6 Non-Relativistic Field Theory

Transcription

1 6 Non-Relativistic Field Theory In this Chapter we will discuss the field theory description of non-relativistic systems. The material that we will be presented here is, for the most part, introductory as this topic is covered in depth in many specialized textbooks, such as Methods of Quantum Field Theory in Statistical Physics by A. Abrikosov, L. Gorkov and I. Dzyaloshinskii (Prentice-Hall, 1963), Statistical Mechanics, A Set of lectures by R. Feynman (Benjamin, 1973) and many others. The discussion of second quantization is very standardand is presented her for pedagogical reasons but can be skipped. 6.1 Non-Relativistic Field Theories, Second Quantization and the Many-Body Problem Let us consider now the problem of a system of N identical non-relativistic particles. For the sake of simplicity I will assume that the physical state of each particle j is described by its position x j relative to some reference frame. This case is easy to generalize. The wave function for this system is Ψ(x 1,...,x N ). If the particles are identical then the probability density, Ψ(x 1,...,x N ) 2,mustbeinvariant (i.e., unchanged) under arbitrary exchanges of the labels that we use to identify (or designate) the particles. In quantum mechanics, the particles do not have well defined trajectories. Only the states of a physical system are well defined. Thus, even though at some initial time t 0 the N particles may be localized at a set of well defined positions x 1,...,x N,theywillbecome delocalized as the system evolves. Furthermore the Hamiltonian itself is invariant under a permutation of the particle labels. Hence, permutations constitute a symmetry of a many-particle quantum mechanical system. In other terms, identical particles are indistinguishable in quantum mechanics. In particular, the probability density of any eigenstate must remain invariant

2 176 Non-Relativistic Field Theory if the labels of any pair of particles are exchanged. If we denote by P jk the operator which exchanges the labels of particles j and k, the wave functions must satisfy P jk Ψ(x 1,...,x j,...,x k,...,x N )=e iφ Ψ(x 1,...,x j,...,x k,...,x N ) (6.1) Under a further exchange operation, the particles return to their initial labels and we recover the original state. This sample argument then requires that φ =0,π since 2φ must not be an observable phase. We then conclude that there are two possibilities: either Ψ is even under permutation and P Ψ=Ψ,or Ψisodd under permutation and P Ψ= Ψ. Systems of identical particles which have wave functions which are even under a pairwise permutation of the particle labels are called bosons. Intheothercase,Ψodd under pairwise permutation, they are Fermions. Itmustbestressedthat these arguments only show that the requirement that the state prioriψbe either even or odd is only a sufficient condition. It turns out that under special circumstances other options become available and the phase factor φ may take values different from 0 or π. Theseparticlesarecalledanyons. For the moment the only cases in which they may exist appears tobeinsituations in which the particles are restricted to move on a line oronaplane. In the case of relativistic quantum field theories, the requirement that the states have well defined statistics (or symmetry) is demanded by a very deep and fundamental theorem which links the statistics of the states of the spin of the field. This is the spin-statistics theorem which we will discuss later on Fock Space We will now discuss a procedure, known as Second Quantization, whichwill enable us to keep track of the symmetry of the states in a simpleway. Let us consider now a system of N particles. The wave functions in the coordinate representation are Ψ(x 1,...,x N )wherethelabelsx 1,...,x N denote both the coordinates and the spin states of the particles in the state Ψ. Forthesakeofdefinitenesswewilldiscussphysicalsystemsdescribable by Hamiltonians Ĥ of the form Ĥ = 2 2m N N 2 j + j=1 j=1 V (x j )+g j,k U(x j x k )+... (6.2) Let {φ 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

