 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/33c/a/solution

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

L_{i} = ε_{ik ℓ} X_{k} P_{ℓ} .

Using

[X_{i}, X_{j}] = [P_{i}, P_{j}] = 0

and

[X_{i}, P_{j}] = i ℏ δ_{ij}

[L_{i}, X_{j}] = ε_{ik ℓ} X_{k} [P_{ℓ}, X_{j}] = - i ℏ ε_{ikj} X_{k} = i ℏ ε_{ijk} X_{k},

and

[L_{i}, P_{j}] = ε_{ik ℓ} [X_{k}, P_{j}] P_{ℓ} = i ℏ ε_{ik ℓ} δ_{kj} P_{ℓ} = i ℏ ε_{ij ℓ} P_{ℓ} .

[L_{i}, X_{j}] = i ℏ ε_{ijk} X_{k}, [L_{i}, P_{j}] = i ℏ ε_{ijk} P_{k} .

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