[eBook] - Science - Physics - Foundations of Quantum Mechanics (35p, Dirac Des Del Principi)

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Foundations of Quantum Mechanics

Dr. H. Osborn1

Michlmas 1997

1LATEXed by Paul Metcalfe comments and corrections to [email protected].


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Revision: 2.5

Date: 1999-06-06 14:10:19+01

The following people have maintained these notes.

date Paul Metcalfe


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Contents

Introduction v

1 Basics 1

1.1 Review of earlier work . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 The Dirac Formalism . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1.2.1 Continuum basis . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2.2 Action of operators on wavefunctions . . . . . . . . . . . . . 5

1.2.3 Momentum space . . . . . . . . . . . . . . . . . . . . . . . . 61.2.4 Commuting operators . . . . . . . . . . . . . . . . . . . . . 7

1.2.5 Unitary Operators . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2.6 Time dependence . . . . . . . . . . . . . . . . . . . . . . . . 8

2 The Harmonic Oscillator 9

2.1 Relation to wavefunctions . . . . . . . . . . . . . . . . . . . . . . . 10

2.2 More comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

3 Multiparticle Systems 13

3.1 Combination of physical systems . . . . . . . . . . . . . . . . . . . . 13

3.2 Multiparticle Systems . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.2.1 Identical particles . . . . . . . . . . . . . . . . . . . . . . . . 14

3.2.2 Spinless bosons . . . . . . . . . . . . . . . . . . . . . . . . . 153.2.3 Spin 12 fermions . . . . . . . . . . . . . . . . . . . . . . . . 16

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

3.4 Observation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

4 Perturbation Expansions 19

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

4.2 Non-degenerate perturbation theory . . . . . . . . . . . . . . . . . . 19

4.3 Degeneracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

5 General theory of angular momentum 23

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

5.1.1 Spin 12 particles . . . . . . . . . . . . . . . . . . . . . . . . . 24

5.1.2 Spin 1 particles . . . . . . . . . . . . . . . . . . . . . . . . . 255.1.3 Electrons . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255.2 Addition of angular momentum . . . . . . . . . . . . . . . . . . . . . 26

5.3 The meaning of quantum mechanics . . . . . . . . . . . . . . . . . . 27

iii


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iv CONTENTS


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Introduction

These notes are based on the course Foundations of Quantum Mechanics given by

Dr. H. Osborn in Cambridge in the Michlmas Term 1997. Recommended books are

discussed in the bibliography at the back.

Other sets of notes are available for different courses. At the time of typing these

courses were:

Probability Discrete Mathematics

Analysis Further Analysis

Methods Quantum MechanicsFluid Dynamics 1 Quadratic Mathematics

Geometry Dynamics of D.E.s

Foundations of QM Electrodynamics

Methods of Math. Phys Fluid Dynamics 2

Waves (etc.) Statistical Physics

General Relativity Dynamical Systems

Physiological Fluid Dynamics Bifurcations in Nonlinear Convection

Slow Viscous Flows Turbulence and Self-Similarity

Acoustics Non-Newtonian Fluids

Seismic Waves

They may be downloaded from

http://www.istari.ucam.org/maths/ or

http://www.cam.ac.uk/CambUniv/Societies/archim/notes.htm

or you can email [email protected] to get a copy of the

sets you require.

v


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AGE.


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Chapter 1

The Basics of Quantum

Mechanics

Quantum mechanics is viewed as the most remarkable development in 20th century

physics. 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 mathematical

basis for these is in Hilbert Spaces but (fortunately!) no knowledge of this area is

required for this course.

1.1 Review of earlier work

This is a brief review of the salient points of the 1B Quantum Mechanics course. If

you anything here is unfamiliar it is as well to read up on the 1B Quantum Mechanics

course. This section can be omitted by the brave.

A wavefunction (x) : R3 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

2

d3x ||2 = 1

then we say that d3x ||2 is the probability of finding the particle in the infinitesimalregion d3x (at x).

Superposition Principle

If 1 and 2 are two wavefunctions representing states of a particle, then so is thelinear combination a11 + a22 (a1, a2 C). This is obviously the statement thatwavefunctions live in a vector space. If = a (with a = 0) then and representthe same physical state. If and are both normalised then a = e. We write e to show that they represent the same physical state.

1


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2 CHAPTER 1. BASICS

For two wavefunctions and we can define a scalar product

(, )

d3x 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 = |(, )|2 and if thewavefunctions are normalised then 0 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 conjugateof the operator Ato be the operator A satisfying (,A) = (A, ). IfA = A then A is Hermitian.Note that ifA is linear then so is A.

In quantum mechanics dynamical variables (such as energy, momentum or angular

momentum) 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 potential V(x) we get the

Hamiltonian H = 22m2 + 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 expanded

as a linear combination of eigenfunctions.

The expectation value for A in a state with wavefunction is A , defined to bei i |ai|2 = (,A). We define the square deviation A2 to be (A A)2

which is in general nonzero.

