 Solution

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

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

With the convention in the question,

T_{r} = exp (\frac{i}{ℏ} r \cdot p)

acts as

(T_{r} ψ) (x) = ψ (x + r) .

Although

V (x + r) = V (x)

for

r \in Λ

, the vector potential changes by

A (x + r) = A (x) + \frac{1}{2} B \times r .

Consequently

T_{r} H T_{r}^{- 1} = \frac{1}{2 m} [p - e A (x + r)]^{2} + V (x) \neq = H

for a nonzero magnetic field in general.

Now let

K = p + e A

. Since

r \cdot K = - i ℏ r \cdot \nabla + \frac{e}{2} r \cdot (B \times x)

and the derivative in the

r

direction annihilates

r \cdot (B \times x)

, the two terms in its exponential commute. The magnetic translation operator therefore acts as

(T_{r} ψ) (x) = exp (\frac{i e}{2ℏ} r \cdot (B \times x)) ψ (x + r) .

Part (a) gives

[K_{i}, p_{j} - e A_{j}] = 0

, so

T_{r}

commutes with the kinetic term. Its phase commutes with

V

, while its translation sends

V (x)

V (x + r) = V (x)

. Hence

[T_{r}, H] = 0 (r \in Λ) .

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