Introduction to Quantum Mechanics

The Manchester Physics Series
General Editors
D. J. SANDIFORD: F. MANDL: A. C. PHILLIPS
Department of Physics and Astronomy,
University of Manchester

Properties of Matter:

B. H. Flowers and E. Mendoza

Statistical Physics:
Second Edition

F. Mandl

Electromagnetism:
Second Edition

I. S. Grant and W. R. Phillips

Statistics:

R. J. Barlow

Solid State Physics:
Second Edition

J. R. Hook and H. E. Hall

Quantum Mechanics:

F. Mandl

Particle Physics:
Second Edition

B. R. Martin and G. Shaw

The Physics of Stars:
Second Edition

A. C. Phillips

Computing for Scientists:

R. J. Barlow and A. R. Barnett

Nuclear Physics:

J. S. Lilley

Introduction to Quantum Mechanics: A. C. Phillips

INTRODUCTION TO
QUANTUM MECHANICS

A. C. Phillips

Department of Physics and Astronomy
University of ManchesterCopyright # 2003 by John Wiley & Sons Ltd,The Atrium, Southern Gate, Chichester,West Sussex, PO19 8SQ, England
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To my sons:
JosephMichaelPatrick
PeterThis page intentionally left blank

Contents
Forewordxi

Editor's preface to the Manchester Physics Seriesxiii

Author's preface

xv

1

PLANCK'S CONSTANT IN ACTION
1.1
1.2
1.3
1.4

2

1
4
7
10
11
13
16
17

È DINGER EQUATIONTHE SCHRO
2.1

2.2

3

Photons
De Broglie Waves
Atoms
Measurement
The uncertainty principle
Measurement and wave±particle duality
Measurement and non-locality
Problems 1

Waves
Sinusoidal waves
Linear superpositions of sinusoidal waves
Dispersive and non-dispersive waves
Particle Wave Equations
A wave equation for a free particle
Wave equation for a particle in a potential energy field
Problems 2

21
21
22
23
26
27
29
31

POSITION AND MOMENTUM
3.1 Probability
Discrete random variables
Continuous random variables
3.2 Position Probabilities
Two-slit interference
The Born interpretation of the wave function

35
35
37
38
38
41

viii

Contents

3.3
3.4
3.5
3.6
4

4.5
4.6

The Hamiltonian Operator
Normal Modes of a String
States of Certain Energy
A Particle in a Box II
A one-dimensional box
A three-dimensional box
States of Uncertain Energy
Basis functions
Energy probability amplitudes
Time Dependence
Problems 4

59
60
63
66
66
69
71
71
73
74
77

SQUARE WELLS AND BARRIERS
5.1

5.2

6

42
44
46
48
49
50
52

ENERGY AND TIME
4.1
4.2
4.3
4.4

5

Momentum Probabilities
A Particle in a Box I
Expectation Values
Operators
Uncertainties
Quantum States
Problems 3

Bound and Unbound States
Bound states
Unbound states
General implications
Barrier Penetration
Stationary state analysis of reflection and transmission
Tunnelling through wide barriers
Tunnelling electrons
Tunnelling protons
Problems 5

83
85
88
93
94
95
97
99
100
103

THE HARMONIC OSCILLATOR
6.1
6.2
6.3
6.4
6.5
6.6

The Classical Oscillator
The Quantum Oscillator
Quantum States
Stationary states
Non-stationary states
Diatomic Molecules
Three-dimensional Oscillators
The Oscillator Eigenvalue Problem
The ground state

109
110
112
112
116
118
121
123
125

Contents

Excited states
Is E0 really the lowest energy?
Mathematical properties of the oscillator eigenfunctions
Pr o b l e m s 6
7

7.3
7.4
7.5

Essential Properties
Position and Momentum
Eigenfunctions for position
Eigenfunctions for momentum
Delta function normalization
Compatible Observables
Commutators
A particle in one dimension
A particle in three dimensions
Constants of Motion
Problems 7

136
138
138
139
140
141
142
143
145
146
148

ANGULAR MOMENTUM
8.1
8.2

8.3

9

126
127
128
128

OBSERVABLES AND OPERATORS
7.1
7.2

8

ix

Angular Momentum Basics
Magnetic Moments
Classical magnets
Quantum magnets
Magnetic energies and the Stern±Gerlach experiment
Orbital Angular Momentum
Classical orbital angular momentum
Quantum orbital angular momentum
Angular shape of wave functions
Spherical harmonics
Linear superposition
Problems 8

155
158
158
159
161
163
163
164
164
169
171
174

THE HYDROGEN ATOM
9.1
9.2
9.3
9.4
9.5
9.6
9.7

Central Potentials
Classical mechanics of a particle in a central potential
Quantum mechanics of a particle in a central potential
Quantum Mechanics of the Hydrogen Atom
Energy levels and eigenfunctions
Sizes and Shapes
Radiative Transitions
The Reduced Mass Effect
Relativistic Effects
The Coulomb Eigenvalue Problem

179
179
182
185
188
191
194
196
198
202

x

Contents

Problems 9
10

IDENTICAL PARTICLES
10.1
10.2
10.3
10.4

11

205

Exchange Symmetry
Physical Consequences
Exchange Symmetry with Spin
Bosons and FermionsP r o b l e m s 10

213
215
219
222
224

ATOMS
11.1
11.2
11.3

Atomic Quantum States
The central field approximation
Corrections to the central field approximation
The Periodic Table
What If ?
P r o b l e m s 11

229
230
234
238
241
246

Hints to selected problems

249

Further reading

262

Index

263

Physical constants and conversion factors

Inside Back CoverForeword
Sadly, Tony Phillips, a good friend and colleague for more than thirty years,
died on 27th November 2002. Over the years, we discussed most topics under
the sun. The originality and clarity of his thoughts and the ethical basis of his
judgements always made this a refreshing exercise. When discussing physics,
quantum mechanics was a recurring theme which gained prominence after his
decision to write this book. He completed the manuscript three months before
his death and asked me to take care of the proofreading and the Index. A
labour of love. I knew what Tony wantedÐand what he did not want. Except
for corrections, no changes have been made.
Tony was an outstanding teacher who could talk with students of all abilities.
He had a deep knowledge of physics and was able to explain subtle ideas in a
simple and delightful style. Who else would refer to the end-point of nuclear
fusion in the sun as sunshine? Students appreciated him for these qualities, his
straightforwardness and his genuine concern for them. This book is a fitting
memorial to him.
Franz Mandl
December 2002

This page intentionally left blankEditors' preface to the
Manchester Physics Series
The Manchester Physics Series is a series of textbooks at first degree level. It
grew out of our experience at the Department of Physics and Astronomy atManchester University, widely shared elsewhere, that many textbooks containmuch more material than can be accommodated in a typical undergraduate
course; and that this material is only rarely so arranged as to allow thedefinition of a short self-contained course. In planning these books we have
had two objectives. One was to produce short books: so that lecturers shouldfind them attractive for undergraduate courses; so that students should not befrightened off by their encyclopaedic size or price. To achieve this, we have beenvery selective in the choice of topics, with the emphasis on the basic physics
together with some instructive, stimulating and useful applications. Our second
objective was to produce books which allow courses of different lengths anddifficulty to be selected with emphasis on different applications. To achieve
such flexibility we have encouraged authors to use flow diagrams showing the
logical connections between different chapters and to put some topics in starredsections. These cover more advanced and alternative material which is not
required for the understanding of latter parts of each volume.
Although these books were conceived as a series, each of them is selfcontained and can be used independently of the others. Several of them aresuitable for wider use in other sciences. Each Author's Preface gives details
about the level, prerequisites, etc., of that volume.
The Manchester Physics Series has been very successful with total sales ofmore than a quarter of a million copies. We are extremely grateful to the many
students and colleagues, at Manchester and elsewhere, for helpful criticisms
and stimulating comments. Our particular thanks go to the authors for all thework they have done, for the many new ideas they have contributed, and for
discussing patiently, and often accepting, the suggestions of the editors.

xiv

Editors' preface to the Manchester Physics Series

Finally we would like to thank our publishers, John Wiley & Sons, Ltd,for their enthusiastic and continued commitment to the Manchester Physics
Series.
D. J. Sandiford
F. Mandl
A. C. Phillips
February 1997

Author's preface
There are many good advanced books on quantum mechanics but there is adistinct lack of books which attempt to give a serious introduction at a levelsuitable for undergraduates who have a tentative understanding of mathematics, probability and classical physics.
This book introduces the most important aspects of quantum mechanics inthe simplest way possible, but challenging aspects which are essential for ameaningful understanding have not been evaded. It is an introduction toquantum mechanics which. motivates the fundamental postulates of quantum mechanics by considering
the weird behaviour of quantum particles. reviews relevant concepts in classical physics before corresponding concepts
are developed in quantum mechanics. presents mathematical arguments in their simplest form
. provides an understanding of the power and elegance of quantum mechanics
that will make more advanced texts accessible.
Chapter 1 provides a qualitative description of the remarkable propertiesof quantum particles, and these properties are used as the guidelines for atheory of quantum mechanics which is developed in Chapters 2, 3 and 4.
Insight into this theory is gained by considering square wells and barriers in
Chapter 5 and the harmonic oscillator in Chapter 6. Many of the concepts usedin the first six chapters are clarified and developed in Chapter 7. Angularmomentum in quantum mechanics is introduced in Chapter 8, but because
angular momentum is a demanding topic, this chapter focusses on the ideasthat are needed for an understanding of the hydrogen atom in Chapter 9,
identical particles in Chapter 10 and many-electron atoms in Chapter 11.
Chapter 10 explains why identical particles are described by entangled quantum
states and how this entanglement for electrons leads to the Pauli exclusion
principle.
Chapters 7 and 10 may be omitted without significant loss of continuity.
They deal with concepts which are not needed elsewhere in the book.

xvi

Author's preface

I would like to express my thanks to students and colleagues at the University of Manchester. Daniel Guise helpfully calculated the energy levels in ascreened Coulomb potential. Thomas York used his impressive computing
skills to provide representations of the position probabilities for particles withdifferent orbital angular momentum. Sean Freeman read an early version of thefirst six chapters and provided suggestions and encouragement. Finally, I
would like to thank Franz Mandl for reading an early version of the bookand for making forcefully intelligent suggestions for improvement.
A. C. Phillips
August 2002

1
Planck's constant in action
Classical physics is dominated by two fundamental concepts. The first is theconcept of a particle, a discrete entity with definite position and momentum
which moves in accordance with Newton's laws of motion. The second is theconcept of an electromagnetic wave, an extended physical entity with a presence at every point in space that is provided by electric and magnetic fields
which change in accordance with Maxwell's laws of electromagnetism. The
classical world picture is neat and tidy: the laws of particle motion account forthe material world around us and the laws of electromagnetic fields accountfor the light waves which illuminate this world.
This classical picture began to crumble in 1900 when Max Planck published atheory of black-body radiation; i.e. a theory of thermal radiation in equilibriumwith a perfectly absorbing body. Planck provided an explanation of the observed properties of black-body radiation by assuming that atoms emit andabsorb discrete quanta of radiation with energy E  hn, where n is the frequencyof the radiation and h is a fundamental constant of nature with valueh  6:626  10ÿ34 J s:
This constant is now called Planck's constant.
In this chapter we shall see that Planck's constant has a strange role oflinking wave-like and particle-like properties. In so doing it reveals that physics
cannot be based on two distinct, unrelated concepts, the concept of a particle
and the concept of a wave. These classical concepts, it seems, are at best
approximate descriptions of reality.

1.1

PHOTONS

Photons are particle-like quanta of electromagnetic radiation. They travel at
the speed of light c with momentum p and energy E given by

2

Planck's constant in action

p

h
l

Chap. 1

and

E

hc
,
l

(1:1)

where l is the wavelength of the electromagnetic radiation. In comparison withmacroscopic standards, the momentum and energy of a photon are tiny. For
example, the momentum and energy of a visible photon with wavelengthl  663 nm arep  10ÿ27 J s

andE  3  10ÿ19 J:

We note that an electronvolt, 1 eV  1:602  10ÿ19 J, is a useful unit for theenergy of a photon: visible photons have energies of the order of an eV and
X-ray photons have energies of the order of 10 keV.
The evidence for the existence of photons emerged during the early yearsof the twentieth century. In 1923 the evidence became compelling whenA. H. Compton showed that the wavelength of an X-ray increases when it isscattered by an atomic electron. This effect, which is now called the Compton
effect, can be understood by assuming that the scattering process is a photon±
electron collision in which energy and momentum are conserved. As illustrated
in Fig. 1.1, the incident photon transfers momentum to a stationary electron sothat the scattered photon has a lower momentum and hence a longer wavelength. In fact, when the photon is scattered through an angle y by a stationary
electron of mass me , the increase in wavelength is given byDl h(1 ÿ cos y):
me c

(1:2)

We note that the magnitude of this increase in wavelength is set by

pf

p

i

q

Pf

Fig. 1.1 A photon±electron collision in which a photon is scattered by a stationary
electron through an angle y. Because the electron recoils with momentum Pf , themagnitude of the photon momentum decreases from pi to pf and the photon wavelength
increases.

1.1

3

Photonsh 2:43  10ÿ12 m,
me ca fundamental length called the Compton wavelength of the electron.
The concept of a photon provides a natural explanation of the Compton
effect and of other particle-like electromagnetic phenomena such as the photoelectric effect. However, it is not clear how the photon can account for thewave-like properties of electromagnetic radiation. We shall illustrate this difficulty by considering the two-slit interference experiment which was first used byThomas Young in 1801 to measure the wavelength of light.
The essential elements of a two-slit interference are shown in Fig. 1.2. Whenelectromagnetic radiation passes through the two slits it forms a pattern ofinterference fringes on a screen. These fringes arise because wave-like disturbances from each slit interfere constructively or destructively when they arrive at
the screen. But a close examination of the interference pattern reveals that it is
the result of innumerable photons which arrive at different points on the screen,
as illustrated in Fig. 1.3. In fact, when the intensity of the light is very low, theinterference pattern builds up slowly as photons arrive, one by one, at random
points on the screen after seemingly passing through both slits in a wave-likeway. These photons are not behaving like classical particles with well-defined
trajectories. Instead, when presented with two possible trajectories, one foreach slit, they seem to pass along both trajectories, arrive at a random pointon the screen and build up an interference pattern.
D
Wave-like entity
incident on two slits

R2

P
X

d

R1

Fig. 1.2 A schematic illustration of a two-slit interference experiment consisting of two
slits with separation d and an observation screen at distance D. Equally spaced brightand dark fringes are observed when wave-like disturbances from the two slits interfere
constructively and destructively on the screen. Constructive interference occurs at the
point P, at a distance x from the centre of the screen, when the path difference R1 ÿ R2 is
an integer number of wavelengths. This path difference is equal to xd=D if d << D.

4Planck's constant in action

Chap. 1

Pattern formed by 100 quantum particles

Pattern formed by 1000 quantum particles

Pattern formed by 10 000 quantum particles

Fig. 1.3 A computer generated simulation of the build-up of a two-slit interference
pattern. Each dot records the detection of a quantum particle on a screen positioned
behind two slits. Patterns formed by 100, 1000 and 10 000 quantum particles areillustrated.

At first sight the particle-like and wave-like properties of the photon are
strange. But they are not peculiar. We shall soon see that electrons, neutrons,
atoms and molecules also behave in this strange way.

1.2

DE BROGLIE WAVESThe possibility that particles of matter like electrons could be both particle-likeand wave-like was first proposed by Louis de Broglie in 1923. Specifically heproposed that a particle of matter with momentum p could act as a wave withwavelengthhl :pThis wavelength is now called the de Broglie wavelength.

(1:3)

1.2

De Broglie Waves

5

It is often useful to write the de Broglie wavelength in terms of the energy ofthe particle. The general relation between the relativistic energy E and themomentum p of a particle of mass m is
E2 ÿ p2 c2  m2 c4 :

(1:4)

This implies that the de Broglie wavelength of a particle with relativistic energy
E is given byhcl  p :(E ÿ mc2 )(E  mc2 )(1:5)

When the particle is ultra-relativistic we can neglect mass energy mc2 and obtain
l

hc
,
E

(1:6)

an expression which agrees with the relation between energy and wavelength fora photon given in Eq. (1.1). When the particle is non-relativistic, we can set
E  mc2  E,
where E  p2 =2m is the kinetic energy of a non-relativistic particle, and obtainhl  p :
2mE

(1:7)

In practice, the de Broglie wavelength of a particle of matter is small anddifficult to measure. However, we see from Eq. (1.7) that particles of lower
mass have longer wavelengths, which implies that the wave properties of thelightest particle of matter, the electron, should be the easiest to detect. The
wavelength of a non-relativistic electron is obtained by substitutingm  me  9:109  10ÿ31 kg into Eq. (1.7). If we express the kinetic energy E
in electron volts, we obtainr
1:5
nm:
l
E

(1:8)

From this equation we immediately see that an electron with energy of 1.5 eV
has a wavelength of 1 nm and that an electron with energy of 15 keV has awavelength of 0.01 nm.
Because these wavelengths are comparable with the distances between atoms
in crystalline solids, electrons with energies in the eV to keV range are diffracted

6

Planck's constant in action

Chap. 1by crystal lattices. Indeed, the first experiments to demonstrate the wave
properties of electrons were crystal diffraction experiments by C. J. Davissonand L. H. Germer and by G. P. Thomson in 1927. Davisson's experiment
involved electrons with energy around 54 eV and wavelength 0.17 nm whichwere diffracted by the regular array of atoms on the surface of a crystal ofnickel. In Thomson's experiment, electrons with energy around 40 keV andwavelength 0.006 nm were passed through a polycrystalline target and diffracted by randomly orientated microcrystals. These experiments showedbeyond doubt that electrons can behave like waves with a wavelength givenby the de Broglie relation Eq. (1.3).
Since 1927, many experiments have shown that protons, neutrons, atoms andmolecules also have wave-like properties. However, the conceptual implicationsof these properties are best explored by reconsidering the two-slit interference
experiment illustrated in Fig. 1.2. We recall that a photon passing through two
slits gives rise to wave-like disturbances which interfere constructively anddestructively when the photon is detected on a screen positioned behind theslits. Particles of matter behave in a similar way. A particle of matter, like aphoton, gives rise to wave-like disturbances which interfere constructively anddestructively when the particle is detected on a screen. As more and moreparticles pass through the slits, an interference pattern builds up on the observation screen. This remarkable behaviour is illustrated in Fig. 1.3.
Interference patterns formed by a variety of particles passing through two
slits have been observed experimentally. For example, two-slit interference patterns formed by electrons have been observed by A. Tonomura, J. Endo,T. Matsuda, T. Kawasaki and H. Exawa (American Journal of Physics, vol. 57,
p. 117 (1989)). They also demonstrated that a pattern still emerges even when thesource is so weak that only one electron is in transit at any one time, confirmingthat each electron seems to pass through both slits in a wave-like way before
detection at a random point on the observation screen. Two-slit interference
experiments have been carried out using neutrons by R. GaÈhler and A. Zeilinger(American Journal of Physics, vol. 59, p. 316 (1991) ), and using atoms by
O. Carnal and J. Mlynek (Physical Review Letters, vol. 66, p. 2689 (1991) ).
Even molecules as complicated as C60 molecules have been observed to exhibit
similar interference effects as seen by M. Arndt et al. (Nature, vol. 401, p. 680(1999) ).
These experiments demonstrate that particles of matter, like photons, are notclassical particles with well-defined trajectories. Instead, when presented withtwo possible trajectories, one for each slit, they seem to pass along bothtrajectories in a wave-like way, arrive at a random point on the screen andbuild up an interference pattern. In all cases the pattern consists of fringes
with a spacing of lD=d, where d is the slit separation, D is the screen distanceand l is the de Broglie wavelength given by Eq. (1.3).
Physicists have continued to use the ambiguous word particle to describe
these remarkable microscopic objects. We shall live with this ambiguity, but we1.3 Atoms

7

shall occasionally use the term quantum particle to remind the reader that theobject under consideration has particle and wave-like properties. We have used
this term in Fig. 1.3 because this figure provides a compelling illustration ofparticle and wave-like properties. Finally, we emphasize the role of Planck's
constant in linking the particle and wave-like properties of a quantum particle.
If Planck's constant were zero, all de Broglie wavelengths would be zero andparticles of matter would only exhibit classical, particle-like properties.

1.3

ATOMS

It is well known that atoms can exist in states with discrete or quantized energy.
For example, the energy levels for the hydrogen atom, consisting of an electron
and a proton, are shown in Fig. 1.4. Later in this book we shall show thatbound states of an electron and a proton have quantized energies given by

Continuumof unboundenergy levels

E` = 0eV
E 3 = − 13.632eV
E 2 = − 13.6
22eV
E 1 = − 13.6
12

Fig. 1.4 A simplified energy level diagram for the hydrogen atom. To a good approximation the bound states have quantized energies given by En  ÿ13:6=n2 eV where n, theprincipal quantum number, can equal 1, 2, 3, . . . . When the excitation energy is above
13.6 eV, the atom is ionized and its energy can, in principle, take on any value in thecontinuum between E  0 and E  1.

8

Planck's constant in action

Chap. 1

En  ÿ

13:6
eV,
n2

(1:9)

where n is a number, called the principal quantum number, which can take on an
infinite number of the values, n  1, 2, 3, . . . . The ground state of the hydrogen
atom has n  1, a first excited state has n  2 and so on. When the excitationenergy is above 13.6 eV, the electron is no longer bound to the proton; the atom
is ionized and its energy can, in principle, take on any value in the continuum
between E  0 and E  1.
The existence of quantized atomic energy levels is demonstrated by theobservation of electromagnetic spectra with sharp spectral lines that arise
when an atom makes a transition between two quantized energy levels. For
example, a transition between hydrogen-atom states with ni and nf leads to aspectral line with a wavelength l given byhc
 jEni ÿ Enf j:lSome of the spectral lines of atomic hydrogen are illustrated in Fig. 1.5.
Quantized energy levels of atoms may also be revealed by scattering processes. For example, when an electron passes through mercury vapour it has ahigh probability of losing energy when its energy exceeds 4.2 eV, which is thequantized energy difference between the ground and first excited state of amercury atom. Moreover, when this happens the excited mercury atoms subsequently emit photons with energy E  4:2 eV and wavelengthl

200 nmhc
 254 nm:E

400 nm

600 nmFig. 1.5 Spectral lines of atomic hydrogen. The series of lines in the visible part of theelectromagnetic spectrum, called the Balmer series, arises from transitions between
states with principal quantum number n  3, 4, 5, . . . and a state with n  2. The seriesof lines in the ultraviolet, called the Lyman series, arises from transitions between stateswith principal quantum number n  2, 3, . . . and the ground state with n  1.

1.3

Atoms

9

But quantized energy levels are not the most amazing property of atoms.
Atoms are surprisingly resilient: in most situations they are unaffected when
they collide with neighbouring atoms, but if they are excited by such encountersthey quickly return to their original pristine condition. In addition, atoms of thesame chemical element are identical: somehow the atomic number Z, thenumber of electrons in the atom, fixes a specific identity which is common to
all atoms with this number of electrons. Finally, there is a wide variation inchemical properties, but there is a surprisingly small variation in size; forexample, an atom of mercury with 80 electrons is only three times bigger thana hydrogen atom with one electron.
These remarkable properties show that atoms are not mini solar systems inwhich particle-like electrons trace well-defined, classical orbits around a nucleus. Such an atom would be unstable because the orbiting electrons would
radiate electromagnetic energy and fall into the nucleus. Even in the absence ofelectromagnetic radiation, the pattern of orbits in such an atom would change
whenever the atom collided with another atom. Thus, this classical picture
cannot explain why atoms are stable, why atoms of the same chemical element
are always identical or why atoms have a surprisingly small variation insize.
In fact, atoms can only be understood by focussing on the wave-like properties of atomic electrons. To some extent atoms behave like musical instruments.
When a violin string vibrates with definite frequency, it forms a standing wave
pattern of specific shape. When wave-like electrons, with definite energy, areconfined inside an atom, they form a wave pattern of specific shape. An atom isresilient because, when left alone, it assumes the shape of the electron wave
pattern of lowest energy, and when the atom is in this state of lowest energy
there is no tendency for the electrons to radiate energy and fall into the nucleus.
However, atomic electrons can be excited and assume the shapes of wave
patterns of higher quantized energy.
One of the most surprising characteristics of electron waves in an atom is thatthey are entangled so that it is not possible to tell which electron is which. As aresult, the possible electron wave patterns are limited to those that are compatible with a principle called the Pauli exclusion principle. These patterns, for an
atom with an atomic number Z, uniquely determine the chemical properties of
all atoms with this atomic number.
All these ideas will be considered in more detail in subsequent chapters, but
at this stage we can show that the wave nature of atomic electrons provides anatural explanation for the typical size of atoms. Because the de Broglie
wavelength of an electron depends upon the magnitude of Planck's constanth and the electron mass me , the size of an atom consisting of wave-like electrons
also depends upon h and me . We also expect a dependence on the strength of
the force which binds an electron to a nucleus; this is proportional to e2 =4pE0 ,where e is the magnitude of the charge on an electron and on a proton. Thus,
the order of magnitude of the size of atoms is expected to be a function of

10

Planck's constant in action Chap. 1

e2 =4pE0 , me and h (or h  h=2p). In fact, the natural unit of length for atomic
size is the Bohr radius which is given by1

a0 

 2
4pE0 h 0:529  10ÿ10 m:
e2 m e(1:10)

Given this natural length, we can write down a natural unit for atomic
binding energies. This is called the Rydberg energy and it is given byER e2 13:6 eV:
8pE0 a0(1:11)

We note, that the binding energy of a hydrogen-atom state with principal
quantum number n is ER =n2 .
The Bohr radius was introduced by Niels Bohr in 1913 in a paper which
presented a very successful model of the atom. Even though the Bohr model isan out-dated mixture of classical physics and ad-hoc postulates, the central ideaof the model is still relevant. This idea is that Planck's constant has a key role inthe mechanics of atomic electrons. Bohr expressed the idea in the following way:
The result of the discussion of these questions seems to be the general
acknowledgment of the inadequacy of the classical electrodynamics indescribing the behaviour of systems of atomic size. Whatever alteration in
the laws of motion of electrons may be, it seems necessary to introduce inthe laws in question a quantity foreign to the classical electrodynamics;i.e., Planck's constant, or as it is often called, the elementary quantum ofaction. By introduction of this quantity the question of the stable configuration of the electrons in atoms is essentially changed, as this constant isof such dimensions and magnitude that it, together with the mass and thecharge of the particles, can determine a length of the order of the magnitude required.
Ten years after this was written, it was realised that Planck's constant has a rolein atoms because it links the particle-like and wave-like properties of atomic
electrons.

1.4

MEASUREMENT

In classical physics, the act of measurement need not affect the object underobservation because the disturbance associated with the measurement can be1

In this dimensional analysis argument we have not included a dependence on the velocity of light c
because electrons in atoms are, to a good approximation, non-relativistic.

1.4 Measurement

11

made arbitrarily small. Accordingly, the properties of a classical object can be
specified with precision and without reference to the process of measurement.
This is not the case in quantum physics. Here measurement plays an active anddisturbing role. Because of this, quantum particles are best described within thecontext of the possible outcomes of measurements. We shall illustrate the roleof measurement in quantum mechanics by introducing the Heisenberg uncertainty principle and then use this principle to show how measurement provides
a framework for describing particle-like and wave-like quantum particles.

The uncertainty principle
We shall introduce the uncertainty principle for the position and momentum ofa particle by considering a famous thought experiment due to Werner Heisenberg in which the position of a particle is measured using a microscope. The
particle is illuminated and the scattered light is collected by the lens of a
microscope as shown in Fig. 1.6.
Because of the wave-like properties of light, the microscope has a finite
spatial resolving power. This means that the position of the observed particle
has an uncertainty given approximately by
Dx 

l
,
sin a

(1:12)

where l is the wavelength of the illumination and 2a is the angle subtended
by the lens at the particle. We note that the resolution can be improved by
reducing the wavelength of the radiation illuminating the particle; visible
light waves are better than microwaves, and X-rays are better than visible lightwaves.
M icroscope
lens

Scatteredradiation

2αIncident
radiation

Observed
particle

Fig. 1.6 A schematic illustration of the observation of a particle using the Heisenberg
microscope. The particle scatters electromagnetic radiation with wavelength l into a lenswith angular aperture 2a and its position is determined with an uncertainty ofDx  l= sin a.

12Planck's constant in action Chap. 1

However, because of the particle-like properties of light, the process ofobservation involves innumerable photon±particle collisions, with the scatteredphotons entering the lens of the microscope. To enter the lens, a scatteredphoton with wavelength l and momentum h=l must have a sideways momentum betweenhÿ sin alandh sin a:l

Thus the sideways momentum of the scattered photon is uncertain to the degreehDp  sin a:l(1:13)The sideways momentum of the observed particle has a similar uncertainty,
because momentum is conserved when the photon scatters.
We note that we can reduce the uncertainty in the momentum of the observed
particle by increasing the wavelength of the radiation illuminating the particle,
but this would result in a poorer spatial resolution of the microscope and anincrease in the uncertainty in the position of the particle. Indeed, by combining
Eq. (1.12) and Eq. (1.13), we find that the uncertainties in the position and in the
momentum of the observed particle are approximately related by
Dx Dp  h:

(1:14)

This result is called the Heisenberg uncertainty principle. It asserts that greater
accuracy in position is possible only at the expense of greater uncertainty inmomentum, and vice versa. The precise statement of the principle is that thefundamental uncertainties in the simultaneous knowledge of the position andmomentum of a particle obey the inequalityh
Dx Dp  ,2

where hh:2p

(1:15)We shall derive this inequality in Section 7.4 of Chapter 7.
The Heisenberg uncertainty principle suggests that a precise determination of
position, one with Dx  0, is possible at the expense of total uncertainty inmomentum. In fact, an analysis of the microscope experiment, which takes into
account the Compton effect, shows that a completely precise determination of
position is impossible. According to the Compton effect, Eq. (1.2), the wavelength of a scattered photon is increased byDl h(1 ÿ cos y),
mc

1.4

Measurement

13

where m is the mass of the observed particle and y is an angle of scatter which
will take the photon into the microscope lens. This implies that, even if we
illuminate the particle with radiation of zero wavelength to get the best possible
spatial resolution, the radiation entering the microscope lens has a wavelengthof the order of h=mc. It follows that the resolution given by Eq. (1.12) is at best
Dx 

l
h

,
sin a mc sin a(1:16)

which means that the minimum uncertainty in the position of an observed
particle of mass m is of the order of h=mc.
Our analysis of Heisenberg's microscope experiment has illustrated the roleof Planck's constant in a measurement: The minimum uncertainties in theposition and momentum of an observed particle are related by DxDp  h, andthe minimum uncertainty in position is not zero but of the order of h=mc.
However, readers are warned that Heisenberg's microscope experiment can bemisleading. In particular, readers should resist the temptation to believe that a
particle can really have a definite position and momentum, which, because of
the clumsy nature of the observation, cannot be measured. In fact, there is noevidence for the existence of particles with definite position and momentum.
This concept is an unobservable idealization or a figment of the imagination ofclassical physicists. Indeed, the Heisenberg uncertainty principle can be considered as a danger signal which tells us how far we can go in using the classicalconcepts of position and momentum without getting into trouble with reality.

Measurement and wave±particle duality
In practice, the particle-like properties of a quantum particle are observed when
it is detected, whereas its wave-like properties are inferred from the random
nature of the observed particle-like properties. For example, in a two-slit
experiment, particle-like properties are observed when the position of a quantum particle is measured on the screen, but the wave-like passage of thequantum particle through both the slits is not observed. It is inferred from apattern of arrival at the screen which could only arise from the interference of
two wave-like disturbances from the two slits.
However, the inferred properties of a quantum particle depend on the experiment and on the measurements that can take place in this experiment. We shallillustrate this subjective characteristic of a quantum particle by considering amodification of the two-slit experiment in which the screen can either be held
fixed or be allowed to move as shown in Fig. 1.7.
When the pin in Fig. 1.7 is inserted, detectors on a fixed screen precisely
measure the position of each arriving particle and an interference pattern builds
with fringes separated by a distance of lD=d.

14

Planck's constant in action Chap. 1
D
Vertical
momentumsensorQuantum particles
incident on two slitsd

Pin whichholds orreleases screen

Fig. 1.7 A modified two-slit experiment in which the screen may move vertically andbecome a part of a detection system which identifies the slit through which each particle
passes.

When the pin is withdrawn, the screen becomes a mobile detection system
which is sensitive to the momentum p  h=l of the particles hitting the screen.
It recoils when a particle arrives and, by measuring this recoil accurately, we
can measure the vertical momentum of the particle detected at the screen andhence identify the slit from which the particle came. For example, near thecentre of the screen, a particle from the upper slit has a downward momentumof pd=2D and a particle from the lower slit has an upward momentum of
pd=2D. In general, the difference in vertical momenta of particles from thetwo slits is approximately Dp  pd=D. Thus, if the momentum of the recoilingscreen is measured with an accuracy of
Dp 

pd
,
D

(1:17)

we can identify the slit from which each particle emerges. When this is the case,a wave-like passage through both slits is not possible and an interference
pattern should not build up. This statement can be verified by considering theuncertainties involved in the measurement of the momentum of the screen.
The screen is governed by the Heisenberg uncertainty principle and an
accurate measurement of its momentum is only possible at the expense of anuncertainty in its position. In particular, if the uncertainty in the vertical
momentum of the screen is Dp  pd=D, so that we can just identify the slit
through which each particle passes, then the minimum uncertainty in thevertical position of the screen is
Dx 

h
hD

:
Dp pd

(1:18)

1.4

Measurement

15

This uncertainty in position can be rewritten in terms of the wavelength of theparticle. Using p  h=l, we obtain
Dx 

lD
:
d

(1:19)

We note that this uncertainty in the vertical position of the screen is sufficient towash out the interference fringes which would have a spacing of lD=d. Hence,
when the pin in Fig. 1.7 is withdrawn so that the recoiling screen can signal theslit from which a particle comes, no interference pattern builds up and no wavelike passage through both slits is inferred.
This thought experiment illustrates how the concepts of measurement anduncertainty can be used to provide a logical and consistent description of thewave-particle properties of quantum particles. In particular, it shows that,when it is possible to identify the slit through which a particle passes, there is
no wave-like passage through both slits, but when there is no possibility ofidentifying the slit, the particle covertly passes through both slits in a wave-likeway. In fact, the wave-like behaviour of quantum particles is always covert.
Unlike a classical electromagnetic wave, the wave describing a quantum particle cannot be directly observed.2
Finally, some readers may find it instructively disturbing to consider afurther variation of the two-slit experiment which was pointed out by Wheeler
in 1978. In this variation, we imagine a situation in which the choice of theexperimental arrangement in Fig. 1.7 is delayed until after the particle has
passed the slits. We could, for example, insert the pin and fix the position of
the screen just before each particle arrives at the screen. In this case aninterference pattern builds up, which is characteristic of wave-like particles
which pass through both slits. Alternatively, just before each particle arrives
at the screen, we could withdraw the pin so as to allow the screen to recoil anddetermine the slit from which the particle comes. In this case, no interference
pattern builds up.
Thus, a delayed choice of the experimental arrangement seems to influence
the behaviour of the particle at an earlier time. The choice of arrangement
seems to make history, either a history in which the particle passes throughboth slits or a history in which it passes through one or other of the slits. Before
you dismiss this as unacceptable behaviour, note that the history created in thisexperiment is not classical history. The particles concerned are not classical2

A real experiment of this kind is described by Greenberger (Reviews of Modern Physics vol. 55,
1983). In this experiment polarized neutrons, that is neutrons with their spin orientated in a
specific direction, pass through two slits. The polarization of the neutrons which pass through oneof the slits is reversed, and then the intensity of neutrons with a specific polarization is measured
at the screen. If the specific polarization at the screen is chosen so that one cannot infer throughwhich slit each neutron passes, an interference pattern builds up. If it is chosen so that one caninfer through which slit each neutron passes, no interference pattern builds up.

16

Planck's constant in action Chap. 1

particles which pass through one slit or the other, nor are they classical waves
which pass through both slits. They are quantum particles which have thecapability to behave in both of these ways, but only one of these ways may
be inferred in a particular experimental arrangement. The history created in adelayed choice experiment is an inferred history of a quantum particle.

Measurement and non-locality
The most important implication of this discussion of measurement is thatquantum mechanics only describes what we can know about the world. For
example, because we cannot know the position and momentum of an electronwith precision, we cannot describe a world in which an electron has both aprecise position and momentum. In the standard interpretation of quantum
mechanics, a precise position or a precise momentum of an electron can bebrought into existence by a measurement, but no attempt is made to explain
how this occurs.
That properties are brought into existence by measurement is not restricted
to the measurement of the position or momentum of a single particle. It applies
to other observable properties of a quantum particle and also to systems ofquantum particles. Amazingly systems of quantum particles exist in which ameasurement at one location can bring into existence a property at a remotelocation. In other words a measurement here can affect things over there. Thus,
measurements can have a non-local impact on our knowledge of the world.
The non-local nature of quantum mechanical measurement is best illustrated
by considering a particular situation in which two photons are emitted byexcited states of an atom. These photons may move off in opposite directions
with the same polarization in the following meaning of the word same: If thephoton moving to the East, say, is observed to have right-hand circular polarization, then the photon moving to the West is certain to be found to haveright-hand circular polarization. But if the photon moving to the East isobserved to have left-hand circular polarization, then the photon moving to
the West is certain to be found to have left-hand circular polarization.
This behaviour would be unremarkable if right and left-hand polarization
were two alternatives which were created at the moment the two photons were
emitted. But this is not the case. At the moment of emission an entangled state is
created in which the photons are simultaneously right and left-handed, but only
one of these two alternatives is brought into existence by a subsequent measurement. Amazingly, when this measurement is performed on one of thephotons, say the photon moving East, there are two outcomes: The observed
photon moving to the East has a specific polarization and the unobserved
photon moving West immediately has the same polarization.
This system of entangled photons is equivalent to having two ambidextrous
gloves separated by a large distance but linked in such a way that if one glove

Problems 1

17

becomes a right-hand glove, the other automatically becomes a right-handglove. The quantum mechanical reasons for this unexpected togetherness of
distant objects are that:(1) the initial state of the gloves is a superposition of right and left-handedness,in much the same way as the state of a quantum particle can be a linear
superposition of two waves passing through two slits; and(2) a measurement not only disturbs what is measured but also brings into
existence what is measured.
Needless to say, it is not possible to fully justify these arguments at the
beginning of a book whose aim is to introduce the theory of quantum particles.
But these arguments form a part of a logically consistent theory and they aresupported by experimental observations, particularly by Alain Aspect and hiscolleagues; see, for example, The Quantum Challenge by G. Greenstein and
A. G. Zajonc, (Jones and Bartlett, 1997).
The topics considered in this chapter provide the guidelines for a theory ofquantum particles. Most importantly, the theory must provide a way of dealing
with the particle and wave-like properties of quantum particles and in so doing it
must involve the constant which links these properties, Planck's constant. In
addition, the theory must recognize that measurement is not a passive act which
has no effect on the observed system, but a way of creating a particular propertyof the system. The basic elements of such a theory will be developed in the next
three chapters and then further developed by application in subsequent chapters.
This development necessarily entails abstract mathematical concepts, but theresults are not abstract because they describe what we can know about the world.
However, we shall limit our dissussion to a world of non-relativistic particles
Relativistic particles, like photons, will not be considered because this presents
the additional challenge of dealing with the creation and destruction of particles.

PROBLEMS 1
1. Compton scattering can be described in terms of a collision between aphoton and an electron, as shown in Fig. 1.1. The energy E and momentum
P of a relativistic electron and the energy E and momentum p of a photon are
related by
E 2 ÿ P2 c2  m2e c4

andE  pc:

Let Ei , Pi and Ef , Pf denote the initial and final energies and momenta of theelectron, and let Ei , pi and Ef , pf denote the initial and final energies andmomenta of the photon. Assume that the electron is initially at rest, so that18

Planck's constant in action Chap. 1

Ei  me c2 and Pi  0, and assume that the photon is scattered through anangle y.
By considering the conservation of momentum show thatE2i ÿ 2Ei Ef cos y  E2f  Ef2 ÿ m2e c4 :Write down the equation describing the conservation of relativistic energy inthe collision and show that it may be rearranged to giveE2i ÿ 2Ei Ef  E2f  Ef2 ÿ 2Ef m2e c4  m2e c4 :By subtracting these equations show thatm e c2(Ei ÿ Ef )
 (1 ÿ cos y):
Ei Ef

Show that this equation implies that the increase in wavelength of thescattered photon is given byDl h(1 ÿ cos y):me c

2. In the photoelectric effect, electromagnetic radiation incident on a metal
surface may eject electrons, but only if the frequency of the radiation exceeds
a threshold value.
Show that this frequency threshold can be understood if the mechanism
for the photoelectric effect involves an interaction between a photon and anelectron, and if the energy needed to eject the electron has a minimum value.
The minimum energy needed to eject an electron from the surface ofmagnesium is 3.68 eV. Show that light with a frequency below 8:89  1014 Hz
cannot produce photoelectrons from magnesium, no matter how intense theillumination may be.
3. Dimensional analysis can provide insight into Stefan±Boltzmann's law for
the radiation from a black body. According to this law the intensity ofradiation, in units of J sÿ1 mÿ2 , from a body at temperature T is
I  sT 4 ,where s is Stefan±Boltzmann's constant. Because black-body radiation
can be considered to be a gas of photons, i.e. quantum particles whichmove with velocity c with typical energies of the order of kT, the intensity
I is a function of h, c and kT. Use dimensional analysis to confirm that I is
proportional to T 4 and find the dependence of s on h and c.

Problems 1

19

4. In Section 1.3 we used dimensional analysis to show that the size of ahydrogen atom can be understood by assuming that the electron in theatom is wave-like and non-relativistic. In this problem we show that, if we
assume the electron in the atom is a classical electron described by the theoryof relativity, dimensional analysis gives an atomic size which is four ordersof magnitude too small.
Consider a relativistic, classical theory of an electron moving in the
Coulomb potential of a proton. Such a theory only involves three physicalconstants: me , e2 =4pE0 , and c, the maximum velocity in relativity. Show that
it is possible to construct a length from these three physical constants, butshow that it too small to characterize the size of the atom.
5. An electron in a circular orbit about a proton can be described by classicalmechanics if its angular momentum L is very much greater than h. Show
that this condition is satisfied if the radius of the orbit r is very much greaterthan the Bohr radius a0 , i.e. ifr >> a0 

4pE0 
h2:
e2 me

6. Assume that an electron is located somewhere within a region of atomic size.
Estimate the minimum uncertainty in its momentum. By assuming that thisuncertainty is comparable with its average momentum, estimate the averagekinetic energy of the electron.
7. Assume that a charmed quark of mass 1:5 GeV=c2 is confined to a volume
with linear dimension of the order of 1 fm. Assume that the average momentum of the quark is comparable with the minimum uncertainty in its momentum. Show that the confined quark may be treated as a non-relativistic
particle, and estimate its average kinetic energy.
8. JJ and GP Thomson, father and son, both performed experiments withbeams of electrons. In 1897, JJ deduced electrons are particles with a definite
value for e=me . In 1927, GP deduced that electrons behave like waves. In JJ's
experiment, electrons with kinetic energies of 200 eV passed through a pairof plates with 2 cm separation. Explain why JJ saw no evidence for wave-likebehaviour of electrons.
9. The wave properties of electrons were first demonstrated in 1925 by Davisson and Germer at Bell Telephone Laboratories. The basic features of theirexperiment are shown schematically below.

20

Planck's constant in action Chap. 1
Incident electron wave

D iffracted
electron wavefSurface atoms with separation D

Electrons with energy 54 eV were scattered by atoms on the surface of acrystal of nickel.3 The spacing between parallel rows of atoms on thesurface was D  0:215 nm. Explain why Davisson and Germer detected
strong scattering at an angle f equal to 50 degrees.
10. The electrons which conduct electricity in copper have a kinetic energy ofabout 7 eV. Calculate their wavelength. By comparing this wavelength with
the interatomic distance in copper, assess whether the wave-like propertiesof conduction electrons are important as they move in copper.
(The density of copper is 8:9  103 kg mÿ3 and the mass of a copper
atom is 60 amu.)
11. Neutrons from a nuclear reactor are brought into thermal equilibrium by
repeated collisions in heavy water at T  300 K. What is the averageenergy (in eV) and the typical wavelength of the neutrons? Explain why
they are diffracted when they pass through a crystalline solid.
12. Estimate the wavelength of an oxygen molecule in air at NTP. Compare
this wavelength with the average separation between molecules in air andexplain why the motion of oxygen molecules in air at NTP is not affectedby the wave-like properties of the molecules.

3

Incidentally, the crystalline structure was caused by accident when they heated the target inhydrogen in an attempt to repair the damage caused by oxidation.

2
The SchroÈdinger equation
The first step in the development of a logically consistent theory of nonrelativistic quantum mechanics is to devise a wave equation which can describe
the covert, wave-like behaviour of a quantum particle. This equation is called
the SchroÈdinger equation.
The role of the SchroÈdinger equation in quantum mechanics is analogous tothat of Newton's Laws in classical mechanics. Both describe motion. Newton's
Second Law is a differential equation which describes how a classical particle
moves, whereas the SchroÈdinger equation is a partial differential equation
which describes how the wave function representing a quantum particle ebbs
and flows. In addition, both were postulated and then tested by experiment.

2.1

WAVES

As a prelude to the SchroÈdinger equation, we shall review how mathematics canbe used to describe waves of various shapes and sizes.

Sinusoidal waves
The most elegant wave is a sinusoidal travelling wave with definite wavelengthl and period t, or equivalently definite wave number, k  2p=l, and angularfrequency, !  2p=t. Such a wave may be represented by the mathematical
functionC(x, t)  A cos (kx ÿ !t)(2:1)

where A is a constant. At each point x, the function C(x, t) oscillates withamplitude A and period 2p=!. At each time t, the function C(x, t) undulateswith amplitude A and wavelength 2p=k. Moreover, these undulations move,

22

The SchroÈdinger equation

Chap. 2

like a Mexican wave, in the direction of increasing x with velocity !=k; forexample, the maximum of C(x, t) corresponding to kx ÿ !t  0 occurs at theposition x  !t=k, and the minimum corresponding to kx ÿ !t  p occurs at
the position x  l=2  !t=k; in both cases the position moves with velocity!=k.
The function sin (kx ÿ !t), like cos (kx ÿ !t), also represents a sinusoidal
travelling wave with wave number k and angular frequency !. Becausesin (kx ÿ !t)  cos (kx ÿ !t ÿ p=2),
the undulations and oscillations of sin (kx ÿ !t) are out of step with those ofcos (kx ÿ !t); the waves sin (kx ÿ !t) and cos (kx ÿ !t) are said to have a phasedifference of p=2. The most general sinusoidal travelling wave with wave
number k and angular frequency ! is the linear superposition
C(x, t)  A cos (kx ÿ !t)  B sin (kx ÿ !t),(2:2)

where A and B are arbitrary constants.
Very often in classical physics, and invariably in quantum physics, sinusoidal
travelling waves are represented by complex exponential functions of the form
C(x, t)  A ei(kxÿ!t) :(2:3)

The representation of waves by complex exponentials in classical physics ismerely a mathematical convenience. For example, the pressure in a sound wave
may be described by the real function A cos (kx ÿ !t), but this real function
may be taken to be the real part of a complex exponential function A ei(kxÿ!t)becauseei(kxÿ!t)  cos (kx ÿ !t)  i sin (kx ÿ !t):Thus, in classical physics, we have the option of representing a real sinusoidal
wave by the real part of a complex exponential. In quantum physics, however,the use of complex numbers is not an option and we shall see that a complex
exponential provides a natural description of a de Broglie wave.

Linear superpositions of sinusoidal waves
Two sinusoidal waves moving in opposite directions may be combined to form
standing waves. For example, the linear superposition
A cos (kx ÿ !t)  A cos (kx  !t)

2.1

Waves

23

gives rise to the wave 2A cos kx cos !t. This wave oscillates with period 2p=!
and undulates with wavelength 2p=k, but these oscillations and undulations do
not propagate; it is a non-Mexican wave which merely stands and waves.
Alternatively, many sinusoidal waves may be combined to form a wave
packet. For example, the mathematical form of a wave packet formed by alinear superposition of sinusoidal waves with constant amplitude A and wave
numbers in the range k ÿ Dk to k  Dk isC(x, t) Z kDkkÿDk

A cos (k0 x ÿ !0 t) dk0 :(2:4)

If k is positive, this wave packet travels in the positive x direction, and in thenegative x direction if k is negative.
The initial shape of the wave packet, i.e. the shape at t  0, may be obtainedby evaluating the integral
C(x, 0) Z kDkkÿDk

A cos k0 x dk0 :This gives
C(x, 0)  S(x) cos kx,

where

S(x)  2ADksin (Dkx)
:
(Dkx)

(2:5)

If Dk << k, we have a rapidly varying sinusoidal, cos kx, with an amplitude
modulated by a slowly varying function S(x) which has a maximum at x  0
and zeros when x is an integer multiple of p=Dk. The net result is a wave packet
with an effective length of about 2p=Dk. Three such wave packets, with different values for Dk, are illustrated in Fig. 2.1. We note that the wave packets
increase in length as the range of wave numbers decreases and that they would
become `monochromatic' waves of infinite extent as Dk ! 0. Similar behaviour
is exhibited by other types of wave packets.
The velocity of propagation of a wave packet, and the possible change ofshape as it propagates, depend crucially on the relation between the angularfrequency and wave number. This relation, the function !(k), is called thedispersion relation because it determines whether the waves are dispersive or
non-dispersive.

Dispersive and non-dispersive waves
The most familiar example of a non-dispersive wave is an electromagnetic wave
in the vacuum. A non-dispersive wave has a dispersion relation of the form
!  ck, where c is a constant so that the velocity of a sinusoidal wave, !=k  c,

24

The SchroÈdinger equation

Chap. 2

(A)

(B)

(C)

Fig. 2.1 The initial shapes of the wave packets given by a linear superposition ofsinusoidal waves with constant amplitude A and wave numbers in the range k ÿ Dk tok  Dk; see Eq. (2.5). The three diagrams show how the length of a wave packet increases
as the range of wave numbers Dk decreases. The value of ADk is constant, but Dk equals
k=8 in diagram (A), Dk equals k=16 in diagram (B) and Dk equals k=32 in diagram (C).
In general, the length of a wave packet is inversely proportional to Dk and becomes
infinite in extent as Dk ! 0:

is independent of the wave number k. A wave packet formed from a linear
superposition of such sinusoidal waves travels without change of shape because
each sinusoidal component has the same velocity.
Non-dispersive waves are governed by a partial differential equation called theclassical wave equation. For waves travelling in three dimensions, it has the form
1 ]2 Cr2 C ÿ 2 2  0,
c ]t

where

r2 

]2
]2
]2


,
]x2 ]y2 ]z2(2:6)

and for waves travelling in one dimension, the x direction say, it has the form
]2 C 1 ]2 Cÿ
 0:]x2 c2 ]t2

(2:7)

2.1

Waves

25

The classical wave equation has an infinite number of solutions corresponding
to an infinite variety of wave forms. For example, the sinusoidal waves,
A cos (kx ÿ !t),A sin (kx ÿ !t),or A ei(kxÿ!t) ,are solutions provided !2  c2 k2 , as may be shown by direct substitution into
Eq. (2.7); solutions with k  !=c describe waves travelling in the positive x
direction and solutions with k  ÿ!=c describe waves travelling in the negative
x direction. Because each term in the classical wave equation is linear in C, alinear superposition of sinusoidal waves is also a solution. For example, asuperposition like Eq. (2.4) is a solution which describes a wave packet which
propagates without change of shape.
However, the majority of waves encountered in classical and in quantum
physics are dispersive waves. A dispersive wave is governed by a partial differential equation which is more complicated than the classical wave equation,
Eq. (2.7).1 The dispersion relation is more complicated than !  ck so that thevelocity of propagation of a sinusoidal wave, !=k, depends upon the wave
number k. Hence a packet of dispersive waves will, in general, change shape as
it propagates. However, if the packet is composed of waves with a narrow rangeof wave numbers, it has a well-defined velocity of propagation. This velocity iscalled the group velocity and it is given byugroup 

d!
,
dk

(2:8)

whereas the velocity of a simple sinusoidal wave, !=k, is called the phasevelocity.
To understand Eq. (2.8), we note that the group velocity describes the motionof a localized disturbance due to constructive interference of many sinusoidal
waves. Let us focus on the point of constructive interference of two sinusoidal
waves with wave numbers k1 and k2 and angular frequencies !1 and !2 which isformed when the waves are in phase; i.e. whenk1 x ÿ !1 t  k2 x ÿ !2 t:
By rearranging this equation, we find that the position of this point of constructive interference is given by
!1 ÿ ! 2t:x
k1 ÿ k2

1

In fact, Eq. (2.7) with the constant c replaced by a frequency-dependent velocity is often used todescribe dispersive waves.

