LECTURES ON STATISTICAL MECHANICS E. MANOUSAKIS

Transcription

1 LECTURES ON STATISTICAL MECHANICS E. MANOUSAKIS February 18, 2011

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3 Contents 1 Need for Statistical Mechanical Description 9 2 Classical Statistical Mechanics Phase Space Statistical Distribution and Equilibrium Ensemble and Macroscopic State Liouville s Theorem Measurement and Averages Micro-canonical Ensemble Statistical Independence-Fluctuations Entropy The Law of Entropy Increase Temperature Problems Derivation of Thermodynamics Reversibility and Adiabatic Process First Law of Thermodynamics Second law of thermodynamics Thermodynamic Functions Quantum Statistical Mechanics The state of a macroscopic system Density Matrix Statistical Averages Micro-canonical Ensemble Entropy Problems Other Ensembles Canonical Ensemble Thermodynamic Quantities Examples from the Canonical Ensemble Distinguishable Free-Particles

4 4 CONTENTS Classical Partition Function Magnetic Dipoles in a Magnetic Field Grand Canonical Ensemble Problems Simulations The general idea Monte Carlo Simulation Solving the equations of motion Examples The Classical gas and liquid The Heisenberg model Problems Quantum Gases Identical Particles Density Matrix of Free Particles Fermi Gas Bose Gas Bose-Einstein Condensation Blackbody Radiation Cosmic Background Radiation Problems Thermodynamics of a Solid Atomic Vibrations A Single Harmonic Oscillator Thermodynamics of a Single Harmonic Oscillator Mono-atomic Chain Classical Treatment Quantization of the phonon field Thermodynamic Behavior of Lattice Vibrations The Fermi Gas at Low Temperature Low temperature expansion in a Fermi gas Magnetism of a Fermi-Gas Relativistic electron gas White Dwarfs Problems The Ising Model The Ising Model The One-dimensional Ising Model Mean-field approximation Bragg-Williams Approximation Bethe-Peierls Approximation Exact solution for the square lattice

5 CONTENTS Problems Critical Phenomena The order parameter Universality-Critical exponents Landau Theory Effective Theory General Approach and Conclusions Role of Long Wave-Length Fluctuations Finite-Size Scaling Problems Path Integrals Path Integrals Trotter Approximation Mapping the quantum to a classical system Path integral for identical particles Quantum spin models Path integral-boson coherent states Path integral-fermion coherent states Perturbation Theory Introduction Green s functions Generating function Free particles Higher order Green s functions Expansion of the partition function Hugenholtz Diagrams Summary of diagram rules Linked Cluster Expansion Averages of operators Zero temperature expansion Finite-density of fermions Frequency Fourier transform of G Perturbation theory for bosons Analytic properties of the Green s function Response Theory Linear Response Density-density response function Normal Fermi Fluid Charged Fermi system Fermi-gas Plasmon mode

6 6 CONTENTS 15.4 Quasiparticles Bose Fluid Green s functions Dyson s Equations Bogoliubov s Trasformation Instabilities of the Fermi fluid Pairing Instability Bogoliubov Transformation Superconductivity Spin Density Waves Charge Density Waves Other Possible Instabilities Problems Renormalization Group Approach Scale Transformation RG Flow-Fixed Points Real Space RG Transformation Momentum Space RG The Gaussian Model One-Component φ 4 Theory-Ising Model The Heisenberg and the non-linear σ Model Renormalization of the σ Model Loop and ɛ Expansion RG Flow n-Component φ 4 Theory Fluctuations in 2D Problems The 2D XY Model-Vortices Role of Spin Waves in 2D KT Phase Transition Renormalization Group approach Results from Simulations A Second Quantization 311 A.1 Hilbert space for identical particles A.2 Operators A.3 Creation and annihilation operators B Coherent States 319 B.1 Coherent States for bosons B.2 Grassmann algebra B.3 Fermion coherent states

