 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2012/s18/1/4/solution

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Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S18 / 1 / 4 / Solution by

Codex  0 2026-10-25

With

A^{μ} = (0, A)

, the lowered components are

A_{μ} = (0, - A)

π = - i \nabla + e A

and

B = \nabla \times A

. This convention describes physical charge

q = - e

. Since

F_{ij} = - ϵ_{ijk} B_{k}

,

σ_{μν} F^{μν} = - 2 (σ \cdot B 0 0 σ \cdot B) .

ψ = e^{- i (m + E) t} ϕ

makes the squared equation

[(m + E)^{2} - m^{2} - π^{2} - e Σ \cdot B] ϕ = 0.

Drop

E^{2}

relative to

2 m E

and write

ϕ = (φ, χ)^{T}

. Both two-component blocks obey the Pauli equation

Eφ = [\frac{( p + e A ) ^{2}}{2 m} + \frac{e}{2 m} σ \cdot B] φ, E χ = [\frac{( p + e A ) ^{2}}{2 m} + \frac{e}{2 m} σ \cdot B] χ .

The gauge-covariant kinetic term contains

p^{2} / (2 m)

,

e (A \cdot p + p \cdot A) / (2 m)

, and

e^{2} A^{2} / (2 m)

: free kinetic, orbital magnetic, and diamagnetic contributions. The spin term is

- μ \cdot B

with

μ = - e σ / (2 m)

, giving spin gyromagnetic factor

g = 2

at this order. In uniform

B

and symmetric gauge, the orbital term is

e B \cdot L / (2 m)

.

Squaring introduces extra formal solutions unless the original first-order equation is retained. In the usual positive-energy Dirac basis it relates the small lower component to the upper one by

χ = (σ \cdot π) φ / (2 m + E) ≃ (σ \cdot π) φ / (2 m)

. Thus these are not two independent nonrelativistic particles. Restoring units places

ℏ

in the spin term,

e ℏ σ \cdot B / (2 m)

.

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