26

The SchroÈdinger equation

Chap. 2Thus our point of constructive interference is located at x  0 when t  0 and it
moves with a velocity given by (!1 ÿ !2 )=(k1 ÿ k2 ), or by Eq. (2.8) if jk1 ÿ k2 j issmall. Of course, with two sinusoidal waves, there are an infinite number ofpoints of constructive interference, but many sinusoidal waves can form a
localized region of constructive interference which moves with a velocity
given by Eq. (2.8).
To illustrate how a group velocity can be derived from Eq. (2.8), we consider
the example of water waves of long wavelength which obey the dispersion
relation!p
gk,

where g is the acceleration due to gravity. The velocity of a sinusoidal water
wave, the so-called phase velocity, is!
uphase  
k

r
g
,
k

and the velocity of a packet of water waves with a narrow range of wave
numbers near k isd! 1ugroup 

dk 2

r
g
:
k

Thus, for water waves, the group velocity is exactly one-half of the phasevelocity. In other words, the sinusoidal waves forming the packet, travel attwice the speed of the region of maximum disturbance formed by the interference of these waves. However, the shape of the disturbance will change as itpropagates; in general it will tend to spread out.

2.2

PARTICLE WAVE EQUATIONSIn classical physics, fundamental laws of physics are used to derive the wave
equations which describe wave-like phenomena; for example, Maxwell's laws ofelectromagnetism can be used to derive the classical wave equation (2.6) whichgoverns electromagnetic waves in the vacuum. In contrast, we shall view the
wave equation governing the wave-like properties of a quantum particle as
a fundamental equation which cannot be derived from underlying basic physical principles. We, like the inventors of quantum theory, can only guess theform of this wave equation and then test for consistency and agreement withexperiment.

2.2

Particle Wave Equations

27

A wave equation for a free particle
We shall construct a possible wave equation for a freely moving non-relativistic
particle by considering the properties of the de Broglie waves describing theparticle.
According to Eq. (1.3) a particle with momentum p has a de Broglie wavelength given by l  h=p. This implies that a de Broglie wave with wave numberk  2p=l describes a particle with momentump
hk,

where h 

h
:
2p

(2:9)

We shall extend this idea by assuming that a de Broglie wave packet with arange of wave numbers between k ÿ Dk and k  Dk describes a particle with anuncertain momentum
Dp  
hDk:(2:10)We shall also assume that the length of this wave packet is a measure of Dx, theuncertainty in the position of the particle. Using Eq. (2.5) and Fig. 2.1 as a
guide, we write
Dx 

2p
:
Dk

(2:11)

If we multiply these uncertainties, we obtainDx Dp  h,in agreement with the Heisenberg uncertainty principle, Eq. (1.14). Thus, a de
Broglie wave packet can account for the uncertainties in the position andmomentum of a quantum particle.
However, we note that a de Broglie wave must be transformed by a measurement. If a precise measurement of the position is made, the new wave packet
describing the particle must be very short, a superposition of sinusoidal waves
with a very wide range of wavelengths. Similarly, if a precise measurement of
the momentum is made, the new wave packet is very long with a sharply
defined wavelength. This implies that the wave packet is a fragile entity which
is transformed by a measurement. No one knows how this happens.
We shall now impose the condition that the wave packet represents a moving
quantum particle. Specifically, we shall require that the group velocity of thepacket is equal to the velocity of a particle with mass m and momentum p  hk;i.e. we shall require that

28

The SchroÈdinger equation

Chap. 2

d! 
hk
 :
dk
m

(2:12)

This equation may be integrated to give the following dispersion relation forthe de Broglie waves describing a freely moving quantum particle of mass m:
!

hk2
:
2m

(2:13)

In obtaining this relation we have set the constant of integration to zero
because this constant gives rise to no observable consequences in non-relativistic quantum mechanics.
Our task is to find a wave equation which has sinusoidal solutions whichobey this dispersion relation. The simplest such wave equation is called theSchroÈdinger equation. For a free particle moving in one dimension, it has the
form
i
h

]C
h2 ]2 Cÿ:2m ]x2]t(2:14)

It is easy to verify that the complex exponential
C(x, t)  A ei(kxÿ!t)(2:15)

is a solution of this equation provided ! and k obey the dispersion relation
Eq. (2.13). If we substitute into Eq. (2.14), the left-hand side yields
ih]Ch!A ei(kxÿ!t) , ih( ÿ i!)A ei(kxÿ!t)  ]t

and the right-hand side yields
ÿ2 ] 2 C hh2 k2
A ei(kxÿ!t) ,2m ]x22m

and we have a solution provided 
h!  
h2 k2 =2m.
Because the sinusoidal solution, Eq. (2.15), describes a wave moving in the x
direction with wave number k and angular velocity !, we shall assume that it
represents a free particle moving in the x direction with a sharply defined
momentum p  hk and energy E  p2 =2m  
h!. There are, of course, manyother solutions of the SchroÈdinger equation which represent other states of
motion of the particle.
We emphasize that in order to accommodate the dispersion relation for de
Broglie waves, Eq. (2.13), we have arrived at a wave equation, the free-particle

2.2

Particle Wave Equations

29

SchroÈdinger equation Eq. (2.14), whose solutions are necessarily complex
functions of space and time. These complex functions are called wave functions.
We recall that classical waves are often represented by complex functions, but
this representation is purely a matter of mathematical convenience; classicalwaves are real functions of space and time. In contrast, SchroÈdinger wave
functions are not real functions of space and time. They are complex functions
which describe the covert wave-like behaviour of a quantum particle.
So far we have only considered sinusoidal solutions of the SchroÈdinger
equation, but given these solutions we can construct other types of solutions.
Because each term in the SchroÈdinger equation is linear in the wave function C,
a superposition of solutions is also a solution. For example,
C(x, t)  A1 ei(k1 xÿ!1 t)  A2 ei(k2 xÿ!2 t) ,with
h! 1 
h2 k212m

andh! 2 


2 k22h,2mwhere A1 and A2 are arbitrary complex constants, is a solution of Eq. (2.14);
this may easily be confirmed by direct substitution. Indeed, the most generalsolution is a superposition of sinusoidal waves with all possible angular frequencies and wave numbers; i.e.
C(x, t) 

Z 1
ÿ1

0

0

A(k0 ) ei(k xÿ! t) dk0

with h!0 h2 k02:2m(2:16)In this superposition, A(k0 ) is an arbitrary complex function of k0 and theintegral represents a sum over all possible values of k0 . If the function A(k0 ) issuch that the sum involves a narrow range of wave numbers around a positive
value k, this superposition yields a wave packet moving in the positive x
direction with a group velocity hk=m. Such a wave packet represents a quantum
particle which moves with velocity 
hk=m, but with position and momentum inaccordance with the Heisenberg uncertainty principle.

Wave equation for a particle in a potential energy field
The interactions of a non-relativistic particle can usually be described in termsof a potential energy field. For example, an electron in a hydrogen atom can bethought of as moving in the potential energy field V (r)  ÿe2 =4pE0 r of a
nucleus. In classical mechanics, this field implies that an electron at a distancer from the nucleus experiences an attractive force of magnitude e2 =4pE0 r2 . In

30

The SchroÈdinger equation

Chap. 2quantum mechanics, it implies that the wave equation for the electron is not thesimple free-particle wave equation given by Eq. (2.14).
In 1926, Erwin SchroÈdinger invented a wave equation for a quantum particle
in a potential energy field which led to a successful description of atoms andother microscopic systems. It is a generalization of the wave equation for a freeparticle given by Eq. (2.14). The SchroÈdinger equation for a particle moving in
the three-dimensional potential energy field V(r) is
"
#
]C
h2 2

r  V (r) C:
i
h
 ÿ
2m
]t

(2:17)

When the particle moves in a one-dimensional potential V(x) the SchroÈdinger
equation simplifies to"#]Ch2 ] 2 V (x) C:ih ÿ2m ]x2]t(2:18)

It is easy to find solutions of the SchroÈdinger equation when the potential
energy is a constant. For example, when a particle moves along the x axis withconstant potential energy V0 , the wave function
C(x, t)  A ei(kxÿ!t)is a solution of Eq. (2.18) providedh! 2 k 2h V0 :
2m

This wave function represents a particle with sharply defined total energy E andmomentum p given by
E

p2
 V0
2m

and

p
hk:

In later chapters we shall find solutions of the SchroÈdinger equation for avariety of potential energy fields, including the most important case of the
Coulomb potential energy of an electron in an atom. But our next task is to
explore the meaning of the SchroÈdinger equation.

Problems 2

31

PROBLEMS 2
1. Waves on the surface of water are dispersive waves. If the wavelength isshort so that surface tension provides the restoring force, the dispersion
relation iss
Tk3
,
!
r
where T is the surface tension and r is the density of water.
Find expressions for the phase velocity of a sinusoidal water wave withwave number k and the group velocity of a packet of water waves with wave
numbers near k.
2. Consider the wave packet represented by
C(x, t) Z kDkkÿDk

A cos (k0 x ÿ !0 t) dk0

where A is a constant and assume that the dispersion relation is!0  ck0 ,where c is a constant. By integrating over k0 show that
C(x, t)  S(x ÿ ct) cos k(x ÿ ct)
whereS(x ÿ ct)  2ADksin [Dk(x ÿ ct)]:
[Dk(x ÿ ct)]

Describe the propagation properties of this wave packet.
3. Verify by direct substitution that the real functions
C  A cos (kx ÿ !t) andC  A sin (kx ÿ !t)are not solutions of the SchroÈdinger equation for a free particle.
4. Verify that the wave function
C(x, t)  A ei(kxÿ!t) ÿ A eÿi(kx!t) ,32

The SchroÈdinger equation

Chap. 2where A is an arbitrary complex constant, is a solution of the SchroÈdinger
equation for a free particle of mass m, Eq. (2.14), if
h! h2 k2:2m

Show that this wave function can be rewritten as
C(x, t)  2iA sin kx eÿi!t :What sort of wave is this?
5. In quantum mechanics it is the convention to represent a free particle with
momentum p and energy E by the wave function
C(x, t)  A ei(pxÿEt)=h :Physicists on another planet may have chosen the convention of representing a free particle with momentum p and energy E by the wave function
C(x, t)  A eÿi(pxÿEt)=h :What is the form of the SchroÈdinger equation on this planet?
6. This question tentatively considers the wave equation for a relativistic
particle. According to the Theory of Relativity, the momentum p and energy
E of a particle of mass m are related byE2 ÿ p2 c2  m2 c4 ,and the velocity of the particle is given by
u

pc2
:
E

(a) Assume that the motion of the particle can be described by a wave
packet in which the angular frequency and wave number are given
by
E
h!

and

p
hk:

Derive an expression for the group velocity of the wave packet and showthat it is identical to the particle velocity.

Problems 233(b) Show that the wave equation
2]2 C
m 2 c4
2] C
ÿ
c

C0
]t2
]x2
h2


has solutions of the form
C  A eÿi(!tÿkx) ,which could possibly describe a relativistic particle of mass m withenergy E  h! and momentum p  
hk.
(At first sight this wave equation, which is called the Klein±
Gordon equation, provides a basis for a theory for relativistic particles which is similar to that provided by the SchroÈdinger equation
for non-relativistic particles. However, a sensible interpretation of
the Klein±Gordon equation cannot be obtained if its solutions aretreated as wave functions. A sensible interpretation only emerges if
the solutions are treated as quantum field operators.)

This page intentionally left blank

3
Position and momentum
The SchroÈdinger equation has an infinite number of solutions which correspond to an infinite number of possible states of motion. These wave functions,being extended and non-localized, can easily describe the wave-like propertiesof a particle, but it is difficult to see how they can describe particle-like
properties. In a two-slit experiment, for example, a wave function can describe
how an electron can pass through two slits, but how can it describe an electronthat arrives as a lump on the screen? This problem may be resolved by adopting
the radical idea that measurements can lead to random outcomes governed by
the laws of probability.
In this chapter we shall focus on how the uncertain outcomes of position andmomentum measurements are described in quantum mechanics. We shall showthat these outcomes are governed by the wave function of a particle. In sodoing, we shall illustrate how a wave function provides a description of theproperties that could emerge from measurements, in other words, a descriptionof potentialities that may become realities.

3.1

PROBABILITY

Because of the importance of probability in quantum measurement, we shallbriefly consider how discrete and continuous random variables are governed byprobability distributions. These general considerations are illustrated by problems at the end of the chapter on the Poisson, Gaussian and exponential
probability distributions.

Discrete random variables
Let us consider a process or experiment with possible outcomes described by adiscrete random variable which can take on the values x0 , x1 , x2 , . . . , with

36

Position and momentum

Chap. 3probabilities p0 , p1 , p2 , . . . . The set of probabilities pn is called a probability
distribution. Because the total probability of all the possible outcomes is equal
to one, the probability distribution pn must satisfy the normalization conditionXpn  1:

(3:1)

all n

For example, if pn is the probability that a reader of this book has n grandparents that are alive and well, thenp0  p1  p2  p3  p4  1:
The probability distribution pn can be used to evaluate the expectation value
for the random variable xn . This is the average value of the many possible
outcomes that may occur when the process or experiment takes place an infinite
number of times. It is given by
hxi 

X

x n pn :

(3:2)

all n

The likely spread in the outcomes about this expectation value is given by
the standard deviation or uncertainty in x. We shall denote this by Dx. The
square of the standard deviation is called the variance and the variance is givenby(Dx)2 X

(xn ÿ hxi)2 pn :

(3:3)

all n

In this expression (xn ÿ hxi) is the deviation of xn from the expected value; thisdeviation may be positive or negative and its average value is zero. However,the variance is the average of the square of this deviation; it is zero when there is
only one possible outcome and it is a positive number when there is more than
one possible outcome.
In many cases it is useful to rewrite Eq. (3.3) in the following way. Using(xn ÿ hxi)2  x2n ÿ 2hxixn  hxi2 ,and bearing in mind that hxi is a number that does not depend on n, we find(Dx)2 X
all n

x2n pn ÿ 2hxi

X
all n

xn pn  hxi2

X

pn :

all n

The first term is equal to hx2 i, the expectation value of the square of xn . The
second term is equal to ÿ2hxihxi  ÿ2hxi2 . The third term is equal to hxi2

3.1

Probability

37

because the probability distribution is normalized. It follows that the variancegiven by Eq. (3.3) may be rewritten as(Dx)2  hx2 i ÿ hxi2 :(3:4)

This equation states that the variance is the difference between the average ofthe square and the square of the average of the random variable.

Continuous random variables
We shall now consider a process or experiment in which the outcomes aredescribed by a continuous variable x. The probability of an outcome between x
and x  dx can be denoted by r(x) dx. The function r(x) is called a probability
density. It satisfies the normalization conditionZ
all xr(x) dx  1,

(3:5)

because the probability of an outcome x anywhere in its range of possible
values must be equal to one. For example, if x is the position of a particle
confined to the region 0  x  a, thenZ a0r(x) dx  1:

The expectation value of x, in analogy with Eq. (3.2), is given by the integral
Z
hxi 

all x

xr(x) dx:

(3:6)

Similarly, the expectation value of x2 is given by2

hx i 

Z
all x

x2 r(x) dx(3:7)

and the standard deviation, or uncertainty, in x is given byDx q
hx2 i ÿ hxi2 :

(3:8)

38

3.2

Position and momentum

Chap. 3POSITION PROBABILITIESWe shall now return to quantum mechanics and consider how the position of aquantum particle is described. This problem was first successfully addressed by
Max Born by introducing an interpretation of the SchroÈdinger wave function
which led to a revolution in the philosophical basis of physics. We shallintroduce Born's interpretation by reconsidering the two-slit interference experiment discussed in Chapter 1. We shall describe how an interference patternarises when a classical wave passes through two slits and then develop a way ofdescribing how a similar pattern is produced by a current of quantum particles.

Two-slit interference
The key elements of a two-slit interference experiment are very simple: a sourceof something with wave-like properties, two slits and an observation screen, as
illustrated in Fig. 3.1.
When a classical wave passes through the slits, two waves emerge which
combine and interfere to form a pattern on the screen. These waves can be
described by real functions of space and time. For example, if the wave number
is k and the angular frequency is !, the combined wave at a point P, at distanceR1 from slit S1 and R2 from slit S2 , may be represented by
C  A1 cos (kR1 ÿ !t)  A2 cos (kR2 ÿ !t),Wave-like entity
incident on two slits

R2

(3:9)

P

S2
R1
S1

Fig. 3.1 A two-slit interference experiment in which a wave-like entity passes throughtwo slits, S1 and S2 , and is detected on a screen. Equally spaced bright and dark fringes
are observed when wave-like disturbances from the two slits interfere constructively anddestructively on the screen. Constructive interference occurs at the point P when thepath difference R1 ÿ R2 is an integer number of wavelengths.

3.2

Position Probabilities

39

where A1 and A2 are amplitudes which are inversely proportional to R1 and R2 ,respectively. The energy density and intensity of the wave at point P areproportional to the square of the wave. If we set A1  A2  A, which is agood approximation when the distances R1 and R2 are large compared with theslit separation, and if we use some simple trigonometry, we find1
C2  2A2 cos2k(R1 ÿ R2 )cos2 !t:2(3:10)

It follows that the intensity has maxima when k(R1 ÿ R2 )=2 is an integer
multiple of p and minima when it is a half-integer multiple of p. If we rewrite
k in terms of the wavelength l, we find that the maxima arise when the path
difference R1 ÿ R2 is equal to an integer number of wavelengths and theminima when R1 ÿ R2 is equal to a half-integer number of wavelengths.
When the effect of the finite width of the slits is taken into account, thesemaxima and minima give rise to an interference pattern like that illustrated inFig. 3.2.
We shall now seek a similar description of how an interference pattern arises
when a current of quantum particles passes through two slits. We shall assume
that the covert passage of a quantum particle through the two slits is

Fig. 3.2 An interference pattern produced by a classical wave and by a current of
quantum particles passing through two slits with a finite width. Constructive interference occurs at points when the difference in paths from the slits equals an integer numberof wavelengths; the central maximum occurs when the path difference is zero. Note thepattern given by a classical wave oscillates with time with a definite frequency whereas
the pattern given by a current of quantum particles builds up gradually as more particles
arrive at the screen.

1

This trigonometry can be avoided by representing classical waves as the real part of a complex
exponential, but we have deliberately not done this because we wish to emphasise that classicalwaves, unlike quantum wave functions, are real functions of space and time.

40

Position and momentum

Chap. 3represented by a wave function, i.e. by a complex function of space and time thatis a solution of the SchroÈdinger equation. If the particle has definite momentump  hk and energy E  h!, the wave function at a point P is a linear superpositionof two terms, a wave from slit S1 and a wave from slit S2 , of the form
C  A1 ei(kR1 ÿ!t)  A2 ei(kR2 ÿ!t) ,(3:11)

where A1 and A2 are complex constants with approximately the same value.
When the particle arrives at the screen a very complicated process occurs. At
each point on the screen there is a measuring device which magnifies microscopic effects due to the particle so that there is a clear signal that a particle hasor has not arrived at that point. In other words, a small-scale event triggers
something that can actually be seen. In practice, the particle may be detected at
any point on the screen and, as more and more particles pass through the slits,
an interference pattern builds up on the screen, as illustrated in Fig. 1.3.
We shall not attempt to understand how this complicated process of detection takes place. Instead, we shall set a more modest objective of describing thepossible outcomes of the process using the SchroÈdinger wave function, C. Thiscan be achieved by making the bold assumption that the probability ofdetecting a particle at a particular location is proportional to the value of theeffective intensity of the complex wave function at that location. In analogy
with classical waves, we shall define this intensity to be a real number given by
jCj2  C*C:

(3:12)The value of jCj2 at a point P on the screen can be found using the wave
function Eq. (3.11). Making the approximation A1  A2  A, we havejCj2  (A* eÿi(kR1 ÿ!t)  A* eÿi(kR2 ÿ!t) )(A ei(kR1 ÿ!t)  A ei(kR2 ÿ!t) ),which gives
jCj2  jAj2  jAj2  jAj2 eik(R1 ÿR2 )  jAj2 eÿik(R1 ÿR2 ) :Using the mathematical relations
cos y  (eiy  eÿiy )=2

andcos y  2 cos2

 
y
ÿ 1,
2

we findk(R1 ÿ R2 ):jCj  2A cos
2
2

2

2

(3:13)

3.2

Position Probabilities

41

If we compare with Eq. (3.10), we see that jCj2 , has maximum and minimum
values on the screen similar to those given by the intensity of a classical wave;
there are maxima when R1 ÿ R2 is an integer number of de Broglie wavelengths
and minima when R1 ÿ R2 is a half-integer number of de Broglie wavelengths.
Hence, if the probability of detection is proportional to jCj2 , an interference
pattern similar to that shown in Fig. 3.2 will build up when many quantum
particles pass through the two slits.
Thus, we have a logical way of describing the interference pattern produced byquantum particles. It is based on two crucial ideas. First, the wave function of theparticle at the screen, C, is a linear superposition of wave functions: a wave fromslit S1 and a wave from slit S2 . Second, the probability of detecting the particle ata particular location is proportional to the value of jCj2 at that location.
But we also need a successful way of describing the outcome of the modified
two-slit experiment shown in Fig. 1.7. In particular, we need to explain why
there is no interference pattern when we can identify the slit through which eachparticle passes. To do so, we shall assume that the process of identification
changes the wave function of the particle, so that it becomes a single wave fromthe slit through which the particle passes. In so doing, we follow the standardpractice of assuming that a measurement can affect a wave function. It is also
standard practice not to explore too deeply how this actually occurs.

The Born interpretation of the wave function
The idea that the wave function can govern the potential outcomes of a position
measurement was first proposed by Max Born in 1926. It is now called the Borninterpretation of the wave function. According to this interpretation, the wave
function C(r, t) is a complex function of the space coordinates whose modulus
squared, jC(r, t)j2 , is a measure of the probability of finding the particle at the
point r at time t. The particle can be found anywhere, but it is more likely to befound at a point where jC(r, t)j2 is large. Specifically(23jC(r, t)j d r 

The probability of finding the particle
at time t in the volume element d3 r:)(3:14)

Thus, jC(r, t)j2 can be thought of as probability density for position. Because ofthis, the wave function C(r, t) is often referred to as a probability amplitude forposition.22Note that the Born interpretation of the wave function implies that the position of a particle canbe determined precisely at the expense of total uncertainty in its momentum, whereas we showedin Section 1.4 that the minimum uncertainty in the position of a particle of mass m is of the orderof h=mc. However, this minimum uncertainty in position is only relevant in relativistic quantum
physics.

42

Position and momentum

Chap. 3

If we integrate the probability density over all possible positions of theparticle, we obtain the probability of finding the particle somewhere in theuniverse. Because the particle is certain to be found somewhere, this probability
must be is equal to unity. Hence the wave function must satisfy the normalization conditionZ

jC(r, t)j2 d3 r  1,

(3:15)

where the integration is over all space.
These equations look less formidable when the particle is restricted to movein one dimension. If the particle moves along the x axis, it may be described bya wave function C(x, t) such that(jC(x, t)j2 dx The probability of finding the particle
at time t between x and x  dx:)(3:14)

Because the particle must be found somewhere between x  ÿ1 and x  1,
the wave function must obey the normalization condition
Z 1ÿ1

jC(x, t)j2 dx  1:

(3:17)

In practice, a wave function is normalized by multiplying a solution of the
SchroÈdinger equation by an appropriate constant. When we encounter a newwave function we shall often normalize it and then explore the potential positionsof the particle by considering the probability density for position. We stress that
this exercise only relies on a schematic description of a position measurement, inwhich a small scale event is amplified and the particle materialises at a specificlocation with probability proportional to jCj2 . This schematic description of ameasurement leads to a powerful theory even though no attempt is made todescribe how the outcome of the measurement emerges. In fact, quantum mechanics is successful because it avoids explaining how events happen.

3.3

MOMENTUM PROBABILITIESWe shall now explain how the momentum properties of a particle can be describedby a SchroÈdinger wave function. If a wave function can represent a particle with arange of potential positions, it is reasonable to expect that it can also represent a
particle with a range of potential momenta. For example, the wave function
C(x, t)  A1 ei(k1 xÿ!1 t)  A2 ei(k2 xÿ!2 t) ,

3.3

Momentum Probabilities

43

could describe a free particle moving in the x direction with two possible
momenta and energies,
p1  hk1 , E1  
h !1

and

p2  
hk2 , E2  h!2 :This idea is also illustrated by the general solution of the free-particle SchroÈdinger equation given by Eq. (2.16),
C(x, t) Z 1
ÿ1

A(k) ei(kxÿ!t) dk,

with

h! 

h2 k2:2m(3:18)

In wave terms, we have a superposition of sinusoidal waves each with adifferent wave number k, and the magnitude of the function jA(k)j2 is a measureof the intensity of the sinusoidal wave with wave number k. In particle terms,
this range of wave numbers corresponds to a range of possible momentap  hk. We shall assume, in analogy with the Born interpretation of the wave
function, that the most probable momenta found in a measurement correspondto the values of hk for which the function jA(k)j2 is large.
More generally, one may treat position and momentum in a symmetrical way
by using Fourier transforms.3 Any wave function for a particle moving in one
dimension can always be written as a Fourier transform
1
C(x, t)  p
2ph

Z 1
ÿ1

e t) eipx=h dp:
C(p,(3:19)

C(x, t) eÿipx=h dx:(3:20)The inverse Fourier transform is1e t)  p

C(p,
2p
h

Z 1
ÿ1

e t) is also normalized; i.e. if
It can be shown, that if C(x, t) is normalized, then C(p,
Z 1
ÿ1

2

jC(x, t)j dx  1,

then

Z 1
ÿ1

e t)j2 dp  1:
jC(p,This symmetry between position and momentum, and the earlier observation
about the interpretation of a superposition of sinusoidal waves, leads us toassume that, since jC(x, t)j2 is the probability density for finding the particle
3

Readers unfamiliar with Fourier transforms should not panic. Eq. (3.19) and Eq. (3.20) are theonly Fourier transform equations we shall be using. If you are willing to accept these equations,
you can appreciate the symmetry that exists in quantum mechanics between position and momentum observables and also gain an understanding of why these observables can be described byoperators.

44

Position and momentum

Chap. 3e t)j2 is the probability density for finding the particle with
with position x, jC(p,momentum p. Further, since the wave function C(x, t) is a probability amplie t) is the probability amplitude fortude for position, its Fourier transform C(p,momentum. This generalization of the Born interpretation of the wave function
can easily be extended to describe the possible momenta of a particle moving inthree dimensions.

3.4

A PARTICLE IN A BOX I

In this section we shall illustrate how position and momentum probability
densities can be calculated by considering one of the simplest systems inquantum mechanics: a particle of mass m confined to a one-dimensional region
0 < x < a. In Section 4.4 we shall show that such a particle has an infinite
number of states with discrete energies labelled by a quantum numbern  1, 2, 3, . . . . A particle in a state with quantum number n has an energy
given by
h2 k2n,2m

En 

where kn np,a

(3:21)

and a wave function given byCn (x, t) N sin kn x eÿiEn t=h0

if 0 < x < aelsewhere.

(3:22)

The constant N, which is called a normalization constant, can be foundby normalizing the position probability density. This is given by jCn (x, t)j2 ;it is zero outside the region 0 < x < a, and inside this region it is givenbyjCn (x, t)j2  N* sin kn x eiEn t=h N sin kn x eÿiEn t=h  jNj2 sin2 kn x:
The total probability of finding the particle at any of its possible locations isgiven by the normalization integral,
Z 1
ÿ1

2

2

jCn (x, t)j dx  jNj

Z aa
sin2 kx dx  jNj2 :20p
By equating this probability to one, we find that N  2=a gives rise to anormalized probability density and to a normalized wave function.
e n (p, t)j2 ,The momentum probability density for the particle is given by jC
where, in accordance with Eq. (3.20),

45

3.4 A Particle in a Box I

0.4

probability density

probability density2(n = 1)

1.5

1

0.5

0
−0.5

0

0.5
position

1

0−6

probability density

probability density

0.1

−4

−2
0
2
momentum

4

6

0.4

1.5

(n = 3)

1

0.5

0
−0.5

0.2

1.5

2

(n = 1)

0.3

0

0.5
position1(n = 3)

0.3

0.2

0.1
0
−6

1.5

−4

−2
0
2
momentum

4

6

Fig. 3.3 The position and momentum probability densities for a particle confined to
the region 0  x  a with normalized wave functions given by Eq. (3.22) with n  1 andn  3. In this figure the units are such that a position equal to one corresponds to x  a
and a momentum equal to one corresponds to p  hp=a. Note that the area under eachcurve is equal to one, signifying that the probability of finding the particle at any of its
possible locations or with any of its possible momenta is equal to one.

Z 11
e n (p, t)  pCCn (x, t) eÿipx=h dx:
2ph ÿ1p
If we use Eq. (3.22) with N  2=a, we obtain1e n (p, t)  p
eÿiEn t=h
C
pha

Z a
0

eÿipx=h sin kn x dx,

and the integral can easily be evaluated using
sin kn x eikn x ÿ eÿikn x:
2i

In Fig. 3.3 we show position and momentum probability densities for aparticle confined to the region 0 < x < a. Two possible states are considered,46

Position and momentum

Chap. 3the ground state with n  1 and the second excited state with n  3. The
position probability densities, shown on the left of the figure, indicate thatthe most likely location is x  a=2 when n  1 and that, when n  3, there arethree most likely locations: at x  a=6, x  a=2 and x  5a=6. The momentum
probability densities, shown on the right of the figure, indicate that the most
likely momentum is zero when n  1 and that there are two most likely
momenta when n  3, at p  ÿ3
hp=a and at p  3hp=a. In fact, a state witha high value for n can be roughly pictured as a particle trapped between x  0
and x  a with two possible momenta, p  ÿnhp=a and p  nhp=a.

3.5

EXPECTATION VALUES

In general, the outcome of a measurement in quantum mechanics is a randomvariable with many possible values. The average of these values is called the
expectation value. In principle, the expectation value can be found by taking
the average result of measurements on an infinite ensemble of identically
prepared systems. Alternatively, we can calculate the expectation value using
the probability distribution which governs the outcomes of the measurement.
To keep the mathematics as simple as possible we will consider a particle
moving along the x axis.
The result of a measurement of the position x is a continuous randomvariable. The wave function C(x, t) is a probability amplitude for the position
observable and jC(x, t)j2 dx is the probability of finding the particle between x
and x  dx at time t. Thus, if a measurement of position is repeated many times
in an identical way on an identical particle in identical circumstances, many
possible outcomes are possible and the expectation value of these outcomes is,according to Eq. (3.6),
hxi 

Z ÿ1
1

x jC(x, t)j2 dx:(3:23)

Similarly, the measured momentum of the particle is also a continuous randome t), is the probability
variable. The Fourier transform of the wave function, C(p,2~amplitude for the momentum observable and jC(p, t)j dp is the probability of amomentum outcome between p and p  dp at time t. Hence, the expectation
value for momentum is
hpi 

Z 1
ÿ1

~ t)j2 dp:
pjC(p,(3:24)~ 2  C*~ C
~ to rewrite theBefore proceeding, we shall use jCj2  C*C and jCj
integrals in Eq. (3.23) and Eq. (3.24) as

3.5

hxi 

Z 1
ÿ1

Expectation Values

47

C*(x, t)x C(x, t) dx(3:25)~~ t) dp:
C*(p,t)p C(p,(3:26)

and
hpi 

Z 1
ÿ1

These expressions and the Fourier transform integrals, Eq. (3.19) and Eq.
(3.20), illustrate the symmetry between position and momentum in quantum
physics. However, this symmetry is often hidden because the expectation value~ t); it is usually calculated byfor momentum is seldom calculated using C(p,using the wave function C(x, t) and the expression
hpi 



]
C(x, t) dx:
C*(x, t) ÿi
h
]x
ÿ1

Z 1

(3:27)

This expression, as we shall soon show, can be derived using Fourier transform
techniques. You can also show that it is plausible. In problem 9, at the end of
the chapter, you are asked to show that Eq. (3.27) predicts that the averagevalues of the uncertain position and momentum of a quantum particle of mass
m obey a classical-like relationship:
hpi  mdhxi:
dt

(3:28)For the benefit of mathematically inclined readers we shall demonstrate thatthe momentum expectation value given by Eq. (3.27) is identical to that given~ t)
by Eq. (3.26). We shall do this by using the relations between C(x, t) and C(p,given by Eq. (3.19) and Eq. (3.20).
Using Eq. (3.19) we write

ÿih





Z 1
]
]
1
~ t) eipx=h dpp
C(p,C(x, t)  ÿi
h
]x
]x
2p
h ÿ1

which gives




Z 1
]
1
~ t) eipx=h dp :pC(p,C(x, t)  pÿih]x2ph ÿ1

48

Position and momentum

Chap. 3

We can now rewrite Eq. (3.27) as


Z 1
1
ipx=
h~
C*(x, t) p
pC(p, t) e
dp dx:
hpi 
2ph ÿ1
ÿ1
Z 1

If the order of integration is interchanged, we obtain
Z 1
Z 1 
1
ipx=h~ t) dp:p
hpi 
C*(x, t) e
dx pC(p,
2p
h ÿ1
ÿ1
We now use Eq. (3.20) to show that the integral in the brackets is equal to~C*(p,t) and obtain
hpi 

Z 1
ÿ1

~
~ t) dp
C*(p,t)pC(p,which is identical to Eq. (3.26).

Operators
We shall now introduce an idea which will become increasingly important as wedevelop the basic elements of quantum mechanics. This is the idea that observables in quantum mechanics can be described by operators. At this stage weshall consider the role of operators in the calculation of position and momentum expectation values using Eq. (3.25) and Eq. (3.27). The recipe for the
calculation is as follows:. First prepare a sandwich with

* and

.

. To find hxi insert x into the sandwich, and to find h pi insert ÿih ]=]x into thesandwich.
. Then integrate over x.
In this recipe, the position observable is represented by x and the momentum
observable is represented by ÿih ]=]x. However, both x and ÿih ]=]x can beconsidered as operators which act on the wave function; the real number x
merely multiplies C(x, t) by a factor x, and the differential expression ÿih ]=]x
differentiates C(x, t) and multiplies it by ÿih. To emphasize the role of operators in quantum physics we rewrite Eq. (3.25) and Eq. (3.27) as
hxi 

Z 1
ÿ1

^ C(x, t) dx
C*(x, t) x

and

hpi 

Z 1
ÿ1

C*(x, t) ^p C(x, t) dx:(3:29)

3.5

Expectation Values

49

The circumflex accent denotes an operator, and the operators for the position
and momentum observables are
^x
x

and

^
p  ÿih]:]x(3:30)The generalization of these ideas to a particle moving in three dimensions isstraightforward: the operators for position and momentum are^r  r and^p  ÿihr(3:31)and the expectation values for position and momentum are given by the threedimensional sandwich integrals,Z
hri C*(r, t) ^r C(r, t) d3 r

Z
and

hpi 

C*(r, t) ^p C(r, t) d3 r:(3:32)

Finally, we emphasise that the order in which functions and operators are
written down is important; for example,
C*(r, t) ^p C(r, t) 6 ^p C*(r, t) C(r, t):

Uncertainties
Operators can also be used to calculate uncertainties in position and momentum. According to Eq. (3.8) these uncertainties are given byDx q
hx2 i ÿ hxi2 andDp q
hp2 i ÿ hpi2 :(3:33)The expectation values of x and p are given by Eq. (3.29), and the expectation
values of x2 and p2 can be found from the generalization of this equation. They
are given by the sandwich integrals
hx2 i 2and hp i 

Z 1
ÿ1
Z 1
ÿ1

^2 C(x, t) dx
C*(x, t) x
(3:34)
2

C*(x, t) ^p C(x, t) dx,^x^ and ^
^2 is equivalent to x
p2 is equivalent to ^p  ^p. Using
where the operator x
Eq. (3.30), we have
^2  x2x^2  ÿ
and p
h2

]2
:
]x2

(3:35)

50

Position and momentum

Chap. 3

The actual values for the uncertainties in position and momentum depend
upon the form of the wave function. For example, in problem 4 at the end of
this chapter, you are asked to show that the uncertainties in the position andmomentum for a particle with the wave function
C(x)  N eÿx=2a

2

p
p
are Dx  a= 2 and Dp  
h=a 2. We note that, in this example, the product of
the uncertainties obey the relationh
Dx Dp  ,2
which is in accordance with the general statement of the Heisenberg uncertainty
principle given by Eq. (1.15). For all other wave functions the product Dx Dp
is larger, as illustrated in problems 5, 6 and 7.

3.6

QUANTUM STATESWe began this chapter with the puzzle of how a wave function can describe
both the wave-like and the particle-like properties of a quantum particle. We
have seen that this puzzle can be resolved if the wave function of a particle
governs the potential outcomes of measurements on the particle. This interpretation of the wave function has transformed the philosophical basis ofphysics. Physics no longer tries to predict exactly what will happen; it is now
content with predicting the probabilities of a range of possible outcomes.
It is not clear whether probability is being used in quantum mechanics, as in
the kinetic theory of gases, to cover up our ignorance of some underlying
specific description of the particle, a description which assigns an exact position
to the particle which is revealed by the measurement, or whether probability is
being used to provide a complete and fundamental description of the particle.
In the latter case, it is pointless or meaningless to speculate on where theparticle is prior to the measurement. Its exact position is not revealed by themeasurement, but brought into existence by the measurement; the particle, likethe experimenter, is surprised by the outcome!
However, these issues, which have been debated since the inception ofquantum mechanics, are not our immediate concern.4 Our main aim is toappreciate the power and elegance of quantum mechanics as a consistent theoryof microscopic phenomena and we shall do this by developing an intuitive
understanding of wave functions.
4

An excellent account of these issues and an extensive list of references are provided by F. LaloÈ,American Journal of Physics, vol. 69, page 655 (2001).

3.6

Quantum States

51

The key idea is that a wave function represents a quantum state of a particle,
a state of motion that bears only a passing resemblance to the well-defined
trajectories of classical physics.5 In the first three chapters of this book we havetouched upon the general properties of quantum states and these properties will
become more understandable as we consider examples of quantum states in thechapters that follow. At this stage, it is useful to set out the following important
properties of quantum states:. In the absence of measurements, a quantum state evolves with time smoothly
and deterministically in accordance with the time-dependent SchroÈdinger
equation, Eq. (2.17).
. A quantum state describes potentialities which can become realities. As
illustrated in this chapter, a quantum state can predict the possible outcomesof position and momentum measurements and the probabilities for theoccurrence of these outcomes. More generally, a quantum state can predictthe possible outcomes of any measurement.
. A quantum state is a linear superposition of other quantum states whichmeans that a particle in one quantum state is also simultaneously in otherquantum states. This property is called the principle of linear superposition.
We used this principle when we wrote down the wave function for a particle
passing through two slits, Eq. (3.11), and when we argued in Section 3.3 thata wave function can describe a particle with a range of possible momenta. It
will be used again in Chapter 4 when we consider a wave function of aparticle with a range of possible energies.
. Lastly, a quantum state is fragile. When a measurement occurs, a quantum
state is destroyed and replaced by a new quantum state which is compatible
with the random outcome of the measurement. This abrupt and non-deterministic process is called the collapse of the wave function. We used this idea
to explain why a two-slit experiment, in which the slit through which eachparticle passes is identified, does not give rise to an interference pattern. To aconsiderable extent, the collapse of the wave function is an arbitrary rule thatbridges the gap between the unobserved quantum system and the observed
quantum system. The underlying mechanism for the collapse is notunderstood.

5

In the most general formulation of quantum mechanics, quantum states are complex vectors.
Some quantum states can also be represented by wave functions and others, like quantum states
describing the spin properties of an electron, by matrices.

52

Position and momentum

Chap. 3

PROBLEMS 3
1. This problem considers the Poisson distribution, a probability distribution
for a discrete random variable which was first used by SimeÂon-Denis Poisson to describe seemingly random criminal events in Paris in 1837. If independent events have a constant tendency to occur and if the average rate ofoccurrence is a, then the probability that n events actually occur is given bypn 

eÿa ann!

withn  0, 1, 2, . . . 1:(a) By noting thatea  1 a a2 a3   ...
1! 2! 3!

show thatnX
1pn  1,n0

thereby verifying that the Poisson distribution is normalized.
(b) By using n=n!  1=(n ÿ 1)! and an  aanÿ1 , show thatn1X

npn  a,n0

thereby verifying that the average rate of occurrence, or the expectation
value hni, is equal to a.
(c) By using similar techniques, find hn2 i and show, using Eq. (3.4), that thestandard deviation of the Poisson distribution is given by
Dn 

p
a:

2. This problem considers a Gaussian distribution, with standard deviation s,
given by1
2
2
r(x)dx  p eÿx =2s dx
2
2pswith ÿ 1 < x < 1:This probability distribution is normalized because

Problems 31p
2ps2

Z 1
ÿ1

2

53

2

eÿx =2s dx  1:

It is famous for describing a random variable which arises from a multitudeof small random contributions, such as the net distance travelled by atottering drunk with very small legs.
(a) By considering the effect of the transformation x ! ÿx on the function
xr(x), show that the expectation value of x is equal to zero.
(b) By using
Z 1
ÿ1


r(x) dx  x r(x)

1
ÿ1

ÿ

Z 1
ÿ1

x

dr
dx,
dx

show that the expectation value of x2 is equal to s2 .
(c) Hence verify that the standard deviation of x is equal to s.
3. The lifetime of an unstable particle is governed by the exponential probability distribution. In particular, the probability that the particle lives for time t
and then decays in the time interval t to t  dt is given byp(t) dt  eÿlt l dt,
where l is a positive decay constant.
(a) Show that the probability that the particle eventually decays is equal toone.
(b) Find an expression for the mean lifetime of the particle.
(c) Find an expression for the probability that the particle lives for at least
time T.
4. In this question you should avoid unnecessary mathematics by using theproperties of the Gaussian probability distribution given in problem 2.
Consider a particle of mass m described by the wave function
2

C(x)  N eÿx =2awhere a is a constant length.

2

54

Position and momentum

Chap. 3(a) Use the properties of the Gaussian probability distribution to confirm thatthe expectation values of the position and the square of the position arehxi  0hx2 i 

and

a2
:
2

(b) Show, without lengthy calculation, that the expectation values of the
momentum and the square of the momentum are
hpi  0hp2 i 

and

2
h
:
2a2

[Hint: I suggest you use your skill at integration by parts to show that
Z 1
ÿ1

C*

d2 C
dx 
dx2

Z 1
ÿ1

dC* dCdx,
dx dx

and also make use of the integrals used in part (a).](c) Hence show that the uncertainty in position, Dx, and the uncertainty inmomentum, Dp, for this particle obey the relationh
Dx Dp  :2
5. A particle is confined to a region 0  x  a and has a wave function of theformc(x)  Nx(a ÿ x),
where N is a constant.
(a) Normalize the wave function and find the average position of theparticle.
(b) Show that the uncertainties in the position and momentum of theparticle are given byr
1
Dx 
a
28

and

Dp 

p h
10 :a

6. A particle of mass m is confined to a region ÿa=2 < x < a=2. Outside thisregion the wave function is zero and inside it is given byProblems 3

55

r
2
px
C
cos :
a
a
(a) Explain the physical significance of the integrals
Z 1
ÿ1

C*xC dx

and


d
C dx:
C* ÿi
h
dx
ÿ1

Z 1

Show that both integrals are zero.
(b) Show that Dx, the uncertainty in the position of the particle, and Dp, theuncertainty in the momentum of the particle, are related byr
p2 ÿ 6h:
Dx Dp 12
(The following integral
Z a=2ÿa=2x2 cos2px
a3 p3 pÿdx  3p 24 4ais useful.)
7. Consider a particle with normalized wave functionC(x) 

Nx eÿax=20

if 0  x < 1
elsewhere,

where a is a positive real constant and N p
a3 =2.

(a) Write down an expression for the probability of finding the particle
between x and x  dx. Illustrate how this probability depends on x
and find the most probable value of x.
(b) Find the expectation values for the position and the square of theposition, hxi and hx2 i.

56

Position and momentum

Chap. 3(c) Find the expectation values for the momentum and the square of the
momentum, hpi and hp2 i.
(d) Show that these expectation values yield uncertainties for position andmomentum which are consistent with the Heisenberg uncertainty relation.
(The mathematical identity
Z 10eÿax xn dx 

n!
an1

for

n > ÿ1is useful.)
8. In this problem the probability density for the position of a particle with wave
function C(x, t) is denoted by r(x, t). In general, the value of r(x, t) in aparticular region will change with time and this change can be attributed to
the flow of probability into and out of the region. Indeed we expect that there
is a probability current density j(x, t) which obeys the continuity equation6
]r
]j
ÿ :
]t
]x
When there are many particles, the actual current of particles is obtained bymultiplying j(x, t) by the actual density of particles.
(a) By noting that the time dependence of the wave function is governed by
the SchroÈdinger equation,
"
#
]C
h2 ]2

 V (x) C, ÿih2m ]x2]t

6

The reader may have come across a similar equation when considering the conservation of charge inelectromagnetism or particles in diffusion. In three-dimensional problems it has the form]r=]t  ÿdivj. It is a book-keeping equation which describes the flow of probability. Its bookkeeping properties can be illustrated by considering a one-dimensional region between x  x1 andx  x2 . The amount of probability in this region, i.e. the probability of finding the particle in theregion at time t is
Z x2
r(x, t) dx:
x1

The rate of change of probability in this region is given by
Z x2
 Z x2Z x2d]r]jr dx 
dx  ÿ
dx  j(x1 , t) ÿ j(x2 , t):
dt x1
x1 ]t
x1 ]x
We see that the rate of change of probability inside the region is the difference between the current
into the region at x  x1 and the current out at x  x2 .

Problems 3

57

derive an expression for the time derivative ofr(x, t)  C*(x, t)C(x, t):(b) Hence show that the probability current is given byj(x, t) 



i
h
]C
]C*
C*
ÿC
:
2m
]x
]x

9. In this problem you are asked to use Eqs. (3.25) and (3.27) to show that the
expectation values of the position and momentum of a particle with mass m
are related bymdhxi
 hpi:
dt

The method is similar to that used in the previous problem.
(a) By noting that the time dependence of the wave function is governed by
the SchroÈdinger equation, show thatd(C*xC)
ih
]2 C
]2 C*
,

xC* 2 ÿ xC
dt
2m
]x
]x2
and show that this can be rewritten asd(C*xC)
ih ]
]C
]C*
i
h
]C
]C*

xC*
ÿ xC
ÿ
C*
ÿC
:
dt
2m ]x
]x
]x
2m
]x
]x
(b) Assuming that the wave function tends to zero sufficiently rapidly atx  1, show that
d
dt

Z 1
ÿ1

i
h
C*xC dx  ÿ2m

Z 1 
ÿ1


]C
]C*
C*
ÿC
dx:
]x
]x

(c) Now integrate by parts and show that
d
m
dt

Z 1
ÿ1

C*xC dx 

Z 1
ÿ1



]
C*(x, t) ÿih
C(x, t) dx:]xThis page intentionally left blank4
Energy and time
This chapter focuses on how the energy observable is described in quantum
mechanics and how it is related to the time evolution of a quantum state. Most
importantly, we shall show that the observable properties of a quantum statewith a sharply defined energy never change.

4.1

THE HAMILTONIAN OPERATORIn quantum mechanics observable quantities are described by operators. In thelast chapter we found that the position and momentum observables are described by^r  r and^p  ÿihr:(4:1)

The energy observable in quantum mechanics is described by an operator called^ We shall assume that thethe Hamiltonian operator and it is denoted by H.
relation between the operators for energy, momentum and position is similar to
the relation between the classical energy, momentum and position. In particular, we shall assume that the Hamiltonian operator for a particle with mass m ina potential energy field V(r) is given by
2

p
^  ^
 V (^r):
H
2m

(4:2)

We can rewrite this using Eq. (4.1) to give2

^  ÿ h r2  V (r):
H
2m

(4:3)

60

Energy and time

Chap. 4^ has a dual role in
We emphasize that the Hamiltonian operator H^
quantum mechanics. First, the operator H describes the energy observable;
for example, the recipe for expectation values given in Section 3.4 implies
that the energy expectation value at time t for a particle with wave function
C(r, t) isZhEi ^ C(r, t) d3 r:
C*(r, t) H

(4:4)

^ governs the time evolution of the wave function
Second, the operator H
because the SchroÈdinger equation, Eq. (2.17), has the form
i
h

]C
^
 HC:
]t

(4:5)

Thus, in quantum mechanics, there is a fundamental connection between
energy and time.
We shall explore this connection by finding solutions to the SchroÈdinger
equation. The procedure used will be identical to that used to solve the classicalwave equation or the diffusion equation: we shall seek a separable solution andthen solve an eigenvalue problem. Because this procedure seems arbitrary in the
abstract, it is best introduced by considering a familiar and simple problem, theproblem of finding the normal modes of a vibrating string.

4.2

NORMAL MODES OF A STRING

Let C(x, t) represent the transverse displacement of a stretched string at the
point x at time t. This displacement is governed by the classical wave equation]2 C 1 ]2 Cÿ
 0,]x2 c2 ]t2(4:6)

where c is the speed of waves on the string. If the ends of the string are fixed atx  0 and x  a, we seek solutions to the wave equation which satisfy theboundary conditions
C(x, t)  0 at x  0 and at x  a, for all time t:(4:7)

There are an infinite number of such solutions because the string can vibrate in
an infinite number of ways.
The normal mode solutions are particularly simple. They correspond tovibrations where all points on the string move with the same time dependence.
They have the separable form

4.2

Normal Modes of a StringC(x, t)  c(x)T(t):

61

(4:8)

The function T(t) describes the common time dependence of each point and thefunction c(x) describes the spatial shape of the vibration. If we substitute theseparable form for C(x, t) into the wave equation, Eq. (4.6), and if we carefully
separate out the functions which depend on t from those which depend on x we
obtain1 d2 T c2 d2 c

:
c dx2
T dt2

(4:9)

The equal sign in this equation asserts that a function of t on the left-hand side isequal to a function of x on the right-hand side, for all x and t. This can be true only
if both functions are equal to the same constant. We shall denote this constant by
ÿ!2 and we shall also set !  ck where k is another constant. By equating both theleft-hand side and the right-hand side of Eq. (4.9) to the constant ÿ!2 , we find thetime dependence T(t) and spatial shape c(x) for each normal mode solution.
The time dependence T(t) is governed by the differential equationd2 T ÿ!2 T:dt2

(4:10)

T(t)  A cos !t  B sin !t,(4:11)The general solution iswhere A and B are arbitrary constants. This solution describes sinusoidal
motion with an angular frequency !, which, as yet, is undetermined.
The normal mode function c(x) is governed by the differential equationd2 c ÿk2 c,dx2

where k !
:c(4:12)

And it also satisfies the boundary conditionsc(0)  c(a)  0,

(4:13)

which follow from Eqs. (4.7) and (4.8). The general solution of the differential
equation (4.12) isc(x)  M cos kx  N sin kx,
where M and N are arbitrary constants. The boundary condition at x  0 gives
M  0 and the boundary condition at x  a restricts the values of k to62

Energy and time

Chap. 4p=a, 2p=a, 3p=a, etc. Thus, there are an infinite number of normal mode solutions with spatial shapes given bycn (x)  N sin kn x,

with kn np,a(4:14)

where n  1, 2, 3, . . . . The spatial shapes of the normal modes of a string withn  1, 2, 3, and 4 are shown in Fig. 4.1.
If these spatial functions are combined with the time-dependent functions,Eq. (4.11), with angular frequencies !n  ckn , we obtain a complete specification of the normal mode solutions. They have the form
Cn (x, t)  [An cos !n t  Bn sin !n t] sin kn x:(4:15)

Because the classical wave equation, Eq. (4.6), is a homogeneous linear
partial differential equation, a linear superposition of normal mode solutions

n=1

1.5

1.5

0.5
0
−0.5
−1
−1.5

0.5
0
−0.5
−1
−1.5

0

0.2

0.4

0.6

0.8

1

0

n=3

1.5

0.2

0.4

0.6

0.8

1

n=4

1.5
1
eigenfunction

1
eigenfunction

n=2

1
eigenfunction

eigenfunction

1

0.5
0
−0.5
−1
−1.5

0.5
0
−0.5
−1
−1.5

0

0.2

0.4
0.6
position x

0.8

1

0

0.2

0.4
0.6
position x

0.8

1Fig. 4.1 The spatial shapes of four eigenfunctions cn (x) of the eigenvalue problem
defined by the differential equation (4.12) and the boundary conditions (4.13) witha  1. In classical physics, these eigenfunctions may describe the shape of a normal
mode of vibration of a string with definite angular frequency. In quantum physics, they
may describe the shape of a wave function of a particle in a box with definite energy.