3 6.1 Non-Relativistic Field Theories, Second Quantization and the Many-Body Problem 177 tensor product of the one-particle states, namely Ψ(x 1,...,x N )= {nj} C(n 1,...n N )φ n1 (x 1 )...φ nn (x N ) (6.3) Thus, if Ψ is symmetric (antisymmetric) under an arbitrary exchange x j x k,thecoefficientsc(n 1,...,n N )mustbesymmetric(antisymmetric)under the exchange n j n k. AsetofN-particle basis states with well defined permutation symmetry is the properly symmetrized or antisymmetrized tensor product Ψ 1,...Ψ N Ψ 1 Ψ 2 Ψ N = 1 ξ P Ψ P (1) Ψ P (N) N! (6.4) where the sum runs over the set of all possible permutation P.Theweight factor ξ is +1 for bosons and 1 forfermions.noticethat,forfermions,the N-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 χ 1,...χ N ψ 1,...,ψ N = 1 ξ P +Q χ N! Q(1) ψ P (1) χ Q(N) ψ P (1) = P,Q = P ξ P χ 1 ψ P (1) χ N ψ P (N) P (6.5) which is nothing but the permanent (determinant) of the matrix χ j ψ k for symmetric (antisymmetric) states, i.e., χ 1 ψ 1... χ 1 ψ N χ 1,...χ N ψ 1,...ψ N =.. (6.6) χ N ψ 1... χ N ψ N ξ In the case of antisymmetric states, the inner product is the familiar Slater determinant. Letusdenoteby{ α } the complete set of one-particle states which satisfy α β = δ αβ α α =1 (6.7) The N-particle states are { α 1,...α N }. Becauseofthesymmetryrequirements, the labels α j can be arranged in the form of a monotonic sequence α 1 α 2 α N for bosons, orintheformofastrict monotonic sequence α 1 <α 2 < <α N for fermions. Letn j be an integer which counts how α

4 178 Non-Relativistic Field Theory many particles are in the j-th one-particle state. The boson states α 1,...α N must be normalized by a factor of the form 1 n1!...n N! α 1,...,α N (α 1 α 2 α N ) (6.8) and n j are non-negative integers. For fermions the states are α 1,...,α N (α 1 <α 2 < <α N ) (6.9) and n j =0> 1. These N-particle states are complete and orthonormal 1 N! α 1,...,α N α 1,...,α N α 1,...,α N = Î (6.10) where the sum over the α s is unrestricted and the operator Î is the identity operator in the space of N-particle states. We will now consider the more general problem in which the number of particles N is not fixed a-priori. Rather, we will consider an enlarged space of states in which the number of particles is allowed to fluctuate. In the language of Statistical Physics what we are doing is to go fromthecanonical Ensemble to the Grand Canonical Ensemble. Thus,letusdenotebyH 0 the Hilbert space with no particles, H 1 the Hilbert space with only one particle and, in general, H N the Hilbert space for N-particles. The direct sum of these spaces H H = H 0 H 1 H N (6.11) is usually called the Fock space.anarbitrarystate ψ in Fock space is the sum over the subspaces H N, ψ = ψ (0) + ψ (1) + + ψ (N) + (6.12) The subspace with no particles is a one-dimensional space spanned by the vector 0 which we will call the vacuum. Thesubspaceswithwelldefined number of particles are defined to be orthogonal to each other in the sense that the inner product in the Fock space χ ψ χ (j) ψ (j) (6.13) j=0 vanishes if χ and ψ belong to different subspaces.

5 6.1 Non-Relativistic Field Theories, Second Quantization and the Many-Body Problem Creation and Annihilation Operators Let φ be an arbitrary one-particle state. Letusdefinethecreation operator â (φ) byitsactiononanarbitrarystateinfockspace â (φ) ψ 1,...,ψ N = φ, ψ 1,...,ψ N (6.14) Clearly, â (φ) mapsthen-particle state with proper symmetry ψ 1,...,ψ N onto the N +1-particle state φ, ψ,..., ψ N,alsowithpropersymmetry. The destruction or annihilation operator â(φ) is defined to be the adjoint of â (φ), χ 1,...,χ N 1 â(φ) ψ 1,...,ψ N = ψ 1,...,ψ N â (φ) χ 1,...,χ N 1 (6.15) Hence χ 1,...,χ N 1 â(φ) ψ 1,...,ψ N ψ = 1,...,ψ N φ, χ 1,...,χ N 1 = ψ 1 φ ψ 1 χ 1 ψ 1 χ N 1 =... ψ N φ ψ N χ 1 ψ N χ N 1 ξ (6.16) We can now expand the permanent (or determinant) to get χ 1,...,χ N 1 â(φ) ψ 1,...,ψ N = N = ξ k 1 ψ k φ k=1 = ψ 1 χ 1... ψ 1 χ N (noψ k )... ψ N χ 1... ψ N χ N 1 N ξ k 1 ψ k φ χ 1,...,χ N 1 ψ 1,... ˆψ k...,ψ N k=1 ξ (6.17) where ˆψ k indicates that ψ k is absent. Thus,thedestructionoperatoris given by â(φ) ψ 1,...,ψ N = N ξ k 1 φ ψ k ψ 1,..., ˆψ k,...,ψ N (6.18) k=1 With these definitions, we can easily see that the operators â (φ) andâ(φ)

