 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/33c/c/solution

Codex  0 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} .

[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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