Time dependence

This is governed by the Schrodinger equation

t= H,

where H is the Hamiltonian. H must be Hermitian for the consistency of quantummechanics:

t(, ) = (, H ) (H , ) = 0

if H is Hermitian. Thus we can impose the condition (, ) = 1 for all time (if isnormalisable).

If we consider eigenfunctions i ofHwith eigenvalues Ei we can expand a generalwavefunction as

(x, t) =

aie Ei

ti(x).

If is normalised then the probability of finding the system with energy Ei is |ai|2.


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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 of

quantum mechanics than that previously discussed. The essential structure of quantum

mechanics is based on operators acting on vectors in some vector space. A wavefunc-

tion corresponds to some abstract vector

|

, a ketvector.

|

represents the state of

some physical system described by the vector space.

If|1 and |2 are ket vectors then | = a1|1 + a2|2 is a possible ket vectordescribing a state this is the superposition principle again.

We define a dual space of bra vectors | and a scalar product |, a complexnumber.1 For any | there corresponds a unique | and we require | = |.We require the scalar product to be linear such that | = a1|1 + a2|2 implies| = a1|1 + a2|2. We see that | = a11| + a22| and so| = a11| + a22|.

We introduce linear operators A| = | and we define operators acting on bravectors to the left |A = | by requiring | = |A| for all . In general, in|A|, A can act either to the right or the left. We define the adjointA ofA suchthat ifA| = | then |A = |. A is said to be Hermitian if A = A.

IfA = a1A1 + a2A2 then A = a1A1 + a2A2, which can be seen by appealing tothe definitions. We also find the adjoint ofBA as follows:

Let BA| = B| = |. Then | = |B = |AB and the resultfollows. Also, if|A = | then | = A|.

We have eigenvectors A| = | and it can be seen in the usual manner that theeigenvalues of a Hermitian operator are real and the eigenvectors corresponding to two

different eigenvalues are orthogonal.

We assume completeness that is any | can be expandedin terms of the basis ketvectors, | = ai|i where A|i = i|i and ai = i|. If| is normalised | = 1 then the expected value of A is A = |A|, which is real ifAis Hermitian.

The completeness relation for eigenvectors ofA can be written as 1 = i |i

i

|,

which gives (as before)

| = 1| =i

|ii|.

We can also rewrite A =

i |iii| and if j = 0 j then we can defineA1 =

i |i1i i|.

We now choose an orthonormal basis {|n} with n|m = nm and the complete-ness relation 1 =

n |nn|. We can thus expand | =

n an|n with an = n|.

We now consider a linear operator A, and then A| = n anA|n = m am|m,with am = m|A| =

n anm|A|n. Further, putting Amn = m|A|n we get

am = n

Amnan and therefore solving A|

= |

is equivalent to solving the

matrix equation Aa = a. Amn is called the matrix representation ofA. We also have| = n ann|, with an = m amAmn, where Amn = Anm gives the Hermitianconjugate matrix. This is the matrix representation ofA.

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 eigenvectors|i. However, in general, in quantum mechanics operators often have continuousspectrum for instance the position operator x in 3 dimensions. x must have eigen-

values x for any point x R3. There exist eigenvectors |x such that x|x = x|x forany x

R3

.As x must be Hermitian we have x|x = xx|. We define the vector space requiredin the Dirac formalism as that spanned by |x.

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

Dirac delta function to get x|x = 3(x x). Completeness givesd3x|xx| = 1.

We can also recover the ket vector from the wavefunction by

| = 1| =

d3x|x(x).

Also x|x| = x(x); the action of the operator x on a wavefunction is multipli-cation by x.

Something else reassuring is

| = |1| =

d3x|xx|

=

d3x |(x)|2 .

The momentum operator p is also expected to have continuum eigenvalues. We

can similarly define states |p which satisfy p|p = p|p. We can relate x and p usingthe commutator, which for two operators A and B is defined by

A, B

= AB BA.

The relationship between x and p is [xi, pj ] = ij . In one dimension [x, p] = .We have a useful rule for calculating commutators, that is:

A, BC

=

A, B

C+ B

A, C

.

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

this to calculate

x, p2

.

x, p2

= [x, p] p + p [x, p]

= 2p.

It is easy to show by induction that [x, pn] = npn1.We can define an exponential by

eap

=

n=0

1

n!

ap

n.


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We can evaluate

x, eap

by

x, e

ap

=

x,

n=0

1

n!

ap

n

=

n=0

1

n!

x, ap

n=

n=0

1

n!

a

n[x, pn]

=

n=1

1

(n 1)! a

npn1

= a

n=1

a

n1pn1

= aeap

and by rearranging this we get that

xeap = e

ap (x + a)

and it follows that eap |x is an eigenvalue of x with eigenvalue x + a. Thus we

see eap

|x = |x + a. We can do the same to the bra vectors with the Hermi-tian conjugate e

ap

to get x + a| = x|e ap . Then we also have the normalisationx + a|x + a = x|x.

We now wish to consider x + a|p = x|e ap |p = e ap x|p. Setting x = 0 givesa|p = e ap N, where N = 0|p is independent ofx. We can determine N from thenormalisation of|p.

