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Photon Statistics 1 • A single photon in the state r has energy r = ħωr. • The number of photons in any state r may vary from 0 to . • The total energy of blackbody radiation is ER = ∑r nrr , where nr is the number of photons in the r’th state, so that Zph(T, V) = ∑R exp (–ER). • The state R of the complete system may be represented by a set of occupation numbers (n1, n2, … nr, …). • We show that ln Zph(T, V) = – ∑r ln [1 – exp(–εr)], • and <nr> = – (1/) ∂(lnZph)/∂εr, • which leads to n(ω) = 1/(e βħω – 1). 1 Photon Statistics 2 2 Photon Statistics 3 This equation shows how to determine the mean number of systems of energy εr. 3 Photon Statistics 4 For a continuous EM (photon) distribution, (ω) = ħω, so that n(ω) = 1/(e βħω – 1). 4 Density of States 1 • For a particle in a cube of side L, the wavefunction Ψ is zero at the walls, so that Ψ(n1,n2,n3) = sink1x sink2y sink3z, where ki = niπ/L (i = 1,2,3), and each k-state is characterized by the set of positive integers (n1, n2, n3). • Neighboring states are separated by Δki = π/L , so that the volume per state in k-space is (π/L)3 = π3/V. In this 2-dimensional figure, each point represents an allowed k-state, associated with an area in k-space of (π/L)2. 5 Density of States 2 • The volume of a spherical shell of radius k is 4πk2dk, which would contain 4πk2dk/(π3/V) = 4V k2dk/π2, where (π3/V) is the volume in k-space associated with each state (n1, n2, n3). • However, since only positive values of ni represent physical situations, the number of k-states in range k to k + dk is (1/8)th of that for the total shell; i.e. g’(k)dk = Vk2dk/(2π2). 6 Density of States 3 • In dealing with g’(k)dk = Vk2dk/(2π2), we transform from the magnitude of the wave vector k to the angular frequency ; i.e. g()d = g’(k)dk = g’(k)(dk/d)d. • Since c = /k, g’(k) = Vk2/(2π2) = V2/(2π2c2), and dk/d = 1/c, g()d = V2/(2π2c3) d. • Thus, the number of photons in the range to + d equals the number of photon states in that range g()d times the occupation of each state n(ω) = 1/(e βħω – 1). • Thus, the energy density u(ω) is given by u(ω)dω = (1/V) ħω n(ω) g(ω) dω. Planck’s Radiation Law • The, the energy density u(ω) is given by u(ω)dω = (1/V) ħω n(ω) g(ω) dω. • Inserting the values g(ω) dω = V ω2 dω/π2c3 , n(ω) = 1/(eħω – 1) , we obtain u(ω,) = ħω3 dω / π2c3(eħω – 1) . • This is Planck’s radiation law, which on integration over all frequencies gives the Stefan-Boltzmann law u(T) = aT4, where a = π2k4/15 ħ3c3 . Note: letting x = ħ, u(,x)dω = (4π2c3ħ3)–1∫x3dx/(ex – 1). 8 Finding the Grand Partition Function ZG 9 Occupation Numbers 1 10 Occupation Numbers 2 11 Bose Einstein and Fermi-Dirac Statistics • The symmetry requirements placed on a system of identical quantum particles depends on their spin. • Particles with integer spin (0, 1, 2,…) follow Bose-Einstein statistics, in which the sign of the total wave function is symmetrical with respect to the interchange of any two particles; i.e. Ψ( ∙ ∙ ∙ Qj ∙ ∙ ∙ Qk∙ ∙ ∙ ) = Ψ( ∙ ∙ ∙ Qk∙ ∙ ∙ Qi ∙ ∙ ∙ ). • Particles with half-integer spin (1/2, 3/2, …) follow Fermi-Dirac statistics, in which the sign of the total wave function is antisymmetrical with respect to the interchange of any two particles; i.e. Ψ( ∙ ∙ ∙ Qj ∙ ∙ ∙ Qk∙ ∙ ∙ ) = – Ψ( ∙ ∙ ∙ Qk∙ ∙ ∙ Qi ∙ ∙ ∙ ). Thus, two particles cannot be in the same state, since Ψ = 0, 12 when particles j and k are in the same state. Two-Particle Systems • Writing the wavefunction for particle j in state A as ψj(qA) etc., we have the following situations: • Maxwell-Boltzmann statistics: Ψ = ψj(qA)ψk(qB). • Bose-Einstein (BE) statistics: Ψ = ψj(qA)ψk(qB) + ψj(qB)ψk(qA). In this case, the total wave function Ψ is antisymmetrical. Examples of bosons are photons and composite particles, such as H1 atoms or He4 nuclei. • Fermi-Dirac (FD) statistics: Ψ = ψj(qA)ψk(qB) – ψj(qB)ψk(qA). In this case, the total wave function Ψ is symmetrical. Ψ = 0, if both particles are placed in the same state – the Pauli exclusion principle. Examples of bosons are electrons, protons, neutrons, and composite particles, such as H2 atoms or He3 nuclei. 