7 CONTENTS 7 B.4 Problems B.5 Gaussian Integrals-Bosons B.6 Gaussian Integrals-Fermions

8 310 CONTENTS

9 Appendix A Second Quantization Here we shall present the formalism of second quantization. There is no additional quantization, the quantization procedure is only one. The name second quantization originates in the history of the problem and the name itself gives no information on what it is it might even be confusing. A more appropriate name is formulation of operators in terms of creation and annihilation operators. This is exactly what we plan on doing. Consider that the particle number is not definite, not because that we necessarily plan to introduce interactions in which particles are destroyed or created. The deeper reason is that in many problems the physical eigenstates of the system, might behave as if they do not respect the particle number definiteness. Sometimes they behave as the definiteness of the particle number is transcended. Using a more technically precise vocabulary, even though the particle conservation is an exact symmetry of the Hamiltonian of the system, the state of the system might show little respect for that symmetry in the limit of N. The symmetry of the Hamiltonian is restored in any finite size system by mixing many states and each one of them does not respect the symmetry but the entire state does. However, there is an energy associated with precession of one such state into another (the mixing of these states) which as N becomes larger and larger this frequency becomes smaller and smaller. In the thermodynamics limit the system might find itself stuck in one such sector where the symmetry is not respected. The way one detects such a tendency in a given system is the development of correlations in the particle creation and in its creation at some very far distance from the creation place. The particle actually has not vanished, it has simply disappeared from the part of space we are looking, thus, creating a nice illusion for us, which has marvelous consequences since this distance gets larger and larger and in the thermodynamic limit it becomes infinite. When one measure disappears by becoming infinite, other quantities also leap into the immeasurable. 311

10 312 APPENDIX A. SECOND QUANTIZATION A.1 Hilbert space for identical particles Let us start by considering the Hilbert space H which has no definite number of particles, symbolically: H = H 0 H 1 H 2...H N... (A.1) Namely, it is thought as the direct sum of the Hilbert subspaces H n with welldefined number of particles n. Now a complete basis of the H N for distinguishable particles can be constructed out of the basis elements of H 1 as follows: γ 1, γ 2,..., γ N = γ 1 1 γ γ N N (A.2) where γ i for fixed i (here γ labels the different basis states) is a complete basis of H 1 and i is the particular particle identity index. This is the same as say that H N = H 1 N (A.3) where the direct product of H 1 with itself is taken N times. For indistinguishable particles the situation is different. In such cases one needs to project out only the symmetric or antisymmetric part of the Hilbert space as discussed in Chapter 7. In the case of bosons or fermions, a basis in the H N is constructed as follows: γ 1, γ 2,..., γ N ) ± = ˆP ( ) ± γ1 1 γ γ N N (A.4) where ˆP ± is a symmetrization (+) or antisymmetrization (-) operator defined as ( ) 1 ˆP ± γ1 1 γ 2... γ N N = (±1) [P ] γ 1 P 1 γ 2 P 2... γ N P N (A.5) N! P and the notation regarding the summation is the same with that used in Chapter 7. Here we have chosen to permute particle indices, however, the same can be done by choosing to permute the indices of the single-particle states. The larger Hilbert space H can be projected to two (dimensionally smaller) spaces each one separately transforming according to the representations of the pair permutations. Symbolically, H ± = ˆP ± H. (A.6) Let us say that we have ordered the single-particle basis states according to our choice so that they are in one to one correspondence with the natural numbers 1,2,3,...,k,... Then, let the set n i, i = 1, 2,..., k,..., give the number of particles occupying each one of these single-particle states using the same order ordering. The numbers n i are called, occupation numbers. Therefore given a state characterized by the set of single particle states γ 1, γ 2,..., γ N there is a corresponding set of occupation numbers n i with at most N non-zero elements. Now it is clear by examining Eq. (A.4) that for fermions each n i can only take

