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6

Non-Relativistic Field Theory

In this Chapter we will discuss the field theory description of non-relativisticsystems. The material that we will be presented here is, for the most part,introductory as this topic is covered in depth in many specialized text-books, such as Methods of Quantum Field Theory in Statistical Physics byA. Abrikosov, L. Gorkov and I. Dzyaloshinskii (Prentice-Hall, 1963), Sta-tistical Mechanics, A Set of lectures by R. Feynman (Benjamin, 1973) andmany others. The discussion of “second quantization” is very standard andis presented her for pedagogical reasons but can be skipped.

6.1 Non-Relativistic Field Theories, Second Quantization andthe Many-Body Problem

Let us consider now the problem of a system of N identical non-relativisticparticles. For the sake of simplicity I will assume that the physical stateof each particle j is described by its position xj relative to some referenceframe. This case is easy to generalize.

The wave function for this system is !(x1, . . . ,xN ). If the particles areidentical then the probability density, |!(x1, . . . ,xN )|2, must be invariant(i.e., unchanged) under arbitrary exchanges of the labels that we use toidentify (or designate) the particles. In quantum mechanics, the particles donot have well defined trajectories. Only the states of a physical system arewell defined. Thus, even though at some initial time t0 the N particles maybe localized at a set of well defined positions x1, . . . ,xN , they will becomedelocalized as the system evolves. Furthermore the Hamiltonian itself isinvariant under a permutation of the particle labels. Hence, permutationsconstitute a symmetry of a many-particle quantum mechanical system. Inother terms, identical particles are indistinguishable in quantum mechanics.In particular, the probability density of any eigenstate must remain invariant

176 Non-Relativistic Field Theory

if the labels of any pair of particles are exchanged. If we denote by Pjk theoperator which exchanges the labels of particles j and k, the wave functionsmust satisfy

Pjk!(x1, . . . ,xj, . . . ,xk, . . . ,xN ) = ei!!(x1, . . . ,xj , . . . ,xk, . . . ,xN )(6.1)

Under a further exchange operation, the particles return to their initial la-bels and we recover the original state. This sample argument then requiresthat ! = 0," since 2! must not be an observable phase. We then concludethat there are two possibilities: either ! is even under permutation andP! = !, or ! is odd under permutation and P! = !!. Systems of iden-tical particles which have wave functions which are even under a pairwisepermutation of the particle labels are called bosons. In the other case, ! oddunder pairwise permutation, they are Fermions. It must be stressed thatthese arguments only show that the requirement that the state priori! beeither even or odd is only a su!cient condition. It turns out that underspecial circumstances other options become available and the phase factor! may take values di"erent from 0 or ". These particles are called anyons.For the moment the only cases in which they may exist appears to be in sit-uations in which the particles are restricted to move on a line or on a plane.In the case of relativistic quantum field theories, the requirement that thestates have well defined statistics (or symmetry) is demanded by a very deepand fundamental theorem which links the statistics of the states of the spinof the field. This is the spin-statistics theorem which we will discuss lateron.

6.1.1 Fock Space

We will now discuss a procedure, known as Second Quantization, which willenable us to keep track of the symmetry of the states in a simple way.Let us consider now a system of N particles. The wave functions in thecoordinate representation are !(x1, . . . ,xN ) where the labels x1, . . . ,xN

denote both the coordinates and the spin states of the particles in the state|!". For the sake of definiteness we will discuss physical systems describableby Hamiltonians H of the form

H = ! !2

2m

N!

j=1

#2j +

N!

j=1

V (xj) + g!

j,k

U(xj ! xk) + . . . (6.2)

Let {!n(x)} be the wave functions for a complete set of one-particle states.Then an arbitrary N -particle state can be expanded in a basis which is the

6.1 Non-Relativistic Field Theories, Second Quantization and the Many-Body Problem177

tensor product of the one-particle states, namely

!(x1, . . . ,xN ) =!

{nj}

C(n1, . . . nN )!n1(x1) . . . !nN (xN ) (6.3)

Thus, if ! is symmetric (antisymmetric) under an arbitrary exchange xj $xk, the coe#cients C(n1, . . . , nN ) must be symmetric (antisymmetric) underthe exchange nj % nk.

A set of N -particle basis states with well defined permutation symmetryis the properly symmetrized or antisymmetrized tensor product

|!1, . . .!N " & |!1"' |!2"' · · ·' |!N " =1(N !

