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1. Second Quantization

In this first part, we shall consider nonrelativistic systems consisting of alarge number of identical particles. In order to treat these, we will introducea particularly efficient formalism, namely, the method of second quantiza-tion. Nature has given us two types of particle, bosons and fermions. Thesehave states that are, respectively, completely symmetric and completely an-tisymmetric. Fermions possess half-integer spin values, whereas boson spinshave integer values. This connection between spin and symmetry (statistics)is proved within relativistic quantum field theory (the spin-statistics theo-rem). An important consequence in many-particle physics is the existence ofFermi–Dirac statistics and Bose–Einstein statistics. We shall begin in Sect.1.1 with some preliminary remarks which follow on from Chap. 13 of Quan-tum Mechanics1. For the later sections, only the first part, Sect. 1.1.1, isessential.

1.1 Identical Particles, Many-Particle States,and Permutation Symmetry

1.1.1 States and Observables of Identical Particles

We consider N identical particles (e.g., electrons, π mesons). The Hamiltonian

H = H(1, 2, . . . , N) (1.1.1)

is symmetric in the variables 1, 2, . . . , N . Here 1 ≡ x1, σ1 denotes the positionand spin degrees of freedom of particle 1 and correspondingly for the otherparticles. Similarly, we write a wave function in the form

ψ = ψ(1, 2, . . . , N). (1.1.2)

The permutation operator Pij , which interchanges i and j, has the followingeffect on an arbitrary N -particle wave function

1 F. Schwabl, Quantum Mechanics, 3rd ed., Springer, Berlin Heidelberg, 2002; insubsequent citations this book will be referred to as QMI.

4 1. Second Quantization

Pijψ(. . . , i, . . . , j, . . . ) = ψ(. . . , j, . . . , i, . . . ). (1.1.3)

We remind the reader of a few important properties of this operator. SinceP 2

ij = 1, the eigenvalues of Pij are ±1. Due to the symmetry of the Hamilto-nian, one has for every element P of the permutation group

PH = HP. (1.1.4)

The permutation group SN which consists of all permutations of N objectshas N ! elements. Every permutation P can be represented as a product oftranspositions Pij . An element is said to be even (odd) when the number ofPij ’s is even (odd).2

A few properties :

(i) If ψ(1, . . . , N) is an eigenfunction of H with eigenvalue E, then the samealso holds true for Pψ(1, . . . , N).Proof. Hψ = Eψ ⇒ HPψ = PHψ = EPψ .

(ii) For every permutation one has

⟨ϕ|ψ⟩ = ⟨Pϕ|Pψ⟩ , (1.1.5)

as follows by renaming the integration variables.(iii) The adjoint permutation operator P † is defined as usual by

⟨ϕ|Pψ⟩ =!P †ϕ|ψ

".

It follows from this that

⟨ϕ|Pψ⟩ =!P−1ϕ|P−1Pψ

"=

!P−1ϕ|ψ

"⇒ P † = P−1

and thus P is unitary

P †P = PP † = 1 . (1.1.6)

(iv) For every symmetric operator S(1, . . . , N) we have

[P, S] = 0 (1.1.7)

and

⟨Pψi|S |Pψj⟩ = ⟨ψi|P †SP |ψj⟩ = ⟨ψi|P †PS |ψj⟩ = ⟨ψi|S |ψj⟩ .(1.1.8)

This proves that the matrix elements of symmetric operators are thesame in the states ψi and in the permutated states Pψi.

2 It is well known that every permutation can be represented as a product of cyclesthat have no element in common, e.g., (124)(35). Every cycle can be written asa product of transpositions,

e.g. (12) oddP124 ≡ (124) = (14)(12) even

Each cycle is carried out from left to right (1 → 2, 2 → 4, 4 → 1), whereas theproducts of cycles are applied from right to left.

1.1 Identical Particles, Many-Particle States, and Permutation Symmetry 5

(v) The converse of (iv) is also true. The requirement that an exchange ofidentical particles should not have any observable consequences impliesthat all observables O must be symmetric, i.e., permutation invariant.Proof. ⟨ψ|O |ψ⟩ = ⟨Pψ|O |Pψ⟩ = ⟨ψ|P †OP |ψ⟩ holds for arbitrary ψ.Thus, P †OP = O and, hence, PO = OP .

Since identical particles are all influenced identically by any physical pro-cess, all physical operators must be symmetric. Hence, the states ψ and Pψare experimentally indistinguishable. The question arises as to whether allthese N ! states are realized in nature.

In fact, the totally symmetric and totally antisymmetric states ψs and ψa

do play a special role. These states are defined by

Pijψ sa(. . . , i, . . . , j, . . . ) = ±ψ s

a(. . . , i, . . . , j, . . . ) (1.1.9)

for all Pij .It is an experimental fact that there are two types of particle, bosons

and fermions, whose states are totally symmetric and totally antisymmetric,respectively. As mentioned at the outset, bosons have integral, and fermionshalf-integral spin.

Remarks:

(i) The symmetry character of a state does not change in the course of time:

ψ(t) = T e− i

!tR

0dt′H(t′)

ψ(0) ⇒ Pψ(t) = T e− i

!tR

0dt′H(t′)

Pψ(0) ,(1.1.10)

where T is the time-ordering operator.3(ii) For arbitrary permutations P , the states introduced in (1.1.9) satisfy

Pψs = ψs (1.1.11)

Pψa = (−1)P ψa , with (−1)P =#

1 for even permutations−1 for odd permutations.

Thus, the states ψs and ψa form the basis of two one-dimensional repre-sentations of the permutation group SN . For ψs, every P is assigned thenumber 1, and for ψa every even (odd) element is assigned the number1(−1). Since, in the case of three or more particles, the Pij do not all com-mute with one another, there are, in addition to ψs and ψa, also statesfor which not all Pij are diagonal. Due to noncommutativity, a com-plete set of common eigenfunctions of all Pij cannot exist. These statesare basis functions of higher-dimensional representations of the permu-tation group. These states are not realized in nature; they are referred to

3 QM I, Chap. 16.


as parasymmetric states.4. The fictitious particles that are described bythese states are known as paraparticles and are said to obey parastatis-tics.

1.1.2 Examples

(i) Two particlesLet ψ(1, 2) be an arbitrary wave function. The permutation P12 leads to P12ψ(1, 2)= ψ(2, 1).From these two wave functions one can form

ψs = ψ(1, 2) + ψ(2, 1) symmetric

ψa = ψ(1, 2) − ψ(2, 1) antisymmetric(1.1.12)

under the operation P12. For two particles, the symmetric and antisymmetric statesexhaust all possibilities.

(ii) Three particlesWe consider the example of a wave function that is a function only of the spatialcoordinates

ψ(1, 2, 3) = ψ(x1, x2, x3).

Application of the permutation P123 yields

P123 ψ(x1, x2, x3) = ψ(x2, x3, x1),

i.e., particle 1 is replaced by particle 2, particle 2 by particle 3, and parti-

cle 3 by particle 1, e.g., ψ(1, 2, 3) = e−x21(x2

2−x23)2 , P12 ψ(1, 2, 3) = e−x2

2(x21−x2

3)2 ,

P123 ψ(1, 2, 3) = e−x22(x2

3−x21)2 . We consider

P13P12 ψ(1, 2, 3) = P13 ψ(2, 1, 3) = ψ(2, 3, 1) = P123 ψ(1, 2, 3)P12P13 ψ(1, 2, 3) = P12 ψ(3, 2, 1) = ψ(3, 1, 2) = P132 ψ(1, 2, 3)(P123)2ψ(1, 2, 3) = P123 ψ(2, 3, 1) = ψ(3, 1, 2) = P132 ψ(1, 2, 3).

Clearly, P13P12 = P12P13 .S3, the permutation group for three objects, consists of the following 3! = 6 ele-ments:

S3 = {1, P12, P23, P31, P123, P132 = (P123)2}. (1.1.13)

We now consider the effect of a permutation P on a ket vector. Thus far we haveonly allowed P to act on spatial wave functions or inside scalar products which leadto integrals over products of spatial wave functions.Let us assume that we have the state

|ψ⟩ =X

x1,x2,x3

direct productz }| {|x1⟩1 |x2⟩2 |x3⟩3 ψ(x1, x2, x3) (1.1.14)

4 A.M.L. Messiah and O.W. Greenberg, Phys. Rev. B 136, 248 (1964), B 138,1155 (1965).

1.1 Identical Particles, Many-Particle States, and Permutation Symmetry 7

with ψ(x1, x2, x3) = ⟨x1|1 ⟨x2|2 ⟨x3|3|ψ⟩. In |xi⟩j the particle is labeled by the num-ber j and the spatial coordinate is xi. The effect of P123, for example, is defined asfollows:

P123 |ψ⟩ =X

x1,x2,x3

|x1⟩2 |x2⟩3 |x3⟩1 ψ(x1, x2, x3) .

=X

x1,x2,x3

|x3⟩1 |x1⟩2 |x2⟩3 ψ(x1, x2, x3)

In the second line the basis vectors of the three particles in the direct product areonce more written in the usual order, 1,2,3. We can now rename the summationvariables according to (x1, x2, x3) → P123(x1, x2, x3) = (x2, x3, x1). From this, itfollows that

P123 |ψ⟩ =X

x1,x2,x3

|x1⟩1 |x2⟩2 |x3⟩3 ψ(x2, x3, x1) .

If the state |ψ⟩ has the wave function ψ(x1, x2, x3), then P |ψ⟩ has the wave functionPψ(x1, x2, x3). The particles are exchanged under the permutation. Finally, wediscuss the basis vectors for three particles: If we start from the state |α⟩ |β⟩ |γ⟩ andapply the elements of the group S3, we get the six states

|α⟩ |β⟩ |γ⟩P12 |α⟩ |β⟩ |γ⟩ = |β⟩ |α⟩ |γ⟩ , P23 |α⟩ |β⟩ |γ⟩ = |α⟩ |γ⟩ |β⟩ ,

P31 |α⟩ |β⟩ |γ⟩ = |γ⟩ |β⟩ |α⟩ ,

P123 |α⟩1 |β⟩2 |γ⟩3 = |α⟩2 |β⟩3 |γ⟩1 = |γ⟩ |α⟩ |β⟩ ,

P132 |α⟩ |β⟩ |γ⟩ = |β⟩ |γ⟩ |α⟩ .

(1.1.15)

Except in the fourth line, the indices for the particle number are not written out,but are determined by the position within the product (particle 1 is the first factor,etc.). It is the particles that are permutated, not the arguments of the states.If we assume that α, β, and γ are all different, then the same is true of the sixstates given in (1.1.15). One can group and combine these in the following way toyield invariant subspaces 5:

Invariant subspaces:

Basis 1 (symmetric basis):

1√6(|α⟩ |β⟩ |γ⟩ + |β⟩ |α⟩ |γ⟩ + |α⟩ |γ⟩ |β⟩ + |γ⟩ |β⟩ |α⟩ + |γ⟩ |α⟩ |β⟩ + |β⟩ |γ⟩ |α⟩)

(1.1.16a)

Basis 2 (antisymmetric basis):

1√6(|α⟩ |β⟩ |γ⟩ − |β⟩ |α⟩ |γ⟩ − |α⟩ |γ⟩ |β⟩ − |γ⟩ |β⟩ |α⟩ + |γ⟩ |α⟩ |β⟩ + |β⟩ |γ⟩ |α⟩)

(1.1.16b)

5 An invariant subspace is a subspace of states which transforms into itself onapplication of the group elements.

Basis 3:8<

:

1√12

(2 |α⟩ |β⟩ |γ⟩ + 2 |β⟩ |α⟩ |γ⟩ − |α⟩ |γ⟩ |β⟩ − |γ⟩ |β⟩ |α⟩− |γ⟩ |α⟩ |β⟩ − |β⟩ |γ⟩ |α⟩)

12 (0 + 0 − |α⟩ |γ⟩ |β⟩ + |γ⟩ |β⟩ |α⟩ + |γ⟩ |α⟩ |β⟩ − |β⟩ |γ⟩ |α⟩)

(1.1.16c)

Basis 4:8<

:

12 (0 + 0 − |α⟩ |γ⟩ |β⟩ + |γ⟩ |β⟩ |α⟩ − |γ⟩ |α⟩ |β⟩ + |β⟩ |γ⟩ |α⟩)1√12

(2 |α⟩ |β⟩ |γ⟩ − 2 |β⟩ |α⟩ |γ⟩ + |α⟩ |γ⟩ |β⟩ + |γ⟩ |β⟩ |α⟩− |γ⟩ |α⟩ |β⟩ − |β⟩ |γ⟩ |α⟩) .

(1.1.16d)

In the bases 3 and 4, the first of the two functions in each case is even underP12 and the second is odd under P12 (immediately below we shall call these twofunctions |ψ1⟩ and |ψ2⟩). Other operations give rise to a linear combination of thetwo functions:

P12 |ψ1⟩ = |ψ1⟩ , P12 |ψ2⟩ = − |ψ2⟩ , (1.1.17a)

P13 |ψ1⟩ = α11 |ψ1⟩ + α12 |ψ2⟩ , P13 |ψ2⟩ = α21 |ψ1⟩ + α22 |ψ2⟩ , (1.1.17b)

with coefficients αij . In matrix form, (1.1.17b) can be written as

P13

„|ψ1⟩|ψ2⟩

«=

„α11 α12

α21 α22

«„|ψ1⟩|ψ2⟩

«. (1.1.17c)

The elements P12 and P13 are thus represented by 2 × 2 matrices

P12 =

„1 00 −1

«, P13 =

„α11 α12

α21 α22

«. (1.1.18)

This fact implies that the basis vectors |ψ1⟩ and |ψ2⟩ span a two-dimensional repre-sentation of the permutation group S3. The explicit calculation will be carried outin Problem 1.2.

1.2 Completely Symmetric and Antisymmetric States

We begin with the single-particle states |i⟩: |1⟩, |2⟩, . . . . The single-particlestates of the particles 1, 2, . . . , α, . . . , N are denoted by |i⟩1, |i⟩2, . . . , |i⟩α,. . . , |i⟩N . These enable us to write the basis states of the N -particle system

|i1, . . . , iα, . . . , iN⟩ = |i1⟩1 . . . |iα⟩α . . . |iN ⟩N , (1.2.1)

where particle 1 is in state |i1⟩1 and particle α in state |iα⟩α, etc. (Thesubscript outside the ket is the number labeling the particle, and the indexwithin the ket identifies the state of this particle.)

Provided that the {|i⟩} form a complete orthonormal set, the productstates defined above likewise represent a complete orthonormal system in the

1.2 Completely Symmetric and Antisymmetric States 9

space of N -particle states. The symmetrized and antisymmetrized basis statesare then defined by

S± |i1, i2, . . . , iN⟩ ≡ 1√N !

$

P

(±1)P P |i1, i2, . . . , iN⟩ . (1.2.2)

In other words, we apply all N ! elements of the permutation group SN of Nobjects and, for fermions, we multiply by (−1) when P is an odd permutation.The states defined in (1.2.2) are of two types: completely symmetric andcompletely antisymmetric.

Remarks regarding the properties of S± ≡ 1√N !

%P (±1)P P :

(i) Let SN be the permutation group (or symmetric group) of N quantities.

Assertion: For every element P ∈ SN , one has PSN = SN .Proof. The set PSN contains exactly the same number of elements as SN and these,due to the group property, are all contained in SN . Furthermore, the elements of

PSN are all different since, if one had PP1 = PP2, then, after multiplication by

P−1, it would follow that P1 = P2.Thus

PSN = SNP = SN . (1.2.3)

(ii) It follows from this that

PS+ = S+P = S+ (1.2.4a)

and

PS− = S−P = (−1)P S−. (1.2.4b)

If P is even, then even elements remain even and odd ones remain odd. IfP is odd, then multiplication by P changes even into odd elements and viceversa.

PS+ |i1, . . . , iN⟩ = S+ |i1, . . . , iN⟩PS− |i1, . . . , iN⟩ = (−1)P S− |i1, . . . , iN ⟩Special case PijS− |i1, . . . , iN⟩ = −S− |i1, . . . , iN⟩ .

(iii) If |i1, . . . , iN⟩ contains single-particle states occurring more than once,then S+ |i1, . . . , iN ⟩ is no longer normalized to unity. Let us assume that thefirst state occurs n1 times, the second n2 times, etc. Since S+ |i1, . . . , iN ⟩contains a total of N ! terms, of which N !

n1!n2!...are different, each of these

terms occurs with a multiplicity of n1!n2! . . . .

⟨i1, . . . , iN |S†+S+ |i1, . . . , iN ⟩ =

1N !

(n1!n2! . . . )2N !

n1!n2! . . .= n1!n2! . . .


Thus, the normalized Bose basis functions are

S+ |i1, . . . , iN ⟩ 1√n1!n2! . . .

=1√

N !n1!n2! . . .

$

P

P |i1, . . . , iN⟩ . (1.2.5)

(iv) A further property of S± is

S2± =

√N !S± , (1.2.6a)

since S2± = 1√

N !

%P (±1)P PS± = 1√

N !

%P S± =

√N !S±. We now consider

an arbitrary N -particle state, which we expand in the basis |i1⟩ . . . |iN⟩

|z⟩ =$

i1,... ,iN

|i1⟩ . . . |iN⟩ ⟨i1, . . . , iN |z⟩& '( )ci1,... ,iN

.

Application of S± yields

S± |z⟩ =$

i1,... ,iN

S± |i1⟩ . . . |iN ⟩ ci1,... ,iN =$

i1,... ,iN

|i1⟩ . . . |iN ⟩S±ci1,... ,iN

and further application of 1√N !

S±, with the identity (1.2.6a), results in

S± |z⟩ =1√N !

$

i1,... ,iN

S± |i1⟩ . . . |iN⟩ (S±ci1,... ,iN ). (1.2.6b)

Equation (1.2.6b) implies that every symmetrized state can be expanded interms of the symmetrized basis states (1.2.2).

1.3 Bosons

1.3.1 States, Fock Space, Creation and Annihilation Operators

The state (1.2.5) is fully characterized by specifying the occupation numbers

|n1, n2, . . .⟩ = S+ |i1, i2, . . . , iN⟩ 1√n1!n2! . . .

. (1.3.1)

Here, n1 is the number of times that the state 1 occurs, n2 the number oftimes that state 2 occurs, . . . . Alternatively: n1 is the number of particles instate 1, n2 is the number of particles in state 2, . . . . The sum of all occupationnumbers ni must be equal to the total number of particles:

∞$

i=1

ni = N. (1.3.2)

1.3 Bosons 11

Apart from this constraint, the ni can take any of the values 0, 1, 2, . . . .The factor (n1!n2! . . . )−1/2, together with the factor 1/

√N ! contained in

S+, has the effect of normalizing |n1, n2, . . .⟩ (see point (iii)). These statesform a complete set of completely symmetric N -particle states. By linearsuperposition, one can construct from these any desired symmetric N -particlestate.

