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.

Foundations of Quantum Mechanics

Dr. H. Osborn1

Michælmas 1997

1LATEXed by Paul Metcalfe – comments and corrections topdm23@cam.ac.uk .

Copyright © 2000 University of Cambridge. Not to be quoted or reproduced without permission.

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. Contents

Introduction v

1 Basics 11.1 Review of earlier work . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 The Dirac Formalism . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2.1 Continuum basis . . . . . . . . . . . . . . . . . . . . . . . . 41.2.2 Action of operators on wavefunctions . . . . . . . . . . . . . 51.2.3 Momentum space . . . . . . . . . . . . . . . . . . . . . . . . 61.2.4 Commuting operators . . . . . . . . . . . . . . . . . . . . . 71.2.5 Unitary Operators . . . . . . . . . . . . . . . . . . . . . . . . 81.2.6 Time dependence . . . . . . . . . . . . . . . . . . . . . . . . 8

2 The Harmonic Oscillator 92.1 Relation to wavefunctions . . . . . . . . . . . . . . . . . . . . . . . 102.2 More comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 Multiparticle Systems 133.1 Combination of physical systems . . . . . . . . . . . . . . . . . . . . 133.2 Multiparticle Systems . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.2.1 Identical particles . . . . . . . . . . . . . . . . . . . . . . . . 143.2.2 Spinless bosons . . . . . . . . . . . . . . . . . . . . . . . . . 153.2.3 Spin12 fermions . . . . . . . . . . . . . . . . . . . . . . . . 16

3.3 Two particle states and centre of mass . . . . . . . . . . . . . . . . . 173.4 Observation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

4 Perturbation Expansions 194.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194.2 Non-degenerate perturbation theory . . . . . . . . . . . . . . . . . . 194.3 Degeneracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

5 General theory of angular momentum 235.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

5.1.1 Spin12 particles . . . . . . . . . . . . . . . . . . . . . . . . . 245.1.2 Spin1 particles . . . . . . . . . . . . . . . . . . . . . . . . . 255.1.3 Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

5.2 Addition of angular momentum . . . . . . . . . . . . . . . . . . . . . 255.3 The meaning of quantum mechanics . . . . . . . . . . . . . . . . . . 26

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Copyright © 2000 University of Cambridge. Not to be quoted or reproduced without permission.

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. Introduction

These notes are based on the course “Foundations of Quantum Mechanics” givenbyDr. H. Osborn in Cambridge in the Michælmas Term 1997. These typeset notes havebeen produced mainly for my own benefit but seem to be officially supported. Recom-mended books are discussed in the bibliography at the back.

A word or two about the philosophy of these notes seem in order. They are based incontent on the lectures given, but I have felt free to expand and contract various details,as well as to clarify explanations and improve the narrative flow. Errorsin content are(hopefully) mine and mine alone but I accept no responsibility for your use of thesenotes.

Other sets of notes are available for different courses. At the time of typing, thesecourses were:

Probability Discrete MathematicsAnalysis Further AnalysisQuantum Mechanics Fluid Dynamics 1Quadratic Mathematics GeometryDynamics of D.E.’s Foundations of QMElectrodynamics Methods of Math. PhysFluid Dynamics 2 Waves (etc.)Applications of QM Dynamical SystemsStatistical Physics

They may be downloaded from

http://pdm23.trin.cam.ac.uk/˜pdm23/maths/ orhttp://www.damtp.cam.ac.uk/

or you can email me onpdm23@cam.ac.uk to get a copy of the sets you require.Even if you download them please email me to let me know, so that I can keep you upto date with the errata and new note sets. The other people who have contributed timeand effort to these note sets are:

Richard Cameron Analysis Hugh Osborn ProofreadingClaire Gough Proofreading Malcolm Perry AccomodationKate Metcalfe Probability David Sanders Proofreading

Although these notes are free of charge anyone who wishes to express their thankscould send a couple of bottles of interesting beer to Y1 Burrell’s Field,Grange Road.

Paul Metcalfe15th December 1997

v

Copyright © 2000 University of Cambridge. Not to be quoted or reproduced without permission.

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. Chapter 1

The Basics of QuantumMechanics

Quantum mechanics is viewed as the most remarkable development in20th centuryphysics. Its point of view is completely different from classical physics. Its predictionsare often probabilistic.

We will develop the mathematical formalism and some applications. We will em-phasize vector spaces (to which wavefunctions belong). These vector spaces are some-times finite-dimensional, but more often infinite dimensional. The pure mathematicalbasis for these is in Hilbert Spaces but (fortunately!) no knowledge ofthis area isrequired for this course.

1.1 Review of earlier work

This is abrief review of the salient points of the 1B Quantum Mechanics course. Ifyou anything here is unfamiliar it is as well to read up on the 1B Quantum Mechanicscourse. This section can be omitted by the brave.

A wavefunction (x) : R3 7! C is associated with a single particle in three di-mensions. represents the state of a physical system for a single particle. If isnormalised, that is k k2 � Z d3x j j2 = 1then we say thatd3x j j2 is the probability of finding the particle in the infinitesimalregiond3x (atx).

Superposition Principle

If 1 and 2 are two wavefunctions representing states of a particle, then so is thelinear combinationa1 1 + a2 2 (a1; a2 2 C ). This is obviously the statement thatwavefunctions live in a vector space. If 0 = a (with a 6= 0) then and 0 representthe same physical state. If and 0 are both normalised thena = e{�. We write � e{� to show that they represent the same physical state.

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2 CHAPTER 1. BASICS

For two wavefunctions� and we can define a scalar product(�; ) � Z d3x�� 2 C :This has various properties which you can investigate at your leisure.

Interpretative Postulate

Given a particle in a state represented by a wavefunction (henceforth “in a state ”) then the probability of finding the particle in state� is P = j(�; )j2 and if thewavefunctions are normalised then0 � P � 1. P = 1 if � �.

We wish to define (linear) operators on our vector space — do the obvious thing.In finite dimensions we can choose a basis and replace an operator with a matrix.

For a complex vector space we can define the Hermitian conjugate of the operatorAto be the operatorAy satisfying(�;A ) = (Ay�; ). If A = Ay thenA is Hermitian.Note that ifA is linear then so isAy.

In quantum mechanics dynamical variables (such as energy, momentum or angularmomentum) are represented by (linear) Hermitian operators, the values of the dynam-ical variables being given by the eigenvalues. For wavefunctions (x), A is usuallya differential operator. For a single particle moving in a potentialV (x) we get theHamiltonianH = � ~22mr2 + V (x). Operators may have either a continuous or dis-crete spectrum.

If A is Hermitian then the eigenfunctions corresponding to different eigenvaluesare orthogonal. We assume completeness — that any wavefunction can be expandedas a linear combination of eigenfunctions.

The expectation value forA in a state with wavefunction is hAi , defined to bePi �i jaij2 = ( ;A ). We define the square deviation�A2 to beh(A� hAi )2i which is in general nonzero.