6 180 Non-Relativistic Field Theory obey the commutation relations â (φ 1 )â (φ 2 )=ξ â (φ 2 )â (φ 1 ) (6.19) Let us introduce the notation Â, ˆB] ξ  ˆB ξ ˆB (6.20) where  and ˆB are two arbitrary operators. For ξ =+1(bosons) wehave the commutator ] â (φ 1 ), â (φ 2 ) ]+1 â (φ 1 ), â (φ 2 ) =0 (6.21) while for ξ = 1 itistheanticommutator { } â (φ 1 ), â (φ 2 ) ] 1 â (φ 1 ), â (φ 2 ) =0 (6.22) Similarly for any pair of arbitrary one-particle states φ 1 and φ 2 we get â(φ 1 ), â(φ 2 )] ξ =0 (6.23) It is also easy to check that the following identity holds ] â(φ 1 ), â (φ 2 ) = φ 1 φ 2 (6.24) ξ So far we have not picked any particular representation. Let us consider the occupation number representation in which the states are labelled by the number of particles n k in the single-particle state k. Inthiscase,wehave n 1,...,n k,... n1 n2 1 n1!n 2!... {}}{{}}{ 1...1, (6.25) In the case of bosons, then j s can be any non-negative integer, while for fermions they can only be equal to zero or one. In general we have that if α is the αth single-particle state, then â α n 1,...,n α,... = n α +1 n 1,...,n α +1,... â α n 1,...,n α,... = n α n 1,...,n α 1,... (6.26) Thus for both fermions and bosons, â α annihilates all states with n α =0, while for fermions â α annihilates all states with n α =1. The commutation relations are â α, â β ]= â α, â β ] =0 ] â α, â β = δ αβ (6.27)

7 6.1 Non-Relativistic Field Theories, Second Quantization and the Many-Body Problem 181 for bosons, and {â α â β } = { } â β, â β =0 { } â α â β = δ αβ (6.28) } for fermions. Here, {Â, ˆB is the anticommutator of the operators  and ˆB {Â, ˆB}  ˆB + ˆB (6.29) If a unitary transformation is performed in the space of one-particle state vectors, then a unitary transformation is induced in the space of the operators themselves, i.e., if χ = α ψ + β φ, then â(χ) =α â(ψ)+β â(φ) â (χ) =αâ (ψ)+βâ (φ) (6.30) and we say that â (χ) transformsliketheket χ while â(χ) transformslike the bra χ. For example, we can pick as the complete set of one-particle states the momentum states { p }. Thisis momentumspace.withthischoicethe commutation relations are ] â (p), â (q) =â(p), â(q)] ξ =0 ξ ] â(p), â (q) ξ =(2π)d δ d (p q) (6.31) where d is the dimensionality of space. In this representation, an N-particle state is p 1,...,p N =â (p 1 )...â (p N ) 0 (6.32) On the other hand, we can also pick the one-particle states to be eigenstates of the position operators, i.e., x 1,...x N =â (x 1 )...â (x N ) 0 (6.33) In position space, the operators satisfy ] â (x 1 ), â (x 2 ) =â(x 1), â(x 2 )] ξ =0 ξ ] â(x 1 ), â (x 2 ) = ξ δd (x 1 x 2 ) (6.34)