13 Grand Partition Function ZG • To obtain the grand partition function ZG , we consider a system in which the number of particles N can vary, which is in contact with a heat reservoir. • The system is a member of a grand canonical ensemble, in which T, V and μ (the chemical potential) are constants. • Assume that there are any number of particles in the system, so that 0 ≤ N ≤ No → , and an energy sequence for each value of N, UN1 ≤ UN2 ≤ … ≤ UNr … in which, Vo = V + Vb, Uo = U + Ub, No = N + Nb, where Vb etc. refer to the reservoir and Vo etc. to the total.14 Comparison of ZG with Z Bath type Heat bath Heat and particle bath Probability pr = exp (– Er)/Z pN,r = exp (μN – EN,r)/Z Statistical Z = r=1 exp (– Er) ZG= N=0 r=1 exp (μN – EN,r) parameter Partition function Grand partition function • The state N,r of the system may be written as a set of particle occupation-numbers (n1, n2,…, nr, …), with ni = (1/)[∂(lnzGi)/∂μ]. • Fermions (particles with half-integer spin): ni = 0 or 1. • Bosons (particles with integer spin): ni = 0,1, 2,…… ∞. 15 Fermi-Dirac Statistics ni is the mean no of spin-1/2 fermions in the i’th state. All values of μ, positive or negative are allowable, since ni always lies in the range 0 ≤ ni ≤ 1. 16 Bose-Einstein Statistics ni is the mean no of bosons in the i’th state. ∑ni = N, the total number of particles in the system of like particles. ni must be positive and finite, i μ for all i. For an ideal gas, i = p2/2m min = 0, so that μ must be negative. For a photon gas, μ = 0. 17 Density of States 4 • For a set of spin-zero bosons, g’(k)dk = V k2dk /(2π2). • For a set of spin-½ fermions, g’(k)dk = 2V k2dk /(2π2), since each set of quantum numbers (n1, n2, n3), has two possible spin states. • The number of states in the range to + d is given by f()d = g’(k)dk = g’(k)(dk/d)d. • Now so that = p2/2m = (kħ)2/2m, k = √(2m)/ħ , dk/d = (1/2ħ)(2m/)1/2. 18 Density of States 5 Bose-Einstein condensation (spin 0 system) • The number of states in the range to + d is given by f()d = g(k)dk = g(k)(dk/d)d, with g(k) = Vk2/(2π2), k = √(2m)/ħ , dk/d = (1/2ħ)(2m/)1/2. • Hence, f()d = Vk2(dk/d)d/2π2 = V[4πm/ħ3](2m/)1/2d; i.e. f()d = (2πV/h3)(2m)3/21/2 d. Free electron theory (spin ½ system) f()d = (4πV/h3)(2m)3/21/2 d. 19 Density of States 6 • Suppose that for an N-particle system with continuous ε, i. the number of states in the range ε to ε + dε is f()d; ii. the mean number of particles of energy is <n()>. • The number of particles with energies in the range ε to ε + dε is dN(ε) = <n()> f()d. • Thus, the total no. of particles is given by N = ∫dN(ε), and the total energy is given by U = ∫ε dN(ε) = ∫ε <n()> f()d, where the integration limits are 0 and ∞. • The values of <n()> for quantum systems are given by <n()> = 1/{exp[β(ε – μ)] ± 1}. • The distribution f()d is called the density of states. 20 Mean number of bosons or fermions If there is a fixed number of particles N, ∑<n(ε)> = N 21 Classical limit • Quantum statistics gives <nr> = 1/{exp[β(εr – μ)] ± 1}. • In the classical limit, the energy states r are infinitesimally close, so that <nr> → 0, <nr>–1 → ∞, exp[β(εr – μ)] » 1 and <nr> → exp[β(μ – εr)] = exp(βμ). exp(– εr) , where z is the single-particle partition function. • The single-particle Boltzmann distribution is pr = <nr>/N = exp(– εr).z . • Thus, N = z exp(βμ). 