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PH2002 Quantum Mechanics

Or Wave-particle duality

by Dennis Dunn

Version date: Tuesday, 16 October, 2007 at 11:45

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Copyright c©2006 Dennis Dunn.

Contents

Contents 3

1 Module Plan 6

1.1 Module Description: PH2002 Quantum Physics 6

1.2 References 8

2 Physics: Experiment and Theory 9

3 Classical Physics: Successes and Failures 11

3.1 The Successes 11

3.2 The Failures 11

4 The beginnings of Quantum Theory 16

4.1 Bohr Model 16

5 Wave-Packets 19

6 Schrodinger’s Equations 22

6.1 Electromagnetic Properties 23

6.2 Energy states 24

6.3 An Example: A Particle in a 1D Box 25

7 Hilbert Space and Eigenvalue Equations in General 29

7.1 Hilbert Space 29

7.2 Operators 30

7.3 Hermitian Operators 31

7.4 Linear Hermitian Operators 31

8 Probability Theory 33

8.1 Tests and Frequencies 34

8.2 A simple example - a die 34

8.3 Continuous outcomes and probability densities 35

9 Quantum Theory 37

4

10 Examples 41

10.1 Experiments measuring particle position at time t=0 41

10.2 Experiments measuring particle position at time t > 0 43

10.3 Measurement of Energy 44

10.4 Expectation Values 45

11 Motion of a Wavepacket 48

11.1 Fourier Analysis 48

11.2 A Gaussian Wavepacket 50

11.3 Two Standard Integrals 50

11.4 Normalization and Uncertainties in Initial Wavefunction 51

11.5 Time Dependence 52

12 Finite Square Well 54

13 Harmonic Oscillator 58

13.1 Energy Eigenstates 58

13.2 Oscillating Wavepacket 61

14 Quantum Dynamics 65

15 3D, Central Forces and Angular Momentum 68

15.1 3D Wavepacket 68

15.2 Central Forces 69

15.3 Angular Momentum in Quantum Theory 70

15.4 Eigenvalues of J2 and Jz 72

15.5 Spherical Polar Co-ordinates 75

16 Hydrogen Atom 78

16.1 Spherically symmetric potential 78

16.2 Hydrogen Atom 78

17 General Uncertainty Relations 82

17.1 Examples 84

18 Many-particle systems 87

19 A Summary of Essential Features 89

19.1 Electron Diffraction 89

19.2 Wave Equation 89

5

19.3 Classical-like solutions 89

19.4 Very non-classical solutions 89

19.5 Meaning of the electron field 90

19.6 General Measurements 90

Index 91

Chapter 1

Module Plan

The module will be organized as follows:

• Physics – experiments & theories• Classical Physics – successes and failures

– Particles & Fields– Planck and Black-body radiation – Planck’s constant– Electron diffraction – de Broglie waves– Photo-electric effect– Stability of atoms– Atomic spectra

• The beginnings of Quantum Theory – de Broglie & Bohr model• Wave packets – brief introduction• Schrodinger’s Equation• Energy eigenstates and the solution to the ’stability of atoms’• Hilbert Space and Eigenvalue Equations• Meaning of the wavefunction and discussion on Probabilities• Proper formulation of Quantum Theory• 1D Examples• Wave-packets again• Uncertainty Relations• The Classical Limit• 3D problems• Angular momentum• Hydrogen Atom• Many-particle wave functions – fermions & bosons

1.1 Module Description: PH2002 Quantum Physics

Module code PH2002Credits 20Terms taught Autumn & SpringModule Convenor Dr Dennis DunnSchool Mathematics, Meteorology and Physics

6

1.1 Module Description: PH2002 Quantum Physics 7

Module prerequisites PH1001 PH1002 and MA111 (or equivalents)Non-modular prerequisitesModule corequisitesExcluded modules

Summary module description An introduction to quantum physics and the solutions toSchrodinger’s wave equations.

Aims

• To provide an introduction to quantum physics and the solutions to Schrodinger’s waveequation

• To provide students with an understanding of the structure of quantum theory and todevelop the mathematical skills required for its implementation

• To provide students with an appreciation of the differences between and similarities ofquantum and classical mechanics.

Assessable Learning Outcomes After the module each student should be able to:

• Discuss both the successes and failures of Classical Physics. In particular, the successes:– Newton’s particle physics;– Maxwell’s electromagnetism;And the failures:– Blackbody radiation;– Photoelectric effect;– Compton effect;– Electron diffraction;– Stability of atoms;– Atomic spectra

• Discuss the Bohr model of the atom;• Discuss the properties of wave packets and explain how a wave can be localized;• Recall Schrodinger’s wave equation;• Recall the expression for the electron charge density of the electron wave;• Explain the important qualitative differences that occur when this charge density is a)

time-dependent and b) time-independent;• Specify the energy states ;• Show that the electron density for the energy states is time-dependent and explain the

importance of this property;• Determine the properties of a 1D particle confined in a range 0 ≤ x ≤ L;• Recall the defining properties of a Hilbert Space;• Define an Hermitian operator;• Specify the properties of the eigenvalues and eigenfields of an Hermitian operator;• Specify the properties of probablities and of probability densities;• Define, and be able to calculate, expectation and uncertainty values;• State the formal rules of quantum theory (at least for a one-electron system);• Solve the energy eigenvalue equations for simple 1D systems.• Determine expectation values and uncertainties.• Apply the methods of Fourier Analysis to the 1D wavepacket.

1.2 References 8

• Determine the quantum dynamical equation and to compare this with Newton’s equation.• Determine the eigenvalues of L2 and Lz.• Explain why angular momentum is important in the solution of the energy eigenvalue

equation for a spherically symmetric potential;• Solve the radial energy eigenvalue equation for hydrogen atom.• Determine the degeneracies of the energy eigenstates of the hydrogen atom.• Explain and apply the general uncertainty relations.• Explain the symmetry properties of many-particle systems.

Outline Content

Quantum Physics theory, applications and interpretation. The mathematical skills requiredwill be introduced as they are required.

Brief description of teaching and learning methods

The core of this course is provided through lectures and detailed printed notes. The lecturesare used to introduce new concepts and to lead students through mathematical derivations andworked examples.

1.2 References

• Quantum Physics, Stephen Gasiorowicz (Wiley 2003, ISBN 0471429457) Tesco £34.15• Introduction to Quantum Mechanics, David J Griffiths (Prentice Hall 2005, ISBN 0-13-

191179-9) Amazon £45.59• Quantum Physics of Atoms, Molecules, Solids, Nuclei and Particle, Robert Eisberg and

Robert Resnick (Wiley 1985) Amazon £73.91 (£45.00 used)• http://www.oberlin.edu/physics/dstyer/StrangeQM/history.html• http://www-history.mcs.st-andrews.ac.uk/HistTopics/

The_Quantum_age_begins.html

Chapter 2

Physics: Experiment and Theory

PHYSICSPHYSICSEXPERIMENT

OBSERVATION

THEORYMODELS PREDICTIONS

Physics begins with observations and experiments – experiments are just more precise obser-vations or, alternatively, observations are qualitative experiments.

However physics is definitely not simply a list or record of the results of experiments.

The main aim of physics is the construction of theories or models.

A theory proposes entities which are the ingredients of the system being modelled and someset of rules which specify the properties and behaviour of these entities. The rules usuallyinvolve mathematics but this is is mathematics with a definite purpose. (Mathematicians oftendon’t appreciate this!) There are no rules about the rules except that they must be consistent.That is one rule should not contradict (even indirectly) another.

The construction of a theory often involves the creation of new mathematics.

9

10

The aim of a theory is to make predictions. A ”theory” which cannot be used to make predic-tions is NOT a theory.

The predictions need to be compared to experimental results. If there are no existing exper-imental results then we need to persuade a friendly experimenter to design a suitable experi-ment and make some measurements!

If the predictions do not agree with the experimental result then the theory fails. (Note thatthis concept of a theory failing does not exist in mathematics!)

When a theory fails we need to invent a new one. Unfortunately there is no prescription fordoing this.

This is a slightly naive view of physics: Problems arise because of the uncertainties involvedin experimental results and, sometimes, in the approximate nature of the quantitative predic-tions. This means that it is not always immediately obvious whether the theoretical predictionand the experimental result agree or disagree. (This explains why there is such an emphasisin physics laboratories on the determination of the uncertainties in the experimental results.)

Uncertainties in the predictions can arise if the equations which result from the theory are toocomplicated to solve exactly and approximations have to be employed – this is usually thecase!

Chapter 3

Classical Physics: Successes and Failures

3.1 The Successes

The main ingredients of classical physics are particles and fields or waves.

Particles are localized and each particle is specified by a position vector ~x (t) and by a velocityvector ~v (t). These quantities change in time according to Newton’s equations.

A field, such as the electromagnetic field or the gravitational field, extends throughout a regionof space (and time): It is not localized.

An important property of fields (or waves) is that they add and subtract. That is we can takea wave f (~x, t) and add it to or subtract it from a wave g (~x, t) to form new waves

f (~x, t) + g (~x, t) f (~x, t)− g (~x, t)

This property is the whole basis of interference and diffraction.

Newton’s theory of particle motion and Maxwell’s theory of electromagnetism were the greatachievements of classical physics.

3.2 The Failures

In spite of the enormous successes of classical physics there were some niggling failures:

3.2.1 Blackbody Radiation

When the Maxwell’s theory of electromagnetism was combined with statistical physics thethermodynamic properties of the electromagnetic waves could be calculated. In particular theenergy per unit volume in a particular frequency range could be calculated. The well-knownRayleigh-Jeans formula for the energy per unit volume for electromagnetic radiation in theangular frequency range ω − (ω + δω) is

ω2

π2 c3kBT (3.1)

However this formula (which is a prediction of classical physics) disagreed, and by a massiveamount, with experiment. (First done – I think – by Lummer and Pringsheim in 1899).

11

3.2 The Failures 12

Max Planck essentially guessed a formula which fitted the experimental results for this energyper unit volume

ω2

π2 c3~ω

exp

(~ωkBT

)− 1

(3.2)

He found that this fitted the whole series of experimental results, for different temperatures,precisely if the constant ~ was chosen to be 1.05× 10−34 kg m2s−1.

Note that these two formulas become identical in the limit of low frequency: ~ω kBT .

He then found that he could justify this formula if a wave of angular frequency ω could beconsidered as composed of many particles each with energy ~ω.

However, at the time, he was not convinced that this had any real significance.

You may derive Planck’s result when you do Statistical Physics in Part 3.

3.2.2 Photoelectric Effect

The photoelectric effect was discovered by Hertz in 1887.

In this experiment a beam of light is directed at a metal surface and electrons are emitted.

In classical terms this could be understood in terms of the energy of the electromagnetic fieldbeing given to electrons. However there was a massive flaw. In classical terms the energyin the electromagnetic wave is proportional to the square of its amplitude. If this amplitudeis made progressively smaller then it should take a progressively longer time for there to beenough energy to emit an electron. No such time delay was (or has been) observed.

It was also found that there was a linear relationship between the energy of the emitted elec-trons and the frequency of the incident electromagnetic wave. There is no classical explanationof this.

Einstein provided a theory of the photoelectric effect by using Planck’s idea of associatingwith the electromagnetic wave (with angular frequency ω) particles with energy ~ω.

3.2 The Failures 13

3.2.3 Compton Effect

Compton discovered that radiation of a given frequency (in the X-ray region) when directedon to a thin metal foil scattered in a way not predicted by classical theory.

According to classical theory the electromagnetic radiation causes the charged particles in thefoil to oscillate (at the same frequency as the wave and in the direction of the electric field)and these oscillating charged particles should produce out-going electromagnetic waves againwith the same frequency and with an intensity which is proportional to 1 + cos2 θ where θ isthe direction relative to the direction of the electric field.

This did not happen in the experiment.

Compton actually found that the frequency of the scattered radiation was different from theincident radiation and that this change in the frequency depended on the angle of emission.

The experimental results could be explained by assuming the radiation (of angular frequencyω) behaved just like a stream of particles – photons – each with momentum

~p = ~~k (3.3)

and energy

E = ~ω = ~ c |~k| (3.4)

where ~k is the wave-vector of the electromagnetic wave and ω is its angular frequency.

That is we are assuming a propagating wave of the form

exp[i(~k • ~x− ω t

)](3.5)

Notice that, just as ~k is generated by operating with −i ~∇, ω is is generated by operating with

id

dt.

Hence there is some relationship between energy E and applying the operator

i ~d

dt

and similarly a relationship between momentum ~P and applying the operator

i ~ ~∇

The properties of the emitted radiation are derived in this approach by considering the colli-sion of two particles - the electron and the photon – and by using the conservation of energyand the conservation of momentum to determine the possible out-going energies and momenta.

That is, the wave is considered as a stream of particles.

3.2 The Failures 14

3.2.4 Electron Diffraction

It was found experimentally that a stream of electrons, each with a particular kinetic energyE, when directed at the surface of crystal were diffracted in exactly the same way as an elec-tromagnetic wave (X-rays) of a particular frequency.