4.3

States of Certain Energy63

is also a solution. Indeed, it can be shown that the general motion of a vibrating
string with fixed ends is given by
C(x, t) Xn1, 2, 3...

[An cos !n t  Bn sin !n t] sin kn x:(4:16)

If the initial displacement and velocity of each point of the string are known,
Fourier series techniques can be used to find the constants, An and Bn , for eachterm in the series.
It is useful to describe what we have done using the mathematical languagethat is used in quantum mechanics. In finding the normal mode solutions of avibrating string we have solved an eigenvalue problem. The eigenvalue problem
is defined by the differential equation (4.12) and the boundary conditions(4.13). We note that the differential equation contains an undetermined parameter k and that solutions to the eigenvalue problem only exist for particularvalues of k given in Eq. (4.14). The function cn (x)  N sin kn x is called theeigenfunction belonging to the eigenvalue kn  np=a. Once we found the eigenfunctions, we wrote down the general motion in terms of a linear superpositionof eigenfunctions, Eq. (4.16).

4.3

STATES OF CERTAIN ENERGY

We shall now see how the mathematics of a vibrating string may be adapted tofind the solutions of the SchroÈdinger equation. We shall begin by showing how
solutions analogous to normal mode solutions may be constructed.
According to Eq. (2.17), the SchroÈdinger equation for a particle of mass m inpotential V(r) is
"
#
]C
h2 2
r  V (r) C:
i
h
 ÿ
2m
]t

(4:17)As in the classical example of a vibrating string, we shall seek separable
solutions of the formC(r, t)  c(r)T(t):(4:18)

If we substitute into the SchroÈdinger equation and carefully separate functions
depending on t from those which depend on r we obtain
"
#
i
h dT
1
h2 2

ÿ
r c  V (r)c :
T dt
c
2m

(4:19)

64

Energy and time

Chap. 4The equal sign in this equation asserts that a function of t on the left-hand sideis equal to a function of r on the right-hand side, for all r and t. This can be true
only if both functions are equal to the same constant. We shall denote thisconstant by E. By equating both the left-hand side and the right-hand side ofEq. (4.19) to the constant E, we can find the time dependence T(t) and spatial
shape c(r) for each separable solution of the SchroÈdinger equation.
The time dependence T(t) is governed by the differential equation

ÿiE
T:
h


(4:20)

T(t)  A eÿiEt=h ,(4:21)dTdt

The general solution iswhere A is an arbitrary constant. The constant E is, as yet, undetermined, butsome readers may have guessed its meaning.
The spatial shape c(r) of the wave function is governed by the differential
equation
"

#
2 2
h
r  V (r) c(r)  Ec(r),ÿ2m(4:22)

which may be rewritten more succinctly as^
Hc(r) Ec(r),(4:23)^ is the Hamiltonian operator. This equation is called the energy
where H^eigenvalue equation and the function c(r) is called the eigenfunction of H
belonging to the eigenvalue E; in practice, there are many eigenvalues andmany eigenfunctions. Equation (4.23) is also called the time-independent SchroÈdinger equation.
An eigenfunction of the Hamiltonian operator is a very special mathematical^ acts upon a function we expect afunction. When the complicated operator H
mess, but when it operates on an eigenfunction it gives the same function
multiplied by a constant, as shown in Eq. (4.23). If the eigenfunction belonging
to the eigenvalue E is combined with the time-dependent function (4.21), we
obtain a special solution to the SchroÈdinger equation:C(r, t)  c(r) eÿiEt=h :(4:24)We shall now show that this wave function represents a state with a sharply
defined energy E.

4.3

States of Certain Energy

65

In general, when the energy is measured, the outcome is uncertain. In
analogy with Eq. (3.33), which gives the uncertainties for position and momentum, the uncertainty in the energy is given byDE q
hE 2 i ÿ hEi2 ,(4:25)

where hEi is the expectation value for the energy and hE 2 i is the expectation
value for the square of the energy. These expectation values, for a particle witha normalized wave function C(r, t), are given by the sandwich integrals
Z
hEi 

^
C*(r, t)HC(r,t) d3 r(4:26)^ 2 C(r, t) d3 r:
C*(r, t)H

(4:27)

andhE 2 i Z

It is easy to evaluate these integrals when the wave function is given by
Eq. (4.24). In this special case, the wave function C(r, t), like c(r), is aneigenfunction of the Hamiltonian operator with eigenvalue E and we can use^
HC(r,t)  EC(r, t)(4:28)

to giveZ
hEi ^C*(r, t)HC(r,t) d3 r  E

Z

C*(r, t)C(r, t) d3 r  E:

^ it is also an eigenfunction of theMoreover, if C is an eigenfunction of H,^^
product of H with H. Using^ 2 C(r, t)  E 2 C(r, t),H(4:29)

we obtainhE 2 i Z^ 2 C(r, t) d3 r  E 2
C*(r, t)HZ

C*(r, t)C(r, t) d3 r  E 2 :When we substitute hEi  E and hE 2 i  E 2 into Eq. (4.25), we find that theuncertainty in energy DE is zero. These results imply that the result of an energy
measurement is certain to be E when the wave function is an eigenfunction of
the Hamiltonian with eigenvalue E. We conclude that an eigenfunction of the
Hamiltonian always describes a state of definite energy.

66

4.4

Energy and time

Chap. 4

A PARTICLE IN A BOX II

One of the key features of quantum physics is that the possible energies of aconfined particle are quantized. Indeed, the familiar quantized energy levels of
atomic, nuclear and particle physics are manifestations of confinement. We
shall illustrate the connection between confinement and quantized energy levels
by considering a particle confined to a box.

A one-dimensional box
We begin by considering a particle moving in one dimension with potential
energy
V (x) 01

if 0 < x < aelsewhere,(4:30)

This infinite square-well potential confines the particle to a one-dimensionalbox of size a, as shown in Fig. 4.2. In classical physics, the particle either lies atthe bottom of the well with zero energy or it bounces back and forth between
the barriers at x  0 and x  a with any energy up to infinity. In quantum
physics, more varied states exist. Each is described by a wave function C(x, t)
which obeys the one-dimensional SchroÈdinger equation"#]Ch2 ] 2ih ÿ V (x) C:
2m ]x2]t(4:31)

2 2p
E4 = 16 h2ma
2

V(x)
2

2

2

2

2

2

p
E3 = 9 h2ma2p
E2 = 4 h2ma2p
E1 = 1 h2ma
2

x=0

x=a

Fig. 4.2 Low-lying energy levels of a particle of mass m confined by an infinite squarewell potential V (x) with width a.

4.4

A Particle in a Box II

67

However, when the particle has a definite value E, the wave function has the
special form
C(x, t)  c(x) eÿiEt=h ,(4:32)

where c(x) satisfies the energy eigenvalue equation
"

#
2 d2
h
ÿ
 V (x) c(x)  Ec(x):2m dx2(4:33)We shall now seek physically acceptable solutions to Eq. (4.33). Because
the potential energy V(x) rises abruptly to infinity at x  0 and at x  a, theparticle is confined to the region 0 < x < a and outside this region the eigenfunction c(x) is zero. Inside this region, the potential energy is zero and theeigenfunction is a solution of Eq. (4.33) with V (x)  0. We shall simplify thisequation by rewriting the energy E as
E

2 k2
h
2m

(4:34)

to gived2 c ÿk2 c:dx2

(4:35)

Physically acceptable solutions of this differential equation are obtained bywriting down the general solutionc(x)  M cos kx  N sin kx,
where M and N are constants, and by imposing boundary conditionsc(0)  c(a)  0,

(4:36)

which ensure that the position probability density of the particle does notchange abruptly at the edge of the box.
Readers should note that the energy eigenvalue problem for a particle in a
one-dimensional box, defined by the differential equation (4.35) and boundary
conditions (4.36), is identical to the eigenvalue problem for a vibrating string
defined by Eqs. (4.12) and (4.13). In both cases, there are an infinite number of
eigenfunctions labelled by an integer n  1, 2, 3, . . . . They are given bycn (x)  N sin kn x,

with

kn 

np
,
a

(4:37)

68

Energy and time

Chap. 4

where N is an arbitrary constant, and they are illustrated in Fig. 4.1. In classicalphysics, the eigenfunctions cn can be used to describe the possible shapes ofnormal modes of vibration of a string. In quantum physics, they can be used todescribe the possible shapes of wave functions of a particle in a box withdefinite energy, labelled by the quantum numbers n  1, 2, 3, . . . .
We conclude that the possible energy levels of a particle in a one-dimensionalbox of width a are given byEn 

n2 p2 h2,
2ma2

with n  1, 2, 3, . . . ,(4:38)and that a particle with energy En has a wave function of the form
Cn (x, t)  N sin kn x eÿiEn t=h :(4:39)

We note the following:. As shown in Fig. 4.2, the separation between the energy levels increases as
the quantum number n increases. However, this separation as a fraction of
the energy decreases; indeed
En1 ÿ En
2
!
En
n

as

n ! 1:

This means that the discrete nature of the energy levels becomes less important when the energy is high.
. The lowest possible energy, in contrast with classical physics, is not zero, but
E1 

2 p2
h
:
2ma2

We can understand this zero-point energy by using the Heisenberg uncertainty principle, Eq. (1.17). If a particle is confined to a region of size a, it has
an uncertain position Dx  a and, hence, an uncertain momentum Dp which
is at least of the order of 
h=2a. Because the average magnitude of the
momentum is always greater than Dp, the average kinetic energy of theparticle is always greater than (Dp)2 =2m which in turn is greater thanh2 =8ma2 .
. The spatial shape of the wave function of a particle in a box with energy En is
identical to the spatial shape of the normal mode of a string with angularfrequency !n . As illustrated in Fig. 4.1, the number of nodes increases as thevalue of n increases.

4.4

A Particle in a Box II

69. The wave function of a particle in a box, unlike the displacement of a string,is not observable, but it can be used to construct properties of the particle
that are observable. The first step is to normalize the wave function so that
Z a0jC(x, t)j2 dx  1:pAs shown in Section 3.4, this condition is satisfied if N  2=a. One canthen calculate probability densities for position and momentum as illustrated
in Fig. 3.3.

A three-dimensional box
We shall now consider the more realistic problem of a particle confined inthree dimensions. If the particle has definite energy E, its wave function has theform
C(x, y, z, t)  c(x, y, z) eÿiEt=h ,(4:40)

where c(x, y, z) satisfies the energy eigenvalue equation"#h2 ]2]2]2
 V (x, y, z) c  Ec:ÿ
2m ]x2 ]y2 ]z2(4:41)We shall choose a potential energy function
V (x, y, z) 0
1

if 0 < x < a, 0 < y < b, 0 < z < celsewhere,(4:42)

which confines the particle to a box with sides a, b and c.
The possible energy eigenfunctions and eigenvalues of the particle may befound by seeking solutions of Eq. (4.41) inside the box which are equal to zeroon all six faces of the box. For example, the function
px py pzc(x, y, z)  N sin
sin
sin
a
b
c
is zero on each of the faces defined byx  0,x  a,y  0,y  b,z0andz  c,and it satisfies Eq. (4.41) inside the box where V (x, y, z)  0, if

70Energy and time

Chap. 4E
2 p 2 1
h
1
1
:


2m a2 b2 c2

In general, there are an infinite set of eigenfunctions and eigenvalues labelled bythree quantum numbers nx  1, 2, 3, . . . , ny  1, 2, 3, . . . , and nz  1, 2, 3, . . . .
The eigenfunctions have the form
cnx , ny , nz (x, y, z)  N sin

n px
xasin

n py n pz
y
z
sin
,
b
c

(4:43)and the energy eigenvalues are given by"#
2 p2 n2x n2y n2zh 
Enx , ny , nz :2m a2 b2 c2(4:44)

Equation (4.44) shows how the quantized energy levels of a particle in a box
depend upon the dimensions of the box, a, b and c. Most importantly, it showsthat some energy levels may coincide when the box has particular dimensions.
We illustrate this in Fig. 4.3 which shows that, for a particle in a cubical box
with a  b  c, energy levels like E1, 1, 2 , E2, 1, 1 , and E1, 2, 1 coincide. When there
are several states, or wave functions, with the same energy, the energy levels are
said to be degenerate. Degenerate energy levels are very important in atomic,
nuclear and particle physics. They arise because the interactions which confine22

E2,2,2 = 16 h2map 222

E1,1,3 = E3,1,1 = E1,3,1 = 11 h2map 2
2

2

2

2

E1,2,2 = E2,2,1 = E2,1,2 = 9 h2map 2

E1,1,2 = E2,1,1 = E1,2,1 = 6 h2map 222pE1,1,1 = 3 h2ma
2Fig. 4.3 Low-lying energy levels of a particle of mass m confined in a cubical box withsides of length a. Note the degeneracy of the first, second and third excited states.

4.5

States of Uncertain Energy

71

electrons in atoms, nucleons in nuclei and quarks in hadrons, have specificsymmetry properties. Indeed, the observed degeneracy of energy levels can beused to deduce these symmetry properties.

4.5

STATES OF UNCERTAIN ENERGY

In Section 4.3 we showed that a quantum state with a sharply defined energy En
is represented by the wave function
Cn (r, t)  cn (r) eÿiEn t=h ,(4:45)

where cn (r) is the energy eigenfunction belonging to the eigenvalue En . We shallnow show that a state with uncertain energy is represented by a wave functionof the formC(r, t) Xn1, 2, ...

cn cn (r) eÿiEn t=h :(4:46)

To do so we need to understand two concepts. The mathematical concept of acomplete set of basis functions and the physical concept of an energy probability
amplitude.

Basis functions
Because the SchroÈdinger equation is a homogeneous linear partial differential
equation, a linear superposition of wave functions with definite energies is alsoa solution. For example, the general wave function for a particle in a onedimensional box isC(x, t) Xn1, 2, 3...:cn Cn (x, t)(4:47)

where, in accord with Eq. (4.39),
Cn (x, t)  N sin kn x eÿiEn t=h
and cn are arbitrary complex constants. This equation is very similar to theequation for the general motion of a vibrating string, Eq. (4.16). In one case, wehave a general wave function expressed as a linear superposition of wave
functions with definite energies and, in the other case, we have a generalvibration expressed as a linear superposition of normal modes. In both thesecases, we have a linear superposition of sine functions, or a Fourier sine series.

72

Energy and time

Chap. 4In more complicated problems, a generalized Fourier series must be used.
Such a series, like an ordinary Fourier series, is based on the idea that the
eigenfunctions of a Hamiltonian form a complete orthonormal set of basis
functions. To illustrate this idea we shall consider a Hamiltonian which gives
rise to an infinite set of energy eigenfunctions and eigenvalues denoted by cn (r)and En with n  1, 2, . . . . It can be shown that these eigenfunctions form acomplete orthonormal set of basis functions. This means three things:. Each eigenfunction Cn (r) may be normalized so thatZ
jcn (r)j2 d3 r  1:

(4:48). The eigenfunctions, cm and cn , belonging to different discrete eigenvalues Em
and En , are orthogonal. As shown in problem 2 at the end of this chapter,they satisfy the condition
Z

cm*(r)cn (r) d3 r  0

if

Em 6 En :(4:49)The eigenfunctions belonging to a degenerate eigenvalue, Em  En , are notuniquely determined. One can utilize this latitude to make them orthogonal.
The orthogonality relation (4.49) then holds generally.
. The eigenfunctions form a complete set because it is always possible toexpress any wave function as a linear superposition of eigenfunctions. This
expression is a generalised Fourier series of the formC(r, t) Xn1, 2, 3...:cn cn (r) eÿiEn t=h ,(4:50)

where the coefficients cn are complex constants.
The coefficients cn of the generalized Fourier series (4.50) can be found if weknow the initial spatial shape of the wave function. For example, to find c3 weconsiderC(r, 0)  c1 c1 (r)  c2 c2 (r)  c3 c3 (r)  . . . ,and multiply both sides by c*(r).
If we integrate over r and use the normaliza3
tion and orthogonality integrals, Eqs. (4.48) and (4.49), we obtainZ0) d3 r  c3 :c*(r)C(r,3

Clearly, the general expression for the coefficients cn is

4.5

Z

States of Uncertain Energy

c*(r)C(r,0) d3 r:
n

cn 

73

(4:51)

Armed with these coefficients, we can use Eq. (4.50) to keep track of thesubsequent evolution of the wave function.1

Energy probability amplitudes
We shall now consider the physical interpretation of a wave function given by
the linear superposition of energy eigenfunctions,C(r, t) Xn1, 2, ...

cn cn (r) eÿiEn t=h :(4:52)

Our first step is to find the condition
R for the wave function (4.52) to benormalized. The normalization integral C*C d3 r has the form
Z

[c*1 c*1 eiE1 t=h  c*2 c*2 eiE2 t=h  . . . ] [c1 c1 eÿiE1 t=h  c2 c2 eÿiE2 t=h  . . . ] d3 r:

Using the normalization and othogonality relations for the eigefunctions cn (r),we find that terms likec*2 c1 ei(E2 ÿE1 )t=h

Z

c*2 c1 d3 r

yield zero and terms likec*1 c1Z

c*1 c1 d3 r

yield jc1 j2 . Hence, we obtainZ

C*C d3 r Xn1, 2, ...

jcn j2 ,and conclude that the wave function (4.52) is normalized if the coefficients cn
satisfy the conditionXn1, 2, ...
1

jcn j2  1:

(4:53)

These equations have to be modified when the potential gives rise to a continuum of energy
eigenvalues. In this case, the normalization and orthogonality of the eigenfunctions involves a
Dirac delta function and the general solution involves an integral over the continuous energy
variable which labels the eigenfunctions. These modifications will be considered in Chapter 7.

74Energy and time

Chap. 4

We shall now calculate the expectation values for the energy and for thesquare of the energy of a particle with wave function (4.52). These, according toEqs. (4.26) and (4.27), are given byZ
hEi ^ C d3 r
C*H

and

2

Z

hE i 

^ 2 C d3 r:
C*H

To evaluate these integrals when the wave function is a superposition of
eigenfunctions cn , we use^ n (r)  En cn (r)Hcand^ 2 cn (r)  E 2 cn (r),Hnand the orthogonality and normalization relations, Eqs. (4.48) and (4.49). We
obtainhEi Xn1, 2, ...

jcn j2 EnandhE 2 i Xn1, 2, ...

jcn j2 En2 :(4:54)

We can now assign a physical meaning to the complex coefficients cn . If werecall the general statements made about probability distributions in
Section 3.1, we see thatpn  jcn j2

withn  1, 2, 3 . . . :(4:55)

is a probability distribution. Eq. (4.53) shows that it is a normalized distribution and Eq. (4.54) shows that pn  jcn j2 is the probability that the energy isequal to En . Because of this, the coefficients cn are called energy probability
amplitudes.
In this section we have illustrated how the possible energies of a quantum
state are described using the principle of linear superposition. According to thisprinciple all quantum states are linear superpositions of other quantum states.
Here we have shown that the state represented by the linear superposition(4.52) describes a particle with uncertain energy and that the possible outcomesof a measurement are E1 , E2 , E3 , . . . with probabilities jc1 j2 , jc2 j2 , jc3 j2 , . . . .

4.6

TIME DEPENDENCE

We shall end this chapter by exploring the fundamental connection between theenergy properties and the time dependence of the observable properties of aquantum state.
We first consider a particle with the wave function
Cn (r, t)  cn (r) eÿiEn t=h

(4:56)

4.6

Time Dependence

75

which represents a quantum state with sharply defined energy En . Even though
the time dependence of this wave function is described by a complex exponential which oscillates with angular frequency En =
h, the observable properties ofthe particle do not change with time. We can illustrate this by showing that theposition probability density is independent of time:
Cn*(r, t)Cn (r, t)  cn*(r) eiEn t=h cn (r) eÿiEn t=h  cn*(r)cn (r):In fact, the wave function Cn (r, t) describes a quantum state with no observabletime dependence; the probabilities and the expectation value for any observable
never change. Such a state is called a stationary state.
In contrast, a quantum state with uncertain energy has observable properties
which change with time. We shall illustrate this by considering a particle withwave function
C(r, t) qqÿiE1 t=h1
1ÿiE2 t=
h
c
(r)
e

:
2 1
2c2 (r) e(4:57)In this case, there are two possible outcomes when the energy is measured: E1
with probability 12 and E2 with probability 12. The energy expectation valuesarehEi  12E1  12E2

andhE 2 i  12E12  12E22 ,and the uncertainty in energy isDE q
hE 2 i ÿ hEi2  12jE1 ÿ E2 j:

The position probability density for this state of uncertain energy is timedependent: it is given by
C*C  12[jc1 j2  jc2 j2  c*1 c2 ei(E1 ÿE2 )t=h  c1 c*2 eÿi(E1 ÿE2 )t=h ],and oscillates with angular frequency jE1 ÿ E2 j=
h, i.e. with period 2ph=jE1 ÿ E2 j.
Thus, Eq. (4.57) represents a quantum state with uncertain energy DE which
has an observable property that oscillates with period ph=DE.
Quantum states of uncertain energy are called non-stationary states because
they have some observable properties which change with time. In general,these properties change more rapidly when the energy is more uncertain. If
dt is the time scale for significant change and DE is the energy uncertainty,
thendt DE  h:(4:58)

76

Energy and time

Chap. 4This general relation is often referred to as the Heisenberg uncertainty
relation for time and energy, but a better name is the time and uncertainty inenergy relation. Time, unlike position, momentum or energy, is not an observable in quantum mechanics. It is a parameter which is used to label a changing
system. The symbol dt in Eq. (4.58) is not an uncertainty in the outcome of ameasurement, but the time scale for change in observable properties of a state.
If this time scale is short, we have a non-stationary state with a large uncertainty in its energy. If this time scale is infinite, we have a stationary state withcertain energy.
In practice, the distinction between stationary and non-stationary states is
subtle. Clearly, the ground state of an atom is a state of definite energy andhence a stationary state with no observable time dependence; accordingly, theelectrons in an undisturbed atom, even though they have kinetic energy, haveno time-dependent properties. At first sight, an atom in an excited state is also
in a state of definite energy; indeed, its wave function is an energy eigenfunctionof the Hamiltonian which describes the interactions between the particles inside
the atom. As such, the atom should be in a stationary state, a state with timelessproperties. But an atom in an excited state changes; it emits electromagnetic
radiation and de-excites. It is, at best, in an almost stationary state and,according to the uncertainty relation (4.58), its energy must have a small
uncertainty.
In fact, an excited state of an atom has uncertain energy because the true
Hamiltonian describes not only the interactions between the particles inside theatom but also an interaction between these particles and fluctuating electromagnetic fields that are always present, even in empty space. These interactions
give rise to an energy uncertainty DE which, in accord with Eq. (4.58), is givenbyh
DE  ,
t(4:59)

where t is the mean lifetime for the excited state to decay. As a consequence, thewavelength of radiation emitted by a decaying atom is uncertain and theobserved spectral line has a natural line width. But, in most situations, naturalline widths are smaller than the widths that arise because atoms move andcollide.
Finally, the reader may be concerned that time and position are treated so
differently in quantum mechanics; time is a label whereas position is an observable which can be measured. To treat them differently is against the spirit of
special relativity which stresses the unity of time and space. Indeed, thisshortcoming in quantum mechanics has to be rectified before the theory canbe effective in describing relativistic phenomena. In relativistic quantum physics neither position nor time are observables; both are labels assigned toquantum field operators.

Problems 4

77

PROBLEMS 4
1. In practice the potential energy function for a particle is a real function. In
this problem you are asked to show that this implies that the energy
eigenvalues of a particle are real.
The energy eigenfunction cn (x) and its complex conjugate cn*(x) satisfy
the equations
"

#
2 d2
h
ÿ
 V (x) cn  En cn ,2m dx2

and
"

#
2 d2
h
ÿ
 V (x) c*n  En*c*:n2m dx2Multiply the first equation by c*n and the second by cn , subtract and show
that


h2 d
dcn
dc*n
ÿ cn
c*
 (En ÿ En*)c*c
ÿn n:
2m dx n dx
dx
By integrating over x and by assuming that cn (x) is zero at x  1, showthat En  En*.
2. In this problem you are asked to show that the eigenfunctions, cm and cn ,belonging to different discrete eigenvalues Em and En , are orthogonal.
For a real potential, the energy eigenfunction cn (x) and the complex
conjugate of cm (x) satisfy the equations
"

#
2 d2
h
ÿ
 V (x) cn  En cn ,2m dx2

and
"

#
2 d2
h
 V (x) cm*  Em cm*:ÿ
2m dx2Multiply the first equation by cm* and the second by cn , subtract and show
that

78

Energy and timeÿChap. 4



h2 d
dc
dc *
cm* n ÿ cn m  (En ÿ Em )cm*cn :
2m dx
dx
dx

By integrating over x and by assuming that cn (x) and cm (x) are zero atx  1, show thatZ 1
ÿ1

cm*(x)cn (x) dx  0 if

Em 6 En :3. What is the energy difference between the lowest and first excited state of aparticle of mass m in a one-dimensional, infinite square-well of width a?
Evaluate this energy:(a) for an electron in a well of atomic size, expressing your answer in eV,(b) for a neutron in a well of nuclear size, expressing your answer in MeV.
4. According to Eq. (4.39), the wave functions for a particle with energy En
confined by an infinite square-well potential are
Cn (x, t)  N sin kn x eÿiEn t=h :Show that these wave functions can be thought of as standing waves formed
by a linear superposition of travelling waves trapped in the region 0 < x < a.
5. Consider a particle of mass m in a two-dimensional box defined by thepotential energy field:
V (x, y) 0
1

if 0 < x < aelsewhere.

and

0<y<b

States of definite energy can be labelled by two quantum numbers, nx and ny ,and they have wave functions of the form
Cnx , ny (x, y, t)  cnx , ny (x, y) eÿiEnx , ny t=h :(a) Find the explicit form of the eigenfunctions cnx , ny (x, y) and eigenvalues
Enx , ny .
(b) Draw diagrams showing the first four energy levels in a box with a  b
and in a box with a  2b. Indicate on the diagrams the degeneracy ofeach level, i.e. the number of independent eigenfunctions that have thesame energy.

Problems 479

6. Two states of a particle with definite energy E1 and E2 are represented by thefollowing normalized, orthogonal solutions of the SchroÈdinger equation:
C1 (x, t)  c1 (x) eÿiE1 t=h

C2 (x, t)  c2 (x) eÿiE2 t=h :and(a) Write down a linear superposition of C1 and C2 which represents thestate for which the expectation value of the energy is 14 E1  34 E2 .
(b) Find the uncertainty in energy for the state written down.
(c) Show, for the state written down, that the probability density oscillateswith time. Find the relation between the period of these oscillations andthe uncertainty in the energy.
7. This problem illustrates the general idea of a complete set of orthonormal
basis functions on the interval 0 < x < a by considering the eigenfunctionsof a particle in a one-dimensional infinite square well of width a. These are
given by Eq. (4.37), i.e. bycn (x)  N sin kn x

for

0 < x < a,

where
kn 

np
a

and

n  1, 2, 3 . . . :(a) Show that the normalization condition
Z a0

jcn (x)j2 dx  1p
is satisfied if N  2=a.
(b) Show that the orthogonality condition
Z a0

cm*cn dx  0 if

m 6 nis satisfied.
(c) Consider the Fourier sine series for the function f(x) on the interval
0<x<af (x) Xn1, 2, 3...

cn cn (x):80Energy and time

Chap. 4Show that the coefficients of this series are given by
2
cn 
a

Z a
0

sin kn x f (x) dx:

8. Consider a particle of mass m in the ground state of an infinite
square-well potential of width a=2. Its normalized wave function at timet  0 is82px
< 2p sinC(x, 0) aa:0

if 0 < x < a=2
elsewhere.

At this time the well suddenly changes to an infinite square-well potentialof width a without affecting the wave function.
By writing C(x, t) as a linear superposition of the energy eigenfunctionsof the new potential, find the probability that a subsequent measurement of
the energy will yield the result
2 p2
h
:
2ma2

E1 

(Hint: A linear superposition of square-well eigenfunctions is a Fourier sineseries, and the coefficients of the series are given by simple integrals.)
9. The general wave function of a particle of mass m in a one-dimensional
infinite square well with width a at time t isC(x, t) 1
Xn1

cn cn (x)eÿiEn t=h ,where cn (x) is an eigenfunction with energy En  n2 p2 h2 =2ma2 . Show
that the wave function returns to its original form in a time T  4ma2 =ph.
10. Consider the electromagnetic radiation of wavelength l which is emitted
when an atom makes a transition from a state with energy E2 to aground state with energy E1 . Assume that the mean lifetime of thestate with energy E2 is t. Show that the uncertainty in the wavelength ofthe emitted radiation, i.e. the natural line width of the spectral line, is given
by
Dl 

l2
:
2pct

Problems 481

11. In particle physics the Z boson is an unstable gauge boson which plays a
key role in mediating the Weak Nuclear Interaction. The fundamental
uncertainty in the mass energy of the Z boson is DE  2:5 GeV. Evaluate
the mean decay lifetime of the Z boson.

This page intentionally left blank5
Square wells and barriers
Insight into how quantum particles can be bound or scattered by potential
energy fields can be obtained by considering models based on square wells andsquare barriers. In these models, the SchroÈdinger equation may be solved easily
using elementary mathematics, the possible energies of a particle may be foundand the properties of the wave functions are self-evident.
We begin by considering the quantum states of a particle in a one-dimensionalsquare-well potential. We shall show that there are unbound states with acontinuous range of energies and that there are, when the well is deep enough,bound states with discrete energies.
We shall then consider a particle incident on a square potential barrier. We
shall see that this is an uncertain encounter with two possible outcomes:
reflection and transmission. Most importantly, we shall show that transmission
is possible even when the particle has insufficient energy to surmount thebarrier. In other words, we shall illustrate how quantum particles can tunnel
through potential barriers.

5.1

BOUND AND UNBOUND STATESIn order to explore the properties of bound and unbound quantum states in a
simple context, we shall consider a particle of mass m in the one-dimensional
potential energy field given by(
V (x) 1
ÿV0
0

if ÿ1 < x < 0
if 0 < x < a
if a < x < 1.

(5:1)

As illustrated in Fig. 5.1, the potential energy changes abruptly at x  0 andx  a. There is an attractive well of depth V0 , which may or may not trap theparticle, and an infinite wall at x  0 which repels the particle.

84

Square wells and barriers

Chap. 5V(x) = −V0

V(x) = + `x=0V(x) = 0x= aFig. 5.1 The potential energy field given by Eq. (5.1) in which there are unbound stateswith a continuous range of energies and, if the well is deep enough, bound states withdiscrete energies.

The behaviour of a classical particle in this potential should be familiar. Theenergy of the particle, E, is given by the sum of its kinetic and potential energies,Ep2
 V (x):2m

When the energy is negative and somewhere between E  ÿV0 and E  0, theparticle is bound or trapped in the well of depth V0 ; it bounces back and forth
between x  0 and x  a with kinetic energy E  V0 . But when the energy ispositive, the particle is unbound. For example, it could approach the well fromx  1 with kinetic energy E, increase its kinetic energy to E  V0 when it
reaches x  a, hit the infinitely-high potential wall at x  0 and then bounceback to x  1.
The behaviour of a quantum particle in this potential is described by a wave
function C(x, t) which is a solution of the SchroÈdinger equationih]Ch2 ]2 C
 V (x)C:ÿ
2m ]x2]t(5:2)

When the particle has definite energy E, the wave function has the form
C(x, t)  c(x) eÿiEt=h ,(5:3)

where c(x) is an eigenfunction satisfying the energy eigenvalue equationÿh2 d2 c
 V (x)c  Ec:
2m dx2

(5:4)

5.1

85

Bound and Unbound States

Once we have solved this eigenvalue equation and found all the possible energy
eigenvalues and eigenfunctions, we can represent any quantum state of theparticle in the potential as a linear superposition of energy eigenfunctions.
To solve the eigenvalue equation, we note that the potential V(x) given by
Eq. (5.1) takes on constant values in three regions of x: ( ÿ 1 < x < 0),(0 < x < a) and (a < x < 1). We shall find solutions of Eq. (5.4) in thesethree regions and then join the solutions together at x  0 and at x  a toobtain physically acceptable eigenfunctions. Because the potential V(x)changes abruptly at x  0 and x  a, we can only require the eigenfunctionsc(x) to be as smooth as possible. In particular, we shall require c(x) to becontinuous at x  0 and x  a in order to avoid unacceptable abrupt changes
in the position probability density. The differential equation (5.4) with potential (5.1) then implies that the first derivative of c(x) is continuous at x  a anddiscontinuous at x  0.1
We are interested in two types of eigenfunctions. The eigenfunctions forbound states and the eigenfunctions for unbound states.

Bound states
If a bound state exists, it has a negative energy somewhere between E  ÿV0and E  0. We shall set E  ÿE, where E is the binding energy, and seek
solutions of Eq. (5.4).
In the region ( ÿ 1 < x < 0), the potential energy is infinite and the only
finite solution of Eq. (5.4) is c(x)  0, signifying that the particle is never foundin the negative x region.
In the region (0 < x < a), the potential energy is V (x)  ÿV0 and Eq. (5.4)
has the formd2 c
 ÿk20 c,
dx2

where

E

2 k20hÿ V0 :
2m(5:5)

The general solution of this second-order differential equation has the formc(x)  C sin (k0 x  g),
where C and g are arbitrary constants. To ensure continuity of c(x) at x  0, weshall set the constant g to zero to givec(x)  C sin k0 x:
1

(5:6)

The infinite change in the potential at x  0 forces a discontinuity in dc=dx at x  0. A more
rigorous approach would be to consider a potential energy with a finite value V1 in the regionx  0, require the continuity of c(x) and dc=dx at x  0 and then take the limit V1 ! 1.

86

Square wells and barriers

Chap. 5In the region (a < x < 1), the potential energy is zero and Eq. (5.4) has the
form
d2 c
 a2 c,
dx2

where

Eÿ

2 a2
h
:
2m

(5:7)

The general solution isc(x)  A eÿax  A0 eax ,where A and A0 are arbitrary constants. To ensure that the eigenfunction is finite
at infinity, we set A0 to zero to give a solution which falls off exponentially with x:c(x)  A eÿax :(5:8)

Our next task is to join the solution given by Eq. (5.6), which is valid in
the region (0 < x < a), onto the solution given by Eq. (5.8), which is valid in theregion (a < x < 1). As mentioned earlier, we shall require the eigenfunction
and its first derivative to be continuous at x  a. Continuity of c(x) gives
C sin k0 a  A eÿaa ,(5:9)

and continuity of dc=dx gives
k0 C cos k0 a  ÿaA eÿaa :(5:10)

If we divide Eq. (5.10) by Eq. (5.9), we obtaink0 cot k0 a  ÿa:(5:11)

Equation (5.11) sets the condition for a smooth join at x  a of the functions
C sin k0 x and A eÿax . It is a non-trivial condition which is only satisfied when
the parameters k0 and a take on special values. And once we have found these
special values, we will be able to find the binding energies of the bound statesfrom E  h2 a2 =2m.
To find these binding energies, we note that a and k0 are not independent
parameters. They are defined byEÿh2 a22m

andE

h2 k20
ÿ V0 ,
2m

which imply thata2  k20  w2 ,where w is given by

V0 

h2 w2:2m(5:12)

5.1

87Bound and Unbound StatesThus, we have two simultaneous equations for a and k0 , Eq. (5.11) andEq. (5.12). These equations may be solved graphically by finding the points
of intersection of the curvesa  ÿk0 cot k0 aanda2  k20  w2 ,as illustrated in Fig. 5.2.
Inspection of Fig. 5.2 shows the number of points of intersection, and hence
the number of bound states, increase as the well becomes deeper. In particular,
there are no bound states for a shallow well with
3
(C)

2.5

α in units of π/a

2
(B)

1.5

1
(A)

0.5

0

0

0.5

1

1.5
k0 in units of π/a

2

2.5

3

Fig. 5.2 Graphical solution of the simultaneous equations a  ÿk0 cot k0 a anda2  k20  w2 . The units of k0 and a are p=a. Three values for the well-depth parameter,w  p=a, w  2p=a and w  3p=a, are labelled by (A), (B) and (C), respectively. For (A)there is one point of intersection and one bound state, for (B) there are two points ofintersection and two bound states and for (C) there are three points of intersection andthree bound states.

88

Square wells and barriers

Chap. 5

w<

p
:
2a

There is one bound state when
p
3p
<w< ,
2a
2a
and two bound states when
3p
5p
<w<
,
2a
2a
and so on.
To illustrate the nature of the bound states, we shall consider a potential withwell-depth parameter w  2p=a which corresponds to a well with depth
V0 2h2 p 2:
ma2

In this case two bound states exist, a ground state and a first excited state withbinding energies
E1  3:26

2 p2
h
2ma2

and

E2  1:17h2 p2:2ma2

The corresponding eigenfunctions are shown in Fig. 5.3. These eigenfunctions
show that a bound quantum particle can be found outside the classical regionof confinement (0 < x < a). Specifically, for x > 0, bound-state eigenfunctions
have non-zero values given byc(x)  A eÿax :Hence the position probability densities for a bound particle fall off exponentially as x penetrates into the classically forbidden region. Because the parameter a is related to the binding energy E via E  
h2 a2 =2m, the degree of quantum
penetration into the classically forbidden region is more pronounced when thebinding energy is low. The phenomenon of quantum penetration will be considered further in Section 5.2.

Unbound states
We shall now consider a particle with positive energy E that approaches thewell, shown in Fig. 5.1, from the right and is then reflected at x  0. It is usefulto write the energy of the particle aseigenfunction

5.1

eigenfunction

89

The ground state

1

0

−1

0

0.5

1

1.5

2

2.5

3

The first excited state

1
0
−1

0

eigenfunction

Bound and Unbound States

0.5

1

1.5

2

2.5

3

An unbound state

1
0
−1

0

0.5

1

1.5
position x

2

2.5

3

Fig. 5.3 The energy eigenfunctions for a particle in a square-well potential with widtha  1 and depth V0  2h2 p2 =ma2 . The eigenfunctions for the ground state, the firstexcited state, and an unbound state with energy E  2h2 p2 =ma2 are shown. The normalization of the eigenfunction of the unbound state is arbitrary.

Eh2 k20
h2 k2
ÿ V0 
2m
2m(5:13)and note that, if the particle were governed by classical physics, its momentum
would be hk0 inside the well and hk outside the well. But the particle is really
governed by quantum mechanics. It has a wave function of the form
C(x, t)  c(x) eÿiEt=hwhere the eigenfunction c(x) satisfies the eigenvalue equation (5.4) for all
values of x. We shall find these eigenfunctions by following the procedure weused to find the bound-state eigenfunctions: we shall find solutions of Eq. (5.4)
in the three regions of x and then join these solutions at x  0 and x  a.
First, the eigenfunction c(x) is zero in the region ( ÿ 1 < x < 0) because thepotential energy is infinite in this region.

90

Square wells and barriers

Chap. 5

Second, in the region (0 < x < a) where the potential energy is ÿV0 , theeigenvalue equation for the unbound state has the same form as the eigenvalue
equation for a bound state, Eq. (5.5), and the solution is given by Eq. (5.6), i.e.
byc(x)  C sin k0 x:(5:14)

Third, in the region (a < x < 1) where the potential energy is zero, theeigenvalue equation has the formd2 c
 ÿk2 c,dx2

(5:15)

c(x)  D sin (kx  d):(5:16)and the solution is

The constant d is called the phase shift.2
Our next task is to join the solution given by Eq. (5.14) onto the solution
given by Eq. (5.16). As mentioned earlier, we shall require the eigenfunction
and its first derivative to be continuous at x  a. Continuity of c(x) gives
C sin k0 a  D sin (ka  d)

(5:17)and continuity of dc=dx gives
k0 C cos k0 a  kD cos (ka  d):(5:18)

If we divide Eq. (5.18) by Eq. (5.17), we obtaink0 cot k0 a  k cot (ka  d):(5:19)

This equation sets the condition for a smooth join of the functions C sin (k0 x)
and D sin (kx  d) at x  a. When we considered the analogous equation for
the eigenfunctions of the bound states, Eq. (5.11), we found that the join was
only smooth when the energy takes on special discrete values. This is not thecase for unbound states. Here we can choose any value of the energy E, find k
2

We recall that the bound-state eigenfunction, Eq. (5.8), decreases exponentially at large x
signifying that a bound particle does not escape to infinity. In contrast, an unbound particle
can escape to infinity and its eigenfunction has a finite value at large distances given by Eq. (5.16).
Readers may worry about whether an eigenfunction for an unbound particle can be normalized.
This can be done by the mathematical trick of placing the entire system, the potential and theparticle, in a big box; to some extent, this box represents the laboratory containing the system. In
practice, the normalization of unbound quantum states is seldom considered. These states areusually characterized by a particle flux or by a probability current density.

5.1

Bound and Unbound States91

and k0 from Eq. (5.13) and use Eq. (5.19) to find the phase shift d. Thus, forany positive energy E, we can always find a smooth eigenfunction of the
form:
8
<0
c(x)  C sin (k0 x):
D sin (kx  d)if ÿ1 < x < 0
if 0 < x < a
if a < x < 1.

(5:20)

We note that there are undulations with wave numbers k0 and k in regions
where a classical particle would have momenta 
hk0 and hk. These undulations
are illustrated in Fig. 5.3 which shows the eigenfunction for an unboundparticle with energy E  2h2 p2 =ma2 in a potential well of depth
2 2
2V0  2h p =ma .
The eigenfunction Eq. (5.20) may be rewritten to reveal a wave being reflected by the potential well. To identify incident and reflected components, werewrite the eigenfunction in terms of complex exponentials. If we use
sin y eiy ÿ eÿiy2i

and if we introduce two new constants
A0  ÿ

C
2i

and A  ÿD eÿ2id,2i

we obtain(c(x) 

A0 eÿik0 x ÿ A0 eik0 x
Ae

ÿikx

ÿAe

2id ikxe

if 0 < x < a
if a < x < 1.

(5:21)

We can now identify a wave travelling back and forth with wave number k0
inside the well, and, outside the well, we can identify incoming and outgoingwaves with wave number k but with a phase difference of 2d. These waves
represent a particle with momentum 
hk being reflected by the potential well
between x  0 and x  a and by the barrier at x  0. Because the incoming andreflected waves have the same intensity jAj2 , the reflection is complete. The
meaning of the phase shift will become evident in a moment.
Even though we have used words like travelling back and forth, incoming
and outgoing, the eigenfunction given by Eq. (5.21) describes a quantum
state which exhibits no observable time dependence because it is a state withdefinite energy E. To describe the dynamics of particle reflection properly,
we need a non-stationary quantum state of uncertain energy. Such a state canbe represented in the region (a < x < 1) by an incoming wave packet of the
form

92

Square wells and barriers

Ci (x, t) Z 1
0

Chap. 5000c(E 0 ) eÿi(E tp x)=h dE 0 ,where E 0 

p2
,
2m

and an outgoing wave packet of the form
Cf (x, t) 

Z 1
0

0

0

0

c(E 0 ) eÿi(E tÿp x)=h e2id(E ) dE 0 :The function c(E 0 ) is an energy probability amplitude and jc(E 0 )j2 dE 0 is theprobability that the particle has energy between E 0 and E 0  dE 0 . If the function
jc(E 0 )j2 is strongly peaked at E 0  E, the wave packets representp an incoming
and outgoing particle with energy E, momentum p  2mE and velocityu  p=m. In this case, one can show that
Ci (x, t) / F (t  x=u)

(5:22)Cf (x, t) / F (t ÿ x=u ÿ 2
hdd=dE),(5:23)

andwhere the function F specifies the shape of the incoming and outgoing wave
packets. These equations show that the energy dependence of the phase shift,d(E), is related to the time delay that occurs during the reflection process; this
delay is approximately given byt  2hdd:
dE(5:24)

This time delay is negative for a particle reflected by the potential shown in
Fig. 5.1 because the particle speeds up as it enters the well at x  a.
For the benefit of more advanced readers, we shall give a brief derivation of theformula for the time delay. If c(E 0 ) is strongly peaked at E 0  E, we can set
p0  p  (E 0 ÿ E)

dp
dE

and
d(E 0 )  d(E)  (E 0 ÿ E)

dd
,
dE

and substitute into Eq. (5.22) and Eq. (5.23). If we note that dp=dE is equal to1=u, where u is the classical velocity of the particle, we obtain5.1

Ci (x, t)  eÿi(Etpx)=h

Z 10

Bound and Unbound States930c(E 0 ) eÿi(E ÿE)(tx=u)=h dE 0

and
Cf (x, t)  eÿi(Etÿpxÿ2hd)=h

Z 1
0

0

c(E 0 ) eÿi(E ÿE)(tÿx=uÿ2hdd=dE)=h dE 0 :We now focus on the x and t dependence that emerges when the integration
over E 0 is carried out. It should be clear that, for a given function c(E 0 ),integration will lead to wave packets of a specific shape, which we denote bya function F, with locations given by
Ci (x, t) / F (t  x=u)
and
Cf (x, t) / F (t ÿ x=u ÿ 2
hdd=dE):Thus, we have an incoming wave packet with velocity u and, after a time delayof 2hdd=dE, an outgoing wave packet with velocity u. In this approximate
calculation the shapes of the incoming and outgoing wave packets are thesame, but in practice wave packets change in shape as they move.

General implications
This section has considered a particle in the potential energy field given in
Fig. 5.1. We have chosen this simple potential because it permits the solution of
the SchroÈdinger equation using elementary methods and because it leads towave functions which are easy to visualize. Clearly, the detail of the results are
only relevant to this particular potential, but the following general features are
relevant in atomic, nuclear and particle physics:. Wave functions undulate in classically allowed regions and fall off exponentially in classically forbidden regions.
. Potentials give rise to bound states with discrete energies when they are
sufficiently attractive.
. When a particle is unbound it can have a continuous range of energies, andwhen it is scattered or reflected by a potential, its wave function experiences aphase shift which can be related to a time delay that occurs during thescattering process.

94

Square wells and barriers

5.2

BARRIER PENETRATION

Chap. 5The ability to penetrate and tunnel through a classically forbidden region is oneof the most important properties of a quantum particle. In the last section, we
discovered that the wave function of a bound particle extends beyond theregion of confinement of a bound classical particle. In this section we shallshow how particles can tunnel through potential barriers. To keep the mathematics as simple as possible we shall consider a particle in a simple potential
energy field of the form:(
V (x) 0
VB0

if ÿ1 < x < 0
if 0 < x < a
if a < x < 1.

(5:25)As shown in Fig. 5.4, we have a square barrier of height VB separating theregions ( ÿ 1 < x < 0) and (a < x < 1).
If a classical particle were to approach this barrier from the left, it would bereflected if its energy is below VB and it would be transmitted if its energy is
above VB . We shall see that when a quantum particle encounters the barrier,the outcome is uncertain; it may be reflected or it may be transmitted. Most
importantly, we shall show that the particle may be transmitted even when its
energy is below VB , and we shall calculate the probability for this to happen.
The behaviour of a particle of mass m in the potential V (x) is described by awave function C(x, t) which is a solution of the SchroÈdinger equationih]Ch2 ] 2 C V (x)C:ÿ
2m ]x2]t

To describe the dynamics of the uncertain encounter with the barrier, we seek awave function C(x, t) which describes an incoming particle and the possibilityof reflection and transmission. This wave function should give rise to an
incoming pulse of probability representing a particle approaching the barrierV(x) = 0

V(x) = VB

x= 0

V(x) = 0x= aFig. 5.4 The potential barrier given by Eq. (5.25) which will be used to illustratequantum mechanical tunnelling.

5.2

Barrier Penetration95

before the encounter. After the encounter, there should be two pulses ofprobability, one representing the possibility of a reflected particle and theother the possibility of a transmitted particle. According to the standard
interpretation of the wave function, reflection and transmission persist as
possible options until the particle is detected. When this happens, the wave
function C(x, t) collapses and one or other of the two options are realized withprobabilities governed by the magnitudes of the reflected and transmitted
pulses.
This dynamical description suggests that we need to solve a problem intime-dependent quantum mechanics. The quantum state that describes anincoming particle and the possibilities of a reflected and transmitted particle
is a non-stationary state. Such a state, as discussed in Section 4.7, is a state ofuncertain energy represented by a linear superposition of energy eigenfunctions. In this case, there is a continuum of possible energies and the wave
function has the form
Z
C(x, t) 0c(E 0 )cE 0 (x) eÿiE t=h dE 0 ,where cE 0 (x) is an eigenfunction with energy E 0 with the appropriate boundary
conditions. The function c(E 0 ) is an energy probability amplitude andjc(E 0 )j2 dE 0 is the probability that the particle has energy between E 0 andE 0  dE 0 . If the function jc(E 0 )j2 is strongly peaked at E 0  E, the wave function C(x, t) consists of a localized wave packet representing a particle of energy
E encountering the barrier.
Even though time-dependent quantum mechanics is essential for a conceptual understanding of an uncertain encounter with a potential barrier, it is notneeded to calculate the probabilities of reflection and transmission. We shallnow show how these probabilities may be found simply by using time-independent quantum mechanics.

Stationary state analysis of reflection and transmission
Provided the uncertainty in the energy of the particle is small compared withthe variations in the potential energy V (x), we can calculate the probabilities ofreflection and transmission by considering a stationary state with definite
energy. Such a state is represented by the wave function
C(x, t)  cE (x) eÿiEt=h ,(5:26)

where cE (x) is an eigenfunction with energy E satisfying the eigenvalue
equation

96

Square wells and barriers

Chap. 5

2 d2 cEh
 V (x)cE  EcE :2m dx2

ÿ(5:27)

Because of the simple nature of the potential V (x) shown in Fig. 5.4, it is easy tofind an eigenfunction which describes incoming, reflected and transmitted
waves. The procedure is to find the solutions of Eq. (5.27) in different regionsof x and then smoothly join the solutions together.3
On the left of the barrier, the potential energy V (x) is zero and the eigenfunction cE (x) satisfies the differential equationd2 cE
 ÿk2 cE ,
dx2

where

E

h2 k2:2m(5:28)The solution representing an incident wave of intensity jAI j2 and a reflectedwave of intensity jAR j2 iscE (x)  AI eikx  AR eÿikx :(5:29)

The form of the eigenfunction inside the barrier depends on whether the energyof the particle is above or below the barrier. When E > VB , the region(0 < x < a) is a classically allowed region and the eigenfunction is governed byd2 cE
 ÿk2B cE ,
dx2

where

E

2 k2Bh VB :
2m(5:30)The general solution involves two arbitrary constants and it undulates withwave number kB as follows
cE (x)  A eikB x  A0 eÿikB x :(5:31)

When E < VB , the region (0 < x < a) is a classically forbidden region. Here,the eigenfunction is governed byd2 cE
 b2 cE ,
dx2

where

Eÿ

2 b2
h
 VB ,
2m(5:32)and the general solution iscE (x)  B eÿbx  B0 ebx ,3(5:33)

A similar procedure is used to describe the reflection and transmission of classical waves. Indeed,
the mathematics describing an electromagnetic wave incident on a thin sheet of metal is almost
identical.

5.2

Barrier Penetration97

where B and B0 are arbitrary constants.
The potential energy is again equal to zero on the right of the barrier. Here,the eigenfunction satisfies Eq. (5.28) and the solution representing a transmitted wave of intensity jAT j2 iscE (x)  AT eikx :(5:34)

If we smoothly join these solutions together at x  0 and at x  a, we canfind the intensities of the reflected and transmitted waves. In particular, we canderive expressions for the ratiosRjAR j2jAI j2andT

jAT j2jAI j2:(5:35)

Because the probability of finding a particle at x is proportional to jcE (x)j2 ,these ratios are probabilities: R is the probability that the particle is reflected
and T is the probability that the particle is transmitted, and the sum of theseprobabilities is equal to one, i.e.
R  T  1:

(5:36)

This interpretation of R and T is confirmed in problem 5 at the end of thischapter.
We shall confine our attention to a particle with an energy below the barrier
and derive an approximate expression for the transmission probability which isuseful when the barrier is wide. In other words, we will find the probability oftunnelling through a wide barrier.