(p p) = p|p = da p|aa|p= |N|2

da e

a(pp)

= |N|2 2 (p p)

So, because we are free to choose the phase of N, we can set N =

12

12 and

thus x|p = 12 12 e xp . We could define |p by|p =

dx |xx|p =

1

2

12

dx |xe 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) = x|. We wish to see whatoperators (the position and momentum operators discussed) do to wavefunctions.


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6 CHAPTER 1. BASICS

Now x|x| = xx| = x(x), so the position operator acts on wavefunctionsby multiplication. As for the momentum operator,

x|p| =

dp x|p|pp|

=

dp px|pp|

=

1

2

12

dppexp p|

= ddx

dp x|pp|

= ddx

x| = ddx

(x).

The commutation relation [x, p] = corresponds to

x, ddx

= (acting on(x)).

1.2.3 Momentum space

|x (x) = x| defines a particular representation of the vector space. It issometimes useful to use a momentum representation, (p) = p|. We observe that

(p) =

dx p|xx|

=

1

2

12

dx exp (x).

In momentum space, the operators act differently on wavefunctions. It is easy to

see that p|p| = p(p) and p|x| = ddp (p).We convert the Schrodinger equation into momentum space. We have the operator

equation H = p2

2m + V(x) and we just need to calculate how the potential operates onthe wavefunction.

p|V(x)| =

dx p|V(x)|xx|

=

1

2

12

dx exp V(x)x|

=1

2

dx dp V(x)(p)e

x(pp)

= dp V(p p)(p),

where V(p) = 12

dx exp V(x). Thus in momentum space,

Hp(p) =p2

2m(p) +

dp V(p p)(p).


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1.2.4 Commuting operators

Suppose A and B are Hermitian and

A, B

= 0. Then A and B have simultaneous

eigenvectors.

Proof. Suppose A

|

=

|

and the vector subspace V is the span of the eigenvec-

tors ofA with eigenvalue . (Ifdim V > 1 then is said to be degenerate.)As A and B commute we know that B| = AB| and so B| V. If is

non-degenerate then B| = | for some . Otherwise we have that B : V Vand we can therefore find eigenvectors of B which lie entirely inside V. We can labelthese as |, , and we know that

A|, = |, B|, = |, .

These may still be degenerate. However we can in principle remove this degener-

acy by adding more commuting operators until each state is uniquely labeled by the

eigenvalues of each common eigenvector. This set of operators is called a completecommuting set.

This isnt so odd: for a single particle in 3 dimensions we have the operators x1, x2and x3. These all commute, so for a single particle with no other degrees of freedomwe can label states uniquely by |x. 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 by |p. To ram the point in more, we could also have taken someweird combination like x1, x2 and p3.

For our single particle in 3 dimensions, a natural set of commuting operators in-

volves the angular momentum operator, L = x p, or Li = ijk xj pk.We can find commutation relations between Li and the other operators we know.

These are summarised here, proof is straightforward.

Li, xl

= iljxj

Li, x2

= 0

Li, pm

= imk pk

Li, p2

= 0

Li, Lj

= ijk Lk

Li, L2 = 0If we have a Hamiltonian H = p

2

2m + V(|x|) then we can also see thatL, H

= 0.

We choose as a commuting set H, L2 and L3 and label states |E, l, m, where theeigenvalue ofL2 is l(l + 1) and the eigenvalue ofL3 is m.


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8 CHAPTER 1. BASICS

1.2.5 Unitary Operators

An operator U is said to be unitary ifUU = 1, or equivalently U1 = U.Suppose U is unitary and U| = |, U| = |. Then | = |U and

| = |. Thus the scalar product, which is the probability amplitude of findingthe state | given the state |, is invariant under unitary transformations of states.

For any operator

A we can define

A =U

A

U. Then |

A| = |

A| andmatrix elements are unchanged under unitary transformations. We also note that if

C = AB then C = AB.The quantum mechanics for the |, |, A, B etc. is the same as for |, |, A,

B and so on. A unitary transform in quantum mechanics is analogous to a canonicaltransformation in dynamics.

Note that ifO is Hermitian then U = eO is unitary, as U = eO

= eO.

1.2.6 Time dependence

This is governed by the Schrodinger equation,

t |(t)

= H

|(t)

.

H is the Hamiltonian and we require it to be Hermitian. We can get an explicitsolution of this if H does not depend explicitly on t. We set |(t) = U(t)|(0),where U(t) = e

Ht . As U(t) is unitary, (t)|(t) = (0)|(0).

If we measure the expectation of A at time t we get (t)|A|(t) = a(t). Thisdescription is called the Schrodinger picture. Alternatively we can absorb the time de-

pendence into the operator A to get the Heisenberg picture, a(t) = |U(t)AU(t)|.We write AH(t) = U(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

t

AH(t) = AH(t), H ,which is easily proven.

