evolve in time.

2.1 Operators
A linear operator on the vector space V is an object Q that transforms
kets into kets in a linear way. That is, if |ψi is a ket, then |φi = Q|ψi isanother ket, and if |χi is a third ket and α and β are complex numbers, wehave
Q α|ψi + β|χi = α(Q|ψi) + β(Q|χi).
(2.1)

Consider now the linear operator

I=

X
i

|iihi|,(2.2)

where {|ii} is any set of basis kets. I really is an operator because if we
apply it to any ket |ψi, we get a linear combination of kets, which must itself
be a ket:X
X(hi|ψi) |ii,
(2.3)
|iihi|ψi =I|ψi =
ii

18

Chapter 2: Operators, measurement and time evolution

where we are able to move hi|ψi around freely because it’s just a complex
number. To determine which ket I|ψi is, we substitute into (2.3) the expansion (1.17) of |ψi and use the orthogonality relation (1.22):
XX
aj |ji|iihi|
I|ψi =
j

i

=

X
i

(2.4)

ai |ii = |ψi.

We have shown that I applied to an arbitrary ket |ψi yields that same ket.
Hence I is the identity operator. We shall make extensive use of this fact.
Consider now the operator
X
Ei |Ei ihEi |.
(2.5)
H=
i

This is the most important single operator in quantum mechanics. It is called
the Hamiltonian in honour of W.R. Hamilton, who introduced its classicalanalogue.1 We use H to operate on an arbitrary ket |ψi to form the ket
H|ψi, and then we bra through by the adjoint hψ| of |ψi. We havehψ|H|ψi =X
i

Ei hψ|Ei ihEi |ψi.

(2.6)

By equation (1.29) hEi |ψi = ai , while by (1.24) hψ|Ei i = a∗i . Thushψ|H|ψi =X
i

Ei |ai |2 =Xipi Ei = hEi .

(2.7)

Here is yet another result of fundamental importance: if we squeeze the
Hamiltonian between a quantum state |ψi and its adjoint bra, we obtain the
expectation value of the energy for that state.
It is straightforward to generalise this result for the expectation valueof the energy to other measurable quantities: if Q is something that we canmeasure (often called an observable) and its spectrum of possible values is{qi }, then we expand an arbitrary ket |ψi as a linear combination of states|qi i in which the value of Q is well defined,Xai |qi i,(2.8)

and with Q we associate the operatorXqi |qi ihqi |.
Q=

(2.9)

|ψi =

i

i

Then hψ|Q|ψi is the expectation value of Q when our system is in the state|ψi. When the state in question is obvious from the context, we shall sometimes write the expectation value of Q simply as hQi.
When a linear operator R turns up in any mathematical problem, it
is generally expedient to investigate its eigenvalues and eigenvectors. An
eigenvector is a vector that R simply rescales, and its eigenvalue is therescaling factor. Thus, let |ri be an eigenvector of R, and r be its eigenvalue,then we have
R|ri = r|ri.
(2.10)
1 William Rowan Hamilton (1805–1865) was a protestant Irishman who was appointed
the Andrews’ Professor of Astronomy at Trinity College Dublin while still an undergraduate. Although he did not contribute to astronomy, he made important contributions to
optics and mechanics, and to pure mathematics with his invention of quaternions, the first
non-commutative algebra.

2.1 Operators19

Box 2.1:

Hermitian Operators

Let Q be a Hermitian operator with eigenvalues qi and eigenvectors |qi i.
Then we bra the defining equation of |qi i through by hqk |, and bra thedefining equation of |qk i through by hqi |:hqk |Q|qi i = qi hqk |qi i

hqi |Q|qk i = qk hqi |qk i.

We next take the complex conjugate of the second equation from the first.
The left side then vanishes because Q is Hermitian, so with equation(1.27)
0 = (qi − qk∗ )hqk |qi i.

Setting k = i we find that qi = qi∗ since hqi |qi i > 0. Hence the eigenvalues
are real. When qi 6= qk , we must have hqk |qi i = 0, so the eigenvectors
belonging to distinct eigenvalues are orthogonal.
What are the eigenvectors and eigenvalues of H? If we apply H to |Ek i, we
findX
Ei |Ei ihEi |Ek i = Ek |Ek i.
(2.11)
H|Ek i =i

So the eigenvectors of H are the states of well defined energy, and its eigenvalues are the possible results of a measurement of energy. Clearly thisimportant result generalises immediately to eigenvectors and eigenvalues ofthe operator Q that we have associated with an arbitrary observable.
Consider the complex number hφ|Q|ψi, where |φi and |ψi are two arbitrary quantum states. After expanding the states in terms of the eigenvectorsof Q, we havehφ|Q|ψi =Xib∗i hqi |

 X
 X
X
b∗i qi ai (2.12)b∗i aj qj δij =aj |qj i =
Q
j

ij

i

P
Similarly, hψ|Q|φi = i a∗i qi bi . Hence so long as the spectrum {qi } of Q
consists entirely of real numbers (which is physically reasonable), then(hφ|Q|ψi)∗ = hψ|Q|φi(2.13)

for any two states |φi and |ψi. An operator with this property is said to
be Hermitian. Hermitian operators have nice properties. In particular,
one can prove – see Box 2.1 – that they have real eigenvalues and mutuallyorthogonal eigenvectors, and it is because we require these properties on
physical grounds that the operators of observables turn out to be Hermitian.
In Chapter 4 we shall find that Hermitian operators arise naturally fromanother physical point of view.
Although the operators associated with observables are always Hermitian, operators that are not Hermitian turn out to be extremely useful. Witha non-Hermitian operator R we associate another operator R† called its Hermitian adjoint by requiring that for any states |φi and |ψi it is true that∗hφ|R† |ψi = hψ|R|φi.

(2.14)

Comparing this equation with equation (2.13) it is clear that a Hermitian
operator Q is its own adjoint: Q† = Q.
By expanding the kets |φi and |ψi in the equation |φi = R|ψi as sums ofbasis kets, we show that R is completely determined by the array of numbers(called matrix elements)
Rij ≡ hi|R|ji.

(2.15)

20

Chapter 2: Operators, measurement and time evolution
Table 2.1 Rules for Hermitian adjoints
ObjectAdjoint

In fact
|φi =

i
−i
X
i

|ψi
hψ|

R
R†

QR
R † Q†

bi |ii = R|ψi =⇒ bi =

X
j

X

R|ψi
hψ|R†

hφ|R|ψihψ|R† |φiaj R|jijaj hi|R|ji =X

Rij aj .

(2.16)

j

If in equation (2.14) we set |φi = |ii and |ψi = |ji, we discover the
relation between the matrix of R and that of R† :† ∗(Rij
) = Rji

†
∗
Rij
= Rji
.

⇔

(2.17)

Hence the matrix of R† is the complex-conjugate transpose of the matrix
for R. If R is Hermitian so that R† = R, the matrix Rij must equal its
complex-conjugate transpose, that is, it must be an Hermitian matrix.
Operators can be multiplied together: when the operator QR operates
on |ψi, the result is what you get by operating first with R and then applying
Q to R|ψi. We shall frequently need to find the Hermitian adjoints of suchproducts. To find out how to do this we replace R in (2.17) by QR:(QR)†ij = (QR)∗ji =X∗
Q∗jk Rki
=

k

X

†
Rik
Q†kj = (R† Q† )ij .

(2.18)

kThus, to dagger a product we reverse the terms and dagger the individual
operators. By induction it is now easy to show that(ABC . . . Z)† = Z † . . . C † B † A† .

(2.19)

If we agree that the Hermitian adjoint of a complex number is its complex conjugate and that |ψi† ≡ hψ| and hψ|† ≡ |ψi, then we can consider thebasic rule (2.14) for taking the complex conjugate of a matrix element to bea generalisation of the rule we have derived about reversing the order anddaggering the components of a product of operators. The rules for taking
Hermitian adjoints are summarised in Table 2.1.
Functions of operators We shall frequently need to evaluate functionsof operators. For example, the potential energy of a particle is a function
V (x̂) of the position operator x̂. Let f be any function of one variable and
R be any operator. Then we define the operator f (R) by the equationf (R) ≡

Xi

f (ri )|ri ihri |,(2.20)

where the ri and |ri i are the eigenvalues and eigenkets of R. This definition
defines f (R) to be the operator that has the same eigenkets as R and theeigenvalues that you get by evaluating the function f on the eigenvalues ofR.
Commutators The commutator of two operators A, B is defined to be[A, B] ≡ AB − BA.

(2.21)

If [A, B] 6= 0, it is impossible to find a complete set of mutual eigenkets of A
and B (Problem 2.19). Conversely, it can be shown that if [A, B] = 0 there
is a complete set of mutual eigenkets of A and B, that is, there is a complete
set of states of the system in which there is no uncertainty in the value thatwill be obtained for either A or B. We shall make extensive use of this fact.

2.2 Time evolution

21

Notice that the word complete appears in both these statements; even in thecase [A, B] 6= 0 it may be possible to find states in which both A and B
have definite values. It is just that such states cannot form a complete set.
Similarly, when [A, B] = 0 there can be states for which A has a definite
value but B does not. The literature is full of inaccurate statements about
the implications of [A, B] being zero or non-zero.
Three invaluable rules are[A + B, C] = [A, C] + [B, C]
AB = BA + [A, B][AB, C] = [A, C]B + A[B, C].

(2.22)All three rules are trivial to prove by explicitly writing out the contents ofthe square brackets. With these rules it is rarely necessary to write out thecontents of a commutator again, so they eliminate a common source of error
and tedium in calculations. Notice the similarity of the third rule to thestandard rule for differentiating a product: d(ab)/dc = (da/dc)b + a(db/dc).
The rule is easily generalised by induction to the rule
[ABC . . . , Z] = [A, Z]BC . . . + A[B, Z]C . . . + AB[C, Z] . . .

(2.23)We shall frequently need to evaluate the commutator of an operator
A with a function f of an operator B. We assume that f has a convergent
Taylor series2 f = f0 +f ′ B+ 21 f ′′ B 2 +· · ·, where f0 ≡ f (0), f ′ ≡ (df (x)/dx)0 ,etc., are numbers. Then
[A, f (B)] = f ′ [A, B] + 21 f ′′ ([A, B]B + B[A, B])
1 ′′′f ([A, B]B 2 + B[A, B]B + B 2 [A, B]) + · · ·+ 3!

(2.24)In the important case in which B commutes with [A, B], this expression
simplifies dramatically[A, f (B)] = [A, B](f ′ + f ′′ B + 21 f ′′′ B 2 + · · ·) = [A, B]

df
.
dB

(2.25)We shall use this formula several times.

2.2 Evolution in time
Since physics is about predicting the future, equations of motion lie at its
heart. Newtonian dynamics is dominated by the equation of motion f =ma, where f is the force on a particle of mass m and a is the resulting
acceleration. In quantum mechanics the analogous dynamical equation is
the time-dependent Schrödinger equation (TDSE):3
ih̄

∂|ψi
= H|ψi.
∂t(2.26)For future reference we use the rules of Table 2.1 to derive from this equation
the equation of motion of a bra:
−ih̄

∂hψ|
= hψ|H,∂t(2.27)

2 If necessary, we expand f (x) about some point x 6= 0, i.e., in powers of x − x , so00
we don’t need to worry that the series about the origin may not converge for all x.
3 Beginners sometimes interpret the tdse as stating that H = ih̄∂/∂t. This is asunhelpful as interpreting f = ma as a definition of f . For Newton’s equation to be usefulit has to be supplemented by a description of the forces acting on the particle. Similarly,
the tdse is useful only when we have another expression for H.

22

Chapter 2: Operators, measurement and time evolution

where we have used the fact that H is Hermitian, so H † = H. The great
importance of the Hamiltonian operator is due to its appearance in the tdse,
which must be satisfied by the ket of any system. We shall see below inseveral concrete examples that the tdse, which we have not attempted tomotivate physically, generates familiar motions in circumstances that permit
classical mechanics to be used.
One perhaps surprising aspect of the tdse we can justify straight away:
while Newton’s second law is a second-order differential equation, the tdse
is first-order. Since it is first order, the boundary data at t = 0 required tosolve for |ψ, ti at t > 0 comprise the ket |ψ, 0i. If the equation were secondorder in time, like Newton’s law, the required boundary data would include∂|ψi/∂t. But |ψ, 0i by hypothesis constitutes a complete set of amplitudes;
it embodies everything we know about the current state of the system. If
mathematics required us to know something about the system in addition to|ψ, 0i, then either |ψi would not constitute a complete set of amplitudes, or
physics could offer no hope of predicting the future, and it would be time totake up biology or accountancy, or whatever.
The tdse tells us that states of well-defined energy evolve in time in an
exceptionally simple wayih̄
which implies that∂|En i= H|En i = En |En i,∂t(2.28)|En , ti = |En , 0ie−iEn t/h̄ .

(2.29)

That is, the passage of time simply changes the phase of the ket at a rate
En /h̄.
We can use this result to calculate the time evolution of an arbitrarystate |ψi. In the energy representation the state is|ψ, ti =Xan (t)|En , ti.

(2.30)n

Substituting this expansion into the tdse (2.26) we find
ih̄

 X

∂|ψi X∂|En i=an H|En i,=ih̄ ȧn |En i + an
∂t
∂t
n
n

(2.31)

where a dot denotes differentiation with respect to time. The right side
cancels with the second term in the middle, so we have ȧn = 0. Since the an
are constant, on eliminating |En , ti between equations (2.29) and (2.30), we
find that the evolution of |ψi is simply given by|ψ, ti =Xnan e−iEn t/h̄ |En , 0i.

(2.32)We shall use this result time and again.
States of well-defined energy are unphysical and never occur in Nature
because they are incapable of changing in any way, and hence it is impossible
to get a system into such a state. But they play an extremely important rolein quantum mechanics because they provide the almost trivial solution (2.32)
to the governing equation of the theory, (2.26). Given the central role of thesestates, we spend much time solving their defining equation
H|En i = En |En i,(2.33)

which is known as the time-independent Schrödinger equation, or
TISE for short.

2.2 Time evolution

23

2.2.1 Evolution of expectation values
We have seen that hψ|Q|ψi is the expectation value of the observable Q
when the system is in the state |ψi, and that expectation values provide anatural connection to classical physics, which is about situations in which theresult of a measurement is almost certain to lie very close to the quantummechanical expectation value. We can use the tdse to determine the rateof change of this expectation value:
ih̄

∂Q
d
hψ|Q|ψi = −hψ|HQ|ψi + ih̄hψ||ψi + hψ|QH|ψi
dt
∂t
∂Q
|ψi,
= hψ|[Q, H]|ψi + ih̄hψ|∂t(2.34)

where we have used both the tdse (2.26) and its Hermitian adjoint (2.27)
and the square bracket denotes a commutator – see (2.21). Usually operators
are independent of time (i.e., ∂Q/∂t = 0), and then the rate of change of an
expectation value is the expectation value of the operator −i[Q, H]/h̄. This
important result is known as Ehrenfest’s theorem.
If a time-independent operator Q happens to commute with the Hamiltonian, that is if [Q, H] = 0, then for any state |ψi the expectation valueof Q is constant in time, or a conserved quantity. Moreover, in these2
circumstances Q2 also commutes with H, so hψ|(∆Q)2 |ψi = Q2 − hQi
is also constant. If initially ψ is a state of well-defined Q, i.e., |ψi = |qi i
for some i, then (∆Q)2 = 0 at all times. Hence, whenever [Q, H] = 0,a state of well defined Q evolves into another such state, so the value of Q
can be known precisely at all times. The value qi is then said to be a goodquantum number. We always need to label states in some way. The label
should be something that can be checked at any time and is not constantly
changing. Good quantum numbers have precisely these properties, so they
are much employed as labels of states.
If the system is in a state of well defined energy, the expectation valueof any time-independent operator is time-independent, even if the operator
does not commute with H. This is true because in these circumstances
equation (2.34) becomesih̄dhE|Q|Ei = hE|(QH − HQ)|Ei = (E − E)hE|Q|Ei = 0,dt(2.35)

where we have used the equation H|Ei = E|Ei and its Hermitian adjoint.
In view of this property of having constant expectation values of all timeindependent operators, states of well defined energy are called stationary
states.
Since H inevitably commutes with itself, equation (2.34) gives for therate of change of the expectation of the energy
d hEi
=
dt



∂H
∂t



.

(2.36)

In particular hEi is constant if the Hamiltonian is time-independent. Thisis a statement of the principle of the conservation of energy since timedependence of the Hamiltonian arises only when some external force is working on the system. For example, a particle that is gyrating in a timedependent magnetic field has a time-dependent Hamiltonian because work
is being done either on or by the currents that generate the field.

24

Chapter 2: Operators, measurement and time evolution

2.3 The position representation
If the system consists of a single particle that can move in only one dimension,
the amplitudes ψ(x) to find the particle at x for x in (−∞, ∞) constitute acomplete set of amplitudes. By analogy with equation (1.29) we have4
Z ∞
|ψi =
dx ψ(x)|xi.
(2.37)−∞

Here an integral replaces the sum because the spectrum of possible valuesof x is continuous rather than discrete. Our basis kets are the states |xi in
which the particle is definitely at x. By analogy with equation (1.30) wehaveψ(x) = hx|ψi.
(2.38)
Notice that both sides of this equation are complex numbers that depend on
the variable x, that is, they are complex-valued functions of x. For historicalreasons, the function ψ(x) is called the wavefunction of the particle. By
the usual rule (1.27) for complex conjugation of a bra-ket we haveψ ∗ (x) = hψ|xi.

(2.39)

The analogue for the kets |xi of the orthogonality relation (1.22) ishx′ |xi = δ(x − x′ ),(2.40)

where the Dirac delta function δ(x − x′ ) is zero for x 6= x′ because when
the particle is at x, it has zero amplitude to be at a different location x′ .
We get insight into the value of δ(x − x′ ) for x = x′ by multiplying equation(2.37) through by hx′ | and using equation (2.38) to eliminate hx′ |ψi:Z′′hx |ψi = ψ(x ) = dx ψ(x)hx′ |xi
Z(2.41)= dx ψ(x)δ(x − x′ ).
Since δ(x − x′ ) is zero for x 6= x′ , we can replace ψ(x) in the integrand byψ(x′ ) and then take this number outside the integral sign and cancel it withthe ψ(x′ ) on the left hand side. What remains is the equationZ
1 = dx δ(x − x′ ).
(2.42)
Thus there is unit area under the graph of δ(x), which is remarkable, giventhat the function vanishes for x 6= 0! Although the name of δ(x) includes
the word ‘function’, this object is not really a function because we cannot
assign it a value at the origin. It is best considered to be the limit of a seriesof functions that all have unit area under their graphs but become more andmore sharply peaked around the origin (see Figure 2.1).
The analogue of equation (1.31) isZ
dx |ψ(x)|2 = 1,
(2.43)
which expresses the physical requirement that there is unit probability offinding the particle at some value of x.
The analogue of equation (2.2) isZ
I = dx |xihx|.
(2.44)
4 The analogy would be clearer if we wrote a(x) for ψ(x), but for historical reasons
the letter ψ is hard to avoid in this context.

2.3 Position representation

25

Figure 2.1 A series of Gaussians of unit area. The Dirac delta function is the limit ofthis series of functions as the dispersion tends to zero.

It is instructive to check that the operator that is defined by the right sideof this equation really is the identity operator. Applying the operator to anarbitrary state |ψi we findZ
I|ψi = dx |xihx|ψi(2.45)
By equations (2.37) and (2.38) the expression on the right of this equationis |ψi, so I is indeed the identity operator.
When we multiply (2.45) by hφ| on the left, we obtain an important
formulaZZ
hφ|ψi = dx hφ|xihx|ψi = dx φ∗ (x)ψ(x),(2.46)
where the second equality uses equations (2.38) and (2.39). Many practicalproblems reduce to the evaluation of an amplitude such as hφ|ψi. The expression on the right of equation (2.46) is a well defined integral that evaluates
to the desired number.
By analogy with equation (2.5), the position operator isZx̂ = dx x|xihx|.
(2.47)

After applying x̂ to a ket |ψi we have a ket |φi = x̂|ψi whose wavefunction
φ(x′ ) = hx′ |x̂|ψi isZ
φ(x′ ) = hx′ |x̂|ψi = dx xhx′ |xihx|ψiZ(2.48)′′′= dx xδ(x − x )ψ(x) = x ψ(x ),where we have used equations (2.38) and (2.40). Equation (2.48) states thatthe operator x̂ simply multiplies the wavefunction ψ(x) by its argument.
In the position representation, operators turn functions of x into otherfunctions of x. An easy way of making a new function out of an old one isto differentiate it. So consider the operator p̂ that is defined byhx|p̂|ψi = (p̂ψ)(x) = −ih̄

∂ψ
.
∂x

(2.49)In Box 2.2 we show that the factor i ensures that p̂ is a Hermitian operator.
The factor h̄ ensures that p̂ has the dimensions of momentum:5 we will find
5 Planck’s constant h = 2πh̄ has dimensions of distance × momentum, or, equivalently,
energy × time, or, most simply, angular momentum.

26

Chapter 2: Operators, measurement and time evolution

Box 2.2: Proof that p̂ is Hermitian
We have to show that for any states |φi and |ψi, hψ|p̂|φi = (hφ|p̂|ψi)∗ . We
use equation (2.49) to write the left side of this equation in the position
representation:Z∂φhψ|p̂|φi = −ih̄ dx ψ ∗ (x) .
∂xIntegrating by parts this becomesZ ∗ ∞∂ψ ∗hψ|p̂|φi = −ih̄ ψ φ −∞ − dx φ(x).
∂x

We assume that all wavefunctions vanish at spatial infinity, so the termin square brackets vanishes, andZ∂ψ ∗= (hφ|p̂|ψi)∗ .
hψ|p̂|φi = ih̄ dx φ(x)∂xthat p̂ is the momentum operator. In Newtonian physics the momentumof a particle of mass m and velocity ẋ is mẋ, so let’s use equation (2.34) tocalculate d hx̂i /dt and see whether it is hp̂i /m.
2.3.1 Hamiltonian of a particle
To calculate any time derivatives in quantum mechanics we need to knowwhat the Hamiltonian operator H of our system is because H appears in thetdse (2.26). Equation (2.5) defines H in the energy representation, but not
how to write H in the position representation. We are going to have to makean informed guess and justify our guess later.
The Newtonian expression for the energy of a particle isE = 21 mẋ2 + V =

p2
+ V,
2m

(2.50)

where V (x) is the particle’s potential energy. So we guess that the Hamiltonian of a particle isp̂2+ V (x̂),(2.51)
H=
2m
where the square of p̂ means the act of operating with p̂ twice (p̂2 ≡ p̂p̂). The
meaning of V (x̂) is given by equation (2.20) with V and x̂ substituted for f
and R. Working from that equation in close analogy with the calculation inequation (2.48) demonstrates that in the position representation the operator
V (x̂) acts on a wavefunction ψ(x) simply by multiplying ψ by V (x). Thatis, hx|V (x̂)|ψi = V (x)ψ(x).
Now that we have guessed that H is given by equation (2.51), the next
step in the calculation of the rate of change of hx̂i is to evaluate the commutator of x̂ and H. Making use of equations (2.22) we findp̂2[x̂, p̂p̂][x̂, H] = x̂,+V =+ [x̂, V (x̂)]
2m2m[x̂, p̂]p̂ + p̂[x̂, p̂].
=2m(2.52)In the last equality we have used the fact that [x̂, V (x̂)] = 0, which follows
because both x̂ and V (x̂) act by multiplication, and ordinary multiplication
is a commutative operation. We now have to determine the value of the2.3 Position representation

27

commutator [x̂, p̂]. We return to the definition (2.49) of p̂ and calculate thewavefunction produced by applying [x̂, p̂] to an arbitrary state |ψi∂ψ ∂(xψ)
hx|[x̂, p̂]|ψi = hx|(x̂p̂ − p̂x̂)|ψi = −ih̄ x−∂x∂x= ih̄hx|ψi.

(2.53)

Since this equation holds for any |ψi, we have the operator equation[x̂, p̂] = ih̄.

(2.54)

This key result, that the commutator of x̂ with p̂ is the constant ih̄, is calleda canonical commutation relation.6 Two observables whose commutator
is ±ih̄ are said to be canonically conjugate to one another, or conjugate
observables.
Finally we have the hoped-for relation between p and ẋ: substituting
equations (2.53) and (2.54) into equation (2.34) we have
d hxi
d
i
i ih̄
= hψ|x̂|ψi = − hψ|[x̂, H]|ψi = −hψ|p̂|ψi
dt
dt
h̄
h̄ m
1
=
hp̂i .
m(2.55)

This result makes it highly plausible that p̂ is indeed the momentum operator.
A calculation of the rate of change of hp̂i will increase the plausibility
still further. Again working from (2.34) and using (2.51) we haveiid hp̂i= − h[p̂, H]i = − h[p̂, V ]i .
dt
h̄
h̄

(2.56)

Since [p̂, x̂] = −ih̄ is just a number, equation (2.25) for the commutator of
one operator with a function of another operator can be used to evaluate[p̂, V (x̂)]. We then haved hp̂idV.
(2.57)=−dtdx̂
That is, the expectation of the rate of change of the momentum is equal
to the expectation of the force on the particle. Thus we have recovered
Newton’s second law from the tdse. This achievement gives us confidence
that (2.51) is the correct expression for H.

2.3.2 Wavefunction for well defined momentum
From the discussion below equation (2.11) we know that the state |pi in whicha measurement of the momentum will certainly yield the value p has to bean eigenstate of p̂. We find the wavefunction up (x) = hx|pi of this importantstate by using equation (2.49) to write the defining equation p̂|pi = p|pi inthe position representation:hx|p̂|pi = −ih̄∂up= phx|pi = pup (x).
∂x(2.58)The solution of this differential equation isup (x) = Aeipx/h̄ .

(2.59)

6 The name that derives from ‘canonical coordinates’ in Hamilton’s formulation ofclassical mechanics.

28

Chapter 2: Operators, measurement and time evolution

Box 2.3:
Consider the integral
I≡

Gaussian integrals
Z ∞

2

2

dx e−(b x +ax) ,

(1)

−∞

where a and b are constants. We observe that b2 x2 + ax = (bx + a/2b)2 −R2
a2 /4b2 −1
a2 /4b2 . Thus we may writeI=ebdz e−z , where z ≡ bx+a/2b.
√
The integral is equal to π. Hence we have the very useful result
√
Z ∞
2 2
π a2 /4b2e.
(2)dx e−(b x +ax) =b−∞
Hence the wavefunction of a particle √of well defined momentum is a plane
wave with wavelength λ = 2π/k = h/ 2mE, where m is the particle’s massand E its kinetic energy; λ is called the particle’s de Broglie wavelength.7
If we try to choose the constant A in (2.59) to ensure that up satisfies
the usual normalisation condition (2.43), we will fail because the integral
over all x of |eipx/h̄ |2 = 1 is undefined. Instead we choose A as follows. By
analogy with (2.40) we require hp′ |pi = δ(p − p′ ). When we use (2.44) toinsert an identity operator into this expression, it becomesZZ′δ(p − p′ ) = dx hp′ |xihx|pi = |A|2 dx ei(p−p )x/h̄ = 2πh̄|A|2 δ(p − p′ ),(2.60)
where we have used equation (C.12) to evaluate the integral. Thus |A|2 =h−1 , where h = 2πh̄ is Planck’s constant, and the correctly normalised wavefunction of a particle of momentum p is1
up (x) ≡ hx|pi = √ eipx/h̄ .
h(2.61)The uncertainty principle It follows from (2.61) that the position of aparticle that has well defined momentum is maximally uncertain: all values ofx are equally probable because |up |2 is independent of x. This phenomenon
is said to be a consequence of the uncertainty principle,8 namely thatwhen an observable has a well-defined value, all values of the canonicallyconjugate observable are equally probable.
We can gain useful insight into the workings of the uncertainty principle
by calculating the variance in momentum measurements for states in whichmeasurements of position are subject to varying degrees of uncertainty. For
definiteness we take the probability density |ψ(x)|2 to be a Gaussian distribution of dispersion σ. So we write
ψ(x) =221
e−x /4σ .
(2πσ 2 )1/4

(2.62)

With equations (2.46) and (2.61) we find that in this state the amplitude tomeasure momentum p is
Z
Z
2
2
1
hp|ψi = dx u∗p (x)ψ(x) = √dx e−ipx/h̄ e−x /4σ .
(2.63)
h(2πσ 2 )1/4
7 Louis de Broglie (1892–1987) was the second son of the Duc de Broglie. In 1924 his
PhD thesis introduced the concept of matter waves, by considering relativistic invarianceof phase. For this work he won the 1929 Nobel prize for physics. In later years he struggled
to find a causal rather than probabilistic interpretation of quantum mechanics.
8 First stated by Werner Heisenberg, Z. Phys., 43, 172 (1927), and consequently often
called ‘Heisenberg’s uncertainty principle’.

2.3 Position representation29

Box 2.3 explains how integrals of this type are evaluated. Setting a = ip/h̄
and b = (2σ)−1 in equation (2) of Box 2.3 we find
√
2 2
2
2 2
2
2σ π
1
hp|ψi = √e−σ p /h̄ =e−σ p /h̄ .
22
1/4(2πh̄ /4σ )h(2πσ 2 )1/4

(2.64)The probability density |hp|ψi|2 is a Gaussian centred on zero with a dispersion σp in p that equals h̄/2σ. Thus, the more sharply peaked the particle’s
probability distribution is in x, the broader the distribution is in p. The
product of the dispersions in x and p is always 12 h̄: σp σ = 21 h̄.
This trade-off between the uncertainties in x and p arises because when
we expand |ψi in eigenkets of p̂, localisation of the probability amplitude
ψ(x) is caused by interference between states of different momenta: in theposition representation, these states are plane waves of wavelength h/p thathave the same amplitude everywhere, and interference between waves of very
different wavelengths is required if the region of constructive interference is
to be strongly confined.

2.3.3 Dynamics of a free particle
We now consider the motion of a free particle – one that is subject to no forces
so we can drop the potential term in the Hamiltonian (2.51). Consequently,
the Hamiltonian of a free particle,
H=

p̂2
,
2m

(2.65)

is a function of p̂ alone, so its eigenkets will be the eigenkets (2.61) of p̂.
By expressing any ket |ψi as a linear combination of these eigenkets, andusing the basic time-evolution equation (2.32), we can follow the motion ofa particle from the initial state |ψi. We illustrate this procedure with thecase in which ψ corresponds to the particle being approximately at the origin
with momentum near some value p0 . Equation (2.64) gives hp|ψi for the casein which p0 vanishes. The amplitude distribution that we require is
hp|ψ, 0i =2221e−σ (p−p0 ) /h̄ .
(2πh̄2 /4σ 2 )1/4

(2.66)

We can now use (2.32) to obtain the wavefunction t units of time later
Z2hx|ψ, ti = dp hx|pihp|ψ, 0ie−ip t/2mh̄
Z(2.67)
2
2
2
2
1
√
=
dp eipx/h̄ e−σ (p−p0 ) /h̄ e−ip t/2mh̄ .
2h(2πh̄ /4σ 2 )1/4
Evaluating the integral in this expression involves some tiresome algebra– you can find the details in Box 2.4 if you are interested. We want theprobability density at time t of finding the particle at x, which is the modsquare of equation (2.67). From the last equation of Box 2.4 we haveσexp|hx|ψ, ti|2 = √2πh̄2 |b|2

−(x − p0 t/m)2 σ 22h̄4 |b|4



.

(2.68)

This is a Gaussian distribution whose centre moves with the velocity p0 /m
associated with the most probable momentum in the initial data (2.66). The
square of the Gaussian’s dispersion is22

σ (t) = σ +



h̄t
2mσ

2

.

(2.69)

30

Chapter 2: Operators, measurement and time evolution

Box 2.4: Evaluating the integral in equation (2.67)
The integral is of the form discussed in Box 2.3. To clean it up we
replace the p2 in the third exponential with (p − p0 )2 + 2p0 p − p20 andgather together all three exponents:2

eip0 t/2mh̄
h(2πh̄2 /4σ 2 )1/4
 2


 
Z
it
σ
p0 t
ip
2
+
x−
− (p − p0 ).
× dp exp
h̄
m
2mh̄
h̄2

hx|ψ, ti = √In Box 2.3 we now set


i
p0 t
a=
x−
h̄
m
and conclude that2hx|ψ, ti = √eip0 t/2mh̄
h(2πh̄2 /4σ 2 )1/4

exp



;

b2 =

ip0
h̄





it
σ2
2 + 2mh̄
h̄

p0 t
x−
m



 √
π −(x−p0 t/m)2 /4h̄2 b2e.
bThis is a complicated result because b is a complex number, but its modsquare, equation (2.68), is relatively simple.

We saw above that in the initial data the uncertainty in p is ∼ σp = h̄/2σ,
which translates to an uncertainty in velocity ∆v ∼ h̄/2mσ. After time t
this uncertainty should lead to an additional uncertainty in position ∆x =∆v t ∼ h̄t/2mσ in perfect agreement with equation (2.69).
These results complete the demonstration that the identification of theoperator p̂ defined by equation (2.49) with the momentum operator, together
with the Hamiltonian (2.51), enable us to recover as much of Newtonian mechanics as we expect to continue valid outside the classical regime. The ideathat in an appropriate limit the predictions of quantum mechanics should
agree with classical mechanics is often called the correspondence principle. The discipline of checking that one’s calculations comply with thecorrespondence principle is useful in several ways: (i) it provides a check on
the calculations, helping one to locate missing factors of i or incorrect signs,(ii) it deepens one’s understanding of classical physics, and (iii) it draws attention to novel predictions of quantum mechanics that have no counterpartsin classical mechanics.
In the process of checking the correspondence principle for a free particle
we have stumbled on a new principle, the uncertainty principle, which implies
that the more tightly constrained the value of one observable is, the more
uncertain the value of the conjugate variable must be. Notice that these
uncertainties do not arise from measurement errors: we have assumed thatx and p can be measured exactly. The uncertainties we have discussed are
inherent in the situation and can only be increased by deficiencies in themeasurement process.
Our calculations have also shown how far-reaching the principle of quantum interference is: equation (2.67), upon which our understanding of the
dynamics of a free particle rests, expresses the amplitude for the particle tobe found at x at time t as an integral over momenta of the amplitude to travelat momentum p. It is through interference between the infinite number ofcontributing amplitudes that classically recognisable dynamics is recovered.
Had we mod-squared the amplitudes before adding them, as classical probability theory would suggest, we would have obtained entirely unphysicalresults.

2.3 Position representation

31

2.3.4 Back to two-slit interference
When we discussed the two-slit interference experiment in §1.2.1, we statedwithout proof that φ1 − φ2 ∝ x, where φi (x) is the phase of the amplitude
Ai (x) for an electron to arrive at the point x on the screen P after passing
through the slit Si . We can now justify this assertion and derive the constantof proportionality. Once the constant has been determined, it is possible toassess the feasibility of the experiment from a practical point of view.
We assume that the quantum state of an electron as it emerges fromthe electron gun can be approximated by a state of well defined momentum|pi. So the wavefunction between the gun and the screen with the slits is aplane wave of wavelength λ = h/p. As an electron passes through a slit weassume that it is deflected slightly but retains its former kinetic energy. So weapproximate its wavefunction after passing through the slit by a wave that is
no longer plane, but still has wavelength λ. Hence the phase of this wave at
position x on the screen P will be the phase at the slit plus 2πD/λ = pD/h̄,
where D(x) is the distance from x to the slit. By Pythagoras’s theoremp
D = L2 + (x ± s)2 ,(2.70)
where L is the distance between the screen with the slits and P, 2s is the
distance between the slits, and the plus sign applies for one slit and the minus
sign to the other. We assume that both x and s are much smaller than L
so the square root can be expanded by the binomial theorem. We then findthat the difference of the phases is
φ1 − φ2 ≃

2psx
.
h̄L

(2.71)The distance X between the dark bands on P is the value of x for which theleft side becomes 2π, sohLX=.
(2.72)
2ps
Let’s put some numbers into this formula. Since h = 6.63 × 10−34 J s is verysmall, there is a danger that X will come out too small to produce observablebands. Therefore we choose L fairly large and both p and s small. Suppose
we adopt 1 m for L and 1 µm for s. From the Hamiltonian (2.65) we have p =√2mE. A reasonable energy for the electrons is E = 100 eV = 1.6 × 10−17 J,
which yields p = 5.5 × 10−24 Ns, and X = 0.057 mm. Hence there should be
no difficulty observing a sinusoidal pattern that has this period.
What do the numbers look like for bullets? On a firing range we canprobably stretch L to 1000 m. The distance between the slits clearly has
to be larger than the diameter of a bullet, so we take s = 1 cm. A bullet
weighs ∼ 10 gm and travels at ∼ 300 m s−1 . Equation (2.72) now yields
X ∼ 10−29 m. So it is not surprising that fire-arms manufacturers findclassical mechanics entirely satisfactory.
2.3.5 Generalisation to three dimensions
Real particles move in three dimensions rather than one. Fortunately, thegeneralisation to three dimensions of what we have done in one dimension isstraightforward.
The x, y and z coordinates of a particle are three distinct observables.
Their operators commute with one another:[x̂i , x̂j ] = 0.

(2.73)

Since these are commuting observables, there is a complete set of mutual
eigenkets, {|xi}. We can express any state of the system, |ψi, as a linear
combination of these kets:ZZ|ψi = d3 x hx|ψi |xi = d3 x ψ(x) |xi,
(2.74)

32

Chapter 2: Operators, measurement and time evolution

where the wavefunction ψ(x) is now a function of three variables, and theintegral is over all space.
The x, y and z components of the particle’s momentum p̂ commute with
one another:[p̂i , p̂j ] = 0.
(2.75)In the position representation, these operators are represented by partial
derivatives with respect to their respective coordinatesp̂i = −ih̄∂∂xiso p̂ = −ih̄∇ .

(2.76)The momenta commute with all operators except their conjugate coordinate,
so the canonical commutation relations are[x̂i , p̂j ] = ih̄δij .

(2.77)

In §4.2 we will understand the origin of the factor δij . Since the three momentum operators commute with one another, there is a complete set ofmutual eigenstates. Analogously to equation (2.61), the wavefunction of thestate with well defined momentum p ishx|pi =

1 ix·p/h̄e.
h3/2(2.78)In the position representation the tdse of a particle of mass m thatmoves in a potential V (x) readsih̄∂hx|ψip̂2= hx|Ĥ|ψi = hx||ψi + hx|V (x̂)|ψi.
∂t2m(2.79)

Now hx|p̂2 |ψi = −h̄2 ∇2 hx|ψi and hx|V (x̂)|ψi = V (x)hx|ψi. Hence using thedefinition ψ(x) ≡ hx|ψi, the tdse becomes
ih̄

∂ψ
h̄2 2
=−
∇ ψ + V (x)ψ.
∂t
2m(2.80)

Probability current Max Born9 first suggested that the mod-square ofa particle’s wavefunction,
ρ(x, t) ≡ |ψ(x, t)|2(2.81)

is the probability density of finding the particle near x at time t. Since
the particle is certain
to be found somewhere, this interpretation implies
R
that at any time d3 x |ψ(x, t)|2 = 1. It is not self-evident that this physicalrequirement is satisfied when the wavefunction evolves according to the tdse
(2.80). We now show that it is in fact satisfied.
We multiple the tdse (2.80) by ψ ∗ and subtract from it the result ofmultiplying the complex conjugate of (2.80) by ψ. Then the terms involving
the potential V (x) cancel and we are left with



h̄2
∂ψ ∗
∗ ∂ψ
ih̄ ψ
=−
ψ ∗ ∇2 ψ − ψ∇2 ψ ∗ .
(2.82)+ψ∂t∂t
2m
The left side of this equation is a multiple of the time derivative of ρ =ψ ∗ ψ. The right side can be expressed as a multiple of the divergence of theprobability currentih̄ ψ∇ψ ∗ − ψ ∗ ∇ψ .
(2.83)
J(x) ≡
2m
9 For this insight Born won the 1954 Nobel Price for physics. In fact the text of the
key paper (Born, M., Z. Physik, 37 863 (1926)) argues that ψ is the probability density,
but a note in proof says “On more careful consideration, the probability is proportional
to the square of ψ”.

2.3 Position representation

33

That is, equation (2.82) can be written∂ρ= −∇ · J.
∂t(2.84)

In fluid mechanics this equation with J = ρv expresses the conservation ofmass as a fluid of density ρ flows with velocity v(x). In quantum mechanics
it expresses conservation of probability. To show that this is so, we simply
integrate both sides of equation (2.84) through a volume V. Then we obtain
d
dt

Z

Z

∂ρ
d x
=−
d xρ =
∂t
V
V
3

3

Z

3

V

d x∇ · J = −I∂V

d2 S · J,(2.85)

where the last equality uses the divergence theorem and ∂V denotes theboundary of
R V. Equation (2.85) states that the rate of increase of the probability P = V d3 x ρ of finding the particle in V is equal to minus the integral
over the volume’s bounding surface of the probability flux out of the volume.
If VR encompasses all space, ψ, and therefore J, will vanish on the boundary,so V d3 x ρ will be constant.
We can gain valuable insight into the meaning of a wavefunction by
explicitly breaking ψ into its modulus and phase:ψ(x) = S(x)eiφ(x) ,(2.86)

where S and φ are real. Substituting this expression into the definition (2.83)
of J, we find
J=

ih̄
h̄
(S∇S − iS 2 ∇φ − S∇S − iS 2 ∇φ) = S 2 ∇φ.
2mm(2.87)

Since S 2 = |ψ|2 = ρ, the velocity v that is defined by setting J = ρv is
v=

h̄∇φ
.
m

(2.88)

Thus the gradient of the phase of the wavefunction encodes the velocity at
which the probability fluid flows. In classical physics, this is the particle’s
velocity. The phase of the wavefunction (2.78) of a particle of well-defined
momentum is φ(x) = x · p/h̄, so in this special case v = p/m as in classicalphysics. Equation (2.88) extends the connection between velocity and thegradient of phase to general wavefunctions.
The virial theorem We illustrate the use of the canonical commutations
relations equations (2.73), (2.75) and (2.77) by deriving a relation between
the kinetic and potential energies of a particle that is in a stationary state.
In §2.2.1 we showed that all expectation values are time-independent whena system is in a stationary state. We apply this result to the operator x̂ · p̂
h
i
d
p̂2
hx̂ · p̂i = hE| x̂ · p̂,+ V (x̂) |Ei
dt
2m
1
2
=
hE|[x̂ · p̂, p̂ ]|Ei + hE|[x̂ · p̂, V (x̂)]|Ei.
2m

0 = ih̄(2.89)

The first commutator can be expanded thus[x̂ · p̂, p̂2 ] =Xjk[x̂j p̂j , p̂2k ] =Xjk[x̂j , p̂2k ]p̂j =X

2ih̄p̂k δjk p̂j = 2ih̄p̂2 . (2.90)

jk

In the position representation the second commutator is simply[x̂ · p̂, V (x̂)] = −ih̄x · ∇V (x).

(2.91)

34

Problems

When we put these results back into (2.89) and rearrange, we obtain thevirial theoremp̂22hE||Ei = hE|(x · ∇V )|Ei.
(2.92)
2m
In important applications the potential is proportional to some power of
distance from the origin: V (x) = C|x|α . Then, because ∇|x| = x/|x|, theoperator on the right is x · ∇V = αC|x|α = αV and the virial theorem
becomesp̂22hE||Ei = αhE|V |Ei.
(2.93)
2m
So twice the kinetic energy is equal to α times the potential energy. For
example, for a harmonic oscillator α = 2, so kinetic and potential energies
are equal. The other important example is motion in an inverse-square force
field, such as the electrostatic field of an atomic nucleus. In this case α = −1,
so twice the kinetic energy plus the potential energy vanishes. Equivalently,
the kinetic energy is equal in magnitude but opposite in sign to the totalenergy.

Problems
2.1 How is a wave-function ψ(x) written in Dirac’s notation? What’s thephysical significance of the complex number ψ(x) for given x?
2.2 Let Q be an operator. Under what circumstances is the complex number ha|Q|bi equal to the complex number (hb|Q|ai)∗ for any states |ai and|bi?
2.3 Let Q be the operator of an observable and let |ψi be the state of oursystem.
a. What are the physical interpretations of hψ|Q|ψi and |hqn |ψi|2 , where|qn i is the nth eigenket of the observable Q and qn is the corresponding
eigenvalue?
P
b. What is the operator n |qnP
ihqn |, where the sum is over all eigenkets
of Q? What is the operator n qn |qn ihqn |?
c. If un (x) is the wavefunction of the state |qn i, write dow an integral that
evaluates to hqn |ψi.
2.4 What does it mean to say that two operators commute? What is thesignificance of two observables having mutually commuting operators?
Given that the commutator [P, Q] 6= 0 for some observables P and Q,
does it follow that for all |ψi =6 0 we have [P, Q]|ψi =6 0?
2.5 Let ψ(x, t) be the correctly normalised wavefunction of a particle ofmass m and potential energy V (x). Write down expressions for the expectation values of (a) x; (b) x2 ; (c) the momentum px ; (d) p2x ; (e) the energy.
What is the probability that the particle will be found in the interval(x1 , x2 )?
2.6 Write down the time-independent (tise) and the time-dependent (tdse)
Schrödinger equations. Is it necessary for the wavefunction of a system tosatisfy the tdse? Under what circumstances does the wavefunction of asystem satisfy the tise?
2.7 Why is the tdse first-order in time, rather than second-order like Newton’s equations of motion?
2.8 A particle is confined in a potential well such that its allowed energies
are En = n2 E, where n = 1, 2, . . . is an integer and E a positive constant.
The corresponding energy eigenstates are |1i, |2i, . . . , |ni, . . .. At t = 0 theparticle is in the state|ψ(0)i = 0.2|1i + 0.3|2i + 0.4|3i + 0.843|4i.

(2.94)

Problems

35

a. What is the probability, if the energy is measured at t = 0 of finding a
number smaller than 6E?
b. What is the mean value and what is the rms deviation of the energy ofthe particle in the state |ψ(0)i?
c. Calculate the state vector |ψi at time t. Do the results found in (a) and(b) for time t remain valid for arbitrary time t?
d. When the energy is measured it turns out to be 16E. After the measurement, what is the state of the system? What result is obtained if
the energy is measured again?
2.9 A system has a time-independent Hamiltonian that has spectrum {En }.
Prove that the probability Pk that a measurement of energy will yield thevalue Ek is is time-independent. Hint: you can do this either from Ehrenfest’s
theorem, or by differentiating hEk , t|ψi w.r.t. t and using the tdse.
2.10 Let ψ(x) be a properly normalised wavefunction and Q an operator on wavefunctions. Let {qr } be the spectrum of Q and {ur (x)} be thecorresponding correctly normalised eigenfunctions. Write down an expression for thePprobability that a measurement of Q will yield the value qr .
Show thatr P (qr |ψ) = 1. Show further that the expectation of Q isR∞
hQi ≡ −∞ ψ ∗ Q̂ψ dx.10

2.11 Find the energy of neutron, electron and electromagnetic waves ofwavelength 0.1 nm.

2.12 Neutrons are emitted from an atomic pile with a Maxwellian distribution of velocities for temperature 400 K. Find the most probable de Broglie
wavelength in the beam.
2.13 A beam of neutrons with energy E runs horizontally into a crystal.
The crystal transmits half the neutrons and deflects the other half verticallyupwards. After climbing to height H these neutrons are deflected through 90◦
onto a horizontal path parallel to the originally transmitted beam. The twohorizontal beams now move a distance L down the laboratory, one distance H
above the other. After going distance L, the lower beam is deflected verticallyupwards and is finally deflected into the path of the upper beam such thatthe two beams are co-spatial as they enter the detector. Given that particles
in both the lower and upper beams are in states of well-defined momentum,
show that the wavenumbers k, k ′ of the lower and upper beams are related
by


mn gH
′
k ≃k 1−.
(2.95)
2E
In an actual experiment (R. Colella et al., 1975, Phys. Rev. Let., 34, 1472)
E = 0.042 eV and LH ∼ 10−3 m2 (the actual geometry was slightly different). Determine the phase difference between the two beams at the detector.
Sketch the intensity in the detector as a function of H.
2.14 A particle moves in the potential V (x) and is known to have energy
En . (a) Can it have well defined momentum for some particular V (x)? (b)Can the particle simultaneously have well-defined energy and position?
2.15 The states {|1i, |2i} form a complete orthonormal set of states for a
two-state system. With respect to these basis states the operator σy has
matrix


0 −i
σy =
.
(2.96)
i 0
Could σ be an observable? What are its eigenvalues and eigenvectors in the{|1i, |2i} basis? Determine the result of operating with σy on the state1|ψi = √ (|1i − |2i).
(2.97)
2

10 In the most elegant formulation of qantum mechanics, this last result is the basic
postulate of the theory, and one derives other rules for the physical interpretation of the qn ,an etc. from it – see J. von Neumann, Mathematical Foundations of Quantum Mechanics.

36

Problems

2.16 A three-state system has a complete orthonormal set of states |1i, |2i, |3i.
With respect to this basis the operators H and B have matrices



1 0
0
H = h̄ω  0 −1 0 
0 0 −1

1 0
B = b0 0
0 1


0
1,
0

(2.98)

where ω and b are real constants.
a. Are H and B Hermitian?
b. Write down the eigenvalues of H and find the eigenvalues of B. Solve forthe eigenvectors of both H and B. Explain why neither matrix uniquely
specifies its eigenvectors.
c. Show that H and B commute. Give a basis of eigenvectors common to
H and B.
2.17 Given that A and B are Hermitian operators, show that i[A, B] is a
Hermitian operator.
2.18 Given a ordinary function f (x) and an operator R, the operator f (R)is defined to beXf (ri )|ri ihri |,(2.99)
f (R) =iwhere ri are the eigenvalues of R and |ri i are the associated eigenkets. Show
that when f (x) = x2 this definition implies that f (R) = RR, that is, that
operating with f (R) is equivalent to applying the operator R twice. Whatbearing does this result have in the meaning of eR ?
2.19 Show that if there is a complete set of mutual eigenkets of the Hermitian operators A and B, then [A, B] = 0. Explain the physical significanceof this result.
2.20 Given that for any two operators (AB)† = B † A† , show that
(ABCD)† = D† C † B † A† .

(2.100)2.21 Prove for any four operators A, B, C, D that
[ABC, D] = AB[C, D] + A[B, D]C + [A, D]BC.

(2.101)

Explain the similarity with the rule for differentiating a product.
2.22 Show that for any three operators A, B and C, the Jacobi identity
holds:[A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0.
(2.102)2.23 Show that a classical harmonic oscillator satisfies the virial equation
2hKEi = αhPEi and determine the relevant value of α.
2.24 Given that the wavefunction is ψ = Aei(kz−ωt) + Be−i(kz+ωt) , where
A and B are constants, show that the probability current density is
J = v |A|2 − |B|2 ẑ,

where v = h̄k/m. Interpret the result physically.

(2.103)3Harmonic oscillators and magnetic
fieldsHarmonic oscillators are of enormous importance for physics because mostof condensed-matter physics and quantum electrodynamics centre on weakly
perturbed harmonic oscillators. The reason harmonic oscillators are so common is simple. The points of equilibrium of a particle that moves in a
potential V (x) are points at which the force −dV /dx vanishes. When weplace the origin of x at such a point, the Maclaurin expansion of V becomesV (x) = constant + 21 V ′′ x2 + O(x3 ), and the force on the particle becomes
F = −V ′′ x + O(x2 ). Consequently, for sufficiently small excursions from thepoint of equilibrium, the particle’s motion will be well approximated by aharmonic oscillator.
Besides providing the background to a great many branches of physics,
our analysis of a harmonic oscillator will introduce a technique that we will
use twice more in our analysis of the hydrogen atom. As a bonus, we will findthat our results for the harmonic oscillator enable us to solve another important, and apparently unrelated problem: the motion of a charged particle ina uniform magnetic field.

3.1 Stationary states of a harmonic oscillator
We can build a harmonic oscillator by placing a particle in a potential that
increases quadratically with distance from the origin. Hence an appropriate
Hamiltonian is given by equation (2.51) with V ∝ x2 .1 For later convenience
we choose the constant of proportionality such that H becomes
H=

1  2
p + (mωx)2 .
2m

(3.1)

In §2.2 we saw that the dynamical evolution of a system follows immediately
once we know the eigenvalues and eigenkets of H. So we now determine
these quantities for the Hamiltonian (3.1).
1 In the last chapter we distinguished the position and momentum operators from theireigenvalues with hats. Henceforth we drop the hats; the distinction between operator andeigenvalue should be clear from the context.

38

Chapter 3: Harmonic oscillators and magnetic fields
We next introduce the dimensionless operatormωx + ip
A≡ √
.
2mh̄ω

(3.2a)

This operator isn’t Hermitian. Bearing in mind that x and p are Hermitian,from the rules in Table 2.1 we see that its adjoint ismωx − ip.
A† = √
2mh̄ω

(3.2b)The product A† A is1(mωx − ip)(mωx + ip)
2mh̄ω
H
1 
(mωx)2 + imω[x, p] + p2 =
− 1,
=
2mh̄ω
h̄ω 2

A† A =(3.3)

where we have used the canonical commutation relation (2.54). This equation
can be rewritten H/(h̄ω) = A† A + 21 , so A is rather nearly the square root of
the dimensionless Hamiltonian H/h̄ω. If we calculate AA† in the same way,
the only thing that changes is the sign in front of the commutator [x, p], so
we have
H
(3.4)
AA† =
+ 1.
h̄ω 2
Subtracting equation (3.4) from equation (3.3) we find that[A† , A] = −1.

(3.5)

We will find it useful to have evaluated the commutator of A† with the
Hamiltonian. Since from equation (3.3) H = h̄ω(A† A + 21 ), we can write[A† , H] = h̄ω[A† , A† A] = h̄ωA† [A† , A] = −h̄ωA† ,(3.6)

where we have exploited the rules of equations (2.22).
We now multiply both sides of the defining relation of |En i, namely
H|En i = En |En i, by A† :
A† En |En i = A† H|En i = HA† + [A† , H] |En i = (H − h̄ω)A† |En i. (3.7)
A slight rearrangement of this equation yields

H(A† |En i) = (En + h̄ω)(A† |En i).

(3.8)

Provided |bi ≡ A† |En i has non-zero length-squared, this shows that |bi is aneigenket of H with eigenvalue En + h̄ω. The length-square of |bi is†

A |En i2†

= hEn |AA |En i = hEn |



H
+1
h̄ω 2



|En i =

En
+ 12 .
h̄ω(3.9)

Now squeezing H between hEn | and |En i we find with (3.1) thatEn = hEn |H|En i =122
p|En i + m2 ω 2 x|En i
≥ 0.
2mω
2

(3.10)

Thus the energy eigenvalues are non-negative, so A† |En i > 0 and by
repeated application of A† we can construct an infinite series of eigenstateswith energy En + kh̄ω for k = 0, 1, . . .
Similarly, we can show that provided A|En i has non-zero length-squared,
it is an eigenket of H for energy En − h̄ω. Since we know that all eigenvalues3.1 Stationary states39

are non-negative, for some energy E0 , A|E0 i must vanish. Equating to zero
the length-squared of this vector we obtain an equation for E0 :


H
E0
2
− 21 |E0 i =− 1.
0 = A|E0 i = hE0 |(3.11)h̄ωh̄ω 2
So E0 = 21 h̄ω and we have established that the eigenvalues of H areh̄ω × ( 21 , 23 , . . . , 2r+1
2 , . . .) that is

Er = (r + 21 )h̄ω.

(3.12)The operators A† and A with which we have obtained these importantresults are respectively called creation and annihilation operators because the first creates an excitation of the oscillator, and the second destroys
one. In quantum field theory particles are interpreted as excitations of thevacuum and each particle species is associated with creation and annihilation
operators that create and destroy particles of the given species. A and A†
are also called ladder operators.
We now examine the eigenkets of H. Let |ri denote the state of energy(r + 12 )h̄ω. In this notation the lowest-energy state, or ground state, is |0i
and its defining equation is A|0i = 0. From equation (3.2a) this equationreads
mωx|0i + ip|0i√
0 = A|0i =.
(3.13)2mh̄ω
We now go to the position representation by multiplying through by hx|.
With equations (2.48) and (2.49) we find that the equation becomes


1
∂
√
hx|0i = 0.
(3.14)mωx + h̄∂x2mh̄ωThis is a linear, first-order differential equation. Its integrating factor is
exp(mωx2 /2h̄), so the correctly normalised wavefunction is2
2
1
hx|0i =
e−x /4ℓ ,(2πℓ2 )1/4

where ℓ ≡

r

h̄
.
2mω

(3.15)

Notice that this solution is unique, so the ground state is non-degenerate.
It is a Gaussian function, so the probability distribution P (x) = |hx|0i|2 forthe position of the particle that forms the oscillator is also a Gaussian: its
dispersion is ℓ.
From equations (2.63) and (2.64) we see that the momentum distributionof the wavefunction (3.15) is
2 2

2P (p) ≡ |hp|0i|2 ∝ e−2ℓ p /h̄ ,(3.16)

which is a Gaussian with dispersion σp = h̄/2ℓ. By inserting x = ℓ andp = σp in the Hamiltonian (3.1) we obtain estimates of the typical kinetic
and potential energies of the particle when it’s in its ground state. We findthat both energies are ∼ 41 h̄ω. In fact one can straightforwardly show that
H(ℓ, σp ) is minimised subject to the constraint ℓσp ≥ h̄/2 when ℓ and σptake the values that we have derived for the ground state (Problem 3.4). In
other words, in its ground state the particle is as stationary and as close to
the origin as the uncertainty principle permits; there is a conflict between theadvantage energetically of being near the origin, and the energetic penalty
that the uncertainty principle exacts for having a well defined position.
Every system that has a confining potential exhibits an analogous zeropoint motion. The energy tied up in this motion is called zero-point
energy. Zero-point motion is probably the single most important predictionof quantum mechanics, for the material world is at every level profoundly
influenced by this phenomenon.

40

Chapter 3: Harmonic oscillators and magnetic fields

We obtain the wavefunctions of excited states by applying powers ofthe differential operator A† to hx|0i. Equation (3.9) enables us to find thenormalisation constant α in the equation |n + 1i = αA† |ni; it implies thatα2 = n + 1. the generalisation of equation (3.11) enables us to determine thenumber β in the equation |n − 1i = βA|ni, and we have finally1A† |ni ;|n + 1i = √n+1

1|n − 1i = √ A|ni.
n(3.17)

It is useful to remember that the normalisation constant is always the squareroot of the largest value of n appearing in the equation. As a specific example




x
∂
∂
1
hx|0i =
hx|0i−ℓmωx − h̄hx|1i = √
∂x
2ℓ
∂x
2mh̄ω
1
x −x2 /4ℓ2
=
e
.
2
1/4
ℓ
(2πℓ )(3.18)

Whereas the ground-state wavefunction is an even function of x, the wavefunction of the first excited state is an odd function because A† is odd in x.
Wavefunctions that are even in x are said to be of even parity, while thosethat are odd functions have odd parity. It is clear that this pattern will
be repeated as we apply further powers of A† to generate the other statesof well-defined energy, so hx|ni is even parity if n is even, and odd parity
otherwise.
Notice that the operator N ≡ A† A is Hermitian. By equations (3.17)
N |ni = n|ni, so its eigenvalue tells you the number of excitations the oscillator has. Hence N is called the number operator.
Let’s use these results to find the mean-square displacement hn|x2 |ni
when the oscillator is in its nth excited state. Adding equations (3.2) we
express x as a linear combination of A and A†
x=

r

h̄
(A + A† ) = ℓ(A + A† ),2mω

(3.19)

where ℓ is defined by (3.15), sohn|x2 |ni = ℓ2 hn|(A + A† )2 |ni.

(3.20)

When we multiply out the bracket on the right, the only terms that contribute are the ones that involve equal numbers of As and A† s. Thus
hn|x2 |ni = ℓ2 hn|(AA† + A† A)|ni = ℓ2 (2n + 1) = ℓ2

2En
,
h̄ω

(3.21)

where we have used equations (3.17) and (3.12). If we use equation (3.15) toeliminate ℓ, we obtain a formula that is valid in classical mechanics (Problem 2.23).

3.2 Dynamics of oscillators

41

3.2 Dynamics of oscillators
By equations (2.29) and (3.12), the nth excited state of the harmonic oscillator evolves in time according to|n, ti = e−i(n+1/2)ωt |n, 0i(3.22)

Consequently, no state oscillates at the oscillator’s classical frequency ω. How
do we reconcile this result with classical physics?
We have seen that we make the link from quantum to classical physics
by considering the expectation values of observables – if classical physics
applies, the measured value of any observable will lie close to the expectation
value, so the latter provides an accurate description of what’s happening.
Equation (2.35) tells us that when a system is in an energy eigenstate, theexpectation value of any time-independent observable Q cannot depend ontime. Equation (3.22) enables us to obtain this result from a different point ofview by showing that when we form the expectation value hQi = hψ|Q|ψi, thefactor e−iEn t/h̄ in the ket |ψ, ti = e−iEn t/h̄ |En i cancels on the corresponding
factor in hψ, t|. Hence energy eigenstates are incapable of motion.2 The
system is capable of motion only if there are non-negligible amplitudes tomeasure more than one possible energy, or, equivalently, if none of the ai inthe sum (2.32) has near unit modulus.
Consideration of the motion of a harmonic oscillator will make this general point clearer. If the oscillator’s state is writtenX|ψ, ti =aj e−iEj t/h̄ |ji,
(3.23)
then the expectation value of x is
X
X
hxi =
a∗k aj ei(Ek −Ej )t/h̄ hk|x|ji =a∗k aj ei(k−j)ωt hk|x|ji.
jk(3.24)

jk

We simplify this expression by using equation (3.19) to replace x with ℓ(A +A† ) and then using (3.17) to evaluate the matrix elements of A and A† :X
hxi = ℓa∗k aj ei(k−j)ωt hk|(A + A† )|ji
jk

=ℓ

X

a∗k aj ei(k−j)ωtjkpp
jhk|j − 1i + j + 1hk|j + 1i.

(3.25)

Since hk|j − 1i vanishes unless k = j − 1, it’s now easy to perform the sum
over k, leaving two terms to be summed over j. On account of the factor√
j we can restrict the first of these sums to j > 0, and in the second sum
we replace j by j ′ ≡ j + 1 and then replace the symbol j ′ by j so we can
combine the two sums. After these operations we haveXphxi = ℓj(a∗j−1 aj e−iωt + a∗j aj−1 eiωt )
j=1

=

X

Xj cos(ωt + φj ),(3.26a)jwhere the real numbers Xj and φj are defined byp
2 jℓ a∗j aj−1 = Xj eiφj .

(3.26b)Thus hxi oscillates sinusoidally at the classical frequency ω regardless of theamplitudes aj . Thus we have recovered the classical result that the frequency
2 If we consider that t is the variable canonically conjugate to energy, this fact becomesa manifestation of the uncertainty principle.

42

Chapter 3: Harmonic oscillators and magnetic fields

Figure 3.1 The potential energy V (x) of an anharmonic oscillator (full curve) and V (x)for the harmonic oscillator obtained by restricting the potential to the first two terms in
its Maclaurin expansion (dashed curve).

at whichp
a harmonic oscillator oscillates is independent of amplitude andequal to k/m, where k is the oscillator’s spring constant.
In the classical regime, the only non-negligible amplitudes aj have indices j that cluster around some large number n. Consequently, a measurement of the energy is guaranteed to yield a value that lies close to E = En ,and from equation (3.21) it follows that the mean value of x2 will lie close
to x2 = 2ℓ2 En /(h̄ω). Classically, the time average of x2 is proportional tothe average potential energy, which is just half the total energy. Hence, averaging the Hamiltonian (3.1) we conclude that classically x2 = E/(mω 2 ),in precise agrees with the quantum-mechanical result. The correspondence
principle requires the classical and quantum-mechanical values of x2 to agree
for large n. That they agree even for small n is a coincidence.
3.2.1 Anharmonic oscillators
The Taylor series of the potential energy V (x) of a harmonic oscillator isvery special: it contains precisely one non-trivial term, that proportional tox2 . Real oscillators have potential-energy functions that invariably deviate
from this ideal to some degree. The deviation is generally in the sense that
V (x) < 12 V ′′ (0)x2 for x > 0 – see Figure 3.1. One reason why deviations fromharmonicity are generally of this type is that it takes only a finite amount ofenergy to break a real object, so V (∞) should be finite, whereas the potential
energy function of a harmonic oscillator increases without limit as x → ∞.
Consider the anharmonic oscillator that has potential energy
a2 V0
V (x) = − 2,
a + x2(3.27)

where V0 and a are constants. We cannot find the stationary states of thisoscillator analytically any more than we can analytically solve its classicalequations of motion.3 But we can determine its quantum mechanics numerically,4 and doing so will help to show which aspects of the results we
3 Murphy’s law is in action here: the dynamics of the pendulum is analytically intractable precisely because it is richer and more interesting than that of the harmonic
oscillator.
4 A good way to do this is to turn the tise into a finite matrix equation and then touse a numerical linear-algebra package to find the eigenvalues of the matrix. Figure 3.2′′ ≃ (ψ2
was obtained using the approximation ψnn+1 + ψn−1 − 2ψn )/∆ , where ψn denotesψ(n∆) with ∆ a small increment in x. With this approximation the tise becomes the
eigenvalue equation of a tridiagonal matrix that has 2b2 /∆2 + Vn /V0 on the leading diagonal and −b2 /∆2 above and below this diagonal, where b2 = h̄2 /2mV0 and Vn = V (n∆).

3.2 Dynamics of oscillators

43

Figure 3.2 The spectrum of theanharmonic oscillator for which thepotential is plotted in Figure 3.1
when the dimensionless variable
2ma2 V0 /h̄2 = 100.

Figure 3.3 Values of aj when there is significant uncertainty in E.

have obtained for the harmonic oscillator are special, and which have generalapplicability.
Figure 3.2 shows the anharmonic oscillator’s energy spectrum. At low
energies, when the pendulum is nearly harmonic, the energies are nearly
uniformly spaced in E. As we proceed to higher energies, the spacing between
levels diminishes, with the consequence that infinitely many energy levels arepacked into the finite energy range between −V0 and zero, where the particle
becomes free. This crowding of the energy levels has the following implication
for the time dependence of hxi. Suppose there are just two energies with nonzero amplitudes, aN and aN +1 . Then hxi will be given by
hxi = a∗N aN +1 ei(EN −EN +1 )t/h̄ hN |x|N + 1i + complex conjugate.

(3.28)

This is a sinusoidal function of time, but its period, T = h/(EN +1 − EN ),depends on N . If we increase the energy and amplitude of the oscillator, we
will increase N and Figure 3.2 shows that T will also increase. Classicallythe period of the oscillator increases with amplitude in just the same way.
Thus there is an intimate connection between the spacing of the energy levels
and classical dynamics.
Consider now the case in which the energy is more uncertain, so thatseveral of the aj are non-zero, and let these non-zero aj be clustered around44

Chapter 3: Harmonic oscillators and magnetic fields

j = N (see Figure 3.3). In this case several terms will occur in the sum forhxi
hxi = · · · + a∗N −1 aN ei(EN −1 −EN )t/h̄ hN − 1|x|N i+ a∗N +1 aN ei(EN +1 −EN )t/h̄ hN + 1|x|N i(3.29)+ a∗N +3 aN ei(EN +3 −EN )t/h̄ hN + 3|x|N i + · · ·where we have anticipated a result of §4.1.4 below that the matrix element
hj|x|ki vanishes if j − k is even. The sum (3.29) differs from the corresponding one (3.26a) for a harmonic oscillator in the presence of matrix elements
hj|x|ki with |j − k| > 1: in the case of the harmonic oscillator these matrix elements vanish, but in the general case they won’t. In consequence
the series contains terms with frequencies (EN +3 − EN )/h̄ as well as termsin ωN ≡ (EN +1 − EN )/h̄. If these additional frequencies were all integer
multiples of a single frequency ωN , the time dependence of hxi would beperiodic with period TN = 2π/ωN , but anharmonic, like that of the classicaloscillator. Now (EN +3 − EN )/h̄ ≃ 3ωN because the spacing between energy
levels changes only slowly with N , so when, as in Figure 3.3, the non-zero
amplitudes are very tightly clustered around N , the additional frequencies
will be integer multiples of ωN to good accuracy, and the motion will indeed
be periodic but anharmonic as classical mechanics predicts.
If we release the oscillator from near some large extension X, the nonnegligible amplitudes aj will be clustered around some integer N as depicted
in Figure 3.3, and their phases will be such that at t = 0 the wavefunctionshx|ji will interfere constructively near X and sum to near zeroP elsewhere,
ensuring that the mod-square of the wavefunction ψ(x, 0) = j aj hx|ji issharply peaked around x = X. At a general time the wavefunction will begiven byX
ei(EN −Ej )t/h̄ aj hx|ji.
(3.30)
ψ(x, t) = e−iEN t/h̄jSince the spacing of the energy levels varies with index j, the frequencies inthis sum will not be precisely equal to integer multiples of ωN = (EN +1 −EN )/h̄, so after an approximate period TN = 2π/ωN most terms in theseries will not have quite returned to their values at t = 0. Consequently,
the constructive interference around x = X will be less sharply peaked thanit was at t = 0, and the cancellation elsewhere will be correspondingly lesscomplete. After each further approximate period TN , the failure of terms inthe series to return to their original values will be more marked, and the peakin |ψ(x, t)|2 will be wider. After a long time t ≫ TN the instants at which
individual terms next return to their original values will be pretty uniformly
distributed around an interval in t of length TN , and |ψ(x, t)|2 will cease to
evolve very much: it will have become a smooth function throughout therange |x| <∼ X.
This behaviour makes perfectly good sense classically. The uncertainty
in E that enables the wavefunction to be highly localised at t = 0 corresponds in the classical picture to uncertainty in the initial displacement X.
Since the period of an anharmonic oscillator is a function of the oscillator’senergy, uncertainty in X implies uncertainty in the oscillator’s period. Aftera long time even a small uncertainty in the period translates into a significant
uncertainty in the oscillator’s phase. Hence after a long time the probability
distribution for the particle’s position is fairly uniformly distributed within|x| ≤ X even in the classical case.

3.3 Motion in a magnetic field45

3.3 Motion in a magnetic field
The formalism we developed for a harmonic oscillator enables us to solve an
important, and you might have thought unconnected, problem: the motionof a particle of mass m and charge Q in a uniform magnetic field of flux
density B.
The first question to address when setting up the quantum-mechanicaltheory of a system is, “what’s the Hamiltonian?” because it is the Hamiltonian that encodes mathematically how the system works, including what
forces are acting. So we have to decide what the Hamiltonian should be fora particle of charge Q and mass m that moves in a magnetic field B(x). The
answer proves to be1H=(p − QA)2 ,(3.31)
2m
where A is the vector potential that generates B through B = ∇ × A. The
most persuasive theoretical motivation of this Hamiltonian involves relativity
and lies beyond the scope of this book. However, since we are exploring a new
and deeper level of physical theory, we can ultimately only proceed by making
conjectures and then confronting the resulting predictions with experimental
measurements. In this spirit we adopt equation (3.31) as a conjecture fromwhich we can try to recover the known behaviour of a charged particle ina magnetic field. In subsequent chapters we will show that this formulaaccounts satisfactorily for features in the spectra of atoms. Hence we can bepretty sure that it is correct.
Since we know the equations of motion of a classical particle in a field B,
let’s investigate the classical limit in the usual way, by finding the equationsof motion of expectation values. With equation (2.34) we have that the rateof change of the expectation value of the ith component of x isih̄

d hxi i1
[xi , (p − QA)2 ] .
= h[xi , H]i =dt
2m(3.32)The rules (2.22) and the canonical commutation relation [xi , pj ] = ih̄δijenable us to simplify the commutator
2mih̄

d hxi i= h[xi , (p − QA)] · (p − QA)idt
+ h(p − QA) · [xi , (p − QA)]i= 2ih̄ hpi − QAi i ,(3.33)

where we have used the fact that x commutes with A because A is a functionof x only. Thus, with this Hamiltonian
hpi = md hxi
+ Q hAi .
dt(3.34)

that is, the momentum is mẋ plus an amount proportional to the vector
potential. It is possible to show that the additional term represents themomentum of the magnetic field that arises because the charge Q is moving(Problem 3.19).
In the classical limit we can neglect the difference between a variable andits expectation value because all uncertainties are small. Then with (3.34)
the Hamiltonian (3.31) becomes just 21 mẋ2 , which makes perfect sense since
we know that the Lorentz force Qẋ × B does no work on a classical particle,
and the particle’s energy is just its kinetic energy.
To show that our proposed Hamiltonian generates the Lorentz force, we
evaluate the rate of change of hpi:
ih̄d hpi i1 = h[pi , H]i =h[pi , (p − QA)] · (p − QA)idt2m
+ h(p − QA) · [pi , (p − QA)]i .

(3.35)

46

Chapter 3: Harmonic oscillators and magnetic fields

We now use equation (2.25) to evaluate the commutator [pi , A] and conclude
that


 
∂A
∂A
Q
d hpi i.
(3.36)· (p − QA) + (p − QA) ·
=
dt
2m
∂xi
∂xi
Notice that we cannot combine the two terms on the right of this equation
because p does not commute with A and its derivatives. In the classicallimit we can replace each operator by its expectation value, and then replace
hp − QAi by m hẋi. Similarly replacing the p on the left, we have in theclassical limit
d2 xi
dAi
∂A
m 2 +Q· ẋ,(3.37)=Qdt
dt∂xi
where we have omitted the expectation value signs that ought to be aroundevery operator. The time derivative on the left is along the trajectory of theparticle (i.e., to be evaluated at hxit ). If A has no explicit time dependence
because the field B is static, its time derivative is just ẋ · ∇Ai . We move thisterm to the right side and have



d2 xi
∂A
m 2 = Q ẋ ·(3.38)− ẋ · ∇Ai , = Q ẋ × (∇ × A) i .
dt∂xi
Thus our proposed Hamiltonian (3.31) yields the Lorentz force in the classicallimit.

3.3.1 Gauge transformations
Any magnetic field is Gauge invariant: A and A′ = A + ∇Λ generate
identical magnetic fields, where Λ(x) is any scalar function. A potential
problem with the Hamiltonian (3.31) is that it changes in a non-trivial way
when we change gauge, which is worrying because H should embody thephysics, which is independent of gauge. We now show that this behaviour
gives rise to no physical difficulty providing we change the phases of all ketsat the same time that we change the gauge in which we write A. The ideathat a change of gauge in a field such as A that mediates a force (in thiscase the electromagnetic force) requires a compensating change in the ketthat is used to describe a given physical state, has enormously far-reachingramifications in field theory.
Suppose ψ(x) = hx|Ei is an eigenfunction of the Hamiltonian for A:

Then we show that(p − QA)2 |ψi = 2mE|ψi.

(3.39)

φ(x) ≡ eiQΛ/h̄ ψ(x)(3.40)

is an eigenfunction of the Hamiltonian we get by replacing A with A′ . We
start by noting thatp − QA′ = p − Q(A + ∇Λ) = (p − Q∇Λ) − QA
and that for any wavefunction χ(x)eiQΛ/h̄ pχ(x) = eiQΛ/h̄ − ih̄∇χ(x)= −ih̄∇ eiQΛ/h̄ χ − Q∇Λ(eiQΛ/h̄ χ)= (p − Q∇Λ) eiQΛ/h̄ χ(x) .

(3.41)

We subtract QAeiQΛ/h̄ χ from each side to obtain

eiQΛ/h̄ (p − QA)χ(x) = (p − QA − Q∇Λ) eiQΛ/h̄ χ(x) ,(3.42)

3.3 Motion in a magnetic field47

and then apply this result to χ ≡ (p − QA)ψ:eiQΛ/h̄ (p − QA)2 ψ(x) = (p − QA − Q∇Λ)eiQΛ/h̄ (p − QA)ψ(x)= (p − QA − Q∇Λ)2 eiQΛ/h̄ ψ(x) ,(3.43)

where the second equality uses (3.42) again, this time with χ put equal toψ. So if1Hψ =(p − QA)2 ψ = Eψ,(3.44)
2m
then



1
H ′ eiQΛ/h̄ ψ =(p − QA − Q∇Λ)2 eiQΛ/h̄ ψ(x) = E eiQΛ/h̄ ψ(x) ,2m(3.45)
In words, we can convert an eigenfunction of the Hamiltonian (3.31) with A
to an eigenfunction of that Hamiltonian with A′ = A + ∇Λ by multiplying
it by eiQΛ/h̄ . Notice that Λ is an arbitrary function of x, so multiplicationby eiQΛ/h̄ makes an entirely non-trivial change to ψ(x).
Given that there is a one-to-one relation between the eigenfunctions of
H before and after we make a gauge transformation, it is clear that the
spectrum of energy levels must be unchanged by the gauge transformation.
What about expectation values? Since both kets and the Hamiltonian undergo gauge transformations, we should be open to the possibility that otheroperators do too. Let R′ be the gauge transform of the operator R. Thenthe expectation value of R is gauge invariant ifZZ
hRi = d3 x ψ ∗ (x)Rψ(x) = d3 x ψ ∗ (x)e−iQΛ/h̄ R′ eiQΛ/h̄ ψ(x).
(3.46)
Clearly this condition is satisfied for R′ = R if R is a function of x only.
From our work above it is readily seen that if R depends on p, the equation
is satisfied if p only occurs through the combination (p − QA), as in the
Hamiltonian.5 We believe that in any physical situation this condition on
the occurrence of p will always be satisfied, so all expectation values are infact gauge-invariant.
3.3.2 Landau Levels
We now find the stationary states of a spinless particle that moves in a
uniform magnetic field. Let the z-axis be parallel to B and choose the gauge
in which A = 21 B(−y, x, 0). Then from equation (3.31) we haveo2
1 npx + 12 QBy + py − 21 QBx)2 + p2z
H=2m(3.47a)p2z
2
2
1,= 2 h̄ω(πx + πy ) +2m
where ω = QB/m is the Larmor frequency and we have defined the dimensionless operators6
πx ≡px + 12 mωy
√
mωh̄

;

πy ≡

py − 21 mωx
√
.
mωh̄

(3.47b)H has broken into two parts. The term p2z /2m is just the Hamiltonian of afree particle in one dimension – in §2.3.3 we already studied motion governedby this Hamiltonian. The part
Hxy ≡ 12 h̄ω(πx2 + πy2 )(3.47c)5 The principle that p and A only occur in the combination p − QA is known as theprinciple of minimal coupling.
6 We are implicitly assuming that QB and therefore ω are positive. It is this assumption that leads to the angular momentum of a gyrating particle never being positive – seeequation (3.58).

48

Chapter 3: Harmonic oscillators and magnetic fields

is essentially the Hamiltonian of a harmonic oscillator because it is the sum ofsquares of two Hermitian operators that satisfy the canonical commutation
relation1[y, py ] − [px , x] = i.
(3.48)[πx , πy ] =2h̄
The ladder operators are1
a = √ (πx + iπy )21a† = √ (πx − iπy )2⇒[a, a† ] = i[πy , πx ] = 1,

(3.49)and in terms of them Hxy is
Hxy = h̄ω(a† a + 21 ).

(3.50)

It follows that the energy levels are E = h̄ω( 21 , 32 , . . .). These discrete energy
levels for a charged particle in a uniform magnetic field are known as Landaulevels.
If particles can move freely parallel to B (which may not be possible incondensed-matter systems), the overall energy spectrum will be continuous
notwithstanding the existence of discrete Landau levels.
In the case of an electron the Larmor frequency is usually called thecyclotron frequency. It evaluates to 176(B/1 T) GHz, so the spacing of
the energy levels is 1.16 × 10−4 (B/1 T) eV. At room temperature electrons
have thermal energies of order 0.03 eV, so the discreteness of Landau levels
is usually experimentally significant in the laboratory only if the system iscooled to low temperatures and immersed in a strong magnetic field. The
strongest magnetic fields known occur near neutron stars, where B ∼ 108 T
is not uncommon, and in these systems electrons moving from one Landaulevel to the next emit or absorb hard X-ray photons.
To find the wavefunction of a given Landau level, we write the groundstate’s defining equation in the position representation
 

∂
∂  1
a|0i = 0 ↔
h̄
+ 2 mω(x + iy) hx|0i = 0.
+i
(3.51)∂x∂y
We transform to new coordinates u ≡ x + iy, v ≡ x − iy. The chain ruleyields7
 ∂
 ∂
∂
∂ 
∂ 
∂
= 12
−i
= 12
+i
;
,
(3.52)
∂u
∂x
∂y
∂v
∂x
∂y
so a and a† can be writtenu 
rB  ∂+ 2
a = −i √2 ∂v 4rB

where;

rB  ∂v a† = −i √− 2 ,2 ∂u 4rB

r

h̄
=
mω

s

(3.53a)h̄
.
QB(3.53b)1∂hx|0i+ 2 uhx|0i = 0.
∂v
4rB

(3.54)rB ≡Equation (3.51) now becomes

7 Aficianados of functions of a complex variable may ask what ∂/∂u can mean since thepartial derivative involves holding constant v, which appears to be the complex conjugate
of u. Use of u, v as independent coordinates requires permitting x, y to take on complex
values. If you are nervous of using this mathematical fiction to solve differential equations,
you should check that the wavefunction of equation (3.58) really is an eigenfunction of H.

3.3 Motion in a magnetic field

49

Solving this first-order linear o.d.e. we find2

hx|0i = g(u) e−uv/4rB(3.55)

where g(u) is an arbitrary function. On account of the arbitrariness of g(u),the ground state of motion in a magnetic field is not unique. This situation
contrasts with the one we encountered when solving for the ground state ofa harmonic oscillator. We obtain the simplest ground state by taking g to
be a suitable normalising constant C – we’ll consider more elaborate choices
below. Our present ground-state wavefunction is22

2

hx|0i = C e−(x +y )/4rB .

(3.56)In classical physics a particle that moves at speed v perpendicular
to a√uniform
magnetic
field
moves
in
circlesofradiusr=
mv/QB=
2mE/QB=p
2E/mω 2 . When E = 21 h̄ω this radius agrees with the dispersion rB in
radius of the Gaussian probability distribution |hx|0i|2 that we have justderived.
A wavefunction in the first excited level ishx|1i ∝ hx|a† |0i ∝

 ∂

∂u

−

2
v  −uv/4rB2∝ ve−uv/4rB .
2 e
4rB(3.57)

It is easy to see that each further application of a† will introduce an additional
power of v, so that we have
2

2

2

2

hx|ni ∝ v n e−uv/4rB = (x − iy)n e−(x +y )/4rB .

(3.58)We shall see in §7.2.3 that the factor (x − iy)n implies that the particle has nunits of angular momentum about the origin.8 We can also show from thisformula that for large n the expectation of the orbital radius increases as
the square root of the energy, in agreement with classical mechanics (Problem 3.20).
Displacement of the gyrocentre A particle in the state (3.58) gyrates
around the origin of the xy plane. Since the underlying physics (unlike the
Hamiltonian 3.31) is invariant under displacements within the xy plane, there
must be a ground-state ket in which the particle gyrates around any given
point. Hence, every energy level associated with motion in a uniform magnetic field is highly degenerate: it has more than one linearly independent
eigenket.
It was our choice of magnetic vector potential A that made the origin
have a special status in H: the potential we used can be written A = 21 B×x.
The choice Λ = − 21 x · (B × a), where a is any vector, makes the gaugetransformation from A to A′ = A − 21 B × a, so if A = 12 B × x, thenA′ = 12 B × (x − a). If we replace A in H with A′ , it will prove expedient
to redefine πx , πy such that the wavefunctions that are generated by theprocedure we used before will describe a particle that gyrates around x = ainstead of the origin. Thus in the gauge A′ , the wavefunction of a groundstate particle that gyrates around x = a is2

2

hx|0′ , ai = C e−(x−a) /4rB .

(3.59)

We can use the theory of gauge transformations that we derived in §3.3.1 totransform this back to our original gauge A. The result is2

2

hx|0, ai = C eiQ(B×a)·x/2h̄ e−(x−a) /4rB .

(3.60)

8 This statement follows because in spherical polar coordinates (x − iy)n = r n e−inφ .

50

Chapter 3: Harmonic oscillators and magnetic fields

This procedure is easily generalised to the determination of the wavefunctionof the nth Landau level for gyration about x = a.
A complete set of mutually orthogonal stationary states is needed if wewant to expand a general state of motion in a magnetic field as a linear
combination of stationary states. Wavefunctions such as (3.56) and (3.60)
that differ only in their gyrocentres are not orthogonal, so it is not convenient
to combine them in a set of basis states. To obtain a complete set of mutuallyorthogonal states we can either return to equation (3.55) and set g(u) =u, u2 , . . ., etc., or we can step still further back to equations (3.47) andnote that we started with four operators, x, y, px and py , but expressed the
Hamiltonian Hxy in terms of just two operators πx and πy , which we thenpackaged into the ladder operators a and a† .
Consider the operators
ξx ≡px − 12 mωy
√
mωh̄

;

ξy ≡

py + 21 mωx
√
.
mωh̄

(3.61)

They differ from the operators πx and πy defined by equations (3.47b) only
in a sign each, and they commute with them. For example
1[px − 21 mωy, px + 12 mωy] = 0mωh̄(3.62)1[ξx , πy ] =[px − 21 mωy, py − 21 mωx] = 0.
mωh̄Consequently they commute with Hxy . On the other hand, [ξx , ξy ] = −i, sofrom these operators we can construct the ladder operators[ξx , πx ] =1
b = √ (ξx − iξy )21b† = √ (ξx + iξy )2⇒[b, b† ] = i[ξx , ξy ] = 1.

(3.63)

Since these ladder operators commute with Hxy , we can find a complete setof mutual eigenkets of b† b and Hxy .
In the position representation the new ladder operators arerB  ∂v u 
rB  ∂
b = −i √+ 2− 2 .
; b† = −i √(3.64)
2 ∂u 4rB
2 ∂v
4rB
2

When we apply b to the ground-state wavefunction hx|0i = Ce−uv/4rB(eq. 3.56), we find2v 
CrB  ∂
(3.65)+ 2 e−uv/4rB = 0.
bhx|0i = −i √2 ∂u 4rB
2

Thus Ce−uv/4rB is annihilated by both a and b. When we apply b† to thiswavefunction we obtain
2
2
CrB  ∂u b† hx|0i = −i √(3.66)− 2 e−uv/4rB ∝ u e−uv/4rB ,2 ∂v 4rB

which is the wavefunction we would have obtained if we had set g(u) = u inequation (3.55). In fact it’s clear from equation (3.66) that every applicationof b† will introduce an additional factor u before the exponential. Therefore
the series of ground-state wavefunctions that are obtained by repeatedly
2
applying b† to e−uv/4rB are all of the form2(b† )n hx|0i ∝ un e−uv/4rB .

(3.67)The only difference between this general ground-state wavefunction9 and the
wavefunction of the nth excited Landau level (eq. 3.57) is that the former has
un rather than v n in front of the exponential. For the physical explanationof this result, see Problem 3.24.
9 The absolute value of the real part of this is shown for n = 4 on the front cover.

3.3 Motion in a magnetic field

51

Figure 3.4 The Aharonov–Bohm experiment3.3.3 Aharonov-Bohm effectImagine a very long, thin solenoid that runs parallel to the z axis. There isa strong magnetic field inside the solenoid, but because the solenoid is long,the field lines are extremely thinly spread as they return from the solenoid’s
north pole to its south pole, and outside the solenoid the magnetic field is
negligibly small. In the limit that the solenoid becomes infinitely thin, a
suitable vector potential for the field isA=Φ  y x 
− 2, 2,0 ,2π
r r(3.68)

pwhere r = x2 + y 2 and Φ is the magnetic flux through the solenoid. To
justify this statement we note that when we integrate A around a circleof radius r, the integral evaluates to Φ independent of r. But by Stokes’
theoremIZZ
2dx · A = d x · ∇ × A = d2 x · B.
(3.69)
Thus Φ units of flux run along the axis r = 0, and there is no flux anywhereelse.
Now we place a screen with two slits in the plane y = 0, with the slits
distance 2s apart and running parallel to the solenoid and on either side ofit. We bombard the screen from y < 0 with particles that have well defined
momentum p = pj parallel to the y axis, and we detect the arrival of theparticles on a screen P that lies in the plane y = L – apart from the presenceof the solenoid, the arrangement is identical to that of the standard twoslit experiment of §2.3.4. Classical physics predicts that the particles areunaffected by B since they never enter the region of non-zero B. Aharonov& Bohm pointed out10 that the prediction of quantum mechanics is different.
Consider the functionΦΛ = − θ,(3.70)
2π
where θ is the usual polar angle in the xy plane. Since θ = arctan(y/x),
y
∂θ
=− 2
∂x
r

and

∂θ
x
= 2,
∂y
r

(3.71)and the gradient of Λ is∇Λ = −

Φ(−y, x, 0)
2πr2(3.72)

which is minus the vector potential A of equation (3.68). So let’s make agauge transformation from A to A′ = A + ∇Λ. In this gauge, the vector
potential vanishes, so the Hamiltonian is just that of a free particle, p2 /2m.
Hence the analysis of §2.3.4 applies, and the amplitude to pass through a
10 Y. Aharonov & D. Bohm, Phys. Rev. 115, 485 (1959)

52

Problems

given slit and arrive at a point on the screen P with coordinate x has a phaseφ that is proportional to x (cf eq. 2.71):
φi = constant ±

psx
,
h̄L

(3.73)

where the plus sign applies for one slit and the minus sign for the other.
Our choice of gauge leads to a tricky detail, however. We require Λ to be
single valued, so we must restrict the polar angle θ to a range 2π in extent.
Consequently, θ and Λ must be somewhere discontinuous. We get aroundthis problem by using different forms of Λ, and therefore different gauges, toderive the amplitudes for arrival at x from each slit. For slit S1 at x = +s,
we take −π < θ ≤ π, and for S2 at x = −s we take −2π < θ ≤ 0. With thesechoices the discontinuity in Λ occurs where the electron does not go, and Λ isalways the same in the region y < 0 occupied by the incoming electron beam.
Consequently, the amplitudes for arrival at a point x on the screen P are thesame as if the solenoid were not there. However, before we can add theamplitudes and calculate the interference pattern, we have to transform to acommon gauge. The easiest way to do this is to transform the amplitude for
S2 to the gauge of S1 . The function that effects the transformation between
the gauges is ∆ ≡ Λ1 − Λ2 , where Λi is the gauge function used for slit Si . At
any point of P the two forms of θ differ by 2π, so ∆ = −Φ. Therefore equation(3.40) requires us to multiplying the amplitude for S2 by exp(−iQΦ/h̄), andthe quantum interference term (1.15) becomes
constant × ei(φ1 −φ2 +QΦ/h̄) ∝ exp 

i 2psx+ QΦ.
h̄L(3.74)The term QΦ in the exponential shifts the centre of the interference patternby an amount ∆x = −LQΦ/2ps, so by switching the current in the solenoidon and off you can change the interference pattern that is generated by particles that never enter the region to which B is confined. This prediction was
first confirmed experimentally by R.G. Chambers.11 Although this effect has
no counterpart in classical mechanics, curiously the shift ∆x is independentof h and does not vanish in the limit h → 0, which is often regarded as theclassical limit.

Problems
3.1 After choosing units in which everything, including h̄ = 1, the Hamiltonian of a harmonic oscillator may be written H = 21 (p2 +x2 ), where [x, p] = i.
Show that if |ψi is a ket that satisfies H|ψi = E|ψi, then2
1 2
2 (p + x )(x ∓ ip)|ψi = (E ± 1)(x ∓ ip)|ψi.

(3.75)

Explain how this algebra enables one to determine the energy eigenvalues ofa harmonic oscillator.
3.2 Given that A|En i = α|En−1 i and En = (n + 12 )h̄ω, where the annihilation operator of the harmonic oscillator ismωx + ip
A≡ √
,
2mh̄ω
show that α =(3.76)√
n. Hint: consider |A|En i|2 .

3.3 The pendulum of a grandfather clock has a period of 1 s and makes
excursions of 3 cm either side of dead centre. Given that the bob weighs
0.2 kg, around what value of n would you expect its non-negligible quantum
amplitudes to cluster?
11 Phys. Rev. Lett. 5, 3 (1960)

Problems

53

Figure 3.5 The wavefunctions hx|2i
and hx|40i of two stationary states ofa harmonic oscillator.

3.4 Show that the minimum value of E(p, x) ≡ p2 /2m + 12 mω 2 x2 withrespect to the real numbers p, x when they are constrained to satisfy xp = 12 h̄,is 21 h̄ω. Explain the physical significance of this result.
3.5 How many nodes are there in the wavefunction hx|ni of the nth excitedstate of a harmonic oscillator?
3.6 Show that in terms of a harmonic oscillator’s characteristic length ℓ ≡ph̄/2mω the ladder operators can be written

∂
x
∂
x
+ℓ
and A† =−ℓ .
(3.77)2ℓ∂x
2ℓ∂x
Hence show that the wavefunction of the second excited state is hx|2i =22constant × (x2 /ℓ2 − 1)e−x /4ℓ and find the normalising constant.
A=

3.7 Explain why the wavefunction hx|ni of the oscillator’s nth stationary
state must have the form22

hx|ni = Hn (x/ℓ) e−x /4ℓ ,th(3.78)

where H√n is an n -order (‘Hermite’)√ polynomial. By casting the equations
A|ni = n|n − 1i and A† |n − 1i = n|ni in the x-representation, show that√√x′Hn′ (x/ℓ) = nHn−1 (x/ℓ) andnHn (x/ℓ) = Hn−1 (x/ℓ) − Hn−1(x/ℓ).
ℓ(3.79)and thus that√√xnHn (x/ℓ) = Hn−1 (x/ℓ) − n − 1Hn−2 (x/ℓ).
(3.80)ℓGiven that H0 = (2πℓ2 )−1/4 and H1 (y) = y/(2πℓ2 )1/4 , use this recurrence
relation to reproduce the plots of the wavefunctions hx|2i and hx|40i shownin Figure 3.5. Explain the physical significance of the vertical arrows. Whyis the amplitude of hx|40i largest near the right arrow?
3.8 Userh̄(A + A† ) = ℓ(A + A† )(3.81)
2mω
to show for a harmonic oscillator that in the energy representation the operator x is
√


1 √0
0
...
√0
2 √0
 1 √0



2 √0
3
···
 0



3 ...




