1 Photon Statistics 1 A single photon in the state r has energy  r = ħω r. The number of photons in any state r may vary from 0 to . The total energy of blackbody radiation is E R = ∑ r n r  r, where n r is the number of photons in the r’th state, so that Z ph (T, V) = ∑ R exp (–  E R ). occupation numbersThe state R of the complete system may be represented by a set of occupation numbers (n 1, n 2, … n r, …). We show that ln Z ph (T, V) = – ∑ r ln [1 – exp(–  ε r )], and = – (1/  ) ∂(lnZ ph )/∂ε r, which leads to  n(ω)  = 1/(e βħω – 1).

2 Photon Statistics 2

3 Photon Statistics 3 This equation shows how to determine the mean number of systems of energy ε r.

4 Photon Statistics 4 For a continuous EM (photon) distribution,  (ω) = ħω, so that  n(ω)  = 1/(e βħω – 1).

5 Density of States 1 For a particle in a cube of side L, the wavefunction Ψ is zero at the walls, so that Ψ(n 1,n 2,n 3 ) = sink 1 x sink 2 y sink 3 z, where k i = n i π/L (i = 1,2,3), and each k-state is characterized by the set of positive integers (n 1, n 2, n 3 ). Neighboring states are separated by Δk i = π/L, so that the volume per state in k-space is (π/L) 3 = π 3 /V. In this 2-dimensional figure, each point represents an allowed k-state, associated with an area in k-space of (π/L) 2.

6 Density of States 2 The volume of a spherical shell of radius k is 4πk 2 dk, which would contain 4πk 2 dk/(π 3 /V) = 4V k 2 dk/π 2, where (π 3 /V) is the volume in k-space associated with each state (n 1, n 2, n 3 ). However, since only positive values of n i 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 = Vk 2 dk/(2π 2 ).

Density of States 3 In dealing with g’(k)dk = Vk 2 dk/(2π 2 ), 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) = Vk 2 /(2π 2 ) = V  2 /(2π 2 c 2 ), and dk/d  = 1/c, g(  )d  = V  2 /(2π 2 c 3 ) 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.

8 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 ω 2 dω/π 2 c 3,  n(ω)  = 1/(e  ħω – 1), we obtain u(ω,  ) = ħω 3 dω / π 2 c 3 (e  ħω – 1). This is Planck’s radiation law, which on integration over all frequencies gives the Stefan-Boltzmann law u(T) = aT 4, where a = π 2 k 4 /15 ħ 3 c 3. Note: letting x =  ħ , u( ,x)dω = (  4 π 2 c 3 ħ 3 ) –1 ∫x 3 dx/(e x – 1).

9 Finding the Grand Partition Function Z G

10 Occupation Numbers 1

11 Occupation Numbers 2

12 Bose Einstein and Fermi-Dirac Statistics spinThe symmetry requirements placed on a system of identical quantum particles depends on their spin. Bose-Einstein statistics,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( ∙ ∙ ∙ Q j ∙ ∙ ∙ Q k ∙ ∙ ∙ ) = Ψ( ∙ ∙ ∙ Q k ∙ ∙ ∙ Q i ∙ ∙ ∙ ). Fermi-Dirac statistics,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( ∙ ∙ ∙ Q j ∙ ∙ ∙ Q k ∙ ∙ ∙ ) = – Ψ( ∙ ∙ ∙ Q k ∙ ∙ ∙ Q i ∙ ∙ ∙ ). Thus, two particles cannot be in the same state, since Ψ = 0, when particles j and k are in the same state.

13 Two-Particle Systems Writing the wavefunction for particle j in state A as ψ j (q A ) etc., we have the following situations: Maxwell-Boltzmann statistics: Ψ = ψ j (q A )ψ k (q B ). Bose-Einstein (BE) statistics: Ψ = ψ j (q A )ψ k (q B ) + ψ j (q B )ψ k (q A ). In this case, the total wave function Ψ is antisymmetrical. bosons Examples of bosons are photons and composite particles, such as H 1 atoms or He 4 nuclei. Fermi-Dirac (FD) statistics: Ψ = ψ j (q A )ψ k (q B ) – ψ j (q B )ψ k (q A ). In this case, the total wave function Ψ is symmetrical. Pauli exclusion principle. Ψ = 0, if both particles are placed in the same state – the Pauli exclusion principle. bosons Examples of bosons are electrons, protons, neutrons, and composite particles, such as H 2 atoms or He 3 nuclei.

