An approximate solution to the ground state hydrogen atom wavefunction
$$\Psi = \begin{bmatrix} e^{iw}e^{r} \\ 0 \\ i\cos\theta e^{iw}e^{r} \\ i\sin\theta e^{i\phi}e^{iw}e^{r}\end{bmatrix}$$
nearly satisfies the Dirac equation
$$i\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1\end{bmatrix}\frac{d\Psi}{dw} + i\begin{bmatrix} 0 & 0 & \cos \theta & \sin \theta e^{-i\phi} \\ 0 & 0 & sin \theta e^{i\phi} & -\cos \theta \\ -\cos \theta & -\sin \theta e^{-i\phi} & 0 & 0 \\ -\sin \theta e^{i\phi} & \cos \theta & 0 & 0\end{bmatrix}\frac{d\Psi}{dr} + \frac{i}{r}\begin{bmatrix} 0 & 0 & -\sin \theta & \cos \theta e^{-i\phi} \\ 0 & 0 & \cos\theta e^{i\phi} & \sin\theta \\ \sin\theta & -\cos\theta e^{-i\phi} & 0 & 0 \\ -\cos\theta e^{i\phi} & -\sin\theta & 0 & 0\end{bmatrix}\frac{d\Psi}{d\theta} + \frac{i}{r\sin\theta}\begin{bmatrix} 0 & 0 & 0 & -ie^{-i\phi} \\ 0 & 0 & ie^{i\phi} & 0 \\ 0 & ie^{-i\phi} & 0 & 0 \\ -ie^{i\phi} & 0 & 0 & 0\end{bmatrix}\frac{d\Psi}{d\phi} \color{red}{-\frac{\alpha}{r}\Psi}= E\Psi$$
where $\alpha$ is some coupling constant, since:
$$i\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1\end{bmatrix}\begin{bmatrix} (i)e^{iw}e^{r} \\ 0\\ (i)(i\cos\theta)e^{iw}e^{r} \\ (i)(i\sin\theta e^{i\phi})e^{iw}e^{r}\end{bmatrix} + i\begin{bmatrix} 0 & 0 & \cos \theta & \sin \theta e^{-i\phi} \\ 0 & 0 & sin \theta e^{i\phi} & -\cos \theta \\ -\cos \theta & -\sin \theta e^{-i\phi} & 0 & 0 \\ -\sin \theta e^{i\phi} & \cos \theta & 0 & 0\end{bmatrix}\begin{bmatrix} e^{iw}e^{r} \\ 0 \\ (i\cos\theta)e^{iw}e^{r} \\ (i\sin\theta e^{i\phi})e^{iw}e^{r}\end{bmatrix} + \frac{i}{r}\begin{bmatrix} 0 & 0 & -\sin \theta & \cos \theta e^{-i\phi} \\ 0 & 0 & \cos\theta e^{i\phi} & \sin\theta \\ \sin\theta & -\cos\theta e^{-i\phi} & 0 & 0 \\ -\cos\theta e^{i\phi} & -\sin\theta & 0 & 0\end{bmatrix}\begin{bmatrix} 0 \\ 0 \\ -(i\sin\theta)e^{iw}e^{r} \\ (i\cos\theta e^{i\phi})e^{iw}e^{r}\end{bmatrix} + \frac{i}{r\sin\theta}\begin{bmatrix} 0 & 0 & 0 & -ie^{-i\phi} \\ 0 & 0 & ie^{i\phi} & 0 \\ 0 & ie^{-i\phi} & 0 & 0 \\ -ie^{i\phi} & 0 & 0 & 0\end{bmatrix}\begin{bmatrix} 0 \\ 0 \\ 0 \\ -(\sin\theta e^{i\phi})e^{iw}e^{r}\end{bmatrix} - \frac{\alpha}{r}\begin{bmatrix} (1)e^{iw}e^{r} \\ (0)e^{iw}e^{r} \\ (i\cos\theta)e^{iw}e^{r} \\ (i\sin\theta e^{i\phi})e^{iw}e^{r}\end{bmatrix}$$
and so
$$=\begin{bmatrix} -e^{iw}e^{r} \\ 0 \\ -(i\cos\theta)e^{iw}e^{r} \\ -(i\sin\theta e^{i\phi})e^{iw}e^{r}\end{bmatrix} + \begin{bmatrix} -e^{iw}e^{r} \\ 0 \\ -(i\cos\theta)e^{iw}e^{r} \\ -(i\sin\theta e^{i\phi})e^{iw}e^{r}\end{bmatrix} + \begin{bmatrix} (-\frac{1}{r})e^{iw}e^{r} \\ 0 \\ 0 \\ 0\end{bmatrix} + \begin{bmatrix} (-\frac{1}{r})e^{iw}e^{r} \\ 0 \\ 0 \\ 0\end{bmatrix} - \frac{\alpha}{r}\begin{bmatrix} (1)e^{iw}e^{r} \\ (0)e^{iw}e^{r} \\ (i\cos\theta)e^{iw}e^{r} \\ (i\sin\theta e^{i\phi})e^{iw}e^{r}\end{bmatrix}$$
$$=\begin{bmatrix} (-2 - \frac{2+\alpha}{r})e^{iw}e^{r} \\ 0 \\ (-2 -\frac{\alpha}{r})i\cos\theta e^{iw}e^{r} \\ (-2 -\frac{\alpha}{r})i\sin\theta e^{i\phi}e^{iw}e^{r}\end{bmatrix}\rightarrow (-2)\begin{bmatrix} e^{iw}e^{r} \\ 0 \\ i\cos\theta e^{iw}e^{r} \\ i\sin\theta e^{i\phi}e^{iw}e^{r}\end{bmatrix} = E\Psi$$
as $ r \rightarrow \infty$, making $\Psi$ an approximate solution to the Dirac equation.
Could anyone show whether or not, for $\gamma = \sqrt{1-\alpha^2}$:
$$ \Psi_1 = \begin{bmatrix} 1 \\ 0 \\ \frac{i(1-\gamma)}{\alpha} \cos\theta \\ \frac{i(1-\gamma)}{\alpha} \sin\theta e^{i\phi}\end{bmatrix}\frac{ke^{iw}e^r}{r^{1-\gamma}}$$
and
$$ \Psi_2 = \begin{bmatrix} 0 \\ 1 \\ \frac{i(1-\gamma)}{\alpha} \sin\theta e^{i\phi} \\ -\frac{i(1-\gamma)}{\alpha} \cos\theta\end{bmatrix} \frac{ke^{iw}e^r}{r^{1-\gamma}}$$
are exact solutions to the Dirac equation above, please?
