Put the Dirac equation in an electromagnetic field and ask what it says about a slow electron. Split the four components into an upper pair $\varphi$ and a lower pair $\chi$ and take out the fast phase of the rest energy. For slow motion the lower pair is small, $\chi\approx(\vec{\sigma}\cdot\vec{\pi})\varphi/2mc$ with the kinetic momentum $\vec{\pi}=\hat{\mathbf{p}}-q\mathbf{A}$, and eliminating it leaves an equation for $\varphi$ alone: $i\hbar\partial\varphi/\partial t=[(\vec{\sigma}\cdot\vec{\pi})^{2}/2m+q\Phi]\varphi$. That is the Pauli equation of the chapter on electromagnetic fields — and $(\vec{\sigma}\cdot\vec{\pi})^{2}$ contains the spin term with $g=2$ exactly. Dirac did not put the electron’s magnetic moment in; his equation produced it. Pushed one order further in $v/c$, the same procedure gives three corrections, which the next lesson uses: the relativistic kinetic energy, the Darwin term, and the spin–orbit coupling, Thomas’s factor of one half included.

A telephoto photograph of a distant car shows only its outline; a closer look shows the wheels turning. The non-relativistic limit is the outline — the Pauli equation — and the next order shows the turning wheels of the fine structure.

For a charge $q$ in the potentials $\Phi,\mathbf{A}$, with $\vec{\pi}=\hat{\mathbf{p}}-q\mathbf{A}$ and , slow motion gives

• $(\vec{\sigma}\cdot\vec{\pi})^{2}=\vec{\pi}^{2}-q\hbar\,\vec{\sigma}\cdot\mathbf{B}$: the Pauli Hamiltonian with exactly.

Write the Dirac equation in the field as two coupled equations for the upper and lower pairs, after removing the rest-energy phase. In the lower equation the rest energy $2mc^{2}$ dominates, which makes $\chi$ small and fixes it from $\varphi$. Putting it into the upper equation leaves $(\vec{\sigma}\cdot\vec{\pi})^{2}/2m$, and the identity for products of Pauli matrices with non-commuting kinetic momenta turns that into the kinetic energy and the spin’s magnetic energy with .

The Dirac representation, after taking out $e^{-imc^{2}t/\hbar}$.

In the second line $2mc^{2}$ dominates $i\hbar\partial_{t}$ and $q\Phi$ for slow motion in weak fields.

Put $\chi$ into the first line.

The Pauli identity, with $\vec{\pi}\times\vec{\pi}=iq\hbar\mathbf{B}$ from the kinetic momenta’s commutator.

$\hat{\mathbf{S}}=\frac{\hbar}{2}\vec{\sigma}$: twice the orbital ratio, .

For a slow electron the lower pair $\chi$ is smaller than the upper $\varphi$ by about $v/2c$; neglecting it gives the Pauli equation.

The kinetic term $(\vec{\sigma}\cdot\vec{\pi})^{2}/2m$ contains the spin’s magnetic energy with exactly: Dirac’s equation produces the magnetic moment.

In the equation for $\chi$ the rest energy $2mc^{2}$ outweighs everything else for slow motion, which makes $\chi$ small and fixes it from $\varphi$.

One order further in $v/c$ the Dirac equation gives three corrections: the relativistic kinetic energy, the Darwin term and the spin–orbit coupling.

Great work! You now know how to:

• reduce the Dirac equation to the Pauli equation for slow electrons
• derive $g=2$ from it
• name the three corrections of the next order