14 Grand Partition Function Z G grand partition functionTo obtain the grand partition function Z G, we consider a system in which the number of particles N can vary, which is in contact with a heat reservoir. grand canonical ensembleThe 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 ≤ N o → , and an energy sequence for each value of N, U N1 ≤ U N2 ≤ … ≤ U Nr … in which, V o = V + V b, U o = U + U b, N o = N + N b, where V b etc. refer to the reservoir and V o etc. to the total.

15 Comparison of Z G with Z The state N,r of the system may be written as a set of particle occupation-numbers (n 1, n 2,…, n r, …), with  n i  = (1/  )[∂(lnz Gi )/∂μ]. FermionsFermions (particles with half-integer spin): n i = 0 or 1. BosonsBosons (particles with integer spin): n i = 0,1, 2,…… ∞. Bath typeHeat bathHeat and particle bath Probability p r = exp (–  E r )/Zp N,r = exp  (μN – E N,r )/Z Statistical parameter Z =  r=1 exp (–  E r ) Partition function Z G =  N=0  r=1 exp  (μN – E N,r ) Grand partition function

16 Fermi-Dirac Statistics  n i  is the mean no of spin-1/2 fermions in the i’th state. All values of μ, positive or negative are allowable, since  n i  always lies in the range 0 ≤  n i  ≤ 1.

17 Bose-Einstein Statistics ∑ n i = N, the total number of particles in the system of like particles.  n i  must be positive and finite,  i  μ for all i. For an ideal gas,  i = p 2 /2m   min = 0, so that μ must be negative. For a photon gas, μ = 0.  n i  is the mean no of bosons in the i’th state.

18 Density of States 4 bosons,For a set of spin-zero bosons, g’(k)dk = V k 2 dk /(2π 2 ). fermions,For a set of spin-½ fermions, g’(k)dk = 2V k 2 dk /(2π 2 ), since each set of quantum numbers (n 1, n 2, n 3 ), 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  = p 2 /2m = (kħ) 2 /2m, so that k = √(2m  )/ħ, dk/d  = (1/2ħ)(2m/  ) 1/2.

19 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) = Vk 2 /(2π 2 ), k = √(2m  )/ħ, dk/d  = (1/2ħ)(2m/  ) 1/2. Hence, f(  )d  = Vk 2 (dk/d  )d  /2π 2 = V[4πm  /ħ 3 ](2m/  ) 1/2 d  ; i.e. f(  )d  = (2πV/h 3 )(2m) 3/2  1/2 d . Free electron theory (spin ½ system) f(  )d  = (4πV/h 3 )(2m) 3/2  1/2 d .

20 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. The number of particles with energies in the range ε to ε + dε is dN(ε) = 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 f(  )d , where the integration limits are 0 and ∞. The values of for quantum systems are given by = 1/{exp[β(ε – μ)] ± 1}. density of statesThe distribution f(  )d  is called the density of states.

21 Mean number of bosons or fermions If there is a fixed number of particles N, ∑ = N

22 Classical limit Quantum statistics gives = 1/{exp[β(ε r – μ)] ± 1}. In the classical limit, the energy states r are infinitesimally close, so that → 0, –1 → ∞, exp[β(ε r – μ)] » 1 and → exp[β(μ – ε r )] = exp(βμ). exp(–  ε r ), where z is the single-particle partition function. The single-particle Boltzmann distribution is p r = /N = exp(–  ε r ).z. Thus, N = z exp(βμ).

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 = p 2 /2m. The number of states in the range ε to ε + dε is given by f(ε) dε = C ε 1/2 dε, where C = (2πV/h 3 )(2m) 3/2 for spin 0 bosons, and (4πV/h 3 )(2m) 3/2 for spin ½ fermions. The occupation numbers of each state are given by the BE and FD distribution functions = 1/{exp[β(ε – μ)] – 1} and 1/{exp[β(ε – μ)] +1} respectively. The distribution of particles is given by dN(ε) = f(  )d .

LOW TEMPERATURE FD DISTRIBUTION As β → 0, exp[β(ε – μ)] behaves as follows: If ε = 1; if ε > μ, exp[β(ε – μ)] → exp(∞) = ∞, so that = 0. Thus dN(ε) = f(ε)dε for ε μ. The Fermi energy The Fermi energy  F is defined as  F ≡ μ(T → 0).

25 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 = T F. 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 T F =  F /k lies in the range 10 4 – 10 5 K for metals with one conduction electron per atom. Below room temperature, T/T F < 0.03, and μ ≈  F.

26 Free-Electron Theory: Heat capacity Simplified calculation for T << T F Assume that only those particles within an energy kT of  F can be excited and have mean energies given by “equipartition”; i.e. N eff  N(kT/  F ) = NT/T F. Thus, U  N eff (3/2)kT = (3/2)NkT 2 /T F, so that C V = dU/dT  3NkT/T F. In a better calculation, 4.9 replaces 3. Thus, for a conductor at low temperatures, with the Debye term included, C V = AT 3 + γT, so that C V /T = AT 2 + γ.

