We consider the simplest case of a particle confined to a potential well that couples like the zeroth component of four-vector; the other option (where we see a scalar potential and effectively the mass becomes position dependent can be dealt with in exactly the same way) $$\hat{H}\psi\equiv c\vec{\alpha}\cdot\hat{\vec{p}}\psi+\beta mc^{2}\psi+V(r)\psi=E\psi.$$ we now look at the symmetries of $\hat{H}$ . Clearly, the angular momentum operator $\hat{L}$ acts nontrivially on $\hat{\vec{p}}$ $$\begin{align*} \left[\hat{L}_{k},\hat{p}_{i}\right] & =\sum_{i'j'}[\epsilon_{i'j'k}\hat{p}_{i'}x_{j'},\hat{p}_{i}]=\sum_{i'j'}\epsilon_{i'j'k}\hat{p}_{i'}[x_{j'},\hat{p}_{i}]\\ & =\sum_{i'j'}\epsilon_{i'j'k}\hat{p}_{i'}\left(-\frac{\hbar}{i}\right)\delta_{j'i}=i\sum_{i'}\hbar\epsilon_{i'ik}\hat{p}_{i'}=-i\sum_{i'}\hbar\epsilon_{kii'}\hat{p}_{i'},\end{align*}$$ and thus $$\left[\hat{\vec{L}}_{k},\hat{H}\right]=c\sum_{i}\alpha_{i}\left[\hat{\vec{L}}_{k},\hat{p_{i}}\right]=-i\hbar c\sum_{i'i}\epsilon_{kii'}\alpha_{i}\hat{p}_{i'}=-i\hbar c(\vec{\alpha}\times\vec{p})_{k}.$$ Fortunately the commutator with $\hat{S}=\frac{\hbar}{2}\vec{\Sigma}$ is minus this result $$\begin{align*} \left[\hat{\vec{S}}_{k},\hat{H}\right] & =\frac{\hbar}{2}c\sum_{i}\left[\left(\begin{array}{cc} \sigma_{k} & 0\\ 0 & \sigma_{k}\end{array}\right),\left(\begin{array}{cc} 0 & \sigma_{i}\\ \sigma_{i} & 0\end{array}\right)\right]\hat{p}_{i}\\ & =i\hbar c\sum_{i'i}\epsilon_{ki'i}\alpha_{i'}\hat{p}_{i}=i\hbar c(\vec{\alpha}\times\vec{p})_{k},\end{align*}$$ and thus $$\hat{\vec{J}}=\hat{\vec{L}}+\hat{\vec{S}}$$ commutes with $\hat{H}$ . [O]  [O] Once again we have been slightly careless: we should say that $\hat{\vec{J}}$ is a matrix of operators, with $\hat{\vec{J}}=\hat{\vec{L}}I_{4}+\hat{\vec{S}}$ . We also need the analogue of space inversion symmetry; the standard inversion operator $\hat{P}$ , which acts on functions of $\vec{r}$ , $$\hat{P}\psi(\vec{r})=\psi(-\vec{r})$$ must of course be used, but we also need to add a matrix part since the symmetry is allowed to act on each component differently. Since the momentum operator $\hat{p}$ is odd under parity, $$\hat{P}\hat{H}\hat{P}\psi=-c\vec{\alpha}\cdot\hat{\vec{p}}\psi+\beta mc^{2}\psi+V(r)\psi,$$ the matrix part of the operation must leave $\beta$ and $I$ invariant, but change the sign of$\vec{\alpha}$ . It is easy to find that the matrix $\beta$ has these properties, and we shall thus define the parity operator as $$\hat{\mathcal{P}}=\beta\hat{P},$$ since $$\mbox{\beta}\vec{\alpha}\beta=\left(\begin{array}{cc} I_{2} & \begin{array}{cc} 0 & 0\\ 0 & 0\end{array}\\ \begin{array}{cc} 0 & 0\\ 0 & 0\end{array} & -I_{2}\end{array}\right)\left(\begin{array}{cc} \begin{array}{cc} 0 & 0\\ 0 & 0\end{array} & \sigma_{i}\\ \sigma_{i} & \begin{array}{cc} 0 & 0\\ 0 & 0\end{array}\end{array}\right)\left(\begin{array}{cc} I_{2} & \begin{array}{cc} 0 & 0\\ 0 & 0\end{array}\\ \begin{array}{cc} 0 & 0\\ 0 & 0\end{array} & -I_{2}\end{array}\right)=\left(\begin{array}{cc} \begin{array}{cc} 0 & 0\\ 0 & 0\end{array} & I_{2}\sigma_{i}(-I_{2})\\ -I_{2}\sigma_{i}I_{2} & \begin{array}{cc} 0 & 0\\ 0 & 0\end{array}\end{array}\right)=-\alpha.$$ From the standard form $$\beta=\left(\begin{array}{cc} I_{2} & 0\\ 0 & -I_{2}\end{array}\right),$$ we see that the lower components of $\psi$ are multiplied with an additional minus sign under a parity transformation, independent of the parity of the function in these components. This additional minus sign is used to state that these components have negative “intrinsic parity”, something you may pick up in your particle physics course as well!