Time dependence

This is governed by the Schrodinger equation{~@ @t = H ;whereH is the Hamiltonian.H must be Hermitian for the consistency of quantummechanics: {~ @@t ( ; ) = ( ;H )� (H ; ) = 0if H is Hermitian. Thus we can impose the condition( ; ) = 1 for all time (if isnormalisable).

If we consider eigenfunctions i ofH with eigenvaluesEi we can expand a generalwavefunction as (x; t) =Xi aie� {Ei~ t i(x):If is normalised then the probability of finding the system with energyEi is jaij2.

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1.2. THE DIRAC FORMALISM 3

1.2 The Dirac Formalism

This is where we take off into the wild blue yonder, or at least a more abstract form ofquantum mechanics than that previously discussed. The essential structure of quantummechanics is based on operators acting on vectors in some vector space. A wavefunc-tion corresponds to some abstract vectorj i, aketvector.j i represents the state ofsome physical system described by the vector space.

If j 1i andj 2i are ket vectors thenj i = a1j 1i+a2j 2i is a possible ket vectordescribing a state — this is the superposition principle again.

We define a dual space ofbra vectorsh�j and a scalar producth�j i, a complexnumber.1 For anyj i there corresponds a uniqueh j and we requireh�j i = h j�i�.We require the scalar product to be linear such thatj i = a1j 1i + a2j 2i impliesh�j i = a1h�j 1i + a2h�j 2i. We see thath j�i = a�1h 1j�i + a�2h 2j�i and soh j = a�1h 1j+ a�2h 2j.

We introduce linear operatorsAj i = j 0i and we define operators acting on bravectors to the lefth�jA = h�0j by requiringh�0j i = h�jAj i for all . In general, inh�jAj i, A can act either to the right or the left. We define theadjoint Ay of A suchthat if Aj i = j 0i thenh jAy = h 0j. A is said to be Hermitian ifA = Ay.

If A = a1A1 + a2A2 thenAy = a�1Ay1 + a�2Ay2, which can be seen by appealing tothe definitions. We also find the adjoint ofBA as follows:

Let BAj i = Bj 0i = j 00i. Thenh 00j = h 0jBy = h jAyBy and the resultfollows. Also, if h jA = h�0j thenj�0i = Ayj�i.

We have eigenvectorsAj i = �j i and it can be seen in the usual manner that theeigenvalues of a Hermitian operator are real and the eigenvectors correspondingto twodifferent eigenvalues are orthogonal.

We assume completeness — that is anyj�i can be expanded in terms of the basisket vectors,j�i = Pi aij ii whereAj ii = �ij ii andai = h ij i. If j i isnormalised —h j i = 1 — then the expected value ofA is hAi = h jAj i, whichis real if A is Hermitian.

The completeness relation for eigenvectors ofA can be written as1 =Pi j iih ij,which gives (as before) j i = 1j i =Xi j iih ij i:

We can also rewriteA = Pi j ii�ih ij and if �j 6= 0 8j then we can defineA�1 =Pi j ii��1i h ij.We now choose an orthonormal basisfjnig with hnjmi = �nm and the complete-

ness relation1 =Pn jnihnj. We can thus expandj i =Pn anjni with an = hnj i.We now consider a linear operatorA, and thenAj i = Pn anAjni = Pm a0mjmi,with a0m = hmjAj i = Pn anhmjAjni. Further, puttingAmn = hmjAjni we geta0m = PnAmnan and therefore solvingAj i = �j i is equivalent to solving thematrix equationAa = �a. Amn is called the matrix representation ofA. We also haveh j = Pn a�nhnj, with a0n� = Pm a�mAymn, whereAymn = A�nm gives the Hermitianconjugate matrix. This is the matrix representation ofAy.

1bra ket. Who said that mathematicians have no sense of humour?

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4 CHAPTER 1. BASICS

1.2.1 Continuum basis

In the above we have assumed discrete eigenvalues�i and normalisable eigenvectorsj ii. However, in general, in quantum mechanics operators often have continuousspectrum — for instance the position operatorx in 3 dimensions.x must have eigen-valuesx for any pointx 2 R3 . There exist eigenvectorsjxi such thatxjxi = xjxi foranyx 2 R3 .

As xmust be Hermitian we havehxjx = xhxj. We define the vector space requiredin the Dirac formalism as that spanned byjxi.

For any statej i we can define a wavefunction (x) = hxj i.We also need to find some normalisation criterion, which uses the 3 dimensional

Dirac delta function to gethxjx0i = �3(x� x0). Completeness givesZ d3xjxihxj = 1:We can also recover the ket vector from the wavefunction byj i = 1j i = Z d3xjxi (x):Also hxjxj i = x (x); the action of the operatorx on a wavefunction is multipli-

cation byx.Something else reassuring ish j i = h j1j i = Z d3xh jxihxj i= Z d3x j (x)j2 :The momentum operatorp is also expected to have continuum eigenvalues. We

can similarly define statesjpi which satisfypjpi = pjpi. We can relatex andp usingthe commutator, which for two operatorsA andB is defined byhA; Bi = AB � BA:

The relationship betweenx andp is [xi; pj ] = {~�ij . In one dimension[x; p] = {~.We have a useful rule for calculating commutators, that is:hA; BCi = hA; Bi C + B hA; Ci :

This can be easily proved simply by expanding the right hand side out.We can use thisto calculate

�x; p2�. �x; p2� = [x; p] p+ p [x; p]= 2{~p:It is easy to show by induction that[x; pn] = n{~pn�1.We can define an exponential bye� {ap~ = 1Xn=0 1n! �� {ap~ �n :

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1.2. THE DIRAC FORMALISM 5

We can evaluatehx; e� {ap~ i byhx; e� {ap~ i = "x; 1Xn=0 1n! �� {ap~ �n#= 1Xn=0 1n! �x;�� {ap~ �n�= 1Xn=0 1n! �� {a~ �n [x; pn]= 1Xn=1 1(n� 1)! �� {a~ �n {~pn�1= a 1Xn=1�� {a~ �n�1 pn�1= ae� {ap~

and by rearranging this we get thatxe� {ap~ = e� {ap~ (x+ a)and it follows thate� {ap~ jxi is an eigenvalue ofx with eigenvaluex + a. Thus weseee� {ap~ jxi = jx + ai. We can do the same to the bra vectors with the Hermi-tian conjugatee {ap~ to gethx + aj = hxje {ap~ . Then we also have the normalisationhx0 + ajx+ ai = hx0jxi.

