fullnotes_phas3226

PHAS3226: Quantum Mechanics

David Bowler(based on notes by D. Tovee)

December 19, 2011


AdministrationOffice hours will be Mondays at 1pm in my office (4C2 in the London Centre for Nanotechnology). Attendance sheetsmust be filled in. They’ll be given out at the start of a lecture and collated over the weeks.

Problem sheets will be given out through the term, roughly every two weeks. The best three problem sheets will countfor 10% of the final course mark. N.B. There will be four sheets during term.

Full sets of lecture notes will be made available approximately one week after the lecture. A complete PDF file willbe available at the end of the course.

PrerequisitesTo have attended and passed the department’s introductory quantum mechanics courses, PHAS2222 or equivalent courses.PHAS2224, the Atomic and Molecular Physics course is desirable. Frequent reference is made to the material inPHAS2222.

The following topics should have been covered previously :

• The time-independent Schrödinger wave equation and its solution for:

1. quantum wells and quantum barriers/steps

2. The harmonic oscillator (classical and quantum)

3. The hydrogen atom including the radial equation as well as the angular equation and its solution with sphericalharmonics.

An understanding of atomic spectroscopic notation (n, l, m quantum numbers) and their physical basis is assumed.

• The expansion postulate; the Born interpretation of the wave function; simple calculations of probabilities andexpectation values.

• For a time-independent Hamiltonian, an understanding of the separability of the full Schrödinger equation into atime-independent wave equation in position space and a time-dependent component is assumed. Familiarity withapplications to eigenstates and/or superpositions thereof is assumed.

• A basic understanding of the postulates of quantum theory is assumed.

Aims & ObjectivesAims This course aims to

• Introduce the basic concepts of Heisenberg’s matrix mechanics. The second year course (PHAS2222) dealt primar-ily with the Schrödinger’s matter wave dynamics. In PHAS3226, matrix mechanics is introduced as an alternativeapproach to quantum dynamics. It is also shown to provide a complementary approach, enabling the treatmentof systems (such as spin systems) where solutions of the non-relativistic Schrodinger wave equation in positioncoordinates is not possible.

• Apply matrix mechanics and operator algebra to the Quantum Harmonic Oscillator and its relation to the 2nd yearwave dynamics solutions using Hermite polynomials.

• Apply matrix mechanics and operator algebra to quantum angular momentum.

• Demonstrate that matrix mechanics predicts and permits solution of spin-1/2 systems using Pauli matrices.

• Develop understanding of fundamental concepts using these new methods. The Heisenberg uncertainty principle isshown to be just one among a family of generalized uncertainty relations, which arise from the basic mathematicalstructure of quantum theory, complementing arguments (e.g. Heisenberg microscope) introduced in the secondyear.

• Explore some concepts of two-particle systems. The addition of two-spins is analysed including fundamentalimplications, exemplified by the Einstein-Podolsky-Rosen paradox and Bell inequalities.

• Introduce approximate methods (time-independent perturbation theory, variational principle) to extend the PHAS2222analytical solution of the hydrogen atom to encompass atoms in weak external electric and magnetic fields and two-electron systems like helium atoms.

• Introduce symmetry requirements and the Pauli Principle .

ii


Objectives After completing the module the student should be able to:

• Formulate most quantum expressions using abstract Dirac notation and understand that it is not simply shorthandnotation for second-year expressions.

• Understand how to formulate and solve simple quantum problems expressing quantum states as vectors and quan-tum operators as matrices.

• Use commutator algebra and creation/annihilation operators to solve for the Quantum Harmonic Oscillator.

• Derive generalized uncertainty relations; calculate variances and uncertainties for arbitrary observables. In thiscontext, have a clear understanding of the relation and difference between operators and observables.

• Use commutator algebra and raising/lowering operators to calculate angular momentum observables.

• Calculate the states and observables of spin-1/2 systems using Pauli matrices.

• Understand and estimate corrections to the energy and properties of low-lying states of helium atoms using approx-imate methods and symmetry.

• Understand time-independent perturbation theory.

• Understand the variational principle.

• Understand the Pauli Principle and symmetrisation requirements on quantum states including for combinations ofspace/spin states.

TextbooksThose which are closest to the material and level of the course are (in alphabetical order)

• B. H. Bransden and C. J. Joachain, Quantum Mechanics, Second Edition (Pearson, 2000)

• S. Gasiorowicz, Quantum Physics, Third Edition (Wiley, 2003).

• A.Rae, Quantum Mechanics, Fifth edition, (Taylor & Francis 2007).

SyllabusThe approximate allocation of lectures to topics is shown in brackets below.

Formal quantum mechanics [7]Revision of year 2 concepts using Dirac bracket notation: Introduction to Dirac notation and application to PHAS2222material including orthonormality of quantum states, scalar products, expansion postulate, Linear Hermitian Operatorsand derivations of their properties; simple eigenvalue equations for energy, linear and angular momentum.

Representations of general operators as matrices and states as vectors using a basis of orthonormal states. Basis settransformations; proof that eigenvalue spectrum is representation independent. Similarity transformations and applica-tions to calculations of observables such as eigenvalues and expectation values.

The Quantum Harmonic Oscillator; Generalised Uncertainty Relations [7]Introduction of creation and annihilation operators: main commutator relations, the number operator. Solution of theQHO eigenstates and spectrum using this method. Relation to wave dynamics and Hermite polynomial solutions obtainedpreviously. The zero-point energy as a consequence of both commutator algebra as well as wave solutions. Uncertaintyrelations for general operators. Solutions of simple examples including Heisenberg Uncertainty Principle.

Generalised Angular Momentum [4]Commutator algebra and raising/lowering operators. Obtaining angular momentum eigenvalue and eigenstates usingraising and lowering operators. Matrix representation and solution of simple problems.

iii


Spin-1/2 systems [8]Introduction to spin-1/2 systems. Matrix representations of spin using eigenstates of z-component of spin: spinors andPauli matrices. Matrix representations of eigentates of spin operators along arbitrary directions. Basis set and similaritytransformations between different basis sets. Addition of two spins. The EPR paradox and derivation of Bell inequalities.

Approximate methods and many-body systems [7]Derivation of time independent Perturbation theory. First and second order theory and examples: helium and atoms inexternal fields. The variational principle; example with helium.

Symmetry, fermions, bosons and the Pauli principle. Two-particle space-spin symmetry. Slater determinants for manybody systems. Exchange corrections in helium spectrum.

iv

CONTENTS

Contents

1 Formal Quantum Mechanics 31.1 Introduction - Review of basic quantum mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.1.1 The Expansion Postulate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Dirac Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.2.1 Dirac notation for state vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 The position (coordinate) representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.4 Basis states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.4.1 Change of basis - change of representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.5 Matrix representation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.5.1 Eigenvalue equation and eigenvalues in matrix representation . . . . . . . . . . . . . . . . . . . 141.5.2 Change of basis - an example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.5.3 Solving for energy eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2 QHO 232.1 Classical mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.2 Schrödinger equation for harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.3 Algebraic operator approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

2.3.1 Eigenvectors of N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.3.2 Eigenvalues of the Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.3.3 Actions of operators a+ and a− . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272.3.4 Spatial wavefunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.3.5 Matrix representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.3.6 Minimum uncertainty for the harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . 302.3.7 Eigenstates of a− . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

2.4 Compatible observables and commuting operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.5 Heisenberg Uncertainty Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

2.5.1 Uncertainty relation for the harmonic oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3 Generalised Angular Momentum 373.1 Orbital angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.2 Eigenvalues and eigenfunctions of Lz and L2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393.3 Central potentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.3.1 Review of hydrogen atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413.4 Generalized angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.4.1 Raising and lowering operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.5 General angular momentum eigenvalue problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

3.5.1 Actions of the J+ and J− . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4 Spin 1/2 Systems 474.1 Useful relations for spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484.2 Representation of spin 1/2 operators and eigenfunctions - Pauli matrices . . . . . . . . . . . . . . . . . . 49

4.2.1 Matrix representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494.2.2 Commutators for the Pauli matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504.2.3 Basis states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504.2.4 Determination of eigenvalues and eigenvectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.2.5 Spin along an arbitrary direction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

4.3 Space-spin wavefunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.3.1 Stern-Gerlach experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4.4 Addition of angular momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

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PHAS3226: Quantum Mechanics CONTENTS

4.4.1 Addition of two spin-1/2 angular momenta . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.4.2 Coupling of spin-1/2 and orbital angular momentum . . . . . . . . . . . . . . . . . . . . . . . . 574.4.3 Addition of orbital and spin-1/2 angular momenta . . . . . . . . . . . . . . . . . . . . . . . . . . 58

5 Approximate methods & Many-body systems 615.1 Time-independent perturbation theory for a non-degenerate energy level . . . . . . . . . . . . . . . . . . 61

5.1.1 First-order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 625.1.2 Second-order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 625.1.3 Observations on the energy corrections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

5.2 The Degenerate case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 655.3 Applications of perturbation theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5.3.1 First-order examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 685.3.2 Second-order example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 725.3.3 Degenerate example - first order . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

5.4 Variational method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 765.4.1 Proof of the variational principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 765.4.2 Excited States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 775.4.3 Variational examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

5.5 Systems of identical particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 825.5.1 Identical particles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 835.5.2 Exclusion principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 845.5.3 N-particle states . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 845.5.4 Helium atom and the exchange force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

A Problem Sheets 91

2

CHAPTER 1. FORMAL QUANTUM MECHANICS

Chapter 1

Formal Quantum Mechanics

1.1 Introduction - Review of basic quantum mechanics

Quantum mechanics is probabilistic and deterministic. The system of N particles has the same set of dynamicalvariables (positions, momenta, kinetic energies, etc.) as in classical mechanics. However it is a fundamental feature ofquantum mechanics that it is impossible to have a precise knowledge of all the dynamical variables at the same timet. The Heisenberg Uncertainty Principle states that it is impossible to know precisely the position and momentum of aparticle at any given time. In particular, the uncertainty in the x-component of position, ∆x, and momentum, ∆px, of aparticle are related by

∆x∆p ≥ 1

2h . (1.1)

In the wave mechanics formulation of quantum mechanics due to Schrödinger, a basic postulate is that for everydynamical system there exists a wavefunction ψ that is a (single-valued) function of the parameters of the system andtime from which all possible information of the physical system can be obtained. In this formulation the wavefunction fora single particle is denoted by ψ(r, t), and by ψ(r1, r2, . . . rN , t) for N particles.The usual interpretation given to ψ is that its square modulus |ψ|2| gives the probability density for finding the particleat specified positions; |ψ(r)|2 d3r is the probability of finding the particle inside a volume element d3r located at vectorposition r . (Recall that the square modulus |ψ(r)|2 = ψ(r)?ψ(r), where ψ∗(r) is the complex conjugate of ψ(r).)

On the other hand, quantum mechanics is deterministic in that given the wavefunction at some time t its value at alater time t′ can be found as ψ(r, t) evolves according to the time-dependent Schrödinger equation

ih∂ψ (r, t)

∂t= Hψ (r, t) , (1.2)

where H is the Hamiltonian operator for the system. For a particle of mass m moving in a potential V (x, t) in one-dimension the Hamiltonian is

H = − h2

2m

d2

dx2+ V (x, t) . (1.3)

If the potential is independent of time, the time-dependent Schrödinger equation separates with solutions of the formφ (x) e−iEt/h and the general solution (by the expansion postulate) is

ψ (x, t) =∑n

cnφn (x) e−iEnt/h (1.4)

with φn (x) satisfying the time-independent Schrödinger equation

− h2

2m

d2φn (x)

dx2+ V (x, t)φn (x) = Enφn (x) (1.5)

with energy eigenvalue En.It is a basic principle of quantum mechanics that if ψ1 and ψ2 are two possible states of a given system then the linear

combination (superposition) ψ = c1ψ1 + c2ψ2 is also a possible state of the system where c1, c2 are constants. Theprobability density associated with ψ (r) becomes

3

PHAS3226: Quantum Mechanics CHAPTER 1. FORMAL QUANTUM MECHANICS

|ψ (r)|2 = |c1ψ1 (r) + c2ψ2 (r)|2 (1.6)= (c∗1ψ

∗1 (r) + c∗2ψ

∗2 (r)) (c1ψ1 (r) + c2ψ2 (r))

P (r) = |c1|2 |ψ1 (r)|2 + |c2|2 |ψ2 (r)|2 + c∗1c2ψ∗1 (r)ψ2 (r) + c1c

∗2ψ1 (r)ψ∗2 (r) (1.7)

= |c1|2 |ψ1 (r)|2 + |c2|2 |ψ2 (r)|2 + 2< [c∗1c2ψ∗1 (r)ψ2 (r)] (1.8)

= |c1|2 P1 (r) + |c2|2 P2 (r) + 2< [c∗1c2ψ∗1 (r)ψ2 (r)] (1.9)

where P1 (r) = |ψ1 (r)|2 and P2 (r) = |ψ2 (r)|2 are the probability densities associated with ψ1 (r) and ψ2 (r). Howeverthe total probability density is not just the weighted sums of P1 (r) and P2 (r) as there is an additional interference term2< [c∗1c2ψ

∗1 (r)ψ2 (r)].

The Hamiltonian operator was mentioned above. It is a postulate of quantum mechanics that every dynamical (vari-able) observable is represented by a linear Hermitian operator. A linear operator A is such that A (c1ψ1 + c2ψ2) =c1Aψ1 + c2Aψ2.

Another postulate is that the only possible results of a measurement of an observable A are the eigenvalues of thecorresponding operator A. That is

Aφn = λnφn, (1.10)

where λn is the eigenvalues for eigenfunction φn. Immediately after such a measurement the wavefunction of the systemwill be the same as the eigenfunction corresponding to the eigenvalue. Examples encountered in the previous course are

p = −ih ∂

∂x; pψ = peikx = −ih∂ψ

∂x= hkeikx, (1.11)

Lz = ih∂

∂φ; LzΦ = Lze

imφ = mheimφ,

L2Y`m (θ, φ) = ` (`+ 1) h2Y`m (θ, φ)

For an operator A, the Hermitian conjugate A† is defined by∫ψ∗1

(Aψ2

)dτ =

∫ (A†ψ1

)∗ψ2dτ =

[∫ψ∗2

(A†ψ1

)dτ

]∗(1.12)

for any functions ψ1 and ψ2. If the operator is Hermitian then A = A†. and so∫ψ∗1

(Aψ2

)dτ =

∫ (Aψ1

)∗ψ2dτ =

[∫ψ∗2

(Aψ1

)dτ

]∗(1.13)

The definition of an Hermitian operator ensures that

1. the eigenvalues are real quantities (a necessity for a measurement).

Since Aφn = anφn then by Hermiticity

an

∫φ∗nφndτ =

∫φ∗nAφndτ =

∫ (Aφn

)∗φndτ = a∗n

∫φ∗nφndτ

an = a∗n

and so an is real.

2. the eigenfunctions φn, φm corresponding to different eigenvalues are orthogonal, i.e.∫φ∗n (r)φm (r) d3r = 1.

If φk and φn are two eigenfunctions corresponding to different eigenvalues ak and an respectively then Aφk =akφk and Aφn = anφn.

Hence ∫φ∗nAφkdτ = ak

∫φ∗nφkdτ∫ (

Aφn

)∗φkdτ = ak

∫φ∗nφkdτ

a∗n

∫φ∗nφkdτ = ak

∫φ∗nφkdτ

(an − ak)

∫φ∗nφkdτ = 0.

4


As an 6= ak then ∫φ∗nφkdτ = 0 (1.14)

i.e. the eigenfunctions are orthogonal.

Note: This proof is not satisfying as it stands as sometimes an operator has more than one eigenfunction for oneeigenvalue (differing by more than a multiplicative factor). The eigenvalue is said to be degenerate. Suppose φn and φmare two such degenerate eigenfunctions with an = am. Then ψ = φm+cφn is also an eigenfunction of A with eigenvaluean (=am);

Aψ = Aφm + cAφn = amφm + canφn = am (φm + cφn) = amψ.

As c is arbitrary it can be chosen to make ψ orthogonal to φn, as∫φ∗nψdτ =

∫φ∗n (φm + cφn) dτ =

∫φ∗nφmdτ + c

∫|φn|2 dτ = 0

ifc = −

∫φ∗nφmdτ /

∫|φn|2 dτ.

Thus if the eigenvalue is degenerate, the eigenfunctions can always be chosen to be orthogonal.

Note on Hermitian operators.If A, B are two Hermitian operators then AB + BA and i(AB − BA) are also Hermitian. From the definition of an

Hermitian operator ∫ψ∗(Aφ)

dτ =

∫ (Aψ)∗φdτ

then ∫ψ∗ABφdτ =

∫ (Aψ)∗ (

Bφ)

dτ =

∫ (BAψ

)∗φdτ.

Similarly ∫ψ∗BAφdτ =

∫ (ABψ

)∗φdτ.

Adding gives ∫ψ∗(AB + BA)φdτ =

∫ [(AB + BA)ψ

]∗φdτ

so AB + BA is Hermitian.Likewise ∫

ψ∗iABφdτ =

∫ (Aψ)∗i(Bφ)

dτ =

∫ (BAψ

)∗iφdτ =

∫ (−iBAψ

)∗φdτ∫

ψ∗iBAφdτ =

∫ (−iABψ

)∗φdτ

so subtracting gives ∫ψ∗i(AB − BA)φdτ =

∫ (i(AB − BA)ψ

)∗φdτ

and so i(AB − BA) is Hermitian, and i[A, B

]is Hermitian.

1.1.1 The Expansion PostulateThe expansion postulate is an essential postulate of quantum mechanics. It says that for every dynamical observable thatcan be measured its eigenfunctions form a complete set, φn i.e. that any wavefunction describing a possible state of thesystem can be expressed as a linear combination of the eigenstates (in same domain and dimensionality). Hence

ψ =∑n

cnφn (1.15)

where the constants, cn, are called the expansion coefficients. Taking the scalar product of equ(1.15) with φk gives∫φ∗kψdτ =

∑n

cn

∫φ∗kφndτ =

∑n

cnδkn = ck.

5


Thus the expansion coefficients are obtained from

cn =

∫φ∗nψdτ. (1.16)

The number of eigenfunctions may be finite or infinite. In some cases the eigenfunctions of an observable form acontinuum, in which case the sum in equ(1.15) is replaced by an integral.

As an example, the functions cos (nπx/2a) and sin (nπx/2a) are eigenfunctions of the Hamiltonian for a particle ina one-dimensional infinite potential well (V (x) = 0, |x| ≤ a, V(x) = 0, |x| > a) and give rise to the Fourier expansionof a function,

f (x) = a0 +

∞∑n=1

[an cos

(nπx2a

)+ bn sin

(nπx2a

)]. (1.17)

A further postulate is that if a measurement of observable A is repeated many times, always with the system in thesame initial state ψ, the average value of A, the expectation value of A is given by

〈A〉 =

∫ψ∗Aψdτ. (1.18)

Using the expansion postulate the expectation becomes

〈A〉 =

∫ ∑k

(c∗kφ∗k) A

∑n

(cnφn) dτ

=

∫ ∑k

∑n

(c∗kφ∗k) (cnanφn) dτ

=∑k

∑n

c∗kcnan

∫φ∗kφndτ =

∑k

∑n

c∗kcnanδkn

〈A〉 =∑n

|cn|2 an =∑n

pnan, (1.19)

where pn is the probability of value an, i.e. pn = |cn|2. Furthermore if ψ is normalized

1 =

∫ψ∗ψdτ =

∫ ∑k

(c∗kφ∗k)∑n

(cnφn) dτ

=∑k

∑n

c∗kcn

∫φ∗kφndτ =

∑k

∑n

c∗kcnδkn =∑n

|cn|2

and ∑n

|cn|2 = 1, (1.20)

as is to be expected if |pn|2 is interpreted as the probability of obtaining value an as a result of a measurement.Many properties of a state are completely independent of how it is represented, e.g.probabilities of obtaining different

measured values of an observable; the expectation value of an observable cannot depend on whether the state was givenas a function of position or some other way. Thus it is desirable to have a way of discussing states which is independent ofthe representation. When the concept of intrinsic angular momentum (spin) of a particle was introduced for electrons toexplain the results of a Stern-Gerlach experiment and the fine structure of spectral lines it was noted that the spin operatorcannot be represented in term of spatial coordinates, nor can the associated eigenfunction.

1.2 Dirac NotationDirac introduced a compact general notation for quantum states. The idea behind it is based on the observation that thewavefunctions behave like vectors in many circumstances. The superposition of two wavefunctions

ψ = c1φ1 + c2φ2 , (1.21)

where c1 and c2 are constants is similar to the addition of two vectors B and C:

D = c1B + c2C . (1.22)

6


Furthermore if φ1 and φ2 are eigenfunctions of the same operator A they are orthonormal, so ψ can be specified by itstwo components (c1, c2) just as the position vector r = xi + yj can be specified by its two components (x, y). For threeeigenfunctions

ψ = c1φ1 + c2φ2 + c3φ3

is analogous to

B = Bxi +Byj +Bzk , (1.23)

where Bx, By and Bz are the Cartesian components of B. The full expansion

ψ =∑n

cnφn (1.24)

just expands the dimensionality of the space.The wavefunction can be represented in terms of different complete orthonormal sets, φ′n as

ψ =∑n

c′nφ′n . (1.25)

Similarly, the vector B can be expressed in an alternative set of Cartesian axes, having unit vectors i′, j′, k′ as

B = B′xi′ +B′yj

′ +B′zk′ . (1.26)

For two vectors B = Bxi +Byj +Bzk and C = Cxi + Cyj + Czk the scalar product is

B ·C = BxCx +ByCy +BzCz.

If two wavefunctions χ and ψ are expanded in terms of the orthonormal set of functionsφn as

χ =∑m

bmφm

ψ =∑n

cnφn

the overlap integral ∫χ∗ψdx =

∑m

∑n

b∗mcn

∫φ∗mφndx =

∑m

∑n

b∗mcnδmn =∑n

b∗ncn (1.27)

is analogous to the scalar product B ·C = BxCx+ByCy+BzCz , but with the difference that (1) it involved the complexconjugate for one of the components, (2) it can involve more dimensions than three (even an infinite number).

Wavefunctions are acted on by linear operators, which map the wavefunctions onto each other e.g.

χ = Aψ . (1.28)

Vectors can also be acted on by operators; a rotation by some chosen angle about a chosen axis; or a reflection in a chosenplane. Under such operations vectors are mapped onto each other.

1.2.1 Dirac notation for state vectorsThe wavefunction ψ and its complex conjugate ψ? are denoted by the symbols |ψ〉 and 〈ψ|, which are called a “ket” anda “bra” respectively so ψ → |ψ〉, and ψ∗ → 〈ψ|. (The reason for these names is that a bra and a ket together make a“bracket”!) The kets are members of a linear vector space whose dimensionality depends on the quantum system, and isoften infinite. The same is true of the bras. The kets are vectors in the sense that

(i) they can be multiplied by (generally complex) numbers c so if |ψ〉 is a member of the space then so is c|ψ〉.

(ii) they can be added so if |φ〉 and |ψ〉 are members of the vector space then so is |φ〉+ |ψ〉.

To each ket, there is a corresponding bra. With 〈ψ| the bra corresponding to |ψ〉, the bra corresponding to c|ψ〉 isc?〈ψ|. The bra corresponding to c1|φ〉+ c2|ψ〉 is c∗1〈φ| + c∗2〈ψ|. The overlap integral of two wavefunctions φ and ψ hasthe properties of a scalar product, ∫

φ?ψ dτ ←→ 〈φ|ψ〉

7


in the Dirac notation. The complex conjugate of such a quantity is[∫φ?ψ dτ

]?=

∫ψ?φdτ (1.29)

〈φ|ψ〉? = 〈ψ|φ〉 (1.30)

and is a key property of the scalar product of a bra and a ket. The normalization condition becomes 〈ψ|ψ〉 = 1 and theorthogonality

∫φ∗mφndτ = δmn (1.31)

〈φm|φn〉 = δmn. (1.32)

The eigenfunctions of an operator A are usually characterised by some quantum number or numbers. For example,the eigenfunctions of a particle in an infinite one-dimensional well are characterised by integers, n. The eigenfunctionsfor a hydrogen atom are specified by knowing the principal quantum number, n, (related to the energy level), the orbitalangular momentum quantum number, `, and the magnetic quantum number, m, so that they are written φn`m (r). InDirac notation these are denoted by φn → |φn〉 → |n〉; ψn`m → |ψn`m〉 → |n, `,m〉. The expansion

ψ =∑n

cnφn

becomes|ψ〉 =

∑cn|φn〉 =

∑n

cn|n〉.

Taking the inner product (= scalar product) with a bra 〈φm| (= 〈m|) gives

〈φm|ψ〉 = 〈m|ψ〉 =∑n

cn〈φm|φn〉 =∑n

cn〈m|n〉 =∑n

cnδmn.

Thuscn = 〈φn|ψ〉 = 〈n|ψ〉. (1.33)

The basis states |φm〉 define a linear vector space. The terminology is adopted that in one representation

wavefunction state functioneigenfunction ⇐⇒ state vector

eigenvector vector space

A closure relation can be developed for a complete set of states since

|ψ〉 =∑n

cn|n〉,

cn = 〈n|ψ〉

and then|ψ〉 =

∑n

〈n|ψ〉|n〉 =∑n

|n〉〈n|ψ〉

so that formally the closure relation ∑n

|n〉〈n| = 1 (1.34)

behaves as an identity operator when the summation is over all eigenstates |n〉.The result of acting with operator A on ket |ψ〉 is another ket written as A|ψ〉. The scalar product of the ket A|ψ〉 with

the bra 〈φ| is written as 〈φ|A|ψ〉. The Hermitian conjugate A† of an operator A was defined by∫φ?(Aψ)

dτ =

∫ (A†φ

)?ψ dτ =

[∫ψ?A†φ dτ

]?. (1.35)

In Dirac notation, this relationship reads:

〈φ|A|ψ〉 = 〈A†φ|ψ〉 = 〈ψ|A†|φ〉? . (1.36)

Thus the bra corresponding to the ket A|ψ〉 can be written as 〈ψ|A†. This means that the quantity 〈φ|A|ψ〉 can beinterpreted in two equivalent ways: (i) it is the scalar product of bra 〈φ| with ket A|ψ〉; (ii) it is the scalar product of bra〈φ|A with ket |ψ〉. This equivalence is an elegant feature of the Dirac notation.

8


If A is Hermitian, A = A† and〈φ|A|ψ〉 = 〈ψ|A|φ〉∗. (1.37)

Hence in Dirac notation the proof that the eigenvalues are real becomes, if |n〉 is an eigenstate of operator A then

〈n|A|n〉 = 〈n|A|n〉∗ = 〈An|n〉∗,an〈n|n〉 = a∗n〈n|n〉

so an is real.The expectation value of observable A is

〈A〉 = 〈ψ|A|ψ〉 =∑m

∑n

c∗mcnan〈m|n〉 =∑n

|cn|2 an. (1.38)

The normalization is

1 = 〈ψ|ψ〉 =∑m

∑n

〈m|c∗mcn|n〉 =∑m

∑n

c∗mcn〈m|n〉,

1 =∑n

|cn|2 .

Hence from the expression for the expectation value it follows that |cn|2 is probability that eigenvalue an is obtainedas a measurement of the observable A. After measurement the eigenfunction is the eigenstate |n〉, i.e. the wavefunction"collapses".

In Dirac notation an operator A transforms a ket |ψ〉 into another ket |χ〉 as

|χ〉 = A|ψ〉.

Taking the scalar product with the bra 〈φ| gives

〈φ|χ〉 = 〈φ|A|ψ〉.

The quantity 〈φ|A|ψ〉 is called the matrix element of the operator A between the states |φ〉 and |ψ〉.

1.3 The position (coordinate) representationSince a particle’s position along the x-axis can be measured there is an Hermitian operator x for this observable. Theresult of the measurement can be any real number, −∞ < x < ∞. If the eigenvalues of operator x are x′ they form acontinuum. The corresponding eigenvector |x′〉 satisfies

x|x′〉 = x′|x′〉. (1.39)

The expansion in discrete states |n〉, |ψ〉 =∑n〈n|ψ〉|n〉 has to become an integral

|ψ〉 =

∫〈x′|ψ〉|x′〉dx′. (1.40)

The components 〈x′|ψ〉 constitute a complex-valued function of the real variable x′. This function is identified with thewavefunction as

ψ (x′) = 〈x′|ψ〉

so establishing the connection between the (abstract) state vector |ψ〉 and the wavefunction ψ (x′). Thus ψ (x′) is acomponent of the state vector in the infinitely dimensional abstract vector space. Thus ψ (x′) is merely just one of manypossible ways of representing the state vector. This representation is called the coordinate or position representation.Equally the momentum eigenvectors of the momentum operator p|p′〉 = p′|p′〉 and the momentum wavefunction φ (p′) =〈p′|ψ〉 would be obtained in the momentum representation.

Note, by taking the scalar product of |ψ〉 with |x〉 gives

〈x|ψ〉 =

∫〈x|x′〉〈x′|ψ〉dx′

so the integral version of the closure relation is ∫|x〉〈x|dx = 1. (1.41)

9


This implies that normalization of the states |x′〉 has to be

〈x|x′〉 = δ (x− x′) (1.42)

where δ is the Dirac δ-function.As an advanced application, begin with the basic commutator

[x, px] = ih. (1.43)

Its matrix element is

〈x′′ |xpx − pxx|x′〉 = ih 〈x′′|x′〉 = ihδ (x′′ − x′)(x′′ − x′) 〈x′′|px|x′〉 = ihδ (x′′ − x′)

〈x′′|px|x′〉 =ihδ (x′′ − x′)

(x′′ − x′). (1.44)

Making use of the property of the derivative of the Dirac delta-function, xδ′ (x) = −δ (x) then

〈x′′|px|x′〉 = −ih ∂

∂x′′δ (x′′ − x′) . (1.45)

Consider px operating on a state |ψ〉 =∫|x′〉〈x′|ψ〉dx′ as

px|ψ〉 =

∫px|x′〉〈x′|ψ〉dx′

and so its expectation value is

〈ψ|px|ψ〉 =

∫ ∫〈ψ|x′′〉dx′′〈x′′|px|x′〉dx′〈x′|ψ〉,

=

∫ ∫〈ψ|x′′〉dx′′

[−ih ∂

∂x′′δ (x′′ − x′)

]dx′〈x′|ψ〉,

=

∫ ∫ψ∗ (x′′) dx′′

[−ih ∂

∂x′′δ (x′′ − x′)

]dx′ψ (x′) . (1.46)

Evaluating the δ (x′′ − x′) dx′′-function over x′′, (i.e. at x′ = x′′) gives

〈ψ|px|ψ〉 =

∫ψ∗ (x′)

[−ih ∂

∂x′

]dx′ψ (x′) (1.47)

as the expectation value of px in the state |ψ〉. Hence the representation of the momentum operator px in the coordinaterepresentation is

px = −ih ∂

∂x(1.48)

as plausibly suggested in an earlier quantum mechanics course!

1.4 Basis statesThe state of a quantum system can be specified by writing the wavefunction as a function of position, ψ (r); ψ (r1, r2 . . . rN ).However there are other ways of specifying the state. For example, ψ can be expressed as a linear combination of thecomplete set of eigenfunctions φn of an operator A, as ψ =

∑n cnφn. In this representation the wavefunction is

specified by the list of coefficients cn instead of being given as an explicit function of position. If ψ is given as a functionof position, essentially this is a representation in terms of the eigenstates of the position operator x. For example,

ψ (x) =∑n

An1√a

sin(nπx

2a

)=∑n

Anφn (x) (1.49)

as φn (x) = 1√a

sin(nπx2a

)are eigenfunctions of the Hamiltonian H for the infinite potential well of width 2a.

The wave function could be specified in terms of the eigenfunctions of the momentum operator

pφ (x) = −ih ∂

∂xφ (x) = pφ (x) ,

φ (p) = eipx/h.

10


Then

ψ (x) =

∫ ∞−∞

φ (p) eipx/h dp (1.50)

(a Fourier transform) and the state of the system is specified by the function φ (p). For discrete momentum values

ψ (x) =∑m

Bm1√2a

eikmx =∑m

Bmφm (x) .

If the energy is measured ψ (x) will reduce to an energy eigenstate φn (x) with probability |A|2; if momentum ismeasured the value p = hkm is obtained with a probability |Bm|2. In the above case, conversion from one representationis straight forward since sin kmx = 1

2i

(eikmx − e−ikmx

). Other cases are less straight forward and require formal

methods.Many properties of a state are completely independent of how it is represented, e.g.probabilities of obtaining different

measured values of an observable cannot depend on whether the state was given as a function of position or some otherway. Thus it is desirable to have a way of discussing states which is independent of the representation. When the conceptof intrinsic angular momentum (spin) of a particle was introduced for electrons to explain the results of the Stern-Gerlachexperiment and fine structure of spectral lines it was noted that the spin operator cannot be represented in terms of spatialcoordinates, or their derivatives, and nor can the associated eigenfunctions.

1.4.1 Change of basis - change of representationTransformation of states

Cold atom physicists can easily adjust the shape of the trapping potential which confines the atoms. Typically this is anharmonic potential VT (x). The Hamiltonian is

H (x, p) =p2

2m+ VT (x) , (1.51)

with VT (x) = 12mω

2x2. Solutions of the time-independent Schrödinger equation are

φn (x) = NnHn (x) e−αx2/2 (1.52)

where Hn (x) is a Hermite polynomial of degree n and α = (mω/h)1/2. The physicist can increase ω (and thus α) by

increasing the strength of the magnetic fields which confine the atom. A tight trap has large ω and α, and hence widelyspaced energy levels En =

(n+ 1

2

)hω and rapid decay of the tail of the probability distribution |ψ (x)|2 showing the

atoms are tightly confined; conversely for a weak potential - small ω and α.Suppose experimenter A prepares atoms in a well defined state by using dissipative laser cooling processes until all

the atoms are in the ground state. The cooling lasers are turned off and the atoms are in state Ψ0 (x) of a potentialV ω0

T (x) = 12mω

20x

2. The confining fields are abruptly weakened and the trap centre is displaced to x = x1. The state

function is still Ψ0 (x) but it is no longer an eigenfunction of the new Hamiltonian of the system H = p2

2m +V ωT (x− x1).A theorist B represents the state in the new basis as

|Ψ0〉 =∑n

cn|ψn〉 (1.53)

with〈x|ψn〉 ≡ ψn (x) = NnHn (x− x1) e−α(x−x1)2/2 (1.54)

with α = (mω/h)1/2. The coefficients cn are calculated by

cn = 〈ψn|Ψ0〉 (1.55)

to get probabilities |cn|2 and expectation values. However suppose the settings in the apparatus are wrong and the trapfrequency is really ω′, β = (mω′/h)

1/2 with eigenstates |χm〉,

〈x|χm〉 ≡ χm (x) = NmHm (x− x1) e−β(x−x1)2/2. (1.56)

The theorist’s expansion should have been|Ψ0〉 =

∑m

dm|χm〉. (1.57)

11


The eigenstates |ψn〉 and |χm〉 both form a complete set of basis functions and mathematically both expansions areequally valid. But only for eq(1.57) can the time evolution be written as

|Ψ0 (t)〉 =∑m

dm|χm〉e−iEmt/h (1.58)

as these states (and energies) correspond to the real apparatus. It is often advantageous to change representations, even inmid-calculation. In the experiment different potentials VT (x) may exist for different times.