8 182 Non-Relativistic Field Theory This is the position space or coordinate representation. A transformation from position space to momentum space is the Fourier transform p = d d x x x p = d d x x e ip x (6.35) and, conversely x = d d p (2π) d p e ip x (6.36) Then, the operators themselves obey â (p) = d d x â (x)e ip x â d d p (x) = (2π) d â (p)e ip x (6.37) General Operators in Fock Space Let A (1) be an operator acting on one-particle states. We can always define an extension of A acting on any arbitrary state ψ of the N-particle Hilbert space as follows:  ψ N ψ 1... A (1) ψ j... ψ N (6.38) j=1 For instance, if the one-particle basis states { ψ j } are eigenstates of A with eigenvalues {a j } we get N  ψ = ψ (6.39) j=1 We wish to find an expression for an arbitrary operator A in terms of creation and annihilation operators. Let us first consider the operator A (1) αβ = α β which acts on one-particle states. The operators A (1) αβ form a basis of the space of operators acting on one-particle states. Then, the N-particle extension of A αβ is  αβ ψ = a j N ψ 1 α ψ N β ψ j (6.40) j=1

9 6.1 Non-Relativistic Field Theories, Second Quantization and the Many-Body Problem 183 Thus  αβ ψ = j N {}}{ ψ 1,..., α,...,ψ N β ψ j (6.41) j=1 In other words, we can replace the one-particle state ψ j from the basis with the state α at the price of a weight factor, the overlap β ψ j.this operator has a very simple expression in terms of creation and annihilation operators. Indeed, â (α)â(β) ψ = N ξ k 1 β ψ k α, ψ 1,...,ψ k 1,ψ k+1,...,ψ N (6.42) k=1 We can now use the symmetry of the state to write ξ k 1 {}}{ α, ψ 1,...,ψ k 1,ψ k+1,...,ψ N = ψ 1,..., α,...,ψ N (6.43) Thus the operator A αβ,theextensionof α β to the N-particle space, coincides with â (α)â(β)  αβ â (α)â(β) (6.44) We can use this result to find the extension for an arbitrary operator A (1) of the form A (1) = α,β k α α A (1) β β (6.45) we find  = α,β â (α)â(β) α A (1) β (6.46) Hence the coefficients of the expansion are the matrix elements ofa (1) between arbitrary one-particle states. We now discuss a few operators of interest. The Identity Operator The Identity Operator ˆ1 oftheone-particlehilbertspace ˆ1 = α α α (6.47) becomes the number operator N N = α â (α)â(α) (6.48)

10 184 Non-Relativistic Field Theory In position and in momentum space we find d N d p = (2π) d â (p)â(p) = d d x â (x)â(x) = where ˆρ(x) =â (x)â(x) istheparticle density operator. The Linear Momentum Operator d d x ˆρ(x) (6.49) In the space H 1,thelinearmomentumoperatoris ˆp (1) d d p j = (2π) d p j p p = d d x x i j x (6.50) Thus, we get that the total momentum operator P j is d P d p j = (2π) d p jâ (p)â(p) = d d x â (x) i jâ(x) (6.51) The one-particle Hamiltonian H (1) has the matrix elements Hamiltonian H (1) = p 2 + V (x) (6.52) 2m x H (1) y = 2 2m 2 δ d (x y)+v (x)δ d (x y) (6.53) Thus, in Fock space we get ) Ĥ = d d x â (x) ( 2 2m 2 +V (x) â(x) (6.54) in position space. In momentum space we can define Ṽ (q) = d d xv(x)e iq x (6.55) the Fourier transform of the potential V (x), and get d d p p 2 Ĥ = 2mâ d d p d d q (2π) d (p)â(p)+ (2π) d (2π) d Ṽ (q)â (p + q)â(p) (6.56) The last term has a very simple physical interpretation. When actingona one-particle state with well-defined momentum, say p, thepotentialterm yields another one-particle state with momentum p + q, whereq is the momentum transfer, withamplitudeṽ (q). This process is usually depicted by the diagram of the following figure.