We now wish to considerhx+ajpi = hxje {ap~ jpi = e {ap~ hxjpi. Settingx = 0 giveshajpi = e {ap~ N , whereN = h0jpi is independent ofx. We can determineN from thenormalisation ofjpi. �(p0 � p) = hp0jpi = Z da hp0jaihajpi= jN j2 Z da e {a(p�p0)~= jN j2 2�~ �(p0 � p)

So, because we are free to choose the phase ofN , we can setN = � 12�~� 12 and

thushxjpi = � 12�~� 12 e {xp~ . We coulddefinejpi byjpi = Z dx jxihxjpi = � 12�~� 12 Z dx jxie {xp~ ;but we then have to check things like completeness.

1.2.2 Action of operators on wavefunctions

We recall the definition of the wavefunction as (x) = hxj i. We wish to see whatoperators (the position and momentum operators discussed) do to wavefunctions.

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6 CHAPTER 1. BASICS

Now hxjxj i = xhxj i = x (x), so the position operator acts on wavefunctionsby multiplication. As for the momentum operator,hxjpj i = Z dp hxjpjpihpj i= Z dp phxjpihpj i= � 12�~� 12 Z dp pe {xp~ hpj i= �{~ ddx Z dp hxjpihpj i= �{~ ddxhxj i = �{~ ddx (x):

The commutation relation[x; p] = {~ corresponds to�x;�{~ ddx� = {~ (acting on (x)).

1.2.3 Momentum spacejxi 7! (x) = hxj i defines a particular representation of the vector space. It issometimes useful to use a momentum representation,~ (p) = hpj i. We observe that~ (p) = Z dx hpjxihxj i= � 12�~� 12 Z dx e� {xp~ (x):

In momentum space, the operators act differently on wavefunctions. It is easy tosee thathpjpj i = p ~ (p) andhpjxj i = {~ ddp ~ (p).

We convert the Schrodinger equation into momentum space. We have the operatorequationH = p22m + V (x) and we just need to calculate how the potential operates onthe wavefunction.hpjV (x)j i = Z dx hpjV (x)jxihxj i= � 12�~� 12 Z dx e� {xp~ V (x)hxj i= 12�~ ZZ dx dp0 V (x) ~ (p0)e {x(p0�p)~= Z dp0 ~V (p� p0) ~ (p0);

where~V (p) = 12�~ R dp e� {xp~ V (x). Thus in momentum space,Hp ~ (p) = p22m ~ (p) + Z dp0 ~V (p� p0) ~ (p0):

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1.2. THE DIRAC FORMALISM 7

1.2.4 Commuting operators

SupposeA andB are Hermitian andhA; Bi = 0. ThenA andB have simultaneous

eigenvectors.

Proof. SupposeAj i = �j i and the vector subspaceV� is the span of the eigenvec-tors ofA with eigenvalue�. (If dimV� > 1 then� is said to be degenerate.)

As A andB commute we know that�Bj i = ABj i and soBj i 2 V�. If � isnon-degenerate thenBj i = �j i for some�. Otherwise we have thatB : V� 7! V�and we can therefore find eigenvectors ofB which lie entirely insideV�. We can labelthese asj�; �i, and we know thatAj�; �i = �j�; �iBj�; �i = �j�; �i:

These may still be degenerate. However we can in principle remove this degener-acy by adding more commuting operators until each state is uniquely labeledby theeigenvalues of each common eigenvector. This set of operators is called acompletecommuting set.

This isn’t so odd: for a single particle in 3 dimensions we have the operatorsx1, x2andx3. These all commute, so for a single particle with no other degrees of freedomwe can label states uniquely byjxi. We also note from this example that a completecommuting set is not unique, we might just as easily have taken the momentum opera-tors and labeled states byjpi. To ram the point in more, we could also have taken someweird combination likex1, x2 andp3.

For our single particle in 3 dimensions, a natural set of commuting operators in-volves the angular momentum operator,L = x ^ p, or Li = �ijkxj pk.

We can find commutation relations betweenLi and the other operators we know.These are summarised here, proof is straightforward.� hLi; xli = {~�ilj xj� hLi; x2i = 0� hLi; pmi = {~�imkpk� hLi; p2i = 0� hLi; Lji = {~�ijkLk� hLi; L2i = 0

If we have a HamiltonianH = p22m +V (jxj) then we can also see thathL; Hi = 0.

We choose as a commuting setH, L2 and L3 and label statesjE; l;mi, where theeigenvalue ofL2 is l(l+ 1) and the eigenvalue ofL3 ism.

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8 CHAPTER 1. BASICS

1.2.5 Unitary Operators

An operatorU is said to beunitary if UyU = 1, or equivalentlyU�1 = Uy.SupposeU is unitary andU j i = j 0i, U j�i = j�0i. Thenh�0j = h�jUy andh�0j 0i = h�j i. Thus the scalar product, which is the probability amplitude of finding

the statej�i given the statej i, is invariant under unitary transformations of states.For any operatorA we can defineA0 = U AU y. Thenh�0jA0j 0i = h�jAj i and

matrix elements are unchanged under unitary transformations. We also notethat ifC = AB thenC 0 = A0B0.The quantum mechanics for thej i, j�i, A, B etc. is the same as forj 0i, j�0i, A0,B0 and so on. A unitary transform in quantum mechanics is analogous to a canonical

transformation in dynamics.Note that ifO is Hermitian thenU = e{O is unitary, asUy = e�{Oy = e�{O .

1.2.6 Time dependence

This is governed by the Schrodinger equation,{~ @@t j (t)i = H j (t)i:H is the Hamiltonian and we require it to be Hermitian. We can get an explicitsolution of this ifH does not depend explicitly ont. We setj (t)i = U(t)j (0)i,whereU(t) = e� {Ht~ . As U(t) is unitary,h�(t)j (t)i = h�(0)j (0)i.

If we measure the expectation ofA at timet we geth (t)jAj (t)i = a(t). Thisdescription is called the Schrodinger picture. Alternatively we can absorbthe time de-pendence into the operatorA to get the Heisenberg picture,a(t) = h jUy(t)AU(t)j i.We writeAH (t) = Uy(t)AU(t). In this description the operators are time dependent(as opposed to the states).AH(t) is the Heisenberg picture time dependent operator.Its evolution is governed by{~ @@tAH(t) = hAH(t); Hi ;which is easily proven.

For a HamiltonianH = 12m p(t)2 + V (x(t)) we can get the Heisenberg equationsfor the operatorsxH andpH ddt xH(t) = 1mpH(t)ddt pH(t) = �V 0(xH (t)):

These ought to remind you of something.

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. Chapter 2

The Harmonic Oscillator

In quantum mechanics there are two basic solvable systems, the harmonic oscillatorand the hydrogen atom. We will examine the quantum harmonic oscillator using al-gebraic methods. In quantum mechanics the harmonic oscillator is governedby theHamiltonian H = 12mp2 + 12m!2x2;with the condition that[x; p] = {~. We wish to solveH j i = Ej i to find the energyeigenvalues.

