rewrite this to

$$H\Psi=(H_{Dirac}+H_{INT})\Psi=E\Psi,
\label{88}$$

where

$$H_{Dirac}=(-i\alpha^k\partial_k+{\beta}m),
\label{89}$$

and

$$H_{Int}=i\left(\frac{2M}{r}\right)^\frac{1}{2}(\partial_r+\frac{3}{4r}).
\label{90}$$

$\Psi$ shall describe an eigenstate to $J^2\ ,J_z$ and $P$ where $J$ describes the norm of the overall angular momentum which means the sum of the orbital angular momentum $L$ and the Spin $S$, $J_z$ its component in $z$-direction and $P$ the parity.

The Parity will be denoted by the quantum number $\omega$ (not to be confused with the Spinor coefficients $\omega_\mu^{mn}$). We set $\omega=+1$ [for states with parity]{} $(-)^{J+\frac{1}{2}}$ and $\omega=-1$ [for states with parity]{} $(-)^{J-\frac{1}{2}}$.

In the following $y_{LJ}^M$ will describe an eigenstate to $J^2$ and $J_z$ where the parity is given by $(-)^L$ which means that $L$ can adopt the two values $L=J+\frac{1}{2}\omega$ and $\bar L=J-\frac{1}{2}\omega$ and that $y_{LJ}^M$ and $y_{\bar LJ}^M$ have opposite parity.

Next we make the following ansatz for $\Psi$

$$\Psi=\frac{1}{r}\left(\begin{array}{cc}
\Phi y_{LJ}^M\\
i \chi y_{\bar LJ}^M\end{array}\right),
\label{66}$$

where $\Phi$ and $\chi$ describe arbitrary radial states. By following a derivation also used in case of the relativistic hydrogen atom (where the free part of the Dirac equation has to be reformulated in spherical coordinates also and which can be found in Ref. [@[book7]] for example) and splitting the two components of the equation 