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

With $A^{\mu }=(0,\mathbf{A})$, the lowered components are $A_{\mu }=(0,-\mathbf{A})$. Define $\mathbf{\pi }=-i\nabla +e\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$. This convention describes physical charge $q=-e$. Since $F_{ij}=-{ϵ}_{ijk}{B}_k$,Writing $\psi =e^{-i(m+E)t}\varphi$ makes the squared equationDrop $E^2$ relative to $2mE$ and write $\varphi =(\phi ,\chi {)}^T$. Both two-component blocks obey the Pauli equationThe gauge-covariant kinetic term contains ${\mathbf{p}}^2/(2m)$, $e(\mathbf{A}\cdot \mathbf{p}+\mathbf{p}\cdot \mathbf{A})/(2m)$, and $e^2{A}^2/(2m)$: free kinetic, orbital magnetic, and diamagnetic contributions. The spin term is $-\mathbf{\mu }\cdot \mathbf{B}$ with $\mathbf{\mu }=-e\mathbf{\sigma }/(2m)$, giving spin gyromagnetic factor $g=2$ at this order. In uniform $\mathbf{B}$ and symmetric gauge, the orbital term is $e\mathbf{B}\cdot \mathbf{L}/(2m)$.

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 $\chi =(\mathbf{\sigma }\cdot \mathbf{\pi })\phi /(2m+E)\simeq (\mathbf{\sigma }\cdot \mathbf{\pi })\phi /(2m)$. Thus these are not two independent nonrelativistic particles. Restoring units places $\hslash$ in the spin term, $e\hslash \mathbf{\sigma }\cdot \mathbf{B}/(2m)$.