A brief hint of the expected answers. Note that on the examination itself you should explain what you are doing, and not simply produce formulae. Notation: subscripts are replaced by suffices, and superscripts are denoted by means of the ^ character. 1a) p1 + p2 + p3 = 1 1b) p1 = 1-2p independent -> probability for both sites in state 1 is p1^2 etc -> E = - 2 p^2 - (1-2p)^2 = - 1 + 4p - 6p^2 per nearest neighbor pair. Per site is z/2 times as much. 1c) S = -k (p1 log p1 + p2 log p2 + p3 log p3) = -k (2p log p + (1-2p)log(1-2p)) 1d) T, p figure as independent parameters in E en S. Can be related by minimizing free energy F = E - T S. Hence eq (1) 1e) For p=1/3 both RHS and LHS vanish. To be expected on basis of permutation symmetry between the three states, since the symmetric point is expected to be a stationary point. 1f) LHS monotonically from +inf at p=0 to -inf at p=1/2 RHS linearly from 1 naar -1/2 Them there must be two more solutions that for T->0 approach p->0 and p->1/2 respectively Last one meets p=1/3 as d/dp of RL and LL of (1) are equal -> kT(-2/(1-2p)-1/p) = -3z (op p=1/3) -> kT = z/3 1g) for high T the solution p=1/3 is unique minimum. for some lower T a new min and max are born at smaller p. They must have higher F . (because path from max to min is always decreasing). for yet lower T (kT=z/3) the max meets the min at p=1/3. At that T the other min must be lower (with argument above). Transition from one min to another, with max in between must be discontinuous -> first order. (note it is also given that at low T the unique pj is highest thus it follows p<1/3) 2a) Distance to transition line is certainly relevant. Since transition left and right of tricritical point are different, the parameter along transition line must also be relevant. yh is expected to be positive at the boundary of ferromagnetic phase. -> All three eigenvalues > 1 (i.e. all three y's positive) 2b) Flowlines all out of tricritical point. Initially tangential to transition line. (because y1>y2) 2c) Singular part of free energy: fs = U1^(d/y1) F[ U2 U1^(-y2/y1), h U1^(-yh/y1) ] differentiate twice w.r.t. U1 -> alphat = 2 - d/y1 2d) equivalent form of fs: fs = U2^(d/y2) Q[ U1 U2^(-y2/y2), h U2^(-yh/y2) ] differentiate once w.r.t. h -> M ~ U2^(d/y2-yh/y2) Since T and Delta analytic in U1 and U2 the singular dependence on Delta has same power. 2e) use (at h=0) fs = U1^(d/y1) F[ U2 U1^(-y2/y1) ] F may be singular as U1->0 for non-zero U2. Singularity determined by fixed point toward which the transition line flows. 2f) use (at h=0) fs = U1^(d/y1) F[ U2 U1^(-y2/y1) ] differentiate twice w.r.t. U1 -> alphat = 2 - d/y1 -> C = U1^(d/y1-2) R[ U2 U1^(-y2/y1) ] (R is combination of F and its derivative) We know that U2 non-zero and that C ~ U1^(-alphac) -> U1^(-alphac) = U1^(d/y1-2) [ U2 U1^(-y2/y1) ]^x -> -alphac = (d/y1-2) - x y2/y1 -> x = (alphac y1 + d - 2 y1)/y2 and A(U2) ~ U2^x