D.74 Elec­tron spin in a mag­netic field

If you are cu­ri­ous how the mag­netic di­pole strength of the elec­tron can just pop out of the rel­a­tivis­tic Dirac equa­tion, this note gives a quick de­riva­tion.

First, a prob­lem must be ad­dressed. Dirac's equa­tion, chap­ter 12.12, as­sumes that Ein­stein's en­ergy square root falls apart in a lin­ear com­bi­na­tion of terms:

$$H = \sqrt{\left(m c^2\right)^2 + \sum_{i=1}^3 \left({\wideh... ...)^2} = \alpha_0 mc^2 + \sum_{i=1}^3 \alpha_i {\widehat p}_i c$$

which works for the 4 $\times$ 4 $\alpha$ ma­tri­ces given in that sec­tion. For an elec­tron in a mag­netic field, ac­cord­ing to chap­ter 13.1 you want to re­place ${\skew 4\widehat{\skew{-.5}\vec p}}$ with ${\skew 4\widehat{\skew{-.5}\vec p}}-q\skew3\vec A$ where $\skew3\vec A$ is the mag­netic vec­tor po­ten­tial. But where should you do that, in the square root or in the lin­ear com­bi­na­tion? It turns out that the an­swer you get for the elec­tron en­ergy is not the same.

If you be­lieve that the Dirac lin­ear com­bi­na­tion is the way physics re­ally works, and its de­scrip­tion of spin leaves lit­tle doubt about that, then the an­swer is clear: you need to put ${\skew 4\widehat{\skew{-.5}\vec p}}-q\skew3\vec A$ in the lin­ear com­bi­na­tion, not in the square root.

So, what are now the en­ergy lev­els? That would be hard to say di­rectly from the lin­ear form, so square it down to $H^2$, us­ing the prop­er­ties of the $\alpha$ ma­tri­ces, as given in chap­ter 12.12 and its note. You get, in in­dex no­ta­tion,

$$H^2 = \left(m c^2\right)^2 I + \sum_{i=1}^3 \Big(({\wideha... ...a_{{\overline{\imath}}}\alpha_{{\overline{\overline{\imath}}}}$$

where $I$ is the four by four unit ma­trix, ${\overline{\imath}}$ is the in­dex fol­low­ing $i$ in the se­quence 123123..., and ${\overline{\overline{\imath}}}$ is the one pre­ced­ing $i$. The fi­nal sum rep­re­sents the ad­di­tional squared en­ergy that you get by sub­sti­tut­ing ${\skew 4\widehat{\skew{-.5}\vec p}}-q\skew3\vec A$ in the lin­ear com­bi­na­tion in­stead of the square root. The com­mu­ta­tor arises be­cause $\alpha_{{\overline{\imath}}}\alpha_{{\overline{\overline{\imath}}}}+\alpha_{{\overline{\overline{\imath}}}}\alpha_{{\overline{\imath}}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, giv­ing the terms with the in­dices re­versed the op­po­site sign. Work­ing out the com­mu­ta­tor us­ing the for­mu­lae of chap­ter 4.5.4, and the de­f­i­n­i­tion of the vec­tor po­ten­tial $\skew3\vec A$,

$$H^2 = \left(m c^2\right)^2 I + \sum_{i=1}^3 \Big(({\wideha... ..._{{\overline{\imath}}}\alpha_{{\overline{\overline{\imath}}}}.$$

By mul­ti­ply­ing out the ex­pres­sions for the $\alpha_i$ of chap­ter 12.12, us­ing the fun­da­men­tal com­mu­ta­tion re­la­tion for the Pauli spin ma­tri­ces that $\sigma_{{\overline{\imath}}}\sigma_{{\overline{\overline{\imath}}}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${\rm i}\sigma_i$,

$$H^2 = \left(m c^2\right)^2 I + \sum_{i=1}^3 \Big(({\wideha... ...array}{cc} \sigma_i & 0\\ 0 & \sigma_i \end{array} \right)$$

It it seen that due to the in­ter­ac­tion of the spin with the mag­netic field, the square en­ergy changes by an amount $-qhc^2\sigma_i{\cal B}_i$. Since $\frac12\hbar$ times the Pauli spin ma­tri­ces gives the spin ${\skew 6\widehat{\vec S}}$, the square en­ergy due to the mag­netic field act­ing on the spin is $-2qc^2{\skew 6\widehat{\vec S}}\cdot\skew2\vec{\cal B}$.

In the non­rel­a­tivis­tic case, the rest mass en­ergy $mc^2$ is much larger than the other terms, and in that case, if the change in square en­ergy is $-2qc^2{\skew 6\widehat{\vec S}}\cdot\skew2\vec{\cal B}$, the change in en­ergy it­self is smaller by a fac­tor $2mc^2$, so the en­ergy due to the mag­netic field is

$$H_{{\cal B}S} = - \frac{q}{m} {\skew 6\widehat{\vec S}}\cdot \skew2\vec{\cal B}$$ (D.52)

which is what was to be proved.