 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/2018/s18/3/c/solution

Use

p = - i ℏ\nabla

and regard

e

as the signed charge in the printed minimal coupling. Multiplying the first Dirac equation by

c γ^{0}

gives

H = c α \cdot π + β m c^{2} + e Φ

, with

π = p - e A / c

. Removing the rest-energy phase yields

i ℏ \partial_{t} \tilde{φ} i ℏ \partial_{t} \tilde{χ} = e Φ \tilde{φ} + c σ \cdot π \tilde{χ}, = c σ \cdot π \tilde{φ} + (e Φ - 2 m c^{2}) \tilde{χ} .

For the alternative equation use its specified field conventions literally. They give

F_{μν} γ^{μ} γ^{ν} = \frac{2}{c} α \cdot E + 2 i Σ \cdot B, Σ = diag (σ, σ) .

Consequently its Hamiltonian is

c α \cdot p + β m c^{2} - κ β Σ \cdot B + (iκ / c) β α \cdot E

. The two shifted equations are

i ℏ \partial_{t} \tilde{φ} i ℏ \partial_{t} \tilde{χ} = - κ σ \cdot B \tilde{φ} + (c σ \cdot p + \frac{iκ}{c} σ \cdot E) \tilde{χ}, = (c σ \cdot p - \frac{iκ}{c} σ \cdot E) \tilde{φ} + (- 2 m c^{2} + κ σ \cdot B) \tilde{χ} .

These electric-field factors depend on the source's convention

E_{i} = c F_{0 i}

; substituting a different Gaussian or SI definition without changing the equation would give incorrect powers of

c

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