I Dirac equation in the hydrogen atom

[Malamala]

Hello! 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!

 

[azntoon]

The spinor above is indeed a valid solution of the Dirac equation and so it does have a definite parity. However, the parity of the two components of the spinor are opposite, which is why the angular part of the wavefunction does not have a definite orbital angular momentum. In the non-relativistic limit, the bottom component is ignored, so the wavefunction has a definite parity and a definite angular momentum. In terms of labeling atomic states, it is common to use the approximate non-relativistic labels, even when considering relativistic effects. This is because in many cases, the relativistic corrections are small and so the non-relativistic labels are still good approximations.