1.1 Creation and Annihilation Operators in Quantum Mechanics

Transcription

1 Chapter Second Quantization. Creation and Annihilation Operators in Quantum Mechanics We will begin with a quick review of creation and annihilation operators in the non-relativistic linear harmonic oscillator. Let a and a be two operators acting on an abstract Hilbert space of states, and satisfying the commutation relation [ a, a ] = (.) where by we mean the identity operator of this Hilbert space. The operators a and a are not self-adjoint but are the adjoint of each other. Let α be a state which we will take to be an eigenvector of the Hermitian operators a a with eigenvalue α which is a real number, a a α = α α (.2) Hence, α = α a a α = a α 2 0 (.3) where we used the fundamental axiom of Quantum Mechanics that the norm of all states in the physical Hilbert space is positive. As a result, the eigenvalues α of the eigenstates of a a must be non-negative real numbers. Furthermore, since for all operators A, B and C we get [AB, C] = A[B, C] + [A, C] B (.4) [ a a, a ] = a (.5) [ a a, a ] = a (.6)

2 2 CHAPTER. SECOND QUANTIZATION i.e., a and a are eigen-operators of a a. Hence, ( a a ) a = a ( a a ) (.7) ( a a ) a = a ( a a + ) (.8) Consequently we find ( a a ) a α = a ( a a ) α = (α )a α (.9) Hence the state a α is an eigenstate of a a with eigenvalue α, provided a α 0. Similarly, a α is an eigenstate of a a with eigenvalue α+, provided a α 0. This also implies that α = α + = α a α (.0) α + a α (.) Let us assume that a n α 0, n Z + (.2) Hence, a n α is an eigenstate of a a with eigenvalue α n. However, α n < 0 if α < n, which contradicts our earlier result that all these eigenvalues must be non-negative real numbers. Hence, for a given α there must exist an integer n such that a n α 0 but a n+ α = 0, where n Z +. Let where But α n = a n α an α a a α n = (α n) α n (.3) a n α n = α n (.4) a α n = a n α an+ α = 0 α = n (.5) In other words the allowed eigenvalues of a a are the non-negative integers. Let us now define the ground state 0, as the state annihilated by a, Then, an arbitrary state n is a 0 = 0 (.6) n = n! (a ) n 0 (.7) which has the inner product n m = δ n,m n! (.8)

3 .. CREATION AND ANNIHILATION OPERATORS IN QUANTUM MECHANICS3 In summary, we found that creation and annihilation operators obey and thus their matrix elements are a n = n + n + (.9) a n = n n (.20) a a n = n n (.2) (.22) m a n = n + δ m,n+ m a n = n δ m,n (.23).. The Linear harmonic Oscillator The Hamiltonian of the Linear Harmonic Oscillator is H = P 2 2m + 2 mω2 X 2 (.24) where X and P, the coordinate and momentum Hermitian operators satisfy canonical commutation relations, [X, P] = i (.25) We now define the creation and annihilation operators a and a as a = [ ] mω 2 X + i P mω a = [ ] mω 2 X i P mω (.26) (.27) which satisfy [ a, a ] = (.28) Since X = P = the Hamiltonian takes the simple form ( H = ω a a + ) 2 ( a + a ) (.29) 2mω ( ) mω a a (.30) 2 i (.3) The eigenstates of the Hamiltonian are constructed easily using our results since all eigenstates of a a are eigenstates of H. Thus, the eigenstates of H are the eigenstates of a a, ( H n = ω n + ) n (.32) 2

4 4 CHAPTER. SECOND QUANTIZATION with eigenvalues ( E n = ω n + ) 2 (.33) where n = 0,,.... The ground state 0 is the state annihilated by a, Since a 0 [ ] mω 2 X + i P 0 = 0 (.34) mω x P φ = i d x φ (.35) dx we find that ψ 0 (x) = x 0 satisfies ( x + ) d ψ 0 (x) = 0 (.36) mω dx whose (normalized) solution is ψ 0 (x) = The wave functions ψ n (x) of the excited states n are ψ n (x) = x n = n! x (a ) n 0 = ( mω n! 2 ( mω ) /4 mω e 2 x2 (.37) π ) n/2 ( x mω ) n d ψ 0 (x) (.38) dx Creation and annihilation operators are very useful. Let us consider for instance the anharmonic oscillator whose Hamiltonian is H = P 2 2m + 2 mω2 X 2 + λx 4 (.39) Let us compute the eigenvalues E n to lowest order in perturbation theory in powers of λ. The first order shift E n is E n = λ n X 4 n + O(λ 2 ) ( ) 2 = λ n (a + a ) 4 n mω ( ) 2 { = λ n a a aa n + other terms with two a s and two a s } mω ( ) 2 ( = λ 6n 2 + 6n + 3 ) +... (.40) 2mω

5 .. CREATION AND ANNIHILATION OPERATORS IN QUANTUM MECHANICS5..2 Many Harmonic Oscillators It is trivial to extend these ideas to the case of many harmonic oscillators, which is a crude model of an elastic solid. Consider a system of N identical linear harmonic oscillators of mass M and frequency ω, with coordinates {Q i } and momenta {P i }, where i =,...,N. These operators satisfy the commutation relations [Q j, Q k ] = [P j, P k ] = 0, [Q j, P k ] = i δ jk (.4) where j, k =,..., N. The Hamiltonian is H = N i= P 2 i 2M i + 2 N V ij Q i Q j (.42) i,j= where V ij is a symmetric positive definite matrix, V ij = V ji. We will find the spectrum (and eigenstates) of this system by changing variables to normal modes and using creation and annihilation operators for the normal modes. To this end we will first rescale coordinates and momenta so as to absorb the particle mass M: x i = M i Q i (.43) p i = U ij = P i Mi (.44) Mi M j V ij (.45) which also satisfy and H = [x j, p j ] = i δ jk (.46) N i= p 2 i N U ij x i x j (.47) i,j= We now got to normal mode variables by means of an orthogonal transformation C jk, i.e. [C ] jk = C kj, N N x k = C kj x j, p k = C kj p j j= N C ki C ji = δ kj, i= j= N C ik C ij = δ kj (.48) Sine the matrix U ij is real symmetric and positive definite, its eigenvalues, which we will denote by ωk 2 (with k =,...,N), are all non-negative, ω2 k 0. The eigenvalue equation is i= N C ki C lj U ij = ωk 2 δ kl, (no sum over k) (.49) i,j=

