Dirac equation for hydrogen atom

Question

I went over a calculation of the hydrogen wavefunction using Dirac equation (this one) and I am a bit confused by the angular part. The final result for the wavefunction based on that derivation is this:

$$ \begin{pmatrix} if(r) Y_{j l_A}^{m_j} \\ -g(r) \frac{\vec{\sigma}\cdot\vec{x}}{r}Y_{j l_A}^{m_j} \end{pmatrix} $$

where $f(r)$ and $g(r)$ are radial functions and $Y_{j l_A}^{m_j}$ are spin spherical harmonics. In the derivation they show that $Y_{j l_A}^{m_j}$ and $-\frac{\vec{\sigma}\cdot\vec{x}}{r}Y_{j l_A}^{m_j}$ differ in their value of orbital angular momentum, $l$ by 1 and they have opposite parities. For example, if $j=1/2$, $Y_{j l_A}^{m_j}$ can have $l=1$ and $-\frac{\vec{\sigma}\cdot\vec{x}}{r}Y_{j l_A}^{m_j}$ would have $l=0$ (or the other way around). This implies (as it is mentioned in that derivation) that $l$ ($L^2$ as an operator) is not a good quantum number for a Dirac spinor.

I am not sure how to think about this. For example the atomic states are usually labeled as $^{2S+1}L_{J}$, which implies that the state has a definite orbital angular momentum, l. Is that just an approximation? Another thing I don't understand is the parity. As we are dealing only with electromagnetism, the wavefunctions should have a definite parity. But the top and bottom part in the spinor above have opposite parities, so it looks like the Dirac spinor doesn't have a defined parity. Can someone explain to me how should I think about these spinors? Should I look only at the top part? I know the bottom part is ignored in non-relativistic limit, but parity should still be a good quantum number even in the relativistic case (where I can't just ignore the bottom part).

Thank you!

Answer 1

(The answer is a bit late, but the article still viewed)

Indeed the eigenstates of the Dirac equation with a spherical potential ($V(\vec{r}) = V(r)$) are only eigenstates of $\vec{J}$ and not of $\vec{L}$ nor $\vec{S}$. The Dirac Hamiltonian only commutes with $\vec{J}=\vec{L}+\vec{S}$. The small and large part each are eigenstates of $\vec{L}$, albeit with a different value of $l$.

The labelling according to the Russel Sauders Coupling Scheme is an approximation. For single electron atoms they only relate to the eigenvalues of the large part of the wave function.

The parity operator on a basis of the Dirac 4-vectors with a basis of large and small states with spin up and spin down is given by the $\beta$ or $\gamma_4$ matrix. See also here https://quantummechanics.ucsd.edu/ph130a/130_notes/node493.html

You can also see this for the solution of a free particle. A particle with spin in the $z$ direction and $\vec{p} = (0,0,p_z)$ is upto a normalisation given by the vector $(1,0,\frac{c p_z}{E+m c^2}, 0 ) \, e^{\mathrm{i} p_z z }$. If you replace $z$ by $-z$ you do not get the eigenstate with $p_z \to -p_z$. For the later you also need to multiply the state by the matrix $\beta = \gamma_4$.