George Paget Thomson was awarded the Nobel prize for his experiments – carried out at theUniversity of Aberdeen – showing this wave property of the electron. His father Joseph JohnThomson had previously been awarded the Nobel prize for discovering – in the CavendishLaboratory, Cambridge – the electron and for determining its particle properties.

de Broglie had previously made the outrageous suggestion that an electron with momentum~p (remember ~p = m~v) should behave like a wave

exp(i~k • ~x

)(3.6)

where the wave-vector ~k is given by

~p = ~~k (3.7)

The wavelength of such a wave is

λ =2π

|~k|= ~

2π

|~p|(3.8)

Note that ~k is the result of applying the differential operator −i ~∇ on the wave. So, in somesense, there is a relationship between

~p and − i ~ ~∇

Remember that

~∇ = ~exd

dx+ ~ey

d

dy+ ~ez

d

dz(3.9)

G P Thomson’s experiments confirmed these predicted wave properties.

3.2.5 Stability of Atoms

The model of the atom, which largely emerged from experiments of Rutherford, is that ofnegatively charged electrons orbiting a positively charged nucleus. The mathematics of thisis exactly the same as that of planets orbiting the Sun.

However electrons are charged and a rapidly oscillating charge emits electromagnetic radia-tion. This is the basis of all radio and television broadcasting. If an atom is emitting radiationthen it must be losing energy. If it is losing energy then its orbital radius will continually

3.2 The Failures 15

be getting smaller. Since objects are made of atoms, the objects will continually be gettingsmaller.

It is not, at first sight, obvious that this is necessarily a problem since if everything is gettingsmaller then so are all the rulers which we use to determine the size of objects.

I will let you think about this!

However it does seem that atoms are stable and there is no doubt that atoms are not observedto be continually emitting radiation. (I am not talking about black-body radiation here whichis a thermal effect. The predicted radiation from atoms would be independent of temperature).

3.2.6 Atomic Spectra

Experimentally it is found that atoms absorb electromagnetic radiation at only selected fre-quencies and that atoms, which have been excited in some way, emit radiation at only selectedfrequencies.

It is as if the electrons in the atoms can only absorb energy in quantized amounts.

There is no explanation for this in classical physics.

3.2.7 Project

I want you to explore these topics in more detail.

There will be an ’essay’-type question on the final examination which will require you toexpand on one, or more, of these topics. You will gain marks from a quantitative discussion.

Chapter 4

The beginnings of Quantum Theory

Suppose that, for whatever reason, with a particle of momentum p and energy E we associatea wave

exp [i (k x− ω t)] (4.1)

where the wave-properties and particle properties are related by

p = ~ k E = ~ω (4.2)

I am, for the moment, ignoring the vector properties.

The wavelength is

λ =2π

|k|=

2π ~|p|

(4.3)

4.1 Bohr Model

Consider a simple atomic model in which one electron is orbiting a proton-nucleus in a circu-lar orbit.

If I equate the centrifugal force to the Coulomb force, I get the relation

mv2

r=

e2

4πε0r2(4.4)

or, re-writing in terms of the momentum,

p2 =me2

4πε0r(4.5)

Now I suppose that the only stable orbits are those for which a wave fits exactly round theorbit. That is, the circumference of the orbit must be a whole number of wavelengths:

2π r = nλ =2π n ~p

(4.6)

16

4.1 Bohr Model 17

where n is a positive integer.

If I use this to substitute for r in equation (4.5) I obtain the following allowed values for r andp

pn =me2

4πε0 ~1

n

rn =4π ε0~2

me2n2

(4.7)

The energy of the orbiting electron is

E =p2

2m− e2

4π ε0 r(4.8)

If I insert the allowed values rn and pn for r and p I get

En = −m2

(e2

4π ε0~

)21

n2(4.9)

Hence, in this simple model, the energy, momentum and radius are all quantized.

The angular momentum is also quantized and in a particularly simple way. For a circular orbitthe angular momentum is L = r p and, using (4.7), the allowed values are

Ln = n ~ (4.10)

The lowest energy is given by n = 1 and in this case the parameters are:

r1 =4π ε0~2

me2= 5.29167× 10−11m

p1 =me2

4πε0 ~= 1.9901× 1024 kg ms−1

E1 = −m2

(e2

4π ε0~

)2

= −2.17969× 10−18 J = −13.605 eV

L1 = ~ = 1.0545× 10−34 Js

(4.11)

Atomic spectra were associated with transitions between two orbits with different allowedenergies and the frequency of the emitted radiation was taken to be

~ωn1n2 =m

2

(e2

4π ε0~

)2(1

n22

− 1

n21

)(4.12)

This formula fitted observed frequencies in the hydrogen spectrum.

4.1 Bohr Model 18

The method I have used was not that originally used by Bohr.

This theory was extended (by Wilson and Sommerfeld - I think) to include elliptical orbits.

However you should realise that this is not a proper theory: It has severe limitations. Althoughwave properties are associated with the electron the wave itself does not actually make anappearance. It is basically an attempt to fix something – classical mechanics – which has toomany flaws to be fixed.

Chapter 5

Wave-Packets

I want to convince you that trying describe a particle by a wave is not as silly as it, at first,seems!

Consider the figure Extended Plane Wave:

−20 −10 0 10 20

−1

−0.5

0

0.5

1

Figure 5.1. Extended Plane Wave

This is the classic wave. It is completely de-localised: It is equally everywhere. It also has awell-defined wavelength. There is no uncertainty in what the wavelength is.

Mathematically such a wave –I am not considering any time variation at present and I haveassumed just one dimension – is simply

A exp (i k0 x) (5.1)

The wavelength is

λ =2π

|k0|(5.2)

I have drawn this with k0 = 6 (in some units) and so λ =2π

6∼ 1 (in those units).

Now consider the wave in the figure Slightly Localized Wave:

This wave is clearly not equally everywhere. There is more of it near x = 0. We could, in fact,identify x = 0 as its average position but clearly there is a large uncertainty in this position.

19

20

−20 −10 0 10 20

−1

−0.5

0

0.5

1

Figure 5.2. Slightly Localized Wave

The wavevector is now no longer exactly defined. We can identify k0 = 6 as the averagewavevector but this will have a small uncertainty associated with it.

Mathematically this wave is

A exp (i k0 x) exp

(−1

2

(xa

)2)

(5.3)

I have drawn this with k0 = 6, as before, and the constant a = 20. a determines the width ofthis wave.

The wave in the final figure (Very Localized Wave) is similar, except that a = 1. Here we seethat the wave has become very localized.

−20 −10 0 10 20−1

−0.5

0

0.5

1

Figure 5.3. Very Localized Wave

The uncertainty in its average position is much less but the wavevector has correspondinglybecome much more uncertain.

If I chose progressively smaller values for a then the uncertainty in the position becomes lessand less. That is, this starts to look more like a particle. However the uncertainty in thewavevector becomes progressively larger.

21

If I denote the uncertainty in the wavevector by ∆k and the uncertainty in the position by ∆xthen using Fourier Analysis I can show that for any wave

∆k ∆x ≥ 1

2(5.4)

This inequality is known as the wavevector-position uncertainty relation. This is somewhatvacuous at the moment because I have not properly defined the two uncertainties: This willbe remedied later.

This uncertainty relation has implications for what we do later in quantum theory but it is hasother applications as well – in electronics for example.

For the waves I have just shown the minimum value of1

2is attained. That is ∆k ∆x =

1

2

Chapter 6

Schrodinger’s Equations

The big flaw with the Bohr attempt to patch up classical mechanics was that although therewas some attempt to introduce wave properties, there was no attempt to specify the wave andno attempt to define a wave equation.

This was remedied by Schrodinger who did specify a wave equation and made, at least, a firstattempt at putting forward a cohesive theory.

Suppose, in classical mechanics, that a single particle is moving under the action of a force~F (~x) and this force is given in terms of a potential

~F (~x) = − ~∇V (~x) (6.1)

The energy of this particle is then

E =~p • ~p2m

+ V (~x) (6.2)

~p is the momentum of the particle and is equal to m~v.

The deliberations of de Broglie and others suggested that there was a wave associated with theparticle and that in some way energy of the particle was associated with the operating on the

wave with the operator i~d

dtand that the particle momentum was associated with the operator

−i ~ ~∇.

If I make these substitutions in equation (6.2) and apply the resulting operator to a waveψ (~x, t), I get the following equation:

i~dψ (~x, t)

dt= − ~2

2m

(~∇ • ~∇

)ψ (~x, t) + V (~x)ψ (~x, t) (6.3)

This is Schrodinger’s Equation for the wave ψ (~x, t).

Notice that the presence of the complex i in this equation means that the wave must be com-plex. In much of what you have done before in your physics course you will have usedcomplex representations of waves as a mathematical convenience but the waves themselveswere considered to be real. Here the wave is inherently complex.

The operator ~∇ • ~∇ in the equation is, in detail,

22

6.1 Electromagnetic Properties 23

~∇ • ~∇ =d2

dx2+

d2

dy2+

d2

dz2(6.4)

You should note that we have not proved Schrodinger’s Equation: This equation has to betaken as one of the assumptions of a new theory.

6.1 Electromagnetic Properties

The electron is charged and so interacts with the electromagnetic field. Hence if the electronfield ψ (~x, t) is supposed to represent the electron it also must interact with the electromagneticfield. This is a two-way process: The electromagnetic field affects the electron field and theelectron field affects the electromagnetic field. The first of these processes I will leave untillater.

One of Maxwell’s equations, for the electric field, is

~∇ • ~E (~x, t) =ρ (~x, t)

ε0(6.5)

ρ (~x, t) is the charge density. This charge density is a source of the electromagnetic field.However this can occur in two quite distinct ways.

6.1.1 Static charge density

If ρ (~x, t), in fact, does not vary with time we say it is static. This produces, correspondingly,a static (time-independent) electric field. There is no flow of electromagnetic energy.

6.1.2 Time-varying charge density

If ρ (~x, t) does indeed vary with time it produces a time-varying electric field (and also atime-varying magnetic field). This then gives rise to radiation which means that there is aflow of electromagnetic energy.

6.1.3 Electron charge density

The charge density associated with the electron field ψ (~x, t) is

ρ (~x, t) = −e ψ∗ (~x, t)ψ (~x, t) (6.6)

where ψ∗ (~x, t) is the complex conjugate of the electron field.

The charge density is a real and negative quantity.

This ρ (~x, t) does, in general, depend on time and so does produce a flow of electromagneticenergy.

6.2 Energy states 24

Only in the exceptional case in which

ψ∗ (~x, t)ψ (~x, t)

is independent of time will there be no electromagnetic energy flow.

6.2 Energy states

I put forward Schrodinger’s Equation on the basis that I replaced energy E by the operator

i~d

dt. Let’s try to put the energy back.

I want to see if I can make a solution of Schrodinger’s Equation also satisfy the equation

i~dψ (~x, t)

dt= Eψ (~x, t) (6.7)

where E is some constant (energy).

The time-dependence of this equation is easily determined:

ψ (~x, t) = ψ (~x, 0) exp

(−iE t

~

)(6.8)

This is a product of a two terms: One which only depends on ~x and one which represents asimple oscillation in time.

If I write the spatially-dependent part as

φ (~x) = ψ (~x, 0) (6.9)

Then the equation for φ can be obtained by inserting (6.8) back into (6.3). The result is

− ~2

2m

(~∇ • ~∇

)φ (~x) + V (~x)φ (~x) = Eφ (~x) (6.10)

This is an eigenvalue equation for the energy and I shall call this Schrodinger’s energy eigen-value equation. You will also find it called as Schrodinger’s Time-Independent Equation.

The main importance of the energy states (or stationary states) is that the electron field is ofthe form

ψ (~x, t) = φ (~x) exp

(−iE t

~

)(6.11)

and the corresponding charge density is

ρ (~x) = −e φ∗ (~x)φ (~x) (6.12)


This is independent of time.

Hence electron fields which have the form of equation (6.11) produce no electromagneticradiation.

We shall see that this is very important when we apply Schrodinger’s Equations to atoms andwill completely solve the problem of the instability of the classical model of atoms.

6.3 An Example: A Particle in a 1D Box

I am going to consider a simple model which does not have much to do with physics but willget you used to the mathematics involved in Quantum Theory.

Consider a model in which there is only one spatial dimension denoted by x and in which theelectron field is confined to the region

0 ≤ x ≤ L (6.13)

The wavefunction (or electron field) ψ (x, t) is chosen to be zero at the boundaries of theregion.

ψ (0, t) = 0 ψ (L, t) = 0 (6.14)

.

In this simple space, in which there is no potential, Schrodinger’s equation becomes

i~dψ (x, t)

dt= − ~2

2m

d2

dx2ψ (x, t) (6.15)

and Schrodinger’s energy eigenvalue equation , which describes the energy states, becomes

− ~2

2m

d2

dx2φ (x) = Eφ (x) (6.16)

where E is some constant energy.