The basis states can be expanded in terms of each other, as each forms a complete set. Explicitly

|ψn〉 =∑m

〈χm|ψn〉 |χm〉 =∑m

Smn|χm〉 (1.59)

and conversely|χm〉 =

∑〈ψn|χm〉 |ψn〉. (1.60)

Hence from eq(1.57) and (1.53) the expansion coefficients are given by

dm = 〈χm|Ψ0〉 = 〈χm|∑n

cn|ψn =∑n

〈χm|ψn〉cn (1.61)

dm =∑n

Smncn. (1.62)

This is an example of matrix multiplication of column vectors d and c and matrix S with Smn = 〈χm|ψn〉,

d = Sc,

d1

d2

...dN

=

〈χ1|ψ1〉〈χ1|ψ2〉 · · · 〈χ1|ψN 〉〈χ2|ψ1〉〈χ2|ψ2〉 · · · 〈χ2|ψN 〉

...... · · ·

...〈χN |ψ1〉〈χN |ψ2〉 · · · 〈χN |ψN 〉

c1c2...cN

. (1.63)

S is the matrix which transforms between the two representations of |Ψ0〉. Noting that cn = 〈ψn|Ψ0〉 and using theexpansion |Ψ0〉 =

∑m dm|χm〉 gives the inverse transformation

c1c2...cN

=

〈ψ1|χ1〉〈ψ1|χ2〉 · · · 〈ψ1|χN 〉〈ψ2|χ1〉〈ψ2|χ2〉 · · · 〈ψ2|χN 〉

...... · · ·

...〈ψN |χ1〉〈ψN |χ2〉 · · · 〈ψN |χN 〉

d1

d2

...dN

(1.64)

c = S†d (1.65)

where S† has elements S∗nm (transpose and complex conjugation of S).From

d = Sc

S−1d = S−1Sc = c = S†d

S−1 = S†

and so S is unitary, i.e.SS† = In. (1.66)

The same results can easily be found using the closure property of a complete set of states. Recall that |Ψ〉 =∑n cn|ψn〉 and cn = 〈ψn|Ψ0〉 so that

|Ψ0〉 =∑n

〈ψn|Ψ0〉 |ψn〉 =∑n

|ψn〉〈ψn|Ψ0〉 (1.67)

so that formally ∑n

|ψn〉〈ψn| = 1. (1.68)

Then sincedm = 〈χm|Ψ〉 , (1.69)

12


using the closure relationdm =

∑n

〈χm|ψn〉〈ψn|Ψ0〉 =∑n

Smncn (1.70)

withSmn = 〈χm|ψn〉 (1.71)

and in matrix formd = Sc. (1.72)

1.5 Matrix representationA state |ψ〉 can be expanded in terms of a complete set of basis states φn as

|ψ〉 =∑n

cn|φn〉. (1.73)

The constants cn can be considered as the elements of a columns vector

c =

c1c2...cn

(1.74)

with cn = 〈φn|ψ〉.The action of an operator A on the wavefunction is to generate another wavefunction |χ〉 = A|ψ〉. The state |χ〉 can

also be expanded in terms of the basis states φn as

|χ〉 =∑m

dm|φn〉 (1.75)

with dm = 〈φm|χ〉. Thus

dm = 〈φm|χ〉 = 〈φm|A|ψ〉 = 〈φm|A|∑n

cn|φn〉 =∑n

〈φm|A|φn〉cn. (1.76)

The quantitiesAmn = 〈φm|A|φn〉 are called the matrix elements of the operator A in the basis φn. The above equationbecomes

dm =∑n

Amncn

and dm are elements of a column vector

d =

d1

d2

...dn

(1.77)

with the matrix equationd = Ac (1.78)

d1

d2

...dn

=

A11 A12 . . . A1n

A21 A22 . . . A2n

......

......

An1 An2 . . . Ann

c1c2...cn

(1.79)

In the coordinate representation

Amn = 〈φm|A|φn〉 =

∫φ∗m (x) Aφn (x) dτ. (1.80)

The scalar product〈χ|ψ〉 =

∑m

∑n

d∗mcn〈φm|φn〉 =∑n

d∗ncn

as 〈φm|φn〉 = δmn so

〈χ|ψ〉 =∑n

d∗ncn = d†c = (d∗1, d∗2, . . . d

∗n)

c1c2...cn

(1.81)

These ideas lead to a matrix representation of quantum systems - a formulation put forward by Heisenberg.

13


1.5.1 Eigenvalue equation and eigenvalues in matrix representationSuppose A|χ〉 = a|χ〉. In matrix representation this is written Ac = ac. Considering the i-th component,∑

n

Aincn1 = aci1 = a∑n

δincn1∑(Ain − δina) cn1 = 0.

This is equivalent to A11 − a A12 A13 . . .A21 A22 − a A23 . . .A31 A32 A33 − a . . .

......

.... . .

c1c2c3...

= 0. (1.82)

This equation has a non-trivial solution for the c’s if

det |Ain − δina| = 0 = det |A− aI| . (1.83)

This is a good way of finding the eigenvalues for operators represented by finite matrices - not so simple for infinitematrices!

Transformation of operators

Recall that the matrix representation of an operator equation such as

φ = Aψ (1.84)

isb = Ac (1.85)

where|φ〉 =

∑n

bn|n〉; |ψ〉 =∑n

cn|n〉 (1.86)

in the same basis |n〉 and the operator is represented by the matrix

Akj =⟨k|A|j

⟩(1.87)

whencebk =

∑j

Akjcj . (1.88)

The operator A in the basis |ψn〉is represented by the matrix⟨ψi|A|ψn

⟩and by

⟨χ`|A|χm

⟩in the basis |χn〉.

Using the closure relation twice⟨χ`|A|χm

⟩=

∑i

∑j

⟨χ`|ψi〉〈ψiA|ψj〉〈ψj |χm

⟩(1.89)(

Aχ

)`m

=∑i

∑j

SliAijS∗jm (1.90)

Aχ = SAψS† = SAψS

−1, (1.91)

and the converseAψ = S−1AχS. (1.92)

Eq(1.91) and (1.92) are called similarity transformations of the matrices representing the operator A.Let

A`m =⟨χ`|A|χm

⟩and

A′in =⟨ψi|A|ψn

⟩But

|χm〉 =∑n

〈ψn|χm〉 |ψn〉 (1.93)

14


and|χ`〉∗ = 〈χ`| =

∑i

〈χ`|ψi〉〈ψi| (1.94)

so that ⟨χ`|A|χm

⟩=

∑i

〈χ`|ψi〉〈ψi|A∑n

|ψn〉〈ψn|χm〉 (1.95)

=∑i

∑n

〈χ`|ψi〉〈ψi|A|ψn〉〈ψn|χm〉 (1.96)

=∑i

∑n

S`iA′inS∗nm (1.97)

A = SA′S† = SA′S−1, (1.98)Aχ = SAψS

−1. (1.99)

Conversely

AS = SA′S−1S = SA′ (1.100)A′ = S−1AS. (1.101)

Eq(1.98) and (1.101) are called similarity transformations of the matrices representing the operator A.Transforming operators, especially the Hamiltonian between bases can be an efficient way to solve the Schrödinger

equation. As an example: represent H with the basis |χn〉 (of harmonic oscillator states) with

H|χn〉 = En|χ〉 =

(n+

1

2

)hω|χ〉, (1.102)

so thatHnn′ =

⟨χn|H|χn′〉

⟩= Enδnn′ (1.103)

and H is a diagonal matrix

H =

E1 0 · · · · · ·0 E2 0 · · ·... 0 E3 · · ·...

......

. . .

. (1.104)

The matrix of an operator in a basis of its own eigenstates is diagonal, i.e.if A|n〉 = λn|n〉, then⟨n|A|n′

⟩= λnδnn′ .

However the matrix H ′ of H in the basis |ψn〉 is not diagonal in this example as |ψn〉 are not the eigenfunctions;the |χm〉 are. However the eigenvalues of H ′ are still

(n+ 1

2

)hω, i.e. the (energy) spectrum is independent of the

representation.

Proof that H and H ′ have the same spectrum To find the eigenvalues of H ′; H ′|n〉 = λn|n〉, the characteristicequation

det (H ′ − λI) = 0

must be solved. But H ′ = S−1HS and λI = λS−1S so and

H ′ − λI = S−1HS − λS−1S = S−1 (H − λI)S

anddet (H ′ − λI) = det

[S−1 (H − λI)S

].

But det (AB) = detAdetB so

det (H ′ − λI) = det(S−1

)det (H − λI) detS

= det(S−1

)detS det (H − λI)

= det (H − λI) ,

since det(S−1

)detS = det

(S−1S

)= det (I). The eigenvalues of H are known and hence so are those of H ′. Also

both the eigenvalues as well as the trace of the matrix of any operator A are independent of the (basis) representation.

15


Demonstration that the expectation value is independent of the (basis) representation

The expectation value of an operator A can be evaluated from⟨A⟩

=⟨

Ψ|A|Ψ⟩. (1.105)

If A is represented in a basis |ψn〉 which is not formed from its eigenfunctions, the matrix representing A is non-diagonal, with elements

A′nn′ =⟨ψn|A|ψn′

⟩. (1.106)

Expanding |Ψ〉 in the same basis gives

|Ψ〉 =∑n

cn|ψn〉 (1.107)

and ⟨Ψ|A|Ψ

⟩=

∑n

〈ψn|c∗nA∑n′

cn′ |ψn′ 〉 =∑n

∑n′

c∗nA′nn′cn′ (1.108)

=(cT)∗A′ψc (1.109)

Using the closure relation ⟨Ψ|A|Ψ

⟩=

∑n

∑n′

⟨Ψ|ψn〉〈ψn|A|ψn′〉〈ψn′ |Ψ

⟩, (1.110)

=∑n

∑n′

c∗nAnn′cn′ , (1.111)

= c†A′ψc.

If |χj〉 is an eigenbasis of A thenA|χj〉 = λj |χj〉 (1.112)

and〈χj |A|χi〉 = λjδji (1.113)

i.e. is diagonal. It could also have been used to expand |Ψ〉 as

|Ψ〉 =∑j

dj |χj〉 (1.114)

then ⟨Ψ|A|Ψ

⟩=

∑j

〈χj |d∗j A∑j

di|χi〉 =∑j

∑j

d∗j 〈χj |A|χi〉di, (1.115)

=(dT)∗Aχd = d†Aχd. (1.116)

⟨Ψ|A|Ψ

⟩= (d∗1, d

∗2, . . .)

λ1 0 · · ·0 λ2 · · ·...

. . .

d1

d2

...

. (1.117)

=∑j

∣∣d2j

∣∣λj . (1.118)

Eq(1.109) and (1.118) give the same result. This follows from SS−1 = 1; d = Sc; d† = c†S† and Aχ = SA′ψS−1, so

d†Aχd = c†S†AχSc = c†S−1AχSc,

= c†A′ψc, (1.119)

and thus〈A〉 =

⟨Ψ|A|Ψ

⟩= c†A′ψc = d†Aχd. (1.120)

16


1.5.2 Change of basis - an exampleN.B. The example below uses ideas from later in the course, but is included for completeness Spin is associated witha magnetic dipole moment µ carried by a spin- 1

2 particle, e.g. an electron where

µ = −gsµBh

s (1.136)

with µB = eh2me

the Bohr magneton, gs the Landé g-factor for the electron (≈ 2). In an external magnetic field B theenergy is

H = E = −µ ·B =gsµBh

s ·B =µes ·B (1.137)

with µe = gsµB/h. In MRI/NMR it is the nuclear proton’s spin that is involved with µN = eh/2mp instead of the Bohrmagneton.

Consider a spin- 12 particle (or system) in a magnetic field B = Bx ı+Bzk. The Hamiltonian is

H = µes ·B

= µe

(BxSx +BzSz

)= µe

h

2(Bxσx +Bzσz)

H = aσz + bσx. (1.138)

Suppose a/h = 4MHz and b/h = 3MHz . The matrix representation of H in the basis functions of σz is

H(z) = a

(1 00 −1

)+ b

(0 11 0

)=

(a bb −a

). (1.139)

The matrix representation of H in the basis functions of σx is

H(x) =

(x〈+|H|+〉x x〈+|H|−〉xx〈−|H|+〉x x〈−|H|−〉x

). (1.140)

But

|+〉x =1√2

(|α〉+ |β〉) =1√2

(11

)and

|−〉x =1√2

(|α〉 − |β〉) =1√2

(1−1

)and hence

x〈+|H|+〉x =1

2(〈α|+ 〈β|) (aσz + bσx) (|α〉+ |β〉) .

But σz|α〉 = |α〉; σz|β〉 = −|β〉; σx|α〉 = |β〉; σx|β〉 = |α〉 so

x〈+|H|+〉x =1

2(〈α|+ 〈β|) (a|α〉 − a|β〉+ b|β〉+ b|α〉)

=1

2(a+ b− a+ b) = b. (1.141)

Using the Pauli matrices

x〈+|H|+〉x =1

2(1, 1)

(a bb −a

)(11

)=

1

2(1, 1)

(a+ bb− a

)= b,

x〈−|H|+〉x =1

2(1,−1)

(a bb −a

)(11

)=

1

2(1,−1)

(a+ bb− a

)= a,

x〈−|H|−〉x =1

2(1,−1)

(a bb −a

)(1−1

)=

1

2(1,−1)

(a− bb+ a

)= −b.

Hence

H(x) =

(b aa −b

)(1.142)

18


H(x) and H(z) are the same Hamiltonian but represented in different bases. It is easy to verify that H(x) and H(z) havethe same spectrum (of eigenvalues). For H(z) one has∣∣∣∣ a− λ b

b −a− λ

∣∣∣∣ = 0

giving − (a− λ) (a+ λ)− b2 = 0, λ2 − a2 − b2 = 0 and λ = ±√a2 + b2 = ±5. For H(x) one has∣∣∣∣ b− λ a

a −b− λ

∣∣∣∣ = 0

so − (b− λ) (b+ λ)− a2 = 0, λ2 − a2 − b2 = 0 and λ = ±√a2 + b2 = ±5.

Suppose that the system is in the ground state |ψ0〉 of H with eigenvalue −√a2 + b2 = −5. What is this state in

terms of the basis of σz? (If a measurement of a sample of these spins along the z-axis is measured the magnetization isMz = µeN 〈Sz〉.) If the eigenvalue is λ and |ψ0〉 = p|α〉+ q|β〉 then

H|ψ0〉 = λ|ψ0〉(aσz + bσx) (p|α〉+ q|β〉) = λ (p|α〉+ q|β〉)

or in terms of matrices (a bb −a

)(pq

)= λ

(pq

)(

4 33 −4

)(pq

)= −5

(pq

)

4p+ 3q = −5p

3p− 4q = −5q

so q = −3p. Normalization requires |p|2 + |q|2 = 1 so p = 1√10

and q = − 3√10

and

|ψ0〉 =1√10|α〉 − 3√

10|β〉. (1.143)

For λ = +5 the eigenstate is

|ψ1〉 =3√10|α〉+

1√10|β〉. (1.144)

The expectation value ⟨Sz

⟩=

h

2〈ψ0|σz|ψ0〉

=h

2

1√10

(1,−3)∗(

1 00 −1

)(13

)1√10⟨

Sz

⟩=

h

2

1

10(−8) = −4

5

(h

2

),

which is just the probability of "up"(

110

(h2

))and "down"

(910

(− h2))

.What is the probability of the spin pointing along the +x−axis? The ground state |ψ0〉 = 1√

10|α〉 − 3√

10|β〉 ≡

c1|α〉+ c2|β〉 needs to be expanded in terms of the eigenstates of Sx as

|ψ0〉 = d1|+〉x + d2|−〉x = c1|α〉+ c2|β〉

Taking the scalar product with |+〉x gives

d1x〈+|+〉x + d2x〈+|−〉x = c1x〈+|α〉+ c2x〈+|β〉

But |+〉x and |−〉x are orthogonal sod1x = x〈+|α〉c1 + x〈+|β〉c2.

Similarly using |−〉xd2x = x〈−|α〉c1 + x〈−|β〉c2.

19


These equations can be represented by the matrix equation

d = Sc (1.145)(d1

d2

)=

(x〈+|α〉 x〈+|β〉x〈−|α〉 x〈−|β〉

)(c1c2

). (1.146)

But |+〉x = 1√2

(|α〉+ |β〉) and |−〉x = 1√2

(|α〉 − |β〉) so x〈+|α〉 = 1√2

etc. so that

S =

(1√2

1√2

1√2− 1√

2

)=

1√2

(1 11 −1

)and

S† =1√2

(1 11 −1

). (1.147)

Thus S is the matrix that transforms representations in the z-basis to those of the x-basis. S† does the reverse. Note that

SS† =1

2

(1 11 −1

)(1 11 −1

)=

1

2

(2 00 −2

)= I2

as it should!If working out S for the transformation between the z-basis and the y-basis care must be taken as there are i’s when

taking transposes and complex conjugations.In this case (

d1

d2

)=

1√2

(1 11 −1

)1√10

(1−3

)=

1√5

(−12

)and so

|ψ0〉 = − 1√5|+〉x +

2√5|−〉x. (1.148)

The probability of measuring spin along the +x-axis is |d1|2 = 15 and the probability of measuring spin along the−x-axis

is |d2|2 = 45 and the magnetization

P+ − P− = 〈ψ0|σx|ψ0〉 = −3

5. (1.149)

The expectation (and magnetization) are measurable quantities and should be independent of the representation. This iseasily verified as in the z-representation

〈ψ0|σx|ψ0〉 =1√10

(1,−3)∗(

0 11 0

)1√10

(1−3

)= −3

5. (1.150)

Transformation of the operators between representations

The Hamiltonian operators H(x) and H(z) are related by a similarity transform

H(x) = SH(z)S−1 = SH(z)S† (1.151)

Explicitly (b aa −b

)=

1√2

(1 11 −1

)(a bb −a

)1√2

(1 11 −1

)(1.152)

For the simple 2× 2 matrix and spin- 12 problems most quantities can be calculated using the transformation of the basis,

e.g.

|+〉x =1√2

(|α〉+ |β〉) ; |−〉x =1√2

(|α〉 − |β〉) (1.153)

and their inverses

|α〉 =1√2

(|+〉x + |−〉x) ; |α〉 =1√2

(|+〉x − |−〉x) (1.154)

But the similarity transforms are best for complex systems of many coupled spins where the matrices can be quite large.Also if the magnet field is constantly changing as in H (t) = µeSzBz (t) + µeSxBx (t). Note that the matrix S isindependent of Bx (t) and Bz (t) so one can switch back and forth thousands of times to calculate the magnetizationsMx, My , Mz at different times.

20


1.5.3 Solving for energy eigenvaluesNowadays most solutions of the time-independent Schrödinger equation are achieved by diagonalizing a matrix such asH ′ rather than solving second-order partial differential equations.

Suppose that the true eigenstates |χ〉 are not known, but that |ψ〉 is a known basis. Then

Hχn〉 = En|χn〉 = H∑j

c(n)j |ψj〉 (1.155)

but also the coefficients c(n)j are unknown (if they were known then so would the |χ〉 be known).

Procedure:

1. The effect of an operator on a state is given by the effect of its matrix on the state’s column vector

2. Represent H in the basis |ψ〉 by H ′

3. Matrix H ′ has the correct energy spectrum, thus

H ′c(n) =

H ′11 H ′12 · · ·H ′21 H ′22 · · ·

......

. . .

c(n)1

c(n)2...

= En

c(n)1

c(n)2...

(1.156)

4. Solve det (H ′ − λI) = 0 to get the eigenvalues En

5. Each eigenvector c(n) contains the coefficients for eq(1.155) so solve

H ′c(n) = Enc(n) (1.157)

Hence now all the eigenvalues and eigenstates are known without solving the wave equation.

21


22

CHAPTER 2. QHO

Chapter 2

Quantum Harmonic Oscillator; GeneralisedUncertainty Relations

2.1 Classical mechanicsThe harmonic oscillator is an important system to study as many physical systems are approximated by it. Many systemshave an interaction potential of this general form, with a quadratic potential.

Expanding the potential function about the minimum at x = x0 as a Taylor series gives

V (x) = V (x0) + (x− x0)

(dV

dx

)x0

+1

2(x− x0)

2

(d2V

dx2

)x0

+1

6(x− x0)

3

(d3V

dx3

)x0

+ · · · (2.1)

At the minimum(

dVdx

)x0

= 0. The force of interaction and equation of motion of a particle of mass m are

F = −dV

dxx = m

d2x

dx2x

md2x

dx2= − (x− x0)

(d2V

dx2

)x0

− 1

2(x− x0)

2

(d3V

dx3

)x0

− · · · .

For small displacements from the equilibrium position x = x0 the leading term is

d2x

dx2= − 1

m

(d2V

dx2

)x0

(x− x0) (2.2)

which is the equation of simple harmonic motion about x = x0.Defining the potentials with respect to the minimum V (x0) and displacements relative to x0 gives

V (x) =1

2kx2 (2.3)

with

k =

(d2V

dx2

)x0

(2.4)

andd2x

dx2= − k

mx = −ω2x, (2.5)

with angular frequency ω =√k/m and solution x (t) = A cos (ωt) + sin (ωt). The turning points are at x =

±(2E/mω2

)1/2.

2.2 Schrödinger equation for harmonic oscillatorExpressing the potential as V (x) = 1

2mω2x2, the time-independent Schrödinger equation is

− h2

2m

d2ψ (x)

dx2+

1

2mω2x2ψ (x) = Eψ (x) . (2.6)

23

PHAS3226: Quantum Mechanics CHAPTER 2. QHO

By introducing the dimensionless variable λ = 2E/hω and ξ = αx where α = (mω/h)1/2 the Schrödinger equation

becomes(

ddx = α d

dξ

)d2ψ (ξ)

dξ2+(λ− ξ2

)ψ (ξ) = 0. (2.7)

This is a second-order differential equation which can be solved by a power series approach. The quantum harmonicoscillator eigenfunctions are

ψn (ξ) = NnHn (ξ) e−ξ2/2, (2.8)

where Hn (ξ) is Hermite polynomial of degree n, parity (−1)n and Nn is a normalisation constant

Nn =

(α√π2nn!

)1/2

. (2.9)

The eigenvalues λn = 2n+ 1 give energy eigenvalues

En =

(n+

1

2

)hω, (2.10)

with n = 1, 2, . . ..In unscaled variables

ψ (x) = Nn (α)Hn (αx) e−α2x2/2. (2.11)

Thus large ω implies a deep potential well, large α and a short tail (rapid fall off) of the wavefunction (and vice versa).The eigenfunctions are orthonormal i.e.∫ ∞

−∞ψ∗m (x)ψn (x) dx =

NmNnα

∫ ∞−∞

e−ξ2

Hm (ξ)Hn (ξ) dξ = δmn, (2.12)

and ∫ ∞−∞

e−ξ2

Hm (ξ)Hn (ξ) dξ =√π2nn!δmn. (2.13)

Recurrence relations exist for the Hermite polynomials

Hn+1 (x)− 2xHn (x) + 2nHn−1 (x) = 0

dHn (x)

dx= 2nHn−1 (x) .

The recurrence relations enable evaluation of the matrix elements, e.g.

xkn =

∫ ∞−∞

ψ∗m (x)xψn (x) dx = 〈ψk |x|ψn〉 =

(h

2mω

)1/2 (√k + 1δk+1,n +

√kδk−1,n

). (2.14)

2.3 Algebraic operator approachIn many physical systems involving harmonic oscillators the interest is in the energy eigenvalues or the matrix elementsand the explicit form of the eigenfunctions is not needed (or is even hard to find). This can be achieved by an elegantoperator approach.

The Hamiltonian can be written as

H =p2

2m+

1

2mω2x2 (2.15)

with the momentum and position operators satisfying

[x, px] = ih. (2.16)

Two new operators a+ and a− are defined by

a+ =

(1

2hmω

)1/2

(mωx− ip) =1√2

[((mωh

)1/2

x− ip

(mhω)1/2

)](2.17)

a+ =1√2

[αx− i

hαp

], (2.18)

a− =

(1

2hmω

)1/2

(mωx+ ip) =1√2

[αx+

i

hαp

]. (2.19)

24


The operators a+ and a− are called raising and lowering operators (also creation and annihilation operators) respectivelyfor reasons that will be clearer later. Since x and p are Hermitian operators, a+ and a− are Hermitian conjugates of eachother,

a†+ = a−; a†− = a+ (2.20)

These operators satisfy the commutator[a−, a+] = 1. (2.21)

Explicitly

[a−, a+] =1

2hωm([mωx+ ip,mωx− ip])

1

2hωmimω ([x,−p] + [p, x])

since [x, x] = [p, p] = 0. But [x, px] = ih so

[a−, a+] =i

2h(−ih− ih) = 1. (2.22)

Introducing the Hermitian operatorN = a+a− (2.23)

then

a+a− =1

2hωm(mωx− ip) (mωx+ ip)

=1

2hωm

(m2ω2x2 + p2 + imω (xp− px)

)=

1

2hωm

(m2ω2x2 + p2 + imωih

)=

1

hω

(1

2mω2x2 +

p2

2m

)− 1

2

=1

hωH − 1

2.

Hence the Hamiltonian can be written as

H = hω

(a+a− +

1

2

)= hω

(N +

1

2

). (2.24)

The operator N = a+a− is called the number operator for reasons that will be more obvious later! It follows thatN = a+a− commutes with the Hamiltonian H and thus they have a common set of eigenvectors.

Using the form a+ = 1√2

[αx− i

hα p]

[a−, a+] =1

2

[(αx+

i

hαp

),

(αx− i

hαp

)]Expanding using commutator algebra and noting that x and p commute with themselves, so

[a−, a+] =1

2

[αx,− i

hαp

]+

[i

hαp, αx

]=

1

2

(− ih

)[x, p] +

i

h[p, x]

=

1

2

(− ih

)ih+

i

h(−ih)

= 1.

Also

a+a− =1√2

[αx− i

hαp

]1√2

[αx+

i

hαp

](2.25)

=1

2

(α2x2 +

p2

h2α2+ αx

i

hαp− i

hαpαx

)=

1

2

(α2x2 +

p2

h2α2+i

h[x, p]

)=

1

2

(α2x2 +

p2

h2α2+i

hih

)=

1

2

(α2x2 +

p2

h2α2− 1

)=

1

2

(mωh

)x2 +

p2

2hωm− 1

2.

=1

hω

(1

2mω2x2 +

p2

2m

)− 1

2=

1

hωH − 1

2, (2.26)

25


and so

H =

(a+a− +

1

2

)hω.

as before.Useful commutators are [

H, a+

]= hωa+;

[H, a−

]= −hωa−. (2.27)

Proof: [H, a+

]= hω

[(a+a− +

1

2

), a+

]= hω [a+a−, a+] = hω a+ [a−, a+] + [a+, a+] a−

= hωa+[H, a−

]= hω

[(a+a− +

1

2

), a−

]= hω [a+a−, a−] = hω a+ [a−, a−] + [a+, a−] a−

= −hωa−.

2.3.1 Eigenvectors of NSuppose |φ〉 is an eigenvector of N with eigenvalue λ, then

N |φ〉 = λ|φ〉

and

〈φ∣∣∣N ∣∣∣φ〉 = λ 〈φ|φ〉 = λ

〈φ|a+a−|φ〉 = λ

〈a−φ|a−φ〉 = λ

since a†+ = a−. Consider the state |φ〉 = a−|φ〉, then from above⟨φ|φ⟩

= λ.

But by necessity 〈φ|φ〉 ≥ 0, and hence λ ≥ 0.Consider

N |φ〉 = (a+a−) a−|φ〉 = (a−a+ − 1) a−|φ〉= a− (a+a−) |φ〉 − a−|φ〉 = a−N |φ〉 − a−|φ〉= a−λ|φ〉 − a−|φ〉

N (a−|φ〉) = (λ− 1) (a−|φ〉) (2.28)

Thus a−|φ〉 is also an eigenvector of N but with eigenvalue λ−1. Hence the reason that a− is called a lowering operator.Note that a+ has the opposite effect, i.e. N (a+|φ〉) = (λ+ 1) (a+|φ〉) and hence is a raising operator and obviouslya+|φ〉 = a+a−|φ〉 = λ|φ〉.

Thus from the state |φ〉 a sequence of eigenstates can be formed by repeated application of a− i.e. a|φ〉, (a−)2 |φ〉,

(a−)3 |φ〉, . . . (a−)

n |φ〉 having eigenvalues λ− 1, λ− 2, λ− 3, . . .λ−n. Hence λ can only have integer values (positiveor zero) since λ ≥ 0. If positive non-integer values of λ existed, the sequence λ, λ− 1, λ− 2, λ− 3, . . .λ− n would beunending with some values which are negative contrary to the requirement that they are all greater or equal to zero. If λis an integer, then λ− n can be zero and its associated eigenstate |φ0〉 satisfies

a−|φ0〉 = 0|φ0〉 = 0,

a+a−|φ0〉 = 0|φ0〉N |φ0〉 = 0|φ0〉, (2.29)

so the eigenvalues of N are positive integers and zero. The corresponding eigenstate is denoted by |φ〉 = |n〉, such that

N |n〉 = n|n〉 (2.30)

and

N (a−|n〉) = (n− 1) (a−|n〉) ,N (a+|n〉) = (n+ 1) (a+|n〉) .

Since N |n〉 = n|n〉 and En =(n+ 1

2

)hω the state |n〉 can be interpreted as containing n quanta each of energy hω.

The state a+|n〉 contains n + 1 quanta and so a+ creates a quantum of energy and is referred to as a creation operator.Likewise a− absorbs a quantum of energy and is referred to as the annihilation operator. The operator N is called theoccupation number operator.

26


2.3.2 Eigenvalues of the Hamiltonian

Since H = hω(N + 1

2

)then

H|φ〉 = hω

(N +

1

2

)|φ〉 =

(n+

1

2

)hω|φ〉 = En|φ〉, (2.31)

so the energy eigenvalues are En =(n+ 1

2

)hω with n = 0, 1, 2, . . . Note that the lowest energy is E0 = 1

2 hω, theso-called "zero-point" energy. In this approach it arises as a consequence of the non-commutability of the position andmomentim operators, and their assocated uncertainty ∆x∆p ≥ 1

2 h.Note also that the deternination of the energy eigenvalues has been done without solving the second-order differential

equation for the harmonic oscillator in the coordinate representation.If |ψE〉 is a state of energy E then

H|ψE〉 = E|ψE〉

and since[H, a±

]= ±hωa± then

(Ha± − a±H

)= ±hωa± and

Ha±|ψE〉 =(a±H ± hωa±

)|ψE〉

= (a±E ± hωa±) |ψE〉= (E ± hω) a±|ψE〉.

Thus a±|ψE〉 is an eigenstate of H with eigenvalue E ± hω, so a+ raise the value of E by hω and a− lowers E by hω.

2.3.3 Actions of operators a+ and a−

Assume that the eigenstates |n〉 of N , are normalised so 〈n|n〉 = 1. Then

N |n〉 = n|n〉; N |n+ 1〉 = (n+ 1) |n+ 1〉 (2.32)

But a+|n〉 is also an eigenstate of N with eigenvalue n+ 1, hence

N (a+|n〉) = (n+ 1) (a+|n〉) (2.33)

butN |n+ 1〉 = (n+ 1) |n+ 1〉 (2.34)

hence (a+|n〉) must be some multiple of |n+ 1〉, such as

a+|n〉 = cn|n+ 1〉. (2.35)

Normalisation gives, since a†+ = a−,

〈n|a−a+|n〉 = |cn|2 〈n+ 1|n+ 1〉 = |cn|2 (2.36)

But [a−, a+] = 1, so a−a+ = a+a− + 1 and

〈n|a+a− + 1|n〉 = |cn|2⟨n|N + 1|n

⟩= |cn|2

n+ 1 = |cn|2 (2.37)

socn =

√n+ 1

and

a+|n〉 =√n+ 1|n+ 1〉. (2.38)

Similarly for a− where a−|n〉 is an eigenstate of N with eigenvalue n− 1. Hence a−|n〉 is a multiple of |n− 1〉, as

a−|n〉 = cn|n− 1〉.

27


Then

a+a+|n〉 = cna+|n− 1〉 = cn√n|n〉

〈n|a+a+|n〉 = cn√n〈n|n〉 = cn

√n

〈n|N |n〉 = n = cn√n

so cn =√n and

a−|n〉 =√n|n− 1〉. (2.39)

If |0〉 is the lowest energy eigenstate thena+|0〉 = |1〉,

(a+)2 |0〉 = a+|1〉 =

√2|2〉,

(a+)3 |0〉 = a+

√2|2〉 =

√2√

3|3〉,(a+)

4 |0〉 = a+

√2√

3|3〉 =√

2√

3√

4|4〉

and by repeated action of a+

(a+)n |0〉 =

√n!|n〉 (2.40)

and hence the state|n〉 =

1√n!

(a+)n |0〉 (2.41)

generates all the eigenstates of N .

2.3.4 Spatial wavefunctionsThe eigenfunctions in the coordinate representation can be generated in a similar manner. The lowest eigenstate satisfies

a−|0〉 = 0,

or equivalently (1

2hmω

)1/2

(mωx+ ip) |0〉 = 0.

In the coordinate representation this is (mωx+ i

(−ih d

dx

))ψ0 (x) = 0 (2.42)

dψ0 (x)

dx+(mωh

)xψ0 (x) = 0. (2.43)

This equation has solutionψ0 (x) = Ae−(mω/2h)x2

(2.44)

satisfying the requirements that ψ0 (x)→ 0 as x→ ±∞. Normalisation requires∫ ∞−∞|A|2 e−(mω/h)x2

dx = 1

and gives A = (mω/hπ)1/4 and

ψ0 (x) =(mωhπ

)1/4

e−(mω/2h)x2

= |0〉. (2.45)

Then ψ1 (x) is obtained from ψ0 (x) using |1〉 = a+|0〉, i.e.

ψ1 (x) =

(1

2hmω

)1/2

(mωx− ip)ψ0 (x) (2.46)

=

(1

2hmω

)1/2(mωx− h d

dx

)ψ0 (x)

and similarly for ψ2 etc. Hence from |n〉 = 1√n!

(a+)n |0〉 comes

ψn (x) =1√n!

((1

2hmω

)1/2(mωx− h d

dx

))nψ0 (x) . (2.47)

28


2.3.5 Matrix representationsSince a+|n〉 =

√n+ 1|n+ 1〉 then taking the scalar product with another state (ket) |k〉 gives

〈k|a+|n〉 =√n+ 1〈k|n+ 1〉 =

√n+ 1δk,n+1 (2.48)

and from a−|n〉 =√n|n− 1〉 comes

〈k|a−|n〉 =√nδk,n−1. (2.49)

Thus the matrix representation of a+ is

a+ =

0 0 0 0 · · ·1 0 0 0 · · ·0√

2 0 0 · · ·0 0

√3 0 · · ·

...... 0

√4

. . .

.

From the definitions of a+ and a−,

a+ =

(1

2hmω

)1/2

(mωx− ip) , (2.50)

a− =

(1

2hmω

)1/2

(mωx+ ip) . (2.51)

adding gives the position operator in terms of a+ and a− as

x =

(h

2mω

)1/2

(a+ + a−) =1√2α

(a+ + a−) , (2.52)

and by subtracting, the momentum operator is

p = i

(mωh

2

)1/2

(a+ − a−) =ihα√

2(a+ − a−) . (2.53)

The matrix elements of x are

〈k|x|n〉 =

(h

2mω

)1/2 √n+ 1δk,n+1 +

√nδk,n−1

(2.54)

for k, n = 1, 2, . . .. Note that there are no diagonal elements, whereas in the basis states |n〉the Hamiltonian H isdiagonal with

H =

⟨

0|H|0⟩ ⟨

0|H|1⟩· · ·⟨

1|H|0⟩ ⟨

1|H|1⟩

......

. . .

= hω

12 0 · · ·0 3

2 · · ·...

.... . .