For a Hamiltonian H = 12m p(t)2 + V(x(t)) we can get the Heisenberg equations

for the operators xH and pH

d

dtxH(t) =

1

mpH(t)

d

dtpH(t) = V(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 oscillator

and the hydrogen atom. We will examine the quantum harmonic oscillator using al-

gebraic methods. In quantum mechanics the harmonic oscillator is governed by the

Hamiltonian

H =1

2mp2 + 12m

2x2,

with the condition that [x, p] = . We wish to solve H| = E| to find the energyeigenvalues.

We define a new operator a.

a =m

2

12

x +

p

m

a = m

2 12

x p

m .a and a are respectively called the annihilation and creation operators. We can

easily obtain the commutation relation

a, a

= 1. It is easy to show that, in terms

of the annihilation and creation operators, the Hamiltonian H = 12

aa + aa

,

which reduces to

aa + 12

. Let N = aa. Then

a, N

= a and

a, N

= a.Therefore Na = a

N 1

and Na = a

N + 1

.

Suppose | is an eigenvector ofN with eigenvalue . Then the commutation rela-tions give that Na| = ( 1) a| and therefore unless a| = 0 it is an eigenvalueofN with eigenvalue 1. Similarly Na| = ( + 1) a|.

But for any |, |N| 0 and equals 0 iffa| = 0. Now suppose we have aneigenvalue of

H, / {0, 1, 2, . . .}. Then n such that a

n

| is an eigenvector of

Nwith eigenvalue n < 0 and so we must have {0, 1, 2, . . .}. Returning to theHamiltonian we get energy eigenvalues En =

n + 12

, the same result as using the

Schrodinger equation for wavefunctions, but with much less effort.

We define |n = Cnan|0, where Cn is such as to make n|n = 1. We can takeCn R, and evaluate 0|anan|0 to find Cn.

9


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10 CHAPTER 2. THE HARMONIC OSCILLATOR

1 = n|n= C2n0|anan|0= C2n0|an1aaan1|0

= C2nC2n1

n 1|aa|n 1

=C2n

C2n1n 1|N + 1|n 1

=C2n

C2n1(n 1 + 1)n 1|n 1

= nC2n

C2n1.

We thus require Cn =Cn1

nand as C0 = 1 we get Cn = (n!)

12 and so we have

the normalised eigenstate (of N)

|n

= 1

n!an

|0

(with eigenvalue n).

|n

is also an

eigenvector of H with eigenvalue

n + 12

. The space of states for the harmonic

oscillator is spanned by {|n}.We also need to ask if there exists a non-zero state | such that a| = 0. Then

0 = |aa| = | + |aa| | > 0.

So there exist no non-zero states | such that a| = 0.

2.1 Relation to wavefunctions

We evaluate

0 = x|a|0 = m2

12

x + m

ddx

x|0and we see that 0(x) = x|0 satisfies the differential equation

d

dx+

m

x

0(x) = 0.

This (obviously) has solution 0(x) = N e12m

x2 for some normalisation con-

stant N. This is the ground state wavefunction which has energy 12.For 1(x) = x|1 = x|a|0 we find

1(x) = m2

12

x

|x

mp

|0

=m

2

12

x

m

d

dx

0(x)

=

2m

12

x0(x).


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2.2. MORE COMMENTS 11

2.2 More comments

Many harmonic oscillator problems are simplified using the creation and annihilation

operators.1 It is useful to summarise the action of the annihilation and creation opera-

tors on the basis states:

a|n = n + 1|n + 1 and a|n = n|n 1.

For example

m|x|n =

2m

12

m|a + a|n

=

2m

12

n m|n 1 + n + 1 m|n + 1

=

2m12

n m,n1 +

n + 1 m,n+1.

This is non-zero only ifm = n1. We note that xr contains terms asars, where0 s r and so m|xr|n can be non-zero only ifn r m n + r.

It is easy to see that in the Heisenberg picture aH(t) = e Ht ae

Ht = eta.

Then using the equations for xH(t) and pH(t), we see that

xH(t) = x cos t +1

mp sin t.

Also, HaH(t) = aH(t)(H+), so if| is an energy eigenstate with eigenvalue

E then aH(t)| is an energy eigenstate with eigenvalue E+ .

1And such problems always occur in Tripos papers. You have been warned.


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12 CHAPTER 2. THE HARMONIC OSCILLATOR


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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 states

and operators, which if Hermitian represent observed quantities.

If we consider two vector spaces V1 and V2 with bases {|r1} and {|s2} withr = 1 . . . dim V1 and s = 1 . . . dim V2. We define the tensor product V1 V2 as thevector space spanned by pairs of vectors

{|r1|s2 : r = 1 . . . dim V1, s = 1 . . . dim V2}.

We see that dim(V1 V2) = dim V1 dim V2. We also write the basis vectors ofV1V2 as |r, s. We can define a scalar product on V1V2 in terms of the basis vectors:r, s|r, s = r|r1s|s2. We can see that if {|r1} and {|s2} are orthonormalbases for their respective vector spaces then {|r, s} is an orthonormal basis for V1V2.

Suppose A1 is an operator on V1 and B2 is an operator on V2 we can define anoperator A1 B2 on V1 V2 by its operation on the basis vectors:

A1 B2 |r1|s2 = A1|r1B2|s2 .