We define a new operatora.a = �m!2~ � 12 �x+ {pm!�ay = �m!2~ � 12 �x� {pm!� :a and ay are respectively called the annihilation and creation operators. We caneasily obtain the commutation relation

�a; ay� = 1. It is easy to show that, in termsof the annihilation and creation operators, the HamiltonianH = 12~! �aay + aya�,which reduces to~! �aya+ 12�. Let N = aya. Then

ha; Ni = a andhay; Ni = �ay.

ThereforeN a = a�N � 1� andN ay = ay �N + 1�.

Supposej i is an eigenvector ofN with eigenvalue�. Then the commutation rela-tions give thatNaj i = (�� 1) aj i and therefore unlessaj i = 0 it is an eigenvalueof N with eigenvalue�� 1. Similarly N ayj i = (�+ 1) ayj i.

But for anyj i, h jN j i � 0 and equals0 iff aj i = 0. Now suppose we havean eigenvalue� =2 f0; 1; 2; : : :g. Then9n such thatanj i is an eigenvector ofNwith eigenvalue� � n < 0 and so we must have� 2 f0; 1; 2; : : :g. Returning to theHamiltonian we get energy eigenvaluesEn = ~! �n+ 12�, the same result as using theSchrodinger equation for wavefunctions, but with much less effort.

We definejni = Cnaynj0i, whereCn is such as to makehnjni = 1. We can takeCn 2 R, and evaluateh0janaynj0i to findCn.

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10 CHAPTER 2. THE HARMONIC OSCILLATOR1 = hnjni= C2nh0janaynj0i= C2nh0jan�1aayayn�1j0i= C2nC2n�1 hn� 1jaayjn� 1i= C2nC2n�1 hn� 1jN + 1jn� 1i= C2nC2n�1 (n� 1 + 1)hn� 1jn� 1i= n C2nC2n�1 :We thus requireCn = Cn�1=pn and asC0 = 1 we getCn = (n!)� 12 and so we

have the normalised eigenstate (ofN ) jni = 1pn! ayj0i (with eigenvaluen). jni is also

an eigenvector ofH with eigenvalue~! �n+ 12�. The space of states for the harmonicoscillator is spanned byfjnig.

We also need to ask if there exists a non-zero statej i such thatayj i = 0. Then0 = h jaayj i = h j i+ h jayaj i � h j i > 0:So there exist no non-zero statesj i such thatayj i = 0.

2.1 Relation to wavefunctions

We evaluate 0 = hxjaj0i = �m!2~ � 12 �x+ ~m! ddx� hxj0iand we see that 0(x) = hxj0i satisfies the differential equation� ddx + m!~ x� 0(x) = 0:This (obviously) has solution 0(x) = Ne� 12 m!~ x2 for some normalisation constantN . This is the ground state wavefunction which has energy12~!.

For 1(x) = hxj1i = hxjayj0i we find 1(x) = �m!2~ � 12 hxj�x� {m! p� j0i= �m!2~ � 12 �x� ~m! ddx� 0(x)= �2m!~ � 12 x 0(x):

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2.2. MORE COMMENTS 11

2.2 More comments

Many harmonic oscillator problems are simplified using the creation and annihilationoperators.1 For examplehmjxjni = � ~2m!� 12 hmja+ ayjni= � ~2m!� 12 �pn hmjn� 1i+pn+ 1 hmjn+ 1i�= � ~2m!� 12 �pn �m;n�1 +pn+ 1 �m;n+1� :

This is non-zero only ifm = n�1. We note thatxr contains termsasayr�s, where0 � s � r and sohmjxrjni can be non-zero only ifn� r � m � n+ r.It is easy to see that in the Heisenberg pictureaH(t) = e{ Ht~ ae�{ Ht~ = e�{!ta.

Then using the equations forxH (t) andpH(t), we see thatxH(t) = x cos!t+ 1m! p sin!t:Also, HayH(t) = ayH(t)(H +~!), so if j i is an energy eigenstate with eigenvalueE thenayH(t)j i is an energy eigenstate with eigenvalueE + ~!.

1And such problemsalwaysoccur in Tripos papers. You have been warned.

Copyright © 2000 University of Cambridge. Not to be quoted or reproduced without permission.

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. Chapter 3

Multiparticle Systems

3.1 Combination of physical systems

In quantum mechanics each physical system has its own vector space of physical statesand operators, which if Hermitian represent observed quantities.

If we consider two vector spacesV1 andV2 with basesfjri1g and fjsi2g withr = 1 : : :dimV1 ands = 1 : : : dimV2. We define the tensor productV1 V2 as thevector space spanned by pairs of vectorsfjri1jsi2 : r = 1 : : :dimV1; s = 1 : : :dimV2g:

We see thatdim(V1 V2) = dimV1 dimV2. We also write the basis vectors ofV1V2 asjr; si. We can define a scalar product onV1V2 in terms of the basis vectors:hr0; s0jr; si = hr0jri1hs0jsi2. We can see that iffjri1g andfjsi2g are orthonormalbases for their respective vector spaces thenfjr; sig is an orthonormal basis forV1V2.

SupposeA1 is an operator onV1 andB2 is an operator onV2 we can define anoperatorA1 � B2 onV1 V2 by its operation on the basis vectors:�A1 � B2� jri1jsi2 = �A1jri1��B2jsi2� :

We writeA1 � B2 asA1B2.Two harmonic oscillators

We illustrate these comments by example. SupposeHi = p2i2m + 12m!x2i i = 1; 2:We have two independent vector spacesVi with basesjnii wheren = 0; 1; : : : andai andayi are creation and annihilation operators onVi, andHijnii = ~! �n+ 12� jnii:For the combined system we form the tensor productV1 V2 with basisjn1; n2i

and HamiltonianH = Pi Hi, soH jn1; n2i = ~! (n1 + n2 + 1) jn1; n2i. There areN + 1 ket vectors in theN th excited state.

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14 CHAPTER 3. MULTIPARTICLE SYSTEMS

The three dimensional harmonic oscillator follows similarly. In general if H1 andH2 are two independent Hamiltonians which act onV1 andV2 respectively then theHamiltonian for the combined system isH = H1 + H2 acting onV1 V2. If fj rigand fj sig are eigenbases forV1 andV2 with energy eigenvaluesfE1rg and fE2sgrespectively then the basis vectorsfjir;sg for V1V2 have energiesEr;s = E1r +E2s .

3.2 Multiparticle Systems

We have considered single particle systems with statesj i and wavefunctions (x) =hxj i. The states belong to a spaceH.Consider anN particle system. We say the states belong toHn = H1 � � � HN

and define a basis of statesj r1i1j r2i2 : : : j rN iN wherefj riiig is a basis forHi.A general stateji is a linear combination of basis vectors and we can define theN particle wavefunction as(x1;x2; : : : ;xN ) = hx1;x2; : : : ;xN ji.The normalisation condition ishji = Z d3x1 : : : d3xN j(x1;x2; : : : ;xN )j2 = 1 if normalised.

