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 Incoming links: Oscillator bilinear commutator

Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 33C / c / Solution Created 2026-09-24 Updated 2026-09-25

For

T_{ij} = a_{i}^{†} a_{j}

, expand the commutator and commute one annihilation operator past one creation operator:

[T_{ij}, T_{k l}] = a_{i}^{†} a_{j} a_{k}^{†} a_{l} - a_{k}^{†} a_{l} a_{i}^{†} a_{j} = δ_{jk} a_{i}^{†} a_{l} - δ_{i l} a_{k}^{†} a_{j} .

Hence the oscillator bilinear commutator is

[T_{ij}, T_{k l}] = δ_{jk} T_{i l} - δ_{i l} T_{kj} .

Since

H = T_{xx} + T_{yy} + 1,

we obtain

[T_{ij}, H] = \sum_{k} (δ_{jk} T_{ik} - δ_{ik} T_{kj}) = 0.

Every

T_{ij}

therefore acts within a fixed energy eigenspace.

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Past exam of the quantum theory course of the University of Oxford / 2021 / 2 / b / ii / Solution 2026-10-25

From part (a),

[N, a_{j}] = - a_{j}

and

[N, a_{j}^{*}] = a_{j}^{*}

. The commutator product rule gives

[N, a_{1} a_{2}^{*}] [N, a_{2} a_{1}^{*}] = [N, a_{1}] a_{2}^{*} + a_{1} [N, a_{2}^{*}] = - a_{1} a_{2}^{*} + a_{1} a_{2}^{*} = 0, = [N, a_{2}] a_{1}^{*} + a_{2} [N, a_{1}^{*}] = - a_{2} a_{1}^{*} + a_{2} a_{1}^{*} = 0.

Consequently

[N, L] = 0.

Each term in

L

transfers one excitation between the two modes, preserving the total number operator eigenvalue. This is an instance of the oscillator bilinear commutator; every number operator eigenspace is an invariant subspace of

L

.

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