We write A1 B2 as A1B2.

Two harmonic oscillators

We illustrate these comments by example. Suppose

Hi =p2i

2m+ 12mx

2i i = 1, 2.

We have two independent vector spaces Vi with bases |ni where n = 0, 1, . . . andai and a

i are creation and annihilation operators on Vi, and

Hi|ni =

n + 12 |ni.

For the combined system we form the tensor product V1 V2 with basis |n1, n2and Hamiltonian H =

i Hi, so H|n1, n2 = (n1 + n2 + 1) |n1, n2. There are

N + 1 ket vectors in the Nth excited state.

13


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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 on V1 and V2 respectively then theHamiltonian for the combined system is H = H1 + H2 acting on V1 V2. If{|r}and {|s} are eigenbases for V1 and V2 with energy eigenvalues {E1r} and {E2s}respectively then the basis vectors {|r,s} for V1V2 have energies Er,s = E1r + E2s .

3.2 Multiparticle Systems

We have considered single particle systems with states | and wavefunctions (x) =x|. The states belong to a space H.

Consider an N particle system. We say the states belong to Hn = H1 HNand define a basis of states |r11|r22 . . . |rN N where {|rii} is a basis for Hi.

A general state | is a linear combination of basis vectors and we can define theN particle wavefunction as (x1,x2, . . . ,xN) = x1,x2, . . . ,xN|.

The normalisation condition is

| =

d3x1 . . . d3xN |(x1,x2, . . . ,xN)|2 = 1 if normalised.

We can interpret d3x1 . . . d3xN |(x1,x2, . . . ,xN)|2 as the probability densitythat particle i is in the volume element d3xi at xi. We can obtain the probabilitydensity for one particle by integrating out all the other xjs.

For time evolution we get the equation t

| = H|, where H is an operatoron HN.

If the particles do not interact then

H =

Ni=1

Hi

where Hi acts on Hi but leaves Hj alone forj = i. We have energy eigenstates in eachHi such that Hi|ri = Er|ri and so | = |r11|r22 . . . |rN N is an energyeigenstate with energy Er1 + + ErN .

3.2.1 Identical particles

There are many such cases, for instance multielectron atoms. We will concentrate on

two identical particles.

Identical means that physical quantities are be invariant under interchange of

particles. For instance if we have H = H(x1, p1, x2, p2) then this must equal thepermuted Hamiltonian H(x2, p2, x1, p1) if we have identical particles. We introduceU such that

Ux1U1 = x2 Ux2U1 = x1

Up1U1 = p2 Up2U1 = p1.

We should also have UHU1 = H and more generally if A1 is an operator onparticle 1 then UA1U1 is the corresponding operator on particle 2 (and vice versa).


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3.2. MULTIPARTICLE SYSTEMS 15

Note that if | is an energy eigenstate of H then so is U|. Clearly U2 = 1 and werequire U to be unitary, which implies that U is Hermitian.

In quantum mechanics we require | and U| to be the same states (for identicalparticles). This implies that U| = | and the requirement U2 = 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 minus

sign then the particles are fermions (which have spin 12 , 32 , . . . ).1

The generalisation to N identical particles is reasonably obvious. Let Uij inter-

change particles i and j. Then UijHU1ij = H for all pairs (i, j).

The same physical requirement as before gives us that Uij| = | 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 and {1} are the two possible representationsof the permutation group in one dimension.

3.2.2 Spinless bosons

(Which means that the only variables for a single particle are x and p.) Suppose

we have two identical non-interacting bosons. Then H = Hi + H2 and we haveH1|ri = Er|ri. The general space with two particles is H1 H2 which hasa basis {|r1|s2}, 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 and H2 is

|r1|r2; 12 (|r1|s2 + |s1|r2) , r = s

.

The corresponding wavefunctions are

r(x1)r(x2) and12

(r(x1)s(x2) + s(x1)r(x2))

and the corresponding eigenvalues are 2Er and Er+Es. The factor of212 just ensures

normalisation and

12

(1r |2s | + 1s |2r |) 12 (|r1|s2 + |s1|r2)

evaluates to rrss + rsrs.For N spinless bosons with H =

Hi we get

1N!

(|r11 . . . |rN N + permutations thereof) ifri = rj1Spin will be studied later in the course.


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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

glue the results back together as above. The hydrogen atom (with a nuclear charge 2e)

has En =2e280n2

, so we get a ground state for the helium atom with energy 2E1 withno degeneracy and a first excited state with energy E1 + E2 with a degeneracy of four.Hopefully these bear some relation to the results obtained by taking the interaction into

account.

3.3 Two particle states and centre of mass

Suppose we have a Hamiltonian H =p21

2m +p22

2m + V(x1 x2) defined on H2. We canseparate out the centre of mass motion by letting

P = p1 + p2 p =1

2(p1 p2)

X =1

2(x1 + x2) x = x1 x2.

Then

Xi, Pj

= ij , [xi, pj] = ij and X, P and x, p commute respectively.