We now combine the states for N = 0, 1, 2, . . . and obtain a completeorthonormal system of states for arbitrary particle number, which satisfy theorthogonality relation6

⟨n1, n2, . . . |n1′, n2

′, . . .⟩ = δn1,n1′δn2,n2′ . . . (1.3.3a)

and the completeness relation$

n1,n2,...

|n1, n2, . . .⟩ ⟨n1, n2, . . .| = 11 . (1.3.3b)

This extended space is the direct sum of the space with no particles (vacuumstate |0⟩), the space with one particle, the space with two particles, etc.; it isknown as Fock space.

The operators we have considered so far act only within a subspace offixed particle number. On applying p,x etc. to an N -particle state, we obtainagain an N -particle state. We now define creation and annihilation operators,which lead from the space of N -particle states to the spaces of N ±1-particlestates:

a†i |. . . , ni, . . .⟩ =

√ni + 1 |. . . , ni + 1, . . .⟩ . (1.3.4)

Taking the adjoint of this equation and relabeling ni → ni′, we have

⟨. . . , ni′, . . .| ai =

*ni

′ + 1 ⟨. . . , ni′ + 1, . . .| . (1.3.5)

Multiplying this equation by |. . . , ni, . . .⟩ yields

⟨. . . , ni′, . . .| ai |. . . , ni, . . .⟩ =

√ni δni

′+1,ni .

Expressed in words, the operator ai reduces the occupation number by 1.Assertion:

ai |. . . , ni, . . .⟩ =√

ni |. . . , ni − 1, . . .⟩ for ni ≥ 1 (1.3.6)

and

ai |. . . , ni = 0, . . .⟩ = 0 .

6 In the states |n1, n2, . . .⟩, the n1, n2 etc. are arbitrary natural numbers whosesum is not constrained. The (vanishing) scalar product between states of differingparticle number is defined by (1.3.3a).


Proof:

ai |. . . , ni, . . .⟩ =∞$

ni′=0

|. . . , ni′, . . .⟩ ⟨. . . , ni

′, . . .| ai |. . . , ni, . . .⟩

=∞$

ni′=0

|. . . , ni′, . . .⟩√ni δni

′+1,ni

=#√

ni |. . . , ni − 1, . . .⟩ for ni ≥ 10 for ni = 0 .

The operator a†i increases the occupation number of the state |i⟩ by 1, and the

operator ai reduces it by 1. The operators a†i and ai are thus called creation

and annihilation operators. The above relations and the completeness of thestates yield the Bose commutation relations

[ai, aj] = 0, [a†i , a

†j ] = 0, [ai, a

†j] = δij . (1.3.7a,b,c)

Proof. It is clear that (1.3.7a) holds for i = j, since ai commutes with itself. Fori = j, it follows from (1.3.6) that

aiaj |. . . , ni, . . . , nj , . . .⟩ =√

ni√

nj |. . . , ni − 1, . . . , nj − 1, . . .⟩= ajai |. . . , ni, . . . , nj , . . .⟩

which proves (1.3.7a) and, by taking the hermitian conjugate, also (1.3.7b).For j = i we have

aia†j |. . . , ni, . . . , nj , . . .⟩ =

√ni

pnj + 1 |. . . , ni − 1, . . . , nj + 1, . . .⟩

= a†jai |. . . , ni, . . . , nj , . . .⟩

and“aia

†i − a†

iai

”|. . . , ni, . . . , nj , . . .⟩ =

`√ni + 1

√ni + 1 −

√ni√

ni

´|. . . , ni, . . . , nj , . . .⟩

hence also proving (1.3.7c).

Starting from the ground state ≡ vacuum state

|0⟩ ≡ |0, 0, . . .⟩ , (1.3.8)

which contains no particles at all, we can construct all states:single-particle states

a†i |0⟩ , . . . ,

two-particle states

1√2!

+a†

i

,2|0⟩ , a†

ia†j |0⟩ , . . .

1.3 Bosons 13

and the general many-particle state

|n1, n2, . . .⟩ =1√

n1!n2! . . .

+a†1

,n1+a†2

,n2

. . . |0⟩ . (1.3.9)

Normalization:

a† |n − 1⟩ =√

n |n⟩ (1.3.10)

--a† |n − 1⟩-- =

√n

|n⟩ =1√n

a† |n − 1⟩ .

1.3.2 The Particle-Number Operator

The particle-number operator (occupation-number operator for the state |i⟩)is defined by

ni = a†iai . (1.3.11)

The states introduced above are eigenfunctions of ni:

ni |. . . , ni, . . .⟩ = ni |. . . , ni, . . .⟩ , (1.3.12)

and the corresponding eigenvalue of ni is the number of particles in the statei.The operator for the total number of particles is given by

N =$

i

ni. (1.3.13)

Applying this operator to the states |. . . , ni, . . .⟩ yields

N |n1, n2, . . .⟩ =

.$

i

ni

/|n1, n2, . . .⟩ . (1.3.14)

Assuming that the particles do not interact with one another and, further-more, that the states |i⟩ are the eigenstates of the single-particle Hamiltonianwith eigenvalues ϵi, the full Hamiltonian can be written as

H0 =$

i

niϵi (1.3.15a)

H0 |n1, . . .⟩ =

.$

i

niϵi

/|n1, . . .⟩ . (1.3.15b)

The commutation relations and the properties of the particle-number opera-tor are analogous to those of the harmonic oscillator.


1.3.3 General Single- and Many-Particle Operators

Let us consider an operator for the N -particle system which is a sum ofsingle-particle operators

T = t1 + t2 + . . . + tN ≡$

α

tα , (1.3.16)

e.g., for the kinetic energy tα = p2α/2m, and for the potential V (xα). For one

particle, the single-particle operator is t. Its matrix elements in the basis |i⟩are

tij = ⟨i| t |j⟩ , (1.3.17)

such that

t =$

i,j

tij |i⟩ ⟨j| (1.3.18)

and for the full N -particle system

T =$

i,j

tij

N$

α=1

|i⟩α ⟨j|α . (1.3.19)

Our aim is to represent this operator in terms of creation and annihilation op-erators. We begin by taking a pair of states i, j from (1.3.19) and calculatingtheir effect on an arbitrary state (1.3.1). We assume initially that j = i

$

α

|i⟩α ⟨j|α |. . . , ni, . . . , nj , . . .⟩

≡$

α

|i⟩α ⟨j|α S+ |i1, i2, . . . , iN ⟩ 1√n1!n2! . . .

= S+

$

α

|i⟩α ⟨j|α |i1, i2, . . . , iN ⟩ 1√n1!n2! . . .

.

(1.3.20)

It is possible, as was done in the third line, to bring the S+ to the front, sinceit commutes with every symmetric operator. If the state j is nj-fold occupied,it gives rise to nj terms in which |j⟩ is replaced by |i⟩. Hence, the effect ofS+ is to yield nj states |. . . , ni + 1, . . . , nj − 1, . . .⟩, where the change in thenormalization should be noted. Equation (1.3.20) thus leads on to

= nj

√ni + 1

1√

nj|. . . , ni + 1, . . . , nj − 1, . . .⟩

= √nj

√ni + 1 |. . . , ni + 1, . . . , nj − 1, . . .⟩

= a†iaj |. . . , ni, . . . , nj , . . .⟩ .

(1.3.20′)

1.3 Bosons 15

For j = i, the i is replaced ni times by itself, thus yielding

ni |. . . , ni, . . .⟩ = a†iai |. . . , ni, . . .⟩ .

Thus, for any N , we have

N$

α=1

|i⟩α ⟨j|α = a†iaj .

From this it follows that, for any single-particle operator,

T =$

i,j

tija†iaj , (1.3.21)

where

tij = ⟨i| t |j⟩ . (1.3.22)

The special case tij = ϵiδij leads to

H0 =$

i

ϵia†iai ,

i.e., to (1.3.15a).In a similar way one can show that two-particle operators

F =12

$

α=β

f (2)(xα,xβ) (1.3.23)

can be written in the form

F =12

$

i,j,k,m

⟨i, j| f (2) |k, m⟩ a†ia

†jamak, (1.3.24)

where

⟨i, j| f (2) |k, m⟩ =0

dx

0dyϕ∗

i (x)ϕ∗j (y)f (2)(x, y)ϕk(x)ϕm(y) . (1.3.25)

In (1.3.23), the condition α = β is required as, otherwise, we would haveonly a single-particle operator. The factor 1

2 in (1.3.23) is to ensure that eachinteraction is included only once since, for identical particles, symmetry im-plies that f (2)(xα,xβ) = f (2)(xβ ,xα).Proof of (1.3.24). One first expresses F in the form

F =12

$

α=β

$

i,j,k,m

⟨i, j| f (2) |k, m⟩ |i⟩α |j⟩β ⟨k|α ⟨m|β .


We now investigate the action of one term of the sum constituting F :$

α=β

|i⟩α |j⟩β ⟨k|α ⟨m|β |. . . , ni, . . . , nj , . . . , nk, . . . , nm, . . .⟩

= nknm1

√nk

√nm

√ni + 1

*nj + 1

|. . . , ni + 1, . . . , nj + 1, . . . , nk − 1, . . . , nm − 1, . . .⟩

= a†ia

†jakam |. . . , ni, . . . , nj , . . . , nk, . . . , nm, . . .⟩ .

Here, we have assumed that the states are different. If the states are identical,the derivation has to be supplemented in a similar way to that for the single-particle operators.

A somewhat shorter derivation, and one which also covers the case offermions, proceeds as follows: The commutator and anticommutator forbosons and fermions, respectively, are combined in the form [ak, aj ]∓ = δkj .

$

α=β

|i⟩α |j⟩β ⟨k|α ⟨m|β =$

α=β

|i⟩α ⟨k|α |j⟩β ⟨m|β

=$

α,β

|i⟩α ⟨k|α |j⟩β ⟨m|β − ⟨k|j⟩& '( )δkj

$

α

|i⟩α ⟨m|α

= a†iaka†

jam − a†i [ak, a†

j ]∓& '( )aka†

j∓a†jak

am

= ±a†ia

†jakam = a†

ia†jamak ,

(1.3.26)

for bosonsfermions.

This completes the proof of the form (1.3.24).

1.4 Fermions

1.4.1 States, Fock Space, Creation and Annihilation Operators

For fermions, one needs to consider the states S− |i1, i2, . . . , iN ⟩ defined in(1.2.2), which can also be represented in the form of a determinant:

S− |i1, i2, . . . , iN⟩ =1√N !

1111111

|i1⟩1 |i1⟩2 · · · |i1⟩N...

.... . .

...|iN ⟩1 |iN ⟩2 · · · |iN ⟩N

1111111. (1.4.1)

1.4 Fermions 17

The determinants of one-particle states are called Slater determinants. If anyof the single-particle states in (1.4.1) are the same, the result is zero. Thisis a statement of the Pauli principle: two identical fermions must not occupythe same state. On the other hand, when all the iα are different, then thisantisymmetrized state is normalized to 1. In addition, we have

S− |i2, i1, . . .⟩ = −S− |i1, i2, . . .⟩ . (1.4.2)

This dependence on the order is a general property of determinants.Here, too, we shall characterize the states by specifying their occupation

numbers, which can now take the values 0 and 1. The state with n1 particlesin state 1 and n2 particles in state 2, etc., is

|n1, n2, . . .⟩ .

The state in which there are no particles is the vacuum state, represented by

|0⟩ = |0, 0, . . .⟩ .

This state must not be confused with the null vector!We combine these states (vacuum state, single-particle states, two-particle

states, . . . ) to give a state space. In other words, we form the direct sumof the state spaces for the various fixed particle numbers. For fermions, thisspace is once again known as Fock space. In this state space a scalar productis defined as follows:

⟨n1, n2, . . . |n1′, n2

′, . . .⟩ = δn1,n1′δn2,n2′ . . . ; (1.4.3a)

i.e., for states with equal particle number (from a single subspace), it is iden-tical to the previous scalar product, and for states from different subspacesit always vanishes. Furthermore, we have the completeness relation

1$

n1=0

1$

n2=0

. . . |n1, n2, . . .⟩ ⟨n1, n2, . . .| = 11 . (1.4.3b)

Here, we wish to introduce creation operators a†i once again. These must

be defined such that the result of applying them twice is zero. Furthermore,the order in which they are applied must play a role. We thus define thecreation operators a†

i by

S− |i1, i2, . . . , iN⟩ = a†i1

a†i2

. . . a†iN

|0⟩

S− |i2, i1, . . . , iN⟩ = a†i2

a†i1

. . . a†iN

|0⟩ .(1.4.4)

Since these states are equal except in sign, the anticommutator is

{a†i , a

†j} = 0, (1.4.5a)

which also implies the impossibility of double occupation+a†

i

,2= 0. (1.4.5b)

The anticommutator encountered in (1.4.5a) and the commutator of twooperators A and B are defined by

{A, B} ≡ [A, B]+ ≡ AB + BA[A, B] ≡ [A, B]− ≡ AB − BA .

(1.4.6)

Given these preliminaries, we can now address the precise formulation. If onewants to characterize the states by means of occupation numbers, one has tochoose a particular ordering of the states. This is arbitrary but, once chosen,must be adhered to. The states are then represented as

|n1, n2, . . .⟩ =+a†1

,n1 +a†2

,n2

. . . |0⟩ , ni = 0, 1. (1.4.7)

The effect of the operator a†i must be

a†i |. . . , ni, . . .⟩ = (1 − ni)(−1)

Pj<i nj |. . . , ni + 1, . . .⟩ . (1.4.8)

The number of particles is increased by 1, but for a state that is alreadyoccupied, the factor (1−ni) yields zero. The phase factor corresponds to thenumber of anticommutations necessary to bring the a†

i to the position i.The adjoint relation reads:

⟨. . . , ni, . . .| ai = (1 − ni)(−1)P

j<i nj ⟨. . . , ni + 1, . . .| . (1.4.9)

This yields the matrix element

⟨. . . , ni, . . .| ai |. . . , ni′, . . .⟩ = (1 − ni)(−1)

Pj<i nj δni+1,ni

′ . (1.4.10)

We now calculate

ai |. . . , ni′, . . .⟩ =

$

ni

|ni⟩ ⟨ni| ai |ni′⟩

=$

ni

|ni⟩ (1 − ni)(−1)P

j<i nj δni+1,ni′ (1.4.11)

= (2 − ni′)(−1)

Pj<i nj |. . . , ni

′ − 1, . . .⟩ni′.

Here, we have introduced the factor ni′, since, for ni

′ = 0, the Kronecker deltaδni+1,ni

′ = 0 always gives zero. The factor ni′ also ensures that the right-hand

side cannot become equal to the state |. . . , ni′ − 1, . . .⟩ = |. . . ,−1, . . .⟩.

To summarize, the effects of the creation and annihilation operators are

a†i |. . . , ni, . . .⟩ = (1 − ni)(−1)

Pj<i nj |. . . , ni + 1, . . .⟩

ai |. . . , ni, . . .⟩ = ni(−1)P

j<i nj |. . . , ni − 1, . . .⟩ .(1.4.12)

1.4 Fermions 19

It follows from this that

aia†i |. . . , ni, . . .⟩ = (1 − ni)(−1)2

Pj<i nj (ni + 1) |. . . , ni, . . .⟩

= (1 − ni) |. . . , ni, . . .⟩ (1.4.13a)

a†iai |. . . , ni, . . .⟩ = ni(−1)2

Pj<i nj (1 − ni + 1) |. . . , ni, . . .⟩

= ni |. . . , ni, . . .⟩ , (1.4.13b)

since for ni ∈ {0, 1} we have n2i = ni and (−1)2

Pj<i nj = 1. On account

of the property (1.4.13b) one can regard a†iai as the occupation-number op-

erator for the state |i⟩. By taking the sum of (1.4.13a,b), one obtains theanticommutator

[ai, a†i ]+ = 1.

In the anticommutator [ai, a†j]+ with i = j, the phase factor of the two terms

is different:

[ai, a†j]+ ∝ (1 − nj)ni(1 − 1) = 0 .

Likewise, [ai, aj ]+ for i = j, also has different phase factors in the two sum-mands and, since aiai |. . . , ni, . . .⟩ ∝ ni(ni−1) = 0, one obtains the followinganticommutation rules for fermions:

[ai, aj]+ = 0, [a†i , a

†j ]+ = 0, [ai, a

†j ]+ = δij . (1.4.14)

1.4.2 Single- and Many-Particle Operators

For fermions, too, the operators can be expressed in terms of creation andannihilation operators. The form is exactly the same as for bosons, (1.3.21)and (1.3.24). Now, however, one has to pay special attention to the order ofthe creation and annihilation operators.The important relation

X

α

|i⟩α ⟨j|α = a†iaj , (1.4.15)

from which, according to (1.3.26), one also obtains two-particle (and many-particle)operators, can be proved as follows: Given the state S− |i1, i2, . . . , iN ⟩, we assume,without loss of generality, the arrangement to be i1 < i2 < . . . < iN . Applicationof the left-hand side of (1.4.15) gives

X

α

|i⟩α ⟨j|α S− |i1, i2, . . . , iN⟩ = S−X

α

|i⟩α ⟨j|α |i1, i2, . . . , iN⟩

= nj(1 − ni)S− |i1, i2, . . . , iN ⟩˛j→i

.

The symbol |j→i implies that the state |j⟩ is replaced by |i⟩. In order to bring thei into the right position, one has to carry out

Pk<j nk +

Pk j.

This yields the same phase factor as does the right-hand side of (1.4.15):

a†iaj |. . . , ni, . . . , nj , . . .⟩ = nj(−1)

Pk<j nka†

i |. . . , ni, . . . , nj − 1, . . .⟩

= ni(1 − ni)(−1)P

kj |. . . , ni + 1, . . . , nj − 1, . . .⟩ .

In summary, for bosons and fermions, the single- and two-particle operatorscan be written, respectively, as

T =$

i,j

tija†iaj (1.4.16a)

F =12

$

i,j,k,m

⟨i, j| f (2) |k, m⟩ a†ia

†jamak, (1.4.16b)

where the operators ai obey the commutation relations (1.3.7) for bosonsand, for fermions, the anticommutation relations (1.4.14). The Hamiltonianof a many-particle system with kinetic energy T , potential energy U and atwo-particle interaction f (2) has the form

H =$

i,j

(tij + Uij)a†iaj +

12

$

i,j,k,m

⟨i, j| f (2) |k, m⟩a†ia

†jamak , (1.4.16c)

where the matrix elements are defined in (1.3.21, 1.3.22, 1.3.25) and, forfermions, particular attention must be paid to the order of the two annihila-tion operators in the two-particle operator.