11 A.1. HILBERT SPACE FOR IDENTICAL PARTICLES 313 two possible values n i = 0, 1, otherwise the state collapses to zero owing to the cancellation of two identical terms (with opposite signs) the following: any term and the corresponding one produced by a permutation of the two particles which are placed in the same state. With the convention that 0! = 1, there is no ambiguity in the case of fermion states. However, in the case of bosons in order guarantee normalization of the state when n i 0, 1, we need to divide by a normalization factor, which the square root of the number of ways by which one is able to re-arrange n 1 states, n 2 states and,... Next we shall examine the normalization of the states. Let us consider the overlap of two states. It is immediately clear that the states are orthogonal if they are states corresponding to different particle numbers. Thus, we consider states having the same particle number (γ 1, γ 2,...γ N γ 1, γ 2,..., γ N ) ± = 1 N! P,P (±1) ([P ]+[P ]) γ P 1 γ P 1 1 γ P 2 γ P γ P N γ P N = 1 (±1) [Q] γ P 1 γ QP N! 1 1 γ P 2 γ QP γ P N γ QP N. (A.7) P Q Here, for convenience we have permuted the state indices. The second part of the equation is obtained by realizing that we can write P = QP where Q another permutation and change the summation variables from P and P to P and Q. The overall sign is determined by the sign of the relative permutation Q. Then the sum over Q is independent of P which can be regarded as the reference permutation. Thus, one find that (γ 1, γ 2,...γ N γ 1, γ 2,..., γ N ) ± = Q (±1) [Q] γ 1 γ Q1 1 γ 2 γ Q γ N γ QN. (A.8) Now, it is clear that, unless the set of states {γ 1, γ 2,...γ N } is the same as the set {γ 1, γ 2,...γ N } the overlap is zero. Therefore they can be different by only a permutation P 0 in which case the overlap is the same as the one with itself multiplied by the factor (±1) [P0] which is required to create the same permutation. Therefore, let us consider (γ 1, γ 2,...γ N γ 1, γ 2,..., γ N ) ± = Q (±1) [Q] γ 1 γ Q1 1 γ 2 γ Q γ N γ QN. (A.9) For Fermions all single-particle states in both N-particle states are different thus, only the identity permutation gives non-zero contribution. Thus for fermions, the states defined by Eq. (A.4) are normalized. For the case of bosons, however, there can be n i > 1, then, there are i=1 n i! permutations corresponding to the same state. Thus, for both fermions (for fermions always n i! = 1) and bosons

12 314 APPENDIX A. SECOND QUANTIZATION one can write: (γ 1, γ 2,...γ N γ 1, γ 2,..., γ N ) ± = Thus an orthonormal set of states can be defined as γ 1, γ 2,...γ N ± = n i!. i=1 1 i=1 n i! γ 1, γ 2,..., γ N ) ±. (A.10) (A.11) Once such a complete set of orthonormal states is constructed, then the completeness relationship in each of the spaces H ± is automatic γ 1, γ 2,..., γ N γ 1, γ 2,..., γ N = 1. (A.12) N=1 {γ i} For simplicity in the notation, we have omitted the label ± of the state which stands for its symmetrization or antisymmetrization. Here 0 stands for the noparticle state. Here, we need to worry about the meaning of the summation. We should sum over all different states. Any permutation of the states γ 1, γ 2,..., γ N does not produce any different state because of the symmetrization (or antisymmetrization). Thus, the summation is over all different sets {γ i }. Two sets with the same elements but differing by a permutation are still the same sets. A.2 Operators An operator Ô which does not change the particle number can be expressed in this orthonormal basis as Ô = γ 1,..., γ N γ 1,..., γ N Ô γ 1,..., γ N γ 1,..., γ N. (A.13) N {γ},{γ } Here, again, we have omitted the anti-symmetrization/symmetrization label for simplicity. Let us, first, consider a one-body operator such as an external field acting on each one of the particles or the kinetic energy operator. In the standard notation: Ô (1) = N i=1 ô (1) i. (A.14) In this case, it is straightforward for one to carry out the evaluation of the matrix elements which enter in (A.13) and show that Ô (1) = N N γ 1,..., γ m,..., γ N γ m ô (1) γ m γ 1,..., γ m,..., γ N. (A.15) m=1 {γ},γ m

13 A.3. CREATION AND ANNIHILATION OPERATORS 315 In the case of two-body operators such as the usual pair interaction Ô (2) = 1 2 N i=1,j=1 ô (2) ij. (A.16) we find Ô (2) = 1 2 N N m=1,n=1 {γ},γ m,γ n γ 1,..., γ m,..., γ n,..., γ N γ m, γ n ô (2) γ m, γ n γ 1,..., γ m,..., γ n,..., γ N. (A.17) In the same way one can define other many-particle operators. We are only going to be concerned with problems involving at the most two-particle operators. A.3 Creation and annihilation operators We, now, wish to define the following operators â γ γ 1, γ 2,..., γ N ) ± γ, γ 1, γ 2,..., γ N ) ±. (A.18) and its application on the orthonormal set of states (A.11) gives â γ γ 1, γ 2,..., γ N ± = n i + 1 γ, γ 1, γ 2,..., γ N ±. (A.19) The operator itself increases the occupation number of that particular state by one. Thus the state on the left-hand-side is an N particle state while the state produced on the right-hand-side is an N +1 state with the same symmetry (symmetric or antisymmetric). It is clear that for the case of fermions the factor ni + 1 will never play any role because if n i = 0 then this factor is 1, while if n i 1 we cannot produce a fermion state with n i fermions in the same single-particle state, thus, the state in the right-hand-side will be zero. The adjoint operator of â i, namely â i (â i ). We wish to find what is the result of the action of â i on any state γ 1, γ 2,..., γ N ± as defined by Eq. (A.4). We obtain: n=1 â i γ 1, γ 2,..., γ N ) ± = 0 â i γ 1, γ 2,..., γ N ± 0 + {γ i,i=1,...,n} γ 1, γ 2,..., γ n â γ γ 1, γ 2,..., γ N ) ± γ 1, γ 2,..., γ n) ± = N (±1) m 1 δ γ,γm γ 1, γ 2,..., ˇγ m, γ N ) ±. (A.20) m=1 where the notation ˇγ P 1 means that the state γ P 1 is missing, thus, the state ˇγ P 1, γ P 2,..., γ P N ± is an N 1 particle state. The last equation is obtained by

