Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 303 / 1 / a / ii

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- Solution ii

Solution

The mean-field approximation replaces

S_{i} S_{j} = m (S_{i} + S_{j}) - m^{2} + (S_{i} - m) (S_{j} - m)

by its first three terms. Since every site has

q

neighbours,

\sum_{⟨ ij ⟩} S_{i} S_{j} ≃ q m \sum_{i} S_{i} - \frac{Nq}{2} m^{2} .

The sites then decouple, and their three possible weights sum to

1 + e^{- β g + β (J q m + B)} + e^{- β g - β (J q m + B)} = 1 + 2 κ cosh (β (J q m + B)) .

Hence

c_{1} = 1

,

c_{2} = 2

, and

Z = e^{- βN J q m^{2} /2} [1 + 2 κ cosh (β (J q m + B))]^{N} .

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