6 6 CHAPTER. SECOND QUANTIZATION Since the transformation is orthogonal, it preserves the commutation relations and the Hamiltonian is now diagonal [ x j, x k ] = [ p j, p k ] = 0, [ x j, p k ] = i δ jk (.50) H = 2 N 2 ( p j + ωj 2 j) x2 i= (.5) We now define creation and annihilation operators for the normal modes ( a j = ωj x j + i ) p j 2 ωj a j = x j = p j = i ( ωj x j i ) p j 2 ωj ( ) a j + a j 2ω j ωj 2 ( ) a j a j (.52) where, once again, [ ] [a j, a k ] a j, a k = 0, [ ] a j, a k = δjk (.53) and the normal mode Hamiltonian takes the standard form H = N j= ( ω j a j a j + ) 2 (.54) The eigenstates of the Hamiltonian are labelled by the eigenvalues of a j a j for each normal mode j, n,...,n N {n j }. Hence, H n,...,n N = N j= ( ω j n j + ) n,...,n N (.55) 2 where N n,...,n N = j= (a j )nj nj! 0,...,0 (.56) The ground state of the system, which we will denote by 0, is the state in which all normal modes are in their ground state, 0 0,..., 0 (.57)

7 .. CREATION AND ANNIHILATION OPERATORS IN QUANTUM MECHANICS7 Thus, the ground state 0 is annihilated by the annihilation operators of all normal modes, and the ground state energy of the system E gnd is The energy of the excited states is a j 0 = 0, j (.58) E gnd = E(n,...,n N ) = N j= 2 ω j (.59) N ω j n j + E gnd (.60) We can now regard the state 0 as the vacuum state and the excited states n,...,n N as a state with n j excitations (or particles) of type j. In this context, the excitations are called phonons. A single-phonon state of type j will be denoted by j, is This state is an eigenstate of H, j= j = a j 0 = 0,...,0, j, 0,...,0 (.6) H j = Ha j 0 = ( ω j + E gnd )a j 0 = ( ω j + E gnd ) j (.62) with excitation energy ω j. Hence, an arbitrary state n,..., n N can also be regarded as a collection of non-interacting particles (or excitations), each carrying an energy equal to the excitation energy (relative to the ground state energy). The total number of phonons in a given state is measured by the number operator N ˆN = a j a j (.63) j= Notice that although the number of oscillators is fixed (and equal to N) the number of excitations may differ greatly from one state to another. We now note that the state n,..., n N can also be represented as n,...,n k,... n n 2 n!n 2!... {}}{{}}{..., (.64) We will see below that this form appears naturally in the quantization of systems of identical particles. Eq.(.64) is symmetric under the exchange of labels of the phonons. Thus, phonons are bosons.

8 8 CHAPTER. SECOND QUANTIZATION.2 The Quantized Elastic Solid We will now consider the problem of an elastic solid in the approximation of continuum elasticity, i.e. we will be interested in vibrations on wavelengths long compared to the inter-atomic spacing. At the classical level, the physical state of this system is determined by specifying the local three-component vector displacement field u( r, t), which describes the local displacement of the atoms away from their equilibrium positions, and by the velocities of the atoms, u t ( r, t). The classical Lagrangian L for an isotropic solid is L = d 3 r ρ 2 ( ) 2 u t 2 d 3 r [K i u j i u j + Γ i u i j u j ] (.65) where ρ is the mass density, K and Γ are two elastic moduli. The classical equations of motion are δl = ( ) δl δu i t δ u i (.66) which have the explicit form of a wave equation ρ 2 u i t 2 K 2 u i Γ i u = 0 (.67) We now define the canonical momentum Π i ( r, t) Π i ( r, t) = δl δ u i = ρ u i (.68) which in this case coincides with the linear momentum density of the particles. The classical Hamiltonian H is H = d 3 rπ i ( r, t) u [ ] Π ( r, t) L = d 3 2 r t 2ρ + K 2 ( i u) 2 + Γ ( ) 2 u 2 (.69) In the quantum theory, the displacement field u( r) and the canonical momentum Π( r) become operators acting on a Hilbert space, obeying the equal-time commutation relations [ u i ( r), u j ( R) ] [ = Π i ( r), Π j ( R) ] [ = 0, u i ( r), Π j ( R) ] = i δ 3 ( r R)δ ij (.70) Due to the translational invariance of the continuum solid, the canonical transformation to normal modes is a Fourier transform. Thus we write ũ i ( p) = d 3 r e i p r u i ( r) (.7) Π i ( p) = d 3 r e i p r Π i ( r) (.72)

9 .2. THE QUANTIZED ELASTIC SOLID 9 Using the representations of the Dirac δ-function δ 3 ( r R) d 3 p = ei p ( r R) (2π) 3 (.73) (2π) 3 δ 3 ( p q) = d 3 r e i( p q) r (.74) we can write u i ( r) = ρ Π i ( r) = ρ d 3 p (2π) 3 ei p r ũ i ( p) (.75) d 3 p (2π) 3 ei p r Πi ( p) (.76) where we have scaled out the density ρ or future convenience. On the other hand, since u i ( r) and Π i ( r) are real (and Hermitian) u i ( r) = u i ( r), Π i( r) = Π i ( r) (.77) their Fourier transformed fields ũ i ( p) and Π i ( p) obey ũ i ( p) = ũ i( p), with equal-time commutation relations [ũ j ( p), ũ k ( q)] = [ Πj ( p), Π k ( q)] = 0, where Π i ( p) = Π i ( p) (.78) [ ũ j ( p), Π ] k ( q) = i (2π) 3 δ 3 ( p + q)δ jk (.79) In terms of the Fourier transformed fields the Hamiltonian has the form H = d 3 p (2π) 3 The 3 3 matrix ωij 2 has two eigenvalues:. 2. ( 2 Π i ( p) Π i ( p) + 2 ω2 ij ( p)ũ i( p)ũ j ( p) ) (.80) ω 2 ij ( p) = K ρ p 2 δ ij + Γ ρ p ip j (.8) with eigenvector parallel to the unit vector p/ p ( ) K + Γ ωl( p) 2 = p 2 (.82) ρ ω 2 T( p) = ( ) K p 2 (.83) ρ with a two-dimensional degenerate space spanned by the mutually orthogonal unit vectors e ( p) and e 2 ( p), both orthogonal to p: e α ( p) p = 0, (α =, 2), e ( p) e 2 ( p) = 0 (.84)