11 6.1 Non-Relativistic Field Theories, Second Quantization and the Many-Body Problem 185 â (p + q) p + q q Ṽ (q) â(p ) p Figure 6.1 One-body scattering. Two-Body Interactions Atwo-particleinteractionisaoperator ˆV (2) which acts on the space of two-particle states H 2,hastheform V (2) = 1 α, β V (2) (α, β) α, β (6.57) 2 α,β The methods developed above yield an extension of V (2) to Fock space of the form V = 1 2 α,β In position space, ignoring spin, we get V = â (α)â (β)â(β)â(α) V (2) (α, β) (6.58) d d x d d y â (x) â (y) â(y) â(x) V (2) (x, y) d d x d d y ˆρ(x)V (2) (x, y)ˆρ(y)+ 1 d d xv (2) (x, x) ˆρ(x) 2 (6.59) while in momentum space we find V = 1 2 d d p (2π) d d d q (2π) d d d k (2π) d Ṽ (k) â (p + k)â (q k)â(q)â(p) (6.60)

12 186 Non-Relativistic Field Theory where Ṽ (k) isonlyafunctionofthemomentumtransfer k.thisisaconsequence of translation invariance. In particular for a Coulomb interaction, for which we have V (2) (x, y) = e2 x y (6.61) Ṽ (k) = 4πe2 k 2 (6.62) â (p + k) p + k q k â (q k) k Ṽ (k) â(p) p q â(q) 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 nonrelativistic field theory. The procedure described in the previous section is commonly known as Second Quantization. If the (identical) particles are bosons, the operators â(φ) obey canonical commutation relations. If the (identical) particles are Fermions, theoperatorsâ(φ) obeycanonical anticommutation relations. Inpositionspace,itiscustomarytorepresentâ (φ) by the operator ˆψ(x) whichobeystheequal-timealgebra ˆψ(x), (y)] ˆψ = ξ δd (x y) ] ˆψ(x), ˆψ(y) = ˆψ (x), ˆψ ] (y) =0 ξ ξ (6.63) In this framework, the one-particle Schrödinger equation becomes the clas-

13 sical field equation 6.3 Non-Relativistic Fermions at Zero Temperature 187 i ] 2 + t 2m 2 V (x) ˆψ =0 (6.64) Can we find a Lagrangian density L from which the one-particle Schrödinger equation follows as its classical equation of motion? The answer is yes and L is given by Its Euler-Lagrange equations are L = i ψ t ψ 2 2m ψ ψ V (x)ψ ψ (6.65) t δl δ t ψ = δl δ ψ + δl δψ (6.66) which are equivalent to the field Equation Eq The canonical momenta Π ψ and Π ψ are and Π ψ = Π ψ = δl = i ψ (6.67) δ t ψ δl δ t ψ = i ψ (6.68) Thus, the (equal-time) canonical commutation relations are ˆψ(x), ˆΠψ (y)] = i δ(x y) (6.69) which require that ˆψ(x), ˆψ (y)] ξ ξ = δd (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-relativistic fermions (with two-body interactions) is (in D = d +1space-time dimensions) S = 1 2 d D x ˆψ i t ˆψ 2 d D x ] 2m ˆψ ˆψ V (x) ˆψ (x) ˆψ(x) d D x ˆψ (x) ˆψ (x )U(x x ) ˆψ(x ) ˆψ (x) ˆψ(x) (6.71)

14 188 Non-Relativistic Field Theory where U(x x )representsinstantaneouspair-interactions, U(x x ) U(x x )δ(x 0 x 0) (6.72) The Hamiltonian Ĥ for this system is Ĥ = d d x 2 2m ˆψ ˆψ + V (x) ˆψ (x)ψ(x)] + 1 d d x d d x 2 ˆψ (x) ˆψ (x )U(x x ) ˆψ(x ) ˆψ(x) (6.73) For Fermions the fields ˆψ and ˆψ satisfy equal-time canonical anticommutation relations while for Bosons they satisfy { ˆψ(x), ˆψ (x)} = δ(x x ) (6.74) ˆψ(x), ˆψ (x )] = δ(x x ) (6.75) In both cases, the Hamiltonian Ĥ commutes with the total number operator ˆN = d d x ˆψ (x) ˆψ(x) sinceĥ conserves the total number of particles. The Fock space picture of the many-body problem is equivalent to the Grand Canonical Ensemble of Statistical Mechanics. Thus, instead offixingthe number of particles we can introduce a Lagrange multiplier µ, thechemical potential, to weigh contributions from different parts of the Fock space. Thus, we define the operator H. H Ĥ µ N (6.76) In a Hilbert space with fixed ˆN this amounts to a shift of the energy by µn. WewillnowallowthesystemtochoosethesectoroftheFockspace but with the requirement that the average number of particles ˆN is fixed to be some number N. Inthethermodynamiclimit(N ), µ represents the difference of the ground state energies between two sectors withn + 1 and N particles respectively. The modified Hamiltonian H is H = d d x σ + 1 d d x 2 ˆψ σ (x) ( 2 2m 2 +V (x) µ d d y σ,σ ) ˆψ σ (x) ˆψ σ (x) ˆψ σ (y)u(x y) ˆψ σ (y) ˆψ σ (x) (6.77)