Tunnelling through wide barriers
The eigenfunction of a particle with energy below the barrier is given by
8< AI eikx  AR eÿikxcE (x)  B eÿbx  B0 ebx:
AT eikx

if ÿ1 < x < 0
if 0 < x < a
if a < x < 1,

(5:37)

with constants AI , AR , AT , B and B0 having values which ensure that cE
and dcE =dx are continuous at x  0 and at x  a. Continuity at x  0
requiresAI  AR  B  B0
and continuity at x  a requires

andikAI ÿ ikAR  ÿbB  bB0 ,98Square wells and barriers

B eÿba  B0 eba  AT eika

Chap. 5and

ÿ bB eÿba  bB0 eba  ikAT eika :Our objective is to find an expression for the tunnelling probability. To do so
we shall express the constants AT and AI in terms of B. First, we use continuity
at x  0 to give
2ikAI  ÿ(b ÿ ik)B  (b  ik)B0 :(5:38)

Second, we use continuity at x  a to give
AT eika 

2b
B eÿba(b ÿ ik)and

B0  B eÿ2ba(b  ik):(b ÿ ik)(5:39)

It is now straightforward to express both AI and AT in terms of B and obtainthe tunnelling probability T  jAT j2 =jAI j2 .
To avoid algebraic tedium, we shall simplify matters by assuming thatthe particle encounters a wide barrier with exp ( ÿ 2ba) << 1. When this is thecase, the constant B0 is much smaller than B and we can approximate Eq. (5.38) by
2ikAI  ÿ(b ÿ ik)B:If we combine this equation with Eq. (5.39), we obtain
AT eika  ÿ

4ikb eÿba
AI :(b ÿ ik)2

This implies that the tunnelling probability is approximately given by"#
16k2 b2Teÿ2ba :
(b2  k2 )2
If we rewrite this equation usingkp
2mE
h


and

b

p
2m(VB ÿ E),h

we obtainT


16E(VB ÿ E) ÿ2bae:
VB2

(5:40)

This approximate expression for the tunnelling probability is valid wheneÿ2ba << 1. When this is the case, the wave function inside the barrier is5.2

Barrier Penetration

99

dominated by the exponentially decaying term B eÿbx and the probability oftunnelling a barrier of width a is proportional to eÿ2ba . In fact, this exponential
dependence on the penetration parameter b and on the width of the barrier a isthe most important, and the most useful, feature of the tunnelling formulaEq. (5.40).
We shall illustrate the importance of barrier penetration by considering two
processes, one involving tunnelling electrons and the other tunnelling protons.

Tunnelling electrons
We know from the photoelectric effect that the minimum energy needed to eject
an electron from the surface of a metal is of the order of a few electron volts.
This energy is needed because electrons in a metal reside in an attractive
potential energy field which increases at the surface of the metal to give apotential step which is a few electron volts above the energy of the mostenergetic electrons in the metal.
When two metal surfaces are placed in close proximity, there are two regions of
low potential energy separated by a potential barrier which is similar to thatshown in Fig. 5.4. But this barrier does not prevent electrons from moving acrossthe gap between the surfaces. Electrons are quantum particles that can tunnel
through the barrier with a probability given by Eq. (5.40), or approximately by
ÿ2ba

T e

with

p
2me (VB ÿ E)
:
b
h


(5:41)In this expression a is the width of the gap between the metal surfaces, me is themass of the electron, VB is the barrier height at the surface of the metal and E isthe energy of the most energetic electrons in the metal.
Equation (5.41) implies that the tunnelling probability is a very sensitive
function of the gap. Indeed, if the gap a changes by a small amount Da, thefractional change in the tunnelling probability is given byDT ÿ2b Da:T

(5:42)

To illustrate the numerical significance of this formula, we shall consider atypical situation in which 4 eV are needed to eject an electron from the metal.
This implies that the height of the barrier VB is about 4 eV higher than theelectron energy E and that the penetration parameter is given byp
2me (VB ÿ E)
 1010 mÿ1 :b
h(5:43)

100

Square wells and barriers

Chap. 5

By substituting this value for b into Eq. (5.42) we can illustrate the incredible
sensitivity of the electron tunnelling probability. For example, there is a measurable 2 % change in the probability when the gap between the surfaces changes
by a mere 0.001 nm!
The extreme sensitivity of electron tunnelling is exploited in a device called
the scanning tunnelling microscope. In this device a sharp metal probe is positioned near to a surface under investigation. The separation is made smallenough to induce the tunnelling of electrons between the probe and the surface,and a potential difference between the probe and the surface is also establishedso that there is a net current of electrons in one direction. As the probe is moved
or scanned across the surface, surface features of atomic dimensions will giverise to measurable changes in the current of tunnelling electrons. In this way thescanning tunnelling microscope can produce a map of the locations of individual atoms on the surface.

Tunnelling protons
The centre of the sun consists of an ionized gas of electrons, protons and lightatomic nuclei at a temperature T of about 107 K. The protons and other lightnuclei collide frequently, occasionally get close and occasionally fuse to releaseenergy which is ultimately radiated from the solar surface as sunshine.
To understand the issues involved in the generation of solar thermonuclear
energy, we shall focus on two protons approaching each other near the centreof the sun. They move in the ionized gas with thermal kinetic energies of theorder ofE  kT  1 keV:
The mutual potential energy of the two protons depends on their separation. As
illustrated in Fig. 5.5, the potential energy at large separation r is dominated by
the repulsive Coulomb potential
V (r) e2:
4pE0 r

But at small separations, when r becomes comparable with the range of nuclear
forces given by rN  2  10ÿ15 m, the potential energy becomes attractive. The
net effect is a Coulomb barrier which rises to a height of about 1 MeV at a
separation of about 2  10ÿ15 m or 2 fm.
Thus, when protons approach each other near the centre of the sun, they do
so with energies of the order of keV and they encounter a Coulomb barrier
measured in MeV. According to classical physics, there is a well-defined distance of closest approach rC , which is given by5.2 Barrier Penetration

101

Coulomb barrier e2/4p 0rV(r)Energy of approach E

rN

rC

r

Fig. 5.5 A schematic representation of the potential energy V(r) of two protons withseparation r. When the separation is less than rN  2  10ÿ15 m, there is a strong nuclear
attraction and the protons may fuse to form a deuteron. Nuclear fusion is hindered by a
Coulomb barrier which is approximately 1 MeV high. The distance rC is the classicaldistance of closest approach for protons with an energy of approach equal to E.

E

e2
:
4pE0 rC(5:44)

Because this distance is three orders of magnitude larger that the range of nuclear
forces rN , the prospect of a close encounter and the possibility of nuclear fusion
look dim. Indeed, at first sight, the sun is not hot enough to shine.
In fact, thermonuclear fusion in the sun, and in other stars, is only possible
because protons are quantum particles that can tunnel through Coulomb
barriers. Two protons with an energy of approach E are described by aneigenfunction c(r) which obeys the time-independent SchroÈdinger equation,
"

#
h2 2
ÿ r  V (r) c(r)  Ec(r),2m(5:45)

where V(r) is the potential shown in Fig. 5.5 and m is equal to mp =2, the reduced
mass of two protons.4 For two protons with low energy, the relevant eigenfunction has no angular dependence and it has the form
4

The reduced mass mp =2 is used because the kinetic energy of two protons approaching each otherwith equal and opposite momentum p is given by
E

p2
p2
p2


:
2mp 2mp mp102

Square wells and barriers

Chap. 5c(r) 

u(r),r(5:46)

where u(r), as shown in problem 2 at the end of the chapter, obeysÿ 2 d2 uh V (r)u  E:
2m dr2(5:47)We shall not attempt to solve this equation. Instead, we shall use the results weobtained for a one-dimensional barrier to write down a plausible form of theeigenfunction and then estimate the probability of tunnelling through a Coulomb barrier.
We begin by considering a three-dimensional barrier with constant height VB
and width rC ÿ rN . In this case, the function u(r) decays exponentially in theclassically forbidden region as r gets smaller and it is given byu(r) / ebr ,where b is given by
Eÿ

2 b2
h
 VB :
2m

The probability that the protons tunnel from r  rC to r  rN is approximately
equal to the ratio of ju(rN )j2 to ju(rC )j2 , and it is given byT  j exp [ ÿ b(rC ÿ rN )] j2 :(5:48)

We now consider the Coulomb barrier shown in Fig. 5.5. In this case, theeigenfunction in the classically forbidden region (rN < r < rC ) is again approximately given byu(r) / ebr ,but b now depends on r because it is given byEÿh2 b2e2
:2m
4pE0 r

The probability that the two protons tunnel from r  rC to r  rN is now
approximately given by a generalization of Eq. (5.48)

Problems 5

 Z rC2
T  exp ÿb dr :
rN103(5:49)

If we assume that rC >> rN and evaluate the integral in Eq. (5.49) by substituting r  rC cos2 y, we find that"   #
EG 1=2T  exp ÿE(5:50)

where E is the relative energy of the protons and EG is defined by

EG 

e2
4pE0 hc

2

2p2 mc2 :(5:51)The energy EG is called the Gamow energy and its value is 493 keV.
We can now estimate the probability that two protons tunnel through the
Coulomb barrier which normally keeps them well apart when they collide near
the centre of the sun. By substituting a typical thermal energy of E  1 keV into
Eq. (5.50), we obtainT  exp [ ÿ 22]  3  10ÿ10 :Thus, with a probability of about one in 3 billion, protons colliding near thecentre of the sun tunnel through the Coulomb barrier. And when they do sothey have a chance of fusing and releasing thermonuclear energy.
In practice, stars evolve slowly by adjusting their temperature so that the
average thermal energy of nuclei is well below the Coulomb barrier. Fusion
then proceeds at a rate proportional to the tunnelling probability. Because thisprobability is very low, fusion proceeds at a slow pace and the nuclear fuel lasts
for an astronomically long time scale.

PROBLEMS 5
1. Consider a particle of mass m in the one-dimensional potential energy field(
V (x) 0
ÿV00

if ÿ1 < x < ÿa
if ÿa < x < a
if a < x < 1.

Because the potential is symmetric about x  0, there are two types ofenergy eigenfunctions. There are symmetric eigenfunctions which obey

104

Square wells and barriers

Chap. 5c(x)  c( ÿ x)
and antisymmetric eigenfunctions which obeyc(x)  ÿc( ÿ x):
Symmetric eigenfunctions are said to have positive parity and antisymmetriceigenfunctions are said to have negative parity.
(a) Show, by considering the energy eigenvalue equation in the three regionsof x, that a positive parity eigenfunction with energy E  ÿh2 a2 =2m has
the form:
8< A eaxc(x)  C cos k0 x
: ÿax
Ae

if ÿ1 < x < ÿa
if ÿa < x < a
if a < x < 1,q
where A and C are constants and k0  2m(E  V0 )=h2 .
(b) Show that the continuity of c(x) and dc=dx at the edges of the potential
well implies thata  k0 tan k0 a:(c) By seeking a graphical solution of the equationsa  k0 tan k0 a and a2  k20  w2 ,q
where w  2mV0 =
h2 , show that there is one bound state if
0<w<

p
,
2a

two bound states if
p
3p
<w< ,
2a
2a
and so on.
(d) Now consider a negative-parity eigenfunction and confirm that thebound state energies are identical to those given by the potential illustrated in Fig. 5.1.

Problems 5

105

2. A particle of mass m moves in three dimensions in a potential energyfieldV (r) ÿV0

if r < R0

if r > R,

where r is the distance from the origin. Its energy eigenfunctions c(r) aregoverned byÿh2 2
r c  V (r)c  Ec,
2m

where, in spherical polar coordinates,
1 ]2 (rc) 1 ]2 c cos y ]c
1 ]2 c
:
 2


r c
r ]r2
r ]y2 sin y ]y sin2 y ]f22

Consider spherically symmetric eigenfunctions with no angular dependenceof the formc(r) u(r):r(a) Show thatÿ

2 d 2 uh V (r)u  E:
2m dr2(b) By solving for u(r) in the regions r < R and r > R and by imposing
appropriate boundary conditions, show that an eigenfunction of abound state with energy E  ÿ
h2 a2 =2m has the formc(r) 

C sin k0 r=r
Aeÿar

if r < R=r

if r > R,q
where C and A are constants and k0  2m(V0  E)=h2 .
(c) Show that there is one bound state of this kind if the depth of the wellobeys
2 p 2
h
9
h2 p2
<
V
<
:
0
8mR2
8mR2

106

Square wells and barriers

Chap. 5(Many of the steps in this question are almost identical to thosecarried out in Section 5.1 for the one-dimensional potential given in
Fig 5.1.)
3. A particle with energy E  
h2 p2 =m is scattered by the potential well shownin Fig. 5.1 with depth V0  2h2 p2 =ma2 . Use Eq. (5.19) to show that the
eigenfunction has a phase shift, in radians, which can be taken asd(E)  3:53:[Note that any integer multiple of p can be added to the phase shift satisfying
Eq. (5.19).]
4. (a) A particle in a potential V (x) has definite energy
Eÿ

2 a2
h
2m

and an eigenfunctionc(x) 

Nx eÿax

if 0  x < 10elsewhere,

where N and a are positive real constants. Verify that the potential isgiven by
V (x) ÿa
h2 =mx

if 0  x < 1

1

elsewhere.

(b) Given that the eigenfunction c(x) given in part (a) describes the groundstate of the particle in the potential V (x), roughly sketch the eigenfunction which describes the first excited state.
5. In problem 8 at the end of Chapter 3, we showed that the function j(x, t),defined by


i
h
]C
]C*
C*
ÿC
,
j(x, t) 2m]x]x
can be interpreted as a current of probability becausedj(x2 , t) ÿ j(x1 , t) dt

 Z x2x1

2

jC(x, t)j dx :Problems 5

107Consider a particle with a stationary state wave function
C(x, t)  c(x) eÿiEt=hincident on the barrier illustrated in Fig. 5.4. Assume that on the left of thebarriercE (x)  AI eikx  AR eÿikx
and on the right of the barriercE (x)  AT eikx :(a) Show that on the leftj(x, t)  jAI j2

k
h
hk

ÿ jAR j2mm

and that on the rightj(x, t)  jAT j2

k
h
:
m

(b) By noting that the position probability density jC(x, t)j2 is constant fora stationary state, show thatjAI j2

k
h
hk

hk

 jAR j2  jAT j2 :mmm

(This implies that the incident probability current equals the sum of thereflected and transmitted probability currents, and that the reflection
and transmitted probabilities satisfy the relation
R  T  1:
As expected, the sum of the probabilities of the only possible twooutcomes of the encounter, reflection and transmission, is unity.)
6. Consider a particle with an energy E above the square potential barrier
illustrated in Fig. 5.4. As in Section 5.2, define wave numbers kB and k byE

2 k2Bhh2 k2
 VB 
2m
2m

108

Square wells and barriers

Chap. 5and show that the transmission probability is given byT

jAT j2jAI j2

1
1  (S sin 2kB a)2

and that the reflection probability is given byRjAR j2(S sin 2kB a)2
,jAI j2 1  (S sin 2kB a)2

where
S

(k2 ÿ k2B ):2kkB

Show that the barrier is completely transparent for certain values of theenergy.
7. Find the classical distance of closest approach for two protons with anenergy of approach equal to 2 keV. Estimate the probability that the protons
penetrate the Coulomb barrier tending to keep them apart. Compare thisprobability with the corresponding probability for two 4 He nuclei with thesame energy of approach.
8. We have seen that the tunnelling through a Coulomb barrier plays a crucialrole in thermonuclear fusion. It also plays a crucial role in the alpha-decay of
nuclei such as 235 U. In the simplest model for alpha-decay, the alphaparticle is preformed and trapped within the nucleus by a potential similar
to that shown in Fig. 5.5. The mean rate of decay, l, is then the product of
the frequency n with which the alpha-particle hits the confining barrier,multiplied by the probability of penetration of the Coulomb barrier; thisprobability is given by Eq. (5.50).
Write down an approximate expression for the decay rate in termsof n, EG and the energy released by alpha-decay, E. The half-life for thealpha-decay of 235 U is 7:1  108 years and the energy released isE  4:68 MeV. Estimate the half-life for the alpha-decay of 239 Pu giventhat the energy released in this decay is 5.24 MeV.

6
The harmonic oscillator
The harmonic oscillator played a leading role in the development of quantum
mechanics. In 1900, Planck made the bold assumption that atoms acted likeoscillators with quantized energy when they emitted and absorbed radiation; in
1905, Einstein assumed that electromagnetic radiation acted like electromagnetic
harmonic oscillators with quantized energy; and in 1907, Einstein assumed thatthe elastic vibrations of a solid behaved as a system of mechanical oscillators withquantized energy. These assumptions were invoked to account for black bodyradiation, the photoelectric effect and the temperature dependence of the specificheats of solids. Subsequently, quantum theory provided a fundamental description of both electromagnetic and mechanical harmonic oscillators.
This chapter deals with the quantum mechanical behaviour of a particle in aharmonic oscillator potential. We shall find the energy eigenvalues and eigenfunctions, and explore the properties of stationary and non-stationary quantum states. These states are very important in molecular and solid state physics,
nuclear physics and, more generally, in quantum field theory.

6.1

THE CLASSICAL OSCILLATOR

The simplest example of a harmonic oscillator is a particle on a spring withelastic constant k. When the particle is displaced from equilibrium by a distancex, there is a force F  ÿkx which opposes the displacement. Because the workneeded to move the particle from x to x  dx is kx dx, the potential energy
stored by displacing the particle by a finite distance x is
V (x) Z x01
kx0 dx0  kx2 :
2

(6:1)

This potential energy is converted into kinetic energy when the particle isreleased.

110

The harmonic oscillator

Chap. 6The equation of motion for a particle of mass m isd2 x
 ÿkx,
dt2(6:2)

  ÿ!2 x,
x

(6:3)

m
and this is usually rewritten as is the acceleration of the particle and ! 
where xisp
k=m. The general solutionx  A cos (!t  a),(6:4)

where A and a are two constants which may be determined by specifying theinitial position and velocity of the particle; for example, if the particle isreleased from rest at position x0 , then A  x0 and a  0.
Eq. (6.4) describes simple harmonic motion with amplitude A, phase a andangular frequency ! or period 2p=!. During the motion, the potential energy
rises and falls as the kinetic energy falls and rises. But the total energy E, thesum of the kinetic and potential energies, remains constant and equal toE  12mx_ 2  12kx2  12m!2 A2 :(6:5)

In the real world of quantum mechanics, simple harmonic motion withdefinite energy, frequency, phase and amplitude never really happens. We
shall see that oscillators either have definite energy and do not oscillate, orthey oscillate with uncertain energy. However, we shall see that, in special
circumstances, the oscillations are almost like simple harmonic motion.

6.2

THE QUANTUM OSCILLATOR

The defining property of a quantum system is its Hamiltonian operator. For aone-dimensional harmonic oscillator the Hamiltonian operator is
p2 1
^  ^
^2 ,
 m!2 x
H
2m 2

(6:6)

h2 ]2
1
^ ÿ
 m!2 x2 :H2m ]x2 2(6:7)

or, if we use Eq. (3.30),

6.2

The Quantum Oscillator

111

The first term represents the kinetic energy operator for a particle of mass mand the second term represents the potential energy operator for a particle in a
potential well which, in classical physics, would give rise to simple harmonic
motion with angular frequency !.
The behaviour of a particle in a harmonic oscillator potential is more variedin quantum physics than in classical physics. There are an infinite number ofquantum states; some are stationary states with definite energy and some are
non-stationary states with uncertain energy. Each of these states is described bya wave function C(x, t) which satisfies the SchroÈdinger equation,
i
h

]C
^
 HC,
]t

(6:8)

^ given by Eq. (6.7).
with the Hamiltonian H
The quantum states with definite energy will be our first concern. As we saw in
Section 4.3, states with energy E are represented by a wave function of the form
C(x, t)  c(x) eÿiEt=h ,(6:9)

where c(x) is an eigenfunction belonging to an energy eigenvalue E. If wesubstitute for C(x, t) into Eq. (6.8) and use Eq. (6.7), we obtain the energy
eigenvalue equation:
"

#
2 d2
h
1
2 2
 m! x c(x)  Ec(x):ÿ
2m dx2 2(6:10)

When we seek solutions to this equation, we shall impose the physicalrequirement that the wave function of the particle is normalizable. To dothis, we shall require the eigenfunctions to go to zero at infinity; i.e.
c(x) ! 0 as

x ! 1:(6:11)

Because the sides of the harmonic oscillator potential, like the walls of theinfinite square-well potential, are infinitely high, we expect an infinite numberof quantized eigenvalues. We shall denote these by En , and the corresponding
eigenfunctions by cn (x), where n is a quantum number. We shall follow theconvention of labelling the ground state by n  0 and the first, second and third
excited states, etc, by n  1, 2, 3, . . . .1. Because we wish to emphasise the physics
of the harmonic oscillator, we shall defer the mathematical problem of finding
the energy eigenvalues and eigenfunctions until Section 6.6.
1Note, the convention for the infinite square-well potential is different. The ground state for thispotential has the label n  1; see Eq. (4.38).

112

The harmonic oscillator

6.3

QUANTUM STATES

Chap. 6

We shall first describe the properties of the stationary states of the harmonic
oscillator and then explore the extent to which the non-stationary states resemble classical simple harmonic motion.

Stationary states
In Section 6.6 we shall show that the energy eigenvalues of a harmonic oscillator with classical angular frequency ! are given by
En  (n  12) h!,withn  0, 1, 2, 3, . . . . . .

(6:12)

As illustrated in Fig. 6.1, the energy levels have equal spacing h! and the lowest
energy level is E0  12 h!.
The wave function of a particle with energy En has the form
C(x, t)  cn (x) eÿiEn t=h

(6:13)

where cn (x) is the energy eigenfunction. The eigenfunctions of the four lowest
states of the oscillator are given in Table 6.1 and they are shown in Fig. 6.2. We
note that the nth eigenfunction has n nodes; i.e. there are n values of x for whichcn (x)  0. In Section 4.4 we showed that the eigenfunctions of an infinite
square well have a similar property. In general, eigenfunctions of excitedbound states always have a number of nodes which increases with the degreeof excitation, just as classical normal modes of oscillation of higher frequency
always have a higher number of nodes.
TABLE 6.1 Normalized eigenfunctions for the four lowest states of a onedimensional harmonic oscillator.
p The parameter which determines the spatial extent of
the eigenfunctions is a  
h=m!
Quantum number

Energy eigenvalue

n0

E0 

1
h!

2

n1

E1 

3
h!

2

n2

E2 

5
h!

2

n3

E3 

7
h!

2

Energy eigenfunction1
1 2 ÿx2 =2a2pec0 (x) a p

12  1
x ÿx2 =2a2p 2
e
a
2a p

12 
x2 
1
2
2
p
2ÿ4
eÿx =2ac2 (x) a8a p
12   x 3 
1
x
2
2
p
12
eÿx =2ac3 (x) 
ÿ8
a
a
48a p
c1 (x) 

113

6.3 Quantum States

7

E3 = 2 h
1V(x) = 2 m

2x2

5

E2 = 2 h

3E1 = 2 h

1E0 = 2 hx=0Fig. 6.1 Low-lying energy levels of a particle of mass m confined by a one-dimensional
harmonic oscillator potential.
n=0

n=1
0.5

eigenfunction

eigenfunction

0.5

0
−0.5
−2

−1

0

1

0
−0.5
−2

2

n=2

0

1

2

0
1
position x

2

n=3
0.5

eigenfunction

0.5

eigenfunction

−1

0
−0.5
−2

−1

0
1
position x

2

0
−0.5
−2

−1

Fig. 6.2 The spatial shapes of the eigenfunctions cn (x) for the pfourloweststates of a

one-dimensional harmonic oscillator with length parameter a  
h=m!  1:

114

The harmonic oscillator

Chap. 6The observable properties of a quantum state with energy En include thefollowing:. The eigenfunction has an observable property called parity, which will be
discussed further in Sections 9.1 and 9.4. If the position coordinate ischanged from x to ÿx, the eigenfunction has a definite symmetry:cn ( ÿ x)  cn (x)if n is even,cn ( ÿ x)  ÿcn (x)if n is odd:

andOscillator states with even n are said to have positive parity and oscillator
states with odd n are said to have negative parity. This property arises
because the Hamiltonian of the harmonic oscillator, Eq. (6.7), is unchanged
under the transformationx ! x0  ÿx:. The position probability density of the particle,
jCn (x, t)j2  jcn (x)j2 ,(6:14)

is time-independent, as expected from Section 4.8, where we showed thatquantum states of definite energy are stationary states, i.e. states with noobservable time dependence. As illustrated in Fig. 6.3, the particle can have
any location between x  ÿ1 and x  1, in marked contrast with aclassical particle which is confined to the region ÿA < x < A, where A isthe amplitude of oscillation.
. The position expectation values are2
hxi  0andÿ
hx2 i  n  12 a2 ,so that the uncertainty in position isDx 
where a 2q q
ÿ
hx2 i ÿ hxi2 n  12 a,p
h=m!.

The expectation values of position and momentum can derived using the results of problems 10
and 11 at the end of this chapter.

0.4

0.3

0.2

0.1

0

probability density

n=0

probability density

0.5

0.5

−2

−1

0

1

n=2

0.3
0.2
0.1
−2

−1
0
1
position x

115

Quantum States

n=1

0.4
0.3
0.2
0.1
0

0.4

0

0.5

2

probability density

probability density

6.3

0.5

−2

−1

0

1

2

−1
0
1
position x

2

n=3

0.4
0.3
0.2
0.1
0

2

−2

Fig. 6.3 The position probability densities jcn (x)j2 for the four
lowest states of a onep
dimensional harmonic oscillator with length parameter a  
h=m!  1:. The momentum expectation values are
hpi  0ÿ h2
and hp2 i  n  12 2 ,aso that the uncertainty in the particle's momentum is
Dp q q
ÿ h
hp2 i ÿ hpi2 n  12 :a. The product of the position and momentum uncertainties isÿh,Dx Dp  n  12 which agrees with the Heisenberg uncertainty principle, Eq. (1.15), whichasserts that in general
Dx Dp  12h:

116

The harmonic oscillator

Chap. 6

We note that when n  0, i.e. when the particle is in the ground state of theoscillator, the product of Dx and Dp is as small as it can be.
. Because the position and momentum of the particle are uncertain, thepotential energy and the kinetic energy are uncertain. The expectation valuesof these uncertain observables are
1
1
m!2 hx2 i  En
2
2

and

hp2 i 1
 En :
2m
2

Not surprisingly, the sum of the expectation values of the uncertain potential
and kinetic energies is equal to En , the sharply defined total energy of thestate.
Finally, it is useful to consider in general terms why the quantum ground stateof the harmonic oscillator is so different from the classical ground state inwhich the particle lies at rest at the bottom of the well with zero kinetic energy
and zero potential energy. When a quantum particle is precisely localized at
the centre of the well, it has a highly uncertain momentum and, hence, ahigh kinetic energy. Similarly, when its momentum is precisely zero, it has ahighly uncertain position and it may be found in regions of high potential
energy. It follows that the sum of the kinetic and potential energies of aquantum particle in a harmonic oscillator potential has a minimum value
when its position and momentum are uncertain, but not too uncertain. This
minimum is called the zero point energy of the harmonic oscillator. A lower
bound for this energy is derived using these ideas in problem 1 at the end of thischapter.

Non-stationary states
The general wave function of a particle in a harmonic oscillator potential has
the form
C(x, t) Xn0, 1, 2...

cn cn (x) eÿiEn t=h :(6:15)

This wave function represents a state of uncertain energy because when theenergy is measured many outcomes are possible: E0  12 h!, E1  32 h!, . . . withprobabilities jc0 j2 , jc1 j2 , . . ..
This wave function also represents a non-stationary state, a state with timedependent observable properties. For example, the position probability amplitude jC(x, t)j2 has time-dependent terms which arise from the interference ofterms involving different energy eigenfunctions, cn (x). In particular, the inter-6.3

Quantum States

117

ference of cm (x) eÿiEm t=h with cn (x) eÿiEn t=h gives rise to a term which oscillateswith angular frequency!m , n 

jEm ÿ En j,h

(6:16)

which is an integer multiple of the classical angular frequency !. Thus, theposition probability density, jC(x, t)j2 , can oscillate with a range of angularfrequencies: !, 2!, 3!, etc.
At first sight, we expect the average position of the particle to oscillate withthe same frequencies, because, when we evaluate the expectation valuehx(t)i Z 1
ÿ1

C*(x, t)xC(x, t) dx,

we obtain terms like
cm*cn xm, n e

i!m, n t,

wherexm, n Z 1
ÿ1

cm*(x)xcn (x) dx:

But most of these terms are zero because, as shown in problem 11, the integral
xm, n is zero when jm ÿ nj is greater than one. It follows that the average position
only oscillates with angular frequency !, and so does the average momentum. In
general, the average position and momentum of a particle of mass m with wave
function (6.15) are given byhx(t)i  A cos (!t  a) andhp(t)i  ÿm!A sin (!t  a):(6:17)

Thus, to some extent, the particle oscillates back and forth like a classicaloscillator. But the resemblance to classical simple harmonic motion may not be
close because the uncertainties in position and momentum may be large andthey may change as the particle oscillates back and forth. Accordingly, it isinstructive to identify quasi-classical states in which the motion most closely
resembles classical simple harmonic motion.
It can be shown that the wave functions of these quasi-classical states are
given by Eq. (6.15) with energy probability amplitudes cn which satisfy the
Poisson probability distribution3
jcn j2 

3nn ÿn
e ,
n!

with


n >> 1:

Quasi-classical states are fully described in Quantum Mechanics, vol. I, C Cohen-Tannoudji, B
Diu and F LaloeÈ, John Wiley & Sons (1977).

118

The harmonic oscillator

Chap. 6

In problem 1 at the end of Chapter 3 we showedthat the mean and standardp
deviation of this distribution are n and n. It follows that the mean andstandard deviation of the energy of a quasi-classical state is given byh!
hEi  (n  12)

and DE pnh!:

When the average excitation 
n is high, the relative uncertainty in energy is given
by
p

n
DE
1

! phEi n  12nso that the uncertainty in the energy becomes less important. Because theuncertainties in the position and momentum also become less important, themotion of the particle approaches the impossible perfection of classical simple
harmonic motion.
At the turn of the twentieth century, Planck and Einstein made the inspired
guess that oscillators which exhibit simple harmonic motion could also havequantized energies. We have followed a logical path in the opposite direction
and indicated how quantum oscillators, which have quantized energies, canalmost exhibit simple harmonic motion.

6.4

DIATOMIC MOLECULES

To a first approximation a diatomic molecule consists of two nuclei held
together in an effective potential which arises from the Coulomb interactionsof the electrons and nuclei and the quantum behaviour of the electrons. Thiseffective potential determines the strength of the molecular bond between the
nuclei and it also governs the vibrational motion of the nuclei. The effective
potential and the vibrational energy levels of the simplest diatomic molecule,the hydrogen molecule, are illustrated in Fig. 6.4.
We note from Fig. 6.4 that an effective potential for a diatomic molecule,
Ve (r), has a minimum and that near this minimum the shape is like a harmonic
oscillator potential. Indeed, if r0 denotes the separation at which the effective
potential has a minimum and x  r ÿ r0 denotes a small displacement from r0 ,we can write
Ve (r)  12kx2
where k is a constant. This equation implies that when the nuclei are displaced a
distance x from their equilibrium separation of r0 , there is a restoring force ofmagnitude kx and the potential energy increases by 12kx2 . The constant k is aneffective elastic constant which characterizes the strength of the molecular bond
between the nuclei in the molecule.

6.4

Diatomic Molecules

119

Continuum ofeVenergy levels

5
Ve(r)

4
3
2

n=2
n=1
n=0

1
0
0

0.1

0.2

nm

Internuclear separation r

Fig. 6.4 The effective internuclear potential Ve (r) and the vibrational energy levels of
the hydrogen molecule. The potential energy near the minimum is approximately
quadratic and acts like a harmonic oscillator potential, and the lowest vibrational levels
are approximately equally spaced and given by the En  (n  12 )h!. The vibrational
levels become more closely spaced as the degree of excitation increases and the dissociation of the molecule gives rise to a continuum of energy levels.

If classical physics were applicable, the nuclei would have energy
Eclassical 

p21
p2
1
 2  kx2 :2m1 2m2 2

where m1 and m2 are the masses of the nuclei and p1 and p2 are the magnitudesof their momenta. In the centre-of-mass frame, we can set p1  p2  p and, byintroducing the reduced mass
m

m1 m2
,
m1  m2

we obtain
Eclassical 

p2 1 2
 kx :
2m 2This energy is the same as the energy of a single particle of mass m on a springwith elastic constant k. Accordingly, we expect the vibrating nuclei in a
diatomicpmolecule to act like a harmonic oscillator with classical frequency!  k=m, where m is the reduced mass of the nuclei and k is an elastic constantcharacterizing the strength of the molecular bond between the nuclei.
The quantum mechanical behaviour of this oscillator is described by a wave
function C(x, t) which satisfies the SchroÈdinger equation

120

The harmonic oscillator

Chap. 6"#]Ch2 ] 2
1 2 2i
h k x C: ÿ
2m ]x2 2]t(6:18)

This equation is almost identical to Eq. (6.8), which formed the starting point
for our discussion of the quantum oscillator. Indeed, if we replace the mass mby the reduced mass m, we can apply all our results to a diatomic molecule.
Most importantly, we can use Eq. (6.12) to write down an expression for thevibrational energy levels of a diatomic molecule with reduced mass m and elastic
constant k:


1
h!,
En  n  
2

where

s
k
!
:
m

(6:19)The quantum number n can take on the values 0, 1, 2, . . ., but, when n is large,
the harmonic oscillator model for molecular vibrations breaks down. Thisoccurs when the vibrational energy becomes comparable with the dissociationenergy of the molecule, as shown for the hydrogen molecule in Fig. 6.4.
A transition from one vibrational level of the molecule to another is often
accompanied by the emission or absorption of electromagnetic radiation, usually in the infrared part of the spectrum. This is particularly so for diatomic
molecules with two different nuclei, i.e. heteronuclear diatomic molecules. For
such molecules, the electrons form an electric dipole which can strongly absorb
or emit electromagnetic radiation. In fact, this mechanism leads to transitions
between adjacent vibrational levels and the emission or absorption of photons
with energy4s
k:E
hm
These photons give rise to a prominent spectral line with wavelength
l
4

hc
 2pc
E

r
m
:
k

(6:20)The probability for transition from cm to cn induced by electric dipole radiation is proportional
to jxm, n j2 whereZ 1xm, n cm*(x)xcn (x) dx:
ÿ1

By using the properties of the harmonic oscillator eigenfunctions, one can show that xm, n  0 if
jm ÿ nj 6 1. (See problem 11 at the end of this chapter.)6.5

Three-Dimensional Oscillators

121

As an example, we consider the carbon monoxide molecule. The reduced
mass of the nuclei is m  6:85 amu, and, when transitions between adjacentvibrational levels occur, infrared radiation with wavelength l  4:6 mm is emitted or absorbed. If we substitute these values for m and l into Eq. (6.20), we findthat the elastic constant, characterizing the strength of the bond in the carbon
monoxide molecule, is k  1908 Nmÿ1 .
In reality, the situation is more complex. First of all, transitions between
adjacent vibrational levels have slightly different wavelengths, because thevibrational energy levels are only approximately equally spaced; as illustrated
in Fig. 6.4, a harmonic oscillator potential does not exactly describe theinteraction between the nuclei in a diatomic molecule. Second, the molecule
may rotate and each vibrational level is really a band of closely spaced levels
with different rotational energies; accordingly, there is a band of spectral lines
associated with each vibrational transition.

6.5

THREE-DIMENSIONAL OSCILLATORSWe shall conclude this chapter by considering a particle of mass m in the threedimensional harmonic oscillator potential
V (r)  12kr2  12k(x2  y2  z2 ):(6:21)

A classical particle at a distance r from the origin would experience a central
force towards the origin of magnitude kr. When displaced from the origin andreleased,pit executes simple harmonic motion with angular frequency!  k=m, but more complicated motion occurs when the particle is displaced
and also given a transverse velocity.
The behaviour of a quantum particle is governed by a Hamiltonian operator^ which is the sum of three one-dimensional Hamiltonians:H^y  H^z^ H^x  HH
where
2
2^ x  ÿ h ]  1 m!2 x2 ,H2m ]x2 222^ y  ÿ h ]  1 m!2 y2 ,H2m ]y2 222^ z  ÿ h ]  1 m!2 z2 :H2m ]z2 2(6:22)

122

The harmonic oscillator

Chap. 6

Stationary states with definite energy are represented by wave functions of
the form
C(x, y, z, t)  c(x, y, z) eÿiEt=h ,(6:23)

where c(x, y, z) and E satisfy the three-dimensional eigenvalue equation^
Hc(x,y, z)  Ec(x, y, z):(6:24)

These states may be found by using the eigenvalue equations for the^ y and H^ z:^ x, H
one-dimensional oscillators governed by Hÿ
^ x cn (x)  nx  1 H
2 h!cnx (x),xÿ^ y cn (y)  ny  1 H
2 h!cny (y),yÿ^ z cn (z)  nz  1 H
2 h!cnz (z),zwhere the quantum numbers nx , ny and nz can take on the values 0, 1, 2 . . . . . .
These three equations imply that the function
cnx , ny , nz (x, y, z)  cnx (x)cny (y)cnz (z)(6:25)

satisfies the three-dimensional eigenvalue equation^ n , n , n (x, y, z)  Enx , ny , nz cn , n , n (x, y, z)
Hc
x
y
z
x
y
z

(6:26)

ÿEnx , ny , nz  nx  ny  nz  32 h!:(6:27)

provided that

Thus, the eigenvalues and the eigenfunctions of the three-dimensional oscillator are labelled by three quantum numbers, nx , ny and nz , each of which cantake on any integer value between zero and infinity. The explicit forms of the
low-lying eigenfunctions can be found by using Table 6.1. When all threequantum numbers are equal to 0, we have the ground state:
3E0, 0, 0  h!
2


and

c0, 0, 0 (x, y, z) 
1 3=2 ÿ(x2 y2 z2 )=2a2
p
e
,
a p

p
where a  h=m!. By changing one of the quantum numbers from 0 to 1, we
obtain three excited states with the same energy:

6.6 The Oscillator Eigenvalue Problem

EnergyDegeneracy

9
2h

10

7
2h

6

5
2h

3

3
2h

1

123

Fig. 6.5 The four lowest energy levels of a particle in a three-dimensional harmonic
oscillator potential. The degeneracy of each level is denoted on the right-hand side.

5
E1, 0, 0  h! and2
5
E0, 1, 0  h! and2
5
E0, 0, 1  h! and2


1 3=2 1=2 x ÿ(x2 y2 z2 )=2a2
p
2
;
e
a
a p


1 3=2 1=2 y ÿ(x2 y2 z2 )=2a2p2;c0, 1, 0 (x, y, z) 
e
a
a p


1 3=2 1=2  z  ÿ(x2 y2 z2 )=2a2pc0, 0, 1 (x, y, z) 
2
:
e
a
a p

c1, 0, 0 (x, y, z) In a similar way we can find six states with energy 7h!=2, ten states with energy
9h!=2, and so on.
The energy levels of the three-dimensional harmonic oscillator are shown in
Fig. 6.5. This diagram also indicates the degeneracy of each level, the degeneracy of an energy level being the number of independent eigenfunctions associated with the level. This degeneracy arises because the Hamiltonian for thethree-dimensional oscillator has rotational and other symmetries.

6.6

THE OSCILLATOR EIGENVALUE PROBLEM

For the benefit of mathematically inclined readers we shall now discuss
the problem of finding the energy eigenfunctions and eigenvalues of a onedimensional harmonic oscillator. The method used is interesting and introduces mathematical methods which are very useful in advanced quantum
mechanics. This section may be omitted without significant loss of continuity.
In order to simplify the task of finding the eigenvalues and eigenfunctions,
we shall clean up the eigenvalue equation (6.10) and give it a gentle massage.
We note that this equation contains three dimensional constants: Planck's
constant h, the classical angular frequency !, and the mass of the confinedparticlepm.
Withthese constants we can construct an energy h! and a

length h=m!. Hence, it is natural to measure the energy E in units of h!

124

The harmonic oscillator

Chap. 6p
h=m!. Accordingly, we shall rewrite Eq. (6.10)
and the length x in units ofusing

E  Eh!

andr
h

:
xq
m!

(6:28)

If we think of the eigenfunction c as a function of q, Eq. (6.10) becomes

ÿ


d2
2
c(q)  2Ec(q):qdq2(6:29)The task of finding the eigenvalues E and eigenfunctions c(q) is made easy bynoting that for any function f (q)



d
d
d2 f (q)df (q) d[qf (q)]qqÿf (q)  q2 f (q) ÿ

:
ÿq
dq
dq
dq2
dq
dq
Using
d[qf (q)]df (q)
 f (q)  q
dqdq
we obtain


d
q
dq



"
#

2
d
d
qÿ
f (q)  q2 ÿ 2  1 f (q):dqdq

It follows that the eigenvalue equation Eq. (6.29) may be written as

q

d
dq


qÿ


d
c(q)  (2E  1)c(q):dq(6:30)

Similar considerations show that we can also write the eigenvalue equation
as

qÿ

d
dq


q


d
c(q)  (2E ÿ 1)c(q):dq(6:31)

Instead of finding all the eigenfunctions at one go, as we did in Section 4.4 forthe infinite square-well, we shall adopt a more elegant approach of first finding
the eigenfunction of the ground state and then use this as a starting point forfinding the eigenfunctions of the excited states.

6.6

The Oscillator Eigenvalue Problem

125

The ground state
Two possible eigenvalues and eigenfunctions are immediately apparent from aninspection of the alternative expressions for the eigenvalue equation given by
Eq. (6.30) and Eq. (6.31).
First, Eq. (6.30) is clearly satisfied if
Eÿ

1
2


and

qÿ


d
c(q)  0:dq

This first-order differential equation for c(q) has the solution
2c(q)  A eq =2
where A is a constant. But this solution must be discarded because it does notsatisfy the boundary conditions, c(q) ! 0 as q ! 1, which are needed for anormalizable wave function.
Second, Eq. (6.31) is clearly satisfied if
1
E
2


and


d
q
c(q)  0:dq

In this case the differential equation for c(q) has the solution
2c(q)  A eÿq =2 ,which is an acceptable eigenfunction because c(q) ! 0 as q ! 1. Later weshall show that this is the eigenfunction of the ground state. Accordingly, we
shall use the quantum number n  0 as a label and take the ground state
eigenvalue and eigenfunction to be
E0   12and2

u0 (q)  A0 eÿq =2 ,(6:32)

where A0 is a normalization constant.
If we use Eq. (6.28) to express the dimensionless variables E and q in terms ofthe dimensional variables E and x, we find that the ground state of a harmonic
oscillator with angular frequency ! has energy
E0  12h!
and that its eigenfuction, as a function of x, is given by(6:33)

126

The harmonic oscillator

c0 (x)  N0 e

Chap. 6

ÿx2 =2a2,

wherer
h
:
a
m!

(6:34)

The constant N0 is a normalization constant.

Excited states
We shall now find the solutions which describe the excited states of the oscillator. The first step is to focus on the nth excited state. Its eigenfunction andeigenvalue satisfy an eigenvalue equation which we may write down using
either Eq. (6.30) or Eq. (6.31). We shall choose Eq. (6.30) and write


d
q
dq



d
qÿ
c (q)  (2En  1)cn (q):dq n

dWe now operate with [q ÿ dq
] on both sides to give



d
qÿ
dq



d
q
dq








d
d
qÿ
c (q)  (2En  1) q ÿ
c (q)
dq n
dq n

and then compare with the form of eigenvalue equation given by Eq. (6.31);
i.e. we compare withdc(q)  (2E ÿ 1)c(q):dqhidThis comparison shows that the function q ÿ dq
cn (q) is an eigenfunction
d
qÿ
dq



q

c(q) with an eigenvalue E given by (2E ÿ 1)  (2En  1); i.e. it is an eigenfunction with eigenvalue E  En  1.
d
We have discovered that the operator [q ÿ dq
] has the role of an energy
raising operator. When it operates on the eigenfunction cn (q) with eigenvalue
En , it gives the eigenfunction cn1 (q) with eigenvalue En1  En  1. It follows
that the ground-state eigenvalue and eigenfunction given by Eq. (6.32) may be
used as the starting point for generating an infinite set of eigenvalues andeigenfunctions which describe excited states of the harmonic oscillator. The
first excited state is described by
E1 

3
2

and



d ÿq2 =2,
c1 (q)  A1 q ÿedq

and the second excited state is described by(6:35)

6.6

E1 

5
2

The Oscillator Eigenvalue Problemd 2 ÿq2 =2
and c2 (q)  A2 q ÿ
e
,
dq

127

(6:36)and so on, ad infinitum. Thus, we can construct an infinite sequence of energy
eigenvalues and eigenfunctions; they may be labelled by the quantum numbersn  0, 1, 2, . . . : and they are given by1
En  n 2
d n ÿq2 =2e:
cn (q)  An q ÿdq

and(6:37)

By using Eq. (6.28) to express E in terms of E, we find that the energy of thenth level of a harmonic oscillator with angular frequency ! isÿEn  n  12 h!:

(6:38)

And by expressing q in terms of x we find, after a little algebra, that the ntheigenfunction has the formcn (x)  Nn Hn

 x
a

e

ÿx2 =2a2,

wherer
h
:
a
m!

(6:39)

The constant Nn is a normalization constant, and the function Hn , a polynomialof order n in x=a, is called a Hermite polynomial. The normalized eigenfunctions for the n  0, 1, 2, 3 and 4 states are listed in Table 6.1.

Is E0 really the lowest energy?
We have one item of unfinished mathematics. We have yet to show thatE0  E0 h!  12 h! is really the lowest energy of a harmonic oscillator withangular frequency !.
In problem 1 at the end of the chapter, we shall show that the Heisenberg
uncertainty principle implies that the energy of the oscillator cannot be less thatE0  12 h!. In addition, in problem 9, we shall show that there is an energy
lowering operator for the harmonic oscillator, but that this operator cannot be
used to construct an eigenfuction with energy less than E0  12 h!. This operatordhas the form [q  dq]. When it operates on the eigenfunction cn (q) with eigenvalue En , it yields the eigenfunction cnÿ1 (q) with eigenvalue Enÿ1 . But when thisoperator acts on the eigenfunction c0 (q) it gives zero, i.e.

q


d
c (q)  0:
dq 0

(6:40)

128

The harmonic oscillator

Chap. 6

Mathematical properties of the oscillator eigenfunctions
The eigenfunctions cn (x), like the eigenfunctions of any Hamiltonian, form acomplete orthonormal set of basis functions. As described in Section 4.6, theysatisfy the condition
Z 1
ÿ1

cm*(x)cn (x) dx  dm, n ,(6:41)

where dm, n  1 if m  n and dm, n  0 if m 6 n. Moreover, they can be used as
basis functions for a generalized Fourier series: any function f (x) may be
represented by the seriesf (x) X
n0, 1, ...

cn cn (x),

where

cn 

Z 1
ÿ1

c*(x)f(x) dx:n(6:42)

This means that the general solution of the time-dependent SchroÈdinger equation for the harmonic oscillator, Eq. (6.8), has the form
C(x, t) Xn0, 1, 2...

cn cn (x) eÿiEn t=h ,(6:43)

where cn are complex constants.

PROBLEMS 6
1. In this question the Heisenberg uncertainty principlehDx Dp  12
is used to derive a lower bound for the energy of a particle of mass m in aharmonic oscillator potential with classical angular frequency !.
(a) Note that the expectation value of the energy of the particle is given byhEi 

hp2 i 1
 m!2 hx2 i
2m 2

and show that, if the average position and momentum of the particle
are both zero, its energy has an expectation value which satisfies theinequality

Problems 6

hEi 

129

2
h
1
 m!2 (Dx)2 :8m(Dx)2 2(b) Show that the minimum value of a function of the form
F (Dx) A2
 B2 (Dx)22(Dx)

is 2AB.
(c) Hence show that the expectation value of the energy of a particle in aharmonic oscillator well satisfies the inequality
hEi  12 h!:
2. By reference to the properties of the Gaussian distribution given in problem
2 at the end of Chapter 2, show that the position probability density for aparticle of mass m in the ground state of a harmonic oscillator with angularfrequency! is a Gaussian probability distribution with standard deviationps  h=2m!.
3. Find the amplitude of oscillation A for a classical particle with the sameenergy as a quantum particle in the ground state of the harmonic oscillator.
Write down an expression for the probability of finding the quantum
particle in the classically forbidden region jxj > A.
4. Consider the potential
V (x) 1

1
2 2
2 m! x

if x < 0
if x > 0,

which describes an elastic spring which can be extended but not compressed.
By referring to the eigenfunctions of the harmonic oscillator potential
shown in Fig. 6.2, sketch the eigenfunctions of the ground and first excitedstates of this new potential.
What are the energies of the ground and first excited states?
5. At time t  0 a particle in a harmonic oscillator potential V (x)  12 m!2 x2
has a wave function
1
C(x, 0)  p [c0 (x)  c1 (x)]2

130

The harmonic oscillator

Chap. 6

where c0 (x) and c1 (x) are real, normalized and orthogonal eigenfunctions
for the ground and first-excited states of the oscillator.
(a) Write down an expression for C(x, t), the wave function at time t.
(b) Show that C(x, t) is a normalized wave function.
(c) Use your knowledge of the energy levels of the harmonic oscillator
potential to show that the probability density jC(x, t)j2 oscillates withangular frequency !.
(d) Show that the expectation value of x has the formhxi  A cos !t,

where

A

Z 1
ÿ1

c0 (x) x c1 (x) dx:

6. Consider the harmonic oscillator quantum state with the wave function
C(x, t) q

1
ÿiE0 t=h c1 (x) eÿiE1 t=h  c2 (x) eÿiE2 t=h ]3[c0 (x) e

where c0 (x), c1 (x) and c2 (x) are taken as real, normalized eigenfunctions of
the harmonic oscillator with energy E0 , E1 and E2 , respectively.
(a) What is the expectation value of the energy?
(b) What is the uncertainty in the energy?
(c) Show that the probability for the position of the particle has the form
jC(x, t)j2  A(x)  B(x) cos !t  C(x) cos 2!t,and find the functions A(x), B(x) and C(x).
7. The transitions between adjacent vibrational levels of the NO molecule giverise to infrared radiation with wavelength l  5:33 mm. Find the elastic
constant k characterizing the strength of the bond between the nuclei in
the NO molecule. (The reduced mass of the NO molecule is 7.46 amu.)
8. In answering this question, you may assume that cn (x) is the energy eigenfunction of a particle of mass m in a one-dimensional harmonic oscillator
potential V (x)  12 m!2 x2 with energy En  (n  12 )h!.
Consider a particle of mass m in a two-dimensional harmonic oscillator
potential with an energy eigenvalue equation of the form:

Problems 6

"

131

#


h2 ]2]2
1
 m!2 (x2  y2 ) cnx , ny (x, y)  Enx , ny cnx , ny (x, y):ÿ2m ]x2 ]y22(a) Verify that
cnx , ny (x, y)  cnx (x)cny (y),with nx  0, 1, 2, . . . and ny  0, 1, 2, . . ., is an energy eigenfunction
with energy Enx , ny  (nx  ny  1)
h!.
(b) Draw an energy level diagram and indicate the degeneracy of theenergy levels.
(c) By expressing c1, 0 and c0, 1 in plane polar coordinates (r, f), findfunctions ca (r, f) and cb (r, f) which obey the equations
ÿih

]ca
 
hca
]f

and

ÿ i
h

]cb
 ÿhcb :]f

The remaining questions are for readers who studied Section 6.6.

hid9. In this question you are asked to show that when the operator q  dq
acts
on the eigenfunction cn (q) with eigenvalue En , it gives the eigenfunction
cnÿ1 (q) with eigenvalue Enÿ1  En ÿ 1.
Consider the nth state of the oscillator with eigenvalue En and eigenfunction cn . Using the form of the eigenvalue equation given by Eq. (6.31),
write



d
d
q
c (q)  (2En ÿ 1)cn (q):
qÿ
dq
dq n
h
i
d
Now operate with q  dq
on both sides, compare with the form of theequation given by Eq. (6.30) and show that the function
heigenvalueidq  dqcn (q) is an eigenfunction c(q) with an eigenvalue E given by
E  En ÿ 1.