From this point on, the development of the theory can be presented simulta-neously for bosons and fermions.

1.5 Field Operators

1.5.1 Transformations Between Different Basis Systems

Consider two basis systems {|i⟩} and {|λ⟩}. What is the relationship betweenthe operators ai and aλ?The state |λ⟩ can be expanded in the basis {|i⟩}:

|λ⟩ =$

i

|i⟩ ⟨i|λ⟩ . (1.5.1)

The operator a†i creates particles in the state |i⟩. Hence, the superposition%

i⟨i|λ⟩ a†i yields one particle in the state |λ⟩. This leads to the relation

a†λ =

$

i

⟨i|λ⟩ a†i (1.5.2a)

1.5 Field Operators 21

with the adjoint

aλ =$

i

⟨λ|i⟩ ai. (1.5.2b)

The position eigenstates |x⟩ represent an important special case

⟨x|i⟩ = ϕi(x), (1.5.3)

where ϕi(x) is the single-particle wave function in the coordinate representa-tion. The creation and annihilation operators corresponding to the eigenstatesof position are called field operators.

1.5.2 Field Operators

The field operators are defined by

ψ(x) =$

i

ϕi(x)ai (1.5.4a)

ψ†(x) =$

i

ϕ∗i (x)a†

i . (1.5.4b)

The operator ψ†(x) (ψ(x)) creates (annihilates) a particle in the positioneigenstate |x⟩, i.e., at the position x. The field operators obey the followingcommutation relations:

[ψ(x), ψ(x′)]± = 0 , (1.5.5a)[ψ†(x), ψ†(x′)]± = 0 , (1.5.5b)

[ψ(x), ψ†(x′)]± =$

i,j

ϕi(x)ϕ∗j (x

′)[ai, a†j ]± (1.5.5c)

=$

i,j

ϕi(x)ϕ∗j (x

′)δij = δ(3)(x − x′) ,

where the upper sign applies to fermions and the lower one to bosons.We shall now express a few important operators in terms of the field

operators.

Kinetic energy7

$

i,j

a†iTijaj =

$

i,j

0d3xa†

iϕ∗i (x)

2− !2

2m∇2

3ϕj(x)aj

=!2

2m

0d3x∇ψ†(x)∇ψ(x) (1.5.6a)

7 The second line in (1.5.6a) holds when the wave function on which the oper-ator acts decreases sufficiently fast at infinity that one can neglect the surfacecontribution to the partial integration.


Single-particle potential$

i,j

a†iUijaj =

$

i,j

0d3xa†

iϕ∗i (x)U(x)ϕj(x)aj

=0

d3xU(x)ψ†(x)ψ(x) (1.5.6b)

Two-particle interaction or any two-particle operator12

$

i,j,k,m

0d3xd3x′ ϕ∗

i (x)ϕ∗j (x

′)V (x,x′)ϕk(x)ϕm(x′)a†ia

†jamak

=12

0d3xd3x′ V (x,x′)ψ†(x)ψ†(x′)ψ(x′)ψ(x) (1.5.6c)

Hamiltonian

H =0

d3x

2!2

2m∇ψ†(x)∇ψ(x) + U(x)ψ†(x)ψ(x)

3+

12

0d3xd3x′ ψ†(x)ψ†(x′)V (x,x′)ψ(x′)ψ(x) (1.5.6d)

Particle density (particle-number density)The particle-density operator is given by

n(x) =$

α

δ(3)(x − xα) . (1.5.7)

Hence its representation in terms of creation and annihilation operators is

n(x) =$

i,j

a†iaj

0d3y ϕ∗

i (y)δ(3)(x − y)ϕj(y)

=$

i,j

a†iajϕ

∗i (x)ϕj(x). (1.5.8)

This representation is valid in any basis and can also be expressed in termsof the field operators

n(x) = ψ†(x)ψ(x). (1.5.9)

Total-particle-number operator

N =0

d3xn(x) =0

d3xψ†(x)ψ(x) . (1.5.10)

Formally, at least, the particle-density operator (1.5.9) of the many-particle system looks like the probability density of a particle in the stateψ(x). However, the analogy is no more than a formal one since the former isan operator and the latter a complex function. This formal correspondencehas given rise to the term second quantization, since the operators, in the cre-ation and annihilation operator formalism, can be obtained by replacing the

1.5 Field Operators 23

wave function ψ(x) in the single-particle densities by the operator ψ(x). Thisimmediately enables one to write down, e.g., the current-density operator(see Problem 1.6)

j(x) =!

2im[ψ†(x)∇ψ(x) − (∇ψ†(x))ψ(x)] . (1.5.11)

The kinetic energy (1.5.12) has a formal similarity to the expectation valueof the kinetic energy of a single particle, where, however, the wavefunction isreplaced by the field operator.Remark. The representations of the operators in terms of field operators that wefound above could also have been obtained directly. For example, for the particle-number density

Zd3ξd3ξ′ ψ†(ξ) ⟨ξ| δ(3)(x − bξ)

˛ξ′¸ψ(ξ′) = ψ†(x)ψ(x), (1.5.12)

where bξ is the position operator of a single particle and where we have made use ofthe fact that the matrix element within the integral is equal to δ(3)(x−ξ)δ(3)(ξ−ξ′).In general, for a k-particle operator Vk:

Zd3ξ1 . . . d3ξkd3ξ1

′ . . . d3ξk′ ψ†(ξ1) . . . ψ†(ξk)

⟨ξ1ξ2 . . . ξk|Vk

˛ξ′1ξ

′2 . . . ξ′

k

¸ψ(ξ′

k) . . . ψ(ξ′1). (1.5.13)

1.5.3 Field Equations

The equations of motion of the field operators ψ(x, t) in the Heisenberg rep-resentation

ψ(x, t) = eiHt/! ψ(x, 0) e−iHt/! (1.5.14)

read, for the Hamiltonian (1.5.6d),

i! ∂

∂tψ(x, t) =

2− !2

2m∇2 + U(x)

3ψ(x, t) +

+0

d3x′ ψ†(x′, t)V (x,x′)ψ(x′, t)ψ(x, t). (1.5.15)

The structure is that of a nonlinear Schrodinger equation, another reason forusing the expression “second quantization”.Proof: One starts from the Heisenberg equation of motion

i! ∂

∂tψ(x, t) = −[H, ψ(x, t)] = −eiHt/! [H, ψ(x, 0)] e−iHt/! . (1.5.16)

Using the relation

[AB, C]− = A[B, C]± ∓ [A, C]±BFermiBose , (1.5.17)

one obtains for the commutators with the kinetic energy:0

d3x′ !2

2m[∇′ψ†(x′)∇′ψ(x′), ψ(x)]

=0

d3x′ !2

2m(−∇′δ(3)(x′ − x) · ∇′ψ(x′)) =

!2

2m∇2ψ(x) ,

the potential energy:0

d3x′ U(x′)[ψ†(x′)ψ(x′), ψ(x)]

=0

d3x′ U(x′)(−δ(3)(x′ − x)ψ(x′)) = − U(x)ψ(x) ,

and the interaction:12

40d3x′d3x′′ ψ†(x′)ψ†(x′′)V (x′,x′′)ψ(x′′)ψ(x′), ψ(x)

5

=12

0d3x′

0d3x′′ [ψ†(x′)ψ†(x′′), ψ(x)]V (x′,x′′)ψ(x′′)ψ(x′)

=12

0d3x′

0d3x′′

6±δ(3)(x′′ − x)ψ†(x′) − ψ†(x′′)δ(3)(x′ − x)

7

× V (x′,x′′)ψ(x′′)ψ(x′)

= −0

d3x′ ψ†(x′)V (x,x′)ψ(x′)ψ(x).

In this last equation, (1.5.17) and (1.5.5c) are used to proceed from the secondline. Also, after the third line, in addition to ψ(x′′)ψ(x′) = ∓ψ(x′)ψ(x′′), thesymmetry V (x,x′) = V (x′,x) is exploited. Together, these expressions givethe equation of motion (1.5.15) of the field operator, which is also known asthe field equation.The equation of motion for the adjoint field operator reads:

i!ψ†(x, t) = −#− !2

2m∇2 + U(x)

8ψ†(x, t)

−0

d3x′ ψ†(x, t)ψ†(x′, t)V (x,x′)ψ(x′, t), (1.5.18)

where it is assumed that V (x,x′)∗ = V (x,x′).If (1.5.15) is multiplied from the left by ψ†(x, t) and (1.5.18) from the rightby ψ(x, t), one obtains the equation of motion for the density operator

n(x, t) =+ψ†ψ + ψ†ψ

,=

1i!

2− !2

2m

3 9ψ†∇2ψ −

:∇2ψ†; ψ

<,

and thus

n(x) = −∇j(x), (1.5.19)

where j(x) is the particle current density defined in (1.5.11). Equation (1.5.19)is the continuity equation for the particle-number density.

1.6 Momentum Representation 25

1.6 Momentum Representation

1.6.1 Momentum Eigenfunctions and the Hamiltonian

The momentum representation is particularly useful in translationally in-variant systems. We base our considerations on a rectangular normalizationvolume of dimensions Lx, Ly and Lz. The momentum eigenfunctions, whichare used in place of ϕi(x), are normalized to 1 and are given by

ϕk(x) = eik·x/√

V (1.6.1)

with the volume V = LxLyLz. By assuming periodic boundary conditions

eik(x+Lx) = eikx, etc. , (1.6.2a)

the allowed values of the wave vector k are restricted to

k = 2π

2nx

Lx,ny

Ly,nz

Lz

3, nx = 0,±1, . . . , ny = 0,±1, . . . , nz = 0,±1, . . . .

(1.6.2b)

The eigenfunctions (1.6.1) obey the following orthonormality relation:0

d3xϕ∗k(x)ϕk′(x) = δk,k′ . (1.6.3)

In order to represent the Hamiltonian in second-quantized form, we needthe matrix elements of the operators that it contains. The kinetic energy isproportional to

0ϕ∗

k′:−∇2; ϕkd3x = δk,k′k2 (1.6.4a)

and the matrix element of the single-particle potential is given by the Fouriertransform of the latter:

0ϕ∗

k′(x)U(x)ϕk(x)d3x =1V

Uk′−k. (1.6.4b)

For two-particle potentials V (x − x′) that depend only on the relative coor-dinates of the two particles, it is useful to introduce their Fourier transform

Vq =0

d3xe−iq·xV (x) , (1.6.5a)

and also its inverse

V (x) =1V

$

q

Vqeiq·x . (1.6.5b)


For the matrix element of the two-particle potential, one then finds

⟨p′,k′|V (x − x′) |p,k⟩

=1

V 2

0d3xd3x′e−ip′·xe−ik′·x′

V (x − x′)eik·x′eip·x

=1

V 3

$

q

Vq

0d3x

0d3x′e−ip′·x−ik′·x′+iq·(x−x′)+ik·x′+ip·x

=1

V 3

$

q

VqV δ−p′+q+p,0V δ−k′−q+k,0.

(1.6.5c)

Inserting (1.6.5a,b,c) into the general representation (1.4.16c) of the Hamil-tonian yields:

H =$

k

(!k)2

2ma†kak +

1V

$

k′,k

Uk′−ka†k′ak +

12V

$

q,p,k

Vqa†p+qa†

k−qakap.

(1.6.6)

The creation operators of a particle with wave vector k (i.e., in the state ϕk)are denoted by a†

k and the annihilation operators by ak. Their commutationrelations are

[ak, ak′ ]± = 0, [a†k, a†

k′ ]± = 0 and [ak, a†k′ ]± = δkk′ . (1.6.7)

The interaction term allows a pictorial interpretation. It causes the annihila-tion of two particles with wave vectors k and p and creates in their place twoparticles with wave vectors k−q and p+q. This is represented in Fig. 1.1a.The full lines denote the particles and the dotted lines the interaction po-tential Vq. The amplitude for this transition is proportional to Vq. This dia-

k p

k − q p + qVq

a)k p

k − q1 p + q1

Vq1

Vq2

b)

k − q1 − q2 p + q1 + q2

Fig. 1.1. a) Diagrammatic representation of the interaction term in the Hamilto-nian (1.6.6) b) The diagrammatic representation of the double scattering of twoparticles

1.6 Momentum Representation 27

grammatic form is a useful way of representing the perturbation-theoreticaldescription of such processes. The double scattering of two particles can berepresented as shown in Fig. 1.1b, where one must sum over all intermediatestates.

1.6.2 Fourier Transformation of the Density

The other operators considered in the previous section can also be expressedin the momentum representation. As an important example, we shall lookat the density operator. The Fourier transform of the density operator8 isdefined by

nq =0

d3xn(x)e−iq·x =0

d3xψ†(x)ψ(x)e−iq·x . (1.6.8)

From (1.5.4a,b) we insert

ψ(x) =1√V

$

p

eip·xap, ψ†(x) =1√V

$

p

e−ip·xa†p , (1.6.9)

which yields

nq =0

d3x1V

$

p

$

k

e−ip·xa†peik·xake−iq·x ,

and thus, with (1.6.3), one finally obtains

nq =$

p

a†pap+q . (1.6.10)

We have thus found the Fourier transform of the density operator in themomentum representation.

The occupation-number operator for the state |p⟩ is also denoted bynp ≡ a†

pap. It will always be clear from the context which one of the twomeanings is meant. The operator for the total number of particles (1.3.13) inthe momentum representation reads

N =$

p

a†pap . (1.6.11)

1.6.3 The Inclusion of Spin

Up until now, we have not explicitly considered the spin. One can think ofit as being included in the previous formulas as part of the spatial degree8 The hat on the operator, as used here for nq and previously for the occupation-

number operator, will only be retained where it is needed to avoid confusion.


of freedom x. If the spin is to be given explicitly, then one has to make thereplacements ψ(x) → ψσ(x) and ap → apσ and, in addition, introduce thesum over σ, the z component of the spin. The particle-number density, forexample, then takes the form

n(x) =$

σ

ψ†σ(x)ψσ(x)

nq =$

p,σ

a†pσap+qσ .

(1.6.12)

The Hamiltonian for the case of a spin-independent interaction reads:

H =$

σ

0d3x(

!2

2m∇ψ†

σ∇ψσ + U(x)ψ†σ(x)ψσ(x))

+12

$

σ,σ′

0d3xd3x′ψ†

σ(x)ψ†σ′ (x′)V (x,x′)ψσ′ (x′)ψσ(x) , (1.6.13)

the corresponding form applying in the momentum representation.For spin- 1

2 fermions, the two possible spin quantum numbers for the zcomponent of S are ±!

2 . The spin density operator

S(x) =N$

α=1

δ(x − xα)Sα (1.6.14a)

is, in this case,

S(x) =!2

$

σ,σ′

ψ†σ(x)σσσ′ψσ′(x), (1.6.14b)

where σσσ′ are the matrix elements of the Pauli matrices.The commutation relations of the field operators and the operators in the

momentum representation read:

[ψσ(x), ψσ′ (x′)]± = 0 , [ψ†σ(x), ψ†

σ′ (x′)]± = 0

[ψσ(x), ψ†σ′ (x′)]± = δσσ′δ(x − x′)

(1.6.15)

and

[akσ, ak′σ′ ]± = 0, [a†kσ, a†

k′σ′ ]± = 0, [akσ, a†k′σ′ ]± = δkk′δσσ′ . (1.6.16)

The equations of motion are given by

i! ∂

∂tψσ(x, t) =

2− !2

2m∇2 + U(x)

3ψσ(x, t)

+$

σ′

0d3x′ ψ†

σ′(x′, t)V (x,x′)ψσ′ (x′, t)ψσ(x, t) (1.6.17)

Problems 29

and

i!akσ(t) =(!k)2

2makσ(t) +

1V

$

k′

Uk−k′ak′σ(t)

+1V

$

p,q,σ′

Vqa†p+qσ′(t)apσ′(t)ak+qσ(t) . (1.6.18)

Problems

1.1 Show that the fully symmetrized (antisymmetrized) basis functions

S±ϕi1(x1)ϕi2(x2) ... ϕiN (xN)

are complete in the space of the symmetric (antisymmetric) wave functionsψs/a(x1, x2, ... , xN).

Hint: Assume that the product states ϕi1(x1) ... ϕiN (xN), composed of thesingle-particle wave functions ϕi(x), form a complete basis set, and express ψs/a

in this basis. Show that the expansion coefficients cs/ai1,...,iN

possess the symmetry

property 1√N!

S±cs/ai1,...,iN

= cs/ai1,...,iN

. The above assertion then follows directly by

utilizing the identity 1√N!

S±ψs/a = ψs/a demonstrated in the main text.

1.2 Consider the three-particle state |α⟩|β⟩|γ⟩, where the particle number is deter-mined by its position in the product.a) Apply the elements of the permutation group S3. One thereby finds six differentstates, which can be combined into four invariant subspaces.b) Consider the following basis, given in (1.1.16c), of one of these subspaces, com-prising two states:

|ψ1⟩ =1√12

2 |α⟩|β⟩|γ⟩ +2 |β⟩|α⟩|γ⟩ −|α⟩|γ⟩|β⟩ −|γ⟩|β⟩|α⟩

−|γ⟩|α⟩|β⟩ −|β⟩|γ⟩|α⟩!

,

|ψ2⟩ =12

0+0 −|α⟩|γ⟩|β⟩ +|γ⟩|β⟩|α⟩ +|γ⟩|α⟩|β⟩ −|β⟩|γ⟩|α⟩

!

and find the corresponding two-dimensional representation of S3.

1.3 For a simple harmonic oscillator, [a, a†] = 1, (or for the equivalent Bose opera-tor) prove the following relations:

[a, eαa†] = αeαa†

, e−αa†a eαa†

= a + α ,

e−αa†eβa eαa†

= eβαeβa , eαa†a a e−αa†a = e−α a ,

where α and β are complex numbers.

Hint:a) First demonstrate the validity of the following relations

ha, f(a†)

i=

∂∂a† f(a†) ,

ha†, f(a)

i= − ∂

∂af(a) .


b) In some parts of the problem it is useful to consider the left-hand side of theidentity as a function of α, to derive a differential equation for these functions, andthen to solve the corresponding initial value problem.c) The Baker–Hausdorff identity

eABe−A = B + [A, B] +12!

[A, [A, B]] + ...

can likewise be used to prove some of the above relations.

