 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-4/34d/a/solution

Codex  0 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

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