Title: ETH_Sem5 / AdvQC / 07_Dirac_H

Q: What is the general procedure to get the energies and wave functions of the Dirac hydrogen atom?
A: 1. Derive the stationary Dirac equation (using BO-approximation)
2. Transform \(h^D\) to spherical coordinates (as is it is central-field problem) by
i) Analyze action of the 2D substructure \(\bm{\sigma}\cdot\mom\) of \(h^D\)
ii) Separation to radial and spin-angular components
iii) Solve the spin-angular part using angular-momentum coupling \(\bm{\alpha}\cdot\mom\)
4. Proceed similarly for the 4D case \(\bm{\alpha}\cdot\mom\)
5. Use product ansatz of radial and angular components for the total spinor
6. Rewrite the Dirac equation to an ordinary differential equation and solve it to obtain the radial functions and the eigenenergies

Q: Write down the general form of the total Hamiltonian and the total wave function for the hydrogen atom! Which assumptions are then used to arrive at the stationary Dirac equation?
A: \[\begin{align}H=h_e^D \otimes \mathbf{1}+\mathbf{1} \otimes h_p^D+V_{e p} \\ \thus \Psi=\psi_e \otimes \psi_p \end{align}\]
1. Assume instantaneous interactions without magnetic effects: \[\begin{align} V_{e p}(r) & \approx q_e \phi=\frac{q_e\,q_{\mathrm{nuc}}}{|r-\bm{R}|}=-\frac{Z e^2}{r} \equiv V_{\mathrm{nuc}}(r) \\ & \text { \ or short: } V(r) \end{align}\]
2. Assume that proton is much heavier than the electron! \(\rightarrow\) Fix origin of the coordinate system as center of the nucleus (BO-approximation) \[\left[c \bm{\alpha} \cdot \bm{p}+\beta m_e c^2+V(r)\right] \Psi(t, \bm{r})=\mathrm{i} \hbar \frac{\partial}{\partial t} \Psi(t, \bm{r})\]
3. The coordinates can now be separated: \[\psi_e(t, \bm{r}) \rightarrow \Psi(t, \bm{r})=\exp (-\mathrm{i} E t / \hbar) \Psi(\bm{r})\]
4. After separation we obtain for the stationary Dirac equation \[\underbrace{\left[c \bm{\alpha} \cdot \bm{p}+\beta m_e c^2+V(r)\right]}_{\equiv h^D} \Psi(\bm{r})=E \Psi(\bm{r})\]

Q: What is the total angular momentum \(\bm{j}_4\)? Which important property does it have?
A: For the solution of the Dirac hydrogen atom it is convenient to introduce the total angular momentum as a 3-vector of \(4\times4\) matrices,
\[\bm{j}_4=\bm{j} \bm{1}_2=\bm{1}_2 \otimes\left(\bm{l} \bm{1}_2+\bm{s}\right)=\bm{l} \bm{1}_4+\bm{1}_2 \otimes \bm{s}=\bm{l} \bm{1}_4+\frac{\hbar}{2} \mathbf{1}_2 \otimes \bm{\sigma}=\bm{l} 1_4+\frac{\hbar}{2} \bm{\Sigma}\]
with
\[\bm{l} \mathbf{1}_4 \equiv\left(l_x \mathbf{1}_4, l_y \mathbf{1}_4, l_z \mathbf{1}_4\right)\]
\[\bm{\Sigma} \equiv\left(\begin{array}{cc}\bm{\sigma} & 0 \\ 0 & \bm{\sigma}\end{array}\right)=\left[\left(\begin{array}{cc}\sigma_x & 0 \\ 0 & \sigma_x\end{array}\right),\left(\begin{array}{cc}\sigma_y & 0 \\ 0 & \sigma_y\end{array}\right),\left(\begin{array}{cc}\sigma_z & 0 \\ 0 & \sigma_z\end{array}\right)\right]\]
The Dirac Hamiltonian for hydrogen-like atoms, \(h^D\), and the square as well as any component of the total angular momentum operator \(\bm{j}_4\) commute:
\[\left[h^D, \bm{j}_4^2\right]=0, \text { and }\left[h^D, j_{4, i}\right]=0 \text {, with } i \in\{x, y, z\}\]

