 Solution

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.