 Solution

ID: past-exam-of-the-theoretical-physics-major-option-course-of-the-university-of-oxford/2016/1/a/solution

Let

β = (k_{B} T)^{- 1}

and sum over all

2^{2 L}

Z_{L} = \sum_{{σ, τ}} e^{- βE (σ, τ)}, f_{L} = - \frac{k _{B} T}{2 L} lo g Z_{L} .

Both spins in each cell count as sites. In particular the denominator is

2 L

, rather than

L

⟨ m ⟩ = \frac{1}{2 L Z _{L}} {σ, τ} \sum [j = 1 \sum L (σ_{j} + τ_{j})] e^{- βE} .

Equivalently introduce a uniform magnetic field through

E_{h} = E - h \sum_{j} (σ_{j} + τ_{j})

and differentiate:

⟨ m ⟩ = \frac{1}{2 L β} \frac{\partial l o g Z _{L} ( h )}{\partial h}_{h = 0} .

This convention absorbs the microscopic magnetic moment into

h

. Simultaneous reversal of every spin leaves

E

invariant and changes the sign of the numerator, so

⟨ m ⟩ = 0

at zero field for finite

L

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