. (2.55)

To obtain matrices for say x2 there are several ways to proceed; by repeated action of x; by expanding x2 in terms of a+,and a−, i.e.

x|n〉 =

(h

2mω

)1/2

(a+ + a−) |n〉 =

(h

2mω

)1/2 (√n+ 1|n+ 1〉+

√n|n− 1〉

),

xx|n〉 =

(h

2mω

)1/2

(a+ + a−)

(h

2mω

)1/2 (√n+ 1|n+ 1〉+

√n|n− 1〉

),

x2|n〉 =

(h

2mω

)[a+

(√n+ 1|n+ 1〉+

√n|n− 1〉

)+ a−

(√n+ 1|n+ 1〉+

√n|n− 1〉

)],

=

(h

2mω

)[√n+ 1

√n+ 2|n+ 2〉+

√n√n|n〉+

√n+ 1

√n+ 1|n〉+

√n√n− 1|n− 2〉

],

x2|n〉 =

(h

2mω

)[√(n+ 2) (n+ 1)|n+ 2〉+ (2n+ 1) |n〉+

√n (n− 1)|n− 2〉

], (2.56)

〈k|x2|n〉 =

(h

2mω

)[√(n+ 2) (n+ 1)δk,n+2 + (2n+ 1) δkn +

√n (n− 1)δk,n−2

]. (2.57)

29


or using [a−, a+] = 1 = a−a+ − a+a−,

x =

(h

2mω

)1/2

(a+ + a−) ,

x2 =

(h

2mω

)1/2

(a+ + a−)

(h

2mω

)1/2

(a+ + a−) ,

=

(h

2mω

)(a+ + a−) (a+ + a−) =

(h

2mω

)(a2

+ + a+a− + a−a+ + a2−),

x2 =

(h

2mω

)(a2

+ + a+a− + a+a− + 1 + a2−)

=

(h

2mω

)(a2

+ + 2N + 1 + a2−

), (2.58)

〈k|x2|n〉 =

(h

2mω

)[√(n+ 2) (n+ 1)δk,n+2 + (2n+ 1) δkn +

√n (n− 1)δk,n−2

]. (2.59)

2.3.6 Minimum uncertainty for the harmonic oscillatorThe uncertainty in the x-position ∆x is given by

(∆x)2

=⟨

(x− 〈x〉)2⟩

=⟨x2⟩− 〈x〉2 .

In terms of the raising and lowering operators a+, a−, the x operator is

x =

(h

2mω

)1/2

(a+ + a−) (2.60)

so that

x|n〉 =

(h

2mω

)1/2

(a+ + a−) |n〉

=

(h

2mω

)1/2 (√n+ 1|n+ 1〉+

√n|n− 1〉

)and hence

〈x〉 = 〈n|x|n〉 = 0. (2.61)

Since x is an Hermitian operator and a†+ = a− and a†− = a+ then

〈n|x x|n〉 =

(h

2mω

)(〈n+ 1|

√n+ 1 + 〈n− 1|

√n) (√

n+ 1|n+ 1〉+√n|n− 1〉

)=

(h

2mω

)(n+ 1) 〈n+ 1|n+ 1〉+ n〈n− 1|n− 1〉 =

((2n+ 1) h

2mω

),

and

(∆x)2

=

(n+

1

2

)h

mω. (2.62)

In terms of a+, a−, the momentum operator p is

p = i

(mhω

2

)1/2

(a+ − a−) (2.63)

and hence

p|n〉 = i

(mhω

2

)1/2

(a+ − a−) |n〉

= i

(mhω

2

)1/2 (√n+ 1|n+ 1〉 −

√n|n− 1〉

)and hence

〈p〉 = 〈n|p|n〉 = 0. (2.64)

30


Also ⟨p2⟩

= 〈n|p p|n〉 =

(mhω

2

)(〈n+ 1|

√n+ 1− 〈n− 1|

√n) (√

n+ 1|n+ 1〉 −√n|n− 1〉

)=

(mhω

2

)(n+ 1 + n) = mhω

(n+

1

2

),

and

(∆p)2

=

(n+

1

2

)mhω. (2.65)

Thus the product

(∆x)2

(∆p)2

=

(n+

1

2

)h

mω

(n+

1

2

)mhω

=

(n+

1

2

)2

h2 (2.66)

and

(∆x) (∆p) =

(n+

1

2

)h. (2.67)

Thus in the lowest state n = 0, the product of uncertainties is (∆x) (∆p) = 12 h. The quantum oscillator satisfies the

lower bound of the generalised uncertainty relation.

2.3.7 Eigenstates of a−Suppose that |α〉 is an eigenstate of a− with eigenvalue α, then

a−|α〉 = α|α〉. (2.68)

Expanding |α〉 in terms of the complete set of states |n〉 as

|α〉 =∑n

cn|n〉.

Thena−|α〉 =

∑n

cna−|n〉 =∑n

cn√n|n− 1〉 = α

∑n

cn|n〉.

Taking the scalar product with the state |k〉 gives

〈k|a−|α〉 =∑n

cn√n〈k|n− 1〉 = α

∑n

cn〈k|n〉

=∑n

cn√nδk,n−1 = α

∑n

cnδkn. (2.69)

Thus for k = 0, 1, 2, . . . in turn,c1 = αc0;

√2c2 = αc1;

√3c3 = αc2; . . . (2.70)

Hence c1 = αc0, c2 = α2√

2c0; c3 = α3

√3√

2c0, etc. and in general

cn =αn√n!c0 (2.71)

so that

|α〉 = c0

∞∑n=0

αn√n!|n〉. (2.72)

Normalisation requires 〈α|α〉 = 1 so

1 = |c0|2∞∑n=0

(αn)∗

√n!

∞∑m=0

αm√m!〈n|m〉

= |c0|2∞∑n=0

|α|2n

n!= |c0|2 e|α|

2

.

31


Hence|c0| = e−|α|

2/2

giving

|α〉 = e−|α|2/2

∞∑n=0

αn√n!|n〉. (2.73)

The state |n〉 contains n quanta each of energy hω. Thus the probability that state |α〉 contains k quanta is (= |ck|2)

Pk = |〈k|α〉|2 =

∣∣∣∣∣e−|α|2/2∞∑n=0

αn√n!〈k|n〉

∣∣∣∣∣2

Pk =

∣∣∣∣e−|α|2/2 αk√k!

∣∣∣∣ =

(|α|2

)kk!

e−|α|2

. (2.74)

This is a Poisson distribution with a mean value |α|2. The average number of quanta in state |α〉 can be found from theexpectation value of the occupation number operator N in the state |α〉, i.e.

〈α|N |α〉 = 〈α |a+a−|α〉 = |α|2 . (2.75)

Alternatively using the probability Pk above, the average number of quanta is

n =

∞∑k=0

kPk =

∞∑k=0

k

(|α|2

)kk!

e−|α|2

=

∞∑k=1

e−|α|2

(|α|2

)k(k − 1)!

=

∞∑k=1

|α|2 e−|α|2

(|α|2

)k−1

(k − 1)!

= |α|2 e−|α|2∞∑k=0

(|α|2

)k(k)!

= |α|2 e−|α|2

e|α|2

= |α|2 .

2.4 Compatible observables and commuting operatorsTwo physical observables A, B are said to be compatible if a measurement of A, giving a value a, followed immediatelyby a measurement of B, giving value b, leaves the system in a state where an immediate remeasurement of A or B givesthe same values a or b again. Statement requires slight modification for a degenerate system. The requirement that thetwo measurements and remeasurement be carried out immediately ensures that between measurements the values of theobservables have not altered owing to the state of the system evolving according to the equation of motion.

Since a definite value an can be assigned to a system only if it is in an eigenstate φn of the operator A, and likewise avalue bn for operator B then φn is an eigenstate of both A and B, i.e. Aφn = anφn and Bφn = bnφn. The observablesA and B have simultaneous eigenfunctions. The two observables are compatible, otherwise they are incompatible. Itfollows that

BAφn = Banφn = bnanφn = anbnφn = ABφn,(AB − BA

)φn = 0. (2.76)

Since any function ψ can be expanded in terms of the complete set φn then(AB − BA

)ψ = 0 and the commutator[

A, B]

= 0, (2.77)

i.e.the two operators commute with each other. Thus compatibility implies commuting operators.The converse is true. If A and B are two commuting observables then a common set of eigenfunctions exits. If φn is

an eigenfunction of A with eigenvalue an then Aφn = anφn. Thus BAφn = anBφn But BA = AB so A(Bφn

)=

an

(Bφn

)and hence Bφn is an eigenfunction of A with eigenvalue an. If an is a non-degenerate value there is only one

eigenfunction having eigenvalue an, hence Bφn must be a multiple of φn, i.e. Bφn = bnφn and φn is an eigenfunctionof B with eigenvalue bn. Thus every eigenfunction of A is an eigenfunction of B.

The argument requires modification if the eigenvalue an is degenerate. Following a measurement of observable Athe system will have an wavefunction corresponding to one of the set of α-degenerate eigenfunctions φns, s = 1, 2, . . . , αfor eigenvalue an. In general this need not be an eigenfunction of observable B and the exact result of a measurementof B is not predictable even though A and B commute. Nevertheless a subsequent measurement of B will result in a

32


wavefunction being equivalent to one of the set of common eigenfunctions so that the results of further measurements ofA or B are completely predictable. If an is α-fold degenerate, φnk, k = 1, 2, . . . , α are α orthogonal eigenfunctions ofA with eigenvalue an. Any linear combination ψn =

∑αk=1 ckφnk is also an eigenfunction. If this is an eigenfunction of

B then Bψn = bnψn and

B

α∑k=1

ckφnk = bn

α∑k=1

ckφnk.

Taking the scalar product with φnm, m = 1, 2, . . . , α gives

α∑k=1

ck

∫φ∗nmBφnkdτ = bn

α∑k=1

ck

∫φ∗nmφnkdτ.

Writing∫φ∗nmBφnk = Bmk gives

α∑k=1

(Bmk − bnδmk) ck = 0. (2.78)

This set of α homogeneous linear equations in the unknowns c1, c2, . . . , cα possess a non-zero (non-trivial) solution ifdet |Bmk − bnδmk| = 0. This is a polynomial equation of degree α in bn having α roots. Corresponding to eachroot b(p)n where p = 1, 2, . . . , α there is a solution for ckp giving the corresponding eigenfunction of B which is also aneigenfunction of A.

The theorem can be extended to more than two observables. If A1, A2,. . .An are commuting observables whoseoperators satisfy [

Ar, As

]= 0 r, s = 1, 2, . . . , n, (2.79)

they possess a complete set of simultaneous eigenfunctions (and vice-versa). This is a powerful result as it is often easierto evaluate commutators, even in complicated situations, when there is no hope of obtaining explicit eigenfunctions.

To exploit this result some properties of commutators are be needed,

[A,B] = − [B,A] (2.80)[A,B + C] = [A,B] + [A,C] (2.81)

[A,BC] = [A,B]C +B [A,C] (2.82)[AB,C] = A [B,C] + [A,C]B (2.83)

[A, [B,C]] + [B, [C,A]] + [C, [A,B]] = 0. (2.84)

The order of the factors must be preserved. From [A,BC] = [A,B]C + B [A,C] it follows that A commutes with A2,and more generally with An and so if fn

(A)

=∑n cnA

n then[A, fn

(A)]

= 0. (2.85)

2.5 Heisenberg Uncertainty RelationsThe most famous pair of non-commuting observables are position and momentum in the same dimension, i.e.

[x, px] = ih. (2.86)

Recall that in the coordinate representation px = −i ∂∂x , so

[x, px]ψ (x) = −ihx∂ψ∂x

+ ih∂ (xψ)

∂x= ih.

For each component[x, px] = [y, py] = [z, pz] = ih (2.87)

but different Cartesian components commute, e.g.

[x, py] = [x, pz] = 0. (2.88)

The product of the uncertainties (∆x) (∆px) forms the Heisenberg Uncertainty Principle

(∆x) (∆px) ≥ 1

2h. (2.89)

33


This is an example of a general principle that the product of the uncertainties of two observables is related to the expecta-tion value of their commutator.

Suppose the state of the quantum system is ψ. In this state measurement of the observable A results in obtaining oneof its eigenvalues of operator A with probability |cn|2 where ψ =

∑n cnφn. The average value is 〈A〉 and the observed

values are scattered about this value. The observableA−〈A〉 will be scattered about zero, so a measure of the uncertaintyis

(∆A)2

=

⟨(A−

⟨A⟩)2

⟩, (2.90)

where⟨A⟩

=∫ψ∗Aψdτ . (This is just the standard deviation of A.) Similarly for another observable B,

(∆B)2

=

⟨(B −

⟨B⟩)2

⟩. (2.91)

Note that

(∆A)2

=

⟨(A−

⟨A⟩)2

⟩=⟨(A−

⟨A⟩)(

A−⟨A⟩)⟩

=

⟨A2 − 2A

⟨A⟩

+⟨A⟩2⟩

=⟨A2⟩− 2

⟨A⟩⟨

A⟩

+⟨A⟩2

(∆A)2

=⟨A2⟩−⟨A⟩2

. (2.92)

Define Hermitian operators α = A −⟨A⟩

, β = B −⟨B⟩

and their linear combination (which is non-Hermitian)

O = α+ iλβ where λ is a real parameter. Clearly ∫ ∣∣∣Oψ∣∣∣2 dτ ≥ 0

and is real and non-negative. Then ∫ (Oψ)∗ (

Oψ)

dτ ≥ 0,∫ [(α+ iλβ

)ψ]∗ [(

α+ iλβ)ψ]

dτ ≥ 0.

From the Hermiticity of α and β, ∫ψ∗(α− iλβ

)(α+ iλβ

)ψdτ ≥ 0.

But from definitions of α and β,(α− iλβ

)(α+ iλβ

)= α2 + λ2β2 + iλ

(αβ − βα

)= α2 + λ2β2 + iλ

[α, β

].

Also [α, β

]=[A−

⟨A⟩, B −

⟨B⟩]

=[A, B

]since

⟨A⟩

and⟨B⟩

are just numbers. Hence∫ψ∗(α2 + λ2β2 + iλ

[A, B

])ψdτ ≥ 0.

Thus the functionf (λ) =

⟨α2⟩

+ λ2⟨β2⟩

+ iλ⟨[A, B

]⟩≥ 0,

and implies that⟨[A, B

]⟩is purely imaginary. This function has a minimum value for

λ0 = −i⟨[A, B

]⟩2⟨β2⟩ .

34


The value of f (λ) at this minimum is

f (λ0) =⟨α2⟩−

⟨[A, B

]⟩2

4⟨β2⟩ +

⟨[A, B

]⟩2

2⟨β2⟩ =

⟨α2⟩

+

⟨[A, B

]⟩2

4⟨β2⟩ ≥ 0.

Thus ⟨α2⟩ ⟨β2⟩≥ −1

4

⟨[A, B

]⟩2

.

But⟨[A, B

]⟩is purely imaginary, say ⟨[

A, B]⟩

= iC

where C is real. Then ⟨(A−

⟨A⟩)2

⟩⟨(B −

⟨B⟩)2

⟩≥ −1

4(iC)

2=

1

4C2,

(∆A)2

(∆B)2 ≥ 1

4C2

(∆A) (∆B) ≥ 1

2C

i.e.(∆A) (∆B) ≥ 1

2

∣∣∣⟨[A, B]⟩∣∣∣ . (2.93)

This is the Generalised Uncertainty Principle.

For the position-momentum commutator [x, px] = ih so that (∆x) (∆px) ≥ 12 h.

35


2.5.1 Uncertainty relation for the harmonic oscillator

The uncertainties are (∆x)2

=⟨

(x− 〈x〉)2⟩

=⟨x2⟩− 〈x〉2 and (∆p)

2=⟨

(p− 〈p〉)2⟩

. In the ground state theeigenfunction is

ψ0 (x) =

(α√π

)1/2

e−α2x2/2 (2.94)

with α = (mω/h)1/2. Then clearly

〈x〉 =

(α√π

)∫ ∞−∞

xe−α2x2

dx = 0

as the integrand is an odd function of x. Then

⟨x2⟩

=

(α√π

)∫ ∞−∞

x2e−α2x2

dx =

(α√π

)1

2α2

√π

α2=

1

2α2,

using the standard integral∫∞−∞ x2 exp

(−ax2

)dx = 1

2a

√πa , hence

(∆x) =1√2α. (2.95)

Note comparing the probability distribution P (x) = |ψ0|2 = α√π

exp(−α2x2

)with the normal (Gaussian) distribution

1√2πσ

exp(−x2/2σ2) the variance σ is 12σ2 = α or σ = 1√

2α.

For the momentum operator,

〈px〉 = −ih α√π

∫ ∞−∞

e−α2x2/2

(d

dx

)(e−α

2x2/2)

dx = −ih α√π

∫ ∞−∞−α2xe−α

2x2

dx = 0,

as integrand is an odd function of x. To evaluate⟨p2x

⟩use integration by parts as

⟨p2x

⟩= −h2 α√

π

∫ ∞−∞

e−α2x2/2

(d2

dx2

)(e−α

2x2/2)

dx

= −αh2

√π

[e−α

2x2/2

(d

dx

)(e−α

2x2/2)]∞−∞−∫ ∞−∞

[(d

dx

)(e−α

2x2/2)]2

dx

=αh2

√π

∫ ∞−∞

[−α2xe−α

2x2/2]2

dx =αh2

√π

∫ ∞−∞

α4x2e−α2x2

dx =α5h2

√π

1

2α2

√π

α2=h2α2

2.

and(∆px) =

hα√2. (2.96)

Two useful sandard integrals are∫ ∞−∞

e−ax2

dx =

√π

a;

∫ ∞−∞

x2e−ax2

dx =1

2a

√π

a. (2.97)

The product of the uncertainties

(∆x) (∆px) =h

2. (2.98)

Thus the ground state of the harmonic oscillator minimises the uncertainty principle. In general for a harmonic oscillatorin energy state En =

(n+ 1

2

)hω, the product of the uncertainties is

(∆x) (∆px) =

(n+

1

2

)h. (2.99)

36

CHAPTER 3. GENERALISED ANGULAR MOMENTUM

Chapter 3

Generalised Angular Momentum

3.1 Orbital angular momentumOne set of non-commuting (and incompatible) operators which have already been met are the components of orbitalmomentum. In general the components of orbital angular momentum cannot be assigned definite values simultaneously(except for Lx = Ly = Lz = 0). A review of some of the properties of orbital angular momentum is given beforemoving on to its generalizations.

In classical mechanics the principle of conservation of angular momentum is a powerful tool in the solution ofrotational problems, e.g. motion of satellites, planets, gyroscopes, etc. The role of angular momentum in quantummechanics is probably even more important. This is due to the fact that the angular momentum properties of a system arelargely independent of the details of the forces acting. (The angular momentum properties are related to the rotationalsymmetry properties.) For example, for a particle moving in a central potential V (r) a part of the wavefunction is fullydetermined by its angular momentum properties, independent of the exact form of the potential. As a consequence, theangular momentum and hence qualitative features of the energy level schemes and spectroscopy of many-electron atomscan be understood without formally solving the Schrödinger equation.

Classically the orbital angular momentum L with respect to the origin of a particle at position r moving with momen-tum p is

L = r× p. (3.1)

This has Cartesian components

Lx = ypz − zpy (3.2)Ly = zpx − xpz (3.3)Lz = xpy − ypx. (3.4)

and square of the magnitude of the orbital angular momentum

L2 = L2x + L2

y + L2z. (3.5)

Using the fundamental position-momentum commutator relations

[x, px] = [y, py] = [z, pz] = ih

and commutator algebra leads to[Lx, Ly

]= ihLz;

[Ly, Lz

]= ihLx;

[Lz, Lx

]= ihLy. (3.6)

Proof: [Lx, Ly

]= [ypz − zpy, zpx − xpz]

= [ypz, zpx]− [zpy, zpx]− [ypz, xpz] + [zpy, xpz]

= ypx [pz, z]− 0− 0 + pyx [z, pz] = ypx (−ih) + pyx (ih)

= ih (xpy − ypx) = ihLz.

Relations for the other components may be obtained by cyclic permutation of the x, y, z as there can be nothing specialabout the choice of x, y, z. These commutation relations mean that two components of the angular momentum can not bemeasure simultaneously. This is very different from classical mechanics. The commutator relations can be expressed as

L× L = ihL. (3.7)

37

PHAS3226: Quantum Mechanics CHAPTER 3. GENERALISED ANGULAR MOMENTUM

The generalized uncertainty relation

(∆A) (∆B) ≥ 1

2|⟨[A, B

]⟩|

gives (∆Lx

)(∆Ly

)≥ 1

2|⟨[Lx, Ly

]⟩=

1

2

∣∣∣⟨ihLz⟩∣∣∣(∆Lx

)(∆Ly

)≥ 1

2h⟨Lz

⟩. (3.8)

So in general all three components can not be known except for Lx = Ly = Lz = 0.The components of L may not commute among themselves but each commutes with L2, as for example[

Lx, L2]

=[Lx, L

2x

]+[Lx, L

2y

]+[Lx, L

2z

]= 0 +

[Lx, Ly

]Ly + Ly

[Lx, Ly

]+[Lx, Lz

]Lz + Lz

[Lx, Lz

]= ihLzLy + ihLyLz − ihLyLz − ihLzLy[

Lx, L2]

= 0. (3.9)

Similarly for the other components so that[Lx, L

2]

=[Ly, L

2]

=[Lz, L

2]

= 0. (3.10)

Thus it is possible to construct simultaneous eigenfunctions of L2 and one component of L. This is by convention chosento be the z-component Lz (owing to its simple representation in spherical polar coordinates). Thus

L2ψ = aψ, (3.11)Lzψ = bψ. (3.12)

Since L2 = L2x + L2

y + L2z then

⟨L2⟩≥⟨L2z

⟩which implies that a ≥ b2.

The differential operators for the components of L are obtained from L = r× p using p = −ih∇ as L = −ihr×∇.In Cartesian form they are

Lx = −ih(y∂

∂z− z ∂

∂y

), (3.13)

Ly = −ih(z∂

∂x− x ∂

∂z

), (3.14)

Ly = −ih(z∂

∂x− x ∂

∂z

). (3.15)

However as orbital angular momentum is intimately related to rotations it is more convenient to express the operators inspherical polar coordinates (r, θ, φ) related to the Cartesian coordinates (x, y, z) by

x = r sin θ cosφ, (3.16)y = r sin θ sinφ, (3.17)z = r cos θ (3.18)

with 0 ≤ r ≤ ∞, 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π. A tedious bit of algebra relating the partial derivatives with respect to(x, y, z) to those with respect to (r, θ, φ) gives

Lx = ih

(sinφ

∂

∂θ+ cot θ cosφ

∂

∂φ

), (3.19)

Ly = ih

(− cosφ

∂

∂θ+ cot θ sinφ

∂

∂φ

), (3.20)

Lz = −ih ∂

∂φ, (3.21)

and

L2 = −h2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2 θ

∂2

∂2φ

]. (3.22)

38


Note as the interest is only in Lz and L2 the following procedure can be used. In spherical polar coordinates

∇ = r∂

∂r+θ

r

∂

∂θ+

φ

r sin θ

∂

∂φ

then L = −ihr×∇ becomes

L = −ih

(− θ

sin θ

∂

∂φ+ φ

∂

∂θ

).

Noting that z = cos θr− sin θθ then Lz = L · z = −ih ∂∂φ . For

L2 = −h2 (r×∇) · (r×∇)

= −h2r · [∇× (r×∇)]

= −h2r · ∇×

(− θ

sin θ

∂

∂φ+ φ

∂

∂θ

)

using the general relation a · (c× d) = c · (d× a). Evaluating the curl

∇×

(− θ

r sin θ

∂

∂φ+ φ

∂

∂θ

)=

1

r2 sin θ

∣∣∣∣∣∣∣r rθ r sin θφ∂∂r

∂∂θ

∂∂φ

0 − rsin θ

∂∂φ r sin θ ∂

∂φ

∣∣∣∣∣∣∣ .The interest is only in the r component so

L2 = −h2r1

r2 sin θ

[∂

∂θ

(r sin θ

∂

∂θ

)− ∂

∂φ

(−r

sin θ

∂

∂φ

)]L2 = −h2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2 θ

∂2

∂φ2

]. (3.23)

3.2 Eigenvalues and eigenfunctions of Lz and L2

The operators Lz and L2 have simultaneous eigenfunctions ψ (θ, φ) satisfying

L2ψ (θ, φ) = aψ (θ, φ) ,

L2zψ (θ, φ) = bψ (θ, φ) .

In polar coordinates

−ih∂ψ (θ, φ)

∂φ= bψ (θ, φ) .

This clearly has solution ψ (θ, φ) = Θ (θ) Φ (φ) = Θ (θ) eibφ/h. It is usual to impose the condition that ψ must be asingle-valued function. In this case that Φ (φ+ 2π) = Φ (φ). This requires eib(φ+2π)/h = eibφ/h i.e. ei2πb/h = ei2mπ

and so b = mh where m = 0,±1,±2, . . ..The single-valued requirement looks reasonable in view of the probability interpretation of |ψ (θ, φ)|2. However this

would only seem to require that |ψ (θ, φ)|2 is single valued. This would admit both integer and half-integer values of m.However in the present context the half-integer values of m can be ruled out because they lead to solutions for Θ (θ) forwhich the probability flux has a physically impossible behaviour (infinite at angles θ = 0 and π). Half-integer values ofm do arise in connection with eigenstates used to describe "spin" of elementary particles.

Normalization is achieved by ∫ 2π

0

Φ∗m (φ) Φm (φ) dφ = 1

so that

Φm (φ) =1√2π

eimφ. (3.24)

The full wavefunction ψ (θ, φ) = Θ (θ) Φ (φ) is such that it satisfies

L2ψ (θ, φ) = −h2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2 θ

∂2

∂φ2

]Θ (θ)

1√2π

eimφ = aΘ (θ)1√2π

eimφ

39


and carrying out the differentiation with respect to φ gives

1

sin θ

∂

∂θ

(sin θ

dΘ (θ)

dθ

)+

(a

h2 −m2

sin2 θ

)Θ (θ) = 0. (3.25)

Putting x = cos θ, ddθ = − sin θ d

dx = −(1− x2

)1/2 ddx , P (cos θ) = Θ (θ) gives

d

dx

[(1− x2

) dP|m|` (x)

dx

]+

[a

h2 −m2

(1− x2)

]P|m|` (x) = 0 (3.26)

which is the associated Legendre differential equation. Physically acceptable solutions which remain finite over the range0 ≤ θ ≤ π exist only if

a

h2 = ` (`+ 1)

where ` = 0, 1, 2, . . ., and m = −`,−` + 1, . . . ` − 1, `. That is ` is a positive integer or zero. The solutions are calledthe associated Legendre polynomials denoted by P |m|` (x). The functions Θ (θ) are normalized so that∫ π

0

Θ∗`′m (θ) Θ`m (θ) sin θdθ = δ`′`

and are given in terms of the Legendre polynomials P |m|` (x) as

Θ`m (θ) = (−1)m

[(2`+ 1) (`−m)!

2 (`+m)!

]1/2

P|m|` (x) , m ≥ 0, (3.27)

= (−1)|m|

Θ`|m| (θ) m < 0. (3.28)

The simultaneous eigenfunctions ψ (θ, φ) = Θ (θ) Φ (φ) are called the spherical harmonics Y`m (θ, φ) given by

Y`m (θ, φ) = (−1)m

[(2`+ 1) (`−m)!

4π (`+m)!

]1/2

P|m|` (x) eimφ m ≥ 0, (3.29)

Y`,−m (θ, φ) = (−1)mY ∗`m (θ, φ) m < 0. (3.30)

The spherical harmonics satisfy the orthonormality relation∫Y ∗`′m′ (θ, φ)Y`m (θ, φ) dΩ =

∫ 2π

0

dφ

∫ π

0

sin θY ∗`′m′ (θ, φ)Y`m (θ, φ) dθ = δ`′`δm′m. (3.31)

The first few spherical harmonics are

Y00 =1√4π,

(3.32)

Y1,±1 = ∓√

3

8πsin θe±imφ, (3.33)

Y10 =

√3

4πcos θ =

√3

4π

(zr

), (3.34)

Y11 + Y1−1 = 2

√3

8πsin θ cosφ = 2

√3

8π

(xr

). (3.35)

The basic eigenvalue equations for L2 and Lz are

L2Y`m = ` (`+ 1) h2Y`m, (3.36)Lz = mhY`m, (3.37)

with ` = 0, 1, 2, . . ., and m = −`,−`+ 1, . . . `− 1, ` . Thus the ` (`+ 1) h2 eigenvalue is (2`+ 1)-fold degenerate. Thequantities `, andm` are called the orbital angular momentum quantum number and the magnetic quantum number respec-tively. A particle is referred to to as being in a state with angular momentum ` - not as one of magnitude h

√` (`+ 1)!

For historical reasons the states with ` = 0, 1, 2, 3 are called s, p, d, and f states. The higher values of ` (= 4, 5, . . .) arereferred to in alphabetical order as g, h, etc. states.

40


3.3 Central potentialsThe Hamiltonian for a particle moving in a potential V (r) is

H = − h2

2m∇2 + V (r) .

For a central potential V depends only on the magnitude of r and is spherically symmetric giving

H = − h2

2m∇2 + V (r) .

In polar coordinates

H = − h2

2m

[1

r2

∂

∂r

(r2 ∂

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

r2 sin2 θ

∂2

∂φ2

]+ V (r) , (3.38)

= − h2

2m

[1

r2

∂

∂r

(r2 ∂

∂r

)− L2

h2r2

]+ V (r) ,

= − h2

2m

1

r2

∂

∂r

(r2 ∂

∂r

)+ V (r) +

L2

2mr2. (3.39)

Since L2 and Lz operate only on the angular coordinates (θ, φ), then[H, L2

]=[H, Lz

]= 0, (3.40)

and hence there are simultaneous eigenfunctions of the total energy E, orbital angular momentum and one component,

ψE`m (r, θ, φ) = RE` (r)Y`m (θ, φ) . (3.41)

The energy level corresponding to the same value of ` are (2`+ 1)-fold degenerate.

3.3.1 Review of hydrogen atomThe time-independent Schrödinger equation is(

− h2

2me∇2 − e2

4πε0r

)ψ (r, θ, φ) = Eψ (r, θ, φ) .

In atomic units where e = me = h = 1/4πε0 = 1 so

H = −1

2∇2 − 1

r,

= f (r) +L2

2r2

and thus

Hψ =

[− 1

2r2

d

dr

(r2 d

dr

)+L2

2r2− 1

r

]ψ = Eψ.

In general since V (r) is a central potential [H, L2

]=[H, Lz

]= 0

and hence there are simultaneous eigenfunctions of H , the orbital angular momentum L2 and one component. Thewavefunction can be written as

ψE`m (r, θ, φ) = RE` (r)Y`m (θ, φ)

and the energy eigenvalues

En`m ≡ En = − 1

2n2, n = 1, 2, 3, . . . .

The radial wavefunctions have the formRn` (r) = Pn` (r) e−r/n

where Pn` (r) is a polynomial of degree n − ` − 1. Note that 0 ≤ ` ≤ n − 1, and m = −`,−` + 1, . . . ` − 1, `. For agiven ` there are 2`+ 1 values of m all giving the same energy. Without a spin-orbit interaction all states of given n havethe same energy regardless of the ` value, i.e.are degenerate. The spin-orbit interaction removes some of this degeneracy.The spectroscopic notation is

` = 0 1 2 3 4 5 · · ·notation s p d f g h · · · .

41


3.4 Generalized angular momentumAn orbiting electron constitutes a current loop which by electromagnetic theory (Biot-Savart law for example) has asso-ciated with it a magnetic moment µL proportional to the electron’s angular momentum L. Thus a hydrogen atom in itsground state (` = 0) would not be expected to have a magnetic moment. However when a beam of such atoms is passedthrough an inhomogeneous magnetic field, the beam splits into two components (Stern-Gerlach type experiment). Thissuggests that the H-atom has indeed a magnetic moment which can have two quantized orientations to the direction ofthe magnetic field. Furthermore this magnetic moment µS is ascribed to an intrinsic property of the electron. Then byanalogy with orbital magnetic moment µL one deduces that the electron possesses an intrinsic angular momentum s inits own rest frame proportional to its magnetic moment µS . This angular momentum is called spin. This spin has twoprojections on the z-axis and this m = 1

2 or − 12 as 2` + 1 = 2 gives ` = 1

2 . The fine structure of spectral lines is alsoexplained in terms of the interaction between the magnetic moments associated with spin and the orbital angular momentaof the electron, via the spin-orbit coupling s · `.

Clearly spin is not an angular momentum of the familiar r × p kind. It has no classical analogue. To discuss it ageneralized definition of angular momentum is needed. Also clearly the eigenfunctions associated with the spin variableare not ordinary functions of position, x, and momentum, p, or time of the system. Thus one has to generalize the idea ofangular momentum in quantum mechanics. The Dirac notation offers a good way to do this.

Orbital angular momentum satisfies the commutator relations[Lx, Ly

]= ihLz

and with cyclic permutations of the components. The will be the starting position for a general definition of angular mo-mentum. As a definition J is an angular momentum if its components are Hermitian operators satisfying the commutationrelations [

Jx, Jy

]= ihJz (3.42)

and with its cyclic permutations., i.e. J× J = ihJ.

3.4.1 Raising and lowering operatorsDefine two operators J+ and J− by

J+ = Jx + iJy, (3.43)

J− = Jx − iJy. (3.44)

These operators are not Hermitian (but they are self-adjoint, or Hermitian conjugates) nor do they represent any physicallymeasurable quantity, but they are very useful in the mathematical developments. Note that⟨

φ|J+ψ⟩

=⟨J−φ|ψ

⟩=⟨ψ|J−φ

⟩∗. (3.45)

Commutator relations for J+ and J− Clearly [J2, J±

]= 0 (3.46)

as J2 commutes with Jx and Jy . The product

J+J− =(Jx + iJy

)(Jx − iJy

)= J2

x + J2y − i

[JxJy − JyJx

]J+J− = J2 − J2

z + hJz (3.47)

SimilarlyJ−J+ = J2 − J2

z − hJz. (3.48)

Subtracting gives (J+J− − J−J+

)=[J+, J−

]= 2hJz.

and in [J±, J∓

]= ±2hJz. (3.49)

42


Finally [Jz, J±

]=

[Jz, Jx ± iJy

]=[Jz, Jx

]± i[Jz, Jy

]= ihJy ± i

(−ihJx

)= ihJy ± hJx = ±h

(Jx ± iJy

)[Jz, J±

]= ±hJ±. (3.50)

3.5 General angular momentum eigenvalue problem

Let the eigenvalues of J2 and Jz be α and β respectively and the corresponding simultaneous eigenfunction be φ. Thus

J2|φ〉 = α|φ〉, (3.51)Jz|φ〉 = β|φ〉. (3.52)

The eigenvalues of J2 and Jz are real as J2 and Jz are Hermitian operators. Moreover the eigenvalues of J2 must bepositive or zero because the expectation value of the square of an Hermitian operator must be positive or zero.

Note⟨ψ∣∣∣J2x

∣∣∣ψ⟩ ≥ 0. Choose |χ〉 = Jx|ψ〉 then⟨ψ∣∣∣Jx∣∣∣χ⟩ =

⟨χ∣∣∣Jx∣∣∣ψ⟩∗ = 〈χ|χ〉∗ = 〈χ|χ〉 ≥ 0, as Jx is

Hermitian. Therefore⟨J2x

⟩,⟨J2y

⟩,⟨J2z

⟩are all ≥ 0. But

⟨J2⟩

=⟨J2x

⟩+⟨J2y

⟩+⟨J2z

⟩≥ 0 and so

⟨J2⟩≥⟨J2z

⟩and hence α ≥ β2.