...
...
.
.
.
 ...

√
xjk = ℓ ... √ 0
n−1
.√. .




n−1
0
n


√
√


n
0
n
+
1
·
·
·


√


n+1
0
···
···
···
···
···
(3.82)
Calculate the same entries for the matrix pjk .
x=

54

Problems

3.9 Show that the momentum operator of a harmonic oscillator can be
expressed in terms of the creation and annihilation operators asih̄
p = (A† − A) where2ℓHence show that2h0|p |0i =



h̄
2ℓ

ℓ≡
2

r

h̄
.
2mω

(3.83).

(3.84)How does this result relate to the physics of a free particle discussed in §2.3.3?
3.10 At t = 0 the state of a harmonic oscillator, mass m frequency ω, is11|ψi = √ |N − 1i + √ |N i.
22(3.85)Show that subsequently√
hxit = N ℓ cos(ωt) where

ℓ≡

r

h̄
.
2mω

(3.86)

Interpret this result physically. What does this example teach us about thevalidity of classical mechanics?
Show that a classical oscillator with energy (N + 21 )h̄ω has amplitude
xmax = 2

q
N + 12 ℓ.

(3.87)To explain the discrepancy between these results, consider the case inwhich initially
N +K−1
X
1
√
|ψi =
|ki
(3.88)
K
k=N√
with N ≫ K ≫ 1. Show that then hxit ≃ 2 N ℓ cos(ωt) consistent withclassical physics.
3.11∗ By expressing the annihilation operator A of the harmonic oscillator
in the momentum representation, obtain hp|0i. Check that your expressionagrees with that obtained from the Fourier transform of
hx|0i =221
e−x /4ℓ ,(2πℓ2 )1/4

where ℓ ≡

r

h̄
.
2mω

(3.89)

3.12 Show that for any two N × N matrices A, B, trace([A, B]) = 0. Comment on this result in the light of the results of Problem 3.8 and the canonicalcommutation relation [x, p] = ih̄.
3.13∗ A Fermi oscillator has Hamiltonian H = f † f , where f is an operator that satisfies
f2 = 0;f f † + f † f = 1.

(3.90)Show that H 2 = H, and thus find the eigenvalues of H. If the ket |0i
satisfies H|0i = 0 with h0|0i = 1, what are the kets (a) |ai ≡ f |0i, and (b)|bi ≡ f † |0i?
In quantum field theory the vacuum is pictured as an assembly of oscillators, one for each possible value of the momentum of each particle type.
A boson is an excitation of a harmonic oscillator, while a fermion in an excitation of a Fermi oscillator. Explain the connection between the spectrumof f † f and the Pauli principle.

Problems

55

3.14 In the time interval (t + δt, t) the Hamiltonian H of some systemvaries in such a way that |H|ψi| remains finite. Show that under thesecircumstances |ψi is a continuous function of time.
A harmonic oscillator with frequency ω is in its ground state when the
stiffness of the spring is instantaneously reduced by a factor f 4 < 1, so its
natural frequency becomes f 2 ω. What is the probability that the oscillator
is subsequently found to have energy 23 h̄f 2 ω? Discuss the classical analogueof this problem.
3.15∗ P is the probability that at the end of the experiment described in
Problem 3.14, the oscillator is in its second excited state. Show that whenf = 21 , P = 0.144 as follows. First show that the annihilation operator of
the original oscillator
A = 12 (f −1 + f )A′ + (f −1 − f )A′† ,(3.91)

where A′ and A′† are the annihilation and creation operators of the final
oscillator. P
Then writing the ground-state ket of the original oscillator as asum |0i = n cn |n′ i over the energy eigenkets of the final oscillator, impose
the condition A|0i = 0. Finally use the normalisation of |0i and the orthogonality of the |n′ i. What value do you get for the probability of the oscillator
remaining in the ground state?
Show that at the end of the experiment the expectation value of theenergy is 0.2656h̄ω. Explain physically why this is less than the original
ground-state energy 12 h̄ω.
This example contains the physics behind the inflationary origin of the
Universe: gravity explosively enlarges the vacuum, which is an infinite collection of harmonic oscillators (Problem 3.13). Excitations of these oscillators
correspond to elementary particles. Before inflation the vacuum is unexcited
so every oscillator is in its ground state. At the end of inflation, there is nonnegligible probability of many oscillators being excited and each excitation
implies the existence of a newly created particle.
3.16∗ In terms of the usual ladder operators A, A† , a Hamiltonian can be
writtenH = µA† A + λ(A + A† ).
(3.92)
What restrictions on the values of the numbers µ and λ follow from therequirement for H to be Hermitian?
Show that for a suitably chosen operator B, H can be rewritten
H = µB † B + constant.

(3.93)

where [B, B † ] = 1. Hence determine the spectrum of H.
3.17∗ Numerically calculate the spectrum of the anharmonic oscillator shownin Figure 3.2. From it estimate the period at a sequence of energies. Compare
your quantum results with the equivalent classical results.
3.18∗ Let B = cA+sA† , where c ≡ cosh θ, s ≡ sinh θ with θ a real constantand A, A† are the usual ladder operators. Show that [B, B † ] = 1.
Consider the HamiltonianH = ǫA† A + 21 λ(A† A† + AA),

(3.94)

where ǫ and λ are real and such that ǫ > λ > 0. Show that whenǫc − λs = Ec;

λc − ǫs = Es

(3.95)

with E a constant, [B, H] = EB. Hence determine the spectrum of H in
terms of ǫ and λ.

56

Problems

3.19 This problem is all classical electromagnetism, but it gives physicalinsight into quantum physics. It is hard to do without a command of Cartesian tensor notation (Appendix B). A point charge Q is placed at the origin
in the magnetic field generated by a spatially confined current distribution.
Given that
Q r
E=
(3.96)
4πǫ0 r3and B = ∇ × A with ∇ · A = 0, show that the field’s momentumZP ≡ ǫ0 d3 x E × B = QA(0).

(3.97)

Write down the relation between the particle’s position and momentum andinterpret this relation physically in light of the result you have just obtained.
Hint: write E = −(Q/4πǫ0)∇r−1 and B = ∇ × A, expand the vector
triple product and integrate each of the resulting terms by parts so as to
2 −1
3exploit in one ∇ · A = 0 and in the other(r). The tensor
R 3 ∇ r =H −4πδ
2form of Gauss’s theorem states that d x ∇i T = d Si T no matter how
many indices the tensor T may carry.
3.20 From equation (3.58) show that the the normalised wavefunction ofa particle of mass m that is in the nth Landau level of a uniform magnetic
field B is22
rn e−r /4rB e−inφ√hx|ni =,
(3.98)
n+1
2(n+1)/2 n! π rB
pwhere rB = h̄/QB. Hence show that the expectation of the particle’s
gyration radius ishrin ≡ hn|r|ni =Show further that

√ (n + 21 )!
2
rB .
n!

(3.99)

δ ln hrin
1
≃
δn
2n

(3.100)√and thus show that in the limit of large n, hri ∝ E, where E is the energyof the level. Show that this result is in accordance with the correspondence
principle.
3.21 Show that in the gauge in which the magnetic vector potential is
A = 21 B × x the wavefunction of the nth Landau level of gyration about the
point a is22

hx|n, ai = eiQ(B×a)·x/2h̄ {(x − ax ) − i(y − ay )}n e−|x−a| /4rB .

(3.101)

3.22 A particle of charge Q is confined to move in the xy plane, withelectrostatic potential φ = 0 and vector potential A satisfying∇ × A = (0, 0, B).

(3.102)

Consider the operators ρx , ρy , Rx and Ry , defined byρ=1êz × (p − QA)QBand

R = r − ρ,(3.103)where r and p are the usual position and momentum operators, and êz is
the unit vector along B. Show that the only non-zero commutators formed
from the x- and y-components of these are2[ρx , ρy ] = irB

and2[Rx , Ry ] = −irB,

(3.104)

Problems

57

2
where rB
= h̄/QB.
The operators a, a† , b and b† are defined via

1a= √(ρx + iρy )
2 rB

and

1
b= √
(Ry + iRx ).
2 rB(3.105)

Evaluate [a, a†] and [b, b† ]. Show that for suitably defined ω, the Hamiltonian
can be written
H = h̄ω a† a + 21 .
(3.106)Given that there exists a unique state |ψi satisfying
a|ψi = b|ψi = 0,

(3.107)

what conclusions can be drawn about the allowed energies of the Hamiltonian
and their degeneracies? What is the physical interpretation of these results?
3.23 Using cylindrical polar coordinates (R, φ, z), show that the probability
current density associated with the wavefunction (3.98) of the nth Landaulevel is22 h̄R2n−1 e−R /2rB
R2
J(R) = − n+1êφ ,(3.108)
n
+
2
2n+2
2rB
2
πn!mrBp
where rB ≡ h̄/QB. Plot J as a function of R and interpret your plot
physically.

3.24 Determine the probability current density associated with the nth
Landau ground-state wavefunction (3.67) (which for n = 4 is shown in Figure Landaufig). Use your result to explain in as much detail as you can why
this state can be interpreted as a superposition of states in which the electron
gyrates around different gyrocentres. Hint: adapt equation (3.108).
Why is the energy of a gyrating electron incremented if we multiply the2wavefunction e−(mω/4h̄)r by v n = (x − iy)n but not if we multiply it bynnu = (x + iy) ?

4Transformations & Observables
In §2.1 we associated an operator with every observable quantity througha sum over all states in which the system has a well-defined value of theobservable (eq. 2.5). We found that this operator enabled us to calculate
the expectation value of any function of the observable. Moreover, from theoperator we could recover the observable’s allowed values and the associated states because they are the operator’s eigenvalues and eigenkets. These
properties make an observable’s operator a useful repository of information
about the observable, a handy filing system. But they do not give the operator much physical meaning. Above all, they don’t answer the question ‘what
does an operator actually do when it operates?’ In this chapter we answer
this question. In the process of doing this, we will see why the canonicalcommutation relations (2.54) have the form that they do, and introduce theangular-momentum operators, which will play important roles in the rest of
the book.

4.1 Transforming kets
When one meets an unfamiliar object, one may study it by moving it around,perhaps turning it over in one’s hands so as to learn about its shape. In §1.3
we claimed that all physical information about any system is encapsulated
in its ket |ψi, so we must learn how |ψi changes as we move and turn thesystem.
Even the simplest systems can have orientations in addition to positions. For example, an electron, a lithium nucleus or a water molecule all
have orientations because they are not spherically symmetric: an electron
is a magnetic dipole, a 7 Li nucleus has an electric quadrupole, and a watermolecule is a V-shaped thing. The ket |ψi that describes any of these objects
contains information about the object’s orientation in addition to its position
and momentum. In the next subsection we shall focus on the location of aquantum system, but later we shall be concerned with its orientation as well,and in preparation for that work we explicitly display a label µ of the system’s orientation and any other relevant properties, such as internal energy.
For the moment µ is just an abstract symbol for orientation information; thedetails will be fleshed out in §7.1.

4.1 Transforming kets

59

Figure 4.1 A spherical wavefunction
and its displaced version.

4.1.1 Translating kets
We now focus on the location of our system. To keep track of this we use acoordinate system Σ0 whose origin is some well-defined point, say the centreof our laboratory. We can investigate |ψi by expanding it in terms of acomplete set of eigenstates |x, µi, where x is the position vector of the centreof mass and µ represents the system’s orientation. The amplitude for finding
the system’s centre of mass at x with the orientation specified by µ is
ψµ (x) ≡ hx, µ|ψi.

(4.1)

If we know all the derivatives of the wavefunction ψµ at a position x, Taylor’s
theorem gives the value of the wavefunction at some other location x − a as
"

#

2
∂
1
∂
ψµ (x − a) = 1 − a ·− . . . ψµ (x)
+
a·
∂x 2!
∂x


∂
ψµ (x)= exp −a ·∂x
 a · p= hx, µ| exp −i
|ψi.
h̄

(4.2)

This equation tells us that in the state |ψi, the amplitude to find the system
at x − a with orientation etc µ is the same as the amplitude to find thesystem with unchanged orientation at x when it is a different state, namely|ψ ′ i ≡ U (a)|ψi

where

U (a) ≡ exp(−ia · p/h̄).

(4.3)

In this notation, equation (4.2) becomesψµ (x − a) = hx, µ|U (a)|ψi = hx, µ|ψ ′ i = ψµ′ (x),(4.4)

so, as Figure 4.1 illustrates, the wavefunction ψµ′ for |ψ ′ i is the wavefunction
we would expect for a system that is identical with the one described by|ψi except for being shifted along the vector a. We shall refer to this new
system as the translated or transformed system and we shall say that thetranslation operator operator U (a) translates |ψi through a even though|ψi is not an object in real space, so this is a slight abuse of language.
The ket |ψ ′ i of the translated system is a function of the vector a. It
is instructive to take its partial derivative with respect to one component ofa, say ax . Evaluating the resulting derivative at a = 0, when |ψ ′ i = |ψi, we
find
∂|ψi
∂|ψi
ih̄
= −ih̄
= px |ψi.
(4.5)
∂ax
∂x
Thus the operator px gives the rate at which the system’s ket changes as
we translate the system along the x axis. So we have answered the question

60

Chapter 4: Transformations & Observables

Box 4.1:

Passive transformations

We can describe objects such as atoms equally well using any coordinate
system. Imagine a whole family of coordinate systems set up throughout
space, such that every physical point is at the origin of one coordinate
system. We label by Σy the coordinate system whose origin coincides
with the point labelled by y in our original coordinate system Σ0 , and we
indicate the coordinate system used to obtain a wavefunction by making
y a second argument of the wavefunction; ψµ (x; y) is the amplitude tofind the system at the point labelled x in Σy . Because the different
coordinate systems vary smoothly with y, we can use Taylor’s theorem
to express amplitudes in, say, Σa+y in terms of amplitudes in Σy . We
have


∂
ψµ (x; a + y) = exp a ·ψµ (x; y).
(1)∂yNow ψµ (x; a) = ψµ (x+a; 0) because both expressions give the amplitude
for the system to be at the same physical location, the point called x in
Σa and called x + a in Σ0 . Then equations (4.4) and (1) givehx, µ; 0|U (−a)|ψi = ψµ (x + a; 0) = ψµ (x; a),(2)where again |x, µ; 0i indicates the state in which the system is locatedat the point labelled by x in Σ0 . This equation tells us that |ψi has
the same wavefunction in Σa that |ψ̄i ≡ U (−a)|ψi has in Σ0 . Therefore, moving the origin of our coordinates through a vector a has thesame effect on an arbitrary state’s wavefunction as moving the systemitself through −a. Physically moving the system is known as an active
transformation, whereas leaving the state alone but changing the coordinate system is called a passive transformation. The infinitesimalvectors required to make logically equivalent active and passive transformations differ in sign. This sign difference reflects the fact that if you
move backwards, the world around you seems to move forwards; hence
moving the origin of one’s coordinates back by δa has the same effect as
moving the system forward by δa. In this book we confine ourselves toactive transformations.
posed above as to what an observable’s operator actually does in the case of
the momentum operators.
Equation (2.78) enables us to expand a state of well-defined position x0
in terms of momentum eigenstates. We have
Z
Z
1
(4.6)
|x0 , µi = d3 p |p, µihp, µ|x0 , µi = 3/2 d3 p e−ix0 ·p/h̄ |p, µi.
hApplying the translation operator, we obtain with (4.3)
Z
1
U (a)|x0 , µi = 3/2 d3 p e−ix0 ·p/h̄ U (a)|p, µihZ1= 3/2 d3 p e−i(x0 +a)·p/h̄ |p, µih= |x0 + a, µi,(4.7)

which is a new state, in which the system is definitely located at x0 + a, as
we would expect.
4.1.2 Continuous transformations and generators
In §1.3 we saw that the normalisation condition hψ|ψi = 1 expresses thefact that when we make a measurement of any observable, we will measure4.1 Transforming kets61

Box 4.2: Operators from expectation values
In this box we show that ifhψ|A|ψi = hψ|B|ψi,(1)

for every state |ψi, then the operators A and B are identical. We set|ψi = |φi+λ|χi, where λ is a complex number. Then equation (1) implies
λ (hφ|A|χi − hφ|B|χi) = λ∗ (hχ|B|φi − hχ|A|φi) .

(2)

Since equation (1) is valid for any state |ψi, equation (2) remains valid as
we vary λ. If the coefficients of λ and λ∗ are non-zero, we can cause theleft and right sides of (2) to change differently by varying the phase of
λ; they can be equal irrespective of the phase of λ only if the coefficientsvanish. This shows that hχ|A|φi = hχ|B|φi for for arbitrary states |φiand |χi, from which it follows that A = B.
some value; for example, if we determine the system’s location, we will findit somewhere. The normalisation condition must be unaffected by any transformation that we make on a system, so the transformation operator1 U
must have the property that for any state |ψi
1 = hψ ′ |ψ ′ i = hψ|U † U |ψi.

(4.8)

From this requirement we can infer by the argument given in Box 4.2 (withA = U † U and B = I, the identity operator) that U † U = I, so U † = U −1 .
Operators with this property are called unitary operators. When we transform all states with a unitary operator, we leave unchanged all amplitudes:hφ′ |ψ ′ i = hφ|ψi for any states |φi and |ψi.
Exactly how we construct a unitary operator depends on the type oftransformation we wish it to make. The identity operator is the unitary
operator that represents doing nothing to our system. The translation operator U (a) can be made to approach the identity as closely as we please bydiminishing the magnitude of a. Many other unitary operators also have aparameter θ that can be reduced to zero such that the operator tends to theidentity. In this case we can write for small δθU (δθ) = I − iδθ τ + O(δθ)2 ,(4.9)

where the factor of i is a matter of convention and τ is an operator. The
unitarity of U implies that
I = U † (δθ)U (δθ) = I + iδθ (τ † − τ ) + O(δθ2 ).

(4.10)

Equating powers of δθ on the two sides of the equation, we deduce that τ
is Hermitian, so it may be an observable. If so, its eigenkets are states inwhich the system has well-defined values of the observable τ .
We obtain an important equation by using equation (4.9) to evaluate|ψ ′ i ≡ U (δθ)|ψi. Subtracting |ψi from both sides of the resulting equation,
dividing through by δθ and proceeding to the limit δθ → 0, we obtaini∂|ψ ′ i= τ |ψ ′ i.
∂θ

(4.11)

Thus the observable τ gives the rate at which |ψi changes when we increase
the parameter θ in the unitary transformation that τ generates. Equation(4.5) is a concrete example of this equation in action.
1 We restrict ourselves to the case in which the operator U is linear, as is every operator
used in this book. In consequence, we are unable to consider time reversal.

62

Chapter 4: Transformations & Observables

A finite transformation can be generated by repeatedly performing an
infinitesimal one. Specifically, if we transform N times with U (δθ) withδθ = θ/N , then in the limit N → ∞ we haveNθU (θ) ≡ lim 1 − i τ= e−iθτ .
N →∞N(4.12)

This relation is clearly a generalisation of the definition (4.3) of the translation operator. The Hermitian operator τ is called the generator of boththe unitary operator U and the transformations that U accomplishes; forexample, p/h̄ is the generator of translations.
4.1.3 The rotation operator
Consider what happens if we rotate the system. Whereas in §4.1.1 we constructed a state |ψ ′ i = U (a)|ψi that differed from the state |ψi only in ashift by a in the location of the centre of mass, we now wish to find a rotation operator that constructs the state |ψ ′ i that we would get if we could
somehow rotate the apparatus on a turntable without disturbing its internal
structure in any way. Whereas the orientation of the system is unaffectedby a translation, it will be changed by the rotation operator, as is physicallyevident if we imagine turning a non-spherical object on a turntable.
From §4.1.2 we know that a rotation operator will be unitary, and have a
Hermitian generator. Actually, we expect there to be several generators, justas there are three generators, px /h̄, py /h̄ and pz /h̄, of translations. Because
there are three generators of translations, three numbers, the components ofthe vector a in equation (4.3), are required to specify a particular translation.
Hence we anticipate that the number of generators of rotations will equal
the number of angles that are required to specify a rotation. Two angles are
required to specify the axis of rotation, and a third is required to specify
the angle through which we rotate. Thus by analogy with equation (4.3), we
expect that a general rotation operator can be obtained by exponentiating
a linear combination of three generators of rotations, and we write
U (α) = exp(−iα · J).

(4.13)

Here α is a vector that specifies a rotation through an angle |α| aroundthe direction of the unit vector α̂, and J is comprised of three Hermitian
operators, Jx , Jy and Jz . In the course of this chapter and the next it will
become clear that the observable associated with J is angular momentum.
Consequently, the components of J are called the angular-momentum operators.
The role that the angular momentum operators play in rotating thesystem around the axis α̂ is expressed by rewriting equation (4.11) withappropriate substitutions asi∂|ψi= α̂ · J|ψi.
∂α(4.14)

4.1.4 Discrete transformations(a) The parity operator
Not all transformations are continuous. In
physics, the most prominent example of a discrete transformation is theparity transformation P, which swaps the sign of the coordinates of all
spatial points; the action of P on coordinates is represented by the matrix


−1 0
0
P ≡  0 −1 0  so Px = −x.
(4.15)
00 −1

4.1 Transforming kets

63

Notice that det P = −1, whereas a rotation matrix has det R = +1. In fact,
any linear transformation with determinant equal to −1 can be written as a
product of P and a rotation R.
Let an arbitrary quantum state |ψi have wavefunction ψµ (x) = hx, µ|ψi,
where the label µ is the usual shorthand for the system’s orientation. Thenthe quantum parity operator P is defined byψµ′ (x) ≡ hx, µ|P |ψi ≡ ψµ (Px) = ψµ (−x) = h−x, µ|ψi.

(4.16)

The wavefunction of the new state, |ψ ′ i = P |ψi, takes the same value at xthat the old wavefunction does at −x. Thus, when the system is in the state
P |ψi, it has the same amplitude to be at x as it had to be at −x when it wasin the state |ψi. The orientation and internal properties of the system areunaffected by P . The invariance of orientation under a parity transformation
is not self evident, but in §4.2 we shall see that it follows from the rules thatgovern commutation of P with x and J.
Applying the parity operator twice creates a state |ψ ′′ i = P |ψ ′ i = P 2 |ψiwith wavefunction
ψµ′′ (x) = hx, µ|P |ψ ′ i = h−x, µ|ψ ′ i = h−x, µ|P |ψi = hx, µ|ψi
= ψµ (x).

(4.17)

Hence P 2 = 1 and an even number of applications of the parity operator
leaves the wavefunction unchanged. It also follows that P = P −1 is its own
inverse.
P is also Hermitian:Z
X
hφ|P |ψi∗ = d3 x
(hφ|x, µihx, µ|P |ψi)∗
=
=
=

Z

Z

Z

µ

d3 x

X

(hφ|x, µih−x, µ|ψi)∗

µ

d3 x

X
µ

d3 x

X
µ

(4.18)
hψ| − x, µihx, µ|P 2 |φihψ| − x, µih−x, µ|P |φi = hψ|P |φi,so P † = P . It now follows that P is unitary2 because P −1 = P = P † . Hencefrom the discussion of §4.1.2 it follows that transforming all states with P
will preserve all amplitudes for the system.
Suppose now that |P i is an eigenket of P , with eigenvalue λ. Then
|P i = P 2 |P i = λP |P i = λ2 |P i, so λ2 = 1. Thus the eigenvalues of P are±1. Eigenstates of P are said to have definite parity, with |+i = P |+ibeing a state of even parity and |−i = −P |−i being one of odd parity.
In §3.1 we found that the stationary-state wavefunctions of a harmonic
oscillator are even functions of x when the quantum number n is even, andodd functions of x otherwise. It is clear that these stationary states are also
eigenstates of P , those for n = 0, 2, 4, . . . having even parity and those forn = 1, 3, 5, . . . having odd parity.
Mirror operators Systems not infrequently exhibit a mirror symmetry ofsome sort. When they do, it can be helpful to define an operator which transforms any state into the corresponding mirror state. Here’s an illustrative
concrete example.
A particle moves in two dimensions, so the amplitudes hx, y|ψi for theparticle to be found at the location (x, y) constitute a complete set of amplitudes. Let the operator M be such that for any state |ψihx, y|M |ψi = hy, x|ψi.

(4.19)

2 Problem 4.9 shows that P is unitary by showing that it has an infinitesimal generator.

64

Chapter 4: Transformations & Observables

Figure 4.2 If the coordinates of thepoint marked with an filled circle are(x, y), then the coordinates of thepoint marked with an open circle are(y, x). The points would be object
and image if a mirror lay along theline y = x.

That is, in the state M |ψi the amplitude to be at the point (x, y) (marked
with a filled dot in Figure 4.2) is the same as the amplitude to be at the
point (y, x) (marked by an open dot) when in the state |ψi. If there were amirror along the line y = x, the image of a light at (x, y) would be locatedat (y, x). Thus the operator M produces the state we get by mirroring al
the amplitudes in the line y = x. We leave as an exercise (Problem 4.12) the
proof that M is a unitary operator, which closely follows the proof we gaveof the unitarity of P .
If we mirror a set of amplitudes twice, we obviously recover the original
amplitudes, so M 2 = 1 and it follows that the eigenvalues of M can only be±1.

4.2 Transformations of operators
When we move an object around, we expect to find it in a new place. Specifically, suppose hψ|x|ψi = x0 for some state |ψi. Since x0 just labels a spatial
point, it must behave under translations and rotations like any vector. For
example, translating a system that is in the state |ψi through a, we obtaina new state |ψ ′ i which has hψ ′ |x|ψ ′ i = x0 + a = hψ|x + Ia|ψi. On the other
hand, from §4.1.1 we know that hψ ′ |x|ψ ′ i = hψ|U † (a)xU (a)|ψi. Since these
expectation values must be equal for any initial state |ψi, it follows from theargument given in Box 4.2 thatU † (a) x U (a) = x + a,

(4.20)

where the identity operator is understood to multiply the constant a. For
an infinitesimal translation with a → δa we have U (a) ≃ 1 − ia · p/h̄. So δa · pδa · px + δa ≃ 1 + i
x 1−i
h̄
h̄
(4.21)i2= x − [x, δa · p] + O(δa) .
h̄
For this to be true for all small vectors δa, x and p must satisfy the commutation relations[xi , pj ] = ih̄δij(4.22)in accordance with equation (2.54). Here we see that this commutation relation arises as a natural consequence of the properties of x under translations.
For a finite translation, we can writeU † (a) x U (a) = U † (a)U (a)x + U † (a)[x, U (a)] = x + U † (a) [x, U (a)] . (4.23)
We use equation (2.25) to evaluate the commutator on the right. Treating
U as the function e−ia·p/h̄ of a · p, we findU † (a) x U (a) = x −i †U (a)[x, a · p]U (a) = x + ah̄(4.24)

4.2 Transformations of operators65

Box 4.3: Rotations in ordinary space
A rotation matrix R is defined by the conditions RT = R−1 anddet(R) = +1. If R(α) rotates around the α̂ axis, it should leave thisaxis invariant so R(α)α̂ = α̂. For a rotation through an angle |α|,
Tr R(α) = 1 + 2 cos |α|.

Let α be an infinitesimal rotation vector: that is,a rotation through α aroundthe axis that is in the direction of the unit vector α̂.
We consider the effect of rotating an arbitrary vector v
through angle α. The component of v parallel to α isunchanged by the rotation.
The figure shows the projection of v into the plane perpendicular toα. The rotated vector is seen to be the vectorial sum of v and the infinitesimal vector α × v that joins the end of v before and after rotation. That isv′ = v + α × v.

as equation (4.20) requires.
Similarly, under rotations ordinary spatial vectors have components
which transform as v → R(α)v, where R(α) is a matrix describing a rotation
through angle |α| around the α̂ axis. The expectation values hψ|x|ψi = x0
should then transform in this way. In §4.1.2 we saw that when a system isrotated through an angle |α| around the α̂ axis, its ket |ψi should be multiplied by U (α) = e−iα·J . If this transformation of |ψi is to be consistent
with the rotation of the expectation value of x, we needR(α)hψ|x|ψi = hψ ′ |x|ψ ′ i = hψ|U † (α) x U (α)|ψi.

(4.25)

x + δα × x ≃ (1 + iδα · J) x (1 − iδα · J)(4.27)

In components, the vector product δα × x can be writtenX(δα × x)i =ǫijk δαj xk ,(4.28)

Since this must hold for any state |ψi, from the argument given in Box 4.2
it follows that
R(α)x = U † (α) x U (α).
(4.26)For an infinitesimal rotation, α → δα and R(α)x ≃ x + δα × x as isshown in Box 4.3, so equation (4.26) becomes= x + i [δα · J, x] + O(δα)2 .

jkwhere ǫijk is the object that changes sign if any two subscripts are interchanged and P
has ǫxyz = 1 (Appendix B). For example, equation (4.28) gives(δα × x)x = jk ǫxjk δαj xk = ǫxyz δαy xz + ǫxzy δαz xy = δαy z − δαz y. The
ith components of equation (4.27) isXX
δαj [Jj , xi ].
(4.29)
ǫijk δαj xk = ijjk

Since this equation holds for arbitrary δα, we conclude that the position
and angular momentum operators xi and Jj must satisfy the commutation
relationX[Ji , xj ] = iǫijk xk .
(4.30)k

In particular, [Jx , y] = iz and [Jz , x] = iy, while [Jx , x] = 0.

66

Chapter 4: Transformations & Observables

In fact, if the expectation value of any operator v is a spatial vector,
then the argument just given in the case of x shows that the components vi
must satisfyX[Ji , vj ] = iǫijk vk .
(4.31)k

For example, since momentum is a vector, equation (4.31) with v = p gives
the commutation relations of p with J,
X[Ji , pj ] = iǫijk pk .
(4.32)k

The product α · J must be invariant under coordinate rotations because
the operator U (α) = e−iα·J depends on the direction α̂ and not on thenumbers used to quantify that direction. Since α is an arbitrary vector, theinvariance of α · J under rotations implies that under rotations the components of J transform like those of a vector. Hence, in equation (4.31) we canreplace v by J to obtain the commutation relationX[Ji , Jj ] = iǫijk Jk .
(4.33)kIn §7.1 shall deduce the spectrum of the angular-momentum operators fromthis relation.
We now show that J commutes with any operator S whose expectation
value is a scalar. The proof is simple: hψ|S|ψi, being a scalar, is not affectedby rotations, sohψ ′ |S|ψ ′ i = hψ|U † (α) S U (α)|ψi = hψ|S|ψi.

(4.34)

Equating the operators on either side of the second equality and using U −1 =U † we have [S, U ] = 0. Restricting U to an infinitesimal rotation givesS ≃ (1 + iδα · J) S (1 − iδα · J) = S + iδα · [J, S] + O(δα)2 .

(4.35)

Since δα is arbitrary, it follows that
[J, S] = 0.

(4.36)

Among other things, this tells us that [J, x · x] = [J, p · p] = [J, x · p] = 0. It
is straightforward to check that these results are consistent with the vector
commutation relations (4.31) (Problem 4.1). It also follows that J 2 = J · J
commutes with all of the Ji ,[J, J 2 ] = 0.

(4.37)

Equations (4.33) and (4.37) imply that it is possible to find a complete setof simultaneous eigenstates for both J 2 and any one component of J (butonly one).
The parity operator Under a parity transform, coordinates behave as
x → Px = −x whereas quantum states transform as |ψi → |ψ ′ i = P |ψi, so−hψ|x|ψi = Phψ|x|ψi = hψ ′ |x|ψ ′ i = hψ|P † xP |ψi,

(4.38)

which implies that P † xP = −x or, since P is a unitary operator,{x, P } ≡ xP + P x = 0.

(4.39)

Two operators A and B for which {A, B} = 0 are said to anticommute,with {A, B} being their anticommutator. The argument we have just given
for x works with x replaced by any vector operator v, so we always have{v, P } = vP + P v = 0.

(4.40)

4.2 Transformations of operators

67

This relation contains important information about the action of P . Suppose|ωi is an eigenstate of a vector operator v with eigenvalues ω such thatv|ωi = ω|ωi. From equation (4.40) we see thatv|ω′ i = v (P |ωi) = −P v|ωi = −ωP |ωi = −ω|ω′ i(4.41)

so the parity-reversed state |ω′ i = P |ωi is also an eigenstate of v, but the
eigenvalue has changed sign.
Let |±i be states of definite parity such that P |±i = ±|±i. Withequation (4.40) we deduce that−h±|v|±i = Ph±|v|±i = h±|P † vP |±i = (±)2 h±|v|±i.

(4.42)

Since zero is the only number that is equal to minus itself, all vector operators
have vanishing expectation value in states of definite parity. More generally,
if |φi and |χi both have the same definite parity, equation (4.40) implies thathφ|v|χi = 0. We’ll use this result in Chapter 9.
We frequently encounter situations in which the potential energy V (x)is an even function of x: V (−x) = V (x). We then say that the potential isreflection-symmetric because the potential energy at −x is the same as it
is at the point x into which −x is mapped by reflection through the origin.
We now show that in such a case the parity operator commutes with the
Hamiltonian. For an arbitrary state |ψi consider the amplitude
hx|P V |ψi = h−x|V |ψi = V (−x)h−x|ψi = V (x)h−x|ψi,(4.43a)where we have used equation (4.16). On the other handhx|V P |ψi = V (x)hx|P |ψi = V (x)h−x|ψi.

(4.43b)Since x and |ψi are arbitrary, it follows that when V is an even function ofx, [P, V ] = 0. This argument generalises to all operators that carry out atransformation that is a symmetry of the potential energy.
Since the momentum p is a vector operator, P p = −pP , so
p2 P =Xk⇒

pk pk P = −2[p , P ] = 0.

X

pk P pk =kXP pk pk = P p2

k(4.44)

Applying these results to the Hamiltonian H = p2 /2m + V (x) of a particleof mass m that moves in a reflection-symmetric potential, we have that
[H, P ] = 0. It follows that for such a particle there is a complete set of
stationary states of well-defined parity. This fact is illustrated by the caseof the harmonic oscillator studied in §3.1, and in Chapter 5 it will enable us
dramatically to simplify our calculations.
In classical physics, a vector product a×b is a pseudovector; it behaves
like an ordinary vector under rotations, but is invariant under parity, sinceboth a and b change sign. We now show that expectation values of theangular momentum operators, hJi, are pseudovectors. If vi are components
of a vector operator, then combining equations (4.31) and (4.40), we obtain{P, [vi , Jj ]} = iXkǫijk {P, vk } = 0.

(4.45)We use the identity (4.76) proved in Problem 4.8 to rewrite the left side ofthis equation. We obtain0 = {P, [vi , Jj ]} = [{P, vi }, Jj ] − {[P, Jj ], vi } = −{[P, Jj ], vi }.

(4.46)

68

Chapter 4: Transformations & Observables

Hence the operator [P, Jj ] anticommutes with any component of an arbitrary
vector. Since P is defined to have precisely this property, [P, Jj ] must be
proportional to P , that is[P, Jj ] = λP,(4.47)
where λ must be the same for all values of j because the three coordinate
directions are equivalent. Under rotations, the left side transforms like a vector, while the right side is invariant. This is possible only if both sides vanish.
Hence the parity operator commutes with all three angular-momentum operators. It now follows thathψ ′ |J|ψ ′ i = hψ|P † JP |ψi = hψ|J|ψi,(4.48)

so hJi is unchanged by a parity transformation, and is a pseudovector.
Mirror operators In §4.1.4 we introduced a typical mirror operator M .
To discover how M interacts with the position operators x and y we argue
that for any state |ψihψ|M † xM |ψi = hψ|y|ψi.

(4.49)

That is, in the state M |ψi the expectation of x must be equal to the expectation value of y in the state |ψi – the truth of this statement follows
immediately from the definition (4.19) of the state M |ψi. Since equation(4.49) holds for arbitrary |ψi, we can infer the operator equation
M † xM = y⇒

xM = M y,(4.50)

where the second equation follows by multiplying both sides of the first equation by M and using the unitarity condition M M † = I. In the same way wecan show that M x = yM , px M = M py and py M = M px .

4.3 Symmetries and conservation laws
Time changes states: in a given time interval t, the natural evolution of thesystem causes any state |ψ, 0i to evolve to another state |ψ, ti. Equation(2.32) gives an explicit expression for |ψ, ti. It is easy to see that with thepresent notation this rule can be written|ψ, ti = e−iHt/h̄ |ψ, 0i,(4.51)

where H is the Hamiltonian. The time-evolution operator
U (t) ≡ e−iHt/h̄

(4.52)

is unitary, as we would expect.3
Now suppose that the generator τ of some displacement (a translation,a rotation, or something similar) commutes with H. Since these operators
commute, their exponentials U (θ) (eq. 4.12) and U (t) also commute. Consequently, for any state |ψi
U (θ)U (t)|ψi = U (t)U (θ)|ψi.

(4.53)

3 The similarity between equations (4.52) and the formula (4.12) for a general unitarytransformation suggests that H is the generator of transformations in time. This is notquite true. If we were to push the system forward in time in the same way that wetranslate it in x, we would delay the instant at which we would impose some given initialconditions, with the result that it would be less evolved at a given time t. The timeevolution operator, by contrast, makes the system older. Hence H is the generator oftransformations backwards in time.

4.3 Symmetries & conservation laws

69

The left side is the state you get by waiting for time t and then displacing,while the right side is the state obtained by displacing first and then waiting.
So the equation says that the system evolves in the same way no matter where
you put it. That is, there is a connection between commuting observables andinvariance of the physics under displacements. Moreover, in §2.2.1 we sawthat when any operator Q commutes with the Hamiltonian, the expectation
value of any function of Q is a conserved quantity, and that in consequence,a system that is initially in an eigenstate |qi i of Q remains in that eigenstate.
So whenever the physics is unchanged by a displacement, there is a conserved
quantity.
If [px , H] = 0, this argument implies that the system evolves in the same
way wherever it is located. We say that the Hamiltonian is translationallyinvariant. It is a fundamental premise of physics that empty space is the same
everywhere, so the Hamiltonian of every isolated system is translationallyinvariant. Consequently, when a system is isolated, the expectation value ofany function of the momentum operators is a conserved quantity, and, if thesystem is started in a state of well-defined momentum, it will stay in thatstate. This is Newton’s first law.
If [Jz , H] = 0, we say that the Hamiltonian is rotationally invariant
around the z axis, and our argument implies that the system evolves in thesame way no matter how it is turned around the z axis. The expectation
value of any function of Jz is constant, and if the state is initially in an
eigenstate of Jz with eigenvalue m, it will remain in that state. Consequently,
m is a good quantum number. In classical physics invariance of a system’s
dynamics under rotations around the z axis is associated with conservationof the z component of the system’s angular momentum. This fact inspires
the identification of h̄J with angular momentum.
Above we used a very general argument to infer that the existence ofa unitary operator that commutes with the Hamiltonian implies that thesystem has a symmetry. In §4.1.4 an explicit calculation (eq. 4.43) showedthat reflection symmetry of the potential energy implied that the potentialenergy operator V commutes with the parity operator P . This argument
generalises to other transformation operators. For example, suppose V (x) isinvariant under some rotation V (R(α)x) = V (x). Then
hx|V U (α)|ψi = V (x)hx|U (α)|ψi = V (x)hR(α)x|ψi,(4.54a)hx|U (α)V |ψi = hR(α)x|V |ψi = V (R(α)x)hR(α)x|ψi,(4.54b)while

so and the operator equation U V = V U follows from the equality of V (R(α)x)
and V (x).
In general, finding all the operators that commute with a given Hamiltonian is a very difficult problem. However, it is sometimes possible to deduce
conserved quantities by direct inspection. For example, the Hamiltonian fora system of n particles that interact with each other, but not with anythingelse, is
n
X
X
p2i
H=
V (xi − xj ),(4.55)+
2mi i<ji=1where the potential-energy function V only depends on the relative positionsof the individual particles. Such a Hamiltonian is invariant under translationsof all particles together (shiftsof the centre of mass coordinate) and thus thePtotal momentum ptot = i pi of this system is conserved.
If the Hamiltonian is a scalar, then [H, J] = 0, [H, J 2 ] = 0 and [H, P ] = 0(Problem 4.10), which implies conservation of angular momentum aroundany axis, conservation of total angular momentum, and conservation of parity. We have already seen that [J, J 2 ] = [J, P ] = 0, so for a scalar Hamiltonian we can find complete sets of simultaneous eigenkets of H, P , J 2 , andany one of the components of J.

70

Chapter 4: Transformations & Observables

The equation [H, P ] = 0 implies that if you set up a system that at
t = 0 is a mirror image of a given system, it will evolve in exactly the same
way as the given system. When the evolution of the mirrored system iswatched, it will appear identical to the evolution of the given system when
the latter is observed in a mirror. Hence, when [H, P ] = 0, it is impossible
to tell whether a system that is being observed, is being watched directly or
through a mirror. One of the major surprises of 20th century physics was anexperiment by Wu et al.4 in 1957, which showed that you can see things in a
mirror that cannot happen in the real world! That is, there are Hamiltonians
for which [H, P ] 6= 0.

4.4 The Heisenberg picture
All physical predictions are extracted from the formalism of quantum mechanics by operating with a bra on a ket to extract a complex number: weeither calculate the amplitude for some event as A = hφ|ψi or the expectation value of a observable through hQi = hψ|Qψi, where |Qψi ≡ Q|ψi. In
general our predictions are time-dependent because the state of our systemevolves in time according to|ψ, ti = U (t)|ψ, 0i,(4.56)

where the time-evolution operator U is defined by equation (4.52).
With every operator of interest we can associate a new time-dependent
operatore t ≡ U † (t)QU (t).
Q
(4.57)

Then at any time t the expectation value of Q can be writtene t |ψ, 0i.
hQit = hψ, t|Q|ψ, ti = hψ, 0|U † (t)QU (t)|ψ, 0i = hψ, 0|Q(4.58)

e t isThat is, the expectation value at time t = 0 of the new operator Q
equal to the expectation value of the original, physical, operator Q at time t.
Similarly, when we wish to calculate an amplitude hφ, t|ψ, ti for something
to happen at time t, we can argue that on account of the unitarity of U (t)it is equal to a corresponding amplitude at time zero:hφ, t|ψ, ti = hφ, 0|ψ, 0i where |φ, ti ≡ U (t)|φ, 0i.

(4.59)

e t , theThus if we work with the new time-dependent operators such as Q
only states we require are those at t = 0. This formalism, is called theHeisenberg picture to distinguish it from the Schrödinger picture inwhich states evolve and operators are normally time-independent.
As we have seen, classical mechanics applies in the limit that it is sufficient to calculate the expectation values of observables, and is concerned withsolving the equations of motion of these expectation values. In the Heisenberg picture quantum mechanics is concerned with solving the equations of
e t , etc. Consequently, there is amotion of the time-dependent operators Q
degree of similarity between the Heisenberg picture and classical mechanics.
e t : we
It is straightforward to determine the equation of motion of Q
simply differentiate equation (4.57)
et
dQ
dU †
dU
=
QU + U † Q
.
dt
dt
dt

(4.60)

4 Wu, C.S., Ambler, E., Hayward, R.W., Hoppes, D.D. & Hudson, R.P., 1957, Phys.
Rev., 105, 1413

4.5 The essence of quantum mechanics

71

But differentiating equation (4.52) we have
dU
iH
=− U
dt
h̄

⇒

dU †
iH †
=
U ,
dt
h̄

(4.61)

where we have taken advantage of the fact that U is a function of H andtherefore commutes with it. Inserting these expressions into equation (4.60)
we obtainetdQ= −HU † QU + U † QU H
ih̄
(4.62)dte= [Qt , H].

This result is similar to Ehrenfest’s theorem (eq. 2.34) as it has to be because
Ehrenfest’s theorem must be recovered if we pre- and post-multiply each sideby the time-independent state |ψ, 0i.
The Heisenberg picture is most widely used in the quantum theory offields. In this theory one needs essentially only one state, the vacuum in the
remote past |0i, which we assume was empty. Excitations of the vacuum
are interpreted as particles, each mode of excitation being associated with adifferent type of particle (photons, electron, up-quarks, etc). The theory isconcerned with the dynamics of operators that excite the vacuum, creating
particles, which then propagate to other locations, where they are detected(annihilated) by similar operators. Sometimes one mode of excitation of thevacuum morphs into one or more different modes of the vacuum, and such an
event is interpreted as the decay of one type of particle into other particles.
The amplitude for any such sequence of events is obtained as a number of thee2 . . . A
en |0i, where the operators A
ei are creation or annihilation
e1 A
form h0|A
operators for the appropriate particles in the Heisenberg picture.

4.5 What is the essence of quantum mechanics?
It is sometimes
P said that commutation relations such as [xi , pj ] = ih̄δij and[Ji , Jj ] = i k ǫijk Jk are inherently quantum mechanical, but this is not true.
Take for example an ordinary classical rotation matrix R(α) which rotates spatial vectors as v → v′ = R(α)v. Define matrices Jx , Jy and Jz
via
exp (−iα · J ) ≡ R(α),(4.63)
where the exponential of a matrix is defined in terms of the power series forex . Clearly, the Ji must be 3 × 3 matrices, and, since R(α) is real and theangles α are arbitrary, the Ji must be pure imaginary. Finally, orthogonalityof R requires
I = RT (α)R(α) = exp(−iα · J )T exp(−iα · J )= exp(−iα · J T ) exp(−iα · J )(4.64)We express α in terms of the angle of the rotation it represents, θ = |α|,and the direction n = α/|α| of the rotation axis, and then we differentiate
equation (4.64) with respect to θ. We obtain
0 = −in · J T exp(−iθn · J T ) exp(−iθn · J )+ exp(−iθn · J T ) exp(−iθn · J )(−in · J )= −in · {JJ

T

(4.65)+ J }.

Since n is an arbitrary unit vector, it now follows that JiT = −Ji , so Ji
is antisymmetric. A pure imaginary antisymmetric matrix is a Hermitianmatrix. Thus the Ji are Hermitian.

72

Chapter 4: Transformations & Observables

For any two vectors α and β, it is easy to show that the product
RT (α)R(β)R(α) is an orthogonal matrix with determinant +1, so it is arotation matrix. It leaves the vector β′ ≡ R(−α)β = RT (α)β invariant:no
RT (α)R(β)R(α) β′ = RT (α)R(β)β = RT (α)β = β′ .
(4.66)
Hence β′ is the axis of this rotation. Therefore

RT (α)R(β)R(α) = R(β′ ) = R(R(−α)β).

(4.67)In Box 4.3 we showed that when |α| is infinitesimal, R(−α)β ≃ β − α × β,
so when β is also infinitesimal, equation (4.67) can be written in terms ofthe classical generators (4.63) as(1 + iα · J ) (1 − iβ · J ) (1 − iα · J ) ≃ 1 − i(β − α × β) · J .

(4.68)The zeroth order terms (‘1’) and those involving only α or β cancel, but theterms involving both α and β cancel only if
αi βj [Ji , Jj ] = iαi βjXkǫijk Jk .

(4.69)

This equation can hold for all directions α and β only if the Ji satisfy[Ji , Jj ] = iXkǫijk Jk ,(4.70)

which is identical to the ‘quantum’ commutation relation (4.33). Our rederivation of these commutation relations from entirely classical considerations
is possible because the relations reflect the fact that the order in which you
rotate an object around two different axes matters (Problem 4.6). This isa statement about the geometry of space that has to be recognised by bothquantum and classical mechanics.
In Appendix D it is shown that in classical statistical mechanics, eachcomponent of position, xi , and momentum, pi , is associated with a Hermitian operator x̂i or p̂i that acts on functions on phase space. The operatorp̂i generates translations along xi , while x̂i generates translations along pi(boosts). The operators L̂i associated with angular momentum satisfy thecommutation relation [L̂x , L̂y ] = iHL̂z , where H is a number with the same
dimensions as h̄ and a magnitude that depends on how x̂i and p̂i are normalised.
If the form of the commutation relations is not special to quantum mechanics, what is? In quantum mechanics, complete information about any
system is contained in its ket |ψi. There is nothing else. From |ψi we canevaluate amplitudes such as hx, µ|ψi for the system to be found at x withorientation µ. If we do not care about µ, the total probability for |ψi to befound at x isX
2
Prob(at x|ψ) =hx, µ|ψi .
(4.71)µ

Eigenstates of the x operator with eigenvalue x0 are states in which thesystem is definitely at x0 , while eigenstates of the p operator with eigenvalueh̄k are states in which the system definitely has momentum h̄k.
By contrast, in classical statistical mechanics we declare at the outsetthat a well defined state is one that has definite values for all measurable
quantities, so it has a definite position, momentum, orientation etc. The
eigenfunctions of p̂ or L̂ do not represent states of definite momentum or
angular momentum, because we have already defined what such states are.
Classical statistical mechanics knows nothing about probability amplitudes, but interprets the functions on phase space on which p̂ or L̂ act

Problems73

as probability distributions. This is possible because, as we show in Appendix D, the integral of such a distribution can be normalised to one and isconserved. We can certainly expand any such distribution in the eigenfunctions of, say p̂. However, as in quantum mechanics the expansion coefficients
will not be positive – in fact, they will generally be complex. Hence theycannot be interpreted as probabilities. What makes quantum mechanics fundamentally different is its reliance on complex quantum amplitudes, and thephysical interpretation that it gives to a functional expansion through thefundamental rule (1.11) for adding quantum amplitudes. Quantum mechanics is therefore naturally formulated in terms of states |ψi that inhabit acomplex vector space of arbitrary dimension – a so called Hilbert space.
These states may always be expanded in terms of a complete set of eigenstates of a Hermitian operator, and the (complex) expansion coefficients havea simple physical interpretation.
Classical statistical mechanics is restricted to probabilities, which haveto be real, non-negative numbers and are therefore never expansion coefficients. Quantum and classical mechanics incorporate the same commutation
relations, however, because, as we stressed in §4.2, these follow from thegeometry of space. From a mathematician’s perspective, the commutation
relations of quantum-mechanical operators and the operators of classical statistical physics have to be the same because both systems of operators provide representations of the ‘Lie algebra’ of the same mathematical group(Appendix E).

Problems
4.1 Verify that [J,Px · x] = 0 and [J, x · p] =P0 by using the commutation
relations [xi , Jj ] = i k ǫijk xk and [pi , Jj ] = i k ǫijk pk .
4.2∗ Show that the vector product a × b of two classical vectors transforms
like a vector under rotations. Hint: A rotation matrix R satisfies
P the relations
R · RT = I and det(R) = 1, which in tensor notation read p Rip Rtp = δitP
and ijk ǫijk Rir Rjs Rkt = ǫrst .
P
4.3∗ We have shown that [vi , Jj ] = i k ǫijk vk for any operator whose
components vi form a vector. The expectation
P value of this operator relationin any state |ψi is then hψ|[vi , Jj ]|ψi = i k ǫijk hψ|vk |ψi. Check that with
U (α) = e−iα·J this relation is consistent under a further rotation |ψi →|ψ ′ i = U (α)|ψi by evaluating both sides separately.
4.4∗

The matrix for rotating an ordinary vector by φ around the z axis is
cos φ − sin φ 0
R(φ) ≡  sin φ cos φ 0 (4.72)
001

By considering the form taken by R for infinitesimal φ calculate from R thematrix J z that √appears in R(φ) = exp(−iJJz φ).
√ Introduce new coordinates
u1 ≡ (−x + iy)/ 2, u2 = z and u3 ≡ (x + iy)/ 2. Write down the matrix M
that appears in u = M · x [where x ≡ (x, y, z)] and show that it is unitary.
Then show that
J z′ ≡ M · J z · M† .
(4.73)
is identical with Sz in the set of spin-one Pauli analogues





0 1 0
0 −i 01
1 
1
Sx = √
1 0 1  , Sy = √  i 0 −i  , Sz =  0
2
2
0 1 0
0 i
0
0


0 0
0 0 .
0 −1(4.74)
Write down the matrix J x whose exponential generates rotations aroundthe x axis, calculate J x′ by analogy with equation (4.73) and check thatyour result agrees with Sx in the set (4.74). Explain as fully as you can themeaning of these calculations.

74

Problems

4.5 Determine the commutator [JJx′ , J z′ ] of the generators used in Problem′
4.4. Show that it is equal to −iJJy , where J y′ is identical with Sy in the set(4.74).
4.6∗ Show that if α and β are non-parallel vectors, α is not invariant under
the combined rotation R(α)R(β). Hence show that RT (β)RT (α)R(β)R(α)
is not the identity operation. Explain the physical significance of this result.
4.7∗

In this problem you derive the wavefunction
hx|pi = eip·x/h̄

(4.75)

of a state of well defined momentum from the properties of the translation operator U (a). The state |ki is one of well-defined momentum h̄k. How would
you characterise the state |k′ i ≡ U (a)|ki? Show that the wavefunctions ofthese states are related by uk′ (x) = e−ia·kuk (x) and uk′ (x) = uk (x − a).
Hence obtain equation (4.75).
4.8 By expanding the anticommutator on the left and then applying thethird rule of the set (2.22), show that any three operators satisfy the identity[{A, B}, C] = {A, [B, C]} + {[A, C], B}.

(4.76)

4.9 Define G in terms of the parity operator P by
G ≡ 21 (1 − P ).

(4.77)Show that G is Hermitian and that Gn = G for positive integer n. Explain
this result in terms of the eigenkets and eigenvalues of G. Show further thatP = U (π) where U (s) ≡ eisG .
4.10 Let P be the parity operator and S an arbitrary scalar operator.
Explain why P and S must commute.
4.11 In this problem we consider discrete transformations other than thatassociated with parity. Let S be a linear transformation on ordinary threedimensional space that effects a a reflection in a plane. Let S be the associated operator on kets. Explain the physical relationship between the kets
|ψi and |ψ ′ i ≡ S|ψi. Explain why we can writeShψ|x|ψi = hψ|S † xS|ψi.

(4.78)

What are the possible eigenvalues of S?
Given that S reflects in the plane through the origin with unit normaln̂, show, by means of a diagram or otherwise, that its matrix is given by
Sij = δij − 2ni nj .

(4.79)√
Determine the form of this matrix in the case that n = (1, −1, 0)/ 2. Show
that in this case Sx = yS and give an alternative expression for Sy.
Show that a potential of the form
V (x) = f (R) + λxy,

where R ≡p
x2 + y 2(4.80)

satisfies V (Sx) = V (x) and explain the geometrical significance of this equation. Show that [S, V ] = 0. Given that E is an eigenvalue of H = p2 /2m + V
that has a unique eigenket |Ei, what equation does |Ei satisfy in addition
to H|Ei = E|Ei?
4.12 Show that the operator defined by hx, y|S|ψi = hy, x|ψi is Hermitian.

5
Motion in step potentials
We follow up our study of the harmonic oscillator by looking at motion ina wider range of one-dimensional potentials V (x). The potentials we study
will be artificial in that they will only vary in sharp steps, but they will
enable us to explore analytically some features of quantum mechanics thatare generic and hidden from us in the classical limit. We start by consideringa particle that is trapped in a potential well and go on to consider a particle
that has a choice of two wells. We find that in this case it can move between
these wells in violation of classical mechanics, and we use this simple system
to mode the operation of an ammonia maser. In §5.3 we ask how potential
wells and barriers affect the motion of a free particle – one that can escapeto infinity. We find that whereas in classical mechanics the particle is neverreflected by a potential well, in quantum mechanics there is generally a nonzero amplitude for such reflection. We find also that particles can “tunnel”
through barriers that classically would certainly reflect them.

5.1 Square potential well
We look for energy eigenstates of a particle that moves in the potential(Figure 5.1)
0
for |x| < aV (x) =
(5.1)
V0 > 0 otherwise.
Since V is an even function of x, the Hamiltonian (2.51) commutes with theparity operator P (page 67). So there is a complete set of energy eigenstatesof well defined parity. The wavefunctions u(x) ≡ hx|Ei of these states will
be either even or odd functions of x, and this fact will greatly simplify thejob of determining u(x).
In the position representation, the governing equation (the tise 2.33)
readsh̄2 d2 u+ V (x)u = Eu.
(5.2)−
2m dx2On account of the step-like nature of V , equation (5.2) reduces to a pair of

76

Chapter 5: Motion in step potentials

Figure 5.1 The dotted line shows the square-well potential V (x). The full curve shows
the ground-state wavefunction.

extremely simple equations,d2 u
2mE=− 2 udx2h̄2m(V0 − E)
d2 u
=
u
dx2
h̄2

for |x| < a(5.3)

otherwise.

We restrict ourselves to solutions that describe a particle that is bound by
the potential well in the sense that E < V0 .1 Then the solution to the second
equation is u(x) = Ae±Kx , where A is a constant andK≡s2m(V0 − E)
.
h̄2

(5.4)

If u is to be normalisable, it must vanish as |x| → ∞. So at x > a we haveu(x) = Ae−Kx , and at x < −a we have u(x) = ±Ae+Kx , where the plus
sign is required for solutions of even parity, and the minus sign is requiredfor odd parity.
For E > 0, the solution to the first of equations (5.3) is either u(x) =B cos(kx) or u(x) = B sin(kx) depending on the parity, where
k≡

r

2mE
.
h̄2

(5.5)

So far we have ensured that u(x) solves the tise everywhere except
at |x| = a. Unless u is continuous at these points, du/dx will be arbitrarily
large, and d2 u/dx2 will be undefined, so u will not satisfy the tise. Similarly,
unless du/dx is continuous at these points, d2 u/dx2 will be arbitrarily large,
so u cannot solve the tise. Therefore, we require that both u and du/dx arecontinuous at x = a, that is)(
B sin(ka) = Ae−Ka
B cos(ka) = Ae−Kaor(5.6)
kB cos(ka) = −KAe−Ka−kB sin(ka) = −KAe−Ka
where the first pair of equations apply in the case of even parity and thesecond in the case of odd parity. It is easy to show that once these equations have been satisfied, the corresponding equations for x = −a will be
automatically satisfied.
1 By considering the behaviour of u near the origin we can prove that E > 0.

5.1 Square potential well77

Figure 5.2 Plots of the left (full) and right (dashed) sides of equation (5.8) for the caseW = 10.

Figure 5.3 A square well inscribed in a general well.

We eliminate A and B from equations (5.6) by dividing the second
equation in each set by the first. In the case of even parity we obtainr
2mV0− k2 .
(5.7)
k tan(ka) = K =h̄2This is an algebraic equation for k, which controls E through (5.5). Before
attempting to solve this equation, it is useful to rewrite it assr
W2
2mV0 a2
−
1
where
W
≡
tan(ka) =
.
(5.8)
(ka)2
h̄2
W and ka are dimensionless variables. The left and right sides of equation(5.8) are plotted as functions of ka in Figure 5.2. Since for ka = 0 the graphsof the two sides start at the origin and infinity, and the graph of the left sideincreases to infinity at ka = π/2 while the graph of the left side terminates
at ka = W , the equation always has at least one solution. Thus no matter
how small V0 and a are, the square well can always trap the particle. The
bigger W is, the more solutions the equation has; a second solution appears
at W = π, a third at W = 2π, etc.
Analogously one can show that for an odd-parity energy eigenstate toexist, we must have W > π/2 and that additional solutions appear when
W = (2r + 1)π/2 for r = 1, 2, . . . (Problem 5.5).
From a naive perspective our discovery that no matter how narrow orshallow it is, a square potential well always has at least one bound state,conflicts with theprinciple: the particle’s momentum cannot√ uncertainty√
exceed pmax = 2mE < 2mV0 and can have either sign, so if the particle
were confined within the well, √the product of the uncertainties in p and x
would be less than 4apmax < 4 2mV0 a2 = 4h̄W , which tends to zero with
W . The resolution of this apparent paradox is that for W ≪ 1 the particle is
not confined within the well; there is a non-negligible probability of finding
the particle in the classically forbidden region |x| > a. In the limit W → 0
the particle is confined by a well in which it is certain never to be found!

78

Chapter 5: Motion in step potentials

Figure 5.4 The wavefunctions of the lowest three stationary states of the infinitely deepsquare well: ground state (full); first excited state (dashed); second excited state (dotted).

Our result that a square well always has a bound state can be extended
to potential wells of any shape: given the potential well U sketched in Figure 5.3, we consider the square well shown by the dashed line in the figure.
Since this shallower and narrower well has a bound state, we infer that thepotential U also has at least one bound state.

5.1.1 Limiting cases(a) Infinitely deep well It is worthwhile to investigate the behaviourof these solutions as V0 → ∞ with a fixed, when the well becomes infinitely
deep. Then W → ∞ and the dashed curve in Figure 5.2 moves higher
and higher up the paper and hits the x axis further and further to theright. Consequently, the values of ka that solve equation (5.8) tend towards
ka = (2r + 1)π/2, so the even-parity energy eigenfunctions become
u(x) =n

A cos[(2r + 1)πx/2a] |x| < a0otherwise.

(5.9)

This solution has a discontinuity in its gradient at x = a because it is thelimit of solutions in which the curvature K for x > a diverges to infinity. The
odd-parity solutions are obtained by replacing the cosine with sin(sπx/a),
where s = 1, 2, . . ., which again vanish at the edge of the well (Figure 5.4).
From this example we infer the principle that wavefunctions vanish at theedges of regions of infinite potential energy.
The energy of any stationary state of an infinite square potential well
can be obtained from 2n2 h̄πEn =, where n = 1, 2, . . .
(5.10)
8m a
The particle’s momentum when it is in the ground state (n = 1) is of orderh̄k = h̄π/2a and of undetermined sign, so the uncertainty in the momentumis ∆p ≃ h̄π/a. The uncertainty in the particle’s position is ∆x ≃ 2a, so∆x ∆p ≃ 2h̄π, consistent with the uncertainty principle (§2.3.2).
(b) Infinitely narrow well In §11.5.1 we will study a model of covalentbonding that involves the potential obtained by letting the width of thesquare well tend to zero as the well becomes deeper and deeper√ in such a
way that the product V0 a remains constant. In this limit W ∝ a V0 (eq. 5.8)
tends to zero, so there is only one bound state and it will be an even-parity
state.
Rather than obtaining the wavefunction and energy of this state fromformulae already in hand, it is more convenient to reformulate the problem
using a different normalisation for the energy: we now set V to zero outside
the well, so V becomes negative at interior points. Then we can write V (x) =−Vδ δ(x), where δ(x) is the Dirac delta function and Vδ > 0. The tise now
readsh̄2 d2 u−− Vδ δ(x)u = Eu.
(5.11)
2m dx2

5.2 A pair of square wells

79

Figure 5.5 Wavefunction of a particle trapped by a very narrow, deep potential well.

Integrating the equation from x = −ǫ to x = ǫ with ǫ infinitesimal, we find
−





ǫ
Z ǫ
du
2m
dx u .
= 2 Vδ u(0) + E
dx −ǫ
h̄
−ǫ

(5.12)

Since u is finite, the integral on the right can be made as small as we pleaseby letting ǫ → 0. Hence the content of equation (5.12) is that du/dx has a
discontinuity at the origin:


ǫ
du
2mVδ
= − 2 u(0).
dx −ǫ
h̄

(5.13)

Since we know that the solution we seek has even parity, it is of theform u(x) = Ae∓Kx , where the minus sign applies for x > 0 (Figure 5.5).
Substituting this form of u into (5.13) and dividing through by 2A we have
K=

mVδ
.
h̄2

(5.14)

Inserting u = e−Kx into equation (5.11) at x > 0 we find that E = −h̄2 K 2 /2m,so the energy of a particle that is bound to a δ-function potential is
E=−

mVδ2
.
2h̄2

(5.15)

Figure 5.5 shows that |ψ(x)|2 is finite in the well, and the well in infinitely
narrow, so the probability of finding the particle in the well is zero – theparticle is certain never to be in the well that traps it! This result is anextreme case of the phenomenon we discussed apropos the application of theuncertainty principle to a shallow well of finite depth.

5.2 A pair of square wells
Some important phenomena can be illustrated by considering motion in apair of potentials that are separated by a barrier of finite height and width.
Figure 5.6 shows the potential
V (x) =(

V00∞

for |x| < afor a < |x| < botherwise.

(5.16)

Since the potential is an even function of x, we may assume that the energy
eigenfunctions that we seek are of well-defined parity.
For simplicity we take the potential to be infinite for |x| > b, and weassume that the particle is classically forbidden in the region |x| < a. Thenin this region the wavefunction must be of the form u(x) = A cosh(Kx)or u(x) = A sinh(Kx) depending on parity, and K is given by (5.4). In
the region a < x < b the wavefunction may be taken to be of the formu(x) = B sin(kx + φ), where B, and φ are constants to be determined and k

80

Chapter 5: Motion in step potentials

Figure 5.6 A double potential well with b/a = 5.

Figure 5.7 Full curves: the left side of equation (5.19) for the case W = 3.5, b = 5a.
Each vertical section is associated with a different value of the integer r. The right side isshown by the dotted curve for even parity, and the dashed curve for odd parity.

is related to the energy by (5.5). From our study of a single square well weknow that u must vanish at x = b, sosin(kb + φ) = 0⇒

φ = rπ − kbwith r = 0, 1, . . .

(5.17)

Again by analogy with the case of a single square well, we require u and its
derivative to be continuous at x = a, so (depending on parity)sinh(Ka) = B sin(ka + φ)cosh(Ka) = B sin(ka + φ)orK cosh(Ka) = kB cos(ka + φ).
K sinh(Ka) = kB cos(ka + φ)
(5.18)
Once these equations have been solved, the corresponding conditions at
x = −a will be automatically satisfied if for −b < x < −a we take u =±B sin(k|x| + φ), using the plus sign in the even-parity case.
Using (5.17) to eliminate φ from equations (5.18) and then proceeding
in close analogy with the working below equations (5.6), we findps
2 − (ka)22even parity
W
coth
W
p

−1=
tan [rπ − k(b − a)]2 tanh(ka)
odd parity,
W 2 − (ka)2(5.19)

where W is defined by equation (5.8).

5.2 A pair of square wells

81

Figure 5.8 The ground state (full curve) and the associated odd-parity state (dashedcurve) of the double square-well potential (shown dotted).

The left and right sides of equation (5.19) are plotted in Figure 5.7; thevalues of ka for stationary states correspond to intersections of the steeply
sloping curves of the left side with the initially horizontal curves of the rightside. The smallest value of ka is associated with the ground state. The values
come in pairs, one for an even-parity state, and one very slightly larger foran odd-parity state. The difference between the k values in a pair increases
with k.
The closeness of the k values in a given pair ensures that in the righthand well (a < x < b) the wavefunctions ue (x) and uo (x) of the even- andodd-parity states are very similar, and that in the left-hand well ue and uo
differ by little more than sign – see Figure 5.8. Moreover, when the k values
are similar, the amplitude of the wavefunction is small in the classicallyforbidden region |x| < a. Hence, the linear combinations1
ψ± (x) ≡ √ [ue (x) ± uo (x)]2(5.20)

are the wavefunctions of a state |ψ+ i in which the particle is almost certain
to be in the right-hand well, and a state |ψ− i in which it is equally certain
to be in the left-hand well.
Consider now how the system evolves if at time 0 it is in the state |ψ+ i,so the particle is in the right-hand well. Then by equation (2.32) at time t
its wavefunction isi
1 hψ(x, t) = √ ue (x)e−iEe t/h̄ + uo (x)e−iEo t/h̄2
(5.21)ih1= √ e−iEe t/h̄ ue (x) + uo (x)e−i(Eo −Ee )t/h̄ .
2
After a time T = πh̄/(Eo − Ee ) the exponential in the square brackets on thesecond line of this equation equals −1, so to √within an overall phase factor
the wavefunction has become [ue (x) − uo (x)]/ 2, implying that the particle
is certainly in the left-hand well; we say that in the interval T the particle hastunnelled through the barrier that divides the wells. After a further period
T it is certainly in the right-hand well, and so on ad infinitum. In classicalphysics the particle would stay in whatever well it was in initially. In fact, theposition of a familiar light switch is governed by a potential that consists of
two similar adjacent potential wells, and such switches most definitely do not
oscillate between their on and off positions. We do not observe tunnelling in
the classical regime because Eo − Ee decreases with increasing W faster thane−2W (Problem 5.16), so the time required for tunnelling to occur increases
faster than e2W and is enormously long for classical systems such as lightswitches.
5.2.1 Ammonia
Nature provides us with a beautiful physical realisation of a system with adouble potential well in the ammonia molecule NH3 . Ammonia contains four

82

Chapter 5: Motion in step potentials

Figure 5.9 The two possible relative locations of nitrogen and hydrogen atoms in NH3 .

nuclei and ten electrons, so is really a very complicated dynamical system.
However, in §11.5.2 we shall show that a useful way of thinking about thelow-energy behaviour of molecules is to imagine that the electrons provide
light springs, which hold the nuclei together. The nuclei oscillate around theequilibrium positions defined by the potential energy of these springs. In
the case of NH3 , the potential energy is minimised when the three hydrogenatoms are arranged at the vertices of an equilateral triangle, while the nitrogen atom lies some distance x away from the plane of the triangle, either‘above’ or ‘below’ it (see Figure 5.9). Hence if we were to plot the molecule’s
potential energy as a function of x, we would obtain a graph that looked like
Figure 5.6 except that the sides of the wells would be sloping rather thanstraight. This function would yield eigenenergies that came in pairs, as inour square-well example.
In many physical situations the molecule would have so little energy thatit could have negligible amplitudes to be found in any but the two lowestlying stationary states, and we would obtain an excellent approximation to
the dynamics of ammonia by including only the amplitudes to be found inthese two states. We now use Dirac notation to study this dynamics.
Let |+i be the state whose wavefunction is analogous to the wavefunction
ψ+ (x) defined above in the case of the double square well; then ψ+ (x) =hx|+i, and in the state |+i the N atom is certainly above the plane containing
the H atoms. The ket |−i is the complementary state in which the N atomlies below the plane of the H atoms.
The |±i states are linear combinations of the eigenkets |ei and |oi of the
Hamiltonian:1|±i = √ (|ei ± |oi).
(5.22)
2
In the |±i basis the matrix elements of the Hamiltonian H are
h+|H|+i = 12 (he| + ho|)H(|ei + |oi) = 12 (Ee + Eo )h+|H|−i = 12 (he| + ho|)H(|ei − |oi) = 12 (Ee − Eo )(5.23)

h−|H|−i = 12 (he| − ho|)H(|ei − |oi) = 12 (Ee + Eo )Bearing in mind that H is represented by a Hermitian matrix, we conclude
that it is


E −A
,
(5.24)H=−A E
where E = 21 (Ee + Eo ) and A ≡ 21 (Eo − Ee ) are both positive.
Now the electronic structure of NH3 is such that the N atom carries
a small negative charge −q, with a corresponding positive charge +q distributed among the H atoms. With NH3 in either the |+i or |−i state there
is a net separation of charge, so an ammonia molecule in these states possesses an electric dipole moment of magnitude qs directed perpendicular tothe plane of H atoms (see Figure 5.9), where s is a small distance.
Below equation (5.21) we saw that a molecule that is initially in thestate |+i will subsequently oscillate between this state and the state |−i ata frequency (Eo − Ee )/2πh̄ = A/πh̄. Hence a molecule that starts in thestate |+i is an oscillating dipole and it will emit electromagnetic radiation

5.2 A pair of square wells

83

Figure 5.10 Energy levels of theammonia molecule as a function of
external electric field strength E.
The quantity plotted, ∆E = E − E.

at the frequency A/πh̄. This proves to be 150 GHz, so the molecule emits
microwave radiation.
The ammonia maser The energy 2A that separates the ground and firstexcited states of ammonia in zero electric field is small, 10−4 eV. Consequently at room temperature similar numbers of molecules are in these twostates. The principle of an ammonia maser2 is to isolate the molecules thatare in the first excited state, and then to harvest the radiation that is emitted as the molecules decay to the ground state. The isolation is achievedby exploiting the fact that, as we now show, when an electric field is applied, molecules in the ground and first excited states develop polarisationsof opposite sign.
We define the dipole-moment operator P by
P |+i = −qs|+i ;P |−i = +qs|−i,(5.25)

so a molecule in the |+i state has dipole moment −qs and a molecule in the|−i state has dipole moment +qs.3 To measure this dipole moment, we canplace the molecule in an electric field of magnitude E parallel to the dipole
axis. Since the energy of interaction between a dipole P and an electric field
E is −P E, the new Hamiltonian is
E + qEs−A
H=.
(5.26)
E − qEs−A
This new Hamiltonian has eigenvaluesp
E± = E ± A2 + (qEs)2 .

(5.27)

These are plotted as a function of field E in Figure 5.10. When E = 0 theenergy levels are the same as p
before. As E slowly increases, E increases
quadratically with E, because A2 + (qEs)2 ≃ A + (qEs)2 /2A, but whenE ≫ A/qs the energy eigenvalues change linearly with E. Notice that in thislarge-field limit, at lowest order the energy levels do not depend on A.
The physical interpretation of these results is the following. In theabsence of an electric field, the energy eigenstates are the states of welldefined parity |ei and |oi, which have no dipole moment. An electric field
breaks the symmetry between the two potential wells, making it energeticallyfavourable for the N atom to occupy the well to which the electric field ispushing it. Consequently, the ground state develops a dipole moment P ,which is proportional to E. Thus at this stage the electric contribution to
the energy of the ground state, which is −P E, is proportional to E 2 . Once
2 ‘maser’ is an acronym for “microwave amplification by stimulated emission of radiation”.
3 The N atom is negatively charged so the dipole points away from it.

84

Chapter 5: Motion in step potentials

this contribution exceeds the separation A between the states of well-definedparity, the molecule has shifted to the lower-energy state of the pair |±i,
and it stays in this state as the electric field is increased further. Thus forlarge fields the polarisation of the ground state is independent of E and theelectric contribution to the energy is simply proportional to E.
While the ground state develops a dipole moment that lowers its energy,
the first excited state develops the opposite polarisation, so the electric field
raises the energy of this state, as shown in Figure 5.10. The response of thefirst excited state is anomalous from a classical perspective.
Ehrenfest’s theorem (2.57) tells us that the expectation values of operators obey classical equations of motion. In particular the momentum of amolecule obeys
d hpx i∂V,
(5.28)=−
dt∂x
where x is a Cartesian coordinate of the molecule’s centre of mass. The
potential depends on x only through the electric field E, so
∂V
∂E
= −P
,
∂x
∂x

(5.29)∂Ed hpx i= hP i.
dt∂x(5.30)

from which it follows that

Since the sign of hP i and therefore the force on a molecule depends onwhether the molecule is in the ground or first excited state, when a jet ofammonia passes through a region of varying E, molecules in the first excitedstate can be separated from those in the ground state.
Having gathered the molecules that are in the excited state, we leadthem to a cavity that resonates with the 150 GHz radiation that is emitted
when molecules drop into the ground state. The operation of an ammonia
maser by Charles Townes and colleagues4 was the first demonstration of
stimulated emission and opened up the revolution in science and technologythat lasers have have since wrought.

5.3 Scattering of free particles
We now consider what happens when a particle that is travelling parallelto the x axis encounters a region of sharply changed potential energy. In
classical physics the outcome depends critically on whether the potential
rises by more than the kinetic energy of the incoming particle: if it does, theparticle is certainly reflected, while it continues moving towards positive x in
the contrary case. We shall find that quantum mechanics predicts that there
are usually non-vanishing probabilities for both reflection and transmission
regardless of whether the rise in potential exceeds the initial kinetic energy.
We assume that each particle has well-defined energy E, so its wavefunction satisfies the tise (5.2). We take the potential to be (Figure 5.11)
V =



V0
0

for |x| < aotherwise,(5.31)

where V0 is a constant. At |x| > a the relevant solutions of (5.2) are
e±ikx

or



sin(kx + φ)
± sin(−kx + φ)

at x > a
at x < −a



with

k=

r

2mE
, (5.32a)h̄2

4 Gordon, J.P., Zeiger, H.J., & Townes, C.H., 1954, Phys. Rev, 95, 282 (1954)

5.3 Scattering of free particles

85

Figure 5.11 A square, classically forbidden barrier and the functional forms for stationary
states of even (top) and odd parity.

where φ is a constant phase. Since the time dependence of these stationary
states is obtained by introducing the factor e−iEt/h̄ , a plus sign in e±ikx
implies that the particle is moving to the right, and a minus sign is associatedwith movement to the left. A wavefunction that is proportional to sin(kx+φ)contains both types of wave with amplitudes of equal magnitude, so it makes
motion in either direction equally likely. At |x| < a the relevant solutions of(5.2) ares
2m(E − V0 )cos(Kx)±iKx
when E > V0eorwith K =sin(Kx)h̄2s2m(V0 − E)cosh(Kx)
e±Kx orwhen E < V0 .
with K =sinh(Kx)h̄2(5.32b)In every case we have a choice between exponential solutions and solutions of well-defined parity. Since our physical problem is strongly asymmetric in that particles are fired in from negative x rather than equally from bothsides, it is tempting to work with the exponential solutions of the tise rather
than the solutions of well-defined parity. However, the algebra involved insolving our problem is much lighter if we use solutions of well-defined parity
because then the conditions that ensure proper behaviour of the solution
at x = a automatically ensure that the solution also behaves properly at
x = −a; if we use exponential solutions, we have to deal with the cases
x = ±a individually. Therefore we seek solutions of the formB sin(k|x| + φ)
for |x| > aψe (x) =cos(Kx) or cosh(Kx) otherwise; ′(5.33)
for x > a
 B sin(kx + φ′ )ψo (x) = A sin(Kx) or A sinh(Kx) for |x| ≤ a,

−B ′ sin(k|x| + φ′ )otherwise,

where A, B, B ′ , φ and φ′ are constants. B, φ and φ′ will be unambiguously
determined by the conditions at x = ±a. These conditions will make B ′proportional to A, which we treat as a free parameter.
In our study of the bound states of potential wells in §5.1, the requirement that the wavefunction vanish at infinity could be satisfied only fordiscrete values of E. These values of E (and therefore k and K) differed
between the even- and odd-parity solutions, so all energy eigenfunctions automatically had well-defined parity. In the case of a free particle, by contrast,
we will be able to construct both an even-parity and an odd-parity solution
for every given value of E. Linear combinations of these solutions ψe (x) andψo (x) of well-defined parity are energy eigenfunctions that do not have welldefined parity. We now show that the sum these solutions of the tise withwell-defined parity can be made to describe the actual scattering problem.
In the solution we seek, there are no particles approaching from theright. Adding the even- and odd-parity solutions, we obtain at x > a asolution of the formψe (x) + ψo (x) = B sin(kx + φ) + B ′ sin(kx + φ′ ) e−ikx  (5.34)′′
eikx  iφ=
Be + B ′ eiφ −Be−iφ + B ′ e−iφ .
2i
2i

86

Chapter 5: Motion in step potentials

The condition that no particles are approaching from the right is′

B ′ = −Bei(φ −φ) ,(5.35)

for then at x > a the solution becomesψe (x) + ψo (x) =′eikx iφBe (1 − e2i(φ −φ) ) (x > a),2i(5.36)

which includes only particles moving to the right. At x < −a the solution isnowψe (x) + φo (x) = B sin(−kx + φ) − B ′ sin(−kx + φ′ ) e−ikx ′′eikx 
−Be−iφ + B ′ e−iφ +Beiφ − B ′ eiφ
=
2i
2i

′
e−ikx iφ 
ikx
−iφ
1 + e2i(φ −φ) .
Be= e iBe+2i(5.37a)In the solution given by equations (5.36) and (5.37a) the incoming amplitude
is iBe−iφ , while the amplitudes for reflection and transmission are

where


B iφ
e 1 + e2i∆φ
2i

B iφ
e 1 − e2i∆φ2i(reflected)(5.37b)(transmitted),∆φ ≡ φ′ − φ(5.37c)is phase difference ∆φ between the odd- and even-parity solutions at |x| >a. From the ratios of the mod-squares of the outgoing amplitudes to thatof the incoming amplitude iB we have that the reflection and transmission
probabilities are
Prefl = cos2 (∆φ)

Ptrans = sin2 (∆φ).

(5.38)

Thus ∆φ determines the reflection and transmission probabilities. Noticethat these formulae for the transmission and reflection probabilities havebeen obtained without reference to the form of the wavefunction at |x| < a.
Consequently, they are valid for any scattering potential V (x) that has evenparity and vanishes outside some finite region, here |x| < a.

The scattering cross section In the case that V0 < 0, so the scattering
potential forms a potential well, the outgoing wave at x > a represents
two physically distinct possibilities: (i) that the incoming particle failed tointeract with the potential well and continued on its way undisturbed, and(ii) that it was for a while trapped by the well and later broke free towards
the right rather than the left. We isolate the possibility of scattering by
writing the amplitude of the outgoing wave as 1 + T times the amplitudeof the incoming wave. Here the one represents the possibility of passing
through undisturbed and T represents real forward scattering. From ourformulae (5.37) for the amplitudes of the incoming and outgoing waves wehave that′
T = 21 e2iφ − e2iφ − 1.
(5.39a)If we similarly write the amplitude of the reflected wave as R times theamplitude of the incoming wave, then from the formulae above we have′
R = − 21 e2iφ + e2iφ .

(5.39b)5.3 Scattering of free particles

87

The total scattering cross section5 is defined to be the sum of the probabilities for forward and backward scattering:σ = |R|2 + |T |2 .

(5.40)

Now |R|2 is just the reflection probability Prefl , and the transmission probability is
Ptrans = |1 + T |2 = 1 + |T |2 + T + T ∗ ,(5.41)so

σ = Prefl + Ptrans − 1 − T − T ∗ = −(T + T ∗ ).

(5.42)

From equation (5.39a) we have an expression for the total scattering crosssection in terms of the phase angles
σ = 2 − cos(2φ′ ) + cos(2φ).

(5.43a)The trigonometric identities 1 + cos 2φ = 2 cos2 φ and 1 − cos 2φ = 2 sin2 φenable us to re-express the cross section as
σ = 2 sin2 φ′ + cos2 φ .
(5.43b)5.3.1 Tunnelling through a potential barrier
Now consider the case V0 > E in which classical physics predicts that all
particles are reflected. From equations (5.33), the conditions for both the
wavefunction and its derivative to be continuous at x = a are
A sinh(Ka) = B ′ sin(ka + φ′ )cosh(Ka) = B sin(ka + φ)orKA cosh(Ka) = B ′ k cos(ka + φ′ ),K sinh(Ka) = Bk cos(ka + φ)
(5.44a)
where
s
r
2mE
2m(V0 − E)k=and K =.
(5.44b)h̄2h̄2
Dividing the equations of each pair into one another to eliminate the constants A, B and B ′ , we obtaintan(ka + φ′ ) = (k/K) tanh(Ka).
(5.45)On account of the fact that for any x, tan(x + π) = tan x, the equations haveinfinitely many solutions for φ and φ′ that differ by rπ, where r is an integer.
From equations (5.39) and (5.43a) we see that these solutions give identicalamplitudes for reflection and transmission and the same value of the total
scattering cross section σ. Hence we need consider only the unique values of
φ and φ′ that lie within ±π/2 of −ka.
Equations (5.38) show that the transmission and reflection probabilities
are determined by the phase difference ∆φ = φ′ − φ. From (5.45) we havetan(ka + φ) = (k/K) coth(Ka) or∆φ = arctan






k
k
tanh(Ka) − arctancoth(Ka) .
K
K(5.46)

Figure 5.12 shows the transmission probability sin2 (∆φ) as a function of
the energy of the incident particle for (2mV0 a2 /h̄2 )1/2 = 0.5, 1 and 1.5. We
see that for the most permiable of these barriers the transmission probability
reaches 50% when the energy is less than a third of the energy, V0 , classicallyrequired for passage. On the other hand, the transmission probability is still
5 This definition of the total scattering cross section only applies to one-dimensional
scattering problems. See §12.3 for the definition of the total scattering cross section thatis appropriate for realistic three-dimensional experiments.

88

Chapter 5: Motion in step potentials

Figure 5.12 The transmission probability for a particle incident on a potential-energy
barrier of height V0 and width 2a as a function of the particle’s energy. The curves arelabelled by the values of the dimensionless parameter (2mV0 a2 /h̄2 )1/2 .

only 80% when E = V0 and classically the particle would be certain to pass.
A barrier of the same height but three times as thick allows the particle topass with only 2% probability when E = V0 /3, and even when E = V0 the
chance of passing this thicker barrier is only a third.
When the barrier is high, Ka ≫ 1 so both t ≡ tanh(Ka) and coth(Ka) =1/t are close to unity:t ≡ tanh(Ka) =eKa − e−Ka
≃ (1 − e−2Ka )2 .
eKa + e−Ka

(5.47)

Consequently, the arguments of the two arctan functions in equation (5.46)
are similar and we can obtain an approximate expression for ∆φ by writing




k
k 2
k
arctant = arctan+(t − 1)K
Kt Kt
(5.48)
1k 2
k+(t − 1),≃ arctanKt
1 + (k/Kt)2 Kt
where we have used the standard formula d arctan x/dx = 1/(1 + x2 ). Using
equations (5.47) and (5.48) in equation (5.46), and we have∆φ ≃ −

4e−2Ka4k≃ − e−2Ka
Kt/k + k/Kt
K(Ka ≫ 1).

(5.49)

Thus the probability of passing the barrier, sin2 (∆φ), decreases like e−4Ka
as the barrier gets higher.

5.3.2 Scattering by a classically allowed region
Now consider the case of scattering by the square potential (5.31) when
E > V0 , so the region of non-zero potential is classically allowed. Physicallythat region could be a classically surmountable barrier (V0 > 0) or a potential
well (V0 < 0). At |x| < a the wavefunctions of well-defined parity are now
either cos(Kx) or A sin(Ka) and from equation (5.33) the conditions forcontinuity of the wavefunction and its derivative at x = a arecos(Ka) = B sin(ka + φ)−K sin(Ka) = Bk cos(ka + φ)



or



A sin(Ka) = B ′ sin(ka + φ′ )KA cos(Ka) = B ′ k cos(ka + φ′ ),(5.50a)5.3 Scattering of free particles

89

Figure 5.13 The probability of reflection by three potential barriers of height V0 andthree half-widths a as functions of E/V0 . The curves are labelled by the dimensionlessparameter (2mV0 a2 /h̄2 )1/2 .

where
k=

r

2mE
h̄2

and K =s

2m(E − V0 ).
h̄2(5.50b)By dividing the second equation of each pair into the first we obtain equations
that uniquely determine the two solutions:tan(ka+φ) = −(k/K) cot(Ka) ortan(ka+φ′ ) = (k/K) tan(Ka). (5.51)

The points in Figure 5.13 at E > V0 were obtained by solving these equations6for φ and φ′ and then calculating the reflection probability cos2 (φ′ −φ), while
the remaining points were obtained from equations (5.46) for E < V0 . We
see that for all three barrier widths the reflection probability obtained for
E > V0 joins smoothly onto that for E < V0 . The reflection probability
tends to zero with increasing E/V0 as we would expect, but its dependence
on the thickness of the barrier is surprising: for E/V0 = 2 the thickest
barrier has the lowest reflection probability. In fact, the reflection probability
vanishes for E slightly larger than 2V0 and then increases at larger energies.
Similarly, the probability for transmission through the next thickest barrier
vanishes near E = 3.5V0 . The cause of this unexpected phenomenon isquantum interference between the amplitudes to be reflected from the front
and back edges of the barrier, which cancel each other when the barrier is ofa particular thickness.
When the constant V0 in the potential (5.31) is negative, there is a potential well around the origin rather than a barrier. In the classical regime
the probability of reflection is zero, but as Figure 5.14 shows, it is in general
non-zero and is large near E/|V0 | ≪ 1. The oscillations in the reflection probability apparent in Figure 5.14 are caused by quantum interference between
reflections from the two edges of the well.

5.3.3 Resonant scattering
In the limit that a barrier becomes very high, the probability that it reflects
an incoming particle tends to unity. Consequently, a particle that encounters
two high barriers (Figure 5.15) can bounce from one barrier to the other agreat many times before eventually tunnelling through one of the barriers and6 An explicit expression (5.78) for the reflection probability in terms of ka and Ka andwithout reference to φ or φ′ can be derived (Problem 5.10). This formula is useful when
limiting cases need to be examined.

90

Chapter 5: Motion in step potentials

Figure 5.14 The probability for reflection by square potential wells of depth |V0 |. The
full curve is for (2m|V |a2 /h̄2 )1/2 = 3 and the dashed curve is for a well only half as wide.

Figure 5.15 Schematic of the potential-energy function V (x) experienced by an α-particle
near an atomic nucleus. The short-range ‘strong’ force causes the particle’s potential
energy to rise extremely steeply at the edge of the nucleus. The long-range electrostatic
repulsion between the nucleus and the alpha particle causes V (x) to drop steadily as the
α-particle moves away from the nucleus.

Figure 5.16 A pair of δ-function potentials form a well within which a particle can betrapped. The forms taken by the wavefunctions of the stationary states of even (top) andodd parity are shown.

escaping to infinity. This situation arises in atomic nuclei because the shortrange ‘strong’ force confines charged particles such as protons and helium
nuclei (α-particles) within the nucleus even though it would be energeticallyadvantageous for them to escape to infinity: the electrostatic energy released
as the positively charged particle recedes from the positively charged nucleus
can more than compensate for the work done on the strong force in moving
beyond its short effective range (Figure 5.15). Some types of radioactivity –the sudden release of a charged particle by a nucleus – are caused by theseparticles tunnelling out of a well that has confined them for up to several
gigayears. We now use a toy model of this physics to demonstrate that there
is an important link between the cross section for scattering by a well and theexistence of long-lived bound states within the well. This connection makes
it possible to probe the internal structure of atomic nuclei and ‘elementary’particles with scattering experiments.
We model the barriers that form the potential well by δ-function potentials, located at x = ±a:
V (x) = Vδ {δ(x + a) + δ(x − a)}with Vδ > 0.

(5.52)

5.3 Scattering of free particles

91

Figure 5.17 The total scattering cross sections of double δ-function barriers as a functionof the wavenumber of the incoming particle. The barriers are located at x = ±a. The fullcurve is for high barriers (2mVδ a/h̄2 = 40) while the dotted curve is for lower barriers(2mVδ a/h̄2 = 10).

By integrating the tise−h̄2 d2 ψ+ Vδ δ(x)ψ = Eψ2m dx2(5.53)

for an infinitesimal distance across the location of the δ-function barriersin equation (5.52), we find that a barrier introduces a discontinuity in thegradient of the wavefunction of magnitude (cf. eq. 5.13)



dψ
= Kψ,
dx

where K ≡

2mVδ
.
h̄2

(5.54)

Hence the energy eigenstates that will enable us to calculate scattering by adouble-δ-function system take the form of sinusoids at |x| < a and at |x| > athat join continuously at x = ±a in such a way that their gradients there
differ in accordance with equation (5.54) (Figure 5.16).
At x = a the requirements on the even-parity solution ψe (x) that it becontinuous and have the prescribed change in derivative, readB sin(ka + φ) = cos(ka)kB cos(ka + φ) = −k sin(ka) + K cos(ka).

(5.55a)Similarly the conditions on the odd-parity solution ψo (x) are
B ′ sin(ka + φ′ ) = A sin(ka)kB ′ cos(ka + φ′ ) = kA cos(ka) + KA sin(ka).

(5.55b)Dividing one equation in each pair by the other to eliminate A, B and B ′ weobtaincot(ka + φ) = K/k − tan(ka)(5.56)
cot(ka + φ′ ) = K/k + cot(ka).
From these expressions and equations (5.38) we can easily recover the probability sin2 (φ′ − φ) that an incoming particle gets past both δ-function barriers. More interesting is the total scattering cross section, which is related
to the phases by equation (5.43b). Figure 5.17 shows as a function of thewavenumber of the incoming particle the cross sections for barriers of twoheights. The height of a barrier is best quantified by the dimensionless number 2mVδ a/h̄2 = Ka. The full curve in Figure 5.17 is for the case that

92

Chapter 5: Motion in step potentials

Ka = 40 and the dotted curve is for the case of a lower barrier such that
Ka = 10. In each case the cross section shows a series of peaks. In the caseof the higher barrier, these peaks lie near ka = nπ/2, with n = 1, 2, . . .. In
the case of the lower barrier the peaks are less sharp and occur at slightly
smaller values of ka.
If the barriers were so high as to be impenetrable, the particle would havebound states with ka = nπ/2, which is the condition for the wavefunction tovanish at x = ±a. Each peak in the scattering cross section is associated with
one of these bound states. Physically, the scattering cross section is largenear the energy of a bound state because at such an energy the particle canbecome temporarily trapped between the barriers, and after a delay escapeeither to the right or the left.
When the barriers have only finite height, the state |trapi in which
the particle is initially trapped in the well is not a stationary state, andits expansion in stationary states will involve states whose energies span a
non-zero range, say (E0 − Γ/2, E0 + Γ/2). For simplicity we assume that
|trapi has even parity, so it can be expressed as a linear combination of theeven-parity stationary states |e; Ei:|trapi =Z E0 +Γ/2

dE a(E)|e; Ei,(5.57)

E0 −Γ/2

where a(E) is the amplitude to measure energy E. Outside the well the
wavefunction of this state isψtrap (x) ∝Z E0 +Γ/2

dE a(E) sin(kx + φ)(x > a).

(5.58)

E0 −Γ/2

Below we shall find that when the well is very deep, φ becomes a sensitive
function of E in the neighbourhood of particular ‘resonant’ energies. Thenthe sines in equation (5.58) cancel essentially perfectly on account of therapidly changing phase φ(E). When the integral is small, there is negligible
probability of finding the particle outside the well.
The evolution of ψtrap with time is obtained by adding the usual factors
e−iEt/h̄ in the integral of equation (5.58):
ψtrap (x, t) ∝Z E0 +Γ/2

dE a(E) sin(kx + φ) e−iEt/h̄

(x > a).