1.4 For independent harmonic oscillators (or noninteracting bosons) described bythe Hamiltonian

H =X

i

ϵia†iai

determine the equation of motion for the creation and annihilation operators in theHeisenberg representation,

ai(t) = eiHt/!aie−iHt/! .

Give the solution of the equation of motion by (i) solving the corresponding initialvalue problem and (ii) by explicitly carrying out the commutator operations in theexpression ai(t) = eiHt/!aie

−iHt/!.

1.5 Consider a two-particle potential V (x′,x′′) symmetric in x′ and x′′. Calculatethe commutator

12

»Zd3x′

Zd3x′′ψ†(x′)ψ†(x′′)V (x′,x′′)ψ(x′′)ψ(x′), ψ(x)

–,

for fermionic and bosonic field operators ψ(x).

1.6 (a) Verify, for an N-particle system, the form of the current-density operator,

j(x) =12

NX

α=1

npα

m, δ(x − xα)

o

in second quantization. Use a basis consisting of plane waves. Also give the formof the operator in the momentum representation, i.e., evaluate the integral, j(q) =R

d3xe−iq·xj(x).(b) For spin- 1

2 particles, determine, in the momentum representation, the spin-density operator,

S(x) =NX

α=1

δ(x− xα)Sα ,

in second quantization.

Problems 31

1.7 Consider electrons on a lattice with the single-particle wave function localizedat the lattice point Ri given by ϕiσ(x) = χσϕi(x) with ϕi(x) = φ(x − Ri). AHamiltonian, H = T + V , consisting of a spin-independent single-particle opera-tor T =

PNα=1 tα and a two-particle operator V = 1

2

Pα=β V (2)(xα, xβ) can be

represented in the basis {ϕiσ} by

H =X

i,j

X

σ

tija†iσajσ +

12

X

i,j,k,l

X

σ,σ′

Vijkla†iσa†

jσ′alσ′akσ ,

where the matrix elements are given by tij = ⟨i | t | j⟩ and Vijkl = ⟨ij | V (2) | kl⟩. Ifone assumes that the overlap of the wave functions ϕi(x) at different lattice pointsis negligible, one can make the following approximations:

tij =

8><

>:

w for i = j ,

t for i and j adjacent sites

0 otherwise

,

Vijkl = Vijδilδjk with Vij =

Zd3x

Zd3y | ϕi(x) |2 V (2)(x,y) | ϕj(y) |2 .

(a) Determine the matrix elements Vij for a contact potential

V =λ2

X

α=β

δ(xα − xβ)

between the electrons for the following cases: (i) “on-site” interaction i = j, and(ii) nearest-neighbor interaction, i.e., i and j adjacent lattice points. Assume asquare lattice with lattice constant a and wave functions that are Gaussians,ϕ(x) = 1

∆3/2π3/4 exp{−x2/2∆2}.(b) In the limit ∆ ≪ a, the “on-site” interaction U = Vii is the dominant con-tribution. Determine for this limiting case the form of the Hamiltonian in secondquantization. The model thereby obtained is known as the Hubbard model.

1.8 Show, for bosons and fermions, that the particle-number operator N =P

i a†iai

commutes with the Hamiltonian

H =X

ij

a†i ⟨i| T |j⟩ aj +

12

X

ijkl

a†i a

†j ⟨ij| V |kl⟩ alak .

1.9 Determine, for bosons and fermions, the thermal expectation value of theoccupation-number operator ni for the state |i⟩ in the grand canonical ensemble,whose density matrix is given by

ρG =1

ZGe−β(H−µN)

with ZG = Tr“e−β(H−µN)

”.


1.10 (a) Show, by verifying the relation

n(x) |φ⟩ = δ(x− x′) |φ⟩ ,

that the state

|φ⟩ = ψ†(x′) |0⟩

(|0⟩ = vacuum state) describes a particle with the position x′.(b) The operator for the total particle number reads:

N =

Zd3x n(x) .

Show that for spinless particles

[ψ(x), N ] = ψ(x) .

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h 3 e 4 @ 5 6³

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2 3 ¢% 9 2 % Â

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2 Identical particles

We are now ready for new material. In this chapter we consider systems which containnot just one or two particles, but any (large) number of particles. After presenting theSchrödinger equation for a general N-particle system, we focus on the case where allN particles are identical. For a classical system this would have no special implications, itwould just mean that all particles have equal properties like charge, mass, etc. For a quan-tum system, however, we show that the situation is completely different: most solutions ofthe Schrödinger equation for a system of identical particles are redundant – they cannotoccur in nature. The physical solutions of the Schrödinger equation are either completelysymmetric with respect to particle exchange (for bosons) or completely antisymmetric(for fermions). Finally, we explain how this allows us to represent any quantum state ofmany identical particles in a very elegant and intuitive way in so-called Fock space – thespace spanned by all physical solutions of the N-particle Schrödinger equation. Table 2.1summarizes the topics covered.

2.1 Schrödinger equation for identical particles

Let us start with some very general considerations about the quantum mechanics of many-particle systems. The wave function of a many-particle state can be written as a functionof the coordinates of all the particles,

ψ(r1, r2, . . . , rN) ≡ ψ({ri}), (2.1)

where N denotes the number of particles. We use the shorthand notation {ri} to denote theset of coordinates r1, r2, . . . , rN . The Schrödinger equation describing the evolution of thiswave function reads generally

H =N!

i=1

"

− h2

2mi

∂2

∂r2i

+ Vi(ri)

#

+ 12

!

i =j

V (2)ij (ri, rj) + 1

6

!

i =j =k

V (3)ijk (ri, rj, rk) + . . . (2.2)

The first term in this equation is just the sum of the one-particle Hamiltonians of all par-ticles. The second and third terms add to this the interactions between the particles: thesecond presents the pairwise interaction between all particles, the third gives the three-particle interactions, and there could be many more exotic interactions up to V (N) givingthe N-particle interactions.

43


In the special case of a system of N identical particles, this Hamiltonian acquires animportant symmetry: it becomes invariant under any permutation of particles. Let usexplain this in more detail. If all particles are identical, they are assumed to all havethe same mass and feel the same potential. More importantly, however, the interactionsbetween the particles also reflect their identity. Quite generally, the interaction energy oftwo particles depends on their coordinates, Eint = V (2)

12 (r1, r2). If the particles are trulyidentical, one can permute them – put particle 1 in the place of particle 2, and particle 2 inthe place of particle 1 – without changing the interaction energy. Thus, for identical parti-cles V (2)

12 (r1, r2) = V (2)21 (r2, r1) should hold. Also the more exotic interactions, that involve

three and more particles, should be symmetric with respect to permutations. For instance,the three-particle interaction satisfies

V (3)123(r1, r2, r3) = V (3)

132(r1, r3, r2) = V (3)231(r2, r3, r1)

= V (3)213(r2, r1, r3) = V (3)

312(r3, r1, r2) = V (3)321(r3, r2, r1),

where we list all six possible permutations of three particles.

Control question. How many permutations are possible for N particles? How manypair permutations are possible?

So, however complicated the structure of a Hamiltonian for identical particles, the mostimportant feature of it is its symmetry. The Hamiltonian does not change upon permutationof the coordinates of any pair of particles. More formally, we can state that for any pair ofparticles i and j

H(. . . , ri, r′i, . . . , rj, r′

j, . . . ) = H(. . . , rj, r′j, . . . , ri, r′

i, . . . ). (2.3)

Here the coordinates r′ refer to the coordinates of the wave function on which theHamiltonian acts, and r to the coordinates of the resulting wave function.

Control question. Do you understand this notation? Can you explain that for a singlefree particle one can write H(r, r′) = − h2

2m∇rδ(r − r′)∇r′?

The many-particle Hamiltonian as presented in (2.2) for the case of identical particles thusreads

H =N!

i=1

"

− h2

2m∂2

∂r2i

+ Vi(ri)

#

+N!

i=2

i−1!

j=1

V (2)(ri, rj) +N!

i=3

i−1!

j=2

j−1!

k=1

V (3)(ri, rj, rk) + . . .

(2.4)

Control question. Do you understand how permutation symmetry is incorporated intothis Hamiltonian?

One can also state the permutation symmetry of the Hamiltonian by saying that Hcommutes with all pair permutation operators, i.e. HPab = PabH, where the permutationoperator Pab effects pair permutation of the particles a and b. This implies that H and Pab

must possess a complete common set of eigenfunctions (see Exercise 3 of Chapter 1). So,

45 2.1 Schrödinger equation for identical particles

by just investigating the symmetries of the permutation operators we thus reveal propertiesof the eigenfunctions of any many-particle Hamiltonian!

Unfortunately, the eigenfunctions of a symmetric Hamiltonian do not have to be of thesame symmetry as the Hamiltonian itself, which you probably already learned in yourelementary course on quantum mechanics. An illustrative example of this fact is given bythe atomic orbitals that are eigenstates of a spherically-symmetric Hamiltonian. While thes-orbitals are also spherically symmetric, the p- and d-orbitals – “hourglasses” and “cloverleafs” – are not. A simpler example concerns a particle confined in a one-dimensionalpotential well, with V(x) = 0 for |x| < L/2 and V(x) = ∞ for x ≥ L/2. The Hamiltoniandescribing this particle is symmetric with respect to reflection, x → −x. Its eigenfunctions,however, found from solving the Schrödinger equation, are either symmetric,

ψ sn(x) =

$2L

cos%

[2n + 1]πxL

&, with n = 0, 1, 2, . . . , (2.5)

or antisymmetric

ψan (x) =

$2L

sin%

2nπxL

&, with n = 0, 1, 2, . . . (2.6)

So indeed, although H(x) = H(−x) in this case, its eigenfunctions can also be antisymmet-ric, ψa

n (x) = −ψan (−x).

The same holds for the permutation symmetric many-particle Hamiltonian. The caseof two identical particles is relatively simple: we know that the pair permutation operatorobeys (P12)2 = 1, so the only possible eigenvalues of P12 are 1 and −1. The two eigen-functions therefore are either symmetric, P12ψs = ψs, or antisymmetric, P12ψa = −ψa,with respect to permutation of the particles. For three and more particles, the symmetryclassification gets more complicated. Different permutation operators do not always com-mute with each other (e.g. P12P23 = P23P12), and therefore there does not exist a completeset of common eigenfunctions for all permutation operators. However, one can still form acompletely symmetric and a completely antisymmetric eigenfunction, i.e. one that remainsunchanged, Pabψs = ψs, and one that flips sign, Pabψa = −ψa, under permutation ofany pair of particles. Further there may also exist states which form partially symmetricsubspaces: any pair permutation performed on a state in a partially symmetric subspaceproduces a linear combination of states in that same subspace. Let us rephrase this in theformal language introduced above. Suppose the states ψ1 and ψ2 form a partially symmet-ric subspace of an N-particle system. Any pair permutation Pab performed on the state ψ1

or ψ2 then produces a linear combination of the same two states: Pabψ1 = α11ψ1 + α21ψ2

and Pabψ2 = α12ψ1 + α22ψ2. Thus, in this partially symmetric subspace all permuta-tion operations are represented by two-dimensional matrices. This is in contrast with theirone-dimensional representations in the fully symmetric and fully antisymmetric case (therethey are represented by a simple number, 1 or −1).

To illustrate this, we now treat the case of three particles explicitly. Using the shorthandnotation |ijk⟩, where i labels the state of the first particle, j of the second, and k of the third,


!Fig. 2.1 All permutations of three particles. The six states obtained by permutation of the particles’ coordinates can berepresented by the vertices of a hexagon. The states connected by a triangle (thick lines) can be obtained from eachother by cyclic permutation. Pair permutations connect states from different triangles and are represented by thinlines: dotted, dashed, and dotted-dashed.

we can make a fully symmetric wave function,

|(123)s⟩ = 1√6

'|123⟩ + |231⟩ + |312⟩ + |132⟩ + |321⟩ + |213⟩

(, (2.7)

which is symmetric under any pair permutation of two particles, meaning that Pab|(123)s⟩= |(123)s⟩, and a fully antisymmetric wave function,

|(123)a⟩ = 1√6

'|123⟩ + |231⟩ + |312⟩ − |132⟩ − |321⟩ − |213⟩

(, (2.8)

which is antisymmetric under any pair permutation, Pab|(123)a⟩ = −|(123)a⟩. This iseasy to see from Fig. 2.1. There are two triangles of states connected by thick lines. Thecomponents of the wave function |(123)a⟩ belonging to different thick triangles come withdifferent sign. Any pair permutation (thin lines) swaps the thick triangles, thus changing thesign of the wave function. The functions |(123)s⟩ and |(123)a⟩ are two linearly independentcombinations of all six permutations of |123⟩. We included a normalization factor of 1/

√6

so that, if |123⟩ is normalized, then also |(123)s,a⟩ are normalized.There exist, however, other linearly independent combinations of the permutations of

|123⟩ which exhibit more complicated symmetry. Let us consider the set of two functions

|(123)+A⟩ = 1√3

'|123⟩ + ε|231⟩ + ε∗|312⟩

(,

|(123)−A⟩ = 1√3

'|132⟩ + ε∗|321⟩ + ε|213⟩

(.

(2.9)

Here we defined ε ≡ ei2π/3 so that ε3 = 1 and ε2 = ε−1 = ε∗. Such functions are trans-formed to themselves by a cyclic permutation of particles, |123⟩ → |231⟩ and so on. Thiscyclic permutation gives Pcyc|(123)+ A⟩ = ε|(123)+ A⟩ and Pcyc|(123)−A⟩ = ε∗|(123)−A⟩.1A pair permutation switches between the two functions |(123)±A⟩. Therefore, any permu-tation of particles in the two-dimensional subspace spanned by these functions producesagain a state within the same subspace. In other words, all permutation operators can be

1 For this reason, these functions are also used in solid state theory to represent the trigonal symmetry of 2π/3rotations. This symmetry is also evident in Fig. 2.1.

47 2.2 The symmetry postulate

represented by two-dimensional matrices in the subspace. The functions are said to form atwo-dimensional irreducible presentation of the permutation group.

There is an alternative set of two functions

|(123)−B⟩ = 1√3

'|123⟩ + ε∗|231⟩ + ε|312⟩

(,

|(123)+B⟩ = 1√3

'|132⟩ + ε|321⟩ + ε∗|213⟩

(,

(2.10)

which have identical transformation properties with respect to all permutations. They aresaid to belong to the same two-dimensional irreducible representation.

Control question. Give the matrix representation of the permutation operator P13 inthe partially symmetric subspaces spanned by (2.9) and (2.10).

In principle these partial symmetries could have been exploited by Nature as well. Thereis no a-priori reason why wave functions must belong to a one-dimensional vector space.One could also create partially symmetric multi-dimensional representations of quantumstates which are eigenstates of a permutation symmetric Hamiltonian. Fortunately, how-ever, it seems that Nature has chosen the most simple option: only fully symmetric andfully antisymmetric states are to be found.

2.2 The symmetry postulate

The general observation that the wave function of a set of identical particles is either fullysymmetric or fully antisymmetric can now be expressed in the form of an exact defini-tion. We put it here as a postulate, which distinguishes all particles as either bosons orfermions.

symmetry postulateThe wave function of N identical bosons is completely symmetric withrespect to particle pair permutations.The wave function of N identical fermions is completely antisymmetricwith respect to particle pair permutations.

One of the implications of this postulate is that quantum particles are more than justidentical, they are truly indistinguishable. Let us illustrate this statement with a comparisonwith classical objects, such as ping-pong balls. We can imagine having two identical ping-pong balls, which we distinguish by putting one ball in our left pocket and the other in ourright pocket (see Fig. 2.2). Then there are two well distinguishable objects: a ball “sittingin our right pocket” and a ball “sitting in our left pocket”. Let us try to follow the samereasoning for two quantum particles, where the two pockets are now two quantum states,|L⟩ and |R⟩. Suppose that the two particles are bosons (but the argument also holds for


!Fig. 2.2 The possible ways to distribute two balls, labeled “0” and “1”, over two bags, so that there is one ball in each bag.(a) If the balls and bags are classical, then there are obviously two possibilities. (b) For quantum balls in “quantumbags” there is only one possibility due to the symmetry postulate.

fermions). We would again like to put each particle in a separate “pocket,” i.e. we wouldlike to create the state |L⟩|R⟩ or |R⟩|L⟩, thereby distinguishing the particles. We then realizethat this pocket trick cannot be implemented. Since the wave function must be symmetricwith respect to permutation, the only permitted state with one ball in each pocket is thesuperposition

ψs = 1√2

(|L⟩|R⟩ + |R⟩|L⟩). (2.11)

Neither |L⟩|R⟩ nor |R⟩|L⟩ can be realized. It is clear that, when the particles are in the stateψs, it is impossible to tell which of them is in state |L⟩ and which in state |R⟩: they both arein both states!

Control question. Why is there a square root in the above formula?

2.2.1 Quantum fields

If one contemplates the implications of the symmetry postulate, it soon gets strange, andone is forced to draw some counterintuitive conclusions. For instance, the postulate seemsto contradict the most important principle of physics, the locality principle. We wouldexpect that objects and events that are separated by large distances in space or time mustbe independent of each other and should not correlate. Suppose Dutch scientists pick up anelectron and confine it in a quantum dot, while a group of Japanese colleagues performs asimilar experiment (Fig. 2.3). The two electrons picked up do not know about each other,have never been in a close proximity, nor do the two research teams communicate with eachother. However, due to the magic of the symmetry postulate, the wave function of these twoidentical electrons must be antisymmetric. Therefore, the electrons correlate! Why?

49 2.2 The symmetry postulate

ee

!Fig. 2.3 Electrons separated by large distances in space and time all originate from the same quantum field. This explains theircorrelations, imposed by the symmetry postulate.

The attempts to solve this paradox have led to the most important theoretical findingof the 20th century: there are no particles! At least, particles are not small separate actorsin the otherwise empty scene of space and time, as they have been envisaged by gen-erations of pre-quantum scientists starting from Epicurus. What we see as particles arein fact quantized excitations of corresponding fields. There is a single entity – the electronfield – which is responsible for all electrons in the world. This field is the same in Japan andin the Netherlands, thus explaining the correlations. The field persists even if no electronsare present: the physical vacuum, which was believed to be empty, is in fact not.

The electron field is similar to a classical field – a physical quantity defined in space andtime, such as an electric field in a vacuum or a pressure field in a liquid. Starting from thenext chapter, most of the material in this book illustrates and explains the relation betweenparticles and fields: a relation which is a dichotomy in classical science, and a unity inquantum mechanics. We see fields emerging from indistinguishable particles, and, con-versely, also see particles emerging when a classical field is quantized. The consequencesof these relations are spelled out in detail.