Operating with J+

J+Jz|φ〉 = βJ+|φ〉.

Using[Jz, J+

]= hJ+ i.e. JzJ+ − J+Jz = hJ+(

JzJ+ − hJ+

)|φ〉 = βJ+|φ〉

Jz

(J+|φ〉

)= (β + h)

(J+|φ〉

). (3.53)

Similarly operating with J− givesJz

(J−|φ〉

)= (β − h)

(J−|φ〉

). (3.54)

Thus(J+|φ〉

)is an eigenfunction of Jz with eigenvalue (β + h) and

(J−|φ〉

)is an eigenfunction of Jz with eigenvalue

(β − h). Thus J+ and J− raise or lower respectively the eigenvalues of Jz (They are called step-up, step-down or raising,lowering, or ladder operators.)

Also from J2|φ〉 = α|φ and using[J2, J±

]= 0

J±J2|φ〉 = αJ±|φ

J2(J±|φ〉

)= α

(J±|φ

)(3.55)

so J±|φ〉 are also eigenfunctions of J2 with eigenvalue α.Hence by repeatedly operating with J+ a sequence of eigenstates of Jz can be constructed with eigenvalues β + h,

β + 2h, . . . β + ph, each eigenstate being an eigenfunction of J2 with eigenvalue α. Similarly repeated application ofJ− produces a sequence of eigenfunctions of Jz with eigenvalues β − h, β − 2h, . . . β − qh. Since α ≥ (β + ph)

2, andα ≥ (β − qh)

2 these sequences must terminate. There will be a maximum value of β + ph, say mT with eigenfunction|φT 〉 such that

J+|φT 〉 = 0 (3.56)

andJz|φT 〉 = mT |φT 〉. (3.57)

Similarly there is a minimum value of β − qh = mB with eigenfunction |φB〉 such that

J−|φB〉 = 0, (3.58)Jz|φB〉 = mB |φT 〉 (3.59)

Furthermore as J+ and J− raise and lower the eigenvalues in steps of h, then

mT −mB = nh (3.60)

43


where n = 0, 1, 2, . . ..Applying J− to J+|φT 〉 = 0 then J−J+|φT 〉 = 0. But J−J+ = J2 − J2

z − hJz so(J2 − J2

z − hJz)|φT 〉 = 0. (3.61)

Thus evaluating using equ(3.51, 3.59) (α−m2

T −mT h)|φT 〉 = 0

andα = mT (mT + h) (3.62)

Similarly from J+J−|φB〉 = 0, but J+J− = J2 − J2z + hJz and so(

α−m2B +mBh

)|φT 〉 = 0

andα = mB (mB − h) (3.63)

Equations (3.62, 3.63) givemT (mT + h) = mB (mB − h) ,

m2T + (mT +mB) h−m2

B = 0.

This equation has two solutions

mT = −mB ,

mT = mB − h.

The second one violates mT ≥ mB thus the only valid solution is mT = −mB and so mT ≥ 0. But mT −mB = nh so

mT = nh

2, n = 0, 1, 2, . . . . (3.64)

Putting n/2 = j means mT = −mB = jh and

α = j (j + 1) h2 (3.65)

with j = 0, 12 , 1,

32 , . . . , i.e. j is an integer or half-integer and

β = mh

with m ranging between −j and j in integer steps., i.e. m has 2j + 1 values for a given j value.The general definition of angular momentum based on the commutator relations has led to a positive integer (or zero)

and half-integer values for j (and form). The half odd-integer values for j are excluded for orbital angular momentum forwhich j = 0, 1, 2, . . .. There is overwhelming evidence that electrons, protons, neutrons and many more particles possessan intrinsic angular momentum, called spin, of 1

2 . For pions s = 0, for photons s = 1, for the Ω− particle s = 32 . Note

properties of electron spin can be derived from Dirac’s relativistic formulation of quantum mechanics.

3.5.1 Actions of the J+ and J−

The state |j,m〉 is an eigenstate of Jz with eigenvalue mh so

Jz|j,m〉 = mh|j,m〉

so

J+Jz|j,m〉 = mhJ+|j,m〉(JzJ+ − hJ+

)|j,m〉 = mh

(J+|j,m〉

)Jz

(J+|j,m〉

)= (m+ 1) h

(J+|j,m〉

), (3.66)

so the state(J+|j,m〉

)is an eigenstate of Jz with eigenvalue (m+ 1) h. But also

Jz|j,m+ 1〉 = (m+ 1) h|j,m+ 1〉 (3.67)

44


46

CHAPTER 4. SPIN 1/2 SYSTEMS

Chapter 4

Spin 1/2 Systems

The theory of the last section provides the formalism for a discussion of the spin angular momentum of a particle. Theusual convention is that s denotes the spin angular momentum number, S the spin angular momentum operator. Thecorresponding quantities for orbital angular momentum are ` and L. The discussion will be confined to spin- 1

2 particles -an important case as electrons and protons have spin- 1

2 . For such particles the only possible values of m (denoted now byms to distinguish it from the orbital magnetic quantum number) are 1

2 h and− 12 h. These are often referred to as "spin-up"

and "spin-down" states (with respect to quantization axis). Since |s| =√s (s+ 1)h =

√34 h and sz = 1

2 h the classicalvector for s makes an angle of 54.7 to the z-axis!

The eigenstates of Sz are denoted by | 12 ,12 〉 ≡ |α〉 for ms = 1

2 h, and | 12 ,−12 〉 ≡ |β〉 for ms = − 1

2 h. Then

S2|α〉 = s (s+ 1) h2|α〉 =3

4h2|α〉, (4.1)

S2|β〉 =3

4h2|β〉, (4.2)

Sz|α〉 =1

2h|α〉, (4.3)

Sz|β〉 = −1

2h|β〉. (4.4)

Since |α〉 and |β〉 belong to different eigenvalues of Sz they are orthogonal, 〈α|β〉 = 〈β|α〉 = 0 and are assumed to benormalized, 〈α|α〉 = 〈β|β〉 = 1.

The angular momentum commutation relations are[Sx, Sy

]= ihSz (4.5)

with cyclic permutations. The action of the raising and lowering operators S+ = Sx + iSy and S− = Sx − iSy are, ingeneral,

S±|s,ms〉 = [s (s+ 1)−ms (ms ± 1)]1/2

h|s,ms ± 1〉, (4.6)

S±|1

2,ms〉 =

[3

4−ms (ms ± 1)

]1/2

h|12,ms ± 1〉. (4.7)

In particular

S+|α〉 = 0; S+|β〉 = h|α〉, (4.8)S−|α〉 = h|β〉; S−|β〉 = 0. (4.9)

SinceSx =

1

2

(S+ + S−

); Sy =

1

2i

(S+ − S−

)(4.10)

this leads to the setSx|α〉 = h

2 |β〉 Sx|β〉 = h2 |α〉

Sy|α〉 = i h2 |β〉 Sy|β〉 = −i h2 |α〉Sz|α〉 = h

2 |α〉 Sz|β〉 = − h2 |β〉. (4.11)

An arbitrary spin- 12 state |χ〉 can be written as a linear superposition of states |α〉 and |β〉 as

|χ〉 = a|α〉+ b|β〉. (4.12)

47

PHAS3226: Quantum Mechanics CHAPTER 4. SPIN 1/2 SYSTEMS

Taking the scalar product with |α〉 and |β〉 in turn gives

a = 〈α|χ〉; b = 〈β|χ〉,

so|χ〉 = |α〉〈α|χ〉+ |β〉〈β|χ〉 (4.13)

which is just an example of the general expansion in eigenstates. If |χ〉 is normalized as 〈χ|χ〉 = 1 then

1 = 〈χ|α〉〈α|χ〉+ 〈χ|β〉〈β|χ〉1 = |a|2 + |b|2 . (4.14)

The expectation value of Sz in the state |χ〉 from

Sz|χ〉 = Sz|α〉〈α|χ〉+ Sz|β〉〈β|χ〉

=h

2|α〉〈α|χ〉 − h

2|β〉〈β|χ〉

〈χ|Sz|χ〉 =h

2|〈α|χ〉|2 − h

2|〈β|χ〉|2 ≡ h

2|a|2 − h

2|b|2 , (4.15)

as would be expected since |a|2 is the probability of finding the particle in state |χ〉 with spin-up and |b|2 the probabilityof spin-down.

4.1 Useful relations for spin

The spin- 12 operators have some other useful properties. One is that S2 is a purely numerical operator since

S2|χ〉 = s (s+ 1) h2|χ〉 =3

4h2|χ〉 (4.16)

for any state |χ〉, so

S2 ≡ 3

4h2.

For spin- 12 only:

S2x|α〉 =

h

2Sx|β〉 =

h2

4|α〉,

S2x|β〉 =

h

2Sx|α〉 =

h2

4|β〉

so

S2x|χ〉 =

h2

4|χ〉. (4.17)

Similar results hold for S2y and S2

z , hence

S2x = S2

y = S2z =

h2

4. (4.18)

Since

S+|α〉 = 0; S+|β〉 = h|α〉,S2

+|α〉 = 0; S2+|β〉 = 0

then for an arbitrary state |χ〉S2

+|χ〉 = 0. (4.19)

Similarly for S−S2−|χ〉 = 0 (4.20)

and thus

0 =(Sx ± iSy

)(Sx ± iSy

),

0 = S2x − S2

y ± i(SxSy + SySx

). (4.21)

48


It follows thatSxSy + SySx = 0 (4.22)

and that Sx and Sy are said to anticommute, Sx, Sy

= SxSy + SyS = 0. (4.23)

Similar anticommutation relations exist between the other components so,Sx, Sy

=Sy, Sz

=Sz, Sx

= 0. (4.24)

Combining the anticommutation relations above with the commutation relations, e.g.

SxSy + SySx = 0

andSxSy − SySx = ihSz

gives

SxSy = ih

2Sz. (4.25)

A similar relation holds for a cyclic permutation of the indices x, y, z so that in general

SxSy = ih

2Sz (4.26)

with cyclic permutation of (x, y, z).This relation is useful because when combined with S2

x = S2y = S2

z = h2

4 it shows that any arbitrary product of spin- 12

operators can be reduced to a spin independent term or a term linear in Sx, Sy, Sz , i.e. the most general spin- 12 operator

isA = A0 + B · S =A0 + BxSx + BySy + BzSz. (4.27)

4.2 Representation of spin 1/2 operators and eigenfunctions - Pauli matricesSo far the discussion of spin has made use solely of the spin operators and their eigenstates |α〉 and |β〉. For somecalculations it is useful to have an explicit representation for the operators and eigenfunctions. It should be obvious fromthe earlier discussion that they cannot be represented as functions of the particle’s position and momentum as in the caseof orbital angular momentum. They can, however, be very conveniently handled by representing the spin operators by2× 2 matrices and the spin states by column vectors.

4.2.1 Matrix representationsSince S2|s,ms〉 = s (s+ 1) h2|s,ms〉 then 〈s′,m′s|S2|s,ms〉 = s (s+ 1) h2δs′sδm′sms

and the matrix representation ofS2 is diagonal with

S2 =3

4h2

(1 00 1

). (4.28)

Since Sz|s,ms〉 = msh|s,ms〉 then 〈s,m′s|Sz|s,ms〉 = mshδm′sms and

Sz =1

2h

(1 00 −1

)=

1

2hσz (4.29)

where σz is a Pauli matrix.The matrix representations of Sx and Sy can be found using

Sx =1

2

(S+ + S−

); Sy =

1

2i

(S+ − S−

)and

S±|s,ms〉 = [s (s+ 1)−ms (ms ± 1)]1/2

h|s,ms ± 1〉.

Hence

Sx|s,ms〉 =1

2[s (s+ 1)−ms (ms + 1)]

1/2h|s,ms + 1〉+ [s (s+ 1)−ms (ms − 1)]

1/2h|s,ms − 1〉. (4.30)

49


Explicity Sx| 12 ,12 〉 = h

2 |12 ,−

12 〉 and Sx| 12 ,−

12 〉 = h

2 |12 ,

12 〉 and so

Sx ≡(〈 12 ,

12 |Sx|

12 ,

12 〉〈 12 ,

12 |Sx|

12 ,−

12 〉

〈 12 ,−12 |Sx|

12 ,

12 〉〈

12 ,−

12 |Sx|

12 ,−

12 〉

)=

(〈α|Sx|α〉〈α|Sx|β〉〈β|Sx|α〉〈β|Sx|β〉

), (4.31)

Sx =h

2

(0 11 0

)=h

2σx. (4.32)

By similar steps

Sy =h

2

(0 −ii 0

)=h

2σy. (4.33)

The three Pauli matrices

σx =

(0 11 0

); σy =

(0 −ii 0

); σz =

(1 00 −1

); (4.34)

and the identity

I =

(1 00 1

)(4.35)

form a complete set in terms of which any 2× 2 matrix may be expanded, possibly with complex coefficients.

4.2.2 Commutators for the Pauli matricesSince Pauli matrices are defined by

S =h

2σ (4.36)

and [Sx, Sy

]= ihSz

then[σx, σy] = 2iσz (4.37)

together with the cyclic permutations of the indices.Also since SxSy = i h2 Sz then

σxσy = iσz. (4.38)

Note also that the traces of all the Pauli matrices vanish,

Trσx = Trσy = Trσz = 0 (4.39)

anddetσx = detσy = detσz = −1. (4.40)

4.2.3 Basis statesThe eigenfunctions |α〉 and |β〉 are represented by the column vectors

|α〉 =

(10

); |β〉 =

(01

). (4.41)

Thus the operator equation

Sz|α〉 =1

2h|α〉

becomes1

2h

(1 00 −1

)(10

)=

1

2h

(10

). (4.42)

The arbitrary spin state |χ〉 = a|α〉+ b|β〉 becomes

|χ〉 = a

(10

)+ b

(01

)=

(ab

). (4.43)

If |η〉 = c|α〉+ d|β〉 then〈η|χ〉 = c∗a+ d∗b. (4.44)

50


This expression can be obtained if the bra 〈η| is represented by a two-component row vector

〈η| = (c∗, d∗) (4.45)

and so

〈η|χ〉 = (c∗, d∗)

(ab

)= c∗a+ d∗b

by the rules of matrix multiplication.As an example consider the expectation value of Sz in the state |χ〉;

〈χ|Ss|χ〉 =h

2(a∗, b∗)

(1 00 −1

)(ab

)=h

2(a∗a− b∗b) =

h

2

(|a|2 − |b|2

)(4.46)

as before.

4.2.4 Determination of eigenvalues and eigenvectorsTo solve, say, Sx|χ〉 = a|χ〉 in the matrix representation. Write a = h

2λ and Sx = h2 σx and express

|χ〉 =

(uv

), (4.47)

so thath

2

(0 11 0

)(uv

)= λ

h

2

(uv

),

then (−λ 11 −λ

)(uv

)= 0. (4.48)

This equation has a non-trivial solution if

det

∣∣∣∣ −λ 11 −λ

∣∣∣∣ = 0 (4.49)

i.e.λ2 − 1 = 0

and hence λ = ±1.For λ = +1,

h

2

(0 11 0

)(uv

)=h

2

(uv

)gives

v = u

u = v.

Normalization of the eigenvector |χ〉 as 1 = 〈χ|χ〉 = |u|2 + |v|2 gives u = v =√

12 . The eigenvalue is thus h

2 and theeigenvector is

|χ+〉 =1√2

(11

)=

1√2

(|α〉+ |β〉) . (4.50)

For λ = −1 one gets u = −v and eigenvector

|χ−〉 =1√2

(1−1

)=

1√2

(|α〉 − |β〉) . (4.51)

In a similar way the eigenvalues and eigenvectors of Sy and Sz can be found. These are summarized below:

eigenvalue eigenstate matrix form

Sx12 h |χ+〉 = |+〉x = 1√

2(|α〉+ |β〉) 1√

2

(11

)− 1

2 h |χ−〉 = |−〉x = 1√2

(|α〉 − |β〉) 1√2

(1−1

)Sy

12 h |+〉y = 1√

2(|α〉+ i|β〉) 1√

2

(1i

)− 1

2 h |−〉y = 1√2

(|α〉 − i|β〉) 1√2

(1−i

)Sz

12 h |+〉z = |α〉

(10

)− 1

2 h |−〉z = |β〉(

01

)

(4.52)

51


The eigenvectors can also be found via the basis states |α〉 and |β〉 Express |χ〉 = a|α〉+ b|β〉 then

Sx|χ〉 = a|χ〉 =1

2hλ|χ〉

1

2

(S+ + S−

)(a|α〉+ b|β) =

1

2hλ|χ〉

bh

2|α〉+ a

h

2|β〉 =

1

2hλ (a|α〉+ b|β〉) (4.53)

Taking the scalar product with |α〉 and |β〉, or noting that |α〉 and |β〉 are linearly independent so that coefficients on eachside may be equated, gives

bh

2=

1

2hλa

ah

2=

1

2hλb.

Thus λ2 = 1, a2 = b2 , λ = ±1. For λ = +1, a = b and normalization gives a = b = 1√2

as before.

4.2.5 Spin along an arbitrary directionThe component of spin S along a direction n is Sn = S · n. Taking the direction

n = (cosφ sin θ, sinφ sin θ, cos θ)

then

Sn =h

2(σx cosφ sin θ + σy sinφ sin θ + σz cos θ)

=h

2

(0 11 0

)cosφ sin θ +

(0 −ii 0

)sinφ sin θ +

(1 00 −1

)cos θ

=

h

2

(cos θ cosφ sin θ − i sinφ sin θ

cosφ sin θ + i sinφ sin θ − cos θ

)Sn =

h

2

(cos θ sin θe−iφ

sin θeiφ − cos θ

). (4.54)

The eigenvalue equation is

Sn|χ〉 =h

2λ|χ〉

with |χ〉 = a|α〉+ b|β〉, or in matrix form(cos θ sin θe−iφ

sin θeiφ − cos θ

)(ab

)= λ

(ab

). (4.55)

Hence for a non-trivial solution ∣∣∣∣ cos θ − λ sin θe−iφ

sin θeiφ − cos θ − λ

∣∣∣∣ = 0

− (cos θ − λ) (cos θ + λ)− sin2 θ = 0

λ2 = 1,

λ = ±1. (4.56)

Thus the eigenvalues are ±h/2 for a spin- 12 system regardless of the axis chosen. To find the eigenvectors, for λ = +1

a cos θ + b sin θe−iφ = a

b2 sinθ

2cos

θ

2e−iφ = a (1− cos θ) = 2a sin2 θ

2

sob

a=

sin θ2eiφ

cos θ2. (4.57)

Choose for λ = +1,

|+〉n =

(cos θ2

sin θ2eiφ

),

52


and for λ = −1,

|−〉n =

(sin θ

2

− cos θ2eiφ

); or |−〉n =

(− sin θ

2

cos θ2eiφ

). (4.58)

In state |+〉n the probability of "spin-up" (along +z) is P+ = cos2 θ2 ; probability of "spin-down" (along -z) is P− =

sin2 θ2 .

4.3 Space-spin wavefunctionsThe wavefunction of a spin- 1

2 particle also has a spatial dependence. The complete wavefunction is

Ψ =

ms=1/2∑ms=−1/2

cmsψms

(r)χms, (4.59)

which for a spin- 12 particle can be written a

Ψ (r, s) = c 12ψ 1

2(r) |α〉+ c− 1

2ψ− 1

2(r) |β〉, (4.60)

Ψ (r, s) =

(c 1

2ψ 1

2(r)

c− 12ψ− 1

2(r)

). (4.61)

Thus a spin- 12 particle is described by a two-component wavefunction (a spinor). The probability that the particle will be

found in a volume element dτ with "spin up"∣∣∣c 1

2ψ 1

2(r)∣∣∣2, and the probability that it is found anywhere with "spin-up" is∣∣∣c 1

2

∣∣∣2provided ψ 12

(r) is normalized, as∫ ∣∣∣ψ 1

2(r)∣∣∣2 dτ = 1. The expectation value

⟨Ψ|Sz|Ψ

⟩=h

2

(∣∣∣c 12

∣∣∣2 − ∣∣∣c− 12

∣∣∣2) (4.62)

Suppose the electron has spin fully aligned along the z-axis, i.e. it’s spin state is |α〉 and it is moving along the x-axis ina particle free region. Then H = − h2

2md2

dx2 and the wavefunction is

ψ (x) = Aeikx|α〉 = Aeikx(

10

)=

(ψ+ (x)ψ− (x)

)(4.63)

with ψ+ (x) = Aeikx and ψ− (x) = 0.

4.3.1 Stern-Gerlach experimentThis is one of the famous experiments that demonstrated the existence of electron spin (1922, Nobel prize 1943). Orig-inally performed with Ag atoms which have an unpaired electron in the outer shell. The beam of Ag atoms is directedalong the x-axis and passes between the poles of a magnet with a very non-uniform field along the z-axis. The potentialenergy is

V = −µ ·B = −(−gs

µBh

s ·B)

= gsµBh

s ·B. (4.64)

and the force is

Fz = −dV

dz= −gs

µBh

s · dB

dzk =− gs

µBhSz

dB

dz. (4.65)

Depending on the sign of Sz (+h/2 or−h/2) the force is upwards or downwards. Hence two "spots" are seen on a screen.Classically s would be orientated at random, so all points on the screen between the two spot limits would be filled in.

Suppose that the atoms are initially polarized with ms = +h/2, i.e. σz = +1. The incoming wavefunction is

ψ↑in (r, t) = f (r− vt) eikx|α〉 (4.66)

where f is an "envelope" function describing the shape of the wave packet - the spatial part of the wave packet at sometime t. Particles are located within the envelope whose maximum advances at speed v = hk/m. When the atoms emergefrom the magnet they will have been deflected by an amount ∆r = ∆zk and the outgoing wavefunction is

ψ↑out (r, t) = f(r− vt+ ∆zk

)eikx|α〉. (4.67)

53


Similarly the "spin-down" atoms have σz = −1 and

ψ↓in (r, t) = f (r− vt) eikx|β〉, (4.68)

ψ↓out (r, t) = f(r− vt−∆zk

)eikx|β〉. (4.69)

Suppose that initially the atoms were fully polarized along the +x-axis, so then

|+〉x =1√2

(|α〉+ |β〉) .

Then before the magnet (ψ↑in (r, t)

ψ↓in (r, t)

)=

1√2

eikx(f (r− vt)f (r− vt)

). (4.70)

After the magnet induce deflections

(ψ↑out (r, t)

ψ↓out (r, t)

)=

1√2

eikx

f(r− vt+ ∆zk

)f(r− vt−∆zk

) . (4.71)

The intensity of each beam will be equal. If the lower beam (state |β〉) is taken through a second magnet withB = Bx (x)the atoms will be found with a 50:50 mix along the +x-axis and the −x-axis since the lower z-beam was in a state|β〉 = 1√

2(|+〉x + |−〉x).

4.4 Addition of angular momentumFor an atom in free space, its total angular momentum is conserved (i.e. it commutes with the Hamiltonian). This meansthat the stationary states (energy eigenstates) of the atom can be chosen so that the square of total angular momentum andthe z-component of angular momentum are precisely defined. But the total angular momentum of an atom is generallythe sum of a number of individual angular momenta. Each electron in the atom has an orbital angular momentum as wellas an intrinsic spin angular momentum. The vector sum of all these angular momenta is the total angular momentum ofthe electrons. In addition, the atomic nucleus generally has an intrinsic angular momentum, which must also be includedto obtain the total angular momentum of the atom. So in order to understand atoms, we need to know how to add angularmomenta. The question is: if two (quantised) angular momenta are add them together, what are the possible values ofthe j and m quantum numbers of the resulting total angular momentum, and how are the eigenvectors of total angularmomentum expressed in terms of the eigenvectors of the individual angular momenta? The general theory of additionof angular momenta is rather complicated, and is presented in the 4th-year quantum mechanics course. Here, only twoimportant special cases are studied: the addition of two spin- 1

2 angular momenta, and the addition of a spin- 12 angular

momentum and an orbital angular momentum. The first case is relevant to the addition of the spin of two electrons inan atom (e.g. the He atom), and the second case is relevant to the addition of the orbital and spin angular momenta of anelectron.

4.4.1 Addition of two spin-1/2 angular momenta

The simplest, but important example, is the addition of the spins of two spin- 12 particles, e.g. two electrons. The vector

operators representing the spin angular momenta of these two particles are denoted by S1 and S2. Then the total spinangular momentum S is the vector sum of S1 and S2:

S = S1 + S2 . (4.72)

Since S is an angular momentum, it is quantized: its square can only have values S(S + 1)h2, with S an integer or ahalf-odd-integer; its z-component Sz can only have the values msh, with ms = −S,−S + 1, . . . S − 1, S.

Let the two eigenvectors of s1z be called α1 and β1 then

S1zα1 =1

2h α1 , S1zβ1 = −1

2h β1 . (4.73)

These are also eigenstates of S21 :

S21α1 =

3

4h2α1 , S2

1β1 =3

4h2β1 . (4.74)

54


Similarly, for particle 2, denote the eigenvectors of S2z by α2, β2:

S2zα2 =1

2h α2 , S2zβ2 = −1

2h β2

S22α2 =

3

4h2α2 , S2

2β2 =3

4h2β2 . (4.75)

Note that the state vectors α1 and β1 on the one hand, and α2 and β2 on the other hand, inhabit completely differentspaces. Linear combinations of α1 and β1, e.g. 1√

2(α1 + β1), can be constructed, describing different quantum states of

particle 1. Similarly, linear combinations of α2 and β2 describe states of particle 2. But α1 + α2 has no meaning.All possible states of the system of two spin- 1

2 particles can be represented as linear combinations of the followingfour basic states:

α1α2, α1β2, β1α2, β1β2. (4.76)

These are eigenstates of both S1z and S2z , for example,

S1zα1α2 =1

2h α1α2, S2zα1α2 =

1

2h α1α2, (4.77)

so that in the state α1α2, S1z has the precisely defined value 12 h, and S2z also has the precisely defined value 1

2 h. Similarlyin the state α1β2, S1z and S2z have the values 1

2 h and− 12 h respectively; in state β1α2 they have values− 1

2 h and 12 h; and

in state β1β2, they have values − 12 h and − 1

2 h. (Note that α1α2 does not mean α1 multiplied by α2 in the normal senseof the word ‘multiplied’. It is just a notation for the state of the system in which S1z and S2z have the values 1

2 h and 12 h.)

The total spin isS = S1 + S2 (4.78)

andSz = S1z + S2z. (4.79)

Denote the general eigenstate by |S,M〉. There are four possible product combinations α1α2, α1β2, β1α2 and β1β2

which are simultaneous eigenvectors of Sz with eigenvalues (M = ms1 +ms2) of 1, 0, 0,−1 respectively. The stateα1α2 with M = 1 must belong to the state S = 1 as

Sz|1, 1〉 =(S1z + S2z

)α1α2 =

h

2α1α2 +

h

2α1α2 = hα1α2. (4.80)

There are two M = 0 states, α1β2, β1α2. From these can be formed two linearly independent combinations. Onecorresponds to the state |1, 0〉 and the other to state |0, 0〉. These can be found by using the lowering operators S− =S1− + S2− and

S−|1, 1〉 =(S1− + S2−

)α1α2 = hβ1α2 + hα1β2 = h (β1α2 + α1β2) . (4.81)

But in general S−|s,ms〉 = [s (s+ 1)−ms (ms − 1)]1/2

h|s,ms − 1〉 so

S−|1, 1〉 =√

2h|1, 0〉

and hence|1, 0〉 =

1√2

(α1β2 + β1α2) (4.82)

Note that S− is symmetric in the label 1, 2 of the two electrons. Hence operating on α1α2 which is also symmetric underthe interchange of labels 1 and 2 must produce a symmetric state. Note also that |1, 0〉 is already normalized. The statewith S = 1 and M = −1 is clearly β1β2. It could be found by operating with S− on |1, 0〉.

The state |0, 0〉 with S = 0, M = 0 must also be a linear combination of α1β2 and β1α2. It is orthogonal to |1, 0〉. As|1, 0〉 is symmetric then |0, 0〉 is antisymmetric, so

|0, 0〉 =1√2

(α1β2 − β1α2) . (4.83)

Formally one can write |0, 0〉 = aα1β2 + bβ1α2 then S+ or S− applied to it must give zero, i.e.

S+ (aα1β2 + bβ1α2) =(S1+ + S2+

)(aα1β2 + bβ1α2) = 0 (4.84)

= bhα1α2 + ahα1α2 = 0 (4.85)

so a = −b. Normalization requires |a|2 + |b|2 = 1 so a = 1√2

.

55


The triplet states with S = 1 are

|1, 1〉 = α1α2; |1, 0〉 =1√2

(α1β2 + β1α2) ; |1,−1〉 = β1β2, (4.86)

and the singlet S = 0 is

|0, 0〉 =1√2

(α1β2 − β1α2) . (4.87)

An alternative approach can be followed that shows that these eigenstates are also eigenstates of

S2 =(S1 + S2

)·(S1 + S2

)(4.88)

= S21 + S2

2 + 2S1 · S2 (4.89)

= S21 + S2

2 + 2(S1xS2x + S1yS2y + S1zS2z

)(4.90)

Since Sx = 12

(S+ + S−

)and Sy = 1

2i

(S+ − S−

)then

S1xS2x + S1yS2y =1

2

(S1+S2− + S1−S2+

)(4.91)

andS2 = S2

1 + S22 + 2S1zS2z +

(S1+S2− + S1−S2+

). (4.92)

Express |1, 0〉 = aα1β2 + bα2β1 then

S2|1, 0〉 = 1 (1 + 1) h2|1, 0〉 = 2aα1β2 + 2bα2β1 (4.93)

and [S2

1 + S22 + 2S1zS2z +

(S1+S2− + S1−S2+

)](aα1β2 + bα2β1) =

+a

[3

4h2α1β2 +

3

4h2α1β2 + 2

(h

2

)(− h

2

)α1β2 + h2β1α2

]+b

[3

4h2α2β1 +

3

4h2α2β1 + 2

(− h

2

)(h

2

)α2β1 + h2β2α1

]

= α1β2

[3

2h2a− 1

2h2a+ h2b

]+ α2β1

[3

2h2b− 1

2h2b+ h2a

]= α1β2 (a+ b) h2 + β1α2 (a+ b) h2. (4.94)

Thus from eq(4.93,4.94) comparing coefficients of α1β2 and β1α2 gives

2a = a+ b; 2b = a+ b

a = b.

Normalization requires |a|2 + |b|2 = 1 so a = b = 1√2

and

|1, 0〉 =1√2

(α1β2 + α2β1) . (4.95)

If a = −b then the eigenvalue is 0 and 1√2

(α1β2 − α2β1) corresponds to the state |0, 0〉.Applying the same method to α1α2:

S2α1α2 =[S2

1 + S22 + 2S1zS2z +

(S1+S2− + S1−S2+

)]α1α2

=

[3

4h2 +

3

4h2 + 2

h

2

h

2+) + 0 + 0

]α1α2 = 2h2α1α2

= S (S + 1) h2α1α2

and so S = 1. For the z-component

Szα1α2 =(S1z + S2z

)α1α2 =

h

2α1α2 +

h

2α1α2

so M = 1 and the state α1α2 = |1, 1〉.

56


4.4.2 Coupling of spin-1/2 and orbital angular momentumIf a spin- 1

2 particle is in a state such that its orbital angular momentum quantum number is `, what are the possible valuesof total angular momentum quantum number j, and how are eigenstates of J2 and Jz , where J is total angular momentumconstructed?

All possible states of the system having orbital angular momentum quantum number ` can be represented as linearcombinations of the state vectors Y`m(θ, φ)α and Y`m(θ, φ)β, where α and β represent spin-up and spin-down, as before.(For a complete specification of the state, a radial function R(r) is needed, but this is of no interest here, since it doesnot affect the angular momentum.) Denote the magnetic quantum number for orbital angular momentum by m`, andthe magnetic quantum number for total angular momentum by mj . Recalling the restriction |m`| ≤ `, there are 2` + 1possible values of m, so that the total number of state vectors Ylm α and Ylm β is 2(2`+ 1).

The total angular momentum J is:J = L + S , (4.96)

where L is the orbital angular momentum and S is the spin angular momentum. The states Y`m(θ, φ)α and Y`m(θ, φ)βare already eigenstates of Jz:

JzY`m(θ, φ)α = LzY`m(θ, φ)α+ SzY`m(θ, φ)α = h

(m+

1

2

)Y`m(θ, φ)α (4.97)

JzY`m(θ, φ)β = LzY`m(θ, φ)β + SzY`m(θ, φ)β = h

(m− 1

2

)Y`m(θ, φ)β . (4.98)

From this, it is clear that the highest value of total magnetic quantum number mj is `+ 12 , and the lowest possible value

is −`− 12 . This means that the highest value of J is `+ 1

2 , and the values of mj are −`− 12 , −`+ 1

2 . . ., `+ 12 . There are

therefore 2(`+ 1

2

)+ 1 = 2`+ 2 states having j = `+ 1

2 . Since the total number of states is 2(2`+ 1), the other 2` statesmust be associated with J = ` − 1

2 . This is conirmed by the fact that there is only a single state for which mj = ` + 12 ,

and a single state for which mj = −`− 12 , but there are two states for all the other values of mj .

Note that most of the states Y`m α and Y`m β, although they are eigenvectors of Jz , are not eigenvectors of J2. Theonly exceptions to this statement are the states Y`` α and Y`−` β. One way to construct all the eigenstates of J2 havingJ = ` + 1

2 is to start with Y`` α and act repeatedly with the total lowering operator J− = L− + S−. The eigenstates ofJ2 for which J = ` − 1

2 can then be obtained by finding the states that are eigenvectors of Jz but are orthogonal to theeigenvectors of Jz having J = `+ 1

2 .As an explicit example, take the case of j = ` + 1

2 and mJ = ` + 12 i.e. the maximal projection state Y``α ≡

|j, j〉. Applying J− to it generates the state of j = ` + 12 with mJ = ` − 1

2 . For any angular momenta, J−|j,m〉 =√j (j + 1)−m (m− 1)h|j,m− 1〉, so

J−|j, j〉 =√

2jh|j, j − 1〉; J−|`+1

2, `+

1

2〉 =

√2

(`+

1

2

)h|`+

1

2, `− 1

2〉.(

L− + S−

)Y``α =

(L− + S−

)|`, `〉|α〉 =

√2`h|`, `− 1〉|α〉+ |`, `〉h|β〉

so √2

(`+

1

2

)h|`+

1

2, `− 1

2〉 =√

2`h|`, `− 1〉|α〉+ |`, `〉h|β〉, (4.99)

and the state

|`+1

2, `− 1

2〉 =

√2`

2`+ 1|`, `− 1〉|α〉+

√1

2`+ 1|`, `〉h|β〉. (4.100)

The state |`− 12 , `−

12 〉 = a|`, `− 1〉|α〉+ b|`, `〉|β〉 is orthogonal to |`+ 1

2 , `−12 〉 so

〈`+1

2, `− 1

2|`− 1

2, `− 1

2〉 = 0, (4.101)

(√2`

2`+ 1〈`, `− 1|〈α|+

√1

2`+ 1〈`, `|〈β|

)(a|`, `− 1〉|α〉+ b|`, `〉|β〉) = 0, (4.102)

a

√2`

2`+ 1+ b

√1

2`+ 1= 0,

b = −a√

2`. (4.103)

57


Since normalization requires |a|2 + |b|2 = 1, then a =√

12`+1 and b = −

√2`

2`+1 , and

|`− 1

2, `− 1

2〉 =

√1

2`+ 1|`, `− 1〉|α〉 −

√2`

2`+ 1|`, `〉|β〉. (4.104)

In summary the total angular momentum J is:

J = L + S , (4.105)

where L is the orbital angular momentum and S is the spin angular momentum and

L2Y`m = ` (`+ 1) h2Y`m, (4.106)S2χ = s (s+ 1) h2χ, (4.107)J2ψ = J (J + 1) h2ψ, (4.108)

alsoJz = Lz + Sz. (4.109)

Four good quantum numbers are J , MJ , `, s where

MJ = m` +ms (4.110)Jzψ = MJ hψ, (4.111)

andJ2|J,MJ , `, s〉 = J (J + 1) h2|J,MJ , `, s〉

with |`− s| ≤ J ≤ `+ s and

Jz|J,MJ , `, s〉 = MJ |J,MJ , `, s〉

=

(m` ±

1

2

)h|J,MJ , `, s〉.