(5.59)

E0 −Γ/2

After a time of order h/Γ the relative phases of the integrand at E0 −Γ/2 andE0 + Γ/2 will have changed by π and the originally perfect cancellation of
the sines will have been sabotaged. The growth of the value of the integral,and therefore the wavefunction outside the well, signals increasing probability
that the particle has escaped from the well. The more rapidly φ changes withE, the smaller is the value of Γ at which a negligible value of the integral inequation (5.58) can be achieved, and the smaller Γ is, the longer the particle
is trapped. Thus sensitive dependence of the phases on energy is associatedwith long-lived trapped states, which are in turn associated with abnormallylarge scattering cross sections. Notice that in Figure 5.17 the peaks are
narrower at small values of k because the smaller the particle’s energy is, thesmaller is its probability of tunnelling through one of the barriers.
The Breit–Wigner cross section We have seen that when particles arescattered by a model potential that contains a well, the total scattering crosssection has narrow peaks. The physical arguments given above suggest that
this behaviour is generic in the sense that it is related to the time it takes
a particle to tunnel out of the well after being placed there. What we have
yet to do is to understand mathematically how the fairly simple formulae(5.43b) and (5.56) generate sharp peaks in the energy dependence of σ. An

5.3 Scattering of free particles

93

Figure 5.18 The values of the phases φ (full curve) and φ′ (dashed curve) from equations(5.56) as functions of the wavenumber of the incoming particle when the latter is scatteredby the double δ-function well (eq. 5.52). These results are for the case 2mVδ a/h̄2 = 40.

understanding of this phenomenon will motivate a simple analytic model ofresonant scattering that is widely used in experimental physics.
Figure 5.18 shows the values of the phases φ and φ′ that solve equation(5.56). For most values of k (and therefore E), the two angles are equal, so
the sum sin2 φ′ + cos2 φ in equation (5.43b) is unity. The peaks in σ occur
where φ and φ′ briefly diverge from one another at the integral-sign featuresin Figure 5.18, which we shall refer to as ‘glitches’.
We are interested in the case K/k ≫ 1. Then for most values of ka
the right sides of equations (5.56) are dominated by the first term, so thecotangent on the left is equal to some large positive value, and its argument
lies close to zero. However each time ka/π approaches (2r + 1)π/2 with r
an integer, the tangent in the first equation briefly overwhelms K/k and theright side changes from a large positive number to a large negative number.
Consequently the argument of the cotangent on the left quickly increases toa value close to π. As ka increases through (2r + 1)π/2, the tangent instantaneously changes sign, and the argument of the cotangent instantaneously
returns to a small value. Examination of the third glitch in Figure 5.18 confirms that φ rises rapidly but continuously by almost π and then suddenly
drops by exactly π as this analysis implies. The abrupt rise in φ is centredon the point at which ka + φ = π/2, at which point φ ≃ −rπ because at a
glitch ka ≃ (2r + 1)π/2. Consequently, glitches in φ are centred on points
at which cos2 (φ) = 1. Meanwhile φ′ ≃ −(2r + 1)π/2, so sin2 (φ′ ) = 1 andequation (5.43b) gives σ = 4. A very similar analysis reveals how the secondof equations (5.56) generates glitches in φ′ .
Putting this argument on a quantitative basis, we Taylor expand tan(ka)around the resonant value of k, kR , at which tan(kR a) = K/kR . Then2
K/k − tan(ka) ≃ −sec2 (kR a)aδk = −(1 + K 2 /kR)aδk,(5.60)

where δk = k − kR . We also observe that glitches in φ occur where ka ≃(2r + 1)π/2, socot(ka + φ) =cos(ka) cos(φ) − sin(ka) sin φ≃ − tan φ.
sin(ka) cos φ + cos(ka) sin(φ)

With these approximations the first of equations (5.56) reads
2aδk.
tan φ ≃ 1 + K 2 /kR(5.61)

94

Chapter 5: Motion in step potentials

Figure 5.19 The total cross section for the scattering of neutrons by 238 U nuclei. (Fromdata published in L.M. Bollinger, et al., Phys. Rev., 105, 661 (1957))

Equation (5.43b) now gives the cross section as



1
2 ′
2 ′
2
σ = 2 sin φ + cos φ = 2 sin φ +1 + tan2 φ


1
2 ′
≃ 2 sin φ +.
2 )2 a2 (δk)2
1 + (1 + K 2 /kR(5.62)

2
Thus in the vicinity of the resonant energy ER = h̄2 kR/2m, whereδk =m(E − ER ),h̄2 kR(5.63)

the cross section has the form
σ = constant +
where
Γ≡

2(Γ/2)2(Γ/2)2 + (E − ER )2,2h̄2 k 32h̄2 kR≃ 2 R.
2
2(1 + K /kR )am
K am(5.64a)(5.64b)Γ has the dimensions of energy and is the characteristic width of the resonance. Experimental data for the energy dependence of cross sections areoften fitted to the functional form defined by equation (5.64a), which is
known as the Breit–Wigner cross section. Figure 5.19 shows a typicalexample.
The dependence on energy of the phase φ and the total scattering crosssection σ in the vicinity of a peak in σ is reminiscent of the behaviour neara resonance of a lightly damped harmonic oscillator (Box 5.1).
By the uncertainty principle, the width Γ of the Breit–Wigner crosssection (5.64a) corresponds to a time scaletR =h̄K 2 am
=
3 .
Γ
2h̄kR

(5.65)A naive calculation confirms that tR is the timescale on which a particle
escapes from the well. When a particle encounters a δ-function barrier, it iseasy to show (Problem 5.11) that its probability of tunnelling through thebarrier is
Ptun =4≃ 4(kR /K)2
4 + (K/kR )2forK ≫ kR .

(5.66)

5.4 How applicable are our results?

95

Box 5.1: Analogy with a damped oscillator
When a weakly damped harmonic oscillator is driven at some angularfrequency ω, the phase of the steady-state response changes sharply inthe vicinity of the oscillator’s resonant frequency ωR . Specifically, if theoscillator’s equation of motion is2ẍ + γ ẋ + ωRz = F cos(ωt),

then the steady-state solution is x = X cos(ωt − φ), where


F
γω
X=p 2
.
and φ = arctan
2 − ω2
ωR(ωR − ω 2 )2 + γ 2 ω 2

As the driving frequency approaches the resonant frequency from below,
the phase lag φ increases from near zero to π/2. As the driving frequency
passes through resonance, φ drops discontinuously to −π/2, and then in2
creases to near zero as ω 2 − ωRbecomes large compared to ωγ. These
results suggest a picture in which a quantum well is an oscillator thatis being driven by the incoming probability amplitude. The oscillator’slevel of damping is set by the well’s characteristic energy Γ of equation (5.64b), and the form of the Breit–Wigner cross section of equation(5.64a) mirrors the Lorentzian form of the oscillator’s amplitude X.
Hence the probability of remaining in the well after bouncing n times off thewalls isPtrap = (1 − Ptun )n .
(5.67)The particle moves from one barrier to the other in a time tf = 2am/h̄kR
and in this time the logarithm of Ptrap changes by ln(1 − Ptun ) ≃ −Ptun , so
d
1
Ptun
=− ,
ln(Ptrap ) ≃ −
dt
tf
tR

(5.68)

where tR is given by equation (5.65). Thus this simple physical argument
confirms that h̄/Γ is the characteristic time for the particle to remain in thewell.

5.4 How applicable are our results?
It seems unlikely that any real system has a discontinuous potential V (x), so
our results are of practical interest only if sufficiently steep changes in V canbe treated as discontinuous. We now investigate how abrupt a change in potential must be for results obtained under the assumption of a discontinuous
potential to be applicable.
When the wavefunction is evanescent (i.e. ∝ e±Kx ) on one side of the
discontinuity, our results carry over to potentials that change continuously:
where E < V (x), the wavefunction is no longer a simple exponential but
its phase remains constant and its amplitude decreases monotonically, whilein the region E > V (x) the sinusoidal dependence ψ(x) ∝ eikx is replacedby some other oscillatory function of similar amplitude. Qualitatively the
results we have obtained for particles confined by a step potential carry overto continuously varying potentials when E < V on one side of a region ofvarying potential.
The situation is less clear when E > V (x) on both sides of the change
in the potential. The relevance to such cases of our solutions for step potentials can be investigated by solving the tise numerically for a potential

96

Chapter 5: Motion in step potentials

Figure 5.20 The full curve shows the probability of reflection when a particle moves fromx = −∞ in the potential (5.69) with energy E = h̄2 k 2 /2m and V0 = 0.7E. The dottedline is the value obtained for a step change in the potential (Problem 5.4).

Figure 5.21 Each curve shows the reflection probability when a particle with kinetic
energy E encounters a region in which the potential V (x) smoothly changes to V0 overa distance 2b, and then smoothly returns to zero; the change in V is given by equation(5.69) with x replaced by x + a, and the fall is given by the same equation with x replacedby −(x − a). The full curves are for ka = 30 and the dashed curves for ka = 15. The
left panel is for barriers of height V0 = 0.7E, while the right panel is for potential wells(V0 = −0.7E).

that changes over a distance that can be varied (Problem 5.15). Consider
for example
V (x) = V0(01
2 [1 + sin(πx/2b)]

1

for x < −bfor |x| < bfor x > b,

(5.69)

which changes from 0 to V0 over a distance 2b centred on the origin. Figure 5.20 shows the probability of reflection when a particle with energyh̄2 k 2 /2m = V0 /0.7 encounters this rise in potential energy as it approaches
from x = −∞. For kb ≪ 1, the probability of reflection is close to that obtained for the corresponding step potential (Problem 5.4), but it falls to very
7much smaller values for kb >∼ 2. Thus treating a rapid change in potential
energy as a discontinuity can lead to a serious over-estimate of the reflected
amplitude.
Figure 5.21 shows reflection probabilities for particles of energy E thatencounter a finite region of elevated or depressed potential energy as a function of the sharpness of the region’s sides – the changes in potential energy
occur in a distance 2b as described by equation (5.69). The left panel is forpotential barriers of height 0.7E and the right panel is for potential wells of
7 The ‘wkbj’ approximation derived in §11.6 provides an analytic approximation to
the solution of the tise when kb is significantly larger than 2π.

5.4 How applicable are our results?

97

depth 0.7E. The full curves are for the case in which the region’s half-widtha satisfies ka = 30, where k is the wavenumber of the incoming and outgoing
wavefunctions, while the dashed curves are for regions only half as wide. The
left panel shows that when kb = 1, the reflection probability generated by asmooth barrier is smaller than that for a sharp step by a factor ∼ 1.7, andwhen kb = 2 it is nearly a factor ten smaller than in the case of an abrupt
barrier. The right panel shows that in the case of a potential well, evensmaller values of kb are required for the assumption of an abrupt change inV (x) to be useful. Thus these results confirm the implication of Figure 5.20
that modelling a change in potential energy by a sharp step is seriously misleading unless the half width of the transition region b satisfies the conditionkb < 1.
The amplitude to be reflected by a region of varying potential energy
decreases rapidly with increasing half width b of the transition region, but
the amplitudes to be reflected at the leading and trailing edges of a regionof varying potential remain comparable. Consequently, destructive interference between these amplitudes is possible for all values of b. Moreover, thephases of the reflected amplitudes depend on b, so the plots of overall reflection probability versus b in Figure 5.21 show there are regular nulls in theprobability for reflection like those we see in Figure 5.14 for the probability
for reflection by an abrupt potential well.
We conclude that many qualitative features of results obtained with step
potentials also hold for continuous potentials, but results obtained for step
potentials with no classically forbidden region are quantitatively misleadingwhen applied to continuous potentials unless the distance over which thepotential changes is small compared to the de Broglie wavelength λ = h/pof the incident particles.
Let’s consider under what circumstances this condition could be satisfied
for a stream of electrons. The de Broglie wavelength of electrons with kinetic
energy E is
−1/2

E
nm.
(5.70)
λ = 1.16
1 eV
For the one-dimensional approximation to apply, we need the beam to be
many λ wide, and for the step approximation to be valid, we require thechange in potential to be complete well inside λ. In practice these conditions
can be simultaneously satisfied only when the potential change is associatedwith a change in the medium through which the electrons are propagating.
If the medium is made of atoms, the change must extend over at least thecharacteristic size of atoms 0.1 nm. Hence we require E <∼ 1 eV.
Realistically step potentials are relevant only for less massive particles,
photons and neutrinos. The propensity for some photons to be transmitted and some reflected at an abrupt change in potential, such as that at a
glass/air interface, plays an important role in optics. By contrast, electrons,
neutrons and protons are unlikely to be partially transmitted and partiallyreflected by a region of varying potential.
These considerations explain why the phenomenon of partial reflection
and partial transmission is unknown to classical mechanics, which is concerned with massive bodies that have de Broglie wavelengths many orders ofmagnitude smaller than an atom at any experimentally accessible energy.

98Chapter 5: Motion in step potentials

5.5 Summary
In this chapter we have examined some highly idealised systems and reachedsome surprising conclusions.
• Any one-dimensional potential well has at least one bound state, andmay have more depending on the size of the value of the dimensionlessparameter W defined by equation (5.8), with V0 and a interpreted as
the well’s characteristic depth and width, respectively.
• A particle trapped by a very narrow or shallow well has negligible probability of being found in the well.
• When solving the tise in the presence of an infinite step in the potential,
we should require the wavefunction to vanish at the base of the step.
• When two identical square potential wells are separated by a barrier,the eigenenergies occur in pairs, and the associated wavefunctions have
either even or odd parity with respect to an origin that is symmetricallyplaced between the wells. The even-parity state of a pair lies slightly
lower in energy than the odd-parity state. A sum of the lowest two
eigenstates is a state in which the particle is certainly in one well, while
the difference gives a state in which the particle is certainly in the other
well. A particle that starts in one well oscillates between the wells with aperiod inversely proportional to the difference between the eigenenergies.
The particle is said to ‘tunnel’ through the barrier that divides the two
wells at a rate that decreases exponentially with the product of thebarrier’s height and the square of its width.
• In an ammonia molecule the nitrogen atom moves in an effective potential that provides two identical wells and the above model explains how
an ammonia maser works.
• A free particle has a non-zero probability to cross a potential barrierthat would be impenetrable according to classical physics. On the other
hand, if the potential changes significantly within one de Broglie wavelength, a particle generally has a non-zero probability of being reflectedby a low barrier that classical physics predicts will be crossed.
• The probabilities for a free particle to be reflected or transmitted by apotential barrier or a well with very steep sides oscillate as functions of
the particle’s energy on account of quantum interference between theamplitudes to be reflected at the front and back edges of the barrier orwell.
• When a free particle is scattered by a region that contains a potential
well, the total scattering cross section peaks in the vicinity of the energiesof the well’s approximately bound states. Longer-lived bound states areassociated with sharper peaks in a plot of scattering cross section versus
energy because the width in energy of a peak, Γ, and the lifetime t0 of
the corresponding bound state are related by the uncertainty relationt0 Γ ∼ h̄.
• The Breit–Wigner formula (5.64a) gives the energy-dependence of a scattering cross-section near a resonance, and the timescale h̄/Γ that appears
in it is the typical time for which a particle is trapped.
• The results we have obtained for discontinuous potentials V (x) are an
accurate guide to what will happen when the potential changes continuously when either (i) the particle is classically forbidden on one sideof the change, or (ii) the change is complete within a fraction of a de
Broglie wavelength. When the potential changes more gradually, theamplitude to be reflected by the region of changing potential is typicallymuch smaller than in our idealised examples, but, as in these examples,
the amplitude to be reflected oscillates as a function of energy.

Problems

99

Problems
5.1 A particle is confined by the potential well
V (x) =

n0 for |x| < a∞ otherwise.

(5.71)

Explain (a) why we can assume that there is a complete set of stationary
states with well-defined parity and (b) why to find the stationary states wesolve the tise subject to the boundary condition ψ(±a) = 0.
Determine the particle’s energy spectrum and give the wavefunctions of
the first two stationary states.
5.2 At t = 0 the particle of Problem 5.1 has the wavefunction
ψ(x) =



√
1/ 2a
0

for |x| < aotherwise.

(5.72)

Find the probabilities that a measurement of its energy will yield: (a)9h̄2 π 2 /(8ma2 ); (b) 16h̄2 π 2 /(8ma2 ).
5.3 Find the probability distribution of measuring momentum p for theparticle described in Problem 5.2. Sketch and comment on your distribution.
Hint: express hp|xi in the position representation.
5.4 Particles move in the potential

V (x) =0
V0

for x < 0for x > 0.

(5.73)

Particles of mass m and energy E > V0 are incident from x = −∞. Show
that the probability that a particle is reflected is


k−K
k+K

2

,

(5.74)

p√where k ≡ 2mE/h̄ and K ≡ 2m(E − V0 )/h̄. Show directly from thetise that the probability of transmission is
4kK(k + K)2(5.75)and check that the flux of particles moving away from the origin is equal tothe incident particle flux.
5.5 Show that the energies of bound, odd-parity stationary states of thesquare potential well
V (x) =



0V0 > 0for |x| < aotherwise,(5.76)are governed bycot(ka) = −s

W2− 1 where
(ka)2

W ≡

r

2mV0 a2h̄2

and k 2 = 2mE/h̄2 .

(5.77)Show that for a bound odd-parity state to exist, we require W > π/2.

100Problems

Figure 5.22 The real part of the wavefunction when a free particle of energy E is scatteredby a classically forbidden square barrier barrier (top) and a potential well (bottom). The
upper panel is for a barrier of height V0 = E/0.7 and half width a such that 2mEa2 /h̄2 = 1.
The lower panel is for a well of depth V0 = E/0.2 and half width a such that 2mEa2 /h̄2 =9. In both panels (2mE/h̄2 )1/2 = 40.

Figure 5.23 A triangle for Problem 5.10

5.6 Show that the correctly normalised wavefunctionof a particle trapped√by the potential V (x) = −Vδ δ(x) is ψ(x) = Ke−K|x| , where K = mVδ /h̄2 .
Show that although this wavefunction makes it certain that a measurementof x will find the particle outside the well where its kinetic energy is negative, the expectation value of its kinetic energy hEK i = 21 mVδ2 /h̄2 is in fact
positive. Reconcile this apparent paradox as follows: (i) show that for anarrow, deep potential well of depth V0 and half-width a, with 2V0 a = Vδ ,ka ≃ W ≡ (2mV0 a2 /h̄2 )1/2 , while Ka ≃ W 2 . (ii) Hence show that the contribution from inside the well to hEK i is |ψ(0)|2 Vδ regardless of the value ofa. Explain physically what is happening as we send a → 0.
5.7 Reproduce the plots shown in Figure 5.22 of the wavefunctions of particles that are scattered by a square barrier and a square potential well. Givephysical interpretations of as many features of the plots as you can.
5.8 Give an example of a potential in which there is a complete set ofbound stationary states of well-defined parity, and an alternative complete
set of bound stationary states that are not eigenkets of the parity operator.
Hint: modify the potential discussed apropos NH3 .
5.9 A free particle of energy E approaches a square, one-dimensional potential well of depth V0 and width 2a. Show that the probability of beingreflected by the well vanishes when Ka = nπ/2, where n is an integer and
K = (2m(E + V0 )/h̄2 )1/2 . Explain this phenomenon in physical terms.
5.10 Show that the phase shifts φ (for the even-parity stationary state)and φ′ (for the odd-parity state) that are associated with scattering by aclassically allowed region of potential V0 and width 2a, satisfy
tan(ka + φ) = −(k/K) cot(Ka) andtan(ka + φ′ ) = (k/K) tan(Ka),Problems

101

where k and K are, respectively, the wavenumbers at infinity and in the
scattering potential. Show that
Prefl = cos2 (φ′ − φ) =(K/k − k/K)2 sin2 (2Ka).
(K/k + k/K)2 sin2 (2Ka) + 4 cos2 (2Ka)(5.78)

Hint: apply the cosine rule for an angle in a triangle in terms of the lengthsof the triangle’s sides to the top triangle in Figure 5.23.
5.11 A particle of energy E approaches from x < 0 a barrier in which thepotential energy is V (x) = Vδ δ(x). Show that the probability of its passing
the barrier is
r
1
2mE
2mVδ
where k =Ptun =, K=.
(5.79)
1 + (K/2k)2h̄2
h̄2
5.12 An electron moves along an infinite chain of potential wells. For
sufficiently low energies we can assume that the set {|ni} is complete, where|ni is the state of definitely being in the nth well. By analogy with ouranalysis of the NH3 molecule we assume that for all n the only non-vanishing
matrix elements of the Hamiltonian are E ≡ hn|H|ni and A ≡ hn ± 1|H|ni.
Give physical interpretations of the numbers A and E.
Explain why we can write
H=

∞
X

n=−∞

E|nihn| + A (|nihn + 1| + |n + 1ihn|) .

Writing an energy eigenket |Ei =P(5.80)n an |ni show thatam (E − E) − A (am+1 + am−1 ) = 0.

(5.81)

Obtain solutions of these equations in which am ∝ eikm and thus find thecorresponding energies Ek . Why is there an upper limit on the values of kthat need be considered?
Initially the electron is in the state1|ψi = √ (|Ek i + |Ek+∆ i) ,2

(5.82)

where 0 < k ≪ 1 and 0 < ∆ ≪ k. Describe the electron’s subsequent motion
in as much detail as you can.
5.13∗ In this problem you investigate the interaction of ammonia moleculeswith electromagnetic waves in an ammonia maser. Let |+i be the state inwhich the N atom lies above the plane of the H atoms and |−i be the state inwhich the N lies below the plane. Then when there is an oscillating electricfield E cos ωt directed perpendicular to the plane of the hydrogen atoms, theHamiltonian in the |±i basis becomesE + qEs cos ωt−AH=.
(5.83)
E − qEs cos ωt−ATransform this Hamiltonian from the |±i basis to the basis provided√ by thestates of well-defined parity |ei and |oi (where |ei = (|+i + |−i)/ 2, etc).
Writing|ψi = ae (t)e−iEe t/h̄ |ei + ao (t)e−iEo t/h̄ |oi,
(5.84)
show that the equations of motion of the expansion coefficients aredae= −iΩao (t) ei(ω−ω0 )t + e−i(ω+ω0 )t
dt


dao
= −iΩae (t) ei(ω+ω0 )t + e−i(ω−ω0 )t ,dt

(5.85)

102

Problems

where Ω ≡ qEs/2h̄ and ω0 = (Eo − Ee )/h̄. Explain why in the case of amaser the exponentials involving ω + ω0 a can be neglected so the equationsof motion becomedae= −iΩao (t)ei(ω−ω0 )t
dt

;

dao
= −iΩae (t)e−i(ω−ω0 )t .
dt

(5.86)

Solve the equations by multiplying the first equation by e−i(ω−ω0 )t and differentiating the result. Explain how the solution describes the decay of a population of molecules that are initially all in the higher energy level. Compare
your solution to the result of setting ω = ω0 in (5.86).
5.14 238 U decays by α emission with a mean lifetime of 6.4 Gyr. Take thenucleus to have a diameter ∼ 10−14 m and suppose that the α particle has
been bouncing around within it at speed ∼ c/3. Modelling the potential
barrier that confines the α particle to be a square one of height V0 and width
2a, give an order-of-magnitude estimate of W = (2mV0 a2 /h̄2 )1/2 . Given
that the energy released by the decay is ∼ 4 MeV and the atomic numberof uranium is Z = 92, estimate the width of the barrier through which the
α particle has to tunnel. Hence give a very rough estimate of the barrier’s
typical height. Outline numerical work that would lead to an improved
estimate of the structure of the barrier.
5.15∗ Particles of mass m and momentum h̄k at x < −a move in thepotential(
0for x < −a(5.87)
V (x) = V0 12 [1 + sin(πx/2a)] for |x| < a1
for x > a,where V0 < h̄2 k 2 /2m. Numerically reproduce the reflection probabilities
plotted Figure 5.20 as follows. Let ψi ≡ ψ(xj ) be the value of the wavefunction at xj = j∆, where ∆ is a small increment in the x coordinate. From
the tise show thatψj ≃ (2 − ∆2 k 2 )ψj+1 − ψj+2 ,(5.88)

pwhere k ≡ 2m(E − V )/h̄. Determine ψj at the two grid points with thelargest values of x from a suitable boundary condition, and use the recurrence
relation (5.88) to determine ψj at all other grid points. By matching thevalues of ψ at the points with the smallest values of x to a sum of sinusoidal
waves, determine the probabilities required for the figure. Be sure to checkthe accuracy of your code when V0 = 0, and in the general case explicitly
check that your results are consistent with equal fluxes of particles towards
and away from the origin.
Equation (11.40) gives an analytical approximation for ψ in the casethat there is negligible reflection. Compute this approximate form of ψ andcompare it with your numerical results for larger values of a.
5.16∗ In this problem we obtain an analytic estimate of the energy difference between the even- and odd-parity states of a double square well. Show
that for large θ, coth θ − tanh θ ≃ 4e−2θ . Next letting δk be the difference
between the k values that solveps cotheven parity
W 2 − (ka)2
W2

p
−
1
=
tan [rπ − k(b − a)] tanh(ka)2W 2 − (ka)2
odd parity,(5.89a)

where

W ≡

r

2mV0 a2h̄2(5.89b)

Problems

103

for given r in the odd- and even-parity cases, deduce that "

−1/2 #


−1 
(ka)2
1
1−
δk
k
W2
i
h p
≃ −4 exp −2 W 2 − (ka)2 .
(5.90)
Hence show that when W ≫ 1 the fractional difference between the energiesof the ground and first excited states is
W2
−1
(ka)2

1/2

+



W2
−1
(ka)2

(b − a) +

√
δE
−8a
≃
e−2W 1−E/V0 .
E
W (b − a)(5.91)

5.17 We consider the scattering of free particles of mass m that move in
one-dimension in the potential V (x) = −W δ(x), with W > 0. (a) For a wellof finite depth V0 and width 2a the condition on the phases φ and φ′ of theeven- and odd-parity wavefunctions ψ ∝ sin(kx + φ), etc, for free particles
aretan(ka + φ) = −kcot(Ka) ;Ktan(ka + φ′ ) = −ktan(Ka)K

Show that in the limit a → 0, V0 = W/2a → ∞ we have tan φ → −h̄2 k/mW
and φ′ → 0. Hence obtain the scattering cross section by the δ-function
potential2
σ=.
(5.92)
1 + (h̄k/mW )2(b) Re-derive the equation above for φ by requiring that ψ = sin(k|x| + φ)satisfy the tise. Convince yourself that ψ = sin(kx) is also consistent withthe tise.
(a) The wavenumber k is constant as we send a → 0 so ka → 0.

6Composite systems
Systems often consist of more than one part. For example a hydrogen atom
consists of a proton and an electron, and a diamond consists of a very large
number of carbon atoms. In these examples of composite systems there issignificant physical interaction between the component parts of the system– the electron moves in the electromagnetic field of the proton, and electromagnetic forces act between the atoms in a diamond. But in principle there
need be no physical interaction between the parts of a composite system: it isenough that we consider the sum of the parts to constitute a single system.
For example ‘quantum cryptography’ exploits correlations between widely
separated photons that are not interacting with each other, and in §7.5 weshall study a system that consists of two completely unconnected gyros that
happen to be in the same box. Even in classical physics specifying the stateof such a system is a complex business because in general there will be correlations between the parts of the system: the probability for obtaining acertain value for an observable of one subsystem depends on the state of theother subsystem. In quantum mechanics correlations arise through quantum
interference between various states of the system, with the result that correlations are sometimes associated with unexpected and sometimes puzzling
phenomena.
In §6.1 we extend the formalism of quantum mechanics to composite
systems. We introduce the concept of ‘quantum entanglement’, which is
how correlations between the different parts of a composite system are represented in quantum mechanics, and we find that subsystems have a propensity to become entangled. In §6.1.4 we discuss a thought experiment withentangled particles that Einstein believed demonstrated that quantum mechanics is merely an incomplete substitute for a deeper theory. Experimentsof this type have since been carried out and the results are inconsistent witha theory of the type sought by Einstein. In §6.2 we introduce the principal
ideas of quantum computing, which is the focus of much current experimental work and has the potential to revolutionise computational mathematics
with major implications for the many aspects of our civilisation that rely
on cryptography. In §6.3 we introduce the operator that enables us to drop
unrealistic assumptions about our level of knowledge of the states of quantum systems and introduce the key concept of entropy. In §6.4 we show that
thermodynamics arises naturally from quantum mechanics. In §6.5 we comeclean about the intellectual black hole that lurks at the heart of quantum
mechanics: the still unresolved problem of measurement.

6.1 Composite systems

105

At several points in the chapter we encounter fundamental questions
about quantum mechanics with which experimental and theoretical physicists are currently wrestling. It is a remarkable feature of quantum mechanics
that already the sixth chapter of an introduction to the subject can bringstudents to the frontier of human understanding.

6.1 Composite systems
Once we understand how to combine two systems A and B to make a composite system AB, we will be in a position to build up systems of arbitrarycomplexity, because we will be able to combine the system AB with someother system C to make a system ABC, and so on indefinitely. So we now
consider what is involved in forming AB out of A and B.
Suppose {|A; ii} and {|B; ji} are sets of states of A and B, respectively.
Then the symbolic product |A; ii|B; ji is used to denote that state of thecomposite system in which A is in the state |A; ii and B is in the state |B; ji:clearly this is a well defined state of AB. We express this fact by writing|AB; i, ji = |A; ii|B; ji,(6.1a)

where the label of the ket before the semicolon indicates what system ishaving its state specified, and the label after the semicolon enumerates thestates. The Hermitian adjoint of equation (6.1a) ishAB; i, j| = hA; i|hB; j|,(6.1b)

and we define the product of a bra and a ket of AB by the rulehAB; i′ , j ′ |AB; i, ji = hA; i′ |A; iihB; j ′ |B; ji.

(6.2)

This rule is well defined because the right side is simply a product of twocomplex numbers. It is a physically sensible rule because it implies that theprobability that AB is in the state i′ j ′ is the product of the probability that
A is in state i′ and B is in state j ′ :p(AB; i′ , j ′ ) = |hAB; i′ , j ′ |AB; i, ji|2 = |hA; i′ |A; ii|2 |hB; j ′ |B; ji|2
= p(A; i′ )p(B; j ′ ).

(6.3)

Any state of AB that like (6.2) can be written as a product of a stateof A and a state of B is rather special. To see this, we consider the simplest
non-trivial example, in which both A and B are two-state systems. Let |+i
and |−i be the two basis states |A; ii of A and let |↑i and |↓i be the twobasis states |B; ji of B – we shall call these the ‘up’ and ‘down’ states of B.
We use these basis states to expand the states of the subsystems:|Ai = a− |−i + a+ |+i ;|Bi = b↓ |↓i + b↑ |↑i,(6.4)

so the state |ABi = |Ai|Bi of AB can be written|ABi = (a− |−i + a+ |+i) (b↓ |↓i + b↑ |↑i, )= a− b↓ |−i|↓i + a− b↑ |−i|↑i + a+ b↓ |+i|↓i + a+ b↑ |+i|↑i.

(6.5)

The coefficients in this expansion are the amplitudes for particular events– for example a− b↓ is the amplitude that A will be found to be minus andB will be found to be down. From them we obtain a relation between theprobabilities of finding A to be in its plus state and B to be either up or
down:
p+↑
|b↑ |2
=
.
(6.6)
p+↓
|b↓ |2

106

Chapter 6: Composite systems

Now by Bayes’ theorem, the probability of finding B to be up given that A
is plus isp(B; ↑|A; +) =

p+↑
1
p+↑
=
.
=
p(A; +)p+↑ + p+↓
1 + p+↓ /p+↑

(6.7)

With equation (6.6) this simplifies top(B; ↑|A; +) =1.
1 + |b↓ /b↑ |2(6.8)

The key thing is that the right side of this expression makes no reference tosubsystem A. Evidently, when the state |ABi of the composite system canbe written as a product |Ai|Bi of states of the subsystems, the probabilityof finding B to be up is independent of the state of A. That is, the two
subsystems are uncorrelated or statistically independent. Usually thestates of subsystems are correlated and then the state of AB cannot be
expressed as a simple product |Ai|Bi.
For example, suppose we have two vertical gear wheels, A with NA
teeth and B with NB teeth. Then the state of A is specified by giving theamplitudes ai that the ith tooth is on top of the wheel. The state of B issimilarly specified by the amplitudes bj for each of its teeth to be uppermost.
However, if both wheels are members of the same train of gears (as in aclock), the probability that the j th tooth of B is on top will depend on which
tooth of A is uppermost. When the orientations of the wheels are correlated
in this way, each of the NA NB configurations of the pair of wheels has anindependent probability, pij . Specifically, when NA = NB , pij will vanish
except when i = j. If these gear wheels are uncorrelated because they are
not meshed together, we need to specify only the NA + NB amplitudes ai andbj . Once the wheels become correlated as a result of their teeth meshing, wehave to specify NA NB amplitudes, one for each probability pij .
We now assume that the sets {|A; ii} and {|B; ji} are complete for theirrespective systems and show that the set of states given by equation (6.1a)
for all possible values of i, j is then a complete set of states for the composite
system. That is, any state |AB; ψi of AB can be written|AB; ψi =X
ij

cij |AB; i, ji =X
ijcij |A; ii|B; ji.

(6.9)

The proof involves supposing on the contrary that there is a state |AB; φi ofAB that cannot be expressed in the form (6.9). We construct the object|AB; χi ≡ |AB; φi −Xijcij |AB; i, ji where cij = hAB; i, j|AB; φi. (6.10)

This object cannot vanish or |AB; φi would be of the form (6.9). But when
AB is in this state, the amplitude for subsystem A to be in any of the states|A; ii vanishes:
X(hA; i|hB; j|) |AB; χi = 0.
(6.11)j

This conclusion is absurd because the set {|A; ii} is by hypothesis complete,
so the hypothesised state |AB; φi cannot exist. Thus we have shown that ageneral state of AB is specified by NA NB amplitudes, just as the argument
about gear wheels suggested.
This result implies that the number of amplitudes required to specify thestate of a composite system grows exceedingly rapidly with the complexityof the subsystems – for example, if NA = NB = 1000, a million amplitudes
are required to specify a general state of AB. By contrast only 2000 amplitudes are required to specify a product state |ABi = |Ai|Bi because the6.1 Composite systems

107

form of such a state automatically sets to zero all correlations between thesubsystems. For a general state a large number of amplitudes are required
to specify these correlations.
Even when a state of AB is given by an expansion of the form (6.9) thatinvolves NA NB amplitudes, the states of A and B may not be correlated. To
PNA
see this let |A; ψi = i=1
ai |A; ii be the state of the subsystem A and let
PNB|B; φi = j=1 bj |B; ji be the state of subsystem B. Then the state of thecomposite system AB isX
ai bj |A; ii|B; ji.
(6.12)|AB; χi ≡ |A; ψi|B; φi =ij

The right side of this equation is identical to the right side of equation (6.9)
except that cij has been replaced with ai bj . Thus equation (6.12) is an
instance of the general expansion (6.9), but it is a very special instance: ingeneral the expansion coefficients cij , which can be thought of as the entries
in an NA ×NB matrix, cannot be written as the product of an NA -dimensional
vector with entries ai and an NB -dimensional vector bj . To see that this isso, consider the ratio cij /cij ′ of the matrix elements in the same row but
different columns. When cij can be expressed as the product of two vectors,
we havecij
ai b jbj==,(6.13)′′
cij
ai b jbj ′so this ratio is independent of i. That is, when the state of AB can be written
as the product of a state of A and a state of B, the expansion coefficients cij
are restricted such that every row of the matrix that they form is a multipleof the top row. Similarly, in this case every column is a multiple of theleftmost column (Problem 6.3).
When the state of AB cannot be written as the product of a state of A
and a state of B, we say that the subsystems A and B are entangled. As wehave seen, the observables of entangled systems are correlated, so we could aswell say that the subsystems are correlated. It is remarkable that correlations
between subsystems, which are as evident in classical physics as in quantum
mechanics, arise in quantum mechanics through the quintessentially quantum
phenomenon of the addition of quantum amplitudes: states of AB in whichsubsystems A and B are correlated are expressed as linear combinations ofstates in which A and B are uncorrelated. The use of the word ‘entanglement’reminds us that correlations arise through an intertwining of states that isinherently quantum-mechanical and without classical analogue.
It may help to clarify these ideas if we apply them to a hydrogen atom.
We work in the position representation, so we require the amplitude
ψ(xe , xp ) = hxe , xp |ψi(6.14)

to find the electron near xe and the proton near xp . Suppose that we havestatesui (xe ) = hxe |ui i and Uj (xp ) = hxp |Uj i(6.15)

that form complete sets for the electron and the proton, respectively. Thenfor any state of the atom, |ψi, there are numbers cij such thatX
cij |ui i|Uj i.
(6.16)|ψi =ij

Multiplying through by hxe , xp | we obtainXψ(xe , xp ) =cij ui (xe )Uj (xp ).

(6.17)

ij

The product of ui and Uj on the right is no longer symbolic: it is an ordinary
product of complex numbers. The quantity cij is the amplitude to find theelectron in the state |ui i and the proton in the state |Uj i.

108Chapter 6: Composite systems

Box 6.1:

Classical correlations

It’s instructive to consider how we can represent correlations between
two classical systems A and B. Let’s assume that each system has afinite number N of discrete states – they might be digital counters on an
instrument panel. Then there are N 2 probabilities cjk to specify.
We can specify the state of A by giving N probabilities aj andsimilarly the state of B can be specified by probabilities bk . We mightchoose to express these in terms of their discrete Fourier transforms âαand b̂β , so
aj =

N
−1
X

âα e2πiαj/N

;

bk =

α=0

N
−1
X

b̂β e2πiβk/N

β=0

If A and B were uncorrelated, so cjk = aj bk , the state of AB could be
writtenXcjk =ĉαβ e2πi(αj+βk)/N ,
(1)
αβ

where

ĉαβ = âα bβ .

(2)

In the presence of correlations we can still represent cjk as the double
Fourier sum (1) but then ĉαβ will not be given by the product of equation(2). Thus the mathematical manifestation of classical correlations canbe very similar to quantum entanglement. The big difference is that in
the classical case the expansion coefficients have no physical interpretation: the basis functions used for expansion (here the circular functions
e2πiαj/N ) and the expansion coefficients âα etc., will not be non-negative
so they cannot be interpreted as probability distributions. In quantum
mechanics these quantities acquire physical interpretations. Moreover,the final probabilities, being obtained by mod-squaring a sum like thatof equation (1), involve quantum interference between different terms inthe sum.
6.1.1 Collapse of the wavefunction
Consider again the composite system we introduced above in which both A
and B are two-state systems, with |−i and |+i constituting a basis for A and|↓i and |↑i constituting a basis for B. Let AB be in the entangled state|ABi = a|+i|↑i + |−i(b|↑i + c|↓i),(6.18a)where b and c are given complex numbers. Then if a measurement of subsystem A is made and it yields +, the state of AB after the system is|ABi = |+i|↑i.

(6.18b)

Conversely, if the measurement of A yields −, the state of AB after themeasurement is
1|ABi = p|−i(b|↑i + c|↓i).
2|b| + |c|2

(6.18c)These rules are extensions of the usual collapse hypothesis, which we introduced in §idealmeasuresec: there we had a single system and we stated thatwhen a measurement is made, the state of the system collapses from a linear
combination of states that are each possible outcomes of the measurement to
the particular state that corresponds to the value of the observable actually
measured. That isXai |ii → |ψi = |3i, say.
(6.19)|ψi =i

6.1 Composite systems

109

The new twist in equations (6.18) is that when we expand the state of acomposite system as a linear combination of states of the subsystem wepropose to measure, the coefficients of those states are states of the othersubsystem rather than amplitudes, and these states are the ones the second
system will be in after the first system has been measured. Consequently, theamplitudes we obtain for a subsequent measurement of the second subsystemdepend on the outcome of the first measurement: if measurement of A yields+, then from (6.18b), a measurement of B is certain to yield ↑, while if themeasurement of A yields −, subsequent measurement of B will yield ↓ withprobability 1/(|b/c|2 + 1).

6.1.2 Operators for composite systems
While the law of multiplication of probabilities leads to the kets of subsystems
being multiplied, we add the operators of subsystems. For example, if A and
B are both free particles, then the Hamiltonian operator of the composite
system isp2p2
HAB = HA + HB = A + B .
(6.20)
2mA
2mB
In this simple example there is no physical interaction between the parts ofthe system, with the consequence that the Hamiltonian splits into a partthat depends only on the operators of A, and a part that depends only on
operators of B. When there is a physical connection between the systems,
there will be an additional part of the Hamiltonian, the interaction Hamiltonian that depends on operators belonging to both systems. For example,
if both particles bear electrostatic charge Q, the interaction Hamiltonian
Hint =

Q2
4πǫ0 |xA − xB |(6.21)

should be added to HA + HB to form HAB . For the rest of this subsection
we assume for simplicity that there is no dynamical interaction between thesubsystems.
When an operator acts on a ket that is a product of one describing A
and one describing B, kets that belong to the other system stand idly by asif they were mere complex numbers. For example
pB |A; ii|B; ji = |A; ii pB |B; ji(6.22)
sohA; i′ |hB; j ′ |(HA +HB )|A; ii|B; ji = hA; i′ |HA |A; iihB; j ′ |B; ji+ hA; i′ |A; iihB; j ′ |HB |B; ji= hA; i′ |HA |A; iiδjj ′ + δii′ hB; j ′ |HB |B; ji.
(6.23)
When we set i′ = i and j ′ = j we obtain the expectation value of HAB when
the system is in the state |A; ii|B; ji. This is easily seen to be just the sum of
the expectation values of the energies of the two free particles, as one would
expect.
We shall several times have to find the eigenvalues and eigenkets of an
operator such as HAB that is the sum of operators HA and HB that belong tocompletely different subsystems. Every operator of subsystem A commuteswith every operator of subsystem B. Consequently when HAB is given byequation (6.20),[HAB , HA ] = [HA + HB , HA ] = 0.
(6.24)
That is, when there is no physical interaction between the subsystems, so
HAB is just the sum of the Hamiltonians of the individual systems, HAB
commutes with both individual Hamiltonians. It follows that in this case

110

Chapter 6: Composite systems

there is a complete set of mutual eigenkets of HAB , HA and HB . Let {|A; ii}
be a complete set of eigenkets of HA with eigenvalues EiA , and let {|B; ii}
be a complete set of eigenkets of HB with eigenvalues EjB . Then it is trivial
to check that the states |AB; i, ji ≡ |A; ii|B; ji are eigenkets of HAB witheigenvalues EiA + EjB . Moreover, we showed above that these product kets
form a complete set. So the states |AB; iji form a complete set of mutual
eigenkets of HAB , HA and HB . In the position representation this result
becomes the statement that the wavefunctions
ABB
ψij(xA , xB ) ≡ hxA , xB |AB; iji = uAi (xA )uj (xB )(6.25)

form a complete set of mutual eigenfunctions for the three operators. Thatis, if we have a composite system with a Hamiltonian that is simply the sumof the Hamiltonians of the parts, we can assume that the eigenfunctions of
the whole system’s Hamiltonian are simply products of eigenfunctions of theindividual component Hamiltonians.
It is instructive to write the tdse for a composite system formed by two
non-interacting subsystems:


∂|ABi
∂
∂|Bi
∂|Ai
ih̄
= ih̄ (|Ai|Bi) = ih̄|Bi + |Ai
∂t
∂t
∂t
∂t
(6.26)= (HA |Ai)|Bi + |Ai(HB |Bi) = (HA + HB )|Ai|Bi= HAB |ABi.
Thus we have been able to derive the tdse for the composite system fromthe tdse for each subsystem. Notice that the physically evident rule foradding the Hamiltonians of the subsystem emerges as a consequence of theket for the whole system being a product of the kets of the subsystems andthe usual rule for differentiating a product.
6.1.3 Development of entanglement
Entangled is an appropriate name because subsystems are as prone to become
entangled as is the line of a kite. To justify this statement, we consider the
dynamical evolution of a composite system AB. Without loss of generality
we can use basis states that satisfy the tdses of the isolated subsystems.
That is, the we may assume that the states |A; ii, etc, satisfyih̄∂|A; ii∂|B; ji= HA |A; ii and ih̄= HB |B; ji.
∂t∂tA general state of the composite system isX
cij |A; ii|B; ji,|ABi =(6.27)(6.28)

ij

where the expansion coefficients cij are all functions of time. The Hamiltonian of the composite system can be written
HAB = HA + HB + Hint ,(6.29)

where the interaction Hamiltonian Hint is the part of HAB that contains
operators belonging to both subsystems (cf eq. 6.21). Substituting this expression for HAB and the expansion (6.28) into the tdse for the composite
system (eq. 6.26), we findX  dcij∂|A; ii∂|B; ji∂|ABi= ih̄
|A; ii|B; ji + cij|B; ji + |A; ii
ih̄
∂t
dt
∂t
∂t
ij
X
cij {(HA |A; ii)|B; ji + |A; ii(HB |B; ji) + Hint |A; ii|B; ji} .
=ij(6.30)

6.1 Composite systems

111

After using equations (6.27) to cancel terms, this simplifies toih̄

X dcijij

dt|A; ii|B; ji =X
ijcij Hint |A; ii|B; ji,(6.31)

which states that the time evolution of the expansion coefficients is entirely
driven by the interaction Hamiltonian. In particular, if there is no coupling
between the systems (Hint = 0), the cij are constant, so if the systems are
initially unentangled, they remain so.
By multiplying equation (6.31) through by hA; k|hB; l| we obtain an
equation that is most conveniently written
ih̄

X
dckl
cij hAB; kl|Hint |AB; iji.
=dt
ij

(6.32)

Let’s suppose that all the matrix elements in this equation vanish except
an element hAB; k0 l0 |Hint |AB; k0 l0 i which lies on the diagonal. Then only
ck0 l0 will have non-vanishing time derivative, so the condition for the subsystems to be unentangled, namely that cij /cij ′ is independent of i, whichis initially satisfied, will soon be violated by the ratio ck0 l0 /ck0 j for j 6= l0 .
Careful consideration of what happens when there are several non-vanishing
matrix elements leads to the same conclusion: almost any coupling between
subsystems will cause them to become entangled from an unentangled initialcondition.
This result is not surprising physically: a coupling makes the motionof one system dependent on the state of the other. So after some time thestate that the second system has reached depends on the state of the firstsystem, which is just to say that the two systems have become correlated orentangled.
6.1.4 Einstein–Podolski–Rosen experiment
In 1935 A. Einstein, B. Podolski and N. Rosen (EPR for short) proposed1
an experiment with entangled particles that they argued would demonstratethat quantum mechanics is an incomplete theory in the sense that to specify
the state of a physical system you need to know the values taken by hidden
variables that quantum mechanics does not consider. In 1964 J.S. Bellshowed2 that for a similar experiment quantum mechanics makes predictions
that are incompatible with the existence of hidden variables. In 1972 anexperiment of this type was successfully carried out3 and its results were
found to vindicate quantum mechanics. We now describe Bell’s formulationof the experiment and discuss its implications.
A nucleus decays from a state that has no spin to another spinless stateby emitting an electron and a positron. The nucleus is at rest both before andafter the decay, so the electron and positron move away in opposite directions
with equal speeds. As we saw in §1.3.5, electrons and positrons are spinningparticles so they each carry some spin angular momentum away from thenucleus. Since the nucleus is at all times without angular momentum, theangular momenta of the electron and positron must be equal and opposite.
At some distance from the decaying nucleus Alice detects the electron andmeasures the component of its spin in the direction of her choice, a. As
we saw in §1.3.5, the result of this measurement will be either + 21 or − 21 .
Meanwhile Bob, who sits a similar distance from the nucleus to Alice, detects
the positron and measures its spin in the direction of his choice, b.
After Alice has obtained + 21 on measuring the spin along a she thinks:“If Bob measures along a too, he must measure − 21 . But if Bob measures
1 E. Einstein, B. Podolski & N. Rosen, Phys. Rev., 47, 777 (1935)
2 J.S. Bell, Phyics, 1, 195 (1964)
3 S.J. Freedman & J.F. Clauser, Phys. Rev. L., 28, 938 (1972)

112

Chapter 6: Composite systems

along some other vector b, I cannot be certain what value he will get, but heisn’t likely to get + 21 if b is only slightly inclined to my vector a, that is, if
1 − a · b ≪ 1.” Alice can see that conservation of angular momentum implies
that the results obtained by Bob and herself must be correlated. Let’s put
this argument on a quantitative basis.
In §7.5.1 we shall see that because the system formed by the electronpositron pair has no net angular momentum, its state can be written1|ψi = √ (|e+i|p−i − |e−i|p+i) .
2(6.33)

Here |e+i is the state in which the component of the electron’s spin along
the z-axis is certain to be + 21 , and similarly for |p±i, etc. We are free toorient the z-axis parallel to Alice’s choice of direction a, so we do this. When
Alice obtains + 21 , she collapses the system’s state into
|ψ ′ i = |e+i|p−i.

(6.34)

Before Alice’s measurement, when the state was given by equation (6.33),
the amplitude
for a measurement of the positron’s spin along a to yield + 12√
was 1/ 2, but after the measurement equation (6.34) shows that it vanishes,
just as Alice reasoned it would. To find the amplitude for Bob to measure forthe positron + 12 along another vector b, we recall equation equation (1.34)afrom §1.3.5:|+, bi = sin(θ/2) eiφ/2 |p−i + cos(θ/2) e−iφ/2 |p+i,(6.35)

where θ and φ are the polar angles that give the orientation of b in a system
in which a is along the z-axis. In particularcos θ = a · b.

(6.36)

Given that after Alice’s measurement the positron is certainly in the state|p−i, it follows from equation (6.35) that the amplitude for Bob to measure+ 21 along his chosen direction is h+, b|p−i = sin(θ/2)e−iφ/2 . Mod-squaring
this amplitude we find that the probability that Bob measures + 21 is
PB (+|A+) = sin2 (θ/2),(6.37)

which is small when a ≃ b as Alice predicted. So quantum mechanics is
consistent with common sense.
We have supposed that Alice measures first, but if the electron andpositron are moving relativistically, a light signal sent to Bob by Alice when
she made her measurement would not have arrived at Bob when he made his
measurement, and vice versa. In these circumstances the theory of relativity
teaches us that the order in which the measurements are made depends onthe velocity of the observer who is judging the matter. Consequently, forconsistency the predictions of quantum mechanics must be independent of
who is supposed to make the first measurement and to collapse the system’s
state. It is easy to see from the discussion above that this condition issatisfied.
What worried EPR was that after Alice’s measurement there is a direction in which Bob will never find + 21 for the positron’s spin, and thisdirection depends on what direction Alice chooses to use. This fact seems
to imply that the positron somehow ‘knows’ what Alice measured for theelectron, and the collapse of the system’s state from (6.33) to (6.34) seems
to confirm this suspicion. Since relativity forbids news of Alice’s work on
the electron from influencing the positron at the time of Bob’s measurement,
EPR argued that the required information must have travelled out with the6.1 Composite systems

113

positron in the form of a hidden variable which was correlated at the timeof the nuclear decay with a matching hidden variable in the electron.
The existence of hidden variables would explain the probabilistic natureof quantum mechanics (which Einstein intensely disliked) because the uncertain outcomes of experiments would reflect our ignorance of the values takenby the hidden variables; the uncertainty would be banished once a better
theory gave us access to these variables.
Bell’s inequality Remarkably, Bell was able to show that any hidden
variable theory will yield a weaker correlation than quantum mechanics between the measurements of Alice and Bob as functions of the angle θ between
their chosen directions. Let’s denote the results of Alice’s and Bob’s measurements by σA = ± 21 and σB = ± 12 and calculate the expectation valueof the product σA σB . There are just four cases to consider, so the desired
expectation value is
hσA σB i = 41 {PA (+)PB (+|A+) + PA (−)PB (−|A−)− PA (+)PB (−|A+) − PA (−)PB (+|A−)},(6.38)

where PA (+) is the probability that Alice obtains σA = + 21 and PB (−|A+)is the probability that Bob finds σB = − 21 given that Alice has measured
σA = + 21 . Since nothing is known about the orientation of the electron
before Alice makes her measurement
PA (+) = PA (−) = 12 .

(6.39)We showed above (eq. 6.37) that PB (+|A+) = sin2 (θ/2), so
PB (−|A+) = 1 − PB (+|A+) = cos2 (θ/2).

(6.40)

Putting these results into equation (6.38) we havehσA σB i = 41 {sin2 (θ/2) − cos2 (θ/2)} = − 14 cos θ = − 41 a · b,

(6.41)

which agrees with Alice’s simple argument when a ≃ ±b.
Consider now the case that the result of measuring the electron’s spinin the direction a is completely determined by the values taken by hidden
variables in addition to a. That is, if we knew the values of these variables,
we could predict with certainty the result of measuring the component of
the electron’s spin in the direction of any unit vector a and Alice is only
uncertain what result she will get because she is ignorant of the values of
the hidden variables. We consider the variables to be the components ofsome n-dimensional vector v, and have that the result of measuring theelectron’s spin along a is a function σe (v, a) that takes the values ± 12 only.
Similarly, the result of measuring the positron’s spin along a unit vector b
is a function σp (v, b) that is likewise restricted to the values ± 12 . As Aliceargued, conservation of angular momentum implies thatσe (v, a) = −σp (v, a).

(6.42)The outcome of a measurement is uncertain because the value of v is uncertain. We quantify whatever knowledge we do have by assigning a probability
density ρ(v) to v, which is such that the probability that v lies in the infinitesimal n-dimensional volume dn v is dP = ρ(v) dn v. In terms of ρ theexpectation value of interest isZ
hσe (a)σp (b)i = dn v ρ(v)σe (v, a)σp (v, b)Z(6.43)= − dn v ρ(v)σe (v, a)σe (v, b),

114

Chapter 6: Composite systems

Figure 6.1 For a family of choicesof the vectors a, b and b′ , quantum
mechanics predicts that the left sideof Bell’s inequality (6.46) is larger
than the right side, contrary to theprediction of any hidden-variable
theory.

where the second equality uses equation (6.42).
Now suppose Bob sometimes measures the spin of the positron parallelto b′ rather than b. Then the fact that σe2 (v, b) = 41 allows us to writeZhσe (a)σp (b)i − hσe (a)σp (b′ )i = − dn v ρ(v)σe (v, a){σe (v, b) − σe (v, b′ )}Z= − dn v ρ(v)σe (v, a)σe (v, b){1 − 4σe (v, b)σe (v, b′ )}.

(6.44)
We now take the absolute value of each side and note that the curly bracket
in the integral is non-negative, while the product σe (v, a)σe (v, b) in front of
it fluctuates between ± 41 . Hence we obtain an upper limit on the value ofthe integral by replacing σe (v, a)σe (v, b) by 41 , and have|hσe (a)σp (b)i − hσe (a)σp (b′ )i| ≤ 14Z

dn v ρ(v){1 − 4σe (v, b)σe (v, b′ )}.

(6.45)
R
We break the right side into two integrals. The first, dn v ρ(v), evaluatesto unity because ρ is a probability density, while changing b → b′ anda → b in equation (6.43) we see that the the second integral evaluates to−4hσe (b)σp (b′ )i. Hence we have that|hσe (a)σp (b)i − hσe (a)σp (b′ )i| ≤ 41 + hσe (b)σp (b′ )i.

(6.46)

This is Bell’s inequality, which must hold for any three unit vectors a, b
and b′ if hidden variables exist. It can be tested experimentally as follows:
for a large number of trials Alice measures the electron’s spin along a while
Bob measures the positron’s spin along b in half the trials and along b′ inthe other half. From the results of these trials the value of the left sideof equation (6.46) can be estimated. The value of the right side is thenestimated from a new series of trials in which Alice measures the electron’s
spin along b and Bob measures the positron’s spin along b′ .
An obvious question is whether Bell’s inequality is consistent with thequantum-mechanical result hσe (a)σp (b)i = − 41 a · b (eq. 6.41). When we
substitute this expression into each side we get
lhs = 41 |a · (b − b′ )|;rhs = 41 (1 − b · b′ ).

(6.47)

Let’s choose a·b = 0 and b′ = b cos φ+a sin φ so as we increase the parameterφ from zero to π/2 b′ swings continuously from b to a. For this choice of b′
we easily find that
lhs = 14 | sin φ|;

rhs = 14 (1 − cos φ).

(6.48)

These expressions for the left and right sides of Bell’s inequality are plottedin Figure 6.1: we see that the inequality is violated for all values of φ other6.1 Composite systems

115

than 0 and π/2. Thus the quantum-mechanical result is inconsistent with
Bell’s inequality and is therefore inconsistent with the existence of hidden
variables.
Inequalities similar to (6.46) can be derived for systems other than spinhalf particles, including pairs of entangled photons. Experiments with photons have produced results that agree with the predictions of quantum mechanics to sufficient precision that they violate the relevant Bell inequalities.4
Consequently, these experiments rule out the possibility that hidden variables
exist.
What general conclusions can we draw from the EPR experiment?
• A measurement both updates our knowledge of a system and disturbs thesystem. Alice’s measurement disturbs the electron but not the positron,and gains her information about both particles.
• Quantum mechanics requires wholistic thinking: when studying the
EPR experiment we must consider the system formed by both particles together rather than treating the particles in isolation. We shallencounter a more spectacular example of this requirement below in connection with ideal gases.
• Many discussions of the EPR experiment generate needless confusion
by supposing that after Alice has measured + 21 for the component of
the electron’s spin parallel to a, the spin is aligned with a. We shallsee in §7.4.2 that the electron also has half a unit of angular momentum
in each of the x and y directions, although the signs of these othercomponents are unknown when we know the value of sz . Hence themost Alice can know about the orientation of the spin vector is that itlies in a particular hemisphere. Whatever hemisphere Alice determines,
she can argue that the positron’s spin lies in the opposite hemisphere.
So if Alice finds the electron’s spin to lie in the northern hemisphere,
she concludes that the positron’s spin lies in the southern hemisphere.
This knowledge excludes only one result from the myriad of possibilities
open to Bob: namely he cannot find sz = + 21 . He is unlikely to find + 21
if he measures the component of spin along a vector b that lies close to
the z axis because the hemisphere associated with this result has a small
overlap with the southern hemisphere, but since there is an overlap, theresult + 12 is not excluded. Contrary to the claims of EPR, the results ofBob’s measurements are consistent with the hemisphere containing thepositron’s spin being fixed at the outset and being unaffected by Alice’s
measurement.
• The experimental demonstration that Bell inequalities are violated establishes that quantum mechanics will not be superseded by a theory
in which the spin vector has a definite direction. In §7.4.1 we shall seethat macroscopic objects only appear to have well defined orientations
because they are not in states of well-defined spin. That is, the ideathat a spin vector points in a well defined direction is a classical notion
and not applicable to objects such as electrons that do have a definite
spin. This idea is an old friend from which we part company as sadly as
after studying relativity we parted company with the concept of universal time. The world we grew accustomed to in playgroup is not the realworld, but an approximation to it that is useful on macroscopic scales.
The study of physics forces one to move on and let childish things go.

4 e.g., W. Tittel et al., PRL, 81, 3563 (1998)

116

Chapter 6: Composite systems

6.2 Quantum computing
There’s an old story about a mathematician at the court of the Chinese emperor. The mathematician had advised the emperor wisely and the emperor,
wishing to express his gratitude in a manner worthy of his greatness, asked
the mathematician to name the reward he would like to receive. “Oh great
Emperor, your offer is too liberal for one who has rendered you such a slight
service. Let a chess board be brought and one grain of rice be placed on thefirst square, two on the second, four on the third, eight on the fourth, andso on till every square of the board has received an allocation of rice.” Theemperor was pleased by the modesty of the mathematician’s proposal andordered it be done. Great was his shock and annoyance the next day when it
was reported to him that all the rice in his great silos had proved insufficient
to pay the mathematician his due. For 264 − 1 ∼ 1019 grains of rice would
be needed to supply the 64 squares on the board. That’s ∼ 1012 tons of riceand vastly more than all the rice on the planet.5
What is the relevance of this old story for quantum mechanics? We haveseen that a system made of two two-state systems has four basis states. If we
add a further two-state system to this four-state composite system, we obtaina system with 2 × 4 = 8 basis states. By the time we have built a systemfrom 64 two-state systems, our composite system will have 264 ∼ 1019 basis
states. Sixty four two-state systems might be constructed from 64 atoms
or even 64 electrons, so could be physically miniscule. But to calculate thedynamics of this miniscule system we would have to integrate the equationsof motion of 1019 amplitudes! This is seriously bad news for physics.
The idea behind quantum computing is to turn this disappointment
for physics into a boon for mathematics. We may not be able to solve
1019 equations of motion, but Nature can evolve the physical system, andappropriate measurements made on the system should enable us to discoverwhat the results of our computations would have been if we had the time tocarry them out. If this approach to computation can be made to work inpractice, calculations will become possible that could never be completed ona conventional computer.
The first step towards understanding how a quantum computer would
work is to map integers onto the basis states of our system. In this context
we refer to a two-state system as a qubit and call its basis states |0i and|1i. A set of N qubits forms a register, which has a complete set of statesof the form |xi|x′ i · · · |x′′ i, where x, x′ , etc., = 0, 1 indicate the states of theconstituent qubits. Now given a number in binary form, such as 7 = 4 + 2 +1 = 111, we associate it with the basis state of the register |0i . . . |0i|1i|1i|1i.
In this way we establish a one to one correspondence between the integers
0 to 2N − 1 and the basis states of a register that comprises N qubits. We
use this correspondence to establish a more compact notation for the basis
states of the register, writing |7i instead of |0i . . . |0i|1i|1i|1i, etc.
This arrangement mirrors the correspondence in a classical computer
between numbers and the states of a classical register formed by N classicaltwo-state systems or bits. The crucial difference between quantum andclassical registers is that whereas a classical register is always in a state thatis associated with a definite number, the generic state |ψi of a quantum
register is a linear combination of states that are associated with different
numbers:
N2X−1
cj |ji.
(6.49)|ψi =j=0Thus nearly all states of a quantum register are not associated with individual numbers but with all representable numbers simultaneously. We shall seethat this ability of a single state of a quantum register to be associated with
5 According to the International Rice Search Institute, in 2007 global rice production
was 650 million tons.

6.2 Quantum computing

117

a huge number of integers enables a quantum computer to conduct massively
parallel computations.
The central processor unit (CPU) of a classical computer is a programmable mechanism that reads a number n from an input register andplaces the number f (n) into the output register, where f is the functionthat the CPU is currently programmed to evaluate. By analogy one might
imagine that a quantum computer would consist of a quantum register anda programmable Hamiltonian H that would cause the state |ni to evolve insome specified time T into the state |f (n)i = e−iHT /h̄ |ni. Unfortunately thisconception is flawed because this machine could not evaluate any functionthat took the same value on different arguments, so f (n) = f (m) = F , say,for some values n 6= m. To see why the computer could not evaluate such afunction recall that the operator U ≡ e−iHT /h̄ is unitary, so it has an inverse
U † . But we have U |ni = U |mi = |F i, and if we apply U † to |F i we mustget both |mi and |ni, which is absurd.
We get around this problem by making our quantum computer slightly
more complex: we let it have two registers, a control register X and adata register Y. The computer then has a basis of states |xi|yi, where x is
the number stored in the control register and y is the number stored in thedata register. We conjecture that we can find a Hamiltonian such that for
any function f the state |xi|yi evolves in time T into the state |xi|y + f (x)i.
Adding the second register solves the problem we encountered above because
applying U to |ni|yi we get |ni|y + F i which is a different state from what we
get when we apply U to |mi|yi, namely |mi|y + F i: adding the extra register
allows the computer to remember the state it was in before the machine
cycle started, and this memory makes it logically possible for U † to restore
the earlier state.
Adding the second register may have demolished an objection to our
original most naive proposal, but is it really possible to construct a timeevolution operator that would enable us to evaluate any function f (x)? This
question is answered affirmatively in two stages. First one defines a handfulof unitary operators U that perform basic bit manipulations on our registers,
and shows that using a sequence of such operators one can perform any of
the standard arithmetical operations, adding, subtracting, multiplying anddividing. Second, for each of these operators U one designs an experimentin which U gives the evolution of a two-state quantum system over sometime interval. Currently many groups use photons as qubits, identifying|0i and |1i with either right- and left-handed circular polarisation, or withlinear polarisation in two orthogonal directions. Other groups use electrons
as qubits, identifying |0i and |1i as states in which the spin in some given
direction is either 21 or − 21 . All such work with real qubits is extremely
challenging and in its infancy, but it has already established that there is noobjection in principle to realising the simple unitary operators that quantum
computing requires. It is too early to tell what physical form qubits will takewhen quantum computing becomes a mature technology. Consequently, we
leave to one side the question of how our operators are to be realised andfocus instead on what operators we require and what could be achieved withthem when they have been realised.
The simplest computer has two one-qubit registers, with a basis of states|0i|0i, |0i|1i, |1i|0i and |1i|1i – we shall refer to basis states of a register withany number of qubits ordered thus by increasing value of the stored number
as the computational basis. In the computational basis of our two-qubit
system, the operator U+ that performs addition (|xi|yi → |xi|y + xi) has the118

Chapter 6: Composite systems

unitary matrix6

1
0
U+ = 
0
0

0
1
0
0

0
0
0
1

To justify this claim we note that


1
0

0
0

0
1
0
0

0
0
0
1


0
0
.
1
0

(6.50)

   0αα0β  β   =  
1
γ
δ
0
δ
γ

(6.51)

so U+ causes the state of the computer|ψi = α|0i|0i + β|0i|1i + γ|1i|0i + δ|1i|1i(6.52)

U+ |ψi = α|0i|0i + β|0i|1i + γ|1i|1i + δ|1i|0i,(6.53)

to evolve into

so the second qubit is indeed incremented by the first modulo 2.
U+ is a simple example of an operator in which the state of the dataregister is changed in a way that depends on the state of the control register
while the state of the control register stays the same. Such operators arecalled controlled-U operators. Another useful operator is the controlledphase operator, which in the computational basis has the matrix
1
0
Uφ = 
0
0

0
1
0
0


0 0
0 0 
.
1 0
iφ
0 e

(6.54)

Uφ has no effect on the first three states of the computational basis, and it
multiplies the phase of the last state by eiφ . It is straightforward to showthat
Uφ |xi|yi = eixyφ |xi|yi(6.55)by checking that the two sides match for all four possible values of (x, y).
It can be shown that any unitary transformation of an n-qubit registercan be simulated if we augment U+ and Uφ with two operators that workon just one qubit. One of these extra operators is the phase operator Uφ1 ,which leaves |0i invariant and increments the phase of |1i by φ:
Uφ1 |0i = |0i

Uφ1 |1i = eiφ |1i)⇔

Uφ1 |xi = eiφx |xi

Uφ1 =

⇔



1
0

0
eiφ



. (6.56)

The other single-qubit operator that we need is the Hadamard operator,
which in the computational basis, |0i |1i, has the matrix
1
UH = √
2



1 1
1 −1



.

(6.57)The Hadamard operator takes a state that represents a number, such as |0i,
and turns it into a state that is a√linear combination of the two representable
numbers: UH |0i = (|0i + |1i)/ 2. Conversely, because UH2 = I so UH is
6 Here x + y must be understood to mean x + y mod 2 because quantum computers
like classical computers do arithmetic modulo one more than the largest number that they
can store.

6.2 Quantum computing

119

Figure 6.2 Schematic diagram to show how two Hadamard operators and two phase shiftoperators suffice to transform |0i into an arbitrary state of a qubit.

Figure 6.3 Evaluating f on every
argument simultaneously. The top
three qubits form the control register, which is initially in the state|0i.

its own inverse, it turns√these linear combinations of numbers into actual
numbers: UH (|0i + |1i)/ 2 = |0i.
Complex operations on qubits can be built up by sequences of phase andHadamard operators and such sequences are conveniently described using
the graphical notation of Figure 6.2. Each qubit is represented by a line
along which the state of the qubit flows from left to right. In the simple
example shown,
the state |0i is converted by the first Hadamard operator to√
1(|0i + |1i)/ 2, and U2θ
converts this to
1√ |0i + e2iθ |1i .
2After the next Hadamard operator this becomes




2iθ
1
(|0i − |1i) = 12 1 + e2iθ |0i + 1 − e2iθ |1i
2 |0i + |1i + e= eiθ {cos θ|0i − i sin θ|1i} .

(6.58a)(6.58b)1
Finally, application of the phase-shift operator Uφ+π/2
converts this to(6.58c)|ψi ≡ eiθ cos θ|0i + eiφ sin θ|1i .

By choosing the values of θ and φ appropriately, we can make |ψi any chosen
state of the qubit. Thus the phase-shift and Hadamard operators form acomplete set of single-qubit operators.
If we apply a Hadamard operator to each qubit of an 2-qubit registerthat is initially in the state |0i|0i, we get(UH |0i) (UH |0i) = 21 (|0i + |1i) (|0i + |1i)

= 12 (|1i|1i + |1i|0i + |0i|1i + |0i|0i)(6.59)= 21 (|3i + |2i + |1i + |0i) .

That is, by setting the register to zero and then applying a Hadamard operator to each of its qubits, we put the register into a linear superposition of
the states associated with each representable number. It is easy to see that
this result generalises to n-qubit registers.7 Using this trick we can simultaneously evaluate a function on every representable argument, simply by
evaluating the function on the state of the control register immediately after
it has been processed by the Hadamard operators. Figure 6.3 illustrates thisprocess, which is described by the equations
7 In fact, applying Hadamard operators to the qubits of an n-qubit register when it is
set to any number will put the register into a linear superposition of states associated withall representable numbers, but if the initial state of the register differs from |0i, exactly
half of the coefficients in the sum will be −2−n/2 and half +2−n/2 (Problem 6.6).

120

Chapter 6: Composite systems

Box 6.2:

Deutsch’s algorithm

Given a function f (x) that takes an n-bit argument and returns either 0
or 1, the exercise is to determine whether f is a constant or ‘balanced’
function. To this end we build a computer with an n-qubit control register and a single qubit data register. We set the control register to |0i andthe data register to |1i and operate on every qubit with the Hadamard
operator UH . Then the computer’s state is
n
2X

−1
1
|xi (|0i − |1i).
(1)2(n+1)/2 x=0
Now we evaluate the function f in the usual way, after which the computer’s state is!
X
1|xi(|f (x)i − |1 + f (x)i) .
(2)2(n+1)/2x

Given that f (x) = 0, 1, it is straightforward to convince oneself that(|f (x)i − |1 + f (x)i) = (−1)f (x) (|0i − |1i) so the computer’s state can be
written
!
X
1
f (x)
(−1)
|xi (|0i − |1i).
(3)
2(n+1)/2x

We now operate on every qubit with UH for a second time. The dataregister returns to |1i because UH is its own inverse, while the control
register only returns to |0i if we can take the factor (−1)f (x) out oftheregister a multiple of
P sum over x, making the state of the controlf (x)
|xi;
if
f
is
‘balanced’,
half
of
the
factors(−1)
are +1 and half −1x
and in this case UH moves the control register to a state |yi for y 6= 0
(Problem 6.6). Hence by measuring the state of the control register, we
discover whether f is constant or balanced: if the control register is set
to zero, f is constant, and if it holds any other number, f is balanced.

n

n

2 −1
2 −1
f
1 X
1 X
→
→
|xi|0i|xi|f (x)i.
|0i|0i2n/2 x=02n/2 x=0UH(6.60)

After the evaluation of f , the computer’s state depends on every possible
value of f . So the state of a 64-qubit computer will depend on the 264 ≃
1019 possible values of f . By exploiting this fact, can we conduct massively
parallel computations with just a pair of quantum registers?
The question is, how can we learn about the values that f takes? An
obvious strategy is to read off a numerical value X from the control registerby collapsing each of its qubits into either the state |0i or the state |1i.
Once this has been done, the state of the composite system |xi|yi will havecollapsed from that given on the right of (6.60) to |Xi|f (X)i, so f (X) canbe determined by inspecting each of the qubits of the data register. The
trouble with this strategy is that it only returns one value of f , and that fora random argument X. Hence if our quantum computer is to outperform aclassical computer, we must avoid collapsing the computer’s state by reading
its registers. Instead we should try to answer questions about f that have
simple answers but ones that involve all the values taken by f .
For example, suppose we know that f (x) only takes the values 0 and 1,
and that it is either a constant function (i.e., either f (x) = 1 for all x, orf (x) = 0 for all x) or it is a ‘balanced function’ in the sense that f (x) = 0 for
half of the possible values of x and 1 for the remaining values. The question
we have to answer is “is f constant or balanced?” With a classical computer
you would have to keep evaluating f on different values of x until either you

6.3 Density operator

121

got two different values (which would establish that f was balanced) or morethan half of the possible values of x had been tried (which would establishthat f was constant). In Box 6.2 we show that from (6.60) we can discoverwhether f is constant or balanced with only a handful of machine cycles.
The algorithm given in Box 6.2 is an extension of one invented by
Deutsch8 , which was an early example of how the parallel-computing potential of a quantum computer could be harnessed. Subsequently algorithms
were developed that dramatically accelerate database searches9 and the decomposition of large numbers into their prime factors. The usefulness of
the internet depends on effective cryptography, which currently relies on
the difficulty of prime-number decomposition. Hence by rendering existing
cryptographic systems ineffective, the successful construction of a quantum
computer would have a big impact on the world economy.
Notwithstanding strenuous efforts around the world, quantum computing remains a dream that will not be realised very soon. Its central idea isthat the the integers up to 2N − 1 can be mapped into the base states of an
N -qubit quantum register, so a general state of such a register is associatedwith all representable integers, and the time evolution of the register involves
massively parallel computing. The field is challenging both experimentallyand theoretically. The challenge for theorists is to devise algorithms thatextract information from a quantum register given that any measurementof the register collapses its state and thus erases much of the informationthat was encoded in it before a measurement was made. Experimentally, thechallenge is to isolate quantum registers from their environment sufficiently
well that they do not become significantly entangled with the environment
during a computation. We discuss the process of becoming entangled withthe environment in the next section.

6.3 The density operator
To this point in this book we have assumed that we know what quantum
state our system is in. For macroscopic objects this assumption is completely
unrealistic, for how can we possibly discover the quantum states of the ∼ 1023
carbon atoms in a diamond, or even the ∼ 105 atoms in a protein molecule?
To achieve this goal for a diamond, at least 1023 observables would haveto be measured, and the number would in reality be vastly greater because
individual atoms would be entangled with one another, making the state ofthe diamond a linear combination of basis states of the form |a1 i|a2 i . . . |aN i,
where |ai i denotes a state of the ith atom. It is time we squared up to thereality of our ignorance of the quantum states of macro- and meso-scopic
objects.
Actually, we need to be cautious even when asserting that we know thequantum state of atomic-scale objects. The claim that the state of a system
is known is generally justified by the assertion that a measurement has justbeen made, with the result that the system’s state has been collapsed intoa known eigenstate of the operator of the given observable. This procedure
for establishing the quantum state of a system is unrealistic in that it makes
no allowance for experimental error, which we all know to be endemic in reallaboratories: real experiments lead to the conclusion that the value of anobservable is x ± y, which is shorthand for “the probability distribution for
the value of the observable is centred on x and has a width of the order y.”
Since the measurement leaves the value of the observable uncertain, it does
not determine the quantum state precisely either.
Let us admit that we don’t know what state our system is in, but conjecture that the system is in one of a complete set of states {|ni}, and for each
8 D. Deutsch, Proc. R. Soc., 400, 97 (1985)
9 L. K. Grover, STOC’96, 212 (1996)

122

Chapter 6: Composite systems

value of n assign a probability pn that it’s in the state |ni.10 It’s important
to be clear that we are not saying that the system is in the state
|φi =

X√
n

pn |ni.

(6.61)

That is a well-defined quantum state, and we are admitting that we don’tknow the system’s state. What we are saying is that the system may be instate |1i, or in state |2i, or state |3i, and assigning probabilities p1 , p2 , . . .
to each of these possibilities.
Given this incomplete information, the expectation value of measuringsome observable Q will be p1 times the expectation value that Q will haveif the system is in the state |1i, plus p2 times the expectation value for thecase that the system is in the state |2i, etc. That isQ=Xn

pn hn|Q|ni,(6.62)

where we have introduced a new notation Q to denote the expectation valueof Q when we have incomplete knowledge. When our knowledge of a system
is incomplete, we say that the system is in an impure state, and correspondingly we sometimes refer to a regular state |ψi as a pure state. Thisterminology is unfortunate because a system in an ‘impure state’ is in a perfectly good quantum state; the problem is that we are uncertain what state
it is in – it is our knowledge of the system that’s impure, not the system’s
state.
It is instructive to P
rewrite equation (6.62) by inserting either side of Q
identity operators I = j |qj ihqj | that are made out of the eigenkets of Q.
Then we have
Q=

X
nkj

pn hn|qk ihqk |Q|qj ihqj |ni =Xnjqj pn |hqj |ni|2 ,(6.63)

where the second equality follows from Q|qj i = qj |qj i and the orthonormalityof the kets |qj i. Equation (6.63) states that the expectation value of Q is thesum of the possible measurement values qj times the probability pn |hqj |ni|2
of obtaining this value, which is the product of the probability of the systembeing in the state |ni and the probability of obtaining qj in the case that itis.
Now consider the density operator
ρ≡Xn

pn |nihn|,(6.64)

where the pn are the probabilities introduced above. This definition is reminiscent of the definitionX
qj |qj ihqj |(6.65)Q=jof the operator associated with an observable (eq. 2.9). In particular, ρ isa Hermitian operator because the pn are real. It should not be considered
an observable, however, because the pn are subjective not objective: theyquantify our state of knowledge rather than hard physical reality. For example, if our records of the results of measurements become scrambled, perhaps
through some failure of electronics in the data-acquisition system, our valuesof the pn will change but the system will not. By contrast the spectrum {qj }10 See Problem 6.9 for a different and more physically plausible physical assumption.

6.3 Density operator123

Box 6.3:

Properties of Tr

The trace operator Tr extracts a complex number from an operator. We
now show that although its definition (6.67) is in terms of a particularbasis {|mi}, its value is independent of the basis used.
P Let {|qj i} be anyother basis. Then we insert identity operators I = j |qj ihqj | either sidePof A in Tr A = n hn|A|ni:
X
X
Tr A =
hn|qj ihqj |A|qk ihqk |ni =hqj |A|qk ihqk |nihn|qj i
njk

=

X
j

kjn

(1)

hqj |A|qj i,Pwhere we have used I = n |nihn| and hqk |qj i = δkj .
Another useful result is that for any two operators A and B,
Tr(AB) = Tr(BA):X
X
Tr(AB) =hn|AB|ni =hn|A|mihm|B|ni
n

=

X
nm

nm

hm|B|nihn|A|mi =X
mhm|BA|mi = Tr(BA).

(2)

By making the substitutions B → C and A → AB in this result we inferthat
Tr(ABC) = Tr(CAB).
(3)
of Q is determined by the laws of nature and is independent of the completeness of our knowledge. Thus the density operator introduces a qualitatively
new feature into the theory: subjectivity.
To see the point of the density operator, we use equations (6.64) and(6.65) to rewrite the operator product ρQ:
ρQ =

X
nj

pn qj |nihn|qj ihqj |.

(6.66)

When this equation is premultiplied by hm| and postmultiplied by |mi andthe result summed over m, the right side becomes the same as the right sideof equation (6.63) for Q. That is,Tr(ρQ) ≡Xmhm|ρQ|mi = Q,

(6.67)

where ‘Tr’ is short for ‘trace’ because the sum over m is of the diagonal elements of the matrix for ρQ in the basis {|ni}. Box 6.3 derives two important
properties of the trace operator.
Equation (6.64) defines the density operator in terms of the basis {|ni}.
What do we get if we P
express ρ in terms of some other basis {|qj i}? To findout we replace |ni by j hqj |ni|qj i and obtain
ρ=

X
njk

=

X
jk

pn hqj |nihn|qk i |qj ihqk |pjk |qj ihqk |

where

pjk ≡

X
n

pn hqj |nihn|qk i.

(6.68)

This equation shows that whereas ρ is represented by a diagonal matrix inthe {|ni} basis, in the {|qj i} basis ρ is represented by a non-diagonal matrix.
This contrast arises because in writing equation (6.64) we assumed that oursystem was in one of the states of the set {|ni}, although we were unsure

124

Chapter 6: Composite systems

which one. In general if the system is in one of these states, it will definitely
not be in any of the states {|qj i} because each |ni will be a non-trivial
linear combination of kets |qj i. Thus when ρ is expanded in this basis, theexpansion does not simply specify a probability
P to be in each state. Instead
it includes complex off-diagonal terms pjk = n pn hqj |nihn|qk i that have noclassical interpretation. When we have incomplete knowledge of the state ofour system, we will generally not know that the system is in some state of agiven complete set, so we should not assume that the off-diagonal elements ofρ vanish. Never the less, we may safely use equation (6.64) because whatever
matrix represents ρ in a given basis, ρ is a Hermitian operator and will havea complete set of eigenkets. Equation (6.64) gives the expansion of ρ interms of its eigenkets. In practical applications we may not know what the
eigenkets |ni are, but this need not prevent us using them in calculations.
The importance of ρ is that through equation (6.67) we can obtain fromit the expectation value of any observable. As the system evolves, these observables will evolve because ρ evolves. To find its equation of motion, we
differentiate equation (6.64) with respect to time and use the tdse. The
differentiation is straightforward because pn is time-independent: if the system was in the state |ni at time t, at any later time it will certainly be inwhatever state |ni evolves into. Hence we have


∂|ni
∂hn|
dρ X
=
hn| + |ni
pn
dt
∂t
∂t
n
X
1
1
=
pn (H|nihn| − |nihn|H) = (Hρ − ρH).
ih̄ n
ih̄

(6.69)

This equation of motion can be written more simplyih̄dρ= [H, ρ].
dt

(6.70)

To obtain the equation of motion of an arbitrary expectation value Q =Tr(ρQ), we expand the trace in terms of a time-independent basis {|ai} anduse equation (6.70):
ih̄

X dρ
X
dQ
= ih̄
ha| Q|ai =ha|(Hρ − ρH)Q|ai = Tr(ρ[Q, H]),
dt
dt
a
a

(6.71)

where the last equality uses equation (3) of Box 6.3. Ehrenfest’s theorem(2.34) states that the rate of change of the expectation value Q for a given
quantum state is the expectation value of [Q, H] divided by ih̄, so equation(6.71) states that when the quantum state is uncertain, the expected rate ofchange of Q is the appropriately weighted average of the rates of change of
Q for each of the possible states of the system.
Notice that the density operator and the operators for the Hamiltonianand other observables encapsulate a complete, self-contained theory of dynamics. If we have incomplete knowledge of our system’s initial state, useof this theory is mandatory. If we do know the initial state, we can still use
this apparatus by assigning our system the density operator
ρ = |ψihψ|(6.72)

rather than using the tdse and extracting amplitudes for possible outcomesof measurements. However, when ρ takes the special form (6.72), the use of
the density operator becomes optional (Problem 6.8).

6.3 Density operator

125

6.3.1 Reduced density operators
We have seen that any physical interaction between two quantum systems
is likely to entangle them. No man is an island and no system is truly
isolated (except perhaps the entire Universe!) Consequently, a real system
is constantly entangling itself with its environment. We now show that even
if our system starts in a pure state, once it has entangled itself with its
environment, it will be in an impure state.
We consider a system that is comprised of two subsystems: A, which
will represent our system, and B, which will represent the environment –the environment consists of anything that is dynamically coupled to oursystem but not observed in sufficient detail for its dynamics to be followed.
Let the density operator of the entire system beXρAB =|A; ii|B; jiρijkl hA; k|hB; l|.
(6.73)ijkl

Let Q be an observable property of subsystem A. The expectation value of
Q is
Q = Tr Qρ
=

X
mn

=

X
mi



hA; m|hB; n|Q 

hA; m|Q|A; ii

X
ijkl

X



|A; ii|B; jiρijkl hA; k|hB; l| |A; mi|B; niρinmn ,

n(6.74)
where the second equality exploits the fact that Q operates only on thestates of subsystem A, and also uses the orthonormality of the states of eachsubsystem: hA; k|A; mi = δkm , etc. We now define the reduced density
operator of subsystem A to be!
X
X
X|A; ii(6.75)
ρA ≡ρinkn hA; k|,hB; n|ρAB |B; ni =
n

im

n

where the second equality uses equation (6.73). In terms of the reduced
density operator, equation (6.74) can be written
Q=XmhA; m|QρA |A; mi = Tr QρA .

(6.76)

Thus the reduced density operator enables us to obtain expectation values ofsubsystem A’s observables without bothering about the states of subsystem
B. It is formed from the density operator of the entire system by taking thepartial trace over the states of subsystem B (eq. 6.75).
Suppose both subsystems start in well-defined states. Then under thetdse the composite system will evolve through a series of pure states |ψ, ti,
and at time t the density operator of the composite system will be (cf. 6.72)
ρAB = |ψ, tihψ, t|.

(6.77)

If the two subsystems have not become entangled, so |ψ, ti = |A, ti|B, ti,then the reduced density operator for A is
X
hB; i|B, tihB, t|B; ii = |A, tihA, t|,(6.78)
ρA = |A, tihA, t|iwhere we have used the fact that the set {|B; ii} is a complete set of states forsubsystem B. Equation (6.78) shows that so long as the subsystems remainunentangled, the reduced density operator for A has the form expected for

126

Chapter 6: Composite systems

a system that is in a pure state. To show that entanglement will generally
lead subsystem A into an impure state, we consider the simplest non-trivial
example: that in which both subsystems are qubits. Suppose they have
evolved into the entangled state1|ψ, ti = √ (|A; 0i|B; 0i + |A; 1i|B; 1i)
2(6.79)

Then evaluating the trace over the two states of B we findρA = 12 hB; 0| (|A; 0i|B; 0i + |A; 1i|B; 1i) (hA; 0|hB; 0| + hA; 1|hB; 1|) |B; 0i+ 21 hB; 1| (|A; 0i|B; 0i + |A; 1i|B; 1i) (hA; 0|hB; 0| + hA; 1|hB; 1|) |B; 1i= 12 (|A; 0ihA; 0| + |A; 1ihA; 1|) ,(6.80)
which is the density operator of a very impure state. Physically this result
makes perfect sense: in equation (6.80) ρA states that subsystem A has
equal probability of being in either |0i or |1i, which is consistent with thestate (6.79) of the entire system. In that state these two possibilities wereassociated with distinct predictions about the state of subsystem B, but
in passing from ρAB to ρA we have lost track of these correlations: if wechoose to consider system A in isolation, we lose the information carriedby these correlations, with the result that we have incomplete information
about system A. In this case system A is in an impure state. So long as we
recognise that A is part of the larger system AB and we retain the ability
to measure both parts of AB, we have complete information, so AB is in apure state.
In this example system A represents the system under study and system
B represents the environment of A, which we defined to be whatever is dynamically coupled to A but incompletely instrumented. If, for example, A isa hydrogen atom, then the electromagnetic field inside the vessel containing
the atom would form part of B because a hydrogen atom, being comprisedof two moving charged particles, is inevitably coupled to the electromagnetic
field. If we start with the atom in its first excited state and the electromagnetic field in its ground state, then atom, field and atom-plus-field are
initially all in pure states. After some time the atom-plus-field will evolve
into the state|ψ, ti = a0 (t)|A; 0i|F; 1i + a1 (t)|A; 1i|F; 0i,(6.81)

where |A; ni is the nth excited state of the atom, while |F; ni is the state of
the electromagnetic field when it contains n photons of the frequency associated with transitions between the atom’s ground and first-excited states.
In equation (6.81), a0 (t) is the amplitude that the atom has decayed to its
ground state while a1 (t) is the amplitude that it is still in its excited state.
When neither amplitude vanishes, the atom is entangled with the electromagnetic field. If we fail to monitor the electromagnetic field, we have todescribe the atom by its reduced density operator
ρA = |a0 |2 |A; 0ihA; 0| + |a1 |2 |A; 1ihA; 1|.

(6.82)

This density operator indicates that the atom is now in an impure state.
In practice a system under study will sooner or later become entangledwith its environment, and once it has, we will be obliged to treat the system
as one for which we lack complete information. That is, we will have to
predict the results of measurements with a non-trivial density operator. The
transition of systems in this way from pure states to impure ones is called
quantum decoherence. Experimental work directed at realising the possibilities offered by quantum computing is very much concerned with arresting
the decoherence process by weakening all couplings to the environment.

6.3 Density operator

127

6.3.2 Shannon entropy
Once we recognise that systems are typically in impure states, it’s natural
to want to quantify the impurity of a state: for example, if in the definition(6.64) of the density operator, p3 = 0.99999999, then the system is almost
certain to be found in the state |3i and predictions made by assuming thatthe system is in the pure state |3i will not be much in error, while if thelargest probability occurring in the sum is 10−20 , the effects of impurity will
be enormous.
A probability distribution {pi } provides a certain amount of information
about the outcome of some investigation. If one probability is close to unity,
the information it provides is nearly complete. Conversely, if all the probabilities are small, no outcome is particularly likely and the missing informationis large. The question we now address is “what is the appropriate measureof the missing information that remains after a probability distribution {pi }has been specified?”
Logic dictates that the required measure s(p1 , . . . , pn ) of missing information must have the following properties:• s must be a continuous, symmetric function of the pi ;• s should be largest when every outcome is equally likely, i.e., whenpi = 1/n for all i. We defines( n1 , . . . , n1 ) = sn

(6.83)

and require that sn+1 > sn (more possibilities implies more missing
information).
• s shall be consistent in the sense that it yields the same missing information when there are different ways of enumerating the possible outcomesof the event.
To grasp the essence of the last requirement, consider an experiment withthree possible outcomes x1 , x2 and x3 to which we assign probabilities p1 , p2
and p3 , yielding missing information s(p1 , p2 , p3 ). We could group the last
two outcomes together into the outcome x23 , by which we mean “either x2
or x3 ”. Then we assign a probability p23 = p2 + p3 to getting x23 , giving
missing information s(p1 , p23 ). To this missing information we have to addthat associated with resolving the outcome x23 into either x2 or x3 . The
probability that we will have to resolve this missing information is p23 , andthe probability of getting x2 given that we have x23 is p2 /p23 , so we argue
that
p
p3 
2
s(p1 , p2 , p3 ) = s(p1 , p23 ) + p23 s.
(6.84),
p23 p23
This equation is readily generalised: we have n possible outcomes x1 , . . . , xn
with probabilities p1 , . . . , pn . We gather the outcomes into r groups and let
y1 be the outcome in which one of x1 , . . . , xk1 was obtained, y2 the outcome inwhich one of xk1 +1 . . . , xk2 was obtained etc, and let wi denote the probabilityof the outcome yi . Then since the probability that we get x1 given that wehave already obtained y1 is p1 /w1 , we haves(p1 , . . . , pn ) = s(w1 , . . . , wr ) + w1 s(p1 /w1 , . . . , pk1 /w1 )+· · · + wr s(pn−kr /wr , . . . , pn /wr ).

(6.85)

Since s is a continuous function of its arguments, it suffices to evaluate
it for rational values of the arguments.
P So we assume that there are integers
ni such that pi = ni /N , wherei ni = N by the requirement that theprobabilities sum to unity. Consider a system in which there are N equally
likely outcomes, and from these form n groups, with ni possibilities in theith group. Then the probability of the group is pi and the probability ofgetting any possibility in the ith group given that the ith group has come up,
is 1/ni . Hence applying equation (6.85) to the whole system we finds(1/N, . . . , 1/N ) = s(p1 , . . . , pn ) +nXipi s(1/ni , . . . , 1/ni )(6.86)

128

Chapter 6: Composite systems

Box 6.4:

Solving Equation (6.88)

Let s(n) ≡ sn . Then equation (6.88) is easily extended tos(mnr · · ·) = s(n) + s(m) + s(r) + · · · ,so with n = m = r = · · · we conclude thats(nk ) = ks(n).

Now let u, v be any two integers bigger than 1. Then for arbitrarily large
n we can find m such that
m
ln v
m+1
≤
<
⇒ um ≤ v n < um+1 .
(1)
n
ln u
n
Since s is monotone increasing,s(um ) ≤ s(v n ) < s(um+1 )⇒⇒ms(u) ≤ ns(v) < (m + 1)s(u)

m
s(v)
m+1
≤
<
.
n
s(u)
n

(2)

Comparing equation (1) with equation (2), we see that
s(v)
ln v
1
−
≤
s(u) ln un⇒s(v) s(u)
−
≤ ǫ,
ln v
ln u

where ǫ = s(u)/(n ln v) is arbitrary small. Thus we have shown thats(v) ∝ ln v.
or with the definition (6.83) of sn ,s(p1 , . . . , pn ) = sN −

n
X

n


X
N=
ni .

pi sni(6.87)

i

i

This equation relates s evaluated on a general argument list to the values
that s takes when all its arguments are equal. Setting all the ni = m we
obtain a relation that involves only sn :sn = snm − sm .

(6.88)

It is easy to check that this functional equation is solved by sn = K ln n,
where K is an arbitrary constant that we can set to unity. In fact, in Box 6.4
it is shown that this is the only monotone solution of equation (6.88). Hencefrom equation (6.87) we have that the unique measure of missing informationiss(p1 , . . . , pn ) = ln N −
=−

X

n
X
i

pi ln ni = −pi ln pi .

Xipi (ln ni − ln N )(6.89)i

Since every probability pi is non-negative and less than or equal to one, s isinherently positive. Claude Shannon (1916–2001) first demonstrated11 thatthe function (6.89) is the only consistent measure of missing information.
Since s(p) turns out to be intimately connected to thermodynamic entropy,
it is called the Shannon entropy of the probability distribution.
The Shannon entropy of a density operator ρ is defined to bes(ρ) = − Tr ρ ln ρ.

(6.90)

11 C.E. Shannon, Bell Systems Technical Journal, 27, 379 (1948). For a much fuller
account, see E.T. Jaynes Probability Theory: the Logic of Science Cambridge University
Press, 2003.

