 Solution

ID: past-exam-of-the-advanced-quantum-physics-course-of-the-university-of-cambridge/2021/section-b/5/a/solution

The components of the momentum operator do not generally commute with a position-dependent magnetic vector potential. Expanding minimal coupling with the ordering intact gives

H = \frac{p ^{2}}{2 m} + V + \frac{e}{2 m} (p \cdot A + A \cdot p) + \frac{e ^{2}}{2 m} A^{2} .

p \cdot A - A \cdot p = - i ℏ\nabla \cdot A

. In the Coulomb gauge the divergence vanishes, so the two cross terms agree. Discarding the quadratic term in the weak radiation amplitude gives

H = H_{0} + H_{1} (t) + O (A^{2}), H_{0} = \frac{p ^{2}}{2 m} + V, H_{1} (t) = \frac{e}{m} p \cdot A (r, t) .

The equality of the two operator orderings follows from the gauge condition, not from treating

p

and

A

as commuting numbers component by component.

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