We shall now try to resolve the solutions to the Dirac equation in states that are both parity and angular momentum eigenstates. We first decompose in upper and lower components, $$\psi=\left(\begin{array}{c} \phi\\ \chi\end{array}\right),$$ where we impose parity, $$\begin{align*} \phi^{\pm}(\pm\vec{r}) & =\pm\phi^{\pm}(\vec{r}),\\ \chi^{\pm}(\pm\vec{r}) & =\mp\chi^{\pm}(\vec{r}).\end{align*}$$

$$\left(\begin{array}{cc} mc^{2}I_{2} & c\vec{\sigma}\cdot\hat{\vec{p}}\\ c\vec{\sigma}\cdot\hat{\vec{p}} & -mc^{2}I_{2}\end{array}\right)\left(\begin{array}{c} a_{j}^{\kappa}(r)\mathcal{Y}_{jm}^{\kappa}(\theta,\phi)\\ ib_{j}^{\kappa}(r)\mathcal{Y}_{jm}^{-\kappa}(\theta,\phi)\end{array}\right)=\left[E-V(r)\right]\left(\begin{array}{c} a_{j}^{\kappa}(r)\mathcal{Y}_{jm}^{\kappa}(\theta,\phi)\\ ib_{j}^{\kappa}(r)\mathcal{Y}_{jm}^{-\kappa}(\theta,\phi)\end{array}\right)$$ We need to evaluate the action of $\vec{\sigma}\cdot\hat{\vec{p}}$ on a vector-spherical harmonic. Start with the identity $$\vec{\sigma}\cdot\vec{r}\vec{\sigma}\cdot\hat{\vec{L}}=\vec{r}\cdot\vec{L}+i\vec{\sigma}.(\vec{r}\times\hat{\vec{L}})=i\vec{\sigma}\cdot(\vec{r}\times\vec{r}\times\hat{\vec{p}})=i\sigma.\vec{r}\,\vec{r}\cdot\hat{\vec{p}}-ir^{2}\vec{\sigma}\cdot\hat{\vec{p}}.$$ Thus we can write $$\vec{\sigma}\cdot\hat{\vec{p}}=i\vec{\sigma}\cdot\vec{e}_{r}\frac{\vec{\sigma}\cdot\hat{\vec{L}}}{r}+\frac{\hbar}{i}\vec{\sigma}.\vec{e}_{r}\frac{\partial}{\partial r}.$$ The attraction of this expression is that anything involving $\hat{\vec{L}}$ or $\vec{e}_{r}$ only acts on the angular components of the wave function, and we shall show below that $$\begin{align} \vec{\sigma}\cdot\vec{e}_{r}\mathcal{Y}_{jm}^{\kappa}(\theta,\phi) & =-\mathcal{Y}_{jm}^{-\kappa}(\theta,\phi),\label{eq:Rel:sigr}\\ \vec{\sigma}\cdot\hat{\vec{L}}\mathcal{Y}_{jm}^{\kappa}(\theta,\phi) & =-(\kappa+1)\hbar\mathcal{Y}_{jm}^{\kappa}(\theta,\phi).