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Past exam of the theoretical physics major option course of the University of Oxford / 2016 / 2 / d / Solution by

Codex  0 2026-10-25

Choose the positive-

x

reference ground state and use the cyclic rotation

S^{z} = S^{x}, S^{x} = S^{y}, S^{y} = S^{z} .

This is a right-handed permutation, so it preserves the spin commutation relations. It puts the ordered direction along the Holstein–Primakoff transformation's quantization axis. To the required order,

S_{j}^{z} = S - n_{j}, S_{j}^{x} = \frac{S}{2} (a_{j} + a_{j}^{†}), S_{j}^{y} = - i \frac{S}{2} (a_{j} - a_{j}^{†}) .

The longitudinal exchange contributes

- L J^{x} S^{2} + 2 S J^{x} \sum_{j} n_{j} + O (S^{0})

. The transverse terms contribute

- \frac{S}{2} \sum_{j} [(J^{y} + J^{z}) (a_{j}^{†} a_{j + 1} + a_{j + 1}^{†} a_{j}) + (J^{y} - J^{z}) (a_{j} a_{j + 1} + a_{j}^{†} a_{j + 1}^{†})] .

Dropping the constant and all terms of order

S^{0}

or smaller gives the stated linear spin-wave approximation, with

A = - \frac{S}{2} (J^{y} + J^{z}), B = - \frac{S}{2} (J^{y} - J^{z}), C = 2 S J^{x} .

The anomalous pair terms are present because the two transverse couplings differ. A different right-handed choice of transverse axes can change the sign of

B

by a boson phase convention, while leaving the dispersion relation unchanged.

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