If I denote the energy operator by

E = − ~2

2m

d2

dx2(6.17)

Then the Schrodinger energy eigenvalue equation can be written as

Eφ = Eφ (6.18)

This is the classic form of an eigenvalue equation.

I can write (6.16) if the form


d2

dx2φ (x) = −k2φ (x) (6.19)

where k2 =2mE

~2. I have implicitly assumed that E is positive. (Note: The classical energy

would certainly be positive. You should be able to explain why.)

The general solutions of the equation are

φ (x) = A sin (kx) +B cos (kx) (6.20)

You can check this simply by differentiating twice.

The requirement that φ (0) = 0 forces the coefficient B to be zero and so

φ (x) = A sin (kx) (6.21)

However I also require φ (L) = 0 and so

A sin (k L) = 0 (6.22)

One solution is, of course, A = 0 but in this case we have no wave so I will ignore thispossibility.

The other possibility is

sin (k L) = 0

This gives certain allowed values for k

kn =nπ

Ln = 1, 2, 3, . . . (6.23)

The corresponding energies and eigenfields are:

φn (x) = A sin(nπ x

L

)En =

~2 π2 n2

2mL2

(6.24)

There are two points to note here:

• The energy is quantized. That is, it has discrete values rather a continuum (which it wouldhave in Classical Mechanics).

• The minimum value is not zero (which it would be in Classical Mechanics).


6.3.1 Orthogonality

The eigenfields have the following interesting property:

L∫0

dx φ∗n1(x)φn2 (x) = 0 n1 6= n2

=A2 L

2n1 = n2

(6.25)

I will leave this as an exercise for you to prove.

The first of these equations is described by saying: φn1 and φn2 are orthogonal.

Notice that in the equations the ∗ is redundant because the functions are real but I will keepit there because we will use similar equations later where the functions are complex and thenthe ∗ will be essential.

If I choose the amplitude A to be

√2

Lthe eigenfield is said to be normalized which means

L∫0

dx φ∗n (x)φn (x) = 1 (6.26)

The equation (6.25) can then be expressed as

L∫0

dx φ∗n1(x)φn2 (x) = δn1,n2 (6.27)

δn1,n2 is the Kronecker Delta Function and is 1 if the two indices are equal and is 0 if the twoindices are different.

6.3.2 φn (x) as Building Blocks

The eigenfields can be used as building blocks. By this I mean that (almost) any functionwhich is defined on the region 0 ≤ x ≤ L can be constructed from linear combinations of theeigenfields. In detail this takes the form:

φ (x) =∞∑n=0

cn φn (x) (6.28)

cn are some set of complex numbers.

The cn can, in fact, be determined explicitly:


cn =

L∫0

dx φ∗n (x)φ (x) (6.29)

This can be proved by using equation (6.27).

Why would anyone want to do this?

Suppose I actually want to solve the Schrodinger Equation for ψ (x, t) and that the initial fieldis

ψ (x, 0) = φ (x)

Then if I have obtained the eigenfield expansion (6.28) for φ (x) I can immediately write downthe solution for ψ (x, t)

ψ (x, t) =∞∑n=0

cn φn (x) exp

(−i Ent

~

)(6.30)

This arises because each solution φn (x) of the time-independent Schrodinger equation gives

rise to a solution φn (x) exp

(−i Ent

~

)of the Schrodinger equation.

Chapter 7

Hilbert Space and Eigenvalue Equations inGeneral

I now want to explore eigenvalue equations in more detail and in more generality.

If I define the energy operator by

E = − ~2

2m

(~∇ • ~∇

)+ V (~x) (7.1)

then time-independent Schrodinger equation can then be expressed in the form

Eφn = En φn (7.2)

This is the classic form of an eigenvalue equation. The solutions for En are known as eigen-values and the corresponding solutions for φn are known as eigenfields.

7.1 Hilbert Space

I will first specify the properties of the fields more precisely. The fields will be elements of aHilbert space H.

A Hilbert space has the following properties:

If f and g are elements of H (that is they are some of our fields) then so are

• f + g• f − g• cf (where c is an arbitrary complex numbex).

These properties are essential for any quantities which exhibit interference and diffraction.

For every pair of elements of H there is a complex number denoted by (f, g) and called theinner product of f and g. In the context of what we have done above this inner product couldbe

(f, g) =

∫dV f ∗ (~x) g (~x) (7.3)

29

7.2 Operators 30

This is the inner product I shall mostly use but there will also be others.

In general, the inner product must satisfy:

(f, g)∗ = (g, f)

(f1 + f2, g) = (f1, g) + (f2, g)

(c1 f, c2 g) = c∗1 c2 (f, g)

(f, f) ≥ 0

(f, f) = 0 if and only if f = 0

(7.4)

In this equation c1 and c2 are arbitrary complex constants and c∗1 is the complex conjugate ofc1.

This inner product can be used to define the size or norm of an element f :

‖ f ‖=√

(f, f) (7.5)

It is implicit in this relation that this size or norm cannot be infinite.

There are some properties which follow from these rules:

| (f, g) | ≤ ‖ f ‖ ‖ g ‖ (7.6)

‖ f + g ‖ ≤ ‖ f ‖ + ‖ g ‖ (7.7)

I will get you to prove these.

7.2 Operators

First let’s look at the concept of an operator. This is very general:

Of = g (7.8)

An operator is simply something which operates on one field, to give another field.

Here are some simple examples of operators

7.3 Hermitian Operators 31

O1f = c1 f

O2f = f 2

O3f =1

f

(7.9)

Here c1 is some arbitrary complex number and f is some arbitrary field.

I shall only consider operators which operate on field in a particular space and produce fieldsin the same space.

7.2.1 Linear Operators

A linear operator satisfies the following rules:

O (f + g) = Of + Og

O (c f) = c O (f)

(7.10)

where f and g are fields and c is an arbitrary complex number.

The operators that we will need in Quantum Mechanics are linear operators.

You should be able to determine which of the 3 operators in (7.9) are linear and which are not.

7.3 Hermitian Operators

An operator O which satisfies (f, Og

)=(Of, g

)(7.11)

for all f and g is called an Hermitian operator.

7.4 Linear Hermitian Operators

The operators we will need in Quantum Mechanics are both linear and hermitian. Suchoperators have very special properties.

If A is a Linear Hermitian Operator then

• The eigenvalues of A are real• Eigenfields correponding to different eigenvalues are orthogonal.• The eigenfields form a complete set.

7.4 Linear Hermitian Operators 32

I will explain what this means.

Suppose I write the eigenvalue equation for A as

A fa = a fa (7.12)

where a is an eigenvalue and fa is the corresponding eigenfield.

The above properties means

• a is real• If fa1 and fa2 are eigenfields corresponding to different eigenvalues a1 and a2 then

(fa1 , fa2) = 0

• Any field f can be constructed from the fields fa. If the eigenvalues are discrete therelation is

f =∑a

cafa

where ca are some set of complex numbers. These complex numbers can be determinedfrom f by

ca = (fa, f) / (fa, fa)

If the eigenvalues are continuous then the sums over a have to be replaced by integrals.

Chapter 8

Probability Theory

Although I have introduced a wave equation (Schrodinger’s Equation) for the electron field, Ihave not discussed any interpretation of this field. Max Born first introduced the idea that thefield was, in some way, related to probabilities. So before we can proceed I need to tell yousomething about Probability Theory.

Consider a event which has a number of possible outcomes. The ’event’ could be throwinga die: In this case there are 6 possible outcomes. The ’event’ could also be conducting anexperiment which has only a certain number of possible results.

With the ith outcome I associate a probability pi. The probability is a measure of the likelihoodof the outcome. The set of probabilities is a model and is set up on the basis of what we know(or don’t know) about the system in question.

By convention a probability zero is assigned to an outcome which is completely impossible;and a probability one is assigned to an outcome which is bound to happen.

The probabilities pi have to satisfy certain rules:

0 ≤ pi ≤ 1∑i

pi = 1(8.1)

If there is some quantity Q which has different values Qi for each outcome of the event thenI define the expectation value or, more simply, the average value of Q by

〈Q〉 =∑i

piQi (8.2)

I also define the uncertainty ∆Q in Q by

∆Q2 =∑i

pi (Qi − 〈Q〉)2 (8.3)

It can be shown that this equation can also be written as

∆Q2 =∑i

piQ2i − 〈Q〉2 (8.4)

33

8.1 Tests and Frequencies 34

8.1 Tests and Frequencies

In some cases precise tests can be made on the model. Suppose the event can be repeated anarbitrary number of times and suppose we choose a large number N . We can then determinethe number of times the ith outcome occurs Ni for all i.

The frequency of the ith outcome is

fi =Ni

N(8.5)

If the model is correct then

fi → pi ;N →∞ (8.6)

That is, as the number of trials N gets larger each measured frequency fi should approach thecorresponding probability pi. Or, in other words, pi is the model’s prediction for the frequency(if N is very large).

However not all events can be repeated an arbitrary number of times.

Consider an event which consists of you crossing Shinfield Road. The 2 outcomes being thatyou cross successfully or you are involved in an accident.

We could, on the basis of traffic statistics for similar sites, form a model for this event. Thatis, we could assign the two probabilities. However you may object to being repeatedly rundown by cars in order to test the model!

This form of probability theory did not exist until 1948 several years after the invention ofquantum theory (1928).

At the time of the invention quantum theory only the frequency concept of probability theoryexisted. This lead people (including Einstein) to regard the wavefunction as describing notone system but an ensemble (that is an infinite set) of identical systems.

With the present view of probability theory we can safely consider the wavefunction describ-ing the one system we are studying.

8.2 A simple example - a die

Consider the event of throwing a die. This has 6 possible outcomes:

• Outcome i corresponds to the upward face of the die showing i dots.

Suppose the quantity Q corresponds to the number of dots shown. That is

Qi = i

If I assume that I know nothing about the die except that it has six faces and six possibleoutcomes then the only sensible choice for the probability model is


pi =1

6

The expectation value of Q is therefore

〈Q〉 =∑i

piQi

= (1 + 2 + 3 + 4 + 5 + 6) /6 = 3.5

and the uncertainty in Q is

∆Q2 =∑i

pi (Qi − 〈Q〉)2

=1

6

((1− 3.5)2 + (2− 3.5)2 + (3− 3.5)2 + (4− 3.5)2 + (5− 3.5)2 + (6− 3.5)2)

= 2.9167

Hence ∆Q = 1.7078.

However you should be alert to the possibility that experiments may show that the model is notvalid. Someone may have tampered with the die (by inserting weights) so that the outcomesare not equally likely.

8.3 Continuous outcomes and probability densities

Not all events have a discrete set of outcomes. In some cases there is a continuum of possibleoutcomes. Suppose that an outcome is labelled by a continuous variable µ. I can assign aprobability density P (µ). This function must satisfy

P (µ) ≥ 0

∞∫−∞

dµP (µ) = 1(8.7)

The probability of an outcome somewhere in the range a ≤ µ ≤ b is

b∫a

dµP (µ) (8.8)

If I have some quantity Q which is a function of the outcome µ, then I can define the averageor expectation value of Q as


〈Q〉 =

∞∫−∞

dµP (µ) Q (µ) (8.9)

Similarly the uncertainty in Q (written as ∆Q) can be defined as

∆Q2 =

∞∫−∞

P (µ) (Q (µ)− 〈Q〉)2 (8.10)

Chapter 9

Quantum Theory

I am now in a position to set out the formal Quantum Theory. I will do this by a set of lawswhich specify the formal theory together with comments on how these specifically relate to asystem involving one electron

Law 1:The state of a physical system is specified by an element ψ, of norm 1, belonging to aHilbert Space H. All the information about the state is contained in ψ.

For a one-electron system the Hilbert Space consists of functions ψ (~x) defined in an infinite3D space. The inner product in this space is

(f, g) =

∫dV f ∗ (~x) g (~x) (9.1)

dV denotes the volume element dx dy dz and ∗ denotes the complex conjugate of the complexfunction. The functions must be such that the integrals exist (ie are not infinite).

Law 2:To every measurable physical quantity A (called an ”observable”) there corresponds alinear Hermitian operator A

The importance of the ’linear hermitian’ is that if I consider the eigenvalue equation for A

A ψa = aψa (9.2)

then the eigenvalues a are real and the eigenvectors ψa are complete. This last statementmeans that any element of the Hilbert Space can be expressed as

ψ =∑a

ca ψa (9.3)

and, if the ψa are normalized, the complex coefficients ca are

37

38

ca = (ψa, ψ) (9.4)

In the one-electron case this is

ca =

∫dv ψ∗a (~x) ψ (~x) (9.5)

The ψa satisfy the orthogonality relations

(ψa, ψa′) = δa, a′ (9.6)

Law 3:A single measurement of the observableA necessarily yields one of the eigenvalues of thecorresponding operator A. Immediately following a measurement of A in state ψ whichyielded a result a the state of the system has become the eigenvector ψa correspondingto a.