We can rewrite the Hamiltonian as H = P2

2M + h, h =p2

m+ V(x), where M = 2m and

we can decompose H2 into HCM Hint. HCM is acted on by X and P and has wave-functions (X). Hint is acted on by x, 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) = eP.X and then

E = P2

2M + Eint. We thus need only to solve the one particle equation h = Eint.

Under the particle exchange operator U we have

(x, s1, s2) (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 potential V(|x|) then we may separate variables to get

(x, s1, s2) = Yl

x

|x|

R(|x|)(s1, s2)

with Yl x|x|

= (1)lYl

x|x|

. For spinless bosons we therefore require l to be even.

3.4 Observation

Consider the tensor product of two systems H1 and H2. A general state | in H1H2can be written as

| =i,j

aij|i1|j2

with |i1 H1 and |j2 H2 assumed orthonormal bases for their respective vectorspaces.


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Chapter 4

Perturbation Expansions

4.1 Introduction

Most problems in quantum mechanics are not exactly solvable and it it necessary to

find approximate answers to them. The simplest method is a perturbation expansion.We write H = H0+H where H0 describes a solvable system with known eigenvaluesand eigenvectors, and H is in some sense small.

We write H() = H0 + H 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 discuss

the time independent Schrodinger equation.

4.2 Non-degenerate perturbation theory

Suppose that H0

|n

= n

|n

for n = 0, 1, . . . . We thus assume discrete energy levels

and we assume further that the energy levels are non-degenerate. We also require Hto be sufficiently non-singular to make a power series expansion possible.

We have the equation H()|n() = En()|n(). We suppose that En()tends to n as 0 and |n() |n as 0. We pose the power seriesexpansions

En() = n + E(1)n +

2E(2)n + . . .

|n() = N|n + |(1)n + . . . ,

substitute into the Schrodinger equation and require it to be satisfied at each power

of. The normalisation constant N is easily seen to be 1 + O(2). The O(1) equationis automatically satisfied and the

O() equation is

H0|(1)n + H|n = E(1)n |n + n|(1)n .

Note that we can always replace |(1)n with |(1)n + |n and leave this equationunchanged. We can therefore impose the condition n|(1)n = 0. If we apply n| to

19


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20 CHAPTER 4. PERTURBATION EXPANSIONS

this equation we get E(1)n = n|H|n the first order perturbation in energy. If we

apply r| where r = n we see that

r|(1)n = r|H|nr n

and therefore

|(1)n = r=n

|rr|H|nr n .

Note that we are justified in these divisions as we have assumed that the eigenvalues

are non-degenerate. On doing the same thing to the O(2) equation we see that

E(2)n = n|H|(1)n

= r=n

r|H|n2r n .

This procedure is valid if r n is not very small when r|H|n = 0.Using these results we can see that d

dEn() = n()|H|n() and

|n() =

r=n

1

Er() En() |r()r()|H|n().

Also H

= H and so

2

2En() = 2n()|H

|n().

Example: harmonic oscillator

Consider H = p2

2m +12

m2x2 + m2x2, which can be viewed as H0 + H, where

H0 is the plain vanilla quantum harmonic oscillator Hamiltonian.Calculating the matrix elements r|x2|n required is an extended exercise in ma-

nipulations of the annihilation and creation operators and is omitted. The results are

E(1)n =

n + 12

E(2)n = 1

2

n +

1

2

.

We thus get the perturbation expansion for En

En =

n + 12

1 + 22 + O(3)

.

This system can also be solved exactly to give En =

n + 12

1 + 2 whichagrees 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 = n. Perturbation theory can beextended to the degenerate case, but we will consider only the first order shift in r.We suppose that the states |n, s, s = 1 . . . N n have the same energy n. Nn is thedegeneracy of this energy level.

As before we pose a Hamiltonian H = H0 + H such that H0|n, s = n|n, sand look for states |() with energy E() n as 0.

The difference with the previous method is that we expand |() as a power seriesin in the basis of eigenvectors ofH0. That is

|() =s

|n, sas + |(1).

As the as are arbitrary we can impose the conditions n, s|(1) = 0 for each s andn. We thus have to solve H|() = E()|() with E() = n + E(1). If we takethe O() equation and apply n, r| to it we get

s

asn, r|H|n, s = arE(1)n

which is a matrix eigenvalue problem. Thus the first order perturbations in n arethe eigenvalues of the matrix n, r|H|n, s. If all the eigenvalues are distinct thenthe perturbation lifts the degeneracy. It is convenient for the purpose of calculation

to choose a basis for the space spanned by the degenerate eigenvectors in which this

matrix is as diagonal as possible.1

1Dont ask...


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22 CHAPTER 4. PERTURBATION EXPANSIONS


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Chapter 5

General theory of angular

momentum

For a particle with position and momentum operators x and p with the commutation

relations[ xi, pj ] =

ijwe define L

=x

p. It can be seen that L is Hermitian and

it is easy to show

Li, Lj

= ijkLk.