\label{eq:Rel:sigL}\end{align}$$ The first step in proving these relations is to show that $$\left[\hat{J},\vec{\sigma}\cdot\hat{\vec{L}}\right]=\left[\hat{J},\vec{\sigma}\cdot\vec{e}_{r}\right]=0,$$ so that neither operator can change the $jm$ values. In the first case (↓) we see that since $\vec{e}_{r}$ has negative parity, we must have $\kappa\rightarrow-\kappa$ on the right-hand side. Using [P]  [P] As usual, we write $d\Omega=d\sin\theta d\phi$ . $$\int d\Omega\left(\vec{\sigma}\cdot\vec{e}_{r}\mathcal{Y}_{jm}^{\kappa}(\theta,\phi)\right)^{\dagger}\vec{\sigma}\cdot\vec{e}_{r}\mathcal{Y}_{jm}^{\kappa}(\theta,\phi)=\int d\Omega\left(\mathcal{Y}_{jm}^{\kappa}(\theta,\phi)\right)^{\dagger}\vec{\sigma}\cdot\vec{e}_{r}\vec{\sigma}\cdot\vec{e}_{r}\mathcal{Y}_{jm}^{\kappa}(\theta,\phi)=\int d\Omega\left(\mathcal{Y}_{jm}^{\kappa}(\theta,\phi)\right)^{\dagger}\mathcal{Y}_{jm}^{\kappa}(\theta,\phi),$$ we see that the magnitude in Eq. (↓) is correct--the phase is purely a matter of an arbitrary choice of phase in the definition of $Y_{LM}$ for different $L$ . The choice made here is the most common one.

$$\begin{align*} \left[\frac{d}{d\rho}-\frac{\kappa}{\rho}\right]G_{j}^{k}(\rho) & -\left(\sqrt{l_{-}/l_{+}}-\alpha'/\rho\right)F_{j}^{\kappa}(\rho)=0,\\ \left[\frac{d}{dr}+\frac{\kappa}{r}\right]F_{j}^{k}(r) & -\left(\sqrt{l_{+}/l_{_{-}}}+\alpha'/\rho\right)G_{j}^{k}(\rho)=0.\end{align*}$$ We now seek solutions of the form of an exponential times a finite polynomial (the fact that the sums for $F$ and $G$ end at the same $N$ is something we can show by performing more analysis) $$F=e^{-\rho}\rho^{\gamma}\sum_{k=0}^{N}b_{k}\rho^{k},\quad G=e^{-\rho}\rho^{\gamma}\sum_{k=0}^{N}c_{k}\rho^{k}.$$ We get the recurrence relation $$\begin{align} (\gamma+m-\kappa)c_{m}-c_{m-1}+\alpha'b_{m}-\sqrt{l_{-}/l_{+}}b_{m-1} & =0,\nonumber \\ (\gamma+m+\kappa)b_{m}-b_{m-1}-\alpha'c_{m}-\sqrt{l_{+}/l_{-}}c_{m-1} & =0.\label{eq:Rel:recurrence}\end{align}$$ For $m=0$ we find $$\begin{align*} (\gamma-\kappa)b_{0} & =\alpha'c_{0},\\ (\gamma+\kappa)c_{0} & =\alpha'b_{0}.\end{align*}$$ Thus $$\gamma=\pm\sqrt{\kappa^{2}-\alpha'^{2}}.$$ If we wish the wavefunction to behave nicely at the origin (integrable singularity), we must have $\gamma>-1/2$ . Need to take positive square root! [R]  [R] For $\alpha Z<1$ , the smallest $\kappa^{2}-\alpha'^{2}$ can be is $1-(Z\alpha)^{2}>0$ , as long as $Z<1/\alpha$ .

