Fine Structure Hydrogen
We wish to use perturbation theory to add some parts missing in the original hydrogen equation. We consider these perturbation to be very small compared to the unperturbed eigenvalues. Recall the hamiltonian for the unperturbed hydrogen which is basically the Bohr model,
\[\begin{split}
\begin{align}
H^0 &= -\frac{\hbar^2}{2m}\nabla^2 - \frac{e^2}{4\pi\varepsilon_0}\frac{1}{r}\\
E_n^0 &= \frac{m}{2\hbar^2}\left(\frac{e^2}{4\pi\epsilon_0}\right)^2\frac{1}{n^2}
\end{align}
\end{split}\]
The correction to the energy for the fine structure hydrogen is on the order of a constant called the fine structure constant \(\alpha\) defined as,
\[ \begin{align}
\alpha \equiv \frac{e^2}{4\pi\epsilon_0\hbar c} \approx \frac{1}{137.036}
\end{align} \]
We may write the \(E_n^0\) in terms of \(\alpha\),
\[ \begin{align}
E_n^0 = \frac{m\alpha^2 c^2}{2}\frac{1}{n^2}
\end{align} \]
The fine structure correction is on the orders of \(E \approx \alpha^2E^0_n\). We will now prove this in the following sections for the fine structure hamiltonian,
\[ H' = K_r' + V_\text{SO}' + V'\]
\(K'\) : Relativistic correction to energy.
\(V'_\text{SO}\) : Spin orbit coupling correction to potential energy.
\(V'\) : Additional correction to potential energy.
The first order correction to the energy eigenvalue of \(H'\) is,
\[\begin{split}
\begin{gather}
E_{nj}^1 = E'_r + E_\text{SO} + E'\\
\boxed{E_{nj}^1 = E_0^1 \frac{\alpha^2}{n^2}\left(\frac{n^2}{j+\frac{1}{2}}-\frac{3}{4}\right)}
\end{gather}
\end{split}\]
Relativistic Correction
An immediately obvious correction is to consider relativistic effects from the moving electron. Recall that,
\[ \begin{align}
K^0 = \frac{p^2}{2m}
\end{align} \]
However special relativity tells us that,
\[\begin{split} \begin{align}
E^2 &= (mc^2)^2 + (pc)^2 \nonumber\\
K &= \sqrt{m^2c^4 + (pc)^2} - mc^2 \nonumber \\
&= mc^2\left(\sqrt{1 + \frac{p^2}{m^2c^2}} - 1\right) \nonumber\\
&\approx mc^2\left(1 + \frac{p^2}{2m^2c^2} - \frac{1}{8}\frac{p^4}{m^4c^4} - 1\right)\nonumber\\
K &\approx \frac{p^2}{2m} - \frac{p^4}{8m^3c^2} \label{eq:kinetic}
\end{align} \end{split}\]
We consider the 2nd term of Equation \(\eqref{eq:kinetic}\) to be the relativistic correction such that,
\[ \begin{align}
\boxed{K'_r = -\frac{p^4}{8m^3c^2}}
\end{align} \]
Because we can come up with an operator that commutes with \(K'\) and \(H^0\) (e.g., \(L^2\), \(L^z\), \(S^2\), \(S_z\)) we may use the nondegenerate perturbation theory and determine the first-order energy correction,
\[\begin{split} \begin{align}
E_r^1 = \langle K' \rangle = \frac{1}{8m^3c^2}\langle \psi | p^4\psi \rangle \nonumber\\
\boxed{E_r^1 = E_n^0\frac{\alpha^2}{n^2}\left(\frac{n}{l +\frac{1}{2}} - \frac{3}{4}\right)}\\
\boxed{E_r^1 = \frac{(E_n^0)^2}{2mc^2}\left(\frac{4n}{l +\frac{1}{2}} - 3\right)}
\end{align} \end{split}\]
Spin Orbit Coupling
We will now consider not only the magnetic field created by the electron spin but also the magnetic field created by the electron orbiting the proton. This creates another term in the hamiltonian \(V'_\text{SO}\),
\[ \begin{align}
V'_\text{SO} = -\boldsymbol{\mu} \cdot \boldsymbol{B}
\end{align} \]
The magnitude of the magnetic field is simplified to a current through a loop (Biot-Savart law),
