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:

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

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,

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

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 $\skew3\vec A$,

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

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

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

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.