6.4 Thermodynamics

129

The right side of this expression involves a function, ln(x) of the operator
ρ. We recall from equation (2.20) that f (ρ) has the same eigenkets as ρ andeigenvalues f (λi ), where λi are the eigenvalues of ρ. Hences = − Tr(ρ ln ρ) = −XX
X Xln(pj )|jihj|ni = −pn ln pn .
pi |iihi|
hn|
n

n

j

i

(6.91)
Hence s is simply the Shannon entropy of the probability distribution {pi }that appears in the definition (6.64) of ρ.

6.4 Thermodynamics
Thermodynamics is concerned with macroscopic systems about which we
don’t know very much, certainly vastly less than is required to define aquantum state. For example, the system might consist of a cylinder fullof fluid and our knowledge be confined to the chemical nature of the fluid(that it is O2 or CO2 , or whatever), the mass of fluid, its volume and thetemperature of the environment with which it is in equilibrium. In the
canonical picture we consider that as a result of exchanges of energy withthe environment, the energy of the fluid fluctuates around a mean U . The
pressure also fluctuates around a mean value P , but the volume V is welldefined and under our control.
Thermodynamics applies to systems that are more complex than bodies
of fluid, for example to a quantity of diamond. In such a case the stress in thematerial is not fully described by the pressure, and thermodynamic relations
involve also the shear stress and the shear strain within the crystal. If thecrystal, like quartz, has interesting electrical properties, the thermodynamic
relations will involve the electric field within the material and the polarisationthat it induces. A fluid is the simplest non-trivial thermodynamic system andtherefore the focus of introductory texts, but the principles that it illuminates
are of much wider validity. For simplicity we restrict our discussion to fluids.
To obtain relations between the thermodynamic variables from a knowledge of the system’s microstructure, we need to assign a probability pi toeach of the system’s zillions of quantum states. We argue that the only rational way to assign probabilities to the stationary states of a thermodynamic
system is to choose them such that (i) they reproduce any measurements
we have of the system, and (ii) they maximise the Shannon entropy. Requirement (ii) follows because in choosing the {pi } we must not specify any
information beyond that included when we satisfy requirement (i) – our probabilities must “tell the truth, the whole truth and nothing but the truth”.
It is straightforward to show (Problem 6.16) that the pi that maximise theShannon entropy for given internal energy
U≡

X

Ei pi

(6.92)

stationary
states i

are given bypi =1 −βEi,eZ(6.93a)where β ≡ 1/(kB T ) is the inverse temperature andZ≡X

e−βEi .

(6.93b)stationary
states i

The quantity Z defined above is called the partition function; it is manifestly a function of T and less obviously a function of the volume V andwhatever other parameters define the spectrum {Ei } of the Hamiltonian. In

130Chapter 6: Composite systems

equation (6.93a) its role P
is clearly to ensure that the probabilities satisfy thenormalisation condition i pi = 1.
Since the probability distribution (6.93a) maximises the Shannon entropy for given internal energy, we take the density operator of a thermodynamic system to be diagonal in the energy representation and to be givenby
1 X(6.94)
e−βEi |iihi|.
ρ=Z stationary
states i

This form of the density operator is called the Gibbs distribution in honourof J.W. Gibbs (1839–1903), who died before quantum mechanics emerged but
had already established that probabilities should given by equation (6.93a).
The sum in equation (6.94) is over quantum states not energy levels. It
is likely that many energyPlevels will be highly degenerate and in this casethe sum simplifies to Z = α gα e−βEα , where α runs over energy levels andgα is the number of linearly independent quantum states in level α.
The expectation of the Hamiltonian of a thermodynamic system isXXX
pi |iihi|ni =pn En = U,
(6.95)
H = Tr(Hρ) =
hn|H
n

n

i

where we have used the definition (6.92) of the internal energy. Thus theinternal energy U of thermodynamics is simply the expectation value of thesystem’s Hamiltonian. Another important expression for U follows straightforwardly from equations (6.92) and (6.93):
U =−∂ ln Z.
∂β(6.96)

We obtain an interesting equation using equation (6.93a) to eliminate
the second occurrence of pn from the extreme right of equation (6.91):Xs=−pn (−βEn − ln Z) = βU + ln Z.
(6.97)nIn terms of the thermodynamic entropy
S ≡ kB s(6.98)and the Helmholtz free energyF ≡ −kB T ln Z

(6.99)

equation (6.97) can be written
F = U − T S,

(6.100)which in classical thermodynamics is considered to be the definition of theHelmholtz free energy. When we substitute our definition of F into equation(6.96), we obtain
U=∂(βF )
∂F
∂F
=F +β
=F −T
.
∂β
∂β
∂T

(6.101)Comparing this equation with equation (6.100) we conclude that
S=−

∂F
.
∂T

(6.102)The difference of equation (6.92) between two similar thermodynamic
states isX(dpi Ei + pi dEi ).
(6.103)dU =i

6.4 Thermodynamics

131P

i pi ln pi of the thermody-

Similarly differencing the definition S = −kB
namic entropy (eqns 6.91 and 6.98), we obtaindS = −kBXi(ln pi + 1) dpi = −kBX

ln pi dpi ,(6.104)iwhere the P
second equality exploits the fact that {pi } is a probability distri= 1 always. By equation (6.93a), ln pi = −Ei /(kB T ) − ln Z,
bution so i piP
so again using i pi = 1, equation (6.104) can be rewritten
T dS =

X

Ei dpi .

(6.105)i

If we heat our system up at constant volume, the Ei stay the same but thepi change because
P they depend on T . In these circumstances the increase ininternal energy, i Ei dpi , is the heat absorbed by the system. Consequently,
equation (6.105) states that T dS is the heat absorbed when the system isheated with no work done. This statement coincides with the definition ofentropy in classical thermodynamics.
Substituting equation (6.105) into equation (6.103) yields
dU = T dS − P dV,(6.106a)X(6.106b)

where
P ≡−

i

pi

∂Ei
.
∂V

If we isolate our system from heat sources and then slowly change its volume, the adiabatic principle (§11.1) tells us that the system will stay inwhatever stationary state it started in. That is, the pi will be constant whilethe volume of the thermally isolated system is slowly changed. In classicalthermodynamics this is an ‘adiabatic’ change. From equation (6.104) we seethat the entropy S is constant during an adiabatic change, just as classicalthermodynamics teaches.
Since dS = 0 in an adiabatic change, the change in U as V is varied
must be the mechanical work done on the system, −P dV, where P is thepressure the system exerts. This argument establishes that the quantity P
defined by (6.106b) is the pressure.
Differentiating equation (6.100) for the Helmholtz free energy and using
equation (6.106a) to eliminate dU , we find thatdF = −SdT − P dV.
From this it immediately follows that


∂F
S=−
;
∂T V

P =−



(6.107)

∂F
∂V



.

(6.108)T

The first of these equations was obtained above but the second one is new.
Equation (6.106a) is the central equation of thermodynamics since itembodies both the first and second laws of thermodynamics. This result establishes that classical thermodynamics is a consequence of applying quantum mechanics to systems of which we know very little. Remarkably, physicists working in the first half of the 19th century discovered thermodynamics
long before quantum mechanics was thought of, using extremely subtle arguments concerning heat engines. Quantum mechanics makes these arguments
redundant. Notwithstanding this redundancy, they continue to feature in undergraduate syllabuses the world over because they are beautiful. But thenso are copperplate writing and slide rules, which have rightly disappeared
from schools.

132

Chapter 6: Composite systems

A possible explanation for the survival of thermodynamics as an independent discipline is as follows. Equations (6.99), (6.100) and (6.108)
establish that any thermodynamic quantity can be obtained from the dependence of the partition function on T and V . Unfortunately, this dependence
can be calculated for only a very few Hamiltonians. In almost all practicalcases we cannot proceed by evaluating Z. However, once we know that Z
and therefore F and S exist, we can determine their functional forms fromexperimental data. For example, by measuring the heat released on cooling
our system
at constant volume to absolute zero, we can determine its entropyR
S = dQ/T . Similarly, we can measure the system’s pressure as a functionof T and V. Then by integrating equation (6.107) we can obtain F (T, V)
and thus infer Z(T, V). In none of these operations is the involvement of
quantum mechanics apparent, so engineers and chemists, who make extensive use of thermodynamics, are generally unaware that it is a consequence ofquantum mechanics. Quantum mechanics provides us with relations between
thermodynamics quantities but does not enable us to evaluate the quantities
themselves. Evaluation must still be done with 19th century technology.
Although thermodynamics systems are inherently macroscopic, quantum mechanics plays a central role in determining their thermodynamic
quantities because it defines the stationary states we have to sum over in(6.93b) to form the partition function. Before quantum mechanics was born,
the thermodynamic properties of an ideal gas – one composed of moleculesthat occupy negligible volume and interact only at very short range – wereobtained by summing over the phase-space locations of each molecule independently. In this procedure there are six distinct states of a three-molecule
gas in which there are molecules at the phase-space locations x1 , x2 and x3 :in one state molecule 1 is at x1 , molecule 2 is at x2 and molecule 3 is at
x3 , and a distinct state is obtained by swapping the locations of molecule
1 and molecule 2, and so forth. Quantum mechanics teaches that the stateof the gas is completely specified by listing the three occupied states, |1i,
|2i and |3i for it is meaningless to say which molecule is in which state.
The classical way of counting states leads to absurd results even for gases atroom temperature (Problem 6.22). At low-temperatures another aspect ofclassical physics leads to erroneous results: the low-lying energy levels of agas are distributed discretely rather than continuously in E, with the resultthat specific heats always vanish in the limit T → 0 (Nernst’s theorem;
Problem 6.23), contrary to the prediction of classical physics.
An important lesson to be learnt from the failure of classical physics topredict the properties of an ideal gas is the importance in quantum mechanics of thinking wholistically: we have to sum over the quantum states of thewhole cylinder of gas, not over the states of individual molecules. This isanalogous to the importance for understanding EPR phenomena of considering the quantum system formed by the entangled particles taken together.
In quantum mechanics the whole is generally very much more than the sumof its parts because there are non-trivial correlations between the parts.12

6.5 Measurement
In §1.4 we asserted that the state of a system ‘collapses’ into one of theeigenstates |qj i of the operator Q the instant we measure the observable Q.
Consequently, the result of measuring Q is to leave the system in the welldefined quantum state |qj i. It’s time to examine this collapse hypothesis
critically.
Superficially the collapse hypothesis is merely an assertion that measurements are reproducible in the sense that if we measure something twice
in quick succession, we will obtain the same result: in §2.1 |qj i was defined
12 The origin of these correlations is the subject of §10.1.

6.5 Measurement133

to be the state in which measurement of Q was certain to yield the value
qj , so if the measurement of Q is to be reproducible, the system has to bein the state |qj i immediately after the measurement. However, our system’squantum state |ψi is supposed to describe the system’s real, physical state,
not just our knowledge of it. So something physical must have happened tomake |ψi shift from the value it had before the measurement to the state|qj i 6= |ψi that it had just after the measurement was completed. Notice thatthe evolution from |ψi to |qj i has not been derived from the tdse, which wehave stated to be the equation that governs the time-evolution of |ψi. So
this Copenhagen interpretation of quantum mechanics implies that every
measurement leads to a momentary suspension of the equations of motion,
so the system can be steered, by forces unspecified, into a randomly chosen
state! This is not serious physics. We need to consider more realisticallywhat is involved in making a measurement.
A first step from Copenhagen towards the real world can be taken byrecognising that since real measurements are associated with error bars, they
will not leave the system in a state in which the result of a subsequent
measurement is certain. It follows that a real measurement of Q will in
general not leave the system in one of the states |qi i in which the result of asubsequent measurement is certain. That is, the collapse hypothesis is false.
The Copenhagen interpretation does, however, contain a crucial insight
into measurement by stressing that any measurement physically disturbs thesystem, so the system’s state after a measurement has been made is different
from what it was earlier. In classical physics we may or may not have toworry about the disturbance of the system by the measuring process – forexample, when we measure the positions of Jupiter’s moons by pointing atelescope at them, we don’t need to worry about disturbance caused by measurement. But when we measure the voltage across a resistor by connecting
a galvanometer in parallel with it, we change the voltage by increasing thecurrent through the circuit either side of the resistor. We minimise this disturbance by buying a galvanometer with the highest affordable impedance,
and we estimate the magnitude of the effect and try to correct for it. Whenmeasurements are made on systems small enough for quantum mechanics
to be relevant, the system will be significantly disturbed because we cannot
make instruments of arbitrary sensitivity – quantum mechanics itself makes
this impossible. So the Copenhagen interpretation is right to stress that
post- and pre-measurement states are significantly different.
Where the Copenhagen interpretation slips up is in supposing that the
disturbance caused by a measurement can be taken into account without
knowing anything about the measuring instrument that was used. Physically
it is obvious that since the disturbance is caused by the instrument, wecannot hope to predict the evolution of the system without knowledge of thephysical principles on which the instrument works, and the configuration it
was in when the measuring process started. In fact, it’s astonishing that
useful predictions can be extracted from a theory that fails to engage withthese key questions!
The Copenhagen interpretation makes progress through two stratagems.
First it assumes that the builder of the measuring instrument has been cleverenough to make an instrument that makes essentially reproducible measurements. This being so, the state in which the measuring instrument leaves
the system must be one of the eigenstates of the observable’s operator. Focusing on instruments that measure reproducibly is a shrewd move because
instruments that do not yield reproducible readings are regarded as ‘noisy’
and tend not to be used. So the Copenhagen interpretation does make anassumption about the nature of the measuring instrument – that it is a good
one, so it steers the system’s state to one of the |qi i – and gets by without
considering the detailed physics that actually does the steering. By declining to consider the physics of the instrument, the theory remains general andable to produce results that apply to any instrument rather than a particularbrand of electrometer, or whatever.

134

Chapter 6: Composite systems

The second stratagem is to abandon causality and assert that the outcome of a measurement is inherently uncertain. It merely supplies probabilities of the various measurement outcomes. So while a differential equation is
supplied with which to calculate with precision the evolution of the system’s
state between measurements, the consequences of measurement are left toblind chance. This stratagem circumvents the failure to consider fully thenature of the measuring equipment, for (as we shall argue) it is the unknown
state of this equipment at the start of the measuring process that makes theoutcome of the measurement uncertain.
A key insight of the Copenhagen interpretation is that the states inwhich the outcome of measuring one observable is certain are generally different from the states in which the outcome of measuring a different observable is certain. That is, there are fundamentally incompatible observablesin the sense that if you are certain what value you will measure for one observable, you cannot be certain what value you will measure for the otherobservable. Since the states in which the outcomes of measurements are
the eigenkets of an observable’s operator, and operators that do not have acomplete set of mutual eigenkets do not commute, incompatible observables
have non-commuting operators. There is nothing deeply mysterious about
incompatible observables – it just happens that the act of measuring one
observable drives the system into different states from those into which thesystem is driven by measuring the other observable. The key thing is to beclear that an observable is not an intrinsic property of the system, but aquestion we can ask of it. In general my particle has neither a position nor
a momentum, but these are questions I can ask of it, and after the question
has been asked, the particle will be (temporarily) in a special state that does
have a well-defined position / momentum.
The probabilistic outcome of a measurement introduces to physics a
new feature of great consequence: irreversibility. After a measurement, it is
impossible to determine what the state of the system was before the measurement was made. This is so because many different initial states of thesystem are consistent with measuring a particular value of the observable Q,
and therefore causing the system to finish in a given state |qi i.
An instrument is itself a dynamical system, and its dynamics is governedby quantum mechanics. We make a measurement by putting the instrument‘into contact’ with our system – that is, we ensure that the instrument andthe system are dynamically coupled by a non-negligible Hamiltonian. Oncein contact, the instrument and our system together form a composite system,
and, like all dynamically coupled subsystems, they soon become entangled.
That is, the state of the instrument becomes correlated with that of thesystem. It is as a consequence of this entanglement that the instrument isable to show a reading that reflects the state of the system being measured.
The instrument must be sufficiently macroscopic that it can be readby a human being – if it were microscopic, an instrument would be needed
to measure it, and so on until eventually the macroscopic scale is reached.
That is, an instrument that is not macroscopic can be considered part of thequantum system being studied and evolved with the tdse; if a measurement
is to be made, at some point the entire quantum system has to interact
with a macroscopic instrument. Anything macroscopic will be in an impure
state (§6.3). Consequently, once interaction with a macroscopic instrument
is established, the outcome of the experiment will be probabilistic, just as
the Copenhagen interpretation asserts.
A measurement will also be irreversible in the sense that one cannot
compute the state that the system was in prior to the interaction becausesuch a computation would require for the initial conditions complete knowledge of the instrument’s state after the measurement.
This discussion shows that the collapse hypothesis is really a clever way
to circumvent our unwillingness to follow the dynamics of system/instrument
interaction. The failure to follow the interaction enables the theory to makeProblems

135

general statements that are valid irrespective of which devices are actuallyused for measurement, but in specific cases it should be possible to obtaina more complete understanding by properly considering the dynamics ofthe measuring instrument. Unfortunately, we probably need an extension
to quantum mechanics to take this step, because a conventional quantummechanical theory of the measuring instrument will require us at some pointto ‘observe’ the instrument using the collapse hypothesis, from which we aretrying to escape: quantum mechanics is a theoretical arena from which the
only exit to the real world is through the turnstile of the collapse hypothesis.
We expect any extension of quantum mechanics that successfully includes the act of measurement to be formulated in terms of density operators, because incomplete knowledge of the state of our instruments certainly
makes a major contribution to the uncertain outcome of measurements, andmay be entirely responsible for it.

Problems
6.1 A system AB consists of two non-interacting parts A and B. The dynamical state of A is described by |ai, and that of B by |bi, so |ai satisfies thetdse for A and similarly for |bi. What is the ket describing the dynamical
state of AB? In terms of the Hamiltonians HA and HB of the subsystems,
write down the tdse for the evolution of this ket and show that it is automatically satisfied. Do HA and HB commute? How is the tdse changed
when the subsystems are coupled by a small dynamical interaction Hint ?
If A and B are harmonic oscillators, write down HA , HB . The oscillating
particles are connected by a weak spring. Write down the appropriate formof the interaction Hamiltonian Hint . Does HA commute with Hint ? Explain
the physical significance of your answer.
6.2 Explain what is implied by the statement that “the physical state ofsystem A is correlated with the state of system B.” Illustrate your answer
by considering the momenta of cars on (i) London’s circular motorway (theM25) at rush-hour, and (ii) the road over the Nullarbor Plain in southern
Australia in the dead of night.
Explain why the states of A and B must be uncorrelated if it is possible
to write the state of AB as a ket |AB; ψi = |A; ψ1 i|B; ψ2 i that is a productof states of A and B. Given a complete set of states for A, {|A; ii} and acorresponding complete set of states for B, {|B; ii}, write down an expression
for a state of AB in which B is possibly correlated with A.
6.3 Given that the state |ABi of a compound system can be written asa product |Ai|Bi
P of states of the individual systems, show that when |ABi
is written as
ij cij |A; ii|B; ji in terms of arbitrary basis vectors for thesubsystems, every column of the matrix cij is a multiple of the leftmost
column.
6.4 Consider a system of two particles of mass m that each move in one
dimension along a given rod. Let |1; xi be the state of the first particle whenit’s at x and |2; yi be the state of the second particle when it’s at y. A
complete set of states of the pair of particles is {|xyi} = {|1; xi|2; yi}. Write
down the Hamiltonian of this system given that the particles attract one
another with a force that’s equal to C times their separation.
Suppose the particles experience an additional potential
V (x, y) = 21 C(x + y)2 .

(6.109)Show that the dynamics of the two particles is now identical with the dynamics of a single particle that moves in two dimensions in a particular potential
Φ(x, y), and give the form of Φ.

136

Problems

6.5 In §6.1.4 we derived Bell’s inequality by considering measurements by
Alice and Bob on an entangled electron-positron pair. Bob measures thecomponent of spin along an axis that is inclined by angle θ to that used by
Alice. Given the expression|−, bi = cos(θ/2) eiφ/2 |−i − sin(θ/2) e−iφ/2 |+i,

(6.110)for the state of a spin-half particle in which it has spin − 12 along the direction
b with polar angles (θ, φ), with |±i the states in which there is spin ± 21 along
the z-axis, calculate the amplitude AB (−|A+) that Bob finds the positron’s
spin to be − 21 given that Alice has found + 12 for the electron’s spin. Henceshow that PB (−|A+) = cos2 (θ/2).
6.6 Show that when the Hadamard operator UH is applied to every qubitof an n-qubit register that is initially in a member |mi of the computational
basis, the resulting state is
|ψi =

1

n
2X
−1

2n/2 x=0

ax |xi,(6.111)

where ax = 1 for all x if m = 0, but exactly half the ax = 1 and the otherhalf the ax = −1 for any other choice of m. Hence show that

X
1
|0i
if all ax = 1
U
a
|xi
=
H
x
|mi 6= |0i if half the ax = 1 and the other ax = −1.
2n/2
x(6.112)

6.7 Show that the trace of every Hermitian operator is real.
6.8 Let ρ be the density operator of a two-state system. Explain why ρ
can be assumed to have the matrix representation


a c
ρ=
,
(6.113)c∗ bwhere a and b are real numbers. Let E0 and E1 > E0 be the eigenenergies ofthis system and |0i and |1i the corresponding stationary states. Show fromthe equation of motion of ρ that in the energy representation a and b aretime-independent while c(t) = c(0)eiωt with ω = (E1 − E0 )/h̄.
Determine the values of a, b√and c(t) for the case that initially the system
is in the state |ψi = (|0i + |1i)/ 2. Given that the parities of |0i and |1i areeven and odd respectively, find the time evolution of the expectation value
x in terms of the matrix element h0|x|1i. Interpret your result physically.
6.9 In this problem we consider an alternative interpretation of the density
operator. Any quantum state can be expanded in the energy basis as|ψ; φi ≡NX√
pn eiφn |ni,

(6.114)n=1

where φn is real and pn is the probability that a measurement of energy will
return En . Suppose we know the values of the pn but not the values of thephases φn . Then the density operator is
Z 2π Nd φ
ρ=|ψ; φihψ; φ|.
(6.115)(2π)N
0PShow that this expression reduces ton pn |nihn|. Contrast the physicalassumptions made in this derivation of ρ with those made in §6.3.
Clearly |ψ; φi can be expanded in some other basis {|qr i} as
Xp(6.116)
Pr eiηr |qr i,|ψ; φi ≡rwhere Pr is the probability of obtaining qr on a measurement of the observable Q and the ηr (φ) are unknown phases. Why does this second expansion
not lead to the erroneous conclusion that ρ is necessarily diagonal in the{|qr i} representation?

Problems

137

6.10 Show that the equation of motion of the density operator ρ is solvedbyρt = U (t)ρ0 U † (t),(6.117)−iHt/h̄where U (t) ≡ e
is the time-evolution operator introduced in §4.3.

6.11∗ Show that when the density operator takes the form ρ = |ψihψ|,the expression Q = Tr Qρ for the expectation value of an observable can be
reduced to hψ|Q|ψi. Explain the physical significance of this result. For the
given form of the density operator, show that the equation of motion of ρ
yields∂|ψi− H|ψi.
(6.118)|φihψ| = |ψihφ| where |φi ≡ ih̄
∂tShow from this equation that |φi = a|ψi, where a is real. Hence determine
the time evolution of |ψi given the at t = 0 |ψi = |Ei is an eigenket of H.
Explain why ρ does not depend on the phase of |ψi and relate this fact tothe presence of a in your solution for |ψ, ti.
P
6.12 The density operator is defined to be ρ =α pα |αihα|, where pα
is the probability that the system P
is in the state α. Given an arbitrarybasis {|ii} and the expansions |αi = i aαi |ii, calculate the matrix elementsρij = hi|ρ|ji of ρ. Show that the diagonal elements ρii are non-negative real
numbers and interpret them as probabilities.
P
6.13 Consider the density operator ρ = ij ρij |iihj| of a system that is ina pure state. Show that every row of the matrix ρij is a multiple of the firstrow and every column is a multiple of the first column. Given that these
relations between the rows and columns of a density matrix hold, show thatthe system is in a pure state. Hint: exploit the real, non-negativity of ρ11
established in Problem 6.12 and the Hermiticity of ρ.
6.14 Consider the rate of change of the expectation of the observable Q
when the system is in an impure state. This isdQ X
d=(6.119)pn hn|Q|ni,dtdtnwhere pn is the probability that the system is in the state |ni. By using
Ehrenfest’s theorem to evaluate the derivative on the right of (6.119), derive
the equation of motion ih̄dQ/dt = Tr(ρ[Q, H]).
6.15 Find the probability distribution (p1 , . . . , pn ) for n possible outcomesthat maximises the Shannon entropy. Hint: use a Lagrange multiplier.
6.16 Use Lagrange multipliers λ and β to extremise the Shannon
Pentropyof the probability
distribution{p}subject
totheconstraints(i)ii pi = 1
P
and (ii) i pi Ei = U . Explain the physical significance of your result.

6.17 Explain why if at t = 0 the density operator of a system is given by
the Gibbs distribution, it remains so at later times.

6.18 A composite system is formed from uncorrelated subsystem A andsubsystem B, both in impure states. The numbers {pAi } are the probabilitiesof the members of the complete set of states {|A; ii} for subsystem A, while
the numbers {pBi } are the probabilities of the complete set of states {|B; ii}
for subsystem B. Show that the Shannon entropy of the composite system isthe sum of the Shannon entropies of its subsystems. What is the relevanceof this result for thermodynamics?
6.19 The |0i state of a qubit has energy 0, while the |1i state has energy ǫ.
Show that when the qubit is in thermodynamic equilibrium at temperature
T = 1/(kB β) the internal energy of the qubit isǫ
U = βǫ
.
(6.120)e +1Show that when βǫ ≪ 1, U ≃ 12 ǫ, while for βǫ ≫ 1, U ≃ ǫe−βǫ . Interpretthese results physically and sketch the specific heat C = ∂U/∂T as a functionof T .

138

Problems

6.20 Show that the time-evolution of the density operator leaves the Shannon entropy s = − Tr ρ log ρ invariant.
6.21 Show that the partition function of a harmonic oscillator of natural
frequency ω ise−βh̄ω/2
Zho =.
(6.121)1 − e−βh̄ωHence show that when the oscillator is at temperature T = 1/(kB β) the
oscillator’s internal energy is
Uho = h̄ω1
12 + eβh̄ω − 1

.

(6.122)

Interpret the factor (eβh̄ω − 1)−1 physically. Show that the specific heatC = ∂U/∂T iseβh̄ω
C = kB βh̄ω
(βh̄ω)2 .
(6.123)(e− 1)2Show that lim T→0 C = 0 and obtain a simple expression for C when kB T ≫h̄ω.
6.22 A classical ideal monatomic gas has internal energy U = 23 N kB T andpressure P = N kB T /V, where N is the number of molecules and V is thevolume they occupy. From these relations, and assuming that the entropy
vanishes at zero temperature and volume, show that in general the entropyisS(T, V) = N kB ( 32 ln T + ln V).
(6.124)
A removable wall divides a cylinder into equal parts of volume V. Initially
the wall is in place and each half contains N molecules of ideal monatomic
gas at temperature T . The wall is removed. Show that equation (6.124)
implies that the entropy of the entire body of fluid increases by 2 ln 2 N kB .
Can this result be squared with the principle that dS = dQ/T , where dQ isthe heat absorbed when the change is made reversibly? What conclusion do
you draw from this thought experiment?
6.23 Consider a ‘gas’ formed by M non-interacting, monatomic moleculesof mass m that move in a one-dimensional potential well V = 0 for |x| < aand ∞ otherwise. Assume that at sufficiently low temperatures all molecules
are either in the ground or first-excited states. Show that in this approximation the partition function is given byln Z = −M βE0 + e−3βE0 − e−3(M+1)βE0

where

E0 ≡

π 2 h̄2.
8ma2(6.125)Show that for M large the internal energy, pressure and specific heat of thisgas are given by
2E0
9E02 −3βE0
M + 3e−3βE0 ; CV =.
eakB T 2(6.126)In what respects do these results for a quantum ideal gas differ from theproperties of a classical ideal gas? Explain these differences physically.
U = E0 (M + 3e−3βE0 ) ; P =7Angular Momentum
In Chapter 4 we introduced the angular-momentum operators Ji as the generP
ators of rotations. We showed that they form a pseudo-vector, so J 2 = i Ji2
is a scalar. By considering the effect of rotations on vectors and scalars, we
showed that the the Ji commute with all scalar operators, including J 2 , andfound that commutator of Ji with a component of vector operator is given
by equation (4.31). From this result we deduced
that the Ji do not commutePwith one another, but satisfy [Ji , Jj ] = i k ǫijk Jk .
Although we have from the outset called the Ji ‘angular-momentum
operators’, the only connection we have established between the Ji and angular momentum is tenuous and by no means justifies our terminology: wehave simply shown that when the Hamiltonian is invariant under rotations
about some axis α̂, and the system starts in an eigenstate of the corresponding angular-momentum operator α̂ · J, it will subsequently remain in thateigenstate. Consequently, the corresponding eigenvalue is then a conserved
quantity. In classical mechanics dynamical symmetry about some axis implies that the component of angular momentum about that axis is conserved,
so it is plausible that the conserved eigenvalue is a measure of angular momentum. This suggestion will be substantiated in this chapter. Anotherimportant task for the chapter is to explain how the orientation of a systemis encoded in the amplitudes for it to be found in different eigenstates ofappropriate angular-momentum operators. We start by using the angularmomentum commutation relations to determine the spectrum of the Ji .

7.1 Eigenvalues of Jz and J 2
Since no two components of J commute, we cannot find a complete set ofsimultaneous eigenkets of two components of J. We can, however, find acomplete set of mutual eigenkets of J 2 and one component of J because[J 2 , Ji ] = 0. Without loss of generality we can orient our coordinates so thatthe chosen component of J is Jz . Let us label a ket which is simultaneously
an eigenstate of J 2 and Jz as |β, mi, whereJ 2 |β, mi = β|β, mi;

Jz |β, mi = m|β, mi.

(7.1)

We now define
J± ≡ Jx ± iJy .

(7.2)

140

Chapter 7: Angular Momentum

These objects clearly commute with J 2 , while their commutation relations
with Jz are[J+ , Jz ] = [Jx , Jz ] + i[Jy , Jz ] = −iJy − Jx = −J+[J− , Jz ] = [Jx , Jz ] − i[Jy , Jz ] = −iJy + Jx = J− .

(7.3)

Since J± commutes with J 2 , the kets J± |β, mi are eigenkets of J 2 witheigenvalue β. Operating with Jz on these kets we findJz J+ |β, mi = (J+ Jz + [Jz , J+ ]) |β, mi = (m + 1)J+ |β, miJz J− |β, mi = (J− Jz + [Jz , J− ]) |β, mi = (m − 1)J− |β, mi.

(7.4)

Thus, J+ |β, mi and J− |β, mi are also members of the complete set of statesthat are eigenstates of both J 2 and Jz , but their eigenvalues with respect to
Jz differ from that of |β, mi by ±1. Therefore we may write
J± |β, mi = α± |β, m ± 1i,(7.5)

where α± is a constant that we now evaluate. We do this by taking the†length-squared of both sides of equation (7.5). Bearing in mind that J+=
J− , we find
|α2+ | = hβ, m|J− J+ |β, mi = hβ, m|(Jx − iJy )(Jx + iJy )|β, mi= hβ, m|(J 2 − Jz2 − Jz )|β, mi = β − m(m + 1).

(7.6)

Similarly, |α− |2 = β − m(m − 1), so
α± =p
β − m(m ± 1).

The Ji are Hermitian operators, so hψ|Ji2 |ψi = Ji |ψi

(7.7)
2

≥ 0. Henceβ = hβ, m|J 2 |β, mi = hβ, m|(Jx2 + Jy2 + Jz2 )|β, mi ≥ m2 .

(7.8)

So notwithstanding equation (7.5), it cannot be possible to create states withever larger eigenvalues of Jz by repeated application of J+ . All that can stop
us doing this is the vanishing of α+ when we reach some maximum eigenvalue
mmax that from equation (7.7) satisfies
β − mmax (mmax + 1) = 0.

(7.9)

Similarly, α− must vanish for a smallest value of m that satisfies
β − mmin (mmin − 1) = 0.

(7.10)

Eliminating β between (7.9) and (7.10) we obtain a relation between mmax
and mmin that we can treat as a quadratic equation for mmin . Solving thisequation we find thatmmin = 21 {1 ± (2mmax + 1)}.

(7.11)The plus sign yields a value of mmin that is incompatible with our requirement that mmin ≤ mmax , so we must have mmin = −mmax . To simplify thenotation, we define j ≡ mmax , so that equation (7.9) becomes β = j(j + 1)
and −j ≤ m ≤ j. Finally, we note that since an integer number of applications of J− will take us from |β, ji to |β, −ji, 2j must be an integer – see
Figure 7.1. In summary, the eigenvalues of J 2 are j(j+1) with 2j = 0, 1, 2, . . .
and for each value of j the eigenvalues m of Jz are (j, j − 1, . . . , −j).
At this point we simplify the labelling of kets by defining |j, mi to bewhat has hitherto been denoted |β, mi with β = j(j + 1) – we clear a great7.1 Eigenvalues of Jz and J 2

141

Figure 7.1 Going from mmin tommax in an integer number of steps
in the cases j = 23 , 2.

deal of clutter from the page by replacing |j(j + 1), mi with |j, mi. The kets’
eigenvalues with respect to J 2 are of course unaffected by this relabelling.
Had we known at the outset that the eigenvalues of J 2 would be of the formj(j + 1), we would have used the new notation all along.
In summary, we can find simultaneous eigenstates of J 2 and one of theJi , conventionally taken to be Jz . The eigenvalues of J 2 are j(j + 1) with
2j = 0, 1, . . ., and for any given j the eigenvalues m of Jz then run from +j
to −j in integer steps:j ≥ m ≥ −j.
(7.12)In order to move from the state |j, mi to the adjacent state |j, m ± 1i we use
the raising or lowering operators J± which act asp
J± |j, mi = α± (m)|j, m ± 1i = j(j + 1) − m(m ± 1)|j, m ± 1i. (7.13)

These operators only change the Jz eigenvalue, so they just realign a given
amount of total angular momentum, placing more (J+ ) or less (J− ) along
the z-axis. So far, we have not discovered how to alter the J 2 eigenvaluej(j + 1).
It is sometimes helpful to rewrite the constants α± (m) of equation (7.13)
in the formp
α+ (m) = (j − m)(j + m + 1)
(7.14)pα− (m) = (j + m)(j − m + 1).

These equations make it clear that the proportionality constants for different
m satisfy
α+ (m) = α+ (−m − 1)
α− (m) = α− (−m + 1)

α+ (m − 1) = α− (m)α− (m) = α+ (−m).

(7.15)For example, when J− lowers the highest state |j, ji, we obtain the same
proportionality constant as when J+ raises the lowest state |j, −ji; consequently, we only need to work out half the constants directly, because we canthen infer the others.
By expressing Jx = 12 (J+ + J− ) in terms of the ladder operators, weobserve that when we apply Jx to a state |j, mi for j > 0 we obtain a ketthat differs from |j, mi, so |j, mi is never an eigenket of Jx . Hence for j > 0
it is impossible to be certain what will be the result of measuring both Jz
and Jx . It is trivial to see that this argument extends to the pair (Jz , Jy ),so for j > 0 it is impossible to be certain of the outcome of more than one of
the Ji . If j = 0 the outcome measuring of any component of J is certainly
0, but a null vector has no direction. Consequently, there are no states inwhich the vector J has a well-defined direction. This situation contrasts withthe case of the momentum vector p, which can have a well defined direction
because its components commute with one another.
In §4.1.2 we discovered that when the system is rotated through anangle α around the z axis, its ket |ψi transforms to |ψ ′ i = U (α)|ψi, where

142

Chapter 7: Angular Momentum

the unitary operator U (α) = exp(−iαJz ). If |ψi = |j, mi is an eigenket ofJz , U (α) simply changes its phase:
U (α)|j, mi = e−iαJz |j, mi = e−iαm |j, mi.

(7.16)

Since 2j is an integer, j (and hence m) must be either an integer or a half integer. Using this information in equation (7.16), we see that, after a rotation
through 2π around the z-axis, we have either|j, mi → |j, mi

for m even

(7.17a)|j, mi → −|j, mi for m odd.

(7.17b)orEquation (7.17a) is as expected; under a 2π rotation, the system returns to
its original state. However, equation (7.17b) says that a system with half
integer angular momentum does not return to its original state after a 2π
rotation – the initial and final states are minus one another! This difference of
behaviour between systems with integer and half-integer angular momentum
is of fundamental importance, and determines many other characteristics ofthese systems. A result of quantum field theory is that ‘spin-half’ fields
never attain macroscopic values: the quantum uncertainty in the value ofa spin-half field is always on the same order as the value of the field itself.
Integer-spin fields, by contrast, can attain macroscopic values: values thatare vastly greater than their quantum uncertainties. Consequently, classicalphysics – physics in the absence of quantum uncertainty – involves integerspin fields (the electromagnetic and gravitational fields are examples) but nospin-half field. Our intuition about what happens when a system is rotatedhas grown out of our experience of classical physics, so we consider thatthings return to their original state after rotation by 2π. If we had hands-on
experience of spin-half objects, we would recognise that this is not generallytrue.
7.1.1 Rotation spectra of diatomic molecules
Knowledge of the spectrum of the angular momentum operators enables us
to understand an important part of the dynamics of a diatomic molecule suchas carbon monoxide. For some purposes a CO molecule can be considered
to consist of two point masses, the nuclei of the oxygen and carbon atoms,
joined by a ‘light rod’ provided by the electrons. In this model the molecule’s
moment of inertia around the axis that joins the nuclei is negligible, while
the same moment of inertia I applies to any perpendicular axis.
In classical mechanics the rotational energy of a rigid body is

 2
Jy2
J2
Jx
E = 21(7.18)
+
+ z ,
Ix
Iy
Iz
where the Ii are the moments of inertia about the body’s three principal
axes and J is the body’s angular-momentum vector. We conjecture thatthe equivalent formula links the Hamiltonian and the angular momentum
operators in quantum mechanics:

Jy2h̄2 Jx2
Jz2
H=.
(7.19)
+
+
2 Ix
Iy
Iz
The best justification for adopting this formula is that it leads us to resultsthat are confirmed by experiments.
In the case of an axisymmetric body, we orient our body such thatthe symmetry axis is parallel to the z axis. Then I ≡ Ix = Iy and the
Hamiltonian can be written1h̄2 J 2
1
2
H=
−
+ Jz
.
(7.20)
2I
Iz
I7.1 Eigenvalues of Jz and J 2

143

From this formula and our knowledge of the eigenvalues of J 2 and Jz , we
can immediately write down the energies that form the spectrum of H:
Ejm =

h̄2
2



j(j + 1)
+ m2
I



1
1
−
Iz
I



,

(7.21)

where j is the total angular-momentum quantum number and |m| < j. In
the case of a diatomic molecule such as CO, Iz ≪ I so the coefficient of m2 isvery much larger than the coefficient of j(j + 1) and states with |m| > 0 will
occur only far above the ground state. Consequently, the states of interest
have energies of the form
Ej = j(j + 1)

h̄2
.
2I

(7.22)For reasons that will emerge in §7.2.2, only integer values of j are allowed.
CO is a significantly dipolar molecule. The carbon atom has a smaller
share of the binding electrons than the oxygen atom, with the result that it
is positively charged and the oxygen atom is negatively charged. A rotating
electric dipole would be expected to emit electromagnetic radiation. Because
we are in the quantum regime, the radiation emerges as photons which, as
we shall see, can add or carry away only one unit h̄ of angular momentum.
It follows that the energies of the photons that can be emitted or absorbed
by a rotating dipolar molecule are
Ep = ± (Ej − Ej−1 ) = ±j

h̄2
.
I

(7.23)

Using the relation E = hν between the energy of a photon and the frequency
ν of its radiation, the frequencies in the rotation spectrum of the moleculeareh̄.
(7.24)
νj = j2πI
In the case of 12 CO, the coefficient of j evaluates to 113.1724 GHz and spectral lines occur at multiples of this frequency (Figure 7.2).
In the classical limit of large j, J = jh̄ is the molecule’s angular momentum, and this is related to the angular frequency ω at which the molecule
rotates by J = Iω. When in equation (7.24) we replace jh̄ by Iω, we discover that the frequency of the emitted radiation ν is simply the frequency
ω/2π at which the molecule rotates around its axis. This conclusion makes
perfect sense physically. Now, because of the form of the Hamiltonian, theenergy eigenstates are also the eigenstates of Jz and J 2 . Therefore in any
energy eigenstate, J 2 = j(j + 1) and for low-lying states with m = 0 andj ∼ O(1), j(j + 1) is significantly larger than j 2 . Therefore νj in (7.24)
is smaller than the frequency at which the molecule rotates when it is in
the upper state of theOn the other hand, νj is larger than thep transition.
h̄rotation frequency (j − 1)j 2πIof the lower state. Hence the frequency atwhich radiation emerges lies between the rotation frequencies of the upperand lower states. Again this makes sense physically. As we approach theclassical regime, j becomes large so j(j + 1) ≃ j 2 ≃ (j − 1)j and the rotation
frequencies of the upper and lower states converge, from above and below,
on the frequency of the emitted radiation.
Measurements of radiation from 115 GHz and the first few multiples ofthis frequency provide one of the two most important probes of interstellar gas.1 In denser, cooler regions, hydrogen atoms combine to form H2
molecules, which are bisymmetric and do not have an electric dipole moment when they are simply rotating. Consequently, these molecules, which
1 The other key probe is the hyperfine line of atomic hydrogen that will be discussed
in Chapter 8.

144

Chapter 7: Angular Momentum

Figure 7.2 The rotation spectrum of CO. The full lines show the measured frequencies
for transitions up to j = 38 → 37, while the dotted lines show integer multiples of thelowest measured frequency. Up to the line for j = 22 → 21 the dotted lines are obscuredby the full lines except at one frequency for which measurements are not available. Forj ≥ 22 the separation between the dotted and full lines increases steadily as a consequenceof the centrifugal stretching of the bond between the molecule’s atoms. Measurements arelacking for several of the higher-frequency lines.

together with similarly uncommunicative helium atoms make up the great
majority of the mass of cold interstellar gas, lack readily observable spectral
lines. Hence astronomers are obliged to study the cold interstellar medium
through the rotation spectrum of the few parts in 106 of CO that it contains.
Important information can be gleaned from the relative intensities oflines associated with different values of j in equation (7.24). The rate at
which molecules emit radiation and thus the intensity of the line2 is proportional to the number nj of molecules in the upper state. As we shall deducein §7.5.3, all states have equal a priori probability, so nj is proportional to
the number of states that have the given energy – the degeneracy or statistical weight g of the energy level. From §7.1 we know that g = 2j + 1
because this is the number of possible orientations of the angular momentum
for quantum number j.
In §6.4 we saw that when a gas is in thermal equilibrium at temperature T , the probability pj that a given molecule is in a state of energy
Ej is proportional to the Boltzmann factor exp(−Ej /kB T ), where kB is theBoltzmann constant (eq. 6.93a). Combining this proportionality with thedependence on the degeneracy 2j + 1 just discussed leads us to expect thatthe intensity of the line at frequency νj will be
Ij ∝ (2j + 1) exp(−Ej /kB T )(j > 0).

(7.25)For E1 < kB T , Ij increases at small j before declining as the Boltzmann
factor begins to overwhelm the degeneracy factor. Fitting this formula, which
has only one free parameter (T ), to observed line intensities enables one bothto measure the temperature of the gas, and to check the correctness of thedegeneracy factor.
Figure 7.2 shows that for large values of the quantum number j, thespacing between lines in the spectrum diminishes in apparent violation of theprediction of equation (7.24). Lines with large j are generated by moleculesthat are spinning very rapidly. The bond between the nuclei is stretched likea spring by the centripetal acceleration of the nuclei. Stretching of the bondincreases the moment of inertia I, and from equation (7.24) this decreases
the frequency of the spectral lines (Problem 7.2).

2 We neglect the absorption of photons after emission, which can actually be an important process, especially for 12 CO.

7.2 Orbital angular momentum

145

7.2 Orbital angular momentum
Let x and p be the position and momentum operators of the system. Then,
inspired by classical mechanics, we define the dimensionless orbital angularmomentum operators by3
L≡1
x × p,h̄that isLi ≡1Xǫijk xj pk .
h̄

(7.26)

jk

From the rules of Table 2.1 and the Hermitian nature of x and p, the Hermitian adjoint of Li is
L†i =

1X
1X
ǫijk p†k x†j =ǫijk xj pk = Li ,
h̄
h̄
jk

(7.27)

jk

where we have used the fact that [xj , pk ] = 0 for j 6= k. Thus the Li are
Hermitian and are likely to correspond to observables. We also define thetotal orbital angular momentum operator by
L2 ≡ L · L = L2x + L2y + L2z ,(7.28)

which is again Hermitian, and calculate a number of commutators. First,
bearing in mind the canonical commutation relation (2.54), we haveX
1X
1X
ǫijl xjǫijk [xj pk , xl ] =ǫijk xj [pk , xl ] = −i
h̄
h̄
j
jk
jk
X
ǫilj xj .
=i(7.29)

X
1X
1X
ǫilj pj .
ǫijk [xj pk , pl ] =ǫijk [xj , pl ]pk = ih̄h̄j(7.30)[Li , xl ] =jSimilarly[Li , pl ] =jkjk

Notice that these commutation relations differ from the corresponding ones
for Ji [equations (4.30) and (4.32)] only by the substitution of L for J. Fromthese relations we can show that Li commutes with the scalars x2 , p2 andx · p. For example[Li , p2 ] =Xj[Li , p2j ] = i

Xǫijk (pk pj + pj pk ) = 0,

(7.31)

jk

where the last equality follows because the ǫ symbol is antisymmetric in jk
while the bracket is symmetrical in these indices (see also Problem 7.3). We
can now also calculate the commutator of one component of L with another.
We have[Lx , Ly ] =1[Lx , (zpx − xpz )] = i(−ypx + xpy ) = iLz .
h̄(7.32)

Clearly each Li commutes with itself, and the other non-zero commutators
can be obtained from equation (7.32) by permuting indices. These commutators mirror the commutators (7.104) of the Ji .
L is a vector operator by virtue of the way it is constructed out of thevectors x and p. It follows that L2 is a scalar operator. Hence the way these3 In many texts L is defined without the factor h̄−1 . By making L dimensionless, thisfactor simplifies many subsequent formulae.

146

Chapter 7: Angular Momentum

operators commute with the total angular momentum operators Ji follows
from the work of §4.2:
X[Ji , Lj ] = iǫijk Lk ; [Ji , L2 ] = 0.
(7.33)k

Although p2 and x2 commute with Li , the total angular momentum
operator J 2 does not:X
X
[Jj2 , Li ] = iǫjik (Lk Jj + Jj Lk ).
(7.34)[J 2 , Li ] =jjk

The right side does not vanish because the final bracket is not symmetric in
jk. The physical significance of [J 2 , Li ] being non-zero is that if our system
is in a state of well-defined total angular momentum, in general there will
be uncertainty in the amount of orbital angular momentum it has about any
axis. We shall explore the consequences of this fact in §7.5.
7.2.1 L as the generator of circular translations
In §4.1.1 we saw that when the system is displaced along a vector a, its ket
is transformed by the unitary operator U (a) = e−ia·p/h̄. We now imagine
successively performing n translations through vectors {a1 , a2 . . . , an }. Since
each translation will cause |ψi to be acted on by a unitary operator, the final
state will be

Y

n
i
exp
U (an ) . . . U (a2 )U (a1 )|ψi =ai · p |ψi
h̄
i=1
! 
(7.35)
ni X
ai · p |ψi,= exph̄ i=1

where the second equality follows because the components of p commute
with one another. Since the exponent
Pn in the last line is proportional to theoverall displacement vector A ≡ i=1 ai , the change in |ψi is independentof the path that the system takes. In particular, if the path is closed, A = 0
and |ψi is unchanged.
Now consider the effect of moving the system in a circle centred on theorigin and in the plane with unit normal n. When we increment the rotation
angle α by δα, we move the system throughδa = δα n × x.

The associated unitary operator is




i
i
U (δa) = exp − δα (n × x) · p = exp − δα n · (x × p)
h̄
h̄
=e

−iδα n·L(7.36)(7.37).

The unitary operator corresponding to rotation through a finite angle α is ahigh power of this operator. Since the exponent contains only one operator,n · L, which inevitably commutes with itself, the product of the exponentials
is simply
U (α) = e−iα·L ,(7.38)
where α ≡ αn.
The difference between the total and orbital angular momentum operators is now apparent. When we rotate the system on a turntable throughan angle α, the system’s ket is updated by e−iα·J . When we move the system around a circle without modifying its orientation, the ket is updatedby e−iα·L . The crucial insight is that the turntable both moves the system
around a circle and reorientates it. The transformations of which J is thegenerator reflects both of these actions. The transformations of which L is
the generator reflects only the translation.

7.2 Orbital angular momentum

147

Figure 7.3 J both swings the particle around the origin and rotates its spin (left), while
L moves the particle, but leaves the direction of the spin invariant (right).

7.2.2 Spectra of L2 and Lz
We have shown that the Li commute with one another in exactly the same
way that the Ji do. In §7.1 we found the possible eigenvalues of J 2 and Jz
from the commutation relations and nothing else. Hence we can without
further ado conclude that the possible eigenvalues of L2 and Lz are l(l + 1)
and m, respectively, with −l ≤ m ≤ l, where l is a member of the set(0, 21 , 1, 32 , . . .).
In the last subsection we saw that L is the generator of translations
on circles around the origin, and we demonstrated that when a complete
rotation through 2π is made, the unitary operator that L generates is simply
the identity. Consider the case in which we move the system right around thez axis when it is in the eigenstate |l, mi of L2 and Lz . The unitary operator
is then e−2πiLz and the transformed ket is|l, mi = e−2πiLz |l, mi = e−2mπi |l, mi.

(7.39)

Since the exponential on the right side is equal to unity only for integer m,
we conclude that Lz , unlike Jz has only integer eigenvalues. Since for given
l, m runs from −l to l, it follows that l also takes only integer values. Thus
the spectrum of L2 is l(l + 1) with l = 0, 1, 2, . . ., and for given l the possible
values of Lz are the integers in the range (−l, l).
7.2.3 Orbital angular momentum eigenfunctions
We already know the possible eigenvalues of the operators L2 and Lz . Now
we find the corresponding eigenfunctions.
In the position representation, the Li become differential operators. For
example1∂∂
Lz = (xpy − ypx ) = −i x(7.40)−yh̄∂y∂x
Let (r, θ, φ) be standard spherical polar coordinates. Then the chain rulestates that
∂
∂x ∂
∂y ∂
∂z ∂
=
+
+
.
(7.41)∂φ∂φ ∂x ∂φ ∂y ∂φ ∂z
Using x = r sin θ cos φ, y = r sin θ sin φ and z = r cos θ we find
 


∂
∂
∂
∂
∂
= x
= iLz .
= r sin θ − sin φ+ cos φ
−y
∂φ
∂x
∂y
∂y
∂x
That isLz = −i∂.
∂φ(7.42)(7.43)

Let |l, mi be a simultaneous eigenket of L2 and m for the eigenvalues l(l + 1)
and m, respectively. Then Lz |l, mi = m|l, mi and the wavefunction ψlm (x) ≡hx|l, mi must satisfy the eigenvalue equation−i∂ψlm= mψlm .
∂φ(7.44)

148

Chapter 7: Angular Momentum

The solution of this equation isψlm (r, θ, φ) = Klm (r, θ)eimφ ,(7.45)

where Klm is an arbitrary function of r and θ. Since m is an integer, ψlm isa single-valued function of position.
In our determination of the spectra of J 2 and Jz in §7.1, important roles
were played by the ladder operators J± = (Jx ± iJy ). If we define
L± ≡ Lx ± iLy ,(7.46)

then by analogy with equation (7.5) we will have that
L± |l, mi = α± |l, m ± 1i,
where
α± (m) =pl(l + 1) − m(m ± 1).

(7.47a)(7.47b)

It will be helpful to express L± in terms of partial derivatives with respect to spherical polar coordinates. We start by deriving a relation between
partial derivatives that we will subsequently require. From the chain rule wehave that

∂
∂
∂
∂ 
− r sin θ .
= r cos θ cos φ+ sin φ(7.48a)
∂θ
∂x
∂y
∂z

Multiplying the corresponding expression (7.42) for φ by cot θ yields


∂
∂
∂
= r cos θ − sin φ+ cos φ.
(7.48b)
cot θ
∂φ
∂x
∂y

Adding or subtracting i times (7.48b) to (7.48a) we obtain


∂
∂
∂
∂
∂
− r sin θ± i cot θ= r cos θ (cos φ ∓ i sin φ)+ (sin φ ± i cos φ)
∂θ
∂φ
∂x
∂y
∂z
 ∂

∂
∂
= r cos θe∓iφ±i− r sin θ .
∂x∂y∂z(7.49)
Multiplying through by e±iφ , we obtain the needed relation:


 ∂
∂
∂
∂
∂ 
±iφ
e
= r cos θ− r sin θ e±iφ .
± i cot θ±i(7.50)
∂θ
∂φ
∂x
∂y
∂z
With this expression in hand we set to work on L+ . In the position representation it is
 


∂
∂
∂
∂
+ z
L+ = −i y
−z
−x
∂z
∂y
∂x
∂z
 ∂
∂
∂ 
(7.51)− (x + iy)
+i
=z
∂x
∂y
∂z
 ∂
∂
∂ 
− r sin θ eiφ ,+i= r cos θ∂x∂y∂z
so with equation (7.50) we can write∂∂ 
L+ = eiφ.
+ i cot θ∂θ∂φSimilarly

 


∂
∂
∂
∂
− z
−z
−x
L− = −i y
∂z
∂y
∂x
∂z
∂
∂
∂ 
= −z
+ (x − iy)
−i
∂x
∂y
∂z



 ∂
∂
∂
− r sin θ e−iφ−i
= − r cos θ
∂x
∂y
∂z
∂

∂
= −e−iφ.
− i cot θ∂θ∂φ(7.52a)(7.52b)7.2 Orbital angular momentum

149

Table 7.1 The first six spherical harmonics

Y0m
Y1m
Y2m

q
q
q

m
±1

0
1
4π
6
8π cos θ∓

10232π (3 cos θ − 1)

∓

q

q

±2

3
±iφ
8π sin θe15±iφ32π sin 2θe

q

2
15
±2iφ
32π sin θe

The state |l, li with the largest permissible value of m for given l mustsatisfy the equation L+ |l, li = 0. Using equations (7.45) and (7.52a), in theposition representation this reads∂Kll− l cot θKll = 0.
∂θ(7.53)

This is Ra first-orderlinear differential equation. Its integrating factor is
exp −l dθ cot θ = sin−l θ, so its solution is Kll = R(r) sinl θ, where R is
an arbitrary function. Substituting this form of Kll into equation (7.45), we
conclude that
ψll (r, θ, φ) = R(r) sinl θ eilφ .
(7.54)
From equation (7.54) we can obtain the wavefunctions ψlm of states withsmaller values of m simply by applying the differential operator L− . For
example∂∂  l ilφψl(l−1) (r, θ, φ) = constant × R(r)e−iφ −sin θ e+ i cot θ∂θ∂φ(7.55)l−1
i(l−1)φ= constant × R(r) sin
θ cos θe.
Hence, the eigenfunctions of L2 and Lz for given l all have the same radial
dependence, R(r). The function of θ, φ that multiplies R in ψlm is conventionally denoted Ylm and called a spherical harmonic. The normalisationof Ylm is chosen such thatZd2 Ω |Ylm |2 = 1

with d2 Ω ≡ sin θ dθdφ(7.56)

the element of solid angle. We have shown thatYll ∝ sinl θeilφ

and Yll−1 ∝ sinl−1 θ cos θei(l−1)φ .

(7.57)The normalising constants can be determined by first evaluating the integralZd2 Ω sin2l θ = 4π 22l(l!)2(2l + 1)!

(7.58)

involved in the normalisation of Yll , and then dividing by the factor α− ofequation (7.47b) each time L− is applied.
The spherical harmonics Ylm for l ≤ 2 are listed in Table 7.1. Figures 7.4
and 7.5 show contour plots of several spherical harmonics. Since sphericalharmonics are functions on the unit sphere, the figures show a series ofballs with contours drawn on them. Since spherical harmonics are complex
functions we had to decide whether to show the real part, the imaginarypart, the modulus or the phase of the function. We decided it was most
instructive to plot contours on which the real part is constant; when the realpart is positive, the contour is full, and when it is negative, the contour isdotted.

150

Chapter 7: Angular Momentumm ) on the unit sphere for m = 15 (left), m = 7 (centre) and
Figure 7.4 Contours of ℜ(Y15m ) = 0 are the heavy curves, while contours
m = 2 (right). The contours on which ℜ(Y15m ) < 0 are dotted. Contours of the imaginary part of Y m would look theon which ℜ(Y15lsame except shifted in azimuth by half the distance between the heavy curves of constantazimuth.

Figure 7.5 Top row: contours of ℜ(Y1m ) for m = 1 (left) and 0 (right) with line styles
having the same meaning as in Figure 7.4. Contours of the imaginary part of Yl1 would
look the same as the left panel but with the circles centred on the y axis. Bottom row:
contours of ℜ(Y2m ) for m = 2 (left), m = 1 (centre) and m = 0 (right).

For large l, Yll is significantly non-zero only where sin θ ≃ 1, i.e., aroundthe equator, θ = π/2 – the leftmost panel of Figure 7.4 illustrates this case.
The first l applications of L− each introduce a term that contains one lesspower of sin θ and an extra power of cos θ. Consequently, as m diminishes
from l to zero, the region of the sphere in which Ylm is significantly non-zero
gradually spreads from the equator toward the poles – compare the leftmost
and rightmost panels of Figure 7.4. These facts make good sense physically:
Yll is the wavefunction of a particle that has essentially all its orbital angularmomentum parallel to the z axis, so the particle should not stray far from thexy plane. Hence Yll , the amplitude to find the particle at θ, should be small
for θ significantly different from π/2. As m diminishes the orbital plane is
becoming more inclined to the xy plane, so we are likely to find the particle
further and further from the plane. This is why Ylm increases away from theequator as m decreases.
For large l the phase of Yll changes rapidly with φ (leftmost panel of
Figure 7.4). This is to be expected, because the particle’s large orbital
angular momentum, lh̄, implies that the particle has a substantial tangential
motion within the xy plane. From classical physics we estimate its tangential
momentum at p = lh̄/r, and from quantum mechanics we know that this7.2 Orbital angular momentum

151

implies that the wavefunction must change its phase at a rate p/h̄ = l/rradians per unit distance. This estimate agrees precisely with the rate ofchange of phase with distance around the equator arising from the factor eilφin Yll . When m is significantly smaller than l (rightmost panel of Figure 7.4),
the rate of change of the wavefunction’s phase with increasing φ is smaller
because the particle’s tangential momentum is not all in the direction ofincreasing φ. Hence Ylm ∝ eimφ .
For any value of m, Lx and Ly both have zero expectation values, as
follows immediately from the relation Lx = 21 (L+ + L− ). So the orientationof the component of the angular momentum vector that lies in the xy plane
is completely uncertain. Because of this uncertainty, the modulus of Ylm
is independent of φ, so there is no trace of an inclined orbital plane whenm < l. An orbital plane becomes defined if there is some uncertainty in Lz ,with the result that there are non-zero amplitudes ψm = hl, m|ψi for several
values of m. In this case quantum interference between states of well-definedLz can generate a peak in |hx|ψi|2 along a great circle that is inclined to theequator.

7.2.4 Orbital angular momentum and parity
In §4.1.4 we defined the parity operator P , which turns a state with wavefunction ψ(x) into the state that has wavefunction ψ ′ (x) ≡ ψ(−x). We now
show that wavefunctions that are proportional to a spherical harmonic Ylm
are eigenfunctions of P with eigenvalue (−1)l .
In polar coordinates the transformation x → −x is effected by θ → π−θ,φ → φ+π. Under this mapping, sin θ = sin(π −θ) is unchanged, while eilφ →eilπ eilφ = (−1)l eilφ . By equation (7.57), Yll ∝ sinl θeilφ , so Yll → (−1)l Yll .
That is, Yll has even parity if l is an even number and odd parity otherwise.
In §4.1.4 we saw that x and p are odd-parity operators: P x = −xP .
From this and the fact that the orbital angular momentum operators Li
are sums of products of a component of x and a component of p, it follows
that both the Li and the ladder operators L± = Lx ± iLy are even-parity
operators. Now Ylm−1 = L− Ylm /α− , where α− is a constant, so applying
the parity operator
P Yll−1 =1
1
L− P Yll = (−1)lL− Yll = (−1)l Yll−1 .
α−α−(7.59)

That is, Yll−1 has the same parity as Yll . Since all the Ylm for a given l canbe obtained by repeated application of L− to Yll , it follows that they all havethe same parity, (−1)l .

7.2.5 Orbital angular momentum and kinetic energy
We now derive a very useful decomposition of the kinetic energy operator
HK ≡ p2 /2m into a sum of operators for the radial and tangential kinetic
energies. First we show that L2 is intimately related to the Laplacian operator ∇2 . From the definition (7.46) of the ladder operators for orbital angularmomentum, we have
L+ L− = (Lx + iLy )(Lx − iLy ) = L2x + L2y + i[Ly , Lx ]= L2x + L2y + Lz .

(7.60)

Hence with equations (7.52) we may write
L2 = L+ L− − Lz + L2z∂  −iφ 
∂ 
∂2
∂
∂
iφ ∂
=e
.
+ i cot θ+ i cot θ
−
e
+i
−
∂θ
∂φ
∂θ
∂φ
∂φ ∂φ2

152

Chapter 7: Angular Momentum

Differentiating out the right side
L2 = −

 ∂
∂
∂2
∂ 
∂
∂2
∂2
2
2
−
i
csc
θ
−
cot
θ
+
cot
θ
−
.
+
i
cot
θ
+
i
−
∂θ2
∂φ2
∂θ
∂φ
∂φ
∂φ ∂φ2

The first-order terms in ∂/∂φ cancel because cot2 θ − csc2 θ = −1. This
identity also enables us to combine the double derivatives in φ. Finally thesingle and double derivatives in θ can be combined so that the equation
becomes




1 ∂
∂
1 ∂2
2
L =−
sin θ
+
,
(7.61)sin θ ∂θ∂θ
sin2 θ ∂φ2which we recognise as −r2 times the angular part of the Laplacian operator∇2 .
Now we ask “what is the operator associated with radial momentum?”.
The obvious candidate is r̂·p, where r̂ is the unit vector in the radial direction.
Unfortunately this operator is not Hermitian:(r̂ · p)† = p · r̂ 6= r̂ · p,

(7.62)

so it is not an observable. This is a particular case of a general phenomenon:
the product AB of two non-commuting observables A and B is never Hermitian. But it is easy to see that 21 (AB + BA) is Hermitian. So we definepr ≡ 21 (r̂ · p + p · r̂),(7.63)

which is manifestly Hermitian. We will need an expression for pr in theposition representation. Replacing p by −ih̄∇ we havepr = −

ih̄
2




1
r · ∇ + ∇ · (r/r) .
r(7.64)

From the chain rule it is straightforward to show that
r

∂
∂
∂
∂
=x
+y
+z
= r·∇ .
∂r
∂x
∂y
∂z

(7.65)

Moreover, ∇ · r = 3, so equation (7.64) can be rewritten
ih̄
2



∂
∂
3
r
+ − 2 +
∂r
r
r
∂r


1
∂
.
+
= −ih̄
∂rr

pr = −

(7.66)

This expression enables us to find the commutator∂[r, pr ] = −ih̄ r,= ih̄.
∂r(7.67)

Squaring both sides of equation (7.66) yields
 2



∂
∂
∂
2 ∂
1
1
1
1
= −h̄2
+
+
+
+
−
∂r
r
∂r
r
∂r2
r ∂r r2
r2


∂
h̄2 ∂
r2
.
=− 2
r ∂r
∂r

p2r = −h̄2(7.68)We recognise this operator as −h̄2 times the radial part of the Laplacian operator ∇2 . Since we have shown that L2 is −r2 times the angular part of the7.2 Orbital angular momentum

153

Table 7.2 The first five Legendre polynomialsl
0 1
Pl (µ)

1 µ

2

3

4

1
1
1
2
3
4
2
2 (3µ − 1) 2 (5µ − 3µ) 8 (35µ − 30µ + 3)

Laplacian (eq. 7.61), it follows that ∇2 = −(p2r /h̄2 + L2 /r2 ). Consequently,
the kinetic-energy operator HK = p2 /2m = −(h̄2 /2m)∇2 can be written
1
HK =
2m



h̄2 L22
pr + 2.
r(7.69)The physical interpretation of this equation is clear: classically, the orbital
angular momentum h̄L is mr × v = mrvt , where vt is the tangential speed,so the term h̄2 L2 /2mr2 = 21 mvt2 is the kinetic energy associated with thetangential motion. On the other hand p2r /2m = 21 mvr2 , so this term represents the kinetic energy due to radial motion, as we would expect. For futurereference we note that the kinetic-energy operator can be also writtenh̄2 1 ∂
L22 ∂
HK = −r− 2 .
2m r2 ∂r∂r
r

(7.70)

7.2.6 Legendre polynomials
The spherical harmonic Yl0 is special in that it is a function of θ only. We
now show that it is, in fact, a polynomial in cos θ. In the interval 0 ≤ θ ≤π of interest, θ is a monotone function of µ ≡ cos θ, so without any lossof generality we may take Yl0 to be proportional to a function Pl (µ). On
this understanding, Pl is an eigenfunction of L2 with eigenvalue l(l + 1).
Transforming the independent variable from θ to µ in our expression (7.61)
for L2 , we find that Pl must satisfy Legendre’s equation:
d
dµ



2 dPl
(1 − µ )+ l(l + 1)Pl = 0.
dµ(7.71)

We look for
Ppolynomial solutions of this equation. Putting in the trial solution Pl = n bn µn , we findXnbn n(n − 1)µn−2 − n(n + 1)µn + l(l + 1)µn = 0.

(7.72)

This equation must be valid for any value of µ in the interval (−1, 1), which
will be possible only if the coefficient of each and every power of µ individuallyvanishes. The coefficient of µk is
0 = bk+2 (k + 2)(k + 1) − bk {k(k + 1) − l(l + 1)} .

(7.73)For k = 0 the expression connects b2 to b0 , while for k = 2 it relates b4 to b2 ,and so on. Thus from this equation we can express bn as a multiple of b0 for
even n, and as a multiple of b1 for odd n. Moreover, if l is an even number,
we know from our discussion of parity that Pl must be an even functionof µ, so in this case bn must vanish for n odd. Finally, bn will vanish forn even and greater than l on account of the vanishing of the curly bracket
in equation (7.73) when k = l. This completes the proof that for even l,
Pl (µ) is a polynomial or order l. An extremely similar argument shows that
Pl (µ) is also a polynomial of order l when l is odd. The first five Legendre
polynomials are listed in Table 7.2.

154

Chapter 7: Angular Momentum

The conventional normalisation of the Legendre polynomial Pl is therequirement that Pl (1) = 1. With this property, the Pl are not orthonormal.
In fact
Z 1
2dµ Pl (µ)Pl′ (µ) =(7.74)
δll′ .
2l + 1−1
From this result it easily follows that the proportionality constant between
Pl (cos θ) and the orthonormal functions Yl0 (θ) is such that
Yl0 (θ) =r

2l + 1Pl (cos θ).
4π(7.75)

7.3 Three-dimensional harmonic oscillator
In this section we discuss the dynamics of a particle that moves in three
dimensions subject to a central force that is proportional to the particle’s
distance from the origin. So the Hamiltonian isH=

p2+ 1 mω 2 r2 .
2m 2

(7.76)

If we use Cartesian coordinates, this Hamiltonian becomes the sum of threecopies of the Hamiltonian of the one-dimensional harmonic oscillator thatwas the subject of §3.1:
H = H x + Hy + Hz ,(7.77)

where, for example, Hx = (p2x /2m)+ 21 mω 2 x2 . These one-dimensional Hamiltonians commute with one another. So there is a complete set of mutual
eigenkets. Let |nx , ny , nz i be the state that is an eigenket of Hx with eigenvalue (nx + 12 )h̄ω eq. 3.12, etc. Then |nx , ny , nz i will be an eigenket of thethree-dimensional Hamiltonian (7.76) with eigenvalue
E = (nx + ny + nz + 23 )h̄ω.

(7.78)

Moreover, in the position representation the wavefunction of this state is justa product of three of the wavefunctions we derived for stationary states of aone-dimensional oscillator
ψ(x) = unx (x)uny (y)unz (z).

(7.79)

In view of these considerations it might be thought that there is nothing
we do not know about the Hamiltonian (7.76). However, it is instructive
to reanalyse the system from a more physical point of view, that recognisesthat the system is spherically symmetric. We have seen that [Li , p2 ] = 0,
and [Li , r2 ] = 0, so [Li , H] = 0 and [L2 , H] = 0. From this result it follows
that there is a complete set of mutual eigenstates of H, L2 and Lz . Veryfew of the eigenstates obtained from the one-dimensional Hamiltonians are
eigenstates of either L2 or Lz . We now show how the eigenvalue problem
associated with (7.76) can be solved in a way that yields mutual eigenkets ofH, L2 and Lz . This exercise is instructive in itself, and some technology that
we will develop along the way will prove extremely useful when we analyse
the hydrogen atom in Chapter 8.
We use equation (7.69) to eliminate p2 from equation (7.76)
H=

p2r
h̄2 L2+ 12 mω 2 r2 .
+2m 2mr2(7.80)

7.3 Three-dimensional harmonic oscillator

155

Figure 7.6 The effective potential(7.82) for (from bottom to top)l = 0, . . . , 5. The beads mark theclassical turning points at the valuesof energy and angular momentum
that quantum mechanics allows.

We can assume that our energy eigenstates are also eigenstates of L2 , so inthis Hamiltonian we can replace L2 by an eigenvalue l(l + 1). Hence we wish
to find the eigenvalues of the radial Hamiltonian
Hl =

p2r
l(l + 1)h̄2+ 21 mω 2 r2 .
+
2m
2mr2

(7.81)

This is the Hamiltonian for a particle that moves in a one-dimensional effective potential
Veff (r) ≡l(l + 1)h̄2+ 21 mω 2 r2 .
2mr2

(7.82)

Hl governs the oscillations of the mass about the minima of this potential,
which is plotted in Figure 7.6. The eigenkets |E, l, mi of H are products ofthe eigenkets |E, li of Hl and the eigenkets |l, mi of L2 and Lz :|E, l, mi = |E, li|l, mi.

(7.83)

Our determination of the allowed energies of a one-dimensional harmonic oscillator exploited the dimensionless operators A and A† , whichrather nearly factorise H/h̄ω. So here we define the operator
1
Al ≡ √2mh̄ω(l + 1)h̄
ipr −+ mωr .
r(7.84)The product of A and its Hermitian adjoint A†l is(l + 1)h̄(l + 1)h̄−ipr −
+ mωr
ipr −
+ mωr
r
r
(

)
2

(l
+
1)h̄
(l
+
1)h̄
1
p2r + −+ mωr + i −+ mωr, pr
=
2mh̄ω
r
r


1
l(l + 1)h̄2
2
2 2 2
=
pr +
+ m ω r − (2l + 3)h̄mω .
2mh̄ω
r2(7.85)
Comparing the right side with equation (7.81), we see thatA†l Al =

1
2mh̄ω

o
n
Hl = h̄ω A†l Al + (l + 23 ) ,(7.86)

which bears a strong similarity to equation (3.3) for the one-dimensional
harmonic oscillator.

156

Chapter 7: Angular Momentum
The commutator of Al and A†l is


 

(l + 1)h̄(l + 1)h̄1
ipr −+ mωr , −ipr −
+ mωr
2mh̄ω
r
r
 

i
(l + 1)h̄=pr , −+ mωr
mh̄ωr(l + 1)h̄
+ 1,
=
mωr2
(7.87)
where we have used equation (7.67) to reach the last line. This result can be
written more usefully in the form
[Al , A†l ] =[Al , A†l ] =Hl+1 − Hl+ 1.
h̄ω(7.88)

From this expression and equation (7.86) we can easily calculate the commutator of Hl with Al :[Al , Hl ] = h̄ω[Al , A†l Al ] = h̄ω[Al , A†l ]Al = (Hl+1 − Hl + h̄ω)Al .

(7.89)

Now let |E, li be an eigenket of Hl with eigenvalue E:
Hl |E, li = E|E, li.

(7.90)

We multiply both sides of the equation by Al and use equation (7.89) toreverse the order of Al and Hl :EAl |E, li = Al Hl |E, li = (Hl Al + [Al , Hl ])|E, li= (Hl+1 + h̄ω)Al |E, li.

(7.91)On rearrangement this yields
Hl+1 (Al |E, li) = (E − h̄ω)(Al |E, li),

(7.92)

which says that Al |E, li is an eigenket of Hl+1 for the eigenvalue E − h̄ω, so
Al |E, li = α− |E − h̄ω, l + 1i,(7.93)

where α− is a normalising constant.
Al creates the radial wavefunction for a state that has more orbital
angular momentum and less energy than the state with which it started.
That is, Al diminishes the radial kinetic energy by some amount and adds asmaller quantity of energy to the tangential motion. If we repeat this process
a sufficient number of times, by following Al with Al+1 and Al+1 with Al+2 ,and so on, there will come a point at which no radial kinetic energy remains– we will have reached the quantum equivalent of a circular orbit. The next
application of Al must annihilate the wavefunction. Hence AL |E, Li = 0,
where L(E) is the largest allowed value of l for energy E. If we operate on|E, Li with HL , we find with equation (7.86) that
E|E, Li = HL |E, Li = h̄ω(L + 23 )|E, Li,so(7.94)

E
− 3.
h̄ω 2
Since L is a non-negative integer, it follows that the ground-state energy is3
2 h̄ω and that the ground state has no angular momentum. In general E/h̄ωis any integer plus 32 . These values of the allowed energies agree perfectlywith what we could have deduced by treating H as a sum of three onedimensional harmonic-oscillator Hamiltonians.
E = (L + 23 )h̄ωand L(E) =7.3 Three-dimensional harmonic oscillator

157

Figure 7.7 Radial probability distributions of circular orbits in the three-dimensionalpharmonic oscillator potential for L = 1 and L = 8. The scale radius ℓ = h̄/2mω.

We shall define a circular orbit to be one that has the maximum angular momentum possible at its energy. We obtain the radial wavefunctionsof these by writing the equation AL |E, Li = 0 in the position representation.
With equations (7.84) and (7.66) this equation becomes∂
1 L + 1 mω+ −+r uL (r) = 0.
∂r r
r
h̄

(7.95)

This is a first-order linear differential equation. Its integrating factor is
exp

Z



 mω 
L mω
r= r−L expr2 ,dr − +rh̄
2h̄

(7.96)

so the solution of equation (7.95) is2

2

uL (r) = constant × rL e−r /4ℓ ,

where

ℓ≡

p
h̄/2mω.

(7.97)

Notice that the exponential factor is simply the product of three exponential
factors from equation (3.15), one in x, one in y and one in z. The wavefunction varies with r, so a circular orbit does have some radial kinetic energy.
In the limit of large L in which classical mechanics applies, the radial kinetic
energy is negligible compared to the tangential kinetic energy, and we neglect
it. But it never really vanishes.
Equation (7.97) gives the radial wavefunction for a circular
R orbit. Themcomplete wavefunction is ψ(x) = uL (r)YL(θ, φ), and since d2 Ω Ylm = 1,2
2
the radial probability density is P (r) = r2 u2L ∝ r2L+2 e−r /2ℓ , where thefactor r2 arises from the expression for the volume element d3 x in sphericalpolar coordinates.
This density is plotted in Figure√√ 7.7 for L = 1 and L = 8.
For r/ℓ ≪ 2L + 2, P rises as r2L+2 . For r/ℓ > 2L + 2 it falls rapidly as
the Gaussian factor takes over. Figure 7.7 shows that the uncertainty in r is
∼ ℓ, which is a small fraction of r when L is not small.
We may obtain the radial wavefunctions of more eccentric orbits byshowing that A†l is a raising operator. Equation (7.89) yields
Al Hl = Hl+1 Al + h̄ωAl(7.98)

Daggering both sides we have
Hl A†l = A†l Hl+1 + h̄ωA†l .

(7.99)

158

Chapter 7: Angular Momentum

Figure 7.8 The (E, l) plane and theaction of Al and A†l−1 .

We now multiply both sides of Hl+1 |E, l + 1i = E|E, l + 1i by A†l :EA†l |E, l + 1i = A†l Hl+1 |E, l + 1i = (Hl − h̄ω)A†l |E, l + 1i.

(7.100)Rearranging
Hl (A†l |E, l + 1i) = (E + h̄ω)(A†l |E, l + 1i).

(7.101)Thus, we have shown thatA†l |E, l + 1i = α+ |E + h̄ω, li,(7.102)where α+ is a normalising constant. By writing A†l in the position representation, we can generate the wavefunctions of all non-circular orbits by
repeatedly applying A†l to the current wavefunction, starting with that ofa circular orbit. We start with the product of rL and a Gaussian factor[equation (7.97)]. From this A†L−1 generates terms proportional to rL+1 andrL−1 times the Gaussian (Problem 7.25). From these two terms A†L−2 thengenerates three terms, rL+2 , rL and rL−2 times the Gaussian, and so on.
Consequently the number of radial nodes – radii at which the wavefunction
vanishes – increases by one with each application of A†l , and the wavefunction oscillates more and more rapidly in radius as A†l invests a larger andlarger fraction of the particle’s kinetic energy in radial motion.
Figure 7.8 helps to organise the generation of radial wavefunctions. Eachdot represents a radial wavefunction. From the dot at (E, l), operating withAl carries one to the next dot up and to the left, while operating with A†l−1
carries one to the next dot down and to the right. At half the energies
only even values of l occur, and only odd values of l occur at the otherhalf of the energies. In Problem 7.22 you can show that, when one bears inmind that each dot gives rise to 2l + 1 complete wavefunctions, the numberof wavefunction with energy E that we obtain in this way agrees with thenumber that we would obtain using wavefunctions of the one-dimensional
harmonic oscillator via equation (7.79).

7.4 Spin angular momentum
In §7.2.1 we saw that the difference between J and L is that J is the generator for complete rotations of the system, while L is the generator for displacements of the system around circles, while leaving its orientation fixed(Figure 7.3). Consequently the difference
S ≡ J−L(7.103)is the generator for changes of orientation that are not accompanied by any
motion of the system as a whole. Since J and L are vector operators, S is
also a vector operator. Its components are called the spin operators.

7.4 Spin159

We saw in §7.2 that L has exactly the same commutation relations as
J with any function of the position and momentum operators only. From
this fact and the definition (7.103), it follows that S commutes with all suchfunctions. In particular [S, x] = [S, p] = [S, L] = 0. This essentially tells usthat S has nothing to do with a system’s location, nor the way in which it
may or may not be moving. S is associated with intrinsic properties of thesystem.
The components Si of the spin operator inherit the usual angular momentum commutation rules from Ji and Li :[Si , Sj ] = [Ji − Li , Jj − Lj ]= [Ji , Jj ] − [Li , Jj ] − [Ji , Lj ] + [Li , Lj ]X=iǫijk (Jk − Lk − Lk + Lk )(7.104)k=iXǫijk Sk .

k

We define S 2 ≡ S · S and then equation (7.104) ensures that
[S, S 2 ] = 0.

(7.105)

Because the Si have exactly the same form of commutation relations as theJi , we know that the possible eigenvalues of S 2 are the numbers 0, 12 , 1, 23 , . . .
and that for given s the eigenvalues m of the Si move in integer steps from −s
to s. Can s take half-integer values? This question is answered affirmatively
by equation (7.103); since [Jz , Lz ] = 0 we can find a complete set of statesthat simultaneously have well-defined values of both Jz and Lz . In general,
the Jz eigenvalue could be either an integer or half-integer, whereas the Lz
eigenvalue must be an integer. The difference Sz = Jz − Lz must then beeither an integer or half-integer.
In the rest of this book we will make extensive use of commutation
relations involving angular momentum operators. In Table 7.3 these havebeen gathered for later reference.

7.4.1 Spin and orientation
We have several times stated without proof that the orientation of the systemis encoded in the amplitudes ajm for the system to be found in states of welldefined angular momentum, |j, mi. We now begin to justify this claim. For
simplicity we consider spin angular momentum because we want to focus on
the orientation of our system without concerning ourselves with its location.
However, what we refer to as ‘spin’ is the total intrinsic angular momentumof the system. If the latter is a hydrogen atom, for example, it may contain acontribution from the orbital angular momentum of the electron in addition
to the contributions from the intrinsic spins of the electron and the proton.
Since the Si are Hermitian operators, any state |ψi may be expanded
in terms of the complete set of eigenstates |s, mi of, say, Sz and S 2 . We
have seen that these states are labelled by an integer or half integer s, with−s ≤ m ≤ s, so the complete expansion is
|ψi =

X

s
X

m=−s
1
s=0, 2 ,1,...

hs, m|ψi|s, mi.

(7.106)

Fortunately, systems for which quantum mechanical effects are significant
rarely have more than a handful of non-zero amplitudes hs, m|ψi in the sumof equation (7.106). In the simplest case we have an object with s = 0, a
spin-zero object, such a pion. The sum in equation (7.106) contains only

160

Chapter 7: Angular Momentum

Table 7.3 Commutators involving angular momentum[Ji , vj ] = i[Ji , s] = 0

P[Li , wj ] = ik ǫijk vk

where v is any vector or pseudovector
where s is any scalar or pseudoscalarPk ǫijk wk

where w is any vector or pseudovector functionof only x, p and constant scalars and pseudoscalars[Li , f ] = 0

where f is any scalar or pseudoscalar functionof only x, p and constant scalars and pseudoscalars[Si , w] = 0

where w is any function of spatial operators

The following are important special cases of the above results[Ji , Jj ] = i

P[Li , Lj ] = i[Ji , J 2 ] = 0

k ǫijk Jk

Pk ǫijk Lk[Ji , Lj ] = i[Si , Sj ] = iPPk ǫijk Lk[Ji , Sj ] = ik ǫijk Sk[Li , Sj ] = 0

P[Ji , L2 ] = 0[Ji , S 2 ] = 0[Li , L2 ] = 0[Li , S 2 ] = 0[Si , L2 ] = 0[Si , S 2 ] = 0[L2 , J 2 ] = 0[S 2 , J 2 ] = 0

k ǫijk Sk

Since J = L + S and therefore J 2 = L2 + S 2 + 2L · S we also have
[Li , J 2 ] = 2iPjk ǫijk Sj Lk[Si , J 2 ] = 2iPjk ǫijk Lj Sk

one eigenstate with s = 0, the state |0, 0i because an object with no spincannot have any spin angular momentum around the z axis.
When we rotate an object about the direction α without translating
it, its state is updated by the operator U (α) = e−iα·S (cf. equation 4.13).
When we apply this operator to a state of a spin-zero object, the state
emerges unchanged:
U (α)|0, 0i = exp (−iα · S) |0, 0i = |0, 0i.

(7.107)

Hence, a spin-zero object, like a perfect sphere, is completely unchanged byan arbitrary rotation. In view of their invariance under rotations, spin zeroparticles are sometimes known as scalar particles.4
Some very important systems require just the two terms in equation(7.106) that are associated with s = 21 . These systems are called spin-half
objects. Electrons, quarks, protons and neutrons fall into this category. For
example, by equation (7.106) the state of an electron can be written1|e− i =

2
X

1
m=− 2

h 12 , m|e− i| 21 , mi.

(7.108)4 Since both J and S commute with the parity operator P , behaviour under rotations
does not tell us about behaviour under reflection, so spin zero particles could also be
pseudoscalars. In fact, pions are pseudoscalar particles.

7.4 Spin

161

Because there are only two terms in this expansion, the quantum uncertainty
in the orientation of a spin-half system is very great. We shall see that themost precise information we can have is that the end of the system’s angularmomentum vector lies in a given hemisphere – for example, we could statethat it lies within the northern rather than the southern hemisphere, or thewestern rather than the eastern hemisphere. Where it lies in the hemisphereis shrouded in quantum uncertainty.
Another important class of systems contains those that have total spinquantum number s = 1. These systems are called spin-one objects. The W
and Z bosons fall in this class. For a spin-one system, the expansion (7.106)
reduces to just (2s + 1) = 3 terms. For example, the state of a Z boson canbe written
1
X
|Zi =
h1, m|Zi|1, mi.
(7.109)m=−1

We will see that we can constrain the end of the angular-momentum vectorof a spin-one system to lie within a chosen polar cap, or in the equatorial
band that lies between opposite polar caps.
The larger a system’s spin s, the more precisely we can constrain theend of its angular momentum vector. It is rather as if systems were subject
to random torques of a certain magnitude, and the faster it is spinning,the more stable its orientation can be in the face of the random torques.
The same physical principle underlies the use of rifling in guns to stabilise
the orientation of the projectile by imparting angular momentum to it as
it flies down the barrel. A few concrete examples will clarify the physicalinterpretation of the quantum states |s, mi.
7.4.2 Spin-half systems
As in equation (7.108), the state of any spin-half system may be expanded
in terms of just two Sz eigenstates | 12 , + 12 i and | 21 , − 12 i which we will call|+i and |−i respectively. Equation (7.108) then reads |ψi = a|+i + b|−i. In
this basis we can write the operators as (cf. equation 2.16)5h+|Sx |+i h+|Sx |−ih+|Sy |+i h+|Sy |−iSx =; Sy =
h−|Sx |+i h−|Sx |−ih−|Sy |+i h−|Sy |−ih+|Sz |+i h+|Sz |−iSz =.
(7.110)h−|Sz |+i h−|Sz |−i

The elements of the matrix Sz are trivially evaluated because |±i are theeigenkets of Sz with eigenvalues ± 21 . To evaluate the other two matrices we
1
notice that Sx = 21 (S+ + S− ), and Sy = 2i
(S+ − S− ), then use the relations
S+ |−i = |+i and S− |+i = |−i which follow from equations (7.5) and (7.7)
for the spin operator. The result of these operations is






0 1
0 −i
1 0
; Sy = 12; Sz = 12.
(7.111)
Sx = 21
1 0
i 0
0 −1
The matrices appearing here are the Pauli matrices,






0 1
0 −i
1 0
σx ≡
; σy ≡; σz ≡,(7.112)
1 0
i 0
0 −1

so we can write S = 12 σ. It is straightforward to verify that the square of
any Pauli matrix is the identity matrix:
σi2 = I.
(7.113)This result implies that for any state Sx2 = Sy2 = Sz2 = 14 , which isconsistent with the fact that the measurement of any component of S canproduce only ± 12 .
The Stern–Gerlach experiment In 1922, Stern and Gerlach6 conducted
5 Here we are again slightly abusing the notation; S are taken to be both the spinioperators and their matrix representations.
6 Gerlach, W. & Stern, O., 1922, Zeit. f. Physik, 9, 349

162

Chapter 7: Angular Momentum

Figure 7.9 Schematic of a Stern–
Gerlach filter. The atomic beam enters from the left. Between the polepieces the magnetic field increases
in intensity upwards, so atoms thathave their spins aligned with B are
deflected upwards and the otheratoms are deflected downwards.

Figure 7.10 Beam split by an SG
filter and then up beam hits a second filter.

some experiments with silver atoms that most beautifully illustrate the degree to which one can know the orientation of a spin-half object. In addition to this interest, these experiments provide clear examples of the standard procedure for extracting experimental predictions from the formalismof quantum mechanics.
A silver atom is a spin-half object and has a magnetic dipole momentµ. which can be used to track the atom’s orientation. In a magnetic field
B, a magnetic dipole experiences a force ∇(µ · B). Consequently, in a fieldthat varies in strength with position, a dipole that is oriented parallel to B
is drawn in to the region of enhanced |B|, whereas one that is antiparallelto B is repelled from this region. Stern and Gerlach exploited this effect toconstruct filters along the lines sketched in Figure 7.9. A powerful magnet
has one pole sharpened to a knife edge while the other forms either a flat
surface (as shown) or is slightly concave. With this geometry the magnetic
field lines are close packed as they stream out of the knife edge, and thenfan out as they approach the flat pole-piece. Consequently the intensity ofthe magnetic field increases towards the knife edge and the Stern–Gerlachfilter sorts particles according to the orientation of their magnetic momentswith respect to B.
The experiments all start with a beam of sliver atoms moving in vacuo,
which is produced by allowing vapourised silver to escape from an oventhrough a suitable arrangement of holes – see Figure 7.10. When the beam
passes into a filter, F1, it splits into just two beams of equal intensity. We
explain this phenomenon by arguing that the operator µi associated with
the ith component of an atom’s magnetic moment is proportional to Si :µi = gSi . Hence the filter has ‘measured’ n · S, where n is the unit vector in
the direction of B; we are at liberty to orient our coordinate system so thatn = ez , and n · S = Sz . We know that for a spin-half system, a measurementof Sz can yield only ± 21 , so the splitting of the beam into two is explained.
Given that there was nothing in the apparatus for producing the beam thatfavoured up over down as a direction for µ, it is to be expected that half of
the atoms return + 21 and half − 21 , so the two sub-beams have equal intensity.
We block the sub-beam associated with Sz = − 21 so that only particles with
Sz = 21 emerge from the filter.
We now place a second Stern-Gerlach filter, F2, in the path of the |+i
sub-beam, as shown in Figure 7.10, and investigate the effect of rotating thefilter’s magnetic axis n in the plane perpendicular to the incoming beam’s
direction. Let this be the yz plane. The incoming particles are definitely inthe state7 |+, zi because they’ve just reported + 12 on a measurement of Sz .
7 We relabel |+i → |+, zi to make clear that this is a state with spin up along the7.4 Spin

163

F2 measures n · S, where n = (0, sin θ, cos θ) with θ the direction betweenn and the z-axis. If |+, θi is the eigenket of n · S with eigenvalue + 21 ,the amplitude that the measurement yields + 21 is h+, θ|+, zi. The defining
equation of |+, θi is 12 n·σ|+, θi = 21 |+, θi or, using the matrix representation(7.111)   cos θ −i sin θaa=,(7.114)i sin θ − cos θbbwhere a ≡ h+, z|+, θi and b ≡ h−, z|+, θi. We have to solve this equation
subject to the normalisation condition |a|2 + |b|2 = 1. From the first row of
the matrix we deduce thatb
1 − cos θ=i.
(7.115)asin θ

From the trigonometric double-angle formulae we have 1 − cos θ = 2 sin2 21 θand sin θ = 2 sin 12 θ cos 21 θ, so

The choices

sin 21 θb=iacos 21 θ.

(7.116)  acos 12 θ=bi sin 21 θ

(7.117)

P2 = |a|2 = cos2 12 θ.

(7.118)satisfy both equation (7.116) and the normalisation condition. The amplitude that a particle with spin up along the z-axis also has spin up along then-axis is a∗ = h+, θ|+, zi, so the probability that an atom will pass F2 isThus, as θ is increased from 0 to π, the fraction of atoms that get throughF2 declines from unity to zero, becoming 21 when θ = π/2 and the magnetic
axes of F1 and F2 are at right angles. Physically it would be surprising if
the fraction that passed F2 when θ = π/2 were not a half since, when themagnetic moments of incoming atoms are perpendicular to the magnetic axisof a filter, there is nothing in the geometry of the experiment to favour theoutgoing particles being parallel to the magnetic axis, rather than antiparallel. When θ = π the magnetic axes of the filters are antiparallel and it isobvious that every atom passed by F1 must be blocked by F2. This agrees
with what we found out about a spin-half object’s orientation in the previous
section; if it is pointing somewhere in the upper z hemisphere, then there issome chance it is also pointing in any other hemisphere apart from the −zone.
We now place a third filter, F3, in the atomic beam that emerges from
F2. Let φ denote the angle between the magnetic axis of this filter andthe z-axis. The atoms that emerge from F2 are in the state |+, θi because
they’ve just returned 12 on a measurement of n · S, so the amplitude thatthese atoms get through F3 is h+, φ|+, θi. The amplitudes a′ ≡ h+, z|+, φiand b′ ≡ h−, z|+, φi can be obtained directly from the formula we alreadyhave for (a, b) with φ substituted for θ. Hence ′ 
acos 21 φ.
(7.119)=i sin 12 φb′and the amplitude to pass F3 isXh+, φ|+, θi =h+, φ|s, zihs, z|+, θis=±=cos 12 φ,= cos 12 (φ − θ).
z-axis.

−i sin 21 φcos 21 θi sin 21 θ



(7.120)164

Chapter 7: Angular Momentum

Thus the amplitude to pass F3 depends only on the angle φ − θ between
the magnetic axes of the filters, and the probability of passing F3 could
have been obtained simply by substituting this angle into equation (7.118).
This conclusion is obvious physically, but it is satisfying to see it emerge
automatically from the formalism.
An especially interesting case is when θ = π/2 and φ = π. In the absenceof F2, F3 would now block every atom that passed F1. But with F2 present
both F2 and F3 allow through half of the atoms that reach them, so a quarterof the atoms that leave F1 with Sz = + 21 pass both filters. These atoms exitfrom F3 with Sz = − 21 . Introducing F2 changes the fraction of atoms thatpass F3 because the measurement that F2 makes changes the states of theatoms. This is a recurring theme in quantum mechanics. No measurement
can be made without slightly disturbing the system that is being measured,and if the system is small enough, the disturbance caused by a measurement
can significantly affect the system’s dynamics.
7.4.3 Spin-one systems
In the case that s = 1, three values of m are possible, −1, 0, 1, and so theSi may be represented by 3 × 3 matrices. The calculation of these matrices
proceeds exactly as for spin half, the main difference being that (7.5) and(7.7) now yield√√
S+ | − 1i = 2|0i ; S+ |0i = 2|1i ;√√(7.121)S− |1i = 2|0i ; S− |0i = 2| − 1i .
The result is
01 
Sx = √
120


1 0
0 1 ;1 01Sz =  00
0 −i 01 
i 0 −i 
Sy = √
2
0 i
0

0 0
0 0 .
0 −1(7.122)

Consider the effect of using Stern-Gerlach filters on a beam of spinone atoms. In the experiment depicted in Figure 7.10 each filter now splits
the incoming beam into three sub-beams, and we block all but the beam
associated with m = +1 along the magnetic axis. One third of the atoms
that emerge from the collimating slits get through the first filter F1 because
each value of m is equally probable at this stage.8 To calculate the fractionof atoms which then pass through F2, the magnetic axis of which is inclined
at angle θ to that of F1, we must calculate the amplitude h1, θ|1, zi. The
defining equation of |1, θi is n · S|1, θi = |1, θi, which with equations (7.122)
can be written   cos θ− √i2 sin θ0aa √i sin θ0− √i2 sin θ (7.123) 2
 b  = b ,cc√i sin θ− cos θ02

where a ≡ h1, z|1, θi, b ≡ h0, z|1, θi, and c ≡ h−1, z|1, θi. The first and third
equations, respectively, yieldi(cos θ − 1)a = √ sin θ b
2

and

i
√ sin θ b = (1 + cos θ)c.
2

(7.124)8 The atoms emerge from the slits in an impure state (§6.3) and we set the probabilities
for each value of m to 13 in order to maximise the Shannon entropy of that state (§6.3.2).
The equality of the probabilities is an instance of the general principle that every quantum
state has equal a priori probability.

