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

In the source's Gaussian electromagnetic units, set $\mathbf{\Pi }=-i\hslash \nabla -e\mathbf{A}/c$ and take $\Phi =0$. ThenThe chosen gauge is independent of $x,z,t$, so simultaneous energy and canonical $p_x,p_z$ eigenstates justify the exponential ansatz. On the remaining spinors use ${\Pi }_x=p_x+eBy/c$, ${\Pi }_y=-i\hslash d/dy$, ${\Pi }_z=p_z$. The coupled equations are $(E-m{c}^2)\phi =c\sigma \cdot \Pi \chi$ and $(E+m{c}^2)\chi =c\sigma \cdot \Pi \phi$.

Eliminate the lower component to get $(E^2-m^2{c}^4)\phi =c^2(\sigma \cdot \Pi {)}^2\phi$. Since $[{\Pi }_x,{\Pi }_y]=ie\hslash B/c$, the Pauli identity gives $(\sigma \cdot \Pi {)}^2={\Pi }^2-e\hslash B{\sigma }_3/c$. Substitute the three momenta and rearrange:The spin-field sign follows from the charge convention and commutator, not an independent assignment.

A simpler physical coordinate is $Y=y+p_{x}c/(eB)$. For ${\sigma }_3$ eigenvalue $s_z=\pm 1$, the equation becomes an oscillator of mass $m$, frequency ${\omega }_c=|eB|/(mc)$, and auxiliary energyIf the source's variable $\xi =(p_{x}c+eBy)/(\hslash c)=b_{B}Y$ is used literally, $b_B=eB/(\hslash c)$, then the same canonical oscillator form has $\boxed{M=m/b_B^2,\omega ={\omega }_c}$ and the same $\epsilon$. Because $\xi$ is an inverse length, this auxiliary $M$ is not a physical particle mass with ordinary mass dimensions; the rescaling must be carried through both kinetic and potential terms.

Using $\epsilon =\hslash {\omega }_c(n+1/2)$ gives the charge-explicit resultFor $B>0$ the two spin signs are equivalently $m^2{c}^4+2m{c}^2[p_z^2/(2m)+\mu B(2n+1)\pm \mu B]$. Both positive and negative relativistic energy branches arise from this squared equation; an actual Dirac spinor must still be reconstructed from the coupled equations, especially at exceptional values where division by $E+m{c}^2$ is unavailable.

Separate $x,z$ plane waves in the spinless equation. The transverse operator is $-{\hslash }^2{d}^2/d{y}^2+(p_x+eBy/c{)}^2$, the same shifted oscillator without a Pauli term. Its eigenvalues are $|e|\hslash |B|(2n+1)/c$. Thuswhich is the printed formula for $B>0$. The absence of the spin splitting is the substantive difference from the Dirac result.

A plausible extension assuming gyromagnetic ratio two iswhere the spin projection is aligned with the charge-field sign. For spin one the lowest aligned mode has $E^2=m^2{c}^4-|e|\hslash |B|c$ at $p_z=0$. Above $|eB|>m^2{c}^3/\hslash$, this is negative: linear time evolution has exponentially growing/decaying modes, signaling an instability of the assumed background.

Spin alone does not fix a magnetic moment. The scalar and Dirac examples cannot establish this formula for every higher-spin particle; the stated instability belongs to the specified $g=2$ field model and its regime of applicability. Interactions and backreaction are needed to describe its eventual outcome.