D.75 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:

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

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,

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

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 $\skew3\vec A$,

\begin{displaymath}
H^2 = \left(m c^2\right)^2 I
+ \sum_{i=1}^3 \Big(({\wide...
...{{\overline{\imath}}}\alpha_{{\overline{\overline{\imath}}}}.
\end{displaymath}

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

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

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

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

which is what was to be proved.