14 316 APPENDIX A. SECOND QUANTIZATION noticing that γ 1, γ 2,..., γ n â γ γ 1, γ 2,..., γ N ) ± = γ 1, γ 2,..., γ N â γ γ 1, γ 2,..., γ n) ± = γ 1, γ 2,..., γ N γ, γ 1, γ 2,..., γ n) ± (A.21) The last part is non-zero only when n = N 1 and the set of single-particle states {γ, γ 1, γ 2,..., γ N 1 } is the same as the set {γ 1, γ 2,..., γ N }. Therefore, there are N different possibilities so that this can happen which are: γ = γ m, and the set γ 1, γ 2,..., γ N 1 is the same as the set γ 1,..., γˇ m,..., γ N. Therefore we have been able to show the following: â γ γ 1, γ 2,..., γ N ± = n i γ, γ 1, γ 2,..., γ N ±, â i {n i } = n i {n i }. (A.22) One can easily show that the following commutation or anticommutation relations are valid respectively for the boson or the fermion creation and annihilation operators. [a γ, a γ ] = 0 (A.23) [a γ, a γ ] = 0 (A.24) [a γ, a γ ] = δ γ,γ (A.25) where the symbol [a, b] = ab ba and the commutator [a, b] is used for the case of bosons and the anticommutator [a, b] + for fermions. These relations can be easily shown by straightforward application to an N-particle state using Eq. (A.18) and Eq. (A.20). Since for identical particles there is no meaning to ask which particle occupies which state, the many-particle state is specified by giving the occupation numbers of all states, namely in the state n 1, n 2,..., n k,..., ± the integers n i with i = 1, 2,... give how many particles occupy the i th state. It can be easily shown that the one-body operator which has been expressed as (A.15) can be written in terms of creation and annihilation operators as: Ô (1) = γ,γ γ ô (1) γ a γa γ. (A.26) Also using the expression (A.17) for the two-body operator we can show that Ô (2) = 1 2 γ 1,γ 2,γ 1,γ 2 γ 1, γ 2 ô (2) γ 1, γ 2 a γ 1 a γ 2 a γ 2 a γ 1. (A.27) As an example of this we consider the case of a uniform system of interacting non-relativistic particles such as the Hamiltonian H = h2 2m N 2 i + 1 u(r ij ), 2 i=1 i j (A.28)

15 A.3. CREATION AND ANNIHILATION OPERATORS 317 for a uniform fluid or electron gas. In the non-interacting plane wave basis this Hamiltonian takes the form H = e(p)a pσ a pσ + 1 p 2 1, p 2 u p 1, p 2 a p pσ p 1, p 2, p 1 σ1a p σ2a p 2 2σ 2 a p1σ 1. 1, p 2,σ1,σ2 (A.29) where the creation operator a pσ adds a particle in an eigenstate of the momentum and spin: a pσ 0 = p σ (A.30) where r p = 1 V e ī h p r (A.31) and σ is the spin-state with spin projection along the z-axis σ. The matrix element can be easily evaluated p 1, p 2 u p 1, p 2 = 1 V ũ( q)δ p 1+ p 2, p 1 + p 2 (A.32) where the Kronecker δ expresses the momentum conservation and ũ( q) is the Fourier transform of u(r) ũ( q) = d 3 ru(r)e ī h q r (A.33) and q = p 1 p 1 = p 2 p 2 is the momentum transfer. The same Hamiltonian can be expressed in terms of creation and annihilation operators ˆψ ( r, σ) which add particles in eigenstates of the position operator: 1 2 σ 1,σ 2 H = h2 d 3 x 2m ˆψ ( r, σ) 2 ˆψ( r, σ) + σ d 3 xd 3 yu( x y ) ˆψ ( x, σ 1 ) ˆψ ( y, σ 2 ) ˆψ( y, σ 2 ) ˆψ( x, σ 1 ). (A.34)