\[\begin{split}
\begin{align*}
B &= \frac{\mu_0I}{2r}\\
I &= \frac{e}{T} = \frac{e}{2\pi mr^2}L\\
B &= \frac{\mu_0e}{4\pi mr^3}L
\end{align*}
\end{split}\]
Plugging the magnitude in for \(\boldsymbol{B}\) which includes \(\boldsymbol{L}\) and the identity \(c=1/\sqrt{\epsilon_0\mu_0}\)
\[ \begin{align}
\boldsymbol{B} &= \frac{1}{4\pi\epsilon_0}\frac{e}{mc^2r^3}\boldsymbol{L} \label{eq:magnetic-field}
\end{align} \]
Now to figure out what the magnitude of \(\boldsymbol{\mu}\) is. Once again for the electron orbiting circularly,
\[\begin{split}\begin{align*}
\mu &= \frac{e\pi r^2}{T}\\
S &= \frac{2\pi m r^2}{T}
\end{align*}\end{split}\]
Reminder that \(S\) is the spin angular momentum (as if the electron was spinning). We combine the two equations above and correct it by a factor of 2 to be,
\[ \begin{align}
\boldsymbol{\mu} = \frac{e}{m}\boldsymbol{S}\label{eq:dipole}
\end{align} \]
Thus we put it all together and write the spin-orbit coupling hamiltonian (ignore bold font for vectors since we know these are operators),
\[
\begin{align}
\boxed{V'_\text{SO} = \left(\frac{e^2}{8\pi\epsilon_0}\right)\frac{1}{m^2c^2}\frac{1}{r^3} S \cdot L}
\end{align}
\]
Unfortunately \(S \cdot L\) does not commute with \(H\) but it does commute with \(L^2, S^2, J^2\). Let’s write \(S \cdot L\) in terms of those 3 operators,
\[\begin{split}
\begin{align*}
J^2 &= (L+S)\cdot(L+S) = L^2 + S^2 + 2L\cdot S\\
L \cdot S &= \frac{1}{2}(J^2 - L^2 - S^2)
\end{align*}
\end{split}\]
\(L \cdot S\) has the eigenvalues,
\[\begin{split}
\begin{gather*}
L \cdot S \; |\phi\rangle = \frac{\hbar^2}{2}\left[j(j+1)-l(l+1)-s(s+1)\right] \; |\phi\rangle\\
s = \frac{1}{2}
\end{gather*}
\end{split}\]
The other operator is \(1/r^3\) which has the expectation value,
\[
\begin{align*}
\avg{\frac{1}{r^3}} = \frac{1}{l(l+\frac{1}{2})(l+1)n^3a^3}
\end{align*}
\]
The energy to \(V'_\text{SO}\) is then \(E_\text{SO}^1\),
\[\begin{split} \begin{align}
E_\text{SO}^1 &= \frac{(E_n^0)^2}{2mc^2}\left(\frac{n \left[j(j+1)-l(l+1)-3/4\right] }{l(l+1/2)(l+1)}\right)\\
E_\text{SO}^1 &= E_n^0 \left(\frac{\alpha^2}{2nl(l+\frac{1}{2})(l+1)}\right)\cdot
\begin{cases}
-l &j = l + \frac{1}{2} \\ l+1 & j=l-\frac{1}{2}
\end{cases}
\end{align}
\end{split}\]
Dirac Potential Correction
While the effects are very small, because of the the magnetic repulsion at the center of the nuclei. This correction will be made into the hamiltonian \(V'\)
\[
\begin{align}
V' = \frac{\pi\hbar^2}{2m^2c^2}\left(\frac{e^2}{4\pi\varepsilon_0}\right)\delta(\vec{r})
\end{align}
\]
Since \(\delta(r)\) only acts on \(r=0\), we don’t care for \(n\) so set all to zero such that,
\[\begin{split}
\begin{align}
\bra{\psi_{nlm}} \delta(\vec{r})\ket{\psi_{nlm}} &= \abs{\psi_{n00}(0)}^2 = \frac{m^3c^3\alpha^3}{\hbar^3}\frac{1}{\pi n^3}\\
E'_p &= \frac{\pi\hbar^4\alpha}{m^2}\abs{\psi_{n00}(0)}^2 \nonumber\\
E'_p &= \frac{1}{2}mc^2\frac{\alpha^4}{\hbar^3} \cdot
\begin{cases}
-E_n^0 \frac{\alpha^2}{n} & l=0\\ 0 & l\neq 0
\end{cases}
\end{align}
\end{split}\]
\[\begin{split}
\begin{align}
H' &= -\vec{d}\cdot \vec{E}\\
\vec{d} &= -q\hat r\\
\vec{E} &= E_0\\
H' &= -qE_0 \hat r
\end{align}
\end{split}\]
\[\begin{split}
\begin{gather}
J_z = L_z^{(1)} + L_z^{(2)} + S_z^{(1)} + S_z^{(2)}\\
\end{gather}
\end{split}\]