10 0 CHAPTER. SECOND QUANTIZATION We can now expand the displacement field ũ( p) into a longitudinal component and two transverse components ũ L i ( p) = ũ L ( p) p i p ũ T i ( p) = α=,2 (.85) ũ T α( p)e α i ( p) (.86) The canonical momenta Π i ( p) can also be expanded into one longitudinal component Π L ( p) and two transverse components Π T α( p). As a result the Hamiltonian can be decomposed into a sum of two terms, H = H L + H T (.87) H L involves only the longitudinal component of the field and momenta H L = d 3 p } { ΠL 2 (2π) 3 ( p) Π L ( p) + ωl 2 ( p) ũ L( p)ũ L ( p) (.88) H T involves only the transverse components of the field and momenta H T = d 3 p } { ΠT 2 (2π) 3 α ( p) Π T α ( p) + ω2 T ( p) ũt α ( p)ũt α ( p) (.89) α=,2 We can now define creation and annihilation operators for both longitudinal and transverse components ( ) ωl i a L ( p) = ( p) ũ L ( p) + Π L ( p) 2 ωl ( p) a L ( p) = a α T( p) = a α T ( p) = ( ωl i ( p) ũ L ( p) 2 ωl ( p) ( ωt ( p) ũ α i T( p) + 2 ωt ( p) ( ωt ( p) ũ α T ( p) i 2 ωt ( p) Π L ( p) Π α T( p) ) ) Π α T ( p) ) which obey standard commutation relations: [a L ( p), a L ( q)] = [ a L ( p), a L ( q) ) ] = 0 [ al ( p), a L ( q) ] = (2π) 3 δ 3 ( p q) ] [a α T ( p), aβt ( q) = [ a α T ( p), aβ T ( q) ] = [ ] a α T ( p), a β T ( q) ) = 0 (2π) 3 δ 3 ( p q)δ αβ [a L ( p), a α T( q)] = [ a L ( p), a α T( q) ) ] = 0 (.90) (.9)

11 .2. THE QUANTIZED ELASTIC SOLID where α, β =, 2. Conversely we also have ũ L ( p) = 2ω L ( p) ( al ( p) + a L ( p) ) Π L ( p) = ũ α T( p) = Π α T ( p) = ωl ( p) ( i al ( p) a L ( p) ) 2 ( a α 2ω T ( p) T ( p) + a α T( p) ) i ωt ( p) 2 ( a α T ( p) a α T ( p) ) (.92) The Hamiltonian now reads H = d 3 p 2 (2π) 3 ω L( p) ( a L ( p) a L ( p) + a L ( p)a L ( p) ) + d 3 p 2 (2π) 3 ω T( p) ( a α T ( p) a α T( p) + a α T( p)a α T( p) ) α=,2 ( d 3 p = E gnd + (2π) 3 ω L ( p) a L ( p) a L ( p) + ) ω T ( p) a α T( p) a α T( p) The ground state 0 has energy E gnd, where we have used that E gnd = V α=,2 (.93) d 3 p (2π) 3 2 ( ω L( p) + 2 ω T ( p)) (.94) lim p 0 δ3 ( p) = V (2π) 3 (.95) As before the ground state is annihilated by all annihilation operators a L ( p) 0 = 0, a α T ( p) 0 = 0 (.96) and it will be regarded as the state without phonons. There are three one-phonon states with wave vector p: One longitudinal phonon state L, p = a L ( p) 0 (.97) with energy E L ( p) E L ( p) = ω L ( p) = v L p (.98)

12 2 CHAPTER. SECOND QUANTIZATION where v L = K + Γ ρ (.99) is the speed of the longitudinal phonon, Two transverse phonon states T, α, p = a α T( p) 0 (.00) one for each polarization, with energy E T ( p) E T ( p) = ω T ( p) = v T p (.0) where v T = K ρ (.02) is the speed of the transverse phonons. The energies of the longitudinal and transverse phonon we found vanish as p 0. These are acoustic phonons and the speeds v L and v T are speeds of sound. Notice that if the elastic modulus Γ = 0 all three states become degenerate. If in addition we were to have considered the effects of lattice anisotropies, the two transverse branches may no longer be degenerate as in this case. Similarly we can define multi-phonon states, with either polarization. For instance a state with two longitudinal phonons with momenta p and q is L, p, q = a L ( p) a L ( q) 0 (.03) This state has energy ω L ( p) + ω L ( q) above the ground state. Finally, let us define the linear momentum operator P, for phonons of either longitudinal or transverse polarization, P = d 3 p (2π) 3 p ( a L ( p) a L ( p) + α=,2 This operator obeys the commutation relations [ P, al ( k) ] = ka L ( k) a α T( p) a α T( p) ) (.04) [ P, a α T ( k) ] = ka α T ( k) [ P, al ( k)] = ka L ( k) [ P, a α T ( k)] = ka α T( k) (.05)

13 .2. THE QUANTIZED ELASTIC SOLID 3 and commutes with the Hamiltonian [ P, H ] = 0 (.06) Hence P is a conserved quantity. Moreover, using the commutation relations and the expressions for the displacement fields it is easy to show that [ P, ui ( x)] = i u i ( x) (.07) which implies that P is the generator of infinitesimal displacements. Hence, it is the linear momentum operator. It is easy to see that P annihilates the ground state P 0 = 0 (.08) which means that the ground state has zero momentum. In other terms, the ground state is translationally invariant (as it should). Using the commutation relations it is easy to show that P L, k = k L, k, P T, α, k = k T, α, k (.09) which allows us to identify the momentum carried by a phonon with k where k is the label of the Fourier transform. Finally, we notice that we can easily write down an expression for the displacement field u i ( r) and the canonical momentum Π i ( r) in terms of creation and annihilation operators for longitudinal and transverse phonons: u i ( r) = ρ + ρ d 3 p (2π) 3 d 3 p (2π) 3 2ω L (p) 2ω T (p) p i ( al ( p)e i p r a L ( p) e i p r) p α=,2 e α i ( p) ( a α T( p)e i p r a α T( p) e i p r) (.0) which is known as a mode expansion. There is a similar expression for the canonical momentum Π i ( r). In summary, after quantizing the elastic solid we found that the quantum states of this system can be classified in terms if a set of excitations, the longitudinal and transverse phonons. These states carry energy and momentum (as well as polarization) and hence behave as particles. For these reason we will regard these excitations as the quasi-particles of this system. We will see that quasi-particles arise generically in interacting many-body system. One problem we will be interested in is in understanding the relation between the properties of the ground state and the quantum numbers of the quasiparticles.

