 Solution

ID: past-exam-of-the-quantum-condensed-matter-field-theory-course-of-the-university-of-cambridge/2015/paper-1/1/a/solution

Use

ℏ = 1

and

S_{n}^{\pm} = S_{n}^{x} \pm i S_{n}^{y}

. Operators on different sites commute, and expanding the two products gives

S_{n}^{+} S_{n + 1}^{-} + S_{n}^{-} S_{n + 1}^{+} = 2 (S_{n}^{x} S_{n + 1}^{x} + S_{n}^{y} S_{n + 1}^{y})

. Therefore the quantum XX chain has

H = - \frac{J}{2} n = 1 \sum N (S_{n}^{+} S_{n + 1}^{-} + S_{n}^{-} S_{n + 1}^{+}) - h n = 1 \sum N S_{n}^{z} .

Each exchange moves a reversed spin between neighbours, conserving the total number of reversed spins. The factor of one half follows from defining

S^{\pm}

without a normalizing square root.

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