We can interpretd3x1 : : : d3xN j(x1;x2; : : : ;xN )j2 as the probability densitythat particlei is in the volume elementd3xi at xi. We can obtain the probabilitydensity for one particle by integrating out all the otherxj ’s.

For time evolution we get the equation{~ @@t ji = H ji, whereH is an operatoronHN .

If the particles do not interact thenH = NXi=1 HiwhereHi acts onHi but leavesHj alone forj 6= i. We have energy eigenstates in eachHi such thatHij rii = Erj rii and soji = j r1i1j r2i2 : : : j rN iN is an energyeigenstate with energyEr1 + � � �+ErN .

3.2.1 Identical particles

There are many such cases, for instance multielectron atoms. We will concentrate ontwo identical particles.

“Identical” means that physical quantities are be invariant under interchange ofparticles. For instance if we haveH = H(x1; p1; x2; p2) then this must equal thepermuted HamiltonianH(x2; p2; x1; p1) if we have identical particles. We introduceU such that U x1U�1 = x2 U x2U�1 = x1U p1U�1 = p2 U p2U�1 = p1:

We should also haveUHU�1 = H and more generally ifA1 is an operator onparticle 1 thenU A1U�1 is the corresponding operator on particle 2 (and vice versa).

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3.2. MULTIPARTICLE SYSTEMS 15

Note that ifji is an energy eigenstate ofH then so isU ji. ClearlyU2 = 1 and werequireU to be unitary, which implies thatU is Hermitian.

In quantum mechanics we requireji andU ji to be the same states (for identicalparticles). This implies thatU ji = �ji and the requirementU2 = 1 gives that� = �1. In terms of wavefunctions this means that(x1; x2) = �(x2; x1). If wehave a plus sign then the particles are bosons (which have integral spin) and if a minussign then the particles are fermions (which have spin12 ; 32 ; : : : ).1

The generalisation toN identical particles is reasonably obvious. LetUij inter-change particlesi andj. ThenUijHU�1ij = H for all pairs(i; j).

The same physical requirement as before gives us thatUij ji = �ji for all pairs(i; j).If we have bosons (plus sign) then in terms of wavefunctions we must have(x1; : : : ; xN ) = (xp1 ; : : : ; xpN );

where(p1; : : : ; pN ) is a permutation of(1; : : : ; N). If we have fermions then(x1; : : : ; xN ) = �(xp1 ; : : : ; xpN );where� = +1 if we have an even permutation of(1; : : : ; N) and�1 if we have an oddpermutation.

Remark for pure mathematicians.1 andf�1g are the two possible one dimensionalrepresentations of the permutation group.

3.2.2 Spinless bosons

(Which means that the only variables for a single particle arex and p.) Supposewe have two identical non-interacting bosons. ThenH = Hi + H2 and we haveH1j rii = Er j rii. The general space with two particles isH1 H2 which hasa basisfj ri1j si2g, but as the particles are identical the two particle state space is(H1 H2)S where we restrict to symmetric combinations of the basis vectors. Thatis, a basis for this in terms of the bases ofH1 andH2 isnj ri1j ri2; 1p2 (j ri1j si2 + j si1j ri2) ; r 6= so :

The corresponding wavefunctions are r(x1) r(x2) and 1p2 ( r(x1) s(x2) + s(x1) r(x2))and the corresponding eigenvalues are2Er andEr+Es. The factor of2� 12 just ensuresnormalisation and1p2 (1h r0 j2h s0 j+ 1h s0 j2h r0 j) 1p2 (j ri1j si2 + j si1j ri2)evaluates to�rr0�ss0 + �rs0�r0s.

For N spinless bosons withH = P Hi the appropriate completely symmetricstates are 1pN !�j r1i1 : : : j rN iN + permutations thereof

�if ri 6= rj

1Spin will be studied later in the course.

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16 CHAPTER 3. MULTIPARTICLE SYSTEMS

3.2.3 Spin 12 fermions

In this case (which covers electrons, for example) a single particle state (or wavefunc-tion) depends on an additional discrete variables. The wavefunctions are (x; s) or s(x). The space of states for a single electronH = L2(R3 ) C 2 has a basis of theform jxijsi � jx; si and the wavefunctions can be written s(x) = hx; sj i. A basisof wavefunctions isf r�(x; s) = r(x)��(s)g, wherer and� are labels for the basis.� takes two values and it will later be seen to be natural to take� = � 12 .

We can also think of the vector�� = ���(1)��(2)�, in which case two possible basis

vectors are

�10� and

�01�. Note that�y�0�� = ���0 .The scalar product is defined in the obvious way:h�r0�0 j�r�i = h r0 j rih��0 j��i,

which equals�rr0���0 if the initial basis states are orthonormal.Thetwoelectron wavefunction is(x1; s1;x2; s2) and under the particle exchange

operatorU we must have(x1; s1;x2; s2) 7! �(x2; s2;x1; s1). The two particlestates belong to the antisymmetric combination(H1 H2)A.

ForN electrons the obvious thing can be done.

Basis for symmetric or antisymmetric 2 particle spin states

There is only one antisymmetric basis state�A(s1; s2) = 1p2 �� 12 (s1)�� 12 (s2)� �� 12 (s1)� 12 (s2)� ;and three symmetric possibilities:�S(s1; s2) =8>><>>:� 12 (s1)� 12 (s2)1p2 �� 12 (s1)�� 12 (s2) + �� 12 (s1)� 12 (s2)� s1 6= s2�� 12 (s1)�� 12 (s2):

We can now examine two non-interacting electrons, withH = H1 + H2 and takeHi independent of spin. The single particle states arej iij�si.The two electron states live in(H1 H2)A, which has a basisj ri1j ri2j�Ai;1p2 (j ri1j si2 + j si1j ri2) j�Ai; r 6= s1p2 (j ri1j si2 � j si1j ri2) j�Si; r 6= s;

with energy levels2Er (one spin state) andEr +Es (one antisymmetric spin state andthree symmetric spin states).

We thus obtain the Pauli exclusion principle: no two electrons can occupy the samestate (taking account of spin).

As an example we can take the helium atom with HamiltonianH = p212m + p222m � 2e24��0 jx1j � 2e24��0 jx2j + e24��0 jx1 � x2j :

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3.3. TWO PARTICLE STATES AND CENTRE OF MASS 17

If we neglect the interaction term we can analyse this as two hydrogen atoms and gluethe results back together as above. The hydrogen atom (with a nuclear charge2e) hasEn = � 2e28��0n2 , so we get a ground state for the helium atom with energy2E1 withno degeneracy and a first excited state with energyE1 +E2 with a degeneracy of four.Hopefully these bear some relation to the results obtained by taking theinteraction intoaccount.

3.3 Two particle states and centre of mass

Suppose we have a HamiltonianH = p212m + p222m +V (x1� x2) defined onH2. We canseparate out the centre of mass motion by lettingP = p1 + p2 p = 12 (p1 � p2)X = 12 (x1 + x2) x = x1 � x2:

ThenhXi; Pji = {~�ij , [xi; pj ] = {~�ij andX, P andx, p commute respectively.

