Chapter 7 Quantum Statistics (Quantenstatistik) 7.1 Density of states of massive particles Our goal: discover what quantum effects bring into the statistical description of systems. Formulate conditions under which QM-specific effects are not essential and classical descriptions can be used. Consider a single particle in a box with infinitely high walls U ε3 ε2 Figure 7.1: Levels of a single particle in a 3D box. At the ε1 box walls the potential is infinite and the wave-function is L zero. A one-particle Hamiltonian has only the momentum p^2 ~2 H^ = = − r2 (7.1) 2m 2m The infinite potential at the box boundaries imposes the zero-wave-function boundary condition. Hence the eigenfunctions of the Hamiltonian are k = sin(kxx) sin(kyy) sin(kzz) (7.2) ~ 2π + k = ~n; ~n = (n1; n2; n3); ni 2 Z (7.3) L ( ) ~ ~ 2π 2 ~p = ~~k; ϵ = k2 = ~n2;V = L3 (7.4) 2m 2m L 4π2~2 ) ϵ = ~n2 (7.5) 2mV 2=3 Let us introduce spin s (it takes into account possible degeneracy of the energy levels) and change the summation over ~n to an integration over ϵ X X X X X X Z X Z 1 ··· = ··· = · · · ∼ d3~n ··· = dn4πn2 ::: (7.6) 0 ν s ~k s ~n s s 45 46 CHAPTER 7. QUANTUM STATISTICS (QUANTENSTATISTIK) (2mϵ)1=2 V 1=3 1 1 V m3=2 n = V 1=3 ; dn = 2mdϵ, 4πn2dn = p ϵ1=2dϵ (7.7) 2π~ 2π~ 2 (2mϵ)1=2 2π2~3 X = gs = 2s + 1 spin-degeneracy (7.8) s Z X g V m3=2 1 ··· = ps dϵϵ1=2 ::: (7.9) 2~3 ν 2π 0 Number of states in a given energy range is higher at high energies than at low energies. helium atom electron photon −18 −14 −4 E111, eV 1:5 × 10 1:1 × 10 1:1 × 10 −14 −11 Teqiuv,K 1:2 × 10 8:7 × 10 0.83 Table 7.1: The lowest energy as computed for a 1 cm3 cubic box. The lowest energy occurs for the state 111 with nx = ny = nz = 1. While the lowest single particle energy is strictly speaking somewhat greater than zero, it is extremely small and, for all practical purposes, can be taken as zero. However, at very low cryogenic temperatures, photons in the lowest energy state, or \ground state", may have a relatively more significant energy. It can normally be assumed that the lowest energy is zero for noninteracting particles in a box at room temperature. In most cases, the spacing between the lowest and second lowest energy is comparable to the lowest energy, and thus similarly negligible (see, however, Bose-Einstein condensation!) Why do we then need quantum mechanics at all? Classical non-quantum physics too would predict a continuum of energies for the particles. And it too would predict the energy to start from zero. The energy of a noninteracting particle is all kinetic energy; classical physics has that zero if the particle is at rest and positive otherwise. There are several reasons quantum mechanics is essential: (i) It predictsp that there are more single- particle states in a given energy range at high energy than at low energy ( ϵ). (ii) The wave function (anti)symmetrization requirements cannot be accommodated using classical physics (bosons and fermions have the same statistics). 7.2 Quantum Gas, Bose-Einstein and Fermi-Dirac Statistics Let us consider N non-interacting identical particles. XN H^ = H^p (7.10) p=1 All H^p have the same set of states. These are given by a one-particle problem H^ jνi = ϵ jνi (7.11) |{z}sp ν single-particle Hamiltonian Any two configurations that differ only by an interchange of two or more identical particles are regarded as one and the same state. Thus the state of the system is determined by giving number of particles nν with energy ϵν . 7.2. QUANTUM GAS 47 Our goal is to calculate the partition function Z with the condition that X nν = N (7.12) ν The values admitted for every nν may be nν = 0; 1; 2; 3; 4 ::: Bose-Einstein case (7.13) nν = 0; 1 Fermi-Dirac case (7.14) Bose-Einstein statistics must be used for particles with integer spin (bosons, for example He4, photons, phonons) and Fermi-Dirac statistics for particles with half-integer spin (fermions, e.g. electrons, neutrons, protons). For Bose particles, any number of particles may occupy a given state. For Fermi particles, however, there can be only one particle at most in each state (Pauli exclusion principle). Total energy of the system is X Etot = |{z}nν ϵν (7.15) ν Xnumber of particles in a state ν nν = N (7.16) ν Hence the canonical partition function ! X0 X X Y Y X Z = exp −β nν ϵν = exp (−βnν ϵν ) = exp (−βnν ϵν ) = (7.17) n1;n2;n38 ν nν ν ν nν < 1 Y 1 + exp(−βϵ ) + exp(−2βϵ ) + ··· = - bosons ν ν 1 − exp(−βϵ ) : ν (7.18) ν 1 + exp(−βϵν ) - fermions P Unfortunately this is NOT correct, since we have an auxiliary condition ν nν = N = const. It is in fact difficult to perform this summation in a canonical ensemble. The trick is to switch to the grandcanonical description: let us connect our system to a \particle-bath" which has chemical potential ! µ such that hNi = N. The grandcanonical partition function removes the restriction on the total number of particles, i.e. all nν are allowed now. Hence ! ! X1 X X X ZG = exp(βµN)Z(N) = exp βµ nν exp −β nν (ϵν − µ) = (7.19) N=0 nν ν ν 8 < 1 Y X Y - Bose-Einstein statistics − − 1 − exp(−β(ϵ − µ)) exp ( β(ϵν µ)nν ) = : ν (7.20) ν nν ν 1 + exp(−β(ϵν − µ)) - Fermi-Dirac statistics 7.2.1 Average occupation numbers The grand potential 8X <> ln [1 − exp(−β(ϵν − µ))] - bosons 1 1 − νX Φ = ln ZG = > (7.21) β β :− ln [1 + exp(−β(ϵν − µ))] - fermions ν 48 CHAPTER 7. QUANTUM STATISTICS (QUANTENSTATISTIK) Recalling thermodynamics Φ = E − TS − µN (7.22) dΦ = T dS − P dV + µdN − (T dS + SdT ) − (µdN + Ndµ) = −P dV − SdT − Ndµ (7.23) 8X > 1 ( ) <> - bosons @Φ exp(β(ϵν − µ)) − 1 h i − νX N = = > 1 (7.24) @µ V;T :>− - fermions exp(β(ϵ − µ)) + 1 ν ν Recalling that * + X hNi = nν (7.25) ν we obtain the average occupation numbers for the energy levels 8 1 <> − − - Bose-Einstein statistics h i exp(β(ϵν µ)) 1 nν = > 1 (7.26) : - Fermi-Dirac statistics exp(β(ϵν − µ)) + 1 7.2.2 The role of the constraint on the number of particles We have now calculated N(µ). For practical reasons we however need µ(N). We will address two points here: (i) when this inversion possible? (ii) if the number of particles is not constrained, can we still use our results? Case 1: The number of particles is known and kept fixed to N. In this case 8X > 1 BE <> = N (µ) - bosons exp(β(ϵ − µ)) − 1 Xν ν N = > 1 (7.27) :> = N FD(µ) - fermions exp(β(ϵ − µ)) + 1 ν ν By inverting these relations we can obtain µ. X 1 Question: For bosons, has many plus-minus infinity points, and therefore the exp(β(ϵ − µ)) − 1 ν ν ! equation N BE(µ) = N has many solutions. Why do we still talk about a unique chemical potential for bosons? To answer this, let us have a look at the number of particles in the ground state, n0. Assuming ϵ0 = 0 1 n = ≥ 0 ) µ ≤ 0 (7.28) 0 exp(−βµ) − 1 BE Hence, for bosons, there is a restriction on the sign of the chemical potential. For fermions, there are no restrictions. Case 2: For some particular types of bosons (e.g. photons, phonons), it is possible to have generation and annihilation of particles. Hence, N is not fixed and, strictly speaking, the grandcanonical description cannot be used. However, we know that if some internal variable is unconstrained, the thermodynamic potential of interest will adjust that variable in such a way that it attains its minimum, e.g. for free energy F = F (T; V; N) - for a system with fixed N (7.29) dF = −SdT − P dV + µdN (7.30) 7.2. QUANTUM GAS 49 if N is not known, it will adjust itself in such a away that F reaches its minimum, i.e. ( ) @F 0 = = µ - for systems with a number of particles which is not fixed (7.31) @N T;V We can still use the grandcanonical result, but with µ = 0. You can also compare this to our \incorrect" treatment at the beginning of the lecture. It was exactly the case with µ = 0. 7.2.3 Equation of state of a 3D quantum gas The grand potential X Φ = −T ln ZG = −T σ ln [1 + σ exp(−β(ϵν − µ))] (7.32) ν 1 nν = − − (7.33) (exp( β(ϵν µ)) + σ +1 fermions σ = (7.34) −1 bosons Changing the sum to an integral Z V m3=2 1 Φ = −T σ p ϵ1=2dϵ ln [1 + σ exp(−β(ϵ − µ))] (7.35) 2 3 2π ~ 0 Integration by part removes the logarithm Z 1 Z 1 3=2 1 3=2 1=2 − − ϵ − − ϵ σβ ϵ dϵ ln [1 + σ exp( β(ϵ µ))] = ln [1 + σ exp( β(ϵ µ))] + dϵ − − 0 3=2 0 0 3=2 σ + exp( β(ϵ µ)) (7.36) Hence Z 3=2 1 X 2 V m 1=2 1 2 2 Φ = − p dϵϵ ϵ = − ϵν nν = − E = −PV (7.37) 2 3 − − |{z} 3 2π ~ 0 σ + exp( β(ϵ µ)) 3 3 ν thermodynamics 2 PV = E (7.38) 3 Let us compare this to the classical ideal gas 2 2 3 2 PV = NT; E = × NT = NT ) PV = E (7.39) 3 3 2 3 The classical expression (in terms of P , V , E) is valid for weakly interacting bosons and fermions, even if the (anti)symmetrization is taken into account! 7.2.4 Classical limit In the classical limit, all states are only \weakly" occupied.