!

P

#P |!P (1)"' · · ·' |!P (N)"

(6.4)where the sum runs over the set of all possible permutation P . The weightfactor # is +1 for bosons and !1 for fermions. Notice that, for fermions, theN -particle state vanishes if two particles are in the same one-particle state.This is the exclusion principle.

The inner product of two N -particle states is

)$1, . . .$N |%1, . . . ,%N " =1

N !

!

P,Q

#P+Q)$Q(1)|%P (1)" · · · )$Q(N)|%P (1)" =

=!

P !

#P!)$1|%P (1)" · · · )$N |%P (N)"

(6.5)

which is nothing but the permanent (determinant) of the matrix )$j |%k" forsymmetric (antisymmetric) states, i.e.,

)$1, . . .$N |%1, . . .%N " =

"""""""

)$1|%1" . . . )$1|%N "...

...)$N |%1" . . . )$N |%N "

""""""""

(6.6)

In the case of antisymmetric states, the inner product is the familiar Slaterdeterminant. Let us denote by {|&"} the complete set of one-particle stateswhich satisfy

)&|'" = (#$!

#

|&")&| = 1 (6.7)

The N -particle states are {|&1, . . .&N "}. Because of the symmetry require-ments, the labels &j can be arranged in the form of a monotonic sequence&1 * &2 * · · · * &N for bosons, or in the form of a strict monotonic sequence&1 < &2 < · · · < &N for fermions. Let nj be an integer which counts how


many particles are in the j-th one-particle state. The boson states |&1, . . .&N "must be normalized by a factor of the form

1(n1! . . . nN !

|&1, . . . ,&N " (&1 * &2 * · · · * &N ) (6.8)

and nj are non-negative integers. For fermions the states are

|&1, . . . ,&N " (&1 < &2 < · · · < &N ) (6.9)

and nj = 0 > 1. These N -particle states are complete and orthonormal

1

N !

!

#1,...,#N

|&1, . . . ,&N ")&1, . . . ,&N | = I (6.10)

where the sum over the &’s is unrestricted and the operator I is the identityoperator in the space of N -particle states.

We will now consider the more general problem in which the number ofparticles N is not fixed a-priori. Rather, we will consider an enlarged spaceof states in which the number of particles is allowed to fluctuate. In thelanguage of Statistical Physics what we are doing is to go from the CanonicalEnsemble to the Grand Canonical Ensemble. Thus, let us denote by H0 theHilbert space with no particles, H1 the Hilbert space with only one particleand, in general, HN the Hilbert space for N -particles. The direct sum ofthese spaces H

H = H0 +H1 + · · ·+HN + · · · (6.11)

is usually called the Fock space. An arbitrary state |%" in Fock space is thesum over the subspaces HN ,

|%" = |%(0)"+ |%(1)"+ · · ·+ |%(N)"+ · · · (6.12)

The subspace with no particles is a one-dimensional space spanned by thevector |0" which we will call the vacuum. The subspaces with well definednumber of particles are defined to be orthogonal to each other in the sensethat the inner product in the Fock space

)$|%" &!!

j=0

)$(j)|%(j)" (6.13)

vanishes if |$" and |%" belong to di"erent subspaces.


6.1.2 Creation and Annihilation Operators

Let |!" be an arbitrary one-particle state. Let us define the creation operatora†(!) by its action on an arbitrary state in Fock space

a†(!)|%1, . . . ,%N " = |!,%1, . . . ,%N " (6.14)

Clearly, a†(!) maps the N -particle state with proper symmetry |%1, . . . ,%N "onto the N + 1-particle state |!,%, . . . ,%N ", also with proper symmetry .The destruction or annihilation operator a(!) is defined to be the adjoint ofa†(!) ,

)$1, . . . ,$N"1|a(!)|%1, . . . ,%N " = )%1, . . . ,%N |a†(!)|$1, . . . ,$N"1"#(6.15)

Hence

)$1, . . . ,$N"1|a(!)|%1, . . . ,%N " =)%1, . . . ,%N |!,$1, . . . ,$N"1"# =

=

"""""""

)%1|!" )%1|$1" · · · )%1|$N"1"...

......