7.4 Spin

165

Eliminating a and c in favour of b yields


sin θ
√i
 
 i

 2 cos θ − 1 − √2 cot(θ/2)
a = b
b = b
.
11√i tan(θ/2)


c
sin θ
2
√i

(7.125)2 cos θ + 1

Thecondition |a|2 + |b|2 + |c|2 = 1 now implies that b =√ normalisation
1
1
2 sin( 2 θ) cos( 2 θ). The coefficient that we need is therefore
h1, θ|1, zi = a = i cos2 ( 12 θ) = 2i (1 + cos θ).

(7.126)

Hence the probability that an atom passes F2 after passing F1 falls from unity
when θ = 0 to zero when θ = π as we would expect. When θ = π/2 theprobability is P3 = 14 , which is substantially smaller than the corresponding
probability of 12 found in (7.118) for the case of spin-half atoms.
From a classical point of view it is surprising that after F1 has selected
atoms that have their angular momentum oriented parallel to the z-axis(in the sense that Sz takes the largest allowed value) there is a non-zeroprobability P3 that the angular momentum is subsequently found to be, inthe same sense, aligned with the y axis. The explanation of this phenomenon
is that for this system, the value of S 2 is s(s+1) = 2 which is twice the largest
allowed value of Sz . Hence, even in the state |1, zi a significant componentof the angular momentum lies in the xy plane. P3 is the probability that
this component is found to be parallel to the y axis. Once the measurementof Sy has been made by F3, the atom is no longer in the state |1, zi and weare no longer certain to obtain 1 if we remeasure Sz .

7.4.4 The classical limit
An electric motor that is, say, 1 cm in diameter and weighs about 10 gm
might spin at 100 revolutions per second. Its angular momentum would thenbe ∼ 10−3 kg m2 s−1 , which is ∼ 1031 h̄. Thus classical physics works withextremely large values of the integers s, m. It is interesting to understandhow familiar phenomena emerge from the quantum formalism when s is nolonger small.
For any value of s we can construct matrices that represent the angularmomentum operators. The matrix for Sz is diagonal with the eigenvaluess, (s−1), . . . , −s down the diagonal. The matrices for Sx and Sy are evaluatedin the usual way from S+ and S− and so are zero apart from strips one
place above and below the diagonal. Using the relations (7.15) between thecoefficients α± (m) of the raising and lowering operators S± we then find0α(s − 1)0...
 α(s − 1)
0
α(s
−
2)


..

.

0
α(s − 2)
0

1
..
.
.
..
..
Sx = 
.
2

.
.
.

..
..
..



0
0
...
0
0
...
...
i
1h
α(m)δm,n−1 + α(m − 1)δm,n+1
=
2


...
...

..

0
0
.
..

.

0
α(s − 2)0α(s − 2)0α(s − 1)

0
0
.
..
..
.















0

α(s − 1) 
0(7.127a)

166

Chapter 7: Angular Momentum

Figure 7.11 The points show the absolute values of the amplitudes am ≡ hs, m, z|s, s, θifor s = 40 and, from left to right, θ = 120◦ , 80◦ , 30◦ . For each value of θ, the vertical line
shows the value of cos θ.
0α(s − 1)0...
 −α(s − 1)
0
α(s
−
2)


..

.

0
−α(s − 2)
0

1 
..
.
.
..
..
Sy =

.
2i 

.
.
.

..
..
..



0
0
...
0
0
...
...
i
1h
=
− α(m)δm,n−1 + α(m − 1)δm,n+1
2i

...
...





s

0

...

0
..
.

0 s −1

 ..
..
.
.

Sz = 
m
0
..
 ..
.
.

..
0
.
0
0
0
... 0
= m δmn ,

...

..

..

.

0
−α(s − 2)0α(s − 2)0−α(s − 1)

0
0
.
..
..
.















0

α(s − 1) 
0(7.127b)



0

0

0
..
.

0 

.. 
. 

0 

.. 
. 

0 
−s

.

...

0
0
.
..

1−s
0

(7.127c)where the α(m) are what were called α+ (m) in (7.15), and the rows andcolumns of the matrix are labelled from +s at the top left to −s at thebottom right. In the same way as for spins s = 21 and s = 1, it is straightforward (for a computer) to determine the amplitudes am ≡ hs, m, z|s, s, θifor measuring Sz to have value m, given that n · S certainly returns value s
when n = (0, sin θ, cos θ) is inclined at angle θ to the z-axis. The points in
Figure 7.11 show the results of this calculation with s = 40 and three valuesof θ. The amplitudes peak around the values of the ordinate, m = s cos θ,that are marked with vertical lines. The larger the spin, the more sharply the
amplitudes are peaked around these lines, so for the extremely large values ofs that are characteristic of macroscopic systems, am is significantly non-zero
only when m differs negligibly from s cos θ. Hence, in the classical limit the
only values of Sz that have non-negligible probabilities of occurring lie ina narrow range around s cos θ, which is just the projection of the classicalangular-momentum vector hSi = sn onto the z-axis. That is, in the classicallimit the probability of measuring any individual value of Sz is small, but weare certain to find a value that lies close to the value predicted by classicalphysics.

7.4 Spin

167

The classical picture implies that when the angular-momentum vector istipped in the yz plane at angle θ to the z axis, the value of Sy should be s sin θ.
So now for the eigenstate of n · S just described, we evaluate the expectation
value hSy i by first multiplying the matrix for Sy on the column vector of
the amplitudes plotted in Figure 7.11, and then multiplying the resulting
vector by the row vector of the complex conjugates of the amplitudes. The
expectation value of Sy in this state is
hSy i = a∗m (Sy )mn ans
1 X=a∗ (α(m)δm,n−1 + α(m − 1)δm,n+1 ) an
2i m,n=−s m≃(7.128)sX

1(α(m)a∗m am+1 + α(m − 1)a∗m am−1 ) ,2i m=−sbearing in mind that α(s) = α(−s − 1) = 0. For a given value of θ, theamplitudes plotted in Figure 7.11 lie on smooth curves so we can use theapproximation |am−1 | ≃ |am | ≃ |am+1 |. The phases of the am increase by
π/2 with successive values of m, so hSy i is real and the two terms (7.128)
add. Finally, we exploit the fact that |am | is small unless m ≃ s cos θ anduse the approximation for large s, m thatα(m) =pps(s + 1) − m(m + 1) ≃ s2 − m2 ≃ s sin θ.

(7.129)Combining these approximations with the normalisation condition on the am
gives
X
hSy i ≃ s sin θ|am |2 = s sin θ(7.130)m

exactly as classical physics leads us to expect.
To determine the uncertainty in Sy we evaluate the expectation of Sy2 .
From equation (7.127b) we find that the matrix Sy2 has elements
− 41

s 
X

p=−s

α(m)δm,p−1 + α(m − 1)δm,p+1
α(p)δp,n−1 + α(p − 1)δp,n+1≃ − 41 {α(m)α(m + 1)δm,n−2 + α(m − 1)α(m − 3)δm,n+2,

− α2 (m) + α2 (m − 1) δmn(7.131)
where in going to the second line we have ignored corrections when m = ±s
because the amplitudes for these are negligible anyway. Using the same
approximations as before, we now find
Sy2 =

X
mn

a∗m (Sy2 )mn an ≃Xm2

α2 (m)|am |2 ≃ (s sin θ)22

2= s sin θ ≃ hSy i .

Xm|am |2(7.132)

1/2
is therefore negligible. A
The uncertainty in Sy , being ∼ Sy2 − hSy i2
similar calculation shows that both hSx i and Sx2 vanish to good accuracy.
Thus in the classical limit it is normal for all three components of S to havesmall uncertainties. However, it should be noted that Sy can be accurately
determined precisely because there is some uncertainty in Sz : our calculation
on hSy i depends crucially on there being several non-zero amplitudes am .
Quantum interference between states with different values of Sz is responsible
for confining the likely values of Sy to a narrow range.
This is the third time we have found that the familiar world re-emerges
through quantum interference between states in which some observable has

168

Chapter 7: Angular Momentum

well-defined values: in §2.3.3 we found that bullets can be assigned positions and momenta simultaneously through interference between states ofwell-defined momentum, in §3.2 we saw that an excited oscillator moves asa result of quantum interference between states of well-defined energy, andnow we find that a gyro has a well defined orientation through quantum
interference between states of well-defined angular momentum. In the classical regime a tiny fractional uncertainty in the value of an observable allows
the vast numbers of states to have non-negligible amplitudes, and interference between these states narrowly defines the value of the variable that is
canonically conjugate to the observable (§2.3.1).
7.4.5 Precession in a magnetic field
A compass needle swings to the Earth’s magnetic north pole because a magnetic dipole such as a compass needle experiences a torque when placed in amagnetic field. Similarly, a proton that is in a magnetic field experiences atorque because it is a magnetic dipole. However, its response to this torquediffers from that of a compass needle because it is a spinning body; instead ofaligning with the magnetic field, it precesses around the field. This precession
forms the basis for nuclear magnetic resonance (NMR) imaging, which has
become an enormously important diagnostic tool for chemistry and medicine.
The theory of NMR is a fine example of the practical application of quantum
mechanics in general and spin operators in particular.
Classically, the potential energy of a magnetic dipole µ in a magnetic
field B is
H = −µ · B,
(7.133)where the minus sign ensures that a dipole aligns with the field because thisis its lowest-energy configuration. We align our coordinate system such thatthe z axis lies along B and assume that the magnetic moment operator µ
is a constant 2µp times the spin operator s. Then the Hamiltonian operator
can be written
H = −2µp Bsz .
(7.134)The stationary states of this Hamiltonian are the eigenstates of sz , which for aspin-half particle such as a proton are the states |±i in which a measurementof sz is certain to yield ± 21 ; the energies of these states are
E± = ∓µp B.

(7.135)The evolution in time of any spin state is|ψ, ti = a− e−iE− t/h̄ |−i + a+ e−iE+ t/h̄ |+i,(7.136)

where the constant amplitudes a± specify the initial condition |ψ, 0i =a− |−i + a+ |+i.
Suppose that initially a measurement of the spin parallel to n ≡ (sin θ, 0, cos θ)
was certain to yield 21 . Then from Problem 7.6 we have that a− = sin(θ/2)
and a+ = cos(θ/2). Hence at time t the proton’s state is|ψ, ti = sin(θ/2)e−iE− t/h̄ |−i + cos(θ/2)e−iE+ t/h̄ |+i= sin(θ/2)eiφ/2 |−i + cos(θ/2)e−iφ/2 |+i,

where

(7.137a)

2E+ t
2µp B
= ωt where ω = −.
(7.137b)h̄h̄
But from Problem 7.6 this is just the state |+, n′ i in which a measurementof the spin parallel to n′ = (sin θ cos φ, sin θ sin φ cos θ) is certain to yield 12 .
Consequently, the direction in which a measurement of the spin is certain toyield 21 rotates around the direction of B at the frequency ω. This mirrors
the behaviour expected in classical physics of a magnetic dipole of magnitude
φ(t) =

7.5 Addition of angular momenta

169µp that has spin angular momentum 21 h̄; its spin axis would precess around
B at angular frequency ω (Problem 7.12).
When material that contains chemically bound hydrogen atoms is immersed in a powerful magnetic field, most of the protons align their spinswith B in order to minimise their energy. Radiation of frequency ω hasjust the energy required kick a proton into the higher-energy state in which
its spin is anti-aligned with B. Consequently, such radiation is readily absorbed by a sample, whereas radiation of neighbouring frequencies is not.
As the analysis above shows, quantum interference between the aligned andanti-aligned states causes the expectation value of the magnetic moment toprecesses at angular frequency ω, and the precessing magnetic moment couples resonantly to the imposed radiation field.
The magnetic field at the location of a proton in a molecule has a contribution from the spins of the electrons that bind the proton, and thiscontribution varies slightly from one location to another. For example, inmethanol, CH3 OH, the magnetic field experienced by the proton that is attached to the oxygen atom differs from those experienced by the protons thatare attached to the carbon atom, and the proton that is on the other sideof the carbon atom from the oxygen atom experiences a different field fromthe protons that are adjacent to the oxygen atom. Since the frequency ω ofthe resonant radiation is proportional to the magnitude of magnetic field at
the location of the proton, methanol has three different resonant frequencies
for a given magnitude of the imposed magnetic field. Consequently, clues tothe chemical structure of a substance can be obtained by determining thefrequencies at which magnetic resonance occurs in a given imposed field.

7.5 Addition of angular momenta
In practical applications of quantum mechanics we can often identify two
or more components of the system that each carry a well defined amountof angular momentum. For example, in a hydrogen atom both the proton
and the electron carry angular momentum 21 h̄ by virtue of their spins, anda further quantity of angular momentum may be present in the orbit of theelectron around the proton. The total angular momentum of the atom isthe sum of these three contributions, so it is important to understand how
to add angular momenta in quantum mechanics. Once we understand how
to add two contributions, we’ll be able to add any number of contributions,because we can add the third contribution to the result of adding the firsttwo, and so on. Therefore in this section we focus the problem of adding theangular momenta of two ‘gyros’, that is two systems that have unvarying total
angular momentum quantum number j but several possible orientations.
Imagine that we have two gyros in a box and that we know that the first
gyro has total angular-momentum quantum number j1 , while the second gyro
has total quantum number j2 . Without loss of generality we may assume
j1 ≥ j2 . A ket describing the state of the first gyro is of the form
|ψ1 i =

j1
X

cm |j1 , mi,(7.138a)j2
Xdm |j2 , mi,(7.138b)m=−j1while the state of the second is|ψ2 i =

m=−j2

and from the discussion in §6.1 it follows that the state of the box is|ψi = |ψ1 i|ψ2 i.

(7.139)

170

Chapter 7: Angular Momentum

The coefficients cm and dm are the amplitudes to find the individual gyros
in particular orientations with respect to the z axis. For example, if bothgyros are maximally aligned with the z axis, we will have |cj1 | = |dj2 | = 1
and cm1 = dm2 = 0 for m1 6= j1 and m2 6= j2 .
The operators of interest are the operators Ji2 , Jiz and Ji± of the ith
gyro and the corresponding operators of the box. The operators Jz and J±
for the box are simply sums of the corresponding operators for the gyros
Jz = J1z + J2z;

J± = J1± + J2± .

(7.140)Operators belonging to different systems always commute, so [J1i , J2j ] = 0
for any values of i, j. The operator for the square of the box’s angularmomentum is(7.141)J 2 = (J1 + J2 )2 = J12 + J22 + 2J1 .J2 .
Now
J1+ J2− = (J1x + iJ1y )(J2x − iJ2y )= (J1x J2x + J1y J2y ) + i(J1y J2x − J1x J2y ).

(7.142)The expression for J1− J2+ can be obtained by swapping the labels 1 and 2,
so9
J1+ J2− + J1− J2+ + 2J1z J2z = 2J1 .J2 .
(7.143)Using this expression to eliminate J1 .J2 from (7.141) we obtain
J 2 = J12 + J22 + J1+ J2− + J1− J2+ + 2J1z J2z .

(7.144)

While the total angular momenta of the individual gyros are fixed, thatof the box is variable because it depends on the mutual orientation of thetwo gyros: if the latter are parallel, the squared angular momentum in thebox might be expected to have quantum number j1 + j2 , while if they are
antiparallel, the box’s angular momentum might be expected to have quantum number j1 − j2 . We shall show that this conjecture is true by explicitly
calculating the values of the coefficients cm and dm for which the box is in an
eigenstate of both J 2 and Jz . We start by examining the state |j1 , j1 i|j2 , j2 i
in which both gyros are maximally aligned with the z axis. It is easy to seethat this object is an eigenket of Jz with eigenvalue j1 + j2 . We use (7.144)
to show that it is also an eigenket of J 2 :J 2 |j1 , j1 i|j2 , j2 i = (J12 + J22 + J1+ J2− + J1− J2+ + 2J1z J2z )|j1 , j1 i|j2 , j2 i
= {j1 (j1 + 1) + j2 (j2 + 1) + 2j1 j2 }|j1 , j1 i|j2 , j2 i,(7.145)where we have used the equation Ji+ |ji , ji i = 0, which follows from equation(7.7). It is straightforward to show that the expression in curly brackets inequation (7.145) equals j(j +1) with j = j1 +j2 . Hence |j1 , j1 i|j2 , j2 i satisfies
both the defining equations of the state |j1 + j2 , j1 + j2 i and we may write|j1 + j2 , j1 + j2 i = |j1 , j1 i|j2 , j2 i.

(7.146)

Now that we have found one mutual eigenket for the box of J 2 and Jz
we can easily find others by applying J− to reorient the angular momentumof the box away from the z axis. Again setting j = j1 + j2 we evaluate thetwo sides of the equation
J− |j, ji = (J1− + J2− )|j1 , j1 i|j2 , j2 i.
Equation (7.7) enables us to rewrite the left sidepp
J− |j, ji = j(j + 1) − j(j − 1) |j, j − 1i = 2j |j, j − 1i.
9 Recall that J

1i commutes with J2j for all ij.

(7.147)(7.148)

7.5 Addition of angular momenta

171

Figure 7.12 The left panel shows states obtained by adding a system of angular momentum j2 = 1 to one with j1 = 2, while the right panel is for j1 = 1 and j2 = 21 .

The right side of (7.147) becomesp
j1 (j1 + 1) − j1 (j1 − 1)|j1 , j1 − 1i|j2 , j2 ip
+ j2 (j2 + 1) − j2 (j2 − 1)|j1 , j1 i|j2 , j2 − 1ipp
= 2j1 |j1 , j1 − 1i|j2 , j2 i + 2j2 |j1 , j1 i|j2 , j2 − 1i.

Putting the two sides back together, we have
s
s
j1
j2
|j, j − 1i =|j1 , j1 − 1i|j2 , j2 i +|j1 , j1 i|j2 , j2 − 1i.
jj(7.149)(7.150)A further application of J− to the left side of this equation and of J1− + J2−
to the right side would produce an expression for |j, j − 2i and so on.
Figure 7.12 helps to organise the results of this calculation. States of thebox with well defined angular momentum are marked by dots. The radiusof each semi-circle is proportional to j ′ , where j ′ (j ′ + 1) is the eigenvalue ofthe kets with respect to J 2 . The height of each ket above the centre of thecircles is proportional to m. The left panel shows the case j1 = 2, j2 = 1,
while the right panel is for j1 = 1, j2 = 21 . The scheme for constructing
eigenstates J 2 and Jz that we have developed so far starts with the state at
the top and then uses J− to successively generate the states that lie on theoutermost semi-circle.
We now seek an expression for the state |j − 1, j − 1i that lies at the topof the first semicircle inwards. It is trivial to verify that |j1 , m1 i|j2 , m2 i is aneigenket of Jz with eigenvalue (m1 + m2 ). We require m1 + m2 = j1 + j2 − 1,so either m1 = j1 − 1 and m2 = j2 , or m1 = j1 and m2 = j2 − 1. Equation(7.150) shows that |j, j − 1i involves precisely these two cases, and must beorthogonal to |j − 1, j − 1i because it has a different eigenvalue with respect
to J 2 . So the ket we seek is the unique (up to an overall phase factor) linear
combination of the kets appearing in (7.150) that is orthogonal to the linear
combination that appears there. That is,
s
s
j2
j1
|j − 1, j − 1i =|j1 , j1 − 1i|j2 , j2 i −|j1 , j1 i|j2 , j2 − 1i. (7.151)
jj
All the kets |j − 1, mi for m = j − 2, . . ., which in Figure 7.12 lie on the first
semicircle in, can be constructed by applying J− to this equation.
Similarly, |j −2, j −2i, which in Figure 7.12 lies at the top of the smallest
semicircle, will be a linear combination of |j1 , j1 −2i|j2 , j2 i, |j1 , j1 −1i|j2 , j2 −1i and |j1 , j1 i|j2 , j2 − 2i and must be orthogonal to |j, j − 2i and |j − 1, j − 2i,which are known linear combinations of these states. Hence we can determine
which linear combination is required for |j − 2, j − 2i, and then generate theremaining kets of the series |j − 2, mi by applying J− to it.
On physical grounds we would expect the box’s smallest total angularmomentum quantum number to be j1 − j2 , corresponding to the case in

172

Chapter 7: Angular Momentum
Table 7.4 Total numbers of statesjj1 + j2
j1 + j2 − 1...
j1 − j2
Total

Number of states2(j1 + j2 ) + 1
2(j1 + j2 ) + 1 − 2...
2(j1 + j2 ) + 1 − 4j2(2j1 + 1)(2j2 + 1)

Figure 7.13 Interpretation of Clebsch–
Gordan coefficients in terms ofpvec3(3 + 1)
tors. The full line has lengthand its vertical component has length
2. The pdotted lines labelled j1 havelength 2(2 + 1) and vertical components of length 2 and 1.

which the two gyros are antiparallel (recall that we have labelled the gyros
such that j1 ≥ j2 ). Does this conjectured smallest value of j allow for thecorrect number of basis states for the box? That is, will there be as manybasis states of the box as there are of the contents of the box? We can easily
evaluate the latter: there are 2j1 + 1 orientations of the first gyro, and foreach of these orientations, the second gyro can be oriented in 2j2 + 1 ways.
So the box’s contents can be in (2j1 + 1)(2j2 + 1) basis states of the form
|j1 , m1 i|j2 , m2 i. The predicted number of basis states of the box is worked
out in Table 7.4. In the main part of the table, the number of states in eachrow is two less than in the row above and there are 2j2 + 1 rows. The sumat the bottom can be obtained by making a third column that is just thesecond column in reverse order and noting that the sum of the numbers inthe second and third columns of a given row is then always 4j1 + 2. Hence
twice the sum of the numbers in the second column is 2j2 + 1 times 4j1 + 2.
Thus we do get the correct number of basis states if the smallest value of j
is j1 − j2 .
The numbers
C(j, m; j1 , j2 , m1 , m2 ) ≡ hj, m|j1 , m1 i|j2 , m2 i

(7.152)that we have been evaluating are called Clebsch–Gordan coefficients. They
have a simple physical interpretation: C(j, m; j1 , j2 , m1 , m2 ) is the amplitude
that, on opening the box when it’s in a state of well defined angular momentum, we will find the first and second gyros to be oriented with amounts m1
and m2 of their spins parallel topthe z axis. For example, equation (7.151)
implies that C(3, 2; 2, 1, 1, 1) = 2/3, so if a box that contains a spin-two
gyro and a spin-one gyro has spin-three, there is a probability 2/3 that on
opening the box the second gyro will be maximally aligned with the z axis
and the second significantly inclined, and only a probability 1/3 of finding
the reverse arrangement. These two possibilities are depicted by the lower
and upper dotted lines in Figure 7.13. The classical interpretation is thatthe two gyros precess around the fixed angular-momentum vector of the box,
and that the two configurations for which the Clebsch–Gordan coefficientsgive amplitudes are two of the states through which the precession carries thesystem. This picture is intuitive and of some value, but should not be taken
too seriously. For one thing, the rules for adding angular momentum are
independent of any statement about the Hamiltonian, and therefore carryno implication about the time evolution of the system. The gyros may or
may not precess, depending on whether they are dynamically coupled.

7.5 Addition of angular momenta173

In §6.1 we saw that the physical significance of the state of a composite
system, such as that formed by two gyros, being a linear combination of
product states such as |j1 , m1 i|j2 , m2 i is that the subsystems are correlated.
The Clebsch–Gordan coefficients encode the correlations between the gyros
required for the box to have well-defined angular momentum. If there is any
uncertainty in the orientation of either gyro, such correlations are essential
if the angular momentum of the box is to be well defined: the angular momentum of the second gyro has to make up a pre-defined total with whatevervalue is measured for the first gyro. This consideration explains why the only
states of the box that are simple products of states of the individual gyros are|j1 + j2 , j1 + j2 i = |j1 , j1 i|j2 , j2 i and |j1 + j2 , −(j1 + j2 )i = |j1 , −j1 i|j2 , −j2 i– so much angular momentum can be aligned with the z-axis only by eachgyro individually straining to align with the axis, and there is then no needfor the gyros to coordinate their efforts.
7.5.1 Case of two spin-half systems
The general analysis we have just given will be clarified by working out someparticular cases. We consider first the case j1 = j2 = 21 , which is relevant, forexample, to a hydrogen atom in its ground state, when all angular momentum
is contributed by the spins of the proton and the electron. The electron hasbase states |±, ei in which Jz returns the value ± 21 , while the proton hascorresponding base states |±, pi. Hence there are four states in all and j
takes just two values, 1 and 0.
Our construction of the states in which the atom has well-defined angularmomentum starts with the state|1, 1i = |+, ei|+, pi(7.153)in which both the electron and the proton have their spins maximally alignedwith the z axis. So the atom has maximum angular momentum, and its
angular momentum is maximally aligned with the z axis. Applying J− =
p
e
J−
+ J−
to this ket we obtain1|1, 0i = √ (|−, ei|+, pi + |+, ei|−, pi) .
2(7.154)

The right side of this equation states that with the atom in this state, measurements of Jz for the electron and proton are certain to find that theyare ‘antiparallel’. This fact is surprising given that the left side states thatthe atom has maximum angular momentum, so you would think that thetwo particles had parallel angular momenta. The resolution of this paradox
is that the z components of the two spins are antiparallel, but the components in the xy plane are parallel, although their direction is unknown to
us. Similarly, when the atom is in the state |1, 1i of equation (7.153), thez components of the electron and proton angular momenta are parallel, but
the components in the xy planealignment in√ are√not well aligned. The poor √2 =the xy plane
explains
why
J
2
for
the
atom
is
less
than
3, which is√p
the sum of J 2 = 3/4 for the electron and the proton.
When we apply J− to |1, 0i we obtain|1, −1i = |−, ei|−, pi.

(7.155)This equation confirms the physically obvious fact that if we want to haveh̄ of angular momentum pointing along the negative z axis, we need to havethe angular momenta of both the proton and the electron maximally alignedwith the negative z axis.
The remaining state of the atom is |0, 0i in which the atom has no angular momentum. This is the unique linear combination of the two compoundstates on the right of equation (7.154) that is orthogonal to |1, 0i:
1|0, 0i = √ (|−, ei|+, pi − |+, ei|−, pi) .
2(7.156)

174

Chapter 7: Angular Momentum

The change of sign on the right of this equation from the right of equation(7.154) for |1, 0i ensures that the spins of the electron and proton are antiparallel in the xy plane as well as along the z axis. We show this by rewriting|1, 0i and |0, 0i in terms of the states in which the electron and proton havewell-defined spin parallel to the x-axis. These states are
1|x+, ei = √ (|+, ei + |−, ei)21|x+, pi = √ (|+, pi + |−, pi)21|x−, ei = √ (|+, ei − |−, ei)2(7.157)1|x−, pi = √ (|+, pi − |−, pi)
2So|0, 0i = |x+, eihx+, e|0, 0i + |x−, eihx−, e|0, 0i= 12 |x+, ei (−|−, pi + |+, pi) − 12 |x−, ei (|−, pi + |+, pi)
1= − √ (|x+, ei|x−, pi + |x−, ei|x+, pi) .
2(7.158)The last line states that when the atom is in the state |0, 0i we are indeed
guaranteed to find the components of the spins of the electron and proton
parallel to x have opposite signs. An analogous calculation starting fromequation (7.154) yields (Problem 7.27)
1|1, 0i = √ (|x+, ei|x+, pi − |x−, ei|x−, pi) ,2(7.159)so when the atom is in the |1, 0i state the two particles have identical components of spin along x .
Notice that all three states in which the atom has j = 1 are unchanged
if we swap the m values of the particles – that is, if we map |±, ei → |∓, ei
and the same for the proton states. The atomic atomic state with j = 0,by contrast, changes sign under this interchange. This fact will prove to beimportant when we consider systems with two electrons (such a helium) or
two protons (such as an H2 molecule).

7.5.2 Case of spin one and spin half
In the first excited state of hydrogen, the electron can have total orbital
angular momentum quantum number l = 1. So we now consider how to
combine angular momentum j = 1 with the electron’s spin, j = 21 . The total
angular momentum quantum number takes two values, j = 32 and j = 12 (seeFigure 7.12). We start with the state| 23 , 32 i = |+i|1, 1i(7.160)in which the spin and orbital angular momenta are both maximally oriented
along the z axis. Applying J− = L− + S− to this equation, we obtain| 23 , 12 i =q1
3 |−i|1, 1i +q

2
3 |+i|1, 0i.

(7.161)

The right side of this equation says that in this state of the atom, the electron
is twice as likely to be found with its spin up as down. A second applicationof J− yieldsqq| 32 , − 12 i =2
3 |−i|1, 0i +1
3 |+i|1, −1i(7.162)

as we would expect from the symmetry between up and down. A final
application of J− yields | 32 , − 23 i = |−i|1, −1i as it must on physical grounds.

7.5 Addition of angular momenta

175

Figure 7.14 Classically the sum
vector J 1 + J 2 can line anywhere on
the sphere of radius |J
J2 | around theend of J 1 .

The state | 21 , 21 i is the linear combination of the states that appear in
the right of equation (7.161) that is orthogonal to | 32 , 21 i. Hence,| 12 , 12 i =q2
3 |−i|1, 1i −q

13 |+i|1, 0i.

(7.163)In this atomic state, the electron’s spin is twice as likely to be down as up.
The last remaining state can be found by applying J− to equation (7.163).
It is
q
q
(7.164)| 12 , − 21 i = 13 |−i|1, 0i − 23 |+i|1, −1i.
7.5.3 The classical limit
In classical physics we identify angular momentum with the vector J ≡ h̄ hJi,
and the angular momentum of the whole system is obtained by vectoriallyadding the angular momenta J 1 and J 2 of the component parts. If θ is theangle between these vectors, thenJ 2 = J12 + J22 + 2J1 J2 cos θ.

(7.165)

If nothing is known about the direction of J 2 relative to J 1 , all points on asphere of radius J2 and centred on the end of J 1 are equally likely locations
for the end of J 2 (Figure 7.14). Consequently, the probability dP that θ liesin the range (θ, θ + dθ) is proportional to the area of the band shown in the
figure. Quantitatively
dP = 12 sin θ dθ,(7.166)
R
where the factor 21 ensures that dP = 1. From equation (7.165) the change
in J when θ changes by dθ is given by
J dJ = −J1 J2 sin θdθ.

(7.167)

Combining equations (7.166) and (7.167), we find that the probability thatthe total angular momentum lies in the interval (J , J + dJ ) is10
dP =

J dJ
.
2J1 J2(7.168)10 We discarded the minus sign in equation (7.167) because we require dP > 0 regardlessof whether J increases or decreases.

176

Problems

In quantum mechanics the fraction of states that have total angularmomentum quantum number j isf=

2j + 1,(2j1 + 1)(2j2 + 1)

(7.169)which in the classical limit of large quantum numbers becomes approximately
j/(2j1 j2 ). If all states were equally likely, this fraction would equal theclassical probability that J ≃ jh̄ lay within h̄ of jh̄. It is easy to checkfrom (7.167) that dP does indeed take the value f when we insert Ji = h̄ji
and dJ = h̄. Thus from consistency with classical mechanics we are led
to the principle of equal a priori probability, namely that when we
have no information relevant to an upcoming measurement, we assign equal
probabilities to the system being in each state of whatever basis we havedecided to work in. This principle is the foundation of all statistical physics.

Problems
7.1 Show that hj, j|Jx |j, ji = hj, j|Jy |j, ji = 0 and that hj, j|(Jx2 +Jy2 )|j, ji =j. Discuss the implications of these results for the uncertainty in the orientation of the classical angular momentum vector J for both small and large
values of j.
7.2 In the rotation spectrum of 12 C16 O the line arising from the transitionl = 4 → 3 is at 461.04077 GHz, while that arising from l = 36 → 35 is at4115.6055 GHz. Show from these data that in a non-rotating CO molecule
the intra-nuclear distance is s ≃ 0.113 nm, and that the electrons provide
a spring between the nuclei that has force constant ∼ 1904 N m−1 . Henceshow that the vibrational frequency of CO should lie near 6.47 × 1013 Hz(measured value is 6.43 × 1013 Hz). Hint: show from classical mechanics
that the distance of O from the centre of mass is 73 s and that the molecule’s
2
moment of inertia is 48
7 mp s . Recall also the classical relation L = Iω.
7.3 Show that Li commutes with x · p and thus also with scalar functionsof x and p.
7.4 Write down the expression for the commutator [σi , σj ] of two Paulimatrices. Show that the anticommutator of two Pauli matrices is{σi , σj } = 2δij .

(7.170)

7.5 Let n be any unit vector and σ = (σx , σy , σz ) be the vector whose
components are the Pauli matrices. Why is it physically necessary that n · σsatisfy (n · σ)2 = I, where I is the 2 × 2 identity matrix? Let m be aunit vector such that m · n = 0. Why do we require that the commutator[m · σ, n · σ] = 2i(m × n) · σ? Prove that that these relations follow from thealgebraic properties of the Pauli matrices. You should be able to show that[m · σ, n · σ] = 2i(m × n) · σ for any two vectors n and m.
7.6 Let n be the unit vector in the direction with polar coordinates (θ, φ).
Write down the matrix n · σ and find its eigenvectors. Hence show that thestate of a spin-half particle in which a measurement of the component of spinalong n is certain to yield 21 h̄ is|+, ni = sin(θ/2) eiφ/2 |−i + cos(θ/2) e−iφ/2 |+i,

(7.171)

where |±i are the states in which ± 12 is obtained when sz is measured.
Obtain the corresponding expression for |−, ni. Explain physically why the
amplitudes in (7.171) have modulus 2−1/2 when θ = π/2 and why one of theamplitudes vanishes when θ = π.

Problems

177

7.7 For a spin-half particle at rest, the rotation operator J is equal to thespin operator S. Use the result of Problem 7.4 to show that in this case therotation operator U (α) ≡ exp(−iα · J) is
U (α) = I cosα
2− iα̂ · σ sin

α
2

,

(7.172)where α̂ is the unit vector parallel to α. Comment on the value this gives
for U (α) when α = 2π.
7.8 Write down the 3 × 3 matrix that represents Sx for a spin-one systemin the basis in which Sz is diagonal (i.e., the basis states are |0i and |±i with
Sz |+i = |+i, etc.)
A beam of spin-one particles emerges from an oven and enters a Stern–
Gerlach filter that passes only particles with Jz = h̄. On exiting this filter,the beam enters a second filter that passes only particles with Jx = h̄, andthen finally it encounters a filter that passes only particles with Jz = −h̄.
What fraction of the particles stagger right through?
7.9∗ Repeat the analysis of Problem 7.8 for spin-one particles coming on
filters aligned successively along +z, 45◦ from z towards x [i.e. along (1,0,1)],and along x.
Use classical electromagnetic theory to determine the outcome in thecase that the spin-one particles were photons and the filters were polaroid.
Why do you get a different answer?
√
7.10 A system that has spin momentum 6h̄ is rotated through an angle φaround the z axis. Write down the 5 × 5 matrix that updates the amplitudes
am that Sz will take the value m.
7.11 Justify physically the claim that the Hamiltonian of a particle thatprecesses in a magnetic field B can be written
H = −2µs · B.

(7.173)

In a coordinate system oriented such that the z axis is parallel to B, aproton is initially in the eigenstate |+, xi of sx . Obtain expressions for the
expectation values of sx and sy at later times. Explain the physical content
of your expressions.
Bearing in mind that a rotating magnetic field must be a source ofradiation, do you expect your expressions to remain valid to arbitrarily latetimes? What really happens in the long run?
7.12 Show that a classical top with spin angular momentum S which is
subject to a torque G = µS×B/|S| precesses at angular velocity ω = µB/|S|.
Explain the relevance of this calculation to magnetic resonance imaging in
general and equation (7.137b) in particular.
7.13∗

Write a computer programme that determines the amplitudes am in|n; s, si =

s
X

m=−s

am |s, miwhere n = (sin θ, 0, cos θ) with θ any angle and |n; s, si is the ket that solves
the equation (n · S)|n; s, si = s|n; s, si. Explain physically the nature of thisstate.
Use your am to evaluate the expectation values hSx i and Sx2 for thisstate
p and hence show that the rms fluctuation in measurements of Sx will
be s/2 cos θ.

178
7.14∗

Problems
We have thatL+ ≡ Lx + iLy = eiφ∂∂θ+ i cot θ

∂ 
.
∂φ

(7.174)

From the Hermitian nature of Lz = −i∂/∂φ we infer that derivative operators
are anti-Hermitian. So using the rule (AB)† = B † A† on equation (7.174),
we infer that


∂
∂
+i
cot θ e−iφ .
L− ≡ L†+ = −∂θ∂φThis argument and the Rresult it leads
to is wrong. Obtain the correct result
Rby integrating by parts dθ sin θ dφ (f ∗ L+ g), where f and g are arbitrary
functions of θ and φ. What is the fallacy in the given argument?
P
7.15∗ By writing h̄2 L2 = (x × p) · (x × p) = ijklm ǫijk xj pk ǫilm xl pm showthat
1 h̄2 L2(7.175)p2 = 2 + 2 (r · p)2 − ih̄r · p .
rr
By showing that p · r̂ − r̂ · p = −2ih̄/r, obtain r · p = rpr + ih̄. Hence obtain
p2 = p2r +h̄2 L2.
r2(7.176)

Give a physical interpretation of one over 2m times this equation.
7.16 The angular part of a system’s wavefunction is√hθ, φ|ψi ∝ ( 2 cos θ + sin θe−iφ − sin θeiφ ).
What are the possible results of measurement of (a) L2 , and (b) Lz , andtheir probabilities? What is the expectation value of Lz ?
7.17 A system’s wavefunction is proportional to sin2 θ e2iφ . What are thepossible results of measurements of (a) Lz and (b) L2 ?
7.18 A system’s wavefunction is proportional to sin2 θ. What are the possible results of measurements of (a) Lz and (b) L2 ? Give the probabilities ofeach possible outcome.
7.19 Consider a stationary state |E, li of a free particle of mass m that has
angular-momentum quantum number l. Show that Hl |E, li = E|E, li, where
1
Hl ≡
2m



l(l + 1)h̄2
2
pr +
.
r2

(7.177)Give a physical interpretation of the two terms in the big bracket. Show thatHl = A†l Al , where1(l + 1)h̄.
(7.178)Al ≡ √
ipr −r
2m
Show that [Al , A†l ] = Hl+1 − Hl . What is the state Al |E, li? Show that for
E > 0 there is no upper bound on the angular momentum. Interpret thisresult physically.
P
7.20∗ Show that [Ji , Lj ] = i k ǫijk Lk and [Ji , L2 ] = 0 by eliminating Li
using its definition L = h̄−1 x × p, and then using the commutators of Ji
with x and p.

Problems

179

7.21∗ In this problem you show that many matrix elements of the position
operator x vanish when states of well defined l, m are used as basis states.
These results will leadPto selection rules for electric dipole radiation. Firstshow that [L2 , xi ] = i jk ǫjik (Lj xk + xk Lj ). Then show that L · x = 0 andusing this result derive[L2 , [L2 , xi ]] = iXjkǫjik Lj [L2 , xk ] + [L2 , xk ]Lj = 2(L2 xi + xi L2 ). (7.179)

By squeezing this equation between angular-momentum eigenstates hl, m|and |l′ , m′ i show that
0 = (β − β ′ )2 − 2(β + β ′ ) hl, m|xi |l′ , m′ i,where β ≡ l(l + 1) and β ′ = l′ (l′ + 1). By equating the factor in front ofhl, m|xi |l′ , m′ i to zero, and treating the resulting equation as a quadratic
equation for β given β ′ , show that hl, m|xi |l′ , m′ i must vanish unless l + l′ =0 or l = l′ ± 1. Explain why the matrix element must also vanish whenl = l′ = 0.
7.22∗ Show that l excitations can be divided amongst the x, y or z oscillators of a three-dimensional harmonic oscillator in ( 21 l + 1)(l + 1) ways. Verify
in the case l = 4 that this agrees with the number of states of well defined
angular momentum and the given energy.
7.23∗

Let
1
Al ≡ √2mh̄ω(l + 1)h̄
ipr −+ mωr .
r

(7.180)be the ladder operator of the three-dimensional harmonic oscillator and |E, li
be the oscillator’s stationary state of energy E and angular-momentum quantum number
l. Show that if we write Al |E, li = α− |E − h̄ω, l + 1i, then√α− = L − l, where L is the angular-momentum quantum number of a circular orbit of energy E. Show similarly that if A†l−1 |E, li = α+ |E +h̄ω, l −1i,√then α+ = L − l + 2.
7.24∗ Show that the probability distribution in radius of a particle thatorbits in the three-dimensional harmonic-oscillator potential on pa circularorbit with angular-momentum quantum number l peaks at r/ℓ = 2(l + 1),
where
r
h̄
.
(7.181)ℓ≡2mω
Derive the corresponding classical result.
7.25∗ A particle moves in the three-dimensional harmonic oscillator potential with the second largest angular-momentum quantum number possible atits energy. Show that the radial wavefunction is
2l + 1 −x2 /4e
u1 ∝ x x −xlwhere x ≡ r/ℓ

with

ℓ≡

r

h̄
.
2mω
(7.182)

How many radial nodes does this wavefunction have?
7.26 A box containing two spin-one gyros A and B is found to have angularmomentum quantum numbers j = 2, m = 1. Determine the probabilities
that when Jz is measured for gyro A, the values m = ±1 and 0 will beobtained.
What is the value of the Clebsch–Gordan coefficient C(2, 1; 1, 1, 1, 0)?

180

Problems

7.27 The angular momentum of a hydrogen atom in its ground state is
entirely due to the spins of the electron and proton. The atom is in the state|1, 0i in which it has one unit of angular momentum but none of it is parallelto the z-axis. Express this state as a linear combination of products of thespin states |±, ei and |±, pi of the proton and electron. Show that the states|x±, ei in which the electron has well-defined spin along the x-axis are1|x±, ei = √ (|+, ei ± |−, ei) .
2(7.183)|1, 0i = |x+, eihx+, e|1, 0i + |x−, eihx−, e|1, 0i,(7.184)By writing

express |1, 0i as a linear combination of the products |x±, ei|x±, pi. Explain
the physical significance of your result.
7.28∗ The interaction between neighbouring spin-half atoms in a crystal isdescribed by the Hamiltonian
H=K

S(1) · S(2)(S(1) · a)(S(2) · a)
−3
a
a3



,

(7.185)where K is a constant, a is the separation of the atoms and S(1) is the firstatom’s spin operator. Explain what physical idea underlies this form of H.
(1) (2)(1) (2)(1) (2)(1) (2)
Show that Sx Sx +Sy Sy = 12 (S+ S− +S− S+ ). Show that the mutual
eigenkets of the total spin operators S 2 and Sz are also eigenstates of H andfind the corresponding eigenvalues.
At time t = 0 particle 1 has its spin parallel to a, while the otherparticle’s spin is antiparallel to a. Find the time required for both spins toreverse their orientations.

8
Hydrogen
Wherever we look, down at ourselves or up into the vastness of the Universe, what we see are atoms. The way atoms interact with themselves andwith electromagnetic radiation structures the world about us, giving colour,
texture, solidity or fluidity to all things, both alive and inanimate. In thewider Universe the way visible matter has aggregated into stars and galaxies
is determined by the interplay between atoms and radiation. In the last two
decades of the twentieth century it emerged that atoms do not really dominate the Universe; on large scales they are rather like icing on the cake. But
they certainly dominate planet Earth, and, like the icing, they are all we cansee of the cosmic cake.
Besides the inherent interest of atomic structure, there is the historical fact that the formative years of quantum mechanics were dominated byexperimental investigations of atomic structure. Most of the principles ofthe subject were developed to explain atomic phenomena, and the stature ofthese phenomena in the minds of physicists was greatly enhanced throughthe rôle they played in revolutionising physics.
It is an unfortunate fact that atoms are complex systems that are noteasily modelled to a precision as good as that with which they are commonlymeasured. The complexity of an atom increases with the number of electrons
that it contains, both because the electrons interact with one another as wellas with the nucleus, and because the more electrons there are, the higher
the nuclear charge and the faster electrons can move. By the middle of theperiodic table the speeds of the fastest electrons are approaching the speedof light and relativistic effects are important.
In this chapter we develop a model of the simplest atom, hydrogen,that accounts for most, but not all, measurements. In Chapter 10 we willtake the first steps towards a model of the second most complex atom, helium, and indicate general trends in atomic properties as one proceeds down
the periodic table. The ideas we use will depend heavily on the model ofhydrogen-like systems that is developed in this chapter. With these applications in view, we generalise from hydrogen to a hydrogen-like ion, in
which a single electron is bound to a nucleus of charge Ze.

182

Chapter 8: Hydrogen

8.1 Gross structure of hydrogen
We start with a rather crude model of a hydrogen-like ion. In this model
neither the electron nor the nucleus has a spin, and the electron moves nonrelativistically under purely electrostatic forces. The structure of an atom orion that is obtained using these approximations is called its gross structure.
The approximations make it easy to write down the model Hamiltonian because they include just three contributions to the energy: the kinetic energiesof the nucleus and the electron, and the bodies’ electrostatic binding energy:
H=

p2n
p2
Ze2
+ e −
,
2mn 2me4πǫ0 |xe − xn |(8.1)

where xe and xn are the position operators of the electron and the nucleus,
respectively, and pe and pn are the corresponding momentum operators. We
wish to solve the eigenvalue equation H|Ei = E|Ei for this Hamiltonian.
In the position representation, the momentum operators become derivative
operators, and the eigenvalue equation becomes a partial differential equationin six variables
Eψ(xn , xe ) = −

h̄2 2
h̄2 2
Ze2 ψ
∇n ψ −∇e ψ −,
2mn
2me
4πǫ0 |xe − xn |(8.2)

where a subscript e or n on ∇ implies the use of derivatives with respect to thecomponents of xe or xn . Remarkably, we can solve this frightening equation
exactly. The key step is to introduce six new variables, the components ofX≡me xe + mn xn
me + mn

and r ≡ xe − xn .

(8.3)

X is the location of the ion’s centre of mass, and r is the vector from thenucleus to the electron. The chain rule yields
∂
∂X ∂
∂r ∂
me
∂
∂
=
·
·
+
=
+
.
∂xe
∂xe ∂X ∂xe ∂rme + mn ∂X ∂r(8.4)

When we take the dot product of each side with itself, we find
∇2e =



me
me + mn2∇2X + ∇2r +

∂2
2me
,
me + mn ∂X · ∂r

(8.5a)

where the subscripts X and r imply that the operator is to be made up ofderivatives with respect to the components of X or r. Similarly
∇2n =



mn
m e + mn2

∇2X + ∇2r −

2mn
∂2
.
me + mn ∂X · ∂r

(8.5b)

We now add m−1
times equation (8.5a) to m−1en times equation (8.5b). The
mixed derivatives cancel leaving
2
−1 2
m−1
e ∇e + mn ∇n =

where
µ≡

1
1
∇2X + ∇2r ,me + mnµme mn
me + mn(8.6a)(8.6b)

is called the reduced mass of the electron. In the case of hydrogen, whenmn = mp = 1836me , the reduced mass differs very little from me (µ =0.99945me), and in heavier hydrogen-like ions the value of µ lies even closer
to me .

8.1 Gross structure

183

Figure 8.1 The effective potential(8.13) for Z = 1 and (from bottom
to top) l = 0, 1, 2, 3, 4.

When we use equation (8.6a) to replace xe and xn in equation (8.2) byr and X, we obtainEψ = −

h̄2
h̄2 2
Ze2
∇2X ψ −∇r ψ −
ψ.
2(me + mn )2µ4πǫ0 r(8.7)

The right side breaks into two parts: the first term is the Hamiltonian HK ofa free particle of mass me + mn , while the second and third terms make up
the Hamiltonian Hr of a particle of mass µ that is attracted to the origin by
an inverse-square law force. Since HK and Hr commute with one another,
there is a complete set of mutual eigenkets. In §6.1.2 we showed (page 109)
that in these circumstances we can assume that ψ is a productψ(xe , xn ) = K(X)ψr (r),

(8.8)

−

h̄2
∇2 K = EK K
2(me + mn ) X

(8.9)

−

Ze2 ψrh̄2 2∇r ψr −
= Er ψr .
2µ4πǫ0 r(8.10)

where

andHere EK and Er are two distinct eigenvalues and their sum is the ion’s totalenergy, E = EK + Er .
From §2.3.3 we know all about the dynamics of a free particle, so equation (8.9) need not detain us. We have to solve equation (8.10). In theinterests of simplicity we henceforth omit the subscript Er .
Equation (7.69) enables us to write the kinetic energy term in equation(8.10) in terms of the radial momentum operator pr and the total orbital
angular momentum operator L2 . Equation (8.10) is then the eigenvalue
equation of the Hamiltonian
H≡

p2r
Ze2
h̄2 L2
−
+
.
2µ
2µr2
4πǫ0 r

(8.11)

L2 commutes with H since the only occurrence in H of the angles θ and φis in L2 itself. So there is a complete set of mutual eigenstates |E, l, mi ofH, L2 and Lz such that L2 |E, l, mi = l(l + 1)|E, l, mi. For these kets theoperator H of equation (8.11) is equivalent to the radial Hamiltonian
Hl ≡

p2r
Ze2
l(l + 1)h̄2
−
+
.
2
2µ
2µr
4πǫ0 r(8.12)

184

Chapter 8: Hydrogen

The operator (8.12) is the Hamiltonian for a particle that moves in one
dimension in the effective potential
Veff (r) ≡Ze2l(l + 1)h̄2−.
2µr2
4πǫ0 r(8.13)

The first term in this expression is the kinetic energy that conservation ofangular momentum requires in tangential motion, while the second term isthe electrostatic potential energy. Veff is plotted in Figure 8.1 for l = 0, . . . , 4.
The radial Hamiltonian Hl governs the oscillations of the reduced massaround the minimum for Veff . By astute exploitation of natural coordinatesand symmetry we have reduced our original intimidating Hamiltonian (8.1),
which contained twelve operators, to a Hamiltonian Hl that contains only
two operators. The eigenkets of the Hamiltonian Hr for the internal structure of the ion are products of the eigenkets |E, li of Hl and eigenkets |l, miof L2 and Lz :|E, l, mi = |E, li|l, mi.
(8.14)
Hl is strikingly similar to the radial Hamiltonian defined by equation(7.81) for which we solved the eigenvalue problem in the course of our studyof the three-dimensional harmonic oscillator. We use essentially the same
technique now, defining the dimensionless ladder operatora0
Al ≡ √
2



l+1
Z
i
pr −
+
h̄
r
(l + 1)a0,(8.15a)where we have identified the Bohr radius1a0 ≡4πǫ0 h̄2.
µe2(8.15b)

The product of Al with A†l is



i
i
l+1
l+1
Z
Z
− pr +
−
−
pr +
2
h̄
(l + 1)a0rh̄(l + 1)a0
r
(
)




2
a2 p2r
Z
l+1
i
l+1
= 0
+
pr ,
.
+
−
2 h̄2
(l + 1)a0rh̄r(8.16)
Equations (2.25) and (7.67) enable us to evaluate the commutator in thisexpression, so we havea2A†l Al = 0

 2

Z2
pr
(l + 1)2
2Z
i l+1
+
+
−
[pr , r]−2 h̄2(l + 1)2 a20r2a0 r h̄ r2
l(l + 1)a2 p2r
2Z
Z2
+
= 0
−
+
2 h̄2
r2
a0 r (l + 1)2 a20
2
2
Z
a µ
.
= 02 Hl +2(l + 1)2h̄

a2
A†l Al = 0(8.17)

If we evaluated the product Al A†l , the sign in front of the commutator in thefirst line of equation (8.17) would be reversed, so we would find
Al A†l =a20 µZ22 Hl+1 + 2(l + 1)2 .
h̄

(8.18)

1 The physical significance of a is clarified by rewriting equation (8.15b) in the form0e2 /(4πǫ0 a0 ) = (h̄/a0 )2 /µ. The left side is the electrostatic potential energy at a0 andthe right side is twice the kinetic energy of zero-point motion (§3.1) of a particle whose
position has uncertainty ∼ a0 . For hydrogen a0 = 5.29177 × 10−11 m.

8.1 Gross structure

185

Taking equation (8.17) from equation (8.18) we obtain the commutator
a20 µ(Hl+1 − Hl ),(8.19)h̄2
a result that recalls equation (7.88) for the three-dimensional harmonic oscillator.
It is useful to rewrite equation (8.17) in the form


Z2
h̄2
†
.
(8.20)
A
A
−
Hl =
l l
µa20
2(l + 1)2
[Al , A†l ] =Commuting each side of this equation with Al and using equation (8.19), we
obtain an expression for the commutator of Hl with Al :[Al , Hl ] =

h̄2
h̄2
†
[A
,
A
A
]
=
[Al , A†l ]Al = (Hl+1 − Hl )Al .
l
l
l
µa20
µa20

(8.21)

This equation simplifies to
Al Hl = Hl+1 Al .

(8.22)We show that Al is a ladder operator by multiplying it into both sidesof the eigenvalue equation Hl |E, li = E|E, li and using equation (8.22):
EAl |E, li = Al Hl |E, li = Hl+1 Al |E, li.

(8.23)

This equation states that Al |E, li is an eigenket of Hl+1 with eigenvalue E.
That is, Al transfers energy from the electron’s radial motion to its tangential
motion. If we repeat this process by multiplying Al |E, li by Al+1 , and so on,
we will eventually arrive at a circular orbit. Let L(E) denote the l value of
this orbit. Then AL must annihilate |E, Li because, if it did not, we would
have a state with even greater angular momentum. Thus with equation(8.17) we can write
2

0 = |AL |E, Li| = hE, L|A†L AL |E, Li =

a20 µ
Z2
2 E + 2(L + 1)2 .
h̄

(8.24)

That is,Z 2 e2Z 2 µe4Z 2 h̄2=−=−,(8.25)
2µa20 n28πǫ0 a0 n22n2 (4πǫ0 h̄)2where we have defined the principal quantum number n ≡ L + 1 and thesecond equality uses the definition (8.15b) of the Bohr radius. The Rydberg
constant R is
 2 2
h̄2
e
e2
1
R≡
=
= 2µ
= 13.6056923 eV,(8.26)
22µa0
8πǫ0 a0
4πǫ0 h̄E=−

where µ = me mp /(me +mp ) is the reduced mass in the case of hydrogen. The
Rydberg constant enables us to give a compact expression for the permitted
values of E and l in hydrogenR(n = 1, 2, . . .) ; 0 ≤ l ≤ n − 1 .
(8.27)
n2
Henceforth we use n rather than E to label kets and wavefunctions. Thus|n, l, mi = |n, li|l, mi (cf. eq. 8.14) is the stationary state of a hydrogen-likeion for the energy given by (8.25) and the stated angular-momentum quantum numbers. The ground state is |1, 0, 0i. The energy level immediately
above the ground state is four-fold degenerate, being spanned by the states|2, 0, 0i, |2, 1, 0i and |2, 1, ±1i. The second excited energy level is 9-fold degenerate, and so on.
This property of our model hydrogen atom, that it has states with different l but the same energy, is unusual and reflects a hidden symmetry of
our model – see Appendix F for details. Atoms with more than one electron
have energy levels that depend explicitly on l even when spin and relativity
are neglected. When our model of hydrogen is upgraded to include spin andrelativity, E becomes weakly dependent on l.
E=−

186

Chapter 8: Hydrogen

Figure 8.2 Schematic diagram of
the Lyman, Balmer and Paschen series of spectral lines in the spectrumof hydrogen.

8.1.1 Emission-line spectra
A hydrogen atom may change its value of n to a smaller value n′ , releasing
the liberated energy as a photon of frequency ν = (En − En′ )/h. Hence theemission spectrum of hydrogen contains lines at the frequencies


1
1
R
.
(8.28)
−
ν=
h n′2
n2
The lines associated with a given lower level n′ form a series of lines ofincreasing frequency and decreasing wavelength. The series associated withn′ = 1 is called the Lyman series, the longest-wavelength member of whichis the Lyman α line at 121.5 nm, followed by the Lyβ line at 102.5 nm,
and so on up to the series limit at 91.2 nm. The series associated withn′ = 2 is called the Balmer series and starts with a line called Hα at
656.2 nm and continues with Hβ at 486.1 nm towards the series limit at
364.6 nm. The series associated with n′ = 3 is the Paschen series, andthat associated with n′ = 4 is the Brackett series. Figure 8.2 shows thefirst three series schematically. Historically the discovery in 1885 by Johann
Balmer (1825–1898), a Swiss schoolmaster, that the principal lines in the
optical spectrum of hydrogen could be fitted by equation (8.28), was crucial
for the development of Niels Bohr’s model atom of 1913, which was the
precursor of the current quantum theory (Problem 8.3).
Equation (8.25) states that, for given n, the energy of an electron scales
as Z 2 . For a many–electron atom electromagnetic interactions between theelectrons invalidate this scaling. However, it holds to a fair approximationfor electrons that have the smallest values of n because these electrons aretrapped in the immediate vicinity of the nucleus and their dynamics is largely
unaffected by the presence of electrons at larger radii. Henry Moseley (1887–
1915) studied the frequencies of X-rays given off when atoms were bombarded
by free electrons. He showed2 that the frequencies of similar spectral lines
from different elements seemed to scale with the square of the atomic number.
At that time the periodic table was something constructed by chemists thatlacked a solid physical foundation. In particular, the atomic numbers of someelements were incorrectly assigned. Moseley’s experiments led to the orderof cobalt and nickel being reversed, and correctly predicted that elementswith atomic numbers 43, 61, 72, 75, 87 and 91 would be discovered.

8.1.2 Radial eigenfunctions
The wavefunctions of hydrogen-like ions are not only important for experiments with atoms and ions that have only one electron, but are also thebuilding blocks from which models of many-electron atoms are built.
The first step in finding any radial eigenfunction for a hydrogen-like ion
is to write the equation An−1 |n, n−1i = 0 (eq. 8.24) as a differential equation
for the radial wavefunction of the circular orbit with angular-momentum
2 Moseley, H.G.J., 1913, Phil. Mag., 27, 703. The lines studied by Moseley wereassociated with transitions n = 2 → 1.

8.1 Gross structure

187

Figure 8.3 The radial wavefunctions un−1(r) of “circular” orbits for n = 1, 2 and 3.
nquantum number l = n − 1. From equations (8.15a) and (7.66) we need to
solve


n−1
Z
∂ n−1
unn−1 = 0,(8.29)
un + −
+
∂r
r
na0
where
uln (r) ≡ hr|n, li.
(8.30)

Equation (8.29) is a first-order linear differential equation. Its integrating
factor is
Z


n−1
Z
exp
dr −
(8.31)= r−(n−1) eZr/na0 ,+r
na0so the required eigenfunction isunn−1 (r) = Crn−1 e−Zr/na0 ,(8.32)

where C is a normalising constant. This wavefunction is very similar to
our expression (7.97) for the wavefunction of a circular orbit in the threedimensional harmonic oscillator potential – the only difference is that theGaussian function has been replaced by a simple exponential. The scalelength in the exponential is (n/Z)a0 , so it increases with energy and decreases
with the nuclear charge. This makes perfect sense physically because it statesthat more energetic electrons can go further from a given nucleus, and that
nuclei with higher electric charge will bind their (innermost) electrons more
tightly.
We choose the normalising constant C in equation (8.32) to ensure thatthe complete wavefunction (eq. 8.14)
hr, θ, φ|(|E, li|l, mi) = hr|n, lihθ, φ|l, mi = uln (r)Ylm (θ, φ)3

2(8.33)
2

normalised. Bearing in mind that d x = r dr d Ω and that
Ris correctly
d2 Ω |Ylm |2 = 1, we find that C must satisfy
Z ∞2n+1 Z ∞
22 na0
2n −2Zr/na0
1=C=Cdρ ρ2n e−ρ
dr r e2Z00(8.34)
 na 2n+1
0= C2(2n)!,
2Z
where we have evaluated the integral with the aid of Box 8.1. The correctly
normalised radial wavefunction is therefore3/2 
n−1
1
2Z
2Zr
unn−1 (r) = p(8.35)
e−Zr/na0 .
na0(2n)! na0

These functions are plotted for n = 1 − 3 in Figure 8.3. For n > 1 thewavefunction rises from zero at the origin to a peak at r = n(n − 1)a0 /Z andfrom there falls exponentially with increasing r.

188

Chapter 8: Hydrogen

Figure 8.4 The probability of finding the electron of a hydrogen atomthat is in its ground state at a radius greater than r. Radii greaterthan 2a0 /Z are classically forbidden.

Box 8.1:

The factorial function

R∞
We often encounter the integral Γ(α + 1) ≡ 0 dt tα e−t . Integrating by
parts we find thatZ ∞∞Γ(α + 1) = − tα e−t 0 + α
dt tα−1 e−t0= αΓ(α).

It is easy to check that Γ(1) = 1. Putting α = 1 in the last equation it
follows that Γ(2) = 1. Setting α = 2 we find Γ(3) = 2, and repeating
this process we see that for any integer n, Γ(n + 1) = n!. We can use thisresult to define the factorial function byZ ∞z! ≡ Γ(z + 1) =dt tz e−t .
(1)0This definition yields a well defined value for z! for any complex numberthat is not a negative integer, and it coincides with the usual definitionof a factorial if z happens to be a non-negative integer.

We obtain the ground-state radial wavefunction by setting n = 1 inequation (8.35):
 3/2Z(8.36)
e−Zr/a0 .
u01 (r) = 2
a0
The complete wavefunction is obtained by multiplying this by Y00 = (4π)−1/2 .
Figure 8.4 shows the probability of finding the electron at a radius greaterthan r. This reaches 13/e4 ≃ 0.24 at r = 2a0 /Z, where the potential energy
is equal to the total energy. In classical physics the probability of finding theelectron at these radii is zero.
It is interesting to calculate the expectation value of r for circular orbits.
We have3Z ∞2(n−1)2Z2Zr
1e−2Zr/na0
dr r3(2n)! na0
na0
0
Z ∞
a0
na0 1
=
dρ ρ2n+1 e−ρ = n(n + 21 ) .
2Z (2n)! 0Z(8.37)In the classical limit of large n, hri ≃ n2 a0 /Z, so E ∝ 1/n2 ∝ 1/ hri asclassical physics predicts. (One can easily show that classical physics yields
hn, n − 1, m|r|n, n − 1, mi =

8.1 Gross structure

189

Figure 8.5 Probability densities forthree orbits in hydrogen. All orbits haven = 3 and m = l. Clockwise from top
left l increases from 0 to 2. The grey
scale shows the log to base 10 of theprobability density.

the correct proportionality constant.) A very similar calculation shows thatr2 = n2 (n + 1)(n + 21 )a20 /Z 2 =

n+1
2
hri ,
n + 12(8.38)

q√2
so the rms uncertainty in r is hr2 i − hri = hri / 2n + 1. Consequently,
as n increases, the uncertainty in r increases as n3/2 , but the fractional
uncertainty in r decreases as n−1/2 .
Our conclusion that the radius of an atom scales as n2 implies that an
atom with n ∼ 100 occupies 1012 times as much volume as an atom in the
ground state. Consequently, only at high-vacuum densities can such highly
excited atoms be considered isolated systems. Radio telescopes detect line
radiation emitted by hydrogen atoms in the interstellar medium that are
reducing their value of n by δn from n ≃ 100. The frequency of such atransition is3
En+δn − En
1
100
R 1
νn =
≃
6.58
δn GHz. (8.39)
−
=
h
h n2
(n + δn)2n
Our analysis of the three-dimensional harmonic oscillator suggests thatapplications of A†l′ to unn−1 should generate the wavefunctions uln for l < n−1.
We show that this is indeed the case by daggering both sides of equation(8.22) to obtain
Hl A†l = A†l Hl+1 .
(8.40)
Consequently, applying A†l to both sides of E|E, l + 1i = Hl+1 |E, l + 1i wehave
E(A†l |E, l + 1i = A†l Hl+1 |E, l + 1i = Hl (A†l |E, l + 1i)

which establishes that A†l |E, l + 1i is an eigenket of Hl as we hoped. Usinga result proved in Problem getAAdaggerprob, we have, in fact, that
−1/2
√ 
2
1
1
|n, li =A†l |n, l + 1i.
(8.41)−
Z(l + 1) n2

190

Chapter 8: Hydrogen

Table 8.1 The first six radial eigenfunctions uln (r) for hydrogen withaZ ≡ a0 /Z. The full wavefunction is uln (r)Ylm (θ, φ).
l012e−r/aZ
3/2

aZ

1
2

n
2
3




−r/2aZ−r/3aZ
2e
r
2r
2r2
2e
1−
1−
+
2aZ
3aZ
27a2Z
(2aZ )3/2(3aZ )3/2
r
25/2 e−r/3aZ re−r/2aZr√
1−6aZ
9(3aZ )3/2 aZ
3(2aZ )3/2 aZ 2
3/2 −r/3aZ
r
2 e
√ √
27 5(3aZ )3/2 aZ

From equations (8.15a) and (7.66) we can writea0A†l = − √

2



∂
l+2
Z
+
−
∂r
r
(l + 1)a0.

(8.42)

Setting l = n − 2 we can apply this operator to unn−1 to obtainunn−2 (r) = constant ×


1−

Zr
n(n − 1)a0

rn−2 e−Zr/na0 .

(8.43)

This wavefunction has a node at r = n(n − 1)a0 /Z. When we apply A†n−3 tothis wavefunction to generate unn−3 , the lowest power of r in the factor thatmultiplies the exponential will be rn−3 , so the exponential will be multiplied
by a quadratic in r and the wavefunction will have two nodes. In our study ofthe three-dimensional harmonic oscillator we encountered the same pattern:
the number of nodes in the radial wavefunction increased by one every timeA† decrements the angular momentum and increases the energy of radial
motion. The radial eigenfunctions for states with n ≤ 3 are listed in Table 8.1
and plotted in Figure 8.5.
Notice that because unn−1 (r) is a real function and A†l is a real operator,
all the radial eigenfunctions are real. Because the probability current J is
proportional to the gradient of the phase of the wavefunction (eq. 2.87), thereality of uln (r) implies that the probability current inside the atom has noradial component. This makes perfectly good sense physically: the electron
moves both inwards and outwards and (unlike in the classical case) at any
given point the electron is as likely to be moving out as in.

8.1.3 Shielding
The electrostatic potential in which a bound electron moves is never exactly proportional to 1/r as we have hitherto assumed. In hydrogen ora single-electron ion the deviations from 1/r proportionality are small but
measurable. In many-electron systems the deviations are large. In all cases
the deviations arise because the charge distribution that binds the electron
is not confined to a point as we have assumed. First, protons and neutrons have non-zero radii – after all there has to be room for three quarks
to move about in there at mildly relativistic speed! Second, even if the nuclear charge were confined to a point, the field it generates would not be aninverse-square field because in the intense electric field that surrounds thenucleus, a non-negligible charge density arises in the vacuum. This charge
density is predicted by quantum electrodynamics, the theory of the interaction of the Dirac field, whose excitations constitute electrons and positrons,
and the electromagnetic field, whose excitations are photons. In a vacuum
the zero-point motions (§3.1) of these fields cause electron-positron pairs to
be constantly created, only to annihilate an extremely short time later. In

8.1 Gross structure191

the strong field near the nucleus, the positrons tend to spend their brief
lives further from the nucleus, which repels them, than do the electrons. In
consequence the charge inside a sphere drawn around the nucleus is slightly
smaller than the charge on the nucleus, the charge deficit being small forboth very small and very large spheres. That is, quantum electrodynamics
predicts that the vacuum is a polarisable dielectric medium, just like an ordinary insulator, in which the electrons and ions move in opposite directions
when a field is applied, giving rise to a net charge density within the medium.
When an atom has more than one electron, the deviation of the electrostatic potential from 1/r proportionality is much larger than in hydrogensince the charge on any electron other than the one whose dynamics we are
studying is distributed by quantum uncertainty through the space aroundthe nucleus, so the charge inside a sphere around the nucleus is comparable
to the charge on the nucleus when the sphere is very small, but falls to ewhen the sphere is large.
Phenomena of this type, in which there is a tendency for a charged
body to gather oppositely charged bodies around it, are often referred to as‘shielding’. A complete treatment of the action of shielding in even singleelectron systems involves quantum field theory and is extremely complex.
In this section we modify the results we have obtained so far to explore an
idealised model of shielding, which makes it clear how shielding modifies theenergy spectrum, and thus the dynamics of atomic species.
The key idea is to replace the atomic number Z in the Hamiltonian witha decreasing function of radius. We adoptaZ(r) = Z0 1 +,(8.44)
r
where Z0 and a are adjustable parameters. For r ≫ a, the nuclear charge
tends to a maximally shielded value Z0 e. For r ≃ a, the charge is largerby ∼ Z0 e. At very small r, the charge diverges, but we anticipate that
this unphysical divergence will not have important consequences because theelectron is very unlikely to be found at r ≪ a0 /Z. With this choice for Z(r),the radial Hamiltonian (8.12) becomes
Hl′ =

p2r
Z0 e2
{l(l + 1) − β}h̄2
−
+
,
2
2µ
2µr
4πǫ0 r(8.45a)Z0 aµe2.
2πǫ0 h̄2(8.45b)where
β≡Because we chose to take the radial dependence of Z to be proportional to
1/r, we have in the end simply reduced the repulsive centrifugal potential
term in the radial Hamiltonian. Let l′ (l) be the positive root of the quadratic
equationl′ (l′ + 1) = l(l + 1) − β.
(8.46)
In general l′ will not be an integer. With this definition, Hl′ is identical with
the Hamiltonian (8.12) of the unshielded case with l′ substituted for l andZ0 replacing Z, that isHl′ = Hl′ (l) .
(8.47)
Consequently, the operator Al′ that is defined by equation (8.15a) with thesame substitutions satisfies (cf. eq. 8.17)
A†l′ Al′ =

a20 µ
Z02
′
2 Hl + 2(l ′ + 1) .
h̄

(8.48)

Moreover by analogy with equation (8.19) we have[Al′ , A†l′ ] =a20 µ(Hl′ +1 − Hl′ ).
h̄2(8.49)

192

Chapter 8: Hydrogen

It follows that Al′ is a ladder operator
EAl′ |E, l′ i = (Hl′ Al′ + [Al′ , Hl′ ])|E, l′ i = Hl′ +1 Al′ |E, l′ i,(8.50)

so Al′ |E, l′ i = α|E, l′ + 1i is an (unnormalised) eigenket of Hl′ +1 just as in
the unshielded case. Applying Al′ +1 to |E, l′ + 1i we argue that eventuallysome maximum value L′ of l′ will be reached, at which point AL′ |E, L′ i = 0.
From the mod square of this equation we conclude thatE=−

Z02 e2,8πǫ0 a0 (L′ + 1)2

where

L′ = l′ (l) + k,

(8.51)

where k is the number of times we have to apply A to achieve annihilation.
Since for a 6= 0, l′ (l) is not an integer, E is given by the formula (8.25)
for the unshielded case with n replaced by a number that is not an integer.
Moreover, the energy now depends on l as well as on n, where n is defined
to be 1 + l plus the number of nodes in the radial wavefunction at r < ∞.
To see this, consider the effect of increasing our initial value of l by one, andcorrespondingly decreasing by one the number of times we have to apply Al′
to achieve annihilation. In an unshielded atom l′ = l, so E is unchanged
when l is increased and k decreased by unity; we have moved between stateswith the same value of n. In the shielded case, increasing l by unity does
not increment l′ (l) by unity, so in equation (8.51) the changes in l′ and k do
not conspire to hold constant L′ . In fact one can show from equation (8.46)
that when l increases by one, l′ increases by more than one (Problem 8.13),
so among states with a given principal quantum number, those with thelargest l values have the smallest binding energies. This makes perfect sense
physically because it is the eccentric orbits that take the electron close to
the nucleus, where the nuclear charge appears greatest.
In 1947 Lamb & Retherford showed3 that in hydrogen the state |2, 0, 0ilies 4.4 × 10−6 eV below the states |2, 1, mi, contrary to naive predictions
from the Dirac equation. This Lamb shift is due to shielding of the proton
by electron-positron pairs in the surrounding vacuum.
8.1.4 Expectation values for r−k
It will prove expedient to have formulae for the expectation value of r−k withhydrogenic wavefunctions and the first three values of k.
The value of r−1 can be obtained from the virial theorem (2.93) sincein hydrogen the potential energy is ∝ r−1 . With α = −1, equation (2.93)
implies thatp22hE|
|Ei = −hE|V |Ei.
(8.52)
2m
On the other hand the expectation of the Hamiltonian yields
p2|Ei + hE|V |Ei = E,2m(8.53)

Z 2 e2
Ze2
hE|r−1 |Ei = 2E = −.
4πǫ04πǫ0 a0 n2(8.54)hE|
so we havehE|V |Ei = −It follows that r−1 = Z/(n2 a0 ).
To obtain r−2 we anticipate a result that we shall prove in §9.1. This
relates to what happens when we add a term βH1 to a system’s Hamiltonian,where β is a number and H1 is an operator. The nth eigenenergy of the3 W.E. Lamb & R.C. Retherford, Phys. Rev. 72, 241

8.2 Fine structure and beyond

193

complete Hamiltonian then becomes a function of β, and in §9.1 we showthatdE
= hE|H1 |Ei.
(8.55)
dβ β=0
We apply this result to a hydrogen-like system with the additional Hamiltonianh̄2.
(8.56)
H1 = −
2µr2
In the last subsection we showed that the exact eigenvalues of this system
are given by equation (8.51). Differentiating the eigenvalues with respect toβ and using equation (8.55) we finddl′d
Z 2 e2Z 2 e2h̄2hE|r−2 |Ei ==−2µdβ β=0 8πǫ0 a0 (l′ + k)24πǫ0 a0 (l + k)3 dβ(8.57)

From equation (8.46) we have dl′ /dβ = −1/(2l + 1), sohE|r−2 |Ei =Z 2 e2 µ
Z2
= 2 3
,
2 3
a0 n (l + 21 )2πǫ0 a0 h̄ n (2l + 1)

(8.58)

where the last equality uses the definition (8.15b) of the Bohr radius.
We determine r−3 by considering the expectation value of the commutator [H, pr ]. As we saw in §2.2.1, in a stationary state the commutatorwith H of any observable vanishes. Hence with equation (8.12) we can write
0 = hE|[H, pr ]|Ei =l(l + 1)h̄2
Ze2
hE|[r−1 , pr ]|Ei (8.59)
hE|[r−2 , pr ]|Ei −2µ4πǫ0Using the canonical commutation relation [r, pr ] = ih̄ [equation (7.67)] toevaluate the commutators in this expression, and the value of r−2 that wehave just established, we findhE|r−3 |Ei =Z3.
a30 n3 l(l + 1)(l + 21 )(8.60)The three values of r−k that we have calculated conform to a pattern.
First the basic atomic scale a0 /Z is raised to the −k th power. Then there isa product of 2k quantum numbers on the bottom, reflecting the tendency forthe atom’s size to grow as n2 . Finally, as k increases, the number of factorsof l increases from zero to three, reflecting the growing sensitivity of r−3
to orbital eccentricity.

8.2 Fine structure and beyond
The model of hydrogen-like ions that we developed in the last section issatisfying and useful, but it is far from complete. We now consider some ofthe physics that is neglected by this model.
In §2.3.5 we saw that when a particle that moves in an inverse-square
force field is in a stationary state, the expectation value of its kinetic energy,
classically 12 mv 2 , is equal in magnitude but opposite in sign to its totalenergy. Equation (8.25) is an expression for the ground-state energy of anelectron in a hydrogen-like ion. When we equate the absolute value of thisexpression to 21 me v 2 , we find that the ratio of v to the speed of light c is
v
= αZ,
c

(8.61)

194

Chapter 8: Hydrogen

where the dimensionless fine structure constant is defined to be
α≡

1
e2
≃
.
4πǫ0 h̄c
137(8.62)

Since relativistic corrections tend to be O(v 2 /c2 ), it follows that in hydrogen
relativistic corrections to the results we have derived may be expected to
be several parts in 105 , but these corrections, being proportional to Z 2 , will
exceed 10% by the middle of the periodic table.
For future reference we note that with the reduced mass µ approximatedby me , equation (8.15b) for the Bohr radius can be writtena0 =h̄λCompton
=
,
αme c
2πα

(8.63)

where we have identified the electron’s Compton wavelength h/me c (thewavelength of a photon that has energy me c2 ). When we use this expression
to eliminate a0 from equation (8.25), we find that the energy levels of ahydrogen-like ion areE=−

Z 2 α2me c2 ,
2n2

so

R = 12 α2 me c2 .

(8.64)

8.2.1 Spin-orbit coupling
Magnetism is a relativistic correction to electrostatics in the sense that a
particle that is moving with velocity v in an electric field E experiences amagnetic field1
B = 2 v × E.
(8.65)
c
If the particle has a magnetic dipole moment µ, it experiences a torqueG = µ × B that will cause its spin S to precess. In the particle’s rest frame
the classical equation of motion of S isdS
1
= µ × B,dth̄

(8.66)

where h̄ appears only because S is the dimensionless spin obtained by dividing the angular momentum by h̄. We assume that the magnetic moment µ is
proportional to the dimensionless spin vector S and write the proportionality
µ=

gQh̄
S,
2m0

(8.67)

where g is the dimensionless gyromagnetic ratio, and Q and m0 are theparticle’s charge and rest mass. In the case of an electron g = 2.002, a value
which is correctly predicted by relativistic quantum electrodynamics, andthe dimensional factor is defined to be the Bohr magneton
µB ≡

eh̄
= 9.27 × 10−24 J T−1 .
2me(8.68)

With this notation, our rest-frame equation of motion (8.66) becomesdS
gQ
S × B.
=
dt
2m0(8.69)The non-zero value of the right side of this classical equation of motion
for S implies that there is a spin-dependent term in the particle’s Hamiltoniansince the operator S commutes with all spatial operators (§7.4) and theright side of the classical equation of motion (8.69) is proportional to the8.2 Fine structure and beyond

195

expectation of [S, H] (cf. eq. 2.57). We want to determine what this term in
H is.
Energy is not a relativistic invariant – it is physically obvious that observers who move relative to one another assign different energies to a given
system. Consequently, when they do quantum mechanics they use different
Hamiltonians. We need the Hamiltonian that governs the dynamics of the
reduced particle in the rest frame of the atom’s centre of mass. So we haveto transform the equation of motion (8.69) to this frame. This is a tricky
business because the reduced particle is accelerating, so the required Lorentz
transformation is time-dependent. Given the delicacy of the required transformation, it is advisable to work throughout with explicitly Lorentz ‘covariant’ quantities, which are explained in Appendix G. In Appendix H these
are used to show that in a frame of reference in which the electron is moving,equation (8.69) becomes
dS
Q
=
dt
2m0 c2h̄ dΦ
−S × L + 2c2 S × B .
m0 r dr(8.70)

It is straightforward to demonstrate (Problem 8.14) from equation (2.34) that
this classical equation of motion of the spin S of an electron (which has charge
Q = −e) arises if we introduce into the quantum-mechanical Hamiltonian(8.1) two spin-dependent terms, namely the spin-orbit Hamiltonian
HSO ≡ −dΦ eh̄2
S · L,
dr 2rm2e c2(8.71)and the Zeeman spin Hamiltonian
HZS ≡

eh̄
S · B.
me(8.72)The Zeeman spin Hamiltonian is just µ · B with equation (8.67) used toreplace the magnetic moment operator by the spin operator. Interestingly,
the spin-orbit Hamiltonian is a factor two smaller than µ · B with µ replacedin the same way and equation (8.65) used to relate B to the electric field
in which the electron is moving. In the 1920s the experimental data clearlyrequired this factor of two difference in the spin Hamiltonians, but its origin
puzzled the pioneers of the subject until, in 1927, L.T. Thomas4 showed that
it is a consequence of the fact that the electron’s rest frame is accelerating(Appendix H). If no torque is applied to a gyro, it does not precess in its
instantaneous rest frame. But if the direction of the gyro’s motion is changing relative to some inertial frame, the sequence of Lorentz transformations
that are required to transform the spin vector into the inertial frame causes
the spin to precess in the inertial frame. This apparent precession of anaccelerated gyro is called Thomas precession.
In a single-electron system such as hydrogen, Φ = Ze/(4πǫ0 r), so
HSO =

Zαh̄3
S · L,
2m2e cr3(8.73)

where the fine-structure constant (8.62) has been used to absorb the 4πǫ0 .
Since the coefficient in front of the operator S · L is positive, spin-orbit
coupling lowers the energy when the spin and orbital angular momenta are
antiparallel.
The operator S · L in equation (8.73) is most conveniently written
S · L = 12 ((L + S)2 − L2 − S 2 ) = 21 (J 2 − L2 − S 2 ),4 Phil. Mag. 3, 1 (1927)(8.74)

196

Chapter 8: Hydrogen

Figure 8.6 The fine structure ofa hydrogen-like ion with Z = 200
that is predicted by equation (8.76a).
The dotted line denotes a break inthe energy scale so that the groundstate can be included.

so HSO is diagonal in a basis made up of mutual eigenkets of J 2 , L2 and S 2 .
In §7.5 we constructed such mutual eigenkets from the eigenkets of S 2 , Sz ,L2 , and Lz . S · L annihilates states with quantum number l = 0 because
then j = s. Hence, there is no spin-orbit coupling in the ground state ofhydrogen. In any excited state, l > 0 is permitted, and from §7.5.2 we knowthat the possible values of j are l ± 21 . The associated eigenvalues of theoperator on the right of equation (8.74) are readily found to be
13
2 {j(j + 1) − l(l + 1) − 4 } =12l− 21 (l + 1)

for j = l + 21 ,for j = l − 21 .

(8.75)

Although S · L commutes with the gross-structure Hamiltonian HGS(eq. 8.1), the other operator in HSO , namely r−3 , does not. So the eigenkets
of HGS + HSO will differ (subtly) from the eigenkets we have found. In§9.1 we shall show that in these circumstances the change in the energy of astationary state can be estimated by replacing the operator by its expectation
value. Equation (8.60) gives this value, and, inserting this with our results
for the spin operators, yields energy shifts∆E ≡ hn, l, m|HSO |n, l, mi ≃ Kn
where
Kn ≡

.

(

(l + 1)(l + 21 )for j = l + 21−l(l + 12 )for j = l − 21

Z 4 α4Z 4 αh̄3=
me c2 ,
4a30 m2e cn3
4n3(l > 0),(8.76a)(8.76b)and the second equality uses equation (8.63) for a0 . The difference between
the energies of states with j = l ± 12 is
El+1/2 − El−1/2 =2Kn.
l(l + 1)

(8.76c)For n = 1 the fine-structure energy scale Kn is smaller than the grossstructure energy (8.64) by a factor Z 2 /2α2 that rises from parts in 105 forhydrogen to more than 10% by the middle of the periodic table.5 In hydrogenfine-structure is largest in the n = 2, l = 1 level, which is split into j = 32 andj = 21 sublevels. According to equations (8.76c), these sublevels are separated
by K2 = 4.53 × 10−5 eV, while the measured shift is 4.54 × 10−5 eV.
Figure 8.6 shows the prediction of equation (8.76a) for the energy levelsof a hydrogen-like ion with Z = 200. With this unrealistically large value
5 Naturally, the fine-structure constant owes its name to its appearance in this ratioof the fine-structure and gross energies.

8.2 Fine structure and beyond

197

of Z the fine structure for n = 2 has comparable magnitude to the grossstructure difference between the n = 2 and n = 3 levels. The levels inthis figure are labelled in an obscure notation that is traditional in atomic
physics and more fully explaine in Box 10.2. The value of n appears first,
followed by one of the letters S, P , D, F to denote l = 0, 1, 2, 3, respectively.6
The value of j appears as a subscript to the letter, and the value of 2s +1 (here always 2) appears before the letter as a superscript. So the level
22 P3/2 has n = 2, s = 1/2, l = 1, and j = 23 . From Figure 8.6 we seethat states in which j is less than l (because the electron’s spin and orbital
angular momenta are antiparallel) are predicted to have lower energies thanthe corresponding states in which the two angular momenta are aligned. The
spin–orbit interaction vanishes by symmetry for s = 0 but otherwise at fixedn the magnitude of the effect decreases with increasing angular momentum
because the electron’s top speed on a nearly circular orbit is smaller than on
an eccentric orbit, so relativistic effects are largest on eccentric orbits.
Equation (8.76a) suggests that states that differ in l but not j shouldhave different energies, whereas they do in fact have extremely similar energies. For example, the 22 S1/2 state lies 4.383 × 10−6 eV above the 22 P1/2
state, while equation (8.76a) implies that this energy difference should be
3−5
eV. This discrepancy arises because the spin-orbit Hamil2 K2 = 6.79×10
tonian does not provide a complete description to order α4 of relativistic
corrections to the electrostatic Hamiltonian. Actually, additional corrections
shift the energy of the 22 S1/2 states into close alignment with the energy ofthe 22 P1/2 states.7 However, in atoms with more than one electron, theelectrostatic repulsion between the electrons shifts the energy of the 22 S1/2
states downwards by much larger amounts. These electrostatic corrections
are hard to calculate accurately, so the much smaller relativistic corrections
are not interesting, experimentally, and the quantities of interest are difference in energy between states with the same values of l but different j. These
difference are correctly given by equation (8.76a).
Relativistic quantum electrodynamics is in perfect agreement with measurements of hydrogen. It uses the Dirac equation rather than classicallyinspired corrections to the electrostatic Hamiltonian. We have devoted significant space to deriving the spin-orbit Hamiltonian not because it plays arole in hydrogen, but because it becomes important as one proceeds down
the periodic table. The other relativistic corrections also become large by the
middle of the periodic table, but outside hydrogen their effects are so masked
by electron-electron interactions that they are of little practical importanceand we shall not discuss them in this book.

8.2.2 Hyperfine structure
A proton is a charged spin-half particle, so like an electron it has a magnetic
moment. By analogy with the definition of the Bohr magneton (eq. 8.68),
we define the nuclear magneton to be
µp ≡

eh̄
= 5.05 × 10−27 J T−1 .
2mp(8.78)

6 These letters are a shorthand for a description of spectral lines that later were foundto involve the various l values: sharp, principal, diffuse, faint.
7 In the lowest order of relativistic quantum electrodynamics, the energy of a hydrogen
atom depends on only n and j: the Dirac equation predicts!)(nα2 Z 2 3R−.
(8.77)
E =− 2 1−
4nn2
j + 21Thus the 22 S1/2 states are predicted to have the same energy as the 22 P1/2 states. The
measured Lamb shift between these states arises in the next order as a consequence ofpolarisation of the vacuum, as described in §8.1.3.

198

Chapter 8: Hydrogen

In terms of µp , the magnetic-moment operator of the proton isµ = g p µp S p ,(8.79)

where gp = 5.58 and Sp is proton’s spin operator, so the proton’s magnetic
moment is smaller than that of an electron by a factor 2.79me /mp ≃ 1.5 ×
10−3 .
The electron in a hydrogen atom can create a magnetic field at thelocation of the proton in two ways: as a moving charge, it generates a current, and it has its intrinsic magnetic moment, so its probability distribution|ψ(x)|2 is a distribution of magnetic dipoles that will generate a magnetic
field just as iron does in a bar magnet.
The ground level of hydrogen is a particularly simple case because inthis state the electron has no orbital angular momentum, so it generates amagnetic field exclusively through its dipole moment. The magnetic vector
potential distance r from a magnetic dipole µe isA=µ µ0 µ e × r
µ0
e
.
=
∇×
3
4π r
4π
r

(8.80)The magnetic field is B = ∇×A, so the hyperfine-structure Hamiltonian
for the ground state isn µ oµ0e.
(8.81)µp · ∇ × ∇ ×
HHFS = µp · B =4πr

Until HHFS is included in the atom’s Hamiltonian, the atom’s lowest
energy level is degenerate because the spins of the electron and the proton
can be combined in a number of different ways. To proceed further we needto evaluate the matrix elements obtained by squeezing HHFS between statesthat form a basis for the ground-level states. The natural basis to use ismade up of the states |j = 0i and |j = 1, mi for m = −1, 0, 1 that can beconstructed by adding two spin-half systems (§7.5.1). In Appendix I we showthat the resulting matrix elements are
2µ0|ψ(0)|2 hs|µp · µe |s′ i32µ0=−|ψ(0)|2 gp µp 2µB hs|Sp · Se |s′ i,3hψ, s|HHFS |ψ, s′ i =(8.82)

where we have replaced the magnetic moment operators by the appropriate multiples of the spin operators. From our discussion of the spin-orbit
Hamiltonian (8.73), which is also proportional to the dot product of two
angular-momentum operators, we know that the eigenstates of the total angular momentum operators are simultaneously eigenstates of Sp · Se witheigenvalues 21 {j(j + 1) − 34 − 43 }, so in this basis the off-diagonal matrix
elements vanish and the diagonal ones arehψ, j, m|HHFS |ψ, j, mi = −2µ0|ψ(0)|2 gp µp µB {j(j + 1) − 32 }.
3(8.83)

In §9.1 we shall show that the diagonal matrix elements provide a good
estimates of the amount by which HHFS shifts the energies of the stationary
states of the gross-structure Hamiltonian.
The total angular-momentum quantum number of the atom can be j = 0
or j = 1 and the two possible values of the curly bracket above differ by two.
Equation (8.36) gives |ψ(0)|2 = 1/(πa30 ), so the energies of these levels differ
by
4µ0
∆E =
5.58µp µB = 5.88 × 10−6 eV.
(8.84)
3πa30
The lower level, having j = 0, is non-degenerate, while the excited state isthree-fold degenerate. Transitions between these levels give rise to radiation

Problems

199

of frequency 1.420 405 7518 GHz. To obtain perfect agreement between thismost accurately measured frequency and equation (8.84), it is necessary tochange the gyromagnetic ratio of the electron from the value of 2 that wehave adopted to the value 2.002319 . . . that is predicted by quantum electrodynamics. The agreement between theory and experiment is then impressive.
The hyperfine line of hydrogen provides the most powerful way of tracing
diffuse gas in interstellar and intergalactic space. Radiation at this frequency
can propagate with little absorption right through clouds of dust and gasthat absorb optical radiation. Consequently it was in radiation at 1.4 GHz
that the large-scale structure of our own galaxy was first revealed in the
1950s. The line is intrinsically very narrow with the consequence that thetemperature and radial velocity of the hydrogen that emits the radiation
can be accurately measured from the Doppler shift and broadening in theobserved spectral line. The existence of 1.4 GHz line radiation from ourgalaxy was predicted theoretically by H.C. van de Hulst as part of his doctoral
work in Nazi-occupied Utrecht. In 1951 groups in the USA and Australia
and the Netherlands, detected the line almost simultaneously. The Dutchgroup used a German radar antenna left over from the war.

Problems
8.1 Some things about hydrogen’s gross structure that it’s important toknow (ignore spin throughout):a) What quantum numbers characterise stationary states of hydrogen?
b) What combinations of values of these numbers are permitted?
c) Give the formula for the energy of a stationary state in terms of the
Rydberg R. What is the value of R in eV?
d) How many stationary states are there in the first excited level and in
the second excited level?
e) What is the wavefunction of the ground state?
f) Write down an expression for the mass of the reduced particle.
g) The wavefunction hx|ni of any state with principal quantum number ncontains an exponential in r = |x|. Write down the scale length of thisexponential in terms of the Bohr radius a0 .
h) We can apply hydrogenic formulae to any two charged particles that areelectrostatically bound. How does the ground-state energy then scalewith (i) the mass of the reduced particle, and (ii) the charge Ze on the
nucleus? (iii) How does the radial scale of the system scale with Z?
8.2 Show, by induction or otherwise, that there are n2 stationary states ofhydrogen with energy E = −R/n2 .

8.3 In the Bohr atom, electrons move on classical circular orbits that haveangular momenta lh̄, where l = 1, 2, . . .. Show that the radius of the first
Bohr orbit is a0 and that the model predicts the correct energy spectrum.
In fact the ground state of hydrogen has zero angular momentum. Why did
Bohr get correct answers from an incorrect hypothesis?
8.4 Show that the speed of a classical electron in the lowest Bohr orbit(Problem 8.3) is v = αc, where α = e2 /4πǫ0 h̄c is the fine-structure constant.
What is the corresponding speed for a hydrogen-like Fe ion (atomic number
Z = 26)? Given these results, what fractional errors must we expect in theenergies of states that we derive from non-relativistic quantum mechanics.
8.5 Show that Bohr’s hypothesis (that a particle’s angular momentum must
be an integer multiple of h̄), when applied to the three-dimensional harmonic
oscillator, predicts energy levels E = lh̄ω with l = 1, 2, . . .. Is there an
experiment that would falsify this prediction?
8.6 Show that the electric field experienced by an electron in the groundstate of hydrogen is of order 5×1011 V m−1 . Why is it impossible to generatecomparable macroscopic fields using charged electrodes. Lasers are available
that can generate beam fluxes as big as 1022 W m−2 . Show that the electricfield in such a beam is of comparable magnitude.