16 318 APPENDIX A. SECOND QUANTIZATION

17 Appendix B Coherent States B.1 Coherent States for bosons Now, we wish to try to construct eigenstates of the annihilation operator, namely a λ {φ µ } = φ λ {φ µ }. (B.1) The states {φ µ }, µ = 1, 2,... are the so-called coherent states. However, we wish to construct them explicitly. Let us expand any such state, in the complete basis {n µ } as {φ µ } = {n µ} C({n µ }) {n µ } (B.2) This equation implies: {m ν } a λ {φ µ } = C({m ν λ }, m λ + 1) m λ + 1. (B.3) Using Eq. (B.1) we find that {m ν } a λ {φ µ } = φ λ C({m ν }). (B.4) Thus, from the last two Eqs. we find C({n µ λ }, n λ + 1) = φ λ nλ + 1 C({n ν λ}, n λ ). (B.5) This equation can be used recursively to yield: C({n λ }) = λ φ n λ λ C(0, 0,...) nλ! (B.6) and normalizing the state such that for the vacuum c(0, 0,...) = 1 we obtain: {φ λ } = (φ λ a λ )n λ 0. (B.7) n λ! {n λ } λ 319

18 320 APPENDIX B. COHERENT STATES which implies that {φ µ } = e λ φ λ a λ 0. (B.8) The state {φ µ } is not normalized and it is straightforward to show that its norm is given by: {φ µ } {φ µ } = e µ φ µ φµ. (B.9) In addition, one can show that {φ µ } {φ µ} = e µ φ µ φ µ, (B.10) which implies that the set of coherent states is an overcomplete set of states. It can be easily shown that the completeness relation is the following: [dφ µ dφ µ ] e λ φ λ φ λ {φ ν } {φ ν } = 1. (B.11) 2πi µ A simple way to show this is by computing the matrix elements of both sides using states in the occupation number representation. If we differentiate both sides of the Eq. (B.8) with respect to φ λ we obtain: a λ {φ µ} = φ λ {φ µ }. (B.12) From Eq.(B.8) we also find that φ a λ = φ φ λ and φ a λ = Therefore for any state Ψ we obtain: φ a λ Ψ = φ λ φ λ φ. Ψ(φ ) (B.13) (B.14) (B.15) and φ a λ Ψ = φ λψ(φ ) (B.16) where Ψ(φ ) = φ Ψ. Using the completeness relation (B.11) we can express the overlap between two states Ψ and Ψ as Ψ Ψ [dφ µ dφ µ ] = e λ φ λ φ λ Ψ {φ ν } {φ ν } Ψ 2πi µ [dφ µ dφ µ ] = e λ φ λ φ λ Ψ ({φ ν })Ψ ({φ 2πi ν}). (B.17) µ

19 B.1. COHERENT STATES FOR BOSONS 321 Therefore we we that the operators a λ and ˆP λ = i ha λ behave as position and momentum operators respectively in the coherent state representation since a λ φ λ ˆP λ = i ha λ i h φ λ (B.18) (B.19) In addition, the standard momentum-position commutation relations hold for these operators also. It is useful to find the analogous relation to the relation x p = 1 V e ī h px (B.20) which gives the overlap between momentum and position eigenstates. Let P be the eigenstates of the momentum operator ˆP λ : ˆP λ {P µ } = P λ {P µ }. (B.21) Where the eigenvalues P λ are i hφ λ. This equation implies: which can be written as follows φ ˆP λ P = P λ φ P. i h φ φ P = P λ φ P λ and by integrating with respect to φ λ we find: φ P = e ī h λ P λφ λ (B.22) (B.23) (B.24) which is the relation we were looking for and it can be identified that it is the same as Eq. (B.10). Using the completeness relation (B.11) we can expand the state Ψ in the basis of coherent states as Ψ = µ [dφ µ dφ µ 2πi ] e λ φ λ φ λ φ Ψ φ (B.25) and, thus, we obtain Ψ(φ ) = µ [dφ µ dφ µ 2πi ] e λ (φ λ φ λ )φ λ Ψ(φ ) (B.26) which can be rewritten in terms of the momentum operator eigenvalues as Ψ(φ [dφ µdp µ ] i ) = e h λ (φ λ φ λ )P λ Ψ(φ ). (B.27) 2π h µ