)%N |!" )%N |$1" · · · )%N |$N"1"

"""""""

#

"

(6.16)

We can now expand the permanent (or determinant) to get

)$1, . . . ,$N"1|a(!)|%1, . . . ,%N " =

=N!

k=1

#k"1)%k|!"

"""""""""

)%1|$1" . . . )%1|$N"1"...

.... . . (no%k) . . .

)%N |$1" . . . )%N |$N"1"

"""""""""

#

"

=N!

k=1

#k"1)%k|!")$1, . . . ,$N"1|%1, . . . %k . . . ,%N "

(6.17)

where %k indicates that %k is absent. Thus, the destruction operator isgiven by

a(!)|%1, . . . ,%N " =N!

k=1

#k"1)!|%k"|%1, . . . , %k, . . . ,%N " (6.18)

With these definitions, we can easily see that the operators a†(!) and a(!)


obey the commutation relations

a†(!1)a†(!2) = # a†(!2)a

†(!1) (6.19)

Let us introduce the notation#A, B

$

""& AB ! # BA (6.20)

where A and B are two arbitrary operators. For # = +1 (bosons) we havethe commutator

#a†(!1), a

†(!2)$

+1&#a†(!1), a

†(!2)$= 0 (6.21)

while for # = !1 it is the anticommutator#a†(!1), a

†(!2)$

"1&%a†(!1), a

†(!2)&= 0 (6.22)

Similarly for any pair of arbitrary one-particle states |!1" and |!2" we get

[a(!1), a(!2)]"" = 0 (6.23)

It is also easy to check that the following identity holds#a(!1), a

†(!2)$

""= )!1|!2" (6.24)

So far we have not picked any particular representation. Let us consider theoccupation number representation in which the states are labelled by thenumber of particles nk in the single-particle state k. In this case, we have

|n1, . . . , nk, . . ." &1(

n1!n2! . . .|

n1' () *1 . . . 1,

n2' () *2 . . . 2 . . ." (6.25)

In the case of bosons, the nj’s can be any non-negative integer, while forfermions they can only be equal to zero or one. In general we have that if|&" is the &th single-particle state, then

a†#|n1, . . . , n#, . . ." =(n# + 1|n1, . . . , n# + 1, . . ."

a#|n1, . . . , n#, . . ." =(n#|n1, . . . , n# ! 1, . . ."

(6.26)

Thus for both fermions and bosons, a# annihilates all states with n# = 0,while for fermions a†# annihilates all states with n# = 1.

The commutation relations are

[a#, a$ ] =#a†#, a

†$

$= 0

#a#, a

†$

$= (#$ (6.27)


for bosons, and

{a#a$} =%a†$ , a

†$

&= 0

%a#a

†$

&= (#$ (6.28)

for fermions. Here,%A, B

&is the anticommutator of the operators A and B

%A, B

&& AB + BA (6.29)

If a unitary transformation is performed in the space of one-particle statevectors, then a unitary transformation is induced in the space of the opera-tors themselves, i.e., if |$" = &|%" + '|!", then

a($) = &#a(%) + '#a(!)

a†($) = &a†(%) + 'a†(!)

(6.30)

and we say that a†($) transforms like the ket |$" while a($) transforms likethe bra )$|.

For example, we can pick as the complete set of one-particle states themomentum states {|p"}. This is “momentum space”. With this choice thecommutation relations are

#a†(p), a†(q)

$

""= [a(p), a(q)]"" = 0

#a(p), a†(q)

$

""= (2")d(d(p! q)

(6.31)

where d is the dimensionality of space. In this representation, an N -particlestate is

|p1, . . . ,pN " = a†(p1) . . . a†(pN )|0" (6.32)

On the other hand, we can also pick the one-particle states to be eigenstatesof the position operators, i.e.,

|x1, . . .xN " = a†(x1) . . . a†(xN )|0" (6.33)

In position space, the operators satisfy#a†(x1), a

†(x2)$

""= [a(x1), a(x2)]"" = 0

#a(x1), a

†(x2)$

""= (d(x1 ! x2)

(6.34)


This is the position space or coordinate representation. A transformationfrom position space to momentum space is the Fourier transform

|p" =+

ddx |x")x|p" =+

ddx |x"eip·x (6.35)

and, conversely

|x" =+

ddp

(2")d|p"e"ip·x (6.36)