Now use $b_{N+1}=c_{N+1}=0$ to derive from $m=N+1$ in Eqs. (↓) that $$\begin{equation} b_{N}=-\sqrt{l_{-}/l_{+}}c_{N}.\label{eq:Rel:finalrecur}\end{equation}$$ Now look at $m=N$ , and eliminate the $N-1$ coefficients $$\begin{align*} (\gamma+N-\kappa)b_{N}+\alpha'c_{N}-b_{N-1}-\sqrt{l_{-}/l_{+}}c_{N-1} & =0\qquad\times l_{+}\\ (\gamma+N+\kappa)c_{N}-\alpha'b_{N}-c_{N-1}-\sqrt{l_{+}/l_{-}}b_{N-1} & =0\qquad\times\sqrt{l_{+}l_{-}}\\ \text{subtract} & \implies\\ {}[(\gamma+N-\kappa)l_{+}+\alpha'\sqrt{l_{+}l_{-}}]b_{N}-[(\gamma+N+\kappa)\sqrt{l_{+}l_{-}}-\alpha'l_{+}]c_{N} & =0,\\ \text{Use Eq. (\ref{eq:Rel:finalrecur})} & \implies\\ -[(\gamma+N-\kappa)\sqrt{l_{-}l_{+}}+\alpha'l_{-}]-[(\gamma+N+\kappa)\sqrt{l_{+}l_{-}}-\alpha'l_{+}] & =0,\\ 2(\gamma+N)\sqrt{l_{-}l_{+}} & =\alpha'(l_{+}-l_{-}).\end{align*}$$ We now use the value of $\gamma$ , $\alpha'$ , $l_{\pm}$ : $$2\sqrt{(mc^{2})^{2}-E^{2}}(\sqrt{\kappa^{2}-\alpha'^{2}}+N)=E\alpha'$$ and solve for $E$ , $$E=\frac{mc^{2}}{\sqrt{1+\frac{a'^{2}}{(\gamma+N)^{2}}}}=\frac{mc^{2}}{\sqrt{1+\frac{Z^{2}\alpha^{2}}{\left(N+\sqrt{(j+\frac{1}{2})^{2}-Z^{2}\alpha^{2}}\right)^{2}}}}.$$ This impressive expression can be Taylor expanded in powers of $\alpha$ ; most conveniently define the “normal” principal quantum number, i.e., the equivalent of the one used in non-relativistic quantum mechanics, $$n=N+|\kappa|=N+j+\frac{1}{2},$$ to find that $$E/mc^{2}=1-\frac{Z^{2}\alpha^{2}}{2n^{2}}+\frac{1}{2}\frac{Z^{4}\alpha^{4}}{n^{4}}\left(\frac{3}{4}-\frac{n}{\left(j+\frac{1}{2}\right)}\right)-\frac{1}{4}\frac{Z^{6}\alpha^{6}}{n^{6}}\left(\frac{5}{4}-3\frac{n}{\left(j+\frac{1}{2}\right)}+\frac{3}{2}\frac{n^{2}}{\left(j+\frac{1}{2}\right)^{2}}+\frac{1}{2}\frac{n^{3}}{\left(j+\frac{1}{2}\right)^{3}}\right)+...$$ This clearly shows that the degeneracy (in $L$ ) for the non-relativistic hydrogen atom is broken by the next-to-leading order term. These corrections are actually quite small for Hydrogen, but not for $Z=100$ , see Fig. 6.8↓.

It is finally instructive to look at the relativistic ground state wave function $$\psi_{0}=\frac{A}{\sqrt{\pi}}\left(\frac{Z}{a_{0}}\right)^{3/2}e^{-Zr/a_{0}}\left(\frac{Zr}{a_{0}}\right)^{\sqrt{1-(Z\alpha)^{2}}-1}\left(\begin{array}{c} \chi^{S}\\ i\frac{1-\sqrt{1-(Z\alpha)^{2}}}{Z\alpha}\vec{\sigma}\cdot\unitvec r\chi^{s}\end{array}\right).$$ For $Z\alpha$ close to 1 we clearly have a very different behaviour near the origin. We can see this in the probability distributions as shown in Fig. 6.9↓, where the probability of the lower component increases dramatically as $Z$ increases, peaking closer and closer to zero.

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