The Hamiltonian of a particle described by the Dirac equation neither commutes with the orbital angular momentum nor with the spin angular momentum. However, it commutes with the sum of the orbital and spin angular momentum. This means that neither the orbital angular momentum nor the spin angular momentum is a good quantum number. Neither of them is conserved. Then how is it meaningful to say that a particle described by the Dirac equation is a spin-$1/2$ particle?
Dirac Equation – Understanding Particle Spin Confusion
angular momentumdirac-equationhamiltonianquantum mechanicsquantum-spin
Best Answer
It is true that the Dirac Hamiltonian does not commute with the spin vector $\vec{S}$. While we have $\left[\vec{J},H_{D}\right]=0$, the related commutator with $\vec{S}$ is nonzero. So the vector components—e.g. $S_{x}=\frac{\hbar}{2}\Sigma_{1}$ in terms of the $4\times4$ Dirac matrix $\Sigma_{1}$—do not represent conserved quantities.* However, the magnitude of the spin does commute with the Hamiltonian, $$\left[\vec{S}\,^{2},H_{D}\right]=0.$$ So the magnitude of the spin is always $\frac{3}{4}\hbar^{2}$, and the spin is always described by a $s=\frac{1}{2}$ representation of the (universal cover of the) rotation group, $SU(2)$.
Note that an analogous commutator formula with $\vec{L}$ does not hold: $$\left[\vec{L}\,^{2},H_{D}\right]\neq0.$$ This leads to the fact that the small and large components of a Dirac spinor describing an energy eigenvalue in a central potential have different eigenvalues of $l$, although they have the same value of $s$ and $j$.
*$H_{D}$ does commute with the helicity $\vec{S}\cdot\hat{p}$, but that is a nonlocal operator, since it involves a unit vector $\hat{p}=\vec{p}/|\vec{p}|$, where $\vec{p}=-i\hbar\vec{\nabla}$. (This is nonlocal because of the $1/|\vec{p}|$; functions of derivatives other than nonnegative integer powers cannot be defined locally.)