Starting with the original constraint equation
$$\hat H\Psi = E\Psi$$
$$i\gamma_w\frac{d\Psi}{dw} + i\gamma_r\frac{d\Psi}{dr} + i\gamma_\theta \frac{d\Psi}{d\theta} + i\gamma_\phi \frac{d\Psi}{d\phi} \color{red}{+A_w \gamma_w \Psi} = E\Psi$$
Multiplying all sides by $\bar \Psi$:
$$i\bar \Psi\gamma_w\frac{d\Psi}{dw} + i\bar \Psi\gamma_r\frac{d\Psi}{dr} + i\bar \Psi\gamma_\theta\frac{d\Psi}{d\theta} + i\bar \Psi\gamma_\phi\frac{d\Psi}{d\phi} + \bar \Psi A_w \gamma_w \Psi = E\bar \Psi\Psi$$
Define $L$ to be
$$L = i\bar \Psi\gamma_w\frac{d\Psi}{dw} + i\bar \Psi\gamma_r\frac{d\Psi}{dr} + i\bar \Psi\gamma_\theta \frac{d\Psi}{d\theta} + i\bar \Psi\gamma_\phi\frac{d\Psi}{d\phi}+\bar \Psi A_w \gamma_w \Psi - E\bar \Psi\Psi$$
Then by the Euler-Lagrange extremization criteria of $L$:
$$\boxed{\frac{dL}{\partial\bar \Psi} = \frac{\partial}{\partial w}\left(\frac{\partial}{d\left(\frac{\partial\bar \Psi}{\partial w}\right)}L\right) + \frac{\partial}{\partial r}\left(\frac{\partial}{\partial\left(\frac{\partial\bar \Psi}{\partial r}\right)}L\right) +\frac{\partial}{\partial\theta}\left(\frac{\partial}{\partial\left(\frac{\partial\bar \Psi}{\partial\theta}\right)}L\right) + \frac{\partial}{\partial\phi}\left(\frac{\partial}{\partial\left(\frac{\partial\bar \Psi}{\partial\phi}\right)}L\right) }$$
we may recover the original constraint equation
$$i\gamma_w\frac{d\Psi}{dw} + i\gamma_r\frac{d\Psi}{dr} + i\gamma_\theta \frac{d\Psi}{d\theta} + i\gamma_\phi \frac{d\Psi}{d\phi} \color{red}{+A_w\gamma_w\Psi} - E\Psi = 0$$
which connects $L$ with $\hat H$.
$A$ may be computed from Coulomb's rule:
$$A_w = - V = \int \frac{kq_{proton}}{r^2} dr = - q\frac{\alpha}{r}$$
giving us
$$i\gamma_w\frac{d\Psi}{dw} + i\gamma_r\frac{d\Psi}{dr} + i\gamma_\theta\frac{d\Psi}{d\theta} + i\gamma_\phi\frac{d\Psi}{d\phi} \color{red}{- q\frac{\alpha}{r}\gamma_w\Psi} = E\Psi$$
Assuming $A_r, A_\theta, A_\phi$ also exist and can be combined linearly, L may be expanded into
$$L = \bar \Psi \left[ \sum_\mu \left(i \gamma_\mu \frac{\partial}{\partial \mu} \Psi + A_\mu \gamma_\mu \Psi \right)- E \Psi\right]$$
$$=i\bar \Psi\gamma_w\frac{d\Psi}{dw} + i\bar \Psi\gamma_r\frac{d\Psi}{dr} + i\bar \Psi\gamma_\theta\frac{d\Psi}{d\theta} + i\bar \Psi\gamma_\phi\frac{d\Psi}{d\phi} + A_w \bar\Psi \gamma_w \Psi+ A_r\bar\Psi \gamma_r \Psi + A_\theta\bar\Psi \gamma_\theta \Psi+ A_\phi \bar\Psi \gamma_\phi \Psi - E\bar \Psi\Psi$$
Synchronizing with electromagnetic L
$$L_{QED} = \bar \Psi \left[ \sum_\mu \left(i \gamma_\mu \frac{\partial}{\partial \mu} \Psi + A_\mu \gamma_\mu \Psi \right)- E \Psi\right] - \frac{1}{2}\left[ (\nabla \times \vec A)^2 + (\nabla \phi + \frac{d \vec A}{dw})^2 \right]$$
where $\{ A_w, A_r, A_\theta, A_\phi \} \rightarrow \{\phi, \vec A\}$
Defining S to be
