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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-81/1/b/solution

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Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 81 / 1 / b / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-07

For occupation much smaller than

2 S

, expand the square root in the Holstein–Primakoff transformation. To leading order

S_{n}^{+} = 2 S a_{n} + O (S^{- 1/2})

S_{n}^{-} = 2 S a_{n}^{†} + O (S^{- 1/2})

and

S_{n}^{z} = S - a_{n}^{†} a_{n}

. Consequently

S_{n} \cdot S_{n + 1} = S^{2} - S (\overset{n}{^}_{n} + \overset{n}{^}_{n + 1}) + S (a_{n}^{†} a_{n + 1} + a_{n + 1}^{†} a_{n}) + O (S^{0}) .

(1)

Periodic boundaries count each occupation in two bonds. The linear spin-wave approximation is therefore

H = - N J S^{2} + J S n \sum [2 a_{n}^{†} a_{n} - a_{n}^{†} a_{n + 1} - a_{n + 1}^{†} a_{n}] + O (S^{0}) .

(2)

The displayed fluctuation Hamiltonian is bilinear and conserves boson number. The omitted terms are magnon interactions, quartic at the first omitted order. With occupation held of order one, they are smaller by

1/ S

than the order-

S

quadratic terms. The polarized vacuum has no zero-point depletion in this ferromagnetic expansion.

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