The discovery of the concept just outlined was revolutionary. It has radically changed theway we understand and perceive the Universe. Theory able to treat particles and fields ina unified framework – quantum field theory – has become a universal language of science.Any imaginable physical problem can be formulated as a quantum field theory (thoughfrequently it is not required, and often it does not help to solve the problem).

Although condensed matter provides interesting and stimulating examples of quantumfield theories (we present some in later chapters), the most important applications of theframework are aimed at explaining fundamental properties of the Universe and concernelementary particles and cosmology. As such, the theory must be relativistic to conformwith the underlying symmetries of the space–time continuum. The proper relativistic the-ory of quantized fields provides a more refined understanding of the symmetry postulate.It gives the relation between spin and statistics: particles with half-integer spin are boundto be fermions while particles with integer spin are bosons.


The quantum field theory corresponding to the Standard Model unifies three of the fourfundamental interactions and explains most experimentally accessible properties of ele-mentary particles. Still the theory is far from complete: the underlying symmetries of thisworld, the relation between fields and geometry, have to be understood better (those beingthe subject of string theory). This is why quantum field theory remains, and will remain inthe foreseeable future, a very active and challenging research area.

2.3 Solutions of the N-particle Schrödinger equation

Let us consider in more detail the solutions of the Schödinger equation for N particles,and try to list all solutions before imposing the symmetry postulate. To simplify the work,we skip the interactions between the particles. The Hamiltonian is then given by the firstline of (2.4), and is just a sum of one-particle Hamiltonians. The Schrödinger equationbecomes

Hψ =N!

i=1

"

− h2

2m∂2

∂r2i

+ V(ri)

#

ψ = Eψ({ri}). (2.12)

We can find the general solution of this equation in terms of one-particle eigenfunctionsϕn(r) that satisfy

"

− h2

2m∂2

∂r2 + V(r)

#

ϕn(r) = Enϕn(r). (2.13)

Here n labels the one-particle quantum states, and to distinguish them from the many-particle states, we call them levels.

Control question. What are the levels ϕn and their energies En for identical particlesin a one-dimensional infinite square well?

These levels ϕn can then be used to construct N-particle solutions. The general solutionto (2.12) can be written as an N-term product of one-particle solutions ϕn,

ψ{ℓi}({ri}) = ϕℓ1 (r1)ϕℓ2 (r2) . . .ϕℓN (rN), (2.14)

where the integer index ℓi gives the level in which the ith particle is situated. The set ofindices {ℓi} = {ℓ1, ℓ2, ℓ3, . . . , ℓN} can be any set of (positive) integers, and to each set ofnumbers corresponds a single wave function (2.14). The set of all ψ{ℓi} we can construct,forms a complete basis of the space of all wave functions of N identical particles.

Control question. Prove that this basis is orthonormal.

To illustrate, with three particles we could create a three-particle state ϕ1(r1)ϕ1(r2)ϕ1(r3), describing the situation where all three particles occupy the first level. Anotherallowed three-particle state, ϕ8(r1)ϕ7(r2)ϕ6(r3), has the first particle in the eighth level, thesecond in the seventh level and the third in the sixth level. We could go on like this forever:every state ϕℓ1 (r1)ϕℓ2 (r2)ϕℓ3 (r3) is formally a solution of the Schrödinger equation.

51 2.3 Solutions of the N-particle Schrödinger equation

The energy corresponding to the many-particle state (2.14) is given by the sum of theone-particle energies,

E =N!

i=1

Eni . (2.15)

It is instructive to re-sum this expression in the following way. We introduce the (integer)occupation numbers nk, which tell us how many times the level k appears in the index {ℓi},or, in other words, how many particles occupy level k. Obviously, we can rewrite

E =∞!

k=1

Eknk. (2.16)

The point we would like to concentrate on now is that there are plenty of many-particlestates which have the same total energy E. We know that in quantum mechanics this iscalled degeneracy. Let us find the degeneracy of an energy eigenvalue that corresponds toa certain choice of occupation numbers {n1, n2, . . . } ≡ {nk}. For this, we have to calculatethe number of ways in which the particles can be distributed over the levels, given thenumbers {nk} of particles in each level. Let us start with the first level: it has to contain n1

particles. The first particle which we put in the first level is to be chosen from N particles,the second particle from N − 1 remaining particles, etc. Therefore, the number of ways tofill the first level is

W1 = 1n1!

N(N − 1) · · · (N − nk + 1) = N!n1! (N − n1)!

. (2.17)

Control question. Do you understand why the factor n1! appears in the denominatorof (2.17)?

Let us now fill the second level. Since we start with N − n1 remaining particles,

W2 = (N − n1)!n2! (N − n1 − n2)!

. (2.18)

The total number of ways to realize a given set of occupation numbers {nk} is

W =∞)

i=1

Wi = N!n1! n2! · · · n∞!

, (2.19)

as illustrated in Fig. 2.4. An infinite number of levels to sum and multiply over should notconfuse us, since for finite N most of the levels are empty, and do not contribute to thesums and products.

These considerations on degeneracy provide a way to connect the seemingly artifi-cial and abstract symmetry postulate with experimental facts. If there were no symmetrypostulate dictating that a wave function be either symmetric or antisymmetric, then thenumber W would indeed give the degeneracy of a state with a certain set of occupationnumbers {nk}, i.e. it would give the number of possible wave functions which realize aspecific energy E = *

n Eknk. Since the degeneracy of E enhances the probability offinding the system at this energy, this degeneracy would manifest itself strongly in statis-tical mechanics. In this case, as shown by Gibbs, the entropy of an ideal gas would not


!Fig. 2.4 The numberW can be understood directly from a simple statistical consideration. Suppose we have a state withoccupation numbers {5, 0, 2, 1, 3, 0, . . . }, as depicted in (a). There are N! ways to fill the levels like this: the firstparticle we put in level 1 can be chosen out of N particles, the second out of N − 1 particles, etc. However, since theparticles are identical, the order in each level does not matter. Indeed, as shown in (b), the factorsϕ5(r1)ϕ5(r2)ϕ5(r3)andϕ5(r3)ϕ5(r1)ϕ5(r2) in the many-particle wave function (2.14) are equal. Therefore N! has to be divided by thenumber of permutations possible in each level, n1! n2! · · ·

be proportional to its volume. This proportionality, however, is an experimental fact, andsuggests that the degeneracy of every energy level is one. Already, before the invention ofquantum theory, Gibbs showed that if one assumes the gas particles to be indistinguishable,the degeneracy of all energy levels reduces to one, resulting in the right expression for theentropy. The symmetry postulate provided the explanation for this absence of degeneracyand for the indistinguishability of identical particles. From the postulate it follows thatthere is only one single quantum state corresponding to each set of occupation numbers.We show this in the two following sections.

2.3.1 Symmetric wave function: Bosons

Let us now explicitly construct wave functions that do satisfy the symmetry postulate. Westart with bosons, so we have to build completely symmetric wave functions. Supposewe have a set of occupation numbers {nk}, and we pick any of the wave functions (2.14)that correspond to this set. All other basis states satisfying the same {nk} can be obtainedfrom this state by permutation of particles. We construct a linear combination of all thesepermuted states, taking them with equal weight C,

)(S){nk} = C

!

P

P'ϕℓ1 (r1)ϕℓ2 (r2) . . .ϕℓN (rN)

(, (2.20)

where the summation is over all different permutations. The result obtained does notchange upon any pair permutation, that is, it presents the completely symmetric wavefunction sought. We are only left to determine the unknown C from the followingnormalization condition,

1 = ⟨)(S)|)(S)⟩ = |C|2!

P,P′⟨P|P′⟩ = |C|2

!

P

⟨P|P⟩

= |C|2 × (number of permutations).

(2.21)


Albert Einstein (1879–1955)Won the Nobel Prize in 1921 for “his services to theoreticalphysics, and especially for his discovery of the law of thephotoelectric effect.”

Albert Einstein is probably the most famous physicist ofthe last three centuries, if not ever. He is most widelyknown for his theory of relativity, but his contributionsto theoretical physics are numerous and cover almost allfields. The work for which he was awarded the Nobel prize

was his explanation of the photoelectric effect, introducing the idea that light consistsof individual discrete quanta – photons. This idea of quantization became a majorinspiration for the development of quantum mechanics in the 1920s.

In 1924 Einstein received a letter from the Indian scientist Satyendra Nath Boseabout the statistical properties of photons. Bose had discovered that the inconsisten-cies between experiments and the theory of radiation could be solved by assumingthat photons were actually indistinguishable. Several journals had rejected Bose’smanuscript on this topic, so that he had decided to write personally to Einstein.Einstein immediately agreed with Bose’s idea. He translated the manuscript to Ger-man, and sent it together with a strong recommendation to the famous Zeitschrift fürPhysik, where it was then accepted immediately. The special statistics resulting fromBose’s idea became known as Bose–Einstein statistics, and the particles obeying itwere called bosons.

After Einstein graduated in 1900 from the Eidgenössische Polytechnische Schule inZurich, he spent almost two years in vain looking for a post as lecturer. Finally, heaccepted a job as assistant examiner at the Swiss patent office in Bern. While beingemployed at the patent office, he worked on his dissertation and started publishingpapers on theoretical physics. The year 1905 became known as his annus mirabilisin which he published four groundbreaking works which would later revolutionizephysics: he explained the photoelectric effect and Brownian motion, he presentedhis theory of special relativity, and he derived the famous relation between mass andenergy E = mc2.

These papers quickly earned him recognition, and in 1908 he was appointedPrivatdozent in Bern. In the following years he moved to Zurich, Prague, and backto Zurich, and in 1914 he finally became Director of the Kaiser Wilhelm Institute forPhysics in Berlin. When Hitler came to power in January 1933, Einstein was visitingthe US to give a series of lectures. He immediately decided not to return to Germany,and accepted a post at Princeton University. Einstein became an American citizen in1940, and stayed there till his death in 1955.


But this number of permutations we know already: it is W, as calculated in (2.19).Therefore,

C =$

n1! n2! · · · n∞!N!

. (2.22)

Let us illustrate this procedure for the case of three particles. Suppose we have a statewhich is characterized by the occupation numbers {2, 1, 0, 0, . . . }: two of the particles arein the first level and the third is in the second level. We pick a wave function satisfyingthese occupations, ϕ1(r1)ϕ1(r2)ϕ2(r3). From this state we construct an equal superpositionof all possible permutations, as described above, and include the prefactor

√2 · 1/

√6,

)(S) = 1√3

'ϕ1(r1)ϕ1(r2)ϕ2(r3) + ϕ1(r1)ϕ2(r2)ϕ1(r3) + ϕ2(r1)ϕ1(r2)ϕ1(r3)

(, (2.23)

which is the desired fully symmetric wave function.Let us emphasize here that the symmetrization procedure creates one unique symmetric

wave function for a given set of occupation numbers. So, indeed, imposing the symmetrypostulate removes all degeneracies in this case.

Control question. Can you construct a three-particle fully symmetric wave functioncorresponding to the occupation numbers {1, 1, 1, 0, . . . }?

2.3.2 Antisymmetric wave function: Fermions

For the case of identical fermions, the symmetry postulate requires a fully antisymmetricwave function. To build such a wave function, we proceed in a way similar to that followedfor bosons.

We first fix the set of occupation numbers {nk}. We must note here, however, that if anyof the nk ≥ 2, there is no chance that we can build a completely antisymmetric function.This follows from the fact that any basis state which satisfies this choice is automaticallysymmetric with respect to permutation of the particles located in the multiply occupiedlevel k. This gives us the celebrated Pauli principle: all occupation numbers for fermionsare either 0 or 1, there is no way that two fermions can share the same level.

We now pick one of the basis states fitting the set of occupation numbers. Again, allother basis states that satisfy the same set are obtained by permutations from this state.To proceed, we have to introduce the concept of parity of a permutation. We define that apermutation is:

• odd if obtained by an odd number of pair permutations, and• even if obtained by an even number of pair permutations,

both starting from the same initial state. We then construct a linear combination of all basisstates, taking those which are even with a weight C and those which are odd with −C. Theresulting linear combination then changes sign upon permutation of any pair of particles,and is therefore the completely antisymmetric state we were looking for,

)(A){nk} = C

!

P

(−1)par(P)P'ϕℓ1 (r1)ϕℓ2 (r2) . . .ϕℓN (rN)

(, (2.24)


Enrico Fermi (1901–1954)Won the Nobel Prize in 1938 for “his demonstrations of theexistence of new radioactive elements produced by neu-tron irradiation, and for his related discovery of nuclearreactions brought about by slow neutrons.”

In 1926, Enrico Fermi was a young lecturer at the Univer-sity of Florence in Italy, and he was studying the statisticalproperties of particles obeying Pauli’s exclusion principle.He showed that assuming a comparable exclusion princi-

ple for the identical particles in an ideal gas, one could derive a theory for the quanti-zation of the gas, which was perfectly consistent with thermodynamics. In the sameyear, Paul Dirac was investigating in a more general context the statistical propertiesof identical quantum particles. He concluded that there are two types of quantum par-ticle: those obeying Bose–Einstein statistics, and those obeying the statistics Fermihad developed for quantizing the ideal gas. The latter became known as Fermi–Diracstatistics, and the particles obeying it as fermions.

In 1927, Fermi gained a professorial position at the University of Rome. He directedhis group there mainly toward the field of nuclear physics, excelling in both theoret-ical and experimental research. During this time, he developed his famous theory ofbeta decay, and made also some important experimental steps toward nuclear fission.In 1938, after receiving the Nobel prize, Fermi moved with his family to New York,mainly because of the threat the fascist regime of Mussolini posed for his Jewishwife. He first accepted a position at Columbia University, and moved in 1946 to theUniversity of Chicago, where he stayed until his death in 1954. During the secondworld war, he was heavily involved in the Manhattan project, the development of theAmerican nuclear bomb.

where par(P) denotes the parity of the permutation P, so that par(P) = 0 for P even, andpar(P) = 1 for P odd. The above superposition can be written in a very compact way usingthe definition of the determinant of a matrix. If we define the elements of the N × N matrixM to be Mij = ϕℓi(rj), then we see that the linear combination (2.24) can be written as

)(A){nk} = C det(M). (2.25)

This determinant is called the Slater determinant.

Control question. In the above equation, i labels the levels and j labels the particles.Why are both dimensions of the matrix N?

We still have to determine C. As in the bosonic case, it follows from the normalizationcondition ⟨)(A)|)(A)⟩ = 1. We find the same result for C, but since every nk is either 0or 1, we can simplify


C =$

n1! n2! · · · n∞!N!

= 1√N!

. (2.26)

Control question. Can you write down the antisymmetrized three-particle wavefunction corresponding to the occupation numbers {1, 1, 1, 0, . . . }?

Following this antisymmetrization procedure, we end up with one unique antisymmetricwave function for a given set of occupation numbers. Therefore, we see that also in the caseof fermions, the symmetrization postulate removes all degeneracies from the many-particlestates.

2.3.3 Fock space

To proceed with the quantum description of identical particles, we need the Hilbert spaceof all possible wave functions of all possible particle numbers. This space is called Fockspace, and in this section we show how to construct it.

In the previous sections we explained that for a system of many identical particles mostsolutions of the Schrödinger equation are actually redundant: only completely symmetricand antisymmetric states occur in Nature. In fact, it turns out that for both bosons andfermions any set of occupation numbers corresponds to exactly one unique many-particlewave function that obeys the symmetry postulate. This suggests that all basis states of themany-particle Hilbert space can be characterized by just a set of occupation numbers. Onthis idea the construction of Fock space is based.

Let us now show formally how to construct Fock space in four steps:

1. We start with the space of all possible solutions of the Schrödinger equation for Nparticles. This space is represented by the basis states in (2.14).

2. We combine all these basis states for all possible N, and introduce a special basis statecorresponding to N = 0. This state is called the vacuum.

3. We construct (anti)symmetric linear combinations of these basis states to obtain statesthat satisfy the symmetry postulate.

4. We throw away all other states since they do not satisfy the postulate.

The resulting set of basis states spans Fock space, and any of these basis states can beuniquely characterized by a set of occupation numbers. Making use of this, we introducethe “ket” notation for the basis states in Fock space,

|{nk}⟩ ≡ |n1, n2, . . .⟩, (2.27)

where the number nk is the occupation number of the level k. Of course, in order to translatea vector in Fock space to an actual many-particle wave function, one needs to know allwave functions ψk(r) corresponding to the levels.

The special state defined in step 2 above, the vacuum, reads

|0, 0, . . .⟩ ≡ |0⟩. (2.28)


With this vacuum state included, we have constructed a complete basis spanning the Hilbertspace of all possible wave functions for any N-particle system, where N can range fromzero to infinity. The basis is orthonormal, so that

⟨{nk}|{n′k}⟩ =

)

k

δnk ,n′k. (2.29)

Control question. Why is there a product over k in this expression?

We hope that it is clear that a vector in Fock space represents in a very intuitive andphysical way the actual many-particle wave function of a system. From the vector we canimmediately see how many particles are present in the system, how they are distributedover the levels, and thus what the total energy of the system is. As we see in the nextchapters, working in Fock space allows for the development of a very convenient tool forsolving many-body problems.


Table 2.1 Summary: Identical particles

Schrödinger equation for many identical particles

H =N!

i=1

"

− h2

2m∂2

∂r2i

+ Vi(ri)

#

+N!

i=2

i−1!

j=1

V (2)(ri, rj) +N!

i=3

i−1!

j=2

j−1!

k=1

V (3)(ri, rj, rk) + . . .

the permutation symmetry of this Hamiltonian imposes restrictions on the allowed symmetriesof the many-particle wave function

Symmetry postulatethere exist only bosons and fermionsbosons: the N-particle wave function is fully symmetric under particle pair permutationfermions: the N-particle wave function is fully antisymmetric under particle pair permutation

Solutions of the N-particle Schrödinger equation, ignoring interactions V (2), V (3), ...general: N-term product of single-particle solutions, ψ{ℓi}({ri}) = ϕℓ1 (r1)ϕℓ2 (r2) . . .ϕℓN (rN ),

where {ℓi} is the set of levels occupied (N numbers)

bosons: symmetric combination )(S){nk} =

$n1! n2! · · · n∞!

N!

!

P

P'ϕℓ1 (r1)ϕℓ2 (r2) . . .ϕℓN (rN )

(,

where {nk} is the set of occupation numbers (maximally N numbers)

fermions: antisymmetric combination )(A){nk} = 1√

N!

!