This is all very different from the classical concept of measurement.

First the theory does not predict the result of the measurement. It only predicts a number ofpossible results – the eigenvalues. It also implies that many values are impossible.

Second, the measurement itself plays an active role: The act of making a measurementchanges the state of the system: ψ → ψa.

Law 4:Probability Interpretation: The probability that the measurement of observable A, instate ψ, yields the eigenvalue an is

Pψ (A→ an) = |(ψan , ψ)|2 (9.7)

This the magnitude squared of the coefficient can in the eigenfunction series for ψ (see equa-tion (9.3).In the one-electron case the probability of a measurement of the electron co-ordinateyielding a value within a volume dV of the space-point ~x is

|ψ (~x)|2 dV (9.8)

This is again very different from classical physics. The predictions of quantum theory areprobabilistic. In classical physics, given a complete knowledge of the state of the system, theprecise results of experiments can be predicted (and the experiments can be done in such a wayas not to intrude on the state of the system). Quantum theory only predicts the probabilitiesof obtaining particular results.

39

This, of course, means that to put quantum theory to the test a very large number of mea-surements need to be made (rather than the one which would have been necessary in classicalphysics).

Law 5:

The expectation (or average) value of observable A in state ψ is

〈A〉ψ =(ψ, Aψ

)(9.9)

This, in fact, follows from the probability interpretation statement.In the one-electron case this becomes

〈A〉ψ =

∫dV ψ∗ (~x) Aψ (~x) (9.10)

Law 6:

Time Evolution: The state of a system necessarily changes with time, that is ψ shouldproperly be denoted by ψ (t) and this ψ (t) must satisfy the equation

i~dψ (t)

dt= E ψ (t) (9.11)

where E is the energy operator. You will also see this called the Hamiltonian.

In the case of a one-electron system this equation is just the Schrodinger equation:

i~dψ (~x, t)

dt= − ~2

2m

(~∇ • ~∇

)ψ (~x, t) + V (~x)ψ (~x, t) (9.12)

Law 7:

40

Basic Observables: The basic operators in Quantum Theory are those corresponding toa particle co-ordinate vector ~Q and to a particle momentum vector ~P . These operatorsare

Qx = x

Qy = y

Qz = z

Px = −i ~ d

dx

Py = −i ~ d

dy

Pz = −i ~ d

dz

(9.13)

and these operators obey the relations

PiQj − QjPi = −i ~ δi,j (9.14)

where i and j are any of x, y or z.

There are some extensions to these laws needed for the cases where the operators have con-tinuous rather than distinct eigenvalues. I will ignore this complication.

Chapter 10

Examples

Consider the 1D example discussed previously in which a particle is confined to a 1D box:0 ≤ x ≤ L. The energy eigenvalues and energy eigenfields were found to be

En =~2 π2 n2

2mL2

φn (x) =

√2

Lsin(nπ x

L

) (10.1)

The factor

√2

Lis to ensure that each eigenfield is normalized. This means that each has norm

(or size) one (as is required by Law 1).

The electron field corresponding to each of these is actually given by

ψn (x, t) = φn (x) exp

(−i Ent

~

)(10.2)

10.1 Experiments measuring particle position at time t=0

I now want to setup two experiments at time t = 0. In this case the time-factors in (10.2)become 1 and so I can deal directly with the fields φn.

10.1.1 Experiment 1

Suppose (at time t=0) the state of the system is specified by one of the eigenfields φn. I thensetup some apparatus which detects whether or not the particle is in the range

0 ≤ x ≤ L

2

That is, in the lower half of the region.

Using equation (9.8), or rather a simplification of this equation to 1D, the probability P of ameasurement of position yielding a value in this lower half is

41

10.1 Experiments measuring particle position at time t=0 42

P =L/2∫0

dx |φn (x) |2

=2

L

L/2∫0

dx sin2(nπxL

) (10.3)

I can simplify the integral by using the trigonometric relation

cos (2θ) = 1− 2 sin2 (θ) (10.4)

Using this relation to substitute for sin2 and performing the integrals gives the result

P =1

2

You should be able to deduce that this is correct by sketching graphs of |φn (x)|2.

10.1.2 Experiment 2

Now I want to repeat this experiment except that now the state of the system is a combinationof two eigenfields:

φ (x) =

√1

2(φn1 (x) + φn2 (x)) (10.5)

The square root factor is to ensure that the new state is normalized. You should check this.

The probability P of a measurement of the particle position yielding a value in the range0 ≤ x ≤ L/2 is in this case

P =1

L

L/2∫0

dx(sin2

(n1π x

L

)+ sin2

(n2π x

L

)+ 2 sin

(n1π x

L

)sin(n2π x

L

))(10.6)

The first two terms in the integrand have already been integrated above and so this reduces to

P =1

2+

2

L

L/2∫0

dx sin(n1π x

L

)sin(n2π x

L

)(10.7)

In this case I can simplify the integrand by using the identity:

2 sin (A) sin (A) = cos (A−B)− cos (A+B) (10.8)

Hence

10.2 Experiments measuring particle position at time t > 0 43

P =1

2+

1

L

L/2∫0

dx

(cos

((n1 − n2)π x

L

)− cos

((n1 − n2)π x

L

))(10.9)

Carrying out the integrals gives

P =1

2+

1

π

(sin ((n1 − n2)π/2)

n1 − n2

− sin ((n1 + n2)π/2)

n1 + n2

)(10.10)

As an example consider n1 = 2 and n2 = 1. The result for this case is

P =1

2+

4

3π= 0.9244 (10.11)

Again you should confirm that this makes sense by sketching |φ|2.

10.2 Experiments measuring particle position at time t > 0

I am now going to repeat the two previous experiments but now at some general (non-zero)time.

10.2.1 Experiment 1a

In this case the time-dependent particle field is

ψ (x, t) = φn (x, t) exp

(−iEnt

~

)(10.12)

However when I work out |ψn|2 the time-dependent exponentials cancel (that is they multiplytogether to give 1). Hence the probability of a measurement of the particle position yielding avalue in the range 0 ≤ x ≤ L/2 is exactly what was determined for the t = 0 case:

P =1

2(10.13)

10.2.2 Experiment 2a

Now if I take the combination of eigenfields as before I have

ψ (x, t) =

√1

2

(φn1 (x, t) exp

(−iEn1t

~

)+ φn2 (x, t) exp

(−iEn2t

~

))(10.14)

However when I now form |ψ|2 it does explicitly depend on time:

10.3 Measurement of Energy 44

|ψ (x, t)|2 =1

2

(φ2n1

+ φ2n2

+ φn1φn2 cos

((En1 − En2) t

~

))(10.15)

The individual integrals (apart from the time factor) have been calculated previously and sothe result for the case considered previously – n1 = 2, n2 = 1 – for the probability is

P =1

2+

4

3πcos

(3 ~ π2 t

2mL2

)(10.16)

This probability varies between a maximum of1

2+

4

3πand a minimum of

1

2− 4

3πas t varies.

10.3 Measurement of Energy

Again I am going to use the ”particle in a 1D box” to illustrate concepts of the theory.

Suppose that at time t = 0 the state of the system is given by

ψ (x, 0) = Ax (L− x) (10.17)

First I need to normalize the function (remember Law 1): That is I need to choose the constantA so that

L∫0

|ψ (x, 0) |2 = 1 (10.18)

I will leave it as an exercise to show that a suitable value for A is

A =

√30

L5(10.19)

Why did I use the words ”suitable value” and not just say ”the value”?

I now want to find an expression for ψ (x, t) for t > 0.

I do this in the following way: The first step is to make use of the completeness of the energyeigenfields in (10.1). Remember this just means that I can use combinations of these to makeany other field. So in particular I write

ψ (x, 0) =

√30

L5x (L− x) =

∑n

cn φn (x) (10.20)

Since the φn are energy eigenvalues I can immediately write down the expression for ψ (x, t):

ψ (x, t) =∑n

cn exp

(−iEnt

~

)φn (x) (10.21)


However for this to be useful I need to find out the coefficients cn.

In order to do this I use the general formula for these coefficients (6.29). In this case itbecomes

cn =2√

15

L3

L∫0

x (L− x) sin(nπ x

L

)(10.22)

This can be evaluated by carrying out two integration by parts: The result is

cn =8√

15

π3 n3n odd

= 0 n even

(10.23)

Now a measurement of energy, in this state, is proposed. According to Law 4 the probabilityof a measurement yielding a value En is

Pn =

∣∣∣∣cn exp

(−iEnt

~

)∣∣∣∣2= 0 n even

=960

π6 n6n odd

(10.24)

Notice that the exponential time factors cancel.

Inserting numerical values gives for the first few energy eigenvalues

P1 = 0.99856 P2 = 0 P3 = 0.00137 P4 = 0

Notice also that we have proved a mathematical relation here. The sum over all the probabili-ties must be 1 and hence

∑n=1,3,5,...

1

n6=

π6

960(10.25)

10.4 Expectation Values

In the ”Laws” section there two (implied) prescriptions for calculating the expectation valueof some observable A.

The specific formula (9.9) is

〈A〉ψ =(ψ, Aψ

)(10.26)


But we are also given the formula for the probability of a measurement of A yielding the ntheigenvalue an:

Pψ (A→ an) = |can|2 (10.27)

where can is a coefficient in the eigenfield expansion of psi

ψ =∑n

can ψan (10.28)

If we know these probabilities then we should be able to use the formula (8.2) to calculate theexpectation value. This yields

〈A〉ψ =∑n

|can|2 an (10.29)

It is not immediately obvious that these two formulas are the same.

In order to prove this I start from (10.26) and substitute the series (10.28) for each of the ψs.This gives

〈A〉ψ =∑n

∑m

c∗amcan

(ψam , Aψan

)(10.30)

The fact that ψan is an eigenfield of A means that this can be written as

〈A〉ψ =∑n

∑m

c∗amcan an (ψam , ψan) (10.31)

Next I use the orthogonality relations (9.6) which gives

〈A〉ψ =∑n

∑m

c∗amcan an δm,n (10.32)

One of the summations, say over m, can be done trivially since the Kronecker-δ ensures thatthe terms in the summation are zero unless m = n.

The final result is exactly equation (10.29).

10.4.1 Example 1

I am now going to test the two formulas for expectation values, (10.26) and (10.29), using the1D box model and the wavefunction:

ψ (x, 0) =

√30

L5x (L− x)

I first use formula (10.26) to work out the energy expectation value in this state. This gives


〈E〉ψ =30

L5

L∫0

dx x (L− x)

(− ~2

2m

d2

dx2

)x (L− x)

= 30~2

mL5

L∫0

dx x (L− x)

= 5~2

mL2

(10.33)

Now I will use the alternative formula (10.29). Using the coefficients cn worked out earlierthis gives:

〈E〉ψ =960

π6

(E1

16+E3

36+E5

56+E7

76+ . . .

)

=960

π6

~2 π2

2mL2

(1 +

1

81+

1

625+

1

2401+ . . .

) (10.34)

Using the terms just up to n = 7 this gives

〈E〉ψ =480

π41.014362

~2

mL2

= 4.9984~2

mL2

(10.35)

This confirms (to a reasonable accuracy) that the two formulas give the same result.

Chapter 11

Motion of a Wavepacket

In this chapter and the following two chapters I will explore some more one-dimensionalexamples but which in these cases have an infinite extent. That is

−∞ < x < +∞

The inner product in this space is

(f, g) =

+∞∫−∞

dx f ∗ (x, t) g (x, t) (11.1)

and so the expectation value of some observable A in a state f , at time t, is given by

〈A〉 (t) =

+∞∫−∞

dx f ∗ (x, t) A f (x, t) (11.2)

I will first consider the motion of a free particle.

By free particle I mean one which has no forces acting on it. The Schrodinger equation forthis case is

i~dψ (x, t)

dt= − ~2

2m

d2ψ (x, t)

dx2(11.3)

Notice that there is no potential term.

11.1 Fourier Analysis

In order to solve this equation, that is determine ψ (x, t) given ψ (x, 0), I will need to call onFourier Analysis (in the spatial variable x).

The basic equations of Fourier Transforms are

48

11.1 Fourier Analysis 49

f (x) =1

2π

+∞∫−∞

dk F (k) exp (ikx)

F (k) =+∞∫−∞

dx f (x) exp (−ikx)

(11.4)

F (k) is called the Fourier Transform of f (x) and f (x) is called the Inverse Fourier Trans-form of F (k).

I am going to use such a Fourier representation for the wavefunction ψ (x, t). That is I amgoing to write this as

ψ (x, t) =1

2π

+∞∫−∞

dkΨ (k, t) exp (ikx) (11.5)

The inverse relation is

Ψ (k, t) =

+∞∫−∞

dxψ (x, t) exp (−ikx) (11.6)

If I insert (11.4) into the Schrodinger equation (11.3) I get the following equation for Ψ (k, t)

i~dΨ (k, t)

dt= −~2k2

2mΨ (k, t) (11.7)

This can be integrated to give the result:

Ψ (k, t) = Ψ (k, 0) exp

(−i~ k

2t

2m

)(11.8)

I now put this back into the Fourier equation to get the corresponding result for ψ (x, t).