Q: Where does the spin-orbit coupling operator \((\bm{\sigma}\cdot\bm{l})\) arise from?
A: The Dirac Hamiltonian for the H-atom in the BO approximation reads
\[h^D=\left(\begin{array}{cc}m_e c^2+V(r) & c \bm{\sigma} \cdot \bm{p} \\ c \bm{\sigma} \cdot \bm{p} & -m_e c^2+V(r)\end{array}\right)\]
We can use a trick and rewrite the off-diagonal elements using Dirac's relation as
\[\begin{aligned} \bm{\sigma} \cdot \bm{p} &=\mathbf{1}_2 \cdot(\bm{\sigma} \cdot \bm{p})\\ &=\left(\frac{\bm{\sigma} \cdot \bm{r}}{r}\right)\left(\frac{\bm{\sigma} \cdot \bm{r}}{r}\right)(\bm{\sigma} \cdot \bm{p}) \\&=\frac{1}{r^2}(\bm{\sigma} \cdot \bm{r})(\bm{\sigma} \cdot \bm{r})(\bm{\sigma} \cdot \bm{p}) \end{aligned}\]
which produces the orbital angular momentum operator \(\bm{l} = \bm{r}\times\bm{p}\)
\[\begin{aligned} \bm{\sigma} \cdot \bm{p}&=\left(\frac{\bm{\sigma} \cdot \bm{r}}{r}\right) \frac{1}{r}[\bm{r} \cdot \bm{p}+\mathrm{i} \bm{\sigma} \cdot(\bm{r} \times \bm{p})] \\&=\frac{1}{r^2}(\bm{\sigma} \cdot \bm{r})[\bm{r} \cdot \bm{p}+\mathrm{i}(\bm{\sigma} \cdot \bm{l})] \end{aligned}\]
As the scalar product \((\bm{r}\cdot\bm{p})\) reads
\[\begin{aligned} \bm{r} \cdot \bm{p}&=-\mathrm{i} \hbar r \frac{\partial}{\partial r} = rp_r + \imag\hbar \\ \text{with}\quad p_r &= -\imag\hbar\frac{1}{r}\pdv{r}r = -\imag\hbar\left(\pdv{}{r}+\frac{1}{r}\right) \end{aligned}\]
in spherical coordinates, all angular variables are contained in \((\bm{\sigma}\cdot\bm{l})\) and a separation of coordinates was achieved \(\rightarrow\) use product ansatz. \((\bm{\sigma}\cdot\bm{l})\) is essentially the spin-orbit coupling operator known from the Pauli elimination.

Q: How are the eigenvectors of the spin-orbit coupling operator \((\bm{s}\cdot\bm{l}) = \hbar/2(\bm{\sigma}\cdot\bm{l})\) obtained? Write the solution down!
A: The eigenvectors can be obtained using angular momentum algebra for the coupling of two angular momenta (here: orbital angular momentum and spin). We expand the eigenvector into the product basis, \(\phi\), of the eigenfunctions of the individual angular momenta: spin eigenvectors \(\ket{s\,m_s}\), \(\alpha\) and \(\beta\), and spherical harmonics \(Y_{lm_l}(\theta,\varphi) = \ket{l\,m_l}\):
\[\chi_{jm_j} = \ket{j\,m_j} = \sum_{m_s = -s}^{s}\sum_{m_l = -l}^{l} \underbrace{\bra{s\,m_s\,l\,m_l}\ket{j\,m_j}}_{C_{sm_slm_l}^{jm_j}}\ \underbrace{\ket{s\,m_s\,l\,m_l}}_{\ket{s\,m_s}\otimes\ket{l\,m_l} = \phi_{lm_l}}\]
And we obtain the Pauli bispinors (note that they can be written in vector form, because of the 2-vector structure of \(\ket{s\,m_s}\))
\[ \begin{aligned} \chi_{\kappa m_j}&=\chi_{j m_j}^{(+)}=+\sqrt{\frac{l+m_j+1 / 2}{2 l+1}} \phi_{l\left(m_j-\frac{1}{2}\right)}+\sqrt{\frac{l-m_j+1 / 2}{2 l+1}} \phi_{l\left(m_j+\frac{1}{2}\right)} \\ \chi_{-\kappa m_j}&=\chi_{j m_j}^{(-)}=-\sqrt{\frac{l-m_j+1 / 2}{2 l+1}} \phi_{l\left(m_j-\frac{1}{2}\right)}+\sqrt{\frac{l+m_j+1 / 2}{2 l+1}} \phi_{l\left(m_j+\frac{1}{2}\right)} \end{aligned}\]