15 6.3 Non-Relativistic Fermions at Zero Temperature The Ground State Let us discuss now the very simple problem of finding the ground statefora system of N spinless free fermions. In this case, the pair-potential vanishes and, if the system is isolated, so does the potential V (x). In general there will be a complete set of one-particle states { α } and, in this basis, Ĥ is Ĥ = α E α â αa α (6.78) where the index α labels the one-particle states by increasing order of their single-particle energies E 1 E 2 E n (6.79) Since were are dealing with fermions, we cannot put more than one particle in each state. Thus the state the lowest energy is obtained by filling up all the first N single particle states. Let gnd denote this ground state gnd = N N â α 0 â 1 â N 0 = {}}{ 1...1, (6.80) α=1 The energy of this state is E gnd with E gnd = E E N (6.81) The energy of the top-most occupied single particle state, E N,iscalledthe Fermi energy of the system and the set of occupied states is called the filled Fermi sea. Astatelike ψ Excited States N 1 {}}{ ψ = (6.82) is an excited state. It is obtained by removing one particle from the single particle state N (thus leaving a hole behind) and putting the particle in the unoccupied single particle state N +1. This is a state with one particle-hole pair, andithastheform The energy of this state is =â N+1âN gnd (6.83) E ψ = E E N 1 + E N+1 (6.84)

16 190 Non-Relativistic Field Theory E Single Particle Energy E F Fermi Energy Fermi Sea Single Particle States Occupied States Empty States Figure 6.3 The Fermi Sea. gnd N N +1 Single Particle States â N+1âN Ψ Single Particle States Figure 6.4 An excited (particle-hole) state. Hence E ψ = E gnd + E N+1 E N (6.85) and, since E N+1 E N,E ψ E gnd.theexcitation energy ε ψ = E ψ E gnd is ε ψ = E N+1 E N 0 (6.86) 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, it is physically more reasonable to use instead the filled Fermi sea gnd as the

17 6.3 Non-Relativistic Fermions at Zero Temperature 191 physical reference state or vacuum state. Thusthisstateisavacuuminthe sense of absence of excitations. Theseargumentsmotivatetheintroduction of the particle-hole transformation. Let us introduce the fermion operators b α such that ˆbα =â α for α N (6.87) Since â α gnd =0(forα N) theoperatorsˆb α annihilate the ground state gnd, i.e., ˆbα gnd =0 (6.88) The following anticommutation relations hold {âα, â α} = {â α, ˆb } β = {ˆbβ, ˆb } { β = â α, ˆb } β =0 } {â α, â α = δ αα, {ˆbβ, ˆb } β = δ ββ (6.89) where α, α >N and β,β N. Thus,relativetothestate gnd, â α and ˆb β behave like creation operators. An arbitrary excited state has the form α 1...α m,β 1...β n ;gnd â α 1...â α mˆb β 1...ˆb β n gnd (6.90) This state has m particles (in the single-particle states α 1,...,α m )andn holes (in the single-particle states β 1,...,β n ). The ground state is annihilated by the operators â α and ˆb β â α gnd = ˆb β gnd =0 (α>nβ N) (6.91) The Hamiltonian Ĥ is normal ordered relative to the empty state 0, i. e. Ĥ 0 =0,butisnotnormalorderedrelativetotheactualgroundstate gnd. The particle-hole transformation enables us to normal order Ĥ relative to gnd. Ĥ = α E α â αâ α = α N E α + α>n E α â αâ α β N E βˆb βˆb β (6.92) where the minus sign in the last term reflects the Femi statistics. Thus Ĥ = E gnd +: Ĥ : (6.93) where N E gnd = E α (6.94) α=1