14 4 CHAPTER. SECOND QUANTIZATION.3 Indistinguishable Particles 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,...,x N ). If the particles are identical then the probability density, Ψ(x,..., x N ) 2, must be invariant (i.e., unchanged) under arbitrary exchanges of the labels that we use to identify (or designate) the particles. In quantum mechanics, 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,..., x N, they will become 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 manyparticle quantum mechanical system. In other terms, identical particles are indistinguishable in quantum mechanics. In particular, the probability density of any eigenstate must remain invariant if the labels of any pair of particles are exchanged. If we denote by P jk the operator that exchanges the labels of particles j and k, the wave functions must change under the action of this operator at most by a phase factor. Hence, we must require that P jk Ψ(x,..., x j,..., x k,..., x N ) = e iφ Ψ(x,...,x j,...,x k,..., x N ) (.) 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 PΨ = Ψ, or Ψ is odd under permutation and PΨ = Ψ. Systems of identical particles which have wave functions which are even under a pairwise permutation of the particle labels are called bosons. In the other case, Ψ odd under pairwise permutation, they are Fermions. It must be stressed that these arguments only show that the requirement that the state Ψ be either even or odd is only a sufficient condition. It turns out that under special circumstances (e.g in one and twodimensional systems) other options become available and the phase factor φ may take values different from 0 or π. These particles are called anyons. For the moment the only cases in which they may exist appears to be in situations 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 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 known as the spin-statistics theorem and it is actually an axiom of relativistic quantum mechanics (and field theory).

15 .4. FOCK SPACE 5.4 Fock Space We will now discuss a procedure, known as Second Quantization, which will enable us to keep track of the symmetry of the states of systems of many identical particles in a simple way. Let us consider a system of N identical non-relativistic particles. The wave functions in the coordinate representation are Ψ(x,...,x N ) where the labels x,..., x N denote both the coordinates and the spin states of the particles in the state Ψ. For the sake of definiteness we will discuss physical systems describable by Hamiltonians Ĥ of the form Ĥ = 2 2m N N 2 j + j= j= V (x j ) + j,k U(x j x k ) +... (.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 tensor product of the one-particle states, namely Ψ(x,..., x N ) = {nj} C(n,... n N )φ n (x )...φ nn (x N ) (.3) Thus, if Ψ is symmetric (antisymmetric) under an arbitrary exchange x j x k, the coefficients C(n,..., n N ) must be symmetric (antisymmetric) under the exchange n j n k. A set of N-particle basis states with well defined permutation symmetry is the properly symmetrized or antisymmetrized tensor product Ψ,... Ψ N Ψ Ψ 2 Ψ N = ξ P Ψ P() Ψ P(N) N! (.4) where the sum runs over the set of all possible permutations P. The weight factor ξ is + for bosons and for fermions. Notice that, for fermions, the N-particle state vanishes if two particles are in the same one-particle state. This is the Pauli Exclusion Principle. The inner product of two N-particle states is χ,... χ N ψ,..., ψ N = ξ P+Q χ Q() ψ P() χ Q(N) ψ P() = N! P,Q = P ξ P χ ψ P() χ N ψ P(N) P (.5) where P = P + Q denotes the permutation resulting from the composition of the permutations P and Q. Since P and Q are arbitrary permutations, P spans the set of all possible permutations as well.

16 6 CHAPTER. SECOND QUANTIZATION It is easy to see that Eq.(.5) is nothing but the permanent (determinant) of the matrix χ j ψ k for symmetric (antisymmetric) states, i.e., χ ψ... χ ψ N χ,... χ N ψ,... ψ N =.. (.6) χ N ψ... χ N ψ N ξ In the case of antisymmetric states, the inner product is the familiar Slater determinant. Let us denote by { α } a complete set of orthonormal one-particle states. They satisfy α β = δ αβ α α = (.7) The N-particle states are { α,... α N }. Because of the symmetry requirements, the labels α j can be arranged in the form of a monotonic sequence α α 2 α N for bosons, or in the form of a strict monotonic sequence α < α 2 < < α N for fermions. Let n j be an integer which counts how many particles are in the j-th one-particle state. The boson states α,...α N must be normalized by a factor of the form α n!... n N! α,..., α N (α α 2 α N ) (.8) and n j are non-negative integers. For fermions the states are α,..., α N (α < α 2 < < α N ) (.9) and n j = 0 >. These N-particle states are complete and orthonormal N! α,...,α N α,...,α N α,..., α N = Î (.20) 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 from the Canonical Ensemble to the Grand Canonical Ensemble. Thus, let us denote by H 0 the Hilbert space with no particles, H 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 H N (.2) is usually called Fock space. An arbitrary state ψ in Fock space is the sum over the subspaces H N, ψ = ψ (0) + ψ () + + ψ (N) + (.22)

17 .5. CREATION AND ANNIHILATION OPERATORS 7 The subspace with no particles is a one-dimensional space spanned by the vector 0 which we will call the vacuum. The subspaces with well defined number of particles are defined to be orthogonal to each other in the sense that the inner product in Fock space χ ψ χ (j) ψ (j) (.23) j=0 vanishes if χ and ψ belong to different subspaces..5 Creation and Annihilation Operators Let φ be an arbitrary one-particle state, i.e. φ H. Let us define the creation operator â (φ) by its action on an arbitrary state in Fock space â (φ) ψ,...,ψ N φ, ψ,...,ψ N (.24) Clearly, â (φ) maps the N-particle state with proper symmetry ψ,..., ψ N to the N + -particle state φ, ψ,..., ψ N, also with proper symmetry. The destruction or annihilation operator â(φ) is then defined as the adjoint of â (φ), χ,...,χ N â(φ) ψ,..., ψ N ψ,..., ψ N â (φ) χ,...,χ N (.25) Hence χ,...,χ N â(φ) ψ,..., ψ N = ψ,..., ψ N φ, χ,..., χ N = ψ φ ψ χ ψ χ N = ψ N φ ψ N χ ψ N χ N ξ (.26) We can now expand the permanent (or determinant) along the first column to get χ,...,χ N â(φ) ψ,..., ψ N = = = N ξ k ψ k φ k= ψ χ... ψ χ N..... (no ψ k )... ψ N χ... ψ N χ N N ξ k ψ k φ χ,...,χ N ψ,... ˆψ k..., ψ N k= where ˆψ k indicates that ψ k is absent. ξ (.27)