Q: Write down the eigenvalue equation and the corresponding eigenvalues of \((\bm{\sigma}\cdot\bm{l})\)!
A: Exploiting the definitions of the spin operator \(\bm{s}=\hbar\bm{\sigma}/2\) and of the total angular momentum \(\bm{j}_2 = \bm{l}\bm{1}_2 + \bm{s}\) we formulate an operator identity
\[\begin{aligned} \bm{j}^2&=\bm{l}^2 \mathbf{1}_2+2(\bm{l} \cdot \bm{s})+\bm{s}^2 \\ &=\bm{l}^2 \mathbf{1}_2+\hbar(\bm{\sigma} \cdot \bm{l})+\bm{s}^2 \end{aligned}\]
which then allows us to deduce an eigenvalue equation
\[\begin{aligned} (\bm{\sigma} \cdot \bm{l}) \chi_{j m_j}^{(\pm)} &=\frac{1}{\hbar}\left(j^2-l^2-s^2\right) \chi_{j m_j}^{(\pm)} \\&=\hbar\left[j(j+1)-l(l+1)-\frac{3}{4}\right] \chi_{j m_j}^{(\pm)} \end{aligned}\]
Taking into account the two cases \(j = l \pm 1/2\) for \(\chi_{j m_j}^{(\pm)}\) we then may write
\[\def\arraystretch{2.5}\begin{array}{>{\displaystyle}r} (\bm{\sigma} \cdot \bm{l}) \chi_{j m_j}^{(+)} =\hbar l \chi_{j m_j}^{(+)}=\hbar\left[-1+\left(j+\frac{1}{2}\right)\right] \chi_{j m_j}^{(+)} \\ (\bm{\sigma} \cdot \bm{l}) \chi_{j m_j}^{(-)}=-\hbar(l+1) \chi_{j m_j}^{(-)}=\hbar\left[-1-\left(j+\frac{1}{2}\right)\right] \chi_{j m_j}^{(-)} \end{array}\]

Q: What is the relativistic analog to the orbital angular momentum quantum number \(\bm{l}\)?
A: The eigenvalue equations of \((\bm{\sigma}\cdot\bm{l})\) suggest that adding \(\hbar\bm{1}_2\) to the operator on the left-hand side of the equations simplifies the eigenvalues \(\rightarrow\) define operator
\[\bm{k} \equiv \bm{\sigma} \cdot \bm{l}+\hbar \mathbf{1}_2\]
that fulfills the eigenvalue equation
\[\bm{k} \chi_{j m_j}^{(\pm)}=\pm \hbar\left(j+\frac{1}{2}\right) \chi_{j m_j}^{(\pm)} \equiv \hbar \kappa^{(\pm)} \chi_{j m_j}^{(\pm)}\]
in which we introduced a new quantum number \(\kappa\). From the equation above we conclude that
\[\kappa^{(+)} = j+\frac{1}{2} \quad\text{and}\quad \kappa^{(-)} = -j-\frac{1}{2}\]
are the eigenvalues to \(\chi_{j m_j}^{(+)}\) and \(\chi_{j m_j}^{(-)}\), respectively.
Hence, we may classify the pair of Pauli spinors for a given set of quantum numbers (\(j,m_j)\) equally well by the new quantum number \(\kappa\),
\[\chi_{\kappa m_j} \equiv \chi_{j m_j}^{(+)} \quad\text{and}\quad \chi_{-\kappa m_j} \equiv \chi_{j m_j}^{(-)}\]
The quantum number \(\kappa\) will turn out to be the analog of \(l\) in non-relativistic theory.