10. In general, the eigenfunctions cn (q) and cn1 (q) are related by

qÿ


d
c (q)  an cn1 (q)dq n

where an is a constant. Show that cn (q) and cn1 (q) have the same
normalization if

132

The harmonic oscillator

Chap. 6

jan j2  2(n  1):[Hint: Write down the normalization integrals, note that integration byparts can be used to show thatZ 1 
ÿ1




Z 1
df (q)dg(q)f (q)g(q) dq  ÿ
dq
dq
dq
ÿ1

if the function f(q)g(q) goes to zero as q ! 1, and use Eq. (6.30).]
Similarly, show that the eigenfunctions cn (q) and cnÿ1 (q), that are
related by

q


d
c (q)  bn cnÿ1 (q),dq nhave the same normalization if
jbn j2  2n:
11. Consider eigenfunctions cn of the harmonic oscillator which obey thenormalization conditionZ 1
ÿ1

jcn (q)j2 dq  1:(a) Show thatZ 1
ÿ1

cm*(q) q cn (q) dq  12(an dm, n1  bn dm, nÿ1 )where an and bn are defined in problem 10.
(b) Show that
Z 1
ÿ1

ÿ

c*(q)
q2 cn (q) dq  n  12n

and thatÿd2
cn (q) dq  n  12 :
c*(q)
ÿ
n
2
dq
ÿ1

Z 1

Problems 6133(c) Now consider the eigenfunction cn as a function of x and normalized
so thatZ 1
ÿ1

jcn (x)j2 dx  1:

Verify thathxi  0andÿ
hx2 i  n  12 a2

hpi  0andÿ
hp2 i  n  12 h=a2 :and that[Hint: Write




1
d
1
d
 qÿ
and
q q
2
dq
2
dq





d
1
d
1
d
 q
ÿ qÿ
,
dq 2
dq
2
dq

d
d
use the fact that [q ÿ dq
] and [q  dq
] are energy raising and lowering
operators, note that the eigenfunctions obey the orthogonality relation
Eq. (6.41), and use Eq. (6.30) and Eq. (6.31).]This page intentionally left blank7
Observables andoperators
Operators have to be used in quantum mechanics to describe observable
quantities because measurements may have uncertain outcomes. In Chapter 3
we used the operators^r  r and^p  ÿihr

to calculate the expectation values and the uncertainties in the position andmomentum of a particle. In Chapter 4 we used the Hamiltonian operator
2

h
^ ÿ
r2  V (r)H2m
to explore the energy properties of a particle. And in the next chapter we shallconsider in detail a fourth operator, the operator describing the orbital angularmomentum of a particle,^  ^r  ^Lp:
In this chapter we shall consider some physical properties of observables inquantum mechanics and link these properties to the mathematical properties ofthe operators which describe observables. In so doing, concepts that were
implicit in the use of operators in earlier chapters will be clarified and developed. This chapter will deal with concepts that are more abstract andmathematical than those encountered elsewhere in this book. It may be omitted
without significant loss of continuity.

136

Observables and operators

7.1

ESSENTIAL PROPERTIES

Chap. 7In Chapter 4 we developed a mathematical description of energy measurement^ We explained why
based upon the properties of the Hamiltonian operator H.
^ belonging to an eigenvalue E represents a quantum statean eigenfunction of H
with sharply defined energy E. We also explained why any wave function C canbe expressed as a linear superposition of energy eigenfunctions. In particular,^ only has eigenfunctions cn with discrete eigenvalues En ,when the operator H
any wave function may be written as
C(r, t) Xncn cn (r) eÿiEn t=h ,where the coefficients cn are probability amplitudes, such that, if the energy
is measured, jcn j2 is the probability of an outcome En . When the Hamiltonian
gives rise to a continuum of energy eigenvalues, the general wave function
involves an integral over the continuous energy variable which labels the eigenfunctions.
We shall use the Hamiltonian operator as a prototype for all operators which
describe observables in quantum mechanics. We shall consider a general ob^ take note of the fundamental concepts
servable A described by an operator A,
discussed in Chapter 4 and highlight the essential mathematical properties of^ They are the following:
the operator A.
. The operator A^ must be a linear operator. This means that, if the action of A^
on the wave functions C1 and C2 is given by^ 1  F1ACand^ 2  F2 ,ACthen the action of A^ on the wave function c1 C1  c2 C2 , where c1 and c2 are
two arbitrary complex numbers, is given by^ 1 C1  c2 C2 )  c1 F1  c2 F2 :
A(c

(7:1)

^ ^r, ^p and by all other
This abstract property is satisfied by the operators for H,
operators which describe observables in quantum mechanics. It ensures thatthese observables are consistent with the principle of linear superposition
which asserts that any quantum state is a linear superposition of other quantum states.
. The operator A^ must be a Hermitian operator which means that it obeys thecondition

7.1

Z

C*1 A^ C2 d3 r ZEssential Properties^ 1 )* C2 d3 r,
(AC

137

(7:2)

where C1 and C2 are any two wave functions; the brackets in the term
^ 1 )* mean that the operator A^ only acts on the wave function C1 and(ACthe complex conjugate of the result is taken. This mathematical property
ensures that the expectation value of the observable,ZhAi 

C* A^ C d3 r,

is real for any wave function C.
It also ensures that the eigenvalues of the operator A^ are real. There may
be eigenfunctions can (r) with discrete eigenvalues an given by^ a  an caAcnn
and=or eigenfunctions ca0 with continuous eigenvalues a0 given by^ 0  a0 c 0 :AcaaIn Chapter 4 we argued that the possible outcomes of an energy measurement are energy eigenvalues. Identical arguments imply that the possible
outcomes of a measurement of the observable A are the eigenvalues of the^ Because the outcomes of all measurements are real numbers, theoperator A.
eigenvalues of the operator A^ must be real numbers.
. Finally, the operator A^ must describe an observable which is always measurable. Specifically, we must be able to predict the outcomes of a measurementof A and the probability of each of these outcomes. This is only possible if theeigenfunctions of the operator A^ form a complete set of basis functions sothat any wave function C(r, t) can be written as
C(r, t) XnZcan (t)can (r) c(a0 , t)ca0 (r) da0 :(7:3)

In this expression can (t) and c(a0 , t) are probability amplitudes for the observable A. If a measurement takes place at time t on a particle with wave
function C, then jcan (t)j2 is the probability of outcome an and jc(a0 , t)j2 da0
is the probability of an outcome between a0 and a0  da0 .1
1

To keep the presentation as simple as possible, we have ignored the complications that arise when
there is degeneracy, i.e. more than one eigenfunction with the same eigenvalue, and we have notaddressed, at this stage, the normalization and orthogonality of eigenfunctions with continuous
eigenvalues.

138

7.2

Observables and operators

Chap. 7POSITION AND MOMENTUM

In Chapter 3 we introduced the operators for the position and momentum of aparticle and described how they can be used to calculate expectation values anduncertainties, but we have not yet explicitly considered the eigenvalues andeigenfunctions of these operators. We shall do so by considering a particle
moving in one dimension. We shall denote an eigenfunction for a particle
with position eigenvalue x0 by cx0 (x) and an eigenfunction for a particle with
momentum eigenvalue p0 by cp0 (x).

Eigenfunctions for position
The position eigenfunction satisfies the eigenvalue equation^cx0 (x)  x0 cx0 (x)
x

(7:4)

^  x to give
which may be rewritten using x
xcx0 (x)  x0 cx0 (x):(7:5)

This equation states that cx0 (x) multiplied by x is the same as cx0 (x) multipliedby the eigenvalue x0 . This is only possible if cx0 (x) is a very peculiar function
which is infinitely peaked at x  x0 . Such a function is normally written as
cx0 (x)  d(x ÿ x0 ),(7:6)

where d(x ÿ x0 ) is a Dirac delta function.
A Dirac delta function can be considered as the limiting case of more familiar
functions. For example, the functiondE (x ÿ x0 ) 



1=2E
0

if jx ÿ x0 j < E
if jx ÿ x0 j > E,

which is illustrated in Fig. 7.1, can behave like a Dirac delta function because it
becomes increasingly high and narrow at x  x0 as E tends to zero. In fact, thedefining property of a Dirac delta function is that, for any function f(x),Z 1
ÿ1

f (x)d(x ÿ x0 ) dx  f (x0 ):(7:7)

This definition is satisfied by dE (x ÿ x0 ) with E ! 0 because, when this limit istaken, the function becomes increasingly high and narrow at x  x0 and thearea under the function remains equal to unity.

7.2

Position and Momentum

139

1/2

d (x − x9)

x
x9
x9 −

x9 +

Fig. 7.1 A function dE (x ÿ x0 ) which behaves like a Dirac delta function d(x ÿ x0 ). As
the parameter E tends to zero, the area under the function dE (x ÿ x0 ) remains equal tounity and the function becomes increasingly high and narrow at x  x0 .

The eigenfunctions of position, like those of any observable, form a complete
set of basis functions. In particular, any wave function C(x, t) may be written as
C(x, t) Z 1
ÿ1c(x0 , t)cx0 (x) dx0(7:8)

where the function c(x0 , t) is a position probability amplitude; i.e. jc(x0 , t)j2 dx0
is the probability of finding the particle between x0 and x0  dx0 at time t. If we use
Eq. (7.6) and the definition of the Dirac delta function, Eq. (7.7), we findC(x, t) Z 1
ÿ1c(x0 , t)d(x0 ÿ x) dx0  c(x, t):This equation confirms the assumption we made in Chapter 3 that a wave
function C(x, t) is a probability amplitude for the position of the particle.

Eigenfunctions for momentum
An eigenfunction for a particle with momentum eigenvalue p0 , cp0 (x), satisfies
the equation^
pcp0 (x)  p0 cp0 (x):This may be rewritten, by using^p  ÿi
h

]
,
]x

(7:9)

140

Observables and operators

Chap. 7

to give the differential equation]cp0 (x) p0 cp0 (x),]x(7:10)

10
cp0 (x)  p eip x=h :2ph(7:11)

ÿihwhich has solutions of the formwave
As expected, an eigenfunction with momentum p0 is a plane p with wave
number k0  p0 =h and wavelength l0  h=p0 . The constant 1= 2ph is a usefulconvention which ensures that the momentum and position eigenfunctions
obey similar normalization conditions.
Because the momentum eigenfunctions form a complete set of basis functions, any wave function C(x, t) may be written as
C(x, t) Z 1
ÿ1c(p0 , t)cp0 (x) dp0 ,(7:12)

where the function c(p0 , t) is a momentum probability amplitude; i.e.
jc(p0 , t)j2 dp0 is the probability that the measured momentum of the particle attime t is between p0 and p0  dp0 . If we use Eq. (7.11), we obtain
1
C(x, t)  p
2p
h

Z 1
ÿ1

0

c(p0 , t) eip x=h dp0 ,which, apart from differences in notation, is identical to the equation
1
C(x, t)  p
2ph

Z 1
ÿ1

e t) eipx=h dp,
C(p,which was introduced in Chapter 3 when we made the assumption that thee t) is the probability amplitude for
Fourier transform of the wave function C(p,momentum; see Eq. (3.19) in Section 3.3. We now see that the assumption wemade in Chapter 3 is consistent with the general description of observables
being developed in this chapter.

Delta function normalization
For the benefit of more mathematically inclined readers we shall briefly discussa minor mathematical problem with the normalization of the eigenfunctions for
position and momentum. This problem arises because both these eigenfunc-

7.3

Compatible Observables

141

tions have continuous eigenvalues; a similar problem afflicts energy eigenfunctions when energy eigenvalues are continuous.
If we use Eq. (7.6) and the definition of the Dirac delta function, Eq. (7.7), we
find that the position eigenfunctions obey the condition
Z 1
ÿ1

0
00
cx*(x)c0x00 (x) dx  d(x ÿ x ):(7:13)

Because the delta function is zero when x0 6 x00 and infinite when x0  x00 , weconclude that position eigenfunctions are mutually orthogonal but that theycannot be normalized to unity.
Similarly, Fourier transform techniques may be used to show that the
momentum eigenfunctions given by Eq. (7.11) obey the condition
Z 1
ÿ1

0
00
cp*(x)c0p00 (x) dx  d(p ÿ p ):(7:14)

Hence momentum eigenfunctions are also mutually orthogonal but not normalizable to unity.
We recall that a wave function must be normalized to unity, as in Eq. (3.17),
if it describes a particle which can always be found somewhere. Thus, strictly
speaking, the position and momentum eigenfunctions, given by Eq. (7.6) andEq. (7.11), cannot be used to describe a physically acceptable wave function.
However, normalizable wave functions can be formed by taking linear superpositions of these eigenfunctions and these wave functions are wave packets
which can be used to describe particles with very small uncertainties in position
or in momentum.
Finally, we note that the use of delta functions can be avoided by imagining
the particle to be in a box of large dimensions. When this approach is adopted,
normalized eigenfunctions of position and momentum can be constructed.

7.3

COMPATIBLE OBSERVABLES

In classical physics it is always possible, in principle, to have precise knowledgeof all the observable properties of a system at a given instant of time. For
example, the specific values of the position and momentum of a particle can beused to find the value of any other dynamical observable. We now know thatthe comprehensive precision of classical physics cannot be realized in practicebecause measurement is an activity which may affect the system. Accordingly,
quantum states of motion may only be specified by data which, in classicalphysics, would be deemed limited or imprecise. If we want precision in quantum physics we have to select a subset of observables which can be determined
without mutual interference or contradiction. Such observables are called

142

Observables and operators

Chap. 7compatible observables. When a complete set of compatible observables is
specified, we can in principle write down a wave function that completely
describes the quantum state. All the information on the complete set of observables and all the information on the probabilities for the uncertain values ofother observables will be contained in this wave function.
To illustrate these general ideas we shall first consider a particle moving in
one dimension and then a particle moving in three dimensions.
A classical state for a particle in one dimension may be defined by specifying
two observables, the position and the momentum. In contrast, a quantum state
for a particle in one dimension is completely defined if one observable isprecisely specified. For example, a quantum state can be defined by a precise
position or by a precise momentum. It could also be defined by precise energy;
such a state is particularly useful because it is a stationary state, a state with notime-dependent observable properties.
More observables are needed to specify a state of a particle moving in three
dimensions. A classical state is defined by six observables, the three position
coordinates and the three components of momentum, but a quantum state canbe defined by only three precisely specified observables. We could, for example,
specify the x, y and z coordinates and have three uncertain components ofmomentum, or we could, for example, specify the x and y coordinates and the z
component of the momentum and live with uncertain momentum in the x and y
directions and an uncertain z coordinate. However, because states with definite
energy are stationary states, it is often most useful to specify the energy and twoother observables.

7.4

COMMUTATORSThe role of compatible and non-compatible observables in quantum mechanics
can be made clearer by introducing the mathematical concept of a commutatorof two operators.
^ is defined by
The commutator of two operators A^ and B^ B]^^  A^ B
^ ÿB
^ A:
[A,

(7:15)

It is a useful concept in the mathematics of operators because, as we shallshow, the order in which two operators act upon a function is important. It
is a useful concept in quantum physics because its value can be used todetermine whether observables are compatible or non-compatible. We shall^ areshow that two observables A and B, described by the operators A^ and B,
non-compatible if^ B]^ 6 0[A,

7.4

Commutators

143

and that they are compatible if^ B]^  0:[A,
This general statement is best understood by reconsidering the quantum states
for a particle in one dimension and in three dimensions.

A particle in one dimension
We can evaluate the commutator of the position and momentum operators fora particle in one dimension by considering^)C(x, t),
(^
x^
pÿ^
px
where C(x, t) is any wave function of the particle. This is non-zero because the^ and ^p matters. Specifically, we haveorder of x


]
^^
x
p C(x, t)  x ÿihC(x, t)
]x
and

^p x
^ C(x, t) 




]
]
x C(x, t)  ÿi
hC(x, t)  x ÿih
C(x, t),
ÿi
h
]x
]x

so that^)C(x, t)  i
(^
x^
pÿ^
px
hC(x, t):(7:16)

Because this is true for any wave function C(x, t), we conclude the operation^ ) is always a multiplication by the number ih; in brief, wedefined by (^x ^p ÿ ^p x
^ and ^
conclude that the commutator of x
p is
[^
x, ^
p]  ih:(7:17)

This relation is so important in quantum mechanics that it is called the canonical commutation relation.
We can reveal the physical significance of the canonical commutation relation by assuming the impossible: the existence of a simultaneous eigenfunctionof position and momentum cx0 p0 (x) satisfying the eigenvalue equations^cx0 p0 (x)  x0 cx0 p0 (x)
x

and

^
pcx0 p0 (x)  p0 cx0 p0 (x):(7:18)

144

Observables and operators

Chap. 7

Such an eigenfunction, if it existed, would represent a quantum state withsharply defined position and momentum, x0 and p0 .
^ and ^p on this hypothetical
Let us consider the action of the commutator of x
simultaneous eigenfunction. If we use the eigenvalue equations (7.18) we obtain^)cx0 p0 (x)  (x0 p0 ÿ p0 x0 )cx0 p0 (x)  0:
x^
pÿ^
px
[^
x, ^p]cx0 p0 (x)  (^If we use the canonical commutation relation, Eq. (7.17), we obtain[^x, ^p]cx0 p0 (x)  ihcx0 p0 (x):These two results imply thatihcx0 p0 (x)  0:
Thus, we have assumed the existence of a simultaneous eigenfunction of position and momentum cx0 p0 (x) and shown that it must be zero for all x. In otherwords, we have shown that a quantum state with definite position and momentum cannot exist. We emphasize that the mathematical reason for the nonexistence of such a state, and hence the non-compatibility of position and^ and ^
momentum, is that the commutator of x
p is non-zero.
Moreover, the degree of non-compatibility of position and momentum, as
expressed by the Heisenberg uncertainty principle Eq. (1.15), may be derived
using the canonical commutation relation (7.17). To fully understand thisderivation, readers need to know about the properties of Hermitian operators
and an inequality called the Schwarz inequality, and they can gain the necessary
understanding by working through problems 5 and 6 at the end of this chapter.
The key steps in the derivation of the Heisenberg uncertainty principle are as
follows:
The square of the uncertainties, or variances, in the position and momentumof a particle with normalized wave function C(x, t) are given by(Dx)2 Z 1
ÿ1c 2 C dx
C* (Dx)and(Dp)2 Z 1
ÿ1c 2 C dx,
C* (Dp)c and Dpc are defined aswhere the operators Dxc x^ ÿ hxi
Dx

and

c^
Dp
p ÿ hpi:

By using the canonical commutation relation (7.17), we can easily show thatthese operators obey the commutation relationc Dp]
c  i
[Dx,h:(7:19)

7.4

Commutators

145

We can now apply the general results derived in problem 4 and 5 to give(Dx)2 (Dp)2 

Z 1

1
4

ÿ1

2

c Dp]
c C dx ,C* [Dx,which can be simplified, using the commutation relation (7.19) and the normalization condition for the wave function C, to give(Dx)2 (Dp)2 

2
h
4

or
h

Dx Dp  :2
Thus, the Heisenberg uncertainty principle, which was introduced in Chapter 1
to illustrate the inherent uncertainties associated with position and momentum
measurements, can be derived by assuming that position and momentum
observables are described by operators that obey the canonical commutation
relation (7.17).

A particle in three dimensions
By considering a particle moving in three dimensions, we can illustrate theconnection between commutators and compatible observables. In this case, aunique quantum state is defined by specifying three compatible observables
which are described by three operators which commute with each other.
For example, we could specify the x and y coordinates and the z componentof the momentum of a particle to define a quantum state. These observables aredescribed by the operators^  x,
x

^
yy

and

p^z  ÿih]:]z

It is easy to show that any two of these operators commute. For example]]^ p^z C  x ÿihC  ÿi
hx C
x
]z
]z
and
^C
p^z x



]
]
xC  ÿihx C:
ÿih
]z
]z

146

Observables and operators

Chap. 7

In fact, we have three commuting operators,y, p^z ]  0,[^x, ^y]  [^
x, p^z ]  [^and simultaneous eigenfunctions of the form
10
cx0 y0 p0z (x, y, z)  d(x ÿ x0 )d(y ÿ y0 ) p eÿipz z :2phMoreover, any wave function C(x, y, z, t) can be expressed as linear superposition of these eigenfunctions as follows:
C(x, y, z, t) 

Z 1
ÿ1

dx0

Z 1
ÿ1

dy0

Z 1
ÿ1

dp0z c(x0 , y0 , p0z , t)cx0 y0 p0z (x, y, z):

In this expression c(x0 , y0 , p0z , t) is a probability amplitude for three compatible
observables. In fact, the probability of finding the particle at time t localized
between x0 and x0  dx0 and between y0 and y0  dy0 , and with momentum in thez direction between p0z and p0z  dp0z , is jc(x0 , y0 , p0z , t)j2 dx0 dy0 dp0z .
This example has illustrated the general procedure of defining a quantum
state of a particle moving in three dimensions by specifying a set of threecompatible observables. This procedure will be used in Chapter 9 when
we construct stationary states of the hydrogen atom by specifying the energy,
the magnitude of the orbital angular momentum and the z component of theorbital angular momentum.

7.5

CONSTANTS OF MOTION

Observables that are compatible with the energy observable have a particularphysical significance. They are constants of the motion. To explain the significance of this statement we consider the expectation value for an observable A
for a particle with wave function C,
Z
hA(t)i 

C*A^ C d3 r:(7:20)

In general, the expectation value hA(t)i will vary with time as the wave
function C(r, t) ebbs and flows in accord with the SchroÈdinger equation
i
h

]C
^
 HC:
]t

(7:21)

7.5

Constants of Motion

147

We can find the rate of change of hA(t)i by differentiating both sides ofEq. (7.20). Using the rules for differentiating a product of functions, we obtain2
dhAi

dt

Z

]C* ^
A C d3 r 
]t

Z

C* A^

]C 3
d r:
]t

If we use the SchroÈdinger equation (7.21) and the complex conjugate of thisequation, we find
dhAi
1
ÿ
dt
ih

Z

1
^
(HC*)
A^ C d3 r 
i
h

Z

^
C* A^ (HC)d3 r:^ like any other operator for an observable in
Because the Hamiltonian H,
quantum mechanics, is a Hermitian operator, we can use Eq. (7.2) to show that
Z

^
(HC*)
A^ C d3 r Z^ A^ C d3 r
C* H

and rewrite the expression for the rate of change of hA(t)i as
dhAi 1

dt
i
h

Z

^ H]
^ C d3 r,
C* [A,(7:22)^ H]^ is a commutator. This equation can be used to determine thewhere [A,time-dependence of the expectation value of any observable.
For an observable A which is compatible with the energy, the commutator^ H]^ is zero and Eq. (7.22) gives
[A,
dhAi
 0:
dt
Such an observable is called a constant of motion because its expectation value
does not change as the wave function evolves with time.
These ideas can be illustrated by considering a particle with the Hamiltonian
2

h
^ ÿ
r2  V (r),H
2mwhere V (r) is a potential energy which only depends on the distance r of theparticle from a fixed origin. For this Hamiltonian, it is easy to show that2Ã does not depend on time. This assumption is possible for any
We are assuming that the operator A
isolated system.

148

Observables and operators^ 6 0,[^r, H]

Chap. 7

^ 6 0,
[^
p, H]

and

^ H]
^  0:[L,

These equations imply that the position and the momentum are not constantsof motion but that the orbital angular momentum is a constant of motion. In
fact, the constants of motion of a system are determined by the symmetry
properties of its Hamiltonian. In this example, the Hamiltonian has rotational
symmetry and this symmetry implies that the orbital angular momentum is aconstant of motion.

PROBLEMS 7
1. Consider a particle moving in a one-dimensional potential energy field V(x).
Show that the operators describing the position, momentum and energy ofthe particle satisfy the following mathematical relations:[^x, ^p] 6 0,^ 6 0,
[^
x, H]

and

^ 6 0:[^p, H]

What is the physical significance of these mathematical relations?
^ and momentum operator ^p
2. (a) Write down the kinetic energy operator T
for a particle of mass m moving along the x axis.
^ and ^(b) Show that T
p satisfy the commutation relation^  0:[^
p, T]
Explain the physical significance of this result.
(c) Show thatc(x)  A cos kx^ but not a eigenfunction
is an eigenfunction of the kinetic operator Tof ^p.
^(d) Are there wave functions which are simultaneously eigenfunctions of Tand ^p? If so, write one down.
3. The canonical commutation relations for a particle moving in three dimensions are[^x, ^px ]  i
h,

[^
y, ^
py ]  ih,and [^z, ^pz ]  ih,Problems 7

149

^, ^
and all other commutators involving x
px , ^y, ^py , ^z, ^pz are zero. These relations can be used to show that the operators for the orbital angular momentum obey the following commutation relations:^ x, L^ y ]  ihL^ z,
[L^ y, L^ z ]  i
^ x,
[L
hL

and

^ z, L^ x ]  ihL
^ y:[L(a) Using^x  ^Ly^pz ÿ ^z^py

and^ y  ^z^
^ ^pz ,
Lpx ÿ xverify that^ y ]  [^^ x, L^^pz ]:
y^pz , ^z^px ]  [^z^py , x[L(b) Using the commutation relations[^z, ^pz ]  i
h

and

[^
y, ^
py ]  ih,verify that^ y ]  i
^ z:
^ x, LhL[L
What is the physical significance of this result?
(c) Using^ x, L^ y ]  ihL^ z,
[L^ y, L^ z ]  i
^ x,
[L
hL

and

^ z, L^ x ]  ihL
^ y,[Lverify that
^ 2, L^ z]  0[L
where^2  L^2  L^ 2:^2  LLxyz
What is the physical significance of this result?
^(Hint: For any two operators A^ and B,^  A^A^ B^ ÿB^ A^A^  A^A^ B^ ÿ A^B^ A^ ÿ B^ A^A^  A^ B
^ A^
[A^ 2 , B]
implies that

150

Observables and operators

Chap. 7

^ A,
^ B]
^ B]
^ )
^  A[^  [A,^ A:[A^ 2 , B]
4. In this problem and in problems 5 and 6 we shall consider some generalmathematical properties of the operators which describe observables inquantum mechanics. To keep the mathematics as simple as possible weshall only consider a particle moving along the x axis.
In general, an operator A^ describing an observable A is a Hermitian
operator which means that it obeys the conditionZ 1ÿ1C1 * A^ C2 dx 

Z 1
ÿ1

^ 1 )* C2 dx,(ACwhere C1 and C2 are any two wave functions; the brackets in the term
^ 1 )* mean that the operator A^ only acts on the wave function C1 and the(ACcomplex conjugate of the result is taken.
By integrating by parts and by assuming that the wave functions go tozero at infinity, verify that the momentum operator ^p  ÿih]=]x is a
Hermitian operator.
5. Consider a particle with wave function C.
(a) Bearing in mind that an observable A is described by a Hermitian
operator, show that its expectation value
hAi 

Z 1
ÿ1

C* A^ C dx

is real.
(b) Show that the expectation value of A2 is given by
Z 1
^
^ dx:
hA2 i 
(AC)*
(AC)
ÿ1

^ that(c) Show for two Hermitian operators A^ and B
Z 1
ÿ1^ C dx 
C* A^ BZ 1
ÿ1Hence, show that
Z 1
ÿ1

*

^ A^ C dx
C* B^ C dx^ B^ A)
C* (A^ B:

Problems 7

151

is real and thatZ 1
ÿ1

^ C dx^ ÿB^ A)
C* (A^ B

is imaginary.
6. In this problem we shall derive an inequality called the Schwarz inequality
and a related inequality which is used to derive the Heisenberg uncertainty
relation.
Let a(x) and b(x) be complex functions of x which give finite values for the
integrals
Z 1
ÿ1

jaj2 dx,

Z 1
ÿ1

jbj2 dx,

Z 1

and

ÿ1

a*b dx,and let f(x) be a complex function given byf(x)  a(x)  lb(x)where l is a complex number. Because
Z 1
ÿ1

jfj2 dx  0,

we haveZ 1
ÿ12

jaj dx  l

Z 1
ÿ1

a*b dx  l*

Z 1
ÿ1

b*a dx  l*l

Z 1
ÿ1

jbj2 dx  0:

Because this inequality is valid for any value of l, it is valid when l is given
by
l

Z 1
ÿ1

2

jbj dx  ÿ

Z 1
ÿ1

b*a dx:(a) Verify the Schwarz inequality
Z 1
ÿ1

jaj2 dx

Z 1
ÿ1

jbj2 dx 

Z 1
ÿ1

2

a*b dx :^ with the function a(x) and BC^ with the function b(x) and(b) Identify AC
use problem 5(b) to show that

152

Observables and operators

2

2

hA ihB i 

Chap. 7

Z 1
ÿ1

2

^ C dx :C* A^ B(c) Now use problem 5(c) to show that hA2 ihB2 i is greater than or equal to
!
2
^ B
^ A^
A^ B
C*
C dx
2
ÿ1

Z 1

!
2
^ ÿB
^ A^
A^ B
C*
C dx :
2
ÿ1

Z 1



7. (a) A particle has a Hamiltonian of the form


h2 ]2

]2
]2
^
 V (x, y, z):Hÿ
2m ]x2 ]y2 ]z2What symmetry property must be satisfied by the potential energy field
V (x, y, z) in order that the x component of the momentum of the particle
is a constant of motion?
(b) A particle has a Hamiltonian of the form

2  2
]
1]
1 ]2
]2
^  ÿ h
 V (r, f, z),H
2m ]r2 r ]r r2 ]f2 ]z2
where (r, f, z) are cylindrical coordinates.
What symmetry properties must be satisfied by the potential energy
field V (r, f, z) in order that the z component of the momentum and the z
component of the orbital angular momentum, two observables described by the operators^pz  ÿi
h

]
]z

]
^ z  ÿi
h ,
L
]f

and

are constants of the motion?
8. In this question you are asked to derive the virial theorem for a particle with
^ T
^ V
^ , with a kinetic energy operator given by
Hamiltonian H2h r^  ÿ
T2m
and a potential energy operator given by^  V (r):V

2

Problems 7

153(a) Show thath 2
^  i
^p[^r  ^p, T]mand that^ ]  ÿihr dV :[^r  ^p, V
dr

^ with eigenvalue E. Bearing in(b) Consider cE (r), an eigenfunction of H^
mind that H is a Hermitian operator, show thatZ

^ cE d3 r  0:p, H]
cE* [^r  ^Hence show thatZ
2

^ cE d3 r 
cE* T

Z
cE* r

dV
c d3 rdr E

which is a statement of the virial theorem.
(c) Show that the expectation values for the kinetic and the potential
energies of a particle in a state with definite energy are related byhTi  hV i

if the potential is

V (r)  12m!2 r2 ,2hTi  hV i

if the potential is

V (r)  ÿand bye2:
4pE0 rThis page intentionally left blank8
Angular momentum
Planck's constant has the units of angular momentum. This suggests that h, orh  h=2p, may be the fundamental unit for angular momentum. It also suggeststhat angular momentum may be a fundamental observable in quantum physics.
Indeed, there are point-like quantum particles which have an intrinsic angularmomentum called spin. The spin of a point particle cannot be related to orbital
motion of constituent parts; it is a fundamental property which has no analogue in classical physics.
The nitty-gritty of spin and orbital angular momentum forms a major part ofadvanced books on quantum mechanics. In this chapter, we shall only set out
and illustrate the most important aspects of this demanding topic. We shallbegin by considering the basic properties of angular momentum and thendescribe how these properties may be revealed by the interaction of magnetic
moments with magnetic fields. Finally, we shall describe how the angular shapeof a wave function is related to the orbital angular momentum of the particle
described by this wave function.

8.1

ANGULAR MOMENTUM BASICS

Quantum particles may possess an orbital angular momentum and an intrinsic
angular momentum, called spin. In appropriate circumstances, the orbital
angular momentum resembles the orbital angular momentum of a classicalparticle; it is a vector, with direction and magnitude describing the inertia ofangular motion. In contrast, spin angular momentum does not have a classicalmanifestation; it is a fundamental quantum property which bears little resemblance to a rotating classical object.
The most important property of angular momentum in quantum mechanics
is that the outcome of measurement is at best a fuzzy vector, a vector with two
defining properties: a definite magnitude and a definite value for just one of its
three Cartesian components. Accordingly, an angular momentum in quantum

156

Angular momentum

Chap. 8physics may be specified using two quantum numbers. Normally, l and ml areused to describe orbital angular momentum, s and ms are used for spin angularmomentum, and the quantum numbers j and mj are used when the angularmomentum arises from a combination of spin and orbital angular momentum,
and when a general angular momentum is being described.1
The only possible precise values for the magnitude of orbital angular momentum are given byLp
l(l  1)
h,

where

l  0, 1, 2, 3, . . . :(8:1)

When the magnitude is fixed by the quantum number l, the orbital angularmomentum in any given direction may have 2l  1 possible values between ÿlhand lh. For example, if we choose to measure the orbital angular momentum
in the z direction, there are 2l  1 possible outcomes given byh,Lz  ml 

where

8
l
>
>
>
(l ÿ 1)>>>>>< (l ÿ 2)
ml  ...
>>> ÿ(l ÿ 2)
>
>
>
>
>
: ÿ(l ÿ 1)
ÿl:

(8:2)

But when this is done, the orbital angular momentum in the x and y directions
are uncertain, in the sense that, if we choose to measure the x or y component,
the possible outcomes will have quantized values somewhere in the range ÿlh tolh. We shall check the validity of all these general statements about orbital
angular momentum in Section 8.3. Readers should note for future reference
that the classification of atomic spectra has led to the spectroscopic notation inwhich the letters s, p, d, f, and g are used to label quantum states withl  0, l  1, l  2, l  3 and l  4. Accordingly, states with l  0 are calleds-states, states with l  1 are called p-states, and so on.
The quantum numbers s and ms are usually used when the angular momentum
is solely due to spin. A particle
is said to have spin s if the magnitude of the spinp
angular momentum is S  s(s  1)
h and if the z components are given bySz  ms 
h,

1

where

8
s
>
>
>
>
< (s ÿ 1)
ms  ...
>>>>: ÿ(s ÿ 1)
ÿs:

(8:3)

In Chapter 11 we shall use capital letters, L, S and J, for the angular momentum quantum
numbers of two or more electrons, but in this chapter these letters will denote the magnitude of anangular momentum.

8.1

Angular Momentum Basics

157

For example, the W boson is a spin-one particle with s  1 and ms  1, 0, ÿ 1
and the electron is a spin-half particle with s  12 and ms   12. Thus, spin
angular momentum can be integer, like orbital angular momentum, but it canalso be half-integer.
Orbital and spin angular momenta may be combined to give a total angularmomentum with magnitude and z component given byJp
j(j  1)handJz  mj h,(8:4)

where, in general, the quantum numbers j and mj may take on integer and halfinteger values given by13
j  0, , 1, , 2, . . . . . . : and
2
2

8 j
>
>
>
> (j ÿ 1)<.
mj  ..
>>>>: ÿ(j ÿ 1)
ÿj:

(8:5)

The actual values of the quantum number j depend on the orbital and spin
angular momenta being combined. It can be shown that, when an orbital
angular momentum with quantum number l is combined with a spin withquantum number s, several total angular momenta may arise with quantum
numbersj  l  s, l  s ÿ 1, . . . jl ÿ sj:(8:6)

For example, we can have j  32 and 12 when l  1 and s  12, and we canhave j  2, 1 and 0 when l  1 and s  1. We note that, in general, twoangular momenta with quantum numbers j1 and j2 may be combined togive an angular momentum with quantum number j which can take on thevaluesj  j1  j2 , j1  j2 ÿ 1, . . . , jj1 ÿ j2 j:Earlier we referred to an angular momentum defined by two quantum
numbers as a fuzzy vector. The fuzziness arises because, when one of its
Cartesian components is sharply defined, the other two components are uncertain but quantized when measured. In view of the uncertainties we have alreadyencountered in position, momentum and energy, uncertain angular momentum
should not be a surprise. Indeed, the uncertainty in orbital angular momentum
can be directly traced to the uncertainties in the position and momentum of aparticle, as indicated in problem 3 at the end of Chapter 7. But it is surprisingthat angular momentum in any given direction can only equal an integer or158

Angular momentum

Chap. 8half-integer multiple of h. We shall see how this surprising property may be
confirmed experimentally in the next section.2

8.2

MAGNETIC MOMENTS

In this section we shall consider magnetic moments and then describe how theinteraction of a magnetic moment with a magnetic field can reveal the properties of an angular momentum.

Classical magnets
The simplest magnetic moment in classical physics consists of an orbiting
charged particle. This magnetic moment is directly proportional to the orbital
angular momentum and it is given by
m

q
L,
2m

(8:7)

where q is the charge and m is the mass of the orbiting particle.
We can check the validity of this relation by considering a particle moving ina circular orbit of radius r with speed u, as shown in Fig. 8.1. Such a particle
gives rise to a circulating electrical current, I, and hence to a magnetic momentwith magnitude IA, where A is the area of the orbit. Because the current is equal
to the charge q divided by the period of the orbit 2pr=u, we have
m

qu
qru
pr2 
,
2pr
2
L = mr 3 v

rv

Fig. 8.1 A particle with charge q and mass m moving in a circular orbit with orbital
angular momentum L  mr  v gives rise to a magnetic moment m  qL=2m.
2

More advanced texts show that the operators for angular momentum also generate rotations. The
geometrical properties of rotations in three dimensions imply that the angular momentum
operators obey commutation relations which require angular momentum to be quantized in
integer and half-integer multiples of h.

8.2

Magnetic Moments

159

which may be rewritten using the angular momentum L  mru as
m

q
L:
2m

The vector version of this equation, Eq. (8.7), follows because the directions of
the magnetic moment and of the orbital angular momentum are both perpendicular to the plane of the orbit.

Quantum magnets
In quantum physics, magnetic moments are also proportional to angularmomenta, but they are now at best fuzzy vectors, with precise values for theirmagnitude and for one Cartesian component. We shall illustrate the relation
between quantum magnets and angular momenta by considering the magnetic
properties of electrons, atoms, protons and neutrons.
The magnetic properties of an electron arise from a spin with magnitude andz component given bySq
ÿ

1 1
h
2 21 andSz  ms h,and from an orbital angular momentum with magnitude and z component
given byLp
l(l  1) handLz  ml h:

The associated magnetic moments are given by formulae similar to Eq. (8.7),
but with minor modifications. In particular, the z components of the magnetic
moments due to electron spin arem(z Spin)  ÿ2eehSz  ÿ2
ms
2me
2me

(8:8)

and those due to orbital angular momentum areÿm(Orbital)
z

e
e
h
Lz  ÿ
ml ,
2me
2me

(8:9)

where me is the mass and ÿe is the charge of an electron. We note thatelectron spin angular momentum is twice as magnetic as orbital angular momentum; the additional factor of 2 in Eq. (8.8) is explained by the Diracequation, a wave equation for relativistic, point-like quantum particles withspin half.

160

Angular momentum

Chap. 8The magnetic moment of an atom arises from the combined spin and orbital
angular momentum of the constituent electrons which can be described byquantum numbers j and mj . In a weak magnetic field, the z component of themagnetic moment of an atom is ÿgm(Atom)zeehJz  ÿg
mj ,
2me
2me

(8:10)

where Jz  mj h is the z component of the angular momentum and g is anumerical factor called the LandeÂ g-factor. In fact, the LandeÂ g-factor for an
atomic state with quantum numbers j, l and s isg1j(j  1) ÿ l(l  1)  s(s  1):
2j(j  1)

The LandeÂ g-factor has the value g  2 if the sole source of magnetism in theatom is due to electron spin, and the value g  1 if due to orbital angularmomentum. Because mj can take on 2j  1 values between ÿj and j, themagnetic moment of the atom has 2j  1 components.
Equations (8.8), (8.9), and (8.10) indicate that the natural unit for magnetic
moments associated with electrons is
mB 

e
h
 9:274  10ÿ24 J Tÿ1 :2me(8:11)

This fundamental constant is called the Bohr magneton.
Protons and neutrons, unlike electrons, are composite objects containing
quarks and gluons. These constituents give rise to angular momenta withquantum numbers j  12 and mj   12. The z components of the associatedmagnetic moments are 2:79m(Proton)
z

e
h
mj
2mp

and

m(Neutron) ÿ1:95
z

eh
mj ,
2mp

(8:12)

where mp is the mass of the proton. We note that the natural unit for thesemagnetic moments, and also for the magnetic moments of nuclei containing
protons and neutrons, is
mN 

eh
 5:05  10ÿ27 J Tÿ1 :2mpThis unit is called the nuclear magneton.

(8:13)

8.2

Magnetic Moments

161

Magnetic energies and the Stern±Gerlach experiment
When a classical magnetic moment m is placed in a magnetic field B, it has anenergy of orientation given by
Emag  ÿm  B:

(8:14)

If we choose the direction of the magnetic field to be the z direction, we have
Emag  ÿmz B,(8:15)

where mz is the z component of the magnetic moment, which can take on any
value between m and ÿm. Hence, for a classical magnet, there is a continuumof energies of orientation between ÿmB and mB.
In marked contrast, the energy of orientation of a quantum magnet in amagnetic field is quantized. For a field B in the z direction, this energy is givenby ÿmz B, where mz is now the quantized z component of the magnetic moment.
For example, we can use Eq. (8.10) to show that the magnetic energy of an
atom in an atomic state with angular quantum numbers j and mj is
Emag  mj gmB B,
where mB is the Bohr magneton and g the LandeÂ g-factor. Thus, for a given
value of j, there are 2j  1 magnetic energy levels given by8
j gm B>>> (j ÿ B1) gm B
>
>
B
<
..
Emag  .
>>>ÿ(j ÿ 1) gmB B>>:
ÿj gmB B:(8:16)

When j  12 there are two energy levels, when j  1 there are three energy levels,
when j  32 there are four energy levels, and so on, as shown in Fig. 8.2.
Indirect evidence for atomic magnetic energy levels is provided by observing
the effect of a magnetic field on spectral lines. The magnetic field splits atomic
energy levels with a given j into 2j  1 magnetic energy levels with different
values for mj , and radiative transitions between states with different values of j
now give rise to several closely spaced spectral lines instead of one. This effect iscalled the Zeeman effect.
However, direct evidence for the quantization of magnetic energies is provided by a Stern±Gerlach experiment. In this experiment individual atoms passthrough a non-uniform magnetic field which separates out the atoms according
the value of their magnetic moment in a given direction.

162

Angular momentum

Chap. 8

Emag

1

j =1

j= 2

3

j= 2

Fig. 8.2 The energy levels in a magnetic field of an atom in states with angularmomentum quantum numbers j  12, 1 and 32. The spacing between levels is given by
gmB B where B is the strength of the magnetic field, mB is the Bohr magneton and g is a
LandeÂ g-factor, a constant which depends on the spin and orbital angular momentum
quantum numbers of the atomic state.

The main features of a Stern±Gerlach experiment are illustrated in Fig. 8.3.
A beam of atoms is passed through a magnetic field produced by speciallyshaped poles of an electromagnet. The direction of the magnetic field is largelyin one direction, the z direction say, but its strength, B(x, y, z), increases
markedly as z increases. In this field, each atom acquires an energy
Emag (x, y, z)  ÿmz B(x, y, z)which depends upon the z component of its magnetic moment mz and on thelocation in the field. Because this magnetic energy varies strongly with z, theatom is deflected by a force in the z direction which is given by

z
z
S

y
x

Collimatedbeam ofatoms

N

Observation
screenM agnetxFig. 8.3 The Stern±Gerlach experiment in which atoms pass through a non-uniform
magnetic field which separates out atoms according to the value of the magnetic momentin the direction of maximum non-uniformity of the field.

8.3

F ÿ

Orbital Angular Momentum

163

]Emag
]B
 mz
:
]z
]z

If the z component of the magnetic moment could take on any value between
m and ÿm, the atomic beam would be smeared out as atoms are dragged up
and down by varying amounts. But for real atoms in states with quantum
number j, the z component of the magnetic moment can only take on 2j  1
discrete values and a beam of such atoms is split into 2j  1 separate beams.
In their original experiment, Stern and Gerlach discovered that a beam ofsilver atoms, in their ground state, is split into two separate beams. This implies
that a measured Cartesian component of the magnetic moment of a silver atom
in its ground state can only take on two possible values and that the angularmomentum quantum numbers for the atom are j  12 and mj  12. They also
showed, by measuring the separation between the two beams of atoms
emerging from the electromagnet, that the magnitude of the magnetic momentof a silver atom is of the order of a Bohr magneton.

8.3

ORBITAL ANGULAR MOMENTUM

In this section we shall remind the reader of the definition of orbital angularmomentum in classical physics, introduce the operators which describe orbital
angular momentum in quantum physics, and then consider how the angularshapes of wave functions are related to orbital angular momentum properties.
In so doing, we shall confirm some of the general statements made about
orbital angular momentum in Section 8.1.

Classical orbital angular momentum
Consider a particle at time t with vector position and momentumr  (x, y, z)andp  (px , py , pz ):The orbital angular momentum about the origin of coordinates is given by thevector productL  r  p,which is a vector with three Cartesian components,
Lx  ypz ÿ zpy ,and a magnitude given by

Ly  zpx ÿ xpz ,Lz  xpy ÿ ypx ,164

Angular momentum

Chap. 8jLj q
L2x  L2y  L2z :Quantum orbital angular momentum
Orbital angular momentum in quantum physics is described by the operator^  ^r  ^Lp  ÿihr  r:

(8:17)

This a vector operator with three Cartesian components,




]
]
^ y  ÿi^ x  ÿih y ] ÿ z ] , L
h z ÿx
,
L
]z
]y
]x
]z


]
]
^
,
h x ÿyLz  ÿi]y]xthat act on wave functions representing possible quantum states of a particle.3
When the wave function C(r, t) is known, expectation values for orbital
angular momentum may be calculated. For example, the integralsZ
hLx i ^ x C d3 r
C* L

and

hL2x i Z^ 2 C d3 r
C* Lxgive the expectation values for the x component and the square of the xcomponent of the orbital angular momentum. And when the wave function is^ x with eigenvalue Lx , i.e. whenan eigenfunction of L
^ x C(r, t)  Lx C(r, t),L
we can follow the procedure outlined in Section 4.3 and showthat hLx i  Lx andq

that hL2x i  L2x . This implies that the uncertainty DLx  hL2x i ÿ hLx i2 is zeroand that the eigenfunction represents a quantum state with a precise value forthe x component of the orbital angular momentum given by the eigenvalue Lx .

Angular shape of wave functions
The wave function of a particle can have an infinite variety of angular shapes.
But any wave function can be expressed in terms of basis wave functions with3^  (S^x , S^y , S^ z ) which acts on a
Spin angular momentum is usually described by an operator S
quantum state which includes a description of the spin properties of the particle. Spin operatorsand spin quantum states are usually represented by matrices.

8.3

165

Orbital Angular Momentum

simpler angular shapes. These basis wave functions are usually taken to be
wave functions with specific orbital angular momentum properties. Accordingly, we shall consider some wave functions with simple angular dependence
and deduce the orbital angular momentum properties of the particle theydescribe. The properties of the following wave functions will be explored: The
spherically symmetric wave function given byc(0, 0)  R(r),(8:18)

where R(r) is any well-behaved function of r 
functionszc(1, 0)  R(r) ,rc(1, 1)  R(r)(x  iy),rp
x2  y2  z2 , and the wavec(1, ÿ1)  R(r)(x ÿ iy):r(8:19)The rationale for the labels (0,0), (1,0) and (1,  1) will become clear after wehave determined the angular momentum properties of the states described bythese wave functions.
The position probability densities for these wave functions,jc(0, 0) j2  jR(r)j2 ,z2
jc(1, 0) j2  jR(r)j2 2
randjc(1, 1) j2  jR(r)j2(x2  y2 ),r2

are illustrated in Fig. 8.4. We note that a particle described by the wave
function c(0, 0) is equally likely to be found at any point on the surface of asphere of radius r, whereas particular regions of the surface are more likely
locations for a particle described by the wave functions c(1, 0) and c(1, 1) . For
the wave function c(1, 0) the North and South poles are more probable(0, 0)(1, 0)(1,61)

Fig. 8.4 The position probability densities on the surface of a sphere for a particle withwave functions c(0, 0) , c(1, 0) and c(1, 1) given by Eqs. (8.18) and (8.19). For futurereference, these wave functions have orbital angular quantum numbers (l, ml ) equal to(0, 0), (1, 0) and (1,  1). (This figure was produced with the permission of Thomas D.
York.)

166

Angular momentum

Chap. 8locations, and for the wave functions c(1, 1) equatorial regions near z  0 are
more likely.
To find the orbital angular momentum properties of the particle described bywave functions Eqs. (8.18) and (8.19), we consider the action of the angularmomentum operator given in Eq. (8.17) on these wave functions.
^ on the function R(r).
We first consider the action of the vector operator LUsing^  ÿiLhr  =

and=R(r)  er

dR
,
dr

where er is a unit vector in the direction of r, we obtaindR^ R(r)  ÿi:L
hr  =R(r)  ÿihr  er
dr^
 0. Hence, the spherically symmetricBecause r  er  0, we deduce that LR(r)wave function c(0, 0)  R(r) satisfies the three equations:^ x c(0, 0)  Lx c(0, 0) ,L^ y c(0, 0)  Ly c(0, 0) ,L^ z c(0, 0)  Lz c(0, 0) ,Lwith Lx  0,with Ly  0,with Lz  0:

It also satisfies the equation^ 2 c(0, 0)  L2 c(0, 0) ,Lwith

L2  0,where^2  L^2  L^ 2:^2  LLxyz
These equations show that any spherically symmetric wave function is a simultaneous eigenfunction of the operators which describe the magnitude and eachof the three Cartesian components of the orbital angular momentum operator,
and that in each case the eigenvalue is equal to zero. We conclude that all
spherically symmetric wave functions describe a particle with zero orbital
angular momentum.
We shall now consider the wave functionc(1, 0)  R(r)

z
r

8.3

Orbital Angular Momentum

167

which describes a quantum particle that is more likely to be found near the
North or South pole and not near the Equator, as shown in Fig. 8.4. To find theorbital angular properties of this particle, we evaluate the action of angularmomentum operators on its wave function. Using the rules for the differentiation of a product, we obtain^ R(r)  R(r) Lz^  R(r) Lz,^^ R(r) z  zL
L
r
r
r
r
which implies that


R(r)
]
]
R(r)
^ x z  ÿi^ x c(1, 0)  R(r) Ly ÿzz  ÿih
y,
h
L
r
r
]z
]y
r


^ y c(1, 0)  R(r) L^ y z  ÿih R(r) z ] ÿ x ] z  ih R(r) x,
L
r
r
]x
]z
r


R(r)
R(r)
]
]
^ z c(1, 0) 
^ z z  ÿiLLx ÿyz  0:hrr]y]x
The first two equations show that the wave function c(1, 0) is not an eigenfunc^ y , but the third equation shows that c(1, 0) is an eigenfunction^ x or of L
tion of L^
of Lz with zero eigenvalue because^ z c(1, 0)  Lz c(1, 0) ,Lwith

Lz  0:(8:20)

By evaluating terms like^ x z  ÿihy,L^ x y  i
L
hz

and

^ 2 z  h2 z,
Lx

it is also easy to show that^ 2 c(1, 0)  
h2 c(1, 0) :LxSimilarly we can easily show that^ 2 c(1, 0)  L
h2 c(1, 0)yand^ 2 c(1, 0)  0:
Lz

When we combine these results we find that
^2  L^ 2 )c(1, 0)  2
^2  Lh2 c(1, 0) :
(L
x
y
z

(8:21)

Equations (8.20) and (8.21) show that the wave function c(1, 0) is a simultan^2  L^2  L^2  L^ 2 and L^ z with eigenvalues
eous eigenfunction of the operators L
x
y
z

168

Angular momentum

Chap. 82
L2 p2h and Lz  0. Therefore, it describes a particle with a precise magnitude
L  2h and precise z component Lz  0, but its orbital angular momentum inthe x and y directions are uncertain.
Clearly, we can construct other wave functions with similar properties. For
example, if we replace z in the expression for c(1, 0) by x or by y, we obtain the
wave functionsc0(1, 0)  R(r)

x
r

and

y
c00(1, 0)  R(r) :r(8:22)

Both these wave functions describe a particle with an orbital angular momenp
tum of magnitude L  2h; but for c0(1, 0) the x component is zero and the yand z components are uncertain, and for c00(1, 0) the y component is zero and thez and x components are uncertain.
We shall finally consider the wave functionsc(1, 1)  R(r)(x  iy)randc(1, ÿ1)  R(r)(x ÿ iy),rboth of which describe a quantum particle which is more likely to be found near
the Equator and never at the North or South poles, as shown in Fig. 8.4. By
evaluating the action of the angular momentum operators on the functions^xx  iy, it is easy to show that these wave functions are not eigenfunctions of L2^^^
or of Ly , but that they are both simultaneous eigenfunctions of Lz and L . Infact,
^ z c(1, 1)  Lhc(1, 1)and^ 2 c(1, 1)  2h2 c(1, 1)L^ z c(1, ÿ1)  ÿL
hc(1, ÿ1)and^ 2 c(1, ÿ1)  2h2 c(1, ÿ1) :
L

and

p
Thus, the wave function c(1, 1) describes a particle with Lz  h and L  p2h,and the wave function c(1, ÿ1) describes a particle with Lz  ÿh and L  2h;in both cases, the x and y components of the orbital angular momentum are
uncertain.
By exploring the properties of these simple wave functions, we have illustrated three general properties of orbital angular momentum in quantum
physics:. Orbital angular momentum in quantum physics is quantized and the naturalunit for angular momentum ish  1:055  10ÿ34 J s:

8.3

Orbital Angular Momentum

169. The orbital angular momentum of a quantum particle is at best a fuzzy
vector. We have only been able to specify precisely the magnitude and justone of the components of orbital angular momentum. This is because thecomponents of angular momentum are non-compatible observables as discussed generally in Chapter 7.
. A quantum particle with specific orbital angular momentum properties has awave function with a specific angular shape. If the orbital angular momentum is zero the wave function is spherically symmetric, and if the orbital
angular momentum is non-zero the wave function has angular dependence.