18 8 CHAPTER. SECOND QUANTIZATION Thus, the destruction operator is given by â(φ) ψ,..., ψ N = N ξ k φ ψ k ψ,..., ˆψ k,...,ψ N (.28) k= With these definitions, we can easily see that the operators â (φ) and â(φ) obey the commutation relations â (φ )â (φ 2 ) = ξ â (φ 2 )â (φ ) (.29) Let us introduce the notation [Â, ˆB] ξ  ˆB ξ ˆB (.30) where  and ˆB are two arbitrary operators. For ξ = + (bosons) we have the commutator [â (φ ), â (φ 2 ) ] + [ â (φ ), â (φ 2 ) ] = 0 (.3) while for ξ = it is the anticommutator [â (φ ), â (φ 2 ) ] { â (φ ), â (φ 2 ) } = 0 (.32) Similarly for any pair of arbitrary one-particle states φ and φ 2 we get [â(φ ), â(φ 2 )] ξ = 0 (.33) It is also easy to check that the following identity generally holds [â(φ ), â (φ 2 ) ] ξ = φ φ 2 (.34) 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. In this case, we have n,..., n k,... n n2 n!n 2!... {}}{{}}{..., (.35) In the case of bosons, the n 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,..., n α,... = n α + n,..., n α +,... â α n,...,n α,... = nα n,..., n α,... (.36) 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 =.

19 .5. CREATION AND ANNIHILATION OPERATORS 9 We conclude that for bosons the following commutation relations hold [ ] [ ] [â α, â β ] = â α, â β = 0 â α, â β = δ αβ (.37) whereas for fermions we obtain instead anticommutation relations { } { } {â α â β } = â β, â β = 0 â α â β = δ αβ (.38) 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., χ = α ψ + β φ â(χ) = α â(ψ) + β â(φ) â (χ) = αâ (ψ) + βâ (φ) (.39) and we say that â (χ) transforms like the ket χ while â(χ) transforms like the bra χ. For example, we can pick as the complete set of one-particle states the momentum states { p }. This is momentum space. With this choice the commutation relations are [â ( p), â ( q) ] ξ = [â( p), â( q)] ξ = 0 [â( p), â ( q) ] ξ = (2π) d δ d ( p q) (.40) where d is the dimensionality of space. In this representation, an N-particle state is p,..., p N = â ( p )...â ( p N ) 0 (.4) On the other hand, we can also pick the one-particle states to be eigenstates of the position operators, i.e., In position space, the operators satisfy x,... x N = â ( x )...â ( x N ) 0 (.42) [â ( x ), â ( x 2 ) ] ξ = [â( x ), â( x 2 )] ξ = 0 [â( x ), â ( x 2 ) ] ξ = δ d ( x x 2 ) (.43) 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 i p x (.44) and, conversely x = d d p (2π) d p e i p x (.45)

20 20 CHAPTER. SECOND QUANTIZATION Then, the operators themselves obey â ( p) = d d x â ( x)e i p x â ( x) = d d p (2π) d â ( p)e i p x (.46).6 General Operators in Fock Space Let A () 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 H N as follows:  ψ N ψ... A () ψ j... ψ N (.47) j= For instance, if the one-particle basis states { ψ j } are eigenstates of  with eigenvalues {a j } we get N  ψ = ψ (.48) j= We wish to find an expression for an arbitrary operator  in terms of creation and annihilation operators. Let us first consider the operator A () αβ = α β which acts on one-particle states. The operators A () αβ form a basis of the space of operators acting on one-particle states. Then, the N-particle extension of A () αβ is N Âαβ ψ = ψ α ψ N β ψ j (.49) Thus j=  αβ ψ = j= a j j N {}}{ ψ,..., α,..., ψ N β ψ j (.50) 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 β ψ k α, ψ,..., ψ k, ψ k+,..., ψ N (.5) k=

21 .6. GENERAL OPERATORS IN FOCK SPACE 2 We can now use the symmetry of the state to write ξ k α, ψ,...,ψ k, ψ k+,..., ψ N = ψ,..., k {}}{ α,...,ψ N (.52) Thus the operator Âαβ, the extension of α β to the N-particle space, coincides with â (α)â(β)  αβ â (α)â(β) (.53) We can use this result to find the extension for an arbitrary operator A () of the form A () = α α A () β β (.54) α,β we find  = α,β â (α)â(β) α A () β (.55) Hence the coefficients of the expansion are the matrix elements of A () between arbitrary one-particle states. We now discuss a few operators of interest.. The Identity Operator: The Identity Operator ˆ of the one-particle Hilbert space ˆ = α α α (.56) in Fock space becomes the number operator ˆN ˆN = α â (α)â(α) (.57) In momentum and in position space it takes the form d ˆN d p = (2π) d â ( p)â( p) = d d x â ( x)â( x) = d d x ˆρ( x) (.58) where ˆρ(x) = â ( x)â( x) is the particle density operator. 2. The Linear Momentum Operator: In the space H, the linear momentum operator is ˆp () j = d d p (2π) d p j p p = d d x x i j x (.59) Thus, we get that the total linear momentum operator ˆP j is ˆP j = d d p (2π) d p jâ ( p)â( p) = d d x â ( x) i jâ( x) (.60)