Q: Fill in the following table!
\[\def\arraystretch{1.5} \begin{array}{l|rrrrrrrrr} \\ \hline \hline \\ \text{Shell} & s_{1 / 2} & p_{1 / 2} & p_{3 / 2} & d_{3 / 2} & d_{5 / 2} & f_{5 / 2} & f_{7 / 2} & g_{7 / 2} & g_{9 / 2} \\ \hline \\ \text{Symbol} & & & & & & & & & \\ \kappa & & & & & & & & & \\ l & & & & & & & & & \\ j=|\kappa|-1/2 & & & & & & & & & \\ \text{Parity }\left([-1]^l\right) & & & & & & & & & \\ \text{Degeneracy} & & & & & & & & & \\ (2|\kappa|=2 j+1) & & & & & & & & & \\ \hline \hline \end{array}\]
A: \[\def\arraystretch{1.75} \begin{array}{l|rrrrrrrrr} \\ \hline \hline \\ \text{Shell} & s_{1 / 2} & p_{1 / 2} & p_{3 / 2} & d_{3 / 2} & d_{5 / 2} & f_{5 / 2} & f_{7 / 2} & g_{7 / 2} & g_{9 / 2} \\ \hline \\ \text{Symbol} & s & \bar{p} & p & \bar{d} & d & \bar{f} & f & \bar{g} & g \\ \kappa & -1 & 1 & -2 & 2 & -3 & 3 & -4 & 4 & -5 \\ l & 0 & 1 & 1 & 2 & 2 & 3 & 3 & 4 & 4 \\ j=|\kappa|-1 / 2 & 1 / 2 & 1 / 2 & 3 / 2 & 3 / 2 & 5 / 2 & 5 / 2 & 7 / 2 & 7 / 2 & 9 / 2 \\ \text{Parity }\left([-1]^l\right) & + & - & - & + & + & - & - & + & + \\ \text{Degeneracy} & 2 & 2 & 4 & 4 & 6 & 6 & 8 & 8 & 10 \\ (2|\kappa|=2 j+1) & & & & & & & & & \\ \hline \hline \\ \end{array}\]

Q: What is the \(4\times4\) analog of \(\bm{k}\)? Why is there a factor of \(\beta\)?
A: \[\bm{K} \equiv \beta\left(\hbar \mathbf{1}_4+\bm{\Sigma} \cdot \bm{l}\right)=\beta\left[\mathbf{1}_2 \otimes\left(\hbar \mathbf{1}_2+\bm{\sigma} \cdot \bm{l}\right)\right] = \beta\left(\mathbf{1}_2 \otimes \bm{k}\right)=\left(\begin{array}{cc}\bm{k} & 0 \\0 & -\bm{k}\end{array}\right)\]
The \(\beta\) factor is introduced to compensate the negative sign of the eigenvalue of \(\bm{k}\chi_{-\kappa m_j}\), in order to yield one quantum number for the angular part of the 4-spinor:
\[\bm{K}\left(\begin{array}{c}\chi_{\kappa m_j} \\ \chi_{-\kappa m_j}\end{array}\right)=\left(\begin{array}{cc}\bm{k} & 0 \\ 0 & -\bm{k}\end{array}\right)\left(\begin{array}{c}\chi_{\kappa m_j} \\ \chi_{-\kappa m_j}\end{array}\right)=\hbar \kappa\left(\begin{array}{c}\chi_{\kappa m_j} \\ \chi_{-\kappa m_j}\end{array}\right)\]

Q: What is the product ansatz for \(\Psi(\pos)\)? Write down the total Dirac Hamiltonian for this ansatz!
A: Because of the general \(2\times2\) block structure of \(h^D\), a suitable ansatz for \(\Psi(\pos)\) is
\[\Psi(\bm{r})=\begin{pmatrix} F_i(r) \chi_{j m_j}(\vartheta, \varphi) \\ \mathrm{i} G_i(r) \chi_{j m_j}^{\prime}(\vartheta, \varphi)\end{pmatrix}\]
Using the fact that
\[\bm{\sigma} \cdot \bm{p}=\frac{1}{r^2}(\bm{\sigma} \cdot \bm{r})[\bm{r} \cdot \bm{p}+\mathrm{i}(\bm{\sigma} \cdot \bm{l})]\]
the Hamiltonian reads
\[h^D=\begin{pmatrix}m_e c^2+V(r) & -\mathrm{i} c\left(\frac{\bm{\sigma} \cdot \bm{r}}{r}\right)\left[\frac{\hbar}{r} \frac{\partial}{\partial r} r-\frac{\bm{k}}{r}\right] \\ -\mathrm{i} c\left(\frac{\bm{\sigma} \cdot \bm{r}}{r}\right)\left[\frac{\hbar}{r} \frac{\partial}{\partial r} r-\frac{\bm{k}}{r}\right] & -m_e c^2+V(r)\end{pmatrix}\]

