Solve the free Dirac equation with plane waves. At rest it is simple: $\beta mc^{2}\psi=E\psi$ gives two solutions with $E=+mc^{2}$, spin up and down, and two with $E=-mc^{2}$. In motion, a positive-energy spinor has a large upper pair of components, a two-component spinor $\chi$, and a small lower pair, $(c\vec{\sigma}\cdot\mathbf{p}/(E+mc^{2}))\chi$, whose size relative to the upper is about $v/2c$ for slow particles. The spin is built in. The orbital angular momentum $\mathbf{L}$ alone does not commute with the Dirac Hamiltonian; neither does the spin; but their sum $\mathbf{J}=\mathbf{L}+(\hbar/2)\vec{\Sigma}$ does. Relativity and a first-order equation together make spin $\tfrac12$ unavoidable, and a moving spinor’s spin along its momentum, its helicity, is conserved for a free particle.

A cyclist’s pedalling and the bicycle’s turning are separate while the bicycle stands still; once it moves they are coupled, and only the total motion keeps a simple rule. Orbit and spin are coupled the same way in a moving Dirac particle, and only their sum is conserved.

The size of the lower pair of components of a positive-energy Dirac spinor relative to the upper pair, $cp/(E+mc^{2})$, against the speed $\beta=v/c$. For slow particles it is about , dashed; it approaches 1 as the speed approaches .

A positive-energy plane wave $\psi=u\,e^{i(\mathbf{p}\cdot\mathbf{r}-Et)/\hbar}$ of the free Dirac equation, with a two-component spinor $\chi$, is

• For each momentum there are four solutions: two spin states at $E>0$ and two at $E<0$.
• The lower components are smaller by $cp/(E+mc^{2})\approx v/2c$ for slow particles: the large and small components.

Compute the two commutators separately. The orbital angular momentum fails to commute with $\vec{\alpha}\cdot\hat{\mathbf{p}}$ because it rotates the momentum; the spin matrix fails to commute because it rotates $$. The two failures are equal and opposite, , and commutes with both.

$\vec{\alpha}$ and $\beta$ commute with $\hat{\mathbf{L}}$; only is rotated.

From $[\sigma_{j},\sigma_{i}]=-2i\epsilon_{ijk}\sigma_{k}$ in each off-diagonal block.

$\beta$ is block-diagonal like $\Sigma_{i}$ and commutes with it.

The two commutators cancel: spin $\tfrac12$ is part of the Dirac particle.

In a positive-energy spinor the lower pair of components is smaller than the upper by $cp/(E+mc^{2})$, about $v/2c$ when slow.

At rest the Dirac equation is $\beta mc^{2}\psi=E\psi$: two solutions at $+mc^{2}$ and two at $$.

For a free Dirac particle neither $\mathbf{L}$ nor the spin is conserved alone; their sum $\mathbf{J}$ is.

The spin term needed to conserve $\mathbf{J}$ comes out of the Dirac equation itself: spin $\tfrac12$ is not added by hand.

Great work! You now know how to:

• write the plane-wave spinors of the Dirac equation
• compare the large and small components
• show that only the total angular momentum is conserved