4.4.3 Addition of orbital and spin-1/2 angular momentaThis section can be ignored If only orbital and one spin angular momentum are to be added the method outlined belowcan be followed:

The linear combination|j,m+

1

2〉 = a|l,m〉α+ b|l,m+ 1〉β (4.112)

is constructed which is an eigenfunction of Jz with eigenvalue(m+ 1

2

)h. The constants a and b have to be found such

that this state is also an eigenstate of J2. Since

J = L + S,

J2 = L2 + S2 + 2L · SJ2 = L2 + S2 + 2LzSz + L+S− + L−S+ (4.113)

Then using

J±|j,m〉 = [j (j + 1)−m (m± 1)]1/2

h|j,m± 1〉,= [(j ∓m) (j ±m+ 1)]

1/2h|j,m± 1〉

(L2 + S2 + 2LzSz + L+S− + L−S+

)(a|l,m〉α+ b|l,m+ 1〉β)

= ah2

(` (`+ 1) +

3

4+ 2m

(1

2

))|`,m〉α

+bh2

(` (`+ 1)

3

4+ 2 (m+ 1)

(−1

2

))|`,m+ 1〉β

+ah2(

[(`−m) (`+m+ 1)]12

)|`,m+ 1〉β

+bh2(

[(`+m+ 1) (`− (m+ 1) + 1)]12

)|`,m〉α.

58


This will be equivalent to

J2|j,m+1

2〉 = j (j + 1) h2 (a|l,m〉α+ b|l,m+ 1〉β)

provided (by comparing coefficients of |l,m〉α and |l,m+ 1〉β) that

ah2

(` (`+ 1) +

3

4+ 2m

(1

2

))+ bh2

([(`+m+ 1) (`− (m+ 1) + 1)]

12

)= j (j + 1) h2a

ah2(

[(`−m) (`+m+ 1)]12

)+ bh2

(` (`+ 1)

3

4+ 2 (m+ 1)

(−1

2

))= j (j + 1) h2b

i.e. [` (`+ 1) +

3

4+m− j (j + 1)

]a+ [(`−m) (`+m+ 1)]

1/2b = 0

[(`−m) (`+m+ 1)]1/2

a+

[` (`+ 1) +

3

4−m− 1− j (j + 1)

]b = 0

These are two simultaneous equations for a and b which have a non-trivial solution if the determinant is zero,[` (`+ 1) +

3

4+m− j (j + 1)

] [` (`+ 1) +

3

4−m− 1− j (j + 1)

]− (`−m) (`+m+ 1) = 0

Putting A = j (j + 1)− ` (`+ 1)− 34 , gives

(A−m) (A+m+ 1)− [` (`+ 1)−m (m+ 1)] = 0,

and

A = −1

2±

√(`+

1

2

)2

so A = ` or A = −`− 1, and j = `+ 12 or j = `− 1

2 . To solve for a and b the matrix equation(` (`+ 1) + 3

4 +m− j (j + 1) [(`−m) (`+m+ 1)]1/2

[(`−m) (`+m+ 1)]1/2

` (`+ 1) + 34 −m− 1− j (j + 1)

)(ab

)= 0

gives for j = `+ 12

(m− `) a+ [(`−m) (`+m+ 1)]1/2

b = 0

a

b=

(`+m+ 1

`−m

)1/2

.

Normalization requires |a|2 + |b|2 = 1, yielding

b =

√`−m2`+ 1

; a =

√`+m+ 1

2`+ 1.

Hence the state

|j,m+1

2〉 ≡ |`+

1

2,m+

1

2〉 =

√`+m+ 1

2`+ 1|l,m〉α+

√`−m2`+ 1

|l,m+ 1〉β. (4.114)

Choosing j = `− 12 gives

|j,m+1

2〉 ≡ |`− 1

2,m+

1

2〉 =

√`−m2`+ 1

|l,m〉α−√`+m+ 1

2`+ 1|l,m+ 1〉β. (4.115)

59


60

CHAPTER 5. APPROXIMATE METHODS & MANY-BODY SYSTEMS

Chapter 5

Approximate methods & Many-body systems

As in classical mechanics there are relatively few interesting problems in quantum mechanics that can be solved ex-actly. Approximate methods are necessary to obtain eigenvalues and eigenfunctions for potentials in the Schrödingerequation. Such methods conveniently divide into two groups according to whether the Hamiltonian of the system istime-independent or time-dependent.

In the first instance the approximate determination of the discrete eigenvalues and eigenfunctions for stationary statesof a time-indpendent Hamiltonian will be discussed.

5.1 Time-independent perturbation theory for a non-degenerate energy level

Suppose that the time-independent Hamiltonian H of a system can be expressed as

H = H0 +H ′ (5.1)

where H0 is the unperturbed Hamiltonian whose corresponding time-independent Schrödinger equation can be solvedexactly, i.e.

H0|ψ(0)n 〉 = E(0)

n |ψ(0)n 〉 (5.2)

where E(0)n is the energy eigenvalue (known) with eigenfunction |ψ(0)

n 〉 (also known). These eigenfunctions form a com-plete orthonormal set, with 〈ψ(0)

k |ψ(0)n 〉 = δkn. The additional term H ′ is in some sense a "small" change to H0 - a

perturbation. In perturbation theory the solution is expanded as a power series in the perturbation. For the theory to bevalid the series must be convergent in a mathematical sense. For the theory to be useful, however, the series must berapidly convergent such that only the first few terms are important. (The convergence of the series will not be discussed.In some cases it can even be shown that the series cannot converge and yet the first few terms do satisfactorily describethe physical system.) To keep track of the order of the perturbation a dimensionless parameter λ is introduced into theHamiltonian via

H = H0 + λH ′ (5.3)

such that λ = 0 gives the unperturbed system and λ = 1 gives the actual system to be solved. It is assumed that H has adiscrete set of eigenvalues which are non-degenerate and satisfy

H|ψn〉 = En|ψn〉, (5.4)(H0 + λH ′

)|ψn〉 = En|ψn〉. (5.5)

Considering a particular unperturbed energy E(0)n it is assumed that λH ′ is small enough that En is closer to E(0)

n thanany other unperturbed level. Thus

limλ→0

En → E(0)n and lim

λ→0|ψn〉 → |ψ(0)

n 〉. (5.6)

The basis idea of perturbation theory is to assume that the eigenvalues and eigenfunctions can be expanded in a powerseries of the perturbation parameter λ as

En =

∞∑j=0

λjE(j)n = E(0)

n + λE(1)n + λ2E(2)

n + . . . (5.7)

and

|ψn〉 =

∞∑j=0

λj |ψ(j)n 〉 = |ψ(0)

n 〉+ λ|ψ(1)n 〉+ λ2|ψ(2)

n 〉+ . . . (5.8)

61


but 〈ψ(0)n |H ′ − E(1)

n |ψ(0)n 〉 = 0 (the first-order correction formula), so

E(2)n = 〈ψ(0)

n |H ′ − E(1)n |

∑k 6=n

a(1)nk |ψ

(0)k 〉 (5.24)

=∑k 6=n

a(1)nk 〈ψ

(0)n |H ′ − E(1)

n |ψ(0)k 〉∑

k 6=n

a(1)nk

(H ′nk − E(1)

n δnk

)=

∑k 6=n

a(1)nk H

′nk

E(2)n =

∑k 6=n

H ′kn(E

(0)n − E(0)

k

)H ′nk, (5.25)

E(2)n =

∑k 6=n

∣∣∣H ′kn∣∣∣2(E

(0)n − E(0)

k

) =∑k 6=n

∣∣∣〈ψ(0)k |H ′|ψ

(0)n 〉∣∣∣2(

E(0)n − E(0)

k

) . (5.26)

and

|ψn〉 = |ψ(0)n 〉+

∑k 6=n

H ′kn(E

(0)n − E(0)

k

) |ψ(0)k 〉 (5.27)

to first-order in the eigenfunction.Note that if the term with k = n is included then

|ψn〉 = (1 + a(1)nn |ψ(0)

n 〉+∑k 6=n

∣∣∣H ′kn∣∣∣2(E

(0)n − E(0)

k

) |ψ(0)k 〉. (5.28)

The coefficients a(1)nn are not determined by these methods. It can be shown that coefficients of this type can not be found

in any order by the method above. Thus one in tempted to conclude that the choice of these quantities has no physicalsignificance, thus could set a(1)

nn = 0!One could attempt to find a(1)

nn = 〈ψ(0)n |ψ(1)

n 〉 by using the normalization condition

1 = 〈ψn|ψn〉 =(〈ψ(0)n |+ λ〈ψ(1)

n |+ λ2〈ψ(2)n |+ . . .

)(|ψ(0)n 〉+ λ|ψ(1)

n 〉+ λ2|ψ(2)n 〉+ . . .

).

This could be done to any order. For example, to first-order in λ

1 =(〈ψ(0)n |+ λ〈ψ(0)

n |)(|ψ(0)n 〉+ λ|ψ(1)

n 〉)

= 1 + λ〈ψ(0)n |ψ(1)

n 〉+ λ〈ψ(0)n |ψ(1)

n 〉

1 = 1 + a(1)nn + a(1)∗

nn

shows that the real parts of a(1)nn vanish, but gives no information on the imaginary parts, so a(1)

nn = iγ say, with γ real. Ifnormalization to second order is carried out one gets

a(2)nn + a(2)∗

nn +∑k

∣∣∣a(1)nk

∣∣∣2 = 0

which again is only an equation for the real parts of a(2)nn . This feature is found for any order. The imaginary parts of

a(j)nn cannot be found from the normalization condition. This remaining arbitrariness corresponds to the fact that one can

multiply a wave function |ψn〉 by an arbitrary phase factor eiα (α real) without changing the normalization. Thus if a(j)nn

is imaginary then(

1 + a(j)nn

)|ψ(0)n 〉 ≡ Aeiα|ψ(0)

n 〉. Therefore without loss of generality one can set the imaginary parts

of the coefficients a(j)nn to zero.

5.1.3 Observations on the energy corrections

1. Unlike the first-order energy shift E(1)n which only involves the state |ψ(0)

n 〉, the second-order correction dependson the complete set of unperturbed states and, in general, leads to formidable calculations.

64


2. If n is the ground state, E(0)n − E(0)

k < 0 (always) and thus E(2)n < 0.

3. The first-order energy shift E(1)n often vanishes exactly on symmetry grounds thus necessitating evaluation of the

second-order correction.

4. The states |ψ(0)k 〉 over which the summation is performed are often called intermediate states. Each term in E(2)

n

is viewed as a succession of two first-order "transitions" n → k; k → n weighted by the energy denominatorE

(0)n − E(0)

k . System appears to leave state |ψ(0)n 〉, propagate to intermediate state |ψ(0)

k 〉 then fall back again tostate |ψ(0)

n 〉.

5. If all matrix elements of H ′ are of the same order of magnitude - a reasonable first guess - then the "nearby" levelshave a bigger effect on E(2)

n than "distant" ones.

6. If n is not the ground state and if an important level k - lying nearby or having H ′kn large - lies above level n thenE

(2)n is <0; if level k lies below the shift is upwards. One speaks of a tendency of levels to repel each other!

7. For perturbation theory to be useful it must give small corrections so that calculations to low order suffice. Forfirst-order changes to |ψ(0)

n 〉 to be small one needs∣∣∣H ′nk∣∣∣ ∣∣∣E(0)

n − E(0)k

∣∣∣ for k 6= n . In general the diagonal

matrix elements H ′nn will be of the same order of magnitude as the non-diagonal elements H ′nk. Thus the condition

implies∣∣∣H ′nn∣∣∣ ∣∣∣E(0)

n − E(0)k

∣∣∣ and therefore∣∣∣H ′nn∣∣∣ min

∣∣∣E(0)n − E(0)

k

∣∣∣. That is the first-order shift E(1)n

must be small compared to the level spacing min∣∣∣E(0)

n − E(0)k

∣∣∣ between E(0)n and the nearest lying energy level.

These conditions break down if the level is degenerate. In this case there are some states |ψ(0)k 〉, k 6= n with

the same energy E(0)k = E

(0)n . Thus some denominators E(0)

n − E(0)k vanish making the expressions for a(1)

nk

meaningless. One can, however, relax the condition that all the unperturbed energy levels are non-degenerate asone only needs that the level whose energy shift is being calculated is non-degenerate. The difficulties only arisefrom the degeneracy of level n, the other levels with E(0)

k 6= E(0)n can be degenerate.

5.2 The Degenerate case

As well as the difficulty referred to above, i.e. E(0)n − E

(0)k being zero for |ψ(0)

k 〉 6= |ψ(0)n 〉 for a degenerate level

there is another difficulty. Suppose level E(0)n is α-fold degenerate. There are α unperturbed wavefunctions |ψ(0)

nµ 〉,(µ = 1, 2, . . . α), corresponding to the level and it is not known a priori to which of these functions, or linear combinations,the perturbed eigenfunction tends as λ→ 0.

The α-unperturbed eigenfunctions |ψ(0)nµ 〉, (µ = 1, 2, . . . α), for levelE(0)

n are orthogonal to those |ψ(0)k 〉 corresponding

to any other levelE(0)k 6= E

(0)n . Although not necessarily orthogonal amongst themselves it is always possible to construct

linear combinations of them that are orthonormal, i.e.⟨ψ

(0)nµ |ψ(0)

nν

⟩= δµν . It will be assumed that this has already been

done so that the setψ

(0)nµ

are orthonormal.

The "correct" unperturbed eigenfunctions |χ(0)nµ〉 that |ψnµ〉 tend to as λ → 0 are some linear combination of the

|ψ(0)nµ 〉, so that

limλ→0|ψnµ〉 = |χ(0)

nµ〉 =

α∑ν=1

cµν |ψ(0)nv 〉. (5.29)

To avoid too many summations, consideration will be given explicitly to a doubly degenerate level (α = 2) as the finalresult is easily generalized to an α-fold degenerate level. In this case the degenerate unperturbed eigenfunctions are |ψ(0)

n1 〉,|ψ(0)n2 〉. The correct zeroth-order eigenfunctions are

|χ(0)n1 〉 = c11|ψ(0)

n1 〉+ c12|ψ(0)n2 〉, (5.30)

|χ(0)n2 〉 = c21|ψ(0)

n1 〉+ c22|ψ(0)n2 〉 (5.31)

Corresponding to eqs(5.7, 5.8) of the non-degenerate case is

|ψnµ〉 = |χ(0)nµ〉+ λ|ψ(1)

nµ 〉+ λ2|ψ(2)nµ 〉+ . . . , (5.32)

Enµ = E(0)n + λE(1)

nµ + λ2E(2)nµ + . . . (5.33)

65


for µ = 1, 2. Thus for first-order eq(5.13) becomes(H0 − E(0)

n

)|ψ(1)nµ 〉 =

(E(1)nµ − H ′

)|χ(0)nµ〉. (5.34)

For a doubly degenerate case this is(H0 − E(0)

n

)|ψ(1)n1 〉 =

(E

(1)n1 − H ′

)|χ(0)n1 〉 (5.35)

=(E

(1)n1 − H ′

)(c11|ψ(0)

n1 〉+ c12|ψ(0)n2 〉)

(5.36)

and (H0 − E(0)

n

)|ψ(1)n2 〉 =

(E

(1)n2 − H ′

)|χ(0)n2 〉 (5.37)

=(E

(1)n2 − H ′

)(c21|ψ(0)

n1 〉+ c22|ψ(0)n2 〉)

(5.38)

Taking the scalar product of eq(5.36)with |ψ(0)n1 〉 and |ψ(0)

n2 〉 in turn gives

〈ψ(0)n1 |(H0 − E(0)

n

)|ψ(1)n1 〉 = 〈ψ(0)

n1 |(E

(1)n1 − H ′

)(c11|ψ(0)

n1 〉+ c12|ψ(0)n2 〉)

(5.39)

〈ψ(0)n2 |(H0 − E(0)

n

)|ψ(1)n1 〉 = 〈ψ(0)

n2 |(E

(1)n1 − H ′

)(c11|ψ(0)

n1 〉+ c12|ψ(0)n2 〉)

(5.40)

The L.H.S is zero just as in the non-degenerate case. Using the orthonormality of |ψ(0)n1 〉 and |ψ(0)

n2 〉 gives

c11

(E

(1)n1 − 〈ψ

(0)n1 |H ′|ψ

(0)n1 〉)− c12〈ψ(0)

n1 |H ′|ψ(0)n2 〉 = 0, (5.41)

−c11〈ψ(0)n2 |H ′|ψ

(0)n1 〉+ c12

(E

(1)n1 − 〈ψ

(0)n2 |H ′|ψ

(0)n2 〉)

= 0. (5.42)

Writing H ′ij = 〈ψ(0)ni |H ′|ψ

(0)nj 〉 these equations are(

H ′11 − E(1)n1

)c11 + H ′12c12 = 0,

H ′21c11 +(H ′22 − E

(1)n1

)c12 = 0

and have a matrix form (H ′11 − E

(1)n1 H ′12

H ′21 H ′22 − E(1)n1

)(c11

c12

)= 0. (5.43)

These simultaneous equations for c11 and c12 have a non-trivial solution if and only if the determinant is zero,∣∣∣∣∣ H ′11 − E(1)n1 H ′12

H ′21 H ′22 − E(1)n1

∣∣∣∣∣ = 0 (5.44)

This is a quadratic equation in E(1)n1 leading to two possible values of E(1)

n1 . One solution gives E(1)n1 , the other gives

E(1)n2 . Substituting these back into eq(5.43) leads to two sets of values for the coefficients, One set for c11 and c12 and

the other set corresponds to c21 and c22. Thus two zero-order eigenfunctions |χ(0)n1 〉 and |χ(0)

n2 〉 are found that are linearcombinations of |ψ(0)

n1 〉 and |ψ(0)n2 〉, and which, in general, correspond to different first-order corrections to the energy

E(1)n1 and E(1)

n2 . These two linear combinations are the correct zeroth-order eigenfunctions and can thus be used alongwith the unperturbed eigenfunctions of the other states to evaluate higher order corrections in a similar way as in thenon-degenerate case.

The generalization to the α-fold degenerate case is straightforward. Since |χ(0)nµ〉 =

∑αν=1 cµν |ψ

(0)nv 〉 then the determi-

nantal equation becomes ∣∣∣∣∣∣∣∣∣∣∣∣

H ′11 − E(1)n1 H ′12 H ′13 . . . H ′1α

H ′21 H ′22 − E(1)n1 H ′23 . . . H ′2α

H ′31 H ′32 H ′33 − E(1)n1 . . . H ′3α

......

......

...H ′α1 H ′α2 H ′α3 . . . H ′αα − E

(1)n1

∣∣∣∣∣∣∣∣∣∣∣∣= 0. (5.45)

66


which has α-roots for E(1)n1 . If all the roots are different the degeneracy is completely removed by the perturbation. If

some of them are the same the degeneracy is only partially removed. The residual degeneracy may be removed in a higherorder of perturbation. In general the determination of the coefficients cµν is a lengthy process.

The determination of the eigenfunctions |χ(0)nµ〉 amounts to finding the correct orthogonal linear combination of the

original zero-order degenerate eigenfunctions |ψ(0)nv 〉 such that the matrix H ′µν =

⟨χ

(0)nµ

∣∣∣H ′∣∣∣χ(0)nν

⟩is diagonal with respect

to the indices µ, ν.A few special cases are worth noting:

1. The matrix of H ′ is already diagonal, as in∣∣∣∣∣ H ′11 − E(1)n1 0

0 H ′22 − E(1)n1

∣∣∣∣∣ = 0 (5.46)

then E(1)n1 = H ′11 and E(1)

n2 = H ′22. Thus the two states |ψ(0)n1 〉 and |ψ(0)

n2 〉 are not connected in first-order. Thedegeneracy is completely removed and the energies are En1 = E

(0)n + H ′11, and En2 = E

(0)n + H ′22. This

happens when the perturbation H ′ commutes with the operator whose eigenvalues the label µ represents. Supposean Hermitian operator A commutes with H0 and H ′, i.e.

[H0, A

]= 0, and

[H ′, A

]= 0. Thus H0 and A possess

a complete set of mutual eigenstates. The eigenstates |ψ(0)nµ 〉, µ = 1, . . . α, belonging to eigenvalue E(0)

n of H0 arealso eigenfunctions of A, with

A|ψ(0)nµ 〉 = anµ|ψ(0)

nµ 〉. (5.47)

Then ⟨ψ(0)nµ

∣∣∣[H ′, A]∣∣∣ψ(0)nν

⟩=

⟨ψ(0)nµ

∣∣∣H ′A− AH ′∣∣∣ψ(0)nν

⟩= 0, (5.48)

=(anν − a∗nµ

) ⟨ψ(0)nµ

∣∣∣H ′∣∣∣ψ(0)nν

⟩= 0. (5.49)

and H ′µν = 0 if anν 6= anµ (note: the eigenvalues are real). Hence if all the α-eigenvalues anµ are different H ′µν is

already diagonal and the |ψ(0)nµ 〉 are already the correct zero-order states.

In the hydrogen atom a degeneracy is associated with the eigenvalues of Lz in that all 2` + 1 m values have thesame energy. Suppose that [

H ′, Lz

]= 0,

then choose |ψ(0)nµ 〉 to be eigenfunctions of Lz i.e. µ ≡ m, so that

Lz|ψ(0)nm〉 = mh|ψ(0)

nm〉. (5.50)

Then ⟨ψ

(0)nm′

∣∣∣[H ′, Lz]∣∣∣ψ(0)nm

⟩=

⟨ψ

(0)nm′ |H

′Lz|ψ(0)nm

⟩−⟨ψ

(0)nm′ |LzH

′|ψ(0)nm

⟩(5.51)

= mh⟨ψ

(0)nm′ |H

′|ψ(0)nm

⟩−m′∗h

⟨ψ

(0)nm′ |H

′|ψ(0)nm

⟩= (m−m′∗) hH ′m′n. (5.52)

But the m values are real and thus if m′ 6= m then H ′m′n = 0, i.e. H ′m′n is a diagonal matrix.

This example illustrates a general result: a degeneracy due to a symmetry of H0 is reduced or removed altogetherby a perturbation of a lower symmetry. In this example, H0 is spherically symmetric but H ′ possesses only axialsymmetry.

The coefficients c11, c12 are easily found from(H ′11 − E

(1)n1 0

0 H ′22 − E(1)n1

)(c11

c12

)= 0. (5.53)

For E(1)n1 = H ′11, c12 = 0 so can choose c11 = 1 and |χ(0)

n1 〉 = |ψ(0)n1 〉. For E(1)

n2 = H ′22, c21 = 0 so can choosec22 = 1 and |χ(0)

n2 〉 = |ψ(0)n2 〉.

67


2. The other interesting case occurs for H ′11 = H ′22 = 0. Then the determinant is∣∣∣∣∣ −E(1)n1 H ′12

H ′21 −E(1)n1

∣∣∣∣∣ = 0 (5.54)

and (E

(1)n1

)2

= H ′12H′21 =

∣∣∣H ′12

∣∣∣2 (5.55)

sinceH ′12 = 〈ψ(0)

n1 |H ′|ψ(0)n2 〉 = 〈ψ(0)

n2 |H ′|ψ(0)n1 〉∗ =

(H ′21

)∗. (5.56)

Hence the first-order energy correction isE

(1)n1 = ±

∣∣∣H ′12

∣∣∣ . (5.57)

The coefficients satisfy−c11E

(1)n1 + c12H

′12 = 0

so thatc11

c12=H ′12

E(1)n1

. (5.58)

For E(1)n1 = +

∣∣∣H ′12

∣∣∣, c11c12 = ±1 depending on the sign of H ′12. For E(1)n1 = −

∣∣∣H ′12

∣∣∣, c11c12 = ∓1. The original

eigenstates |ψ(0)n1 〉 and |ψ(0)

n2 〉are said to be fully mixed with the correct zeroth-order eigenfunctions being

|χ(0)n1 〉 =

1√2

(|ψ(0)n1 〉 ± |ψ

(0)n2 〉), (5.59)

|χ(0)n2 〉 =

1√2

(|ψ(0)n1 〉 ∓ |ψ

(0)n2 〉). (5.60)

5.3 Applications of perturbation theory

5.3.1 First-order examplesThe anharmonic oscillator To calculate the energy eigenvalues of an anharmonic oscillator represented by the Hamil-tonian

H =p2

2m+

1

2kx2 + γx4 . (5.61)

The presence of the anharmonic term γx4 makes this problem difficult to solve exactly, but if the term γx4 is small (in asense to be clarified below), first-order perturbation theory can be used. The unperturbed Hamiltonian H0 is the harmonicoscillator,

H0 =p2

2m+

1

2kx2 , (5.62)

and the perturbation Hamiltonian is:H ′ = γx4. (5.63)

Consider the change of energy of the ground state. The unperturbed ground-state energy is E(0)0 = 1

2 hω, where ωis the frequency of the harmonic oscillator ω = (k/m)1/2. According to eqn (5.14), the change of ground-state energycaused by the term γx4 is:

E(1)0 = 〈φ0|H ′|φ0〉 , (5.64)

where |φ0〉 is the ground-state eigenvector. This quantity could be evaluated using the operator methods for the harmonicoscillator but here the normal wavefunction methods are used .

The normalized ground-state wave-function of the harmonic oscillator is

φ0(x) =(mωπh

)1/4

exp(−mωx2/2h) . (5.65)

Hence

E(1)0 =

(mωπh

)1/2∫ ∞−∞

γx4 exp(−mωx2/h)dx . (5.66)

68


In order to calculate this integral note that ∫ ∞−∞

e−αx2

dx = π1/2α−1/2. (5.67)

Differentiating this with respect to α once gives

− d

dα

∫ ∞−∞

e−αx2

dx =

∫ ∞−∞

x2e−αx2

dx =1

2π1/2α−3/2 (5.68)

and differentiating again

− d

dα

∫ ∞−∞

x2e−αx2

dx =

∫ ∞−∞

x4e−αx2

dx =3

4π1/2α−5/2. (5.69)

It follows that ∫ ∞−∞

x4 exp(−mωx2/h)dx =3

4π1/2 (mω/h)

−5/2 (5.70)

and hence that

E(1)0 =

3

4γ(mωπh

)1/2(πh

mω

)1/2(h

mω

)2

=3h2γ

4m2ω2. (5.71)

Thus the ground-state energy is raised by an amount proportional to γ, as should be expected.It is useful to consider how the total ground-state energyE0 ' E(0)

0 +E(1)0 deviates from the unperturbed ground-state

energy E(0)0 ;

E0/E(0)0 '

(E

(0)0 + E

(1)0

)/E

(0)0 = 1 + E

(1)0 /(

1

2hω) = 1 +

3hγ

2m2ω3. (5.72)

The strength of the perturbation can therefore be characterized by the dimensionless quantity g = 2hγ/m2ω3. The tableshows how the approximation for E0/E

(0)0 from first-order perturbation theory compares with essentially exact numerical

predictions. This shows that the energy shift is given very accurately for small values of g, but becomes less accurate as gincreases.

g = 2hγ/m2ω3 E0/E(0)0 (pert. th.) E0/E

(0)0 (exact)

0.01 1.00750 1.007350.10 1.075 1.0650.2 1.15 1.12

Table 5.1: The ground-state energy E0 of the anharmonic oscillator divided by the ground-state energy of the unperturbedharmonic oscillator E(0)

0 for different values of γ. The size of γ is specified by the dimensionless quantity g = 2hγ/mω3.Second and third columns give E0/E

(0)0 from first-order perturbation theory and from almost exact numerical solution of

the Schrödinger equation.

Harmonic oscillator As an example to demonstrate the validity of the perturbation approach consider a harmonicoscillator where the perturbation is itself harmonic. The unperturbed Hamiltonian

H0 =p2

2m+

1

2kx2 (5.73)

has energy eigenvalues

E(0)n =

(n+

1

2

)hω0,

where ω0 =√k/m. Suppose that the force constant k is changed slightly, k → k′ such that k′ − k = ε with 0 < ε k.

The Hamiltonian is now

H =p2

2m+

1

2kx2 +

1

2εx2 = H0 +

1

2εx2 (5.74)

=p2

2m+

1

2(k + ε) x2

69


and the new force constant is k + ε so the new frequency of oscillation

ω =

(k + ε

m

)1/2

=

(k

m

)1/2 (1 +

ε

k

)1/2

= ω0

(1 +

ε

k

)1/2

(5.75)

and the exact energy levels are

En =

(n+

1

2

)hω0

(1 +

ε

k

)1/2

En =

(n+

1

2

)hω0

(1 +

1

2

ε

k− 1

8

( εk

)2

+ . . .

). (5.76)

The same problem will now be solved using time-independent perturbation theory. The perturbation is H ′ = 12εx

2.The energy levels are non-degenerate, so the first-order correction to E(0)

n =(n+ 1

2

)hω0 is

E(1)n = 〈ψ(0)

n |H ′|ψ(0)n 〉 =

1

2ε〈ψ(0)

n |x2|ψ(0)n 〉 =

1

2ε(x2)nn.

The matrix elements may be evaluated several ways; using the eigenfunctions for ψ(0)n ; or using the operator approach.

The latter is easier. The matrix elements(x2)nn

were calculated earlier when considering the generalized uncertaintyrelation for the harmonic oscillator, i.e.

〈n|x2|n〉 =2n+ 1

2mω0h =

(n+

1

2

)h

mω. (5.77)

For the second-order correction matrix elements between different states will be needed. In anticipation of this recall thatx =

(h

2mω0

)(a+ + a−) and

x|n〉 =

(h

2mω0

)1/2

(a+ + a−) |n〉 =

(h

2mω0

)1/2 (√n+ 1|n+ 1〉+

√n|n− 1〉

)so

〈n|xx|k〉 =

(h

2mω0

)(√n+ 1〈n+ 1|+

√n〈n− 1|

) (√k + 1|k + 1〉+

√k|k − 1〉

),

=

(h

2mω0

)( √(n+ 1) (k + 1)〈n+ 1|k + 1〉+

√(n+ 1) k〈n+ 1|k − 1〉

+√n (k + 1)〈n− 1|k + 1〉+

√nk〈n− 1|k − 1〉

),

=

(h

2mω0

)(√(n+ 1) (k + 1)δnk +

√(n+ 1) kδn,k−2 +

√n (k + 1)δn,k+2 +

√nkδnk

),

〈n|x2|k〉 =

(h

2mω0

)((2n+ 1) δnk +

√(n+ 1) (n+ 2)δn,k−2 +

√n (n− 1)δn,k+2

). (5.78)

The first-order correction is

E(1)n =

1

2ε(x2)nn

=1

2ε

(n+

1

2

)h

mω0. (5.79)

This is the same as the first term in the expansion equ(5.76) since(n+

1

2

)hω0

1

2

ε

k=

(n+

1

2

)hω0

1

2

ε

mω20

=

(n+

1

2

)hε

2mω0. (5.80)

The second-order correction is obtained from

E(2)n =

∑k 6=n

∣∣∣〈k|H ′|n〉∣∣∣2(E

(0)n − E(0)

k

) =∑k 6=n

∣∣〈k| 12εx2|n〉∣∣2((

n+ 12

)hω0 −

(k + 1

2

)hω0

)=

∑k 6=n

(1

2ε

)2∣∣〈k|x2|n〉

∣∣2(n− k) hω0

=1

4

ε2

hω0

∑k 6=n

∣∣〈k|x2|n〉∣∣2

(n− k). (5.81)

70


But from equ(5.78) the only non-zero matrix elements are for k = n ± 2; k = n already being excluded from thesummation. Hence

E(2)n =

1

4

ε2

hω0

[∣∣〈n+ 2|x2|n〉∣∣2

(n− (n+ 2))+

∣∣〈n− 2|x2|n〉∣∣2

(n− (n− 2))

], (5.82)

=1

4

ε2

hω0

( h

2mω0

)2

∣∣∣√(n+ 1) (n+ 2)∣∣∣2

−2+

(h

2mω0

)2

∣∣∣√n (n− 1)∣∣∣2

2

,=

1

4

ε2

hω0

(h

2mω0

)2 [ |(n+ 1) (n+ 2)|−2

+|n (n− 1)|

2

],

E(2)n =

(hε2

32m2ω30

)(−4n− 2) = −

(hε2

8m2ω30

)(n+

1

2

). (5.83)

The perturbation result is in agreement with the exact result equ(5.76), since from the expansion the second-order correc-tion is (

n+1

2

)hω0

(−1

8

( εk

)2)

= −1

8

(n+

1

2

)hε2

m2ω30

.

Ground state of helium atom The neutral He atom consists of two electrons bound to a nucleus having charge Z|e|,with Z = 2. Because of the electrostatic repulsion between the electrons, the Schrödinger equation for this system cannotbe solved exactly. One way of tackling it is to treat the inter-electron repulsion by perturbation theory.

The Hamiltonian for the two-electron system is:

H (r1, r2) = − h2

2m∇2

1 −h2

2m∇2

2 −Ze2

4πε0

(1

r1+

1

r2

)+

e2

4πε0

1

r12. (5.84)

Here, r1 and r2 are the positions of the two electrons, and r12 = |r1 − r2| is the distance between the electrons; ∇21 and

∇22 are the Laplacian operators for the two electron positions. Since the electrons have spin- 1

2 the full space-spin wave-function should describe both the spin state and the space state. The electrons are also fermions so the total wave-functionmust be antisymmetric with respect to interchange of electrons. However, the ground state is a spin singlet so that the spinpart of the two-electron wave-function is (α1 β2 − β1 α2)/

√2. This is already antisymmetric, so that the spatial part of

the wave-function Ψ(r1, r2) is symmetric. Since the Hamiltonian does not depend on spin, Schrödinger’s equation is:(− h2

2m∇2

1 −h2

2m∇2

2 −Ze2

4πε0

(1

r1+

1

r2

)+

e2

4πε0

1

r12

)Ψ(r1, r2) = EΨ(r1, r2) . (5.85)

The inter-electron repulsion e2/(4πε0r12) makes it impossible to solve Schrödinger’s equation exactly, and perturba-tion theory offers one way forward. The unperturbed Hamiltonian H0 omits the repulsion

H0 = − h2

2m∇2

1 −h2

2m∇2

2 −Ze2

4πε0

(1

r1+

1

r2

), (5.86)

and is a separable Hamiltonian withH0 (r1, r2) = H0 (r1) + H0 (r1) . (5.87)

In atomic unitsH0 (r) = −1

2∇2 − Z

r. (5.88)

The unperturbed Schrödinger equation,

H0 (r1, r2) Ψ(0)0 (r1, r2) = E0Ψ

(0)0 (r1, r2) , (5.89)

is easy to solve, because it represents two independent electrons. The ground state wavefunction of H0 is just the productof the ground-state wavefunctions of the individual electrons,

Ψ(0)0 (r1, r2) = ψ100(r1)ψ100(r2) , (5.90)

where ψ100(r) is the ground-state wavefunction for a single electron bound to a nucleus of charge Z|e|. This can beobtained from the usual hydrogenic wave-function by replacing e2 by (Ze)

2 and the energy for each electron is

E(0)n = − Z2

2n2, (5.91)

71


withψ100(r) = Ae−Zr/a0 , (5.92)

and with a0 = 4πε0h2/me2 ' 0.529 Å the usual Bohr radius. The ground-state energy E(0)

0 of the He atom in this ap-proximation is twice the ground-state energy of a single electron bound to a nucleus of chargeZ|e|, which is−Z2/2 a.u, sothat E(0)

0 = −Z2 = −4 a.u. ("Hartrees"). Not surprisingly, this is a poor approximation compared with the experimentalvalue of the He ground-state energy, which is −2.905 a.u. .