Spherical harmonics
So far we have considered wave functions to be functions of the Cartesian
coordinates x, y and z. In practice, it is more useful to consider wave functions
to be functions of the spherical polar coordinates r, y and f illustrated Fig. 8.5.
This figure shows that the Cartesian and spherical coordinates of the point P
are related byx  r sin y cos f,y  r sin y sin f,andz  r cos y:

When a quantum state is represented by a wave function C(r, y, f), thedependence on y and f specifies an angular shape that determines the orbital
angular momentum properties of the state. In fact, all possible orbital angularmomentum properties can be described using simultaneous eigenfunctions of^ 2 and L^ z . These eigenfunctions are called spherical harmonics. They areLdenoted Yl , ml (y, f) and they satisfy the eigenvalue equations:
z
P
r

q
y
f

x

Fig. 8.5

The spherical polar coordinates (r, y, f) of the point P.

170

Angular momentum

Chap. 8^ 2 Yl , m  l(l  1)
L
h2 Yl , mlland

^ z Yl , m  ml hYl , m ,Lll(8:23)

where the quantum numbers l and ml can take on the values l  0, 1, 2, . . . andml  ÿl, . . . ,  l.
These eigenfunctions are orthogonal because they satisfyZ
0(8:24)
Y *l 0 , m0 Yl , ml dV  0 if l 0 6 l and ml 6 mlland they are usually normalized so thatZ
jYl , ml j2 dV  1:

(8:25)In these integrals dV is the solid angledV  sin y dy df
and the limits of integration are from y  0 to y  p and from f  0 to f  2p.
Explicit forms of the spherical harmonics with l  0, l  1, and l  2 aregiven in Table 8.1. If we compare these with the wave functions given by
Eqs. (8.18) and (8.19), we see thatc(0, 0) / Y0, 0 (y, f)
and thatc(1, 0) / Y1, 0 (y, f) andTABLE 8.1

c(1, 1) / Y1, 1 (y, f):

Spherical harmonics with l  0, 1 and 2

Spherical harmonics as functionsof y and fr
1
Y0, 0 4p

Spherical harmonics as functionsof x,y and zr
1
Y0, 0 4pr
3
cos y
Y1, 0 4pr
3
sin y eif
Y1, 1  
8pr
5
Y2, 0 (3 cos2 y ÿ 1)
16p

r
3 z
Y1, 0 4prr
3 x  iy
Y1, 1  
8p rr
5 3z2 ÿ r2
Y2, 0 16p r2r
15
Y2, 1  
sin y cos y eif8pr
15
Y2, 2 sin2 y e2if
32pr
15 (x  iy)zY2, 1  
8p
r2r
2
15 x ÿ y2  2ixyY2, 2 
32p
r2

8.3

Orbital Angular Momentum

171

We also note that spherical harmonics have a simple dependence on theazimuthal angle f, given by
Yl , ml (y, f)  Fl , ml (y) eiml f ,(8:26)

but that the y dependence becomes increasingly complicated as l increases.
The angular shape of the position probability density for a particle withangular momentum quantum numbers l and ml is given by jYl , ml (y, f)j2 . The
angular shapes for l  0 and l  1 were shown in Fig. 8.4 and the more
complex shapes for l  2 and l  3 are shown in Figs. 8.6 and 8.7. We notethat there is no dependence on the azimuthal angle f, but the dependence on
the angle y becomes more complex as l increases.

Linear superposition
We have already emphasized that each orbital angular momentum eigenfunction has a specific angular shape. We shall now describe how these shapes form
a complete set of angular shapes. To illustrate this idea in the simplest possible
context, we shall focus exclusively, for the moment, on the f dependence of the
wave function and suppress any reference to the r and y coordinates.
Any complex function c(f) in the interval 0  f  2p can be expressed as
the Fourier seriesc(f) 

Xncn einf

where n is an integer that runs from ÿ1 to 1, and where the coefficients cn
are given by(2, 0)(2,61)(2,62)

Fig. 8.6 The position probability densities on the surface of a sphere for a particle
with quantum numbers (l, ml ) equal to (2, 0), (2,  1) and (2,  2). (This figure wasproduced with the permission of Thomas D. York.)

172

Angular momentum

Chap. 8(3, 0)(3,61)(3,62)(3,63)

Fig. 8.7 The position probability densities on the surface of a sphere for a particle withquantum numbers (l, ml ) equal to (3, 0), (3,  1), (3,  2), and (3,  3). (This figure wasproduced with the permission of Thomas D. York.)

1
cn 
2p

Z 2p
0

eÿinf c(f) df:

To bring the notation into line with the conventions of quantum physics, weshall rewrite this Fourier series asc(f) Xml

cml Zml (f),eiml fwhere Zml (f)  p ,2p(8:27)

where ml is an integer that runs from ÿ1 to 1.
In problem 5, we shall show that the basis functions Zml (f) are eigenfunc^ z with eigenvalues ml 
h. Thus Eq. (8.27) is yet another example of thetions of L
principle of linear superposition in quantum mechanics, which states that any
quantum state is a linear superposition of other quantum states; in this case, alinear superposition of quantum states with definite values for Lz . The coefficients cml are probability amplitudes for Lz , because jcml j2 is the probability that

8.3

Orbital Angular Momentum

173

the measured value of the z component of orbital angular momentum is equal
to ml h.
In a similar way, the y and f dependence of any wave function can be
expressed as a generalized Fourier series involving basis functions which are^ z . These eigenfunctions form a complete set of three^ 2 and L
eigenfunctions of L
dimensional angular shapes so that any wave function c(r, y, f) can be expressed asc(r, y, f) lX
1 mlX
l0 ml ÿlcl , ml (r)Yl , ml (y, f):(8:28)

By using the orthogonality and normalization conditions for spherical harmonics, Eqs. (8.24) and (8.25), we can show that the coefficients cl , ml (r) of this series
are given byZ
cl , ml (r) Y l*, ml (y, f)c(r, y, f) dV:(8:29)

These coefficients are probability amplitudes for orbital angular momentum;in fact, the probability that the particle
is foundbetween r and r  dr withp

orbital angular momentum L  (l(l  1)
h and Lz  ml h is given by
jcl , ml (r)j2 r2 dr.
0
As an example, let us consider the wave function c(1, 0) given by Eq. (8.22).
By using Table 8.1, we findr
2p R(r)[Y1, 1 (y, f)  Y1, ÿ1 (y, f)]:c(1, 0)  ÿ
3 r
0

Because this is a linear superposition of spherical harmonics withl  1, ml  1 and l  1, ml  ÿ1, a measurement of the magnitude andz component
pof the orbital angularp momentum can have two possible outcomes: L  2h, Lz  h or L  2h, Lz  ÿh. Because the magnitudes of thecoefficients of the superposition are the same, each of these outcomes has thesame probability.
The linear superposition given by Eq. (8.28) provides a useful representationof the wave function of a scattered particle. In this case, the function cl , ml (r) is
called a partial wave. It can be decomposed into an incoming spherical wave
and an outgoing spherical wave, and the effect of scattering is to cause a shift in
the phase of the outgoing spherical wave. The analogous phase shift in a onedimensional scattering process was considered in Section 5.1, but in a threedimensional scattering process, there is a phase shift for each orbital angularmomentum. These phase shifts can be used to calculate the scattering crosssection.

174

Angular momentum

Chap. 8PROBLEMS 8
1. A particle has orbital angular momentum given by the quantum numberl  3 and spin angular momentum given by the quantum number s  1.
(a) How many distinct states are there with different values for the z
components of the orbital and spin angular momenta?
(b) What are the possible values for the quantum number j that describes
the total angular momentum of the particle?
(c) How many distinct states are there with different values for the magnitude and z component of total angular momentum?
(Note that the rules for the addition of angular momenta given by Eq. (8.6)
are such that, when angular momenta with quantum numbers l and s are
combined to give total angular momenta with quantum numbers
j  l  s, l  s ÿ 1, . . . jl ÿ sj, then the number of distinct states with different values for ml and ms is equal to the number of distinct states withdifferent values for j and mj .)2. A classical electron moves in a circle of radius 1 mm with velocity 1 mm sÿ1 .
(a) What is the value of the quantum number l which gives a quantizedangular momentum close to the angular momentum of this classicalelectron?
(b) How many discrete values are possible for the z component of thisorbital angular momentum?
(c) How closely spaced are these values as a fraction of the magnitude of theorbital angular momentum?
3. The ground state of the hydrogen atom consists of an electron and a protonwith zero orbital angular momentum and with magnetic moments given by
Eq. (8.8) and Eq. (8.12). The atom is placed in a magnetic field of 0.5 T.
(a) Explain why, if the effect of the proton magnetic moment can beignored, the ground state energy is split into two energy levels. What is
the spacing between these energy levels in eV?
(b) Explain why each of these magnetic energy levels consists of two closely
spaced energy levels if the effect of the proton magnetic moment is taken
into account, and if the external magnetic field is large compared with

Problems 8

175

any internal field. What is the spacing between these closely spaced
levels in eV?
4. Two particles of mass m are attached to the ends of a massless rod of lengtha. The system is free to rotate in three dimensions about its centre of mass.
(a) Write down an expression for the classical kinetic energy of rotation of
the system, and show that the quantum rotational energy levels are givenbyEl l(l  1)h2
ma2

with

l  0, 1, 2, . . . :(b) What is the degeneracy of the lth energy level?
(c) The H2 molecule consists of two protons separated by a distance of
0.075 nm. Find the energy needed to excite the first excited rotational
state of the molecule.
5. (a) By considering the relation between Cartesian and spherical polar coordinates,x  r sin y cos f,y  r sin y sin f,andz  r cos y,and the chain rule]c ]c ]x ]c ]y ]c ]z,]f ]x ]f ]y ]f ]z ]fshow that the operator for the z component of the orbital angularmomentum of a particle,]]b z  ÿi,h x ÿyL]y]x
can be rewritten asb z  ÿih ] :L]f(b) Verify thateiml fZml (f)  p
2p

176

Angular momentum

Chap. 8^ z with eigenvalue ml 
is an eigenfunction of Lh.
(c) Explain why it is not unreasonable to assume that any wave function
satisfies the conditionc(r, y, f)  c(r, y, f  2p):Show that this condition implies that ml is an integer.
(d) Show that the integral
Z 2p0

Zm*0l (f)Zml (f) df

is equal to one if m0l  ml and zero if m0l 6 ml .
(e) Show that, ifc(r, y, f) Xmlcml (r, y)Zml (f),thencml (r, y) Z 2p0

Z m*l (f)c(r, y, f) df:

6. Consider a wave function with azimuthal dependencec(r, y, f) / sin 2f cos f:
What are the possible outcomes of a measurement of the z component of theorbital angular momentum and what are the probabilities of these outcomes?
(This question can be tackled using the expression for cml given in the lastpart of the preceding problem, but the simplest approach is to rewrite sin 2f
and cos f in terms of complex exponentials and tidy up.)
7. Consider a wave function with azimuthal dependencec(r, y, f) / cos2 f:
What are the possible outcomes of a measurement of the z component of theorbital angular momentum and what are the probabilities of these outcomes?

Problems 8

177

8. In spherical polar coordinates the operator for the square of the orbital
angular momentum is^ 2  ÿ
L
h2

 2

]
cos y ]1 ]2:
]y2 sin y ]y sin2 y ]f2

b z have the formb 2 and LShow that the simultaneous eigenfunctions of L
Yl , ml (y, f)  Fl , ml (y) eiml f ,where the function Fl:ml (y) satisfies the differential equation
d2 Fl , ml
dy2





m2
cos y dFl , ml
 l(l  1) ÿ 2l Fl , ml  0:
sin y dysin y

It can be shown that finite solutions of this differential equation, in the range
0  y  p, only exist if the quantum numbers l and ml take on the values
given by l  0, 1, 2,:: and ml  ÿl, . . . : ,  l. Find, by substitution, thevalues of l and ml for which the following functions are solutions:Fa (y)  A,

Fb (y)  B cos yand

Fc (y)  C sin y,where A, B and C are constants.
9. Reconsider the energy eigenfunctions for a three-dimensional harmonic
oscillator given in Section 6.5. Note that, by using Table 8.1, it is possible
to form linear combinations of these eigenfunctions to give simultaneous
eigenfunctions of energy, L2 and Lz .
(a) Verify that the eigenfunction with energy 32 
h! has orbital angular momentum quantum numbers l  0 and ml  0.
(b) Construct eigenfunctions with energy 52 
h! with quantum numbers l  1
and ml  ÿ1, 0  1.
(c) Construct eigenfunctions with energy 72 
h! with quantum numbers l  2
and ml  ÿ2, ÿ 1, 0,  1,  2, and one eigenfunction with energy 72 h!
with orbital quantum numbers l  0 and ml  0.

This page intentionally left blank9
The hydrogen atom
Just as the solar system provided the first meaningful test of the laws of classicalmechanics, the hydrogen atom provided the first meaningful test of the laws ofquantum mechanics. The hydrogen atom is the simplest atom, consisting of anelectron with charge ÿe and a nucleus with charge e. To a first approximation, the nucleus, with a mass much larger than the electron mass, can be taken
as a fixed object. This means that it should be possible to understand theproperties of the hydrogen atom by solving a one-particle quantum mechanicalproblem, the problem of an electron in the Coulomb potential energy field
V (r)  ÿe2:
4pE0 r

(9:1)

In this chapter we shall find the energy eigenvalues and eigenfunctions for such
an electron and use these results to describe the main features of the hydrogenatom. We will then show how this description can be improved by including
small effects due to relativity and the motion of the nucleus.

9.1

CENTRAL POTENTIALSWe shall begin by considering the general problem of a particle in a central
potential which is a potential, like the Coulomb potential, that only depends onthe distance of the particle from a fixed origin.

Classical mechanics of a particle in a central potential
Let us consider a classical particle of mass m with vector position r, momentum p
and orbital angular momentum L  r  p with respect to a fixed origin. If theparticle moves in a central potential V (r), it is subject to a force which is given by

180

The hydrogen atom

Chap. 9Area swept out

dA =

dr

1
r 3 dr2

rFig. 9.1 The radius vector r of a particle with angular momentum L  mr  dr=dt
sweeps out a vector area dA  (L=2m) dt in time dt.

Fÿ

dV
er
dr

where er is a unit vector in the direction of r. Because this force acts along theradius vector r, the torque acting on the particle, N  r  F, is zero and theparticle moves with constant angular momentum L.
The geometrical implications of constant angular momentum can be understood by considering Fig. 9.1 which shows that the vector area swept by theradius vector r in time dt is given by
dA 

L
dt:
2m

This implies that, when the angular momentum L is a constant vector, theparticle moves in a fixed plane with a radius vector which sweeps out area at aconstant rate of L=2m.
The momentum p of a particle moving in a plane has two independent
components which may be conveniently taken to be the radial and transverse
componentspr  m

dr
dt

and

pt 

L
:
m

By writing the kinetic energy p2 =2m on terms of pr and pt , we find that theconstant total energy of the particle is given by
E

p2r
L2

 V (r):2m 2mr2(9:2)

We note that the energy of the particle can be viewed as the sum of two
terms, the radial kinetic energy p2r =2m and an effective potential energy of the
form

9.1

Ve (r) 

Central Potentials

L2
 V (r):
2mr2

181

(9:3)

The effective force corresponding to this effective potential acts in the radial
direction and has magnitude
Fe  ÿ

dVe
L2
dV
 3ÿ
:
dr
mr
dr

The term L2 =mr3 , which equals mu2 =r for a particle moving with speed u incircle of radius r with angular momentum L  mru, represents an outward
centrifugal force. Thus, the term L2 =2mr2 in the effective potential (9.3) canbe thought of as either a centrifugal potential energy or as a transverse kinetic
energy.
The most important example of classical motion in a central potential isplanetary motion. A planet of mass m moves around the sun with gravitational
potential energy
V (r)  ÿ

GmM
rwhere M is the solar mass and G is Newton's fundamental constant of gravity.
Books on classical mechanics show that planets in the solar system move in
elliptic orbits. If a planet could shed excess energy its orbit would eventually
become circular. If it could acquire energy so that it could just escape from thesun, its orbit would become a parabola, and when the energy is higher still its
orbit would be a hyperbola. Circles, ellipses, parabolas and hyperbolas areconic sections which are best studied by taking a ripe, conical pear and slicing itup.
Planetary motion provided the first extra-terrestrial test of the laws ofclassical mechanics. These classical laws passed the test with flying colours
because the orbital angular momentum of a planet is many orders of magnitude
greater than the fundamental quantum unit of angular momentum h; forexample, the orbital angular momentum of planet earth is a stupendous
3  1074 h.
The hydrogen atom provides another example of motion in a central
potential. In this case, an electron with charge ÿe moves around a nucleus
with charge e in the Coulomb potential Eq. (9.1). When the electron has
an angular momentum much greater than h, classical mechanics can beused and the electron traces an orbit which is a conic section. But, when
the angular momentum is comparable with h, quantum mechanics must
be used and the electron is described by a quantum state with uncertain
properties.

182

The hydrogen atom

Chap. 9Quantum mechanics of a particle in a central potential
The quantum states of a particle in a central potential are described by awave function C(r, y, f, t). We shall focus on quantum states with sharply
defined energy E which, according to Section 4.3, have wave functions of theformC(r, y, f, t)  c(r, y, f) eÿiEt=h ,(9:4)

where c(r, y, f) is an energy eigenfuction satisfying the eigenvalue equation
"

#
h2 2
ÿ
r  V (r) c  Ec:
2m

(9:5)

This partial differential equation in three independent variables r, y and f
may be greatly simplified if we assume that the quantum state has, inaddition to definite energy E, definite angular momentum properties of thetype described in Chapter 8. In particular, if
assume that the magnitudepwe
of the orbital angular momentum is L  l(l  1)h and its z-componentis Lz  ml h, where l and ml are quantum numbers which could take on
the values l  0, 1, 2 . . . : and ml  ÿl, . . . , l, the eigenfunctions have the formc(r, y, f)  R(r)Yl , ml (y, f):(9:6)^ 2 and L^z
In this equation, Yl , ml (y, f) is a simultaneous eigenfunction of L
satisfying Eq. (8.23) and R(r) is an unknown function of r. If we substituteEq. (9.6) into Eq. (9.5), use the identities
r2 c 
1 ]2 (rc) 1 ]2 c cos y ]c
1 ]2 c
r ]r2
r2 ]y2 sin y ]y sin2 y ]f2and^ 2  ÿ
h2
L

 2

]
cos y ]1 ]2,
]y2 sin y ]y sin2 y ]f2

and also use Eq. (8.23), we obtain the following ordinary differential equation
for R(r):"#h2 d2 (rR)l(l  1)h2ÿ V (r) R  ER:
2mr dr2
2mr2

(9:7)

9.1

Central Potentials

183

By introducing a radial function u(r), defined byR(r) 

u(r)
,
r

(9:8)

we obtain"#h2 d2 ul(l  1)h2
 V (r) u  Eu:
ÿ
2m dr22mr2(9:9)

This important equation is called the radial
SchroÈdinger equation. It describesp
a particle with angular momentum L  l(l  1)
h which behaves like a particle in a one-dimensional effective potential of the form
Ve (r) l(l  1)h2
 V (r):2mr2(9:10)

If we compare this potential with the analogous effective potential in classicalmechanics given in Eq. (9.3), we see that the first term, l(l  1)h2 =2mr2 , caneither be thought of as a kinetic energy associated with transverse motion or asa centrifugal potential that arises from the orbital angular momentum of theparticle.
When solutions of the radial SchroÈdinger equation (9.9) are sought, theboundary conditionu(r)  0

at r  0

must be imposed to ensure that the function R(r)  u(r)=r, and hence the actual
three-dimensional eigenfunction given by Eq. (9.6), is finite at the origin. In
addition, bound state solutions, which describe a particle that cannot escape toinfinity, must also satisfy the boundary conditionu(r) ! 0

as r ! 1:

Bound states only exist if the effective potential Ve (r), Eq. (9.10), is sufficiently attractive. We shall label these states by a quantum number
nr  0, 1, 2, . . . which will be shown to be equal to the number nodes of theradial eigenfunction u(r) between r  0 and r  1. This means a bound state ofa particle in a central potential can always be specified by three quantum
numbers nr , l and ml and that the eigenfunction has the form
cnr , l , ml (r, y, f) 

unr , l (r)Yl , ml (y, f):r(9:11)

184

The hydrogen atom

Chap. 9By using the normalization condition (8.25) for spherical harmonics, we caneasily show that the eigenfunction cnr , l , ml (r, y, f) is normalized if the radial
eigenfunction unr , l (r) obeys the condition
Z 1
0

junr , l j2 dr  1:

(9:12)The energy of these bound states will be denoted Enr , l . By thinking of thisenergy as the sum of three terms, the average radial kinetic energy, the averagetransverse kinetic energy and the average Coulomb energy, we can see thatstates with higher l and higher nr have higher energies. The energy Enr , lincreases with l because the average transverse kinetic energy is given by
Z 1
0

un*r , l (r)l(l  1)h2
2mr2!
unr , l (r) dr;

and the energy Enr , l also increases with nr because the average radial kinetic
energy, which is given by
Z 1
0

!
2 d2
h
un*r , l (r) ÿ
unr , l (r) dr,
2m dr2increases as the number of radial nodes increases.1
Before considering the explicit expressions for the energy levels and eigenfunctions of a particle in a Coulomb potential, we pause and reconsider how we
obtained the radial SchroÈdinger equation Eq. (9.9). The crucial step was to seeka quantum state with definite E, L and Lz for a particle in a central potential.
Such states must exist because this step has had a successful outcome: it has led
to Eq. (9.9) which can be solved to give sensible energy eigenvalues andeigenfunctions. Readers who have studied compatible observables in Chapter
7 should not be surprised by this success. In general, E, L and Lz can be taken
as three compatible observables which uniquely define a quantum state of aparticle in a central potential.
In addition, a quantum state of a particle with definite energy in a central
potential has another observable property with a definite value. It is called
parity. Eigenfunctions with the propertyc( ÿ r)  c(r),1(9:13)

We encountered a similar behaviour when we considered the energy levels of one-dimensional
potential wells in Chapters 4, 5 and 6. In general, more nodes in a wave function means higher
energy in quantum mechanics, just as more nodes in a classical normal mode means
higher frequency.

9.2

Quantum Mechanics of the Hydrogen Atom185

are said to have even parity, and eigenfunctions with the propertyc( ÿ r)  ÿc(r)(9:14)

are said to have odd parity. By using Table 8.1, we can easily show that aneigenfunction with definite orbital angular momentum given by Eq. (9.6), i.e. byc(r, y, f)  R(r)Yl , ml (y, f),
has even parity when l  0 and l  2 and odd parity when l  1. It can beshown that, in general, the parity of a particle in a central potential is evenwhen l is even and odd when l is odd.
Parity is one of the simplest observables in quantum mechanics, but, because
it is an observable with no classical analogue, it is often perceived as mysterious. Energy and parity are compatible observables whenever the Hamiltonianis unchanged when the coordinates undergo a reflection through the origin.
Because this is true for all Hamiltonians which do not involve the weak nuclear
interaction, parity has an important role in classifying quantum states inatomic, nuclear and particle physics.

9.2

QUANTUM MECHANICS OF THE HYDROGEN ATOMIn this section we shall give a brief description of the hydrogen atom using thequantum mechanical concepts introduced in the last section.
Because the hydrogen atom is essentially an electron in a Coulomb potential
and the Coulomb potential is a central potential, bound states of atomic
hydrogen maypbe
taken to have definite orbital angular momentum properties
h. These states have wave functions of thegiven by L  l(l  1)h and Lz  ml 
form
Cnr , l , ml (r, y, f, t)  cnr , l , ml (r, y, f) exp( ÿ iEnr , l t=h)
with
cnr , l , ml (r, y, f) 

unr , l (r)Yl , ml (y, f),rwhere unr , l (r) is an eigenfunction given by the radial SchroÈdinger equation (9.9)
for an electron in a Coulomb potential
V (r)  ÿe2:
4pE0 r

186

The hydrogen atom

Chap. 9Specifically, the radial eigenfunction unr , l (r) is a solution of the differential
equation"#h2 d2 unr , ll(l  1)h2
e2
un , l  Enr , l unr , l ,ÿ
2me dr2
2me r2
4pE0 r r(9:15)

which satisfies the boundary conditions
unr , l (r)  0

at r  0 and atr  1:(9:16)The qualitative features of the energy levels given by the eigenvalue problem
defined by Eqs. (9.15) and (9.16) may be deduced by considering the effective
potential that occurs in Eq. (9.15),
Ve (r) l(l  1)
h2
e2
:
ÿ
2
2me r
4pE0 r

(9:17)The shape of this potential for electrons with different values for the orbital
angular momentum quantum number l are shown in Fig. 9.2. We see that, for
non-zero values of l, the effective potential is attractive at large r and repulsiveat small r. By setting dVe =dr to zero, we can easily show that Ve (r) has aminimum value ofVe (r)  ÿERl(l  1)at r  l(l  1)a0 ,(9:18)

where a0 and ER are the natural units of length and energy in atomic physics,
defined as follows: the Bohr radius

a0 

 2
4pE0 h 0:529  10ÿ10 m
e2 m e(9:19)e2
 13:6 eV:
8pE0 a0(9:20)and the Rydberg energy
ER 

The minimum givenpby
Eq. (9.18) implies that bound states with angular
momentumL  l(l  1)
hhave
energies
somewhere
betweenE  ÿER =l(l  1) and E  0. It also implies that the spatial extent of boundstate eigenfunctions with low angular momentum is of the order of a0 and thatthe eigenfunctions extend to larger distances when the angular momentum
increases. When the angular momentum greatly exceeds h, we expect many

187

9.2 Quantum Mechanics of the Hydrogen Atom
0.5

Energy in units of the Rydberg energy

(1)

(2)

(3)

0

−0.5

(0)

−1

−1.5

0

5
10
D istance in units of the Bohr radius

15

Fig. 9.2 The effective potential energy Ve (r) for an electron in a hydrogen atom withorbital angular momentum quantum numbers l  0, 1, 2 and 3. The unit of distance is
the Bohr radius a0 , defined in Eq. (9.19), and the unit of energy is the Rydberg energy
ER , defined in Eq. (9.20). The effective potential for an electron with quantum number lhas a minimum value of ÿER =l(l  1) at r  l(l  1)a0 which, in classical physics,
corresponds to theenergy and radius of a circular orbit of an electron with angularp
momentum L  l(l  1)
h.

bound states with closely spaced energy levels that correspond to the circularand elliptic orbits of classical mechanics.
We also see from Fig. 9.2 that the effective potential is purely attractive foran electron with zero angular momentum. In classical mechanics, such an
electron simply plunges towards the proton and there are no stable boundstates. But there are stable bound states with zero angular momentum inquantum mechanics, whose existence can be understood by using the uncertainty principle.
The uncertainty principle implies that an electron localized in a region of sizer has an uncertain momentum of the order of 
h=r and an average kinetic energy
which is at least of the order of 
h2 =2me r2 . This means that the least energy of anelectron with zero orbital angular momentum in a region of size r near a proton
is roughly given by
E

h2
e2
:
ÿ
2
2me r
4pE0 r

188

The hydrogen atom

Chap. 9As the region of localization decreases, this energy decreases because thepotential energy decreases, but eventually the kinetic energy of localization,
h2 =2me r2 , increases more rapidly. As a result the total energy has a minimum
value of aboutE  ÿERatr  a0 :This minimum provides an estimate of the lowest possible energy of an electronwith zero angular momentum in a Coulomb potential and suggests that there
are quantum bound states with energies in the range E  ÿER and E  0.

Energy levels and eigenfunctions
The energy levels and eigenfunctions for an electron bound in a Coulomb
potential are found by solving the eigenvalue problem defined by Eqs.(9.15)
and (9.16). In order to focus on the physical properties of the hydrogen atom,
we shall consider the results of this mathematical problem before describing
how these results are obtained in Section 9.7.
Wepshallshow in Section 9.7 that an electron with angular momentum
L  l(l  1)h in a Coulomb potential has an infinite number of bound stateswith energies given by
E nr , l  ÿER(nr  l  1)2

withnr  0, 1, 2, 3, . . .

(9:21)The quantum number nr is called the radial quantum number. These energy
levels are illustrated in Fig. 9.3.
As expected, there are bound states with zero and non-zero angular momentum. Also the energy levels are very closely spaced if the angular momentum isvery large, indicating a correspondence with a continuum of classical boundstate energies.
Unexpectedly, many of the energy levels in Fig. 9.3, those with the samevalue for nr  l, have the same energy. Because of this degeneracy, the energy
levels of the hydrogen atom are usually given as
En  ÿ

ER
n2(9:22)

where ER is the Rydberg energy and n is a quantum number defined byn  nr  l  1:

(9:23)

This quantum number is called the principal quantum number and can take on
the values n  1, 2, 3, . . .

9.2

Quantum Mechanics of the Hydrogen Atom
l= 0

l= 1

l= 2

E

nr = 2nr = 1nr = 0ER

nr = 1nr = 0

Energy

l= 3

189

l= 4

0
−

ER
42

− 32R

−

22

E

− 12R

nr = 0Fig. 9.3 The energy levels given by Eq. (9.21) for the bound states of an electron ina hydrogen atom. For each value of the orbital angular momentum quantum numberl  0, 1, 2, , there is an infinite stack of energies Enr , l with radial quantum number nr  0, 1, 2, . . . . Note that, because bound states with the same value for nr  lhave the same energy, these energy levels are often given as En  ÿER =n2 where ER is
the Rydberg energy and n is defined by n  nr  l  1; the quantum number n is called
the principal quantum number and can take on the values n  1, 2, 3, . . . . There is also acontinuum of unbound or ionized states with positive energy.

We shall also show in Section 9.7 that the radial eigenfunction unr , l (r)belonging to the energy eigenvalue Enr , l has three characteristics:(1) In Section 5.1 we found that the eigenfunction of a particle in a onedimensional square well with binding energy E  
h2 a2 =2m falls off exponenÿax
tially like e . In a similar way, the radial eigenfunction of an electron in a
Coulomb potential with binding energy
E

ER
h2 1


n2
2me n2 a20

falls off exponentially at large r like
unr , l (r) / eÿr=na0 :(2) Because of the singular nature at r  0 of the centrifugal potentiall(l  1)h2 =2me r2 , the behaviour of the eigenfunction at small r is governedby the orbital angular momentum and is given by
unr , l (r) / r l1 :(3) Finally, because the radial quantum number nr denotes the number of nodes
between r  0 and r  1, the eigenfunction unr , l (r) is proportional to apolynomial with nr zeros. If this polynomial is denoted by pnr , l (r), we have

190

The hydrogen atom

Chap. 9unr , l (r) / pnr , l (r):By combining these three characteristics, we arrive at a radial eigenfunction of
the form
unr , l (r)  Npnr , l (r)rl1 eÿr=na0 ,(9:24)

where N is a constant which ensures that the normalization condition (9.12) issatisfied.
Explicit expressions for the radial eigenfunctions with low values for the
angular momentum quantum number l and low values for the radial quantum
number nr are given in Table 9.1, and some of these eigenfunctions are illustrated in Fig. 9.4. To conform with the conventions of atomic physics these
eigenfunctions are labelled using spectroscopic notation. This notation employs
the principal quantum number n  nr  l  1 and a letter to designate the valueof l; the letter s is used for l  0, p for l  1, d for l  2, and f for l  3. The
historical origin of this notation dates back to the early days of atomic physics
when spectral lines were labelled s for sharp, p for principal, d for diffuse and f
for fundamental.
TABLE 9.1
atom

Normalized radial eigenfunctions for low-lying states of the hydrogen

Spectroscopic
notation

Radial eigenfunction
unr , l (r) 2r ÿr=a0
u0, 0 (r)  p
e
a0 a0

  1
1 r
r ÿr=2a0
u1, 0 (r)  p 1 ÿ
e
2 a0
a0
2a0
"
 
  # 
2
2 r
2 r 2
r ÿr=3a0eu2, 0 (r)  p 1 ÿ
3 a0
27 a0
a0
3 3a0
"
 
 
 3 # 
1
3 r
1 r 2 1r
r ÿr=4a0
eu3, 0 (r)  p 1 ÿ
ÿ
4 a0
4 a0
8 a0
192 a0a0
 2
1
r
u0, 1 (r)  p
eÿr=2a0
2 6a0 a0"
 # 2
8
1 r
r
u1, 1 (r)  p 1 ÿ
eÿr=3a0
6 a0
a0
27 6a0s" 
  # 2
1
5
1 r
1 r 2
r
u2, 1 (r) 
eÿr=4a01ÿ
16 3a0
4 a0
80 a0
a0
 3
4
r
eÿr=3a0
u0, 2 (r)  p
81 30a0 a0  3
1
1 r
r
u1, 2 (r)  p 1 ÿ
eÿr=4a012aa
64 5a0
0
0

1s
2s
3s
4s
2p
3p
4p
3d
4d

9.3
1

1

1s radial eigenfunction

2p radial eigenfunction

0.5

0

0.5

0

0.5

2

4

6

8

10

0

2

4

6

8

10

3p radial eigenfunction

0

0

0.5

5

10

15

20

−0.5

0

5

10

15

20

0.5

3s radial eigenfunction

4p radial eigenfunction

0

−0.5

0
0.5

2s radial eigenfunction

0

−0.5

191

Sizes and Shapes

0

0

10
20
30
D istance in Bohr radii

40

−0.5

0

10
20
30
D istance in Bohr radii

40

Fig. 9.4 Radial eigenfunctions unr , l (r) for an electron in the hydrogen atom with radialquantum numbers nr  0, 1, and 2 and with angular momentum quantum numbersl  0 and 1. The eigenfunctions have been labelled using spectroscopic notation ns andnp, where n is the principal quantum number n  nr  1  l and s denotes l  0 and p
denotes l  1. Note that the unit of distance is the Bohr radius a0 and that different
scales are used for eigenfunctions with different values of nr .

9.3

SIZES AND SHAPES

The size and shape of a quantum state of atomic hydrogen can be determinedby considering the most probable locations of the electron in the atom. For astate with eigenfunction cnr , l , ml (r, y, f), the probability of finding the electronwith coordinates (r, y, f) in a volume element d3 r is
jcnr , l , ml (r, y, f)j2 d3 r:
We can easily find the radial probability distribution for the electron. To doso, we use d3 r  r2 dr dV, where dV  sin y dy df is an element of solid angle,
and express the eigenfunction as

192

The hydrogen atom

Chap. 9cnr , l , ml (r, y, f) 

unr , l (r)Yl , ml (y, f),rwhere Yl , ml (y, f) is a spherical harmonic which obeys the normalization condition Eq. (8.25). The probability of finding the electron at a distance betweenr and r  dr from the nucleus is then given byZ

unr , l (r) 2 2r drr

jYl , ml (y, f)j2 dV  junr , l (r)j2 dr:Thus, the radial shape of the quantum state is described by a radial probability
density junr , l (r)j2 .
Radial probability densities for states with nr  0, 1 and 2 and with l  0 and
1 are shown in Fig. 9.5. We note that the radial extent increases as nr increases
and as l increases. This may be confirmed by considering the mean radius of astate with quantum numbers nr and l which is given by

1s

0.4

0.2

0

0.2

0

2

4

6

8

10

0.2

0

2

4

6

8

10

0.1

0.05

0.05

0

0.1

5

10

15

3p

0.15

0.1

20

0

0

0.1

3s

0.05

0

0

0.2
2s

0.15

0

2p

0.4

5

10

15

20

4p

0.05

0

10
30
20
D istance in Bohr radii

40

0

0

10
20
30
D istance in Bohr radii

40

Fig. 9.5 Radial probability densities for the 1s, 2s, 3s, 2p, 3p and 4p states of thehydrogen atom. Note that the unit of distance is the Bohr radius a0 and that a different
scale is used for states with a different number of radial nodes.

9.3

hrinr , l Z 10

Sizes and Shapes

193

rjunr , l (r)j2 dr:This integral may be evaluated by using the mathematical properties of theeigenfunctions unr , l (r) to give an expression,
hrinr , l  12[3n2r  6nr (l  1)  (l  1)(2l  3)]a0 ,(9:25)

which explicitly shows how the radial extent increases with nr and l.
By using explicit forms for unr , l (r) and Yl , ml (y, f) we can evaluate theprobability density jcnr , l , ml (r, y, f)j2 and explore both the radial and angularshapes of a hydrogen atom state with quantum numbers nr , l, ml . First, we notethat the angular shape does not depend on the azimuthal angle f because, as
indicated by Eq. (8.26), the spherical harmonic Yl , ml (y, f) has the f dependence
eiml f . Thus, the probability density of the state does not change if it is rotatedabout the z axis. This means that the size and shape of the state may be fully
specified by showing the most likely positions of the electron on any verticalplane that passes through the z axis, as demonstrated in Figs. 9.6 and 9.7.
In Figs. 9.6 and 9.7 we show the sizes and shapes of the 3p and 3d states ofatomic hydrogen. The 3p states in Fig. 9.6 have one radial node and an angulardependence given by jY1, ml (y, f)j2 with ml  1, 0 and ÿ1. The 3d states inFig. 9.7 have no radial nodes and the angular shape of a state with l  2.
<

48 Bohr radii

3p state with m = 0><

48 Bohr radii>

3p state with m = + 1 or −1

Fig. 9.6 The size and shape of the 3p states of the hydrogen atom with a z componentof orbital angular momentum equal to m
h. These states have rotational symmetry about
the z axis. The density of dots is proportional to the probability of finding an electron ona vertical plane passing through the z axis. These pictures were produced by selecting apoint on the plane at random and deciding to plot or not to plot in accordance with thevalue of jcnr , l , ml (r, y, f)j2 at that point.

194<

The hydrogen atom
48 Bohr radii><

Chap. 948 Bohr radii>

3d state with m= + 1 or −1

3d state with m= 0<

48 Bohr radii>

3d state with m= + 2 or −2

Fig. 9.7 The size and shape of the 3d states of the hydrogen atom with a z componentof orbital angular momentum equal to m
h.

9.4

RADIATIVE TRANSITIONSWhen a hydrogen atom interacts with an electromagnetic field, quantum stateswith quantum numbers n, l and ml are, in general, no longer stationary stateswith definite energy, and radiative transitions between these states may takeplace in which electromagnetic energy is either absorbed or emitted.
The most probable radiative transitions are called electric dipole transitions.
They are caused by an interaction of the electric field component E of theelectromagnetic field with the operator describing the electric dipole moment of
the electron±nucleus system. The electric dipole operator is d  ÿer, where r is
the vector position operator for the electron in the atom, and the interaction is
given by^ I  ÿd  E:
H

(9:26)In the presence of this interaction, the probability for a transition between
states with quantum numbers ni , li , mli and nf , lf , mlf is proportional toZj^ I cn , l , m (r) d3 r j2 :c*nf , lf , mlf (r) Hi ili(9:27)

We can easily prove that electric dipole transitions always involve a change inparity by showing that the integral in Eq. (9.27) is zero if the initial and final
states have the same parity. We show this by considering the effect of changing^ I  ÿd  E changes sign,
the integration variable from r to ÿr. The interaction H
but the sign of the eigenfunction, cni , li , ml (r) or cnf , lf , ml (r), is unchanged if theifeigenfunction has even parity and it is changed if the eigenfunction has odd
parity, as shown by Eqs. (9.13) and (9.14). Thus, when both eigenfunctions havethe same parity, the integrand in Eq. (9.27) changes sign when the integration
variable r is changed to ÿr and this implies that the integral must be zero.

9.4

Radiative Transitions

195

It can also be shown, by noting that the angular dependence of theeigenfunctions cnf , lf , ml (r) and cni , li , ml (r) are given by spherical harmonics,ifthat the integral in Eq. (9.27), and hence the probability of transition, iszero unless the difference Dl  lf ÿ li is 1 or ÿ1. This means that all electric
dipole transitions in the hydrogen atom also obey the selection ruleDl  1:(9:28)The electric dipole transitions between low-lying states of the hydrogen atom
are shown as dotted lines in Fig. 9.8, where spectroscopic notation, 1s, 2s, 2p,
etc. has been used to label the levels corresponding to states with different
values for the principal quantum number n and orbital angular momentum
quantum number l; for example, 2s corresponds to n  2 and l  0 and 2p
corresponds to n  2 and l  1.
The transitions shown in Fig. 9.8 may be induced or spontaneous. Induced
transitions between states with energy Eni and Enf occur strongly when theatom interacts with an external electromagnetic field which oscillates with anangular frequency ! which satisfies the resonant conditionh!  jEnf ÿ Eni j;


l= 0

l= 1

l= 2

3s

3p

3d

2s

2p

l= 3

n= 4
n= 3
n= 2

n= 1

1s

Fig. 9.8 Electric dipole radiative transitions between low-lying energy levels of thehydrogen atom with different values for the quantum numbers n and l. Spectroscopic
notation, 1s, 2s, 2p, etc. has been used to label the energy levels; for example, 2s
corresponds to n  2 and l  0 and 2p corresponds to n  2 and l  1. We note thatthe electric dipole transitions shown by the dotted lines obey the Dl  1 selection rule
given in Eq. (9.28).

196

The hydrogen atom

Chap. 9electromagnetic energy is absorbed when Enf > Eni and it is emitted whenEnf < Eni . Spontaneous transitions are seemingly not caused by anything, butthey are really caused by the interaction of the atom with a quantized radiation
field which is always present, even when the atom is entirely isolated. As a resultof this interaction, an atom with energy Eni decays to an atom with energy Enf
by emitting a photon with energy E  Eni ÿ Enf . Such transitions give rise tospectral lines with wavelengths l given by!
hc
11 ER 2 ÿ 2 :lnf ni(9:29)As mentioned in Section 1.3, these lines form a series of lines in the ultra-violet
called the Lyman series, with wavelengths given by


hc
1
1
 ER 2 ÿ 2 ,l1niwith ni  2, 3, 4 . . . ,a series of lines in the visible called the Balmer series, with wavelengths given by


hc
1
1
 ER 2 ÿ 2 ,l2niwith ni  3, 4, 5 . . . ,and other series with longer wavelengths.
Inspection of Fig. 9.8 shows that the 2s state of the hydrogen atom cannot
decay by electric dipole radiation because a 2s ! 1s transition would violate
the Dl  1 rule given in Eq. (9.28). In fact, the 2s state is a metastable stablestate with a long lifetime which eventually decays to the 1s state by a mechanism which is less probable than an electric dipole transition. Whereas the meantime for a spontaneous 2p ! 1s transition is 1:6  10ÿ9 seconds, the mean time
for a spontaneous 2s ! 1s transition is a lengthy 0.14 seconds.

9.5

THE REDUCED MASS EFFECT

So far we have assumed that the mass of the nucleus of the hydrogen atom issufficiently large so that its motion can be ignored. In fact, this motion gives
rise to small but important effects which can be incorporated by introducing
the concept of the reduced mass of the electron±nucleus system.2

2

We encountered the reduced mass of the proton-proton system at the end of Section 5.2 and the
reduced mass of a diatomic molecule in Section 6.4.

9.5

The Reduced Mass Effect197

To understand the reduced mass effect in atomic hydrogen, we consider theclassical energy of an electron of mass me and a nucleus of mass mN interacting
via a Coulomb potential
E

p2e
p2
e2
,
 N ÿ
2me 2mN 4pE0 r

where pe and pN are the magnitudes of momenta of the electron and thenucleus, and r is the distance between the electron and the nucleus. In thecentre-of-mass frame we can set pe  pN  p and obtain an expression for theenergy of the form
E

p2
e2
ÿ
,
2m 4pE0 r

where m me mN:me  mN

This expression shows that a classical electron±nucleus system in the centre ofmass frame acts like a single particle of mass m. This mass is called the reduced
mass of the electron±nucleus system.
In a similar way a quantum electron±nucleus system also acts like a single
particle with reduced mass m. This means that we can take account of the
motion of the nucleus in the hydrogen atom by simply replacing the electronmass me by the reduced mass m. In particular, the length and energy scales givenby the Bohr radius a0 , Eq. (9.19), and the Rydberg energy ER , Eq. (9.20), aremodified slightly and become
a00 



 2
4pE0 h

me

a0
e2 m
m

and
ER0 

e2
m

ER ,
8pE0 a00 me

and all the quantitative results given in Eqs. (9.21), (9.22), (9.24) and (9.25), in
Table 9.1, and in Figs. 9.3, 9.4 and 9.5 are modified accordingly. Most importantly, the modified energy levels of the hydrogen atom are given by
En0  ÿ

ER0
m ER

:
n2
m e n2(9:30)The reduced mass effect in the hydrogen atom is apparent when the spectral
lines of ordinary hydrogen are compared with the spectral lines of heavyhydrogen. For an atom of ordinary hydrogen, the nucleus is a proton with

198

The hydrogen atom

Chap. 9mass 1836me and the reduced mass is (1836=1837)me , whereas for an atom ofheavy hydrogen the nucleus is a deuteron with mass 3671me and the reduced
mass is (3671=3672)me . The tiny difference between these two reduced masses
gives rise to an observable difference in the wavelengths of the spectral lines
emitted by these atoms. For example, the Ha line of the Balmer series, whicharises from a transition from a state with n  3 to one with n  2, has awavelength l given by


hc
m
1
1

ER 2 ÿ 2 :lme
23For ordinary hydrogen this gives l  656:4686 nm and, for heavy hydrogen, it
gives l  656:2899 nm. In fact, Harold Urey discovered heavy hydrogen in1934 in an experiment which revealed that each line in the Balmer series was
accompanied by a faint line due to the small admixture of heavy hydrogen thatis present in naturally occurring hydrogen.

9.6

RELATIVISTIC EFFECTS

In this section we shall confirm that, to a good approximation, the electron in ahydrogen atom is non-relativistic, but that relativistic effects give rise to smalland significant corrections to the energy levels.
We can confirm that the electron in the hydrogen atom is approximately
non-relativistic by estimating its average momentum. Because the uncertainty
in the position of the electron is of the order of the Bohr radius a0 , theuncertainty in its momentum, and hence its average momentum, is of theorder of p0  h=a0 . Using Eq. (9.19) we find thatp0  ame c,(9:31)

where a is a dimensionless constant, called the fine structure constant,given by
a

e2
1

:
4pE0 
hc 137:035 989 5(9:32)The small numerical value of the fine structure constant has an important
implication for atomic physics. It implies that atomic electrons, like the electron
in a hydrogen atom, have momenta which are small compared with me c andthat non-relativistic physics is a good approximation.
We can estimate the magnitude of relativistic corrections to the hydrogen
atom by considering the relation between the relativistic energy E and momentum p of an electron,

9.6

E

Relativistic Effects

199q
m2e c4  p2 c2 :If p << me , we can use the binomial theorem to give the approximate expression


p2
1 p 4
ÿ
m e c2 :E  me c 
2me 8 me c
2

The first term is the rest-mass energy, the second is the non-relativistic kinetic
energy and the third term is the leading relativistic correction to the kinetic
energy. We note that, because the average momentum of an electron in ahydrogen atom is of the order of p0  ame c, the relativistic correction to its
kinetic energy is of the order ofhErel i  ÿ
1 p0 41
me c2  ÿ a4 me c2 :8 me c8(9:33)

Corrections of a similar magnitude also arise from the interaction of the spin
magnetic moment of the electron with a magnetic field caused by the relative
motion of the nucleus and the electron. This interaction is called the spin±orbit
interaction.
The magnitude of the spin±orbit interaction can be estimated by considering aclassical electron moving around the nucleus in a circle of radius r. If the electron
has velocity u and orbital angular momentum L  me ru, it will take timet

2pr 2pme r2Lu

to complete one orbit. Just as we perceive the sun to move around the earth, theelectron perceives a nucleus which appears to move in a circle of radius r withperiod t. Because the nucleus has charge e, an electric current I  e=t circles theelectron constituting a current loop. Using the standard formula for the magnetic field at the centre of a circular current loop of radius r, B  m0 I=2r, andp
the relation c2  1= E0 m0 , we find that the electron perceives a magnetic field ofmagnitudeBeL:
4pE0 me c2 r3(9:34)

Because the electron has a spin magnetic moment, which according to Eq. (8.8)
is given bym  ÿ2

e
S,
2me

200

The hydrogen atom

Chap. 9there is an energy of interaction given by
Emag  ÿm  B e2
L  S:
4pE0 m2e c2 r3This estimate has given the right order of magnitude for the spin±orbit
interaction. A more careful calculation, which takes account of the accelerationof the electron, introduces a factor of one-half and gives
Emag e2
L  S:
8pE0 m2e c2 r3(9:35)The quantitative effects of the spin±orbit interaction on the energy levels of
the hydrogen atom can be obtained by evaluating expectation values of Emag .
However, two general aspects of the spin±orbit interaction are worth mentioning here:(1) The spin±orbit interaction is really a relativistic effect and we can confirm
this by showing that it leads to a correction which is comparable with therelativistic correction to the kinetic energy given by Eq. (9.33)
1
hErel i  ÿ a4 me c2 :8Using the fine structure constant a defined in Eq. (9.32), we rewrite
Eq. (9.35) for the spin±orbit energy as
Emag  a LSh:2m2e c r3

For a low-lying state of the hydrogen atom, we can assume that the
expectation value of L  S=r3 is of the order of h2 =a30 . Rewriting the Bohrradius a0 defined in Eq. (9.19) in terms of the fine structure constant as
a0 

1 
h
,
a me cwe find that the spin±orbit energy of a low-lying hydrogen atom state is of
the order ofhEmag i  a4 me c2 :(2) The spin±orbit interaction implies that quantum states of the hydrogen
atom with definite energy have definite values for J 2 , L2 and S2 , where9.6

Relativistic Effects

201

J  L  S is the total angular momentum due to the orbital and spin
angular momenta of the electron. We can confirm this by using
J  J  (L  S)  (L  S)  L2  S 2  2L  S
to show that the spin±orbit interaction is proportional toLS

J 2 ÿ L2 ÿ S 2:
2

As a quantitative illustration of the combined effect in the hydrogen atom of
the relativistic correction to the kinetic energy and of the relativistic spin±orbit
interaction, we consider quantum states with principal quantum number n  2.
Because of the spin±orbit interaction, these states will have definite energy ifthey have definite magnitudes for spin, orbital and total angular momenta
given bySp
s(s  1)h,Lp
l(l  1)
h

and

J

p
j(j  1)h,

where l, s and j are quantum numbers. The rule for the addition of spin andorbital angular momenta, which is given by Eq. (8.6), implies that, in general, j
can take the values
j  l  s, l  s ÿ 1, . . . , jl ÿ sj:
For hydrogen atom states with n  2, the orbital quantum number l can be 0 or
1. The n  2 states with l  0 and s  12 have j  12 and are denoted by 2s1=2 , andthe n  2 states with l  1 and s  12 have j  12 or j  32 and are denoted by 2p1=2
and 2p3=2 .
If the spin±orbit interaction were absent, all states of the hydrogen atom withn  2 would have the same energy
E2  ÿ

ERa2
me c2 ,

ÿ
22
8

where ER is the Rydberg energy and a is the fine structure constant. Whenrelativistic and spin±orbit corrections are calculated using a theory called
perturbation theory, these states are found to have energies that depend on
the quantum numbers l, s and j. These energies are given by
5 4
E(2s1=2 )  E2 ÿ 64
a m e c2 ,5 4
a m e c2 ,E(2p1=2 )  E2 ÿ 64

and

202

The hydrogen atom

Chap. 91 4
E(2p3=2 )  E2 ÿ 64
a m e c2 :We note that the difference in energies of the 2p3=2 and 2p1=2 states can beverified by observing a small difference in the wavelengths of the radiation
emitted by the transitions 2p3=2 ! 1s1=2 and 2p1=2 ! 1s1=2 . We also note thatthe 2s1=2 and 2p1=2 states are predicted to have the same energy, but that thisdegeneracy is removed by a small effect called the Lamb shift which arises fromthe quantum field properties of the electromagnetic field.