ψ (x, t) =1

2π

+∞∫−∞

dkΨ (k, 0) exp

(−i~ k

2t

2m

)exp (i k x) (11.9)

where Ψ (k, 0) is determined by the initial wave function ψ (x, 0):

Ψ (k, 0) =

+∞∫−∞

dxψ (x, 0) exp (−ikx) (11.10)

These two equations completely specify how the wavefunction changes with time.

11.2 A Gaussian Wavepacket 50

The procedure is:

• Specify the initial function ψ (x, 0);• use equation (11.10) to determine Ψ (k, 0);• insert Ψ (k, 0) into (11.9) to get ψ (x, t).

11.2 A Gaussian Wavepacket

I will now explore this in more detail by taking a particular initial wavefunction:

ψ (x, 0) = A exp (i k0 (x− x0)) exp

(−(x− x0)

2

4σ20

)(11.11)

The first exponential factor, by itself, represents a plane wave with wavelength2π

|k0|. The

second factor, which is a Gaussian function, effectively confines the wave to a neighbourhoodof the point x0. I take the amplitude A to be real. This type of free-particle wavefunction iscalled a wavepacket

The probability density ψ∗ ψ is simply

ψ∗ (x, 0)ψ (x, 0) = A2 exp

(−(x− x0)

2

2σ20

)(11.12)

11.3 Two Standard Integrals

Throughout this chapter I shall make use of the following two standard integrals

+∞∫−∞

dζ exp(−α (ζ − ζ0 − iβ)2

)=

√π

α

+∞∫−∞

dζ (ζ − ζ0 − iβ)2 exp(−α (ζ − ζ0 − iβ)2

)=

1

2

√π

α3

(11.13)

In these equations ζ0 and β are arbitrary real constants and α is a complex constant which hasa positive real part.

These integrals will be used with a variety of particular values for these constants.

11.4 Normalization and Uncertainties in Initial Wavefunction 51

11.4 Normalization and Uncertainties in Initial Wavefunction

The wavefunction must, of course, be normalized. This means that the integral of ψ∗ ψ overall x must be 1. Using the standard integrals to evaluate the integral of (11.12) reveals that Amust be

A =1√

σ0

√2π

(11.14)

The expectation value of an observable B in a state ψ (t) is

〈B〉 (t) =(ψ (t) , Bψ (t)

)(11.15)

The uncertainty in an observable B in a state ψ (t) can be written as

∆B2 =(ψ (t) , B2ψ (t)

)−(ψ (t) , Bψ (t)

)2

=(Bψ (t) , Bψ (t)

)−(ψ (t) , Bψ (t)

)2(11.16)

You should be able to explain why these two forms are equivalent.

I am going to use these formulas first with B replaced by x and second with B replaced by

P = −i~ d

dx.

The results, making use of the standard integrals, are

〈x〉 (0) = x0

〈P 〉 (0) = ~k0

(11.17)

∆x(0) = σ0

∆P (0) =~

2σ0

(11.18)

I will leave the details for you to complete!

These uncertainty results show that

• The (initial) uncertainty in position is σ0.

• The (initial) uncertainty in momentum is~

2σ0

.

• The uncertainty in position is inversely proportional to the uncertainty in momentum.• The uncertainty product is

∆x(0) ∆P (0) =~2


We will see later that this product of position-momentum uncertainties is the smallest valuepossible.

11.5 Time Dependence

I will now return to finding the electron wave at a general time t.

The Fourier transform of this initial wavefunction is

Ψ (k, 0) =1√

σ0

√2π

+∞∫−∞

dx exp (i k0 (x− x0)) exp

(−(x− x0)

2

4σ20

)exp (−ikx) (11.19)

If I multiply inside the integrand by exp (ikx0) and outside the integrand by exp (−ikx0), theresult is, of course, unchanged. However the integral can now be put into the form (11.13).

The resulting expression for Ψ (k, 0) is

Ψ (k, 0) =

√2σ0

√2π exp (−i k x0) exp

(−σ2

0 (k − k0)2) (11.20)

The integral expression (11.9) for ψ (x, t) is therefore

ψ (x, t) =

√(2σ0√2π

) +∞∫−∞

dk exp (i k (x− x0)) exp(−σ2

0 (k − k0)2) exp

(−i~ k

2t

2m

)(11.21)

I can manipulate this integral into the following form:

√(2σ0√2π

) +∞∫−∞

dk exp

(−(σ2

0 +i~t2m

)(k − k0)

2 + i (k − k0)

(x− x0 −

~ k0 t

m

))(11.22)

I then ”complete the square” involving the variable k − k0 to get the integral finally into theform (11.13).

The resulting expression for ψ (x, t) is

ψ (x, t) =1√

√2π

(σ0 +

i~t2mσ0

) exp (ik0 (x− x0)) exp

−(x− x0 − v0t)2

4

(σ2

0 +i~t2m

) exp

(−i~k

20t

2

2m

)

(11.23)


where I have defined the velocity

v0 =~ k0

m

The probability density associated with this wave is

ψ∗ (x, t)ψ (x, t) =1√2π

1√σ2

0 +

(~t

2mσ0

)2exp

− (x− x0 − v0t)2

2

(σ2

0 +

(~t

2mσ0

)2) (11.24)

If we compare this to the original probability density, we see the following changes:

• x0 → x0 + v0t;

• σ0 →

√σ2

0 +

(~t

2mσ0

)2

That is, the position expectation value has changed from x0 to x0 + v0t and the position

uncertainty has changed from σ0 to

√σ2

0 +

(~t

2mσ0

)2

The time-dependent position uncertainty

√σ2

0 +

(~t

2mσ0

)2

has some interesting properties:

In the long-time limit it is approximately

~t2mσ0

This is inversely proportional to the initial uncertainty (σ0). So the smaller the initial uncer-tainty the faster the wave spreads out.

The time t0 at which the spreading out starts to become significant is roughly when the two

terms in the expression

√σ2

0 +

(~t

2mσ0

)2

are equal. That is when t0 =2mσ2

0

~.

Consider two simple examples:

• Supposem is the mass of an electron 9.1×10−31 kg and σ0 is roughly the size of an atom,say, 1.0× 10−10 m. In this case the time t0 is 1.7× 10−16 s.

• Suppose m is the mass of an snooker ball 0.2 kg (This is a guess!) and σ0 is 0.01 mm. Inthis case the time t0 is 3.8× 1023 s. That is, 1.2× 1015 years.

These two simple numerical examples show clearly that for small objects, like an electron,confined to a small region quantum effects are important: For a macroscopic object like asnooker ball they are completely negligible.

Chapter 12

Finite Square Well

I want to consider a system which is similar to the ’particle in a 1D box’. However in thiscase x has an infinite extent

−∞ < x +∞

and the potential is given by

V (x) = 0 − L

2≤ x ≤ +

L

2

= V0 |x| > L

2

(12.1)

I am going to attempt to find the energy eigenstates for this system.

In a classical system if the energy E is larger than the limiting values of the potential V (x)as x→ ±∞ then the motion is unbounded. If, however, the energy E is less than the limitingvalues of the potential then the motion is bounded. The particle is restricted to the regiondefined by

V (x) ≤ E

This has some, but not quite the same, significance in quantum theory.

If, in a quantum system, the energy E is larger than the limiting values of the potential V (x)as x→ ±∞ then the energy eigenvalues form a continuum.

If the energy E is less than the limiting values of the potential then the energy is quantized.That is there are only discrete solutions to the energy eigenvalue equation.

In this system the different cases occur according to whether the energy is greater or less thanV0.

I am only going to consider the case for E < V0 which corresponds to discrete energy eigen-values.

The energy eigenvalue equation can be written as

54

55

E φ (x) = − ~2

2m

d2φ (x)

dx2|x| ≤ L

2

= − ~2

2m

d2φ (x)

dx2+ V0φ (x) |x| > L

2

(12.2)

I can write these two equations as

d2φ (x)

dx2= −2mE

~2φ (x) = −α2φ (x) |x| ≤ L

2

d2φ (x)

dx2=

2m (V0 − E)

~2φ (x) = β2φ (x) |x| ≥ L

2

(12.3)

where I have defined

α =

√2mE

~2

β =

√2m(V0 − E)

~2

(12.4)

I can write the general solution to these equations as

φ (x) = A exp (β x) x ≤ −L2

φ (x) = B cos (αx) + C sin (αx) − L

2≤ x ≤ L

2

φ (x) = D exp (−β x) x ≥ L

2

(12.5)

Note: Exponential terms which would diverge as x → ±∞ are not allowed and have beenexcluded. You should be able to explain why.

The four coefficients have to be chosen so that the function φ (x) and its derivative are con-

tinuous at x = ±L2

.

This gives rise to four equations:

56

A exp

(−β L

2

)= B cos

(αL

2

)− C sin

(αL

2

)

Aβ exp

(−β L

2

)= B α sin

(αL

2

)+ C α cos

(αL

2

)

D exp

(−β L

2

)= B cos

(αL

2

)+ C sin

(αL

2

)

Dβ exp

(−β L

2

)= −B α sin

(αL

2

)+ C α cos

(αL

2

)(12.6)

I solve the first two equations for B and C in terms of A and the second pair I solve for B andC in terms of D. This gives

B =

(cos

(αL

2

)+β

αsin

(αL

2

))A exp

(−β L

2

)

C =

(− sin

(αL

2

)+β

αcos

(αL

2

))A exp

(−β L

2

)

B =

(cos

(αL

2

)+β

αsin

(αL

2

))D exp

(−β L

2

)

C =

(sin

(αL

2

)− β

αcos

(αL

2

))D exp

(−β L

2

)(12.7)

If I consider the B-equations I get D = A but if I consider the C-equations I get D = −A.How can this be?

The resolution is that either

• D = A and C = 0• or• D = −A and B = 0

In the former case the eigenfunctions are even and in the latter case odd.

Even solutions.

In this case C = 0 and so the eigenvalue equation is

tan

(αL

2

)=β

α

tan

(L

2

√2mE

~2

)=

√V0 − E

E

(12.8)

57

Odd solutions.

In this case B = 0 and the eigenvalue equation is

tan

(αL

2

)= −α

β

tan

(L

2

√2mE

~2

)= −

√E

V0 − E

(12.9)

These equations have to be solved numerically or graphically. The graphical method involvingplotting each side of the equation as a function of E. The eigenvalues are then determined bythe crossing points of the two curves.

The number of eigenvalues depends on the size of V0. The greater V0 the more eigenvalues.

There is always at least one solution to (12.8) however small V0 but (12.9) may have nosolutions.

There is only one eigenvalue when

V0 <π2 ~2

2mL2

The parts of the eigenfunctions for |x| > L2

exhibit what is known as tunnelling. The eigen-function penetrates a region from which the particle would be classically forbidden.

Chapter 13

Harmonic Oscillator

I am now going to consider an electron moving in a 1D harmonic potential. In this case theSchrodinger equation for the electron field is

i~dψ (x, t)

dt= − ~2

2m

d2ψ (x, t)

dx2+

1

2mΩ2

0x2ψ (x, t) (13.1)

In classical mechanics Ω0 would be the angular frequency of the motion which would have a

period2π

Ω0

13.1 Energy Eigenstates

I first consider the determination of the energy eigenstates. These are states in which the wavehas the form

ψ (x, t) = φ (x) exp

(−iE t

~

)(13.2)

The energy E has to be determined from the eigenvalue equation:

− ~2

2m

d2φ (x)

dx2+

1

2mΩ2

0x2φ (x) = E φ (x) (13.3)

If we can find solutions of this equation then E is the energy eigenvector and φ is the energyeigenstate.

Remember a significance of such states is that the probability density is independent of timeand that no electromagnetic radiation is emitted.

13.1.1 Finding the Eigenstates

The solutions we are looking for must be normalizable. This means that the integral

+∞∫−∞

dx φ∗ (x)φ (x) (13.4)

58


must be finite.

We will ultimately choose the integral to be one. This can easily be done. If we firstly evaluatethe integral and find the value to be N (this has to be positive), then if we divide the originalfunction φ by

√N the integral changes to one.

However this normalization procedure fails if the integral is infinite.

In order for the integral to be finite the function φ must tend to zero as x gets large.

13.1.2 Choosing the units

In order to simplify the equation, I choose to use units which relate to this harmonic oscillator.

In particular I write the space variable as

x =

√~

mΩ0

ζ (13.5)

√~

mΩ0

has the dimension of a length and so ζ is the space variable in units of√

~mΩ0

.

Similarly I choose to write the energy as

E =~Ω0

2ε (13.6)

~Ω0

2has the dimension of energy and so ε is the energy variable in units of

~Ω0

2.

