Quantum Grand Canonical Ensemble

Chapter 16 Quantum Grand Canonical Ensemble How do we proceed quantum mechanically? For fermions the wavefunction is antisymmetric. An N particle basis function can be constructed in terms of single-particle wavefunctions as follows: ψ(r 1,r 2,...,r N ) = 1 ( 1) P Pφ 1 (r 1 )φ 2 (r 2 ) φ N (r N ), (16.1) N! P where P is the permutation operator, and ( 1) P = ±1 depending on whether P is an even or odd permutation. This result can be written as a (Slater) determinant ψ(r 1,r 2,...,r N ) = 1 detφ i (r ). (16.2) N! For bosons, the wavefunction must be symmetric: 1 1 ψ(r 1,r 2,...,r N ) = Pφ 1 (r 1 ) φ N (r N ), (16.3) N1! N l! N! where N is the number of occurrences of φ. The extra combinatorial factor comes from the fact that you get a distinct wavefunction N 1!N 2! N l! times. A subsystem consists of N particles, with total energy E. It is described by a state vector E, N, k, where k are the other quantum numbers necessary to specify the state. The system consists of subsystems which do not interact with each other. For fermions, the system is described by a state vector E, N, k = P P ( 1) P P 1 n E, N, k N!, (16.4) N! where the outer product is over vectors in different Hilbert spaces, and P permutes particle between the different spaces. For bosons, the ( 1) P factor would not be present. =1 89 Version of April 26, 21

9 Version of April 26, 21CHAPTER 16. QUANTUM GRAND CANONICAL ENSEMBLE The microcanonical distribution for the entire system is described by the density operator E+ǫ/2 1 ρ ǫ (E) = de δ(e H) Ω ǫ (E, N) E ǫ/2 1 = E, N, k E, N, k, (16.5) Ω ǫ (E, N) k where the sum is over all states having energies beween E ǫ/2 and E + ǫ/2. Here, as before, energies are measured in units of ǫ, so all these energies are considered the same. The averaged structure function is Ω ǫ (E, N) = k E, N, k E, N, k, (16.6) which is the number of states in the energy interval [E ǫ/2, E + ǫ/2] and with occupation number N. The total degeneracy of the entire system is again given by the convolution law Ω ǫ (E, N) = δ E, Eδ N, Ω ǫ (E, N ). (16.7) N {N }{E } Note that there are no N!s because they are included in the definitions of the physical states. The single-subsystem distribution function is P (1) E 1,N 1 = Ω(n 1) ǫ (E E 1, N N 1 ), (16.8) Ω ǫ (n) (E, N) which is the ratio of the number of states for which subsystem 1 has energy E 1 and occupation number N 1 to the total number of states. Again, there are no N!s. The grand structure function is here defined by W ǫ (E, z) = z N Ω ǫ (E, N). (16.9) N= For the composite system W ǫ (E, z) = δ E, Eδ N, z N Ω ǫ (E, N ) N N N {E } = {E } δ E, The grand partition function is E W ǫ (E, z). (16.1) X ǫ (α, z) = E e αe W ǫ (E, z) = N z N χ(α, N) = X ǫ (α, z). (16.11)

91 Version of April 26, 21 Once again, we do an asymptotic evaluation using δ E,E = ǫ dα e α(e e ), (16.12) 2πi from which we deduce C P (1) E 1,N 1 = e β(e1 µn1) X 1 (β, µ), (16.13) and in turn, by recognizing E 1, N 1 as the eigenvalues of the Hamiltonian and number operator for the single subsystem, we deduce the density operator ρ = e β(h µn), (16.14) X(β, µ) where H is the Hamiltonian operator, whose eigenvalues are the possible energy states of the system, and N is the number operator, whose eigenvalues are the number of particles in the system. Again note, in contradistinction with the classical probability distribution, there is no N! because the combinatorical factors are taken care of in the definition of the quantum state vectors. This holds whether the particles are bosons or fermions. Because the grand partition function is Trρ = 1, (16.15) X(β, µ) = Tre β(h µn) = E, N, k e β(h µn) E, N, k E,N,k = E,N e β(e µn) g E,N, (16.16) where g E,N is the degeneracy of the state with energy E and number of particles N. Now ln X(β, µ) = H µn U µn, β (16.17) 1 ln X(β, µ) = N = N, β µ (16.18) where U and N are the thermodynamic quantities. The pressure is so therefore p = F V = H V = 1 β V ln X(β, µ), (16.19) d ln X(β, µ, V ) = (U µn)dβ + βn dµ + βp dv, (16.2)