P

(−1)par(P)P'ϕℓ1 (r1)ϕℓ2 (r2) . . .ϕℓN (rN )

(,

where nk is either 0 or 1

Fock spaceeach set of {nk} corresponds to exactly one allowed many-particle wave functionthe basis of |{nk}⟩ ≡ |n1, n2, . . .⟩ is thus complete

59 Exercises

Exercises

1. Pairwise interaction for two bosons (solution included). Consider particles in the systemof levels labeled by n, corresponding to the states |n⟩, with real wave functions ψn(r)and energies En. There is a weak pairwise interaction between the particles,

Hint = U(r1 − r2) = Uδ(r1 − r2).

This interaction is called contact interaction since it is present only if two particles areat the same place.a. Assume that there are two bosonic particles in the system. Write down the wave

functions of all possible stationary states in terms of two-particle basis states |nm⟩ ≡ψn(r1)ψm(r2).

b. Express the matrix element ⟨kl|Hint|nm⟩ of the contact interaction in terms of theintegral

Aklmn =+

drψk(r)ψl(r)ψm(r)ψn(r).

Note that this integral is fully symmetric in its indices k, l, m, and n.c. Compute the correction to the energy levels to first order in U. Express the answer

in terms of Aklmn.d. Now assume that there are N bosonic particles in one level, say |1⟩. Compute the

same correction.2. Particle on a ring (solution included). Consider a system of levels in a ring of radius

R. The coordinates of a particle with mass M are conveniently described by just theangle θ in the polar system (Fig. 2.5a). The Hamiltonian in this coordinate reads

H = − h2

2MR2

,∂

∂θ

-2

,

if there is no magnetic flux penetrating the ring. Otherwise, the Hamiltonian is given by

H = h2

2MR2

,−i

∂

∂θ− ν

-2

,

!Fig. 2.5 (a) Particle on a ring penetrated by a flux ν . The polar angle θ parameterizes the particle’s coordinate. (b) Twostrongly repelling particles on a ring go to opposite points on the ring, forming a rigid molecule. (c) A moleculeformed by N = 4 repelling particles.


ν expressing the flux in the ring in dimensionless units: ν = ,/,0. Here, ,0 = 2π h/eis the flux quantum, where e is the electric charge of the particle. The effect of thisflux is an extra phase φ = ν(θ1 − θ2) acquired by the particle when it travels from thepoint θ1 to θ2. Indeed, the transformation )(θ ) → )(θ ) exp(−iνθ ) eliminates the fluxfrom the last Hamiltonian.

Owing to rotational symmetry, the wave functions are eigenfunctions of angularmomentum, |m⟩ = (2π )−1/2eimθ . An increase of θ by a multiple of 2π brings the parti-cle to the same point. Therefore, the wave functions must be 2π -periodic. This restrictsm to integer values.a. Let us first carry out an exercise with the levels. Compute their energies. Show that

at zero flux the levels are double-degenerate, meaning that |m⟩ and |−m⟩ have thesame energy. Show that the energies are periodic functions of ν.

b. We assume ν = 0 and concentrate on two levels with m = ±1. Let us denotethe levels as |+⟩ and |−⟩. There are four degenerate states for two particles inthese levels, |++⟩, |−+⟩, |+−⟩, and |−−⟩. How many states for boson parti-cles are there? How many for fermion particles? Give the corresponding wavefunctions.

c. Compute the contact interaction corrections for the energy levels of two bosons.Does the degeneracy remain? Explain why.

d. Explain why the results of the previous exercise cannot be immediately applied forpoint (b) although the wave functions of the levels can be made real by transformingthe basis to |a⟩ = {|+⟩ − |−⟩}/i

√2 and |b⟩ = {|+⟩ + |−⟩}/

√2.

3. Pairwise interaction for two fermions. Let us take the setup of Exercise 1 and considerelectrons (fermions with spin 1

2 ) rather than bosons. The levels are thus labeled |nσ ⟩,with σ = {↑, ↓} labeling the spin direction and n indicating which “orbital” the electronoccupies.a. Put two electrons into the same orbital level n. Explain why only one state is permit-

ted by the symmetry postulate and give its wave function. Compute the correctionto its energy due to contact interaction.

b. How does the wave function change upon permutation of the electrons’coordinates?

c. Now put two electrons into two different orbital levels n and m. How many statesare possible? Separate the states according to the total electron spin into singlet andtriplet states and give the corresponding wave functions.

d. Compute again the corrections to the energies due to the contact interaction.4. Pairwise interaction for three fermions.

a. Consider three electrons in three different orbital levels n, m, and p. We assumethat there is one electron in each orbital level. How many states are then possible?Classify these states according to their transformation properties with respect topermutation of the particles’ coordinates.Hint. Use (2.7–2.10).

b. Classify the states found in (a) according to total spin. Compute the matrix ele-ments of the contact interaction (assuming real orbital wave functions) and find theinteraction-induced splitting of these otherwise degenerate states.

61 Solutions

c. Consider now the case where two electrons are in the same orbital level n while thethird one is in m. Classify the states again. Find the interaction-induced splitting.

5. Molecule on a ring. Let us return to the ring setup described in Exercise 2. We nowput two particles in the ring and assume a strong long-range repulsion between theparticles. In this case, at low energy the particles are situated in opposite points of thering: they form a sort of rigid molecule. The resulting two-particle wave function isagain a function of a single coordinate θ (see Fig. 2.5(b)).a. Assume that the particles are distinguishable while having the same mass. Write

down the wave functions corresponding to discrete values of momentum, andcompute the corresponding energies.

b. What changes if the particles are indistinguishable bosons?Hint. A rotation of the molecule by an angle π permutes the particles.

c. Characterize the energy spectrum assuming the particles are fermions.d. Consider a molecule made of N strongly repelling particles (N = 4 in Fig. 2.5(c)).

Give the energy spectrum for the cases where the particles are bosons and fermions.6. Anyons. In 1977, Jon Magne Leinaas and Jan Myrheim figured out that the symmetry

postulate would not apply in two-dimensional space. In 1982, Frank Wilczek proposeda realization of this by constructing hypothetical particles that he called anyons. Thoseare obtained by attributing a magnetic flux to the usual fermions or bosons. When oneanyon makes a loop around another anyon, it does not necessarily end up in the samestate it started from, but it acquires an extra phase shift due to the “flux” present in theloop. It is believed that elementary excitations of the two-dimensional electron gas inthe Fractional Quantum Hall regime are in fact anyons.a. Consider the two-particle molecule from Exercise 5. Suppose the molecule is

rotated over an angle α. By what angle is particle 1 then rotated with respect toparticle 2?

b. Attach a flux ν to each particle. What is the phase change acquired by the moleculeafter rotation over an angle α? What is the phase change acquired in the course of aπ -rotation that permutes the particles?

c. Show that the effect of this flux can be ascribed to particle statistics. The same effectis achieved if one considers fluxless particles of which the wave function acquiresa phase factor eiβ upon permutation. Express β in terms of ν.

d. Find the energy spectrum of the molecule.

Solutions

1. Pairwise interaction for two bosons. This is a relatively simple problem.a. The symmetrized wave functions read

if n = m|nm⟩ + |mn⟩√

2, with E = En + Em,

if n = m |nn⟩, with E = 2En.

b. The matrix element between two non-symmetrized states reads

⟨kl|Hint|nm⟩ = U+

dr1dr2 ψk(r1)ψl(r2)ψm(r1)ψn(r2)δ(r1 − r2)

= UAklmn.

c. Substituting the symmetrized states found at (a), we get δE = U2Annmm if n = m.One would guess that δE = U2Annnn for n = m, yet this is incorrect, in fact δE =UAnnnn.

d. One has to recall that the interaction is between each pair of particles,

U =!

1≤i<j≤N

U(ri − rj).

Each term in the above sum equally contributes to the correction. Therefore,δE(N) = 1

2 N(N − 1)UA1111.2. Particle on a ring. This is a more complicated problem.

a. Just by substituting the wave function ∝ exp(imθ ) into the Schrödinger equationone obtains

Em = h2

2MR2 (m − ν)2.

Plotting this versus ν yields an infinite set of parabolas which is periodic in ν withperiod 1.

b. Three states are symmetric with respect to particle exchange,

|++⟩, |−+⟩ + |+−⟩√2

, and |−−⟩,

and therefore suitable for bosonic particles. For fermions only one single antisym-metric state 1√

2{|−+⟩ − |+−⟩} is suitable.

c. The corrections read

δE++ = U(2π )2

+dφ1dφ2δ(φ1 − φ2) = U/2π ,

δE−− = U/2π ,

δE+− = U/π ,

and the degeneracy is lifted. Indeed, |++⟩ and |−−⟩ have angular momentum ±2,so they have the same energy, while the momentum of the third state is 0.

d. In the new basis, the interaction has non-diagonal matrix elements. While thosecan be disregarded in the case of non-degenerate levels, they become important fordegenerate ones.

3 Second quantization

In this chapter we describe a technique to deal with identical particles that is called secondquantization. Despite being a technique, second quantization helps a lot in understand-ing physics. One can learn and endlessly repeat newspaper-style statements particles arefields, fields are particles without grasping their meaning. Second quantization brings theseconcepts to an operational level.

Since the procedure of second quantization is slightly different for fermions and bosons,we have to treat the two cases separately. We start with considering a system of identicalbosons, and introduce creation and annihilation operators (CAPs) that change the numberof particles in a given many-particle state, thereby connecting different states in Fock space.We show how to express occupation numbers in terms of CAPs, and then get to the heartof the second quantization: the commutation relations for the bosonic CAPs. As a nextstep, we construct field operators out of the CAPs and spend quite some time (and energy)presenting the physics of the particles in terms of these field operators. Only afterward,do we explain why all this activity has been called second quantization. In Section 3.4 wethen present the same procedure for fermions, and we mainly focus on what is different ascompared to bosons. The formalism of second quantization is summarized in Table 3.1.

3.1 Second quantization for bosons

We now use the framework of identical particles we built up earlier. We begin with a basisof single-particle states – the levels. As explained in the previous chapter, these levels arethe allowed one-particle states in the system considered. For instance, for a square wellwith infinitely high potential walls they are the single-particle eigenfunctions we knowfrom our introductory course on quantum mechanics, and for particles in a large volumethe levels are the plane wave solutions we discussed in Section 1.3.

In any case, we can label all available levels with an index k, and then use state vectors inFock space to represent all many-particle states. In this section we focus on bosons, so themany-particle basis states in Fock space read |{nk}⟩, where the numbers {nk} ≡ {n1, n2, . . . }give the occupation numbers of the levels k. Since we are dealing with bosons, the numbersn1, n2, . . . can all be any positive integer (or zero of course).

We now introduce an annihilation operator in this space. Why introduce? Usually, anoperator in quantum mechanics can be immediately associated with a physical quantity,the average value of this quantity in a certain state to be obtained with the operator acting

63


on the wave function. Would it not be simpler to first discuss the corresponding physicalquantity, instead of just “introducing” an operator? Well, you must believe the authors thatthe annihilation operator is no exception: there indeed is a physical quantity associatedwith it. However, let us wait till the end of the section for this discussion.

Instead, we define the bosonic annihilation operator bk by what it does: it removes a par-ticle from the level k. Therefore, when it acts on an arbitrary basis state |{nk}⟩, it transformsit into another basis state that is characterized by a different set of occupation numbers. Asexpected from the definition of bk , most occupation numbers eventually remain the same,and the only change is in the occupation number of the level k: it reduces by 1, from nk tonk − 1. We thus write formally

bk |. . . , nk, . . .⟩ = √nk|. . . , nk − 1, . . .⟩. (3.1)

As we see, it does not simply flip between the states, it also multiplies the state with anugly-looking factor

√nk. We will soon appreciate the use of this factor, but for now just

regard it as part of the definition of bk .Let us first consider the Hermitian conjugated operator, which is obtained by transpo-

sition and complex conjugation, b†nm = (bmn)∗. It has a separate meaning: it is a creation

operator,1 and its action on a basis state is defined as

b†k |. . . , nk, . . .⟩ =

.nk + 1|. . . , nk + 1, . . .⟩, (3.2)

that is it adds a particle into the level k.

Control question. Can you show how the factor√

nk + 1 in (3.2) directly followsfrom the definition of the annihilation operator (3.1)?

3.1.1 Commutation relations

Let us now come to the use of the ugly square-roots in (3.2) and (3.1). For this, we considerproducts of CAPs, bk b†

k and b†k bk . We now make use of the definitions given above to write

b†k bk |. . . , nk, . . .⟩ = b†

k√

nk|. . . , nk − 1, . . .⟩ =.

(nk − 1) + 1√

nk|. . . , nk, . . .⟩= nk|. . . , nk, . . .⟩.

(3.3)

We see that the basis states in Fock space are eigenstates of the operator b†k bk with eigen-

values nk. In other words, when this operator acts on some state |g⟩, it gives the number ofparticles in the level k,

⟨g|b†k bk |g⟩ = nk. (3.4)

Therefore, the operator b†k bk is also called the number operator for the level k, and it is

sometimes denoted nk ≡ b†k bk . It is now clear why the square-root prefactors in (3.2) and

1 Some people have a difficulty remembering that b annihilates and b† creates. A confusion might come fromthe association of † with the sign of the cross, which may symbolize death in the Christianity-based cultures.The following mnemonics could help: death is a new birth or just + adds a particle.

65 3.1 Second quantization for bosons

(3.1) are a very convenient choice: they provide us with a very simple operator which wecan use to retrieve the number of particles in any of the levels!

Control question. Can you interpret the above equation for the case where the state|g⟩ is not one of the basis functions |{nk}⟩?

Using this number operator we can construct another useful operator: the operator of thetotal number of particles in the system,

N =!

k

nk =!

k

b†k bk . (3.5)

Let us now look at another product of CAPs, bk b†k . Doing the same algebra as above, we

obtain

bk b†k |. . . , nk, . . .⟩ = bk

.nk + 1|. . . , nk + 1, . . .⟩ =

.nk + 1

.nk + 1|. . . , nk, . . .⟩

= (nk + 1)|. . . , nk, . . .⟩.(3.6)

We see that this product is also a diagonal operator in the basis of |{nk}⟩, and it has eigen-values nk + 1. It is important to note that the difference of the products bk b†

k and b†k bk ,

i.e. the commutator of bk and b†k , is just 1,

bk b†k − b†

k bk = [bk , b†k] = 1. (3.7)

This property does not depend on a concrete basis, it always holds for CAPs in the samelevel. Since CAPs that act on different levels always commute, we can now formulate thegeneral commutation rules for bosonic CAPs.

commutation relations for boson CAPs

[bk , b†k′ ] = δkk′

[bk , bk′ ] = [b†k , b†

k′ ] = 0

These commutation rules are not only simple and universal, but they could serve as analternative basis of the whole theory of identical particles. One could start with the com-mutation rules and derive everything else from there. Besides, they are very convenient forpractical work: most calculations with CAPs are done using the commutation rules ratherthan the ugly definitions (3.1) and (3.2).

3.1.2 The structure of Fock space

We see that the CAPs provide a kind of navigation through Fock space. One can use themto go step by step from any basis state to any other basis state. Let us illustrate this byshowing how we can create many-particle states starting from the the vacuum |0⟩. First wenote that we cannot annihilate any particle from the vacuum, since there are no particles init. We see from the definition (3.1) that for any level k

bk |0⟩ = 0. (3.8)


!Fig. 3.1 There are many different ways to create the state |1, 1⟩ from the state |0, 0⟩ using creation and annihilationoperators: b†

2 b†1 (gray arrows), b

†1 b

†2 (dotted arrows), or why not b1 b2 b

†1 b

†1 b

†2 b

†2 (black arrows)?

However, we are free to create particles in the vacuum. We can build any basis state out ofthe vacuum by applying the creation operators b†

k as many times as needed to fill all levelswith the desired number of particles. Therefore,

|{nk}⟩ = 1./

k nk!

)

k

%b†

k

&nk |0⟩. (3.9)

The order of the operators in the above formula is not important since all b†k commute with

each other. The prefactor ensures the correct normalization given the square-root in thedefinition (3.1).

Control question. Prove the above formula.

In this way the CAPs give us insight into the structure of Fock space. One could saythat the space looks like a tree: a basis state is a node, and from this node one can makesteps in all directions corresponding in all possible b(†)

k . But actually the space looks morelike a grate. In distinction from an ideal tree, there are several different paths – sequencesof bks and b†

ks – connecting a pair of given nodes. Let us illustrate this with an examplein the simple Fock space spanned by only two levels, 1 and 2. The basis states can thenbe presented as the nodes of a two-dimensional square grate labeled by n1 and n2, such asdepicted in Fig. 3.1. We see that there are multiple paths from any basis state to another. Forinstance, to get from the vacuum |0, 0⟩ to the state |1, 1⟩, one could apply either of the oper-ators b†

2b†1 (as indicated with gray arrows) and b†

1b†2 (dotted arrows), or even b1b2b†

1b†1b†

2b†2

(black arrows).

3.2 Field operators for bosons

We now introduce another type of operator, the so-called field operators. To construct fieldoperators from CAPs, we first note that the CAPs as presented in (3.1) and (3.2) are definedin the convenient framework of Fock space. The basis spanned by the eigenstates of thesingle-particle Hamiltonian, i.e. levels with a definite energy. By defining the CAPs in

67 3.2 Field operators for bosons

this way, we have made a very specific choice for the basis. However, as we have alreadyemphasized in Chapter 1, the choice of a basis to work in is always completely democratic:we can formulate the theory in any arbitrary basis. Let us now try to transform the CAPs,as defined in Fock space, to a representation in the basis of eigenstates of the coordinateoperator r.

To start, we recall how to get a wave function in a certain representation if it is known inanother representation. Any one-particle wave function in the level representation is fullycharacterized by the set of its components {ck}. This means that the wave function can bewritten as the superposition

ψ(r) =!

k

ckψk(r), (3.10)

ψk(r) being the wave functions of the level states. This wave function can be regarded inthe coordinate representation as a function ψ(r) of coordinate, or, equivalently, in the levelrepresentation as a (possibly discrete) function ck of the level index k. In Section 1.3 weillustrated this correspondence of different representations in more detail by comparing thecoordinate representation with the plane wave representation of a free particle.

The idea now is that we transform the CAPs to another basis, in a similar way to thewave functions discussed above. We replace ck → bk in (3.10), which brings us to thedefinition of the field operator,

ψ(r) =!

k

bkψk(r). (3.11)

Since b(†)k annihilates (creates) a particle in the level k, the field operator ψ (†)(r) annihilates

(creates) a particle at the point r.Let us look at some simple properties of field operators. We note that the commutation

relations take the following elegant form:

commutation relations for boson field operators

[ψ(r), ψ†(r′)] = δ(r − r′)[ψ(r), ψ(r′)] = [ψ†(r), ψ†(r′)] = 0

So, bosonic field operators commute unless they are at the same point (r = r′). These com-mutation relations can be derived from the definition (3.11) if one notes the completenessof the basis, that is expressed by

!

k

ψ∗k (r)ψk(r′) = δ(r − r′). (3.12)

Control question. Are you able to present this derivation explicitly?

