 Solution

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

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

Part (a) gives

[K_{i}, K_{j}] = - i e ℏ ε_{ijk} B_{k} .

For

M = \frac{i}{ℏ} r \cdot K, N = \frac{i}{ℏ} r^{'} \cdot K,

[M, N] = \frac{i e}{ℏ} (r \times r^{'}) \cdot B .

It therefore commutes with both

M

and

N

. Applying the stated Baker--Campbell--Hausdorff identity in both orders yields the magnetic translation algebra

T_{r} T_{r^{'}} = exp [\frac{i e}{ℏ} (r \times r^{'}) \cdot B] T_{r^{'}} T_{r} .

Φ = B \cdot (r \times r^{'})

, the translations commute exactly when their phase is one:

\frac{e Φ}{ℏ} = 2 πn, n \in Z .

Thus the cell flux must be an integral multiple of the magnetic flux quantum:

Φ = \frac{2 π ℏ}{e} n = \frac{h}{e} n .

 New to topics? Read the docs here!