The electron-electron mutual repulsion is taken as the perturbation part

H ′ =e2

4πε0r12≡ 1

r12, (5.93)

so that by first-order perturbation theory, the total energy is shifted by:

E(1)0 =

⟨Ψ

(0)0

∣∣∣∣ e2

4πε0r12

∣∣∣∣Ψ(0)0

⟩≡⟨

Ψ(0)0

∣∣∣∣ 1

r12

∣∣∣∣Ψ(0)0

⟩. (5.94)

This can be evaluated exactly, but it is long and tedious as 1r12

= 1|r1−r2| depends on the angular coordinates. The result is

E(1)0 = 1.25 a.u., so that the corrected ground-state energy is−4+1.25 = −2.75 a.u. This is still not in perfect agreement

with the experimental value of−2.905 a.u, but it is much better than the value obtained by omitting the repulsion betweenelectrons.

5.3.2 Second-order exampleStark effect in hydrogen To illustrate the application of perturbation theory to a real physical problem consider theeffect of a time-independent, spatially uniform external electric field on the energy levels of an atom - the so-calledStark effect. The Stark effect is the shift of the optical adsorption and emission frequencies of atoms induced by theapplication of an external electric field. This shift arises from the shift of the atomic energy levels. To keep the discussionas manageable as possible it will be restricted to hydrogen-like atoms, i.e.to one electron atoms.

The unperturbed Hamiltonian H0 is, as usual,

H0 = − h2

2m∇2 − e2

4πε0r, (5.95)

whose zero-order eigenfunctions are the hydrogenic wavefunctions. It is useful to recall some of their properties:(1) The eigenfunctions are ψn`m (r, θ, φ) = Rn` (r)Y`m (θ, φ), with Rn` (r) the radial wavefunction and Y`m (θ, φ)

the spherical harmonic,(2) the parity of ψn`m (r, θ, φ) is (−1)

`,(3) the eigenfunctions are orthonormal,

∫∞0

∫ π0

∫ 2π

0ψ∗n′`′m′ψn`mr

2dr sin θdθdφ = δn′nδ`′`δm′m

(4) the unperturbed energy levels are En = − m2h2

Ze2

4πε01n2 = −mc

2

2

(Zαn

)2where n = 1, 2, 3, . . . and α = e2

4πε0hc'

1137 is the fine structure constant

(5) for a given `, m has 2`+ 1 values, −`,−`+ 1, . . . `− 1, `(6) for a given n, 0 ≤ ` ≤ n− 1.(7) each level is (without considering spin)

∑n−1`=0 (2`+ 1) = n2 degenerate; including spin doubles this value.

The applied electric field E is uniform over the atomic dimensions and assumed to be directed along the the positivez-axis, so that the additional potential energy of the electron is

δV (r) = eEz = −p · E (5.96)

where p = −ez is the electric dipole moment of the electron. This interaction energy must be added to the Coulombpotential energy of the electron in the electric field of the nucleus.

Consider first the ground state (n = 1, ` = 0, m = 0). The first-order shift E(1)100 is

E(1)100 = 〈ψ100|δV |ψ100〉 =

∫drψ?100(r)δV (r)ψ100(r) . (5.97)

The ground-state wave-function ψ100(r) is:

ψ100(r) =1

π1/2a3/20

e−r/a0 ≡ 1√π

e−r, (5.98)

72


so that

E(1)100 = eE

∫|ψ100 (r)|2 zdr . (5.99)

This is exactly zero, since |ψ100 (r)|2 is an even function under parity (r → −r) whereas the perturbation is an oddfunction as z→ −z and hence overall the integrand is odd. Positive values of the integrand cancel negative values. Thereason for this is that the value of the integrand at any point r is exactly equal and opposite to its value at −r. This isthe result of symmetry; the charge distribution in the unperturbed atom is spherically symmetric, whereas the perturbationHamiltonian δV changes sign when r → −r. Thus the first-order shift is zero. Hence for the ground state there is noenergy shift linear in the electric field strength E . Classically a system with electric dipole moment p will experience anenergy shift −p · E hence it is concluded that the atom has, in the ground state, no permanent dipole moment.

Since the first-order shift is zero a second-order calculation must be done. For the ground state the second-ordercorrection is given by

E(2)100 =

∞∑n>1

n−1∑`=0

m=∑m=−`

∣∣∣〈ψ(0)n`m|H ′|ψ

(0)100〉

∣∣∣2(E

(0)100 − E

(0)n`m

) ,

= e2E2∞∑n>1

n−1∑`=0

m=∑m=−`

∣∣∣〈ψ(0)n`m|z|ψ

(0)100〉

∣∣∣2(E

(0)100 − E

(0)n`m

) .Since E(0)

100 − E(0)n`m < 0, then E(2)

n < 0 i.e. the ground state energy is lowered by the interaction. To evaluate the matrixelements put z = r cos θ then

〈ψ(0)n`m|z|ψ

(0)100〉 =

∫ ∫Rn` (r)Y ∗`m (θ, φ) r cos θR10Y00r

2 dr dΩ. (5.100)

But Y00 = 1/√

4π and cos θ =√

4π3 Y10 so that

〈ψ(0)n`m|z|ψ

(0)100〉 =

∫Rn` (r)R10r

3 dr

∫Y ∗`m (θ, φ)

1√3Y10 dΩ.

From the orthonormality of the spherical harmonics this integral vanishes unless ` = 1 and m = 0. Thus only the states|ψ(0)n10〉 contribute to the integral and

E(2)100 = e2E2

∞∑n=2

∣∣∣〈ψ(0)n10|z|ψ

(0)100〉

∣∣∣2(E

(0)100 − E

(0)n10

) . (5.101)

Explicit evaluation of∣∣∣〈ψ(0)

n10|z|ψ(0)100〉

∣∣∣2 is very difficult but clearly it is proportional to some length squared and dependson n. Thus one can write ∣∣∣〈ψ(0)

n10|z|ψ(0)100〉

∣∣∣2 =(a0

Z

)2

f (n) , (5.102)

with a0 the Bohr radius, Actually∣∣∣〈ψ(0)

n10|z|ψ(0)100〉

∣∣∣2 = 13

28n7(n−1)2n−5

(n+1)2n+5 a20 ! so that

E(2)100 = −e2E2

(a0

Z

)2 ∞∑n=2

f (n)mc2

2 (Zα)2 (

1− 1n2

) = − 2e2E2a20

α2Z4mc2

∞∑n=2

n2f (n)

(n2 − 1). (5.103)

Hence E(2)100 ∝ e2E2 as is to be expected from electromagnetic theory as it is due to an induced dipole moment p = eχE

which has an energy 12pE ∝ E

2.

5.3.3 Degenerate example - first orderStark effect in hydrogen - excited states The first excited state has n = 2. If fine-structure (i.e. spin-orbit) effects areneglected this level is 4-fold degenerate. The degenerate unperturbed eigenfunctions are for n = 2 , ` = 0, |ψ200〉, andn = 2, ` = 1, m = 1, 0,−1, |ψ211〉 , |ψ210〉 , |ψ21−1〉 . Consequently to find even the first-order correction for the excitedstates, degenerate perturbation theory needs to be used. At first sight it appears that one needs to calculate the 16 matrixelements `′m′H ′`m = 〈ψ2`′m′ |eEz|ψ2`m〉 (in a slight deviation from the usual notation to emphasise the states involved)in the table below

73


`′m′ \ `,m 0, 0 1, 0 1, 1 1,−1

0, 0 00H′00 − E

(1)2µ 00H

′10 00H

′11 00H

′1−1

1, 0 10H′00 10H

′10 − E

(1)2µ 10H

′11 10H

′1−1

1, 1 11H′00 11H

′10 11H

′11 − E

(1)2µ 11H

′1−1

1,−1 1−1H′00 1−1H

′10 1−1H

′11 1−1H

′1−1 − E

(1)2µ

and evaluate a 4×4 determinant. However an examination of the form of these elements can result in a dramatic reductionin the number (and effort) of them to evaluate explicitly.

1. The perturbation H ′ = eEz has odd parity and the eigenstates |ψ2`m〉 have parity (−1)`. Thus⟨

ψ2`′m′

∣∣∣H ′∣∣∣ψ2`m

⟩= 0 if `′ = `. (5.104)

Thus all the diagonal elements of H ′`′m′`m vanish, i.e. 00H′00 = 10H

′10 = 11H

′110 = 1−1H

′1−1 = 0. In addition

the elements 10H′11, 10H

′1−1, 11H

′10, 11H

′1−1, 1−1H

′10, and 1−1H

′11 are also zero - resulting in a reduction of 10

elements that do no need to be evaluated explicitly, leaving

`′m′ \ `,m 0, 0 1, 0 1, 1 1,−1

0, 0 −E(1)2µ 00H

′10 00H

′11 00H

′1−1

1, 0 10H′00 −E(1)

2µ 0 0

1, 1 11H′00 0 −E(1)

2µ 0

1,−1 1−1H′00 0 0 −E(1)

2µ

.

2. The perturbationH ′ = eEz commutes with Lz whose eigenvalues result in the degeneracy indexm, i.e.[Lz, H

′]

=

eE[Lz, z

]= 0. Hence

0 =⟨ψ2`′m′

∣∣∣[Lz, H ′]∣∣∣ψ2`m

⟩=⟨ψ2`′m′

∣∣∣[LzH ′ − H ′Lz]∣∣∣ψ2`m

⟩,

0 = h (m′ −m)⟨ψ2`′m′

∣∣∣H ′∣∣∣ψ2`m

⟩. (5.105)

Thus if m′ 6= m then⟨ψ2`′m′

∣∣∣H ′∣∣∣ψ2`m

⟩= 0. Hence elements 00H

′11, 00H

′1−1, 10H

′11, 10H

′1−1, 11H

′00, 11H

′10,

11H′1−1, 1−1H

′00, 1−1H

′10, 1−1H

′11 are all zero, giving

`′m′ \ `,m 0, 0 1, 0 1, 1 1,−1

0, 0 −E(1)2µ 00H

′10 0 0

1, 0 10H′00 −E(1)

2µ 0 0

1, 1 0 0 −E(1)2µ 0

1,−1 0 0 0 −E(1)2µ

.

Hence only two non-zero matrix elements exist, 00H′10 =

⟨ψ200

∣∣∣H ′∣∣∣ψ210

⟩and 10H

′00 =

⟨ψ210

∣∣∣H ′∣∣∣ψ200

⟩. But

as H ′ is Hermitian then 00H′10 =10 H

′∗00 so only one matrix element need be calculated! As a general policy it is

always advisable to examine the symmetries and origin of the degeneracy before using brute force to evaluate allthe matrix elements

The 4× 4 matrix00H

′00 − E

(1)2µ 00H

′10 00H

′11 00H

′1−1

10H′00 10H

′10 − E

(1)2µ 10H

′11 10H

′1−1

11H′00 11H

′10 11H

′11 − E

(1)2µ 11H

′1−1

1−1H′00 1−1H

′10 1−1H

′11 1−1H

′1−1 − E

(1)2µ

c1c2c3c4

= 0 (5.106)

reduces to −E(1)

2µ 00H′10 0 0

10H′00 −E(1)

2µ 0 0

0 0 −E(1)2µ 0

0 0 0 −E(1)2µ

c1c2c3c4

= 0 (5.107)

74


and only the secular equation ∣∣∣∣∣ −E(1)2µ 00H

′10

10H′00 −E(1)

2µ

∣∣∣∣∣ = 0 (5.108)

needs to be solved. This has two roots E(1)2µ = ±

∣∣∣00H′10

∣∣∣. Explicit evaluation

00H′10 = 〈ψ200 |eEz|ψ210〉 = eE 〈ψ200 |z|ψ210〉 = eE

∫ ∫R∗20Y

∗00 (r cos θ)R21Y10r

2 dr dΩ. (5.109)

of 00H′10 is now needed. Using expressions for the hydrogenic wavefunctions,

R20 (r) = 2

(Z

2a0

)3/2(1− Zr

2a0

)e−Zr/2a0 , (5.110)

R21 (r) =1√3

(Z

2a0

)3/2Zr

a0e−Zr/2a0

with Z = 1, and z = r cos θ

00H′10 = eE

∫2

(Z

2a0

)3/2(1− Zr

2a0

)e−Zr/2a0r3 1√

3

(Z

2a0

)3/2Zr

a0e−Zr/2a0dr

∫Y ∗00 (cos θ)Y10 dΩ. (5.111)

Evaluating the angular integral first (always a wise move!), and noting that Y00 =(

14π

)1/2and Y10 =

(3

4π

)1/2cos θ,∫

Y ∗00 cos θY10 dΩ =

∫1√4π

√4π

3Y10Y10dΩ =

1√3

(5.112)

using the orthonormality of the spherical harmonics. Hence

00H′10 =

2

3eE(Z

2a0

)3 ∫ ∞0

r3

(Zr

a0

)(1− Zr

2a0

)e−Zr/a0dr

=2

3eE(Z

2a0

)3 (a0

Z

)4∫ ∞

0

ρ4(

1− ρ

2

)e−ρdρ

setting ρ = Zr/a0. Then

00H′10 =

1

12eE(a0

Z

)∫ ∞0

ρ4e−ρdρ− 1

2

∫ ∞0

ρ5e−ρdρ

=

1

12eE(a0

Z

)4!− 1

25!

= −3eE a0

Z, (5.113)

using the standard integral,∫∞

0xne−axdx = n!/an+1. Hence the first-order energy correction is

E(1)2µ = ±

∣∣∣00H′10

∣∣∣ = ±3eE a0

Z, for µ = 1, 2, (5.114)

= 0 for µ = 3, 4. (5.115)

Thus the energy shift is proportional to E , i.e. a linear Stark effect. The linear Stark effect is only non-zero because then = 2 state is degenerate.

The correct linear combination of the unperturbed eigenfunctions |ψ2`m〉 can be found by solving for the coefficientsc, (

−E(1)2µ 00H

′10

10H′00 −E(1)

2µ

)(c1c2

)= 0

soc1c2

=00H

′10

E(1)2µ

(5.116)

Since 00H′10 = −3eE a0Z then for E(1)

21 = +∣∣∣00H

′10

∣∣∣ = 3eE a0Z , c1c2 = −1 and

|χ(0)21 〉 = c1|ψ200〉+ c2|ψ210〉 =

1√2

(|ψ200〉 − |ψ210〉) .

75


For E(1)22 = −

∣∣∣00H′10

∣∣∣ = −3eE a0Z , c1c2 = +1 and

|χ(0)22 〉 = c1|ψ200〉+ c2|ψ210〉 =

1√2

(|ψ200〉+ |ψ210〉) .

For completeness, E(1)23 = 0 and |χ(0)

23 〉 = |ψ211〉 and E(1)24 = 0 and |χ(0)

24 〉 = |ψ21−1〉.The states in the presence of the electric field are no longer eigenstates of L2 or parity as |χ(0)

21 〉 and |χ(0)22 〉 mix these

states and so neither ` nor parity are "good" quantum numbers. The magnetic quantum number m is still a good quantumnumber (as H ′ commutes with Lz) and the system is invariant under rotations about the z-axis. Physically the externalelectric field defines a preferred direction in space so the system cannot be invariant under under arbitrary rotations (i.e.L2) but still invariant for rotations about this preferred direction (i.e. Lz).

The degeneracy of the states |ψ211〉 and |ψ21−1〉 is not removed in first-order. The 4-fold degenerate n = 2 level splitssymmetrically into three sub-levels.

This means that the hydrogen atom (in its unperturbed first-excited state) behaves as if it had a permanent electricdipole moment of magnitude 3ea0 which can be orientated in three different ways,

1. antiparallel to electric field E for |χ(0)21 〉 = 1√

2(|ψ200〉 − |ψ210〉) (Note energy in electric field E of a dipole µ is

W = −µ ·E.)

2. parallel to electric field E for |χ(0)22 〉 = 1√

2(|ψ200〉+ |ψ210〉)

3. two states with zero component along electric field E for |χ(0)23 〉 = |ψ211〉 and |χ(0)

24 〉 = |ψ21−1〉.

5.4 Variational method

There are many stationary state problems which cannot be solved exactly and for which the perturbation approach is un-satisfactory because the first-order is not sufficiently accurate and higher orders involve enormous (complex) calculations.Basically the perturbation expansion converges too slowly. The variational method - also known as the Rayleigh-Ritzmethod - does not presuppose a knowledge of the simpler problem and is, therefore, very versatile. The method is partic-ularly suitable for calculating the ground state energies of a system.

The quantum ground state of a system has a special importance. For systems of electrons, in particular, one is ofteninterested only in the ground state. For example, when quantum mechanics is used to calculate the energies of chemicalreactions, or the properties of materials, excited states are usually irrelevant. Because of this, methods for calculating theground-state energy of quantum systems are important. In developing such methods, the variational principle of quantummechanics usually plays a key role. This principle states that

“The expectation value of the Hamiltonian, evaluated in an arbitrary state, is always greater than or equalto the ground-state energy.”

This principle forms the basis for practical approximate methods of calculating the ground-state energy of a complexsystem. The idea is to guess an approximate ground-state wave-function, called the “trial wave-function”, which isused to calculate the expectation value of the Hamiltonian. To improve the estimate of the ground-state energy, thetrial wave-function contains variable parameters. The expectation value of the Hamiltonian is minimized with respect tothese parameters, and the minimum energy provides the best estimate of the true ground-state energy, within the givenparameterized ‘family’ of wavefunctions.

5.4.1 Proof of the variational principle

The Hamiltonian H possesses a complete set of orthonormal eigenstates φn , φ0, φ1, φ2, . . . , 〈φn|φm〉 = δnm andcorresponding energies E0, E1, E2, . . . with the ground state energy E0 ≤ E1 ≤ E2 . . . .and

H|φn〉 = En|φn〉. (5.117)

Let |ψ〉 be an approximate wavefunction which satisfies the correct boundary conditions. Expand |ψ〉 in terms of thecomplete set φn as

|ψ〉 =∑n

cn|φn〉 (5.118)

76


then the expectation value of the energy in the state |ψ〉 is

E [ψ] = 〈E〉ψ =〈ψ∣∣∣H∣∣∣ψ〉〈ψ|ψ〉

,

=

∑n

∑m c∗mcn〈φm|H|φn〉∑

n

∑m c∗mcnEn〈φm|φn〉

,

〈E〉ψ =

∑n |cn|

2En∑

n |cn|2 .

and

〈E〉ψ − E0 =

∑n |cn|

2En∑

n |cn|2 − E0 =

∑n |cn|

2(En − E0)∑

n |cn|2 . (5.119)

Since every term on the R.H.S. of equ(5.119) is positive then it follows that

〈E〉ψ ≥ E0,

〈ψ∣∣∣H∣∣∣ψ〉〈ψ|ψ〉

≥ E0. (5.120)

and gives an upper bound on the ground state energy. Equality will occur if the trial function |ψ〉 should be the same as

the correct ground state eigenfunction |φ0〉 in which case |c0| = 1 and cn = 0 for all n 6= 0. If〈ψ|H|ψ〉〈ψ|ψ〉 = E0 but cn 6= 0

for more than one n then the ground state is degenerate.In practice a trial function is chosen which depends on some parametersα1 α2, . . . αs, i.e. |ψT 〉 = |ψT (α1, α2, . . . αs)〉

and one calculates

〈E〉ψ =〈ψT

∣∣∣H∣∣∣ψT 〉〈ψT |ψT 〉

= E (α1, α2, . . . αs) . (5.121)

This energy is minimized with respect to the variational parameters α1, α2, . . . αs by solving the equations

∂E (α1, α2, . . . αs)

∂αi= 0, i = 1, 2, . . . , s. (5.122)

The resulting minimum value of E (α1, α2, . . . αs) represents the best estimate of the ground state energy with thatparticular trial function.

Clearly the success of the variational method depends on choosing a trial function that incorporates the correct featuresof the ground state. Symmetry and other physical properties of the system are useful guides. The use of powerful com-puters allows trial functions of great complexity and hence great flexibility leading to very accurate results. One shouldbear in mind that the variational method optimizes the energy. The trial function is not necessarily a good approximationto the true wavefunction and may give poor results when used in calculations of quantities other than energy.

If |ψ〉 is almost the true eigenfunction for the ground state |φ0〉 but with a bit of |φ1〉 mixed in so that |ψ〉 = |φ0〉 +a1|φ1〉 then the approximation to E0 is as good as

〈E〉ψ =E0 + |a|2E1

1 + |a|2= E0

(1 + |a|2E1/E0

1 + |a|2

)(5.123)

〈E〉ψ − E0 = |a|2 (E1 − E0)

1 + |a|2. (5.124)

and the error δE is quadratic in the coefficients. This means that accurate values of the ground-state energy can often beobtained with rather simple guesses for the approximate ground-state wavefunction.

5.4.2 Excited StatesThe variational method can be used to obtain an upper bound for the energy of an excited state provided the trial functionis made orthogonal to all the energy eigenfunctions corresponding to states having an energy lower than the energy levelbeing considered. Unfortunately often the lower eigenfunctions are not known exactly and one has only approximations(obtained perhaps from a variational calculation) for these functions. In this case the orthogonality conditions cannot beachieved exactly. Consequently 〈E〉φ1 may not necessarily provide an upper bound on the energy E1. These difficultiesdo not arise if one looks for an excited energy level whose symmetry properties differ from those of lower-lying levels.Choosing a trial function with the correct symmetry automatically ensures orthogonality to the lower-lying states and thevariational method gives an upper bound to the energy of the selected excited state.

77


5.4.3 Variational examplesGround state of hydrogen - actually an unrealistic example as it can be solved exactly! However it is useful to seehow to choose a trial function and perform the integrals and the variation.

The ground state is an s-state with ` = 0. Thus |ψ〉 depends only on r with |ψ (0)〉 6= 0. At large distances a boundstate wavefunction must vanish. Thus take the trial function as ψ (r) = Ce−αr. Normalization requires

〈ψ|ψ〉 = 4πC2

∫ ∞0

e−2αrr2dr = 1. (5.125)

Using the standard integral∫∞

0rne−ardr = n!/an+1,

1 = 4πC2 2!

(2α)3

and

C =

(α3

π

)1/2

and

ψ (r) =

(α3

π

)1/2

e−αr. (5.126)

The Hamiltonian is H = − h2

2m∇2 − e2

4πε0r. The expectation value of the potential energy is

〈ψ| − e2

4πε0r|ψ〉 = −

(α3

π

)e2

4πε04π

∫ ∞0

e−2αrrdr, (5.127)

= −(α3

π

)e2

4πε04π

1

(2α)2 ,

= − e2

4πε0α. (5.128)

The following "trick" frequently simplifies the kinetic energy calculation. For any function ψ (r) a vector identity is∫V

∇· (ψ∗∇ψ) dτ =

∫ψ∗(∇2ψ

)dτ +

∫∇ψ∗· (∇ψ) dτ,

and using Gauss’ theorem on the L.H.S.∫V

∇· (ψ∗∇ψ) dτ =

∫S

(ψ∗∇ψ) · ndS =

∫ψ∗(∇2ψ

)dτ +

∫|(∇ψ)|2 dτ.

As ψ → 0 sufficiently rapidly as r →∞ the surface integral vanishes (certainly the case for the exponential function) sothat ∫

ψ∗(∇2ψ

)dτ = −

∫|(∇ψ)|2 dτ. (5.129)

Using this result on the expectation value of the kinetic energy term

〈ψ| − h2

2m∇2|ψ〉 =

h2

2m

∫|(∇ψ)|2 dτ =

h2

2m

∫|(∇ψ)|2 4πr2dr.

But for the trial function

|(∇ψ)|2 =

∣∣∣∣∂ψ∂r∣∣∣∣2 = α2 |ψ|2 .

Hence

〈ψ| − h2

2m∇2|ψ〉 =

h2α2

2m

∫|ψ|2 4πr2dr =

h2α2

2m(5.130)

Hence the expectation value of the energy is

E (α) =h2α2

2m− e2

4πε0α (5.131)

and∂E (α)

∂α=h2α

m− e2

4πε0

78


and

α =e2m

4πε0h=

1

a0, (5.132)

where a0 is the Bohr radius. Consequently the minimum energy is

E

(1

a0

)= −1

2

e2

4πε0a0= −1

2a.u. (5.133)

Harmonic oscillator As a simple illustration of how the variational principle works consider the harmonic oscillator,with the usual Hamiltonian:

H = − h2

2m

d2

dx2+

1

2kx2 . (5.134)

For this system the exact ground-state energy E0 = 12 hω is known, where ω is the frequency ω = (k/m)1/2; also the

exact ground-state wavefunction which is φ0(x) = (mω/πh)14 exp(−mωx2/2h). However, to show that the variational

principle really works suppose that this wave function is not known and a trial function is constructed. Because thepotential is symmetric about x = 0 and the wavefunction must satisfy ψ (x) → 0 as x → ∞ then take as a trialwavefunction as

ψT(x) = (2α/π)1/4e−αx2

, (5.135)

with α (positive) variational parameter. To demonstrate that the expectation value of H evaluated with this wavefunctionis always greater than or equal to the exact ground-state energy.

First, verify that the ψT(x) as given by equ (5.135) is already correctly normalized,∫ ∞−∞|ψT(x)|2 dx =

(2α

π

)1/2 ∫ ∞−∞

e−2αx2

dx . (5.136)

For any positive number β,∫ ∞−∞

e−βx2

dx =

(π

β

)1/2

; and

∫ ∞−∞

x2e−βx2

dx =1

2β

(π

β

)1/2

. (5.137)

In the present case, β = 2α, so that∫∞−∞ exp(−2αx2) dx = (π/2α)1/2. This confirms that∫ ∞

−∞|ψT(x)|2 dx = 1 . (5.138)

Now to evaluate the expectation value of H , work out separately the expectation values of the kinetic energy T andthe potential energy V ,

〈T 〉 = 〈ψT|T |ψT〉 =

∫ ∞−∞

ψT(x)?(− h2

2m

d2

dx2

)ψT(x) dx (5.139)

In general by integrating by parts∫ ∞−∞

ψT(x)?∇2ψT(x)dx = [ψT(x)?∇ψT(x)]∞−∞ −

∫∇ψ∗T(x)∇ψT(x)dx

= −∫|∇ψT(x)|2 dx (5.140)

Thus

〈T 〉 = 〈ψT|T |ψT〉 =h2

2m

∫ ∞−∞

d

dx

[(2α

π

)1/4

e−αx2

]2

dx,

=h2

2m

(2α

π

)1/2 ∫ ∞1/4

−∞4α2x2e−αx

2

dx, (5.141)

=h2

2m

(2α

π

)1/2

4α2 1

2 (2α)

( π2α

)1/2

, (5.142)

〈T 〉 =h2α

2m. (5.143)

79


The expectation value of the potential energy is

〈V 〉 = 〈ψT|V |ψT〉 =1

2k

(2α

π

)1/2 ∫ ∞−∞

x2e−2αx2

dx, (5.144)

=1

2k

(2α

π

)1/21

2 (2α)

( π2α

)1/2

, (5.145)

〈V 〉 =k

8α. (5.146)

Adding 〈T 〉 and 〈V 〉 the expectation value of the Hamiltonian is

〈H〉 =h2α

2m+

k

8α. (5.147)

Now note that as α → 0, 〈H〉 → k/8α → +∞, and as α → ∞, 〈H〉 → h2α/2m → ∞. Clearly, 〈H〉 must have aminimum for some value of α. To find this minimum, differentiate with respect to α,

d

dα〈H〉 =

h2

2m− k

8α2= 0, (5.148)

so that

α2 =mk

4h2 =m2ω2

4h2 . (5.149)

This means that for α > 0 there is only a single stationary point, which must be a minimum. The value of α that givesthis minimum is

αmin = mω/2h. (5.150)

Hence the trial wave-function that gives the minimum value of 〈H〉 is:

ψmin(x) = (mω/πh)1/4 exp(−mωx2/2h) . (5.151)

This is exactly the correct ground-state wave-function.From equ (5.147), the value of 〈H〉 given by ψmin(x) is:

〈H〉min =h2

2m· mω

2h+k

8· 2h

mω=

1

4hω +

1

4hω =

1

2hω, (5.152)

which (not surprisingly) is exactly the correct result.In this case, the variational principle gives exactly the correct result, because the trial wavefunction used is capable of

giving the correct result. If some other kind of trial wavefunction had been taken, this would not have happened.

Ground state of helium atom The Schrödinger equation for the helium atom is impossible to solve exactly because ofthe electrostatic repulsion between the two electrons. The Hamiltonian is:

H = − h2

2m∇2

1 −h2

2m∇2

2 −Ze2

4πε0

(1

r1+

1

r2

)+

e2

4πε0

1

r12, (5.153)

= −1

2∇2

1 −1

2∇2

2 −Z

r1− Z

r2+

1

r12. (5.154)

Here, Z is the charge on the nucleus in units of e, which for the helium atom is Z = 2.This was treated earlier as an example of perturbation theory, where H was separated into an unperturbed part H0,

in which electron-electron repulsion is ignored, and H ′ = e2/4πε0r12, which is the electron-electron repulsion. Areasonably good estimate of the ground-state energy can be obtained by using first-order perturbation theory to calculatethe energy shift due to H ′. The zero-th order wavefunctions used in the perturbation calculation assume that each electronmoves in the Coulomb field of the nucleus, with Z = 2. In fact the effect of the mutual repulsion of the electrons is toreduce the effect of the nuclear filed experienced by each electron. This screening can be accounted for by assuming aneffective nuclear charge λ (1 ≤ λ < 2). The ground-state wavefunction of a single electron bound to a nucleus of chargeZ is

ψ100(r) =1√π

(Z

a0

)3/2

e−Zr/a0 ≡ 1√πZ3/2e−Zr, (5.155)

80


and the single-electron ground-state energy is

E1 = − Z2e2

8πε0a0= −Z

2h2

2ma20

= −Z2

2a.u. (5.156)

The two-electron ground-state wave-function is

Ψ0(r1, r2) = ψ100(r1)ψ100(r2) =1

π

(Z

a0

)3

e−Z(r1+r2)/a0 ≡ 1

πZ3e−Z(r1+r2) (5.157)

i.e.the product of the two single-electron wave-functions.This form of the approximate wave-function immediately suggests how to do a variational calculation. This suggests

that Z is replaced in Ψ0(r1, r2) by a variable parameter λ, an effective charge, so that the trial wave-function is:

ΨT(r1, r2) =1

π

(λ

a0

)3

e−λ(r1+r2)/a0 , (5.158)

≡ λ3

πe−λ(r1+r2). (5.159)

This trial function is already normalized since ψ100(r1) and ψ100(r2) were normalized.

The Hamiltonian can be written as

H = −1

2∇2

1 −1

2∇2

2 −Z

r1− Z

r2+

1

r12(5.160)

= −1

2∇2

1 −Z

r1− 1

2∇2

2 −Z

r2+

1

r12

= H1 + H2 + Vee, (5.161)

whereH1 = −1

2∇2

1 −Z

r1(5.162)

and similarly for H2. H1 can be re-written as

H1 = −1

2∇2

1 −λ

r1− (Z − λ)

r1,

= hλ −(Z − λ)

r1, (5.163)

where hλ is the Hamiltonian for a hydrogenic atom of nuclear charge λ. Thus by analogy with the hydrogen wavefunctions

φ100 (r1) =(λ3

π

)1/2

e−λr1 is an eigenfunction of hλ with eigenvalue E1 = − λ2

2n2 . Hence

〈H1〉 =⟨hλ

⟩+

⟨− (Z − λ)

r1

⟩,

〈H1〉 = −λ2

2+

⟨− (Z − λ)

r1

⟩(5.164)

since ⟨hλ

⟩=

⟨φ100 (r1)φ100 (r2) |hλ|φ100 (r1)φ100 (r2)

⟩=⟨φ100 (r1) |hλ|φ100 (r1)〉〈φ100 (r2)φ100 (r2)

⟩= −λ

2

2× 1.

Now ⟨− (Z − λ)

r1

⟩= (Z − λ)

λ3

π

∫ ∞0

4π1

r1e−2λr1r2

1dr1 = (Z − λ)λ3

π

∫ ∞0

4πe−2λr1r1dr1

= (Z − λ) 4λ3 1

(2λ)2 = −λ (Z − λ) , (5.165)

noting that 〈φ100 (r2)φ100 (r2)〉 = 1 and∫∞

0re−2λrdr = 1/ (2λ)

2 and hence

〈H1〉 = −λ2

2− λ (Z − λ) =

λ2

2− λZ. (5.166)

81


By the interchange r1 ↔ r2 it follows that

〈H2〉 =λ2

2− λZ

and

〈H1〉+ 〈H2〉 = λ2 − 2λZ. (5.167)

The electron-electron interaction expectation value is much more difficult to evaluate as it involves the relative anglesbetween the position vectors r1, r2 as r12 = |r1 − r2| = r2

1 + r22 − 2r1r2 cos θ12. The result is

〈Vee〉 =

(λ3

π

)2 ∫ ∞0

dr1

∫ ∞0

dr21

r12e−2λ(r1+r2)

=5

8λ. (5.168)

Adding the three terms, we have the expectation value of the total Hamiltonian

〈H〉 = λ2 − 2λZ +5

8λ. (5.169)

Since this is a quadratic form that goes to∞ for large λ, it has a single minimum. To find this, differentiating with respectto λ gives

d〈H〉dλ

= 2λ− 2Z +5

8= 0, (5.170)

so that

λ = Z − 5

16=

27

16. (5.171)

The term −5/16 reduces λ from the value of λ = Z = 2 that it would have in the absence of the electron-electronrepulsion. The value of the total energy for this value of λ is:

ET = λ2 − 2λ

(λ+

5

16

)+

5

8λ = −λ2 = −

(27

16

)2

, (5.172)

and

ET = −(27/16)2 ' −2.85 au. (5.173)

The experimental value of the total energy of the He atom is −2.905 au. The variational estimate of −2.85 au which isslightly above the correct value (as it must be!), but the accuracy is good; the error is less than 2 %.

An example of a more elaborate trial function with two screening constants is

ψ (r1, r2) = C[e−(αr1+βr2)/a0 + e−(αr2+βr1)/a0

]and gives E = −2.875 au. With improved variational wavefunctions, which include explicitly the correlation betweenthe electrons, essentially exact values of the ground-state energy can be obtained.

5.5 Systems of identical particles

A wide range of physical systems, e.g. electrons in atoms or molecules; protons and neutrons in nuclei; cold atoms in aBose-Einstein condensate (BEC) consist of many identical particles. If they interact strongly solutions of the Hamiltonianare very difficult due to the complexity. Generally the difficulty increases "exponentially" with the number of particles.Even for "non-interacting" particles attention must be paid to the fact that the particles are indistinguishable.

In classical dynamics particles move in well defined trajectories and their positions can be tracked with arbitrary preci-sion. Hence they can be distinguished. In quantum mechanics if their wavefunctions overlap they cannot be distinguished,especially if the Hamiltonian does not "label" the particles in some way.

82


5.5.1 Identical particlesIdentical particles are often called indistinguishable particles to emphasize that they cannot be distinguished by anyphysical measurement. For example, two electrons may be at different positions, have different momenta and even havedifferent z-component of spin as these are dynamical parameters of the system but they have the same charge, mass andintrinsic magnetic moment.