22 Summary • Consider N particles of an ideal quantum gas, with closely spaced states, which may be taken as a continuum. • Such is the case for the energy of a molecule of an ideal monatomic gas ε = p2/2m. • The number of states in the range ε to ε + dε is given by f(ε) dε = C ε1/2 dε, where C = (2πV/h3)(2m)3/2 for spin 0 bosons, and (4πV/h3)(2m)3/2 for spin ½ fermions. • The occupation numbers of each state are given by the BE and FD distribution functions <n(ε)> = 1/{exp[β(ε – μ)] – 1} and 1/{exp[β(ε – μ)] +1} respectively. The distribution of particles is given by dN(ε) = <n()> f()d. LOW TEMPERATURE FD DISTRIBUTION • As β → 0, exp[β(ε – μ)] behaves as follows: • If ε < μ, exp[β(ε – μ)] → exp(-∞) = 0, so that <n(ε)> = 1; if ε > μ, exp[β(ε – μ)] → exp(∞) = ∞, so that <n(ε)> = 0. • Thus dN(ε) = f(ε)dε for ε < μ, and dN(ε) = 0 for ε > μ. The Fermi energy F is defined as F ≡ μ(T → 0). Free-Electron Theory: Fermi Energy 1 The Fermi energy F ≡ μ(T=0) At T=0, the system is in the state of lowest energy, so that the N lowest single-particle states are filled, giving a sharp cut-off in n() at T = TF. At low non-zero temperatures, the occupancies are less than unity, and states with energies greater than μ are partially occupied. Electrons with energies close to μ are the ones primarily excited. The Fermi temperature TF = F/k lies in the range 104 – 105 K for metals with one conduction electron per atom. 25 Below room temperature, T/TF < 0.03, and μ ≈ F. Free-Electron Theory: Heat capacity Simplified calculation for T << TF Assume that only those particles within an energy kT of F can be excited and have mean energies given by “equipartition”; i.e. Neff N(kT/F) = NT/TF. Thus, U Neff(3/2)kT = (3/2)NkT2/TF, so that CV = dU/dT 3NkT/TF. In a better calculation, 4.9 replaces 3. Thus, for a conductor at low temperatures, with the Debye term included, CV = AT3 + γT, so that CV/T = AT2 + γ. 26 Free-Electron Theory: Fermi Energy 2 The number of electrons with energies between and + d is given by dN() = n() f() d, where n() = 1/{[exp( – μ)] + 1}, . 27 Free-Electron Theory: Calculation of <n>. Now U = 0 n()f() d. 28 LOW TEMPERATURE B-E DISTRIBUTION • The distribution of particles dN(ε) = <n()> f()d cannot work for BE particles at low temperatures, since all the particles enter the ground-state, while the theoretical result indicates that the density of states [f(ε) = C ε1/2] is zero at ε = 0. • The thermodynamic approach to Bose-Einstein Condensation shows the strengths and weaknesses of the statistical method: mathematical expressions for the phenomena are obtained quite simply, but a physical picture is totally lacking. • The expression dN(ε) = <n()> f()d works down to a phase transition, which occurs at the Bose or condensation temperature TB, above which the total number of particles is given by N = ∫ dN(ε). • Below TB , appreciable numbers of particles are in the ground state. Bose-Einstein Condensation 1 • The number of particles with energies in the range ε to ε + dε is dN(ε) = <n()> f()d, with <n()> = 1/{exp[β(ε – μ)] – 1} and f()d = (2πV/h3)(2m)3/21/2 d. • Thus, dN(ε) = (2πV/h3)(2m)3/21/2 d /{exp[β(ε – μ)] – 1}, and the total number of particles N is given by N = (2πV/h3)(2m)3/2 ∫1/2 d /{exp[β(ε – μ)] – 1}, which is integrated from = 0 to ∞. 