Q: What is the so called equivalence restriction?
A: If one exploits the fact that we have identified the eigenfunctions of \(\bm{k}\), one can rewrite the ansatz for the spinor as
\[\Psi_{n \kappa m_j}(\bm{r})=\begin{pmatrix} F_i(r) \chi_{j m_j}^{(\pm)}(\vartheta, \varphi) \\ \mathrm{i} G_i(r) (\bm{\sigma} \cdot \hat{\bm{r}}) \chi_{j m_j}^{(\pm)}(\vartheta, \varphi)\end{pmatrix} = \begin{pmatrix} F_{n \kappa}(r) \chi_{\kappa m_j}(\vartheta, \varphi) \\ \mathrm{i} G_{n \kappa}(r) \chi_{-\kappa m_j}(\vartheta, \varphi)\end{pmatrix}\]
because of
\[(\bm{\sigma} \cdot \hat{\bm{r}}) \chi_{\kappa m_j}(\vartheta, \varphi)=-\chi_{-\kappa m_j}(\vartheta, \varphi)\]
Without loss of generality we assign here a single pair of radial functions, \(F_{n \kappa}(r)\) and \(G_{n \kappa}(r)\), to a given set of \((2j+1)\)-degenerate spherical spinors \(\chi_{j m_j}^{(\pm)}(\vartheta, \varphi)\) with given \((j,m_j)\), which is called equivalence restriction.

Q: Which ansatz for the spinor is used to solve the Dirac equation? Write down the Dirac radial equation!
A: In close analogy to the non-relativistic theory, it is advantageous to introduce a second set of radial functions,
\[\Psi_i(\bm{r}) \rightarrow \Psi_{n_i \kappa_i m_j(i)}(\bm{r})=\begin{pmatrix} \frac{P_{n_i \kappa_i}(r)}{r} \chi_{\kappa_i m_{j(i)}}(\vartheta, \varphi) \\ \mathrm{i} \frac{Q_{n_i \kappa_i}(r)}{r} \chi_{-\kappa_i m_{j(i)}}(\vartheta, \varphi) \end{pmatrix}\]
as with this ansatz
1. The volume element \(\mathrm{d}^3r = \mathrm{d} x\,\mathrm{d} y\,\mathrm{d} z\) in spherical coordinates reads \(r^2\,\mathrm{d} r\, \sin\vartheta\,\mathrm{d}\vartheta\,\mathrm{d}\varphi\) and \(r^2\) drops out in \(\left\langle F_{n_i \kappa_i} \mid F_{n_j \kappa_j}\right\rangle+\left\langle G_{n_i \kappa_i} \mid G_{n_j \kappa_j}\right\rangle\).
2. The Dirac radial equation takes the simple form \[\begin{aligned} {\left[V(r)+m_e c^2\right] P_i(r)+c \hbar\left[-\frac{\mathrm{d}}{\mathrm{d} r}+\frac{\kappa_i}{r}\right] Q_i(r) } &=E_i P_i(r) \\ c \hbar\left[\frac{\mathrm{d}}{\mathrm{d} r}+\frac{\kappa_i}{r}\right] P_i(r)+\left[V(r)-m_e c^2\right] Q_i(r) &=E_i Q_i(r) \end{aligned}\]
3. and as these are ordinary differential equations, they can be solved analytically.

