MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Statistical Physics I Spring Term Solutions to Problem Set #10

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1 MASSACHUSES INSIUE OF ECHNOLOGY Physics Department Statistical Physics I Spring erm 203 Problem : wo Identical Particles Solutions to Problem Set #0 a) Fermions:,, 0 > ɛ 2 0 state, 0, > ɛ 3 0,, > ɛ 2 + ɛ 3 b) Bosons: 2, 0, 0 > 0 0 state,, 0 > ɛ 2, 0, > ɛ 3 0, 2, 0 > 2ɛ 2 0,, > ɛ 2 + ɛ 3 0, 0, 2 > 2ɛ 3 c) Let β /k. Z F ( ) e ɛ 2β + e ɛ 3β + e (ɛ 2+ɛ 3 )β Z ( ) + e ɛ2β + e ɛ 3β 2 B + e ɛ2β + e (ɛ 2+ɛ 3 )β + e 2ɛ 3β d) Z F ( ) e ɛ 2β + e ɛ 3β Z F < E > F ZF β β ɛ2e ɛ2β + ɛ 3 e ɛ 3 ɛ 2 + ɛ e e ɛ 2β + e ɛ 3β + e (ɛ 3 ɛ 2 )β [ ] [ ] ɛ + ɛ e (ɛ 3 ɛ 2 )β 2 3 e (ɛ 3 ɛ 2 )β 2 (e (ɛ 3 ɛ 2 )β ) 3 (ɛ 3 ɛ 2 )β ɛ + (ɛ ɛ )e (ɛ 3 ɛ 2 )β ɛ 3 e 2(ɛ 3 ɛ 2 )β e 2(ɛ 3 ɛ 2 )β ɛ 2 + (ɛ 3 ɛ 2 )e (ɛ 3 ɛ 2 )β

2 his result shows a finite < E > at 0 and energy gap behavior with ɛ 3 ɛ 2. ZB < E > B β Z B ɛ2e ɛ 2β + e ɛ 2β ɛ e ɛ 2β 2 his result shows < E > 0 at 0 and energy gap behavior with ɛ 2. Problem 2: A Number of wo-state Particles he two single-particle available are indicated below. ψ ɛ ψ 0 ɛ 0 a) Use the number of particles in the upper single particle state as the index for the manyparticle : n 0, n > N n, n > and E n n n 0,, 2, N N + many-particle b) Let β /k. ( ) e n β e n N Z(N, ) N n 0 n0 β ( ) n ( ) n e β e β n 0 n N+ ( ) e β n ( ) e β N+ ( ) e m n 0 m0 e (N+) β e β e β β e (N+) β e β 2

3 c) n /k e p(n ) e /k Z e (N+) /k e n /k d) If the particles are distinguishable, similar, and non-interacting, then N Zd(N, ) ( Z one particle ) ( + e ) /k N Problem 3: Identical Particle Effects in Rotational Raman Scattering a) Since the Cl nuclei have half-integer spin, I 3/2, they are Fermions and their wavefunction must be anti-symmetric under interchange of the two particles. he two-particle wavefunction consists of two factors, one representing the spatial motion (rotation) of the particles and one representing their spins. he spatial wavefunctions are eigenfunctions of the angular momentum with quantum numbers L and m L. he wavefunctions with even L are symmetric; those with odd L are anti-symmetric. hus the of even L must be paired with spin that are anti-symmetric and those with odd L must be paired with symmetric spin. he nuclear spin are formed by adding two angular momenta, each of magnitude 3/2. he resulting total spin can have the values I 3, 2, and 0. he state with the highest magnitude of the total spin is always symmetric, and the symmetry alternates as the total magnitude is stepped down, by each time. So there are a total of 0 symmetric spin ((2 3 + ) + (2 + )), and 6 anti-symmetric spin ((2 2 + ) + (2 0 + )). he rotational energy levels increase monotonically with L, so the spin, and the associated degeneracy, must alternate between the symmetric and anti-symmetric values. For the spin degeneracies that would be 6, 0, 6, 0,. hus the ratio of the two separate intensity envelopes is 6/0 3/ b) reating L as a continuous variable gives the expression for the envelope of the series of delta functions in the Raman spectrum. We find the spot where that envelope is a maximum. d (2L + ) exp[ L(L + )ΘR / ] (2 dl 0 (Θ R / )(2L + ) 2 ) exp[ L(L + )Θ R / ] (2L + ) 2 2(/Θ R ) L 2Θ R 3