We can rewrite the Hamiltonian asH = P22M + h, h = p2m +V (x), whereM = 2m andwe can decomposeH2 intoHCM Hint. HCM is acted on byX andP and has wave-functions�(X). Hint is acted on byx, p and any spin operators. It has wavefunctions (x; s1; s2). We take wavefunctions(x1; s1;x2; s2) = �(X) (x; s1; s2) in H2.

This simplifies the Schrodinger equation, we can just have�(X) = e{P:X~ and thenE = P22M +Eint. We thus need only to solve the one particle equationh = Eint .Under the particle exchange operatorU we have (x; s1; s2) 7! (�x; s2; s1) = � (x; s1; s2);

with a plus sign for bosons and a minus sign for fermions. In the spinless case then (x) = (�x).If we have a potentialV (jxj) then we may separate variables to get (x; s1; s2) = Yl� xjxj�R(jxj)�(s1; s2)

with Yl�� xjxj� = (�1)lYl� xjxj�. For spinless bosons we therefore requirel to be even.

3.4 Observation

Consider the tensor product of two systemsH1 andH2. A general stateji inH1H2can be written as ji =Xi;j aij j ii1j�ji2with j ii1 2 H1 andj�ji2 2 H2 assumed orthonormal bases for their respective vectorspaces.

Suppose we make a measurement on the first system leaving the second systemunchanged, and find the first system in a statej ii1. Then1h iji = Pj aij j�ji2,

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18 CHAPTER 3. MULTIPARTICLE SYSTEMS

which we write asAij�i2, wherej�i2 is a normalised state of the second system. WeinterpretjAij2 as the probability of finding system 1 in statej ii1. After measurementsystem2 is in a statej�i2.

If aij = �i�ij (no summation) thenAi = �i and measurement of system 1 asj ii1determines system 2 to be in statej�ii2.

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. Chapter 4

Perturbation Expansions

4.1 Introduction

Most problems in quantum mechanics are not exactly solvable and it it necessary tofind approximate answers to them. The simplest method is a perturbationexpansion.We writeH = H0+H 0 whereH0 describes a solvable system with known eigenvaluesand eigenvectors, andH 0 is in some sense small.

We write H(�) = H0 + �H 0 and expand the eigenvalues and eigenvectors inpowers of�. Finally we set� = 1 to get the result. Note that we do not necessarilyhave to introduce�; the problem may have some small parameter which we can use.This theory can be applied to the time dependent problem but here we will only discussthe time independent Schrodinger equation.

4.2 Non-degenerate perturbation theory

Suppose thatH0jni = �njni for n = 0; 1; : : : . We thus assume discrete energy levelsand we assume further that the energy levels are non-degenerate. We also require H 0to be sufficiently non-singular to make a power series expansion possible.

We have the equationH(�)j n(�)i = En(�)j n(�)i. We suppose thatEn(�)tends to�n as� ! 0 and j n(�)i ! jni as� ! 0. We pose the power seriesexpansions En(�) = �n + �E(1)n + �2E(2)n + : : :j n(�)i = N jni+ �j (1)n i+ : : : ;

substitute into the Schrodinger equation and require it to be satisfied at each powerof �. The normalisation constantN is easily seen to be1+O(�2). TheO(1) equationis automatically satisfied and theO(�) equation isH0j (1)n i+ H 0jni = E(1)n jni+ �nj (1)n i:

Note that we can always replacej (1)n i with j (1)n i + �jni and leave this equationunchanged. We can therefore impose the conditionhnj (1)n i = 0. If we applyhnj to

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20 CHAPTER 4. PERTURBATION EXPANSIONS

this equation we getE(1)n = hnjH 0jni — the first order perturbation in energy. If weapplyhrj wherer 6= n we see thathrj (1)n i = �hrjH 0jni�r � �nand therefore j (1)n i = �Xr 6=n jrihrjH 0jni�r � �n :Note that we are justified in these divisions as we have assumed that theeigenvaluesare non-degenerate. On doing the same thing to theO(�2) equation we see thatE(2)n = hnjH 0j (1)n i= �Xr 6=n ���hrjH 0jni���2�r � �n :

This procedure is valid if�r � �n is not very small whenhrjH 0jni 6= 0.Using these results we can see thatdd�En(�) = h n(�)jH 0j n(�)i and@@� j n(�)i = �Xr 6=n 1Er(�) �En(�) j r(�)ih r(�)jH 0j n(�)i:Also @H@� = H 0 and so@2@�2En(�) = 2h n(�)jH 0 @@� j n(�)i:

Example: harmonic oscillator

ConsiderH = p22m + 12m!2x2 + �m!2x2, which can be viewed asH0 + H 0, whereH0 is the plain vanilla quantum harmonic oscillator Hamiltonian.Calculating the matrix elementshrjx2jni required is an extended exercise in ma-

nipulations of the annihilation and creation operators and is omitted. The results areE(1)n = ~! �n+ 12�E(2)n = � 12~! �n+ 12� :We thus get the perturbation expansion forE0nE0n = ~! �n+ 12� �1 + �� 12�2 +O(�3)� :This system can also be solved exactly to giveE0n = ~! �n+ 12�p1 + 2� which

agrees with the perturbation expansion.

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4.3. DEGENERACY 21

4.3 Degeneracy

The method given here breaks down if�r = �n for r 6= n. Perturbation theory can beextended to the degenerate case, but we will consider only the first order shift in �r.We suppose that the statesjn; si, s = 1 : : :Nn have the same energy�n. Ns is thedegeneracy of this energy level.

As before we pose a HamiltonianH = H0 + �H 0 such thatH0jn; si = �njn; siand look for statesj (�)i with energyE(�) ! �n as�! 0.

The difference with the previous method is that we expandj (�)i as a power seriesin � in the basis of eigenvectors ofH0. That isj (�)i =Xs jn; sias + �j (1)i:

As theas are arbitrary we can impose the conditionshn; sj (1)i = 0 for eachs andn. We thus have to solveH j (�)i = E(�)j (�)i with E(�) = �n+�E(1). If we taketheO(�) equation and applyhn; rj to it we getXs ashn; rjH 0jn; si = arE(1)rwhich is a matrix eigenvalue problem. Thus the first order perturbations in �n arethe eigenvalues of the matrixhn; rjH 0jn; si. If all the eigenvalues are distinct thenthe perturbation “lifts the degeneracy”. It is convenient for the purpose of calculationto choose a basis for the space spanned by the degenerate eigenvectors in which thismatrix is “as diagonal as possible”.1

1Don’t ask...

Copyright © 2000 University of Cambridge. Not to be quoted or reproduced without permission.

