MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department 8.044 Statistical Physics I Final Exam, Solutions Problem 1 (20 points) Binary Alloy a) Find the number of different ways of choosing the n α-sites to be vacated and occupied by β atoms. N! #α = n!(N − n)! b) Find the number of different ways of choosing the n β-sites from which to take the β atoms. N! #β = n!(N − n)! c) Find the entropy of the system as a function of n. S(n) = kB ln Ω = kB ln(#α × #β ) = 2kB ln N! n!(N − n)! d) Find U (T, N ). S(n) = 2kB [N ln N − n ln n − (N − n) ln(N − n) − N + n + (N − n)] 1 = T ∂S ∂U = N ∂S ∂n ∂n 2kB = [−1 − ln n + 1 + ln(N − n)] ∂U N N | {z } 1/ = − ln 2kB T n N −n → n = e−/2kB T N −n Solving for n gives n= N 1 + e/2kB T → 1 U (T, N ) = N 1 + e/2kB T Problem 2 (20 points) DNA Model a) This is a classical system with N non-degenerate states with energies En = n. Z1 = N X e−n/kB T n=0 b) When kT one need consider only the lowest two energy states; this becomes an energy gap dominated situation. N X hni = n n=0 e−n/kB T Z 0 e−0 + 1 e−/kB T e−/kB T = e−0 + e−/kB T 1 + e−/kB T ≈ ≈ e−/kB T c) When kB T one can approximate sums over n by integrals. For example Z= N X e −n/kB T = n=0 kB T X N −n/kB T e kB T n=0 kB T ≈ ∞ Z |0 e−x dx = {z } 1 d) In a similar manner hni = N X n n=0 e−n/kB T Z N N X n −n/kB T kB T X n −n/kB T = e = e k T k T kB T B B n=0 n=0 Z kB T ∞ x ≈ xe dx 0 | {z } 1 ≈ kB T Alternatively use Z1 = 1/(β). 1 hni = = −1 ∂Z Z ∂β 2 1 = −1 Z −Z β = kB T kB T Problem 3 (20 points) Spin Waves a) ~ = Lx Ly Lz = V D(k) 2π 2π 2π (2π)3 b) ~ #(ω) = (volume of sphere in k-space) × D(k) = = D(ω) = D(ε) 4 3 V πk (ω) use k = (ω/a)1/2 3 3 (2π) V ω 3/2 6π 2 a d#(ω) V = a−3/2 ω 1/2 2 dω (2π) c) 0 ε ∞ Z h(ω)i D(ω) dω U = 0 Z ∞ ~ω V −3/2 ω 1/2 dω + Z.P. contribution = a ~ω/k 2 B T − 1) (e (2π) 0 3/2 Z ∞ 3/2 x V 1 5/2 = (k T ) dx + Z.P. contribution B (2π)2 ~a ex − 1 0 | {z } ≡ I CV (T, V ) = ∂U ∂T V 5 = 2 kB V 8π kB T ~a 3/2 I d) There is no energy gap behavior (CV ∝ T n e−∆/kB T ) because of the integration over a continuous distribution of gaps (∆ = ~ω), some of which are less than kB T for any physical T. 3 Problem 4 (20 points) Graphene a) ~ = Lx Ly = A D(k) 2π 2π (2π)2 b) ~ #() = 2 × (area of disk in k-space) × D(k) = 2 × πk 2 () A = 2π 1 ~v A use k = 2 (2π) ~v 2 d#() A Dc () = = d π 2 1 ~v 2 c) D(ε) 0 ε d) µ(T = 0) rests at the last filled state at T = 0 which is at the top of the valence band, so µ(T = 0) = 0 . D() is symmetric about = 0. If µ stays at = 0 the symmetry of hn(, T )i assures that as T increases the number of electrons lost from the valence band is exactly equal to the number of electrons appearing in the conduction band. Thus µ(T ) = 0 for all T covered by this model. 4 e) ∞ Z hn(, T )i D() d U = −∞ Z ∞ hn(, T )i Dc () d = 2 0 A = π A = π A = π 1 ~v 2 Z 1 ~v 2 1 ~v 2 ∞ 1 (e/kB T 0 3 Z (kB T ) |0 ∞ + 1) 2 d x2 dx (ex + 1) {z } ≡ I (kB T )3 I f) CA (T ) = (∂U/∂T )A so CA (T ) will be proportional to T 2 , that is the temperature exponent n = 2. 5 Problem 5 (20 points) BEC a) ∞ Z N = < n > D() d 0 ∞ Z 1 = e(−µ)/kB T 0 V = 4π 2 = V 4π 2 n = V 4π 2 2m ~2 " V − 1 (2π)2 3/2 Z √ ∞ 2m ~2 3/2 √ # d d e(−µ)/kB T − 1 3/2 Z ∞ √ 2mkB T x dx 2 (x − y) ~ e −1 0 | {z } ≡ I(y) 3/2 2mkB T I(y) ~2 0 b) Bose-Einstein condensation begins when the above condition is satisfied with µ = 0 which also means our dimensionless parameter y = 0. 3/2 V 2mkB Tc nc = 2 I(y = 0) 4π ~2 c) dU = T dS − P dV + µdN ; change independent variables to T, V, N : dS = ∂S ∂T dT + T,V ∂S ∂V dV + T,N ∂S ∂N dN T,V So ∂U/∂N |V,T = µ + T ∂S/∂N |T,V . Now use Maxwell relation derivable from dF = ... on the information sheet: ∂S/∂N |T,V = −∂µ/∂T |T,N . so ∂U ∂N =µ−T V,T ∂µ ∂T = −T 2 N,V ∂ µ ∂T T = N,V ∂(βµ) ∂β N,V In the Bose condensed phase µ = 0 and is independent of the temperature, so both terms in ∂U/∂N are zero. 6 d) From answer to part a) 1 n= 2 4π 2mk ~2 3/2 T 3/2 Z ∞ 0 √ x dx e(x−y) − 1 N C × β 3/2 = V Z 0 ∞ √ x dx e(x−y) − 1 where C is a collection of constants. Differentiate this equation implicitly w.r.t. β and y = µβ, Z ∞ √ 3 N 1/2 e(x−y) dy xdx (x−y) × β = 2 2 V (e − 1) dβ 0 Both the term on the left and the term in () are positive definite. Thus dy/dβ > 0. 7

8.044 Final Exam (Solutions)

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