92 Version of April 26, 21CHAPTER 16. QUANTUM GRAND CANONICAL ENSEMBLE or from Eq. (14.26) d[β(u µn) + ln X] = β(du d(µn)) + βn dµ + βp dv = β[du + p dv µ dn] = βδq = ds k, (16.21) so, up to a constant, or S k = β(u µn) + ln X, (16.22) TS = U µn + kt ln X. (16.23) This suggests defining still another kind of free energy, the grand potential, J = F µn = U TS µn = kt ln X, (16.24) which is analogous to F = kt ln χ. Note that dj = T ds p dv + µ dn d(ts) d(µn) = p dv S dt N dµ, (16.25) which says that J(T, µ, V ) is a function of the indicated variables, that is, ( ) J T µ,v = S, ( ) J µ T,V = N, ( ) J = p. (16.26) V T,µ The last two equations are ust those given in Eqs. (16.19) and (16.18), while the last is J T = k ln X kt T ln X = k ln X + 1 T β ln X = 1 [ U + µn kt ln X] = S. (16.27) T 16.1 Bose-Einstein and Fermi-Dirac Distributions The grand structure function for a gas of noninteracting particles is (no N!) X = N z N χ(α, N), (16.28) where χ(α, N) = {n } δ N, n e βnε, (16.29)

16.1. BOSE-EINSTEIN AND FERMI-DIRAC DISTRIBUTIONS93 Version of April 26, 21 where n is the number of particles in the single-particle energy state ε. Thus X = z n e βnε {n } = e β[µn nε], (16.3) n which also immediately follows from X = e β(e µn) = e β nε+βµ n. (16.31) N,E {n } For particles obeying Bose-Einstein statistics, bosons, the sum on n ranges from to, so 1 z n e βnε =, (16.32) βε 1 ze n y= while for particles obeying Fermi-Dirac statistics, fermions, the sum on n ranges only from to 1: 1 z n e βnε = 1 + ze βε. (16.33) so in general n = n z n e βnε = (1 ± ze βε ) ±1, (16.34) where the upper sign refers to fermions, and the lower to bosons. Thus the grand partition function is X = (1 ± ze βε ) ±1, (16.35) and ln X = ± ln(1 ± ze βε ) = ± ln(1 ± e βµ e βε ). (16.36) Then, the total number of particles is N = 1 β µ ln X = e β(µ ε) 1 ± e β(µ ε) = 1 e β(ε µ) ± 1, (16.37) and U µn = β ln X = ε µ e β(ε µ) ± 1, (16.38)

94 Version of April 26, 21CHAPTER 16. QUANTUM GRAND CANONICAL ENSEMBLE which implies that the thermodynamic energy is U = ε e β(ε µ) ± 1. (16.39) The mean number of particles in the l energy level is so as expected n l = 1 ln X = eβ(µ εl) β ε l 1 ± e β(µ ε l) 1 = e β(ε l µ) ± 1, (16.4) N = l n l, U = l n l ε l. (16.41) These results coincide with those found in Sec. 12.1, with ζ = e βµ. 16.2 Photons For photons, there is no restriction on the number of particles, so we can set z = 1 or µ = : X = e βnε = (1 e βε ) 1, (16.42) {n } U = β ln X = n = 1 β ε e βε 1, (16.43) 1 ln X = ε e βε 1. (16.44) The fluctuation in the individual level occupation numbers is (n l n l ) 2 = n 2 l n l 2 = 1 β 2 1 X = 1 β 2 2 ε 2 l 2 ε 2 l 16.3 Planck Distribution X 1 β 2 ln X = 1 β ( 1 X ) 2 X ε l ε l n l e βε l = (e βε l 1) 2 = n l + n 2 l = n l (1 + n l ). (16.45) For a photon gas in a volume V, the number of states in a wavenumber interval (dk) is 2 V (dk), hk = p, E = pc, (16.46) (2π) 3

16.3. PLANCK DISTRIBUTION 95 Version of April 26, 21 where the factor of 2 emerges because photons are helicity 2 particles; that is, there are two polarization states for each momentum state. Then the logarithm of the grand partition function is ln X = = 8πV (2π) 3 ln ( 1 e βε) dk k 2 ln ( 1 e β hck) = V [ 1 π 2 3 k3 ln ( 1 e β hck) 1 3 = V β hc 3π 2 e β hck dk k 3 1 e β hkc β hc dk k 3 1 e β hck 1 = F kt, (16.47) where F is either the Helmholtz free energy or the grand potential (the distinction disappears when µ = ). The internal energy is U = β ln X = l ε l e βε = V l 1 π 2 hc dk k 3 e β hck 1. (16.48) We see here the characteristic Planck distribution. In the classical limit, h, V V π 2 kt dk k 2. (16.49) which exhibits the Rayleigh-Jeans law, and exhibits the famous ultraviolet catastrophe. This breakdown of classical physics led Planck to the introduction of the quantum of light, the photon. Note that here F = 1 U, (16.5) 3 and so the pressure is p = F V = 1 U 3 V, (16.51) the characteristic law for a radiation gas. To determine the total energy, recall that from Eq. (7.16) dx xn 1 e x = Γ(n)ζ(n), (16.52) 1 where ζ(n) is the Riemann zeta function. For integer argument the zeta function may be expressed as a Bernoulli number, ] In this way we find ζ(2n) = (2π)2n 2(2n)! B n. (16.53) dx x3 e x 1 = π4 15, (16.54)

96 Version of April 26, 21CHAPTER 16. QUANTUM GRAND CANONICAL ENSEMBLE and then we recover Stefan s law: and the specific heat for the photon gas, U = 3F = π2 k 4 15 ( hc) 3 T 4 V, (16.55) c v = U T = 4π2 k 4 15 ( hc) 3 T 3 V. (16.56)