18 192 Non-Relativistic Field Theory is the ground state energy, and the normal ordered Hamiltonian is : Ĥ := α>n E α â aâα β N ˆb βˆb β E β (6.95) The number operator N is not normal-ordered relative to gnd either. Thus, we write N = â αâ α = N + â αa α ˆb βˆb β (6.96) α α>n β N We see that particles raise the energy while holes reduce it. However, if we deal with Hamiltonians that conserve the particle number N (i.e., N,Ĥ] = 0) for every particle that is removed a hole must be created. Hence particles and holes can only be created in pairs. Aparticle-holestate α, β gnd is α, β gnd â αˆb β gnd (6.97) It is an eigenstate with an energy ( ) Ĥ α, β gnd = E gnd +: Ĥ : â αˆb β gnd =(E gnd + E α E β ) α, β gnd (6.98) and the excitation energy is E α E β 0. Hence the ground state is stable to the creation of particle-hole pairs. This state has exactly N particles since N α, β gnd =(N +1 1) α, β gnd = N α, β gnd (6.99) Let us finally notice that the field operator ˆψ (x) inpositionspaceis ˆψ (x) = α x α â α = α>n φ α (x)â α + β N φ β (x)ˆb β (6.100) where {φ α (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 in the thermodynamic limit N and V at constant density ρ. In this limit, the space of Hilbert space is the set of states which is obtained by acting with creation and annihilation operators finitely on the ground state. The spectrum of states that results from this approach consists on the set of states with finite excitation energy. Hilbert spaces which are built on reference states with macroscopically different number of particles are effectively disconnected from each other. Thus, the normal ordering of a Hamiltonian of asystemwithaninfinitenumberofdegreesoffreedomamountsto a choice

19 6.3 Non-Relativistic Fermions at Zero Temperature 193 of the Hilbert space. This restriction becomes of fundamental importance when interactions are taken into account 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 = d d x ˆψ σ(x) 2 2m 2 µ] ˆψ σ (x) (6.101) σ=, where the label σ =, indicates the z-projection of the spin of the electron. The value of the chemical potential µ will be determined once we satisfy that the electron density is equal to some fixed value ρ. In momentum space, we get ˆψ σ (x) = d d p (2π) d ˆψ p x i σ (p) e (6.102) where the operators ˆψ σ (p) and ˆψ σ(p) satisfy { ˆψσ (p), ˆψ σ (p)} =(2π) d δ σσ δ d (p p ) { ˆψ σ (p), ˆψ σ (p)} = { ˆψ(p), ˆψ σ (p )} =0 (6.103) The Hamiltonian has the very simple form d H d p = (2π) d (ε(p) µ) ˆψ σ (p) ˆψ σ (p) (6.104) where ɛ(p) isgivenby σ=î, ɛ(p) = p2 2m (6.105) For this simple case, ε(p) is independent of the spin orientation. It is convenient to measure the energy relative to the chemical potential (or Fermi energy) µ = E F.TherelativeenergyE(p) is E(p) =ε(p) µ (6.106) Hence, E(p) istheexcitationenergymeasuredfromthefermienergye F = µ. TheenergyE(p) doesnothaveadefinitesignsincetherearestateswith