Q: What are the main steps to solve the Dirac radial equations?
A: 1. Investigate the long-range behavior, \(r\rightarrow\infty\), which leads to the ansatz \[\begin{aligned} P(r) &= p(r)\exp(-\rho(r)) \\ Q(r) &= q(r)\exp(-\rho(r)) \end{aligned}\]
2. Study the short-range behavior, \(r\rightarrow0\), and expand in a Taylor series: \[\begin{aligned} p(r)&=r^\alpha\sum_{i=0}^\infty p_ir^i \\ q(r)&=r^\alpha\sum_{i=0}^\infty q_ir^i \end{aligned}\]
3. Determine the coefficients \(\rho\), \(p_i\) and \(q_i\): \[\begin{aligned} \rho &= \frac{\sqrt{m_e^2c^4-E^2}}{c\hbar} \\ q_i &\approx\frac{2\rho}{i}q_{i-1} \\ p_i &\approx\frac{2\rho}{i}p_{i-1} \\ &\text{for large }i \end{aligned}\]
4. Realize that this would yield \[\begin{aligned} p(r)& \xrightarrow{\text{for large }i} \exp(2\rho r) \\ q(r) & \xrightarrow{\text{for large }i}\exp(2\rho r) \end{aligned}\] and to recover square-integrable spinors (and radial functions), the series expansions must be truncated at \((i-1)=n_r\).

Q: How is the radial density calculated? What is its characteristic?
A: The radial density, \[ \rho_i(r)=P_{n_i \kappa_i}^2(r)+Q_{n_i \kappa_i}^2(r)\] features no longer any nodes!

Q: What are the energies and in particular the ground state energy of the Dirac hydrogen atom?
A: The ground state energy in Hartree atomic units is given by
\[\begin{aligned}E_{n\abs{\kappa}} &= c^2\left\{1+\left(\frac{Z/c}{\sqrt{\kappa^2-Z^2/c^2}+n-\abs{\kappa}}\right)^2\right\}^{-1/2} \\ \thus & {E_0\equiv E_{11}=c^2 \sqrt{1 - \frac{Z^2}{c^2}}} \end{aligned}\]
The ground state is doubly degenerate with the two eigensolutions
\[\begin{aligned}\Psi_{1(-1) 1/2} &= \frac{\left[2 m_e(Z \alpha)\right]^{3 / 2}}{\sqrt{4 \pi}} \sqrt{\frac{1+\alpha_1}{2 \Gamma\left(1+2 \alpha_1\right)}}\left[2 m_e(Z \alpha) r\right]^{\alpha_1-1} \, e^{-m_e(Z \alpha) r}\begin{pmatrix} 1 \\ 0 \\ \frac{\mathrm{i}\left(1-\alpha_1\right)}{Z \alpha} \cos \vartheta \\ \frac{\mathrm{i}\left(1-\alpha_1\right)}{Z \alpha} \sin \vartheta e^{\mathrm{i} \varphi}\end{pmatrix} \\ \Psi_{1(-1)(-1/2)} &= \frac{\left[2 m_e(Z \alpha)\right]^{3 / 2}}{\sqrt{4 \pi}} \sqrt{\frac{1+\alpha_1}{2 \Gamma\left(1+2 \alpha_1\right)}}\left[2 m_e(Z \alpha) r\right]^{\alpha_1-1} \, e^{-m_e(Z \alpha) r}\begin{pmatrix} 0 \\ 1 \\ \frac{\mathrm{i}\left(1-\alpha_1\right)}{Z \alpha} \sin \vartheta e^{\mathrm{i} \varphi} \\ \frac{-\mathrm{i}\left(1-\alpha_1\right)}{Z \alpha} \cos \vartheta\end{pmatrix}\end{aligned}\]
where \(\alpha_1 = \sqrt{1-Z^2\alpha^2}\). Here, \(\alpha\) is the fine-structure constant, \(\alpha = e^2/(\hbar c)\). \(\Gamma(x)\) is the Gamma function resulting from the normalization condition.

Q: Sketch the \(P_{1s}(r)\) and \(Q_{1s}(r)\) radial functions for \(Z = \{40,80,130\}\) !
A: By convention, the large components \(P_i(r)\) are chosen to start with positive values from the origin of the \(r\) coordinate. The radial functions \(P_i(r)\) and \(Q_i(r)\) take zero values at the origin because of the \(r\)-weighting. Consequently, the weak singularity of \(F_i(r)\) and \(G_i(r)\) at the origin is not visible. The size of the small component \(Q_{1 s}(r)\) increases with nuclear charge \(Z_e\). For super-heavy atoms (\(Z>100\)) the small component is no longer small.