4 c) For N 2 L he N 2 spectrum in the notes peaks near the 8 th line, which arrises from transitions from the L 7 level. For N 2 L so one expects the rotational Raman spectrum to peak near the 22 nd line. d) Since the two nuclei are two different isotopes they are not identical particles. All 6 nuclear spin are allowed with each of the rotational. he spectrum will evolve smoothly with L. he intensities of the lines will not alternate. Problem 4: Langmuir Isotherm a) he number of ways one can place N identical atoms on M N distinct sites, given that each site can hold no more than one atom is equal to the number of ways of choosing N objects from M when order does not count: M! (M N)!N! b) Z M! exp[ E /k B ] state g n exp[ E n /k B ] exp[nɛ 0 /k B ] (M N)!N! n c) F k B ln Z ln Z Nɛ 0 /k B + M ln M M (M N) ln(m N) + (M N) N ln N + N µ ( ) F N k B [ɛ 0 /k B + + ln(m N) ln N + ] ( ) M N ɛ 0 k B ln N 4

5 d) From the lecture we have the chemical potential of a classical non-interacting gas in 3 dimensions. µ k ln nλ 3 ( bulk B ) In equilibrium the two chemical potentials must be equal. ( ) f ɛ 0 k B ln f ( ) f ln f µ surface µ bulk ( f ) f k B ln nλ 3 ( ) ɛ 0 /k B + ln nλ 3 ( ) λ 3 ( ) e ɛ0 /k B n λ 3 ( ) eɛ 0 /k B n f + λ 3( ) eɛ 0 /kb n λ( ) h/ 2πmk B so if h 0 then f 0. e) Use the ideal gas law P (N/V )k B nk B to change the variable representing the bulk gas from number density n to pressure P. αp f + αp where the parameter α depends on temperature through the expression ( ) 3 ɛ 0 /k B 3 α λ ( )/k ɛ /k B e h (2πm) 3/2 (k B ) 5/2 e 0 B α 5

6 Problem 5: Information heory Approach to Statistical Mechanics a) p state and Ω () Ω ( ) ( ) S I k B ln Ω Ω ( ) k B ln(ω () Ω S I k B ln Ω S microcanonical b) e βe state p state, Z e βe state and F (, V ) k B ln Z Z S I k B p state ln ( ) e βe state Z k B p state ( βe state ln Z) < E > +k B ln Z p state }{{} (U + k B ln Z) S I (U F ) (U (U S)) S canonical 6

7 c) [S I + λ () + λ U U + λ NN] [ kb p N,ν ln p N,ν + λ p N,ν + λ U E N,ν p N,ν + λ NNp N,ν ] p N,ν p N,ν N,ν ( ) pn,ν k B + ln p N,ν + λ + λ U E N,ν + λ p NN] 0 N,ν ln p N,ν λ /k B + (λ U /k B ) E N,ν + (λ N/k B) N d) k B [pn,ν ln p N,ν ] [(λ k B )p N,ν + λ U E N,ν p N,ν + λ N Np N,ν ] N,ν N,ν S I (λ k B ) + λ U U + λ N N We can solve the given expression for the grand potential for S and find Φ U µ S + N If the information theory entropy is to equal the thermodynamic entropy then we must make the following assignments. Φ (λ k B ) λ U µ λ N Substituting these back into the expression we found for the probabilities we found in c) gives Φ E N ln p N,ν kb,ν µn + k B k B p N,ν exp[(µn E N,ν )/k B ] exp[φ/k B ] 7

8 Recall that the grand partition function Z is related to the grand potential Φ by the relation Φ(, V, µ) k B ln Z Using this brings us to the final form for the probability density in the grand canonical ensemble. p N,ν exp[(nµ E N,ν )/k B ] /Z Problem 6: Maxwell Relations Begin with the internal energy U. du ds P dv + µdn Holding V constant and recognizing that the cross derivatives of and µ must be equal since U is a state function gives ( ) ( ) µ S V,N N S,V Inverting the partial derivatives on both sides gives us ( ) ( ) S N µ V,N S,V Now examine the Helmholtz free energy F U S. df Sd P dv + µdn Hold the temperature constant and equate the cross derivatives of µ and P. ( ) ( ) µ P V,N N,V Continuing alphabetically, let s examine the Gibbs free energy G U S + P V. dg Sd + V dp + µdn Hold P fixed and equate the cross derivatives of µ and ( ) ( ) µ S P,N N,P 8

9 Finally examine the enthalpy H U + P V. dh ds + V dp + µdn Hold S constant and equate the cross derivatives of mu and V. ( ) ( ) µ V P N S,N S,P 9

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