20 194 Non-Relativistic Field Theory ε(p) >µas well as states with ε(p) <µ.letusdefinebyp F the value of p for which E(p F )=ε(p F ) µ =0 (6.107) This is the Fermi momentum.thus,for p <p F,E(p) isnegative while for p >p F,E(p) ispositive. We can construct the ground state of the system by finding the state with lowest energy at fixed µ. SinceE(p) isnegativefor p p F,wesee that by filling up all of those states we get the lowest possible energy. It is then natural to normal order the system relative to a state in which all one-particle states with p p F are occupied. Hence we make the particlehole transformation ˆbσ (p) = ˆψ σ(p) for p p F â σ (p) = ˆψ (6.108) σ (p) for p >p F In terms of the operators â σ and ˆb σ,thehamiltonianis H = d d p (2π) d E(p)θ( p p F )â σ(p)â σ (p)+θ(p F p )E(p)ˆb σ (p)ˆb σ(p)] σ=, where θ(x) isthestepfunction θ(x) = { 1 x>0 0 x 0 (6.109) (6.110) Using the anticommutation relations in the last term we get H = d d p (2π) d E(p)θ( p p F )â σ (p)â σ(p) θ(p F (p) )ˆb σ (p)ˆb σ (p)]+ẽ gnd σ=, (6.111) where Ẽ gnd,thegroundstateenergymeasuredfromthechemicalpotential µ, isgivenby Ẽ gnd = σ=, d d p (2π) d θ(p F p )E(p)(2π) d δ d (0) = E gnd µn (6.112) Recall that (2π) d δ(0) is equal to (2π) d δ d (0) = lim (2π) d δ d (p) = lim p 0 p 0 d d xe ip x = V (6.113) where V is the volume of the system. Thus, Ẽ gnd is extensive Ẽ gnd = V ε gnd (6.114)

21 6.3 Non-Relativistic Fermions at Zero Temperature 195 and the ground state energy density ε gnd is ε gnd =2 p p F d d p (2π) d E(p) =ε gnd µ ρ (6.115) where the factor of 2 comes from the two spin orientations. Putting everything together we get d d ( ) p p 2 pf ( ) ε gnd =2 p p F (2π) d 2m µ =2 dp p d 1 S d p 2 0 (2π) d 2m µ (6.116) where S d is the area of the d-dimensional hypersphere. Our definitions tell us that the chemical potential is µ = p2 F 2m E F where E F,istheFermienergy. Thus the ground state energy density ε gnd (measured from the empty state) is equal to S d E gnd = 1 m (2π) d pf 0 pd+2 F S d dp p d+1 = m(d +2) (2π) d =2E F p d F S d (d +2)(2π) d (6.117) How many particles does this state have? To find that out we need tolook at the number operator. The number operator can also be normal-ordered with respect to this state ˆN = = d d p (2π) d d d p (2π) d σ=, ˆψ σ(p) ˆψ σ (p) = {θ( p p F )â σ (p)â σ(p)+θ(p F p )ˆb σ (p)ˆb σ (p)} σ=, Hence, ˆN can also be written in the form (6.118) ˆN =: ˆN :+N (6.119) where the normal-ordered number operator : ˆN :is d : ˆN d p := (2π) d θ( p p F )â σ(p)â σ (p) θ(p F p )ˆb σ(p)ˆb σ (p)] (6.120) σ=, and N, thenumberofparticlesinthereferencestate gnd, is d d p N = (2π) d θ(p F p )(2π) d δ d (0) = 2 S d d pd F (2π) d V (6.121) σ=,

22 196 Non-Relativistic Field Theory Therefore, the particle density ρ = N V is S d ρ = 2 d (2π) d pd F (6.122) This equation determines the Fermi momentum p F in terms of the density ρ. Similarly we find that the ground state energy per particle is E gnd N = 2d d+2 E F. The excited states can be constructed in a similar fashion. The state +,σ,p +,σ,p =â α(p) gnd (6.123) is a state which represents an electron with spin σ and momentum p while the state,σ,p,σ,p = ˆb σ(p) gnd (6.124) representsa hole with spinσ and momentum p.fromourpreviousdiscussion we see that electrons have momentum p with p >p F while holes have momentum p with p <p F.Theexcitationenergyofaone-electronstate is E(p) 0(for p >p F ), while the excitation energy of a one-hole state is E(p) 0(for p <p F ). Similarly, an electron-hole pair is a state of the form σp,σ p =â σ(p)ˆb σ (p ) gnd (6.125) with p >p F and p <p F.ThisstatehasexcitationenergyE(p) E(p ), which is positive. Hence, states which are obtained from the ground state without changing the density, can only increase the energy. This proves that gnd is indeed the ground state. However, if the density is allowed to change, we can always construct states with energy less than E gnd by creating a number of holes without creating an equal number of particles.