Any operator representing a physical measurement on the system must remain unchanged if the labels assigned to theparticles are interchanged. For example, if one writes the Hamiltonian of two identical particles as H (1, 2) then one musthave

H (1, 2) = H (2, 1) . (5.174)

Such a Hamiltonian is that for the helium atom

H (r1, r2) = − h2

2m∇2

1 −Ze2

4πε0r1− h2

2m∇2

2 −Ze2

4πε0r2+

e2

4πε0 |r1 − r2|, (5.175)

which is symmetric in the electron coordinates, i.e. invariant under the interchange of the labels 1⇔ 2. For such systemsit is meaningless to ask for the probability that helium in its ground state has electron 1 in a volume element dVa atposition a and electron 2 in a volume element dVb at position b. One can only ask for the probability that one electron isin dVa and the other in dVb.

An immediate consequence of the symmetry in equ(5.174) is that the eigenvalues of H are degenerate. Suppose that

H (1, 2)ψ (1, 2) = Eψ (1, 2) . (5.176)

Interchanging 1↔ 2 givesH (2, 1)ψ (2, 1) = Eψ (2, 1)

but H (1, 2) = H (2, 1) soH (1, 2)ψ (2, 1) = Eψ (2, 1) . (5.177)

Thus if ψ (1, 2) is an eigenfunction with energy eigenvalue E, then so also is ψ (2, 1). Hence the eigenvalue is two-folddegenerate. This is known as exchange degeneracy. Thus ψ (1, 2), ψ (2, 1) or any linear combination of them is also aneigenfunction. Two particular combinations are

ψS =1√2

[ψ (1, 2) + ψ (2, 1)] (5.178)

andψA =

1√2

[ψ (1, 2)− ψ (2, 1)] , (5.179)

which are, respectively, symmetric and antisymmetric under the interchange 1↔ 2.It follows from the Schrödinger equation that the symmetry of the wavefunction is a constant of the motion. In fact

physically acceptable wavefunctions representing identical particles must be either symmetric or antisymmetric. Definea particle exchange operator P12 such that P12 acting on any function of the variables of the two particles interchangestheir labels. Then

P12H (1, 2) = H (2, 1)

also

P12

[H (1, 2)ψ (1, 2)

]= H (2, 1)ψ (2, 1) ,

= H (1, 2)P12ψ (1, 2) ,

so [P12, H (1, 2)

]ψ (1, 2) = 0. (5.180)

But ψ (1, 2) is any two-particle wavefunction so [P12, H (1, 2)

]= 0. (5.181)

Thus P12 and H (1, 2) are compatible and have a common set of eigenfunctions. If φ (1, 2) is such an eigenfunction,then to be an eigenfunction of P12 it satisfies

P12φ (1, 2) = pφ (1, 2) ,

φ (2, 1) = pφ (1, 2) .

83


Operate with P12 againP12φ (2, 1) = φ (1, 2) = pP12φ (1, 2) = p2φ (1, 2) (5.182)

so

p2 = 1,

p = ±1. (5.183)

Henceφ (1, 2) = ±φ (2, 1)

and is either symmetric or antisymmetric under the interchange 1 ↔ 2. Not only the Hamiltonian but any operatorrepresenting a physical property of the system must be symmetric under 1 ↔ 2 and must therefore commute with P12.Thus whatever measurement is made in the system the resulting wavefunction will be an eigenfunction of P12. No loss ofgenerality occurs if one assumes that the wavefunction always has this property.

This property can be extended to a many-body wavefunction for identical particles. The wavefunction must be eithersymmetric or antisymmetric with respect to interchange of any pair of particles. It follows that every particle belongsto one of two classes depending on whether a number of them is symmetric or antisymmetric under particle exchange.Experimental evidence is that for a given kind of particle the symmetry is always of one kind only. For example, electrons,positrons, protons, neutrons, neutrinos are described by antisymmetric wavefunctions; photons, pions, K-mesons, 4He(alpha particles) have symmetric wavefunctions. Particles with antisymmetric wavefunctions are called fermions, whilethose with symmetric wavefunctions are bosons. The choice of symmetry is closely related with the value of the total spinof the particle. It an empirical fact that fermions have spin s = 1

2 ,32 ,

52 , . . ., i.e. odd half-integer values, and bosons have

s = 0, 1, 2, 3, . . ., i.e. integral values. This one-to-one correspondence between spin and the interchange symmetry canbe shown to be a necessary consequence of a relativistic quantum field theory of identical particles (and the spin-statisticstheorem).

5.5.2 Exclusion principleConsider two identical particles which interact very weakly with one another. To a first approximation one may neglecttheir mutual interaction and hence write the Hamiltonian H (1, 2) as the sum of the two single-particle Hamiltonians as

H (1, 2) = H1 (1) + H2 (2) . (5.184)

If H1 (1) has a complete set of eigenfunctions ψa (1) with eigenvalues Ea, such that H1 (1)ψa (1) = Eaψa (1) andsimilarly for H2 (2)ψb (2) = Ebψb (2) then

H (1, 2)ψa (1)ψb (2) =(H1 (1) + H2 (2)

)ψa (1)ψb (2) = (Ea + Eb)ψa (1)ψb (2) . (5.185)

The symmetric and antisymmetric eigenfunctions of equ(5.184) are

ΨSA (1, 2) =

1√2

[ψa (1)ψb (2)± ψa (2)ψb (1)] . (5.186)

For fermions, the antisymmetric wavefunction vanishes identically if states a = b, i.e.two identical fermions cannot exitsin the same single-particle state. This is known as the Pauli Exclusion Principle.

The Exclusion Principle actually states that each single-particle eigenfunction can only be used once in constructingproducts, linear combinations of which form the total wavefunction.

This obviously generalizes to more than two fermions; such that in a quantum system at most one particle can occupyany one single-particle state. In this form it is restricted to non-interacting particles. The requirement of antisymmetricwavefunctions for identical particles with odd half-integer spin always holds and represents a general statement of thePauli Exclusion Principle.

No such restriction applies to bosons. In this case states a = b are allowed and

Ψ (1, 2) = ψa (1)ψa (2) (5.187)

and any number of bosons can occupy the same single-particle state. The helium atom is a good example to illustrate theideas of symmetric and antisymmetric wavefunctions.

5.5.3 N-particle statesThe discussion can be extended to N -particle Hamiltonians for non-interacting particles,

H = − h2

2m

N∑i=1

∇2i +

N∑i=1

Vi (ri) =

N∑i=1

hi (5.188)

84


where hi = − h2

2m∇2i + Vi (ri) is a single particle Hamiltonian. The physics becomes more interesting - genuine many-

body - if there are particle-particle interactions∑i 6=j V (rij) e.g. the Coulomb interaction

V (rij) =Z2

|ri − rj |, (5.189)

or a short-range spin exchange as in a spin chain,

Vij = KSi · Sjδi,j±1. (5.190)

For the i-th particlehiφni

(ri) = εiφni(ri) (5.191)

with 〈φn (ri) |φm (ri)〉 = δnm. If the total Hamiltonian is separable the solutions are the products

Ψn1n2...nN(r1, r2, . . . rN ) = φn1 (r1)φn2 (r2) . . . φnN

(rN ) , (5.192)

and

HΨn1n2...nN(r1, r2, . . . rN ) = EjΨn1n2...nN

(r1, r2, . . . rN ) (5.193)(N∑i=1

hi

)φn1

(r1)φn2(r2) . . . φnN

(rN ) =

(N∑i=1

ε(j)i

)φn2

(r2) . . . φnN(rN ) (5.194)

The j-th energy eigenvalue of the system is

Ej =

N∑i=1

ε(j)i . (5.195)

Owing to the separability of H then if

HΨn1n2...nN(r1, . . . rk, . . . r`, . . . rN ) = EjΨn1n2...nN

(r1, . . . rk, . . . r`, . . . rN ) ; (5.196)Ej = ε1 + . . . εk . . . ε` . . . εN (5.197)

the interchange of the position of two particles (e.g.rk ↔ r` ) leaves the energy unchanged,

HΨn1n2...nN(r1, . . . r`, . . . rk, . . . rN ) = EjΨn1n2...nN

(r1, . . . r`, . . . rk, . . . rN ) ; (5.198)Ej = ε1 + . . . ε` . . . εk . . . εN . (5.199)

This will hold for Hamiltonians which are symmetric with respect to exchange whether with or without interaction, e.g.

H (r1, r2) = h1 (r1) + h2 (r2) + V (r1, r2)

if V (r1, r2) = V (r2, r1).Let Pk` be the operator which interchanges particle k with particle `. Then applied to equ(5.196) gives

Pk`HΨn1n2...nN(r1, . . . rk, . . . r`, . . . rN ) = EjPk`Ψn1n2...nN

(r1, . . . rk, . . . r`, . . . rN )

Pk`HΨn1n2...nN(r1, . . . rk, . . . r`, . . . rN ) = EjΨn1n2...nN

(r1, . . . r`, . . . rk, . . . rN ) , (5.200)

but equ(5.198) can be re-written as

HPk`Ψn1n2...nN(r1, . . . rk, . . . r`, . . . rN ) = EjΨn1n2...nN

(r1, . . . r`, . . . rk, . . . rN ) . (5.201)

But the R.H.S. of equ(5.200, 5.201) are the same so from the L.H.S. one has(HPk` − Pk`H

)Ψn1n2...nN

(r1, . . . rk, . . . r`, . . . rN ) = 0,(HPk` − Pk`H

)= 0,[

H, Pk`

]= 0, (5.202)

so Pk` and H commute and there are simultaneous eigenstates of Pk` and H for all k, ` combinations.Also two operations of exchange of particles k and `

P 2k`Ψn1n2...nN

(r1, . . . rk, . . . r`, . . . rN ) = Ψn1n2...nN(r1, . . . rk, . . . r`, . . . rN ) (5.203)

85


leaves the wavefunction unchanged. Hence the eigenvalue of P 2k` is 1 and that of Pk` are ±1. Hence the states are either

symmetric

Pk`Ψ(S)k` = +Ψ

(S)k` (5.204)

or antisymmetricPk`Ψ

(A)k` = −Ψ

(A)`k . (5.205)

Explicitly for two non-interacting indistinguishable particles the symmetric wavefunction is

Ψ(S) (r1, r2) =1√2

[φ1 (r1)φ2 (r2) + φ1 (r2)φ2 (r1)] (5.206)

and the antisymmetric one is

Ψ(A) (r1, r2) =1√2

[φ1 (r1)φ2 (r2)− φ1 (r2)φ2 (r1)] . (5.207)

With interactions the individual wavefunctions are more complex (not necessarily products) but the same symmetryapplies, e.g.

Ψ(A) (r1, r2) =1√2

[Φ (r1, r2)− Φ (r2, r1)] (5.208)

and oftenΦ (r1, r2) =

∑m,n

cnmφm (r1)φn (r2) . (5.209)

Symmetrizing for N > 2 particles

Anti-symmetrizing the wavefunction when there are more than two particles is important when doing molecular or atomicstructure calculations, e.g."Hartree-Fock", ab-initio calculations in quantum chemistry. The generalization for N > 2 isthe Slater Determinant,

Ψ(A)n1n2...nN

(r1, r2, . . . rN ) =1√N !

∣∣∣∣∣∣∣∣∣φn1 (r1) φn1 (r2) · · · φn1 (rN )φn2 (r1) φn2 (r2) · · · φn2 (rN )

......

......

φnN(r1) φnN

(r2) · · · φnN(rN )

∣∣∣∣∣∣∣∣∣ (5.210)

which makes use of the rules for evaluating determinants. For example if N = 3 the antisymmetric state is

Ψ(A)n1n2n3

(r1, r2, r3) =1√6

[φn1(r1)φn2

(r2)φn3(r3)− φn1

(r2)φn2(r1)φn3

(r3)

+φn1(r2)φn2

(r3)φn3(r1)− φn1

(r3)φn2(r2)φn3

(r1) (5.211)+φn1

(r3)φn2(r1)φn3

(r2)− φn1(r1)φn2

(r3)φn3(r2)]

or equivalently

Ψ(A)n1n2n3

(r1, r2, r3) =1√6

∑perm

(−1)PPφn1

(r1)φn2(r2)φn3

(r3) (5.212)

where the summation is over all permutations of 1, 2, 3.The symmetric wavefunction is

Ψ(S)n1n2n3

(r1, r2, r3) =1√6

∑perm

Pφn1(r1)φn2

(r2)φn3(r3) (5.213)

which replaces all the − signs in equ(5.212) with + signs.Consider a two-electron state ψ. Its wavefunction must be antisymmetric with respect to interchange of both position

and spin coordinates, i.e.exchange of the particles. The wavefunction is

Ψn1n2(r1, s1, r2, s2) =

1√2

[φn1(r1, s1)φn2

(r2, s2)− φn1(r2, s2)φn2

(r1, s1)] (5.214)

where s1, s2 denote "spin-up" or "spin-down" states. If the space-spin states are the same, i.e. n1 = n2, s1 = s2,φn1

(r, s1) = φn2(r, s2) then Ψ = 0 for all r. If s1 6= s2 there is no problem, nor if n1 6= n2. The same applies to the

Slater determinant. If two rows of a determinant are identical it evaluates to zero. To include spin, replace n by n, s.

86


5.5.4 Helium atom and the exchange force

The Hamiltonian for the helium atom is

H = − h2

2m∇2

1 −−h2

2m∇2

2 −Ze2

4πε0r1− Ze2

4πε0r2+

e2

4πε0 |r1 − r2|

if small corrections due to spin-spin and spin-orbit interactions are neglected. Since H is spin-independent, spin operatorsS2 = (s1 + s2)

2 and Sz commute with H and so are constants of the motion. Operators L2, Lz also commute with H ,so the simultaneous eigenfunctions of H , L2, Lz , S2, Sz can be written as products of functions. The two electrons canbe in spin state S = 0, a singlet,

|0, 0〉 =1√2

(α1β2 − β1α2) , (5.215)

or S = 1, a triplet,

|1, 1〉 = α1α2, (5.216)

|1, 0〉 =1√2

(α1β2 + β1α2) , (5.217)

|1,−1〉 = β1β2. (5.218)

The singlet is antisymmetric with respect to particle exchange, 1 ↔ 2; the triplet is symmetric. Since the overallwavefunction must be antisymmetric with respect to particle exchange, 1↔ 2 the products are spatially symmetric wave-functions Ψ(S) (r1, r2) with antisymmetric spin wavefunctions, χ(A) (s1, s2), i.e. of the form Ψ(S) (r1, r2)χ(A) (s1, s2)or spatially antisymmetric wavefunctions Ψ(A) (r1, r2) with antisymmetric spin wavefunctions, χ(S) (s1, s2), i.e. of theform Ψ(A) (r1, r2)χ(S) (s1, s2). The lowest energy single particle state would have n1 = 1, n2 = 1 and be the spatiallysymmetric state

ψ(S)11 (r1, r2) = φ100 (r1)φ100 (r2) (5.219)

which is combined with an antisymmetric singlet spin state |0, 0〉 to give

Ψ(Sing) (r1, r2) = φ100 (r1)φ100 (r2)1√2

(α1β2 − β1α2) (5.220)

as the ground state eigenfunction.For the first excited state n1 = 1, (` = 0,m = 0) and n2 = 2, (` = 0, 1) so there are two possibilities,

ψ(S) (r1, r2) =1√2

[φ100 (r1)φ2`m (r2) + φ2`m (r1)φ100 (r2)] , (5.221)

ψ(A) (r1, r2) =1√2

[φ100 (r1)φ2`m (r2)− φ2`m (r1)φ100 (r2)] . (5.222)

The symmetric spatial wavefunction ψ(S) must combine with the antisymmetric spin function |0, 0〉 to give

Ψ(Sing) (r1, r2) = ψ(S) (r1, r2) |0, 0〉 (5.223)

and the antisymmetric spatial function ψ(A) combines with the symmetric triplet spin functions |1, 1〉, |1, 0〉, |1,−1〉 togive

Ψ(Trip) (r1, r2) =

ψ(A) (r1, r2) |1, 1〉ψ(A) (r1, r2) |1, 0〉ψ(A) (r1, r2) |1,−1〉

. (5.224)

A perturbation calculation of the first excited state should use degenerate perturbation theory. However(a) the perturbation e2

4πε0|r1−r2| does not involve spin so all perturbation matrix elements between different spin states

vanish as⟨s,ms| 1

r12|s′,ms′

⟩= δss′δmsms′ , i.e.S is a constant.

(b) the total angular momentum cannot change as there are no external torques, so matrix elements between different`-values are also zero.

(c) the perturbation is symmetric in exchange 1 ↔ 2 so matrix elements between states of different symmetry alsovanish.

87


The expectation value of the inter-electron repulsion for the singlet and triplet states are

∆E(S,T ) =

⟨Ψ(Sing,Trip) (r1, r2) | 1

r12|Ψ(Sing,Trip) (r1, r2)

⟩(5.225)

=〈 1√

2[φ100 (r1)φ2`m (r2)± φ2`m (r1)φ100 (r2)]

∣∣∣ 1r12

∣∣∣1√2

[φ100 (r1)φ2`m (r2)± φ2`m (r1)φ100 (r2)]〉, (5.226)

= 12⟨φ100 (r1)φ2`m (r2)

∣∣∣ 1r12

∣∣∣φ100 (r1)φ2`m (r2)⟩

+⟨φ2`m (r1)φ100 (r2)

∣∣∣ 1r12

∣∣∣φ2`m (r1)φ100 (r2)⟩

±〈φ2`m (r1)φ100 (r2)∣∣∣ 1r12

∣∣∣φ100 (r1)φ2`m (r2)〉

±⟨φ100 (r1)φ2`m (r2)

∣∣∣ 1r12

∣∣∣φ2`m (r1)φ100 (r2)⟩

. (5.227)

The first two terms are symmetric in 1 ↔ 2 and so equal, similarly the last two terms are symmetric in 1 ↔ 2 and soequal so that

∆E(S,T ) =

⟨φ100 (r1)φ2`m (r2)

∣∣∣∣ 1

r12

∣∣∣∣φ100 (r1)φ2`m (r2)

⟩±⟨φ100 (r1)φ2`m (r2)

∣∣∣∣ 1

r12

∣∣∣∣φ2`m (r1)φ100 (r2)

⟩(5.228)

∆E(S,T ) =

∫|φ100 (r1)|2 1

r12|φ2`m (r2)|2 dr1dr2

±∫φ∗100 (r1)φ∗2`m (r2)

1

r12φ2`m (r1)φ100 (r2) dr1dr2. (5.229)

The first integral

J =

∫|φ100 (r1)|2 1

r12|φ2`m (r2)|2 dr1dr2 (5.230)

is called the direct integral. It is clearly positive and represents the mutual electrostatic energy of two charge densities|φ100 (r1)|2 and |φ2`m (r2)|2 separated by a distance r12, i.e. the Coulomb repulsive energy and thus has a classicalanalogue.

The second termK =

∫φ∗100 (r1)φ∗2`m (r2)

1

r12φ2`m (r1)φ100 (r2) dr1dr2 (5.231)

has no classical analogue. Its origin lies in the Pauli Principle and is referred to as the exchange integral. Because of thisexchange contribution the singlet and triplet states are no longer degenerate, with

∆E(S) = J +K. (5.232)∆E(T ) = J −K. (5.233)

The J term is obviously positive. For the K term, it may be evaluated for the hydrogenic wavefunctions and is alsopositive. For ` = n− 1 this is (obvious) because the wavefunctions appearing in K have no nodes. That the triplet statehas lower energy than the singlet state can be seen on qualitative grounds. For the triplet state the spatial wavefunctionis antisymmetric and so the electrons tend to stay away from each other (ψ(A) ∼ 0; r ' 0; 〈r12〉Trip > 〈r12〉Sing); theprobability of finding the two electrons at the same place is low. This makes the repulsion between the electrons in thetriplet state less than in the singlet., 〈V12〉Trip < 〈V12〉Sing. An interesting feature of the singlet-triplet energy splitting

is that although the perturbing interaction e2

4πε0|r1−r2| is spin-independent the symmetry of the wavefunction make thepotential behave as if it were spin-dependent.

Note that S2 = (s1 + s2) = s21 + s2

2 + 2s1 · s2 so that

〈2s1 · s2〉 =⟨S2⟩−⟨s2

1

⟩−⟨s2

2

⟩= S (S + 1) h2 − 3

4h2 − 3

4h2,

= S (S + 1) h2 − 3

2h2. (5.234)

Thus as S = 0 for the singlet and S = 1 for the triplet,

2

h2 〈s1 · s2〉 =1

2〈σ1 · σ2〉 = S (S + 1)− 3

2=

− 3

2 sin glet12 triplet

, (5.235)

88


Hence the energy shift is

∆E = J − 1

2(1 + σ1 · σ2)K. (5.236)

An effective "exchange interaction" term appears having a similar magnitude to the electrostatic Coulomb repulsion.Note that the magnetic dipole interaction which gives rise to the spin-orbit coupling is not responsible for the spin-spincoupling - it is too weak. The exchange force arises owing to symmetry effects alone. It gives rise to ferromagnetism viaan interaction of the form V =

∑Ni=1Aσi · σi+1 between neighbouring atoms in a ferromagnet. (In ferromagnetism spins

tend to align because −Kσ1 · σ2 lowers the energy if the spins are aligned and K > 0).For helium the direct integral depends on the average separation of the electrons. Since the probability density of the

electrons in the 100 state is spherically symmetric, the average distance between the electrons depends only on the radialprobability of the two particles, i.e. ∝ r2R2

n` (r). Thus the average separation is greater for ` = 0 , ` = 0 than ` = 0,` = 1 (see radial density plots, e.g. Eisberg p 306), so that 〈r12〉00 > 〈r12〉01, and hence 〈V12〉00 < 〈V12〉01 and thep-state energy is increased more than the s-state. Singlets lie above triplets in a multiplet - an example of Hund’s rule;other things being equal, state of highest spin has lowest energy.

89


90

APPENDIX A. PROBLEM SHEETS

Appendix A

Problem Sheets

The problem sheets and model answers for the course are reproduced here

91

PHAS3226: Quantum Mechanics APPENDIX A. PROBLEM SHEETS

92

PHAS3226: Quantum Mechanics Problem Sheet 1

Quantum Mechanics: PHAS3226, Winter 2011Problem Sheet 1

Hand in your answers by Tuesday 25 October, either at the lectures or Dr Bowler’s pigeonhole in Physics. Marks persection are shown in square brackets.

1. (a) In quantum theory, physical observables are represented by Hermitian operators. Using Dirac notation, definethe meaning of the term “Hermitian”. [2]

(b) A Hermitian operator A has eigenvalues an and normalised eigenvectors |φn〉. Demonstrate that if two eigen-values am and an are different then the associated eigenvectors are orthogonal: 〈φm|φn〉 = 0. [2]

(c) A general normalised state vector |ψ〉 is a linear superposition of the normalised eigenstates |φn〉:

|ψ〉 =∑n

cn|φn〉

When a measurement of the observable A is made in the state |ψ〉, the probability of finding the value an isdenoted by pn. Explain why pn is equal to |cn|2. (Assume that the expectation value of A in the given state is〈A〉 = 〈ψ|A|ψ〉.) [2]

2. The observable A is represented by operator A. The eigenvectors of A are |φ1〉 and |φ2〉 which are orthonormal.The corresponding eigenvalues are +1 and −1 respectively. The system is prepared in a state |ψ〉 given by

|ψ〉 =4

10|φ1〉+ c|φ2〉.

(a) If the state |ψ〉 is normalized what is the value of |c|? [2]

(b) Calculate the numerical value of the expectation value of A in the state |ψ〉. [4]

(c) With the system in the state |ψ〉 a measurement of observable A is made. What is the probability of obtainingthe eigenvalue −1? [2]

(d) Immediately after the measurement is made, with the result −1, what is now the expectation value of A?Explain your answer briefly. [2]

3. An electron in a hydrogen atom is described by the wavefunction:

ψ(r) ∝ (ψ100(r) + 2ψ210(r)− 3ψ32−2(r)− 4ψ411(r))

where ψnlmlare the orthonormal eigenfunctions of the hydrogen atom with n, l,ml the usual quantum numbers.

(a) Normalise this wavefunction [2]

(b) Evaluate the probability that the electron is measured to be in the ground state [2]

(c) Evaluate the expectation value of the energy (you may assume that the energy levels of the hydrgen atom aregiven by En = −1/2n2 a.u.) [2]

(d) Evaluate the expectation value of Lz [3]

(e) Evaluate the probability that the electron is found to have orbital angular momentum l = 1 [2]

(f) Evaluate the probability that, having obtained l = 1, a subsequent measurement of Lz obtains the valueml = 0 [3]

4. A bound quantum system has a complete set of orthonormal, non-degenerate energy eigenfunctions un with differ-ent energy eigenvalues En. The operator B corresponds to some other observable and is such that:

Bu1 = u2;Bu2 = u1;Bun = 0, n ≥ 3

(a) FInd the complete orthonormal set of eigenfunctions of the operator B [Hint: expand out the eigenvectors ofB in terms of ui and do not neglect any solutions] [6]

(b) If B is measured and found to have the eigenvalue +1, what is the expectation value of the energy in theresulting state ? [4]

93


Quantum Mechanics: PHAS3226, Winter 2011Problem Sheet 1 Worked Answers

1. (a) Definition of “Hermitian”: an operator A is Hermitian if, for any two arbitrary state vectors |ψ1〉 and |ψ2〉:

〈ψ1|A|ψ2〉? = 〈ψ2|A|ψ1〉 (A.1)

(b) For two eigenvectors |φm〉 and |φn〉:

〈φm|A|φn〉 = an〈φm|φn〉 (A.2)

Take complex conjugate and use the fact that A is Hermitian:

〈φm|A|φn〉? = 〈φn|A|φm〉 = a?n〈φm|φn〉? = an〈φn|φm〉 (A.3)

since a?n = an and 〈φm|φn〉? = 〈φn|φm〉 by definition. But we also have: φn|A|φm〉 = am〈φn|φm〉 Itfollows that:

an〈φn|φm〉 = am〈φn|φm〉 (A.4)

Since am 6= an, we must have 〈φn|φm〉 = 0.

(c) From meaning of expectation value:〈A〉 =

∑n

pnan (A.5)

But the expectation value is also given by:

〈A〉 = 〈ψ|A|ψ〉 =∑m,n

c?mcn〈φm|A|φn〉 =∑m,n

c?mcnan〈φm|φn〉 =∑n

|cn|2an (A.6)

Hence: for any state |ψ〉, we must have pn = |cn|2.

2. (a) Normalization requires 1 = 16100 + |c|2, so c =

√84

10 ' 0.9165.

(b) The expectation values of A in state |ψ〉 is⟨ψ|A|ψ

⟩=

(4

10〈φ1|+ c∗〈φ2|

)A

(4

10|φ1〉+ c|φ2〉

)(A.7)

Using the eigenvalue equations A|φ1〉 = +|φ1〉 and A|φ2〉 = −|φ2〉⟨ψ|A|ψ

⟩=

(4

10〈φ1|+ c∗〈φ2|

)(4

10|φ1〉 − c|φ2〉

)(A.8)

=16

100− |c|2 =

16

100− 84

100= − 68

100(A.9)

Alternatively, and far quicker, can calculate the expectation value from∑n |cn|

2an i.e. 16

100 (1)+ 84100 (−1) =

− 68100 .

(c) The probability is the modulus squared of the coefficient of the state |φ2〉, i.e. 84100 .

(d) Immediately after the measurement the system is in state |φ2〉. Hence the expectation value for A is −1.

3. (a) The wavefunction is an expansion in terms of orthonormal eigenfunctions for the hydrogen atom of the form∑n cnψn. Normalization requires

∑n |cn|

2= 1. In this case

∑n |cn|

2= 30, so the normalized wavefunc-

tion isψ (r) =

1√30

(ψ100 + 2ψ210 − 3ψ32−2 − 4ψ411) . (A.10)

(b) The probability that the system is measured to be in an energy state given by wavefunction ψn`m is |cn|2.

Therefore, the probability that the electron is measured to be in the ground state, ψ100 , is∣∣∣ 1√

30

∣∣∣2 = 130 .

(c) The expectation value of the energy is

〈E〉 =∑n

|cn|2En =1

30

(−1

2

)+

4

30

(−1

8

)+

9

30

(− 1

18

)+

16

30

(− 1

32

)= − 1

15au = −0.066667au.

(A.11)

94


(d) The expectation value of Lz , is

〈Lz〉 =∑n

|cn|2mnh =1

30(0) +

4

30(0) +

9

30(−2) +

16

30(1) =

1

15h. (A.12)

(e) Measurement of ` = 1 select states ψ210 and ψ411. Thus the probability is∣∣∣ 2√

30

∣∣∣2 +∣∣∣ 4√

30

∣∣∣2 = 23

(f) Having obtained ` = 1 the resulting state ψ′ ∝ 2ψ210 − 4ψ411. Normalization of this state gives the wave-function after the measurement as ψ′ = 1√

20(2ψ210 − 4ψ411), so the probability that a measurement of Lz

obtains the value m` = 0 is∣∣∣ 2√

20

∣∣∣2 = 15 .

4. (a) The eigenvalues, φi for B will satisfy Bφi = biφi. We can expand these eigenvalues in terms of the orthonor-mal energy eigenfunctions un:

φi =∑n

cinun (A.13)

Substituting this in, and using the information given for B operating on un gives:

B∑n

cinun = ci1u2 + ci2u1 = bi∑n

cinun (A.14)

Take the scalar product on the left with uk; I will shift to Dirac notation for simplicity:

ci1〈uk|u2〉+ ci2〈uk|u1〉 = bi∑n

cin〈uk|un〉 (A.15)

Now, we know that 〈ui|uj〉 = δij , so

ci2 = bici1, k = 1 (A.16)ci1 = bici2, k = 2 (A.17)

0 = bicik, k ≥ 3 (A.18)

We can now solve for bi and cin. If we add Eq. (A.16) and Eq. (A.17), we get:

ci1 + ci2 = bi (ci1 + ci2) (A.19)⇒ bi = 1, ci1 = ci2, cin = 0n ≥ 3 (A.20)

Setting i = 1, we have b1 = 1 and φ1 = (u1 + u2)/√

2 (following normalisation). If we subtract Eqs. (A.16)and (A.17), then we find instead:

ci1 − ci2 = bi (ci2 − ci1) (A.21)⇒ bi = −1, ci1 = −ci2, cin = 0n ≥ 3 (A.22)

Taking i = 2, we have b1 = −1 and φ1 = (u1 − u2)/√

2 (following normalisation). Is that all ? No. We alsohave a solution for bi = 0, which yields ci1 = ci2 = 0. From the conditions of the question, we must haveci3 = 1 for i ≥ 3, which gives φi = ui, i ≥ 3.

(b) If the eigenvalue +1 has been obtained, then we know that the system has state φ1. The expectation value ofthe energy is then:

〈E〉 = 〈φ1|H|φ2〉 (A.23)

= 〈(u1 + u2)/√

2|H|(u1 + u2)/√

2〉 (A.24)

=1

2(〈u1|H|u1〉+ 〈u2|H|u2〉 =

1

2(E1 + E2) (A.25)

as we have that 〈ui|uj〉 = δij and the ui are energy eigenfunctions.

95


96



Hand in your answers by Tuesday 8 November, either at the lectures or Dr Bowler’s pigeonhole in Physics. Marks persection are shown in square brackets.

1. Starting from the basic commutator of position and momentum, [x, px] = ih, show that

(a)[x, p2

x

]= 2ihpx, [1]

(b)[x, p3

x

]= 3ihp2

x, [1]

(c) Prove by induction that [x, pnx ] = ihnpn−1x where n is a positive integer. [2]

(d) By expressing the position operator x and momentum operator p within a harmonic oscillator in terms of theraising and lowering operators a+ and a− respectively, show that in any harmonic oscillator state |n〉 that〈x〉 = 〈p〉 = 0 and

⟨x2⟩

=(n+ 1

2

) (hmω

). (You may assume the properties a+|n〉 =

√n+ 1|n + 1〉 and [2]

[2]a−|n〉 =√n|n− 1〉.)

(e) Using the Hamiltonian for the harmonic oscillator, or otherwise, show that the expectation value of the kineticenergy is equal to the expectation value of the potential energy and half that of the total energy. [2]

2. An harmonic oscillator is in the state |ψ〉 = 1√6|0〉+ 2√

6|1〉+ 1√

6|2〉.

(a) Evaluate in this state |ψ〉 the expectation values of x, px. [4]

(b) Show, by direct substitution into the eigenvalue equation a−|α〉 = α|α〉, that the state

|α〉 = e−|α|2/2

∞∑n=0

αn√n!|n〉

is the eigenfunction of the lowering operator a− with eigenvalue α. [4]

(c) Show that the expectation value of the position operator x in this state is given by (you may assume that thestate |α〉 is normalized, as it is!) [2]

〈x〉 =

(2h

mω

)1/2

Re (α) .

3. The ground state wave function for a particle of mass m moving with energy E in a one-dimensional harmonicoscillator potential with classical frequency ω is:

u0(x) = N0e−α2x2/2

where N0 is some normalisation constant and α =√mω/h.

(a) For the ground state, show explicitly that the quantum mechanical expectation values 〈x〉 and 〈p〉 are bothzero. [3]

(b) If the uncertainties ∆x and ∆p are given by:

(∆x)2

= 〈x2〉 − 〈x〉2; (∆p)2

= 〈p2〉 − 〈p〉2

obtain an expression for the expectation value of the energy in terms of the uncertainties. [3]

(c) If ∆x∆p = c for c a constant, deduce a value for c by minimising the ground state energy. What significancedoes the value of c have in light of the uncertainty principle ? [4]

97



1. (a) The property of commutator algebra to use is

[A,BC] = B [A,C] + [A,B]C (A.26)

Then [x, p2

x

]= px [x, px] + [x, px] px = pxih+ ihpx = 2ihpx. (A.27)

(b) Similarly [x, p3

x

]= px

[x, p2

x

]+ [x, px] p2

x = px2ihpx + ihp2x = 3ihp2

x. (A.28)

(c) If we assume that the relation is true for n, i.e.:

[x, pnx ] = ihnpn−1x (A.29)

then we can write:[x, pn+1

x

]= px [x, pnx ] + [x, px] pnx = pxihnp

n−1x + ihpnx = (n+ 1)ihpnx , (A.30)

so it is also true for n + 1. But we know it’s true for n = 1, and we have shown it for n = 2, 3 above, so byinduction it is true for all n.

(d) Raising and lowering operators are a+ =(

12hmω

)1/2(mωx− ip) and a− =

(1

2hmω

)1/2(mωx+ ip). By

adding and then re-arranging we have

x =

(h

2mω

)1/2

(a+ + a−) (A.31)

and by subtracting we have

p = i

(mhω

2

)1/2

(a+ − a−) . (A.32)

In the harmonic oscillator state |n〉

〈x〉 =

(h

2mω

)1/2

〈n|a+ + a−|n〉 . (A.33)

But a+|n〉 =√n+ 1|n+ 1〉 and a−|n〉 =

√n|n− 1〉 so

〈n|a+ + a−|n〉 = 〈n|√n+ 1|n+ 1〉+ 〈n|

√n|n− 1〉 = 0 (A.34)

since the states |n〉 are orthonormal, 〈k|n〉 = δkn.Similarly

〈p〉 = i

(mhω

2

)1/2

〈n|a+ − a−|n〉 = 0. (A.35)

For⟨x2⟩

=(n+ 1

2

) (hmω

)we use the action of x on the state |n〉 is

x|n〉 =

(h

2mω

)1/2

(a+ + a−) |n〉 =

(h

2mω

)1/2 (√n+ 1|n+ 1〉+

√n|n− 1〉

)(A.36)

Since x is Hermitian, the Hermitian conjugate of x|n〉 is 〈n|x so that

〈n|xx|n〉 =

(h

2mω

)〈n+ 1|

√n+ 1 + 〈n− 1|

√n√

n+ 1|n+ 1〉+√n|n− 1〉

⟨n|x2|n

⟩=

(h

2mω

)n+ 1 + n =

(n+

1

2

)(h

mω

). (A.37)

98


(e) The Hamiltonian for the harmonic oscillator is

H =p2

2m+

1

2mω2x2 = T + V (A.38)

and ⟨H⟩

= 〈T 〉+ 〈V 〉 =1

2m

⟨p2⟩

+1

2mω2

⟨x2⟩. (A.39)

But⟨H⟩

= En =(n+ 1

2

)hω and

⟨x2⟩

=(n+ 1

2

) (hωmω2

)so

〈V 〉 =1

2mω2

(n+

1

2

)(hω

mω2

)=

(n+

1

2

)1

2hω =

1

2En.