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. Chapter 5

General theory of angularmomentum

For a particle with position and momentum operatorsx andp with the commutationrelations[xi; pj ] = {~�ij we defineL = x ^ p. It can be seen thatL is Hermitian and

it is easy to showhLi; Lji = {~�ijkLk.

5.1 Introduction

More generally we define Hermitian angular momentum operatorsJ with the commu-tation relation[Ji; Jj ] = {~�ijkJk.1 We ask on what space of states can this algebra ofoperators be realised, or alternatively, what are the representations?

We choose one component ofJ whose eigenvalues label the states; in accordancewith convention we chooseJ3. Then we haveJ3jmi = m~jmi for m 2 R (discrete)and we also havehm0jmi = �m0m.

By applying the commutation relation we get (easily) that�J3;J2� = 0, so asJ2

is Hermitian we can choose a simultaneous eigenbasis. That is,J2jmi = �~2jmi. Weknow that� � 0 sinceJ2 is the sum of the squares of Hermitian operators.

At this stage we choose an alternate basis; that is we splitJ into J+, J� andJ3,with J� = J1 � {J2. Note thatJy� = J�. The commutation relations forJi give that[J3; J�] = �{~J� and[J+; J�] = 2~J3. It will be useful later to note thatJ2 = 12 (J+J� + J�J+) + J23 = (J�J+ + J23 + ~J3J+J� + J23 � ~J3 :

Proof of this is immediate. We can now rewrite the[J3; J�] commutation rela-tion asJ3J� = J� (J3 � ~) and so we see thatJ�jmi is an eigenvector ofJ3 witheigenvalue(m + 1)~ and soJ�jmi = ~N�mjm � 1i, with some normalisation con-stantN�m. By evaluating the norm ofJ�jmi, and noting thathmjmi = 1 we see that(N�m)2 = ��m2 �m.

We can now define statesjm�ni for n = 0; 1; 2; : : : . We can pin them down moreby noting that(N�m)2 � 0 for positive norms. However the formulae we have are,

1Because we are now grown up we will omit the hats if they do not add to clarity.

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24 CHAPTER 5. GENERAL THEORY OF ANGULAR MOMENTUM

given�, negative for sufficiently largejmj and so to avoid this we must havemmax = jsuch thatJ+jji = 0 and so

�N+j �2 = �� j2 � j = 0 and so� = j(j + 1).We can perform a similar trick withJ�; there must existmmin = �j0 such thatJ�j � j0i = 0; thus� = j0(j0 + 1). Soj0 = j and as�j0 = �j = j � n for somen 2 f0; 1; 2; : : :g we havej = 0; 12 ; 1; 32 ; : : : .In summary the states can be labelled byjj mi such thatJ2jj mi = ~2j(j + 1)jj miJ3jj mi = ~mjj miJ�jj mi = ~ ((j �m) (j �m+ 1)) 12 jj m+ 1i

with m 2 f�j;�j+1; : : : ; j � 1; jg andj 2 f0; 12 ; 1; 32 ; : : : g. There are2j+1 stateswith differentm for the samej. jj mi is the standard basis of the angular momentumstates.

We have obtained a representation of the algebra labelled byj. If J = L = x ^ pwe must havej an integer.

Recall that if we haveA we can define a matrixA�0� by Aj�i =P�0 j�0iA�0�.Note that(BA)�0� = P�B�0�A��. Given j, we have(J3)m0m = ~m�m0m and(J�)m0m = ~p(j �m) (j �m+ 1) �m0;m�1, giving us(2j + 1)�(2j + 1) matricessatisfying the three commutation relations[J3; J�] = �~J� and[J+; J�] = 2~J3.

If J are angular momentum operators which act on a vector spaceV and we havej i 2 V such thatJ3j i = ~kj i andJ+j i = 0 then is a state with angularmomentumj = k. The other states are given byJn�j i, 1 � n � 2k. The conditionsalso giveJ2j i = ~2k (k + 1) j i5.1.1 Spin 12 particles

This is the simplest non-trivial case. We havej = 12 and a two dimensional state spacewith a basisj 12 12 i andj 12 � 12 i. We have the relationsJ3j 12 � 12 i = � 12~j 12 � 12 i andJ+j 12 12 i = 0 J�j 12 12 i = ~j 12 � 12 iJ�j 12 � 12 i = 0 J+j 12 � 12 i = ~j 12 12 i:

It is convenient to introduce explicit matrices� such thatJj 12 mi =Xm0 j 12 m0i 12~ (�)m0m :The matrices� are2� 2 matrices (called the Pauli spin matrices). Explicitly, they

are �+ = �0 20 0� �� = �0 02 0� �3 = �1 00 �1��1 = 12 (�+ + ��) = �0 11 0� �2 = � {2 (�+ � ��) = �0 �{{ 0 � :Note that�21 = �22 = �23 = 1 and�y = �. These satisfy the commutation relations[�i; �j ] = 2{�ijk�k (a slightly modified angular momentum commutation relation)

and we also have�2�3 = {�1 (and the relations obtained by cyclic permutation), so

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5.2. ADDITION OF ANGULAR MOMENTUM 25�i�j + �j�i = 2�ij1. Thus if n is a unit vector we have(�:n)2 = 1 and we see that�:n has eigenvalues�1.We define the angular momentum matricess = 12~� and sos2 = 34~21.

The basis states are� 12 = �10� and�� 12 = �01�.

5.1.2 Spin 1 particles

We apply the theory as above to getS3 = 0@1 0 00 0 00 0 �11A S+ = 0@0 p2 00 0 p20 0 0 1AandS� = S+y.5.1.3 Electrons

Electrons are particles with intrinsic spin12 . The angular momentumJ = x ^ p + s,wheres are the spin operators for spin12 .

The basic operators for an electron arex, p ands. We can represent these operatorsby their action on two component wavefunctions: (x) = X�=� 12 �(x)��:

In this basisx 7! x, p 7! �{~r ands 7! 12~�. All other operators are constructedin terms of these, for instance we may have a HamiltonianH = p22m + V (x) + U(x)�:LwhereL = x ^ p.

If V andU depend only onjxj then[J; H ] = 0.

5.2 Addition of angular momentum

Consider two independent angular momentum operatorsJ(1) andJ(2) with J(r) actingon some spaceV (r) andV (r) having spinjr for r = 1; 2.

We now define an angular momentumJ acting onV (1)V (2) byJ = J(1)+J(2).Using the commutation relations forJ(r) we can get[Ji; Jj ] = {~�ijkJk.