Then, the operators themselves obey

a†(p) =

+ddx a†(x)eip·x

a†(x) =

+ddp

(2")da†(p)e"ip·x

(6.37)

6.1.3 General Operators in Fock Space

Let A(1) be an operator acting on one-particle states. We can always definean extension of A acting on any arbitrary state |%" of the N -particle Hilbertspace as follows:

,A|%" &N!

j=1

|%1" ' . . .'A(1)|%j" ' . . .' |%N " (6.38)

For instance, if the one-particle basis states {|%j"} are eigenstates of A witheigenvalues {aj} we get

,A|%" =

-

.N!

j=1

aj

/

0 |%" (6.39)

We wish to find an expression for an arbitrary operator A in terms of creationand annihilation operators. Let us first consider the operator A(1)

#$ = |&")'|which acts on one-particle states. The operators A(1)

#$ form a basis of thespace of operators acting on one-particle states. Then, the N -particle exten-sion of A#$ is

,A#$ |%" =N!

j=1

|%1" ' · · ·' |&" ' · · ·' |%N ")'|%j" (6.40)


Thus

,A#$ |%" =N!

j=1

|%1, . . . ,

j'()*& , . . . ,%N " )'|%j" (6.41)

In other words, we can replace the one-particle state |%j" from the basiswith the state |&" at the price of a weight factor, the overlap )'|%j". Thisoperator has a very simple expression in terms of creation and annihilationoperators. Indeed,

a†(&)a(')|%" =N!

k=1

#k"1)'|%k" |&,%1, . . . ,%k"1,%k+1, . . . ,%N " (6.42)

We can now use the symmetry of the state to write

#k"1|&,%1, . . . ,%k"1,%k+1, . . . ,%N " = |%1, . . . ,k'()*& , . . . ,%N " (6.43)

Thus the operator A#$, the extension of |&")'| to the N -particle space,coincides with a†(&)a(')

,A#$ & a†(&)a(') (6.44)

We can use this result to find the extension for an arbitrary operator A(1)

of the form

A(1) =!

#,$

|&")&|A(1)|'" )'| (6.45)

we find

,A =!

#,$

a†(&)a('))&|A(1)|'" (6.46)

Hence the coe#cients of the expansion are the matrix elements of A(1) be-tween arbitrary one-particle states. We now discuss a few operators of in-terest.

The Identity Operator

The Identity Operator 1 of the one-particle Hilbert space

1 =!

#

|&")&| (6.47)

becomes the number operator ,N,N =

!

#

a†(&)a(&) (6.48)


In position and in momentum space we find

,N =

+ddp

(2")da†(p)a(p) =

+ddx a†(x)a(x) =

+ddx )(x) (6.49)

where )(x) = a†(x)a(x) is the particle density operator.

The Linear Momentum Operator

In the space H1, the linear momentum operator is

p(1)j =

+ddp

(2")dpj |p")p| =

+ddx |x" !

i*j )x| (6.50)

Thus, we get that the total momentum operator ,Pj is

,Pj =

+ddp

(2")dpj a

†(p)a(p) =

+ddx a†(x)

!

i*j a(x) (6.51)

Hamiltonian

The one-particle Hamiltonian H(1)

H(1) =p 2

2m+ V (x) (6.52)

has the matrix elements

)x|H(1)|y" = ! !2

2m#2 (d(x! y) + V (x)(d(x! y) (6.53)

Thus, in Fock space we get

,H =

+ddx a†(x)

1! !2

2m#2 +V (x)

2a(x) (6.54)

in position space. In momentum space we can define

3V (q) =

+ddx V (x)e"iq·x (6.55)

the Fourier transform of the potential V (x), and get

,H =

+ddp

(2")dp 2

2ma†(p)a(p)+

+ddp

(2")d

+ddq

(2")d3V (q)a†(p+q)a(p) (6.56)

The last term has a very simple physical interpretation. When acting on aone-particle state with well-defined momentum, say |p", the potential termyields another one-particle state with momentum p + q, where q is themomentum transfer, with amplitude V (q). This process is usually depictedby the diagram of the following figure.


a(p )

a†(p+ q)

p+ q

p

q

3V (q)

Figure 6.1 One-body scattering.

Two-Body Interactions

A two-particle interaction is a operator V (2) which acts on the space oftwo-particle states H2, has the form

V (2) =1

2

!