3.2.1 Operators in terms of field operators

We now show how we can generally express the one- and two-particle operators for amany-particle system in terms of field operators. To this end, we first consider a simple


Vladimir Fock (1898–1974)

Vladimir Fock received his education from the Universityof Petrograd in difficult times. In 1916, shortly afterenrolling, he interrupted his studies to voluntarily join thearmy and fight in World War I. He got injured (causing aprogressive deafness which he suffered from the rest ofhis life), but survived and returned to Petrograd in 1918.By then, a civil war had broke out in Russia resulting innationwide political and economical chaos. Fortunately,

to ensure, despite the situation, a good education for the brightest students, a newinstitute was opened in Petrograd, where Fock could continue his studies.

After receiving his degree in 1922, Fock stayed in Petrograd, closely followingthe development of quantum theory. When Schrödinger published his wave the-ory in 1926, Fock immediately recognized the value of it and started to developSchrödinger’s ideas further. When he tried to generalize Schrödinger’s equation toinclude magnetic fields, he independently derived the Klein–Gordon equation (seeChapter 13), even before Klein and Gordon did. In the subsequent years, Fock intro-duced the Fock states and Fock space, which later became fundamental conceptsin many-particle theory. In 1930 he also improved Hartree’s method for findingsolutions of the many-body Schrödinger equation by including the correct quantumstatistics: this method is now know as the Hartree–Fock method (see e.g. Chapter 4).

In 1932 Fock was finally appointed professor at the University of Leningrad (as thecity was called since 1924). He stayed there until his retirement, making importantcontributions to a wide range of fields, including general relativity, optics, theory ofgravitation, and geophysics. He was brave enough to confront Soviet “philosophers”who stamped the theory of relativity as being bourgeois and idealistic: a stand thatcould have cost him his freedom and life.

system consisting of only a single particle. In the basis of the levels – the allowed singleparticle states ψk – we can write any operator acting on a this system as

f =!

k,k′fk′k|ψk′ (r)⟩⟨ψk(r)|, (3.13)

where the coefficients fk′k are the matrix elements fk′k = ⟨ψk′ (r)|f |ψk(r)⟩. For a many-particle system, this operator becomes a sum over all particles,

F =N!

i=1

!

k,k′fk′k|ψk′(ri)⟩⟨ψk(ri)|, (3.14)

where ri now is the coordinate of the ith particle. Since the particles are identical, thecoefficients fk′k are equal for all particles.


We see that, if in a many-particle state the level k is occupied by nk particles, the termwith fk′k in this sum will result in a state with nk −1 particles in level k and nk′ +1 particlesin level k′. In Fock space we can write for the term fk′k

N!

i=1

fk′k|ψk′ (ri)⟩ ⟨ψk(ri)| . . . , nk′ , . . . , nk, . . .⟩ = Cfk′k|. . . , nk′ + 1, . . . , nk − 1, . . .⟩,

(3.15)where the constant C still has to be determined. It follows from the following reasoning.The original state as we write it in Fock space is a normalized symmetrized state: it consistsof a sum over all possible permutations of products of single particle wave functions (2.20)with a normalization constant Cold = √

n1! · · · nk′ ! · · · nk! · · · /N!. The operator in (3.15)then produces a sum over all permutations obeying the new occupation numbers. Since ineach term of this sum, any of the nk′ + 1 wave functions ψk′ can be the one which wascreated by the operator |ψk′ ⟩⟨ψk|, there are formed nk′ + 1 copies of the sum over all per-mutations. The state vector |. . . , nk′ + 1, . . . , nk − 1, . . .⟩ with which we describe this newstate comes by definition with a constant Cnew = √

n1! · · · (nk′ + 1)! · · · (nk − 1)! · · · /N!.Combining all this, we see that

C = (nk′ + 1)Cold

Cnew=.

nk′ + 1√

nk. (3.16)

This allows for the very compact notation

F =!

k,k′fk′k b†

k′ bk , (3.17)

where we again see the use of the ugly square root factors in the definition of the CAPs. Ofcourse, we can write equivalently

F =+

dr1dr2fr1r2 ψ†(r1)ψ(r2), (3.18)

where we have switched to the representation in field operators. The coefficient fr1r2 is nowgiven by the matrix element fr1r2 = ⟨r1|f |r2⟩.

Let us illustrate how one can use this to present very compactly single-particle operatorsin a many-particle space. Suppose we would like to find the many-particle density operatorρ. We know that for a single particle this operator is diagonal in the coordinate basis,ρ = |r⟩⟨r|. We now simply use (3.18) to find

ρ =+

dr1dr2 ρr1r2 ψ†(r1)ψ(r2) = ψ†(r)ψ(r). (3.19)

We can check this result by calculating the operator of the total number of particles

N =+

dr ρ(r) =+

dr ψ†(r)ψ(r) =!

k,k′

+drψ∗

k′ (r)ψk(r)b†k′ bk

=!

k,k′δkk′ b†

k′ bk =!

k

b†k bk .

(3.20)

Comparing this with (3.5) we see that we have indeed found the right expression.


Control question. Which property of the level wave functions has been used to getfrom the first to the second line in the above derivation?

3.2.2 Hamiltonian in terms of field operators

The true power of field operators is that they can provide a complete and closed descrip-tion of a dynamical system of identical particles without invoking any wave functions orthe Schrödinger equation. Since the dynamics of a quantum system is determined by itsHamiltonian, our next step is to get the Hamiltonian in terms of field operators.

Let us start with a system of non-interacting particles. Since in this case the many-particle Hamiltonian is just the sum over all one-particle Hamiltonians, we can write it asin (3.18),

H(1) =+

dr1dr2

0r1

1111p2

2m+ V(r)

1111 r2

2ψ†(r1)ψ(r2)

=+

dr ψ†(r)

"

− h2∇2

2m+ V(r)

#

ψ(r),(3.21)

where we have used the fact that ⟨r1| 12m p2 + V(r)|r2⟩ =

3− h2

2m∇2r1

+ V(r1)4δ(r1 − r2).

The same Hamiltonian can also be expressed in terms of CAPs that create and annihilateparticles in the levels ψk. Since the levels are per definition the eigenstates of the singleparticle Hamiltonian 1

2m p2 + V(r), we can substitute ψ(r) = *k bkψk(r) in (3.21) and

write

H(1) =!

k,k′

+drψ∗

k (r)

"

− h2

2m∇2 + V(r)

#

ψk′ (r)b†k bk′

=!

k,k′δkk′εk′ b†

k bk′ =!

k

εknk,(3.22)

where εk is the energy of the level ψk. We see that the representation in CAPs reduces toa very intuitive form: the operator H(1) counts the number of particles in each level andmultiplies it with the energy of the level. It is easy to see that any basis state in Fock spaceis an eigenstate of this operator, its eigenvalue being the total energy of the state.

The following step is to include also interaction between particles into the Hamiltonian.We consider here pairwise interaction only. We have seen that in the Schrödinger equationfor N identical particles the pairwise interaction is given by

H(2) = 12

!

i =j

V (2)(ri − rj), (3.23)

where the summation runs over all particles. We assume here that the interaction betweentwo particles depends only on the distance between the particles, not on the precise location


of the pair in space. Let us rewrite this Hamiltonian in a form where the number of particlesis of no importance. We do so by making use of the particle density defined as

ρ(r) =!

i

δ(r − ri). (3.24)

This provides us a handy way to deal with the sums over particle coordinates. For anyfunction A(r) of a coordinate, or function B(r, r′) of two coordinates, we can now write

!

i

A(ri) =+

dr ρ(r)A(r), (3.25)

!

i,j

B(ri, rj) =+

dr dr′ ρ(r)ρ(r′)B(r, r′). (3.26)

So we rewrite the Hamiltonian as

H(2) = 12

+dr1dr2V (2)(r1 − r2)ρ(r1)[ρ(r2) − δ(r1 − r2)]. (3.27)

The δ-function in the last term is responsible for the “missing” term i = j in the doublesum in (3.23): a particle does not interact with itself! Now we use ρ(r) = ψ†(r)ψ(r) toobtain

H(2) = 12

+dr1dr2V (2)(r1 − r2)ψ†(r1)ψ†(r2)ψ(r2)ψ(r1). (3.28)

Control question. How does the order of the field operators in (3.28) incorporate thefact that a particle does not interact with itself?Hint. Use the commutation relations for field operators.

Another instructive way to represent the pairwise interaction in a many-particle systemis in terms of the CAPs, which create and annihilate particles in the states with certainwave vectors (or momenta). This will reveal the relation between the pairwise interac-tion and scattering of two particles. We rewrite the Hamiltonian (3.28) in the wave vectorrepresentation, substituting ψ(r) = 1√

V*

k bkeik·r, to obtain

H(2) = 12V

!

k,k′,q

V (2)(q) b†k+qb†

k′−qbk′ bk. (3.29)

Control question. Do you see how to derive (3.29) from (3.28)?Hint. Switch to the coordinates R = r1 − r2 and r2, and then perform the integrationsin (3.28) explicitly. Remember that

5dr = V .

Let us now focus on the physical meaning of (3.29). We concentrate on a single term inthe sum, characterized by the wave vectors k, k′, and q. Suppose that this term acts on aninitial state which contains two particles with wave vectors k and k′. The term annihilatesboth these particles and creates two new particles, with wave vectors k+q and k′−q. If wenow compare the initial and final states, we see that effectively the particles have exchangedthe momentum q with each other. In other words, this term describes a scattering event inwhich the total momentum is conserved. We can depict this interaction in a cartoon styleas in Fig. 3.2.


!Fig. 3.2 A term in a pairwise interaction Hamiltonian pictured as a scattering event. Two particles with wave vectors k and k′

can interact and thereby exchange momentum q. After this interaction the particles have wave vectors k+ q andk′ − q. The amplitude of the process is proportional to the Fourier component V(2)(q) of the interaction potential.

3.2.3 Field operators in the Heisenberg picture

Let us now investigate the time dependence of the CAPs and field operators when con-sidering them in the Heisenberg picture. In this picture the wave function of a system ofidentical particles is “frozen,” and all operators move (see Section 1.4). Assuming non-interacting particles and making use of the Hamiltonian (3.22), we see that the Heisenbergequation for CAPs becomes

ih∂ bk(t)∂t

= [bk, H(1)] = εkbk(t), (3.30)

its solution being bk(t) = e− ih εktbk(0). If we rewrite this in coordinate representation, we

get the evolution equation for field operators

ih∂ψ(r, t)∂t

="

− h2∇2

2m+ V(r)

#

ψ(r, t). (3.31)

This equation is linear and therefore still easy to solve. If we also include interactions, theequations for the field operators are still easy to write down. Importantly, the equations areclosed: to solve for the dynamics of field operators, one does not need any variables otherthan the field operators. This is why a set of time-evolution equations for field operatorstogether with the commutation relations is said to define a quantum field theory. In thepresence of interactions, however, the equations are no longer linear. The general solutionsare very difficult to find and in most cases they are not known.

3.3 Why second quantization?

Let us now add some explanations which we have postponed for a while. We have learneda certain technique called second quantization. Why? From a practical point of view, theresulting equations for systems with many particles are much simpler and more compactthan the original many-particle Schrödinger equation. Indeed, the original scheme operateswith a wave function of very many variables (the coordinates of all particles), which obeys

73 3.3 Why second quantization?

Pascual Jordan (1902–1980)Never won the Nobel Prize, supposedly because of hisstrong support of the Nazi party.

Shortly after receiving his Ph.D. degree at the Universityof Göttingen in 1924, Pascual Jordan was asked bythe 20 years older Max Born to help him understandHeisenberg’s manuscript on the formulation of quantummechanics in terms of non-commuting observables. Bornhad immediately suspected a relation with matrix algebra,but he lacked the mathematical background to work out a

consistent picture. Within a week Jordan showed that Born’s intuition was correct,and that it provided a rigid mathematical basis for Heisenberg’s ideas. In September1925, Jordan was the first to derive the famous commutation relation [x, p] = ih.Later that year, he wrote a paper on “Pauli statistics” in which he derived the Fermi–Dirac statistics before either Fermi or Dirac did. The only reason why fermions arenowadays not called jordanions is that Born delayed publication by forgetting aboutthe manuscript while lecturing in the US for half a year.

In the years 1926–27, Jordan published a series of papers in which he introduced thecanonical anti-commutation relations and a way to quantize wave fields. These worksare probably his most important, and are commonly regarded as the birth of quantumfield theory. After moving to Rostock in 1928, Jordan became more interested in themathematical and conceptual side of quantum mechanics, and he lost track of mostof the rapid developments in the field he started. In 1933 he joined the Nazi party (heeven became a storm trooper), which damaged his relations with many colleagues.Despite his strong verbal support of the party, the Nazis also did not trust him fully:quantum mechanics was still regarded as a mainly Jewish business. Jordan thereforeended up practically in scientific isolation during the war.

After the war, his radical support of the Nazis caused him severe trouble. Only in1953 did he regain his full rights as professor at the University of Hamburg, wherehe had been since 1951 and stayed till his retirement in 1970.

a differential equation involving all these variables. In the framework of second quantiza-tion, we reduced this to an equation for an operator of one single variable, (3.31). Apartfrom the operator “hats,” it looks very much like the Schrödinger equation for a singleparticle,

ih∂ψ(r, t)∂t

= Hψ ="

− h2∇2

2m+ V(r)

#

ψ(r, t), (3.32)

and is of the same level of complexity.


From an educational point of view, the advantage is enormous. To reveal it finally, letus take the expectation value of (3.31) and introduce the (deliberately confusing) notationψ(r, t) = ⟨ψ(r, t)⟩. Owing to the linearity of (3.31), the equation for the expectation valueformally coincides with (3.32). However, let us put this straight: this equation is not for awave function anymore. It is an equation for a classical field.2 Without noticing, we havethus got all the way from particles to fields and have understood that these concepts arejust two facets of a single underlying concept: that of the quantum field.

Another frequently asked question is why the technique is called second quantization.Well, the name is historic rather than rational. A quasi-explanation is that the first quan-tization is the replacement of numbers – momentum and coordinate – by non-commutingoperators, whereas the wave function still is a number,

[px, x] = 0, [ψ∗,ψ] = 0. (3.33)

As a “second” step, one “quantizes” the wave function as well, so that finally

[ψ†, ψ] = 0. (3.34)

Once again, this is an explanation for the name and not for the physics involved. The wavefunction is not really quantized, it is not replaced with an operator! The field operator isthe replacement of the field rather than that of a wave function.

3.4 Second quantization for fermions

As we see below, the second quantization technique for fermions is very similar to the onefor bosons. We proceed by repeating the steps we made for bosons and concentrate on thedifferences.

We again start from a basis in the single-particle space – the basis of levels – and labelthe levels with an index k. The first difference we note is that fermions are bound to havehalf-integer spin,3 whereas bosons have integer spin. For bosons the simplest case is givenby spinless (s = 0) bosons. We did not mention spin in the section on bosons, since wetacitly assumed the bosons to be spinless. For fermions, however, the simplest case is givenby particles with two possible spin directions (s = 1

2 ). A complete set of single particlebasis states now includes an index for spin. For example, the set of free particle planewaves now reads

ψk,↑ = 1√V

exp(ik · r),

10

-,

ψk,↓ = 1√V

exp(ik · r),

01

-,

(3.35)

2 The field at hand is a complex scalar field. By this it differs from, say, an electromagnetic field which is a vectorfield. However, we will quantize the electromagnetic field soon and see that the difference is not essential.

3 Provided they are elementary particles. There are examples of excitations in condensed matter that are naturallyregarded as spinless fermion particles.

75 3.4 Second quantization for fermions

where the two components of the wave functions correspond to the two different directionsof spin. One thus has to remember always that all levels are labeled by a wave vector kand a spin index σ = {↑, ↓}. In the rest of the section we just use the labels k, which areshorthand for the full index, k ≡ {k, σ }.

Now we can construct a basis in Fock space as we did before. We again label the basisstates in Fock space with sets of occupation numbers {nk} ≡ {n1, n2, . . . }, and express theorthonormality of the basis as

⟨{nk}|{n′k}⟩ =

)

k

δnk ,n′k. (3.36)

We note that for fermions these occupation numbers can take only two values, 0 or 1. A setof occupation numbers therefore reduces to just a long sequence of zeros and ones.

This defines the basis in Fock space. Let us note, however, that at this moment still noth-ing in the basis reflects the most important difference between bosons and fermions: thefact that a bosonic wave function is symmetric, whereas a fermionic one is antisymmetric.We could use the same basis presented above to describe some fancy bosons that for somereason strongly dislike sharing the same level.4 So the question is: how do we incorporatethe information about the antisymmetry of the wave function into the second quantizationscheme?

3.4.1 Creation and annihilation operators for fermions

The only place to put the information that a fermionic many-particle wave function isantisymmetric, is in the structure of the CAPs. To understand this, let us do a step backand consider identical distinguishable particles that prefer not to share the same level. Inthis case, the number of possible (basis) states is no longer restricted by the symmetrypostulate. So we must consider a space of all possible products of basis states, not onlythe (anti-)symmetrized ones. For instance, a state where a first particle is put in level 5, asecond particle in level 3, and a third one in level 2, is different from a state with the sameoccupation numbers but a different filling order. Using elementary creation operators c†

kwe can rephrase this statement as

c†2c†

3c†5|0⟩ = c†

3c†2c†

5|0⟩, (3.37)

where |0⟩ represents the vacuum. We require here that the order of application of the oper-ators corresponds to the order of the filling of the levels. We see that if the particles weredistinguishable, the order of the operators determines which particle sits in which level.The two states created in (3.37) are thus in fact the very same state, but with the particlesin levels 2 and 3 permuted. For the indistinguishable fermions, such a permutation shouldresult in a minus sign in front of the many-particle wave function.

