 Solution

ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2022/3/b/solution

z = 4

. The mean-field decoupling is

S_{i} S_{j} ≃ S (S_{i} + S_{j}) - S^{2} .

Accounting for the factor

1/2

in the directed-neighbour sum gives the Hamiltonian in the mean-field approximation:

H_{MF} = 2 J N S^{2} - (4 J S + B) \sum_{i} S_{i} .

The constant term is necessary to correct the double counting of the interaction. Let

h = 4 J S + B

Z_{1} = \sum_{s = - 1, 0, 1} e^{β h s} = 1 + 2 cosh (β h),

Z_{MF} = e^{- 2 β J N S^{2}} Z_{1}^{N}

F_{MF} (S, T, B) = N [2 J S^{2} - \frac{1}{β} lo g {1 + 2 cosh [β (4 J S + B)]}] .

The equilibrium mean-field approximation is obtained by minimizing this expression over the allowed

S

. The

s = 0

state supplies the extra

1

Z_{1}

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