9.7

THE COULOMB EIGENVALUE PROBLEM

In this section we shall find the energy levels and eigenfunctions of an electron
in a Coulomb potential by solving the eigenvalue problem defined by thedifferential equation (9.15) and the boundary conditions (9.16). This section
may be omitted without significant loss of continuity.
As a first step we shall tidy up Eq. (9.15) by settingr  qa0andE  ÿg2 ER ,(9:36)

so that q is a dimensionless measure of distance and g2 is a dimensionlessmeasure of the binding energy. If we use the definitions for a0 and ER ,Eqs. (9.19) and (9.20), we find that the radial eigenfunction u(q), when expressed as a function q, satisfies the differential equation
d2 u
2 l(l  1)
u  g2 u
ÿ

dq2
q
q2

(9:37)and the boundary conditions
u(q)  0

at q  0 andat q  1:(9:38)

Our next step is to find the behaviour of u(q) at large q and at small q. At
large q the differential equation (9.37) becomesd2 u g2 u:dq2
The general solution isu(q)  A eÿgq  B egq ,where A and B are constants, but to satisfy the boundary condition u(q) ! 0 as
q ! 1 we set B  0 to give9.7

The Coulomb Eigenvalue Problem

u(q)  A eÿgq at large q:

203(9:39)

At small q, Eq. (9.37) becomes
d2 u l(l  1)u
ÿ
 0:
dq2
q2
This equation has solutions of the form ql1 and qÿl , as can be shown by
substitution, but to ensure that the boundary condition u(0)  0 is satisfied wemust haveu(q) / ql1

at small q:(9:40)

Our third step is to seek an eigenfunction of the formu(q)  f (q)ql1 eÿgq ,(9:41)

where the function f (q) does not invalidate the small q and large q behaviour
given by Eqs. (9.40) and (9.39). If we substitute this expression into Eq. (9.37)
and usedudf
 [(l  1)ql ÿ gql1 ] eÿgq f  ql1 eÿgqdq
dq
to gived2 u
 [l(l  1)qlÿ1 ÿ 2g(l  1)ql  g2 ql1 ] eÿgq fdq2
 2[(l  1)ql ÿ gql1 ] eÿgqdf
d2 f
 ql1 eÿgq 2 ,dq
dq

we find that the function f(q) satisfies the differential equation
q

d2 f
df
 2[(l  1) ÿ gq]  2[1 ÿ g(l  1)] f  0:dq2
dq

(9:42)

Acceptable solutions of Eq. (9.42) only exist for special values of the energy
parameter g. The simplest is
f (q)  constantwhich is clearly a solution if g  1=(l  1). If we use Eq. (9.36) to express q interms of r and g in terms of E, we find that this solution gives eigenfunctions
and energy eigenvalues with radial quantum number nr  0 of the form

204

The hydrogen atom

Chap. 9u0, l (r)  Nrl1 eÿr=(l1)a0

andE0, l  ÿER:(l  1)2

To find the other eigenfunctions and eigenvalues, we shall express f(q) as an
infinite power series,Xf (q) as qs ,
(9:43)s0, 1, 2...
and substitute into Eq. (9.42). After some tedious tidying we obtainXs0, 1, 2,...

f[s(s  1)  2(s  1)(l  1)]as1 ÿ [2g(s  l  1) ÿ 2]as gqs  0:

Thus, the power series (9.43) is a solution of Eq. (9.42) if the coefficients of theseries satisfy the relation
2g(s  l  1) ÿ 2
as :as1 
s(s  1)  2(s  1)(l  1)

(9:44)

We note that this relation implies that
 
2g
as
as1 !
sas

s ! 1:

By comparing this behaviour with the behaviour of the function
q e2gq  q

1X(2gq)ss0s!

,

we conclude that, at large q, the function f (q) is proportional to q e2gq andthat, at large q, the function u(q)  f (q)ql1 eÿgq is proportional to ql2 egq .
Because this form for u(q) does not satisfy the boundary condition (9.39), theinfinite power series (9.43) is not an acceptable solution of Eq. (9.42).
However, the series does give rise to an acceptable solution when the series
terminates at a finite power of q to give a polynomial in q. To find the condition
for a polynomial solution of degree nr , we consider the relation between thecoefficients anr 1 and anr given by Eq. (9.44),
2g(nr  l  1) ÿ 2an :anr 1 
nr (nr  1)  2(nr  1)(l  1) r
This relation shows that anr 1 is zero and all subsequent coefficients are also
zero, if the energy parameter g takes on the special value

Problems 9

g

205

1
:
nr  l  1

Thus, when g has this value, the power series terminates at anr qnr to give apolynomial solution of Eq. (9.42) which we will denote by pnr , l (q).
This polynomial gives rise to a solution of Eq. (9.37) of the formu(q)  pnr , l (q)qq1 eÿgqthat obeys the boundary conditions (9.38). If we express q in terms of r and g interms of E, we obtain the eigenfunctions and eigenvalues,
unr , l (r)  Npnr , l (r)rl1 eÿr=(nr l1)a0

and

Enr , l  ÿER,(nr  l  1)2

which formed the basis of our discussion of the properties of atomic hydrogenin Section 9.2.

PROBLEMS 9
1. Consider an electron in a Coulomb potential
V (r)  ÿe2
4pE0 rwith wave function
c(r)  N eÿr=awhere a is a constant.
(a) What is the orbital angular momentum of the electron?
(b) Show that the expectation values of the potential energy and of the
kinetic energy of the electron are given by
hV i  ÿe2
4pE0 a

and

hTi 

h2
:
2me a2(c) Show that the expectation value of the total energy of the electron is aminimum when a is equal to the Bohr radius a0 . Find the value of thisminimum.

206

The hydrogen atom

Chap. 9(The integral
Z 10

rk eÿar dr 

k!
ak1

is useful. It is also useful in problems 3, 4 and 10.)
2. Consider the effective potential Ve (r) given by Eq. (9.3) for a classicalparticle in a Coulomb potential
V (r)  ÿe2:
4pE0 r(a) Show that at
L2 4pE0r  rc m e2
Ve (r) has a minimum value ofVe (rc )  ÿ

L2
:
2mr2c

Hence, show by inspection of Eq. (9.2), that a particle with angularmomentum L has a minimum energy when it is in a circular orbit withradius rc .
(b) Show that if the energy is given byEÿe2
8pE0 awhere a is a constant, then the particle
has anin
which thep
 distance orbit pfrom the origin ranges from a ÿ a2 ÿ arc to a  a2 ÿ arc .
3. The radial eigenfunction for the quantum
state of lowest energy forp
an electron with angular momentum L  l(l  1)h in a Coulomb potentialisu0, l (r)  Nrl1 eÿr=(l1)a0
where N is a constant.

Problems 9207(a) Show that the eigenfunction satisfies the normalization condition
Eq. (9.12) if
N2 



2l3
2
1
:
(l  1)a0(2l  2)!

(b) Show that the most probable radius of the state isrmost probable  (l  1)2 a0 :(c) Show that the average radius of the state ishri (2l  3)(l  1)a0:
2(d) Show that the average of the square of the radius is
hr2 i (2l  4)(2l  3)(l  1)2 a20
4

and that the uncertainty in the radius Dr becomes small compared withhri as l ! 1.
(e) Show that in the limit l ! 1, the most probable radius and the averageradius are both equal to rc , the radius derived in problem 2 for a circularorbit of a classical particle with angular momentum L.
4. The eigenfunction for the ground state of the hydrogen atom has the formc1 (r)  N1 eÿr=a0
where a0 is the Bohr radius, N1 is a constant and r is the radial distance ofthe electron from the nucleus.
(a) By normalizing the integral in accordance withZ 1
0

jc1 (r)j2 4pr2 dr  1,

find the constant N1 .
(b) Given that the eigenfunction of an excited state has the form

208

The hydrogen atom

Chap. 9c2 (r)  N2 (1  lr) eÿr=2a0 ,use the orthogonality condition for c1 and c2 to find the constant l, andsketch the eigenfunction c2 (r).
5. In this problem you are asked to verify that gn  n2 gives the number of
independent hydrogen states with energy En  ÿER =n2 , i.e. gn is the degeneracy of the level En . To do so you need to note that for a given value of n, thequantum number l can equal n ÿ 1, n ÿ 2, . . . 0, and that for a given value ofl, the quantum number ml can have 2l  1 values between l and ÿl.
(a) Verify that gn  n2 is valid when the principal quantum number takes on
the values 1 and 2.
(b) Verify that, if gn  n2 is valid when n  k, then it is also valid whenn  k  1.
Note (a) and (b) imply that gn  n2 is valid for all the possible values ofthe principal quantum number.
6. Rewrite Eq. (9.25) for the mean radius of hydrogen atom states in terms of
the principal quantum number n and orbital angular momentum quantum
number l. Find the mean radii for the 1s, 2s, 2p, 3s, 3p and 3d states.
7. Use the virial theorem, derived in problem 8 at the end of Chapter 7, to findthe expectation values of the kinetic and potential energies of an electron ina hydrogen atom state with principal quantum number n.
8. Use the three-dimensional equivalent of Eq. (3.20),
~
c(p)


1
(2p
h)3=2

Z

eÿpr=h c(r) d3 r,

to show that the momentum probability amplitude for an electron in the
ground state of the hydrogen atom with wave function
1c(r)  q eÿr=a0
pa30
is given byp
5=2
2 2 p0~,c(p) 
p (p2  p20 )2

Problems 9

209

where p0  h=a0 .
Hence showpthat
 the most probable magnitude of the momentum of theelectron is p0 = 3 and that its mean value is 8p0 =3p.
9. According to Eq. (9.27), the probability for an electric dipole transition
between hydrogen states with quantum numbers ni , li , mli and nf , lf , mlf is
proportional to the modulus squared of the integralZ^ I cn , l , m (r) d3 r:
c*nf , lf , mlf (r) Hi iliShow that this integral is zero for a 2s ! 1s transition, but that it is notzero for a 2p ! 1s transition.
10. Show that the expectation values of the spin±orbit energy given by
Eq.(9.35) for a 2p3=2 and a 2p1=2 state of a hydrogen atom are
hEmag i2p3=2 

1 4
a m e c296

andhEmag i2p1=2  ÿ

2 4
a m e c2 :96

(Hint: use the eigenfunctions given in Table 9.1 to find the expectation
value of 1=r3 and note thath(J 2 ÿ L2 ÿ S 2 )i  j(j  1) ÿ l(l  1) ÿ s(s  1)
for a state with quantum numbers j, l and s.)
Find the energy difference of the 2p3=2 and 2p1=2 states and showthat the difference in wavelengths for the transitions 2p3=2 ! 1s1=2 and
2p1=2 ! 1s1=2 is 5:4  10ÿ4 nm.
11. Positronium is a bound state of an electron with charge ÿe and its antiparticle, a positron with charge e.
(a) Given that the mass of the positron is equal to the mass of the electron,
what is the reduced mass of the electron±positron system?
(b) Write down an expression for the energy levels of positronium.
(c) Given that the mean radius of the ground state of the hydrogen
atom is 32 a0 , what is the mean radius of the ground state of positronium?
12. A muonic hydrogen atom is a bound state of a muon with charge ÿe and a
proton with charge e.

210

The hydrogen atom

Chap. 9(a) Given that the mass of the muon is 207me , what is the reduced mass ofthe muon±proton system?
(b) Write down an expression for the energy levels of muonic hydrogen.
(c) Given that the mean radius of the ground state of an ordinary hydrogen atom is 32 a0 , what is the mean radius of the ground state of amuonic hydrogen atom?
13. It is possible for the electron in a hydrogen atom to get very close to the
nucleus. To explore this possibility, consider a radius R which is muchsmaller than a0 and show that an electron in a 1s state has a probability of
about 4(R=a0 )3 of being found between r  0 and r  R.
Given that the nucleus is a proton with a radius of about 2  10ÿ15 m,
make a rough estimate of the probability that a 1s electron is inside theproton. Make a similar estimate for a 1s muon in a muonic hydrogen atom,and explain why the energy levels of a muonic hydrogen atom are moresensitive to the size of the proton than the energy levels of an ordinary atomof hydrogen.
14. By modifying the formulae for the hydrogen atom given in this chapter,
write down the energy En of an electron with principal quantum number nin the Coulomb potential due to a point charge Ze,
V (r)  ÿ

Ze2:
4pE0 r

By inspection of Table 9.1, write down the 1s, 2s, and 2p radial eigenfunctions for this electron.
What are the ionization energies of the one-electron ions He and2
Li ?
Explain why relativistic corrections are more important for He and2
Li than they are for the hydrogen atom.
15. Before you tackle this problem, you should work through problem 8 at theend of Chapter 4.
Consider the electron in an atom of the heavy isotope of hydrogen,tritium. The nucleus has charge e, and, apart from a small correction due
to the reduced mass effect, the electron has energies and eigenfunctions thatare identical to those of an ordinary hydrogen atom. However, the nucleusof the atom, a triton, is unstable and decays by beta-decay to form anucleus of 3 He. When it does so, the electron in the tritium atom suddenly
finds itself in a new Coulomb potential, the potential due a nucleus withcharge 2e.

Problems 9

211

Assume that the electron is initially in the ground state of the tritium atomand show that
P

128
a60

Z 1
0

r2 eÿ3r=a0 dr

2

is the probability that the electron is, after the decay, in the ground state of
the He ion.
Evaluate the integral and verify that this probability is 0.702.

This page intentionally left blank10
Identical particles
In classical physics we can, in principle, keep track of particles as they moveand maintain a record of which one is which, even if they are identical. This is
not possible in quantum physics. Quantum particles with uncertain position
and momentum cannot be tracked, and, if they are identical, they are truly
indistinguishable. The indistinguishability of identical particles is a fundamental principle in quantum physics which gives rise to phenomena which have noclassical analogue. We shall show how this principle leads to the importantconcept of exchange symmetry and to a classification of quantum particles into
two types called bosons and fermions.
This chapter, like Chapter 7, will deal with concepts that are more abstract
than those encountered elsewhere in this book. Even though these concepts arefundamental, they need not be studied in detail in order to understand atoms
which is the topic covered in Chapter 11.

10.1

EXCHANGE SYMMETRY

We shall primarily develop an understanding of identical particles by considering a system of two particles p and q. All the properties of such a system may be
extracted from a two-particle wave function C(rp , rq , t). For example, the joint
probability for finding particle p in a volume element d3 a at rp  a and particle
q in volume element d3 b at rq  b is given byjC(a, b, t)j2 d3 a d3 b:(10:1)

When the particles are identical, the wave function must give rise to identicalproperties for each particle. In particular, the probability density for finding
particle p at a and particle q at b must be the same as the probability density forfinding particle q at a and particle p at b. This requirement is met by a wave
function which obeys the condition

214

Identical particles

Chap. 10jC(a, b, t)j2  jC(b, a, t)j2 :(10:2)

When this condition for an acceptable wave function is satisfied,jC(a, b, t)j2 d3 a d3 b is the probability that one or other of the particles isfound at a and the other is found at b.1
Condition (10.2) implies that the function C(a, b, t) differs from the function
C(b, a, t) by at most a phase factor; i.e. we must haveC(a, b, t)  eid C(b, a, t):(10:3)

However, the phase factor eid can only take on two possible values. We canshow this by noting that the exchange of a and b in Eq. (10.3) gives
C(b, a, t)  eid C(a, b, t)
which, when Eq. (10.3) is used again, gives
C(b, a, t)  eid eid C(b, a, t):This equation shows that eid eid must be equal to one and that the phase factor
eid has two possible values:eid  1oreid  ÿ1:

(10:4)

Thus, any wave function for two identical particles must either be a symmetricfunctionC(a, b, t)  C(b, a, t)(10:5)or an antisymmetric function
C(a, b, t)  ÿC(b, a, t):(10:6)

These functions are said to have definite exchange symmetry, a property which
ensures that identical particles cannot be distinguished.

1

Some readers may consider our treatment of identical quantum particles to be contrived, because
we label the identical particles and then we only accept wave functions for which the labelling has
no physical consequences. If so desired, the contrived logic of using labels which cannot be
observed can be avoided, but to do so one must use quantum field theory.

10.2 Physical Consequences

10.2

215

PHYSICAL CONSEQUENCES

We have shown that identical quantum particles must be described by wave
functions that are either symmetric or antisymmetric when the particles areexchanged. We shall explore the physical consequences of this exchange symmetry by considering the simple example of two identical particles of mass m ina one-dimensional harmonic oscillator potential. We shall assume that theparticles have no direct interaction with each other so that their Hamiltonian
operator is given byh2 ] 2h2 ] 2

1
1^ p , xq )  ÿ ÿ m!2 x2p  m!2 x2q :H(x2m ]x2p 2m ]x2q 2
2(10:7)

In problem 2 at the end of this chapter, we show that each of the particles may
occupy single-particle states represented by eigenfunctions cn with energies
given by En  (n  12 )h! with n  0, 1, 2 . . ..
We first consider two particles in the same single-particle state. If this state
has quantum number n, the total energy of the two particles ish!
E  En  En  (2n  1)
and the two-particle wave function with this energy isC(S) (xp , xq , t)  cn (xp )cn (xq ) eÿi(En En )t=h ,(10:8)

where cn is an eigenfunction for a particle with energy En in a harmonic oscillator
potential; explicit expressions for the eigenfunctions with low energy are given in
Table 6.1. We have labelled the two-particle wave function with the superscript
S to indicate that it is symmetric under particle exchange which means that it is anacceptable wave function for two identical particles. We note that it is impossible
to construct an antisymmetric wave function if both particles are in the same
single-particle state. This implies that two identical particles with an antisymmetric wave function cannot occupy the same single-particle state.
We now consider two particles in two different single-particle states. Whenthe particles occupy states with quantum number n and n0 , their energy ish!:E  En  En0  (n  n0  1)
The wave function for two distinguishable particles with this energy may be
given byÿi(En En0 )t=h,
C(D)
1 (xp , xq , t)  cn (xp )cn0 (xq ) eby(10:9)

216

Identical particles

Chap. 10ÿi(En En0 )t=h
C(D),2 (xp , xq , t)  cn (xq )cn0 (xp ) e(10:10)or by a linear combination of the form
(D)
C(D) (xp , xq , t)  c1 C(D)1 (xp , xq , t)  c2 C2 (xp , xq , t),(10:11)

where c1 and c2 are constants. The label D has been used to denote that thesewave functions can be used to describe distinguishable particles. We note thatEq. (10.11) specifies a strange wave function in which both particles areassociated with both single-particle states. Indeed, provided the wave functions
2DCD
1 and C2 are normalized, jc1 j gives the probability that particle p is in staten and particle q is in state n0 and jc2 j2 gives the probability that particle q is instate n and particle p is in state n0 . Because quantum states with wave functions
like (10.11) are called entangled states, we shall occasionally refer to states
described by wave functions like (10.9) and (10.10) as untangled states. Whether
or not distinguishable particles are in an entangled or untangled quantum state
depends on the process that led to the formation of the state.
The wave function for two identical particles must have definite exchange
symmetry, and for particles with energy En  En0 this wave function is necessarily an entangled wave function. It can be a symmetric wave function of the form
1
C(S) (xp , xq , t)  p[cn (xp )cn0 (xq )  cn (xq )cn0 (xp )] eÿi(En En0 )t=h2(10:12)or an antisymmetric wave function of the form
1
C(A) (xp , xq , t)  p[cn (xp )cn0 (xq ) ÿ cn (xq )cn0 (xp )] eÿi(En En0 )t=h :2(10:13)

Both these wave functions represent states in which each particle is equallyassociated with both single-particle states.
We shall now use these wave functions to show that exchange symmetry
leads to a surprising tendency for identical particles to either huddle together or
to avoid each other. To do so, we shall set xp  xq  x0 and compare the valuesof the wave functions for distinguishable and for identical particles when theparticles have the same location x0 . For distinguishable particles with untangled wave functions given by Eqs. (10.9) and (10.10), we obtain
ÿi(En En0 )t=h
C(D):1, 2 (x0 , x0 , t)  cn (x0 )cn0 (x0 ) e

For identical particles with a symmetrical entangled wave function (10.12) we
obtain10.2 Physical Consequences

C(S) (x0 , x0 , t) 

217

p
0
2cn (x0 )cn0 (x0 ) eÿi(En En )t=h ,and for identical particles with an antisymmetrical entangled wave function(10.13) we obtain
C(A) (x0 , x0 , t)  0:
These equations show that, all other things being the equal, two identicalparticles with a symmetric wave function are twice as likely to be found at
the same location as two distinguishable particles with an untangled wave
function, and that two identical particles with an antisymmetric wave function
are never found at the same location. The wave-mechanical origin of thisbehaviour is interference, either constructive in a symmetrical wave function
or destructive in an antisymmetrical wave function.
The tendency of identical particles for togetherness or for avoidance may beillustrated more fully by considering an example where the particles occupy
harmonic oscillator states with n  0 and n0  1. If we use the eigenfunctions
listed in Table 6.1 and use the coordinatesx  xp ÿ xq

and

X

xp  xq,2

we obtain the following expressions for the symmetric and antisymmetric wave
functions given by Eqs. (10.12) and (10.13):
C(S) (x, X , t) 

2
2
2
2
2
p eÿx =4a X eÿX =a eÿi(E0 E1 )t=h
a2 p

C(A) (x, X , t) 

ÿ1
2
2
2
2
p x eÿx =4a eÿX =a eÿi(E0 E1 )t=h :
a2 p

and

The square modulus of each of these wave functions gives a probability
density for the particles to have a separation x and a centre of mass locatedat X. By integrating over all possible values of X, we can find the probability fora separation x. Straightforward calculus shows that the probability for aseparation with a magnitude between jxj and jx  dxj is
2
2
2
P(S) (x) dx  p eÿx =2a dx
a 2p
for the symmetric wave function and(10:14)

218

Identical particles

Chap. 102 x2
2
2
P(A) (x) dx  p 2 eÿx =2a dx
aa 2p(10:15)for the antisymmetric wave function. The corresponding probability for two
distinguishable particles with an untangled wave function like that given by
Eq. (10.9) or by Eq. (10.10) is


1
x2
2
2
P(D) (x) dx  p 1  2 eÿx =2a dx:
aa 2p(10:16)

The probability densities P(S) (x), P(A) (x) and P(D) (x) are plotted in Fig. 10.1.
The graphs on the left show that identical particles huddle together if they havea symmetric wave function and the graphs on the right show that identical

1

1

0.8

0.8

probability density

probability density

(S)

0.6
(D)

(D)

0.6

0.4

0.4

0.2

0.2

0

0

(A)

0

1
2
two -particle separation |x|

3

0

1
2
two -particle separation |x|3

Fig. 10.1 The separation of two identical particles in a one-dimensional harmonic
oscillator potential. We have assumed that the particles are in single-particle stateswith quantum numbers n  0 and n0  1 and that the harmonic oscillator length parameter a is equal to one. The probability densities P(S) (x) and P(A) (x) for symmetric andantisymmetric wave functions are labelled (S) and (A), and they are compared with theprobability density labelled (D) for two distinguishable particles with an untangled wave
function given by Eq. (10.9) or by Eq. (10.10).

10.3 Exchange Symmetry with Spin

219

particles avoid each other if they have an antisymmetric wave function. Similar
behaviour could be demonstrated by distinguishable particles, but only whenparticular physical processes lead to the formation of symmetrical or antisymmetrical entangled states.
We emphasise that these tendencies for togetherness or avoidance are not aconsequence of attractive or repulsive forces between the particles; indeed in
our illustrative example, the particles are governed by a Hamiltonian,Eq. (10.7), in which there is no direct interaction between the particles. Thesetendencies are a result of the exchange symmetry that always exists for identicalquantum particles. We also note from Fig. 10.1, that the effects of exchange
symmetry become less important as the separation of the particles increases.
This reflects the fact that if identical particles can keep their distance they
become almost distinguishable.

10.3

EXCHANGE SYMMETRY WITH SPIN

We have shown that the indistinguishability of identical particles leads to wave
functions with definite exchange symmetry. We have yet to consider the circumstances that determine whether the wave function is symmetric or antisymmetric.
These circumstances depend on the spin of the particles, but to understand therole of spin in determining exchange symmetry, we need to extend our representation of a quantum state so that it includes a description of particle spin.
As stated in Section 8.1, a particle is said to have spin s if the magnitude andthe z component of its intrinsic angular momentum are given by8s>>> (s ÿ 1)><p
S  s(s  1)
h and Sz  ms h, where ms  ...
>>>>: ÿ(s ÿ 1)
ÿs:For example, electrons, protons and neutrons have spin s  12, the deuteronnucleus 2 H has spin s  1 and the helium nucleus 4 He has spin s  0.
From Section 8.3 we recall that all the orbital angularmomentum propertiesp
of a particle p with orbital angular momentum l(l  1)h can be described
all theusing 2l  1 eigenfunctions Yl , ml (yp , fp ). Similarly,p
 intrinsic angularmomentum properties of a particle p with spin s(s  1)h can be describedusing 2s  1 eigenvectors xs, ms (p) with ms  ÿs, ÿ s  1, . . . ,  s. In particular, in analogy with Eq. (8.28), the most general spin state has the formx(p) sX
ms ÿscms xs, ms (p)220

Identical particles

Chap. 10where jcms j2 is equal to the probability that the particle has a z component ofspin equal to ms h. (See footnote 2 below.)
When the spatial and spin properties of a particle are independent of eachother, the quantum state may be represented by a productF(p)  c(rp )x(p):For example, the first term c(rp ) describing the spatial properties of the particle
could be a hydrogen±like wave function with quantum numbers n, l, ml and thesecond term w(p) could be a spin state with quantum numbers s  12 andms   12. When this is the case, we have a single-particle quantum state of
the form
Fn, l , ml , ms (p)  cn, l , ml (rp )xs, ms (p):We can now write down expressions for quantum states which describe thespatial and spin properties of two identical particles.
When both particles occupy the same single-particle state, say one withspatial and spin quantum numbers n, l, ml , ms , we can construct a symmetricalstate for two identical particles of the form
F(S) (p, q)  Fn, l , ml , ms (p)Fn, l , ml , ms (q):(10:17)

But an antisymmetric two-particle state for two identical particles cannot be
constructed when both particles occupy the same single-particle state. Thisimplies that, when identical particles have antisymmetric exchange symmetry,
two or more particles cannot occupy the same single-particle state.
When the particles are associated with two different single-particle states, it ispossible to construct both symmetric and antisymmetric two-particle states.
For example, we can have a symmetric state of the form
1
F(S) (p, q)  p [Fn, l , ml , ms (p)Fn0 , l 0 , m0 , m0 (q)sl2 Fn, l , ml , ms (q)Fn0 , l 0 , m0 , m0 (p)]l2

(10:18)s

In general, the probability amplitude cms depends on time, but to keep the notation simple we shall
ignore time dependence. The spin eigenvectors xs, ms (p) are less abstract if they are represented bycolumn matrices with 2s  1 components; for example, a particle with spin s  12 can be describedusing the matrices
 
 
1
0
x 12, 12 (p) and x12, ÿ12 (p) 
,
0 p
1 p
where the subscript p is necessary because a matrix representing particle p must be distinguishedfrom a matrix representing another particle. The mathematics of the representation of spinquantum states is covered in more advanced books, but this mathematics will not be needed here.

10.3 Exchange Symmetry with Spin

221

and an antisymmetric state of the form
1
F(A) (p, q)  p [Fn, l , ml , ms (p)Fn0 , l 0 , m0 , m0 (q)sl2ÿ Fn, l , ml , ms (q)Fn0 , l 0 , m0 , m0 (p)]:l(10:19)s

Alternatively, we can construct symmetric and antisymmetric two-particle
states by combining two-particle wave functions and two-particle spin stateswith the appropriate symmetry. We shall illustrate this procedure by considering two spin-half particles.
Spin states for two spin-half particles may be labelled by the quantum
numbers S and MS , indicating that the magnitude andz component of thep
combined spin angular momentum of the state are
S(S  1)h and MS h,respectively.3 We have already encountered the rule for the addition of theorbital and spin angular momenta of a single particle in Section 9.6. The rule
for the addition of two spins is very similar. In particular, two spins withquantum number s  12 can be combined to give a combined spin with quantum
numberS  12  12  1

orS  12 ÿ 12  0:This rule is merely a quick way of saying that two spin-half states, x12 , m s (p) and
x12 , m s (q), may be combined to give two-particle spin states with quantum
numbers S  1 and S  0. Explicitly, if we use a simplified notation in which
x12 , m s is denoted by x when ms   12 and xÿ when ms  ÿ 12, we have threesymmetric spin states,x(S)1, 1 (p, q)  x (p)x (q)

9
>
>
>
>
=

1
x(S)
1, 0 (p, q)  p [x (p)xÿ (q)  x (q)xÿ (p)]
>
2
>
>
>
(S)
;
x1, ÿ1 (p, q)  xÿ (p)xÿ (q),(10:20)which correspond to S  1 and MS  1, 0 and ÿ1, and one antisymmetricspin state1x(A)0, 0 (p, q)  p [x (p)xÿ (q) ÿ x (q)xÿ (p)]2

(10:21)

which corresponds to S  0 and MS  0.
3

We follow the convention of using the capital letters S and MS to denote the spin angularmomentum of two or more electrons, even though the symbol S was also used to denote themagnitude of a spin angular momentum in Chapter 8.

222

Identical particles

Chap. 10

These two-particle spin states may be combined with two-particle wave
functions to produce a two-particle quantum state with definite exchange
symmetry. For example, we can construct antisymmetric quantum states fortwo electrons, in two ways. We can combine a symmetric spin state withquantum numbers S  1 and MS  1, 0, ÿ 1 and an antisymmetric wave function c(A) (rp , rq ) to giveF(A) (p, q)  c(A) (rp , rq )x(S)S , MS (p, q),(10:22)or we can combine an antisymmetric spin state with quantum numbers S  0
and MS  0 and a symmetric wave function c(S) (rp , rq ) to giveF(A) (p, q)  c(S) (rp , rq )x(A)S , Ms (p, q):(10:23)We have chosen this example because, as we shall discover in the next
section, electron quantum states are always antisymmetric. This meansthat the symmetry of the wave function of two electrons is fixed by the valueof the combined spin of the electrons in accordance with Eqs. (10.22) and(10.23).
Finally, in order to keep this and the preceding section as simple as possible,
we have only discussed quantum states for two identical particles. When several
identical particles are considered, similar conditions for acceptable quantum
states emerge. They are that the quantum state of a system of identical particles
must be either symmetric or antisymmetric when any two of the particles areexchanged.

10.4

BOSONS AND FERMIONS

According to the arguments presented in the last section, a quantum state
describing the spatial and spin properties of identical particles must havedefinite exchange symmetry and there are two possible options for this symmetry: symmetric or antisymmetric. But, in the real world, identical particles do
not have the freedom to choose between these options. In the real world there
are two types of quantum particles called bosons and fermions with thefollowing characteristics:. Bosons are particles with integer spin and a system of identical bosons musthave quantum states which are symmetric when two particles are exchanged.
. Fermions are particles with half-integer spin and systems of identical fermions must have quantum states which are antisymmetric when two particles
are exchanged.

10.4 Bosons and Fermions

223

This surprising connection between the exchange symmetry and the spin ofidentical particles is called the spin-statistics theorem. Its far-reaching consequences for the properties of matter were identified in the early days ofquantum mechanics and it was subsequently shown to be a theoretical consequence of relativistic quantum field theory. The spin-statistics theorem meansthat in applied quantum mechanics we encounter two ways of being indistinguishable, the boson way and the fermion way.
Because electrons have spin half, they are indistinguishable in the fermion
way. They never have symmetric states like Eq. (10.18) but only antisymmetricstates like Eq. (10.19). These antisymmetric states for electrons are very relevant to our understanding of the world around us because the properties ofatoms and solids are largely determined by the quantum mechanics of electrons. The three most important consequences of the antisymmetry of electron
quantum states are the following:First, a single-particle state can be occupied by at most one electron, because,
if it were occupied by more than one electron, the multi-electron quantum state
would be symmetric and not antisymmetric when the electrons were exchanged.
This characteristic of electrons is called the Pauli exclusion principle.4 The Pauliprinciple plays a governing role in determining the physical and chemical
properties of atoms and leads, as we shall show in the next chapter, to anunderstanding of the Periodic Table of the Elements. It also has a crucial role indetermining the electrical and thermal properties of electrons in solids.
Second, when the spin part of the quantum state of two electrons is symmetric, the spatial part, i.e. the wave function, is necessarily antisymmetric as in Eq.
(10.22). When this is the case, the electrons, like the two identical particles in aharmonic oscillator potential described by the graph on the right of Fig. 10.1,
have a strong tendency to avoid each other. This tendency for avoidance isultimately responsible for the rigidity of ordinary matter; a solid resists compression because identical electrons avoiding each other prevent atoms fromgetting closer.
Third, when the spin state of two electrons is antisymmetric, the wave
function is necessarily symmetric as in Eq. (10.23). In this case, the electrons,
like the two identical particles described by the graph on the left of Fig. 10.1,
have a tendency to huddle together. When this happens between adjacentatoms, a covalent bond between the atoms may be created and a molecule
formed.
Protons and neutrons, like electrons, are also spin-half fermions, and thefermion-like indistinguishability of protons and of neutrons plays a governing
role in the shell model of the nucleus. In this model protons and neutrons
occupy single-particle states, but a single-particle state can be occupied by at
4

We note with admiration that W Pauli discovered the exclusion principle empirically in 1924 andthat he also proved the spin-statistics theorem some 17 years later using relativistic quantum field
theory.

224

Identical particles

Chap. 10

most one proton and by one neutron. Similarly, theoretical models of protons,
neutrons and other hadrons are governed by the idea that quarks of a specificflavour and colour also act like systems of identical fermions with antisymmetric quantum states.
The boson way of being indistinguishable also leads to important physicalphenomena. Because bosons are described by symmetric quantum states, manybosons may occupy the same single-particle state and when this happens
quantum-mechanical behaviour on a macroscopic scale may arise.
The most important example of boson togetherness is the coherent light of alaser. This coherence arises because photons, bosons with spin one, have a highprobability to have the same energy and momentum, in much the same way as
two particles with a symmetric wave function have a high probability of beingat the same location.
Boson togetherness is also responsible for the superfluidity of liquid helium attemperatures below 2.2 K. Liquid helium consists of a system of weakly interacting helium atoms which behave like bosons because they consist of a 4 He nucleus
with spin zero and two electrons with a combined spin of zero. At low temperatures, a considerable fraction of the atoms in liquid helium `condense' into thesame lowest-energy state. They form a Bose±Einstein condensate in which theatoms have wave functions which are coherent with each other and movecollectively without friction. Recently almost pure Bose±Einstein condensates
have been produced by cooling atoms in magnetic traps; indeed the 2001 Nobel
Prize in Physics was awarded to Eric Cornell, Wolfgang Ketterle and Carl Wieman for their work in producing the first pure Bose±Einstein condensate in 1995.
Surprisingly, boson-like togetherness also occurs in situations where fermion-like behaviour is expected. It occurs in the superconductivity of metals
at low temperatures because pairs of electrons act like indistinguishable bosons.
It also probably occurs when liquid helium-3 becomes a superfluid at very low
temperatures. Helium-3 atoms, unlike the normal helium atoms, are fermionsbecause the 3 He nucleus has spin half, but pairs of helium-3 atoms can act like asystem of indistinguishable bosons and give rise to collective motion with nofriction in liquid helium-3.

PROBLEMS 10
1. In Section 10.1 we explained why the wave function of two identical particles
has a definite exchange symmetry. In this problem we show that this exchange symmetry remains unchanged as the wave function evolves.
Given that the time evolution of the wave function for two particles isgoverned by the SchroÈdinger equation,ih]C(rp , rq , t)
^ p , rq )C(rp , rq , t),
 H(r
]t

Problems 10225^ p , rq ) is the Hamiltonian operator for particles p and q, show that
where H(rthe relation between the wave functions at time t and at time t  dt, isi ^C(rp , rq , t  dt)  1 ÿ H(rp , rq )dt C(rp , rq , t):hWhat general symmetry condition must be satisfied by the Hamiltonian^ p , rq ) in order that the exchange symmetry of the wave function
operator H(ris the same at times t and t  dt? Explain why this condition is always
satisfied when the particles are identical.
(Hint: Eq. (10.7) is an example of a two-particle Hamiltonian operator.)
2. The wave function C(xp , xq , t) of two particles with mass m and total energy
E in a one-dimensional harmonic oscillator potential has the form
C(xp , xq , t)  c(xp , xq ) eÿiEt=h ,where c(xp , xq ) satisfies the two-particle eigenvalue equation"#
h2 ]2h2 ] 2

1
1
2 2
2 2
ÿ
 m! xp  m! xq c  Ec:ÿ
2m ]x2p 2m ]x2q 2
2

Show by substitution that the function
c(xp , xq )  cn (xp )cn0 (xq ),where cn and cn0 are one-particle harmonic oscillator eigenfunctions withenergies En and En0 , satisfies these equations if E  En  En0 .
3. Consider the following wave function for two distinguishable particles
C(D) (xp , xq , t)  cn (xp )cn0 (xq ) eÿi(En En0 )t=h ,and the following wave functions for two identical particles
1
C(S) (xp , xq , t)  p [cn (xp )cn0 (xq )  cn (xq )cn0 (xp )] eÿi(En En0 )t=h2
and1
C(A) (xp , xq , t)  p [cn (xp )cn0 (xq ) ÿ cn (xq )cn0 (xp )] eÿi(En En0 )t=h :2

226

Identical particles

Chap. 10(a) Show that all of these wave functions are normalized if the singleparticle eigenfunctions cn and cn0 are normalized and orthogonal.
(b) Suppose that there is a weak, repulsive interaction between the particles
given by the potential V (jxp ÿ xq j) which causes the energy of particles
with wave function C(xp , xq , t) to shift by
DE 

Z 1
ÿ1

dxp

Z 1
ÿ1

dxq C*(xp , xq , t)V (jxp ÿ xq j)C(xp , xq , t):

How does the energy shift for distinguishable particles compare with theenergy shifts for identical particles?
4. Consider two non-interacting particles p and q each with mass m in a cubicalbox of size a. Assume that the energy of the particles isE

3h2 p2 6
h2 p2

:
2
2ma
2ma2

Using the eigenfunctions
cnx , ny , nz (xp , yp , zp ) and

cnx , ny , nz (xq , yq , zq )given by Eq. (4.43) in Section 4.4, write down two-particle wave functions
which could describe the system when the particles are:(a) distinguishable, spinless bosons;(b) identical, spinless bosons;(c) identical spin-half fermions in a symmetric spin state;(d) identical spin-half fermions in the antisymmetric spin state.
5. Suppose that five non-interacting particles are placed in a three-dimensional
harmonic oscillator potential described in Section 6.5, for which the singleparticle energy isÿEn  nx  ny  nz  32 h!:What is the lowest energy of the five-particle state when the particles are:(a) distinguishable, spinless bosons;(b) identical, spinless bosons;(c) identical fermions each with spin s  12 ;(d) identical fermions each with spin s  32 ?

Problems 10

227

6. The hydrogen molecule contains two identical, spin-half nuclei which canvibrate and rotate. We discussed the vibrational energies in Section 6.4 andthe rotational energies in problem 4 at the end of Chapter 8. We shall label
the nuclei by p and q, and describe their spin properties using the spin states
wS, MS (p, q) with S  1, MS  1, 0, ÿ 1 and S  0, MS  0. When the molecule is in a rotational state with orbital angular momentum quantum
numbers l and ml , the spatial separation of the nuclei r  rp ÿ rq is governedby a wave function of the form
cl , ml (r)  R(r)Yl , ml (y, f):Bearing in mind the symmetry properties of the spin states given in
Eqs. (10.20) and (10.21) and the symmetry properties of the sphericalharmonics given in Table 8.1, explain why the rotational states of thehydrogen molecule with odd l have nuclei with a combined spin given byS  1 and those with even l have nuclei with a combined spin given by S  0.
(Because the spins of the nuclei in a hydrogen molecule are seldom
affected by a collision, a gas of hydrogen acts as if it consists of two typesof molecules, ortho-hydrogen with rotational states with odd l and parahydrogen with rotational states with even l. It takes days for a gas ofortho-hydrogen to reach thermal equilibrium with a gas of para-hydrogen.)This page intentionally left blank

11
Atoms
As Richard Feynman once said, why do chemists count funny? Instead ofsaying 1, 2, 3, 4, . . . , they say hydrogen, helium, lithium, beryllium, . . . . The
reason is remarkable. An atom with Z electrons is identical to all other atoms
with Z electrons. They have the same size, the same ionization energy and thesame tendency to react or not to react. Atomic sameness may be labelled by one
number, the atomic number Z, and there are only 100 or so different types.
Moreover, this sameness is resilient. When an atom is excited by an encounter
with a photon, an electron or another atom, it will always return to its original
pristine condition.
The resilient sameness of atoms with atomic number Z cannot be understoodin terms of classical physics. It is a property of the quantum states of Z
electrons. The essential features of these quantum states may be understoodby combining, in an approximate way, the concepts we used to describe thehydrogen atom with the implications that arise from the fact that electrons areindistinguishable fermions. Even though we have to be content with approximate representations for the quantum states of an atom with atomic number Z,
these states in principle provide a complete description which is unique to all
atoms with atomic number Z.

11.1

ATOMIC QUANTUM STATESAn atom with atomic number Z consists of Z electrons held together in thepotential energy field due to the Coulomb attraction of a nucleus and the
Coulomb repulsion between each pair of electrons. In the helium atom, forexample, two electrons move in the potential energy field,
V (rp , rq )  ÿ

2e2
2e2
e2
,
ÿ

4pE0 rp 4pE0 rq 4pE0 jrp ÿ rq j

(11:1)

230

Atoms

Chap. 11and the energy levels and eigenfunctions can be found by solving the eigenvalue
equation"#h2ÿ(r2  r2q )  V (rp , rq ) c(rp , rq )  Ec(rp , rq ):2me p(11:2)

Accurate numerical solutions of this equation may be obtained, but for atoms
with many electrons, approximate methods based on the central field approximation must be used. These methods, which were first developed by E. Fermi,
D. R. Hartree and L. H. Thomas, lead to the following qualitative descriptionof atomic quantum states.

The central field approximation
In the central field approximation, each electron in an atom moves independently in a central potential due to the Coulomb attraction of the nucleus and theaverage effect of the other electrons in the atom. If the electron under consideration has coordinate rp , it will see, when rp is large, a nucleus with charge Ze
screened by inner electrons with total charge ÿ(Z ÿ 1)e and we expect a
potential energy of the form
VB (rp )  ÿ

e2:
4pE0 rp(11:3)

But when rp is small, the electron sees a bare, unscreened nucleus and we expect
a potential energy of the form
VC (rp )  ÿ

Ze2:
4pE0 rp(11:4)

These considerations imply that the rough form of the central potential for eachindependent electron in an atom is similar to the curve (A) shown Fig. 11.1. At
large distances this potential approaches the Coulomb potential due to a point
charge e, curve (B), and at small distances it approaches the Coulomb potential
due to a point charge Ze, curve (C). A very simple model for this central
potential is given by
VA (rp )  ÿz(rp )e2,
4pE0 rpwhere z(rp )  (Z ÿ 1) eÿr=a  1:

(11:5)

This equation with a equal to half a Bohr radius and Z  6 was used to
calculate curve (A) in Fig. 11.1.

11.1 Atomic Quantum States

231

0

Potential energy in units of the Rydberg energy

−5

−10
(B)

−15

(A)

(C)

0.5

1
1.5
2
Distance in units of the Bohr radius

−20

−25

−30

−35

−40

0

2.5

3

Fig. 11.1 A simple model for the central potential for an electron in a carbon atom withatomic number Z  6. Curve (A) gives the screened potential energy VA (r) given by
Eq. (11.5) in units of the Rydberg energy ER as a function of r in units of the Bohrradius, a0 , with a screening radius a  a0 =2. Curves (B) and (C) give the potentials due topoint charges e and 6e, VB (r) and VC (r) given by Eqs. (11.3) and (11.4).

Given a central potential V (r) one can find the energy levels and eigenfunctions for each electron in the atom by solving the single-particle energy eigenvalue equation



r2 2
ÿ
r  V (r) c(r)  Ec(r):2me(11:6)

This equation is identical to Eq. (9.5) which was the starting point for our
discussion in Chapter 9 of the quantum mechanics of a particle in a central
potential and of the hydrogen atom. This means that the eigenfunctions forthese single-particle states may be labelled by the same quantum numbers used
to label the eigenfunctions of the hydrogen atom: the principal quantum
number n, the orbital angular quantum numbers l and ml , and the electron
spin quantum number ms . But in contrast with Eq. (9.22) for the hydrogenatom, the energy levels depend on two of these quantum numbers, n and l.

232

Atoms

Chap. 11

The energy of these levels may be labelled Enl but we will use the spectroscopic
notation 1s, 2s, 2p, etc. We shall also follow the atomic physics convention ofreferring to these single-particle states as orbitals.
The effect of screening on the energies of the orbitals in a carbon atom is
illustrated in Fig. 11.2. The 1s, 2s and 2p energy levels for the unscreened
potential VC (r) are shown on the left and those for the screened potential
VA (r) are shown on the right. We note that for the screened potential, theenergy of an orbital increases when the principal quantum number n increases,
and that, for a given value of n, the energy increases when the orbital angularmomentum quantum number l increases. This increase with l can be understood by recalling from Section 9.1 that the effective potential in the radial
SchroÈdinger equation includes the centrifugal potential, l(l  1)h2 =2me r2 . Because the centrifugal potential becomes more repulsive as l increases, an electron with a higher value for l has a lower probability of penetrating close to the
nucleus and a higher probability of being located at larger distances where thenucleus is well screened by inner electrons. The net effect of higher l is a weakerattraction by the nucleus and a higher energy.
The eigenfunctions for these orbitals may be used to construct approximate
multi-electron quantum states for the atom as a whole. As discussed in Chapter
Energy levels for the
Coulomb potential VC
E= 0

E = −9ER

2s

Energy levels for thescreened potential VA
2p
2s

2p

1s

E = −36ER

1s

Fig. 11.2 The energies of the 1s, 2s and 2p orbitals in the carbon atom. The effect ofelectron screening is to raise the energy levels. This can be seen by comparing the energy
levels on the left, given by an unscreened Coulomb potential VC (r) due a point charge
with Z  6, with those on the right given by the screened potential VA (r) with Z  6 anda equal to half a Bohr radius; the energies for the screened potential were calculated
numerically by Daniel Guise.

11.1 Atomic Quantum States

233

10, quantum states which describe a system of indistinguishable electrons must
be antisymmetric whenever two electrons are exchanged. This can only beachieved if electrons are assigned to orbitals in accordance with the Pauli
exclusion principle; i.e. not more than one electron may occupy an orbitalwith the same quantum numbers n, l, ml , ms . This means that at most twoelectrons can be assigned to 1s orbitals, one with quantum numbers n  1,l  0, ml  0, and ms   12, and one with quantum numbers n  1, l  0,
ml  0, and ms  ÿ 12. Similarly, not more than two electrons can be assigned
to 2s orbitals, but up to six electrons can be assigned to 2p orbitals because there
are six of these orbitals with quantum numbers n  2, l  1, ml  1, 0, ÿ 1,
and ms   12. When the 1s, 2s and 2p orbitals are fully occupied, additional
electrons may only be assigned to orbitals with principal quantum numbers ngreater than 2. These orbitals have higher energy and also a limited capacity.
We can illustrate how to use this construction kit for atomic states byconsidering a carbon atom containing six electrons which may occupy energy
levels similar to those shown on the right-hand-side of Fig. 11.2. The groundstate is obtained by assigning six electrons to orbitals with the lowest possibleenergy; a maximum of two electrons can have the energy E1s , a maximum of
two can have the energy E2s and the minimum energy for each of the tworemaining electrons is E2p . These assignments give an electron configuration
denoted by (1s)2 (2s)2 (2p)2 with energy
E  2E1s  2E2s  2E2p :The first excited state of the carbon atom is obtained by assigning only one
electron to a 2s orbital and three electrons to 2p orbitals. This gives rise to theelectron configuration (1s)2 (2s)(2p)3 with energy
E  2E1s  E2s  3E2p :If the energy levels shown in Fig. 11.2 for the screened potential VA (r) are used
as a rough guide, the energy of the ground state is ÿ41:1ER and the energy ofthe first excited state is ÿ40:6ER . Clearly, states of higher excitation may beobtained by assigning more electrons to 2p orbitals or by assigning electrons to
3s, 3p, 3d, . . . orbitals.
In the preceding paragraph we have followed custom and given the wrong
impression that particular electrons are in particular orbitals. This is not thecase. Because all the electrons in the atom are indistinguishable, each electronis equally associated with each of the occupied orbitals. In fact, like thetwo-electron state given by Eq. (10.19), the multi-electron quantum state is
antisymmetric when any two of the electrons are exchanged.
We have also wrongly given the impression that the central potential which
represents the effect of the attraction of the nucleus and of the average effects ofelectron±electron repulsion, is easy to find. In fact, the central potential and the

234

Atoms

Chap. 11eigenfunctions for independent electrons in an atom are like the chicken and theegg; it is unclear which comes first. This problem may be resolved by a sequenceof calculations in which a central potential gives rise to eigenfunctions andelectron probabilities which in turn can be used to find an improved approximation to the central potential. Ultimately a self-consistent central potential
emerges which accurately describes independent electrons in the atom.

Corrections to the central field approximation
There are two important corrections to the central field approximation. The
first is due to a residual electron±electron repulsion which arises because the
Coulomb repulsion between the electrons in the atom cannot be describedaccurately by a central potential. The second is due to a spin±orbit interaction
which is similar to that considered for the hydrogen atom in Section 9.6. But
before we can understand these corrections, we need to develop a more precise
description of atomic quantum states which takes into account how the angularmomenta of atomic electrons may be combined. Readers who are content with
an approximate description based solely on electron configurations should proceed
directly to Section 11.2.
Because the angular momenta of atomic electrons involve unfamiliar notation, we shall begin by recalling some basic facts and stating some conventions
about angular momentum.
. The magnitude and z component of an angular momentum can have the valuesp
j(j  1)
h

and

mj 
h,

where j is a quantum number which can equal 0, 12 , 1, 32 , . . . and mj is aquantum number which can equal j, j ÿ 1, . . . , ÿ j.
. Two angular momenta with quantum numbers j1 and j2 may be combined togive an angular momentum with quantum number j which can take on thevaluesj  j1  j2 , j1  j2 ÿ 1, . . . , jj1 ÿ j2 j:. Lower case letters are used in atomic physics for the angular momentum of asingle electron: l and ml for orbital angular momentum, s and ms for spin
angular momentum, and j and mj for a combined orbital and spin angularmomentum. But capital letters are used for the angular momentum of two or
more electrons: L and ML for orbital angular momentum, S and MS for spin
angular momentum, and J and MJ for a combined orbital and spin angularmomentum.