#,$

|&,'"V (2)(&,'))&,'| (6.57)

The methods developed above yield an extension of V (2) to Fock space ofthe form

,V =1

2

!

#,$

a†(&)a†(')a(')a(&) V (2)(&,') (6.58)

In position space, ignoring spin, we get

,V =1

2

+ddx

+ddy a†(x) a†(y) a(y) a(x) V (2)(x,y)

& 1

2

+ddx

+ddy )(x)V (2) (x,y))(y) +

1

2

+ddx V (2)(x,x) )(x)

(6.59)

while in momentum space we find

,V =1

2

+ddp

(2")d

+ddq

(2")d

+ddk

(2")d3V (k) a†(p + k)a†(q ! k)a(q)a(p) (6.60)


where 3V (k) is only a function of the momentum transfer +k. This is a conse-quence of translation invariance. In particular for a Coulomb interaction,

V (2)(x,y) =e2

|x! y| (6.61)

for which we have

3V (k) =4"e2

k 2(6.62)

a(p) p

a†(p+ k) p+ k

k

3V (k)

a(q)q

a†(q ! k)

q ! k

Figure 6.2 Two-body interaction.

6.2 Non-Relativistic Field Theory and Second Quantization

We can now reformulate the problem of an N -particle system as a non-relativistic field theory. The procedure described in the previous section iscommonly known as Second Quantization. If the (identical) particles arebosons, the operators a(!) obey canonical commutation relations. If the(identical) particles are Fermions, the operators a(!) obey canonical anti-commutation relations. In position space, it is customary to represent a†(!)by the operator %(x) which obeys the equal-time algebra

#%(x), %†(y)

$

""= (d(x! y)

#%(x), %(y)

$

""=#%†(x), %†(y)

$

""= 0

(6.63)

In this framework, the one-particle Schrodinger equation becomes the clas-

6.3 Non-Relativistic Fermions at Zero Temperature 187

sical field equation4i!*

*t+

!2

2m#2 !V (x)

5% = 0 (6.64)

Can we find a Lagrangian density L from which the one-particle Schrodingerequation follows as its classical equation of motion? The answer is yes andL is given by

L = i!%†*t% !!2

2m!%† ·#% ! V (x)%†% (6.65)

Its Euler-Lagrange equations are

*t(L(*t%†

= !! · (L(!%†

+(L(%†

(6.66)

which are equivalent to the field Equation Eq. 10.17. The canonical momenta$% and $†

% are

$% =(L(*t%†

= !i!% (6.67)

and

$†% =

(L(*t%

= i!%† (6.68)

Thus, the (equal-time) canonical commutation relations are#%(x), $%(y)

$

""= i!((x! y) (6.69)

which require that#%(x), %†(y)

$

""= (d(x! y) (6.70)

6.3 Non-Relativistic Fermions at Zero Temperature

The results of the previous sections tell us that the action for non-relativisticfermions (with two-body interactions) is (in D = d + 1 space-time dimen-sions)

S =

+dDx

4%†i!*t% !

!2

2m!%† ·!% ! V (x)%†(x)%(x)

5

! 1

2

+dDx

+dDx$%†(x)%†(x$)U(x! x$)%(x$)%†(x)%(x)

(6.71)


where U(x! x$) represents instantaneous pair-interactions,

U(x! x$) & U(x! x$)((x0 ! x$0) (6.72)

The Hamiltonian H for this system is

H =

+ddx [

!2

2m!%† ·!% + V (x)%†(x)%(x)]

+1

2

+ddx

+ddx$%†(x)%†(x$)U(x! x$)%(x$)%(x)

(6.73)

For Fermions the fields % and %† satisfy equal-time canonical anticommu-tation relations

{%(x), %†(x)} = ((x ! x$) (6.74)

while for Bosons they satisfy

[%(x), %†(x$)] = ((x! x$) (6.75)

In both cases, the Hamiltonian H commutes with the total number operatorN =

6ddx%†(x)%(x) since H conserves the total number of particles. The

Fock space picture of the many-body problem is equivalent to the GrandCanonical Ensemble of Statistical Mechanics. Thus, instead of fixing thenumber of particles we can introduce a Lagrange multiplier µ, the chemicalpotential, to weigh contributions from di"erent parts of the Fock space.Thus, we define the operator H.