30 Bose-Einstein Condensation 2 • Since the number of particles are fixed, with N = (2πV/h3)(2m)3/2 ∫1/2 d /{exp[β(ε – μ)] – 1}, the integral • ∫1/2 d /{exp[β(ε – μ)] – 1} must be positive and independent of temperature. • Since min= 0 for an ideal gas, the chemical potential μ ≤ 0, the factor exp(β |μ|) – 1 = exp(|μ|/kT) – 1, must be constant, so that (|μ|/T) is independent of temperature. • This can happen only down to the Bose Temperature TB, at which μ becomes 0. What happens below TB? 31 Bose-Einstein Condensation 3 • At the Bose (or condensation) temperature TB, μ ≈ 0, so that N = (2πV/h3)(2m)3/2 ∫1/2 d /{exp(ε/kTB) – 1}, which yields on integration, TB = (h2/2πmk)(N/2.612V)2/3. • The expression f()d = K1/2 d, with K = (2πV/h3)(2m)3/2, indicates that for T > TB, the number of particles in the ground state ( = 0) is negligible, since f() = K1/2 → 0. Behavior of μ above TB. Bose-Einstein Condensation 4 33 Bose-Einstein Condensation 5 For T >TB, the number of particles in the ground state (N) is zero. 34 Bose-Einstein Condensation 6 35 Bose-Einstein Condensation 7 36 Bose-Einstein Condensation 8 37 Bose-Einstein Condensation 9 Classical high-temperature value 38 Bose-Einstein Condensation 10 39 Appendix Alternative Approach to Quantum Statistics PHYS 4315 R. S. Rubins, Fall 2008 Lagrange Method of Undetermined Multipliers 1 Simple example • How to find an extremum for a function f(x,y), subject to the constraint φ(x,y) = constant. • Suppose f(x,y) = x3 +y3, and φ(x,y) = xy = 4. Method 1 Eliminating y, f(x,y) = x3 +(4/x)3, so that df/dx = 3x2 - 3(4/x)4 When df/dx = 0, x6 = 64, x = 2, y = 2. Method 2 (Lagrange method) and = α, so that In this example, α is a Lagrange undermined multiplier. 41 Lagrange Method of Undetermined Multipliers 2 Suppose that the function of is needed. This occurs when df = (∂f/∂x1)dx1 + … + (∂f/∂xn)dxn = 0. Let there be two constraints = N, = U, where in the calculations of the mean number of particles in the state j. Lagrange’s method of undetermined multipliers gives the following set of equations: In the calculations that follow, the function f equals ln(ω), 42 where ω is the thermodynamic degeneracy. Alternate Fermi-Dirac Calculation 1 If the j’th state has degeneracy gj, and contains Nj particles, Nj ≤ gj for all j, since the limit is one particle per state; e.g. The number of ways of dividing N indistinguishable particles into two groups is In the Fermi-Dirac case, . 43 Alternate Fermi-Dirac Calculation 2 The total no. of microstates is obtained by summing over all j; i.e. Therefore, Using Stirling’s theorem, ln N! ≈ N ln N – N, we obtain 44 Alternate Fermi-Dirac Calculation 3 Since the constraints are , we let φ(N1…Nj…) = N, and ψ(N1 …Nj…) = U, so that , where α and β are Lagrange multipliers. Inserting the expression for ln ωFD and remembering that we obtain 45 . Alternate Fermi-Dirac Calculation 4 reduces to Thus <nj> = . The constraint α has been replaced by μ/kT, where μ is the chemical potential, and β by –1/kT. 46 Alternate Bose-Einstein Calculation 1 The j’th energy level has gj quantum states, and contains a total of Nj identical particles, with up to Nj particles in each state. All possible microstates can be obtained by rearranging (gj – 1) partitions and Nj dots, in a diagram like that shown below. The number of microscopes for a given Nj and gj is . 47 Alternate Bose-Einstein Calculation 2 The total no. of microstates is obtained by summing over all j; i.e. Therefore, Using Stirling’s theorem, ln N! ≈ N ln N – N, we obtain 48 Alternate Bose-Einstein Calculation 3 Using the method of Lagrange multipliers as before, where α and β are Lagrange multipliers. Inserting the expression for ln ωFD, we obtain hence 49 . Alternate Bose-Einstein Calculation 4 reduces to Thus <nj> = . The constraint α has been replaced by μ/kT, where μ is the chemical potential, and β by –1/kT. 50