22 22 CHAPTER. SECOND QUANTIZATION â ( p + q) p + q q Ṽ ( q) â( p) p Figure.: One-body scattering. 3. Hamiltonian: The one-particle Hamiltonian H () has the matrix elements H () = p 2 + V ( x) (.6) 2m x H () y = 2 2m 2 δ d ( x y) + V ( x)δ d ( x y) (.62) Thus, in Fock space we get ( ) Ĥ = d d x â ( x) 2 2m 2 +V ( x) â( x) (.63) in position space. In momentum space we can define Ṽ ( q) = d d x V ( x)e i q x (.64) the Fourier transform of the potential V (x), and get Ĥ = d d p p 2 2mâ (2π) d ( p)â( p) + d d p (2π) d d d q (2π) d Ṽ ( q)â ( p + q)â( p) (.65) The last term has a very simple physical interpretation. When acting on a one-particle state with well-defined momentum, say p, the potential term yields another one-particle state with momentum p + q, where q is the momentum transfer, with amplitude Ṽ ( q). This process is usually depicted by the diagram of Fig... Two-Body Interactions: A two-particle interaction is an operator ˆV (2) which acts on the space of

23 .6. GENERAL OPERATORS IN FOCK SPACE 23 two-particle states H 2, which has the form V (2) = α, β V (2) (α, β) α, β (.66) 2 α,β The methods developed above yield an extension of V (2) to Fock space of the form ˆV = â (α)â (β)â(β)â(α) V (2) (α, β) (.67) 2 α,β In position space, ignoring spin, we get ˆV = d d x d d y â ( x) â ( y) â( y) â( x) V (2) ( x, y) 2 d d x d d y ˆρ( x)v (2) ( x, y)ˆρ( y) + d d x V (2) ( x, x) ˆρ( x) 2 2 (.68) where we have used the commutation relations. In momentum space we find instead ˆV = d d p d d q d d k 2 (2π) d (2π) d (2π) d Ṽ ( k) â ( p + k)â ( q k)â( q)â( p) (.69) where Ṽ ( k) is only a function of the momentum transfer k. This is a consequence of translation invariance. In particular for a Coulomb interaction, for which V (2) ( x, y) = Ṽ ( k) = 4πe2 k 2 e2 x y (.70) (.7) â ( p + k) p + k q k â ( q k) kṽ ( k) â( p) 4. p q â( q) Figure.2: Two-body interaction.

24 24 CHAPTER. SECOND QUANTIZATION.7 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 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, the operators â(φ) obey canonical anticommutation relations. In position space, it is customary to represent â (φ) by the operator ˆψ( x) which obeys the equal-time algebra [ ˆψ( x), ( y)] ˆψ = δ d ( x y) ξ [ ] [ ˆψ( x), ˆψ( y) = ˆψ ( x), ˆψ ( y)] = 0 ξ ξ (.72) In this framework, the one-particle Schrödinger equation becomes the classical field equation [ i ] t + 2 2m 2 V ( x) ψ = 0 (.73) 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 L = i ψ 2 tψ ψ 2m ψ V ( x)ψ ψ (.74) Its Euler-Lagrange equations are t δl δ t ψ = δl δ ψ + δl δψ (.75) which are equivalent to the field Equation Eq..73. The canonical momenta Π(x) and Π (y) are Π ψ = δl = i ψ (.76) δ t ψ and Π ψ = δl δ t ψ = i ψ (.77) Thus, the (equal-time) canonical commutation relations are [ ] ˆψ( x), ˆΠ( y) = i δ( x y) (.78) which require that [ ˆψ( x), ˆψ ( y)] ξ ξ = δd ( x y) (.79)

25 .8. NON-RELATIVISTIC FERMIONS AT ZERO TEMPERATURE 25.8 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 + space-time dimensions) S = 2 [ d D x ˆψ i t ˆψ 2 ] 2m ˆψ ˆψ V ( x) ˆψ (x) ˆψ(x) d D x d D x ˆψ (x) ˆψ (x )U(x x ) ˆψ (x) ˆψ(x)] where U(x x ) represents instantaneous (two-body) interactions, The Hamiltonian Ĥ for this system is (.80) U(x x ) U( x x )δ(x 0 x 0 ) (.8) Ĥ = d d x [ 2 2m ˆψ ˆψ + V ( x) ˆψ ( x)ψ( x)] + d d x d d x 2 ˆψ (x) ˆψ (x )U(x x ) ˆψ(x ) ˆψ(x) (.82) For Fermions the fields ˆψ(x) and ˆψ (x) satisfy equal-time canonical anticommutation relations { ˆψ( x), ˆψ ( x)} = δ D ( x x ) (.83) while for Bosons they satisfy [ ˆψ( x), ˆψ ( x )] = δ D ( x x ) (.84) In both cases, the Hamiltonian Ĥ commutes with the total number operator ˆN = d d x ˆψ (x) ˆψ(x) since Ĥ 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 of fixing the number of particles we will introduce a Lagrange multiplier µ, the chemical potential, to weigh contributions from different parts of the Fock space. Thus, we define the operator H, H Ĥ µ N (.85) In a Hilbert space with fixed N this amounts to a shift of the energy by µn. We will now allow the system to choose the sector of the Fock space but with the requirement that the average number of particles N is fixed to be some number N. In the thermodynamic limit (N ), µ represents the difference

26 26 CHAPTER. SECOND QUANTIZATION of the ground state energies between two sectors with N + and N particles respectively. The modified Hamiltonian H is (for spinless fermions) H = d d x ˆψ ( x)[ 2 2m 2 +V ( x) µ] ˆψ( x) + d d x d d y 2 ˆψ ( x) ˆψ ( y)u( x y) ˆψ( y) ˆψ( x) (.86).9 The Ground State of a System of Free Fermions E single particle energy E F Fermi Energy occupied states Fermi Sea empty states single particle states Figure.3: The Fermi Sea. Let us discuss now the very simple problem of finding the ground state for a system of N spinless free fermions. In this case, the interaction potential vanishes and, if the system is isolated, so does the one-body 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 (.87) where the index α labels the one-particle states by increasing order of their single-particle energies E E 2 E n (.88)

27 .0. EXCITED STATES 27 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 â N 0 = {}}{..., (.89) α= The energy of this state is E gnd with E gnd = E + + E N (.90) The energy of the top-most occupied single particle state, E N, is called the Fermi energy of the system and the set of occupied states is called the filled Fermi sea..0 Excited States A state like ψ N {}}{ ψ = (.9) 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 +. This is a state with one particle-hole Ground N N + single particle states â N+âN Ψ single particle states Figure.4: An excited (particle-hole) state. pair, and it has the form = â N+âN gnd (.92) The energy of this state is E ψ = E + + E N + E N+ (.93)