27 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},.

28 Free-Electron Theory: Calculation of. Now U =  0   n(  )  f(  ) d .

LOW TEMPERATURE B-E DISTRIBUTION The distribution of particles dN(ε) = 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 ε 1/2 ] is zero at ε = 0. Bose-Einstein CondensationThe 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. Bose or condensation temperatureThe expression dN(ε) = f(  )d  works down to a phase transition, which occurs at the Bose or condensation temperature T B, above which the total number of particles is given by N = ∫ dN(ε). Below T B, 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) = f(  )d , with = 1/{exp[β(ε – μ)] – 1} and Thus, dN(ε) = (2πV/h 3 )(2m) 3/2  1/2 d  /{exp[β(ε – μ)] – 1}, and the total number of particles N is given by N = (2πV/h 3 )(2m) 3/2 ∫  1/2 d  /{exp[β(ε – μ)] – 1}, which is integrated from  = 0 to ∞. 30 f(  )d  = (2πV/h 3 )(2m) 3/2  1/2 d .

Bose-Einstein Condensation 2 Since the number of particles are fixed, with N = (2πV/h 3 )(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. Bose TemperatureThis can happen only down to the Bose Temperature T B, at which μ becomes 0. What happens below T B ? 31

Bose-Einstein Condensation 3 At the Bose (or condensation) temperature T B, μ ≈ 0, so that N = (2πV/h 3 )(2m) 3/2 ∫  1/2 d  /{exp(ε/kT B ) – 1}, which yields on integration, T B = (h 2 /2πmk)(N/2.612V) 2/3. The expression f(  )d  = K  1/2 d , with K = (2πV/h 3 )(2m) 3/2, indicates that for T > T B, the number of particles in the ground state (  = 0) is negligible, since f(  ) = K  1/2 → 0. Behavior of μ above T B.

33 Bose-Einstein Condensation 4

34 Bose-Einstein Condensation 5 For T >T B, the number of particles in the ground state (N) is zero.

35 Bose-Einstein Condensation 6

Bose-Einstein Condensation 7 36

37 Bose-Einstein Condensation 8

Bose-Einstein Condensation 9 38 Classical high-temperature value

39 Bose-Einstein Condensation 10

Appendix Alternative Approach to Quantum Statistics PHYS 4315 R. S. Rubins, Fall 2008

41 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) = x 3 +y 3, and φ(x,y) = xy = 4. Method 1 Eliminating y, f(x,y) = x 3 +(4/x) 3, so that df/dx = 3x 2 - 3(4/x) 4 When df/dx = 0, x 6 = 64,  x = 2, y = 2. Method 2 (Lagrange method) and = α, so that In this example, α is a Lagrange undermined multiplier.

42 Lagrange Method of Undetermined Multipliers 2 Suppose that the function ofis needed. This occurs when df = (∂f/∂x 1 )dx 1 + … + (∂f/∂x n )dx n = 0. Let there be two constraints = N, = U, where in the calculations of the Lagrange’s method of undetermined multipliers gives the following set of equations: mean number of particles in the state j. In the calculations that follow, the function f equals ln(ω), where ω is the thermodynamic degeneracy.

43 Alternate Fermi-Dirac Calculation 1 If the j’th state has degeneracy g j, and contains N j particles, N j ≤ g j 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,.

44 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

45 Alternate Fermi-Dirac Calculation 3 Since the constraints are, we let φ(N 1 …N j …) = N, and ψ(N 1 …N j …) = U, so that where α and β are Lagrange multipliers. Inserting the expression for ln ω FD and remembering that, we obtain.

46 Alternate Fermi-Dirac Calculation 4 reduces to Thus =. The constraint α has been replaced by μ/kT, where μ is the chemical potential, and β by –1/kT.

47 Alternate Bose-Einstein Calculation 1 The j’th energy level has g j quantum states, and contains a total of N j identical particles, with up to N j particles in each state. All possible microstates can be obtained by rearranging (g j – 1) partitions and N j dots, in a diagram like that shown below.. The number of microscopes for a given N j and g j is

48 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

49 Alternate Bose-Einstein Calculation 3 where α and β are Lagrange multipliers. Inserting the expression for ln ω FD, we obtain hence. Using the method of Lagrange multipliers as before,

50 Alternate Bose-Einstein Calculation 4 reduces to Thus =. The constraint α has been replaced by μ/kT, where μ is the chemical potential, and β by –1/kT.