11.1 Atomic Quantum States

235. Spectroscopic notation uses the letters s, p, d, f . . . to denote the orbital
angular momentum of a single electron with l  0, 1, 2, 3 . . .. But capitalletters S, P, D, F . . . are used to denote the orbital angular momentum of two
or more electrons with L  0, 1, 2, 3 . . ..
With these preliminaries out of the way, we can describe how the angularmomenta of atomic electrons may be combined. Two ways of combining
angular momenta are used in atomic physics. They are called L ÿ S coupling(also called Russell-Saunders coupling) and j ÿ j coupling.
In L ÿ S coupling, the orbital angular momenta of the electrons are coupled
to give a combined orbital angular momentum described by a quantum number
L, and the spin angular momenta of the electrons are coupled to give a
combined spin angular momentum described by a quantum number S. These
combined orbital and spin angular momenta are then coupled to give a total
angular momentum described by a quantum number J. In j ± j coupling, theorbital and spin angular momenta of each electron are coupled to give a
combined angular momentum described by a quantum number j. The combined angular momenta of each of the electrons are then coupled to give a total
angular momentum described by a quantum number J. The L ± S coupling
scheme is most useful when the residual electron±electron repulsion is larger
than the spin±orbit interaction, and j ± j coupling is most useful in the converse
situation. In practice, j ± j coupling is used for atoms with high atomic numbers
and L ± S coupling is used for atoms with low atomic numbers.
We shall illustrate these general ideas by considering how L ± S coupling canbe used to describe the carbon atom. We shall focus on the low-lying quantum2
states with the electron configuration (1s)2 (2s)2 (2p) . As stressed in Section
10.4, these quantum states must be antisymmetric when any two electrons areexchanged.
The two electrons in the 1s orbitals, which necessarily have a symmetric wave
function, are in an antisymmetric quantum state if they have an antisymmetricspin state. Such a spin state, according to Eq. (10.21), has zero spin. Thus, inaddition to having a zero orbital angular momentum, the two 1s electrons havea combined spin angular momentum equal to zero. This is also the case for thetwo 2s electrons.
The two electrons in the 2p orbitals, each with orbital angular momentum
quantum number l  1, can have a combined orbital angular momentum witha quantum number L which can take on the values 2, 1 or 0, and a combined
spin angular momentum with a quantum number S which can take on thevalues 1 or 0. As shown in Eqs. (10.20) and (10.21), the spin state with S  1 issymmetric and the spin state with S  0 is antisymmetric. It can be also be
shown that the exchange symmetry of the orbital angular momentum
wave functions is symmetric when L  2, antisymmetric when L  1 andsymmetric when L  0. Hence, acceptable quantum states with antisymmetricexchange symmetry only exist for the two 2p electrons if the spin and orbital

236

Atoms

Chap. 11angular momentum quantum numbers S and L occur in one of the combinations(i) S  1, L  1;(ii) S  0, L  2;(iii) S  0, L  0:

These states are usually denoted by the spectroscopic notation 3 P, 1 D and 1 S
where P denotes L  1, D denotes L  2 and S denotes L  0, and the superscript denotes the number of possible values for the z component of the spin,
2S  1. (States with 2S  1  1 and 2S  1  3 are called singlet and triplet
states respectively.) Because there are also 2L  1 possible values of the z
component of the orbital angular momentum, there are 3  3 3 P states,
1  5 1 D states and 1  1 1 S states, giving a total of 15 possible states with theelectron configuration (1s)2 (2s)2 (2p)2 . If the electrons in an atom really moved
in a central potential, all these 15 states would have the same energy
E  2E1s  2E2s  2E2p :But this degeneracy does not occur because not all of the effects of electron±
electron repulsion can be included in the central potential. There is a residual
electron±electron repulsion which causes the quantum states 3 P, 1 D and 1 S tohave different energy levels, as shown in Fig. 11.3. These energy differences
arise because the effect of the residual electron±electron repulsion is different inwave functions with different exchange symmetry and different orbital angularmomentum.
In addition to the residual electron±electron repulsion, a spin±orbit interaction, similar to that given by Eq. (9.35) for the hydrogen atom in Section 9.6,
also affects the energy levels of an atom. It gives rise to a fine structure in which
the energy depends, not only on the configuration and on the values L and S,
but also on the value of the total angular momentum quantum number J, which
can take on values between
J LSandJ  jL ÿ Sj:

For example, if we consider the 3 P states of the carbon atom we can couple theorbital and spin angular momenta with L  1 and S  1 to give states with a
total angular momentum with J equal to 2, 1 or 0. These states are usuallydenoted by 3 P2 , 3 P1 and 3 P0 , where the subscript denotes the value of thequantum number J, and they have slightly different energies, as illustrated on
the right-hand side of Fig.11.3. The energy levels of the 1 D and 1 S states are notsplit by the spin±orbit interaction because only one value of J is possible forthese states, J  2 and J  0 respectively.
To sum up, we have discussed how the energy levels of the carbon atom may
be described with increasing precision. To a first approximation, the energy ofthe level is determined by an electron configuration which describes independent electrons in a central potential which represents the Coulomb attraction of11.1 Atomic Quantum States
(1)

(2)

(3)

1S

1S
0

237

1.4 eV1D

1D

1.3 eV(1s)2 (2s)2 (2p)2

3P

2

3P

2

3P

1

3P

0

3.4 meV
2.0 meVFig. 11.3 Three approximations for the lowest energy levels of the carbon atom.
(1) A single level given by the energy of the configuration (1s)2 (2s)2 (2p)2 . (2) Threelevels split by the residual electron±electron repulsion, labelled by the quantum numbers
L and S. (3) Five levels depending on the spin±orbit interaction, labelled by the quantum
numbers L, S and J; the energy differences of the 3 P0 , 3 P1 and 3 P2 levels, which are of theorder of milli-electron volts, have not been drawn to scale.

the nucleus and the average effect of the Coulomb repulsion between electrons
in the atom. To a second approximation, the energy depends upon the residual
electron±electron repulsion and the level is labelled by the orbital and spinquantum numbers L and S. To a third approximation, the energy depends
upon the spin±orbit interaction and the level is labelled by L, S and by the total
angular momentum quantum number J.
Finally, we note that most of our knowledge about atomic energy levels isderived from the spectral lines that arise when radiative transitions occur
between the levels. The most probable transitions for emission or absorption
in the visible and the ultra-violet region of the spectrum are electric dipole
transitions, which were discussed in the simpler context of the hydrogen atom
in Section 9.4. For atoms more complex than the hydrogen atom, the selection
rules for electric dipole transitions are more complicated than the Dl  1 rule
given in Eq. (9.28). The change in the total angular momentum quantum
number, DJ  Jf ÿ Ji , obeys the ruleDJ  0,  1,but

Ji  0 ! Jf  0 is strictly forbidden:(11:7)

When L ÿ S coupling provides an accurate description of the quantum states,
the following rules are also obeyed

238

Atoms

Chap. 11DL  0,  1 andDS  0:(11:8)

There is an additional selection rule, as mentioned in Section 9.4: the parityof an atomic state must change in an electric dipole transition. The parity isdetermined by the behaviour of the quantum state when all the coordinates
undergo a reflection through the origin. If the state is unchanged it has evenparity and if the state is changed by a factor of ÿ1 it has odd parity. As stated at
the end of Section 9.1, the parity of a one-electron state is even when the orbital
angular momentum quantum number l is even and it is odd when l is odd. Moregenerally, if an atomic state has a single configuration (n1 l1 ) (n2 l2 ) . . . (nZ lZ ), its
parity is even if the sum of the orbital angular quantum numbers
l1  l2  . . .  lZ is even and it is odd if this sum is odd.

11.2

THE PERIODIC TABLE

Atoms with different atomic numbers Z have different chemical properties, butthese properties vary periodically with Z. For example, lithium, sodium, potassium and rubidium with Z  3, 11, 19 and 37, respectively, are called alkali
metals because they have similar metallic properties; fluorine, chlorine andbromine with Z  9, 17, and 35, respectively, are called halogens because
they have similar reactive properties; and helium, neon, argon and krypton
with Z  2, 10, 18 and 36, respectively, are called noble gases because theyshow little tendency to react. In fact, every chemical element belongs to a groupof elements with similar properties and each of these groups contains elementswith a well-defined sequence of atomic numbers.
Quantum mechanics provides a physical explanation for why the propertiesof the chemical elements vary periodically. Only two quantum mechanical ideas
are involved. The first is that each electron in an atom occupies a single-particle
state, or orbital, with definite energy. The second is that electrons occupy theseorbitals in accordance with the Pauli exclusion principle.
The orbitals occupied by atomic electrons have energy levels which depend
upon the value of the atomic number Z of the atom; with minor variations foratoms with atomic numbers greater than 20, the sequence of the energy levels is
the same for all atoms. This sequence is shown in Fig. 11.4. We note that, as in
Fig.11.2, the energy of an orbital increases when the principal quantum number nincreases, and that, for a given value of n, the energy increases when the orbital
angular momentum quantum number l increases. An important consequence of
the increase of energy with l is that the energies of orbitals with different values ofn may be close together. In particular, the 3d and 4s energies are close, the 4d and5s energies are close and the 4f, 5d and 6s energies are close.
Given the sequence of energy levels shown in Fig. 11.4 and the constraints
imposed by the Pauli exclusion principle, we can write down electron configurations for any chemical element. The lowest-energy electron configurations for11.2 The Periodic Table
5f (14)4f (14)E

4d (10)5d (10)5p (6)
5s (2)

6d (10)6p (6)
6s (2)239

7s (2)

4p (6)
3p (10)4s (2)

3p (6)
3s (2)
2s (6)

2s (2)

1s (2)

Fig. 11.4 A schematic representation of the sequence of energy levels for electron
orbitals in an atom. The number in the brackets is the degeneracy of the level whichequals the maximum number of electrons that can be assigned to the level. The order ofthe levels depends, to a limited extent, on the atomic number Z of the atom. For
example, the 4s level is lower than the 3d level for calcium with Z  20, but the 4s
level is higher than the 3d level for scandium with Z  21:

the first 30 chemical elements, together with their ionization energies, are given
in Table 11.1. This table shows how a many-electron atom, in its ground state,
has a specific set of occupied orbitals. Because an orbital with higher energy has
a greater spatial extent, the atom has several shells of negative charge. For
example, an argon atom at Z  18 has a K shell formed by occupied 1s orbitals,
an L shell formed by occupied 2s and 2p orbitals and an M shell formed byoccupied 3s and 3p orbitals.
As we move down Table 11.1, we periodically encounter atoms with common
properties for their least bound electrons.
Most importantly, we encounter, at Z  2, 10, and 18, helium, neon andargon atoms with a closed-shell structure. These are very tightly bound systems
with small sizes and high ionization energies. They are also difficult to excite;
the minimum excitation energy is about 20 eV for helium, 16 eV for neon and
10 eV for argon. These atoms are almost chemically inert.
At Z  3, 11 and 19 we encounter lithium, sodium and potassium atoms witha structure consisting of tightly bound closed shells of electrons plus one loosely

240

Atoms

Chap. 11TABLE 11.1. The lowest-energy electron configurations and ionization energies EI of
the first 30 elements
Element

Z

Electron configuration

EI (eV)

HydrogenHelium

H
He

1
2

(1s)
(1s)2

13.6
24.6

Lithium
Beryllium
Boron
CarbonNitrogen
OxygenFluorine
Neon

Li
Be
B
C
N
O
F
N

3
4
5
6
7
8
9
10

(1s)2 (2s)(1s)2 (2s)2(1s)2 (2s)2 (2p)(1s)2 (2s)2 (2p)2(1s)2 (2s)2 (2p)3(1s)2 (2s)2 (2p)4(1s)2 (2s)2 (2p)5(1s)2 (2s)2 (2p)6

5.4
9.3
8.3
11.3
14.5
13.6
17.4
21.6

SodiumMagnesium
Aluminium
SiliconPhosphorus
SulfurChlorine
Argon

Na
Mg
Al
Si
P
S
Cl
A

11
12
13
14
15
16
17
18

(1s)2 (2s)2 (2p)6 (3s)(1s)2 (2s)2 (2p)6 (3s)2(1s)2 (2s)2 (2p)6 (3s)2 (3p)(1s)2 (2s)2 (2p)6 (3s)2 (3p)2(1s)2 (2s)2 (2p)6 (3s)2 (3p)3(1s)2 (2s)2 (2p)6 (3s)2 (3p)4(1s)2 (2s)2 (2p)6 (3s)2 (3p)5(1s)2 (2s)2 (2p)6 (3s)2 (3p)6

5.1
7.6
6.0
8.1
10.5
10.4
13.0
15.8

Potassium
Calcium
Scandium
Titanium
Vanadium
Chromium
Manganese
Iron
CobaltNickelCopper
Zinc

K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn

19
20
21
22
23
24
25
26
27
28
29
30

(1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (4s)(1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (4s)2(1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)(4s)2(1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)2 (4s)2(1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)3 (4s)2(1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)5 (4s)(1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)5 (4s)2(1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)6 (4s)2(1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)7 (4s)2(1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)8 (4s)2(1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)10 (4s)(1s)2 (2s)2 (2p)6 (3s)2 (3p)6 (3d)10 (4s)2

4.3
6.1
6.5
6.8
6.7
6.8
7.4
7.9
7.9
7.6
7.7
6.7

bound electron. These atoms are called alkali metal atoms and they react
readily with other atoms, normally by transferring their loosely bound, or
valence electron.
At Z  9 and 17 we encounter fluorine and chlorine atoms with a structure
which needs one more electron to form a closed shell. These atoms have highionization energies but they are reactive because the acquisition of an extra
electron is energetically favourable; the energy released in binding an extra
electron, the electron affinity, is 3.5 eV for fluorine and 3.6 eV for chlorine.
Finally, we note that the periodicity of the periodic table becomes somewhat
erratic for atoms with higher atomic numbers. This behaviour sets in at Z  19
because the 3d and 4s energy levels are very close together and for some values

11.3

What if ?

241

of Z the 3d level is higher than the 4s level and for other values of Z it is not, as
demonstrated by the electron configurations of elements between potassium
and zinc given in Table 11.1. Similar complications arise when the 4d and 5s
orbitals are filled, when the 4f, 5d and 6s orbitals are filled, and when the 5f, 6d
and 7s orbitals are filled, but we will leave these complications to chemists tosort out.

11.3

WHAT IF?

It should be clear from the last section that the periodic variation of theproperties of atoms with atomic number is largely a consequence of the Pauli
exclusion principle, a principle that arises because electrons are indistinguishable quantum particles which conceal their identity by having antisymmetricquantum states. Because the Pauli principle is a symmetry principle which has
no analogue in classical physics, we shall end this chapter by reflecting on whatatoms would be like without it. We shall do so by using an approximate andsimple model which is based on the intuition we should have acquired byworking through 240 or so pages of quantum mechanics. We shall introduce
the model by considering two atoms whose ground states are unaffected by thePauli principle, the hydrogen and helium atoms.
The hydrogen atom consists of one electron and a nucleus of charge e. If theatom is in a quantum state with spatial extent R, the average potential energy isof the order ofV ÿe2
4pE0 Rand the average kinetic energy is at least of the order of
T

2
h
:
2me R2This expression for the kinetic energy can be thought of as the minimum kinetic
energy of an electron localized in a region of size R. It can be derived by notingthat the average momentum of the electron is comparable with the uncertainty
in its momentum, Dp, and that, in accordance with the uncertainty principle,
the minimum value of this uncertainty is of the order of
Dp 

of

h
:
R

The sum of the kinetic and potential energies gives a total energy of the order

242

Atoms

Chap. 11

E

2
h
e2
:
ÿ
2
2me R
4pE0 R(11:9)For future reference, we will rewrite this energy as
E

A1 B1
ÿ ,
R2 R

(11:10)where A1 and B1 are the constants
A1 

h2
2me

and

B1 

e2
,
4pE0

(11:11)

which determine the kinetic energy and the potential energy for a state of thehydrogen atom with spatial extent R. We note that for a state with large R, theenergy is dominated by the potential energy ÿB1 =R, that for a state with smallR, the energy is dominated by the kinetic energy of localization A1 =R2 , and thata balance between the attractive effect of the potential energy and the repulsive
effect of the kinetic energy gives rise to a minimum energy whendE
A1 B1
 ÿ2 3  2  0,RR
dR
i.e. when R  2A1 =B1 . We conclude that this model predicts a ground statewith an energy and size given byE1  ÿ

B21
4A1

and

R1 

2A1
:
B1

(11:12)

When we substitute for A1 and B1 and use the expressions for the Bohr radius
a0 and Rydberg energy ER , Eqs. (9.19) and (9.20), we find that Eq. (11.12) gives
the correct energy and radius for the ground state of the hydrogen atom,
E1  ÿER

andR1  a0 :(11:13)We shall now show that the model, with minor adjustments, can also describe
the energy and size of the ground state of the helium atom. In this atom there
are two electrons and a nucleus of charge 2e. If both electrons are in the same
single-particle state, the energy of a two-electron quantum state of size R isroughly given by
E2

h2
e2
e2

ÿ
4
,
2me R2
4pE0 R 4pE0 Ree(11:14)

11.3

What if ?

243

where the last term represents the potential energy due to Coulomb repulsion
between two electrons with separation Ree . Because the typical electron±electron separation is of the order of R, we shall approximate the potential energyof repulsion by settingRee  fR

(11:15)where f is an electron±electron avoidance parameter; we expect f to be of theorder of one, but greater than one if the electrons are good at avoiding eachother. With this approximation, the energy of the two-electron atom becomes

 2
2
h
1
e
:
E2
ÿ 4ÿ
2me R2f 4pE0 R(11:16)

In analogy with our description of the hydrogen atom, we rewrite this expression as
A2 B2
E 2ÿ ,RR

with

A2  2A1

and



1
B2  4 ÿ B1 ,f

(11:17)

where A1 and B1 are the constants given by Eq. (11.11) that determine theenergy of the hydrogen atom.
Thus, in the helium atom, a balance is struck between the attractive andrepulsive effects of the potential energy ÿB2 =R and the kinetic energy of
localization A2 =R2 to give a quantum state with minimum energy
E2  ÿB22 =4A2 and size R2  2A2 =B2 . If we substitute for A2 and B2 we obtain
E2  ÿ

4 ÿ f1
2

2
ER

and

R2  

2
4 ÿ 1f

 a0 :(11:18)This formula for E2 gives the correct binding energy of the ground state ofhelium, 5:8ER , if we take the electron±electron avoidance parameter to be 1.67,
which in turn gives a spatial extent of R2  0:6a0 , that is in agreement with themeasured radius of 0:6a0 .
Having established an acceptable model for the ground states of hydrogenand helium, we will use the model to describe the ground state of a hypotheticalatom containing Z electrons which are not constrained by the Pauli exclusion
principle. When all of the Z electrons are in the same single-particle state, thetotal energy is given by
EZ

2
h
e2
Z(Z ÿ 1) e2

ÿ Z2
:
2
2me R
4pE0 R
4pE0 Ree2(11:19)

244

Atoms

Chap. 11

If we assume that the separation of each of the Z(Z ÿ 1)=2 pairs of electrons isgiven by Ree  fR, we obtain
EZ


 2
2
h
Z(Z ÿ 1)
e
2
,
ÿ
Z
ÿ
2me R2
4pE0 R2f(11:20)which can be rewritten as
E

AZ BZ
ÿ
,
R2
R


with

AZ  ZA1

and BZ Z2 ÿ
Z(Z ÿ 1)B1 : (11:21)
2f

By finding the minimum of this energy, we obtain the following expressions forthe energy and size of the ground state of our hypothetical atom:EZ  ÿ

Z 2 ÿ Z(Zÿ1)
2f

2

Z

ER

and

Z
 a0 :
RZ  
2
Z ÿ Z(Zÿ1)
2f(11:22)

To find the ionization energy of the atom, i.e. the energy needed to remove
one electron, we need to know the energy of Z ÿ 1 electrons in a negative ion
with a nucleus of charge Ze. By working through problem 8 at the end of thischapter, you can easily show that this energy is given byEZ ÿ  ÿ

2Z(Z ÿ 1) ÿ (Zÿ1)(Zÿ2)
2f(Z ÿ 1)

ER :(11:23)The ionization energy may then be found from
EI  EZÿ ÿ EZ :(11:24)

But before we can do so, we need to specify the value of the electron±electron
avoidance parameter f. Even though we expect f to decrease with Z, we shall for
simplicity use the same value, f  1:67, which worked for the helium atom.
Straightforward algebra then gives
EI  (0:07Z 2  0:57Z  0:36)ER


and

RZ 


1
a0 :
0:7Z  0:3(11:25)

We can now explicitly illustrate the role of the Pauli exclusion principle inatoms. We shall do this by comparing the predictions of Eq. (11.25) for ahypothetical atom with atomic number Z with the measured ionization energy
and radius of the real atom with atomic number Z. The results of this compari-11.3

What if ?

245

son for the first 11 elements are shown in Fig 11.5. The solid circles giveenergies and radii for real atoms in which the Pauli principle has a governingrole, and the open circles give the energies and radii for hypothetical atoms inwhich the Pauli principle plays no role.
We see that without the Pauli principle, the ionization energies steadily
increase and the radii steadily decrease with atomic number Z. In particular,
the periodicity of chemical properties of real atoms is replaced by a chemistry inwhich atoms steadily become less reactive than helium. A world without thePauli exclusion principle would be very different. One thing is for certain: it
would be a world with no chemists.
Ionizationenergy (E R )
14
12
10
8
6
4
2
2

4

6

8

10

R adius (a 0)
3
2
1
2

4
6
8
10
Atomic number Z

Fig. 11.5 The effect of the Pauli exclusion principle on the ionization energies and radiiof the first 11 elements. The solid circles correspond to atoms in which the Pauliprinciple constrains the behaviour of the electrons and the hollow circles correspondto atoms in which the constraints of the Pauli principle are not imposed. The Pauliprinciple has no effect on the ground states of the first elements, hydrogen and helium.
The Rydberg energy and the Bohr radius have been used as units for the ionizationenergies and radii.

246

Atoms

Chap. 11

PROBLEMS 11
1. Assume that Z ÿ 1 electrons around a nucleus with charge Ze have a charge
densityr(r)  ÿ(Z ÿ 1)e eÿr=a
:
r=a
4pa3

(a) Show that the mean charge radius of this distribution of electrons isequal to 2a.
(b) Find the charge inside a radius r.
(c) Show that an electron moving in the electrostatic field of the nucleus
and of this distribution of electrons has a potential energy given by
Eq. (11.5), i.e.
V (r)  ÿz(rp )e2,
4pE0 rwherez(r)  (Z ÿ 1) eÿr=a  1:

(Hint: For part (c) use Gauss' theorem to find the magnitude of theelectric field E(r) and find the electrostatic potential f(r) usingE  ÿdf=dr.)
2. Given that the energy needed to remove the two electrons from the groundstate of the helium atom is 79 eV, show that the energy needed to remove justone electron from the ground state of the helium atom is 24.6 eV.
(In this problem and in the next problem you may use without derivation
the expression you obtained in problem 14 at the end of Chapter 9 for theenergy of an electron moving in the Coulomb potential of a point charge.)
3. Consider what chemists call the valence electron in a sodium atom. The
other 10 electrons partially screen the charge 11e on the nucleus so that, to afirst approximation, the valence electron sees an effective point charge Z*ewhich is less than 11e. The binding energy of the valence electron in sodium
is 5.12 eV for the 3s state, 2.10 eV for the 3p state and 1.52 eV for the 3dstate. What are the magnitudes of the effective point charge, Z*, of thescreened nucleus as seen by an electron in a 3s, in a 3p and in a 3d state?
4. Explain why up to 10 electrons may be assigned to 3d orbitals in an atom
and why 14 can be assigned to 4f orbitals.
5. Consider the following electron configurations of the helium atom:

Problems 11(1s)2 , (1s)(2s), (1s)(2p)

and

247

(1s) (3d):Write down the possible values for the total orbital angular momentum
quantum number L and the possible values for the total spin quantum
number S. For each value for L and S write down the possible values forthe total angular momentum quantum number J.
6. Consider the following electron configurations of the carbon atom:(1s)2 (2s)2 (2p)(3s)and(1s)2 (2s)2 (2p)(3p):

Write down the possible values for the total orbital angular momentum
quantum number L and the possible values for the total spin quantum
number S. For each value of L and S write down the possible values forthe total angular momentum quantum number J.
7. Explain why there is a 3 S state of the carbon atom with the electron
configuration (1s)2 (2s)2 (2p)(3p) but not one with the configuration(1s)2 (2s)2 (2p)2 .
8. Consider an ion with Z ÿ 1 electrons and a nucleus with charge Ze. Using
the model described in Section 11.3, write down an expression for the energyof a quantum state of the ion with spatial extent R. Show that the minimumof this energy isEZÿ  ÿ
2
Z(Z ÿ 1) ÿ (Zÿ1)(Zÿ2)
2f(Z ÿ 1)ER :9. According to the model used in Section 11.3, the energy of a two-electron
quantum state with spatial extent R of an ion with a nucleus with charge Ze
is
E2

h2
e2
e2

ÿ 2Z
,
2
2me R
4pE0 R 4pE0 Ree

with

Ree  fR:(a) Consider the two-electron ions with the experimental ionization energies, EIexpt in units of the Rydberg energy, given below. Using theelectron±electron avoidance parameter that worked for the helium
atom, f  1:67, show that the model predicts the theoretical ionizationenergies given by EItheory in the following Table.

248

Atoms

Chap. 11

Ion

Li

Be2

B3

C4

N5

O6

F7

EIexpt

5.57

11.32

19.08

28.84

40.61

54.39

70.12

EItheory

5.6

11.4

19.2

29.0

40.8

54.6

70.4

(b) A hydrogen atom is capable of binding a second electron to form aHÿ ion. Show that the model gives the experimental binding energy ofthis ion, 0.75 eV, if the electron±electron avoidance parameter f is equal
to 1.8.

Hints to selected problems
CHAPTER 1
1. Rewrite the equations for the conservation of momentum and of energy in theform
jpi ÿ pf j  jPf ÿ Pi jandEi ÿ Ef  Ef ÿ Ei ,and make use of the relations between the energy and momentum of a particle.
Finally, make use of the relation between the energy of a photon and its wavelength.
2. If the minimum energy needed to eject an electron is W, the photon must have afrequency which is at least W=h.
3. Because kT is an energy and hc is an energy-length, (kT)4 =(hc)3 is an energy per unit
volume and c(kT)4 =(hc)3 is an energy per second per unit area. Hence, theenergy radiated from unit area of a black-body with temperature T is expected tobeI  sT 4  fc(kT)4(hc)3where f is a dimensionless constant.
4. In the theory of relativity me c2 is the rest energy of an electron and in electromagnetism e2 =4pE0 R is a potential energy if R is a length. It follows thatR

e2
4pE0 me c2

250

Hints to selected problems
is a fundamental length in a relativistic theory of the electron. Show that this lengthis a2 a0 where a0 is the Bohr radius and a  e2 =4pE0 
hc. The constant a is called thefine structure constant and it is approximately equal to 1=137.

5. The force between the electron and the proton, e2 =4pE0 r2 , causes a centripetal acceleration equal to me u2 =r. The orbital angular momentum of the electron is L  mur.
6. Make use of the fact that the magnitude of the momentum of a particle is at least as
big as the uncertainty in its momentum and use the uncertainty principle Eq. (1.15).
7. Use the uncertainty principle to show that the uncertainty in the momentum of thequark, and hence the minimum value of its average momentum, is small comparedwith mc where m is the mass of the quark.
8. Evaluate the de Broglie wavelength of electrons with kinetic energy 200 eV andconsider the condition for strong diffraction by a slit.
9. Show that the de Broglie wavelength of a 54 eV electron is l  0:166 nm. The
condition for constructive interference of waves scattered by atoms on the surface
is D sin f  nl. Show that this condition is satisfied when D  0:215 nm,
f  50 degrees and n  1.
10. Show that the wave due to a conduction electron in copper is strongly diffracted by
the lattice of atoms. To do this show that the de Broglie wavelength of a 7 eV electronis 0.46 nm and that this is comparable with the distance between atoms in copper.
11. Show that the de Broglie wavelength of a neutron with thermal energy 32 kT iscomparable with the distance between atoms in a solid if T  300 K.
12. Estimate the thermal energy of an oxygen molecule at T  273 K and show that thede Broglie wavelength is much smaller than the typical distance between molecules inair.

CHAPTER 2
1. The phase and group velocities are given byuphase !
kandugroup d!
:dkVerify that ugroup  32 uphase .
2. UseZcos k0 (x ÿ ct) dk0 

sin k0 (x ÿ ct)(x ÿ ct)

Hints to selected problems

251

and

 

AB
AÿB
sin A ÿ sin B  2 cos
sin
:
2
2
3. Use relations like] cos (kx ÿ !t) ! sin (kx ÿ !t)]t
and]2 cos (kx ÿ !t) ÿk2 cos (kx ÿ !t):]x2
4. The expression
C(x, t)  2iA sin kx eÿi!tis obtained by using eiy ÿ eÿiy  2i sin y. It is a complex standing wave with wave
number k and angular frequency !.
5. Verify that the wave function is a solution of the differential equation
ÿi
h

]C
h2 ]2 C:ÿ2m ]x2]t

6. (a) Useugroup d!
:dk(b) Show that C  A eÿi(!tÿkx) is a solution if E2 ÿ p2 c2  m2 c4 .

CHAPTER 3
1. Make use of
eÿa

1 n
X
a
n0

n!

 eÿa ea  1:

2. (a) Evaluate the integral
Z 1ÿ1xr(x) dx

252

Hints to selected problemsby sketching the integrand xr(x).
(b) Show that
dr
x
ÿ 2r
dx
s
and find hx2 i by usingZ 1
ÿ1x2 r(x) dx  ÿs2

Z 1
ÿ1

x

dr
dx
dx

and the identity given in the question.
q
(c) Use s  hx2 i ÿ hxi2 .
3. (a) Evaluate the integral
Z 10p(t) dt:(b) Evaluate the integral
Z 1

tp(t) dt:0(c) The probability for living for at least time T is equal to the probability for decay
between t  T and t  1 and this probability is given by
Z 1
Tp(t) dt:

4. (a) Show that the position probability density is
jC(x)j2  N 2 eÿx =2s ,22where 2s2  a2 :Use the properties of the Gaussian distribution to evaluate the sandwich integrals for hxi and hx2 i.
(b) Use the properties of the Gaussian distribution to evaluate the sandwich integrals for hpi and hp2 i.
(c) UseDx q
hx2 i ÿ hxi2

and

Dp 

q
hp2 i ÿ hpi2 :Hints to selected problems

253

5. (a) Show that
Z a0

jC(x)j2 dx N 2 a5:30(b) Show that
a
2

and

hx2 i hpi  0 and

hp2 i 

hxi 

2a2
:
7

Show that
10
h2
:
2
a

7. (a) Show that jC(x)j2 has a maximum at x  2=a.
(b) Show that
3
a

and

hx2 i hpi  0 and

hp2 i 

hxi 

12
:
a2

(c) Show that
2 a2
h
:
4

(d) Show thatDx Dp 

r
3
h:

4

CHAPTER 4
3. (a) Assume the well width a is equal to the Bohr radius a0 and note that the
Rydberg energy is given by
ER 

2
h
 13:6 eV:
2me a20(b) Assume the well width a is equal to 1 fm  1  10ÿ15 m.
4. Rewrite using
sin kn x eikn x ÿ eÿikn x
:
2i

254

Hints to selected problems

5. Use the two-dimensional equivalents of Eqs. (4.43) and (4.44).
6. Consider the illustrative example provided by Eq. (4.57) and modify.
8. The wave function may be expressed as
C(x, 0) 1
Xn1

cn cn (x)where cn (x) is the eigenfunction of the particle with energy En in an infinite squarewell of width a. As in problem 7
cn 

2
a

Z a
0

sin kn x C(x, 0) dx:

Find jc1 j2 .
9. By showing that eÿiEn T=h  1 verify that C(t  T)  C(t).
10. The wavelength l of the emitted radiation, given byhc E2 ÿ E1 ,lis uncertain if E2 has uncertainty DE2  
h=t.

CHAPTER 5
1. (a) The eigenvalue equation is given by Eq. (5.5) when jxj < a and by Eq. (5.7) when
jxj > a.
(b) and (c) Modify the mathematics leading to Eq. (5.11) and the graphical solution
shown in Fig. 5.2.
4. (a) Substitute the expressions for c and E into the eigenvalue equationÿ2 d2 ch
 V (x)c  Ec
2m dx2and find the form of V(x) which ensures that the equation is satisfied.
(b) The eigenfunction of the first excited state has a node between x  0 andx  1.

Hints to selected problems

255

6. The wave function for x < 0 is given by Eq. (5.29), for 0 < x < a by Eq. (5.31) andfor x > a by Eq. (5.34). Impose the conditions that c and dc=dx are continuous atx  0 and x  a.
7. Use Eqs. (5.44), (5.50) and (5.51).
8. The half-life is inversely proportional to the decay rate.

CHAPTER 6
3. Show that
if the energy of a classical particle with amplitude A equals 12 h! thenp
Aa 
h=m!.
Show that the probability for finding the quantum particle in the region jxj > A is
2
p
a p

Z 1
a

2

2

eÿx =a dx:

4. The potential is identical to that of the harmonic oscillator for 0 < x < 1, but
presents an infinite barrier which prevents the particle from entering the region
ÿ1 < x < 0. The energies are 32 
h!, 72 h!, etc because the eigenfunctions c(x) havethe following properties:(i) c(x)  0 for ÿ1 < x < 0;(ii) c(x) is identical to a harmonic oscillator eigenfunction for 0 < x < 1;(iii) c(x) is continuous at x  0 but this is only satisfied if c(x) in the region
0 < x < 1 is identical to a harmonic oscillator eigenfunction withn  1, 3, 5, . . .
5. (a) According to Eq. (6.15)
1
C(x, t)  p [c0 (x) eÿiE0 t=h  c1 (x) eÿiE1 t=h ]:2(b) Follow the steps that led to Eq. (4.53).
(c) Use E1 ÿ E0  h!.
6. (a) and (b) Use Eq. (4.54).
(c) Use eiy  eÿiy  2 cos y.
7. Use Eq. (6.20) to show that k  1548 N mÿ1 .
8. (a) and (b) Modify the arguments set out in Section 6.5.
(c) Consider linear superpositions of the form c1, 0  ic0, 1 .

256Hints to selected problems

CHAPTER 7
1. Show that
2

h ]
^ 
[^
x, H]
m ]x

[^
x, ^
p]  ih,anddV^  ÿi
[^
p, H]
h
:
dx

^ and ^p.
2. (d) Show that the functions A eikx are simultaneous eigenfunctions of T
5. (a) The complex conjugate of the expectation value is
hAi* 

Z 1
ÿ1^C (AC)*
dx 

Z 1
ÿ1

^
(AC)*
C dx:

^ is Hermitian
Because A
Z 1
ÿ1

^
(AC)*
C dx 

Z 1
ÿ1^ C dx  hAi:
C* A

Hence
hAi*  hAi:(b) If we rewrite
hA2 i 

Z 1
ÿ1

^A
^ C dx,
C* A^ we obtainusing C2  AC
hA2 i Z 1
ÿ1

^ C2 dx:
C* A^ is Hermitian we have
Because A
hA2 i 

Z 1
ÿ1

^
(AC)*
C2 dx 

Z 1
ÿ1

^
^
(AC)*
(AC)
dx:

^ is Hermitian(c) Because A
Z 1
ÿ1

^B
^ C dx 
C* A^ is Hermitian
Because B

Z 1
ÿ1

^
^ C dx:(AC)*
B

Hints to selected problems
Z 1
ÿ1

^
^ C dx 
(AC)*
B

Z 1
ÿ1

^
^ AC)*(B
C dx: 

Z 1
ÿ1

257


*
^
^
C* BA C dx :h]=]x if the potential energy field
7. (a) The Hamiltonian commutes with ^px  ÿi
satisfies the condition]V 0 for all x, y and]xz:This means that the potential energy field is unchanged under the translationx ! x  a.
(b) See hint for part (a).
8. (a) Use^r  r and^p  ÿihr:(b) Using the definition of a commutator we haveZ^ ] cE d3 r 
cE* [ ^r  ^p, H

Z^ ÿ H^^r^cE* ( ^r  ^pHp ) cE d3 r:^ is Hermitian we have
Assuming that HZ^ ] cE d3 r 
cE* [ ^r  ^p, HZ^ cE d3 r ÿ
cE* ^r  ^pHZ^ E )* ^r  ^(Hc
p cE d3 r:^ with a real eigenvalue E we can use
Because cE is an eigenfunction of H
^ E  EcE
Hcand^ E )*  EcE *(Hc

to obtainZ^ ] cE d3 r  0:cE* [ ^r  ^p, H

CHAPTER 8
ÿ33
2. (a) Thepclassical angular momentum, mur  9:1  10 J s, is approximately equal
to l(l  1)
h if l  86.

3. (a) The energy splitting arises from the interaction of the electron magnetic
moment with the magnetic field. According to Eq. (8.16), the splitting is
2  mB B  5:8  10ÿ5 eV.

258

Hints to selected problems(b) Additional splitting equal to 2  2:79mN B  8:8  10ÿ8 eV arises from the interaction of the proton magnetic moment with the magnetic field.

4. (a) The classical rotational energy is L2 =2I where L is the angular momentum and I
is the moment of inertia about the centre of mass. The moment of inertia is
I  ma2 and, in quantum mechanics, the eigenvalues of L2 are l(l  1)
h2 withl  0, 1, 2, : : :(b) For every value of l, Lz can have 2l  1 values given by Lz  ml 
h withml  ÿl, . . .  l. Hence, there are 2l  1 independent eigenfunctions with rotational energy El  l(l  1)
h2 =ma2 .
(c) For a hydrogen molecule E1 ÿ E0  1:5  10ÿ2 eV. Note that the rotational
states of the hydrogen molecule are excited at room temperature becausekT  1=40 eV.
5. (a) Show that^ z Zm  ml LhZml :l(b) A point with coordinates (r, y, f) also has coordinates (r, y, f  2p).
6. Usesin 2f ei2f ÿ eÿi2f
2i

andcos f eif  eÿif,2and show thatc(r, y, f) / ei3f  eif ÿ eÿif ÿ eÿi3f :Comparison with Eq. (8.27) shows that a measurement of Lz can yield four possible
values 3
h,  h, ÿ 
h and ÿ3
h with equal probabilities of 1/4.
8. (a) The eigenfunction with energy 32 
h! has l  0 and ml  0 because it is sphericallysymmetric.
(b) Form linear combinations of eigenfunctions with energy 52 
h! which are proportional to x  iy, z and x ÿ iy.

CHAPTER 9
1. (a) A spherical symmetric wave function implies zero orbital angular momentum.
(b) Evaluate the integrals

Hints to selected problems
hV i Z 1
0N eÿar

259




ÿe2
N eÿar 4pr2 dr
4pE0 r

and
hTi 

Z 1
0

Ne

!
ÿ
h2 d2
ÿar
4pr2 dr:rN e
2me r dr2ÿar(c) Find the minimum of hEi  hTi  hV i by setting dhEi=da  0.
2. (a) The minimum of Ve (r) is found using
dVe
L2
e2
 3ÿ
 0:
dr
mr
4pE0 r2(b) The maximum and minimum distances r, which occur when pr  0, are given by
e2
L2
e2
ÿ

:
8pE0 a 2mr2 4pE0 r
3. Use the integral given in problem 1 to evaluate
N2

Z 1
0

r 2l2 eÿ2r=(l1)a0 dr:(b) Find the maximum of r2l2 eÿ2r=(l1)a0 .
(c) Use the integral given in problem 1 to evaluate
Z 10

u*0, l (r)ru0, l (r) dr

and

Z 1
0

u*0, l (r)r2 u0, l (r) dr:(e) For l >> 1, the orbital angular momentum L tends to lh and rmost probable and hri
both tend to l 2 a0 or L2 a0 =h2 .
q
4. (a) The eigenfunction is normalized if N1  1= pa30 .
(b) Show that
Z 1
0

2
c*(r)c2
1 (r)r dr  0

if l  ÿ1=2a0 .
8. Choose k to be along the z axis so that eikr  eikr cos y and write d3 r  r2 dr d( cos y) df.
Integrate from f  0 to 2p, from cos y  ÿ1 to 1 and from r  0 to 1.

260

Hints to selected problems
2~
To find the most probable momentum, locate the maximum of 4pp2 jc(p)j. Tofind the average momentum evaluate the sandwich integral

hpi 

Z 1
0

~
~
c*(p)pc(p)4pp2 dp:

9. The simplest way is to write out the integral using Cartesian coordinates.
11. See Section 9.5.
14. When an electron is in the Coulomb potential due to a point charge Ze, the size of abound state is proportional to 1=Z and the binding energy is proportional to Z2 .

CHAPTER 10
1. Exchange symmetry is a constant of motion if the Hamiltonian operator is unchanged when the particles are exchanged.
3. (a) Show that
Z 1
ÿ1

dxp

Z 1
ÿ1

dxq jC(xp , xq , t)j2  1:(b) Show that
DE (S)  DE (D)  K

DE (A)  DE (D) ÿ K,

andwhere K is given by
K

Z 1
ÿ1

dxp

Z 1
ÿ1

dxq c*(x0p )V (jxp ÿ xq j)cn (xp )cn0 (xq ):n q )cn*(xCHAPTER 11
1. (a) The mean charge radius is given by
Z 1
0

rr(r) 4pr2 dr

Z 10r(r) 4pr2 dr:(b) The charge inside radius r is given by
q(r) 

Z r
0

r(r0 ) 4pr02 dr0 :Hints to selected problems

261

3. The effective point charges seen by the valence electron of a sodium atom when it isin a 3s, 3p and 3d state are
Z*(3s)e  1:84e,

Z*(3p)e  1:18e andZ*(3d)e  1:003e:

Note the shielding by the inner electrons increases with l.
8. The energy of a quantum state with spatial extent R of an ion with Z ÿ 1 electrons
and a nucleus with charge Ze is
E  (Z ÿ 1)2he2
(Z ÿ 1)(Z ÿ 2) e2
ÿ Z(Z ÿ 1)
:
2
2me R
4pE0 R
4pE0 Ree2

Further reading
Below is a short list of books on quantum mechanics for further reading. They vary inapproach, topics covered and level of treatment, but all should be accessible to a
student of this book.

Feynman, R.P., Leighton, R.B. and Sands, M., 1965, The Feynman lectures on physics,
vol. III: Quantum mechanics, Addison-Wesley, Reading, MA.
Gasiorowicz, S., 1995, Quantum physics, 2nd edn, Wiley, New York.
Griffiths, D.J., 1995, Introduction to quantum mechanics, Prentice-Hall, Englewood
Cliffs, NJ.
Liboff, R.L., 2002, Introduction to quantum mechanics, 4th edn, Addison-Wesley, Reading, MA.
Mandl, F., 1992, Quantum mechanics (Manchester Physics Series), Wiley, Chichester,
England.
Park, D., 1992, Introduction to quantum theory, 3rd edn, McGraw-Hill, New York.
Peebles, P.J.E., 1992, Quantum mechanics, Princeton University Press, Princeton, NJ.
Rae, A.I.M., 2002, Quantum mechanics, 4th edn, Adam Hilger, Institute of Physics,
Bristol, England.
Townsend, J.S., 1992, A modern approach to quantum mechanics, McGraw-Hill, NewYork.

Index
References to sections are printed in italics
Alpha decay 108
Angular momentum 8.1
addition 157, 221, 234
commutation relations 149
operators 164, 175, 177orbital 156, 8.3
quantum numbers 156, 157, 221,234
spin 156
total 157
Angular momentum eigenfunctions
166±169, 172, 177parity of 184±185
Avoidence parameter 243
Balmer series 8, 196
Barrier penetration 5.2 see also
Tunnelling
Basis functions 71
Bohr magneton 160
Bohr radius 10, 186
Born's probability interpretation of wave
function 41
Bose±Einstein condensate 224
Boson 222
Boundary conditions 63, 67, 85, 183
Canonical commutation relations 143,
148
Carbon atom 232±233, 235±237
Central field approximation 230
corrections to 234±238
Central potential, particle in 9.1
angular momentum properties 184
energy eigenfunctions 183

boundary conditions 183
parity of 184±185energy levels 184, 232
quantum numbers 184, 231
Centrifugal potential 183, 232
Collapse of wave function 51
Commutation relations forangular momentum 149
position and momentum 143, 145±146,
148
Commutator 7.4
Compatible observables 7.3, 143, 145,
147, 184
Complete set of functions 71, 72, 137
Compton effect 2, 12, 17Compton wavelength 3
Configuration of electrons 233
Constant of motion 7.5, 260
Continuity equation 56
Coulomb barrier 100, 108
Coulomb eigenvalue problem 9.7
de Broglie wave 1.2, 27
de Broglie wavelength 4
Decaying state 53, 76
Degeneracy 123
Delta function normalization 140
Diatomic molecules 6.4 see also
Hydrogen molecule
Dirac delta function 138
Dispersion relation 23, 28
Eigenfunctions 63, 64
complete set of 72, 137
expansion in 72, 137

264

Index

Eigenfunctions (contd.)normalization of 72
orthogonality of 72, 77
orthonormality of 72
simultaneous 146, 177
Eigenvalue problem 63
Eigenvalues 63, 64
continuous 141
degenerate 123
reality of 77
Eigenvector 219
Electric dipole operator 194
Electric dipole transitions 194±195
selection rules in hydrogen 194±195
selection rules in many-electron
atoms 237±238
Electronmagnetic moment 159
spin 157
Electron affinity 240
Electron diffraction 6
Energy eigenvalue problem 64
Entangled states 9, 16, 216
Exchange symmetry 10.1, 222 see also
Identical particles
of states with spin 10.3
Exclusion principle see Pauli exclusion
principle
Expectation value 36, 37, 3.5of energy 65
Fermion 222
Fine structure 201, 236
Fine structure constant 198
Gamow energy 103
Gaussian probability distribution 52
Group velocity 25, 27
Hamiltonian operator 4.1
Harmonic oscillator in one dimension
6.2
eigenvalue problem 6.6
non-stationary states 116
quasi-classical states 117
stationary states 112±116
Harmonic oscillator in three
dimensions 6.5, 177
degeneracy of energy levels 123
Heisenberg microscope 11

Heisenberg uncertainty principle 11
relation for position andmomentum 11, 27, 144±145
relation for time and uncertainty inenergy 76
Hermite polynomial 127
Hermitian operator 136, 150
completeness of eigenfunctions 137
reality of eigenvalues 137
Hydrogen atom 9.2
boundary conditions 183, 186
eigenfunctions 188
eigenvalue problem 9.7
energy levels 7, 188
degeneracy of 188, 208
fine structure 201
principal quantum number 188
radial eigenfunctions 186, 189±191
radial quantum number 188
radiative transitions 9.4
selection rules 194±195
reduced mass effect 9.5
relativistic effects 9.6, 201
spectral lines 8, 195±196, 197
spin±orbit interaction 199
Hydrogen molecule 119
rotational energy levels 175, 227
vibrational energy levels 119
Identical particles
exchange symmetry 10.1, 222of particles with spin 10.3
symmetry of wave functions 214,
216
Indistinguishable particles 213
j±j-coupling 235
K shell 239
L shell 239
Lamb shift 202
LandeÂ g-factor 160
Lifetime 53, 76
Line width 76
Linear superposition 22, 29, 51, 71, 74,
136, 171
Lowering operator 127
L±S-coupling 235
Lyman series 8, 196

IndexM shell 239
Magnetic energies 161
Magnetic moments 8.2
Measurement 1.4, 27, 41, 42
and non-locality 16
and wave±particle duality 13
Metastable state 196
Molecules see Diatomic molecules
Momentum operator 49
eigenfunctions for 139
Muonic hydrogen atom 209, 210
Normalization ofprobability distribution 36, 37
wave function 42, 73
Nuclear magneton 160
Observables 48, 136, 7.1
compatible 7.3, 143, 145, 147
complete set of 142
non-compatible 142
Operators 48, 7.1
commuting 145
Hermitian 136, 150
linear 136
momentum 49
position 49
Orbital 232
Orthogonality 72
Ortho-hydrogen 227
Orthonormality 72
Para-hydrogen 227
Parity 104, 114, 184±185, 238
of spherical harmonics 185
Partial wave 173
Particle in a box
one-dimensional 3.4, 66
three-dimensional 69
Pauli exclusion principle 9, 223, 233,
11.3
Periodic table 11.2
Phase velocity 25
Phase shift 90, 173
Photoelectric effect 18
Photons 1.1
Planck's constant 1
Poisson distribution 52
Position operator 49
eigenfunctions for 138

265

Positronium 209
Potential barrier 5.2 see also Tunnelling
Principal quantum number 8, 188
Probability 3.1
amplitude 137
for angular momentum 172, 173
for energy 71, 74
for momentum 44, 140
for position 41, 139
current density 56
density 37, 41, 44
interpretation of wave function 40
Probability distribution forcontinuous random variable 37
discrete random variable 35
p-state 156
Quantized energy levels 66, 68
Quantum numbers 44, 68 see also
Angular momentum, Principal,
RadialQuantum particle 4, 7, 13, 38
Quantum states 3.6
Quasi-classical states 117
Radial function 183
Radial quantum number 188
Radial SchroÈdinger equation 183
Radiative transitions 120, 9.4, 237
Raising operator 126
Reduced mass 119, 196
Reflection probability 97, 107
Residual electron±electron repulsion 234,
236
Russell-Saunders coupling 235
Rydberg energy 10, 186
Scanning tunnelling microscope 100
SchroÈdinger equation 21, 28, 30
time-independent 64
Schwarz inequality 151
Selection rules for electric dipole
transitions
angular momentum 195, 237±238
parity 194, 238
Shell structure of atomic electrons 239
Single-particle states 231
Singlet spin state 236
Spectroscopic notation 156, 190, 232, 235,
236

266

Index

Spherical harmonics 169
completeness of 173
linear superposition of 171, 173
normalization 170
orthonormality 170
Spin 155
eigenvectors 219
quantum numbers 156, 221
Spin±orbit interaction 199, 234, 236
Spin-statistics theorem 223
Square-well potential
bound states 85
infinite see Particle in a box
unbound states 88
s-state 156
Standard deviation 36, 37
State
non-stationary 75, 91
of certain energy 4.3of uncertain energy 4.5, 91
stationary 75
Stern±Gerlach experiment 161
Superposition principle see Linear
superposition
Time-dependence 4.6, 147
Transmission probability 97, 107
Transverse kinetic energy 181, 183
Triplet spin state 236
Tunnelling 94

alpha particles 108
electrons 99
protons 100
through wide barriers 97
Two-slit interference 3, 6, 13, 38
Uncertainty 12, 36, 37, 49, 157
Uncertainty relations see Heisenberg
uncertainty principle
Untangled states 216
Valence 240
Variance 36
Virial theorem 153
Wave function 29
boundary conditions for 63, 67, 85, 183
complex 22, 29
continuity conditions for 85
normalization 42, 73
probability interpretation 40
Wave packet 23, 27, 29, 91±93
Wave±particle duality 13
Waves 2.1
dispersive and non-dispersive 23
sinusoidal 21
superposition of 22, 29
Zeeman effect 161
Zero-point energy 68, 116

PHYSICAL CONSTANTS AND CONVERSION FACTORS
SymbolDescription

Numerical Value

c
m0
E0

velocity of light in vacuum
permeability of vacuump
permittivity of vacuum where c  1= E0 m0299 792 458 m sÿ1 , exactly
4p  10ÿ7 N Aÿ2 , exactly
ÿ2
8:854  10ÿ12 C2 Nÿ1 mhhPlanck constanth=2p

6:626  10ÿ34 J s
1:055  10ÿ34 J s

Ggravitational constant

6:674  10ÿ11 m3 kgÿ1 sÿ2eeVaelementary charge
electronvolt
hcfine structure constant, e2 =4pE0 

1:602  10ÿ19 C
1:602  10ÿ19 J
1=137:0

me
me c2
mB

electron masselectron rest-mass energy
Bohr magneton, eh=2me

9:109  10ÿ31 kg
0.511 MeV
9:274  10ÿ24 J Tÿ1

ER
a0
Ê
A

Rydberg energy a2 me c2 =2
Bohr radius (1=a) (h=me c)angstrom

13.61 eV
0:5292  10ÿ10 m
10ÿ10 mmp
mp c 2mn c 2mN
fmb

proton mass
proton rest-mass energy
neutron rest-mass energy
nuclear magneton, e
h=2mp
femtometre or fermi
barn

1:673  10ÿ27 kg
938.272 MeV
939.566 MeV
5:051  10ÿ27 J Tÿ1
10ÿ15 m10ÿ28 m2uNA1
atomic mass unit, 12m(12 C atom)
Avogadro constant, atoms in gram mol

1:661  10ÿ27 kg
6:022  1023 molÿ1

Tt
k
R
s

triple-point temperature
Boltzmann constantmolar gas constant, NA k
Stefan±Boltzmann constant, (p2 =60) (k4 =
h3 c2 )

273.16 K
1:381  10ÿ23 J Kÿ1
8:315 J molÿ1 Kÿ1
5:670  10ÿ8 W mÿ2 Kÿ4