34B (Applications of Quantum Mechanics)

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

(1)

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,

(2)

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} .

(3)

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} .

(4)

These are the commutators of the kinetic momentum and magnetic pseudomomentum.

b

 0  0

Solution

 0  0

With the convention in the question,

T_{r} = exp (\frac{i}{ℏ} r \cdot p)

(1)

acts as

(T_{r} ψ) (x) = ψ (x + r) .

(2)

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 .

(3)

Consequently

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

(4)

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)

(5)

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) .

(6)

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) .

(7)

c

 0  0

Solution

 0  0

Part (a) gives

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

(1)

For

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

(2)

their commutator is the scalar

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

(3)

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} .

(4)

Writing

Φ = B \cdot (r \times r^{'})

, the translations commute exactly when their phase is one:

\frac{e Φ}{ℏ} = 2 πn, n \in Z .

(5)

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 .

(6)

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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 34B (Applications of Quantum Mechanics)

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