Hence

En = 〈T 〉+ 〈V 〉 = 〈T 〉+1

2En, (A.40)

〈T 〉 =1

2En. (A.41)

2. (a) Since x =(

h2mω

)1/2(a+ + a−) then its expectation value in the state |ψ〉 = 1√

6|0〉+ 2√

6|1〉+ 1√

6|2〉 is given

by

〈ψ|x|ψ〉 =

(1√6〈0|+ 2√

6〈1|+ 1√

6〈2|)(

h

2mω

)1/2

(a+ + a−)

(1√6|0〉+

2√6|1〉+

1√6|2〉). (A.42)

Using the actions of a+ and a− on states |n〉, i.e. a+|n〉 =√n+ 1|n + 1〉 and a−|n〉 =

√n|n − 1〉 and the

orthonormality of the basis states, 〈n|m〉 = δnm we get

〈ψ|x|ψ〉 =

(h

2mω

)1/2(1√6〈0|+ 2√

6〈1|+ 1√

6〈2|)(

2√6|0〉+

1 +√

2√6|1〉+

2√

2√6|2〉+

√3√6|3〉

),

=

(h

2mω

)1/21

6

(2〈0|0〉+ (1 +

√2)〈0|1〉+ 2

√2〈0|2〉+

√3〈0|3〉+ 4〈1|0〉+ 2(1 +

√2)〈1|1〉

+4√

2〈1|2〉+ 2√

3〈1|3〉+ 2〈2|0〉+ (1 +√

2)〈2|1〉+ 2√

2〈2|2〉+√

3〈2|3〉

),

=

(h

2mω

)1/21

6

(2 + 2(1 +

√2) + 2

√2)

=2 + 2

√2

3

(h

2mω

)1/2

.

For px,

〈ψ|px|ψ〉 =

(1√6〈0|+ 2√

6〈1|+ 1√

6〈2|)i

(mhω

2

)1/2

(a+ − a−)

(1√6|0〉+

2√6|1〉+

1√6|2〉)

(A.43)

= i

(mhω

2

)1/21

6(〈0|+ 2〈1|+ 〈2|)

(−2|0〉+ (1−

√2)|1〉+ 2

√2|2〉+

√3|3〉

), (A.44)

= i

(mhω

2

)1/21

6

(−2 + 2(1−

√2) + 2

√2)

= 0 (A.45)

(b) Since a−|n〉 =√n|n− 1〉 then applying a− to |α〉 as given above leads to

a−|α〉 = e−|α|2/2

∞∑n=0

αn√n!a−|n〉 = e−|α|

2/2∞∑n=0

αn√n!

√n|n− 1〉 (A.46)

= e−|α|2/2

∞∑n=1

ααn−1√(n− 1)!

|n− 1〉 = αe−|α|2/2

∞∑n=0

αn√n!|n〉 = α|α〉. (A.47)

(c) Express x in terms of a+ and a− as

x =

(h

2mω

)1/2

(a+ + a−) . (A.48)

Since a−|α〉 = α|α〉, and the state |α〉 is normalized already, then

a−|α〉 = α|α〉, 〈α|a−|α〉 = α〈α|α〉 = α.

99


Also as a+ and a− are Hermitian conjugates of each other then taking the Hermitian conjugate

(a−|α〉)† = (α|α〉)† , (A.49)〈α|a+ = 〈α|α∗, (A.50)

〈α|a+|α〉 = 〈α|α∗|α〉 = α∗. (A.51)

Hence

〈x〉 =

(h

2mω

)1/2

〈α| (a+ + a−) |α〉 =

(h

2mω

)1/2

(〈α|a+|α〉+ 〈α|a−|α〉) , (A.52)

=

(h

2mω

)1/2

(α∗ + α) =

(h

2mω

)1/2

2Re (α) =

(2h

mω

)1/2

Re (α) . (A.53)

3. (a) Expectation values are found from:

〈A〉 =

∫ ∞∞

u?0(x)Au0(x)dx =

∫ ∞∞|u0(x)|2Adx (A.54)

For x we find 〈x〉 = 0 as the integrand is an odd function. For p we find:

〈p〉 =

∫ ∞∞

u?0(x)

(−ih d

dx

)u0(x)dx = |N0|2

∫ ∞∞

e−α2x2/2

(−ih d

dx

)e−α

2x2/2 (A.55)

= −ih|N0|2∫ ∞∞

e−α2x2/2(−α2x)e−α

2x2/2dx = 0 (A.56)

as the integrand is odd again.

(b) As 〈x〉 = 0, then (∆x)2

= 〈x2〉 and similarly (∆p)2

= 〈p2〉. The energy E = p2/2m+mω2x2/2 so:

〈E〉 =〈p2〉2m

+1

2mω2〈x2〉 =

(∆p)2

2m+

1

2mω2 (∆x)

2 (A.57)

(c) We have that ∆x∆p = c, so we can write ∆p = c/∆x and so:

〈E〉 =c2

2m (∆x)2 +

1

2mω2 (∆x)

2 (A.58)

Minimising with respect to (∆x)2 gives:

∂〈E〉∂ (∆x)

2 = − 1

2m

c2

(∆x)4 +

1

2mω2 = 0 (A.59)

(∆x)2

=c

mω(A.60)

Emin =cω

2+cω

2= cω (A.61)

But we know that the ground state energy of the harmonic oscillator is E = 12 hω, so c = 1

2 h and therefore∆x∆p = 1

2 h, which is consistent with the Uncertainty Principle.

100



Hand in your answers by Tuesday 29 November, either at the lectures or Dr Bowler’s pigeonhole in Physics. Marks persection are shown in square brackets.

1. (a) The operators Jx, Jy and Jz are Cartesian components of the angulare momentum operator, obeying the usual

commutation relations ([Jx, Jy

]= ihJz etc). Use these commutation relations to show that the operators

J+ = Jx + iJy and J− = Jx − iJy satisfy the following commutation relations: [3][Jz, J+

]= hJ+ and

[Jz, J−

]= −hJ−

(b) Let |jm〉 be a simultaneous eigenvector of J2 and Jz having eigenvalues j(j + 1)h2 and mh, respectively.Show that J−|jm〉 is an eigenvector of Jz , and that its eigenvalue is (m− 1)h. [3]

(c) With |jm〉 correctly normalised, obtain a formula for 〈jm|J−J+|jm〉 in terms of j and m. For a given valueof j, what value must m have if 〈jm|J−J+|jm〉 = 0 ? What value must m have if 〈jm|J+J−|jm〉 = 0 ? [4]

2. The raising and lowering angular momentum operators, J+, J− are defined in terms of the Cartesian componentsJx, Jy , Jz of angular momentum J by J+ = Jx + iJy and J− = Jx − iJy.

(a) Obtain the matrix representation of Jy for the state with j = 1 in terms of the set of eigenstates of Jz . [6]

(b) Solve the eigenvalue equation, i.e.Jyc = λc, using the matrix representation of Jy and the basis states

|j,m〉 ≡ |1, 1〉 =

100

; |1, 0〉 =

010

; |1,−1〉 =

00−1

; (A.62)

(i) to find ALL the eigenvalues λ, [2](ii) and ANY ONE of the eigenvectors c . (Make the first element of the eigenvector real and positive.) [2]

3. (a) Calculate explicitly the commutator [σz, σx] and show how it is related to σy [2]

(b) The eigenvectors of Sz are denoted by | 12 〉 and | − 12 〉. A system is in a spin state represented by the state

vector:

|ψ〉 =1√2

(|12〉+ | − 1

2〉)

(A.63)

(i) If a measurement is made of Sz , what are the possible outcomes of the measurement, and what are theprobabilities of these outcomes ? [2]

(ii) If a measurement is made on the same state of Sx, what are the possible outcomes of the measurementand the probabilities of these outcomes ? [2]

(c) The basis states |α〉 and |β〉 are the eigenstates of Sz with eigenvalues h/2 and −h/2 respectively. Show thatthe state |χ+

n 〉 = cos (θ/2) |α〉 + sin (θ/2) |β〉 is an eigenstate of the spin operator Sn = S · n = Sx sin θ +

Sz cos θ, the component of spin S of a spin- 12 particle in the direction of the unit vector n = (sin θ, 0, cos θ)

lying in the x-z plane, with eigenvalue +h/2. [2]

(d) If a beam of electrons polarised with spin +h/2 along the z-axis moves through a Stern-Gerlach magnetoriented along a direction n = (sin θ, 0, cos θ), expand the initial state in terms of the state |χ+〉 and theequivalent state |χ−〉. How many beams emerge, and what are their relative intensities ? [2]

101



1. (a) We will need to expand out explicitly. First J+:[Jz, J+

]= (Jz

(Jx + iJy

)−(Jx + iJy

)Jz (A.64)

= JzJx − JxJz + iJzJy − iJyJz =[Jz, Jx

]+ i[Jz, Jy

](A.65)

= ihJy + i(−ihJx

)(A.66)

= h(Jx + iJy) = hJ+ (A.67)

as required. Now J−:[Jz, J−

]= (Jz

(Jx − iJy

)−(Jx − iJy

)Jz (A.68)

= JzJx − JxJz − iJzJy + iJyJz =[Jz, Jx

]+ i[Jy, Jz

](A.69)

= ihJy + i(ihJx

)(A.70)

= −h(Jx − iJy) = −hJ− (A.71)

again, as required.

(b) We will need to act on J−|jm〉 with Jz; we will use the commutator relations just derived as well:

JzJ−|jm〉 =(−hJ− + J−Jz

)|jm〉 (A.72)

= −hJ−|jm〉+ J−mh|jm〉 (A.73)= (m− 1)hJ−|jm〉 (A.74)

which shows that J−|jm〉 is an eigenvector of Jz with eigenalue (m− 1)h, as required.

(c) We will expand out J+ and J−:

〈jm|J−J+|jm〉 = 〈jm|(Jx − iJy

)(Jx + iJy

)|jm〉 (A.75)

= 〈jm|J2x + J2

y + iJxJy − iJyJx|jm〉 (A.76)

= 〈jm|J2 − J2z + i

[Jx, Jy

]|jm〉 (A.77)

= 〈jm|J2 − J2z + i(ihJz)|jm〉 (A.78)

= 〈jm|J2 − J2z − hJz|jm〉 (A.79)

= j(j + 1)h2 −m2h2 −mh2 = (j(j + 1)−m(m+ 1)) h2 (A.80)

where we have used: J2|jm〉 = j(j + 1)h2|jm〉 and Jz|jm〉 = mh|jm〉. Then if m = j we will have〈jm|J−J+|jm〉 = 0. The only change if we look at 〈jm|J+J−|jm〉 will be the sign of the commutator afterexpansion, so we will have 〈jm|J+J−|jm〉 = (j(j + 1)−m(m− 1)) h2 so if m = −j then we will find〈jm|J+J−|jm〉 = 0.

2. By adding, we have

Jy =1

2i

(J+ − J−

). (A.81)

The action of J+, J− on |1,m〉 are

J+|1,m〉 =√

2−m (m+ 1)h|1,m+ 1〉 (A.82)

J−|1,m〉 =√

2−m (m− 1)h|1,m− 1〉 (A.83)

and so ⟨1,m′|J+|1,m

⟩=

√2−m (m+ 1)h〈1,m′|1,m+ 1〉 =

√2−m (m+ 1)hδm′m+1, (A.84)⟨

1,m′|J−|1,m⟩

=√

2−m (m− 1)h〈1,m′|1,m− 1〉 =√

2−m (m− 1)hδm′m−1, (A.85)⟨1,m′|Jy|1,m

⟩=

1

2i

√2−m (m+ 1)hδm′m+1 −

√2−m (m− 1)hδm′m−1

. (A.86)

102


(a) The matrix can be constructed by choosing m′ = 1, 0,−1 and m = 1, 0,−1 in turn in eq(A.86) and notingthat the Kronecker delta δkn = 1 if k = n and δkn = 0 if k 6= n. Clearly the diagonal elements are zero andthe table below can be built up,

m = 1 m = 0 m = −1

m′ = 1 0√

22i h 0

m′ = 0 −√

22i h 0

√2

2i h

m′ = −1 0 −√

22i h 0

and so the matrix representation of Jy is Jy = h√2

0 −i 0i 0 −i0 i 0

.

(b) We have to solve the matrix equation Jyc = λc for the eigenvectors c. This equation has a non-trivial solutionfor c if

det∣∣∣Jy − λI∣∣∣ = 0. (A.87)

Writing λ = h√2µ we have to solve

h√2

0 −i 0i 0 −i0 i 0

c1c2c3

= λ

c1c2c3

=h√2µ

c1c2c3

, (A.88)

h√2

−µ −i 0i −µ −i0 i −µ

c1c2c3

= 0 (A.89)

This equation has a non-trivial solution for c if

det

∣∣∣∣∣∣−µ −i 0i −µ −i0 i −µ

∣∣∣∣∣∣ = 0. (A.90)

−µ∣∣∣∣ −µ −i

i −µ

∣∣∣∣+ i

∣∣∣∣ i −i0 −µ

∣∣∣∣ = −µ(µ2 − 1

)+ i(−iµ) = −µ3 + 2µ = −µ

(µ2 − 2

)= 0. (A.91)

Thus µ = ±√

2, 0 and the eigenvalues are λ = +h, 0h, −h (As is to be expected as there is not differencebetween the y-axis and the z-axis for a free particle.)To find the eigenvectors we have to find the components c1, c2, c3.For λ = h, (µ = +

√2)

h√2

0 −i 0i 0 −i0 i 0

c1c2c3

= h

c1c2c3

, (A.92)

0 −i 0i 0 −i0 i 0

c1c2c3

=√

2

c1c2c3

. (A.93)

Multiplying out gives

−ic2ic1 − ic3ic2

=√

2

c1c2c3

so

c2 = i√

2c1; c1 − c3 = −i√

2c2; c2 = −i√

2c3, (A.94)

so c1 = −c3 = c2/(i√

2). Since the eigenvector must be normalized, i.e. |c1|2 + |c2|2 + |c3|2 = 1 thenusing the above |c2|2/2 + |c2|2 + |c2|2/2 = 1, and c2 = ±1/

√2. Taking c2 = 1/

√2 then the eigenvector for

eigenvalue +h is

c+1 =

−i/21/√

2i/2

=1

2

−i√2i

. (A.95)

103


For eigenvalue λ = −h then we have

h√2

0 −i 0i 0 −i0 i 0

c1c2c3

= −h

c1c2c3

, (A.96)

0 −i 0i 0 −i0 i 0

c1c2c3

= −√

2

c1c2c3

. (A.97)

andic2 =

√2c1; i(c1 − c3) = −

√2c2; ic2 = −

√2c3, (A.98)

so as before c1 = −c3 = ic2/√

2. Normalizing as before, c2 = ±1/√

2. Taking c2 = 1/√

2 then theeigenvector for eigenvalue −h is

c−1 =

i/2

1/√

2−i/2

=1

2

i√2−i

. (A.99)

Finally for λ = 0 we have

h√2

0 −i 0i 0 −i0 i 0

c1c2c3

= 0. (A.100)

Then−ic2 = 0; i(c1 − c3) = 0; ic2 = 0,

so c1 = c3 and normalization gives c1 = 1√2

. Hence the eigenvector for eigenvalue 0h is

c0 =

1√2

01√2

. (A.101)

3. (a) We have:

[σz, σx] =

(1 00 −1

)(0 11 0

)−(

0 11 0

)(1 00 −1

)(A.102)

=

(0 1−1 0

)−(

0 −11 0

)=

(0 2−2 0

)(A.103)

This last matrix is, of course, 2iσy so we can write:

[σz, σx] = 2iσy (A.104)

(b) (i) The eigenvalues of Sz are mh, so the possible outcomes of a measurement are ± 12 h. The probabilities

are 50% for each (as the components of the spin state are equal (N.B. a probability of 12 is also a good

way to say this)).(ii) Using matrix notation, we can write:

|ψ〉 =1√2

(10

)+

1√2

(01

)=

1√2

(11

)(A.105)

But this is an eigenvector of σx with eigenvalue 1. We have that Sx = 12 hσx so the outcome of measuring

Sx in this state is certainly 12 h

(c) The easiest way to show that Sn|χ+n 〉 = h

2 |χ+n 〉 with Sn = Sx sin θ + Sz cos θ and |χ+

n 〉 = cos (θ/2) |α〉 +

sin (θ/2) |β〉 is to substitute. Noting that Sx = 12

(S+ − S−

)then(

1

2

(S+ − S−

)sin θ + Sz cos θ

)(cos (θ/2) |α〉+ sin (θ/2) |β〉)

=h

2sin θ cos (θ/2) |β〉+

h

2sin θ sin (θ/2) |α〉+

h

2cos θ cos (θ/2) |α〉 − h

2cos θ sin (θ/2) |β〉,

=h

2(cos θ cos (θ/2) + sin θ sin (θ/2)) |α〉+

h

2(sin θ cos (θ/2)− cos θ sin (θ/2)) |β〉,

=h

2cos (θ/2) |α〉+

h

2sin (θ/2) |β〉 =

h

2(cos (θ/2) |α〉+ sin (θ/2) |β〉) =

h

2|χ+

n 〉. (A.106)

104


(d) Initially the beam is in the state |α〉. The Stern-Gerlach measures the component of spin along the direction nin the x-z plane, i.e. it measures Sn. Thus expand |α〉 in terms of the eigenstates |χ+

n 〉, and |χ−n 〉 of Sn witheigenvalues +h/2 and −h/2 respectively, as

|α〉 = a|χ+n 〉+ b|χ−n 〉. (A.107)

with |a|2 + |b|2 = 1 for normalization. The magnet splits the beam into 2 components.The relative intensities are |a|2 and |b|2. But a = 〈χ+

n |α〉 = (〈α| cos (θ/2) + 〈β| sin (θ/2)) |α〉 = cos (θ/2).Hence the relative probabilities are |a|2 = cos2 (θ/2) and |b|2 = 1− |a|2 = sin2 (θ/2).

105


106



Hand in your answers by Tuesday 13 December, either at the lectures or Dr Bowler’s pigeonhole in Physics. Marks persection are shown in square brackets.

1. The vector operator S representing the total spin angular momentum of two spin-1/2 particles is: S = S1 + S2,where S1 and S2 are spin angular momenta of particle 1 and 2. The x−, y− and z−components of S1 and S2 aredenoted by S1x, S1y , S1z and S2x, S2y , S2z , respectively, and the components of S are Sx = S1x + S2x, etc...The state vectors α1α2, α1β2, β1α2 and β1β2 represent the states in which both particles are spin-up, particle 1 isspin-up and particle 2 is spin-down, etc...

(a) With S1+ = S1x + iS1y and S1− = S1x − iS1y being the raising and lowering operators for particle 1 and asimilar notation being used for particle 2, show that: [2]

S1+S2− + S1−S2+ = 2(S1xS2x + S1yS2y

)(A.108)

(b) Hence, show that the scalar product of S1 and S2is: [2]

S1 · S2 =1

2

(S1+S2− + S1−S2+

)+ S1zS2z (A.109)

(c) Use the fact that S2 = S21 + S2

2 + 2S1 · S2 to examine the effect of S2 acting on the state vector α1α2, andshow that α1α2 is an eigenvector of S2. What is the eigenvalue? [3]

(d) Using the same method as in part (c), show that the state vector 1√2

(α1β2 + β1α2) is an eigenvector of S2,

and find the eigenvalue. If a measurement of S1 · S2 is made, what will be the outcome ? [3]

2. A particle of mass m is confined to move in a one-dimensional "infinite" potential well defined by V (x) = 0, |x| ≤a, V (x) = ∞ otherwise. The energy eigenvalues are En = n2π2h2

8ma2 , with n = 1, 2, 3, . . . and the orthonormaleigenfunctions are the even and odd functions

ψn (x) =

1√a

cos(nπx2a

)for n = 1, 3, 5 . . .

1√a

sin(nπx2a

)for n = 2, 4, 6 . . .

.

The potential is modified between −a < x < a to V (x) = ε π2h2

8ma2 sin(

3πx2a

)with ε < 1.

(a) Determine the ground state energy to first-order in ε. [4](b) Determine the ground state energy to second-order in ε. [6]

[HINT: (a) Note 2 sinA cosB = sin(A+B) + sin (A−B), (b) the eigenfunctions ψn (x) are orthonormal. ]

3. Solve the above problem with ε = 12 , using the variational method. With b a variational parameter, use a trial

function: [10]

φT (x) =1√a

cos(πx

2a

)+ b

1√a

sin(πxa

)≡ ψ1 (x) + bψ2 (x)

[HINT: (a) When evaluating 〈H〉 with the trial function, make use of the fact that ψ1 (x) and ψ2 (x) are eigenfunc-tions of the one-dimensional infinite potential well, (b) the eigenfunctions ψn (x) are orthonormal.]

4. A system consists of two indistinguishable spin-1 fermions, both confined inside the same box of length L centredon the origin. The particles do not interact with each other.

(a) What is the total energy of the ground state of this system ? (Use the standard formulas for the energyeigenvalues of a single particle in a box.) [1]

(b) What is the spin angular momentum of the system in the ground state? [2](c) How many distinct and mutually orthogonal space-spin energy eigenstates of the whole system are there in

which there is one particle in the single-particle ground state and one particle in the single-particle first excitedstate? Choosing these eigenstates so that they are also eigenvectors of the square and the z-component of totalspin angular momentum, write explicit formulas for all these space-spin eigenstates. (Use α, β notation forspin-up and spin-down states.) [4]

(d) With no interaction between the particles, the energy eigenstates described in part (c) all have exactly the sameenergy (they are degenerate). We now introduce a repulsive interaction between the two particles. Describethe qualitative effect that this interaction has on the relative energies of the singlet and triplet eigenstates. [3]

107



1. (a) We have:

S1+S2− =(S1x + iS1y

)(S2x − iS2y

)= S1xS2x + S1yS2y + i

(S1yS2x − S1xS2y

)S1−S2+ =

(S1x − iS1y

)(S2x + iS2y

)= S1xS2x + S1yS2y + i

(S1xS2y − S1yS2x

)Hence:

S1+S2− + S1−S2+ = 2(S1xS2x + S1yS2y

)(b) Writing out, we find:

S1 · S2 = S1xS2x + S1yS2y + S1zS2z

=1

2

(S1+S2− + S1−S2+

)+ S1zS2z

(c) Expanding out, we have:S2α1α2 = S2

1α1α2 + S22α1α2 + 2S1 · S2α1α2

We know that:

S21α1 = s(s+ 1)h2α1 =

3

4h2α1

S22α2 = s(s+ 1)h2α2 =

3

4h2α2

since s = 1/2 in both cases. Also:

2S1 · S2α1α2 = 2S1zS2zα1α2 +(S1+S2− + S1−S2+

)α1α2 = 2(

1

2h)2α1α2 + 0

since anything with S1+ or S2+ acting on α1α2 is zero. Hence:

S2α1α2 =′(

2× 3

4+ 2× 1

4

)h2α1α2 = 2h2α1α2

So α1α2 is an eigenvector of S2 with eigenvalue 2h2.

(d) In this case:

S2 1√2

(α1β2 + β1α2) =(S2

1 + S22 + 2S1 · S2

) 1√2

(α1β2 + β1α2)

Now α1β2 and β1α2 are eigenvectors of S21 and S2

2 with the eigenvalues in both cases being 34 h

2. Turning toS1S2, we see that:

2S1 · S2α1β2 =(

2S1zS2z + (S1+S2− + S1−S2+))α1β2 = −1

2h2α1β2 + h2β1α2

Similarly:

2S1 · S2β1α2 =(

2S1zS2z + (S1+S2− + S1−S2+))α1β2 = −1

2h2β1α2 + h2α1β2

Hence:

S2 1√2

(α1β2 + β1α2) =

(2× 3

4h2 − 1

2h2 + h2

)1√2

(α1β2 + β1α2) = 2× 1√2

(α1β2 + β1α2)

so the given vector is an eigenvector of S2 with eigenvector 2h2. If a measurement of S1 · S2 is made then wewill find:

S1 · S21√2

(α1β2 + β1α2) =

(−1

4h2 +

1

2

)1√2

(α1β2 + β1α2)

so the result will be definite, with value 14 h

2.

108


2. (a) The ground state is non-degenerate and has n = 1, E1 = π2h2

8ma2 and the unperturbed eigenfunction is |ψ(0)1 〉 =

1√a

cos(πx2a

)for |x| < a, and |ψ(0)

1 〉 = 0, for |x| > a. The first-order correction to the ground state energy,

E(1)1 is obtained from

E(1)1 = 〈ψ(0)

1 |επ2h2

8ma2sin

(3πx

2a

)|ψ(0)

1 〉 = επ2h2

8ma2〈ψ(0)

1 | sin(

3πx

2a

)|ψ(0)

1 〉, (A.110)

= επ2h2

8ma2

∫ a

−aψ

(0)∗1 (x) sin

(3πx

2a

)1√a

cos(πx

2a

)dx,

= επ2h2

8ma2

∫ a

−aψ

(0)∗1 (x)

1

2

1√a

sin

(4πx

2a

)+

1√a

sin

(2πx

2a

)dx,

= επ2h2

8ma2

∫ a

−aψ

(0)∗1

1

2

ψ

(0)4 + ψ

(0)2

dx = ε

π2h2

16ma2

〈ψ(0)

1 |ψ(0)4 〉+ 〈ψ(0)

1 |ψ(0)2 〉

= 0,(A.111)

since the states |ψ(0)1 〉, |ψ

(0)2 〉, |ψ

(0)4 〉 are mutually orthogonal. Thus there is no change to first-order in ε.

This could also be seen immediately from the form of

E(1)1 = 〈ψ(0)

1 |επ2h2

8ma2sin

(3πx

2a

)|ψ(0)

1 〉 (A.112)

since ψ(0)1 (x) is an even function of x and the perturbation term sin

(3πx2a

)is an odd function of x.

(b) The second-order correction E(2)1 is obtained from the expression

E(2)1 =

∞∑n=2

∣∣∣〈ψ(0)n |H ′|ψ(0)

1 〉∣∣∣2

E(0)1 − E(0)

n

=

∞∑n=2

∣∣∣〈ψ(0)n

∣∣∣ε π2h2

8ma2 sin(

3πx2a

)∣∣∣ψ(0)1 〉∣∣∣2

E(0)1 − E(0)

n

. (A.113)

The matrix elements are

H ′n1 = 〈ψ(0)n

∣∣∣∣ε π2h2

8ma2sin

(3πx

2a

)∣∣∣∣ψ(0)1 〉 = ε

π2h2

8ma2〈ψ(0)n

∣∣∣∣sin(3πx

2a

)∣∣∣∣ψ(0)1 〉. (A.114)

Since the ground state has even parity, the perturbation has odd parity, the only non-zero elements are for theodd parity states, |ψ(0)

n 〉 i.e. for n = 2, 4, 6 . . .. Consider

〈ψ(0)n


2a

)∣∣∣∣ψ(0)1 〉 =

∫ a

−aψ(0)∗n (x) sin

(3πx

2a

)1√a

cos(πx

2a

)dx, (A.115)

=

∫ a

−aψ(0)∗n (x)

1

2

1√a

sin

(4πx

2a

)+

1√a

sin

(2πx

2a

)dx,

=

∫ a

−aψ(0)∗n

1

2

ψ

(0)4 + ψ

(0)2

dx =

1

2

〈ψ(0)n |ψ

(0)4 〉+ 〈ψ(0)

n |ψ(0)2 〉,

=1

2δn4 + δn2 . (A.116)

Hence the only terms that contribute to the summation in the expression for E(2)1 are for n = 2 and n = 4.

Thus

E(2)1 =

∣∣∣〈ψ(0)2 |H ′|ψ

(0)1 〉∣∣∣2

E(0)1 − E(0)

2

+

∣∣∣〈ψ(0)4 |H ′|ψ

(0)1 〉∣∣∣2

E(0)1 − E(0)

4

(A.117)

=

(επ2h2

8ma2

)2 (

12

)2π2h2

8ma2 (1− 22)+

(12

)2π2h2

8ma2 (1− 42)

=ε2

4

π2h2

8ma2

1

−3+

1

−15

=

ε2

4

π2h2

8ma2

−2

5

= − ε

2

10

(π2h2

8ma2

). (A.118)

and the energy to second-order in ε is

E1 =

(π2h2

8ma2

)(1− ε2

10

). (A.119)

109


3. The variational method says that for any trial function φT satisfying the boundary conditions⟨H⟩

=〈φT |H|φT 〉〈φT |φT 〉

≥ E0 (A.120)

where E0 is the energy of the ground state.

The trial function is φ (x, b) = ψ1 (x) + bψ2 (x) where ψ1 (x) and ψ2 (x) are the eigenfunctions for a particle inthe infinite potential well defined above. Normalisation gives, since |ψ1〉 and |ψ2〉 are orthonormal.

〈φT |φT 〉 = (〈ψ1|+ b〈ψ2|) (|ψ1〉+ b|ψ2〉) = 〈ψ1|ψ1〉+ b2〈ψ2|ψ2〉+ b〈ψ1|ψ2〉+ b〈ψ2|ψ1〉,= 1 + b2. (A.121)

The Hamiltonian for the system H can be written as H = H0 + H ′ where H0 is a particle in the infinite potentialwell defined above, and H ′ = V (x) = ε π

2h2

8ma2 sin(

3πx2a

). Then

〈φT |H|φT 〉 = 〈φT |H0|φT 〉+ 〈φT |V (x) |φT 〉, (A.122)= 〈ψ1 + bψ2|H0|ψ1 + bψ2〉+ 〈φT |V (x) |φT 〉. (A.123)

But H0|ψ1〉 = E(0)1 |ψ1〉 and H0|ψ2〉 = E

(0)2 |ψ2〉 so

〈φT |H|φT 〉 = E1 + b2E2 + 〈φT |V (x) |φT 〉. (A.124)

Expand 〈φT |V (x) |φT 〉 as

〈φT |V (x) |φT 〉 = 〈ψ1 + bψ2|V (x) |ψ1 + bψ2〉,= 〈ψ1|V (x) |ψ1〉+ b2〈ψ2|V (x) |ψ2〉+ b 〈ψ1|V (x) |ψ2〉+ 〈ψ2|V (x) |ψ1〉 .(A.125)

But V (x) andψ2 (x) are odd functions of x, whereasψ1 (x) is even. Thus the terms 〈ψ1|V (x) |ψ1〉 and 〈ψ2|V (x) |ψ2〉are zero and 〈φT |V (x) |φT 〉 reduces to

〈φT |V (x) |φT 〉 = b 〈ψ1|V (x) |ψ2〉+ 〈ψ2|V (x) |ψ1〉 = 2b〈ψ2|V (x) |ψ1〉 (A.126)

= 2b〈ψ2|επ2h2

8ma2sin

(3πx

2a

)|ψ1〉 = 2bε

π2h2

8ma2〈ψ2| sin

(3πx

2a

)|ψ1〉. (A.127)

and

〈ψ(0)2


2a

)∣∣∣∣ψ(0)1 〉 =

∫ a

−aψ

(0)∗2 (x) sin

(3πx

2a

)1√a

cos(πx

2a

)dx, (A.128)

=

∫ a

−aψ

(0)∗2 (x)

1

2

1√a

sin

(4πx

2a

)+

1√a

sin

(2πx

2a

)dx,

=

∫ a

−aψ

(0)∗2

1

2

ψ

(0)4 + ψ

(0)2

dx =

1

2

〈ψ(0)

2 |ψ(0)4 〉+ 〈ψ(0)

2 |ψ(0)2 〉

(A.129)

=1

2(A.130)

giving

〈φT |V (x) |φT 〉 = bεπ2h2

8ma2= bεE1. (A.131)

Hence ⟨H⟩

=E1 + b2E2 + bεE1

1 + b2= E1

(1 + 4b2 + bε

1 + b2

)= E (b) . (A.132)

To find the minimum value, differentiate E (b) with respect to the variational parameter, b, and set the result to zero

∂E (b)

∂b= E1

(6b− εb2 + ε

(1 + b2)2

)= 0, (A.133)

giving b = −6±√

36+4ε2

−2ε . Taking ε = 12 , then b = 6 ± 6

√(1 + 1

36

)So b1 = −0.08 276 or b2 = 12. 08276. Either

by differentiating again or noting that ε+ 6b− εb2 = − (b− b1) (b− b2) shows that b1 = −0.08 276 correspondsto a minimum. With this value of b the best upper bound on the ground state energy with this trial function is

E ≤ 0.9793E1. (A.134)

Note the correct eigenstate |ψ〉 for the ground state can be expanded in terms of the complete set of eigenstatesof the infinite potential well as |ψ〉 =

∑n=1 cn|ψ

(0)n 〉. This calculation has taken only the first two terms. The

estimate could be improved by taking more terms.

110


4. (a) The ground-state energy of a single particle of mass m in a box of length L is h2/8mL2. The ground state forthe 2-fermion system has both particles in the single-particle spatial ground state, one with spin-up and onewith spin-down. Hence, total energy is h2/4mL2.

(b) Since the two particles are in the same spatial state, the space part of the two-particle wave-function is sym-metric. Therefore the spin part is antisymmetric. But an antisymmetric spin state for two spin- 1

2 particles hastotal spin = 0.

(c) Call single-particle spatial ground-state ψ1(x), spatial first-excited state ψ2(x). There are two kinds of 2-particle states:

• Spatial state exchange symmetric, spin state exchange anti-symmetric• Spatial state exchange anti-symmetric, spin state exchange symmetric

For the first kind, there is only one anti-symmetric spin state:

1√2

(α1β2 − β1α2)

So the state of the first kind is:

1√2

(ψ1(x1)ψ2(x2) + ψ2(x1)ψ1(x2))1√2

(α1β2 − β1α2)

For the second kind, there are three symmetric spin states:

α1α2,1√2

(α1β2 + β1α2) , β1β2

So there are three states of the second kind:

1√2

(ψ1(x1)ψ2(x2)− ψ2(x1)ψ1(x2))α1α2

1√2

(ψ1(x1)ψ2(x2)− ψ2(x1)ψ1(x2))1√2

(α1β2 + β1α2)

1√2

(ψ1(x1)ψ2(x2)− ψ2(x1)ψ1(x2))β1β2

The total number of distinct, orthogonal space-spin energy eigenstates is four.

(d) A repulsive interaction between the particles raises the energy of the system. But it raises it by differentamounts for singlet and triplet states. In the triplet spin states, the spatial state is antisymmetric, so that theprobability of finding the particles at the same position is zero. So the spatial antisymmetry keeps the particlesaway from each other. But in the singlet spin states, the spatial state is symmetric, and the particles are notkept away from each other. Hence, the repulsive interaction raises the energy of the triplet state less than thatof the singlet state. The repulsive interaction breaks the degeneracy, so that the triplet energy is below thesinglet energy.

111

fullnotes_phas3226

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