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 Incoming links: Symmetric gauge

Past exam of the mathematics course of the University of Cambridge / 2003 / ia / Paper 4 / 9E / Solution Created 2026-09-24 Updated 2026-09-24

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Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 4 / 34D / a / Solution Created 2026-09-24 Updated 2026-09-25

With the magnetic minimal coupling convention for charge

e

,

H = \frac{1}{2 m} (- i ℏ\nabla - e A)^{2} .

Here

\nabla \cdot A = 0

,

A^{2} = B^{2} (x^{2} + y^{2}) /4

, and

A \cdot \nabla = \frac{B}{2} (- y \frac{\partial}{\partial x} + x \frac{\partial}{\partial y}) .

Therefore in Cartesian coordinates

H = - \frac{ℏ ^{2}}{2 m} \nabla^{2} + \frac{i e ℏ B}{2 m} (- y \partial_{x} + x \partial_{y}) + \frac{e ^{2} B ^{2}}{8 m} (x^{2} + y^{2}) .

In cylindrical coordinates this is

H = - \frac{ℏ ^{2}}{2 m} [\frac{1}{r} \partial_{r} (r \partial_{r}) + \frac{1}{r ^{2}} \partial_{θ}^{2} + \partial_{z}^{2}] + \frac{i e ℏ B}{2 m} \partial_{θ} + \frac{e ^{2} B ^{2} r ^{2}}{8 m} .

This choice of

A

is the symmetric gauge.

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Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 34B / a / Solution Created 2026-09-24 Updated 2026-09-25

In the symmetric gauge,

\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} .

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

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