20 322 APPENDIX B. COHERENT STATES Eq. (B.27) is a consequence of the fact that dpλ 2π h e ī h P λ(φλ φ λ ) = δ(φ λ φ λ ). (B.28) The next question is how one evaluates the matrix elements, using the coherent states as a basis set, of an an operator Ô({â λ, â λ}) which is a function of creation and annihilation operators. Let us further assume that the operator is in the so-called normal ordered form, which is a polynomial of products of creation and annihilation operators in which all the creation operators are on the left of the product of the annihilation operators. The matrix element of an operator which is in a normal ordered form can be calculated using Eqs. (B.1), (B.13) and (B.10) {φ λ } Ô({â λ, â λ}) {φ λ} = O({ ˆφ λ, φ λ λ})e φ λ φ λ (B.29) where the exponential factor is due to the overlap of the coherent states Eq. (B.10). B.2 Grassmann algebra In order to introduce fermion states which are analogous to the boson coherent states it is necessary to introduce the so-called Grassmann algebra. We introduce a set of generators ψ 1, ψ 2,..., ψ n of the algebra which are anticommuting numbers: ψ λ ψ λ = ψ λ ψ λ (B.30) thus ψλ 2 = 0. We also need to define the operation of conjugation on one such Grassmann variable: ψ λ ψ λ (B.31) In other words, every Grassmann number has a counterpart in the algebra to which is related by conjugation operation. This can be written more simply as (ψ λ ) = ψ λ (B.32) (ψ λ) = ψ λ. (B.33) The properties of the conjugation operation are the following: (ψ λ1 ψ λ2...ψ λn ) = ψ λ n ψ λ n 1...ψ λ 1 (B.34) (cψ λ ) = c ψ λ, (B.35) where c is an arbitrary complex number, not a Grassmann number. We shall consider the case of an algebra with two generators ψ and ψ to illustrate the properties of the algebra. The generalization to an arbitrary number of generators is straightforward. All possible powers of them are 1, ψ, ψ and ψψ. The most general function f(ψ) of ψ is a linear function: f(ψ) = f 0 + f 1 ψ (B.36)

21 B.2. GRASSMANN ALGEBRA 323 because higher powers of ψ vanish. The most general function O(ψ, ψ ) of both ψ and ψ is the following: O(ψ, ψ ) = O 0 + O 1 ψ + Ō1ψ + O 12 ψ ψ. (B.37) Having defined the most general functions of Grassmann variables we can define the operations of the derivative and of the definite integral on these functions. We introduce the derivative as follows: ψ ψ = 1 (B.38) (ψψ ) = ψ ψ (B.39) (ψ ψ) = ψ ψ (B.40) where the last equation follows from the fact that these variables anticommute. Therefore using these definitions we can compute the derivative of any function f(ψ) or of O(ψ, ψ ) as follows: which imply that f(ψ) ψ = f 1 (B.41) O(ψ, ψ ) = O 1 O 12 ψ ψ (B.42) O(ψ, ψ ) ψ = Ō1 + O 12 ψ (B.43) ψ ψ O(ψ, ψ ) = O 12 (B.44) ψ ψ O(ψ, ψ ) = O 12 (B.45) ψ ψ = ψ ψ (B.46) which implies that the derivatives with respect to Grassmann variables anticommute. Notice that the derivative has no similar meaning with the definition of the derivative of an analytic function. In this case we use the name, namely derivative of a function of Grassmann variables, to define the class of operations on such functions which have been defined before. We also need to define the operation of definite integration of a function of Grassmann variables. We define: dψ1 = 0 (B.47) dψψ = 1, (B.48)