Choosing the units to relate to the system being studied (rather than simply using the standardSI units) almost always results in simpler equations.

In terms of the new variables ζ and ε the energy eigenvalue equation is

−d2φ (ζ)

dζ2+ ζ2φ (ζ) = ε φ (ζ) (13.7)

13.1.3 Removal of large term

For large ζ the term ζ2φ (ζ) in the equation is much larger than the term ε φ (ζ). I want tomodify the equation to remove this potentially large term.

I can do this by writing φ (ζ) as

φ (ζ) = h (ζ) exp

(−ζ

2

2

)The hope in doing this is that the function h (ζ) is (relatively) simple and certainly much moreslowly varying than the exponential factor.


With this modification the equation for h (ζ) becomes

−d2h (ζ)

dζ2+ 2ζ

d h (ζ)

dζ= (ε− 1) h (ζ) (13.8)

13.1.4 Series solution

Since h (ζ) is hoped to be slowly varying, I try to find a solution by writing this as a simplepower series

h (ζ) =∞∑n=0

cnζn (13.9)

where cn are some, as yet unknown, coefficients.

If I insert the series expression (13.9) into (13.8) I get the following relations between thecoefficients:

cn+2 = cn2n− ε+ 1

(n+ 2) (n+ 1)n = 0, 1, 2, . . . (13.10)

c0 and c1 can be chosen arbitrarily. I use this freedom to choose either c1 or c0 to be zero: Thisresults in even and odd function respectively.

Consider this recurrence relation for large values of n. In this case the relation is

cn+2 = cn2

n

The solution of this equation (for say the even case) is c2n =1

(2n)!and this is the series for

exp (ζ2).

This function goes to infinity at large x and so is not integrable.

How can we avoid this? Clearly large values of n in the series (13.10) must not occur. This isthe case if one of the coefficients vanishes and this happens if the factor (2n− ε+1) vanishesfor some value of n. That is, the dimensionless energy must be in the form:

ε = εN = 2N + 1 N = 0, 1, 2, . . . (13.11)

In this case cN is the last non-zero term in the series: all the higher terms are zero.

The recurrence relation for a given εN is

cn+2 = −2 cnN − n

(n+ 2) (n+ 1)n < N (13.12)

Using the recurrence I can get the first few complete (but unnormalized) solutions:


N εN φN (ζ)

0 1 exp(−ζ2

2

)1 3 ζ exp

(−ζ2

2

)2 5

(1− 2 ζ2

)exp

(−ζ2

2

)3 7 ζ

(1− 2

3ζ2

)exp

(−ζ2

2

)

If, at this point, I revert back to the original variables in the original units, the energy eigen-values are

EN =

(n+

1

2

)~Ω0 n = 0, 1, 2, . . . (13.13)

The first few (unnormalized) eigenfunctions are:

N EN φN (x)

012

~Ω0 exp(−x2mΩ0

2 ~

)1

32

~Ω0 x exp(−x2mΩ0

2 ~

)2

52

~Ω0

(1− 2 x2mΩ0

~

)exp

(−x2mΩ0

2 ~

)3

72

~Ω0 x

(1− 2 x2mΩ0

3~

)exp

(−x2mΩ0

2 ~

)

13.2 Oscillating Wavepacket

In the first part of this chapter I have concentrated on finding the energy eigenstates of theharmonic oscillator. However in such states the electron probability density is independent oftime: Nothing is moving!

If we take a combination of two energy eigenstates there is a periodic time dependence butthe system does not resemble a simple classical pendulum.

I now want to investigate some genuine oscillatory motion reminiscent of the simple pendu-lum. I can construct an oscillating wavepacket, similar to that studied for a free particle.

The exact solution for a wavepacket moving in this harmonic well is

ψ(x, t) = A(t) exp

[−(x− x(t))2

4a(t)+ i

mv(t)x

~− ib(t)

](13.14)

where x(t) and v(t) are the mean particle position and velocity at time t; A(t) and a(t)determine the amplitude and pulse width. These quantities are given by


A(t) =1√√2π

exp(−iΩ0t

2)√

σ0 cos2(Ω0t) +~

2mΩ0σ0

sin2(Ω0t) +i

2(

~2mΩ0σ0

− σ0) sin(2Ω0t)

(13.15)

a(t) =σ2

0 cos(Ω0t) + i~

2mΩ0

sin(Ω0t)

cos(Ω0t) + iσ2

02mΩ0

~sin(Ω0t)

(13.16)

x(t) = x0 cos(Ω0t) +v0

Ω0

sin(Ω0t) (13.17)

v(t) = v0 cos(Ω0t)− x0Ω0 sin(Ω0t) (13.18)

b(t) =m

2~

[v0x0(cos(2Ω0t)− 1) +

1

2(v2

0

Ω0

− x20Ω0) sin(2Ω0t)

](13.19)

x0 and v0 are the initial mean position and velocity.

Don’t panic! I don’t expect you to remember or to be able to prove these results.

However there are some interesting aspects of this wavefunction. First the position expecta-tion value, which is x (t), and the velocity expectation value, which is v (t), change with timein exactly the same way as the co-ordinate and velocity of a classically oscillating particle.That is (13.17) and (13.18) are exactly what you would get by solving Newton’s equations fora harmonic oscillator.

The pulse width, defined by ∆x2 =< x2 > − < x >2, also changes with time:

∆x(t) =

√σ2

0 cos2(Ω0t) +

(~

2mΩ0σ0

)2

sin2(Ω0t) (13.20)

This oscillates between σ0 (when sin(Ω0t) = 0) and ~/(2mΩ0σ0) (when cos(Ω0t) = 0).

You should compare these expressions with those for a free wave packet. The above resultshould be the same as the free wavepacket in the limit Ω0 → 0.

In the special case in which σ0 is chosen to be√

~2mΩ0

the position uncertainty is independent

of time. This is probably the state which looks most like the classical pendulum.

You should note that the state represented by the wavepacket (13.14) is definitely not anenergy eigenstate.

If I apply the energy operator (13.1) to this function I just get a different function and certainlynot a constant times the original function.

The figures below illustrate an example in which the particle starts on the right-hand side andis initially moving to the left. The wave exhibits an oscillation with period 2π/Ω0.

The first pair of figures illustrates the real part of the (complex) wave at t = 0 and at one-quarter period.


You should notice that the typical wavelengths are smaller at one quarter period than at zero:this means the particle is moving faster.

The horizontal axis is ζ defined in equation (13.5).

The second pair show the corresponding probability densities.

You should notice that the spread of the wavepacket changes.

−6 −4 −2 0 2 4 6

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

Figure 13.1. Real Part t = 0

−6 −4 −2 0 2 4 6

−0.6

−0.4

−0.2

0

0.2

0.4

0.6

0.8

Figure 13.2. Real Part t =π

2 Ω0


−6 −4 −2 0 2 4 6

0

0.1

0.2

0.3

0.4

Figure 13.3. Probability Density t = 0

−6 −4 −2 0 2 4 6

0

0.2

0.4

0.6

0.8

Figure 13.4. Probability Density t =π

2 Ω0

Chapter 14

Quantum Dynamics

Much of the course has emphasized energy eigenstates. This might be called quantum staticssince these are states in which the electron wave is not propagating. That is nothing is moving.I am now going to explicitly investigate motion.

The position expectation value is defined as⟨~X

⟩t

=

∫dV ψ∗ (~x, t) ~xψ (~x, t) (14.1)

This depends on time because ψ∗ (~x, t) and ψ (~x, t) depend on time.

I can evaluate the derivative of this expectation value by using the Schrodinger equation (andits complex conjugate!).

This gives

m

d

⟨~X

⟩t

dt=

⟨~P

⟩t

(14.2)

where ~P is the momentum operator.

I will let you prove this!

I can now repeat this procedure and differentiate⟨~P

⟩t

which is the momentum expectation

value. This gives

d

⟨~P

⟩t

dt=

⟨~F

⟩t

(14.3)

~F is the vector operator corresponding to the force. This is usually just a function of ~X theco-ordinate operator. That is

~F = ~F

(~X

)(14.4)

65

66

However in general the force may depend on momentum as well – magnetic force for example.⟨~F

⟩t

is the expectation value of the force.

If I combine equations (14.2) and (14.3) I get

m

d2

⟨~X

⟩t

dt2=

⟨~F

⟩t

(14.5)

This is the quantum dynamical equation.

In the case of a force which does not depend on velocity this becomes

m

d2

⟨~X

⟩t

dt2=

⟨~F

(~X

)⟩t

(14.6)

I now want to compare this with the classical dynamical equation that is Newton’s equation.This is

md2~Xcl (t)

dt2= ~F

(~Xcl (t)

)(14.7)

where I have called the classical particle co-ordinate ~Xcl (t) and, for simplicity, I have as-sumed that the force does not depend on velocity.

I now identify the classical particle co-ordinate with the quantum position expectation valueand I compare the right-hand sides of the quantum and classical dynamical equations. Theseare: ⟨

~F

(~X

)⟩t

~F

(⟨~X

⟩t

) (14.8)

You will notice that these are very similar but nevertheless different. The latter, the classicalterm, is the force worked out at the position expectation value. In the former, the properquantum version, it is the expectation value of the force.

In order to see the similarities and the differences, I will look at some simple examples.

• A Constant Force~F

(~X

)= ~F0

In this case you should be able to show that the two right-hand sides are identical. Thismeans that Newton’s equation is quantum mechanically correct.

• A Linear Force

67

An example of such a force is the harmonic oscillator

~F

(~X

)= −k ~X

Again you should be able to show that the two right-hand sides are identical and, again,this means that Newton’s equation is quantum mechanically correct.

• A Quadratic ForceThe most general case of a quadratic force has the form

~Fi

(~X

)=

3∑j,k=1

Aijk Xj Xk

In this case the two right-hand sides are

3∑j,k=1

Aijk

⟨Xj Xk

⟩t

in the quantum case and

3∑j,k=1

Aijk

⟨Xj

⟩t

⟨Xk

⟩t

in the classical case.These are now clearly different.

In practice the differences between the classical and quantum dynamics will only be signifi-cant if the spread of the particle field is large compared to the typical distance involved in thevariation of the force.

Chapter 15

3D, Central Forces and Angular Momentum

I am now going to look at three dimensional systems. In this case the electron field will be afunction of x, y and z and the inner product will be

(ψ1, ψ2) =

∫dV ψ∗1 (~x)ψ2 (~x) (15.1)

The Schrodinger wave equation, for this case, is

i~dψ (~x, t)

dt= − ~2

2m~∇ • ~∇ψ (~x, t) + V (~x)ψ (~x, t) (15.2)

V is the potential which gives rise to the force

~F (~x) = − ~∇V (~x) (15.3)

15.1 3D Wavepacket

Chapter 11 on the motion of a 1D wavepacket can essentially be repeated in 3D. The methodis the same.

Suppose we start (ie at t = 0) with a wavepacket

ψ (~x, 0) = A exp(i ~k0 • (~x− ~x0)

)exp

(−(~x− ~x0) • (~x− ~x0)

4σ20

)(15.4)

This represents an electron wave which is confined to a (3D) region around the point ~x0.

σ0 represents the ’width’ of the region around ~x0.

The wave is moving in a direction specified by the wavevector ~k0. In the absence of theconfining Gaussian term, the wavevector would correspond to a wave with wavelength λ0 =2π

|~k0|. However the presence of the confining terms means that the wave is made up of many

wavelengths spread around λ0.

If σ0 is small enough this represents something which confined to a very small 3D region –very like a particle.

68

15.2 Central Forces 69

The form of the wavepacket at a later time can be determined by using 3D Fourier analysis.The result for the moving wavepacket is

ψ (~x, t) =1(√

2π

(σ0 +

i~t2mσ0

))3/2exp

(i~k0 • (~x− ~x0)

)

exp

−(~x− ~x0 − ~v0t) • (~x− ~x0 − ~v0t)

4

(σ2

0 +i~t2m

)

exp

(−i~

~k0 • ~k0 t2

2m

)(15.5)

where ~v0 is defined as ~~k0/m and is the velocity (speed and direction) at which the wavepacketis propagating.

This 3D wavepacket spreads out in time in exactly the same way as the 1D wavepacket

σ0 →

√σ2

0 +

(~t

2mσ0

)2

Again, just as in the 1D case, if we apply quantum theory to macroscopic objects the spreadingout of the wavepackets can be exceedingly slow – too slow too measure!

But for small particles, like the electron, the spreading out can be quite rapid.

15.2 Central Forces

Let’s revert briefly to classical mechanics. If the potential is a function only of the distancefrom the origin

V → V (r) r =√x2 + y2 + z2 (15.6)

then the force is in the radial direction

~F = −~x

|~x|d V (r)

dr(15.7)

This is the sort of force exerted by the earth by the Sun. Such a potential is said to be spheri-cally symmetric.