$$S = \int_{r=0}^{\infty}\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}\int_{t=-\infty}^{\infty} L dr d\theta d\phi dt$$ $$= \int\int\int\int \bar \Psi \left[ \sum_\mu \left(i \gamma_\mu \frac{\partial}{\partial \mu} \Psi + \color{red}{A_\mu \gamma_\mu \Psi} \right)- E \Psi\right] - \frac{1}{2}\left[ (\nabla \times \vec A)^2 + (\nabla \phi + \frac{d \vec A}{dw})^2 \right]$$
we may consider only the $A, \Psi$ interacting terms as $S_1$:
$$\boxed{S_1 = \int\int\int\int A_w \bar\Psi \gamma_w \Psi+ A_r\bar\Psi \gamma_r \Psi + A_\theta\bar\Psi \gamma_\theta \Psi+ A_\phi \bar\Psi \gamma_\phi \Psi }$$
Then given 2 ground state solutions of different spin
$$ \Psi_1 = \begin{bmatrix} e^{it}e^{-r} \\ 0 \\ i\cos\theta e^{it}e^{-r} \\ i \sin\theta e^{i\phi}e^{it}e^{-r} \end{bmatrix}, \Psi_2 = \begin{bmatrix} 0 \\ e^{it}e^{-r} \\ i \sin\theta e^{i\phi}e^{it}e^{-r} \\ -i\cos\theta e^{it}e^{-r} \end{bmatrix}, \gamma_w = \begin{bmatrix} 1&0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1\end{bmatrix}$$
$$A_w = \frac{\alpha}{r}, A_r = 0, A_\theta = 0, A_\phi = 0$$
we obtain
$$S_1 = \int_{r=0}^{\infty}\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}\int_{t=-\infty}^{\infty} A_w \bar\Psi_1 \gamma_w \Psi_1 dr d\theta d\phi dt$$ $$\small{= \int \frac{\alpha}{r} \begin{bmatrix} e^{-it}e^{-r} & 0 & -i\cos\theta e^{-it}e^{-r} & -i \sin\theta e^{-i\phi}e^{-it}e^{-r} \end{bmatrix} \begin{bmatrix} 1&0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1\end{bmatrix} \begin{bmatrix} e^{it}e^{-r} \\ 0 \\ i\cos\theta e^{it}e^{-r} \\ i \sin\theta e^{i\phi}e^{it}e^{-r} \end{bmatrix}}$$ $$=\int_{r=0}^{\infty}\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}\int_{t=-\infty}^{\infty}\frac{\alpha}{r} \left(e^{-2r} - \cos^2 \theta e^{-2r} - \sin^2 \theta e^{-2r}\right) dr d\theta d\phi dt$$
$$=\int_{r=0}^{\infty}\int_{\theta=0}^{\pi}\int_{\phi=0}^{2\pi}\int_{t=-\infty}^{\infty}\frac{\alpha}{r} \left(e^{-2r} -e^{-2r}\right) dr d\theta d\phi dt$$ $$=0$$
Similarly,
$$\boxed{\int \bar\Psi_1 \gamma_w \Psi_1 = \int \bar\Psi_2 \gamma_w \Psi_2 = \int \bar\Psi_1 \gamma_w \bar\Psi_1 = \int \bar\Psi_1 \gamma_w \bar\Psi_2 = \int \bar\Psi_2 \gamma_w \bar\Psi_1 = \int \bar\Psi_2 \gamma_w \bar\Psi_2 = 0}$$
$$\boxed{\int \frac{\alpha}{r} \bar\Psi_1 \gamma_w \Psi_2 = \int \frac{\alpha}{r} \bar\Psi_2 \gamma_w \Psi_1 = \int \frac{1}{137}\frac{2i\sin\theta\cos\theta\sin\phi e^{-2r}}{r} \approx 0.0000000000018115914381782422 i}$$
Suppose an operator exists that flips $\Psi_1$ into $\Psi_2$. 4 repeated applications of the same operator will run through the spin and helicity states before returning the solution to its original state:
$$\begin{bmatrix} 1\\ 0 \\ a \\ b\end{bmatrix} \rightarrow \begin{bmatrix} 0\\ 1 \\ b \\ -a\end{bmatrix} \rightarrow \begin{bmatrix} 1\\ 0 \\ -a \\ -b\end{bmatrix}\rightarrow \begin{bmatrix} 0\\ 1 \\ -b \\ a\end{bmatrix}\rightarrow \begin{bmatrix} 1\\ 0 \\ a \\ b\end{bmatrix} $$