200

Problems

8.7 Positronium consists of an electron and a positron (both spin-half andof equal mass) in orbit around one another. What are its energy levels? Bywhat factor is a positronium atom bigger than a hydrogen atom?
8.8 The emission spectrum of the He+ ion contains the Pickering series ofspectral lines that is analogous to the Lyman, Balmer and Paschen series inthe spectrum of hydrogen.
Balmer i = 1, 2, . . . 0.456806 0.616682 0.690685 0.730884
Pickering i = 2, 4, . . . 0.456987 0.616933 0.690967 0.731183
The table gives the frequencies (in 1015 Hz) of the first four lines of the Balmer
series and the first four even-numbered lines of the Pickering series. The
frequencies of these lines in the Pickering series are almost coincident withthe frequencies of lines of the Balmer series. Explain this finding. Provide aquantitative explanation of the small offset between these nearly coincidentlines in terms of the reduced mass of the electron in the two systems. (In 1896
E.C. Pickering identified the odd-numbered lines in his series in the spectrumof the star ζ Puppis. Helium had yet to be discovered and he believed thatthe lines were being produced by hydrogen. Naturally he confused the evennumbered lines of his series with ordinary Balmer lines.)
8.9 Tritium, 3 H, is highly radioactive and decays with a half-life of 12.3
years to 3 He by the emission of an electron from its nucleus. The electron
departs with 16 keV of kinetic energy. Explain why its departure can betreated as sudden in the sense that the electron of the original tritium atom
barely moves while the ejected electron leaves.
Calculate the probability that the newly-formed 3 He atom is in an excited state. Hint: evaluate h1, 0, 0; Z = 2|1, 0, 0; Z = 1i.
8.10∗

A spherical potential well is defined by
V (r) =

0
V0

for r < aotherwise,(8.85)

where V0 > 0. Consider a stationary state with angular-momentum quantum
number l. By writing the wavefunction ψ(x) = R(r)Ylm (θ, φ) and using
p2 = p2r + h̄2 L2 /r2 , show that the state’s radial wavefunction R(r) mustsatisfy2d1
l(l + 1)h̄2
h̄2
R + V (r)R = ER.
(8.86)
+
R+
−
2m dr
r
2mr2
Show that in terms of S(r) ≡ rR(r), this can be reduced to
d2 S
S
2m
− l(l + 1) 2 + 2 (E − V )S = 0.
dr2
r
h̄

(8.87)

Assume that V0 > E > 0. For the case l = 0 write down solutions to thisequation valid at (a) r < a and (b) r > a. Ensure that R does not diverge
at the origin. What conditions must S satisfy at r = a? Show that these
conditions can be simultaneously satisfied if and only if a solution can befound to k cot ka = −K, where h̄2 k 2 = 2mE and h̄2 K 2 = √2m(V0 − E).
Show graphically that the equation can only be solved when 2mV0 a/h̄ >π/2. Compare this result with that obtained for the corresponding onedimensional potential well.
The deuteron is a bound state of a proton and a neutron with zeroangular momentum. Assume that the strong force that binds them produces
a sharp potential step of height V0 at interparticle distance a = 2 × 10−15 m.
Determine in MeV the minimum value of V0 for the deuteron to exist. Hint:
remember to consider the dynamics of the reduced particle.

Problems

201

8.11 Let the wavefunction of the stationary states of the gross-structure
Hamiltonian of hydrogen be hx|n, l, mi = uln (r)Ylm (θ, φ). Show thatZ ∞0

dr r2 uln (r)uln′ (r) = δnn′ .

(8.88)

By considering an appropriate Sturm-Liouville equation, or otherwise, show
further that
Z ∞
′

0

dr uln (r)uln (r) = Cl δll′ .

(8.89)

8.12 Show that for hydrogen the matrix element h2, 0, 0|z|2, 1, 0i = −3a0 .
On account of the non-zero value of this matrix element, when an electricfield is applied to a hydrogen atom in its first excited state, the atom’s energy
is linear in the field strength (§9.1.2).
q
8.13∗ From equation (8.46) show that l′ + 21 = (l + 12 )2 − β and that theincrement ∆ in l′ when l is increased by one satisfies ∆2 +∆(2l′ +1) = 2(l+1).
By considering the amount by which the solution of this equation changes
when l′ changes from l as a result of β increasing from zero to a small number,show that2β∆=1+ 2+ O(β 2 ).
(8.90)
4l − 1
Explain the physical significance of this result.

8.14 Show that Ehrenfest’s theorem yields equation (8.70) with B = 0
as the classical equation of motion of the vector S that is implied by thespin–orbit Hamiltonian (8.71).
8.15∗ (a) A particle of mass m moves in a spherical potential V (r). Show
that according to classical mechanicsddV der(p × Lc ) = mr2
,
dt
dr dt

(8.91)

where Lc = r × p is the classical angular-momentum vector and er is theunit vector in the radial direction. Hence show that when V (r) = −K/r,
with K a constant, the Runge-Lenz vector Mc ≡ p × Lc − mKer is aconstant of motion. Deduce that Mc lies in the orbital plane, and that for
an elliptical orbit it points from the centre of attraction to the pericentre of
the orbit, while it vanishes for a circular orbit.
(b) Show that in quantum mechanics (p × L)† − p × L = −2ip. Hence
explain why in quantum mechanics we take the Runge-Lenz vector operator
to be
M ≡ 21 h̄N − mKer where N ≡ p × L − L × p.
(8.92)P
Explain why we can write down the commutation relation [Li , Mj ] = i k ǫijk Mk .
(c) Explain why [p2 , N ] = 0 and why [1/r, p × L] = [1/r, p] × L. Henceshow that


1
1 2
2
[1/r, N] = i 3 (r p − x x · p) − pr − p · x x 3 .
(8.93)rr(d) Show that
 

 X
1 1
xj
xj 
[p , er ] = ih̄ − p + p +pj 3 x + x 3 pj .
r
r
r
r
j
2

(8.94)(e) Hence show that [H, M] = 0. What is the physical significance of
this result?

202

Problems

P(f) Show that (i) [Mi , L2 ] = i jk ǫijk (Mk Lj + Lj Mk ), (ii) [Li , M 2 ] = 0,
where M 2 ≡ Mx2 + My2 + Mz2 . What are the physical implications of theseresults?
(g) Show thatX
[Ni , Nj ] = −4iǫiju p2 Lu(8.95)uand that[Ni , (er )j ] − [Nj , (er )i ] = −and hence that

[Mi , Mj ] = −2ih̄2 mH

4ih̄ Xǫijt Ltr t

Xǫijk Lk .

kWhat physical implication does this equation have?

(8.96)(8.97)

9
Perturbation theory
It is rarely possible to solve exactly for the dynamics of a system of experimental interest. In these circumstances we use some kind of approximation
to tweak the solution to some model system that is as close as possible tothe system of interest and yet is simple enough to have analytically solvable
dynamics. That is, we treat the difference between the experimental systemand the model system as a ‘perturbation’ of the model. Perturbation theory
in this sense was an important part of mathematical physics before quantum
mechanics appeared on the scene – in fact the development of Hamiltonianmechanics was driven by people who were using perturbation theory to understand the dynamics of the solar system. Interestingly, while perturbation
theory in classical mechanics remains an eclectic branch of knowledge that isunderstood only by a select few, perturbation theory in quantum mechanics
is a part of main-stream undergraduate syllabuses. There are two reasonsfor this. First, analytically soluble models are even rarer in quantum than inclassical physics, so more systems have to be modelled approximately. Second, in quantum mechanics perturbation theory is a good deal simpler andworks rather better than in classical mechanics.

9.1 Time-independent perturbations
Let H be the Hamiltonian of the experimental system and H0 the Hamiltonian of the model system for which we have already solved the eigenvalue
problem. We hope that ∆ ≡ H − H0 is small and define
Hβ = H0 + β∆.

(9.1)

We can think of Hβ as the Hamiltonian of an apparatus that has a knob on
it labelled ‘β’; when the knob is turned to β = 0, the apparatus is the model
system, and as the knob is turned round to β = 1, the apparatus is gradually
deformed into the system of experimental interest.
We seek the eigenkets |Ei and eigenvalues E of Hβ as functions of β.
Since the Hamiltonian of the apparatus is a continuous function of β, we
conjecture that the |Ei and E are continuous functions of β too. In fact, we204

Chapter 9: Perturbation theory

conjecture that they are analytic functions1 of β so they can be expanded as
power series|Ei = |ai + β|bi + β 2 |ci + · · ·E = Ea + βEb + β 2 Ec + · · · ,

;

(9.2)

where |ai, |bi, etc., are states to be determined and Ea , Eb , etc., are appropriate numbers. When we plug our conjectured forms (9.2) into the eigenvalue
equation H|ψi = E|ψi, we have
(H0 + β∆) |ai + β|bi + β 2 |ci+) = Ea + βEb + β 2 Ec + |ai + β|bi + β 2 |ci + .
(9.3)
Since we require the equality to hold for any value of β, we can equate thecoefficient of every power of β on either side of the equation.
β0 :
β1 :
β2 :

H0 |ai = Ea |ai
H0 |bi + ∆|ai = Ea |bi + Eb |ai
H0 |ci + ∆|bi = Ea |ci + Eb |bi + Ec |ai.

(9.4)

The first equation simply states that Ea and |ai are an eigenvalue and eigenket of H0 . Physically, |ai is the state that we will find the system in if we
slowly turn the knob back to zero after making a measurement of the energy.
Henceforth we shall relabel Ea with E0 and relabel |ai with |E0 i, the zeroreminding us of the association with β = 0 rather than implying that |E0 i
is the ground state of the unperturbed system.
To determine Eb we multiply the second equation through by hE0 |:hE0 |H0 |bi + hE0 |∆|E0 i = E0 hE0 |bi + Eb .

(9.5)

Now from Table 2.1, hE0 |H0 |bi = (hb|H0 |E0 i)∗ = E0 hE0 |bi. Cancelling thiswith the identical term on the right, we are left with
Eb = hE0 |∆|E0 i.

(9.6)

Thus the first-order change in the energy is just the expectation value ofthe change in the Hamiltonian when the system is in its unperturbed state,
which makes good sense intuitively. This is the result that we anticipated in§§8.2.1 and 8.2.2 to estimate the effects on the allowed energies of hydrogenof the spin-orbit and hyperfine Hamiltonians.
To extract the second-order change in E we multiply the third of equations (9.4) by hE0 |. Cancelling hE0 |H0 |ci on E0 hE0 |ci by strict analogy withwhat we just did, we obtain
Ec = hE0 |∆|bi − Eb hE0 |bi.

(9.7)

To proceed further we have to determine |bi, the first-order change in thestate vector. Since the eigenkets |En i of H0 form a complete set of states,
we can write |bi as the sumX|bi =bk |Ek i.
(9.8)
kIn the second of equations (9.4) we replace |bi by this expansion and multiply
through by hEm | 6= hE0 | to findbm =hEm |∆|E0 i.
E0 − Em(9.9)

1 Much interesting physics is associated with phenomena in which a small change in
one variable can produce a large change in another (phase changes, narrow resonances,caustics, . . . ). In classical physics perturbation theory is bedevilled by such phenomena.
In quantum mechanics this conjecture is more successful, but still untrustworthy as weshall see in §9.1.2.

9.1 Time-independent perturbations

205

Box 9.1: Ensuring that ha|bi = 0
Since the perturbed eigenket should be properly normalised, we have
1 = hE|Ei = (hE0 | + βhb| + · · ·) (|E0 i + β|bi + · · ·)= 1 + β (hE0 |bi + hb|E0 i) + O(β 2 ).

Equating the coefficient of β on each side of the equation we conclude
that hE0 |bi + hb|E0 i = 0, from which it follows that hE0 |bi is pure imaginary. The phase of |Ei is arbitrary, and we are free to choose this phaseindependently for each model Hamiltonian Hβ . In particular, instead ofusing |Ei we can use |E ′ i ≡ eiαβ |Ei, where α is any real constant: |E ′ i
is our original eigenket but with its phase shifted by a linear function of
β. When we expand |E ′ i in powers of β we have
|E ′ i = |E0 i + β|b′ i + · · · ,where |b′ i is the derivative of |E ′ i with respect to β evaluated at β = 0.
This isd|E ′ id|b′ i ==
= iα|E0 i + |bi.
eiαβ |Eidβ β=0dβ
β=0Consequently,

hE0 |b′ i = iα + hE0 |bi.

Since hE0 |bi is known to be pure imaginary, it is clear that we can choose
α such that hE0 |b′ i = 0. This analysis shows that the phases of theperturbed eigenkets can be chosen such that the first order perturbation|bi is orthogonal to the unperturbed state |E0 i and one generally assumes
that this choice has been made.
This expression determines the coefficient of all kets in (9.8) that have energies that differ from the unperturbed value E0 . For the moment we assume that E0 is a non-degenerate eigenvalue of H0 , so there is only one
undetermined coefficient, namely that of |E0 i. Fortunately we can argue
that this coefficient can be taken to be zero from the requirement that
|Ei = |E0 i + β|bi + O(β 2 ) remains correctly normalised. The complete argument is given in Box 9.1 but we can draw a useful analogy with changing
a three-dimensional vector so that the condition |r| = 1 is preserved; clearly
we have to move r on the unit sphere and the first-order change in r is necessarily perpendicular to the original value of r. The quantum-mechanicalnormalisation condition implies that as β increases |Ei moves on a hypersphere in state space and hE0 |bi = 0. So we exclude |E0 i from the sum in(9.8) and have that the first-order change to the stationary state is|bi =X hEm |∆|E0 i

m6=0

E0 − Em|Em i.

(9.10)

When this expression for |bi is inserted into equation (9.7), we have that thesecond-order change in E isEc =X hE0 |∆|Ek ihEk |∆|E0 i

k6=0

E0 − Ek.

(9.11)

9.1.1 Quadratic Stark effectLet’s apply the theory we’ve developed so far to a hydrogen atom that has
been placed in an electric field E = −∇Φ. An externally imposed electricfield is small compared to that inside an atom for field strengths up to E ≃206

Chapter 9: Perturbation theory

5 × 1011 V m−1 (Problem 8.6) so perturbation theory should yield a good
estimate of the shifts in energy level that ordinary fields effect. By thedefinition of the electrostatic potential Φ, the field changes the energy of theatom byδE = e{Φ(xp ) − Φ(xe )},(9.12)
where xp and xe are the position vectors of the proton and electron, respectively. We assume that the field changes very little on the scale of the atom,
and, as in §8.1, we define r ≡ xe − xp . Then we may writeδE ≃ −er · ∇Φ = er · E.

(9.13)

We orient our coordinate system so that E is parallel to the z axis and use
the notation E = |E|. Then it is clear that the effect of imposing an external
electric field is to add to the unperturbed Hamiltonian a term
∆ = eEz.

(9.14)

Suppose the atom is in its ground state |100i, where the digits indicate thevalues of n, l and m. Then from equation (9.6) the first-order energy change
in E isEb = eEh100|z|100i.
(9.15)
In §4.1.4 we saw that the expectation value of any component of x vanishes
in a state of well-defined parity. Since the ground-state ket |100i has welldefined (even) parity, Eb = 0, and the change in E is dominated by thesecond-order term Ec . For our perturbation to the ground state of hydrogen,equation (9.11) becomes
Ec = e2 E 2∞ X
Xh100|z|nlmihnlm|z|100i

n=2

l<n
|m|≤l

E1 − En

.

(9.16)

Symmetry considerations make it possible to simplify this sum dramatically.
First, since [Lz , z] = 0 (Table 7.3), z|nlmi is an eigenfunction of Lz witheigenvalue m, and therefore orthogonal to |100i unless m = 0. Therefore
in equation (9.16) only the terms with m = 0 contribute. Second, we candelete from the sum over l all even values of l because, as we saw in §4.1.4,
the matrix elements of an odd-parity operator between states of the same
parity vanish. In fact, a result proved in Problem 7.21 shows that the termswith l = 1 are the only non-vanishing terms in the sums over l in (9.16).
Thus∞
X
h100|z|n10ihn10|z|100i
Ec = e2 E 2.
(9.17)
E1 − Enn=2

It is easy to understand physically why the change in E is proportional
to E 2 . In response to the external electric field, the probability density of theatom’s charge changes by an amount that is proportional to the coefficientsbk , and these coefficients are proportional to E. That is, the field polarises
the atom, generating a dipole moment P that is ∝ E. The dipole’s energy is−P · E, so the energy change caused by the field is proportional to E 2 .
9.1.2 Linear Stark effect and degenerate perturbation theory
Consider now the shift in the energy of the n = 2, l = 0 state of Hydrogen
when an electric field is applied. The sum over k in (9.16) now includes thetermh200|z|210ih210|z|200i,
E20 − E21
which is infinite if we neglect the very small Lamb shift (§8.1.3), because
the top is non-zero (Problem 8.12) and the difference of energies on the9.1 Time-independent perturbations

207

bottom vanishes. It hardly seems likely that a negligible field will produce
an arbitrarily large change in the energy of the first excited state of hydrogen.
So what did we do wrong?
Our error was to assume at the outset that a small stimulus produces asmall response as we did when we wrote equations (9.2). Our infinite contribution to Ec can be traced to our expression (9.9) for bm , which diverges
as Em → E0 . That is, the change in the wavefunction that a given field produces is inversely proportional to the energy difference between the originalstate |E0 i and the state |Em i we are pushing the system towards. This is anentirely reasonable result, analogous to what happens as we push a marble
that lies at the bottom of a bowl: the distance the marble moves before
coming into equilibrium depends on the curvature of the bowl. In the limitthat the curvature goes to zero, and the bottom of the bowl becomes flat, an
infinitesimal force will move the marble arbitrarily far, because all locations
have the same energy. So we conclude that when the system’s initial energy
is a degenerate eigenvalue Ed of H0 , a tiny stimulus is liable to produce a bigchange in the state (but not the energy) of the system. Disaster will attend
an attempt to calculate this abrupt change of state by the approach we havebeen developing.
So must we just give up in despair? No, because we can see that the
only states that are going to acquire a non-negligible amplitude during theabrupt change are ones that have the same energy as Ed . That is, the state
to which the system abruptly moves can be expressed as a linear combinationof the kets belonging to Ed . In many cases of interest there are only a small
number of these (four in the problem of hydrogen on which we are working).
What we have to do is to diagonalise the matrix ∆ij formed by ∆ squeezed
between all pairs of these kets. The eigenkets of ∆ in this small subspace
will be states of well-defined energy in the slightly perturbed system. As β
is ramped up from zero to unity their energies will diverge from Ed . We
conjecture that in the instant that β departs from zero, the system’s state
jumps to the eigenket with the lowest energy, and subsequently stays in thisstate as β increases. If this conjecture is correct, we should be able to use theperturbation theory we have developed provided we use as basis kets ones
that diagonalise ∆ as well as H0 .
So let’s diagonalise eEz in the 4-dimensional subspace of Hydrogen ketswith n = 2. When we list the kets in the order |200i, |210i, |211i, |21−1i,
the matrix of ∆ looks like this

0 a 0 0
 a∗ 0 0 0 ∆ij = eE 
(9.18)
 where a = h200|z|210i.
0 0 0 00 0 0 0From Problem 8.12 we have that h200|z|210i = −3a0 . It is now easy to showthat the eigenvalues of ∆ are ±3eEa0 and 0, while appropriate eigenkets are
2−1/2 (1, ∓1, 0, 0), (0, 0, 1, 0) and (0, 0, 0, 1). We conclude that as soon as theslightest perturbation is switched on, the system is in the state of lowest
energy, |ψi = 2−1/2 |200i + |210i , and we use this state to determine Eb .
We find
Eb = 21 eE h200| + h210| z |200i + |210i(9.19)= −3a0 eE.
From our discussion of the quadratic Stark effect, we know that a change
in E that is proportional to E requires the dipole moment P of an atom to be
independent of E. Since Eb is proportional to E we conclude that a hydrogen
atom in the n = 2 state has a permanent electric dipole.
In classical physics this result is to be expected because the orbit of theelectron would in general be elliptical, and the time-averaged charge density
along the ellipse would be higher at the apocentre than at the pericentre,2 because the electron lingers at the apocentre and rushes through the pericentre.
2 An orbit’s apocentre is the point furthest from the attracting body, while thepericentre is the point nearest that body.

208

Chapter 9: Perturbation theory

Figure 9.1 The charge distribu-√
tion of the state (|200i + |210i)/ 2
is axisymmetric. Here we plot the
distribution in the (R, z) plane ofcylindrical polar coordinates.

Hence the centre of charge would lie on the opposite side of the geometricalcentre of the ellipse from the focus, where the proton’s cancelling charge lies.
Thus, if the electron’s orbit were a perfect Kepler ellipse, the atom would
have a permanent electric dipole moment parallel to the orbit’s major axis.
Any deviation of the radial force field from F ∝ r−2 will cause the major
axis of the ellipse to precess, and therefore the time-averaged polarisation of
the atom to be zero. In hydrogen the force-field deviates verify little froman inverse-square law, so the precession occurs very slowly in the classicalpicture. Consequently, even a weak external field can prevent precession andthus give rise to a steady electric dipole.
In the quantum-mechanical picture, shielding shifts the energy of the S
state below that of the P states, thus ensuring that, in the absence of an
imposed field, the atom is spherical and has no dipole moment. An electricfield deprives L2 of its status as a constant of motion because the field canapply a torque to the atom. Shielding is a very weak effect in hydrogen(because it relies on the vacuum’s virtual electrons and positrons), so the
S state lies very little below the P states and in even a weak electric field
this offset√ becomes irrelevant. The lowest-energy state becomes (|200i +|210i)/ 2. This is not an eigenket of L2 but it is an eigenket of Lz witheigenvalue zero. Thus its angular momentum is perpendicular to the field,
as we expect from the classical picture of a Kepler ellipse with its major axis
parallel to E. Figure 9.1 shows that in this state the charge distribution
comprises a dense cloud around the origin and an extended cloud centred on
R = 0, z ≃ −3a0 . We can think of these clouds as arising from pericentre
and apocentre, respectively, of eccentric orbits that have
R their major axes
roughly aligned with the negative z axis. The integral d3 x, z|ψ|2 = −3a0 ,so in this state the atom has dipole moment P = +3ea0 .
9.1.3 Effect of an external magnetic field
When an atom is placed in a magnetic field, the wavelengths of lines in
its spectrum change slightly. Much of quantum mechanics emerged fromattempts to understand this phenomenon. We now use perturbation theory
to explain it.
In §3.3 we discussed the motion of a free particle in a uniform magnetic
field. Our starting point was the Hamiltonian (3.31), which governs the motion of a free particle of mass m and charge Q in the magnetic field produced

9.1 Time-independent perturbations

209

by the vector potential A. This is the Hamiltonian of a free particle, p2 /2m,with p replaced by p − QA. Hence we can incorporate the effects of a magnetic field on a hydrogen atom by replacing pn and pe in the gross-structure
Hamiltonian (8.1) with pp − eA and pe + eA, respectively. With Z = 1 thekinetic energy term in the Hamiltonian then becomes(pe + eA)2
(pp − eA)2
+
2mp
2me




2
2
pp
pe
pp
pp
pe
pe
1
=
·A+A·
+ O(A2 )++ 2e−−
2mp 2meme
mp
me
mp(9.20)
We neglect the terms that are O(A2 ) on the grounds that when the field isweak enough for the O(A) terms to be small compared to the terms in the
gross-structure Hamiltonian, the O(A2 ) terms are negligible.
Equation (8.4) and the corresponding equation for ∂/∂xp imply that
HKE ≡

pe =

me
pX + prme + mpand pp =mp
pX − pr ,me + mp(9.21)

where pX is the momentum associated with the centre of mass coordinate
X, while pr is the momentum of the reduced particle. From the algebra that
leads to equation (8.6a) we know that the first two terms on the right ofthe second line of equation (9.20) reduce to the kinetic energy of the centreof-mass motion and of the reduced particle. Using equations (9.21) in theremaining terms on the right of equation (9.20) yields
HKE =

p2X
p2
e
+ r +
(pr · A + A · pr ),2(me + mp ) 2µ 2µ(9.22)

where µ is the mass of the reduced particle (eq. 8.6b). It follows that an
external magnetic field adds to the gross-structure Hamiltonian of a hydrogen
atom a perturbing HamiltonianHB =e(pr · A + A · pr ).
2µ(9.23)On the scale of the atom the field is likely to be effectively homogeneous, so
we may take A = 12 B × r (page 49). Then HB becomes
HB =e(p · B × r + B × r · p),
4me(9.24)

where we have approximated µ by me and dropped the subscript on p.
The two terms in the bracket on the right can both be transformed into
B · r × p = h̄B · L because (i) these scalar triple products involve only
products of different components of the three vectors, and (ii) [xi , pj ] = 0
for i 6= j. Hence, we do not need to worry about the order of the r andp operators and can exploit the usual invariance of a scalar triple productunder cyclic interchange of its vectors.
If an atom has more than one unpaired (‘valence’) electron, each electron
will contribute a term of this form to the overall Hamiltonian. We can foldthese separate contributions into a single contribution HB by interpreting L
as the sum of the angular-momentum operators of the individual electrons.
In §8.2.1 we discussed terms that must be added to hydrogen’s grossstructure Hamiltonian to account for the effects of the electron’s intrinsic
dipole moment. We found that the coupling with an external field is generated by the Zeeman spin Hamiltonian (8.72). Adding this to the value of HB
that we have just computed, and orienting our coordinate system so that thez axis is parallel to B, we arrive at our final result, namely that a uniform
magnetic field introduces a perturbation
HBs =

eh̄
B(Lz + 2Sz ) = µB B(Jz + Sz ),2me(9.25)

210

Chapter 9: Perturbation theory

Figure 9.2 Eigenvalues of the spin-dependent Hamiltonian AL · S + B(Lz + 2Sz ) as
functions of B/A for the case l = 1, s = 21 . The right side of the diagram (field strong
compared to spin-orbit coupling) quantifies quantifies the Paschen–Back effect, while theleft side of the diagram quantifies the Zeeman effect (weak field). The top and bottomlines on the extreme right show the energies of the states |1, 1i|+i and |1, −1i|−i, which
are eigenstates of the full Hamiltonian for all values of B/A.

where S is the sum of the spin operators of all the valence electrons.
The Hamiltonian formed by adding HBs to the gross-structure Hamiltonian (8.1) commutes with L2 , Lz , S 2 and Sz . Its eigenkets are simply the
eigenkets of the gross-structure Hamiltonian upgraded to include eigenvaluesof S 2 and Sz . The only difference from the situation we studied in §8.1 isthat the energies of these eigenkets now depend on both Lz and Sz . Hence,each energy level of the gross-structure Hamiltonian is split by the magnetic
field into as many sub-levels as ml + 2ms can take. For example, if l = 0
and s = 12 , there are two sublevels, while when l = 1 and s = 21 there are
five levels in which ml + 2ms ranges between ±2.
In practice the perturbation HBs always acts in conjunction with thespin-orbit perturbation HSO of equation (8.73).3 The general case in which
HBs and HSO are comparable, requires numerical solution. The extreme
cases in which one operator is larger than the other can be handled analytically.
Paschen–Back effect In a sufficiently strong magnetic field, HSO affects
the atom much less than HBs , so HSO simply perturbs the eigenkets of the
Hamiltonian formed adding HBs to the gross-structure Hamiltonian. Thechange in the energy of the state |n, l, ml , s, ms i is
Eb = hn, l, ml , s, ms |HSO |n, l, ml , s, ms i
= ζhn, l, ml , s, ms |L · S|n, l, ml , s, ms i,(9.26)

where ζ is a number with dimensions of energy that is independent of mland ms . By writing L · S = 21 (L+ S− + L− S+ ) + Lz Sz (eq. 7.143) we see thathL · Si = ml ms . So in a strong magnetic field the eigenenergies are
Egross + µB B(ml + 2ms ) + ζml ms .

(9.27)The levels on the extreme right of Figure 9.2 show the energies described bythis formula in the case that l = 1 and s = 21 . The fact that in a strong
magnetic field an atom’s energies depend on ml and ms in this way is known
as the Paschen–Back effect.
Zeeman effect In a sufficiently weak magnetic field, HSO affects the atom
more strongly than HBs . Then spin-orbit coupling assigns different energies
3 There is no spin-orbit coupling for an S state, but an allowed spectral line from an S
state will connect to a P state for which there is spin-orbit coupling. Hence the frequencies
of allowed transitions inevitably involve spin-orbit coupling.

9.1 Time-independent perturbations

211

to states that differ in j. Consequently, when we use perturbation theory
to calculate the smaller effect of an imposed magnetic field, the degenerate
eigenspace in which we have to work is that spanned by the states that havegiven values of j, l and s but differ in their eigenvalues m of Jz . Fortunately,
HBs is already diagonal within this space because [Jz , Sz ] = 0. So the shift
in the energy of each state is simplyEb = hj, m, l, s|HBs |j, m, l, si = µB m + hj, m, l, s|Sz |j, m, l, si .

(9.28)

As we saw in §7.5, our basis states do not have well-defined values of Sz– in general they are linear combinations of eigenstates of Lz and Sz :|j, m, l, si =sXm′ =−scm′ |l, m − m′ i|s, m′ i,(9.29)

where the coefficients cm′ are Clebsch–Gordan coefficients (eq. 7.152). In any
concrete case it is straightforward to calculate the required expectation valueof Sz from this expansion. However, a different approach yields a generalformula that was important historically.
In the classical picture, spin-orbit coupling causes the vector S to precess
around the invariant vector J. Hence, in this picture the expectation value of
S is equal to the projection of S onto J.4 The classical vector triple productformula enables us to express S in terms of this projection:
J × (S × J) = J 2 S − (S · J)J

so J 2 S = (S · J)J + J × (S × L). (9.30)

In the classical picture, the expectation value of the vector triple producton the right side vanishes. If its quantum expectation value were to vanish,
the expectation value of the z component of the equation would relate hSz i,
which we require, to the expectation values of operators that have the states|j, m, l, si as eigenstates, so our problem would be solved. Motivated by theseclassical considerations, let’s investigate the operator
G ≡ J × (S × L)

so Gi ≡

Xǫijk Jj ǫklm Sl Lm .

(9.31)

jklm

It is straightforward to check that its components commute with the angularmomentum operators Ji in the way we expect the components of a vector todo:X[Ji , Gj ] = iǫijk Gk .
(9.32)k

From equation (9.31) it is also evident that J · G = 0. In Problem 7.21
identical conditions on the operators L and x suffice to prove that hxi = 0 in
any state that is an eigenket of L2 . So the steps of that proof can be retraced
with L replaced by J and x replaced by G to show that for the states ofinterest hGi = 0.
Now that we have established that the quantum-mechanical expectation
value of G does indeed vanish, we reinterpret equation (9.30) as an operator
equation, and, from the expectation value of its z component, deduce
hj, m, l, s|Sz |j, m, l, si =hJ · Si m.
j(j + 1)

(9.33)

From equation (8.74) we have
J · S = L · S + S 2 = 21 (J 2 − L2 + S 2 ),4 This heuristic argument is often referred to as the vector model.

(9.34)

212

Chapter 9: Perturbation theory

so we findj(j + 1) − l(l + 1) + s(s + 1)
EB = mgL µB B where gL ≡ 1 +. (9.35)
2j(j + 1)
The factor gL is called the Landé g factor. In the early days of quantum
theory, when the Bohr atom was taken seriously, people expected the magnetic moment of an electron to be ±µB and therefore thought a magnetic
field would shift energy levels by ±µB B. Equation (9.35) states that theactual shift is mgL times this. When this factor differed from unity, theyspoke of an anomalous Zeeman effect.
The left hand side of Figure 9.2 shows the energy levels described byequation (9.35) in the case l = 1, s = 12 . The possible values of j are 32 and1
2 , and the magnetic field splits each of these spin-orbit levels into 2j + 1
components.

9.2 Variational principle
We now describe a method of estimating energy levels, especially a system’s
ground-state energy, that does not involve breaking the Hamiltonian down
into a part that has known eigenkets and an additional perturbation. In
Chapter 10 we shall show that this method yields quite an accurate value
for the ionisation energy of helium.
Let H be the Hamiltonian for which we require the eigenvalues En andtheassociatedeigenkets |ni. We imagine expanding an arbitrary state |ψi =
P
a
|ni
as
a
linear combination of these eigenkets, and then calculate thennexpectation value of H in this state asP|ai |2 Ei
hHi = hψ|H|ψi = Pi,2
j |aj |

(9.36)

where we have included the sum of the |aj |2 on the bottom to cover the
possibility that |ψi is not properly normalised. hHi is manifestly independentof the phase of ai . We investigate the stationary points of hHi with respect
to the moduli |ai | by differentiating equation (9.36) with respect to them:P
2|ak | i |ai |2 Ei∂ hHi
2|ak |Ek
−
= P
P
2 .
2
∂|ak |
2
j |aj ||a|jj

(9.37)

Equating this derivative to zero, we find that the conditions for a stationary
point of hHi are
0 = |ak | Ek

Xi2|ai | −

Xi

2

|ai | Ei!

(k = 0, 1, . . .)(9.38)

These equations are trivially solved by setting ak = 0 for every k, but then|ψi = 0 so the solution is of no interest. For any value of k for which ak 6= 0,
we must haveP2
i |ai | Ei
Ek = P.
(9.39)
2
i |ai |

Since the right side of this equation does not depend on k, the equation canbe satisfied for at most one value of k, and it clearly is satisfied if we set ai = 0
for i 6= k and ak = 1, so |ψi = |ki. This completes the proof of Rayleigh’s
theorem: The stationary points of the expectation value of an Hermitian
operator occur at the eigenstates of that operator. Moreover, all eigenstates
provide stationary points of the operator. That is, for general |ψi the number

9.3 Time-dependent perturbation theory

213of hψ|H|ψi isn’t equal to an eigenvalue of H, but if hψ|H|ψi is stationary
with respect to |ψi in the sense that it doesn’t change when |ψi is changed
by a small amount, Rayleigh’s theorem tells us that the number hψ|H|ψi
is an eigenvalue of H. Problem 9.10 gives a geometrical interpretation ofRayleigh’s theorem.
The stationary point associated with the ground state is a minimum of
hHi. To see that this is so, we subtract the ground-state energy E0 fromboth sides of equation (9.36) and haveP|ai |2 (Ei − E0 ).
(9.40)
hHi − E0 = i P
2
j |aj |
Both the top and bottom of the fraction on the right are non-negative, so
hHi ≥ E0 . The stationary points of hHi associated with excited states are
saddle points (Problem 9.14).
The practical use of Rayleigh’s theorem is this. We write down a trial
wavefunction ψa (x) that depends on a number of parameters a1 , . . . , aN .
These might, for example, be the coefficients in an expansion of ψa as alinear combination of some convenient basis functions ui (x)ψa (x) ≡NX

ai ui (x).

(9.41)

i=1

More often the ai are parameters in a functional form that is motivatedby some physical argument. For example, in Chapter 10 we will treat thevariable Z that appears in the hydrogenic wavefunctions of §8.1.2 as one of
the ai . Then we use ψa to calculate hHi as a function of the ai and findthe stationary points of this function. The minimum value of hHi that we
obtain in this way clearly provides an upper limit on the ground-state energy
E0 . Moreover, since hHi is stationary for the ground-state wavefunction,hHi − E0 increases only quadratically in the difference between ψa and theground-state wavefunction ψ0 . Hence, with even a mediocre fit to ψ0 thisupper limit will lie close to E0 . This approach to finding eigenvalues andeigenfunctions of the Hamiltonian is called the variational principle.
In Problems 9.12 and 9.13 you can explore how the variational principle
works in a simple case.

9.3 Time-dependent perturbation theory
We now describe a way of obtaining approximate solutions to the tdse (2.26)
that we shall use to study both scattering of particles and the emission andabsorption of radiation by atoms and molecules.
Consider the evolution of a system that is initially in a state that isnearly, but not quite, in a stationary state. Specifically, at t = 0 it is in the
N th eigenstate of a Hamiltonian H0 that differs by only a small, possibly
time-dependent, operator V from the true Hamiltonian H:
H = H0 + V.

(9.42)

Inspired by (2.32) we expand the solution to the tdse for this H in the formX|ψi =an (t)e−iEn t/h̄ |En i,(9.43)nwhere |En i is a (time-independent) eigenket of H0 with eigenvalue En . This
expansion doesn’t restrict |ψi because {|En i} is a complete set and the functions an (t) are arbitrary. Substituting it into the tdse we have

X
∂|ψi
ih̄
En |En i + V |En i an e−iEn t/h̄= (H0 + V )|ψi =∂tn(9.44)
X
ih̄ȧn + En an e−iEn t/h̄ |En i.
=
n

214

Chapter 9: Perturbation theory

We simplify this by multiplying through by hEk |:ih̄ȧk e−iEk t/h̄ =Xnan e−iEn t/h̄ hEk |V |En i.

(9.45)

This constitutes a set of linear ordinary differential equations for the an (t)which must be solved subject to the boundary conditions aN (0) = 1 andan (0) = 0 for n 6= N . Hence, at the earliest times the term on the right of(9.45) with n = N will dominate the equation of motion of ak with k 6= N ,and we have the approximationiȧk ≃ − e−i(EN −Ek )t/h̄ hEk |V |EN i.
h̄(9.46)

We now assume that any time dependence of V takes the form V (t) = V0 eiωt ,where V0 is a time-independent operator. This assumption is in practice notvery restrictive because the theory of Fourier analysis enables us to express
any operator of the form V0 f (t), where f is an arbitrary function, as a linear
combination of sinusoidally varying operators. Replacing V by V0 eiωt inequation (9.46) and integrating from t = 0 we findak (t) =hEk |V0 |EN i h −i(EN −Ek −h̄ω)t′ /h̄ ite,EN − Ek − h̄ω0(9.47)

so the probability that after t the system has made the transition to the kth
eigenstate of H0 is
Pk (t) = |ak |2|hEk |V0 |EN i|2=(EN − Ek − h̄ω)2= 4|hEk |V0 |EN i|2(EN − Ek − h̄ω)t2 − 2 cosh̄

(9.48)

2sin ((EN − Ek − h̄ω)t/2h̄).
(EN − Ek − h̄ω)2For a time of order h̄/(EN − Ek − h̄ω) this expression grows like t2 . Subsequently it oscillates.
9.3.1 Fermi golden rule
In many applications of equation (9.48), there are a large number of stationary states of H0 with energies Ek that lie within h̄/t ofEout ≡ EN − h̄ω,(9.49)

and we are interested in the probability that the system has made the transition to any one of these states. Hence we sum the Pk over k. Let therebe g(Ek ) dEk eigenvalues in the interval (Ek + dEk , Ek ). Then the totaltransition probability isXkPk (t) = 4

Z

2
=
h̄

Z

dEk g(Ek ) |hEk |V |EN i|2

sin2 ((Eout − Ek )t/2h̄)(Eout − Ek )2

sin2 (xt),dx g(Eout − 2h̄x) |hEout − 2h̄x|V |EN i|2x2(9.50)

where we’ve introduced a new variable, x = (Eout − Ek )/2h̄. For given t,
the function ft (x) ≡ sin2 (xt)/x2 is dominated by a bump around the originthat is of height t2 and width 2π/t. Hence, the area under the bump is
proportional to t and in the limit of large t,
sin2 (xt)∝ tδ(x).
x2(9.51)

9.3 Time-dependent perturbation theory

215

R
We find the constant of proportionality by differentiating dx ft with respect
to t:
Z ∞
Z
d ∞
sin(2xt)
sin2 (xt)dx== π,(9.52)
dx
2
dt −∞
x
x
−∞
R
where we have used a result, dx sin x/x = π, from the theory of contour
integration in the complex plane. Hencesin2 (xt)= πtδ(x).
t→∞
x2
lim

(9.53)

Inserting this relation in (9.50) and integrating over x, we have finally
X
k

Pk =

2πt
g(Eout ) |hout|V |ini|2 .
h̄(9.54)

This equation establishes Fermi’s golden rule5 of perturbation theory: aperturbation V eiωt causes a system to transition to a new state lower inenergy by h̄ω at a rate equal to 2π/h̄ times the mod-square of the matrix
element of V between the initial and final states times the density of statesat the final energy. It is easy to see that if the time-dependence of theperturbation were e−iωt , it would cause transitions at the same rate to stateshigher in energy by h̄ω.
9.3.2 Radiative transition rates
We now use equation (9.48) to calculate the rate at which a electromagnetic
waves induce an atom to make radiative transitions between discrete stationary states. Our treatment is valid when the quantum uncertainty in theelectromagnetic field may be neglected, and the field treated as a classicalobject. This condition is satisfied, for example, in a laser, or at the focus of
the antenna of a radio telescope.
Whereas in our derivation of Fermi’s golden rule, we took the frequency
ω of the perturbation to be fixed and assumed a continuum of final states,
now that we are considering the case of a discrete final state, we argue
that the electromagnetic field is a superposition of plane waves of various
frequencies, and that we should sum the transition probability (9.48) thateach wave independently contributes. Thus we write
X

waves

Pk (t) = 4

X

waves

|hEk |δV0 |EN i|2

sin2 ((EN − Ek − h̄ω)t/2h̄),(EN − Ek − h̄ω)2(9.55)

where δV0 and ω relate to an individual wave.
In vacuo the electric field of an electromagnetic wave is divergence free,being entirely generated by Faraday’s law, ∇ × E = −∂B/∂t. It follows thatthe whole electromagnetic field of the wave can be described by the vector
potential A through the equations
B = ∇ × A and E = −∂A/∂t.

(9.56)We are considering a superposition of plane waves, which individually havea potential
δA(x, t) = δA0 cos(k · x − ωt),
(9.57)
where δA0 is a constant vector and k is the wavevector. From equations(9.56) and (9.57) we have that the wave’s contribution to the electric field isδE(x, t) = −ωδA0 sin(k · x − ωt),(9.58)

5 The golden rule was actually first given by P.A.M. Dirac, Proc. Roy. Soc. A, 114,
243 (1927)

216

Chapter 9: Perturbation theory

so δE is parallel to δA0 . From ∇ · δE = 0 it follows thatk · δA0 = 0,(9.59)

so k · δE = 0 and the wave is transverse.
In §9.1.3 we saw that an external electromagnetic field adds to an atom’s
Hamiltonian the perturbing term (9.23) for each electron. In the present casethe perturbation ise{p · δA0 cos(k · x − ωt) + cos(k · x − ωt)δA0 · p}.
2meδV (x, t) =(9.60)

By virtue of equation (9.59), δA0 ·p commutes with k·x because a componentof momentum always commutes with a perpendicular component of position.
Since δA0 is a constant, it commutes with p. So we can simplify δV toe
δA0 · p cos(k · x − ωt)
me


e
=
δA0 · p ei(k·x−ωt) + e−i(k·x−ωt) .
2me

δV (x, t) =(9.61)

We now make the approximation that the electromagnetic wavelengthis much bigger than the characteristic size of the atom or molecule. This isa good approximation providing the atom or molecule moves between statesthat are separated in energy by much less than αme c2 (Problem 9.20), as
will be the case for waves with frequencies that are less than those of soft
X-rays. In this case we will have k · x ≪ 1 for all locations x in the atom
or molecule at which there is significant probability of finding an electron.
When this condition is satisfied, it makes sense to expand the factors e±ik·x
in equation (9.61) as a power series and discard all but the constant term.
We then haveeδV (x, t) =
δA0 · p e−iωt + eiωt ,(9.62)
2me
where we have retained the exponentials in time because large values oft cannot be excluded in the way that we can exclude large values of x.
Finally, we note that in the gross-structure Hamiltonian H0 , p occurs onlyin the term p2 /2me , so [H0 , x] = −i(h̄/me )p. When we use this relation toeliminate p from equation (9.62), we haveδV (x, t) = ieδA0[H0 , z](e−iωt + eiωt ),2h̄

(9.63)

where we have chosen to make the z axis parallel to A0 . Thus a plane electromagnetic wave gives rise to perturbations with both positive and negative
frequencies. Above we derived the frequency condition ω = (EN − Ek )/h̄for transitions from |EN i to |Ek i, so the negative frequency perturbation
is associated with excitation of the system (Ek > EN ), while the positivefrequency perturbation is associated with radiative decays.
We identify the time-independent part of δV as the operator δV0 that
occurs in equation (9.55) and then have that the net transition probability
is
X

e2 X
sin2 ((EN − Ek − h̄ω)t/2h̄)(δA0 )2 |hEk |[H0 , z]|EN i|22(EN − Ek − h̄ω)2h̄ wavesX
sin2 (xt)e2,(δA0 )2= 4 (Ek − EN )2 |hEk |z|EN i|2
x2
4h̄
waves
(9.64a)Pk (t) =

waves

where
x ≡ (EN − Ek − h̄ω)/2h̄.

(9.64b)

9.3 Time-dependent perturbation theory

217

Even though the expressionρ=1{(E/c)2 + B 2 }2µ0(9.65)

for the energy density of an electromagnetic field is quadratic in the field
amplitudes E and B, the volume-averaged energy density of a superpositionof plane waves is just the sum of the energy densities of each individual
wave. Moreover, the electric and magnetic energy densities of a plane wave
are equal, so the energy density contributed by our plane wave is just twice
its electric energy density, and from equations (9.58) and (9.65) we infer that
our wave contributes the time-averaged energy densityδρ =ω 2 (δA0 )2= 12 ω 2 ǫ0 (δA0 )2 ,2µ0 c2(9.66)

where the second equality uses µ0 c2 = 1/ǫ0 . Using this expression to eliminate δA0 from equation (9.64a), we obtainX

Pk (t) =waves

X
sin2 (xt)
e2
2
2
(E
−
E
)
|hE
|z|E
i|
δρ
.
k
N
k
N
ω 2 x22ǫ0 h̄4waves(9.67)

Let ρ(ω) be the power contained in all waves that have frequencies less thanω. In symbols
X
ρ(ω) ≡
δρω′ .
(9.68)
waves with ω ′ <ω

Then

X

waves

δρ =

Z

dω

dρ
.
dω

(9.69)

When we use this expression to replace the sum on the right side of equation(9.67) by an integral, and we use equation (9.64b) to replace dω with −2dx,
we obtainZXdρ sin2 (xt)
e2
2
2
(E
−
E
)
|hE
|z|E
i|
dx
.
(9.70)Pk (t) =kNkNdω ω 2 x2
ǫ0 h̄4waves
We now let t become large and exploit equation (9.53) to evaluate the integral. The result isX

Pk (t) = t

waves

πe2
2 dρ
.
2 |hEk |z|EN i| dωǫ0 h̄
ω=(EN −Ek )/h̄(9.71)

The coefficient of t on the right of this equation gives the rate R at whichtransitions occur. When we express the cumulative energy density of thewave-field in terms of frequency ν rather than angular frequency ω and use
equation to eliminate ǫ0 in favour of the Bohr radius, the rate becomes
R=

2π
dρ
|hEk |z|EN i|2.
a 0 medν ν=|EN −Ek |/h(9.72)

When Ek > EN , the negative-frequency term in equation (9.62) gives
rise to excitations at an identical rate. Thus we have recovered from a dynamical argument Einstein’s famous result that stimulated emission of photons
occurs, and that the coefficient B that controls the rate of stimulated emission is equal to the absorption coefficient (Box 9.2). Einstein’s predictionof stimulated emission led 38 years later to the demonstration of a maser(§5.2.1) and 44 years later to the construction of the first laser by Theodore

218

Chapter 9: Perturbation theory

Box 9.2: Einstein A and B Coefficients
In 1916, when only the merest fragments of quantum physics were known,
Einstein showed (Verh. Deutsch. Phys. Ges. 18, 318) that systems must
be capable of both spontaneous and stimulated emission of photons, andthat the coefficient of stimulated emission must equal that for absorptionof a photon. He obtained these results by requiring that in thermal
equilibrium there are equal rates of absorption and emission of photons
of a given frequency ν by an ensemble of systems. He considered afrequency ν for which hν = ∆E, the energy difference between two states|1i and |2i of the systems. The rate of absorptions he assumed to be
Nabs = Ba N1 (dρ/dν), where Ba is the absorption coefficient, N1 is thenumber of systems in the state |1i, and (dρ/dν) is the energy density
in radiation of frequency ν. The rate of emissions he assumed to be
Nem = Be N2 (dρ/dν) + AN2 , where Be is the coefficient for induced
emission and A is that for spontaneous emission. Equating Nabs to Nem
yieldsdρ
0 = (Be N2 − Ba N1 )+ AN2 .
dνIn thermal equilibrium N1 = N2 ehν/kT and dρ/dν is given by the Planck
function. Using these relations to eliminate N1 and dρ/dν and thencancelling N2 , we find8πhν 30 = (Be − Ba ehν/kT ) 3 hν/kT
− A.
c (e
− 1)

In the limit of very large T , ehν/kT → 1, so the factor multiplying thebracket with the Bs becomes large, and the contents of this bracket tends
to Be − Ba . It follows that these coefficients must be equal. We therefore
drop the subscripts on them, take B out of the bracket, cancel the factors
with exponentials, and finally deduce thatA = 8πh(ν/c)3 B.

(1)Maiman.6 In view of this history, it’s a remarkable fact that a laser operates
in the regime in which the electromagnetic field can be treated as a classicalobject, as we have done here. Emission of light by a humble candle, by contrast, is an inherently quantum-mechanical phenomenon because it occurs
through spontaneous emission. Our treatment does not include spontaneousemission because we have neglected the quantum uncertainty in the electromagnetic field. This uncertainty endows the field with zero-point energy(§3.1), and spontaneous emission can be thought of as emission stimulatedby the zero-point energy of the electromagnetic field.
Using the argument given in Box 9.2, Einstein was able to relate thecoefficient A of spontaneous emission to B. Einstein’s argument does notyield a numerical value for either A or B. Our quantum mechanical treatment
has yielded a value for B, and with Einstein’s relation (eq. 1 in Box 9.2)
between B and A we can infer the value of A:A=

16π 2 hν 3
|hEk |z|EN i|2 .
c3 a 0 me(9.73)

From this we can estimate the typical lifetime for radiative decay from an
excited state of an atom.
When the radiation density ρ is very small, the number N2 of atoms inan excited state obeys Ṅ2 = −AN2 (Box 9.2), so N2 decays exponentially
6 The word ‘laser’ is an acronym for “light amplification by stimulated emission”.
Curiously Maiman’s paper (Nature, 187, 493 (1960)) about his laser was rejected by the
Physical Review.

9.3 Time-dependent perturbation theory219

with a characteristic time A−1 . Unless some symmetry condition causes
the matrix element in equation (9.73) to vanish, we expect the value of thematrix element to be ∼ a0 . So the characteristic radiative lifetime of a stateisme c2 λ 1.
(9.74)
τ = A−1 =hν a0 16π 2 ν
For an optical transition, hν ∼ 2 eV, λ ∼ 650 nm ∼ 1.2 × 104 a0 , and ν ∼4.6 × 1014 Hz, so τ ∼ 4 × 10−8 s. It follows that ∼ 107 oscillations of theatom occur before the radiation of energy causes the atom to slump into thelower state.
9.3.3 Selection rules
Equation (9.72) states that the rate of radiative transitions is proportional
to the mod-square of the electric dipole operator ez. For this reason theapproximation we made, that k · x ≪ 1, is called the electric dipole approximation.
There are important circumstances in which symmetry causes the matrix
element of the dipole operator to vanish between the initial and final states.
Transitions between such states are said to be forbidden in contrast to allowed transitions, for which the matrix element does not vanish. Someapproximations were involved in our derivation of equation (9.72), so thetransition rate does not necessarily vanish completely when the matrix element is zero. In fact, forbidden transitions often do occur, but at rates thatare much smaller than the characteristic rate of allowed transitions (eq. 9.74)
because the rate of a forbidden transition is proportional to terms that we
could neglect in the derivation of equation (9.72). We now investigate relations between the initial and final states that must be satisfied if the states
are to be connected by an allowed transition. Such relations are called selection rules. The slower rate of forbidden transitions must be determinedby either including the next term of the Taylor expansion of eik·x , or taking
into account the perturbation µB S · B that arises from the interaction of the
intrinsic magnetic moment of an electron with the wave’s magnetic field.
We are interested in matrix elements between states that are eigenstatesof operators that commute with the Hamiltonian H that the atom would
have if it were decoupled from electromagnetic waves. The Hamiltonian
should include spin-orbit coupling as well as interaction with whatever steadyexternal electric or magnetic fields are being applied. The operator in thematrix element is the component of the position operator parallel to theelectric field of the radiation that is being either absorbed or emitted.
Even in the presence of an external field, the angular-momentum parallelto the field, which we may call Jz , commutes with H, so the kets of interest
are labelled with m. Since [Jz , z] = 0, the ket z|E, mi is an eigenket of Jz
with eigenvalue m. It follows that hE, m|z|E ′ , m′ i = 0 unless m = m′ . This
gives us the first selection rule listed in Table 9.1, namely that when theelectric vector of the radiation is parallel to the imposed field, the quantum
number m is unchanged by radiation.
If we define x± = x ± iy, we have
[Jz , x± ] = iy ± i(−ix) = ±x± .

(9.75)

It follows that x± |E, mi is an eigenket of Jz with eigenvalue m ± 1, so
hE, m|x|E ′ , m′ i = 12 hE, m|(x+ + x− )|E ′ , m′ i
= 0 unless m′ = m ± 1.

(9.76)

Obviously the same result applies to the matrix element for y. Hence wehave the second selection rule listed in Table 9.1: when the electric vector of
the radiation is perpendicular to the imposed field, the quantum number mchanges by ±1. If the direction of observation is along the imposed field, the220Problems
Table 9.1 Selection rulesjmls|j − j ′ | ≤ 1 but j = 0 6→ j ′ = 0|m − m′ | ≤ 1; m = m′ for E parallel to an external field;|m − m′ | = 1 for photon emitted parallel to an external field|l − l′ | = 1
s = s′

electric vector of the radiation must be perpendicular to the field. Hence,in this case m must change by ±1. In fact, m increases when a left-handcircularly-polarised photon is emitted in the positive z direction, and conversely for the emission of a right-hand polarised photon. When the directionof observation is perpendicular to the imposed field, the electric vector of theradiation can be either perpendicular to the field, in which case m changes
by ±1, or parallel to the field, and then m does not change.
When there is no imposed field, m may be unchanged or change by ±1,
and we can observe photons associated with any of these changes in m whenobserving from any direction.
When there is no imposed field, J 2 commutes with H and the ketsof interest are labelled with E, j and m. The selection rule for j can beobtained from the rules for adding angular momentum that were discussed
in §7.5: hE, j, m|xk |E ′ , j ′ , m′ i vanishes unless it is possible to make a spin-j
object by adding a spin-one object to a spin-j ′ object. For example, thematrix element vanishes if j = j ′ = 0 because spin-one is all you can get byadding a spin-one system to a spin-zero one. Subject to the selection rules onm just discussed, the matrix element does not vanish if j = 0 and j ′ = 1, or if
j = 1 and j ′ = 1, because both a spin-zero system and a spin-one one can beobtained by adding two spin-one subsystems. The matrix element vanishes
if j = 1 and j ′ = 3 because by adding spin-one and spin-three the smallest
spin you can get is spin-two. In summary, the selection rule is |j − j ′ | ≤ 1
except that j = 0 → j ′ = 0 is forbidden.
The selection rules for j that we have just given follow from a powerfulresult of group theory, the Wigner-Eckart theorem. Unfortunately, asignificant amount of group theory is required to prove this theorem. In
Appendix J we give a proof of the selection rule for j that builds on thecalculation involved in Problem 7.21.
When spin-orbit coupling is weak, the total orbital angular momentum
L2 and the total spin angular momentum S 2 are constants of motion, so
their quantum numbers l and s are likely to appear as labels in the kets.
Since [x, S] = 0, it is clear that the selection rule for s is that it should notchange. The selection rule for l was derived in Problem 7.21: |l − l′ | = 1.

Problems
9.1 A harmonic oscillator with mass m and angular frequency ω is perturbed by δH = ǫx2 . (a) What is the exact change in the ground-stateenergy? Expand this change in powers of ǫ up to order ǫ2 . (b) Show thatthe change given by first-order perturbation theory agrees with the exact
result to O(ǫ)√ (c) Show that the first-order change in the ground state is|bi = −(ǫℓ2 / 2h̄ω)|E2 i. (d) Show that second-order perturbation theory
yields energy change Ec = −ǫ2 h̄/4m2 ω 3 in agreement with the exact result.
9.2 The harmonic oscillator of Problem 9.1 is perturbed by δH = ǫx. Show
that the perturbed Hamiltonian can be written
1
H=
2m



ǫ2
p +m ω X − 2
ω
2

2

2

2



,

where X = x+ǫ/mω 2 and hence deduce the exact change in the ground-stateenergy. Interpret these results physically.

Problems

221

What value does first-order perturbation theory give? From perturbation theory determine the coefficient b1 of the unperturbed first-excited statein the perturbed ground state. Discuss your result in relation to the exact
ground state of the perturbed oscillator.
9.3 The harmonic oscillator of Problem 9.1 is perturbed by δH = ǫx4 .
Show that the first-order change in the energy of the nth excited state is2h̄δE = 3(2n2 + 2n + 1)ǫ
.
(9.77)
2mω
Hint: express x in terms of A + A† .
9.4 The infinite square-well potential V (x) = 0 for |x| < a and ∞ for|x| > a is perturbed by the potential δV = ǫx/a. Show that to first order inǫ the energy levels of a particle of mass m are unchanged. Show that evento this order the ground-state wavefunction is changed to
X
16ǫ
1
n
ψ1 (x) = √ cos(πx/2a) + 2 √sin(nπx/2a),(−1)n/2 2a
π E1 a n=2,4,(n − 1)3where E1 is the ground-state energy. Explain physically why this wavefunction does not have well-defined parity but predicts that the particle is more
likely to be found on one side of the origin than the other. State with reasons but without further calculation whether the second-order change in the
ground-state energy will be positive or negative.
9.5 An atomic nucleus has a finite size, and inside it the electrostatic
potential Φ(r) deviates from Ze/(4πǫr). Take the proton’s radius to beap ≃ 10−15 m and its charge density to be uniform. Then treating the difference between Φ and Ze/(4πǫ0 r) to be a perturbation on the Hamiltonianof hydrogen, calculate the first-order change in the ground-state energy ofhydrogen. Why is the change in the energy of any P state extremely small?
Comment on how the magnitude of this energy shift varies with Z in hydrogenic ions of charge Z. Hint: exploit the large difference between ap and a0
to approximate the integral you formally require.
9.6 Evaluate the Landé g factor for the case l = 1, s = 21 and relate yourresult to Figure 9.2.
9.7 A particle of mass m moves in the potential V (x, y) = 21 mω 2 (x2 + y 2 ),where ω is a constant. Show that the Hamiltonian can be written as thesum Hx + Hy of the Hamiltonians of two identical one-dimensional harmonic
oscillators. Write down the particle’s energy spectrum. Write down ketsfor two stationary states in the first-excited level in terms of the stationary
states |nx i of Hx and |ny i of Hy . Show that the nth excited level is n + 1
fold degenerate.
The oscillator is disturbed by a small potential H1 = λxy. Show that
this perturbation lifts the degeneracy of the first excited level, producing
states with energies 2h̄ω ± λh̄/2mω. Give expressions for the correspondingkets.
The mirror operator M is defined such that for any state |ψi, hx, y|M |ψi =hy, x|ψi. Explain physically the relationship between the states |ψi and
M |ψi. Show that [M, H1 ] = 0. Show that M Hx = Hy M and thus that
[M, H] = 0. What do you infer from these commutation relations?
9.8∗

The Hamiltonian of a two-state system can be written
A1 + B1 ǫ B2 ǫ
H=
,
B2 ǫ
A2

(9.78)

where all quantities are real and ǫ is a small parameter. To first order in ǫ,
what are the allowed energies in the cases (a) A1 6= A2 , and (b) A1 = A2 ?
Obtain the exact eigenvalues and recover the results of perturbation
theory by expanding in powers of ǫ.

222

Problems

Figure 9.3 The relation of input
and output vectors of a 2 × 2 Hermitian matrix with positive eigenvalues
λ1 > λ2 . An input vector (X, Y ) on
the unit circle produces the output
vector (x, y) that lies on the ellipsethat has the eigenvalues as semiaxes.

9.9∗ For the P states of hydrogen, obtain the shift in energy caused by aweak magnetic field (a) by evaluating the Landé g factor, and (b) by use
equation (9.28) and the Clebsch–Gordan coefficients calculated in §7.5.2.
9.10 The 2 × 2 Hermitian matrix H has positive eigenvalues λ1 > λ2 . The
vectors (X, Y ) and (x, y) are related by
H·



X
Y



 
x
=
.
y

Show that the points (λ1 X, λ2 Y ) and (x, y) are related as shown in Figure 9.3.
How does this result generalise to 3 × 3 matrices? Explain the relation ofRayleigh’s theorem to this result.
9.11 We find an upper limit on the ground-state energy of the harmonic
oscillator from the trial wavefunction ψ(x) = (a2 + x2 )−1 . Using the substitution x = a tan θ, or otherwise, show that
Z ∞
0

Z ∞

dx |ψ|2 = 41 πa−30Z ∞dx x2 |ψ|2 = 41 πa−10dx ψ ∗ p2 ψ = 81 πa−5(9.79)
Hence show that hψ|H|ψi/hψ|ψi is minimised by setting a = 2l/4 ℓ, where ℓ isthe characteristic
length of the oscillator. Show that our upper limit on E0√
is h̄ω/ 2. Plot the the final trial wavefunction and the actual ground-state
wavefunction and (a) say whether you consider it a good fit, and (b) how it
might be adapted into a better trial wavefunction.
9.12 Show that for the unnormalised spherically-symmetric wavefunction
ψ(r) the expectation value of the gross-structure Hamiltonian of hydrogen is
hHi =

h̄2
2me

Z

dr r

2

dψ
dr

2

e2
−
4πǫ0

Z

dr r|ψ|2!,Z

dr r2 |ψ|2 .

(9.80)For the trial wavefunction ψb = e−br show that
hHi =

e2 b
h̄2 b2
−
,
2me
4πǫ0

and hence recover the definitions of the Bohr radius and the Rydberg constant.
9.13∗ Using the result proved in Problem 9.12, show that the trial wave2 2function ψb = e−b r /2 yields −8/(3π)R as an estimate of hydrogen’s groundstate energy, where R is the Rydberg constant.
9.14 Show that the stationary point of hψ|H|ψi associated with an excitedstate of H is a saddle point. Hint: consider the state |ψi = cos θ|ki + sin θ|li,
where θ is a parameter.

Problems

223

9.15 At early times (t ∼ −∞) a harmonic oscillator of mass m and natural
2
2angular frequency ω is in its ground state. A perturbation δH = Exe−t /τ
is then applied, where E and τ are constants.
a. What is the probability according to first-order theory that by late times
the oscillator transitions to its second excited state, |2i?
b. Show that to first order in δH the probability that the oscillator transitions to the first excited state, |1i, isP =πE 2 τ 2 −ω2 τ 2 /2
e
,
2mh̄ω

(9.81)

c. Plot P as a function of τ and comment on its behaviour as ωτ → 0 andωτ → ∞.
9.16 A particle of mass m executes simple harmonic motion at angularfrequency ω. Initially it is in its ground state but from t = 0 its motion is
disturbed by a steady force F . Show that at time t > 0 and to first order inF the state is|ψ, ti = e−iE0 t/h̄ |0i + a1 e−iE1 t/h̄ |1i
wherei
a1 = √
2mh̄ω

Z t

′

dt′ F (t′ )eiωt .

0Calculate hxi (t) and show that your expression coincides with the classicalsolutionZtx(t) =

0dt′ F (t′ )G(t − t′ ),where the Green’s function is G(t − t′ ) = sin[ω(t − t′ )]/mω. Show that asuitable displacement of the point to which the oscillator’s spring is anchored
could give rise to the perturbation.
9.17∗ A particle of mass m is initially trapped by the well with potential
V (x) = −Vδ δ(x), where Vδ > 0. From t = 0 it is disturbed by the timedependent potential v(x, t) = −F xe−iωt . Its subsequent wavefunction canbe written
Z−iE0 t/h̄|ψi = a(t)e|0i + dk {bk (t)|k, ei + ck (t)|k, oi} e−iEk t/h̄ ,(9.82)
where E0 is the energy of the bound state |0i and Ek ≡ h̄2 k 2 /2m and |k, ei
and |k, oi are, respectively the even- and odd-parity states of energy Ek (seeProblem 5.17). Obtain the equations of motion


Z


ih̄ ȧ|0ie−iE0 t/h̄ + dk ḃk |k, ei + ċk |k, oi e−iEk t/h̄Z
= v a|0ie−iE0 t/h̄ + dk (bk |k, ei + ck |k, oi) e−iEk t/h̄ .

(9.83)

Given that the free states are normalised such that hk ′ , o|k, oi = δ(k − k ′ ),show that to first order in v, bk = 0 for all t, and thatiF
sin(Ωk t/2)
hk, o|x|0i eiΩk t/2
,
h̄
Ωk /2

Ek − E0− ω.
h̄(9.84)
Hence show that at late times the probability that the particle has become
free is
2πmF 2 t |hk, o|x|0i|2.
(9.85)
Pfr (t) =
k
h̄3
Ωk =0
ck (t) =where Ωk ≡

224

Problems

Given that from Problem 5.17 we havehx|0i =

√ −K|x|Ke

where K =show thathk, o|x|0i =

mVδ
h̄2

r

1
and hx|k, oi = √ sin(kx),π(9.86)4kK
K.
π (k 2 + K 2 )2(9.87)

Hence show that the probability of becoming free isp
Ef /|E0 |8h̄F 2 t
Pfr (t) =,mE02 (1 + Ef /|E0 |)4(9.88)

where Ef > 0 is the final energy. Check that this expression for Pfr isdimensionless and give a physical explanation of the general form of theenergy-dependence of Pfr (t)9.18∗ A particle travelling with momentum p = h̄k > 0 from −∞ encounters the steep-sided potential well V (x) = −V0 < 0 for |x| < a. Use the Fermi
golden rule to show that the probability that a particle will be reflected by
the well is
V2
Preflect ≃ 0 2 sin2 (2ka),
4E
where E = p2 /2m. Show that in the limit E ≫ V0 this result is consistent
with the exact reflection probability derived in Problem 5.10. Hint: adopt
periodic boundary conditions so the wavefunctions of the in and out states
can be normalised.
9.19∗ Show that the number states g(E) dE d2 Ω with energy in (E, E+dE)
and momentum in the solid angle d2 Ω around p = h̄k of a particle of mass
m that moves freely subject to periodic boundary conditions on the walls ofa cubical box of side length L isg(E) dE d2 Ω =



L
2π

3

m3/2 √
2E dE dΩ2 .
h̄3(9.89)

Hence show from Fermi’s golden rule that the cross section for elastic scattering of such particles by a weak potential V (x) from momentum h̄k into
the solid angle d2 Ω around momentum h̄k′ is
dσ =

m2
d2 Ω
(2π)2 h̄4Z′2

d3 x ei(k−k )·x V (x) .

(9.90)

Explain in what sense the potential has to be “weak” for this Born approximation to the scattering cross section to be valid.
9.20 Given that a0 = h̄/(αme c) show that the product a0 k of the Bohrradius and the wavenumber of a photon of energy E satisfies
a0 k =

E.
αme c2(9.91)

5
Hence show that the wavenumber kα of an Hα photon satisfies a0 kα = 72αand determine λα /a0 . What is the connection between this result and ourestimate that ∼ 107 oscillations are required to complete a radiative decay.
Does it imply anything about the way the widths of spectral lines fromallowed atomic transitions varies with frequency?

9.21 Equation (9.75) implies that x± act as ladder operators for Jz . Why
did we not use these operators in §7.1?

Problems

225

9.22 Given that a system’s Hamiltonian is of the formH=

p2
+ V (x)2me(9.92)

show that [x, [H, x]] = h̄2 /me . By taking the expectation value of this expression in the state |ki, show thatXn6=k|hn|x|ki|2 (En − Ek ) =

h̄2
,
2me

(9.93)

where the sum runs over all the other stationary states.
The oscillator strength of a radiative transition |ki → |ni is defined
to be
2me
fkn ≡ 2 (En − Ek )|hn|x|ki|2(9.94)h̄PShow that n fkn = 1. What is the significance of oscillator strengths forthe allowed radiative transition rates of atoms?

10Helium and the periodic table
In this chapter we build on the foundations laid by our study of hydrogen in
Chapter 8 to understand how the atoms of heavier elements work. Most of
the essential ideas already emerged in Chapter 8. In fact, only one important point of principle needs to be introduced before we can move down the
periodic table understanding why elements have the spectra and chemistrythat they do. This outstanding issue is the remarkable implications thatquantum mechanics carries for the way in which identical particles interact
with one another. We shall be concerned with the case in which the particles are electrons, but the principles we elucidate apply much more widely,for example, to the dynamics of the three quarks that make up a proton or
neutron, or the two oxygen atoms that comprise an oxygen molecule.

10.1 Identical particles
Consider a system that contains two identical spinless particles. Then acomplete set of amplitudes is given by a function ψ of the coordinates x andx′ of the particles: the complex number ψ(x, x′ ) is the amplitude to find one
particle at x and the other particle at x′ . What’s the physical interpretationof the number ψ(x′ , x)? It also gives the amplitude to find particles at x andx′ . If the particles were not identical – if one where a pion the other a kaon,for example – finding the pion at x and the kaon at x′ would be a physicallydistinct situation from finding the kaon at x and the pion at x′ . But if
both particles are pions, ψ(x, x′ ) and ψ(x′ , x) are amplitudes for identicalphysical situations. Does it follow that ψ(x, x′ ) = ψ(x′ , x)? No, because
experimentally we can only test the probabilities to which these amplitudes
give rise. So we can only safely argue that the probability |ψ(x, x′ )|2 must
equal the probability |ψ(x′ , x)|2 , or equivalently that
ψ(x, x′ ) = eiφ ψ(x′ , x),(10.1)

where φ is a real number. This equation must hold for any x and x′ . So thefunction ψ has the property that if you swap its arguments, you increment
its phase by φ. Specificallyψ(x′ , x) = eiφ ψ(x, x′ ).

(10.2)

10.1 Identical particles

227

Substituting this equation into the right side of equation (10.1), it followsthatψ(x, x′ ) = e2iφ ψ(x, x′ ),(10.3)
which implies that either φ = 0 or φ = π. Thus we have shown that the
wavefunction for a system of two identical spinless particles has to be either
symmetric or antisymmetric under interchange of the particles’ coordinates.
Consider now the case of two spin-s particles, which might, for example,
be electrons (s = 12 ) or photons (s = 1). A complete set of amplitudes would
be the amplitude for one particle to be at x in the state that has eigenvalue
m of Sz , and the other particle to be at x′ with eigenvalue m′ . Let thecomplex number ψmm′ (x, x′ ) denote this amplitude – that is, let the firstsubscript on ψ give the orientation of the spin of the particle that is foundat the position given by the first argument of ψ. Then the possibly different
amplitude ψm′ m (x′ , x) is for the identical physical situation. Henceψm′ m (x′ , x) = eiφ ψmm′ (x, x′ ).

(10.4)

This equation must hold for all m, m′ and x, x′ . So swapping the subscripts
on ψ at the same time as swapping the arguments, is equivalent to multiplying through by eiφ . Swapping a second time leads toψmm′ (x, x′ ) = e2iφ ψmm′ (x, x′ ),(10.5)

so, as in the case of spin-zero particles, either φ = 0 or φ = π. It turns outthat there is no change of sign if the particles are bosons (s = 0, 1, 2, . . .), andthere is a change of sign if the particles are fermions (s = 12 , 23 , . . .). That is′

ψmm′ (x, x ) =−ψm′ m (x′ , x)+ψm′ m (x′ , x)

for fermionsfor bosons.

(10.6)

These relations between amplitudes are said to arise from the principle ofexchange symmetry between identical particles.
Generalisation to the case of N identical particles If our system
contains N identical fermions, the wavefunction will change its sign when we
swap the arguments (both spin quantum numbers and spatial coordinates)
associated with any two slots in the wavefunction. Similarly, if the system
contains N bosons, the wavefunction will be invariant when we swap thearguments associated with any two slots.

10.1.1 Pauli exclusion principle
An immediate consequence of the wavefunction of fermions being antisymmetric under a swap of its arguments, is that there is zero probability offinding two fermions with their spins oriented in the same way at the same
location: since ψmm (x, x) = −ψmm (x, x), the amplitude ψmm (x, x) mustvanish. Since wavefunctions are continuous functions of position, and their
spatial derivatives are constrained in magnitude by the particles’ momenta,
ψmm (x, x′ ) can vanish at x = x′ only if it is small whenever the two arguments are nearly equal. Hence, fermions with similarly oriented spins avoid
each other; they are anticorrelated. This fact has profound implications
for atomic and condensed-matter physics.
If the particles’ spins have different orientations, there can be a nonzero amplitude of finding them at the same location: from ψmm′ (x, x) =−ψm′ m (x, x) it does not follow that the amplitude ψmm′ (x, x) vanishes.
The principle of exchange symmetry arises as a constraint on amplitudes,
but we now show that it has implications for the structure of the underlying
states. Let {|ni} be a complete set of states for a single fermion – so thelabel n carries information about both the electron’s motion in space and228

Chapter 10: Helium and the periodic table

its orientation (spin state). Then from §6.1 we have that any state of anelectron pair can be expanded in the form1
|ψi =

X
nn′

ann′ |ni|n′ i.

(10.7)

Multiplying through by hx, x′ , m, m′ | we obtain the amplitude ψmm′ (x, x′ )that is constrained by exchange symmetry:ψmm′ (x, x′ ) = hx, x′ , m, m′ |ψi =X
nn′ann′ hx, m|nihx′ , m′ |n′ i.

(10.8)

We now swap x, m with x′ , m′ throughout the equation and add the result
to our existing equation. Then by exchange symmetry the left side vanishes,
and we have
0=

X
nn′

ann′ hx, m|nihx′ , m′ |n′ i +Xnn′ann′ hx′ , m′ |nihx, m|n′ i.

(10.9)In the second sum we may swap the labels n and n′ (since they are being summed over), and we may also reverse the order of the amplitudes
hx′ , m′ |n′ i and hx, m|ni (because they are mere complex numbers). Then we
have
0=

X
nn′

=

X
nn′

ann′ hx, m|nihx′ , m′ |n′ i +X
n′ n

an′ n hx, m|nihx′ , m′ |n′ ihx, m|nihx′ , m′ |n′ i(ann′ + an′ n )= hx, x′ , m, m′ |Xnn′

(10.10)|ni|n′ i(ann′ + an′ n ).

Since this equation holds for arbitrary x, x′ , m, m′ it follows that the sum
vanishes, and from the linear independence of the basis kets |ni|n′ i it follows
that the coefficient of each such ket vanishes. Hence we haveann′ = −an′ n .

(10.11)

In particular ann vanishes so there is zero amplitude to find that both
fermions are in the same single-particle state |ni. This result is known as
the Pauli exclusion principle.
The Pauli exclusion principle ensures that any expansion of the form
(10.7) involves at least two terms. When √there are only two terms, equation(10.11) ensures that ann′ = −an′ n = ±1/ 2, so equation (10.7) reduces to1|ψi = ± √ (|ni|n′ i − |n′ i|ni).
2(10.12)In §6.1 we saw when the wavefunction of a pair of particles is a non-trivial sum
over products of wavefunctions for each particle, the particles are correlated.
Hence the Pauli exclusion principle implies that identical fermions are always
correlated.
1 Equation (10.7) implies that we can distinguish the two electrons – electron “1” isin the state |ni, while electron “2” is in state |n′ i. Physically this is meaningless. What
we are doing is writing down states of distinguishable particles that are consistent withthe restrictions imposed on states of pairs of indistinguishable particles.

10.1 Identical particles

229

10.1.2 Electron pairs
We now specialise to the important case of identical spin-half particles, suchas electrons. For complete specification of the quantum state of a single
electron we must give the values of two functions of x, which can be theamplitudes ψ± (x) to find the electron at x and oriented such that Sz returns± 21 . These functions form a two-component wavefunction:hx|ψi ≡



ψ+ (x)
ψ− (x)

.

(10.13)

Similarly, to specify completely the state of a pair of electrons, four functionsof two locations are required, namely ψmm′ (x, x′ ) for m, m′ = ±. Thus anelectron pair has a four-component wavefunctionψ++ (x, x′ )′
 ψ (x, x ) hx, x′ |ψi ≡  −+.
ψ+− (x, x′ )′ψ−− (x, x )(10.14)

We often wish to consider the states of an electron pair in which it is the pairrather than its individual members that has well-defined spin – in §7.5.1 weinvestigated the states of a hydrogen atom in which the atom rather thanits constituent particles has well-defined spin. Our derivation of the results
obtained there relied only on the properties of the spin raising and lowering
operators S± , so they are valid for any pair of spin-half particles, including
an electron pair. Multiplying equation (7.153) through by hm, m′ | we seethat when the pair has unit spin and Sz yields 1, the only non-vanishing
amplitude ψmm′ is ψ++ . Hence in this state of the pair the wavefunction is
 
1
′
1
′ 0
hx, x |ψ, 1, 1i = ψ1 (x, x )   ,0
0(10.15)where ψ11 ≡ ψ++ is required by exchange symmetry to be an antisymmetricfunction of x and x′ :
ψ11 (x′ , x) = −ψ11 (x, x′ ).
(10.16)
Similarly, from equation (7.154) it follows that when the pair has s = 1
but m = 0, there are equal amplitudes to find the individual spins −+ and+−, so 01hx, x′ |ψ, 1, 0i = ψ10 (x, x′ )   ,(10.17)10

where ψ10 (x, x′ ) ≡ ψ−+ (x, x′ ) = ψ+− (x, x′ ). Swapping the labels x and x′
on both sides of the equivalence and using first equation (10.6) and thenequation (10.17), yields
ψ10 (x′ , x) = ψ−+ (x′ , x) = −ψ+− (x, x′ ) = −ψ10 (x, x′ ).

(10.18)Thus ψ10 like ψ11 is an antisymmetric function of its arguments.
Similarly, from equation (7.155) we infer that when the pair has s = 1
and m = −1 its wavefunction is
 
0
−1
′ 0
′
hx, x |ψ, 1, −1i = ψ1 (x, x )   ,0
1(10.19)

230

Chapter 10: Helium and the periodic table

where ψ1−1 is an antisymmetric function of its arguments.
Finally we must consider the spin-zero state of the pair. By equation(7.156) it is associated with values of the amplitudes ψ−+ and ψ+− that are
equal in magnitude and opposite in sign. So we can write



0
 1 
hx, x′ |ψ, 0, 0i = ψ0 (x, x′ ) .
−1
0(10.20)In this case use of the exchange principle yields
ψ0 (x, x′ ) ≡ ψ−+ (x, x′ ) = −ψ+− (x′ , x) = ψ0 (x′ , x)(10.21)

so ψ0 , in contrast to the previous functions ψ1m , is a symmetric function of
its arguments.
The spin-one states of an electron pair are generally called triplet
states while the spin-zero state is called the singlet state. We have seenthat the wavefunction of a triplet state is an antisymmetric function of x
and x′ , while wavefunction of the singlet state is a symmetric function of x
and x′ . We saw above that electrons that have equal components of angularmomentum parallel to the z axis avoid each other. We now see that thismutual avoidance is a general characteristic of all the triplet states.
One way of constructing a function of two variables is to take the productu(x)v(x′ ) of two functions u and v of one variable. Unless u = v, this productis neither symmetric nor antisymmetric under interchange of x and x′ , so itcannot be proportional to the wavefunction of either a triplet or a singlet
state. To achieve such proportionality, we must extract the symmetric orantisymmetric part of the product. That is, for appropriate u and v we may
have1ψ1m (x, x′ ) = √ {u(x)v(x′ ) − u(x′ )v(x)} (m = 1, 0, −1)21
ψ0 (x, x′ ) = √ {u(x)v(x′ ) + u(x′ )v(x)}.
2(10.22)In the case u = v, the triplet wavefunctions are identically zero but the singlet wavefunction can be non-vanishing; that is, two distinct single-particle
wavefunctions are required for the construction of a triplet state, while just
one single-particle wavefunction is all that is required for a singlet state.
Wavefunctions of the form (10.22) are widely used in atomic physics
but one should be clear that it is an approximation to assume that a twoelectron wavefunction can be written in terms of just two single-particle
wavefunctions; any wavefunction can be expanded as a sum of products ofsingle-particle wavefunctions, but the sum will generally contain more than
two terms.

10.2 Gross structure of helium
About a quarter of the ordinary matter in the Universe is in the form ofhelium, the second simplest element. The tools that we now have at our
disposal enable us to build a fairly detailed model of these important atoms.
This model will illustrate principles that apply in all many-electron atoms.
We seek the stationary states of the Hamiltonian that describes theelectrostatic interactions between the two electrons and the alpha particle
that make up a helium atom. This Hamiltonian is (cf. eq. 8.1)


p2n
p21
p22
e2
2
1
2
H=
,
+
+
−
+
−
2mn
2me
2me
4πǫ0 |x1 − xn | |x2 − xn | |x1 − x2 |(10.23)

10.2 Gross structure of helium

231

where xi and xn are the position operators of the ith electron and the nucleus,
respectively, and pi and pn are the corresponding momentum operators. We
shall work in the atom’s centre-of-mass frame and neglect the small displacement from the origin and kinetic energy that the nucleus has in this frame.
With this approximation, H can be written as the sum of two hydrogenic
Hamiltonians with Z = 2 (cf. eq. 8.10) and the term that describes themutual electrostatic repulsion of the electrons
H = HH (p1 , x1 ) + HH (p2 , x2 ) +
where

e2
,
4πǫ0 |x1 − x2 |

2e2
p2
−
.
2me
4πǫ0 |x|

HH (p, x) ≡(10.24a)(10.24b)

We cannot determine the eigenkets of H exactly, so we resort to the approximate methods developed in the previous chapter.
10.2.1 Gross structure from perturbation theory
Our first approach is to use the perturbation theory of §9.1. In §8.1 we foundthe eigenfunctions of HH . These proved to be products uln (r)Ylm (θ, φ) of theradial eigenfunctions uln derived in §8.1.2 and the spherical harmonics Ylm
derived in §7.2.3. From the work of §6.1 it follows that the eigenfunctions of
the operator
H0 ≡ HH (p1 , x1 ) + HH (p2 , x2 )(10.25)are products
′

′

′

m
m
l
l
m
Ψm
nl (x1 )Ψn′ l′ (x2 ) ≡ un (r1 )Yl (θ1 , φ1 )un′ (r2 )Yl′ (θ2 , φ2 ),(10.26)

where n and n′ are any positive integers. From equation (8.27) the corresponding eigenvalues are
E0 ≡ −4R



1
1
+ ′2
n2
n



.

(10.27)

The ground-state wavefunction of H0 will be a product of the ground-state
eigenfunctions (4π)−1/2 u01 (r) of HH , where the function u01 is given by equation (8.36) with Z = 2. From equation (10.27) the ground-state energy of
H0 is
E0 = −8R = −108.8 eV.
(10.28)The Hamiltonian (10.23) commutes with all spin operators because it
makes no reference to spin. Therefore we are at liberty to seek eigenfunctionsof H that are simultaneously eigenfunctions of the total spin operators S 2 andSz . We have seen that these eigenfunctions are either singlet or triplet states
and are either symmetric or antisymmetric functions of the spatial variables.
The ground-state wavefunction of H0 is an inherently a symmetric functionof x1 and x2 , so the ground-state is a singlet. The first-order contribution
to the ground-state energy is the expectation value of the perturbing part of
the Hamiltonian (10.24.) This expectation value is
e2
∆E =
D0
4πǫ0

where D0 ≡Z

d3 x1 d3 x2|Ψ010 (x1 )|2 |Ψ010 (x2 )|2. (10.29)|x1 − x2 |Box 10.1 describes the evaluation of the six-dimensional integral D0 . We
find that ∆E = 25 R, so our estimate of the ground-state energy of helium is
E = E0 + ∆E = − 11
2 R = −74.8 eV. The experimentally measured value is−79.0 eV.

232

Chapter 10: Helium and the periodic table

Box 10.1: Evaluating the
integral D0 in equation (10.29)
We express the two position vectors in spherical polar coordinates. Since
x1 is a fixed vector during the integration over x2 , we are at liberty
to orient our pz axis for the x2 coordinate system parallel to x1 . Then|x1 − x2 | = |r12 + r22 − 2r1 r2 cos θ2 | is independent of φ2 . The modsquare of Ψ010 does not depend on φ, so the integrand is independent of
φ2 and we can trivially integrate over φ2 . What remains isZZ
r2 sin θ2 e−2r2 /aZ
2
302
dr2 dθ2 p 2
d x1 |Ψ10 (x1 )|D0 = 3(1)
aZ|r12 + r22 − 2r1 r2 cos θ2 |where aZ ≡ a0 /2. Nowsosin θ21 dp=
2
2
r1 r2 dθ2
|r1 + r2 − 2r1 r2 cos θ2 |Z π
0q|r12 + r22 − 2r1 r2 cos θ2 |,|r1 + r2 | − |r1 − r2 |sin θ2 dθ2
p
=
r1 r2
|r12 + r22 − 2r1 r2 cos θ2 |

2/r1 for r1 > r2=
2/r2 for r1 < r2 .

After using this expression in equation (1), we have to break the integral
over r2 into two parts, and have

Z r1
Z ∞
Z
4
r22 −2r2 /aZ−2r2 /aZ302
D0 = 3dr2 r2 e+dr2 e
d x1 |Ψ10 (x1 )|
aZ
r1
r1
0
Z
2
1 
=
2 − e−ρ1 (2 + ρ1 ) ,d3 x1 |Ψ010 (x1 )|2aZ
ρ1

where ρ1 ≡ 2r1 /aZ . The integral over x1 is relatively straightforward:
given the normalisation of the spherical harmonics, we simply have to
integrate over r1 . We transform to the scaled radius ρ1 and find
Z

1
5
5
D0 =
=
.
dρ1 ρ1 e−ρ1 2 − e−ρ1 (2 + ρ1 ) =
aZ
8aZ
4a0

10.2.2 Application of the variational principle to helium
We can use the variational principle (§9.2) to refine our estimate of helium’s
ground-state energy. Our estimate is based on the assumption that the electrons’ wavefunctions are those that would be appropriate if the electrons did
not repel one another. Suppose we could somehow switch off this repulsion
without affecting the attraction between each electron and the alpha particle.
Then the electrons would settle into the wavefunctions we have assumed. If
we then turned the electric repulsion back on, it would push the electrons
apart to some extent, and the atom would become bigger. This thought
experiment suggests that we might be able to obtain a better approximation
to the electrons’ wavefunctions by increasing the characteristic lengthscalethat appears in the exponential of a hydrogenic ground-state wavefunction(eq. 8.36) from a0 to some value a. The variational principle assures us thatthe minimum value with respect to a of the expectation value of H that we
obtain with these wavefunctions will be a better estimate of the ground-stateenergy than the estimate we obtained by first-order perturbation theory.
Consider, therefore, the expectation value of the Hamiltonian (10.24a)for the case in which the electronic wavefunction is a product of hydrogenic
wavefunctions with a0 replaced by a. From the work of the last subsection wealready know the value taken when a = a0 . Moreover, the expectation value
is made up of five terms, two kinetic energies, and three potential energies,10.2 Gross structure of helium233

and by dimensional analysis it is clear how each term must scale with a: thekinetic energies scale as a−2 because p is proportional to the gradient of the
wavefunction, which scales like a−1 , while the potential energy contributions
scale as a−1 since they explicitly have distances in their denominators. We
know that when a = a0 and the wavefunctions are hydrogenic, the expectation value of the sum of hydrogenic Hamiltonians in equation (10.24a) is−8R, and we know from the virial theorem (eq. 2.93) that this overall energy
is made up of 8R kinetic energy and −16R of potential energy. We saw abovethat when a = a0 the electrostatic repulsion of the electrons contributes 52 Rof potential energy. Bearing in mind the way that these kinetic and potential
energies scale with a it follows that for general a, the expectation value of
helium’s Hamiltonian is
hHia = R{8x2 − (16 − 52 )x}

where

x≡

a0
.
a

(10.30)27. When we
The derivative of hHia with respect to x vanishes when x = 32
insert this value of x into equation (10.30) we find our improved estimate ofhelium’s ground-state energy is − 21 (3/2)6 R = 77.4 eV. As was inevitable,
this value is larger than the experimentally measured value of −79.0 eV. But
it is significantly closer than the value we obtained by first-order perturbation
theory.
An important indicator of the chemical nature of an element is themagnitude of the energy required to strip an electron from an atom, whichis called the element’s ionisation energy. In the case of hydrogen, thisenergy is simply the binding energy, 13.6 eV. In the case of helium it is thedifference between the binding energies of the atom and the ion He+ thatremains after one electron is stripped away. Since the He+ ion is hydrogenic
with Z = 2, its binding energy is 4R = 54.4 eV, so the ionisation energyof helium is 79.0 − 54.4 = 24.6 eV. This proves to be the largest ionisationenergy of any atom, which makes helium perhaps the least chemically active
element there is.

10.2.3 Excited states of helium
We now consider the low-lying excited states of helium. Given our successin calculating the ground-state energy of helium with the aid of hydrogenic
wavefunctions, it is natural to think about the excited states using the samehydrogenic language. Thus we suppose that the electronic wavefunction ismade up of products of single-particle wavefunctions. We recognise that thesingle-particle wavefunctions that should be used in these products will differ
slightly from hydrogenic ones, but we assume that they are similar to thehydrogenic ones that carry the same orbital angular momentum and havethe same number, n − 1, of radial nodes. Hence we can enumerate the singleparticle wavefunctions by assigning the usual quantum numbers n and l toeach electron. We expect to be able to obtain reasonable estimates of theenergies of excited states by taking the expectation value of the Hamiltonian
for hydrogenic states.
In the first excited state it is clear that one of the electrons will have beenpromoted from its n = 1 ground state to one of the n = 2 states. From our
discussion of shielding in §8.1.3, we expect that the state with l = 0 will haveless energy than any other n = 2 state. Thus we seek to construct the firstexcited state from a product of the hydrogenic ground-state wavefunction
Ψ1 (x) and the wavefunction Ψ2 (x) for n = 2, l = 0, which is also sphericallysymmetric. Since we are working with distinct single-particle states, we canconstruct both singlet and triplet states as described in §10.1.3. The spatial
part of the wavefunction (ψ1m of ψ0 ) will be constructed from the productΨ1 Ψ2 symmetrised as described by equation (10.22). Since the two possible
ways of symmetrising the product differ only in a sign, we defer choosing
between them and make our formulae valid for either case, putting the sign

234

Chapter 10: Helium and the periodic table

for the singlet state on top. We now have to calculateZhHi = 21 d3 x d3 x′ {Ψ∗1 (x)Ψ∗2 (x′ ) ± Ψ∗1 (x′ )Ψ∗2 (x)}(10.31)× H {Ψ1 (x)Ψ2 (x′ ) ± Ψ1 (x′ )Ψ2 (x)} .

