= Solution
For a point, stationary proton, the leading hydrogen <fine structure> comes from three order-$1/c^2$ contributions to the nonrelativistic electron <Hamiltonian>:
$$
\boxed{H_{\rm fs}=-\frac{p^4}{8m_e^3c^2}+\frac1{2m_e^2c^2r}\frac{dV}{dr}\,\mathbf L\cdot\mathbf S+\frac{\hbar^2}{8m_e^2c^2}\nabla^2V,\qquad V=-\frac{e^2}{4\pi\epsilon_0r}.}
$$
These are the <relativistic kinetic-energy correction>, <spin-orbit interaction> and <Darwin term>. The spin-orbit coefficient includes the <Thomas precession> factor of one half; $\langle\mathbf L\cdot\mathbf S\rangle=\hbar^2[j(j+1)-l(l+1)-3/4]/2$. The Coulomb Darwin contribution is a positive contact term, relevant for S states. These corrections distinguish total <angular momentum> $j=l\pm1/2$, while leaving magnetic substates degenerate without an external field.
Writing $\alpha_{\rm fs}=e^2/(4\pi\epsilon_0\hbar c)$, the combined <hydrogen fine-structure correction> is
$$
\boxed{\Delta E_{nj}=-\frac{m_ec^2\alpha_{\rm fs}^4}{2n^3}\left[\frac1{j+1/2}-\frac3{4n}\right].}
$$
Thus the splitting is of order $\alpha_{\rm fs}^2$ times the principal binding energy. In the $n=2$ shell, $2S_{1/2}$ and $2P_{1/2}$ remain degenerate in this approximation, while $2P_{3/2}$ is higher by $m_ec^2\alpha_{\rm fs}^4/32$. The <Lamb shift> lifts the remaining S–P degeneracy; <hyperfine structure> is a separate nuclear-spin effect. Neither is part of the displayed correction. Nuclear recoil and finite nuclear size have also been neglected. <Topics in Quantum Mechanics by David Tong>[§§3.1.2–3.1.5, printed pp.68–79, relativistic motion, spin-orbit/Darwin terms and their sum]{ref}
\Image[/past-exam-of-the-advanced-quantum-physics-course-of-the-university-of-cambridge/2012/paper-1-aqp2012-hydrogen-fine-structure.png]
{title=Leading hydrogen n=2 fine-structure corrections retain the 2S1/2–2P1/2 degeneracy}
{source=paper-1-aqp2012-hydrogen-fine-structure.py}
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