28 28 CHAPTER. SECOND QUANTIZATION Hence E ψ = E gnd + E N+ E N (.94) and, since E N+ E N, E ψ E gnd. The excitation energy ǫ ψ = E ψ E gnd is ǫ ψ = E N+ E N 0 (.95). 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 physical reference state or vacuum state. Thus this state is a vacuum in the sense of absence of excitations. These arguments motivate the introduction of the particle-hole transformation. Let us introduce the fermion operators b α such that ˆbα = â α for α N (.96) Since â α gnd = 0 (for α N) the operators ˆb α annihilate the ground state gnd, i.e., ˆbα gnd = 0 (.97) The following anticommutation relations hold } {â α, â α} = {â α,ˆb β = {ˆbβ,ˆb β} = } {â α, â α = δ αα {ˆbβ,ˆb β } = δ ββ { } â α,ˆb β = 0 (.98) where α, α > N and β, β N. Thus, relative to the state gnd, â α and ˆb β behave like creation operators. An arbitrary excited state has the form α...α m, β... β n ; gnd â α... â α mˆb β...ˆb β n gnd (.99) This state has m particles (in the single-particle states α,...,α m ) and n holes (in the single-particle states β,..., β n ). The ground state is annihilated by the operators â α and ˆb β â α gnd = ˆb β gnd = 0 (α > N β N) (.200) The Hamiltonian Ĥ is normal ordered relative to the empty state 0, i.e. Ĥ 0 = 0, (.20) but it is not normal ordered relative to the physical ground state 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 (.202)

29 .. CONSTRUCTION OF THE PHYSICAL HILBERT SPACE 29 Thus where Ĥ = E gnd + :Ĥ : (.203) E gnd = N E α (.204) α= is the ground state energy, and the normal ordered Hamiltonian is :Ĥ := α>n E α â aâ α β N ˆb βˆb β E β (.205) The number operator ˆN is not normal-ordered relative to gnd either. Thus, we write ˆN = â αâα = N + â α a α ˆb βˆb β (.206) 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 ˆN (i.e., [ ˆN, Ĥ] = 0) for every particle that is removed a hole must be created. Hence particles and holes can only be created in pairs. A particle-hole state α, β; gnd is It is an eigenstate with an energy Ĥ α, β; gnd = This state has exactly N particles since α, β; gnd â αˆb β gnd (.207) (E gnd + :Ĥ : ) â αˆb β gnd = (E gnd + E α E β ) α, β; gnd (.208) ˆN α, β; gnd = (N + ) α, β; gnd = N α, β; gnd (.209) Let us finally notice that the field operator ˆψ (x) in position space is ˆψ ( x) = α x α â α = α>n φ α ( x)â α + β N φ β ( x)ˆb β (.20) 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 a system with an infinite number of degrees of freedom amounts to a choice of the Hilbert space. This restriction becomes of fundamental importance when interactions are taken into account.

30 30 CHAPTER. SECOND QUANTIZATION.2 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 ˆψσ ( x) (.2) 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 ) ˆψ d σ ( p) e i p x (.22) 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 (.23) The Hamiltonian has the very simple form d H d p ( = (2π ) d ǫ( p) µ) ˆψ σ ( p) ˆψ σ ( p) (.24) where ε( p) is given by σ=î, ε( p) = p2 (.25) 2m For this simple case, ε( p) is independent of the spin state. It is convenient to define the energy of the one-particle states measured relative to the chemical potential (or Fermi energy) µ = E F, E( p) as E( p) = ε( p) µ (.26) Hence, E( p) is the excitation energy measured from the Fermi energy E F = µ. The energy E( p) does not have a definite sign since there are states with ε( p) > µ as well as states with ε( p) < µ. Let us define by p F the value of p for which E(p F ) = ε(p F ) µ = 0 (.27) This is the Fermi momentum. Thus, for p < p F, E( p) is negative while for p > p F, E( p) is positive. We can construct the ground state of the system by finding the state with lowest energy at fixed µ. Since E( p) is negative for p p F, we see that by

31 .2. THE FREE FERMI GAS 3 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 particle- hole transformation ˆbσ ( p) = ˆψ σ( p) for p p F â σ ( p) = ˆψ (.28) σ ( p) for p > p F In terms of the operators â σ and ˆb σ, the Hamiltonian is 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) is the step function (.29) θ(x) = { x > 0 0 x 0 (.220) 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) +Ẽgnd σ=, (.22) where Ẽgnd, the ground state energy measured from the chemical potential µ, is given by Ẽ gnd = σ=, d d p (2π ) d θ(p F p )E( p)(2π) d δ d (0) = E gnd µn (.222) 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 x e i p x/ = V (.223) where V is the volume of the system. Thus, Ẽ gnd is extensive and the ground state energy density ε gnd is ε gnd = 2 p p F Ẽ gnd = V ε gnd (.224) d d p (2π ) d E( p) = ε gnd µ ρ (.225) 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 S d p 2 0 (2π ) d 2m µ (.226)

32 32 CHAPTER. SECOND QUANTIZATION 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, is the Fermi energy. Thus the ground state energy density ε gnd ( measured from the empty state) is equal to S d E gnd = m (2π ) d pf 0 pd+2 F S d dp p d+ = m(d + 2) (2π ) d = 2E F p d F S d (d + 2)(2π ) d (.227) How many particles does this state have? To find that out we need to look 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=, σ=, ˆψ σ ( p) ˆψ σ ( p) = Hence, ˆN can also be written in the form { } θ( p p F )â σ( p)â σ ( p) + θ(p F p )ˆb σ ( p)ˆb σ( p) (.228) ˆN =: ˆN : +N (.229) where the normal-ordered number operator : ˆN : is : ˆN d d p [ ] := (2π ) d θ( p p F )â σ ( p)â σ( p) θ(p F p )ˆb σ ( p)ˆb σ ( p) σ=, and N, the number of particles in the reference state gnd, is N = d d p (2π ) d σ=, Therefore, the particle density ρ = N V is θ(p F p )(2π ) d δ d (0) = 2 d pd F S d (.230) S d (2π ) d V (.23) ρ = 2 d (2π ) d pd F (.232) 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 = d d+2 E F. The excited states can be constructed in a similar fashion. The state +, σ, p +, σ, p = â α ( p) gnd (.233) is a state which represents an electron with spin σ and momentum p while the state, σ, p, σ, p = ˆb σ ( p) gnd (.234)