When we substitute for H from equation (10.24a), integrals over terms suchas Ψ∗1 (x)Ψ∗2 (x′ )HH Ψ1 (x′ )Ψ2 (x) arise, where HH is the hydrogenic operator
that appears in equations (10.24). The orthogonality of Ψ1 and Ψ2 causes′
these integrals to vanish, because HH contains either x or
not bothR x3, butoperators so there is always an integral of the form 0 = d x Ψ∗1 (x)Ψ2 (x).
The integral over Ψ∗1 (x)Ψ∗2 (x′ )HH Ψ1 (x)Ψ2 (x′ ) evaluates to either −4R or−R depending on whether HH contains x or x′ . HencehHi = −5R +e2{D ± E},4πǫ0(10.32a)where D and E are, respectively, the direct and exchange integrals:Z|Ψ1 (x)Ψ2 (x′ )|2|x − x′ |ZΨ∗ (x)Ψ2 (x)Ψ∗2 (x′ )Ψ1 (x′ ).
E ≡ d3 x d3 x′ 1
|x − x′ |D≡

d3 x d3 x′(10.32b)

Since both Ψ1 and Ψ2 are spherically symmetric functions of their arguments,
both integrals can be evaluated by the technique described in Box 10.1. Aftera good deal of tedious algebra one discovers that hHi = −(56.6 ∓ 1.2) eV,
where the upper sign is for the singlet state. The experimentally measured
values are −(58.8 ∓ 0.4) eV. Hence perturbation theory correctly predicts
that the triplet state lies below the singlet state.
The differences between our perturbative values and the experimental
results arise because the hydrogenic wavefunctions we have employed are notwell suited to helium. The deficiency is particularly marked in the case of
the n = 2 wavefunction because the nuclear charge is significantly shieldedfrom the outer electron, so the n = 2 wavefunction should extend to larger
radii than the hydrogenic wavefunction we have employed, which assumes
that the electron sees the full nuclear charge. Consequently, we have overestimated the overlap between the two wavefunctions: the extent to which
the wavefunctions permit the electrons to visit the same place. Because ourwavefunctions have unrealistically large overlap, they yield values for both D
and E that are too large. The exchange integral is particularly sensitive tooverestimation of the overlap because it vanishes when there is no overlap,
which D does not. Thus it is entirely understandable that our treatment
yields binding energies that are insufficiently negative, and a singlet-triplet
splitting that is too large.
The sensitivity of the singlet-triplet splitting to wavefunction overlap
leaves a clear imprint on the energy-level structure of helium that is shownin Figure 10.1: the separations of corresponding full (singlet) and dotted(triplet) lines diminishes as one goes up any column (increasing n) or fromleft to right (increasing l). Quantitatively, the singlet-triplet splitting whenthe excited electron is in a n = 2, l = 1 state (bottom of second column),
rather than the n = 2, l = 0 state that we have just investigated (bottom of
the first column), is 0.2 eV rather than 0.8 eV because, as we saw in §8.1.2,
the l = 1 state has smaller amplitudes at the small radii at which the n = 1
state has large amplitudes.
We have discussed the splitting between singlet and triplet states in thecase in which both the single-particle wavefunctions employed are sphericallysymmetric, so the wavefunctions are entirely real. The analysis for wavefunctions that have l 6= 0 is significantly more involved, but the essential result10.2 Gross structure of helium

235

Figure 10.1 Excited states of helium for n ≤ 6 and l ≤ 3. Energies are given with respect
to the ground-state energy, and the line at the top shows the ionisation energy. Full lines
show singlet states and dotted lines show triplet states. Fine structure splits the triplet
states with l ≥ 1 but the splittings are much too small to show on this scale – the largest
is 0.00012 eV.

is the same because the exchange integral E is always real (Problem 10.3)
and positive. We can see that E is positive as follows. The exchange integral
is dominated by the region x ≃ x′ in which the denominator is small. In
this region the numerator does not differ much from |Ψ1 (x)Ψ2 (x)|2 , so it ispositive. Hence E is positive. Thus it is quite generally true that the triplet
states lie below the corresponding singlet state.
In our discussion of spin-orbit coupling in §8.2.1 we saw that the energy
scale of that coupling is ∼ 41 Z 2 α4 me c2 (eq. 8.76b). For helium this evaluatesto ∼ 0.006 eV, which is two orders of magnitude smaller than the singlettriplet splitting. Moreover, we found that the coupling vanishes for stateswith l = 0, so it should vanish in the first excited state of helium. The
singlet-triplet splitting is large because it has an electrostatic origin, rather
than being a mere relativistic effect: a triplet state has less energy because
in it the electrons are anticorrelated (§10.1.1).
It is commonly stated that on account of this anticorrelation the energyof electrostatic repulsion between the electrons is smaller in triplet than insinglet states. This is false: the inter-electron potential energy is larger inthe triplet than the singlet state.2 The reason the triplet has lower energy
is because it places the electrons closer to the nucleus than the singlet does.
Moving the electrons towards the nucleus and thus towards one another naturally increases the energy of electron-electron repulsion, but this increaseis outweighed by the lowering of the negative electron-nucleus energy. The
quantitative results we obtained above should not be used to evaluate theinter-electron energy because they are based on hydrogenic wavefunctions,which provide a poor approximation to the true wavefunction. As we saw
in §9.2, even a poor approximation to the wavefunction of a stationary stateyields a useful approximation to the energy of that state because the expectation value of H is stationary in the neighbourhood of a stationary state.
But the expectation value of a single term in H, such as the inter-electron
potential energy, is not extremised by a stationary state, so the error in it
will be of order the error in the wavefunction. In particular, to obtain a valuethat is accurate to first order in the perturbation, it is mandatory to use awavefunction that is correct to first order, whereas we used the zeroth-orderwavefunctions. Because the electrons do a better job of keeping out of eachother’s way in the triplet state, in that state they can cohabit in a smaller
volume, where the attraction of the nucleus is stronger. On account of this
2 B. Schiff, H. Lifson, C.L. Pekeris & P. Rabinowitz, Phys. Rev., 140, A1104, (1965)
find the inter-electron energy to be 6.80 eV in the singlet state and 7.29 eV in the triplet
state.

236

Chapter 10: Helium and the periodic table

Box 10.2: Spectroscopic Notation
Standard spectroscopic notation presumes that l and s, the total orbitaland spin angular momenta, are good quantum numbers. The electronic
configuration is a specification of the principal (n) and orbital angular momentum (l) quantum numbers of the individual electrons of theoutermost shell. Within a configuration a spectroscopic term specifies definite values for the total orbital l and spin s angular momenta of
the outer electrons. Within each term a fine-structure level specifies
a definite value for the total electronic angular momentum j. Within
a fine-structure level may be distinguished different hyperfine levels
that differ in total angular momentum f . The letters S, P , D, F denotel = 0, 1, 2, 3, respectively.
A typical configuration is denoted 2s2p3 meaning one electron has n = 2,l = 0, and three electrons have n = 2, l = 1.
Terms are denoted by (2s+1) lj ; for example 4 P1/2 means s = 23 , l = 1,
j = 21 .
effect, the true singlet and triplet wavefunctions differ by more than justa change of sign; in equation (10.31) the functions Ψ1 and Ψ2 should alsochange between the singlet and triplet cases.
The singlet–triplet splitting in helium reflects destructive interference
between the amplitudes for the two electrons to be simultaneously at thesame place, and it is very much a quantum-mechanical effect. Throughout the periodic table, this mechanism gives rise to large energy differences
between atomic states that differ only in their spin. These differences makeferromagnetism possible, and thus provide us with the dynamos, power transformers and electric motors that keep our civilisation on the move.
10.2.4 Electronic configurations and spectroscopic terms
The ground state of helium has neither spin nor orbital angular momentum.
In conventional spectroscopic notation (Box 10.2) it is designated 1s2 , which
implies that it has two electrons in the n = 1 S state. A related notation isused to indicate the spin, orbital and total angular momentum of the entire
atom. In this system the ground state is designated 1 S0 . The superscript
1 implies that the state is a spin-singlet because there is zero spin angularmomentum. The S implies that there is no orbital angular momentum, andthe subscript 0 implies that there is zero total angular momentum.
The lowest dotted line in Figure 10.1 represents a triplet of excitedstates. These have the electronic configuration 1s2s because there is anelectron in the n = 1, l = 0 state and one in the n = 2, s = 0 state.
They form the spectroscopic term 3 S1 because the angular momenta of
the whole atom are given by s = 1, l = 0 and j = 1.
Just above this triplet of states comes the singlet state that has the same
electronic configuration 1s2s but which forms the distinct spectroscopic term
1
S0 .
Next come four spectroscopic terms that both have the electronic configuration 1s2p: the most energetic of these terms is the singlet 1 P1 , which isa set of three quantum states that have exactly the same energy but different orientations of the one unit of total angular momentum. Below this are
three terms that have very similar energies: 3 P0 , 3 P1 and 3 P2 . These terms
differ from one another in the degrees of alignment of the spin and orbital
angular momenta. In the 3 P0 term the angular momenta are anti-parallel,with the result that the atom has zero angular momentum overall, while inthe 3 P2 term the angular momenta are parallel, so the atom has two units ofangular momentum. There is just one quantum state in the 3 P0 term, and10.3 Periodic table

237

five quantum states in the 3 P2 term. The small energy differences between
the 3 Pj terms are due to spin-orbit coupling.
Spectrum of helium The selection rules listed in Table 9.1 include ∆s =0, so in Figure 10.1 transitions between full and dotted levels are forbidden.
Hence, an atom which is excited into one of the upper triplet states will
cascade down through triplet states until it reaches the 3 S1 level at thebottom of the triplet hierarchy. The states in this level are metastablebecause they can decay radiatively only by making the forbidden transition
to the 1 S0 ground state, which takes appreciable time. Table 9.1 includes
the rule ∆l = ±1, so transitions are only allowed between states that lie inadjacent columns, and the excited singlet state that is designated 1 S0 is also
metastable.

10.3 The periodic table
The understanding of atomic structure that we have gained in our studies ofhydrogen and helium suffices to explain the structure of the periodic tableof the elements.
10.3.1 From lithium to argon
Imagine that we have a helium atom in its first excited state and that we
simultaneously add a proton to the nucleus and an electron to the vacancy
with principal quantum number n = 1 that arose when the atom was put
into its excited state. After making these changes we would have a lithium
atom in its ground state. The effects on the outermost electron of adding thepositively charged proton and the negatively charged electron might be expected to largely cancel, so we would expect the ionisation energy of lithium
to be similar to that of a helium atom that’s in its first excited state. This
expectation is borne out by experimental measurements: the ionisation energy of once excited helium is 4.77 eV while that of lithium in its groundstate is 5.39 eV. Thus the energy required to strip an electron from lithium
is smaller than that required to take an electron from hydrogen or helium
by factors of 2.5 and 4.6, respectively. The comparative ease with which
an electron can be removed from a lithium atom makes compounds such as
LiH stable (Problem 10.5). It also makes lithium is a metal by making it
energetically advantageous for each atom in a lithium crystal to contributeone electron to a common pool of delocalised electrons.
In their ground states atoms of hydrogen and helium cannot absorb
radiation at optical frequencies because the first excited states of these atoms
lie rather far above the ground state (10.2 and 19.8 eV, respectively). The
first excited state of lithium is obtained by promoting the n = 2 electronfrom l = 0 to l = 1. This change in quantum numbers only increases theelectron’s energy by virtue of shielding (§8.1.3), so the energy difference is amere 1.85 eV, the quantity of energy carried by photons of wavelength 671 nm
that lie towards the red end of the optical spectrum. Elements that lie beyond
helium in the periodic table, the so-called heavy elements, feature very
prominently in astronomical measurements even though they are present intrace amounts compared to hydrogen and helium because their absorption
spectra contain lines at easily observed optical wavelengths.
There is a useful parallel between a lithium atom and a hydrogen atom
in its first excited state: the lithium nucleus, shielded by the two n = 1
electrons, appears to have the same net charge as the proton in hydrogen,so the n = 2 electron moves in a similar electric field to that experienced byan electron with n = 2 in hydrogen. We can test this parallel quantitatively
by comparing the ionisation energy of lithium (5.39 eV) with the energy of
H with n = 2 (3.40 eV). This agreement is not terribly good because then = 2, l = 0 wavefunction that forms the ground state of lithium overlaps

238

2

S1=2
Li3

2

He2
1s2

H1
1s1

s
S1=2
2 1

1

Be4

B5

s
S0

2 2

1

2

p
P1=2
2 1

C6

p
P0

2 2

3

Mg12

Al13

Si14

2

1

2

3

s
S1=2
3 1

s
S0

3 2

K19

Ca20

S 21

Ti22

V23

2

1

2

3

4

s
S1=2
4 1

s
S0

4 2

Rb37

Sr38

2

1

s
S1=2
5 1

s
S0

5 2

d
D3=2
3 1

Y39

2

d
D3=2
4 1

d
F2

3 2

Zr40

d
F2

4 2

3

Cr24

d s d
F3=2 7 S3

3 3

4 13 5

Nb41

Mo42

Mn25

Fe26

Co27

Ni28

6

5

4

3

d
S5=2

3 5

T 43

d
D4

3 6

Ru44

d
F9=2

3 7

Rh45

P15
4

p
S3=2

3 3

3

p
P2

2

p
P3=2
2 5

S16

Cl17

3

2

p
P2

3 4

p
P3=2
3 5

Zn30

Ga31

Ge32

As33

Se34

Br35

2

1

2

3

4

3

2

3 8

d
F4

4 1 3 10

Pd46

Ag47

Cd48

In49

Sn50

Sb51

Te52

2

1

2

3

4

3

s d s d s d s d s d s d
D1=2 7 S3 6 D9=2 5 F5 4 F9=2 1 S0
Figure 9.2

p
P0

3 2

p
S3=2

2 4

F9

Cu29

s d
S1=2

s d
S1=2

5 1 4 4 5 1 4 5 5 1 4 6 5 1 4 7 5 1 4 8 5 0 4 10 5 1 4 10
6

p
P1=2
3 1

4

2 3

O8

d
S0

3 10

d
S0

4 10

The rst ve rows of the periodi table.

p
P1=2
4 1

p
P1=2
5 1

p
P0

4 2

p
P0

5 2

p
S3=2

4 3

p
S3=2

5 3

p
P2

4 4

p
P2

5 4

p
P3=2
4 5
I53

2

p
P3=2
5 5

S0

Ne10
2p6
1

S0

Ar18
3p6
1

S0

Kr36
4p6
1

S0

Xe54
5p6
1

S0

Chapter 10: Helium and the periodic table

Na11

N7

10.3 Periodic table239

significantly with the n = 1 wavefunction, and therefore has exposure to
the full nuclear charge. There is a more satisfying parallel between the firstexcited state of lithium, in which the n = 2 electron has l = 1 and thecorresponding state of hydrogen: in this state lithium has ionisation energy
3.54 eV.
Consider now the effect of transmuting lithium into beryllium by simultaneously increasing the nuclear charge by one unit and adding a second
electron to the n = 2, l = 0 state. The parallel that we have just describedsuggests that this operation will be analogous to moving up from hydrogento helium, and will significantly increase the ionisation energy of the atom.
Experiment bears out this expectation, for the ionisation energy of beryllium is 9.32 eV, 1.7 times that of lithium. As in helium, the ground stateof beryllium has total spin zero, while the first excited states have spin one.
However, whereas in the excited states of helium the two electrons have different values of n, in beryllium they both have n = 2, and they differ in theirvalues of l. Consequently, the overlap between the single-electron states thatform the beryllium triplet is significantly larger than the corresponding overlap in helium. This fact makes the exchange integral in equations (10.32)
large and causes the singlet excited state to lie 2.5 eV above the triplet of
excited states.
If we add a unit of charge to the nucleus of a beryllium atom, we create
an atom of singly ionised boron. The four electrons with l = 0 that envelop
the ion’s nucleus screen the nuclear charge to a considerable extent from theperspective of the lowest-energy unfilled single-particle state, which is a 2p
state (n = 2, l = 1). The screening is far from complete, however, so the
nuclear charge Z that the outermost electron perceives is greater than unity
and the dynamics of the outermost electron of boron is similar to that of theelectron in a hydrogenic atom with Z > 1. The ionisation energy from then = 2 level of hydrogen is 41 Z 2 R = 3.40Z 2 eV, while that of boron is 8.30 eV,
so Z ∼ 1.6.
Spin-orbit coupling causes the ground state of boron to form the 2 P1/2
term in which the electron’s spin and orbital angular momenta are antiparallel. At this early stage in the periodic table, spin-orbit coupling is weak,so the excited states of the 2 P3/2 term lie only 0.0019 eV above the groundstate. C+ ions have the same electronic configuration as boron atoms, andin interstellar space are more abundant by factors of several thousand. Evenat the low temperatures (∼ 20 K) that are characteristic of dense interstellar clouds, collisions carry enough energy to lift C+ ions into the low-lying
excited states of the 2 P3/2 term, so such collisions are often inelastic, incontrast to collisions involving the very much more abundant hydrogen andhelium atoms and hydrogen molecules. At the low densities prevalent in interstellar space, an excited C+ ion usually has time to return to the groundstate by emitting a photon before it is involved in another collision. So C+
ions cool the interstellar gas by radiating away its kinetic energy. As a result of this physics, the temperature of interstellar gas depends sensitively
on the abundances in it of the commonest heavy elements, carbon, nitrogenand oxygen. The propensity of interstellar gas to collapse gravitationally
into stars depends on the temperature of the gas, so the formation of stars
depends crucially on the existence of low-lying excited states in boron andthe next few elements in the periodic table.
When we add another unit of charge to the nucleus of a boron atom,
the binding energy of the outermost electron increases by a factor of order(6/5)2 = 1.44. Adding a further electron, which can go into another 2p statealongside the existing outer electron, offsets this increase in binding energy
to some extent, so we expect the ionisation energy of carbon to lie somewhere between the 8.30 eV of boron and 1.44 times this value, 12.0 eV. The
experimental value is 11.3 eV, which lies at the upper end of our anticipatedrange, implying that the mutual repulsion of the two 2p electrons is not very
important energetically. This is to be expected because the ground state of240

Chapter 10: Helium and the periodic tablecarbon belongs to the triplet term3 3 P0 , so the electrons keep out of eachother’s way. As in boron, the first excited states lie very close to the groundstate – they form a 3 P1 term 0.0020 eV above the ground state, and there is
a 3 P2 term 0.0033 eV above that.
Adding a unit of charge to the nucleus of carbon and then dropping
an electron into another 2p single-particle state creates a nitrogen atom.
The ionisation energy increases (to 14.5 eV) for exactly the same reason thatit did when we transmuted boron into carbon. The spin of all three 2p
electrons are aligned to ensure that the wavefunction is antisymmetric in theelectrons’ spatial coordinates. Hence the ground state of nitrogen belongs
to a quadruplet of states. The total orbital angular momentum proves to be
zero (Problem 10.7), so all states in this quadruplet have the same energy,
and there are actually four distinct ground states that together comprise theterm 4 S3/2 – it is sometimes rather confusingly said that the ground ‘state’of nitrogen is four-fold degenerate. The lowest excited states form the 2 D3/2
term, and they lie 2.3835 eV and 2.3846 eV above the ground state.
Since there are only three single-particle spatial wavefunctions available
with l = 1, namely the wavefunctions for m = ±1, 0, when we add an electron to nitrogen to form oxygen, the overall wavefunction cannot be totally
antisymmetric in the spatial coordinates of the electrons. The result is thatin oxygen the electrons are less effective in keeping out of each other’s way
than are the electrons in carbon and nitrogen, and the ionisation energy ofoxygen (13.6 eV) is slightly smaller than that of nitrogen. The ground states
form the term 3 P2 , while states in the 3 P1 and 3 P0 terms lie 0.020 eV and
0.028 eV above the ground state. In these terms three of the electrons havecancelling orbital angular momenta as in nitrogen, so the orbital angularmomentum of the atom is just the single unit introduced by the fourth electron: hence the P in the ground-state term. Spin-orbit interaction causes
the ground state to have the largest available value of j, whereas in carbon
the reverse was the case.
The easiest way to understand fluorine, the element that follows oxygenin the periodic table, is to skip ahead two places from oxygen to neon, in
which a full house of six electrons is packed into the 2p states. Every spin ispaired and every value of the orbital angular momentum quantum number m
is used twice, so both the spin and the orbital angular momenta sum to zero.
Each of the six 2p electrons is exposed to a large fraction of the ten unitsof charge on the nucleus, so the ionisation energy is large, 21.6 eV, second
only to helium of all the elements. There are no low-lying excited states.
This fact together with the large ionisation potential makes neon chemicallyinactive.
We transmute neon into fluorine by taking away a unit of nuclear charge
and one of the 2p electrons. The ‘hole’ we have left in the otherwise complete
shell of 2p electrons behaves like a spin-half particle that carries one unit oforbital angular momentum. Hence the ground state of fluorine has s = 12
and l = 1. Spin-orbit interaction causes the 2 P3/2 term to lie 0.050 eV below
the 2 P1/2 term that also arises when spin-half is combined with one unit oforbital angular momentum. In the case of oxygen we encountered a similar
maximisation of j, and it turns out that the ground states of atoms withshells that are more than half full generally maximise j, while j is minimisedin the ground state of an atom with a shell that is less than half full.
We have now reached the end of the first long period of the table. The
second long period, from sodium to argon, is an almost perfect copy of theperiod we have just covered. Figure 10.3 illustrates this statement by showing the ionisation energies of the elements in the first three periods. Thereis an abrupt drop in the ionisation energy as we move from neon to sodium,from an inert noble gas to a highly reactive alkali metal. Then the ionisation
3 See Problem 10.6 for an explanation of why the ground state of carbon has l = 1
rather than l = 0 or l = 2, which are the other possible results of combining two electrons,each of which has l = 1.

10.3 Periodic table

241

Figure 10.3 Ionisation energies of the first nineteen elements.

Figure 10.4 The lowest-lying energy
levels of carbon silicon and germanium. Along this sequence the finestructure splitting between the threelowest-lying levels increases dramatically. In the case of germanium the
spread in the energies of the triplet
states is no longer negligible compared to the energy gap between thelowest-lying singlet state and the top
triplet state.

energy creeps up as one moves along the period, with two small setbacks, between magnesium and aluminium, and between phosphorus and sulfur, thatare associated with the start and the half-way point of the 3p, respectively.
10.3.2 The fourth and fifth periods
After reaching argon with its full 3p shell, one might expect to start fillingthe 3d shell. Actually the 4s states prove to be lower-lying because theirvanishing angular momentum allows much greater penetration of the cloudof negative charge associated with the electrons that have n ≤ 3. But oncethe 4s shell has been filled, filling of the 3d shell commences with scandium
and continues unbroken through zinc. Once the 3d shell is full, filling of the
4p shell commences, finishing with the noble gas krypton at Z = 39.
In the next period, filling of the 5s shell takes precedence over filling of
the 4d shell, and when, after cadmium at Z = 48, the 4d shell is full, fillingthe 5p shell takes precedence over filling the 4f shell. The last two periods
are very long and have complex patterns due to the availability of shells withlarge l that tend to be filled much later than shells with the same n but smalll.
It is instructive to compare the pattern of energy levels as we move down
one column of the periodic table. Figure 10.4 shows the lowest energy levelsof carbon and the elements, silicon and germanium, that lie beneath it in
the periodic table. As we saw above, carbon has at the bottom of its energylevel diagram a cluster of three very closely spaced energy levels that form theterms 3 Pj for j = 0, 1, 2. As we proceed through silicon and germanium thespacing within this cluster grows markedly because it is caused by spin-orbit
coupling, which scales like Z 4 (§8.2.1). By the time we reach silicon theenergy differences created by spin-orbit coupling are no longer very small
compared to the energy difference between the triplet and singlet states,
which P
we know is of electrostatic origin. The total electron spin operator
2
which isS 2 ≡ ( i Si ) does not commute with the term in the HamiltonianP
generated by spin-orbit interaction, which is proportional to i Si · Li . Solong as this term is small compared to the terms in the Hamiltonian that do
commute with S 2 , total electron spin is a good quantum number and it ismeaningful to describe the atom with a spectroscopic term such as 3 P0 . In

242

Problems

reality the atom’s wavefunction will include contributions from states thathave l 6= 1, but the coefficients of these terms will be very small, and for most
purposes they can be neglected. As the contribution to the Hamiltonian fromspin-orbit coupling grows, the coefficients in the ground-state wavefunctionof terms with l 6= 1 grows, and the designation 3 P0 becomes misleading. The
lowest-lying three levels of germanium can for most purposes be treated as
3
Pj terms. In the case of tin, which lies under germanium, the designation
3
Pj is highly questionable, and in lead, which lies under tin, it is valueless.

Problems
10.1 Show that when the state of a pair of photons is expanded asX|ψi =(10.33)bnn′ |ni|n′ i,nn′where {|ni} is a complete set of single-photon states, the expansion coefficients satisfy bnn′ = bn′ n .
10.2 By substituting from equation (10.22) for ψ0 into equation (10.20),
express the singlet state of an electron pair |ψ, 0, 0i as a linear combination of
products of the single-particle states |u, ±i and |v, ±i in which the individual
electrons are in the states associated with spatial amplitudes u(x) and v(x)with Sz returning ± 21 . Show that your expression is consistent with the Pauli
condition ann′ = −an′ n .
Given the four single-particle states |u, ±i and |v, ±i, how many linearly
independent entangled states of a pair of particles can be constructed if
the particles are not identical? How many linearly independent states are
possible if the particles are identical fermions? Why are only four of thesestates accounted for by the states in first excited level of helium?
10.3 Show that the exchange integral defined by equation (10.32b) is real
for any single-particle wavefunctions Ψ1 and Ψ2 .
10.4 The H− ion consists of two electrons bound to a proton. Estimate
its ground-state energy by adapting the calculation of helium’s ground-stateenergy that uses the variational principle. Show that the analogue for H− ofequation (10.30) isa0hHia = R(2x2 − 11(10.34)4 x) where x ≡ a .
Hence find that the binding energy of H− is ∼ 0.945R. Will H− be a stableion?
10.5∗ Assume that a LiH molecule comprises a Li+ ion electrostaticallybound to an H− ion, and that in the molecule’s ground state the kinetic
energies of the ions can be neglected. Let the centres of the two ions be
separated by a distance b and calculate the resulting electrostatic binding
energy under the assumption that they attract like point charges. Given thatthe ionisation energy of Li is 0.40R and using the result of Problem 10.4,
show that the molecule has less energy than that of well separated hydrogenand lithium atoms for b < 4.4a0 . Does this calculation suggest that LiH is astable molecule? Is it safe to neglect the kinetic energies of the ions within
the molecule?
10.6∗ Two spin-one gyros are a box. Express that states |j, mi in which
the box has definite angular momentum as linear combinations of the states|1, mi|1, m′ i in which the individual gyros have definite angular momentum.
Hence show that1|0, 0i = √ (|1, −1i|1, 1i − |1, 0i|1, 0i + |1, 1i|1, −1i)3
By considering the symmetries of your expressions, explain why the groundstate of carbon has l = 1 rather than l = 2 or 0. What is the total spin
angular momentum of a C atom?

Problems

243

10.7∗ Suppose we have three spin-one gyros in a box. Express the state|0, 0i of the box in which it has no angular momentum as a linear combinationof the states |1, mi|1, m′ i|1, m′′ i in which the individual gyros have welldefined angular momenta. Hint: start with just two gyros in the box, giving
states |j, mi of the box, and argue that only for a single value of j will it bepossible to get |0, 0i by adding the third gyro; use results from Problem 10.6.
Explain the relevance of your result to the fact that the ground state ofnitrogen has l = 0. Deduce the value of the total electron spin of an N atom.
10.8∗ Consider a system made of three spin-half particles with individual
spin states |±i. Write down a linear combination of states such as |+i|+i|−i(with two spins up and one down) that is symmetric under any exchange ofspin eigenvalues ±. Write down three other totally symmetric states and saywhat total spin your states correspond to.
Show that it is not possible to construct a linear combination of productsof |±i which is totally antisymmetric.
What consequences do these results have for the structure of atoms suchas nitrogen that have three valence electrons?

11
Adiabatic principle
We often need to understand the quantum mechanics of systems that havea large number of degrees of freedom. We might. for example, be interested
in the speed at which sound waves propagate through a macroscopic crystalof diamond. This depends on the deformability of the bonds between the
crystal’s carbon atoms, which is in turn determined by the orbits of theatoms’ electrons. Also relevant is the inertia of a block of crystal, which
is mostly contributed by the carbon nuclei. These nuclei are dynamicalsystems, in which protons and neutrons move at mildly relativistic speed.
Each proton or neutron is itself a dynamical systems in which three quarks
and some gluons race about relativistically. When a sound wave passes
through the crystal, each nucleus experiences accelerations that must affectits internal dynamics, and the dynamics of its constituent quarks. Is there
any chance that a sound wave will induce a nucleus to transition to an excitedstate? Could a sound wave cause an atom to become electronically excited?
So long as such transitions are realistic possibilities, it is going to be
extremely difficult to calculate the speed of sound, because the calculationis going to involve atomic physics, nuclear physics and quantum chromodynamics – the theory strong interactions and quarks, which governs theinternal structure of protons and neutrons. The adiabatic approximation,
which is the subject of this chapter, enables us to infer that such transitions
are exceedingly unlikely to occur. Consequently, in this case and a vast number of similar situations, the adiabatic approximation greatly simplifies our
problem by permitting us to neglect phenomena, such as electron or nuclear
excitation, that have energy scales that are significantly larger than the characteristic energy scale of the phenomenon under investigation, even though
the different degrees of freedom are dynamically coupled. Moreover, we shallsee that the adiabatic approximation enables us to calculate quantities suchas the spring constant of the bonds that bind a crystals’s atoms from the
dynamics of the electrons that form these bonds. It also provides the theoretical underpinning for the kinetic theory of gases, for most of condensedmatter physics and much of chemistry. It is enormously important for thedevelopment of quantum field theory and our understanding of quantum
chromodynamics. Hence, the adiabatic approximation is an extraordinarily
important tool with applications that span the natural sciences.
We start by deriving the adiabatic approximation. Then we study inturn elementary applications of it to kinetic theory, to thermodynamics, tocondensed-matter physics, and to chemistry.

11.1 Derivation245

11.1 Derivation of the adiabatic principle
In §2.2 we stressed that the tdse (2.26) is valid even when H is timedependent. However, we have mostly confined ourselves to Hamiltonians
that are time-independent. In §9.3 we did consider a time-dependent Hamiltonian, but we assumed that the time-dependent part of H was small. Now
we consider the case in which H can change by an amount that is large, solong as the time T over which this change takes place is long in a sense thatwill be specified below.
We consider the dynamics of a system that has a slowly varying Hamiltonian H(t). At any instant, H(t) has a complete set of eigenkets |En (t)iand eigenvalues En (t). For the case of vanishing time dependence, equation (9.43) provides an appropriate trial solution of the tdse (2.26). Aftermodifying this solution to allow for time-variation of the En , we have|ψ, ti =Xan (t) exp

n



i
−
h̄

Z t
0


dt En (t ) |En (t)i.
′′(11.1)

Since for each t the set {|En (t)i} is complete and the numbers an (t) arearbitrary, no assumption is involved in writing down this expansion of thesystem’s ket. When we substitute the expansion (11.1) into the tdse, we
find


Z
X
i t ′
∂|ψi
= H|ψi =an exp −dt En (t′ ) H(t)|En (t)i
ih̄
∂t
h̄ 0
n


X
∂|En i={ih̄ȧn + an En (t)}|En (t)i + ih̄an(11.2)
∂t
a


Z
i t ′
′
× exp −
dt En (t ) .
h̄ 0Exploiting the fact that |En (t)i is an eigenket of H(t) we can cancel a termfrom each side and are left with


Z
X
i t ′
∂|En i ′
exp −dt En (t ) .
(11.3)
ȧn |En (t)i + an
0=
∂t
h̄ 0
n
Now we use the perturbation theory developed in §9.1 to expand |En (t+δt)i as a linear combination of the complete set {|En (t)i}. That is, we write|En (t + δt)i − |En (t)i =

X

m6=n

bnm |Em (t)i,(11.4)

where from (9.9) we havebnm =hEm (t)|δH|En (t)i
En (t) − Em (t)(11.5)

with δH the change in H between t and t + δt. Dividing equation (11.4) byδt and substituting the result into (11.3) we find
 i Z t
X hEm (t)|Ḣ|En (t)iX|Em i exp −ȧn |En (t)i+an
dt′ En (t′ ) .
0=
En (t) − Em (t)
h̄ 0
n
m6=n

(11.6)

246

Chapter 11: Adiabatic principle

Figure 11.1 A plot of sin(kx) times the slowly varying function 1/(1 + x2 ). As k → ∞and the wavelength of the oscillations becomes shorter, the negative contribution fromeach section of the curve below the x axis more nearly cancels the positive contribution
from the preceding upward section.

When we multiply through by hEk (t)| this yieldsȧk = −Xn6=kan (t)

 i
hEk (t)|Ḣ|En (t)iexp −En (t) − Ek (t)
h̄

Z t
0


dt′ {En (t′ ) − Ek (t′ )} .

(11.7)
Although we have used first-order perturbation theory, our working so far
has been exact because we can make δH as small as we please by taking
δt to be small. Now we introduce an approximation by supposing that H
is a slowly varying function of time in the sense that it changes by very
little in the time h̄/ min(|En − Ek |), which is the time required for significant
motion to occur as a result of interference between the stationary states withenergies En and Ek (§3.2). In this approximation, the right side of equation(11.7) is a product of a slowly varying function of time and an approximately
sinusoidal term that oscillates much more rapidly. When we integrate thisexpression to get the change in ak , the integral vanishes rather precisely
because the contributions from adjacent half-periods of the oscillating factor
nearly cancel (Figure 11.1). Hence, if initially ak = 1 for some k, it willremain unity throughout the evolution. This completes the derivation of theadiabatic approximation: if a system is initially in the k th state of welldefined energy, it will stay in this state when the Hamiltonian is changed
sufficiently slowly.

11.2 Application to kinetic theory
Consider air that is being compressed in the cylinder of a bicycle pump.
The air resists the compression by exerting pressure on the cylinder andits piston, and it grows hot as we drive the piston in. This phenomenon
is usually explained by treating the air molecules as classical particles thatbounce elastically off the cylinder walls. In this section we use the adiabatic
principle to interpret the phenomenon at a quantum-mechanical level.
We proceed by first imagining that there is only one molecule in thecylinder, and then making the assumption that when there are a large number N of molecules present, the pressure is simply N times the pressure we
calculate for the single-particle case. The Hamiltonian that governs our basic
system, a particle in a box, is
H(t) =p2+ V (x, t),2m(11.8)

where the potential V (x, t) is provided by the walls of the box. The simplest
model is
0 for x in the cylinder
V (x, t) =(11.9)∞ for x in a wall or the piston.
The time dependence of V arises because the piston is moving. We needto find the eigenvalues En and eigenkets |En i of the Hamiltonian (11.8).

11.2 Kinetic theory247

We work in the position representation, in which the eigenvalue equation
becomesh̄2 2−∇ un + V un = En un(11.10)2mwith un (x) ≡ hx|En i. From §5.1.1(a) we have that un should vanish on
the walls of the cylinder and the piston. For x inside the cylinder, thesecond term on the left of equation (11.10) vanishes, so En and un (x) are
the solutions to−h̄2 2∇ un = En un
2m

with

un = 0 on the boundary.

(11.11)

We assume that the cylinder’s cross section is rectangular or circular, so
coordinates exist such that (i) the cylinder’s walls are all surfaces on which
one coordinate vanishes and (ii) the Laplacian operator separates. That is,
we can write∂2∇2 = ∇22 + 2 ,(11.12)∂zwhere ∇22 is an operator that depends on the two coordinates, x and y,that specify location perpendicular to the cylinder’s axis, and z is distancedown that axis. In this case, we can find a complete set of solutions toequation (11.11) for eigenfunctions that are products un (x) = X(x, y)Z(z)of a function X of x and y, and a function of z alone. Substituting thisexpression for un into equation (11.11) and rearranging, we findZ∇22 X +

2mEn
d2 Z
XZ
=
−X
.
dz 2
h̄2

(11.13)

When we divide through by XZ, we find that the left side does not dependon z while the right side does not depend on x or y. It follows that neitherside depends on any of the coordinates. That is, both sides are equal tosome constant, which we may call 2mEz /h̄2 . This observation enables us toseparate our original wave equation into two equations
2m(En − Ez )
X
h̄2
2mEz
d2 Z
− 2 =
Z.
dz
h̄2

−∇22 X =(11.14)The physical content of these equations is clear: Ez is the kinetic energy
associated with motion along the cylinder’s axis, so motion perpendicular to
the axis carries the remaining energy, En − Ez . As we push in the piston,neither the equation governing X and En − Ez nor its boundary conditions
change, so En − Ez is invariant. What does change is the boundary condition
subject to which the equation for Z has to be solved.
We place one end of the cylinder at z = 0 and the piston at z = L.
Then it is easy to see that the required solution for Z is [cf. §5.1.1(a)]Z(z) ∝ sin(kπz/L) withk = 1, 2, . . . ,(11.15)and the possible values of Ez are
Ez =

h2 2
k .
8mL2

(11.16)

The adiabatic principle assures us that if we let the piston out slowly,
the particle’s value of the quantum number k will not change, and its energy
Ez will evolve according to equation (11.16). By conservation of energy, theenergy lost by the particle when L is increased by dL must equal the workthat the particle does on the piston, which is P dV, where P is the pressure

248

Chapter 11: Adiabatic principle

it exerts and dV is the increase in the cylinder’s volume. Let A be the areaof the piston. Then conservation of energy requires that−dEz = 2Ez

dL= P A dL,
L

(11.17)from which it follows that
P =

Ez
2Ez
=2 .
AL
V

(11.18)N
hEz i ,
V

(11.19)When we sum the contributions to the pressure that arise from a large
number, N , of molecules in the cylinder, equation (11.18) yields
P =2

where the angle brackets mean the average over all molecules. At this point
we have to take into account collisions between the N molecules. Colliding
molecules change the directions of their momenta and thus transfer energy
between motion in the z direction and motion in the plane perpendicular toit. Collisions do not satisfy the adiabatic approximation, so they do change
the quantum numbers of particles. Their overall effect is to ensure that thevelocity distribution remains isotropic even though the piston’s motion ischanging Ez and not the energy of motion in the plane of the piston, En − Ez .
So we may assume that hEi = 3 hEz i. Let U ≡ N hEi be the internal energyof the gas. Then eliminating hEz i from equation (11.19) in favour of U , weobtainP V = 23 U.
(11.20)This result is identical with what we obtain by combining the equation ofstate of an ideal gas, P V = N kB T , with the expression for the internalenergy of such a gas, U = 23 N kB T . Actually our result is more generalthan the result for an ideal gas because we have not assumed that the gas
is in thermal equilibrium: the only assumption we have made about the
distribution of kinetic energy among particles is that it is isotropic.

11.3 Application to thermodynamics
In §6.4 we saw that when a system is in thermodynamic equilibrium, we
do not know what quantum state it is in but can assign a probability pi ∝e−Ei /kB T that it is in its ith stationary state (eq. 6.93a). The energy Ei
of this state depends on the variables, such as volume, electric field, shear
stress, etc., that quantify the system’s environment. In the simplest nontrivial case, that in which the system is a fluid, the only relevant variable isthe volume V and we shall consider only this case. Hence we consider theenergy of each stationary state to be a function Ei (V).
In an adiabatic compression of our system, we slowly change V while
isolating the system from heat sources. From the adiabatic principle it follows
that during such a compression the system remains in whatever stationary
state it was in when the compression started. Consequently, the probabilities
pi of its being
P in the various stationary states are constant, and the entropy
S = −kB i pi ln pi (eq. 6.91) is constant during an adiabatic change, justas classical thermodynamics teaches.
P During an adiabatic compression, the change in the internal energy U =i pi Ei is
dU =

X
i

pi

∂Ei
= −P dV
dV

where P ≡ −

X
i

pi

∂Ei
.
∂V

(11.21)Since there is no heat flow, the increment in U must equal the work done,
which is the pressure that the system exerts times −dV, so the quantity P
defined by equation (11.21) is indeed the pressure.

11.4 Compressibility of condensed matter

249

11.4 The compressibility of condensed matter
As a second application of the adiabatic principle, we estimate the compressibility of solids and liquids. In condensed matter atoms touch one anotherand the volume of the bulk material can be reduced only if every atom ismade smaller. If an atom’s electrons are to be confined to a smaller volume,by the uncertainty principle, their momenta and therefore their kinetic energies must increase. We estimate the compressibility of matter by equating
the work done in compressing it to the resulting increase in the energy of
the atom. The adiabatic approximation tells us that during slow compression, the atom remains in its ground state. Hence the compressibility can be
deduced if we can calculate the ground-state energy E0 as a function of theatom’s volume V.
Compressibility χ is defined to be the fractional change in volume perunit applied pressure P :
1 dV
χ=−
.
(11.22)V dP
Conservation of energy implies that −P dV, the work done by the compressor,
is equal to the increase in the ground-state energy dE0 , so P = −dE0 /dVand 2 −1
d E0
χ= V.
(11.23)dV 2
E0 (V) can be obtained by solving for the atom’s stationary states with theelectronic wavefunction required to vanish on the surface of a sphere of volume V. A highly simplified version of such a calculation enables us to obtaina rough estimate of the compressibility of condensed matter.
We assume that when the atom is confined in a sphere of radius a,its wavefunction hx|ai is the same as the wavefunction for the confining
sphere of radius a0 with all distances rescaled by a/a0 and the appropriate
adjustment in the normalisation. In this case, we can argue as in §9.2 thatthe expectation value ha|K|ai of the atom’s kinetic energy operator K scalesas (a0 /a)2 , while the expectation value of the potential-energy operator V
scales as a0 /a. HencedE0
dha|K|ai ha|V |ai=(ha|K|ai + ha|V |ai) ≃ −2
−
.
da
da
a
a

(11.24)Equation (8.52) states that 2ha|K|ai = −ha|V |ai, so the right side of thisequation vanishes.1 Differentiating again, we findd2 E0
ha|K|aiha|V |aiE0≃6+2= −2 2 ,
da2
a2
a2
a

(11.25)where equation (8.52) has been used again to simplify the right side. Since
V ∝ a3 , dV/da = 3V/a and bearing in mind our result that dE0 /da = 0 wefind
 a 2 d2 E
d2 E0
E0
0
≃
= − 92 2 .
(11.26)
dV 2
3V
da2
V

Using this result in equation (11.23), we conclude that the compressibility isχ ≃ 92V.
|E0 |(11.27)

Some care is required in the application of this result to many-electron
atoms. Our assumption that ha|V |ai scales as a−1 is valid only if the wavefunction is simultaneously rescaled in the coordinates of all the atom’s electrons. Unfortunately, it is physically obvious that, at least for small fractional
1 Equation (8.52) was actually only derived for hydrogen, but the result applies to thegross structure of any atom.

250

Chapter 11: Adiabatic principle

changes in volume, only the outermost shell of electrons will be significantly
affected by the confining sphere. So realistically we should assume that thesystem formed by the inner electron shells remains fixed and the wavefunction
is rescaled in its dependence on the coordinates of electrons in the outermost
shell. In this spirit we shall replace |E0 | by N times the atom’s ionisationenergy, where N is the number of electrons in the outermost shell. Since theelectrostatic potential produced by the nucleus and the fixed inner shells ofelectrons does not vary with radius as r−1 , ha|V |ai will not scale as a−1 andthe factor 29 in equation (11.27) will be in error. None the less, the equation
should correctly predict the order of magnitude of an atom’s compressibility.
For lithium we take V = 34 π(2a0 )3 and |E0 | = 5.39 eV to find χ =2.6×10−11 Pa−1 . The measured value varies with temperature and is of order
10−10 Pa−1 , which is in excellent agreement with our quantum-mechanicalestimate given the sensitivity of the latter to the adopted value of the rather
ill-defined parameter V.

11.5 Covalent bonding
The air we breathe, the living tissue of our bodies, and the plastics in theclothes, chairs and carpets that surround us, are held together by covalentbonds. These are bonds between two atoms of similar electronegativity, suchas two oxygen atoms, two carbon atoms or a carbon atom and an oxygen
atom. In this section we explain how they arise through the sharing by theatoms of one or more electrons. Unlike the ionic bonds that hold together acrystal of common salt, which are crudely electrostatic in nature, a covalentbond is intrinsically quantum-mechanical.

11.5.1 A model of a covalent bond
To show how covalent bonding works, we study a one-dimensional modelthat is not at all realistic but it is analytically tractable.2 We imagine aparticle of mass m that moves along the x axis in a potential V (x) that is
made up of two δ-function potentials of the type we introduced in §5.1.1(b).
The wells are separated by distance 2a:
V (x) = −Vδ {δ(x + a) + δ(x − a)}.

(11.28)We have placed the origin at the midpoint between the two wells, which we
can do without loss of generality. This placement ensures that the Hamiltonian commutes with the parity operator, and we can seek solutions of thetise that have well-defined parity. There are three distinct regions in which
V (x) = 0, namely x < −a, −a < x < a and x > a and in these regions the
wavefunction u(x) must be a linear combination of the exponentials e±kx ,where k is related to E byk=√
−2mE /h̄.

(11.29)With an eye to the construction of solutions of definite parity we let oursolutions in these regions be ekxu(x) ∝ cosh(kx) or sinh(kx) −kxefor x < −a,for −a < x < a,for x > a,(11.30)2 Physicists call a model that lacks realism but nonetheless captures the physicalessence of a phenomenon, a toy model.

11.5 Covalent bonding

251

Figure 11.2 Even- and odd-parity solutions to the two delta-function problem.

Bearing in mind that the wavefunction has to be continuous across the potential wells, we see from Figure 11.2 that solutions of each parity must beof the form
 ek(x+a)
for x < −a,u+ (x) = A × cosh(kx)/ cosh(ka) for −a < x < a,
 −k(x−a)e
for x > a,
(11.31)k(x+a)
 −efor x < −a,u− (x) = B × sinh(kx)/ sinh(ka) for −a < x < a,
 −k(x−a)e
for x > a,where the constants on each segment of the real line have been chosen to
ensure that the wavefunctions equal A and B, respectively, at x = a.
On account of the symmetry of the problem, it suffices to choose k foreach parity such that the equation (5.13) is satisfied at x = a. From thisequation we have that2K =

k{1 + tanh(ka)} for even parityk{1 + coth(ka)} for odd parity(11.32)where K is defined by equation (5.14). By expressing the hyperbolic functions in terms of eka , we can rearrange the equations intok
− 1 = ±e−2ka ,K(11.33)where the upper sign is for even parity. In the upper panel of Figure 11.3
the left and right sides of these equations are plotted; the solution k is theordinate at which the straight line of the left side intersects with the decaying
exponential plot of the right side. The value of k+ that we obtain fromthe upper curve associated with the even-parity case is always larger thanthe value k− obtained for the odd-parity case. By equation (11.29), theparticle’s binding energy increases with k, so the even-parity state is themore tightly bound. If we increase a, the exponential curves in the top panelof Figure 11.3 become more curved and approach the x-axis more rapidly.
Hence k+ diminishes, and k− grows. In the limit a → ∞, the exponentials
hug the axes ever more tightly and k+ and k− converge on the point k = K
at which the sloping line crosses the x-axis. This value of k is precisely thatfor an isolated well as we would expect, since in the limit a → ∞ the wells
are isolated from one another. The lower panel of Figure 11.3 shows theenergies associated with k± from equation (11.29).
Suppose our particle is initially in the ground state of two wells thatare distance 2a apart, and imagine slowly moving the two wells towards one
another. By the adiabatic principle, the particle stays in the ground state,

252

Chapter 11: Adiabatic principle

Figure 11.3 Graphical solution of equation (11.33). In the top panel, the exponential isdrawn for the case Ka = 1. Bottom panel: binding energy versus inverse separation. The
scale energy E0 = −h̄2 K 2 /2m is the energy of the bound state of an isolated δ-function
potential well.

which, as we have seen, moves to lower energies. Hence the particle loses
energy. Where does this energy go? It can only go into the mechanism that
positions the wells. A little thought reveals that if work is done on thismechanism, it must be resisting the mutual attraction by the holes. Hence
we have arrived at the conclusion that two potential wells that are jointly
binding a particle, can experience a mutual attraction that would not bepresent if the bound particle were absent.

11.5.2 Molecular dynamics
The toy model just presented describes an inherently quantum-mechanicalmechanism by which atoms experience mutual attraction or repulsion throughsharing a bound particle. An essentially identical calculation, in which theenergy Ee of the two shared electrons on an H2 molecule is studied as afunction of the separation b of the protons, enables one to understand thestructure of the H2 molecule. Analogously with the toy model, the energyof the shared electrons decreases monotonically with b, so in the absence ofthe mutual electrostatic repulsion of the protons, which had no analogue inour model, the electrons would draw the protons ever closer together. In
reality there is a particular value b0 of b at which the rate at which Ee (b)decreases equals the rate at which the electrostatic energy Ep (b) of the protons increases with decreasing b. The classical intranuclear separation of an
H2 molecule is b0 .

11.6 The wkbj approximation

253

A more complete theory is obtained by considering V (b) ≡ Ee + Ep to
be a dynamical potential in which the two nuclei move. The analysis of thisproblem proceeds in close analogy with our treatment of the hydrogen atomin §8.1: we introduce a reduced particle and observe that the Hamiltonianthat governs its dynamics commutes with the reduced particle’s angularmomentum operator; this observation enables us to derive a radial wave
equation for each value of the angular-momentum quantum number l. This
radial wave equation describes oscillations that are governed by the effective
potential V (b). The rotation-vibration spectrum of H2 may be understoodas arising from transitions between states that are characterised by l and thequantum number n associated with the oscillations in b.
Similar principles clearly apply to studies of the dynamics of many othermolecules: one starts by determining the energy of shared electrons for agrid of fixed locations of the nuclei. The resulting energies together withthe energy of electrostatic repulsion between the nuclei yields an effective
potential, that can then be used to study the quantum dynamics of thenuclei. The essential approximation upon which this kind of work depends
is that the frequencies at which the nuclei oscillate are low compared to any
difference between energy levels of the electronic system, divided by h̄. Sinceelectrons are so much lighter than nuclei, this approximation is generally an
excellent one when the molecular rotations and vibrations are not strongly
excited. The approximation is guaranteed to break down during dissociationof a molecule, however. We return to the toy model to explain why this isso.
11.5.3 Dissociation of molecules
In the model of §11.5.1, the force provided by the particle that is shared
between the potential wells is not always attractive: if the particle in the
excited, odd-parity bound-state the energy of the particle increases as theseparation of the wells 2a is diminished, so the positioning mechanism mustbe pushing the two wells together as it resists the mutual repulsion of thewells. Consider now a two-well molecule that is held together by the attractive force provided by the shared particle in its ground state when a photon
promotes the particle to its excited state. Then the force provided by theparticle becomes repulsive, and the wells will begin to move apart. As they
move, much of the energy stored in the excitation of the particle is converted
into kinetic energy of the wells, and soon there is one bare well and one wellwith a trapped particle.
As the wells move apart, the energy difference between the ground andexcited states decreases, while the rate of increase of a increases. Hencethe adiabatic approximation, which requires that a/ȧ ≪ (Eo − Ee )/h̄ must
break down. In a more complex system, such as a real CO molecule, thisbreakdown can cause some of the energy stored in the particle’s excitation
being transferred to excitation of one or both of the final atoms rather thanbeing converted to the kinetic energy of their motion apart.

11.6 The WKBJ approximation
In §5.4 we learnt from a numerical solution of the tise that when a particle
encounters a modest ramp in its potential energy V , the amplitude for reflection is small unless the distance over which V changes is small comparedto the particle’s de Broglie wavelength. This result is closely related to theadiabatic approximation in the sense that in the particle’s rest frame, thepotential that it experiences changes slowly compared to the time taken to
cover a de Broglie wavelength. Now we given an important analytical argument that leads to the same conclusion and allows us to determine theevolution of the wave as it moves over the barrier.

254

Chapter 11: Adiabatic principle
The equation we wish to solve isd2 ψ= −k 2 ψ
dx22m(E − V ),h̄2(11.34)p
E − V (x′ ),(11.35)where k 2 (x) ≡and V (x) is some real function. We define
φ(x) ≡Z x′′

dx k(x ) =

√

2m
h̄

Z x

dx′

so k = dφ/dx. Then without loss of generality we write
ψ(x) = S(x)eiφ ,(11.36)

where S is a function to be determined. When we substitute this expressionfor ψ into the tise (11.34), cancel the right side with one of the terms on
the left and then divide through by eiφ , we obtain
d2 S
dS
dk
+ 2ik
+ iS
= 0.
dx2
dx
dx

(11.37)Now we reason that when k is a slowly varying function of x, S will be
also. In fact, if k changes on a lengthscale L ≫ 1/k that is greater than
the wavelength 2π/k, we will have |dk/dx| ∼ |k|/L, |dS/dx| ∼ |S|/L and|d2 S/dx2 | ∼ |S|/L2 . In these circumstances it follows that the first termin equation (11.37) is negligible compared to the other two, and we may
approximate the equation byd ln kd ln S= − 12dx
dx(11.38)Integrating both sides from x1 to x2 we find that√√(S k) x1 = (S k) x2 .

(11.39)The particle flux implied by ψ(x) is proportional to the probability density|S|2 times the particle speed k/m. Hence equation (11.39) states that theparticle flux at x1 is equal to that at x2 . In other words, when the wavenumber changes very little in a wavelength, the reflected amplitude is negligible
and the wavefunction is approximately

ψ(x) ≃ constant ×h̄22m(E − V )

1/4

eiφ

(11.40)where φ(x) is given by equation (11.35). This solution is known as the
WKBJ approximation.3 The wkbj approximation guarantees conservation of particle flux in the classical limit of very small de Broglie wavelengths.
It also has innumerable applications outside quantum mechanics, including
the working of ear trumpets, tidal bores and Saturn’s rings.

3 The wkbj approximation is named after Wentzel, Kramers, Brillouin and Jeffreys.
This is frequently abbreviated to ‘wkb approximation’.

Problems

255

Problems
11.1 In §9.1 we obtained estimates of the amount by which the energyof an atom changes when an electric or magnetic field is applied. Discuss
whether the derivation of these results implicitly assumed the validity of theadiabatic principle.
11.2 In §11.2 we assumed that the potential energy of air molecules isinfinitely large inside a bicycle pump’s walls. This cannot be strictly true.
Give a reasoned order-of-magnitude estimate for the potential in the walls,
and consider how valid it is to approximate this by infinity.
11.3 Explain why E/ω is an adiabatic invariant of a simple harmonic oscillator, where ω is the oscillator’s angular frequency. Einstein proved thisresult in classical physics when he was developing the “old quantum theory”,
which involved quantising adiabatic invariants such as E/ω and angular momentum. Derive the result for a classical oscillator by adapting the derivationof the wkbj approximation to the oscillator’s equation of motion ẍ = −ω 2 x.
11.4 Consider a particle that is trapped in a one-dimensional potential
well V (x). If the particle is in a sufficiently highly excited state of thiswell, its typical de Broglie wavelength may be sufficiently smaller than thecharacteristic lengthscale of the well for the wkbj approximation to be valid.
Explain why it is plausible that in this case
1
h̄

Z x2
x1

dx′

p
2m{E − V (x′ )} = nπ,(11.41)

where the E − V (xH i ) = 0 and n is an integer. Relate this condition to thequantisation rule dx px = nh used in the “old quantum theory”.

11.5 Show that the “old quantum theory” (Problem 11.4) predicts thatthe energy levels of the harmonic oscillator are nh̄ω rather than (n + 21 )h̄ω.
Comment on the dependence on n of the fractional error in En .

11.6 Suppose the charge carried by a proton gradually decayed from its
current value, e, being at a general time f e. Write down an expression forthe binding energy of a hydrogen atom in terms of f . As α → 0 the binding
energy vanishes. Explain physically where the energy required to free theelectron has come from.
When the spring constant of an oscillator is adiabatically weakened bya factor f 4 , the oscillator’s energy reduces by a factor f 2 . Where has theenergy gone?
In Problems 3.14 and 3.15 we considered an oscillator in its groundstate when the spring constant was suddenly weakened by a factor f = 1/16.
We found that the energy decreased from 21 h̄ω to 0.2656h̄ω not to h̄ω/512.
Explain physically the difference between the sudden and adiabatic cases.
11.7 Photons are trapped inside a cavity that has perfectly reflecting walls
which slowly recede, increasing the cavity’s volume V. Give a physical motivation for the assumption that each photon’s frequency ν ∝ V −1/3 . Using
this assumption, show that the energy density of photons u ∝ V −4/3 andhence determine the scaling with V of the pressure exerted by the photons
on the container’s walls.
Black-body radiation comprises an infinite set of thermally excited harmonic oscillators – each normal mode of a large cavity corresponds to a new
oscillator. Initially the cavity is filled with black-body radiation of temperature T0 . Show that as the cavity expands, the radiation continues tobe black-body radiation although its temperature falls as V −1/3 . Hint: useequation (6.121).

256

Problems

11.8 Show that when a charged particle gyrates with energy E in a uniform magnetic field of flux density B, the magnetic moment µ = E/B isinvariant when B is changed slowly. Hint: recall Problem magmomentprob.
By applying the principle that energy must be conserved when the magnetic
field is slowly ramped up, deduce whether a plasma of free electrons forms apara- or dia-magnetic medium.

12
Scattering Theory
In this chapter we study situations in which a free particle approaches aregion of enhanced potential, is deflected, and moves away in a new direction.
Different potentials lead to different probabilities for a particle to be scatteredin a particular direction, so by carefully measuring the outcomes of repeatedscattering experiments, we can infer the potential that was responsible.
In fact, most of what we know about the small-scale structure of matterhas been learnt this way. For example, Rutherford discovered that atoms
have dense, compact nuclei by studying the distribution of α-particles scattered by gold foil, while nowadays we study the sub-atomic structure of matter by scattering extremely fast-moving electrons or protons off one anotherin high-energy accelerators.
The task of scattering theory is to build a bridge between the Hamiltonians that govern the evolution of states and the quantities – cross sections andbranching ratios – that are actually measured in the laboratory. In §5.3 weinvestigated the scattering of particles that are constrained to move in one
dimension, and found that quantum mechanics predicts qualitatively new
scattering phenomena. We expect the freedom to move in three dimensions
rather than one to be fundamental for the physics of scattering, so in thischapter we investigate three-dimensional scattering. We shall find that the
new phenomena we encountered in §5.3 do carry over to physically realistic
situations.

12.1 The scattering operator
Let |ψi be the state of a particle in a scattering experiment. The evolutionof |ψi is governed by the tdseih̄∂
|ψi = H|ψi∂t⇔

|ψ; ti = U (t)|ψ; 0i,(12.1a)where U (t) = e−iHt/h̄ is the time evolution operator introduced in §4.3. We
break the Hamiltonian into a sum H = HK +V of the kinetic-energy operator
HK = p2 /2m and the potential V that causes scattering. If |ψi represents amoving particle – one that approaches or leaves some collision – it must be a
non-trivial superposition of the eigenstates of H – see §2.3.3. Unfortunately,258

Chapter 12: Scattering Theory

when V 6= 0 we may not know what these eigenstates are, and it may beprohibitively difficult to find them.
A crucial physical insight allows us to make progress: in a scatteringexperiment, long before the particle reaches the interaction region it is approximately free. The evolution of a free particle |φi is governed by HKih̄∂|φi = HK |φi or |φ; ti = UK (t)|φ; 0i,∂t(12.1b)where UK (t) = e−iHK t/h̄ , so the statement that |ψi behaves like a free particle
in the asymptotic past is the requirement thatlim U (t)|ψ; 0i = lim UK (t)|φ; 0i

t→−∞

t→−∞

(12.2a)for some free state |φi. Implicit in this equation is the assumption that theorigin of time is chosen such that the interaction takes place at some finite
time. For example, it might be in full swing at time t = 0, which we shallsometimes refer to as ‘the present’.
In the asymptotic future the scattered particle will have moved far awayfrom the interaction region and will again be approximately free. Hence wealso requirelim U (t)|ψ; 0i = lim UK (t)|φ′ ; 0i(12.2b)t→+∞t→+∞where |φ′ i is another free state.
Equations (12.2) allow us to relate the real state |ψi to both |φi and|φ′ i at the present time via|ψ; 0i = lim U † (t)UK (t)|φ; 0i = ′ lim U † (t′ )UK (t′ )|φ′ ; 0it→−∞′= Ω+ |φ; 0i = Ω− |φ ; 0i,t →+∞(12.3)

where the operators Ω± are defined by1Ω± ≡ lim U † (t)UK (t).
t→∓∞(12.4)The origin of the irritating choices of sign in this definition will be explainedin §12.2 below. In terms of the Ω± operators, the scattering operator S
is defined asS ≡ Ω†− Ω+ .
(12.5)
Here’s what the scattering operator does: first, S evolves a free-particle stateback to the distant past, then matches it onto a real state which has the same
past asymptotic behaviour. Next, S evolves this real state forwards – all theway through the scattering process to the far future. There, the real particle
again behaves like a free particle, and S matches their states before finally
evolving the free state back to the present. If the real particle is in somestate |ψi, and looks like a free-particle state |φi well before the interaction
occurs, then the amplitude for it will look like some other free state |λi longafter the interaction is hλ|S|φi. Hence the probability to find the particle in
the free state |λi is just |hλ|S|φi|2 . The scattering operator is useful because
it always acts on free states, so if we use it we do not need to know the
eigenstates of the full Hamiltonian H.
Notice that S is defined as a product of four unitary evolution operatorsand is therefore itself unitary.
When V = 0, the particle isn’t scattered, and its future state is thesame as its past state. In such circumstances, the scattering operator mustbe just the identity operator S = 1, and we can check this is indeed true by
1 It is not self-evident that the limits as t → ±∞ that appear in equation (12.4) exist.
Appendix K derives a condition on V that ensures that Ω± is well defined.

12.1 The scattering operator

259

putting H = HK in equations (12.3) to (12.5). The operator that describes
a genuine interaction is the transition operator
T ≡S −1(12.6)and the probabilities for actual transitions are given by
Prob(|φi → |λi) = |hλ|T |φi|2 = |hλ|S|φi − hλ|φi|2 .

(12.7)

Since S is unitary, we have
1 = S†S = 1 + T † + T + T †T .

(12.8)

Squeezing this equation between hφ| and |φi we obtain−2ℜe (hφ|T |φi) = hφ|T † T |φi.

(12.9)We also have that22|hφ|S|φi| = |1 + hφ|T |φi| = 1 + 2ℜe(hφ|T |φi) + |hφ|T |φi|2 .

(12.10)Rearranging and using equation (12.9) yields
1 − |hφ|S|φi|2 =X|ψi i6=|φihφ|T † |ψi ihψi |T |φi =X|ψi i6=|φi|hψi |T |φi|2 ,(12.11)

where {|ψi i} is a complete set of states that includes |φi. The left side ofequation (12.11) is one, minus the probability that at t = +∞ the particle
is still in the state it was in at t = −∞, while the right side is the sum of
the probabilities that the particle has made the transition to some state |ψi idifferent from the original state |φi.
12.1.1 Perturbative treatment of the scattering operator
The definition S ≡ Ω†− Ω+ is difficult to use in practical calculations, because
the true Hamiltonian H (whose eigenstates we do not know) is buried rather
deep inside. To get at it we first differentiate Ω(t) ≡ U † (t)UK (t) with respect
to t, findingdiiΩ(t) = eiHt/h̄ (H − HK )e−iHK t/h̄ = eiHt/h̄ V e−iHK t/h̄ ,
dt
h̄
h̄

(12.12)where we have been careful to preserve the order of the operators. We now
re-integrate this equation between t′ and t to reachZi tΩ(t) = Ω(t ) +dτ eiHτ /h̄ V e−iHK τ /h̄
h̄ t′
Z
i t
′
dτ U † (τ )V UK (τ ).
= Ω(t ) +h̄ t′′

(12.13)

Taking the Hermitian adjoint of this equation we have
Z
i t
†
dτ UK
(τ )V U (τ )
h̄ t′
Z
i t
†
= Ω† (t′ ) −dτ UK(τ )V UK (τ )Ω† (τ ).
h̄ t′Ω† (t) = Ω† (t′ ) −(12.14)

260

Chapter 12: Scattering Theory

The integrand itself contains Ω† (τ ), but suppose we use this equation to
replace it by Ω(t′ ) plus an integral that involves Ω† (τ ′ ), and then repeat thisprocess once more. The result isZi t†Ω† (t) = Ω† (t′ ) −dτ UK(τ )V UK (τ ) Ω† (t′ )h̄ t′
Z t Z τ1†(τ )V UK (τ − τ ′ )V UK (τ ′ )Ω† (t′ )− 2 dτ dτ ′ UK
′
′
h̄ t
t
Z Z τ Z τi t†
+ 3 dτ dτ ′ dτ ′′ UK(τ )V UK (τ − τ ′ )V UK (τ ′ − τ ′′ )V UK (τ ′′ )Ω† (τ ′′ ).
h̄ t′ t′t′′(12.15)
Through repeated use of equation (12.14) we can push the operator Ω† (τ )for τ > t′ off into an integral that contains as many powers of V as we please.
For sufficiently small V , the magnitude of the term in which Ω(τ ) occurs will
be negligible, and we will be able to drop it. Then multiplying the equationby Ω(t′ ), and taking the limits t → ∞ and t′ → −∞, we obtain an expansionof S in powers of V . Since Ω† (t′ )Ω(t′ ) = 1, this expansion is
Z
i ∞
S =1−
dτ U † (τ )V UK (τ )h̄ −∞ KZ ∞ Z τ1†− 2
dτ dτ ′ UK(τ )V UK (τ − τ ′ )V UK (τ ′ )h̄ −∞ −∞Z ∞ Z τ Z τ′i†dτ dτ ′ dτ ′′ UK(τ )V UK (τ − τ ′ )V UK (τ ′ − τ ′′ )V UK (τ ′′ )+ 3h̄ −∞ −∞ −∞+ ···(12.16)The virtue of equation (12.16) is that all the evolution operators involve only
the free Hamiltonian HK – information about scattering has been encodedin the expansion in powers of V .2
Equation (12.16) has an intuitive physical interpretation. The zerothorder term is the identity operator and represents no scattering; its presence
was anticipated by equation (12.6). The term S (1) with one power of V acts
on a free particle as
hλ; 0|S

(1)

Z
i ∞
†
|φ; 0i = −dτ hλ; 0|UK(τ )V UK (τ )|φ; 0i
h̄ −∞
Z ∞
i
dτ hλ; τ |V |φ; τ i.
=−
h̄ −∞

(12.17)The integrand hλ; τ |V |φ; τ i is the amplitude for a particle in the free state |φito be deflected by the potential V at time τ , transferring it into another freestate |λi. Since we only observe the initial and final states, we do not know
when the interaction took place, so we add the amplitudes for the deflection
to have occurred at any time. Similarly, the second-order term S (2) gives theamplitudehλ; 0|S(2)

 2 Z ∞ Z τi|φ; 0i =dτ dτ ′ hλ; τ |V UK (τ − τ ′ )V |φ; τ ′ i
h̄
−∞ −∞

(12.18)for an incoming particle in the free state |φi to be deflected by the potential
at time τ ′ , then to propagate freely for a further time τ − τ ′ , and finallyto be deflected again by V into the final state |λi. Since we do not know
2 This expansion is reminiscent of the perturbation theory developed in §9.1. However,that theory hinged on the assumption that the response of the system to changes in its
Hamiltonian is analytic in the parameter β. Here we need no such assumption. Instead
we guess that certain integrals become small.

12.2 The S-matrix

261

when either deflection occurred, we integrate over all τ and τ ′ , subject tothe condition τ ′ < τ that the first deflection happens earlier. Higher-orderterms describe trajectories that involve larger numbers of deflections.
If the potential is sufficiently weak, we might hope to approximate equation (12.16) by its lowest-order terms
Z
i ∞
S ≃1−
dτ U † (τ )V UK (τ ).
(12.19)h̄ −∞ K
This drastic curtailing of the series for S is known as the Born approximation. Whether it is a good approximation in a given physical situation must
be checked, often by estimating the order-of-magnitude of the second-orderterm S (2) , and checking it is acceptably smaller than the Born term.

12.2 The S-matrix
It is impossible to put any physical particle into a pure energy eigenstate,because such states are not localised in time. Nonetheless, energy eigenstates
are useful as mathematical tools, being simpler to handle than realistic superpositions. Calculating with energy eigenstates presents special problems
in scattering theory, because the idea that the particle moves towards thepotential is central to our entire formalism, but a particle that is in an energy
eigenstate goes nowhere.
12.2.1 The iǫ prescription
To get to the root of the problem, notice that equation (12.4) implies thatde−iHt/h̄ Ω±
dt t=0d−iHt/h̄ †lim eU (τ )UK (τ )= ih̄
dt t=0 τ →∓∞


d
†
= ih̄
lim U (τ − t)UK (τ − t)UK (t)dt t=0 τ →∓∞de−iHK t/h̄ = Ω± HK .
= ih̄Ω±
dt t=0HΩ± = ih̄

(12.20)

Therefore, if the interacting state |ψi initially resembles some eigenstate|E; freei of the free Hamiltonian, then equations (12.2a) and (12.20) imply
that
H|ψi = HΩ+ |E; freei = Ω+ HK |E; freei = EΩ+ |E; freei = E|ψi,(12.21)so |ψi must actually be an eigenstate |E; truei of the true Hamiltonian, withthe same energy E. The trouble with this is that energy eigenstates look thesame at all times, so |E; truei and |E; freei would always look like each other– a state of affairs that only makes sense if V = 0 and there is no scattering.
Our argument shows that the initial and final states cannot have welldefined energy – they must be non-trivial superpositions of energy eigenstates. Nonetheless, because energy eigenstates often simplify otherwise difficult calculations, we are reluctant to forego them. Instead, we seek a way
to avoid the problem. From equation (12.13), write Ω± in the formΩ± = 1 +

i
h̄

i
=1+
h̄

Z ∓∞

dτ U † (τ )V UK (τ )

0

Z ∓∞
0

dτ

Z

(12.22)
3

3 ′

†

′

′

d x d x U (τ )|xihx|V |x ihx |UK (τ ) .

262

Chapter 12: Scattering Theory

When Ω± is applied to a real scattering state |φi, the integrand vanishes forlarge τ , because hx|UK |φi is non-negligible only far from the scattering centre,
where hx|V |xi′ ≃ 0. Hence, if we include a convergence factor e−ǫ|τ |/h̄ in the
integrand, for sufficiently small ǫ > 0 we make a negligible difference to theaction of Ω± on a real scattering state: for finite τ this factor approximatese thatunity to arbitrary accuracy as ǫ → 0. Consider therefore the operator Ω
has this harmless factor. Taking the limit that the constant ǫ approaches
zero from above we obtain


Z
i ∓∞
dτ U † (τ )V e−ǫ|τ |/h̄UK (τ ) |φi.
1 + lim+ǫ→0 h̄ 0(12.23)e ± on a state |φi that is a non-trivial superposition of enThe action of Ω
ergy eigenstates is identical to that of Ω± . However, Problem 12.1 shows thate ± satisfies an equation that differs crucially from equation
the product H Ω(12.21):e± = Ω
e ± (HK ± iǫ) ∓ iǫ.
HΩ(12.24)e ± |φi =Ω± |φi −→ Ωe ± |E; freei, where |E; freei is anConsequently, when we apply H to |ψi ≡ Ωeigenstate of HK , we finde ± |E; freei = (E ± iǫ)|ψi ∓ iǫ|E; freei,
H|ψi = H Ω(12.25)so |ψi is an eigenstate of H only when V = 0 and |ψi = |E; freei. Therefore,e ± to generate ‘interacting’ states from eigenstates of HK ,when we use Ωwhich will henceforth be simply labelled |Ei, our interacting states are notstationary states of the true Hamiltonian, and thus can describe scattering.
The crucial point that makes the whole procedure consistent is that for any
physically realistic superposition, it makes no difference whether we constructe ±.
interacting states with Ω± or Ωe
We can simplify Ω± a little: since |τ | = τ for τ ≥ 0 and |τ | = −τ forτ < 0, with |φi = |Ei equation (12.23) becomes
e ± |Ei =
Ω



Z
i ∓∞
†
−i(E±iǫ)τ /h̄|Ei.
dτ U (τ )V e
1 + lim+ǫ→0 h̄ 0(12.26)

Therefore, our modification merely supplements the energy eigenvalue Ewith a small imaginary piece +iǫ for initial states and −iǫ for final states– the sign on iǫ corresponds to the subscript on Ω± and is historically theorigin of the naming of the Ω operators. This procedure is known as theiǫ prescription. In practice the prescription is implemented by using theoriginal Ω± operators, but pretending that all eigenstates of HK satisfy
HK |Ei = (E + iǫ)|Ei for initial kets when acted on by Ω+
HK |E ′ i = (E ′ − iǫ)|E ′ i for final kets when acted on by Ω− .

(12.27a)

Similarly, the Hermitian adjoints of the modified operators (12.23) imply
that we should likewise pretend thathE|HK = hE|(E − iǫ) for initial bras when acted on by Ω†+

hE ′ |HK = hE ′ |(E ′ + iǫ) for final bras when acted on by Ω†− .

(12.27b)

In no way do we mean that the Hermitian operator HK actually has a complex eigenvalues E; equations (12.27) are merely useful fictions that enable
us to carry out the iǫ prescription.

12.2 The S-matrix

263

12.2.2 Expanding the S-matrix
Since the incoming and outgoing states are free states, and the momentum
operators commute with the free Hamiltonian HK , the scattering operator isconveniently studied in the momentum representation. We then work with
the S-matrix
S(p, p′ ) ≡ hp′ |S|pi.
(12.28)where we must use the iǫ prescription of equations (12.27) to interpret theaction of Ω+ on |pi that is implicit in this definition. From equation (12.16),
the lowest-order contribution to the S-matrix is then
Z
i ∞
′
′
hp |S|pi ≃ hp |pi −dτ hp′ |eiHK τ /h̄ V e−iHK τ /h̄ |pi
h̄ −∞
Z ∞
(12.29)i ′−i(Ep −Ep′ )τ /h̄′dτ e.
= hp |pi − hp |V |pih̄−∞
Here, we used the rules (12.27) to find that the argument of the exponential
in the integrand is actually independent of ǫ. We recognise the integral as
2πh̄δ(Ep − Ep′ ).
Potentials that depend only on position are diagonal in the x representation, so the momentum-space elements hp′ |V |pi are
hp′ |V |pi =Z

d3 x d3 x′ hp′ |x′ ihx′ |V |xihx|piZ,13−iq·x/h̄
=
d
x
e
V
(x)
(2πh̄)3(12.30)where we have used our expression (2.78) for the wavefunction of a state ofwell-defined momentum and defined the momentum transfer
q ≡ p′ − p.

(12.31)

Therefore, the Born approximation to the S-matrix just depends on the
Fourier transform of V (x):hp′ |S (1) |pi = −2πi
δ(Ep − Ep′ )(2πh̄)3Z

d3 x e−iq·x/h̄ V (x).

(12.32)

From the theory of Fourier transforms, we see that potentials which vary
rapidly with x lead to S-matrices that contain significant amplitudes for large
momentum transfers. Turning this around, if a particle suffers a large change
in momentum when it is scattered by V (x), we infer that V (x) has sharp
features. Arguing along these lines (albeit more classically), Rutherford wasable to deduce the existence of nuclei from the occasional back-scatteringof α-particles off gold foil. More recently, a team of physicists3 working
at SLAC in Stanford scattered high-energy electrons off protons; the electrons sometimes suffered large-angle scattering, providing evidence for the
existence of quarks inside the nucleons.
The second-order term in the scattering operator can be treated in asimilar manner. From equation (12.16) we find1
hp′ |S (2) |pi = − 2h̄

Z ∞ Z τ†dτ dτ ′ hp′ |UK(τ )V UK (τ − τ ′ )V UK (τ ′ )|pi. (12.33)−∞−∞

The free-evolution
operators can be evaluated by inserting the identity opR
erator 1 = d3 k |kihk| anywhere between the two V operators. Bearing in
3 D.H. Coward, et. al., Phys. Rev. Lett. 20, 292, (1968).

264

Chapter 12: Scattering Theory

mind that HK |pi = (Ep + iǫ)|pi and hp′ |HK = hp′ |(Ep′ + iǫ) in accordancewith the iǫ prescription, we find
′

hp |S

(2)


d k hp′ |V |kihk|V |pi
Z ∞ Z τ′ i(Ep′ +iǫ−Ek )τ /h̄ −i(Ep −Ek +iǫ)τ ′ /h̄× dτ dτ ee.

1|pi = lim − 2
ǫ→0+
h̄

Z

3

−∞

−∞

(12.34)The integral over τ ′ isZ τ′e−i(Ep −Ek +iǫ)τ /h̄,dτ ′ e−i(Ep −Ek +iǫ)τ /h̄ = ih̄Ep − Ek + iǫ−∞

(12.35)so the second-order contribution to the S-matrix is
Z
Z ∞
′
i
(2)
′
3 hp |V |kihk|V |pi
S (p , p) = lim+ −d kdτ e−i(Ep −Ep′ )τ /h̄h̄Ep − Ek + iǫ
ǫ→0−∞Z
hp′ |V |kihk|V |pi.
= −2πi δ(Ep − Ep′ ) lim
d3 k
Ep − Ek + iǫ
ǫ→0+(12.36)
The numerator in the integrand is the amplitude for the particle to scatterfrom the |pi state into the |ki state, and then from the |ki state into the |p′ istate. The denominator arose from the integration over τ ′ , which in turn waspresent because the particle travelled freely for some time τ − τ ′ in between
the two interactions. Since it comes from this free propagation, the factor(Ep −Ek +iǫ)−1 is known as the propagator, written here in the momentum
representation. Finally, because we do not measure the intermediate state,
equation (12.36) adds up the amplitudes for scattering via any state.
Higher order terms are handled in a similar way: V occurs n times in
S (n) (p′ , p), so there are n − 1 intermediate evolution operators, leading to
n − 1 propagators. Similarly, there are n − 1 sets of intermediate states, all ofwhich are integrated over. S (n) (p′ , p) may be represented diagrammaticallyas in Figure 12.1. These Feynman diagrams are an order-by-order bookkeeping system for calculating contributions to the S-matrix: each term inthe series for the S-matrix corresponds to a diagram, and Feynman rules
can be defined that enable the algebraic expression for the term to be inferredfrom the diagram. Thus Feynman diagrams summarise complicated integrals
in an intuitive way.
The Feynman rules required here are extremely simple: (i) each vertex
has just two lines going into it and is associated with a factor V ; (ii) each
‘internal line’ (one that has a vertex at each end) is associated with the propagator (Ep − Ek + iǫ)−1 , where k, which is integrated over, is the momentum
carried by that line; (iii) there is an overall prefactor −2πiδ(Ep − Ep′ ), where
p and p′ are the ingoing and outgoing momenta, respectively. With theserules we can only construct one diagram with a given number n of vertices,
and it’s a simple chain. Feynman diagrams become much more interestingand valuable when one recognises that when an electron is scattered by anelectrostatic potential V (x), for example, it really collides with a photon, andone needs to include the coupled dynamics of the photons. In this more sophisticated picture, V (x) is replaced by the electromagnetic vector potential
A, which becomes a quantum-mechanical object, and our diagrams include
propagators for both photons and electrons. Moreover, the vertices become
points at which three or more lines meet, two for the incoming and outgoing
electron, and one or more for photons. With a richer set of lines and vertices
on hand, many different diagrams can be constructed that all have the same
number of vertices, and therefore contribute to the S-matrix at the sameorder.

12.2 The S-matrix

265

Figure 12.1 Feynman diagramsfor the scattering process to lowest
orders in V .

12.2.3 The scattering amplitude
Both the first- and second-order approximations to the S-matrix are proportional to an energy-conserving delta function. This result is not limited
to the series expansion for S, but actually holds for the exact S-matrix as
we now demonstrate. Equation (12.20) and its Hermitian adjoint state thatHΩ± = Ω± HK and Ω†± H = HK Ω†± . Now S ≡ Ω†− Ω+ , so
HK S = HK Ω†− Ω+ = Ω†− HΩ+ = Ω†− Ω+ HK = SHK ,(12.37)that is, [S, HK ] = 0. Sandwiching this commutation relation between momentum eigenstates and using the iǫ prescription of equations (12.27) gives
the relation
0 = hp′ |[S, HK ]|pi = (Ep + iǫ − Ep′ − iǫ)S(p, p′ ) = (Ep − Ep′ )S(p, p′ ),(12.38)so the S-matrix vanishes unless the initial and final states have the same
(real) energy. This tells us that the exact S-matrix must have the form
S(p, p′ ) ∝ δ(Ep − Ep′ ).
In equation (12.6) we broke S into the sum S = 1 + T to isolate thescattering amplitude, and it is clear that hp′ |T |pi is also proportional toδ(Ep −Ep′ ). Motivated by this insight we define the scattering amplitude
f (p → p′ ) by
hp′ |T |pi =i
δ(Ep − Ep′ )f (p → p′ ),2πh̄m(12.39)where the factor of i/(2πh̄m) is included for later convenience. On accountof the delta function, f (p → p′ ) depends on p′ only through its direction p̂′ .
To understand the significance of the scattering amplitude, consider thefollowing argument. According to the discussion in §12.1, long after theinteraction, a particle that scattered from the free state |φi can be describedby the free state |λi = S|φi. Therefore, in the idealised case that the initialstate was a momentum eigenstate |pi, the wavefunction of the final state isZ
hr|λi = hr|S|pi = hr|pi + d3 p′ hr|p′ ihp′ |T |pi
Z(12.40)i= hr|pi +d3 p′ hr|p′ iδ(Ep − Ep′ )f (p → p′ ).
2πh̄mSince the states |p′ i in the integrand are final states, the iǫ prescription tells
2
us to take p′ /2m = (Ep′ − iǫ), so in spherical polar coordinates4
2

d3 p′ = p′ dp′ dΩ = m

q2m(Ep′ − iǫ) dEp′ dΩ.

(12.41)

Using this in equation (12.40) and integrating over Ep′ using the delta function givespZ√
i 2m(Ep − iǫ)
ir 2m(Ep −iǫ) p̂′ ·r̂/h̄hr|λi = hr|pi+f (p → p′ ). (12.42)dΩe(2πh̄)5/2
4 In this chapter it is convenient to define dΩ = sin θ dθ dφ rather than d2 Ω =sin θ dθ dφ as in earlier chapters.

266

Chapter 12: Scattering Theory

This free-particle wavefunction only looks like the true wavefunction of thescattered particle long after the collision, so equation (12.42) will only correspond to the physical wavefunction as r → ∞. In this limit, the phase of
the exponential in equation (12.42) varies extremely rapidly as a functionof the variables θ and φ that define the direction of p̂, over which we areintegrating. The different contributions to the integral will therefore cancel
each other out except where the phase of the integrand is stationary withrespect to angle. For sufficiently large r the sensitivity of the exponential toangle will exceed that of f (p → p′ ). Hence the dominant contribution to
the integral arises when∂∂ ′(p̂ · r̂) =cos θ = 0
∂θ
∂θ

and

∂
cos θ = 0,∂φ

(12.43)where we have aligned the polar axis with the (fixed) direction r̂ . These
conditions are satisfied when θ = 0, π, independent of φ. When θ = π, andp̂′ · r̂ = −1, the integrand of equation (12.42) is exponentially suppressed as
r → ∞ by the iǫ prescription. Therefore the integral over the unit sphere isdominated by the contribution from a small disc centred on the direction r̂.
This insight justifies the approximationZdΩ eir√2m(Ep −iǫ) p̂′ ·r̂/h̄f (p → p′ )Z√′
≃ 2πf (p → p ) d cos θ eir 2m(Ep −iǫ) cos θ/h̄

(12.44)√f (p → p′ )≃ 2πh̄ p
eir 2m(Ep −iǫ)/h̄ .
ir 2m(Ep − iǫ)Using this expression in equation (12.42) we have finally√!
1
eir 2m(Ep −iǫ)/h̄
lim hr|φi = lim hr|pi +f (p → pr̂)
r→∞
r→∞
r
(2πh̄)3/2

eipr/h̄1ip·r/h̄f (p → pr̂) ,
e
+
= lim
r→∞ (2πh̄)3/2
r(12.45)where in the last line we have taken the limit ǫ → 0+ . Equation (12.45) showsthat a particle that was initially in a momentum eigenstate will emerge fromthe scattering process in a superposition of its original state (no scattering)
and a wave travelling radially outwards. The scattering amplitude f (p → pr̂)is just the amplitude of this outgoing wave.
In equation (12.45) the time-dependence is suppressed by our convention
that the S-matrix generates the wavefunction at the generic time 0. We now
restore explicit time dependence by introducing a factor e−iEp t/h̄ and replace
the incoming state |pi by a realistic superposition of such states. Then theoutgoing wavefunction becomes
hr|φ; ti =Z

d3 p φ(p)(2πh̄)3/2


ei(pr−Ep t)/h̄f (p → pr̂) ,ei(p·r−Ep t)/h̄ +r(12.46)

which is the sum of the incoming wave packet plus a wave packet that travels
radially outwards from the scattering centre.

12.3 Cross-sections & experiments

267

12.3 Cross-sections and scattering experiments
Children sometimes test their skill by taking turns to throw pebbles at a distant target, perhaps a rock. If a pebble hits, it will bounce off in a different
direction, whereas a pebble that misses will simply continue undisturbed.
Each throw will not be repeated exactly, and after a long time we might
imagine that the children have thrown pebbles randomly, such that the distribution of throws per unit area is uniform over a region surrounding thetarget. If so, we can estimate the area of the target that the children see bysimply counting the number of pebbles that hit it – if Nin pebbles are thrownin per unit area, and Nsc of them hit the rock, the rock has cross-sectionalarea
A ≃ Nsc /Nin .
(12.47)
With more care, we can measure the angle through which throws are deflected. Pebbles that strike nearby points of a smooth rock will bounce off
in roughly the same direction, whereas a jagged rock may deflect pebbles
that hit closely spaced points very differently. Hence, counting the numberof pebbles that end up going in a given direction gives us information about
the rock’s shape. We define the differential cross-section δσ to be thearea of the target that deflects pebbles into a small solid angle δΩ. If there
are N (θ, φ)δΩ such pebbles, thenδσ ≡N (θ, φ)δΩ
Nin

or

δσ
N (θ, φ)
,
=
δΩ
Nin

and the total cross-section is
Z
Z
Z
dσ
Nsc
N (θ, φ)σtot ≡ dσ = dΩ
=
= dΩ
dΩ
Nin
Nin

(12.48)(12.49)as above.
This may seem a rather baroque manner in which to investigate rocks,
but when you go out on a dark night with a torch, you probe objects in a very
similar way by throwing photons at them. A more complete analogy can be
drawn between pebble-throwing children and physicists with particle accelerators: a beam containing a large number Nb of particles is fired towards
a target, and detectors measure the number of particles that scatter off into
each element of solid angle δΩ. Long before the collision, a typical particle
in the beam looks like a free state |φi, so the probability density of eachparticle is |hx|φi|2 and the number of particles per unit area perpendicular
to the beam direction isZnin (x⊥ ) = Nb dxk |hx|φi|2 ,(12.50)where the integral is along the beam direction.
When |φi is expanded in terms of momentum eigenstates, equation(12.50) becomesZ′Nbnin (x⊥ ) =dxk d3 p d3 p′ ei(p−p )·x/h̄ φ(p)φ∗ (p′ )3(2πh̄)Z′Nb
d3 p d3 p′ δ(pk − p′k )ei(p⊥ −p ⊥ )·x⊥ /h̄ φ(p)φ∗ (p′ ),=(2πh̄)2(12.51)
where the integral over xk produced the delta function of momentum along
the beam direction. Experimental beams are highly collimated, so φ(p)vanishes rapidly unless the momentum is near some average value p̄. In
particular, they contain only small amounts of momentum perpendicular toˆ ≡ p̄/|p̄|. Consequently, throughout a region of nonthe beam direction p̄
negligible extent near the centre of the beam, at x⊥ = 0, we have eip⊥ ·x⊥ /h̄ ≃268

Chapter 12: Scattering Theory

1. With this approximation, the number of particles incident per unit area
perpendicular to the beam is uniform near the beam centre, soZNbnin (x⊥ ) ≃(12.52)d3 p d3 p′ δ(pk − p′k )φ(p)φ∗ (p′ ).
(2πh̄)2
Equation (12.52) may seem a bizarre way to rewrite the intuitively clear
expression (12.50), but it will soon prove its worth.
We must now calculate N (θ, φ). At large distances, we know that the
wavefunction of particles scattered from the state |φi is hr|T |φi. If we hadplaced detectors at some large distance r0 from the scattering centre, overtime they would have detected any particle that has the same values of θ, φand is predicted to lie at r > r0 . Thus the total number of particles that aredetected in the element of solid angle δΩ isZ ∞dr r2 Nb δΩ |hr|T |φi|2 .
(12.53)N (θ, φ)δΩ =r0

Equation (12.46) gives hr|φi = hr|(1 + T )|φi, so
N (θ, φ)δΩ = Nb δΩ

Z ∞

dr

r0

Z

2

d3 p φ(p) ipr/h̄ef (p → pr̂) .
(2πh̄)3/2(12.54)So long as the scattering amplitude is reasonably smooth, the collimationof the beam allows us to replace f (p → pr̂) by its value at the average
momentum f (p̄ → p̄r̂), which givesZ ∞Z 3 3 ′′
d pd p∗′dr ei(p−p )r/h̄ .
φ(p)φ
(p
)
N (θ, φ)δΩ = Nb δΩ|f (p̄ → p̄r̂)|23(2πh̄)
r0(12.55)
Explicitly writing the incoming particle’s momentum in terms of the average
momentum p̄ of the beam and its deviation δp from this value, we findpp̄ · δpˆ · δp,(12.56)= p̄ + p̄
p = (p̄ + δp) · (p̄ + δp) ≃ p̄ 1 +p̄2
so the argument of the exponential in equation (12.55) involves
ˆ = (δpk − δp′k ) = (pk − p′k ).
(p − p′ ) ≃ (δp − δp′ ) · p̄(12.57)Since r > r0 is very large, the phase of this exponential oscillates rapidly, so
again the integral is dominated by contributions for which pk = p′k giving
N (θ, φ)δΩ ≃ Nb δΩ|f (p̄ → p̄r̂)|2= nin |f (p̄ → p̄r̂)|2 δΩ,Z

d3 pd3 p′
φ(p)φ∗ (p′ )δ(pk − p′k )(2πh̄)2(12.58)where we have used equation (12.52).
Combining this with the definition (12.48) of the differential cross-section,
we find dσ/dΩ for scattering from momentum p (now relabelled) into a different momentum p′ of the same magnitude:5dσ= |f (p → p′ )|2 ,dΩ(12.59)where p̂′ points towards the centre of the element of solid angle dΩ. The
total scattering cross-section isZ
σtot = dΩ |f (p → p′ )|2 .
(12.60)5 Our language here is loose: neither the incoming nor the outgoing states are strictly
states of well-defined momentum.

12.3 Cross-sections & experiments

269

The two remarkably simple formulae (12.59) and (12.60) form crucial
links between experiment and theory. If the scattering potential is sufficiently
weak that the Born approximation is valid, equation (12.32) tells us that the
scattering amplitude is f (p → p′ ) = −4π 2 h̄mhp′ |V |pi, and the differential
cross-section ismdσ= (4π 2 h̄m)2 |hp′ |V |pi|2 =
dΩ
2πh̄2

Z

2

d3 x e−iq·x/h̄ V (x) ,(12.61)

where q = p′ − p. The integral in equation (12.61) is just the Fouriertransform Ve (q) of the potential, so the equation can be rewrittenm2dσ=
P (q),dΩ
4π 2 h̄4(12.62)where P (q) = |Ve (q)|2 is the power spectrum of V (x). Thus, by measuring
the number of particles that are scattered into a given direction, we candetermine the power spectrum of the interaction potential.
If we could complement this information by measuring the phases of theFourier transform, we could reconstruct V (x) from the scattering data. The
obvious way to measure the phases is to observe interference between the
scattered and incident amplitudes – interference of this type is what generates holograms, from which the three-dimensional structure of the scattering
object can be reconstructed. A high-energy accelerator does not produce
sufficiently pure quantum states (in the sense of §6.3) for interference between the incident and scattered amplitudes to be observable. Moreover, in
realistic circumstances, experiments in which this interference was observed
would be of limited interest because in reality the potential V (x) fluctuatesin time. For example, in §12.4 below we discuss scattering of electrons byatoms, and in this case the electrostatic potential V varies in time as theelectrons that partly generate it whizz about the atom. These internal motions cause rapid variability in the phases of Ve (q), while affecting the power
spectrum of V to a much smaller extent: the latter depends on the number
and structure of the lumps associated with the electrons and nucleus, rather
than on their locations. Thus scattering experiments enable us to unveil asmuch of the structure of matter as we are in practice interested in. For thisreason they are one of the most powerful tools we can deploy in our efforts
to understand nature.

12.3.1 The optical theorem
The simple connection between the power spectrum of V (x) and the scattering cross-section established above relies on the Born approximation. Thisapproximation is certainly not always valid, so it is interesting to see what
we can say about cross-sections in general.
From equation (12.8) we have thatT + T † = −T † T = −Z

d3 p′′ T † |p′′ ihp′′ |T .

(12.63)

Squeezing this equation between hp′ | and |pi and using equation (12.39), we
find that the scattering amplitude f (p → p′ ) satisfiesδ(Ep − Ep′ ) f (p → p′ ) − f ∗ (p′ → p)Zid3 p′′ δ(Ep′′ − Ep′ )f ∗ (p′ → p′′ ) δ(Ep′′ − Ep )f (p → p′′ ).
=2πh̄m(12.64)

270

Chapter 12: Scattering Theory

Figure 12.2 The differential cross-section for neutron-proton scattering at two values of
the centre-of-mass energy. Data obtained from M. Kreisler et. al., Phys. Rev. Lett., 16,
1217, (1966). The diffraction peak at θ = 0 can be understood in terms of the opticaltheorem.

The second delta function in the integral ensures that Ep′′ = Ep , so we canreplace the first delta function by δ(Ep − Ep′ ), and then bring it outside theintegral since it no longer depends on p′′ . Then we havef (p → p′ )−f ∗ (p′ → p) =

i
2πh̄m

Z

d3 p′′ δ(Ep′′ −Ep )f ∗ (p′ → p′′ )f (p → p′′ ).

(12.65)
When p′ = p (equal directions as well as magnitudes), the left side becomesf (p → p) − f ∗ (p → p) = 2iℑmf (p → p) and, after changing variables in thedelta function to obtain δ(Ep′′ − Ep ) = (m/p′′ ) δ(p′′ − p), equation (12.65)
reduces toZ
1
1ℑmf (p → p) =d3 p′′ ′′ δ(p − p′′ )f ∗ (p → p′′ )f (p → p′′ )4πh̄
pZ(12.66)p′′ 2=
dΩ |f (p → p )| .
4πh̄In this last expression we recognise from equation (12.60) the total crosssection for scattering from a state of initial momentum p. We have derived
the relation
4πh̄
σtot (p) =ℑmf (p → p).
(12.67)pThis equation is known as the optical theorem, and relates the total crosssection to the imaginary part of the scattering amplitude in the forward
direction.PIt is at heart a re-expression of equation (12.9) with an identity
operator|ψi ihψi | inserted after T † on the right. The forward scattering
gives the probability a particle is removed from the original beam, and this isassociated with the total probability the particle is deflected into some otherdirection.
When neutrons are scattered from protons, the differential cross-section
has a peak in the forward direction, as shown in Figure 12.2. As the centreof-mass energy is raised, this peak increases in height and decreases in width.
This behaviour is explained by the optical theorem. Experimentally, the total
cross-section becomes roughly constant as p → ∞. Equation (12.67) thenimplies that ℑmf (p → p) rises roughly in proportion to p, so from equation(12.59) the differential cross-section in the forward direction grows at least′ 2
as fast as p2 . Conversely,
R since |f (p → p )| is necessarily positive, the total
cross section σtot = dΩ |f (p → p′ )|2 is never less than the cross-section
for scattering into any solid angle ∆Ω < 4π. Choosing ∆Ω to be the region

12.4 Scattering electrons off hydrogen271

around the forward direction in which |f (p → p′ )|2 is falling from its peakat p′ = p, but still greater than 21 |f (p → p)|2 , givesZdΩ |f (p → p′ )|2 ≥ 12 ∆Ω|f (p → p)|2 ≥ 12 ∆Ω|ℑmf (p → p)|2 .
σtot ≥∆Ω(12.68)Hence, from the optical theorem, the fwhm of the peak around the forward
direction is bounded by
32π 2∆Ω ≤ 2(12.69)p σtotand therefore shrinks as p−2 as p → ∞. This diffraction peak is familiarfrom optics: collimated light can be diffracted by two slits, and the resulting
intensity in the Fraunhofer region is peaked in the forward direction, with afwhm that shrinks as the frequency of the light is increased.