In such systems it is useful to consider the angular momentum. This is a vector defined by

~L = m~x ∧ ~v = ~x ∧ ~P (15.8)


I leave it as an exercise for you to show that as a consequence of ~F being parallel to ~x

d ~L

dt= 0 (15.9)

That is ~L is constant.

This means that ~x and ~v are in a plane perpendicular to ~L and hence the motion is two-dimensional.

The energy can be expressed as

E =1

2mv2

r +~L • ~L2mr2

+ V (r) (15.10)

where vr is the velocity in the radial direction. There is a clearly identifiable contribution tothe energy from the angular motion.

15.3 Angular Momentum in Quantum Theory

In quantum theory, as in the classical case, the angular momentum is important in a system inwhich the potential V is spherically symmetric.

The angular momentum operators are defined by

Lx =(yPz − zPy

)= −i~

(yd

dz− z

d

dy

)

Ly =(zPx − xPz

)= −i~

(zd

dx− x

d

dz

)

Lz =(xPy − yPx

)= −i~

(xd

dy− y

d

dx

)(15.11)

If I consider the energy eigenvalue equation corresponding to this spherically symmetric thenenergy operator can be expressed as

E =P 2r

2m+

1

2mr2~L • ~L + V (r) (15.12)

So that the square of the angular momentum operators appears in the energy operator.

I shall simply write this as

L2 = ~L • ~L = L2x + L2

y + L2z (15.13)

In order to solve the energy eigenvalue problem it is important to first determine the eigenval-ues of L2.


The individual angular momentum components satisfy the following commutation rules:

LxLy − LyLx = i~Lz

LyLz − LzLy = i~Lx

LzLx − LxLz = i~Ly

(15.14)

These can be derived directly from the definitions.

It can also be shown from the definitions that L2 commutes with all of the components. Thatis

LxL2 − L2Lx = 0

LyL2 − L2Ly = 0

LzL2 − L2Lz = 0

(15.15)

In order to simplify what follows I am going to define a dimensionless angular momentum

~J =1

~~L (15.16)

This quantity ~J then satisfies the slightly simpler relations:

JxJy − JyJx = iJz

JyJz − JzJy = iJx

JzJx − JxJz = iJy

(15.17)

JxJ2 − J2Jx = 0

JyJ2 − J2Jy = 0

JzJ2 − J2Jz = 0

(15.18)

I want to find the eigenvalues of these Js.

15.3.1 Simultaneous Eigenvalues

Suppose I have two hermitian operators A and B which commute:

AB − BA = 0 (15.19)


Then it is possible to find φmn which are eigenvectors of both A and B. That is,

Aφmn = amφmn

Bφmn = bnφmn

(15.20)

where am are the eigenvalues of A and bn are the eigenvalues of B.

However if

AB − BA 6= 0 (15.21)

this is not possible.

15.4 Eigenvalues of J2 and Jz

If I apply the result of the previous section to the J operators I see that I can, at the same time,find the eigenvalues of J2 and Jz.

I shall attempt to do this.

I first introduce combinations of Jx and Jy.

J+ = Jx + iJy

J− = Jx − iJy

(15.22)

J+ and J− are called raising and lowering operators – you will see why shortly. These are notHermitian operators but each is the Hermitian conjugate of the other.

This means

(ψ1, J+ψ2

)=(J−ψ1, ψ2

)(ψ1, J−ψ2

)=(J+ψ1, ψ2

) (15.23)

The commutation properties of these operators are:

J+J− − J−J+ = 2Jz

JzJ+ − J+Jz = J+

JzJ− − J−Jz = −J−

(15.24)


J+J2 − J2J+ = 0

J−J2 − J2J− = 0

JzJ2 − J2Jz = 0

(15.25)

I will also need the following operators

J+J− = J2 − J2z + Jz

J−J+ = J2 − J2z − Jz

(15.26)

These two operators must have eigenvalues which are non-negative.

Suppose I start from some eigenvector φJm which satisfies

J2φJm = J2φJm

JzφJm = mφJm

(15.27)

I should point out that J , m and φJm are, as yet, unknown.

Now I form a new state by operating on φJm with J+:

φ(+)Jm = J+φJm (15.28)

I shall leave it as an exercise for you to show that

J2φ(+)Jm = J2φ

(+)Jm

Jzφ(+)Jm = (m+ 1)φ

(+)Jm

(15.29)

Hence φ(+)Jm is an eigenvector of J2 with the same eigenvalue (as that of φJm) and it is also an

eigenvector of Jz but with an eigenvalue (m+ 1) (that is one greater than that of φJm).

This is the origin of the name raising operator for J+.

So if we have found one eigenvector φJm I can repeatedly operate with J+ to get a sequenceof eigenvectors each of which has has an eigenvalue (of Jz) which is raised by 1.

m, m+ 1, m+ 2, m+ 3, . . .

If I start this process again but now using the lowering operator I obtain a sequence of neweigenvectors each of which has has an eigenvalue (of Jz) which is lowered by 1.

m, m− 1, m− 2, m− 3, . . .


At this point it looks like the Jz eigenvalues continue indefinitely to ±∞. However the oper-ators in (15.26) save us from this.

Each of the operators must have non-negative eigenvalues. If I apply each of these to φJm Iget

J+J−φJm =(J2 − J2

z + Jz

)φJm = (J2 −m2 +m)φJm

J−J+φJm =(J2 − J2

z − Jz

)φJm = (J2 −m2 −m)φJm

(15.30)

Hence for any eigenvalue m I must have

(J2 −m2 +m) ≥ 0

(J2 −m2 −m) ≥ 0(15.31)

I will leave this as an exercise to show that this implies m must lie in the range

−

(1

2+

√1

2+ J2

)≤ m ≤

(1

2+

√1

2+ J2

)(15.32)

This seems to conflict with the previous statements about the raising and lowering operators.This is resolved as follows: If we apply the raising operator to the eigenvector with the maxi-

mum m =

(1

2+

√1

2+ J2

)we get exactly zero and not another eigenvector. Similarly for

lowering operator and the minimum eigenvalue.

Now suppose that we start from the eigenvector corresponding to the minimum eigenvalueand repeatedly operate with the raising operator. After n steps, where n is, of course, aninteger we get to the maximum eigenvalue. That is

−

(1

2+

√1

2+ J2

)+ n =

(1

2+

√1

2+ J2

)(

1

2+

√1

2+ J2

)=n

2

(15.33)

I am going to call` =

n

2

Equation (15.33) then gives the eigenvalues as

J2 = ` (`+ 1)

m : −`, −`+ 1, −`+ 2, . . . , `− 2, `− 1, `(15.34)


In what we have done ` is any half-integral value

` = 0,1

2, 1,

3

2, 2,

5

2, 3,

7

2. . .

This is indeed the case in the general angular momentum theory and is important in the prop-erties of elementary particles. However in the present case only integer values of `, and hencem, are allowed. (We shall see why later).

` = 0, 1, 2, 3, . . .

The eigenvalues of L2 and Lz can be got from those of J2 and Jz by simply multiplying by~2 and ~ respectively.

15.5 Spherical Polar Co-ordinates

In the previous section I obtained the eigenvalues and eigenvectors of L2 and Lz in a waywhich avoided using the particular forms of the operators. This is rather elegant but does notreally bring out the detail of the eigenfunctions.

In order to investigate this detail I am going to introduce spherical polar co-ordinates r, θ andφ:

x = r sin θ cos φ

y = r sin θ sin φ

z = r cos θ

(15.35)

In terms of these variables the angular momentum operators are

L+ = ~ exp (i φ)

(d

d θ+ i cot θ

d

d φ

)

L− = −~ exp (−i φ)

(d

d θ− i cot θ

d

d φ

)

Lz = −i~ d

d φ

(15.36)

Notice that neither the radial co-ordinate nor its derivatives are involved.

The corresponding operator for L2 is

L2 = −~2

[1

sin θ

d

d θ

(sin θ

d

d θ

)+

1

sin2 θ

d2

d φ2

](15.37)


Remember that we already know the eigenvalues of these operators. What I want to do is toderive the eigenvectors explicitly.

I denote the joint eigenvector of L2 and Lz by Y`m (θ, φ):

L2 Y`m (θ, φ) = ~2` (`+ 1)Y`m (θ, φ)

Lz Y`m (θ, φ) = −~mY`m (θ, φ)

(15.38)

If first I use the form (15.36) for Lz then the dependence on φ can easily be obtained

Y`m (θ, φ) = P`m (θ) exp (imφ) (15.39)

Since φ = 0 and φ = 2π represent the same place the wave must have the same value hence

exp (i 2πm) = 1

This implies that m (and hence ` ) must be an integer. It is this condition which rules out thehalf-integral values found above. However these half-integral values do re-appear in elemen-tary particle physics and are the origin of spin.

The equation for P`m (θ) is

[1

sin θ

d

d θ

(sin θ

d

d θ

)− m2

sin2 θ+ ` (`+ 1)

]P`m (θ) = 0 (15.40)

Although this equation is relatively easy to solve I shall simply quote a few of the simplersolutions.

` m Y`m (θ, φ)

0 0

√14π

1 1

√38π

sin θ exp (iφ)

1 0

√34π

cos θ

1 -1

√38π

sin θ exp (−iφ)

2 2

√1532π

sin2 θ exp (i 2φ)

2 1

√154π

sin θ cos θ exp (i φ)

2 0

√5

16π

(3 cos2 θ − 1

)2 -1

√154π

sin θ cos θ exp (−i φ)

2 -2

√1532π

sin2 θ exp (−i 2φ)


These can be checked by inserting into equation (15.40).

The normalization condition for these functions which only have angular (and not radial)dependence is

π∫0

sin θ dθ

π∫−π

dφ |Y`m (θ, φ)|2 = 1 (15.41)

.

Chapter 16

Hydrogen Atom

16.1 Spherically symmetric potential

If the potential V is a function of only r then the operators L2 and Lz commute with theenergy operator E. Hence we can find eigenfunctions Φn`m which are eigenfunctions of L2,Lz and E.

n is some, as yet unknown, parameter which specifies the energy eigenvalues.

Since we know the form of the angular parts the energy eigenfunctions, in this case, can bewritten (in spherical polar co-ordinates) as

Φn`m (r, θ, φ) = Rn` (r)Y`m (θ, φ) (16.1)

Rn` (r) is called the radial part of the energy eigenfunction and this satisfies the radial energyeigenvalue equation:

− ~2

2me

(d2

dr2+

2

r

d

dr

)Rn` (r) +

(V (r) +

~2` (`+ 1)

2me r2

)Rn` (r) = En`Rn` (r) (16.2)

I have called the electron mass me in order to distinguish it from the eigenvalue m~ (theeigenvalue of Lz). Notice that latter eigenvalue does not appear in the radial equation.

16.2 Hydrogen Atom

I am now going to consider the hydrogen atom or rather an approximation to the hydrogenatom in which the nucleus (proton) is considered to infinitely massive compared to the elec-tron.

In this case the potential is the Coulomb potential:

V (r) = − e2

4π ε0 r(16.3)

78


16.2.1 Units

I am going to choose units which are appropriate to the situation. The unit of length I chooseto be:

a =~2 4π ε0me e2

(16.4)

and the unit of energy to be

~2

2me a2(16.5)

I now introduce a dimensionless radius ρ and a dimensionless energy ε by

ρ =r

a

εn ` = En `

(2me a

2

~2

) (16.6)

In terms of these variables the radial equation is

d2Rn `

dρ2+

2

ρ

dRn `

dρ+

(εn ` +

2

ρ− ` (`+ 1)

ρ2

)Rn ` = 0 (16.7)

I explore the nature of the solutions by considering first ρ very small and then ρ very large.You should be able to show that for small ρ the solution is like

Cρ`

and for large ρ it is like

D exp(−√|ε| ρ

)I then choose to look for a solution in the form

Rn` (ρ) = Aρ` un` (ρ) exp(−√|ε| ρ

)(16.8)

This incorporates the correct behaviour for small and large ρ. The hope in writing the solutionin this way is that the remaining unknown function un` (ρ) is simple.

The resulting equation for un` (ρ) is

d2un` (ρ)

dρ2+ 2

(`+ 1

ρ−√|ε|)dun` (ρ)

dρ+ 2

(1− (`+ 1)

√|ε|

ρ

)un` (ρ) = 0 (16.9)


I now write un` (ρ) as a simple power series in ρ:

un` (ρ) =∞∑k=0

ckρk (16.10)

I will leave it for you, as an exercise, to show that these coefficients ck satisfy the relation

ck+1 = 2ck

((`+ k + 1)

√|ε| − 1

(k + 1) (k + 2`+ 2)

)(16.11)

For large values of k this relation approximates to

ck+1 = ck2√|ε|k

which has a solution ck = C

(2√|ε|)k

k!. The function which results from such a series in ρ

is C exp(2√|ε|)

. This does not lead to an eigenfunction whose magnitude squared can beintegrated. It is therefore not a suitable function for quantum mechanics.