3H & ,H ! µ ,N (6.76)

In a Hilbert space with fixed N this amounts to a shift of the energy byµN . We will now allow the system to choose the sector of the Fock spacebut with the requirement that the average number of particles )N " is fixedto be some number N . In the thermodynamic limit (N $,), µ representsthe di"erence of the ground state energies between two sectors with N + 1and N particles respectively. The modified Hamiltonian H is

3H =

+ddx

!

&

%†&(x)

1! !2

2m#2 +V (x)! µ

2%&(x)

+1

2

+ddx

+ddy

!

&,&!

%†&(x)%

†&!(y)U(x! y)%&!(y)%&(x)

(6.77)


6.3.1 The Ground State

Let us discuss now the very simple problem of finding the ground state for asystem of N spinless free fermions. In this case, the pair-potential vanishesand, if the system is isolated, so does the potential V (x). In general therewill be a complete set of one-particle states {|&"} and, in this basis, H is

H =!

#

E#a†#a# (6.78)

where the index & labels the one-particle states by increasing order of theirsingle-particle energies

E1 * E2 * · · · * En * · · · (6.79)

Since were are dealing with fermions, we cannot put more than one particlein each state. Thus the state the lowest energy is obtained by filling up allthe first N single particle states. Let |gnd" denote this ground state

|gnd" =N7

#=1

a†#|0" & a†1 · · · a†N |0" = |

N' () *1 . . . 1, 00 . . ." (6.80)

The energy of this state is Egnd with

Egnd = E1 + · · · + EN (6.81)

The energy of the top-most occupied single particle state, EN , is called theFermi energy of the system and the set of occupied states is called the filledFermi sea.

6.3.2 Excited States

A state like |%"

|%" = |N"1' () *1 . . . 1 010 . . ." (6.82)

is an excited state. It is obtained by removing one particle from the singleparticle state N (thus leaving a hole behind) and putting the particle in theunoccupied single particle state N +1. This is a state with one particle-holepair, and it has the form

|1 . . . 1010 . . ." = a†N+1aN |gnd" (6.83)

The energy of this state is

E% = E1 + · · ·+ EN"1 + EN+1 (6.84)


EF

E

Fermi Energy

Single Particle Energy

Occupied States Empty States

Single Particle StatesFermi Sea

Figure 6.3 The Fermi Sea.

Single Particle States

Single Particle States|gnd"

|!"

N N + 1

a†N+1aN

Figure 6.4 An excited (particle-hole) state.

Hence

E% = Egnd + EN+1 ! EN (6.85)

and, since EN+1 - EN , E% - Egnd. The excitation energy ,% = E% ! Egnd

is

,% = EN+1 ! EN - 0 (6.86)

6.3.3 Normal Ordering and Particle-Hole transformation:Construction of the Physical Hilbert Space

It is apparent that, instead of using the empty state |0" for reference state, itis physically more reasonable to use instead the filled Fermi sea |gnd" as the


physical reference state or vacuum state. Thus this state is a vacuum in thesense of absence of excitations. These arguments motivate the introductionof the particle-hole transformation.

Let us introduce the fermion operators b# such that

b# = a†# for & * N (6.87)

Since a†#|gnd" = 0 (for & * N) the operators b# annihilate the ground state|gnd", i.e.,

b#|gnd" = 0 (6.88)

The following anticommutation relations hold8a#, a

$#

9=%a#, b$

&=%b$, b

$$

&=%a#, b

†$

&= 0

%a#, a

†#!

&= (##! ,

%b$, b

†$!

&= ($$!

(6.89)

where &,&$ > N and ','$ * N . Thus, relative to the state |gnd", a†# and b†$behave like creation operators. An arbitrary excited state has the form

|&1 . . .&m,'1 . . . 'n; gnd" & a†#1. . . a†#m

b†$1 . . . b†$n|gnd" (6.90)

This state has m particles (in the single-particle states &1, . . . ,&m) and nholes (in the single-particle states '1, . . . ,'n). The ground state is annihi-lated by the operators a# and b$

a#|gnd" = b$|gnd" = 0 (& > N ' * N) (6.91)

The Hamiltonian H is normal ordered relative to the empty state |0", i. e.H|0" = 0, but is not normal ordered relative to the actual ground state |gnd".The particle-hole transformation enables us to normal order H relative to|gnd".

