 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/2017/s18/3/2/solution

Eliminate the lower component to get

(E^{2} - m^{2} c^{4}) φ = c^{2} (σ \cdot Π)^{2} φ

. Since

[Π_{x}, Π_{y}] = i e ℏ B / c

, the Pauli identity gives

(σ \cdot Π)^{2} = Π^{2} - e ℏ B σ_{3} / c

. Substitute the three momenta and rearrange:

[\frac{d ^{2}}{d y ^{2}} - \frac{( p _{x} c + e B y ) ^{2}}{ℏ ^{2} c ^{2}} + \frac{E ^{2} - m ^{2} c ^{4}}{ℏ ^{2} c ^{2}} - \frac{p _{z}^{2}}{ℏ ^{2}} + \frac{e B}{ℏ c} σ_{3}] φ = 0.

The spin-field sign follows from the charge convention and commutator, not an independent assignment.

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