Q: Sketch the energy levels of H-like atoms for the Schrödinger and Dirac case!
A: Shells with the same \(l\) but different \(j\) quantum numbers are now split:
spin-orbit splitting
Note that each drawn line actually represents two degenerate states (spin-up or spin-down) as the wavefunctions \(\Psi_{n_i \kappa_i m_j(i)}(\bm{r})\) are degenerate for different \(m_j\) but same \(n_i\) and \(\kappa_i\).
This means, that for example in \(1s\) or \(1s_{1/2}\) the electron can either be spin-up or spin-down.

Q: Does it make a difference in case of the Dirac hydrogen, if the nuclei are described as finite or point-like? And if this is the case, what changes?
A: The electron-nucleus potential can be written in a series expansion as
\[\begin{aligned} & V_\mathrm{nuc} = v_{-1} r^{-1} + v_0 + \mathcal{O}(r) \\ &\mathrm{with}\qquad v_{-1} = \left\{\begin{array}{rl} -Ze^2 & \text{for point-like nuclear charges} \\ 0 & \text{for finite-size nuclei}\end{array} \right. \end{aligned}\]
This affects the coefficients of the \(r^{\alpha-1}\) term of radial equation which yields:
\[ \alpha=\sqrt{\kappa^2-v_{-1}^2 /\left(\hbar^2 c^2\right)}=\left\{\begin{array}{rl} \sqrt{\kappa^2-Z^2 e^4 /\left(\hbar^2 c^2\right)} & \text{point-like nuclei} \\ |\kappa| & \text{finite-size nuclei} \end{array}\right.\]
Hence, it makes an important qualitative difference whether a singular Coulomb potential or a finite attraction potential is employed in the description of a relativistic Dirac atom.
While for a point-like nuclei the first exponent of the series expansion is not integral, it is well-behaved in the case of the potential of a finite nucleus.

Q: Does it make a difference in case of the Schrödinger hydrogen atom, if the nuclei are described as finite or point-like? And if this is the case, what changes?
A: The qualitative difference of the short-range behavior of the radial functions observed for the Dirac hydrogen does not occur in the case of non-relativistic Schrödinger atoms. The Schrödinger equation is a differential equation of second order. Therefore, the lowest powers in the radial function's series expansion are solely determined by the second derivative, and by the centrifugal potential if \(l>0\), rather than by the nuclear potential. Recall that in the Dirac hydrogen the short range behavior \(r\rightarrow0\) was examined to determine the unknown exponent \(\alpha\). In this case \(\dv{}{r} r^{\alpha+i}\) and \(r^{\alpha+i}/r\) lead to the same power in \(r\) and to non-vanishing coefficients \(q_0\) and \(p_0\) for \(r\rightarrow0\). However, in the case of the radial Schrödinger equation insertion of a series expansion of \(R(r)=(r^\alpha\sum_{i=0}^\infty p_i r^i )\exp{(-\rho r)}\) leads to an equation of the form
\[\begin{aligned} - & \frac{\hbar^2}{2m_e}\left[\sum_{i=0}^\infty (\alpha+i)\right. (\alpha+i-1)p_ir^{\alpha+i-2} \\ & - \rho \sum_{i=0}^\infty (\alpha+i)p_ir^{\alpha+i-1} + \rho^2\sum_{i=0}^\infty p_ir^{\alpha+i} \\ &- \left. \sum_{i=0}^\infty p_ir^{\alpha+i-2}l(l+1)\right] + \sum_{i=0}^\infty p_ir^{\alpha+i}V_\mathrm{nuc}(r) \\ & + \sum_{i=0}^\infty p_ir^{\alpha+i}E = 0 \end{aligned}\]
By multiplying with \(r^{2-\alpha}\) and taking \(r\rightarrow0\) one can see that only the terms of the second derivative and the centrifugal potential are non-vanishing and thus determine the exponent \(\alpha\). As any potential is given by
\(V_\mathrm{nuc} = v_{-1} r^{-1} + v_0 + \mathcal{O}(r)\)
without terms of \(\mathcal{O}(r^{-2})\) it does not contribute to the short-range exponent determination.