,H =!

#

E#a†#a# =

!

#%N

E# +!

#>N

E#a†#a# !

!

$%N

E$ b†$ b$ (6.92)

where the minus sign in the last term reflects the Femi statistics.Thus

,H = Egnd+ : ,H : (6.93)

where

Egnd =N!

#=1

E# (6.94)


is the ground state energy, and the normal ordered Hamiltonian is

: ,H :=!

#>N

E#a†aa# !

!

$%N

b†$ b$ E$ (6.95)

The number operator ,N is not normal-ordered relative to |gnd" either. Thus,we write

,N =!

#

a†#a# = N +!

#>N

a†#a# !!

$%N

b†$ b$ (6.96)

We see that particles raise the energy while holes reduce it. However, if wedeal with Hamiltonians that conserve the particle number ,N (i.e., [ ,N, H ] =0) for every particle that is removed a hole must be created. Hence particlesand holes can only be created in pairs. A particle-hole state |&,' gnd" is

|&,' gnd" & a†#b†$|gnd" (6.97)

It is an eigenstate with an energy

,H|&,' gnd" =:Egnd+ : ,H :

;a†#b

†$|gnd"

= (Egnd + E# ! E$) |&,' gnd"(6.98)

and the excitation energy is E# ! E$ - 0. Hence the ground state is stableto the creation of particle-hole pairs.

This state has exactly N particles since

,N |&,' gnd" = (N + 1! 1)|&,' gnd" = N |&,' gnd" (6.99)

Let us finally notice that the field operator %†(x) in position space is

%†(x) =!

#

)x|&"a†# =!

#>N

!#(x)a†# +

!

$%N

!$(x)b$ (6.100)

where {!#(x)} are the single particle wave functions.The procedure of normal ordering allows us to define the physical Hilbert

space. The physical meaning of this approach becomes more transparent inthe thermodynamic limit N $, and V $, at constant density ). In thislimit, the space of Hilbert space is the set of states which is obtained byacting with creation and annihilation operators finitely on the ground state.The spectrum of states that results from this approach consists on the set ofstates with finite excitation energy. Hilbert spaces which are built on refer-ence states with macroscopically di"erent number of particles are e"ectivelydisconnected from each other. Thus, the normal ordering of a Hamiltonian ofa system with an infinite number of degrees of freedom amounts to a choice


of the Hilbert space. This restriction becomes of fundamental importancewhen interactions are taken into account.

6.3.4 The Free Fermi Gas

Let us consider the case of free spin one-half electrons moving in free space.The Hamiltonian for this system is

H =

+ddx

!

&=&,'

%†&(x)[!

!2

2m#2 !µ]%&(x) (6.101)

where the label - =., / indicates the z-projection of the spin of the electron.The value of the chemical potential µ will be determined once we satisfythat the electron density is equal to some fixed value ).

In momentum space, we get

%&(x) =

+ddp

(2")d%&(p) e

"ip·x! (6.102)

where the operators %&(p) and %†&(p) satisfy

%%&(p), %

†&!(p)

&= (2")d(&&!(

d(p! p$)%%†&(p), %

†&!(p)

&= {%(p), %&!(p$)} = 0

(6.103)

The Hamiltonian has the very simple form

H =

+ddp

(2")d

!

&=i,'

(,(p)! µ) %†&(p)%&(p) (6.104)

where .(p) is given by

.(p) =p2

2m(6.105)

For this simple case, ,(p) is independent of the spin orientation.It is convenient to measure the energy relative to the chemical potential

(or Fermi energy) µ = EF . The relative energy E(p) is

E(p) = ,(p)! µ (6.106)

Hence, E(p) is the excitation energy measured from the Fermi energy EF =µ. The energy E(p) does not have a definite sign since there are states with


,(p) > µ as well as states with ,(p) < µ. Let us define by pF the value of|p| for which

E(pF ) = ,(pF )! µ = 0 (6.107)

This is the Fermi momentum. Thus, for |p| < pF , E(p) is negative while for|p| > pF , E(p) is positive.

We can construct the ground state of the system by finding the statewith lowest energy at fixed µ. Since E(p) is negative for |p| * pF , we seethat by filling up all of those states we get the lowest possible energy. Itis then natural to normal order the system relative to a state in which allone-particle states with |p| * pF are occupied. Hence we make the particle-hole transformation

b&(p) = %†&(p) for |p| * pF

a&(p) = %&(p) for |p| > pF(6.108)

In terms of the operators a& and b&, the Hamiltonian is

H =!