33 .3. FREE BOSE AND FERMI GASES AT T > 0 33 represents a hole with spin σ and momentum p. From our previous discussion we see that electrons have momentum p with p > p F while holes have momentum p with p < p F. The excitation energy of a one-electron state 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 σ p, σ p = â σ ( p)ˆb σ ( p ) gnd (.235) with p > p F and p < p F. This state has excitation energy E( 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..3 Free Bose and Fermi Gases at T > 0 We will now quickly review the properties of free Fermi and Bose gases and Bose condensation, and their thermodynamic properties. We ll work in the Grand Canonical Ensemble in which particle number is not exactly conserved, i.e. we couple the system to a bath at temperature T and fixed chemical potential µ. The Grand Partition Function at fixed T and µ is Z G = e βω β(ĥ µ ˆN) = tr e (.236) where Ĥ is the Hamiltonian, ˆN is the total particle number operator, β = (kt), where T is the temperature and k is the Boltzmann constant, and Ω is the thermodynamic potential Ω = Ω(T, V, µ). For a free particle system the total Hamiltonian H is a sum of one-particle Hamiltonians, H = ε l â lâl (.237) l where l = 0,,... labels the states of the complete set of single-particle states { l }, and {ε l } are the eigenvalues of the single-particle Hamiltonian. The Grand Partition Function Z G can be computed easily in the occupation number representation of the eigenstates of H: Z G = n...n k... n... n k... e β l (µ ε l)n l n...n k... (.238) where n l is the occupation number of the l-th single particle state. The allowed values of the occupation numbers n l for fermions and bosons is n l = 0, n l = 0,,... fermions bosons (.239)

34 34 CHAPTER. SECOND QUANTIZATION Thus, the Grand Partition Function is Z G = l=0 n l n l e β(µ εl)nl n l (.240) For fermions we get Z F G = e βωf G = l=0 n l =0, e β(µ ε l) = l=0 [ + e β(µ ε l) ] (.24) or, what is the same, the thermodynamic potential for a system of free fermions Ω F G at fixed T and µ is given by an expression of the form Ω F G = kt For bosons we find instead, l=0 [ ] ln + e β(µ ε l) Fermions (.242) Z B G = e βωb G = l=0 n l =0 e β(µ ε l) = l=0 e β(µ ε l) (.243) Hence, the thermodynamic potential Ω B G for a system of free bosons at fixed T and µ is [ ] Ω B G = kt ln e β(µ ε l) Bosons (.244) l=0 The average number of particles N is N = tr ˆNe β(ĥ µ ˆN) tre β(ĥ µ ˆN) = β Z G Z G µ = β µ lnz G (.245) N = ρv = β µ [ βω G] = Ω G µ where ρ is the particle density and V is the volume. For bosons N is (.246) N = β l=0 e β(µ ε l) ( ) eβ(µ ε l) β (.247) Hence, N = l=0 e β(ε l µ) Bosons (.248)

35 .3. FREE BOSE AND FERMI GASES AT T > 0 35 Whereas for fermions we find, N = l=0 e β(ε l µ) + Fermions (.249) In general the average number of particles N is N = n l = e β(ε l µ) ± l=0 l=0 where + holds for fermions and holds for bosons. The internal energy U of the system is (.250) where Hence, H µn = Z G U = H = H µn + µ N (.25) Z G β = lnz G β = β (βω G) (.252) U = β (βω G) + µ N (.253) As usual, we will need to find the value of the chemical potential µ in terms of the average number of particles N (or in terms of the average density ρ) at fixed temperature T and volume V. Once this is done we can compute the thermodynamic variables of the systems, such as the pressure P and the entropy S by using standard thermodynamic relations, e.g. Pressure: P = ΩG V µ,t Entropy: S = ΩG T µ,v The following integrals will be helpful below and I(a, b) = + e a 2 x2 bx n! a (n+)/2 for n even I n (a) = (n/2)! 2n/2 0 for n odd dx = e b2 2a (.254) 2π a (.255).3. Bose Case Let ε 0 = 0, ε l 0, and µ 0. The thermodynamic potential for bosons Ω B G is Ω B G = kt l=0 ln [ e β(µ ε l) ] (.256)

36 36 CHAPTER. SECOND QUANTIZATION The number of modes in the box of volume V with momenta in the infinitesimal region d 3 p is sv d3 k (2π ), and for bosons of spin S we have set s = 2S +. Thus 3 for spin- bosons there are 3 states. (However, in the case of photons S = but have only 2 polarization states and hence s = 2.) The single-particle energies ε( p) are ε( p) = p2 2m (.257) where we have approximated the discrete levels by a continuum. In these terms, Ω B G becomes Ω B G = ktsv d 3 [ ] p p2 β(µ (2π ) 3 ln e 2m ) (.258) and ρ = N /V is given by ρ = N V = V Ω µ = s e βp2 2m e βµ e βp2 2m +βµ It is convenient to introduce the fugacity z and the variable x d 3 p (2π ) 3 (.259) z = e βµ x 2 = βp2 2m in terms of which the volume element of momentum space becomes (.260) d 3 p (2π ) 3 4πp2 (2π ) 3 dp = x2 2π 2 3 ( 2m β ) 3/2 dx (.26) where we have carried out the angular integration. The density ρ is [ ze x2 x 2 ( ) ] 3/2 2m ρ = s o ze x2 2π 2 3 dx β ( ) 3/2 s 2m + = 4π 2 3 dx z n+ e (n+)x2 x 2 β n=0 ( ) 3/2 s 2m + = 4π 2 3 dx z n x 2 e nx2 = β = = s 4π 2 3 s 4π 2 3 n= ( ) 3/2 2m β ( 2m β n= ) 3/2 n= z n n 3/2 z n n 3/2 + π 2 dy y 2 e y2 We now introduce the (generalized) Riemann ζ-function, ζ r (z) ζ r (z) = n= (.262) z n n r (.263)