 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/34b/a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

\partial_{i} A_{j} + \partial_{j} A_{i} = 0, \partial_{i} A_{j} - \partial_{j} A_{i} = ε_{ijk} B_{k} .

Together with

[p_{i}, f (x)] = - i ℏ \partial_{i} f

, this gives

[p_{i} + e A_{i}, p_{j} - e A_{j}] = - e [p_{i}, A_{j}] + e [A_{i}, p_{j}] = i e ℏ (\partial_{i} A_{j} + \partial_{j} A_{i}) = 0,

whereas

[p_{i} + e A_{i}, p_{j} + e A_{j}] = e [p_{i}, A_{j}] + e [A_{i}, p_{j}] = i e ℏ (\partial_{j} A_{i} - \partial_{i} A_{j}) = - i e ℏ ε_{ijk} B_{k} .

Thus

[p_{i} + e A_{i}, p_{j} - e A_{j}] = 0, [p_{i} + e A_{i}, p_{j} + e A_{j}] = - i e ℏ ε_{ijk} B_{k} .

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