5.1 Introduction

We want to find out if there are other Hermitian operators J which satisfy this com-

mutation relation.1 We ask on what space of states can this algebra of operators be

realised, or alternatively, what are the representations?

We want

[Ji, Jj ] = ijkJk.

We will choose one component ofJ whose eigenvalues label the states. In accor-

dance with convention we choose J3. Note thatJ2, J3

= 0, so we can simultaneously

diagonalise J2 and J3. Denote the normalised eigenbasis by | , so that

J2| = | and J3| = | .We know that 0 since J2 is the sum of the squares of Hermitian operators.

Now define J = J1 J2. These are not Hermitian, but J+ = J.It will be useful to note that

[J, J3] = J[J+, J] = 2J3

J

2

=1

2 (J+J + JJ+) + J2

3

= J+J J3 + J23= JJ+ + J3 + J23 .

1The is taken outside - you can put it back in if you want, it is inessential but may or may not appear in

exam questions. Since we are now grown up we will omit the hats if they do not add to clarity.

23


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24 CHAPTER 5. GENERAL THEORY OF ANGULAR MOMENTUM

Proof of this is immediate.

Using the [J, J3] relation we have J3J = JJ3 J3, so that

J3J| = ( 1) J|

and J| is an eigenstate of J3 with eigenvalue 1. By similar artifice wecan see that J| is an eigenstate ofJ2 with eigenvalue .

Now evaluate the norm ofJ| , which is

|JJ| = 2 0.

We can now define states | ( n) for n = 0, 1, 2, . . . . We can pin them downmore by noting that 2 2 0 for positive norms. However the formulae wehave are, given , negative for sufficiently large || and so to avoid this we must havemax = j such that J+|j = 0: hence j2 j = 0 and so = j(j + 1).

We can perform a similar trick with J; there must exist min = j such thatJ| j = 0: thus = j(j + 1). So j = j and as j = j = j n for somen {0, 1, 2, . . .} we have j = 0,

1

2 , 1,3

2 , . . . .In summary the states can be labelled by |j m such that

J2|j m = j(j + 1)|j mJ3|j m = m|j m

J|j m = ((j m) (j m + 1))12 |j m 1

with m {j, j + 1, . . . , j 1, j} and j {0, 12 , 1, 32 , . . . }. There are 2j + 1states with different m for the same j. |j m is the standard basis of the angularmomentum states.

We have obtained a representation of the algebra labelled by j. IfJ = L = x pwe must havej an integer.Recall that if we have A we can define a matrix A by A| =

|A.

Note that (BA) =

BA. Given j, we have (J3)mm = m mm and

(J)mm =

(j m) (j m + 1) m,m1, giving us (2j + 1) (2j + 1) matricessatisfying the three commutation relations [J3, J] = J and [J+, J] = 2J3.

IfJ are angular momentum operators which act on a vector space V and we have| V such that J3| = k| and J+| = 0 then is a state with angularmomentum j = k. The other states are given by Jn|, 1 n 2k. The conditionsalso give J2| = k (k + 1) |

5.1.1 Spin 12

particles

This is the simplest non-trivial case. We have j = 12 and a two dimensional state space

with a basis | 12 12 and |12 12 . We have the relations J3|12 12 = 12 |12 12 and

J+|12 12 = 0 J|12 12 = |12 12J+|12 12 = |12 12 J|12 12 = 0.


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5.1. INTRODUCTION 25

It is convenient to introduce explicit matrices such that

J|12 m =m

| 12 m12 ()mm .

The matrices are 2 2 matrices (called the Pauli spin matrices). Explicitly, theyare

+ =

0 20 0

=

0 02 0

3 =

1 00 1

1 =1

2(+ + ) =

0 11 0

2 =

2

(+ ) =

0 0

.

Note that 21 = 22 =

23 = 1 and

= . These satisfy the commutation relations[i, j ] = 2ijkk (a slightly modified angular momentum commutation relation)and we also have 23 = 1 (and the relations obtained by cyclic permutation), soij + ji = 2ij1. Thus ifn is a unit vector we have (.n)

2= 1 and we see that

.n has eigenvalues 1.We define the angular momentum matrices s = 1

2 and so s2 = 3

421.

The basis states are 12

=

10

and 1

2=

01

.

5.1.2 Spin 1 particles

We apply the theory as above to get

S+ =

0

2 0

0 0

20 0 0

S3 =

1 0 00 0 0

0 0 1

and S

= S+.

5.1.3 Electrons

Electrons are particles with intrinsic spin 12 . The angular momentum J = x p+ s, sare the spin operators for spin 1

2.

The basic operators for an electron are x, p and s. We can represent these operators

by their action on two component wavefunctions:

(x) =

= 12

(x).

In this basis x x, p and s 2. All other operators are constructedin terms of these, for instance we may have a Hamiltonian

H =p2

2m+ V(x) + U(x).L

where L = x p.IfV and U depend only on |x| then [J, H] = 0.


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5.2 Addition of angular momentum

Consider two independent angular momentum operatorsJ(1) and J(2) with J(r) acting

on some space V(r) and V(r) having spin jr for r = 1, 2.