22 324 APPENDIX B. COHERENT STATES and based on these we also have dψ = 0 (B.49) dψ ψ = 1. (B.50) Using these definitions we can integrate any function of Grassmann variables: dψf(ψ) = f 1 (B.51) dψo(ψ, ψ ) = O 1 O 12 ψ (B.52) dψ O(ψ, ψ ) = Ō1 + O 12 ψ (B.53) dψ dψo(ψ, ψ ) = O 12 (B.54) dψdψ O(ψ, ψ ) = O 12 (B.55) (B.56) which imply that dψ dψo(ψ, ψ ) = dψdψ O(ψ, ψ ) (B.57) The overlap integral of two functions f(ψ) and g(ψ) of Grassmann variables is defined as follows: f g dψ dψe ψ ψ f (ψ)g(ψ ) (B.58) in analogy with Eq. (B.10) for bosons. Using the definition of the integration and the most general linear functions such as (B.36) we can easily show that f g = f 0 g 0 + f 1 g 1 (B.59) which coincides with the definition of the inner product in a vector space. B.3 Fermion coherent states We shall use these Grassmann variables to define fermion coherent states. These states are not physical states but when we shall express the observables, such as the trace of e β(ĥ µ ˆN) in terms of coherent states they should give the same result as that computed in the physical Fock space. In order to construct fermion coherent states as a linear combination of fermion antisymmetric states using Grassmann variables as coefficients we first

23 B.3. FERMION COHERENT STATES 325 need to associate Grassmann variables ψ λ and ψ λ to each fermion annihilation and creation operator a λ and a λ respectively. Having done that one needs to assume that the Grassmann variables anticommute with the fermion creation or annihilation operators. Namely, {ψ λ, a λ } = {ψ λ, a λ } = 0 (B.60) {ψ λ, a λ } = {ψ λ, a λ } = 0. (B.61) In addition the adjoint operation is related to the conjugation of the Grassmann variables as follows: (ψ λ a λ ) = a λ ψ λ (B.62) (ψ λ a λ ) = a λ ψ λ (B.63) (ψ λa λ ) = a λ ψ λ (B.64) (ψ λa λ ) = a λ ψ λ. (B.65) It is straightforward to show that the state defined as follows ψ = e λ ψ λa λ 0 (B.66) which has the same form as the boson coherent state. In fact the introduction of the Grassmann algebra was made to achieve this goal. Since ψ λ a λ commutes with ψ λ a λ we can write ψ = µ (1 ψ µ a µ) 0 (B.67) and we can prove that (B.66) is an eigenstate of the fermion annihilation operator by considering the action of a λ on such state a λ ψ = µ λ(1 ψ µ a µ)a λ (1 ψ λ a λ ) 0. (B.68) And since the following statements are true a λ (1 ψ λ a λ ) 0 = a λ 0 + ψ λ a λ a λ 0 = ψ λ 0 = ψ λ (1 ψ λ a λ ) 0, we find that a λ ψ = ψ λ ψ. One can show in a similar way that ψ = 0 e λ ψ λ a λ. (B.69) (B.70) In addition, with straightforward application of the properties of Grassmann variables and Grassmann derivatives one can show that ψ a λ = ψ ψ λ a ψ = ψ λ ψ ψ a λ = ψ ψ. (B.71)

24 326 APPENDIX B. COHERENT STATES The overlap of two fermion coherent states has the same form as the one for boson coherent states, ψ ψ = e λ ψ λ ψ λ (B.72) This can be shown in a straightforward way by writing the left and right coherent states as a product (B.67) and by evaluating the vacuum to vacuum expectation values of the terms for each set of single-fermion quantum numbers λ. Finally one needs to use that (1 + ψλ ψ λ ) = eψ λ ψ λ. Again as in the case of boson coherent states the completeness relation is given as: dψλdψ λ e λ ψ λ ψ λ ψ ψ = 1. (B.73) λ In order to show that, we shall consider two fermion determinants γ 1, γ 2,..., γ n = a γ 1 a γ 2...a γ n 0 (B.74) α 1, α 2,..., α m = a α 1 a α 2...a α m 0, (B.75) and we shall compute the matrix elements with these two states of both sides of Eq. (B.73). Thus, we need to evaluate γ 1, γ 2,..., γ n ψ = 0 a γn a γn 1...a γ1 (1 + ψ λ a λ ) 0 Similarly = 0 a γn a γn 1...a γ1 (ψ γ1 a γ 1 )(ψ γ2 a γ 2 )...(ψ γn a γ n ) 0 λ = ( 1) n ψ γn ψ γn 1...ψ γ1. (B.76) ψ α 1, α 2,..., α m = ( 1) m ψ α 1 ψ α 2...ψ α m. (B.77) Therefore we find that the left-hand-side of (B.73) when evaluated in between the states (B.74,B.75) is given by ( 1) m+n λ [ dψ λ dψ λ (1 ψ λψ λ ) ] ψ γn ψ γn 1...ψ γ1 ψ α 1 ψ α 2...ψ α m. (B.78) Let us ask when this is not zero. For this lets not worry at first for the signs that we need to keep track when permuting Grassmann variables. There are the following two possibilities for each different single-fermion set of quantum numbers λ. dψλdψ λ (1 ψλψ λ ) = 1 (B.79) dψλdψ λ (1 ψλψ λ )ψλψ λ = 1 (B.80) If there is only ψ λ or only ψ λ the integral is zero and if there is ψn or (ψ ) n with n > 1 is also zero. This implies that if a ψ λ exists in the product, then ψ λ must also exist. This means that the states γ 1, γ 2,..., γ n and α 1, α 2,...α m are

