Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 4 / 35A / b / ii

Table of contents
- Solution ii

Solution

At

B = 0

,

M = (a b b a), a = e^{β J}, b = e^{- β J},

with eigenvalues

λ_{+} = a + b = 2 cosh (β J)

and

λ_{-} = a - b = 2 sinh (β J)

. Diagonalizing gives

M^{p} = \frac{1}{2} (λ_{+}^{p} + λ_{-}^{p} λ_{+}^{p} - λ_{-}^{p} λ_{+}^{p} - λ_{-}^{p} λ_{+}^{p} + λ_{-}^{p}) .

Substitution in part (i), together with

Z = λ_{+}^{N} + λ_{-}^{N}

, gives

⟨ s_{1} s_{i} ⟩ = \frac{λ _{-}^{i - 1} λ _{+}^{N - i + 1} + λ _{+}^{i - 1} λ _{-}^{N - i + 1}}{λ _{+}^{N} + λ _{-}^{N}} .

Since

λ_{-} / λ_{+} = tanh (β J)

, the required One-dimensional Ising correlation function is

⟨ s_{1} s_{i} ⟩ = \frac{tanh ^{i - 1} ( β J ) + tanh ^{N - i + 1} ( β J )}{1 + tanh ^{N} ( β J )} .

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