The electron moves at a hundredth of the speed of light, so relativity enters as a correction of order α² — small enough to ignore for eight chapters and large enough to measure.

Two loose ends. §9.6 asks how good the non-relativistic approximation was and finds the corrections — small, but they split levels that §9.2 left degenerate. §9.7 finally solves Eq. (9.15), which every section so far has quoted the answer to.

How non-relativistic is the electron?

Its momentum is uncertain by about $\hbar/a_0$, so that is roughly its average momentum:

where $\alpha$ is the fine structure constant fine structure constant α = e²/4πε₀ħc = 1/137.0, dimensionless. It is the electron's orbital speed in hydrogen as a fraction of c, so α² ≈ 5 × 10⁻⁵ measures how badly non-relativistic quantum mechanics is wrong. defined in the toolkit — open in glossary :

Expanding the relativistic energy $E = \sqrt{m_e^2c^4 + p^2c^2}$ for $p \ll m_ec$ gives the rest mass, the familiar $p^2/2m_e$, and then a correction

Spin–orbit coupling

A second correction of the same size comes from magnetism. In the electron’s frame the nucleus appears to orbit, and a circulating charge is a current loop, producing a field at the electron

The electron’s spin moment $\boldsymbol\mu = -2(e/2m_e)\mathbf S$ (Eq. 8.8) then has an orientation energy $-\boldsymbol\mu\cdot\mathbf B$. A careful treatment that accounts for the electron’s acceleration introduces a factor of one-half, giving the spin–orbit interaction spin–orbit interaction The coupling ∝ L·S between the electron's spin magnetic moment and the magnetic field it experiences from the nucleus's apparent motion. Of order α⁴m_ec², the same size as the relativistic kinetic-energy correction, and it forces states to be labelled by j. defined in ch. 9 — open in glossary :

Two consequences matter.

First, it is relativistic. Rewriting Eq. (9.35) with $a_0 = \hbar/\alpha m_ec$ and $\langle L\cdot S/r^3\rangle \sim \hbar^2/a_0^3$ gives $\langle E_{\mathrm{mag}}\rangle \sim \alpha^4m_ec^2$ — the same order as Eq. (9.33). The two corrections are not independent effects that happen to be comparable; both are relativity showing up at order $\alpha^4$.

Second, it forces a new labelling. Since $\mathbf J = \mathbf L + \mathbf S$,

so a state of definite energy must have definite $J^2$, $L^2$ and $S^2$ — that is, definite $j$, $l$ and $s$. The good quantum numbers have changed, and that is why atomic states are labelled $n\,l_j$ from here on.

Fine structure at n = 2

Without spin–orbit coupling all $n = 2$ states share $E_2 = -E_R/4 = -(\alpha^2/8)m_ec^2$. With $l = 0$ or $1$ and $s = \tfrac12$, Eq. (8.6) gives $j = \tfrac12$ for the s-state and $j = \tfrac12$ or $\tfrac32$ for the p-state — three levels, $2s_{1/2}$, $2p_{1/2}$ and $2p_{3/2}$.

Perturbation theory then splits them — the fine structure fine structure The splitting of hydrogen's levels at order α⁴m_ec² ≈ 10⁻⁴ eV, from the relativistic correction to the kinetic energy and the spin–orbit interaction together. It separates 2p_{3/2} from 2p_{1/2} but leaves 2s_{1/2} and 2p_{1/2} degenerate. defined in ch. 9 — open in glossary of hydrogen — and this is where the site parts company with the printed text.

§9.7 — actually solving it

The book marks this section “may be omitted without significant loss of continuity.” This site builds it, because it is the only place the energy levels are derived rather than quoted, and because the argument is a good one: the answer comes from demanding that a power series stop.