We want to construct statesjJ Mi forming a standard angular momentum basis,that is such that: J3jJ Mi = ~M jJ MiJ�jJ Mi = ~N�J;M jJ M�1iwith N�J;M = p(J �M) (J �M + 1). We look first for states inV which satisfyJ+jJ Ji = 0 andJ3jJ Ji = ~J jJ Ji. The maximum value ofJ we can get isj1+ j2;

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26 CHAPTER 5. GENERAL THEORY OF ANGULAR MOMENTUM

andjj1+j2 j1+j2i = jj1 j1i1jj2 j2i2. ThenJ+jj1+j2 j1+j2i = 0. Similarly this isan eigenvector ofJ3 with eigenvalue~ (j1 + j2). We can now applyJ� repeatedly toform all thejJ Mi states. ApplyingJ� we getjJ M�1i = �jj1 j1�1i1jj2 j2i2 + �jj1 j1i1jj2 j2�1i2:

The coefficents� and� can be determined from the coefficentsN�a;b, and we musthave�2 + �2 = 1. If we choosej i a state orthogonal to thisj i = ��jj1 j1�1i1jj2 j2i2 + �jj2 j2i1jj2 j2�1i2:J3j i can be computed and it shows thatj i is an eigenvector ofJ3 with eigenvalue~ (j1 + j2 � 1). Now0 = h jj1+j2 j1+j2�1i / h jJ�jj1+j2 j1+j2iand soh jJ�j�i = 0 for all statesj�i in V . ThusJ+j i = 0 and hence we havej i =jj1+j2�1 j1+j2�1i. We can then construct the statesjj1+j2�1 Mi by repeatedlyapplyingJ�.

For eachJ such thatjj1 � j2j � J � j1 + j2 we can construct a statejJ Ji. Wedefine the Clebsch-Gordan coefficientshj1 m1 j2 m2jJ Mi, and sojJ Mi = Xm1;m2hj1 m1 j2 m2jJ Mijj1 m1ijj2 m2i:The Clebsch-Gordan coefficients are nonzero only whenM = m1 +m2.

We can check the number of states;j1+j2XJ=jj1�j2j (2J + 1) = j1+j2XJ=jj1�j2jn(J + 1)2 � J2o = (2j1 + 1) (2j2 + 1) :Electrons

Electrons have spin12 and we can represent their spin states with�� 12 (s). Using thisnotation we see that two electrons can form a symmetric spin 1 triplet�m(s1; s2) = 8>><>>:� 12 (s1)� 12 (s2)1p2 �� 12 (s1)�� 12 (s2) + �� 12 (s1)� 12 (s2)��� 12 (s1)�� 12 (s2)

and an antisymmetric spin 0 singlet;�0(s1; s2) = 1p2 �� 12 (s1)�� 12 (s2)� �� 12 (s1)� 12 (s2)� :5.3 The meaning of quantum mechanics

Quantum mechanics deals in probabilities, whereas classical mechanics is determin-istic if we have complete information. If we have incomplete information classicalmechanics is also probabilistic.

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5.3. THE MEANING OF QUANTUM MECHANICS 27

Inspired by this we ask if there can be “hidden variables” in quantum mechanicssuch that the theory is deterministic. Assuming that local effects have local causes, thisis not possible.

We will take a spin example to show this. Consider a spin12 particle, with twospin statesj"i andj#i which are eigenvectors ofS3 = 12~�3. If we choose to use twocomponent vectors we have�" = �10� �# = �01� :

Supposen = (sin �; 0; cos �) (a unit vector) and let us find the eigenvectors of�:n = �cos � sin �sin � � cos �� :As (�:n)2 = 1 we must have eigenvalues�1 and an inspired guess gives�";n and�#;n as�";n = cos �2�" + sin �2�# and �#;n = � sin �2�" + cos �2�#:Thus (reverting to ket vector notation) if an electron is in a statej"i then the prob-

ability of finding it in a statej";ni is cos2 �2 and the probability of finding it in a statej#;ni is sin2 �2 .Now, suppose we have two electrons in a spin0 singlet state;j�i = 1p2 fj"i1j#i2 � j#i1j"i2g :Then the probability of finding electron 1 with spin up is12 , and after making this

measurement electron 2 must be spin down. Similarly, if we find electron 1 with spindown (probability12 again) then electron 2 must have spin up. More generally, supposewe measure electron 1’s spin along directionn. Then we see that the probability forelectron 1 to have spin up in directionn (aligned) is12 and then electron 2 must be inthe statej#;ni2.

If we have two electrons (say electron 1 and electron 2) in a spin 0 state we mayphysically separate them and consider independent experiments on them.

We will consider three directions as sketched. For electron 1 there are three vari-ables which we may measure (in separate experiments);S(1)z = �1, S(1)n = �1 andS(1)m = �1. We can also do this for electron 2.

We see that if we find electron 1 hasS(1)z = 1 then electron 2 hasS(2)z = �1 (etc.).If there exists an underlying deterministic theory then we could expectsome prob-

ability distributionp for this set of experiments;0 � p�S(1)z ; S(1)n ; S(1)m ; S(2)z ; S(2)n ; S(2)m � � 1which is nonzero only ifS(1)

dirn = �S(2)dirn andXfsg p(fsg) = 1:

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28 CHAPTER 5. GENERAL THEORY OF ANGULAR MOMENTUM

Bell inequality

Suppose we have a probability distributionp(a; b; c) with a; b; c = �1. We definepartial probabilitiespbc(b; c) =Pa p(a; b; c) and similarly forpac(a; c) andpab(a; b).Then pbc(1;�1) = p(1; 1;�1) + p(�1; 1;�1)� p(1; 1;�1) + p(1; 1; 1) + p(�1; 1;�1) + p(�1;�1;�1)� pab(1; 1) + pac(�1;�1):

Applying this to the two electron system we getP�S(1)n = 1; S(2)m = 1� � P�S(1)z = 1; S(2)n = �1�+ P�S(1)z = �1; S(2)m = 1� :We can calculate these probabilities from quantum mechanicsP�S(1)z = 1; S(2)n = �1� = P�S(1)z = 1; S(1)n = 1� = cos2 �2P�S(1)z = �1; S(2)m = 1� = cos2 �+�2 andP�S(1)n = 1; S(2)m = 1� = sin2 �2 :The Bell inequality givessin2 �2 � cos2 �2 + cos2 �+�2 which is not in general true.

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. References� P.A.M. Dirac,The Principles of Quantum Mechanics, Fourth ed., OUP, 1958.

I enjoyed this book as a good read but it is also an eminently suitable textbook.� Feynman, Leighton, Sands,The Feynman Lectures on Physics Vol. 3, Addison-Wesley, 1964.

Excellent reading but not a very appropriate textbook for this course. I read it as acompanion to the course and enjoyed every chapter.� E. Merzbacher,Quantum Mechanics, Wiley, 1970.

This is a recommended textbook for this course (according to the Schedules). I wasn’tparticularly impressed but you may like it.

There must be a good, modern textbook for this course. If you know of one pleasesend me abrief review and I will include it if I think it is suitable. In any case, withthese marvellous notes you don’t need a textbook, do you?

Related courses

There are courses onStatistical PhysicsandApplications of Quantum MechanicsinPart 2B and courses onQuantum Physicsand Symmetries and Groups in QuantumPhysicsin Part 2A.

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