 Solution

ID: past-exam-of-the-theoretical-physics-major-option-course-of-the-university-of-oxford/2017/1/d/solution

Insert the decorating-spin observable into the one bond that contains it. The decorated-spin occupation insertion uses the matrix:

Q_{s t} = \sum_{t_{0} = - 1}^{1} t_{0}^{2} e^{K t_{0} (s + t)} = 2 cosh [K (s + t)] = (2 cosh 2 K 2 2 2 cosh 2 K)_{s t} .

The partition function still contains

L

decorating-spin sums; one of them has been replaced by this weighted sum. Hence

⟨ τ_{1}^{2} ⟩_{L} = \frac{tr ( Q V ^{L - 1} )}{Z _{L}} = \frac{q _{+} λ _{+}^{L - 1} + q _{-} λ _{-}^{L - 1}}{λ _{+}^{L} + λ _{-}^{L}}, q_{+} = 2 cosh 2 K + 2, q_{-} = 2 cosh 2 K - 2.

Taking the dominant eigenvector gives

L \to \infty lim ⟨ τ_{1}^{2} ⟩_{L} = \frac{q _{+}}{λ _{+}} = \frac{cosh 2 K + 1}{cosh 2 K + 2} = \frac{2 cosh ^{2} K}{1 + 2 cosh ^{2} K} .

This is the probability that the decorating site is in either nonzero state. It tends to

2/3

at high temperature, when its three states are equally likely, and to one at low temperature, when the nonzero aligned states gain bond energy. Unlike the magnetization, this even observable is not forced to vanish by spin reversal.

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