&=&,'

+ddp

(2")d[E(p)/(|p|!pF )a†&(p)a&(p)+/(pF ! |p|)E(p)b&(p)b

†&(p)]

(6.109)where /(x) is the step function

/(x) =

<1 x > 00 x * 0

(6.110)

Using the anticommutation relations in the last term we get

H =!

&=&,'

+ddp

(2")dE(p)[/(|p|!pF )a†&(p)a&(p)!/(pF!|(p)|)b†&(p)b&(p)]+Egnd

(6.111)where Egnd, the ground state energy measured from the chemical potentialµ, is given by

Egnd =!

&=&,'

+ddp

(2")d/(pF ! |p|)E(p)(2")d(d(0) = Egnd ! µN (6.112)

Recall that (2")d((0) is equal to

(2")d(d(0) = limp(0

(2")d(d(p) = limp(0

+ddxeip·x = V (6.113)

where V is the volume of the system. Thus, Egnd is extensive

Egnd = V ,gnd (6.114)


and the ground state energy density ,gnd is

,gnd = 2

+

|p|%pF

ddp

(2")dE(p) = ,gnd ! µ) (6.115)

where the factor of 2 comes from the two spin orientations. Putting every-thing together we get

,gnd = 2

+

|p|%pF

ddp

(2")d

1p2

2m! µ

2= 2

+ pF

0dp pd"1 Sd

(2")d

1p2

2m! µ

2

(6.116)where Sd is the area of the d-dimensional hypersphere. Our definitions tell us

that the chemical potential is µ =p2F2m & EF where EF , is the Fermi energy.

Thus the ground state energy density ,gnd (measured from the empty state)is equal to

Egnd =1

m

Sd

(2")d

+ pF

0dp pd+1 =

pd+2F

m(d+ 2)

Sd

(2")d= 2EF

pdFSd

(d+ 2)(2")d

(6.117)How many particles does this state have? To find that out we need to lookat the number operator. The number operator can also be normal-orderedwith respect to this state

N =

+ddp

(2")d

!

&=&,'

%†&(p)%&(p) =

=

+ddp

(2")d

!

&=&,'

{/(|p|! pF )a†&(p)a&(p) + /(pF ! |p|)b&(p)b†&(p)}

(6.118)

Hence, N can also be written in the form

N =: N : +N (6.119)

where the normal-ordered number operator : N : is

: N :=

+ddp

(2")d

!

&=&,'

[/(|p|!pF )a†&(p)a&(p)!/(pF!|p|)b†&(p)b&(p)] (6.120)

and N , the number of particles in the reference state |gnd", is

N =

+ddp

(2")d

!

&=&,'

/(pF ! |p|)(2")d(d(0) = 2

dpdF

Sd

(2")dV (6.121)


Therefore, the particle density ) = NV is

) =2

d

Sd

(2")dpdF (6.122)

This equation determines the Fermi momentum pF in terms of the density ).Similarly we find that the ground state energy per particle is

Egnd

N = 2dd+2EF .

The excited states can be constructed in a similar fashion. The state|+,-,p"

|+,-,p" = a†#(p)|gnd" (6.123)

is a state which represents an electron with spin - and momentum p whilethe state |!,-,p"

|!,-,p" = b†&(p)|gnd" (6.124)

represents a hole with spin - and momentum p. From our previous discussionwe see that electrons have momentum p with |p| > pF while holes havemomentum p with |p| < pF . The excitation energy of a one-electron stateis E(p) - 0(for |p| > pF ), while the excitation energy of a one-hole state is!E(p) - 0 (for |p| < pF ).

Similarly, an electron-hole pair is a state of the form

|-p,-$p$" = a†&(p)b†&!(p

$)|gnd" (6.125)

with |p| > pF and |p$| < pF . This state has excitation energy E(p)!E(p$),which is positive. Hence, states which are obtained from the ground statewithout changing the density, can only increase the energy. This provesthat |gnd" is indeed the ground state. However, if the density is allowedto change, we can always construct states with energy less than Egnd bycreating a number of holes without creating an equal number of particles.

non-relativistic field theory - university of...

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