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PHYSICS 140B : STATISTICAL PHYSICS MIDTERM EXAM SOLUTIONS Consider a four-state ferromagnetic Ising model with the Hamiltonian X X Si Sj − H Si , Ĥ = −J i hiji where the first sum is over all links of a lattice of coordination number z. The spin variables Si take values in the set {−1 , 0 , 0 , +1}. Note that there are two distinct states, each with Si = 0, and a total of four possible states on each site. Taking the trace for a single site means we sum over the four independent states, one with S = +1, two with S = 0, and one with S = −1. (a) Making the mean field Ansatz Si = m + (Si − m), where m = hSi i is presumed independent of i, derive the mean field Hamiltonian ĤMF . [15 points] Solution : As usual, we neglect fluctuations and obtain X ĤMF = 21 N zJm2 − (zJm + H) Si . i (b) Find the mean field free energy F (m, T, H). [15 points] Solution : The free energy is obtained from the partition function, N −βF − 21 N zβJm2 β(zJm+H)S Z=e =e Tr e S " =e − 21 N zβJm2 2 + 2 cosh zJm + H kB T " #N . Thus, F (m, T, H) = 12 N zJm2 − N kB T ln 2 + 2 cosh zJm + H kB T # . (c) Adimensionalize, writing θ = kB T /zJ and h = H/zJ. Find the dimensionless free energy per site f = F/N zJ. [15 points] Solution : We have " # m + h f (m, θ, h) = 12 m2 − θ ln 2 + 2 cosh θ " # m + h = 21 m2 − 2θ ln 2 cosh . 2θ 1 (d) What is the self-consistent mean field equation for m? [15 points] ∂f Solution : Setting ∂m = 0 we obtain the mean field equation m+h m = tanh . 2θ (e) Find the critical temperature θc . Show that when h = 0 the graphical solution to the mean field equation depends on whether θ < θc or θ > θc . [15 points] Solution : We set h = 0 and ask when the RHS of the above equation has slope unity. This occurs for θ = θc , where θc = 12 . The graphical solution is depicted in fig 6.4 of the lecture notes. (f) For θ > θc , find m(h, θ) assuming |h| 1. [15 points] Solution : For θ > θc , if |h| 1 then |m| 1 and we can replace the tanh function by the first term in its Taylor series. Thus, m' m+h 2θ =⇒ m(h) = h . 2θ − 1 (g) What is the mean field result for hSi2 i? Hint : We don’t neglect fluctuations from the same site. [10 points] Solution : We have (−1)2 · e−(m+h)/θ + 2 · (0)2 · e0 + (+1)2 · e(m+h)/θ e−(m+h)/θ + 2 · e0 + e(m+h)/θ cosh m+h θ . = 1 + cosh m+h θ hS 2 i = Note that as θ → ∞ we have hS 2 i → 12 , since all four states are equally probable and two of them have S 2 = +1 and the other two have S 2 = 0. As θ → 0 the ground state configurations are selected. These are two completely polarized states, one with Si = +1 ∀ i and the other with Si = −1 ∀ i. Thus hS 2 i → 1 in this limit. 2