We now define an angular momentum J acting on V(1) V(2) by J = J(1) +J(2).Using the commutation relations for J(r) we can get [Ji, Jj ] = ijkJk.

We want to construct states |J M forming a standard angular momentum basis,that is such that:

J3|J M = M|J MJ|J M = NJ,M|J M 1

with NJ,M =

(J M) (J M + 1). We look for states in V which satisfyJ+|J J = 0 and J3|J J = J|J J. The maximum value ofJ we can get is j1 + j2;and |j1 +j2 j1 +j2 = |j1 j11|j2 j22. Then J+|j1 +j2 j1 +j2 = 0. Similarly thisis an eigenvector ofJ3 with eigenvalue (j1 +j2). We can now apply J repeatedly toform all the

|J M

states. Applying J

we get

|J M 1 = |j1 j1 11|j2 j22 + |j1 j11|j2 j2 12.

The coefficents and can be determined from the coefficents Na,b, and we musthave 2 + 2 = 1. If we choose | a state orthogonal to this;

| = |j1 j1 11|j2 j22 + |j1 j11|j2 j2 12.

J3| can be computed and it shows that | is an eigenvectorofJ3 with eigenvalue (j1 +j2 1). Now

0 =

|j1 +j2 j1 +j2

1

|J

|j1 +j2 j1 +j2

and so |J| = 0 for all states | in V. Thus J+| = 0 and

| = |j1 +j2 1 j1 +j2 1.

We can then construct the states |j1 +j2 1 M by repeatedly applying J.For each J such that |j1 j2| J j1 + j2 we can construct a state |J J. We

define the Clebsch-Gordan coefficients j1 m1 j2 m2|J M, and so

|J M =

m1,m2

j1 m1 j2 m2|J M|j1 m1|j2 m2.

The Clebsch-Gordan coefficients are nonzero only when M = m1 + m2.We can check the number of states;

j1+j2J=|j1j2|

(2J + 1) =

j1+j2J=|j1j2|

(J + 1)2 J2

= (2j1 + 1)(2j2 + 1) .


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5.3. THE MEANING OF QUANTUM MECHANICS 27

Electrons

Electrons have spin 12 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) = 1

2(s1) 1

2(s2)

12 12 (s1) 12 (s2) + 12 (s1) 12 (s2) 12 (s1) 12 (s2)

and an antisymmetric spin 0 singlet;

0(s1, s2) =1

2

1

2(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 classical

mechanics is also probabilistic.

Inspired by this we ask if there can be hidden variables in quantum mechanics

such that the theory is deterministic. Assuming that local effects have local causes, this

is not possible.

We will take a spin example to show this. Consider a spin 12 particle, with two

spin states | and | which are eigenvectors ofS3 = 23. If we choose to use twocomponent vectors we have

=

10

=

01

.

Suppose n = (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 eigenvalues1 and an inspired guess gives ,n and

,n as

,n = cos2 + sin

2 and ,n = sin 2 + cos 2.

Thus (reverting to ket vector notation) if an electron is in a state | then the prob-ability of finding it in a state |,n is cos2 2 and the probability of finding it in a state|,n is sin2 2 .

Now, suppose we have two electrons in a spin 0 singlet state;

|

=

1

2 {|1

|2

|1

|2

}.

Then the probability of finding electron 1 with spin up is 12

, and after making this

measurement electron 2 must be spin down. Similarly, if we find electron 1 with spin

down (probability 12 again) then electron 2 must have spin up. More generally, suppose

we measure electron 1s spin along direction n. Then we see that the probability for


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electron 1 to have spin up in direction n (aligned) is 12 and then electron 2 must be in

the state |,n2.If we have two electrons (say electron 1 and electron 2) in a spin 0 state we may

physically 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 and

S(1)m = 1. We can also do this for electron 2.We see that if we find electron 1 has S

(1)z = 1 then electron 2 has S

(2)z = 1 (etc.).

If there exists an underlying deterministic theory then we could expect some prob-

ability distribution p 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

1

which is nonzero only ifS(1)dirn = S(2)dirn and

{s}p({s}) = 1.

Bell inequality

Suppose we have a probability distribution p(a,b,c) with a,b,c = 1. We definepartial probabilitiespbc =

ap(a,b,c) and similarly for pac and pab. 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 get

P

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 mechanics

P

S(1)z = 1, S(2)n = 1

= P

S(1)z = 1, S

(1)n = 1

= cos2 2

P

S(1)z = 1, S(2)m = 1

= cos2 +2 and

P

S(1)n = 1, S(2)m = 1

= sin2 2 .

The Bell inequality gives sin2 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 a

companion 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 wasnt

particularly impressed but you may like it.

There must be a good, modern textbook for this course. If you know of one please

send me a brief review and I will include it if I think it is suitable. In any case, with

these marvellous notes you dont need a textbook, do you?

Related courses

There are courses on Statistical Physics and Applications of Quantum Mechanics in

Part 2B and courses on Quantum Physics and Symmetries and Groups in Quantum

Physics in Part 2A.

29

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