The three corrections from the Dirac equation shift hydrogen’s levels at order $\alpha^{4}mc^{2}$, a few hundred-thousandths of the binding energy, and they can be computed with first-order perturbation theory in the hydrogen states. The kinetic correction depends on $l$; the spin–orbit coupling depends on $l$ and $j$ and acts only for $l\ge1$; the Darwin term, a smearing of the electron over its Compton wavelength, acts only on s states, which reach the nucleus. Each depends on $l$, yet their sum does not: the result depends only on $n$ and $j$, $\Delta E=-(\alpha^{4}mc^{2}/2n^{4})(n/(j+\tfrac12)-\tfrac34)$. So $2s_{1/2}$ and $2p_{1/2}$ stay degenerate, and $2p_{3/2}$ lies above them by $4.5\times10^{-5}$ eV, a line split by 11 GHz. The Dirac equation, solved exactly, gives the same formula to this order. The one thing it misses, the small Lamb shift pushing $2s_{1/2}$ above $2p_{1/2}$, belongs to the quantised electromagnetic field.

Three accountants each round a bill differently, and the rounding errors cancel so neatly that the total depends only on the sum paid, not on the items. The three corrections depend on $l$, and their sum does not.

Hydrogen’s $n=2$ level without and with fine structure, to scale in units of $\alpha^{4}mc^{2}/100$: $2p_{3/2}$ moves down by and with by , leaving them apart. The Lamb shift, from the quantised field, would lift a tenth of the way back up.

To first order in the three corrections, hydrogen’s levels depend on $n$ and $j$ only:

• The kinetic, spin–orbit and Darwin corrections each depend on $l$; their sum does not: levels of equal $n$ and $j$ stay degenerate, such as $2s_{1/2}$ and .

Average each correction over a hydrogen state. The kinetic term is rewritten with $p^{2}=2m(E_{n}-V)$, which needs $⟨$ and . The spin–orbit term needs and the value of in a state of given . The Darwin term needs the density at the nucleus. Adding them for , and for , the -dependence cancels.

$p^{2}=2m(E_{n}-V)$ in the state; $⟨V⟩=$ and as stated.

For $l\ge1$, with $\braket{1/r^{3}}$ and $\widehat{\mathbf{L}}\cdot \widehat{\mathbf{S}}=\frac{{\mathrm{\hslash }}^2}{2}[j(j+1)-l($.

$\nabla^{2}V=\frac{e^{2}}{\varepsilon_{0}}\delta^{3}(\mathbf{r})$ for the Coulomb potential; only s states reach .

For both signs the $l$ in the kinetic term trades for $j$; for $l=0$ the Darwin term plays the part of the spin–orbit one.

To order $\alpha^{4}$, hydrogen’s levels depend only on $n$ and $j$.

The three corrections each depend on $l$, but their sum depends only on $n$ and $j$: $2s_{1/2}$ and $$ keep the same energy.

The Darwin term involves the Laplacian of the Coulomb potential, a delta function at the nucleus, so only s states feel it.

Adding the kinetic and spin–orbit shifts for $j=l\pm\tfrac12$, the $l$ in each trades for $$; the Darwin term does the same job for .

Great work! You now know how to:

• average the three relativistic corrections over hydrogen states
• combine them into the fine-structure formula
• say what the formula leaves to the quantised field