25 B.3. FERMION COHERENT STATES 327 identical, thus, m = n and α i = γ i, i = 1, 2,..., n. Thus, in order to separate the integrals one needs to simultaneous move pairs ψ λ ψ λ. Namely, one moves first ψ γ 1 ψ γ1 (α 1 = γ 1 ), next one moves ψ γ 2 ψ γ2 (which now become consecutive), etc. Moving such pairs of Grassmann variables produces no minus signs which we would need to keep track. Thus, when the two states (B.74,B.75) are identical we find 1 otherwise zero, which means that we have evaluated the matrix elements of the unit operator. This property of the fermion coherent states demonstrates the utility of the Grassmann variables. Namely, even though the coherent states are not states in the physical Fock space, their algebra allows construction of the unit operator in the physical Fock space. This enables us to calculate the trace of an operator Ô as follows. We start from the standard definition of trace, using a basis n of the physical Fock space: trô = = = n n Ô n = ] λ λ[ dψ λ dψ λ e λ ψ λ ψ λ n n ψ ψ Ô n ] λ[ dψ λ dψ λ e λ ψ λ ψ λ n ψ Ô n n ψ ] λ[ dψ λ dψ λ e λ ψ λ ψ λ ψ Ôψ. (B.81) Notice that when we permuted the n ψ and ψ Ô n we picked up a minus sign in the coherent state. Namely, the following is true: f ψ ψ g = ψ g f ψ. (B.82) Thus, the trace of any operator in the fermion Fock space using coherent states is given by Eq. (B.81), thus, the Grassmann variables as defined in the previous section are useful because one can cast the trace and as we shall see the path integral over fermion coherent states in the same form as for bosons. Of course the price that we have to pay is that we are dealing not with complex numbers but with Grassmann variables. In the fermion coherent basis, any state Ψ is expressed as [ ] Ψ = dψ λ dψ λ e λ ψ λ ψ λ Ψ(ψ ) ψ. (B.83) λ where Ψ(ψ ) = ψ Ψ. In addition, we find that the representation of the operators a λ and a λ is the following: ψ a λ Ψ = ψ Ψ(ψ ) ψ a λ Ψ = ψ λψ(ψ ). (B.84) (B.85) On the other hand matrix elements of an operator Ô(a λ, a λ), which is a normal ordered form with respect to the operators a and a, are given as follows: ψ Ô(a λ, a λ) ψ = e λ ψ λ ψ λ O(ψλ, ψ λ). (B.86)

26 328 APPENDIX B. COHERENT STATES B.4 Problems 1. As we have discussed the coherent states are not physical states, for example the expectation value of the particle number operator ˆN = λ a λ a λ is given by ψ ˆN ψ ψ ψ = λ ψ λψ λ (B.87) which has no physical interpretation. However, this is because the coherent states are not forming basis in a physical space. We saw how to use them to express physical quantities. For example, take a state n from the physical Fock space, such as (B.74) and using coherent states express the expectation value n ˆN n in terms of integrals over coherent states. Show that one finds the same particle number as in the state n. 2. Show that the following behaves as a δ function among functions of Grassmann variables: δ(ψ, ψ ) = dξe ξ(ψ ψ ), (B.88) Namely, show that given any function of Grassmann variables: f(ψ) dψ δ(ψ, ψ )f(ψ ) = f(ψ). (B.89) 3. For Grassmann variables show that: f ψ ψ g = ψ g f ψ. (B.90)