Clearly large values of k must not occur. The possibility of large k is completely eliminated ifone of the coefficients, say cn+1, is zero. If this is the case then all the coefficients with k > nwill be zero.

If we inspect equation (16.11) we see that cn+1 = 0 only if

(`+ n+ 1)√|ε| = 1 (16.12)

This equation can be re-written as

εn` = − 1

(n+ `+ 1)2 n = 0, 1, 2, . . . (16.13)

There are two things to take note of here: First the negative sign occurs because |ε| = −ε; andsecond, this has identified the parameter n used to specify εn`.

If I use the relation (16.11) the recurrence relation for the coefficients becomes:

ck+1 = ck2 (k − n)

(k + 1) (k + 2`+ 2) (n+ `+ 1)(16.14)

Some of the lower energy eigenvalues and eigenvectors are:


n εn` un` (ρ)

0 − 1(` + 1)2

1

1 − 1(` + 2)2

1− ρ

(` + 1) (` + 2)

2 − 1(` + 3)2

1− 2ρ

(` + 1) (` + 3)+

2ρ2

(` + 1) (2` + 3) (` + 3)2

Of course, in order to get the complete energy eigenfunction the other factors in (16.8) needto be included and the resulting function must be normalized.

It is worth emphasizing, yet again, that these energy eigenstates have no counterparts in classi-cal mechanics and that, in this case, lead to atomic states which do not radiate electromagneticenergy and so can exist indefinitely.

You should contrast this with the classical orbital model of an atom which does radiate elec-tromagnetic energy and whose orbital radius progressively decreases.

There are time-dependent quantum states of the hydrogen atom (which we will not explore)which however do look very like the classical model of an orbiting electron.

16.2.2 Degeneracy

The energy eigenvalue is

εn` = − 1

(n+ `+ 1)2

This does not depend on the Lz quantum number m. However the eigenvector does.

For a given ` there are 2`+1 values of m. That is there are 2`+1 different energy eigenfunc-tions which have the same energy.

In such cases we say the the energy is degenerate and has a degeneracy equal to 2`+ 1. Thatis the degeneracy is the number of different eigenfunctions which have the same energy.

Since the energy eigenvalue only depends on the sum n+ ` there are additional degeneracies.

For example n = 0, ` = 1 and n = 1, ` = 0 have the same energy but different eigen-functions.

You should explore other degeneracies.

Chapter 17

General Uncertainty Relations

What I intend to show is that for two Hermitian operators, corresponding to two physicalquantities:

∆Aψ ∆Bψ ≥1

2

∣∣∣〈(AB − BA)〉ψ∣∣∣ (17.1)

The expectation value of any operator is defined by

〈C〉ψ =

∫dV ψ∗ (~x, t) Cψ (~x, t) (17.2)

and the uncertainty is defined by

∆C2ψ = 〈C2〉ψ − 〈C〉2ψ (17.3)

It is very important that the expectation values and uncertainty are defined with respect to thestate ψ.

It is the failure to realise this that leads to spurious arguments about the supposed violationsof uncertainty relations.

I am going to derive (17.1) using only very general properties of quantum systems. In factjust general properties of what I called the inner product.

The properties of inner products are specified in (7.4). In particular I shall need

(φ, φ) ≥ 0 (17.4)

This is true for any φ and, in fact, the quantity is only zero if φ = 0.

I am going to use (17.4) with φ in the form

φ =[(A− 〈A〉ψ

)+ iλ

(B − 〈B〉ψ

)]ψ (17.5)

I then form (φ, φ) and take advantage of the fact that the operators are hermitian. This meansthat they can be operated on either side of an inner product without changing the result.

The inner product is then given by

82

83

(φ, φ) =

(ψ,(A− 〈A〉ψ

)2

ψ

)

+λ2

(ψ,(B − 〈B〉ψ

)2

ψ

)

+λ(ψ, i

[AB − BA

]ψ)

(17.6)

I temporarily use the notation

C = i[AB − BA

](17.7)

You should note that C is an Hermitian operator and its expectation value is real: This wouldnot be true without the factor i.

If I use the above definitions of uncertainties and expectation values I get

(φ, φ) = λ2∆B2ψ + λ〈C〉ψ + ∆A2

ψ (17.8)

This quantity must never be negative for any value of λ and so if I plot the right-hand side asa function of λ the curve must never cross the horizontal axis.

Suppose I ignore this and try to solve the equation

λ2∆B2ψ + λ〈C〉ψ + ∆A2

ψ = 0 (17.9)

That is I am trying to find where it does cross zero.

The solutions to this quadratic equation are

λ = − 1

2∆B2ψ

(〈C〉ψ ±

√〈C〉2ψ − 4∆A2

ψ∆B2ψ

)(17.10)

So, on one hand, I have the statement that the right-hand side of (17.8) should never cross theaxis; and, on the other hand, I have equation (17.10) which says where it does cross the axis.

In order to resolve this apparent paradox you should realise that a quadratic equation alwayshas solutions but that these solutions need not be real.

Clearly the ”solutions” (17.10) must be complex. Hence the term inside the square root mustbe negative:

〈C〉2ψ ≤ 4∆A2ψ∆B2

ψ (17.11)

If I take the square root of both sides and insert the definition of C I get exactly the generaluncertainty relation (17.1).

17.1 Examples 84

17.1 Examples

17.1.1 Position and Momentum

In general the properties of position and momentum operators are specified by (9.13) and(9.14).

Suppose first I consider the case where A is replaced by the j-component of the momentum Pjand B is replaced by the k-component of the postion operator Qk where j and k are different.

In this case there is no lower limit to the product

∆ (Pj)ψ ∆ (Qk)ψ

because the two operators commute.

Now suppose that the components are the same (j = k). There is now a restriction on theproduct

∆ (Pj)ψ ∆ (Qj)ψ ≥~2

(17.12)

This is the original Heisenberg uncertainty relation.

If you search the literature you will find references to ”violations” of this relation. These”violations” always originate from people calculating the uncertainties using different wave-functions.

Quantum theory makes no general predictions about the product

∆ (Pj)ψ1 ∆ (Qj)ψ2

when the two states represented by ψ1 and ψ2 are different.

17.1.2 Angular Momentum

Suppose I consider the operators corresponding to the x and y-components of angular mo-mentum Lx and Ly. The commutation relation for these two operators is

LxLy − LyLx = i~Lz

Therefore applying the general formalism to this case gives

∆ (Lx)ψ ∆ (Ly)ψ ≥~2|〈Lz〉ψ| (17.13)

If the expectation value of Lz is non-zero there is a restriction on the product of the uncertain-ties in Lx and Ly. So if one is very small the other must be correspondingly large.

17.1 Examples 85

If I choose ψ to be an eigenfunction of Lz with eigenvalue m~ (see chapter 15) then thisbecomes simply

∆ (Lx)ψ ∆ (Ly)ψ ≥~2

2|m| (17.14)

Hence in a state in which Lz has precise value, if the uncertainty in Lx is small then theuncertainty in Ly must be large.

17.1.3 Energy-Time Uncertainty

You will often see, in elementary physics texts, an uncertainty relation written as

∆E ∆t ≥ ~2

This is often ”derived” from the Heisenberg momentum-position uncertainty relation by sim-ply quoting ”relativity”. In relativity momentum and energy are related as are space and time.

But, in its simple form, ∆t makes no sense. Time is a parameter in Quantum Theory it is notan observable: there is no uncertainty associated with t.

However the formula can be made to have some sense.

Suppose I go back to the original formula and replace A with the energy operator E but leaveB arbitrary. The general formula now reads

∆Eψ ∆Bψ ≥1

2

∣∣∣〈(EB − BE)〉ψ∣∣∣

The right-hand side is related to the time-derivative of the expectation value of B:

d 〈B〉ψdt

=−i~〈(EB − BE

)〉ψ (17.15)

Hence the uncertainty relation can be written as

∆Eψ ∆Bψ ≥~2

∣∣∣∣∣d 〈B〉ψdt

∣∣∣∣∣ (17.16)

I can define a time

∆tB ψ =∆Bψ∣∣∣∣∣d 〈B〉ψdt

∣∣∣∣∣(17.17)

This is (roughly) the time it takes for the expectation value in B to change by an amount equalto the uncertainty in B. Using this time definition the uncertainty relation becomes

17.1 Examples 86

∆Eψ ∆ tB ψ ≥~2

(17.18)

which is the energy-time uncertainty but now the ∆ tB ψ has a precise meaning and is relatedto some physical quantity B.

Chapter 18

Many-particle systems

Up to this point I have discussed a system comprising one particle (and in particular oneelectron). This is rather limiting: There are a lot of particles in the universe.

Suppose I consider now a system with N identical particles. The field for such a system canbe written as

Ψ (~x1, ~x2, . . . , ~xN , t) (18.1)

The first variable in the argument list relates to particle 1 and the second variable relates toparticle 2 etc. So if I write

Ψ (~x2, ~x1, . . . , ~xN , t)

then ~x1 is referring to particle 2.

The quantity

|Ψ (~x1, ~x2, . . . , ~xN , t)|2 dV1dV2 . . . dVN (18.2)

is the probability that the detection of N particles yields one within a volume dV1 of ~x1, onewithin a volume dV2 of ~x2 and so on.

Since the particles have been assumed to be identical there should be no observable effect ininterchanging 2 particles. Hence

|Ψ (~x1, ~x2, . . . , ~xN , t)|2 = |Ψ (~x2, ~x1, . . . , ~xN , t)|2 (18.3)

This implies that

Ψ (~x2, ~x1, . . . , ~xN , t) = exp (iα) Ψ (~x1, ~x2, . . . , ~xN , t) (18.4)

where exp (iα) is some arbitrary phase factor.

However experimental evidence suggests that that the phase factor can only be either 1 or−1.

It is 1 for Bose particles – such as photons – and −1 for Fermi particles – such as electrons.

Hence for electrons we have anti-symmetric wavefunctions

87

88

Ψ (~x2, ~x1, . . . , ~xN , t) = −Ψ (~x1, ~x2, . . . , ~xN , t) (18.5)

This has the important consequence that if two of the co-ordinates are the same the functionmust vanish

Ψ (~x, ~x, . . . , ~xN , t) = 0 (18.6)

Hence there is zero probability of a detection of particle positions yielding two particles at thesame place.

Suppose, for simplicity, I consider just two Fermi particles.

Ψ → Ψ (~x1, ~x2, t) (18.7)

I could construct such a function out of single-particle functions (like the ones we have studiedin the rest of the module). For example

Ψ (~x1, ~x2, t) =1√2

(ψa (~x1, t)ψb (~x2, t)− ψb (~x1, t)ψa (~x2, t)) (18.8)

The factor1√2

is to ensure proper normalization.

In this case we would say that one particle is in state ψa and one is in state ψb. Of course, wecan’t say which particle is in which state because they are identical.

However suppose we try to put both particles in the same state. That is we let ψa = ψb: Inthis case the whole function is zero. Hence it is not possible for two Fermi particles to be inthe same state.

This is known as the Pauli exclusion principle.

Chapter 19

A Summary of Essential Features

Here I summarize the essential features of Quantum Mechanics. It is very easy to get lost inthe mathematical complexity.

19.1 Electron Diffraction

Experiments on electrons suggests that these have wave properties and hence that there is anelectron field ψ (~x, t).

19.2 Wave Equation

Schrodinger proposed a wave equation for the electron field

i~dψ

dt= − ~2

2m∇2ψ + V ψ

19.3 Classical-like solutions

There are solutions of this wave equation in which the electron field is very localized in spaceand moves around very like a classical particle. Such waves (usually) do spread out with time.

If the Schrodinger model is applied to macroscopic particles (billiard balls; earth; etc) witha much larger mass than the electron the spreading time of these wave-packets is extremelylong.

19.4 Very non-classical solutions

There are some solutions of the wave equation which have no counterpart in classical me-chanics. These are the energy eigenstates.

ψ(~x, t) = φ(~x) exp

(−iEt

~

)89

19.5 Meaning of the electron field 90

− ~2

2m

d2φ

dx2+ V φ = Eφ

In such states the magnitude of the wave is independent of time. Nothing is moving – that isthe wave is not propagating – and the electron does not emit any electromagnetic radiation.

19.5 Meaning of the electron field

ψ∗ (~x, t)ψ (~x, t) dV is the probability that a measurement of the position of the particle, attime t, yields a value in region dV surrounding the point ~x.

19.6 General Measurements

In a measurement of some quantity Q at time t, the probability of the result being eigenvalueqn is

∣∣∣∣∫ dV ψ∗n (~x)ψ (~x, t)

∣∣∣∣2ψn (~x) is the eigenvector of Q which is the operator corresponding to the physical quantity Q:

Qψn = qnψn

If a particular measurement yields the value qm then immediately after the measurement hasbeen made the electron field has changed to

ψm (~x)

where this is the eigenvector corresponding to the measured value. So (normally) performinga measurement changes the state of the system.

You ought to be able to say under what circumstances the measurement does not change thestate.

19.6 General Measurements 91

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