12.4 Scattering electrons off hydrogen
We now apply our scattering formalism to a physical problem, namely scattering of electrons by a hydrogen atom that is in its ground state |1, 0, 0i(§8.1). Taking the proton to be a pointlike object at the centre of the atom,
the atom’s charge distribution is2

ρ(r) = eδ 3 (r) − e |hr|1, 0, 0i| .

(12.70)From §8.1.2 we have that |hr|1, 0, 0i|2 = e−2r/a0 /πa30 where a0 is the Bohrradius (eq. 8.15b). Hence, the atom is the source of an electric field E =−∇Φ, whereZ−2r ′ /a0eeρ(r′ )
3 ′ e
d
r
=
−
|r − r′ |4πǫ0 r 4πǫ0 πa30
|r − r′ |Z′e2e
r′2 sin θ e−2r /a0=−
dr′ dθ 2.
3
4πǫ0 r 4πa0 ǫ0(r + r′ 2 − 2rr′ cos θ)1/2

Φ(r) =

1
4πǫ0

Z

d3 r′

(12.71)

The integral differs only trivially from that evaluated in Box 10.1. Adapting
the result obtained there we conclude that


1
1
e
(12.72)e−2r/a0 .
+Φ(r) =
4πǫ0 ra0
Notice how the ground-state electron shields the pure 1/r Coulomb potentialof the proton, causing the overall potential to decline exponentially at large
distances. This potential will scatter a passing charged particle such as anelectron. It will turn out that our calculations only apply to electrons thathave enough energy to excite or even ionise the atom. Never the less, weshall consider only the case of elastic scattering, in which the atom remains
throughout in its ground state.
Equation (12.61) gives the Born approximation for the differential crosssection in terms of the Fourier transform of the interaction potential V (r) =−eΦ(r). By equation (12.72) V is a function of distance r only, and for any
such function it is straightforward to show thatZZ
 qr 
4πh̄ ∞
3−iq·r/h̄d reV (r) = −V (r).
(12.73)dr r sinq 0h̄Substituting for V (r) = −eΦ(r) from equation (12.72) we findZ

d3 r e−iq·r/h̄ V (r) =4πh̄2 8 + (qa0 /h̄)2.
me (4 + (qa0 /h̄)2 )2(12.74)

272

Chapter 12: Scattering Theory

p’
θ

q
p

Figure 12.3 Trigonometry of theisosceles triangle tells us the magnitude of momentum transfer, since|p′ | = |p|.

Plugging this result into equation (12.61), we have finallydσ= 4a20dΩ

8 + (qa0 /h̄)2(4 + (qa0 /h̄)2 )2

2.

(12.75)

Now q = |p′ − p| = 2p sin(θ/2) (see Figure 12.3), so q is smallest and thecross-section is greatest for forward scattering (θ = 0). Quantitatively,dσ= a20 ,dΩ θ=0(12.76)

independent of the incoming electron’s energy. When the electron’s momentum is large, the cross-section drops sharply as we move away from theforward direction. This behaviour is in rough agreement with the opticaltheorem, although we should not expect equation (12.67) to hold exactly
because we have used the Born approximation.
We now check the validity of the Born approximation. The potentialof equation (12.72) has a characteristic range a0 . When an electron with
momentum ∼ p̄ is aimed at the atom, it is within this range for a time oforder δt ≃ a0 m/p̄. Averaged over that time, the potential it experiences isof orderZ1e2
V ≡ 3 dr r2 V (r) = −= −R,(12.77)a0
8πǫ0 a0
where we have used the definition (8.15b) of a0 and R is the Rydberg constant(eq. 8.26). From the tdse the fractional change that V effects in its ket
during this interval is of order δ|ψi/|ψi ∼ V δt/h̄. We expect the Born
approximation to be is valid if this fractional change is small, that is, provided
a0 m |V |
1≫
=
p̄ h̄

p
Rme /2.
p̄

(12.78)Hence the inequality holds for electrons with energiesp̄2≫ 14 R.
2me(12.79)Since R ∼ 13.6 eV, while the rest-mass energy of the electron is me c2 ∼511 keV, there is a wide range of energy that is high enough for the Born
approximation to be valid, yet small enough for the electron to be nonrelativistic. In Figure 12.4 we plot the experimentally measured differential
cross section alongside our estimate (12.75) from the Born approximation
for three electron energies: 4.9, 30 and 680 eV. At the lowest energy the
Born approximation is useless. At 30 eV ∼ 2R the approximation works
moderately well for back-scattering but seriously underpredicts the crosssection for forward scattering. At 680 eV the approximation works well forall scattering angles.

12.5 Partial wave expansions

273

Figure 12.4 Elastic e− H scattering at electron kinetic energies E = 4.9, 30 and 680 eV.
The curves show the predictions of the Born approximation (eq. 12.75) while the points
show experimental data from J.F. Williams, J. Phys. B, 8 no. 13 (1975). The accuracy of
the Born approximation increases with energy.

12.5 Partial wave expansions
In §12.2 we introduced the S-matrix by squeezing the scattering operator
between states |pi of definite momentum. This allowed us to evaluate theaction of the free evolution operators UK (τ ), because {|pi} is a complete setof eigenstates of HK . In §7.2.5 we saw that states |E, l, mi of definite angularmomentum also form a complete set of eigenstates of HK , so we could justas well consider the matrix hE ′ , l′ , m′ |S|E, l, mi.
From equation (12.37) we have that [HK , S] = 0, from which it follows that hE ′ , l′ , m′ |S|E, l, mi vanishes unless E ′ = E. Also, if the scattering potential is spherically symmetric, it follows from the work of §4.2 that
[L, S] = 0, so[Lz , S] = 0;[L± , S] = 0 ;[L2 , S] = 0.

(12.80)From the first and last of these commutators it follows that hE ′ , l′ , m′ |S|E, l, mivanishes unless m′ = m and l′ = l. Moreover, the second commutator implies
that6
0 = hE, l, m|[S, L+ ]|E, l, m − 1i
∝ hE, l, m|S|E, l, mi − hE, l, m − 1|S|E, l, m − 1i,(12.81)

so not only is S diagonal in the |E, l, mi basis, but hE, l, m|S|E, l, mi isactually independent of m. We can summarise these constraints on the Smatrix of a spherically symmetric potential by writinghE ′ , l′ , m′ |S|E, l, mi = δ(E − E ′ )δl′ l δm′ m sl (E),(12.82)where sl (E) is a number that depends on E and l. Finally, since the S-matrix
is unitary, sl (E), must have unit modulus, sohE ′ , l′ , m′ |S|E, l, mi = δ(E − E ′ )δl′ l δm′ m e2iδl (E) ,(12.83)where all the remaining information is contained in the real phase shifts
δl (E). This reduction of the whole scattering process to a mere set of phases
makes the angular-momentum basis invaluable for scattering problems.
Equation (12.83) implies that
S|E, l, mi = e2iδl (E) |E, l, mi,
6 Recall from equations (7.15) that α (m − 1) = α (m).
+−(12.84)

274

Chapter 12: Scattering Theory

Box 12.1: The amplitude hp|E, l, mi
We require the amplitude hp|E, l, mi that a particle that has well-definedenergy E and angular momentum will be found to have momentum p
and energy p2 /2M . Since the quantity we seek is the momentum-space
wavefunction of a particle of well-defined angular momentum, we simply
repeat the work of §7.2.3 in the momentum representation. We have


1
∂
∂
Lz = (py x − px y) = i py
,
− px
h̄
∂px
∂py
where the operators are written in the momentum representation. We
now introduce polar coordinates (p, ϑ, ϕ) for momentum space, and, inexact analogy to the derivation of equation (7.43), show that Lz = i∂/∂ϕ.
This is simply minus the corresponding real-space result (7.43). It is easy
to see that proceeding in this way we would obtain momentum-space
representations of L± that differ from their real-space analogues (7.52)
only in the substitutions θ → ϑ and φ → ϕ and an overall change ofsign. Consequently the momentum-space wavefunction of a state of welldefined angular momentum must be ψ(p) = g(p)Ylm∗ (ϑ, ϕ), where g(p)is as yet undetermined and the complex conjugate spherical harmonic is
required because Lz = +i∂/∂ϕ. If we require E to equal p2 /2M , it isclear that g = Gδ(E − p2 /2M ). The constant of proportionality, G, isdetermined by the normalization condition
δ(E − E ′ ) = hE, l, m|E ′ , l, lmi
ZZ= G2 dp p2 δ(E − p2 /2M )δ(E ′ − p2 /2M ) dΩ|Ylm |2Zp2= G M dEp 2M Ep δ(E − Ep )δ(E ′ − Ep )√= G2 M 2M E δ(E − E ′ ) = G2 M pδ(E − E ′ ).
Thus G = (M p)−1/2 .
so if prior to scattering the particle is in the state |E, l, mi, it will emerge fromthe scattering region in a state that differs only by the acquisition of an extra
phase 2δl (E). This fact mirrors our finding in §5.3 that a one-dimensional
scattering process is entirely determined by the phase shifts of the even- andodd-parity solutions to the tise – which are the one-dimensional analoguesof states of well-defined angular momentum.
The above discussion generalises straightforwardly to the case of particles with non-zero spin, provided we replace the orbital angular momentum
operator L with the total angular momentum operator J, and relabel thestates and phase shifts accordingly. For simplicity, we will confine ourselvesto scalar particles for the rest of this section.
To relate the S-matrix in the form of equation (12.83) to experimental
cross-sections, we must calculate the scattering amplitude f (p → p′ ). Using
equations (12.6) and (12.83) we obtain
X Z
hp′ |T |pi =dEdE ′ hp′ |E ′ , l′ , m′ ihE ′ , l′ , m′ |T |E, l, mihE, l, m|pil′ lm′ m=XZlmdE hp′ |E, l, mihE, l, m|pi(e2iδl (E) − 1).
(12.85)

In Box 12.1 we show that
hp|E, l, mi = (M p)−1/2 δ(E − Ep )Ylm∗ (ϑ, ϕ),where (ϑ, ϕ) are the polar coordinates of p.

(12.86)

12.5 Partial wave expansions

275

If we align the z-axis with the beam direction, the initial state is unchanged by rotations around the z-axis. Equation (7.38) with α = ẑ thentells that the initial state has m = 0, so hpẑ|E, l, mi vanishes unless m = 0.
Thus for a beam in this direction
hE, l, m|pi =δ(Ep − E)√
δm0 Yl0 (0).
Mp

(12.87)

In §7.2.3 we saw that Yl0 (ϑ) is a real lth -order polynomial in cos ϑ:
Yl0 (ϑ) =r

2l + 1Pl (cos ϑ).
4π(12.88)We have, moreover, that Pl (1) = 1. Consequently, equation (12.87) yields
hE, l, m|pi = δ(Ep − E)δm0s

2l + 1.
4πM p(12.89)Using equation (12.86) again to eliminate hp′ |E, l, mi from equation (12.85),
and then using equation (12.88) to eliminate Yl0 , we obtain
XZ

hp′ |T |pi =ldE hp′ |E, l, 0ihE, l, 0|pi e2iδl (Ep ) − 1

X 2l + 1 Z dE
√ ′ δ(E − Ep′ )δ(Ep − E)Pl (cos ϑ′ ) e2iδl (Ep ) − 1
=
4πM
pp
l
X 2l + 1iδ(Ep − Ep′ )h̄Pl (cos ϑ′ ) e2iδl (E) − 1 ,
=
2πh̄M
2ip
l
(12.90)where ϑ′ is the angle between p and p′ . Comparing this equation with thedefinition (12.39) of the scattering amplitude, we see finally thatf (p → p′ ) =X(2l + 1)Pl (cos ϑ′ )fl (Ep ),(12.91a)l

where the partial-wave amplitude is defined to befl (E) ≡ h̄

e2iδl (E) − 1h̄= eiδl (E) sin δl (E).
2ip
p(12.91b)

The differential cross-section is just the mod square of the scattering
amplitude, and because the spherical harmonics are orthonormal when integrated over all angles, the total cross-section is
σtot =

ZdΩ |f (p → p′ )|2

Z
Xp
∗
′
= 4π
(2l + 1)(2l + 1)fl′ (E) fl (E) dΩ Yl0′ (ϑ′ )Yl0 (ϑ′ )
l′ l

= 4π

X
l

(2l + 1)|fl (E)|2 = 4πh̄2

X 2l + 1
p2lThis equation is often written as σtot =section of order l,P

σl ≡ 4π(2l + 1)|fl (E)|2 = 4πh̄2(12.92)sin2 δl (E).

l σl , where the partial

cross-

2l + 1
sin2 δl (E),p2

(12.93)

276

Chapter 12: Scattering Theoryis the cross-section for scattering a particle that has total squared angularmomentum l(l + 1)h̄2 . Clearly, the partial cross-sections are restricted by
0 ≤ σl ≤ 4πh̄22l + 1,p2(12.94)

with σl only vanishing when the phase shift δl = nπ. Notice from equations(12.91a) thatℑmf (p → p) =X

(2l + 1)Pl (1)ℑmfl (E) =l

X 2l + 1lph̄ sin2 δl (E). (12.95)

Comparison of this with equation (12.92) shows that the optical theorem(12.67) is explicitly satisfied in this basis. This fact follows from conservationof angular momentum – we have treated the incoming beam as a superposition of states of well-defined angular momentum; since the potential isspherically symmetric, it cannot change the particle’s angular momentum,
so each state of well-defined angular momentum scatters separately, and does
so in conformity with the optical theorem.
In the classical picture of scattering, the angular momentum of a particleof energy p2 /2M is determined by its impact parameter b, which is the
distance between the scattering centre and the straight line tangent to theincoming trajectory. Quantitatively, the angular momentum has magnitude
L = bp. Large b corresponds to a glancing collision and a small scattering
angle, while at small b the encounter is nearly head-on and the particle isliable to back-scatter. Thus we expect the differential cross section fl (p →p′ ) to be largest for ϑ′ ≈ 0 when l is large, and for ϑ′ ≈ π to be largest whenl ≈ 0. The partial cross section σl is expected to decrease as l, and thereforeb, increases.

12.5.1 Scattering at low energy
At low energy, p is small and for l > 0 the classical impact parameter b = L/pbecomes large. Hence we expect low-energy scattering to be dominated by
the partial wave with l = 0. In this subsection we show that this naive
expectation is borne out by our quantum-mechanical formulae.
To discover how a given particle is actually scattered, we must relate
the phase shifts δl (E) to the scattering potential V (r). Since the free state|E, l, mi is an eigenstate of HK , L2 and Lz , from equation (7.70) it follows
that7


 
1 ∂
l(l + 1) 2mE
2 ∂hr|E, l, mi.
(12.96)r− 2
hr|E, l, mi =
r2 ∂r
∂r
r2
h̄
When l 6= 0, the angular momentum term dominates the right side near
the origin, and one can easily show that hr|E, l, mi ∼ rl for small r (Problem 12.8). Consequently, there is only a very small probability of finding aparticle that has high angular momentum near the origin. This reasoning
breaks down when the second term on the right side becomes important. Forenergy E = p2 /2m this occurs at r ≈ lh̄/p, which for large l coincides withthe classical impact parameter. Suppose that the scattering potential acts
over some characteristic length R, beyond which it is negligible – for example, in the case of the potential (12.72), R is of the order of a few Bohr radii.
If R ≪ lh̄/p, then V (r) is only strong in a region where a free wavefunction
would be very small. In this case, the lth partial wave will scarcely be affected, so δl ≃ 0. Roughly, the only states that suffer significant scattering
are those with angular momenta in the range lh̄ <∼ pR, so for low incoming
7 In fact, hr|E, l, mi = j (kr)Y m (θ, φ), where k =ll√

2M E/h̄ and jl is the lth sphericalBessel function, but we do not need this result here (see Problem 12.8).

12.5 Partial wave expansions

277

momenta the total scattering amplitude can be well approximated by the
first few terms in the infinite sum (12.91a).
In fact, we can see quite generally that only the lowest l states makesignificant contributions to the low-energy cross-section. The scattering amplitude f (p → p′ ) should be a smooth function of p and p′ as p, p′ → 0,
because at low energies the incoming particle has a large wavelength and cannot resolve any sharp features in the potential. Since in equation (12.91a)
Pl is an lth -order polynomial in cos ϑ = p′ · p/p′ p, we see that if f (p → p′ )is to be an analytic function of the Cartesian components of p and p′ at low
energies, the partial wave amplitude fl (E) must vanish with p at least as
fast aslim fl (E) ∼ (p′ p)l = p2l .
(12.97)p→0

The total cross section (12.92) then behaves asXlim σtot = 4π
a2l p4lp→0(12.98)lin terms of some constants al , and can be well approximated by just thelowest few terms. In the extreme low energy limit, the only non-vanishing
amplitude is f0 (Ep → 0) = a0 and the differential cross section
lim

p→0

dσ
≃ a20 P0 = a20dΩ(12.99)is isotropic.
An eigenstate of the true Hamiltonian H with the same energy andangular-momentum quantum numbers as |E, l, mi has a radial wavefunctionul (r) that satisfies a version of equation (12.96) modified by the inclusion of
a potential V (r). Writing ul (r) = Ul (r)/r we findd2 Ul (r)2mE
= − 2 Ul (r) + Veff Ul (r),
dr2
h̄
where

(12.100a)2mV (r) l(l + 1).
(12.100b)+r2
h̄2
For a general potential, we typically have to solve this equation numerically, and then find the phase shifts by comparing our solution with equation(12.45) in the large r limit. However, we can obtain a heuristic understandingof the behaviour of the phase shift as follows. If the potential is attractive(V (r) < 0), Ul (r) will have a greater curvature, and hence oscillate more
rapidly in the presence of V that it would have done if the particle were free.
A potential with finite range R ≪ lh̄/p only acts over a small part of a radial
oscillation, so when V < 0, the wavefunction emerges from the interaction
region slightly further along its cycle than a free wavefunction. On the other
hand, when the potential is repulsive, Ul (r) has smaller curvature, so oscillates more slowly than a free wavefunction, emerging from the interaction
region slightly behind. Equation (12.84) tells us that states emerge from thescattering process changed only in phase; we now see that the sign of thephase shift δl will typically be opposite to that of the potential.
As the magnitude of V increases, so does the difference in oscillation
rates between interacting and free eigenstates, and hence at fixed energy|δl (E)| likewise increases. When the potential is sufficiently strong, the interacting wavefunction can oscillate precisely half a cycle more (or less if
V > 0) in the interaction region than it would do if the state were free, andthen |δl (E)| = π with the consequence that fl (E) → 0. In these circumstances this angular-momentum state suffers no scattering at all.
In §10.3, we saw that atoms of a noble gas such as argon are chemicallyinert because in their ground states they have spherically-symmetric distributions of electron charge, and they have no low-lying excited states that
Veff (r) ≡278

Chapter 12: Scattering Theory

Figure 12.5 An exponential attractive force combined with centrifugal
repulsion generates a minimum inthe effective potential.

can be mixed in by a perturbation to generate a less symmetrical charge distribution. As a consequence, these atoms generate negligible electric fields
beyond some limiting radius R ≃ a0 that contains nearly all the probability
density of the outermost shell of electrons. At r ≤ R there is a significant
electric field, and any particle that penetrates this region will be appreciably
scattered, but particles that have larger impact parameters will be negligibly scattered. At energies low enough that Rp ≪ h̄, scattering from stateswith l > 0 is negligible, while the considerations of the previous paragraphsuggest that there could be an energy at which there is also no scatteringfrom states with l = 0. Then an electron is not scattered at all. Exactly
this Ramsauer–Townsend effect was observed before the development of
quantum mechanics8 when electrons of energy ∼ 0.7 eV were discovered topass undeflected past noble-gas atoms.

12.6 Resonant scattering
Nuclear physics involves a combination of short- and long-range forces: thestrong interaction that binds protons and neutrons into nuclei has only ashort range, while the electrostatic repulsion between protons has a longrange. Figure 12.5 illustrates the fact that when a short-range attractive
force is combined with a long-range repulsive force, the overall effective potential Veff of equation (12.100b) is likely to have a local minimum. In §5.3.3
we studied scattering by a one-dimensional potential that contained such a
potential well and demonstrated that a plot of the scattering cross-section
versus energy would show sharp peaks at the energies that allow the particle
to be trapped in the well for an extended period of time. The method ofpartial waves allows us to consider the physics of temporary entrapment in
the much more realistic case of scattering in three dimensions. We shall find
not only that the results of §5.3.3 largely carry over to realistic scattering
potentials, but we are able to extend them to include a quantitative accountof the delay between the particle reaching the potential well and its making good its escape. Physicists have learnt much of what is known about
the structure of both nuclei and baryons such as protons and neutrons by
exploiting the connections between bound states and anomalous scatteringcross sections that emerges from this section.
Equation (12.46) gives the wavefunction at late times of a particle thatwas initially in the free state |φi. It breaks the wavefunction into two parts.
The first is a sum of plane waves φ(p)ei(p·r−Et)/h̄ . If |φ(p)|2 has a well-definedpeak at momentum p̄, from §2.3.3 we know that the amplitude of this firstˆ · r = p̄t/m. To determine the location at
contribution peaks on the plane p̄
8 C. Ramsauer, Ann. Physik, 4, 64 (1921); V.A. Bailey & J.S. Townsend, Phil. Mag.,42, 873 (1921).

12.6 Resonant scattering279

which the amplitude of the second part peaks, we observe that as t → ∞,the phase of the exponential in equation (12.46) varies extremely rapidly
as a function of momentum, causing the sign of the integrand to oscillate
quickly. If φ(p) is a smooth function, the integral will be dominated by thecontribution from momenta at which the phase is stationary. To find thesepoints, we must take into account the phase of f (p → pr̂). By equations(12.91), the scattering amplitude for each partial wave is real except fora factor eiδl (Ep ) . Hence the dominant contribution to the second term inequation (12.46) arises when∂{pr − Ep t + h̄δl (Ep )} = 0∂pi.e. when r =p∂t − h̄ δl (Ep ). (12.101)m∂p

We see that if the phase shift δl (Ep ) increases sharply for momenta near the
average momentum of the initial state, the amplitude of the wave will beconcentrated at a smaller radius than the incoming wave would have been.
Consequently, there will be two distinct peaks in the probability of a particle
reaching a detector at some distance from the scattering centre. The firstis associated with the possibility that the particle misses its target, and thesecond with the possibility that it hits and is temporarily trapped by it
before making good its escape. Thus unstable bound states are associatedwith rapid increases with Ep in the phase shift of the scattered particle. Thisconclusion mirrors our finding in §5.3.3 that temporary trapping of particles
by a one-dimensional well is associated with rapid variations with energy inthe phases φ and φ′ of the states of well-defined parity.
We can model a dramatic increase of δl (E) by postulating that, forenergy near some value ER , the phase shift behaves as


−Γ/2
−1
δl (E) ≃ tan,(12.102)
E − ER
where the fixed energy scale Γ is included to ensure that the argument of theinverse tangent is dimensionless, and must be positive if we want ∂δl /∂p > 0.
In this model the phase shift rapidly increases by π as the energy increases
through ER . Using the model in equation (12.101), we find that the timedelay between the two peaks in the probability density of the particle hitting
a given detector is


mh̄ ∂
−Γ/2
mh̄ ∂E ∂∆t =δl (E) =tan−1
p ∂p
p ∂p ∂E
E − ER(12.103)h̄Γ/2=.
(E − ER )2 + (Γ/2)2
We infer from this delay that the lifetime of the quasi-bound state is ≈ h̄/Γ inagreement with the much less rigorous conclusion that we reached in §5.3.3.
Calculating Γ for a physically realistic potential usually requires numerical analysis. However, since the lifetime of the quasi-bound state increases
as Γ decreases, we anticipate that smaller values of Γ correspond to deeper
minima in the potential: a deeper well traps the particle for longer. The
limiting case Γ → 0+ implies that the delay in emergence becomes infinite.
We interpret this to mean that the dip in V is just deep enough to genuinely
bind an incoming particle.
If V is so deep that there is a state that has a strictly negative energy,
the final state may not resemble the initial free state. For example, theincoming particle may get trapped for good in the potential, or it may knockout another particle that is already trapped (as in ionisation of an atom).
In such cases, the scattering is said to be inelastic and the methods of thischapter must be extended9 .
9 The difficulty is not too severe. True bound states have energy E < 0 whereas all
free states must have energy E ≥ 0. Hence, if |bi is bound and |φi is free, hb|φi = 0 so H
acts on a larger Hilbert space than does HK . Including these extra states carefully allows
us to treat bound states. (See also Problem 12.6.)

280

Chapter 12: Scattering Theory

Returning to the case Γ > 0, we now investigate how the cross-section isaffected by the delayed emergence of our particle. From equations (12.6) and(12.46) the wavefunction of the scattered particle in the asymptotic future isZd3 pei(pr−Et)/h̄f (p → pr̂)
φ(p)
3/2
r
(2πh̄)ZX
ei(pr−Et)/h̄
d3 p
φ(p)
=
(2l + 1)Pl (p̂ · r̂)fl (E),3/2r(2πh̄)
l(12.104)
where the second line uses equation (12.91a) to relate f (p → p′ ) to thepartial-wave amplitudes fl (E). To derive the cross-section (12.59), we assumed that the scattering amplitudes are more slowly varying functions ofp than the wavepacket φ(p) – see the discussion after equation (12.54). In
the presence of a resonance, this approximation may break down. Indeed,
when one of the phase shifts δl (E) has the form of equation (12.102), fromequation (12.91b) the corresponding partial wave amplitude is
hr|T |φ; ti =fl (E) ≡Γ/2h̄h̄ iδl (E)esin δl (E) = −,p
p E − ER + iΓ/2

(12.105)

and for small Γ this varies rapidly when E ≈ ER . We assume that theincoming wavepacket φ(p) contains states |pi that are restricted in energy toa range of width ∆ and consider first the case ∆ ≪ Γ in which the resonanceis broader than the uncertainty in the energy of the incoming particle.

12.6.1 Breit–Wigner resonances
If ∆ ≪ Γ, then fl (E) varies slowly with energy in comparison to φ(p),and equation (12.93) for the partial cross-sections σl applies. Again using
equation (12.102), we have (cf. eq. 5.64a)σl (E) = 4π(2l + 1)|fl |2 ≃(2l + 1)(Γ/2)2
4πh̄2
.
2
p (E − ER )2 + (Γ/2)2(12.106)

A peak in the cross-section that follows this famous formula is said arise
from a pure Breit–Wigner resonance. Breit–Wigner resonances are easily detected in plots of a cross-section versus energy and are the experimental
signature of quasi-bound states in the scattering potential. Figure 12.6 is aplot of equation (12.106). Notice that the energy dependence is a combination of the slow decline with E associated with the factor p−2 and thepeak that arises from the Lorentzian final factor – such factors are familiarfrom the theory of a damped harmonic oscillator (Box 5.1). If Γ ≪ ER ,the factor p−2 changes very little over the width of the bump, and the resonance curve falls to half its maximum height when E ≃ ER ± Γ/2. Thus
the resonance lifetime h̄/Γ can be determined from the fwhm of the peakin the cross-section. This result explains why we needed to restrict ourselves
to superpositions with ∆ ≪ Γ: in order to resolve the Breit–Wigner curve
experimentally, there had better be a good chance that our particle’s energy
lies within Γ of ER , where all the action lies.

12.6.2 Radioactive decay
The width Γ of a very long-lived resonance may be so small that our experimental apparatus cannot generate incoming particles with sufficiently small
uncertainty ∆ in the energy to resolve the curve of Figure 12.6. Then, using
equation (12.86), we expand the momentum amplitudes φ(p) of the initial12.6 Resonant scattering

281

Figure 12.6 The Breit–Wigner formula for a scattering cross section in the presence of aresonance. Here ER = 10Γ and the cross section is normalised by s ≡ 2(2l + 1)h̄2 /mΓ.

state asφ(p) =

XZ
lm

=

X
lm

dE ′ hp|E ′ , l, mihE ′ , l, m|φihE, l, m|φiYlm∗ (p̂) √
,
Mp

p2
.
where E ≡2M

(12.107)

Suppose for simplicity that, near some energy ER , only one partial wave
amplitude has the form of equation (12.105), the others all being negligible by comparison. Then, ignoring any angular dependence, the final-state
wavefunction (12.104) of the scattered particle contains a factor
Zp2
hE, l, m|φi Γh̄ ei(pr−Et)/h̄(2πh̄)3/2 (M p)1/2 2pr E − ER + iΓ/2sZei(pr−Et)/h̄Γh̄ ∞ dE
M
hE,
l,
m|φi
.
=
2r 0 (2πh̄)3/2
p
E − ER + iΓ/2

hr|T |φ; ti ∝dp

(12.108)

If the initial state |φi has average energy around hHK i ≃ ER , but is a superposition of states with different energies, smooth over a range ∆ ≫ Γ, we may−1/2
approximate p−1/2 hE, l, m|φi by its value at resonance, pR hER , l, m|φi andbring it outside the integral. Similarly, near resonance we approximate p byp ≃ pR +dp
E − ER,(E − ER ) = pR +dE
vR

(12.109)

where vR = pR /M , soei(pr−Et)/h̄ ≃ ei(pR r−ER t)/h̄ ei(E−ER )(r/vR −t)/h̄ .

(12.110)Substituting these approximations into equation (12.108), we findZ ∞ei(E−ER )(r/vR −t)/h̄.
dE
E − ER + iΓ/2
0R
(12.111)
The remaining integral can be done by contour integration. Since the denominator is large except near E = ER , we can extend the range of integrationto −∞ < E < ∞, without drastically affecting the integral. If r > vR t, weclose the contour in the upper half complex E plane. Since the only pole isin the lower half-plane, the integral evaluates to zero. If r < vR t, we closeΓh̄ hER , l, m|φi ei(pR r−ER t)/h̄
hr|T |φ; ti ∼ 1/2r(2πh̄)3/2
2v

282

Problems

the contour in the lower half-plane. Evaluating the residue of the pole at
E = ER − iΓ/2, we conclude thathr|T |φ; ti ∼ei(pR r−ER t)/h̄ −Γ(t−r/vR )/2h̄
iΓ
hE
,
l,
m|φi
e
. (12.112)Rr2(2πh̄vR )1/2

Consequently,|hr|T |φ; ti|2 ∼





0

if r > vR t|hER , l, m|φi|2 −Γ(t−r/vR )/h̄Γe
8πh̄vRr2
2otherwise.

(12.113)

The physical interpretation of this equation is the following. The probability
density |hr|T |φ; ti|2 is zero before time t′ = r/vR because the particle travels
radially outwards at speed ≃ vR . Subsequently, the probability of finding the′particle anywhere on a sphere of radius r decays exponentially as e−Γ(t−t )/h̄ .
This result provides a remarkable explanation of the law of radioactive
decay: we interpret the emission of a neutron by an unstable nucleus as theendpoint of a scattering experiment that started months earlier in a nuclear
reactor, where the nucleus was created by absorption of a neutron. Moredramatic is the case of 238 U, which decays via emission of an α-particle to234
Th with a mean life h̄/Γ ≃ 6.4 Gyr. Because Γ/h̄ is tiny, the probability(12.113) is nearly constant over huge periods of time. Our formalism tells
us that if we were to scatter α-particles off 234 Th, they would all eventually
re-emerge, but only after a delay that often exceeds the age of the universe!
Thus 238 U is really a long-lived resonance of the (α,234 Th) system, rather
than a stationary state. It is only because the timescale h̄/Γ is so longthat we speak of 238 U rather than a resonance in the (α, 234 Th) system. In
fact, 234 Th is itself a resonance, ultimately of Pb. The longevity of 238 U is
inevitably associated with a very small probability that 238 U will be formed
when we shoot an α-particle at a 234 Th nucleus. To see this notice thatthe final-state wavefunction hr|S|φ; ti = hr|φ; ti + hr|T |φ; ti, also involves an
unscattered piece. On account of the smallness of Γ, the ratio of probabilities
Prob(α is trapped)Γ2 m|hER , l, m|φi|2≈
Prob(α unscattered)h̄pR(12.114)is extremely small. Hence it is exceptionally difficult to form 238 U by firing αparticles at 234 Th nuclei. Naturally occurring 238 U was formed in supernovae,
where the flux of α-particles and neutrons was large enough to overcome thissuppression.

Problems
e ± defined by equation (12.23) obey
12.1 Show that the operators Ω
e± = Ω
e ± (HK ± iǫ) ∓ iǫ.
HΩ(12.115)

12.2 Obtain the first and second order contributions to the S-matrix fromthe Feynman rules given in §12.3.
12.3 Derive the Lippmann–Schwinger equation|±i = |Ei +1
V |±i,
E − H K ± iǫ

(12.116)

where |±i are in and out states of energy E and |Ei is a free-particle state of
the same energy. In the case that the potential V = V0 |χihχ| for some state|χi and constant V0 , solve the Lippmann–Schwinger equation to find hχ|±i.

Problems283

12.4 A certain potential V (r) falls as r−n at large distances. Show thatthe Born approximation to the total cross-section is finite only if n > 2. Is
this a problem with the Born approximation?
12.5 Compute the differential cross section in the Born approximation forthe potential V (r) = V0 exp(−r2 /2r02 ). For what energies is the Born approximation justified?
12.6 When an electron scatters off an atom, the atom may be excited (oreven ionised). Consider an electron scattering off a hydrogen atom. The
Hamiltonian may be written as H = H0 + H1 where
H0 =

p̂21
p̂2
e2
+ 2
−
2m 4πǫ0 r1
2m(12.117)is the Hamiltonian of the hydrogen atom (whose electron is described by coordinate r1 ) together with the kinetic Hamiltonian of the scattering electron(coordinate r2 ), and
H1 =

e2
4πǫ0



1
1
−
|r1 − r2 | r2



(12.118)

is the interaction of the scattering electron with the atom.
By using H0 in the evolution operators, show that in the Born approximation the amplitude for a collision to scatter the electron from momentum p2 to p′2 whilst exciting the atom from the state |n, l, mi to the state|n′ , l′ , m′ i isf (p2 ; n, l, m → p′2 ; n′ , l′ , m′ )Z
4π 2 h̄m=−
d3 r1 d3 r2 e−iq2 ·r2 hn′ , l′ , m′ |r1 ihr1 |n, l, miH1 (r1 , r2 ),(2πh̄)3(12.119)
where q2 is the momentum transferred to the scattering electron. (Neglect
the possibility that the two electrons exchange places. You may wish to
perform the d3 r1 integral by including a factor e−αr1 and then letting α → 0.)Compute the differential cross-section for the |1, 0, 0i → |2, 0, 0i transition and show that at high energies it falls as cosec12 (θ/2).
12.7 Use the optical theorem to show that the first Born approximation is
not valid for forward scattering.
12.8 A particle scatters off a hard sphere, described by the potential
V (r) =n

∞ for |r| ≤ a
0 otherwise.

(12.120)

By considering the form of the radial wavefunction u(r) in the region r > a,showthat the phase shifts are given by tan δl = jl (ka)/nl (ka), where k =√2mE/h̄ and jl (kr) and nl (kr) are spherical Bessel functions and Neumannfunctions, which are the two independent solutions of the second-order radial
equation


 
1 d
l(l + 1) 2mE
2 d
u(r).
(12.121)
r
−
u(r)
=
r2 dr
dr
r2
h̄2
In the limit kr → 0, show that these functions behave asjl (kr) →(kr)l2l + 1nl (kr) → −2l − 1.
(kr)l+1(12.122)

Use this to show that in the low-energy limit, the scattering is sphericallysymmetric and the total cross-section is four times the classical value.

284

Problems

12.9 Show that in the Born approximation the phase shifts δl (E) for scattering off a spherical potential V (r) are given byδl (E) ≃ −2mkh̄2

Z ∞dr r2 V (r) (jl (kr))2 .

(12.123)

0

When is the approximation valid?
12.10 Two α-particles collide. Show that when the α-particles initiallyhave equal and opposite momenta, the differential cross-section isdσ= |f (θ) + f (θ − π)|2 .
dΩ(12.124)

Using the formula for f (θ) in terms of partial waves, show that the differential
cross-section at θ = π/2 is twice what would be expected had the α-particles
been distinguishable.
A moving electron crashes into an electron that is initially at rest. Assuming both electrons are in the same spin state, show that the differential
cross-section falls to zero at θ = π/4.

The laws of probability

285

Appendices
Appendix A: The laws of probability
Events are frequently one-offs: Pretty Lady will run in the 4.20 at Sandown Park
only once this year, and if she enters the race next year, her form and the field will
be different. The probability that we want is for this year’s race. Sometimes events
can be repeated, however. For example, there is no obvious difference between one
throw of a die and the next throw, so it makes sense to assume that the probabilityof throwing a 5 is the same on each throw. When events can be repeated in thisway we seek to assign probabilities in such a way that when we make a very large
number N of trials, the number nA of trials in which event A occurs (for example
5 comes up) satisfiesnA ≃ pA N.
(A.1)
In any sequence of throws, the ratio nA /N will vary with N , while the probability
pA does not. So the relation (A.1) is rarely an equality. The idea is that we shouldchoose pA so that nA /N fluctuates in a smaller and smaller interval around pA as
N is increased.
Events can be logically combined to form composite events: if A is the eventthat a certain red die falls with 1 up, and B is the event that a white die falls
with 5 up, AB is the event that when both dice are thrown, the red die shows 1
and the white one shows 5. If the probability of A is pA and the probability of B
is pB , then in a fraction ∼ pA of throws of the two dice the red die will show 1,
and in a fraction ∼ pB of these throws, the white die will have 5 up. Hence thefraction of throws in which the event AB occurs is ∼ pA pB so we should take theprobability of AB to be pAB = pA pB . In this example A and B are independent
events because we see no reason why the number shown by the white die could
be influenced by the number that happens to come up on the red one, and viceversa. The rule for combining the probabilities of independent events to get theprobability of both events happening, is to multiply them:p(A and B) = p(A)p(B)(independent events).

(A.2)

Since only one number can come up on a die in a given throw, the event A
above excludes the event C that the red die shows 2; A and C are exclusive events.
The probability that either a 1 or a 2 will show is obtained by adding pA and pC .
Thus, for classical probabilityp(A or C) = p(A) + p(C)(exclusive events).

(A.3)

In the case of reproducible events, this rule is clearly consistent with the principlethat the fraction of trials in which either A or C occurs should be the sum of thefractions of the trials in which one or the other occurs. If we throw our die, thenumber that will come up is certainly one of 1, 2, 3, 4, 5 or 6. So by the rule justgiven, the sum of the probabilities associated with each of these numbers coming
up has to be unity. Unless we know that the die is loaded, we assume that no286

Appendix B: Cartesian tensors

number is more likely to come up than another, so all six probabilities must be
equal. Hence, they must all equal 61 . Generalising this example we have the rules
With just N mutually exclusive outcomes,NX

pi = 1.

i=1

If all outcomes are equally likely, pi = 1/N.

(A.4)

Expectation values A random variable x is a quantity that we can measure
and the value that we get is subject to uncertainty. Suppose for simplicity that
only discrete values xi can be measured. In the case of a die, for example, x could
be the number that comes up, so x has six possible values, x1 = 1 to x6 = 6. If pi
is the probability that we shall measure xi , then the expectation value of x isXhxi ≡pi xi .
(A.5)
i

If the event is reproducible, it is easy to show that the average of the values that
we measure on N trials tends to hxi as N becomes very large. Consequently, hxi
is often referred to as the average of x.
Suppose we have two random variables, x and y. Let pij be the probability
that our measurement returns xi for the value of x and yj for the value of y. Thenthe expectation of the sum x + y isXXX
hx + yi =pij (xi + yj ) =pij xi +pij yj
(A.6)
P

ij

ij

ij

But j pij is the probability that we measure xi regardless of what we measureP
for y, so it must equal pi . Similarly i pij = pj , the probability of measuring yj
irrespective of what we get for x. Inserting these expressions in to (A.6) we findhx + yi = hxi + hyi .

(A.7)

That is, the expectation value of the sum of two random variables is the sum of
the variables’ individual expectation values, regardless of whether the variables are
independent or not.
A useful measure of the amount by which the value of a random variablefluctuates from trial to trial is the variance of x:(x − hxi)2 = x2 − 2 hx hxii + hxi2 ,(A.8)

where we have made use of equation (A.7). The expectation hxi is not a randomvariable, but has a definite value. Consequently hx hxii = hxi2 and hxi2 = hxi2 ,so the variance of x is related to the expectations of x and x2 by∆2x ≡ (x − hxi)2 = x2 − hxi2 .

(A.9)

Appendix B: Cartesian tensors
Vector notation is very powerful, but sometimes it is necessary to step outside
it and work explicitly with the components of vectors. This is especially true inquantum mechanics, because when operators are in play we have less flexibility
about the order in which we write symbols, and standard vector notation can beprescriptive about order. For example if we want p to operate on a but not b,
we have to write b to the left of p and a on the right, but this requirement isincompatible with the vectorial requirements if the classical expression would be
p × (a × b). The techniques of Cartesian tensors resolve this kind of problem.
Even in classical physics tensor notation enables us to use concepts that cannot be
handled by vectors. In particular, it extends effortlessly to spaces with more than
three dimensions, such as spacetime, which vector notation does only in a limitedway.
Instead ofPwriting a, we write ai for the ith component of the vector. Thena · b becomes i ai bi . When a subscript is used twice in a product, as i is here,
it is generally summed over and we speak of the subscript on a being contracted
on the subscript on b.
The ij th component of the 3 × 3 identity matrix is denoted δij and called theKronecker delta: so
1 if i = j
δij =
(B.1)
0 otherwise.

Cartesian tensors

287

P
The equation ai =matrix times aj δij aj expresses the fact that the identityPequals a. The scalar product often appears in the form a · b = ij δij ai bj . To seethat this is equivalent to the usual expression, we do the sum over j. ThenP thedelta vanishes except when j = i, when it is unity, and we are left with i qi bi .
Notice that
P it does not matter in what order the symbols appear; we have alsoa · b = ij ai δij bj , etc. – when using Cartesian tensors, the information that invector notation is encoded in the positions of symbols is carried by the way thesubscripts on different symbols are coupled together.
To make the vector product we need to introduce the alternating symbol
or Levi–Civita symbol ǫijk . This is a set of 27 zeros and ones defined such thatǫ123 = 1 and the sign changes if any two indices are swapped. So ǫ213 = −1,
ǫ231 = 1, etc. If we cyclically permute the indices, changing 123 into first 312 andthen 231, we are swapping two pairs each time, so there are two cancelling changesof sign. That is, ǫ123 = ǫ312 = ǫ231 = 1 and ǫ213 = ǫ321 = ǫ132 = −1. All theremaining 21 components of the alternating symbol vanish, because they have atleast two subscripts equal, and swapping these equal subscripts we learn that thiscomponent is equal to minus itself, and therefore must be zero.
We now haveXǫijk aj bk .
(B.2)
(a × b)i =jk

To prove this statement, we explicitly evaluate thePright side for i = 1, 2 and 3.
For example, setting i = 1 the right side becomes jk ǫ1jk aj bk . In this sum ǫ1jk
is non-vanishing only when j is different from k and neither is equal one. So there
are only two terms:
ǫ123 a2 b3 + ǫ132 a3 a2 = a2 b3 − a3 b2(B.3)

which is by definition the third component of a × b.
A few simple rules enable us to translate between vector and tensor notation.
1. Fundamentally we are writing down the general component of some quantity,
so if that quantity is a vector, there should be one subscript that is “spare”
in the sense that it is not contracted with another subscript. Similarly, if thequantity is a scalar, all indices should be contracted, while a tensor quantity
has two spare indices.
2. Each scalar product is expressed by choosing a letter that has not alreadybeen used for a subscript and making it the subscript of both the vectors
involved in the product.
3. Each vector product is expressed by choosing three letters, say i, j and k andusing them as subscripts of an ǫ. The second letter becomes the subscriptthat comes before the cross, and the third letter becomes the subscript of thevector that comes after the cross.
We need a lemma to handle vector triple products:
Xǫijk ǫirs = δjr δks − δkr δjs(B.4)
i

Before we prove this identity (which should be memorised), notice its shape: onthe left we have two epsilons with a contracted subscript. On the right we have
two products of deltas, the subscripts of which are matched “middle to middle,
end to end” and “middle to end, end to middle”. Now the proof. For the sum on
the left to be non-vanishing, both epsilons must be non-vanishing for some valueof i. For that value of i, the subscripts j and k must take the values that i doesnot. For example, if i is 1, j and k must between them be 2 and 3. For the same
reason r and s must also between them be 2 and 3. So either j = r and k = s
or j = s and k = r. In the first case, if ijk is an even permutation of 123, thenso is irs, or if ijk is an odd permutation, then so is irs. Hence in the first caseeither both epsilons are equal to 1, or they are both equal to −1 and their productis guaranteed to be 1. The first pair of deltas on the right cover this case. If, on
the other hand, j = s and k = r, irs will be an odd permutation of 123 if ijk isan even one, and vice versa if ijk is an odd permutation. Hence in this case one
epsilon is equal to 1 and the other is equal to −1 and their product is certainly
equal to −1. The second product of deltas covers this case. This completes theproof of equation (B.4) because we have shown that the two sides always take thesame value no matter what values we give to the subscripts.
Besides enabling us to translate vector products into tensor notation, thealternating symbol enables us to form the determinant of any 3 × 3 matrix. In fact,
this is the symbol’s core role and its use for vector products is a spinoff from it.

288

Appendix C: Fourier series and transforms

The simplest expression for det(A) is
Xǫijk A1i A2j A3k .
det(A) =

(B.5)

ijk

A more sophisticated expression that is often invaluable isXǫijk Ari Asj Atk .
det(A)ǫrst =

(B.6)

ijk

These expressions are best treated as the definition of det(A) and used to derive
the usual rule for the evaluation of a determinant by expansion down a row orcolumn. This popular rule is actually a poor way to define a determinant, and adreadful way of evaluating one. It should be avoided whenever possible.

Appendix C: Fourier series and transforms
The amplitude for a particle to be located at x and the amplitude for the particle
to have momentum p are related by Fourier transforms, so they play a significant
role in quantum mechanics. In this appendix we derive the basic formulae. Like
Fourier1 himself we start by considering a function of one variable, f (x), that isperiodic with period L: that is, f (x + L) = f (x) for all x. We assume that f canbe expressed as a sum of sinusoidal waves with wavelength L:∞X
Fn e2πinx/L ,(C.1)
f (x) =n=−∞where the Fn are complex numbers to be determined. At this stage this is justa hypothesis – only 127 years after Fourier introduced his series did Stone2 provethat the sum on the right always converges to the function on the left. However
numerical experiments soon convince us that the hypothesis is valid because it isstraightforward to determine what the coefficients Fn must be, so we can evaluate
them for some specimen functions f and see whether the series converges to thefunction. To determine the Fn we simply multiply both sides of the equation by
e−2πimx/L and integrate from −L/2 to L/2:3
Z L/2
Z L/2
∞
X
Fn
dx e2πi(n−m)x/L
dx e−2πimx/L f (x) =
(C.2)
−L/2
−L/2
n=−∞
= LFm ,

where the second equality follows because for n 6= m the integral of the exponential
on the right vanishes, so there is only one non-zero term in the series. Thus the
expansion coefficients have to be
Z
1 L/2
Fm =
dx e−2πimx/L f (x).
(C.3)
L −L/2In terms of the wavenumbers of our waves,
2πn
,
kn ≡
L
our formulae become
Z
∞
X
1 L/2
dx e−ikm x f (x).
Fn eikn x ; Fm =f (x) =
L
−L/2
n=−∞

(C.4)

(C.5)

At this stage it proves expedient to replace the Fn with rescaled coefficientsfe(kn ) ≡ LFn .
(C.6)
so our equations become∞X
1ef (x) =f (kn ) eikn x
L
n=−∞

;

fe(km ) =Z L/2

dx e−ikm x f (x).

(C.7)

−L/2

1 After dropping out from a seminary Joseph Fourier (1768–1830) joined the Auxerre
Revolutionary Committee. The Revolution’s fratricidal violence led to his arrest but heavoided the guillotine by virtue of Robespierre’s fall in 1794. He invented Fourier series
while serving Napoleon as Prefect of Grenoble. His former teachers Laplace and Lagrange
were not convinced.
2 Marshall Stone strengthened a theorem proved by Karl Weierstrass in 1885.
3 You can check that the integration can be over any interval of length L. We havechosen the interval (− 21 L, 21 L) for later convenience.

Operators in classical statistical mechanics

289

Now we eliminate L from the first equation in favour of the difference dk ≡ kn+1 −kn = 2π/L and have
Z L/2
∞
X
dk e
f (kn ) eikn x ; fe(km ) =dx e−ikm x f (x).
(C.8)
f (x) =
2π
−L/2
n=−∞

Finally we imagine the period getting longer and longer without limit. As L grows
the difference dk between successive values of kn becomes smaller and smaller, so
kn becomes a continuous variable k, and the sum in the first equation of (C.8)
becomes an integral. Hence in the limit of infinite L we are left with
Z ∞
Z ∞
dk e
f (x) =
f (k) eikx ; fe(k) =dx e−ikx f (x).
(C.9)
−∞ 2π
−∞

These are the basic formulae of Fourier transforms. The original restriction toperiodic functions has been lifted because any function can be considered to repeat
itself after an infinite interval. The only restriction on f for these formulae to bevalid is that it vanishes sufficiently fast at infinity for the integral
R ∞in the second of
equations (C.9) to converge: the requirement proves to be that −∞ dx |f |2 exists,
which requires that asymptotically |f | < |x|−1/2 . Physicists generally don’t worry
too much about this restriction.
Using the second of equations (C.9) to eliminate the Fourier transform fe fromthe first equation, we have
Z ∞
Z ∞
′
dk
dx′ eik(x−x ) f (x′ ).
(C.10)
f (x) =−∞−∞ 2π
Mathematicians stop here because our next step is illegal.4 Physicists reverse theorder of the integrations in equation (C.10) and writeZ ∞Z ∞dk ik(x−x′ )f (x) =
dx′ f (x′ )e.
(C.11)
2π−∞−∞
Comparing this equation with equation (2.41) we see that the inner integral on theright satisfies the defining condition of the Dirac delta function, and we haveZ ∞dk ik(x−x′ )e.
(C.12)
δ(x − x′ ) =−∞ 2πAppendix D: Operators in classical statistical mechanics
In classical statistical mechanics we are interested in the dynamics of a system with
N degrees of freedom. We do not know the system’s state, which would be quantified by its position q = (q1 , . . . , qN ) and the canonically conjugate momentum
p. Our limited knowledge is contained in the probability density ψ(q, p), which isdefined such that the probability of the system being in the elementary phase-spacevolume dτ = dN q dN p is ψ(q, p) dτ .
Over time q and p evolve according to Hamilton’s equations
q̇ =

∂H
∂p

;

ṗ = −

∂H
,
∂q

(D.1)

and ψ evolves such that probability is conserved:∂∂∂ψ+· (q̇ψ) +· (ṗψ)
∂t
∂q
∂p
∂H ∂ψ
∂H ∂ψ
∂ψ
(D.2)
+
·
−
·
=
∂t
∂p ∂q
∂q ∂p
∂ψ
=
+ {ψ, H},∂twhere the second equality follows from substituting for q̇ and ṗ from Hamilton’s
equations, and the last line follows from the definition of a Poisson bracket: if
0=

4 It is legitimate to reverse the order of integration only when the integrand is absolutely convergent, i.e., the integral of its absolute value is finite. This condition is clearly
not satisfied in the case of eikx . By breaking the proper rules we obtain an expression
for an object, the Dirac delta function, that is not a legitimate function. However, it isextremely useful.

290

Appendix D: Operators in classical statistical mechanics

Box D.1: Classical operators for a single particle
In the simplest case, our system consists of a single particle with Hamiltonian
H = 12 p2 /m + V (x). Then the operators associated with px , x, H and Lz are∂∂p̂x = −ih̄{·, px } = −ih̄
; x̂ = −ih̄{·, x} = ih̄
∂x
∂px
p
∂ 
· ∇ − ∇V ·Ĥ = −ih̄{·, H} = −ih̄m∂p(1)L̂z = −ih̄{·, Lz } = −ih̄{·, xpy − ypx }
 ∂
∂
∂ 
∂
− py
−y
+ px
.
= −ih̄ x
∂y
∂x
∂py
∂px
d
2 ) 6= (p̂)2 . The commutators of these operators are
Notice that (p[x̂, p̂x ] = 0 ; [L̂x , L̂y ] = ih̄L̂z .
(The easiest way to derive the second of these results is to apply (D.8).)(2)

F (q, p) and G(q, p) are any two functions on phase space, the Poisson bracket{F, G} is defined to be
X ∂F ∂G
∂F ∂G
−
.
(D.3)
{F, G} ≡
∂q
∂p
i ∂pi
i ∂qi
i
We use the Poisson bracket to associate with F an operator F̂ on other functions of phase space: we define the operator F̂ by its action on an arbitrary functionψ(p, q):F̂ ψ ≡ −ih̄{ψ, F }.
(D.4)
Here h̄ is some constant with the dimensions of the product p · q – i.e., the inverseof the dimensions of the Poisson bracket – and is introduced so the operator F̂ has
the same dimensions as the function F. The factor i is introduced to ensure thatwith the obvious definition of an inner product, the operator F̂ is Hermitian:
Z

Z
Z
∂ψ ∂F
∂ψ ∂F
dN p dN q φ∗dτ φ∗ F̂ ψ = −ih̄·− dN q dN p φ∗
·
∂q ∂p
∂p ∂q
Z

Z
∗
∗
∂φ
∂F
∂φ
∂F
(D.5)
dN p dN q= ih̄·ψ − dN q dN p
·
ψ
∂q ∂p
∂p ∂q
Z
= dτ (F̂ φ)∗ ψ.
When written in terms of the classical Hamiltonian operator Ĥ, the classical evolution equation (D.2) takes a familiar form∂ψ
= Ĥψ.
(D.6)∂t
It is straightforward (if tedious) to show that Poisson brackets, like a commutators, satisfies the Jacobi identity (cf. 2.102)
ih̄{A, {B, C}} + {B, {C, A}} + {C, {A, B}} = 0.

(D.7)

We use this identity to express the commutator of two such operators in terms of
the Poisson bracket of the underlying functions:[Â, B̂]ψ = −h̄2 {{ψ, B}, A} − {{ψ, A}, B}= h̄2 {ψ, {A, B}}

(D.8)

d
= ih̄{A,B}ψ.

dwhere {A,B} denotes the operator associated with the function {A, B}.
Let A(p, q) be some function on phase space. Then the rate of change of thevalue of A along a phase trajectory is
∂A
∂A
dA
=
· ṗ +· q̇ = {A, H}.
dt
∂p
∂q

(D.9)

Consequently A is a constant of motion if and only if 0 = {A, H}, which by (D.8)
requires its operator to commute with the Hamiltonian operator: as in quantum
mechanics, A is a constant of the motion if and only if [Â, Ĥ] = 0.

Lie groups and Lie algebras291

It is instructive to repeat the analysis of §4.2 with the classical operatorsof a single-particle system (Box D.1). If we displace the system through a, its
probability density becomes
"
#

2
1
∂
∂
′
+
ψ (x, p) ≡ ψ(x − a, p) = 1 − a ·a·− . . . ψ(x, p)∂x
2!
∂x(D.10)
a · p̂
∂ψ(x, p) = exp −iψ(x, p).
= exp −a ·∂xh̄Thus p̂/h̄ is the generator of displacements, as in quantum mechanics. Displacement of the system by δa clearly increases the expectation of x by δa, so withdτ ≡ d3 x d3 pZZδa · p̂hxi + δa = dτ xψ ′ (x, p) = dτ x 1 − iψ(x, p) + O(δa2 ).
(D.11)h̄This equation will hold for an arbitrary probability density ψ if and only ifZZZih̄δij = dτ xi p̂j ψ = dτ (p̂j xi )∗ ψ = ih̄ dτ {xi , pj }ψ,(D.12)

where the second equality uses the fact that p̂j is Hermitian. Thus equation (D.11)
holds if and and only if the Poisson brackets {xi , pj } rather than the commutators[x̂i , p̂j ] satisfy the canonical commutation relations. This crucial difference between
the quantum and classical cases arises from the way we calculate expectation values:
in classical physics the quantum rule hQi = hψ|Q|ψi is replaced byZ
hQi = dN q dN p Qψ,(D.13)

where (i) Q is the function not the associated operator, and (ii) ψ occurs once nottwice because it is a probability not a probability amplitude. On account of thesedifferences, whereas equation (4.21) yields [xi , pj ] = ih̄δij , its classical analogue,(D.11) yields {xi , pj } = δij .

Appendix E: Lie groups and Lie algebras
A group is a set of objects that is equipped with an associative product and an
identity member: given any two members a, b of a group G, there is a member ofG, c = ab, and there is a member 1 such that 1a = a for any a in G. For any athere must be an inverse a−1 such that aa−1 = 1. The associativity of the productmeans that for any a, b, c ∈ G, (ab)c = a(bc). If the multiplication is commutative(ab = ba always) then the group is Abelian.
Some groups, such as that of the rotations that turn a square into itself, havea finite number of (discrete) members, but others, such as the group SO(3) of all
three-dimensional rotations have an infinite number of members.1 Moreover, in agroup such as SO(3) some members are almost identical to other members because
they are rotations through almost the same angle around almost the same axis. A
Lie group is such a continuous group.2
The linear transformations of vector spaces provide representations of groups,that is, concrete realisations of an abstract group in which each group member isrepresented by a linear transformation in such a way that the product of groupmembers is faithfully represented by the compounding of transformations. We generally quantify the linear transformations of an n-dimensional vector space withn × n matrices, so a representation of a group consists of a rule associating eachgroup member a with a matrix Ma such that if ab = c, then Ma · Mb = Mc .
In Chapter 4 we saw how translations and rotations of a system are mirrored by unitary transformations U (a) and U (α) of the system’s ket |ψi. These
unitary transformations form representations of the groups of translations and rotations. So group theory, and especially results relating to possible representationsof groups, are important for quantum mechanics. In particular, quantum mechanics enormously extends the range of representations that are physically significant
by introducing the usually infinite-dimensional (Hilbert) space of possible kets. By
1 The group formed by orthogonal rotations in n-dimensional space is called SO(n).
Each group member is most naturally represented by a n × n real orthogonal matrix withunit determinant.
2 Technically, a Lie group is a group that is also a differentiable manifold.

292

Appendix F: The hidden symmetry of hydrogen

contrast, in classical physics groups generally only have finite-dimensional representations.
The structure of a Lie group is largely determined by the group members thatlie near the identity, because by left-multiplying any member of a neighbourhoodof the identity by a, we can map this neighbourhood into a neighbourhood of a.
Moreover, if we repeatedly multiply the identity by a nearby group member, we
can generate members that lie far from the identity, and if the structure of thegroup is simple, we can generate any group member. This logic becomes veryclear when we work with a quantum-mechanical representation: a member near
the identity of SO(3) is represented by the matrix I − iδα · J, where |δα| ≪ 1 andthe result of repeatedly multiplying the identity by this member is represented by
U (α) = e−iα·J (eq. 4.12).
The infinitesimal generators of the group – in the case of SO(3) the angularmomentum operators Ji – form a Lie algebra because they have the following
properties: (i) they can be multiplied by (complex) numbers and added, and (ii)they are equipped with an antisymmetric product (the commutator) which (a)satisfies the Jacobi identity[A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0,

(E.1)

and (b) evaluates to a linear combination of the original
generators. That is, if
P
I1 , . . . , In are n operators which satisfy [Ii , Ij ] = i k ckij Ik , where ckij is a set ofcomplex numbers, then the set {Ii } forms the basis of an n-dimensional Lie algebra.
In the case of SO(3), n = 3 and ckij = iǫijk .
The group SU(2) formed by 2 × 2 unitary matrices with unit determinant
plays a big role in quantum mechanics because it is an important subgroup of thegroup of Lorentz transformations. Its Lie algebra is identical to that of SO(3), sohere we have a manifestation of the fact that a Lie group is not entirely determinedby its Lie algebra, although it nearly is. In fact, SU(2) is essentially an extension
of SO(3).
The group SU(3) of 3 × 3 unitary matrices with unit determinant is crucial
for nuclear physics because the quantum field whose excitations we call quarks
provides representations of SU(3).

Appendix F: The hidden symmetry of hydrogen
The gross structure of hydrogen is degenerate with respect to both l and m; there
are states with different (l, m) that have the same energy. Degeneracy with respect
to m is to be expected whenever the Hamiltonian is rotationally symmetric, but
degeneracy with respect to l is special to the Coulomb potential and indicates thepresence of additional symmetry.
To uncover this symmetry, we define the Hermitian Runge-Lenz operator(Problem 8.15)
mZe2 r.
(F.1)
M ≡ 12 h̄(p × L − L × p) −4πǫ0 r
The important algebraic properties of this operator are derived (after lengthy
computations) in Problem 8.15. Crucially it commutes with the Hamiltonian,[M, H] = 0, so each component of M is associated with a conserved quantity(§2.2.1) and the unitary transformations that M generates, U (θ) ≡ exp(−iθ · M)
where θ is a triple of real numbers, transform stationary states into stationary
states of the same energy because these transformations are dynamical symmetriesof H (§4.3).
The components of M have the commutation relations
Xǫijk Lk .
(F.2)[Mi , Mj ] = −2ih̄2 mH
kConsequently, the action of U (θ) on stationary states of energy E < 0 can beobtained by using the generators
M′ ≡

1
M,
(2h̄2 m|E|)1/2

which have the commutation relations[Mi′ , Mj′ ] = iXkǫijk Lk ,

(F.3)

(F.4)

The hidden symmetry of hydrogen

293

Since M′ is an (axial) vector operator, from §4.2 we know that the commutators
between its components and the orbital-angular momentum operators areXǫijk Mk′ .
(F.5)[Li , Mj′ ] = ikFrom equations (F.4) and (F.5) it follows that the components of
M± ≡ 12 (L ± M′ )(F.6)

commute with each other because[Mi+ , Mj− ] = 41 ([Li , Lj ] − [Li , Mj ] + [Mi , Lj ] − [Mi , Mj ])Xǫijk (Lk − Mk + Mk − Lk ) = 0
= 4i

(F.7)

[Mi± , Mj± ] = 14 ([Li , Lj ] ± [Li , Mj ] ± [Mi , Lj ] + [Mi , M − j])XXǫijk Mk± .
ǫijk (Lk ± Mk ± Mk + Lk ) = i
= 4i

(F.8)

ijk

while

k

k

Thus the components of M+ and M− satisfy the commutation relations we firstencountered in §4.2 in connection with the rotation operators. These relations
define the Lie Algebra of the group SU(2) formed by 2×2 unitary matrices with unit
determinant – for a very brief account Lie groups and Lie algebras, see Appendix E.
The group SO(4) formed by the orthogonal rotations of 4-dimensional vectors
has just the Lie algebra that is formed by the operators M± , namely two sets ofthree operators that commute with each other like the angular-momentum operators, with any operator from one set commuting with any in the other set. This
coincidence of the Lie algebras establishes that the invariance group of hydrogen isSO(4). The manifest spherical symmetry of the atom is just the SO(3) sub-groupof SO(4). The extra dimension of symmetry is obscure because in the classicallimit it is not a point transformation – a mapping of spatial points into spatial
points – but involves momentum in an essential way, so if (x′ , p′ ) is the image ofthe phase-space point (x, p), x′ depends on p as well as on x.
The operators Mi′ that generate the hidden symmetries of hydrogen do notcommute with L2 , so a general symmetry transformation U (θ) will turn an eigenstate |li of L2 into a state that is not an eigenstate of L2 (Problem 8.15). For a
judicious choice of θ it is possible move |li into a different eigenstate |l ± 1i of L2 .
This is almost what the ladder operators Al and A†l of §8.1 accomplish – almost
because Al and A†l only modify the radial part of the wavefunction, and we havem
to change the angular part from Ylm to Yl±1
by hand.
From the results above and our work with the angular-momentum operators,
we know that a complete set of commuting operators for hydrogen will consist ofH, one component of each of M± , say Mz+ and Mz− , and the operators (Mx+ )2 +(My+ )2 + (Mz+ )2 and (Mx− )2 + (My− )2 + (Mz− )2 . So we could label the members−+−of a complete set of hydrogen’s stationary states, |E, m+
z , mz , m , m i, with theireigenvalues with respect to these operators. These five eigenvalues correspond to
the five classical constants of motion of Kepler orbits. Five constraints on the six
phase-space coordinates confine the orbit to a one-dimensional sub-manifold (anellipse) of phase space. Indeed, in any fixed spherical potential, the direction of
the angular momentum vector confines the orbit to a plane, and the magnitudeof the angular momentum and the energy together determine the radial scale andeccentricity of the orbit. The fifth constant of motion, provided by the hidden
symmetry of the inverse-square law, fixes the point of closest approach to thecentre of attraction (the periapsis). The precession of the periapsis of Mercury’s
orbit demonstrates that the planet moves in a force field that deviates from theinverse-square law. About 10% of this deviation is caused by general relativity andthe rest arises from the gravitational fields of the other planets, predominantly thatof Jupiter.

294

Appendix G: Lorentz covariant equations

Appendix G: Lorentz covariant equations
Special relativity is about determining how the numbers used by moving observers
to characterise a given system are related to one another. All observers agree onsome numbers such as the electric charge on a particle – these numbers are called
Lorentz scalars. Other numbers, such as energy, belong to a set of four numbers,
called a four-vector. If you know the values assigned by some observer to all four
components of a four-vector, you can predict the values that any other observer
will assign. If you do not know all four numbers, in general you cannot predict
any of the values that a moving observer will assign. The components of every
ordinary three-dimensional vector are associated with the spatial components of afour-vector, while some other number completes the set as the ‘time component’of the four-vector. We use the convention that Greek indices run from 0 to 3 while
Roman ones run from 1 to 3; the time component is component 0, followed by thex component, and so on. All components should have the same dimensions, so, forexample, the energy-momentum four vector is(p0 , p1 , p2 , p3 ) ≡ (E/c, px , py , pz ).

(G.1)

The energies and momenta assigned by an observer who moves at speed v parallelto the x axis are found by multiplying the four-vector by a Lorentz transformation
matrix. For example, if the primed observer is moving at speed v along the x axis,
then she measures ′  E/c
γ−βγ 0 0E /c′ px   −βγγ0 0  px  , ′ =
(G.2)
 py   0
0
1 0   py 
p′z

pz
p
where β ≡ v/c and the Lorentz factor is γ = 1/ 1 − β 2 . The indices on thefour-vector p are written as superscripts because it proves helpful to have a formof p in which the sign of the time component is reversed. That is we define
0

0

0

1

(p0 , p1 , p2 , p3 ) ≡ (−E/c, px , py , pz ),(G.3)

and we write the indices on the left as subscripts to signal the difference in the timecomponent. It is straightforward to verify that in the primed frame the componentsof the down vector are obtained by multiplication with a matrix that differs slightly
from the one used to transform the up vector


 

−E/c
γ βγ 0 0−E ′ /c
 p′x   βγ γ 0 0   px 
.

 

(G.4)
 p′y  =  0
0 1 0   py 
p′z

0

0

0

pz

1

The Lorentz transformation matrices that appear in equation (G.2) and (G.4) areinverses of one another. In index notation we write these equations
X µ
X ν µ
(G.5)
Λν pµ .
Λ µp
and p′ν =
p′ν =
µ

µ

Notice that we sum over one down and one up index; we never sum over two down
or two up indices. Summing over an up and down index is called contraction of
those indices.
The dot product of the up and down forms of a four vector yields a Lorentz
scalar. For example
Xµpµ pµ = −E2
+ p2x + p2y + p2z = −m20 c2 ,
c2

(G.6)

where m0 is the particle’s rest mass. Observers in relative motion will typicallyassign different values to all
P four components of p and yet from them they willcompute the same value of µ pµ pµ . The dot product of two different four vectors
PP
is also a Lorentz scalar: the value of µ pµ hµ = µ pµ hµ is the same in any frame.
The time and space coordinates form a four-vector(x0 , x1 , x2 , x3 ) ≡ (ct, x, y, z).

(G.7)

In some interval dt of coordinate time, a particle’s position four-vector x incrementsby dx and the Lorentz scalar associated with dx is −c2 times the square of theLorentz covariance295

proper-time interval associated with dt:
1 
1 X
dxµ dxµ = (dt)2 − 2 (dx)2 + (dy)2 + (dz)2(dτ )2 = − 2
c µ
c


v2
= (dt)2 1 − 2 = (dt)2 (1 − β 2 ).
c

(G.8)

The proper time dτ is just the elapse of time in the particle’s instantaneous rest
frame; it is the amount by which the hands move on a clock that is tied to theparticle. From the last equality above it follows that dτ = dt/γ, so moving clocks
tick slowly.
The four-velocity of a particle is




dxµ
dct dx dy dz
dx dy dz
uµ =
=
,
,
,
,
,
= γ c,,(G.9)
dτdτ dτ dτ dτdt dt dt
where γ is the particle’s Lorentz factor. In a particle’s rest frame the four velocity
points straight into the future: uµ = (1, 0, 0, 0). In any frameuµ uµ = −c2 .

(G.10)The electrostatic potential φ and the magnetic vector potential A form a four
vectorAµ = (φ/c, Ax , Ay , Az ).
(G.11)
Some numbers are members of a set of six numbers that must all be known
in one frame before any of them can be predicted in an arbitrary frame. The six
components of the electric and magnetic fields form such a set. We arrange themas the independent, non-vanishing components of an antisymmetric four by four
matrix, called the Maxwell field tensor

0−Ex /c −Ey /c −Ez /c
 Ex /c
0
Bz
−By 
.
(G.12)
Fµν ≡  Ey /c
−Bz
0
Bx 
Ez /c

By

−Bx

0

The electric and magnetic fields seen by a moving observer are obtained by preand post-multiplying this matrix by an appropriate Lorentz transformation matrix
such as that appearing in equation (G.4).
The equation of motion of a particle of rest mass m0 and charge Q isX
duλFλν uν .
(G.13)=Qm0dτνThe time component of the four-velocity u is γc, and the spatial part is γv, so,using our expression (G.12) for F, the spatial part of this equation of motion is

dγv
dv
= γm0
+ O(β 2 ),(G.14)dτdτwhich shows the familiar electrostatic and Lorentz forces in action.
The great merit of establishing these rules is that we can state that the dynamics of any system can be determined from equations in which both sides areof the same Lorentz-covariant type. That is, both sides are Lorentz scalars, orfour-vectors, or antisymmetric matrices, or whatever. Any correct equation that
does not conform to this pattern must be a fragment of a set of equations that do.
Once a system’s governing equations have been written in Lorentz covariant form,
we can instantly transform them to whatever reference frame we prefer to work in.
γQ(E + v × B) = m0

Lagrangian and Hamiltonian for motion in an electromagnetic field Every particle traces a path x(τ ) through space-time. This path is called the particle’s
world line. The particle’s action S[x(τ )] is a functional of the world line – anumber that depends on the whole line. In fact S[x] is given byZ
S[x] = dτ s[x(τ )],(G.15)

where s is a function of x(τ ) and its derivatives. Quantum mechanics ensures thatthe world line taken by a particle as it moves from a given event x1 to a second
event x2 is the one that extremises S. Thus the particle’s dynamics is determinedby the action S, which is in turn determined by the function s[x].
The action S must be a Lorentz scalar since there is no obvious higher n-tuple
into which we can fit it. The proper time τ is a Lorentz scalar, so s must be ascalar too, and to determine the form of s we have only to ask what scalars we can296

Appendix G: Lorentz covariant equations

construct from the world-line x(τ ) and quantities such as A, F associated with theelectromagnetic field in which the particle moves.
First we note that S shouldn’t depend on our choice of origin, so only derivatives of x(τ ) such as dx/dτ should occur in s, not x itself. Furthermore, theEuler-Lagrange equation, which is used to extract equations of motion from a Lagrangian, involves differentiation with respect to the variable that parametrises
position along the extremal path, in this case τ . So we will get a second-orderequation of motion, if s depends on dx/dτ , but not on higher derivatives of x(τ ).
Similarly, the Euler-Lagrange equation involves differentiation with respect to theposition vector x, so if the equation of motion is to depend on F and not its derivatives, s should depend on A but not F. So the Lorentz scalars to consider are (i)|dx/dτ |2 = −c2 and (ii) (dx/dτ ) · A – we exclude |A|2 from consideration since
its contribution to S proves to be both gauge- and world-line dependent. So thesimplest thing to try isZdx
S = dτ −m0 c2 + Q·A ,(G.16)dτ
where we’ve included the rest mass m0 for future convenience and Q is a constantscalar, which will turn out to be the charge.
Now that we have chosen a form for s, we should find the differential equationof the world line that extremises S between given events x1 and x2 . Unfortunately,
this is a non-standard problem because the elapse of proper time between these
events depends on the world line taken between them, and the Euler-Lagrange
equations require the integration variable to have fixed values at the end-points.
By contrast, the coordinate time t does have fixed values at x1 and x2 , so wechange integration variable from τ to t. Since dt/dτ = u0 /c = γ, we obtain
Z


p
dx
·A .
(G.17)
S = dt − m0 c2 1 − v 2 /c2 + Q
dt
If we are willing to restrict ourselves to non-relativistic motion, we can simplify
equation (G.17) by expanding the square root and discarding terms smaller thanv 2 /c2 . Then we have


Z


v2
dx
S = dt − m0 c2 1 − 21 2 + Q· A + QcA0 ,(G.18)cdt

where the term with A0 arises because the vectors x and A are now three- rather
than four-dimensional so we have to add explicitly the contributions of the timecomponents to the contraction of two four-vectors. The term in the integrand−m0 c2 yields a contribution to S that is world-line-independent and therefore
plays no role in determining the world line. Hence we may delete this term andarrive at our final expression for the action of a charged, non-relativistic particleZdx· A − Qφ ,(G.19)
S = dt 21 m0 v 2 + Q
dt

where φ = −A0 /c is the electrostatic potential.
In classical mechanics the Lagrangian is a function L(x, ẋ) of position andvelocity ẋ ≡ dx/dt that yields the action throughZdx
S[(x(t)] = dt L x,.
(G.20)
dt
Hence equation (G.19) states that the Lagrangian of a charged particle is
L (x, ẋ) = 12 m0 |ẋ|2 + Qẋ · A − Qφ.
The classical momentum is defined to be∂L= m0 ẋ + QA,p≡∂ ẋand the classical Hamiltonian is defined to be
H(x, p) ≡ p · ẋ − L|p − QA|2
p − QAp − QA−−Q· A + Qφ
m0
2m0
m0
2
|p − QA|
=
+ Qφ.
2m0

=p·

(G.21)(G.22)(G.23)

Thomas precession

297

Appendix H: Thomas precession
In this appendix we generalise the equation of motion of an electron’s spin (eq. 8.69)
gQ
dS
S×B
(H.1)
=
dt
2m0
from the electron’s rest frame to a frame in which the electron is moving. We do
this by writing equation (H.1) in Lorentz covariant form (Appendix G).
The first step in upgrading equation (H.1) to Lorentz covariant form is toreplace S and B with covariant structures. We hypothesise that the numbers Si
comprise the spatial components of a four vector that has vanishing time component
in the particle’s rest frame. Thus
(s0 , s1 , s2 , s3 ) = (0, Sx , Sy , Sz )(rest frame),(H.2)

and we can calculate sµ in an arbitrary frame by multiplying this equation by anappropriate Lorentz transformation matrix. Since in the rest frame sµ is orthogonal
to the particle’s four-velocity uµ , the equation
uµ sµ = 0

(H.3)

holds in any frame. In equation (8.69) B clearly has to be treated as part of theMaxwell field tensor Fµν (eq. G.12). In the particle’s rest frame dt = dτ and  
−S · E/c00−Ex /c −Ey /c −Ez /c
X






Ex /c
0
Bz
−By   Sx  
 (H.4)=Fµν sν =  Ey /c
S×B 
Sy
−Bz
0
Bx
ν
Sz
Ez /c
By
−Bx
0
so equation (H.1) coincides with the spatial components of the covariant equation
dsµ
gQ X
=
Fµν sν .
(H.5)
dτ
2m0 ν

This cannot be the correct equation, however, because it is liable to violate thecondition (H.3). To see this, consider the case in which the particle moves atconstant velocity and dot equation (H.5) through by the fixed four-velocity uν .
Then we obtain
X d µ
gQ X
(u sµ ) =Fµν uµ sν .
(H.6)
dτ
2m
0
µ
µν

The left side has to be zero but there is no reason why the right side should vanish.
We can fix this problem by adding an extra term to the right side, so that
!
X λ
uν uµ
gQ X
dsµ
ν
s Fλν 2=Fµν s −
.
(H.7)
dτ
2m0
c
ν
λν
µ

When this equation is dotted throughP by u , and equation (G.10) is used, theright side becomes proportional to µν Fµν (sµ uν + sν uµ ), which vanishes because
F is antisymmetric in its indices while the bracket into which it is contracted issymmetric in the same indices.1
If our particle is accelerating, equation (H.7) is still incompatible with equation(H.3), as becomes obvious when one dots through by uµ and includes a non-zeroterm duµ /dτ . Fortunately, this objection is easily fixed by adding a third term to
the right side. We then have our final covariant equation of motion for s
!
X λ
dsµ
uν uµ
gQ X
1 X λ duλs Fλν 2s
Fµν sν −
=
uµ .
(H.8)
+ 2
dτ
2m0
c
c
dτ
ν
λν

λ

In the rest frame the spatial components of this covariant equation coincide withthe equation (8.69) that we started from because ui = 0. The two new terms on
the right side ensure that s remains perpendicular to the four-velocity u as it must
do if it is to have vanishing time component in the rest frame.
The last term on the right of equation (H.8) is entirely generated by theparticle’s acceleration; it would survive even in the case g = 0 of vanishing magnetic
moment. Thus the spin of an accelerating particle precesses regardless of torques.
This precession is called Thomas precession.2

1 Here’s a proof that the contraction of tensors S and A that are respectively symPP
metric and antisymmetric in their indices vanishes.
µν Sµν Aµν =µν Sνµ Aµν =P− µν Sνµ Aνµ . This establishes that the sum is equal to minus itself. Zero is the only
number that has this property.
2 L.T. Thomas, Phil. Mag. 3, 1 (1927). For an illuminating discussion see §11.11 ofClassical Electrodynamics by J.D. Jackson (Wiley).

298

Appendix H: Thomas precession

If the particle’s acceleration is entirely due to the electromagnetic force thatit experiences because it is charged, its equation of motion is (G.13). Using this inequation (H.8), we find
!
X λ
X
uν uµ
Q
dsµ
s Fλν 2Fµν sν − (g − 2)
g
.
(H.9)
=
dτ
2m0
c
ν
λν

For electrons, g = 2.002 and to a good approximation the extra terms we haveadded cancel and our originally conjectured equation (H.5) holds after all. We now
specialise on the unusually simple and important case in which g = 2.
From our equation of motion of the covariant object s we derive the equationof motion of the three-vector S whose components are the expectation values ofthe spin operators. We choose to work in the rest frame of the atom. By equation(H.2), S is related to s by a time-dependent Lorentz transformation from this frameto the electron’s rest frame. We align our x axis with the direction of the required
boost, so
 0
  
s
γ
−βγ 0 00
  s1  Sx   −βγ
γ
0
0
 .
 =
(H.10)
 Sy   0
0
1 0   s2 
s3
Sz
0
0
0 1

The time equation implies that s0 = βs1 , so the x equation can be written
Sx = γ(s1 − βs0 ) = γ(1 − β 2 )s1 =s1= s1 (1 − 12 β 2 + · · ·).
γ(H.11)The y and z components of equation (H.10) state that the corresponding components of S and s are identical. Since s1 is the projection of the spatial part of s
onto the particle’s velocity v, we can summarise these results in the single equationv·s
S=s−
v + O(β 4 )(H.12)
2c2
as one can check by dotting through with the unit vectors in the three spatial
directions. Differentiating with respect to proper time and discarding terms oforder β 2 , we find


dv
ds
1
v · s dv
dS
=
− 2
·s v−
+ O(β 2 ).
(H.13)dτdτ
2cdτ
2c2 dτ
Equation (H.9) implies that with g = 2



ds
Q s0
Q v · s=
E+s×B =
E+s×B ,
2
dτ
m0 c
m0
c

(H.14)

where the second equality uses the relation s0 = βs1 = (v · s)/c. We now use thisequation and equation (G.14) to eliminate ds/dτ and dv/dτ from equation (H.13).
dS
Q(v · s)E − (E · s)v + 2c2 s × B + O(β 2 )=dτ
2m0 c2(H.15)

Q
2
2
S
×
(E
×
v)
+
2c
S
×
B
+
O(β
).
=
2m0 c2Since we are working in the atom’s rest frame, B = 0 unless we are applying an
external electric field. The difference between the electron’s proper time τ and theatom’s proper time t is O(β 2 ), so we can replace τ with t. We assume that E isgenerated by an electrostatic potential Φ(r) that is a function of radius only. Then
E = −∇Φ = −(dΦ/dr)r/r points in the radial direction. Using this relation inequation (H.15) we find


dS
Q
1 dΦ
2
=
−
S
×
(r
×
v)
+
2c
S
×
B
.
(H.16)
dt
2m0 c2r drWhen r × v is replaced by h̄L/m0 , we obtain equation (8.70). The factor of two
difference between the coefficients of S in the spin-orbit and Zeeman Hamiltonians(8.71) and (8.72) that so puzzled the pioneers of quantum mechanics, arises because the variable in equation (H.5) is not the usual spin operator but a Lorentz
transformed version of it. The required factor of two emerges from the derivativesof v in equation (H.13). Hence it is a consequence of the fact that the electron’srest frame is accelerating.

Dipole-dipole interaction

299

Appendix I: Matrix elements for a dipole-dipole
interaction
We calculate the matrix elements obtained by squeezing the hyperfine Hamiltonian(8.81) between states that would all be ground states if the nucleus had no magnetic
dipole moment. We assume that these are S states, and therefore have a sphericallysymmetric spatial wavefunction ψ(r). They differ only in their spins. In practicethey will be the eigenstates |j, mi of the total angular momentum operators J 2 andJz that can be constructed by adding the nuclear and electron spins. We use thesymbol s as a shorthand for j, m or whatever other eigenvalues we decide to use.
The matrix elements areZ(I.1a)
Mss′ ≡ hψ, s|HHFS |ψ, s′ i = d3 x ρ(r)hs|HHFS |s′ i,whereρ(r) ≡ |ψ(r)|2 ,′(I.1b)′

and for given s, s hs|HHFS |s i is a function of position x only. Substituting for
HHFS from equation (8.81) we haveZn
 µ oµ0eMss′ =|s′ i.
(I.2)
d3 x ρhs|µN · ∇ × ∇ ×
4πr

We now use tensor notation (Appendix B) to extract the spin operators from theintegral, findingµ0 X
Mss′ =
ǫijk ǫklm hs|µNi µem |s′ iI,(I.3a)
4π ijklm

whereZ∂ 2 r −1.
(I.3b)∂xj ∂xl
The domain of integration is a large sphere centred on the origin. On evaluating the derivatives of r −1 and writing the volume element d3 x in spherical polar
coordinates, the integral becomes


Z
Z
xj xl
δjl
I = ρ(r)r 2 dr d2 Ω 3 5 − 3 .
(I.4)
r
r
I≡

d3 x ρ(r)We integrate over polar angles first. If j 6= l, the first term integrates to zero
because the contribution from a region in which xj is positive is exactly cancelled
by a contribution from a region in which xj is negative. When j = l, we orient ouraxes so that xj is the z axis. Then the angular integral becomes


Z
Z
2π
δjl
xj xl
dθ sin θ(3 cos2 θ − 1) = 0.
(I.5)
dΩ 3 5 − 3 = 3rrr

The vanishing of the angular integral implies that no contribution to the integral ofequation (I.3b) comes from the entire region r > 0. However, we cannot conclude
that the integral vanishes entirely because the coefficient of ρ in the radial integralof (I.4) is proportional to 1/r, so the radial integral is divergent for ρ(0) 6= 0.
Since any contribution comes from the immediate vicinity of the origin, we
return to our original expression but restrict the region of integration to an infinitesimal sphere around the origin. We approximate ρ(r) by ρ(0) and take it outof the integral. Then we can invoke the divergence theorem to state thatZZ
∂ 2 r −1∂r −1
I = ρ(0) d3 x
= d2 Ω rxj,(I.6)∂xj ∂xl∂xl

where we have used the fact that on the surface of a sphere of radius r the infinitesimal surface element is d2 S = d2 Ω rx. We now evaluate the surviving derivative of
1/r:
Z
4π
xj xl
(I.7)
I = −ρ(0) d2 Ω 2 = − ρ(0)δjl ,r3
where we have again exploited the fact that the integral vanishes by symmetry if
j 6= l, and that when j = l it can be evaluated by taking xj to be z. Inserting thisvalue of I in equation (I.3a), we haveX
µ0ǫijk ǫklm δjl hs|µNi µem |s′ i.
(I.8)
Mss′ = − ρ(0)
3
ijklm

Now

X

ijklm

ǫijk ǫklm δjl =

X

ijkm

ǫijk ǫkjm = −

X

ijkm

ǫijk ǫmjk

(I.9)

300

Appendix J: Selection rule for j

This sum must be proportional to δim because if i 6= m, it is impossible for both(ijk) and (mjk) to be permutations of (xyz) as they must be to get a non-vanishing
contribution to the sum. We can determine the constant of proportionality bymaking a concrete choice for i = m. For example, when they are both x we haveXǫxjk ǫxjk = ǫxyz ǫxyz + ǫxzy ǫxzy = 2.
(I.10)
jk

When these results are used in equation (I.8), we have finallyX
2µ0µNi · µei |s′ i.
|ψ(0)|2 hs|
Mss′ =
3i(I.11)

Appendix J: Selection rule for j
In Problem 7.21 the selection rule on l is derived by calculating [L2 , [L2 , xi ]] andthen squeezing the resulting equation between states hl, ml | and |l′ , m′l i.P
The algebra uses only two facts about the operators L and x, namely [Li , xj ] = i k ǫijk xk ,and L · x =P0. Now if we substitute J for L, the first equation carries over (i.e.,[Ji , xj ] = i k ǫijk xk ) but the second does not (J · x = S · x). To proceed, we define
the operator
X ≡ J × x − ix.
(J.1)
Since
P X is a vector operator, it will satisfy the commutation relations [Ji , Xj ] =i k ǫijk Xk , as can be verified by explicit calculation. Moreover X is perpendicular
to J:
X
X
Jm xm
ǫklm Jk Jl xm − i
J·X=
m

klm

= 21

X

ǫklm [Jk , Jl ]xm − i

klm

= 12

X

ǫklm i

klm

X
p

X
m

ǫklp Jp xm − i

Jm xm
X

(J.2)

Jm xm = 0,mwhere the last equality uses equation (I.10). We can now argue that the algebraof Problem 7.21 will carry over with J substituted for L and X substituted for x.
Hence the matrix elements hjm|Xk |j ′ m′ i satisfy the selection rule |j − j ′ | = 1.
Now we squeeze a component of equation (J.1) between two states of welldefined angular momentum
X
X
hjm|Js |j ′′ m′′ ihj ′′ m′′ |xt |j ′ m′ i − ihjm|xr |j ′ m′ iǫrsthjm|Xr |j ′ m′ i =j ′′ m′′st=

Xǫrst hjm|Js |jm′′ ihjm′′ |xt |j ′ m′ i − ihjm|xr |j ′ m′ i,

m′′ st(J.3)
where the sum over j ′′ has been reduced to the single term j ′′ = j because [J 2 , Js ] =0. The left side vanishes for |j−j ′ | =6 1. Moreover, since J·x is a scalar, it commuteswith J 2 and we have that hjm|J · x|j ′ m′ i = 0 unless j = j ′ , orX
(J.4)
hjm|Jt |jm′′ ihjm′′ |xt |j ′ m′ i ∝ δjj ′
m′′ t

′

Let |j − j | > 1, then in matrix notation we can write equations (J.3) for r = x, yand (J.4) as
0 = Jy z − Jz y − ix
0 = Jz x − Jx z − iy
(J.5)
0 = Jx x + Jy y + Jz z,

where Jx etc are the (2j +1)×(2j +1) spin-j matrices introduced in §7.4.4 and x etc
are the (2j+1)×(2j ′ +1) arrays of matrix elements that we seek to constrain. These
are three linear homogeneous simultaneous equations for the three unknowns x, etc.
Unless the 3 × 3 matrix that has the J matrices for its elements is singular, theequations will only have the trivial solution x = y = z = 0. One can demonstratethat the matrix is non-singular by eliminating first x and then y. Multiplying thefirst equation by iJx and then subtracting the third, and taking the second fromiJz times the first, we obtain
0 = (iJx Jy − Jz )z − (iJx Jz + Jy )y
0 = (iJz Jy + Jx )z − (iJ2z − i)y.

(J.6)

Restrictions on scattering potentials

301

Eliminating y yields
0 = {i(J2z − 1)(iJx Jy − Jz ) − (iJx Jz + Jy )(iJz Jy + Jx )}z= {−J2z Jx Jy − iJ3z + Jx Jy + iJz(J.7)+ Jx J2z Jy − i(Jx Jz Jx + Jy Jz Jy ) − Jy Jx }z.
We can simplify the matrix that multiplies z by working Jz to the front. In fact,using
Jx J2z Jy = (Jz Jx − iJy )Jz Jy = Jz (Jz Jx − iJy )Jy − i(Jz Jy + iJx )Jy= J2z Jx Jy − 2iJz J2y + Jx Jy
and
Jx Jz Jx = Jz J2x − iJy Jx
Jy Jz Jy = Jz J2y + iJx Jy)⇒ Jx Jz Jx + Jy Jz Jy = Jz (J2x + J2y ) − Jz

(J.8)

(J.9)

equation (J.7) simplifies to{iJz (3 − J2 − 2J2y ) + Jx Jy }z = 0.

(J.10)The matrix multiplying z is not singular, so z = 0. Given this result, the secondof equations (J.9) clearly implies that y = 0, which in turn implies that x = 0.
This completes the demonstration that the matrix elements of the dipole operator
between states of well defined angular momentum vanish unless |j − j ′ | ≤ 1.

Appendix K: Restrictions on scattering potentials
The Ω± operators of equation (12.3) require us to evaluate eiHt/h̄ e−iH0 t/h̄ as t →∓∞. Since e±i∞ is not mathematically well defined, we must check we really knowwhat Ω± actually means.
We can determine Ω± from equation (12.13) if it is possible to take the limitt → ∓∞ in the upper limit of integration. Hence the Ω± operators will make sense
so long as this integral converges when acting on free states.
Let’s concentrate for a while on Ω− , with |ψ; 0i = Ω− |φ; 0i telling us that |ψi
and |φi behave the same way in the distant future. Using equation (12.13), we
have
Z
i ∞
|ψ; 0i = Ω(t′ )|φ; 0i +dτ U † (τ )V U0 (τ )|φ; 0i.
(K.1)
h̄ t′

To decide if the integral converges, we ask whether its modulus is finite (as it must
be if |ψi can be normalized). The triangle inequality |v1 + v2 | ≤ |v1 | + |v2 | tells
us that the modulus of an integral is no greater than the integral of the modulus
of its integrand, so
Z ∞Z ∞dτ U † (τ )V U0 (τ )|φ; 0i ≤dτ U † (τ )V U0 (τ )|φ; 0i .
(K.2)
t′

t′

Since U (τ ) is unitary, the integrand simplifies to |V U0 (τ )|φ; 0i| = |V |φ; τ i| where|φ; τ i is the state of the free particle
at time τ . If the potential depends only on
Rposition, it can be written V = d3 x V (x)|xihx|, and the integrand on the right ofequation (K.2) becomes
1/2
Z
d3 x V 2 (x)|hx|φ; τ i|2.
(K.3)
V |φ; τ i = hφ; τ |V 2 |φ; τ i1/2 =What does this expression mean? At any fixed time, |hx|φ; τ i|2 d3 x is the probability that we find our particle in a small volume d3 x. Equation (K.3) instructs us to
add up these probabilities over all space, weighted by the square of the potential –in other words (with the square root) we calculate the rms V (x) felt by the particle
at time τ . As time progresses, the particle moves and we repeat the process, adding
up the rms potential all alongRthe line of flight.
Now 1 = hφ; τ |φ; τ i = d3 x |hx|φ; τ i|2 , we can be confident that for any
given value of τ the integral over x on the right of (K.3) will be finite. Since theintegrand of the integral over τ is manifestly positive, convergence of the integral
over τ requires that
1/2
Z
< O(τ −1 ).
(K.3b)
d3 x V 2 (x) |hx|φ; τ i|2
lim
τ →∞

302

Appendix K: Restrictions on scattering potentials

We began our discussion of scattering processes by claiming that the real particle
should be free when far from the target, so it’s not surprising that we now find acondition which requires that the particle feels no potential at late times.
If we neglect dispersion, |hx|φ; τ i|2 is just a function of the ratio ξ ≡ x/τ
as the particle’s wavepacket moves around. Assuming that the potential varies assome power r −n at large radii, we have for large τ
Z
Z
(K.4)
d3 x V 2 (x) |hx|φ; τ i|2 ≃ τ 3−2n d3 ξ V 2 (ξ)f (ξ).
Hence, at late times the rms potential varies as ∼ τ −n+3/2 and Ω± is certainly welldefined for potentials that drop faster than r −5/2 . When dispersion is taken into
account, we can sometimes strengthen this result to include potentials that drop
more slowly – see Problem 12.4.
Unfortunately, the Coulomb potential does not satisfy our condition. We will
not let this bother us too greatly because pure Coulomb potentials never arise – if
we move far enough away, they are always shielded by other charges.