We can now build the Fock space for fermions. To conform to the antisymmetry require-ment, we impose that two states with the same occupation numbers but a different filling

4 Such models of impenetrable boson particles have been actually put forward.


order must differ by a minus sign if the number of permutations needed to go from the onestate to the other is odd. For example,

c†2c†

3c†5|0⟩ = −c†

2c†5c†

3|0⟩ and c†2c†

3c†5|0⟩ = c†

3c†5c†

2|0⟩. (3.38)

This imposes a specific requirement on the commutation relations of the fermionic CAPs:they must anticommute. Indeed, for two particles we can write analogously to (3.38)

c†nc†

m|0⟩ = −c†mc†

n|0⟩, (3.39)

which is satisfied provided

c†nc†

m + c†mc†

n ≡ {c†n, c†

m} = 0. (3.40)

The antisymmetry is crucial here. If we run these procedures for the fancy impenetrablebosons mentioned above, we have to require c†

nc†m|0⟩ = c†

mc†n|0⟩ and again end up with

commuting creation operators.It is convenient to require that a product of a creation and annihilation operator again

presents the occupation number operator, as it does for bosons. This leads us to a uniquedefinition of CAPs for fermions. A word of warning is appropriate here: the fermionicCAPs are in fact simple and nice to work with, but this does not apply to their definition.So brace yourself!

The annihilation operator ck for a fermion in the level k is defined as

ck |. . . . . . , 1k, . . .⟩ = (−1)*k−1

i=1 ni |. . . . . . , 0k, . . .⟩, (3.41)

ck |. . . . . . , 0k, . . .⟩ = 0. (3.42)

Here the zeros and ones refer to the occupation number of the kth level, as indicated withthe index k. The annihilation operator thus does what we expect: it removes a particlefrom the level k if it is present there. Besides, it multiplies the function with a factor ±1.Inconveniently enough, this factor depends on the occupation of all levels below the levelk. It becomes +1(−1) if the total number of particles in these levels is even (odd). Theconjugated operator to ck creates a particle in the level k as expected,

c†k |. . . . . . , 0k, . . .⟩ = (−1)

*k−1i=1 ni |. . . . . . , 1k, . . .⟩. (3.43)

Control question. Can you find the values of the matrix elements ⟨1, 1, 1|c1|1, 1, 1⟩,⟨1, 0, 1|c2|1, 1, 1⟩, and ⟨1, 1, 1|c†

3|1, 1, 0⟩?

What is really strange about these definitions is that they are subjective. They explicitlydepend on the way we have ordered the levels. We usually find it convenient to organizethe levels in order of increasing energy, but we could also choose decreasing energy orany other order. The point is that the signs of the matrix elements of fermionic CAPsare subjective indeed. Therefore, the CAPs do not purport to correspond to any physicalquantity. On the contrary, the matrix elements of their products do not depend on the choiceof the level ordering. Therefore, the products correspond to physical quantities.

77 3.4 Second quantization for fermions

Let us show this explicitly by calculating products of CAPs. We start with c†k c†

k′ fork = k′. We apply this product to a state |. . . , 0k, . . . , 0k′ , . . .⟩ since any other state produceszero. From the definition (3.43) we find

c†k c†

k′ |. . . , 0k, . . . , 0k′ , . . .⟩ = c†k(−1)

*k′−1i=1 ni |. . . , 0k, . . . , 1k′ , . . .⟩

= (−1)*k−1

i=1 ni(−1)*k′−1

i=1 ni |. . . , 1k, . . . , 1k′ , . . .⟩.(3.44)

Changing the order of the two operators, we get

c†k′ c

†k |. . . , 0k, . . . , 0k′ , . . .⟩ = c†

k′ (−1)*k−1

i=1 ni |. . . , 1k, . . . , 0k′ , . . .⟩

= (−1)*k−1

i=1 ni(−1)1+*k′−1i=1 ni |. . . , 1k, . . . , 1k′ , . . .⟩ = −c†

k c†k′ |. . . , 0k, . . . , 0k′ , . . .⟩.

(3.45)

So, the creation operators anticommute as expected.For the product c†

k ck′ we can proceed in the same way to find the expected result

c†k ck′ |. . . , 0k, . . . , 1k′ , . . .⟩ = −ck′ c

†k |. . . , 0k, . . . , 1k′ , . . .⟩. (3.46)

In a similar way one can find for the case k = k′ the relations

c†k ck |{n}⟩ = nk|{n}⟩, (3.47)

ck c†k |{n}⟩ = (1 − nk)|{n}⟩, (3.48)

so c†k ck + ck c†

k = 1. Conveniently, the representation of the occupation numbers in termsof CAPs is the same as for bosons, nk = c†

k ck . The operator of the total number of particlesreads

N =!

k

nk =!

k

c†k ck . (3.49)

We summarize the commutation relations for the fermionic CAPs as follows:

commutation relations for fermion CAPs

{ck , c†k′} = δkk′

{ck , ck′} = {c†k , c†

k′} = 0

Let us conclude this section with a few comments on the structure of the Fock space forfermions. We have again one special state, the vacuum |0⟩. By definition, ck |0⟩ = 0 for allk since there are no particles to be annihilated in the vacuum. We have already noted thatall states can be obtained from the vacuum state by creating particles one by one in it, thatis, by applying the proper sequence of creation operators,

|{nk}⟩ =)

k ∈ occupied

c†k |0⟩. (3.50)

The product runs over the occupied levels only, those for which nk = 1. A distinction fromthe similar formula for bosons is that the order in the sequence is now in principle impor-tant: fermionic creation operators do not commute. However, while they anticommute, the


difference between different sequences of the same operators is minimal: it is just a factor±1. Another important distinction from bosons is that the fermionic Fock space spannedby a finite number of levels M is of a finite dimension 2M (where M is taken to include thespin degeneracy).

Control question. Can you explain why the dimension is such? If not, count thenumber of basis states in the Fock space for M = 1, 2, . . .

3.4.2 Field operators

The field operators, the Hamiltonian in terms of field operators, and the dynamic equationsgenerated by this Hamiltonian almost literally coincide with those for bosons. One of thedifferences is spin: there are (at least) two sorts of field operator corresponding to particlesof different spin directions,

ψσ (r) =!

k

ckσψk(r). (3.51)

The most important difference, however, is in the commutation relations. Since the fieldoperators are linear combinations of CAPs, they anticommute as well:

commutation relations for fermion field operators

{ψσ (r), ψ†σ ′ (r′)} = δ(r − r′)δσσ ′

{ψσ (r), ψσ ′ (r′)} = {ψ†σ (r), ψ†

σ ′ (r′)} = 0

Remarkably (and quite conveniently) this change in commutation relations hardlyaffects the form of the Hamiltonians. Quite similarly to their boson counterparts, thenon-interacting Hamiltonian and the pairwise interaction term read

H(1) =!

σ

+dr ψ†

σ (r)

"

− h2∇2

2m+ V(r)

#

ψσ (r), (3.52)

H(2) = 12

!

σ ,σ ′

+dr1dr2 V (2)(r1 − r2)ψ†

σ (r1)ψ†σ ′ (r2)ψσ ′ (r2)ψσ (r1). (3.53)

In the wave vector representation, the interaction term now reads

H(2) = 12V

!

k,k′,q

!

σ ,σ ′V (2)(q) c†

k+q,σ c†k′−q,σ ′ ck′,σ ′ ck,σ . (3.54)

This can still be interpreted in terms of two-particle scattering, such as depicted in Fig. 3.2.A minor difference arises from the presence of spin indices, since the total spin of bothparticles is conserved in the course of the scattering.

79 3.5 Summary of second quantization

The Heisenberg equations for the fermionic CAPs, although derived from differentcommutation rules, are also almost identical to those for the boson CAPs,

ih∂ ck(t)∂t

= [ck, H(1)] = εkck(t), (3.55)

so that ck(t) = e− ih εktck(0). Transforming this time-evolution equation to the coordinate

representation, we obtain

ih∂ψσ (r, t)∂t

="

− h2∇2

2m+ V(r)

#

ψσ (r, t), (3.56)

indeed the same equation as for bosons. However, there is a physical difference. As dis-cussed, for bosons we can take the expectation value of the above equation and come upwith an equation that describes the dynamics of a classical field. It is a classical limit of theabove equation and is relevant if the occupation numbers of the levels involved are large,nk ≫ 1. Formally, we could do the same with (3.56). However, the resulting equation isfor a quantity ⟨ψ(r, t)⟩, the expectation value of a fermionic field. This expectation value,however, is zero for all physical systems known. One can infer this from the fact that thematrix elements of single fermionic CAPs are in fact subjective quantities (see the dis-cussion above). The resulting equation does therefore not make any sense. This is usuallyformulated as follows: the fermionic field does not have a classical limit. Another way ofseeing this is to acknowledge that for fermions, occupation numbers can never be large, nk

is either 0 or 1, and therefore there is no classical limit.

3.5 Summary of second quantization

The most important facts about second quantization are summarized in Table 3.1 below.Some of the relations turn out to be identical for bosons and fermions: in those cases we usethe generic notation a for the annihilation operator. Otherwise we stick to the convention bfor bosons and c for fermions.

Let us end this chapter with a few practical tips concerning CAPs. In all applications ofthe second quantization technique, you sooner or later run into expressions consisting ofa long product of CAPs stacked between two basis vectors of the Fock space. In order toderive any useful result, you must be able to manipulate those products of CAPs, i.e. youmust master the definition and commutation relations of the CAPs at an operational level.Any expression consisting of a long string of CAPs can be handled with the followingsimple rules:

1. Do not panic. You can do this, it is a routine calculation.2. Try to reduce the number of operators by converting them to occupation number

operators a†l al |{nk}⟩ = nl|{nk}⟩.

3. In order to achieve this, permute the operators using the commutation rules.


4. Do not calculate parts which reduce obviously to zero. Use common sense and thedefinition of the CAPs to guess whether an expression is zero before evaluating it.

Let us illustrate this by evaluating the following expression involving fermionic operators:

⟨{nk}|c†l1

c†l2

cl3cl4

|{nk}⟩ = . . .?

The last rule comes first. The bra and ket states are identical, and this allows us to establishrelations between the level indices l1, l2, l3, and l4. The two annihilation operators in theexpression kill particles in the ket state in the levels l3 and l4. In order to end up in |{nk}⟩again and thus have a non-zero result, the particles in these levels have to be re-created bythe creation operators. So we have to concentrate on two possibilities only,5

l1 = l3, l2 = l4, l1 = l2,

or l1 = l4, l2 = l3, l1 = l2.

We now focus on the first possibility, and try to reduce the CAPs to occupation numberoperators,

⟨{nk}|c†l1

c†l2

cl1cl2

|{nk}⟩ (permute the second and the third)

= − ⟨{nk}|c†l1

cl1c†

l2cl2

|{nk}⟩ (use reduction rule)

= − ⟨{nk}|c†l1

cl1|{nk} ⟩nl2 (once again)

= − nl1 nl2 .

Treating the second possibility in the same way, we find

⟨{nk}|c†l1

c†l2

cl2cl1

|{nk}⟩ (permute the third and the fourth)

= − ⟨{nk}|c†l1

c†l2

cl1cl2

|{nk}⟩ (permute the second and the third)

= + ⟨{nk}|c†l1

cl1c†

l2cl2

|{nk} (use reduction rule twice)

= + nl1 nl2 .

5 The case l1 = l2 = l3 = l4 gives zero since all levels can only be occupied once. The term cl3 cl4 then alwaysproduces zero.

81 3.5 Summary of second quantization

Table 3.1 Summary: Second quantization

Bosons FermionsMany-particle wave function: ψ({ri}, t)

fully symmetric fully antisymmetricFock space, where the basis states are labeled by the sets {nk} of occupation numbersoccupation numbers can be: non-negative integers 0 or 1Creation and annihilation operators: commute anticommute

[bk , bk′ ] = 0 {ck , ck′ } = 0[b†

k , b†k′ ] = 0 {c†

k , c†k′ } = 0

[bk , b†k′ ] = δkk′ {ck , c†

k′ } = δkk′

Occupation number operatornumber of particles in the level k: nk ≡ a†

k aktotal number of particles: N = *

k nk = *k a†

k ak

Hamiltonianwith particle–particle interactions: H =

!

k

εkb†kbk the same, with spin

+ 12V

!

k,k′ ,q

V(q) b†k+qb†

k′−qbk′ bk

Field operators: ψ(r) = *k akψk(r)

dynamics are (without interactions) the same as the Schrödinger equation for the wave function

Heisenberg equation: ih∂

∂tψ(r, t) =

"

− h2∇2

2m+ V(r)

#

ψ(r, t)


Exercises

In all Exercises, we use the following notation for the CAPs: the operators b and c arerespectively bosonic and fermionic operators, while a is an operator that can be eitherbosonic or fermionic. Matrices Mkl with k, l labeling the levels are denoted as M.

1. Applications of commutation rules (solution included).a. Use the commutation relations to show that

[a†k al , am] = −δmkal,

for both bosonic and fermionic CAPs.b. Given the Hamiltonian of non-interacting particles

H =!

k

εknk =!

k

εka†k ak,

use the Heisenberg equations of motion to show that

ddt

ak = − ihεkak.

c. Compute for an ideal Bose-gas of spinless particles the matrix element

⟨g|bi1bi2b†i3b†

i4|g⟩,where |g⟩ is a number state in Fock space, |g⟩ ≡ |{nk}⟩. The level indices i1 . . . i4can be different or coinciding. Compute the matrix element considering all possiblecases.

d. Do the same for ⟨g|b†i1bi2b†

i3bi4|g⟩.e. Compute a similar matrix element, ⟨g|c†

i1ci2c†i3ci4|g⟩ for the fermion case.

2. Fluctuations. Any one-particle operator A can be written using creation and annihilationoperators as

A =!

i,j

Aija†i aj .

a. Calculate the expectation value ⟨g|A|g⟩ ≡ ⟨A⟩, |g⟩ being a number state, |g⟩ ≡|{nk}⟩. Is there a difference for fermions and bosons?

b. Calculate the variance ⟨A2⟩ − ⟨A⟩2 for fermions and bosons, making use of theresults found in 1(d) and 1(e).

3. Linear transformations of CAPs. Let us consider a set of bosonic CAPs, bk and b†k . A

linear transformation A, represented by the matrix L, transforms the CAPs to anotherset of operators,

b′m =

!

k

Lmkbk.

a. Find the conditions on the matrix L representing the transformation A under whichthe resulting operators are also bosonic CAPs, that is, they satisfy the bosoniccommutation rules.

83 Solutions

A more general linear transformation B can also mix the two types of operator,

b′m =

!

k

Lmkbk +!

k

Mmkb†k .

b. Find the conditions on the matrices L and M of the more general transform B underwhich the resulting operators are bosonic CAPs.

c. Consider the transformations A and B in application to fermionic CAPs and find theconditions under which the transformed set remains a set of fermionic CAPs.

4. Exponents of boson operators.a. Find

[bn, b†].

b. Employ the result of (a) to prove that for an arbitrary sufficiently smooth functionF it holds that

[F(b), b†] = F′(b).

c. Use the result of (b) to prove that

F(b) exp%αb†

&= exp

%αb†

&F(b + α).

d. Show that

exp%αb† + βb

&= exp

%αb†

&exp

%βb&

eαβ/2.

Hint. Consider the operator

O(s) = exp%−sαb†

&exp

%−sβb

&exp

%s(αb† + βb)

&,

and derive the differential equation it satisfies as a function of s.5. Statistics in disguise.

a. Consider the operator Z ≡ exp%−*k,l c†

kMklcl

&, where M is a Hermitian matrix.

Show that

Tr{Z} = det[1 + exp(−M)].

Hint. Diagonalize M.b. Compute Tr

3exp

%−*k,l b†

kMklbl

&4.

Solutions

1. Applications of commutation rules.a. For bosons,

[b†k bl , bm] = b†

k bl bm − bmb†k bl = b†

k bmbl − (δkm + b†k bm)bl = −δkmbl ,


where after the second equality sign we make use of commutation relations for twoannihilation operators to transform the first term, and that for a creation and anannihilation operator to transform the second term. For fermions,

[c†k cl , cm] = c†

k cl cm − cmc†k cl = −c†

k cmcl − (δkm − c†k cm)cl = −δkmcl ,

where we make use of anticommutation rather than commutation relations.b. The Heisenberg equation of motion for the operator ak reads

ddt

ak = ih

[H, ak].

We compute the commutator using the relation found in (a),6!

l

εla†l al , ak

7

= −εkak,

which proves the equation of motion given in the problem.c. Since the numbers of particles are the same in ⟨g| and |g⟩, the matrix element is

non-zero only if each creation of a particle in a certain level is compensated by anannihilation in the same level. This gives us three distinct cases:

I. i1 = i3, i2 = i4, and i1 = i2. In this case, we start by commuting the secondand third operator,


i2|g⟩ = ⟨g|bi1b†i1bi2b†

i2|g⟩ = ⟨g|bi1b†i1|g⟩(1+ni2) = (1+ni2)(1+ni1),

where we have made use of bk b†k |g⟩ = (1 + nk)|g⟩.

II. i1 = i4, i2 = i3, and i1 = i2. In this case, the first and fourth operators have tobe brought together. We commute the fourth operator with the middle two,


i1|g⟩ = ⟨g|bi1b†i1bi2b†

i2|g⟩ = ⟨g|bi1b†i1|g⟩(1+ni2) = (1+ni2)(1+ni1),

thus getting the same as in the previous case.III. All indices are the same, i1 = i2 = i3 = i4. The matrix element can then be

computed directly giving


i1|g⟩ = (2 + ni1)(1 + ni1),

where we use the definitions (3.1) and (3.2).Collecting all three cases, we obtain


i4|g⟩ = (δi1,i3δi2,i4 + δi1,i3δi2,i4)

×'(1 + ni1)(1 + ni2) − 1

2δi1,i2ni1(1 + ni2)(.

d. We proceed in the same way considering the casesI. i1 = i2, i3 = i4, and i1 = i3 gives ni1ni3.

II. i1 = i4, i3 = i2, and i1 = i3 gives ni1ni3.III. i1 = i2 = i3 = i4 also gives ni1ni3.Collecting all cases again, we now obtain

⟨g|b†i1bi2b†

i3bi4|g⟩ = (δi1,i2δi3,i4 + δi1,i4δi2,i3)(1 − 12δi1,i3)ni1ni3.

85 Solutions

e. Here we have the same cases as in (d). In the cases I and III, we do not have tocommute any operators and the matrix element is readily reduced to ni1ni3. Forcase II we derive

⟨g|c†i1ci3c†

i3ci1|g⟩ = ⟨g|c†i1ci1ci3c†

i3|g⟩ = ni1(1 − ni3).

Collecting the three cases thus yields

⟨g|c†i1ci2c†

i3ci4|g⟩ = δi1,i2δi3,i4ni1ni3 + δi1,i4δi2,i3ni1(1 − ni3).

It might seem that this expression does not account for case III, but in fact it does.If we set all indices equal on the right-hand side, we get nini + ni(1 − ni). But inany number state, ni(1 − ni) = 0 for all i, dictated by fermionic statistics.

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