 Solution

In the ideal point-nucleus Dirac-Coulomb problem, hydrogen levels depend on $n$ and $j$, so $2S_{1/2}$ and $2P_{1/2}$ are degenerate despite their different orbital angular momenta. The Lamb shift removes this degeneracy: the $2S_{1/2}$ level lies about $h\times 1.06\mathrm{GHz}$ above $2P_{1/2}$, an energy of roughly $4.4\mu \mathrm{eV}$. This is distinct from the much larger $2P_{3/2}-2P_{1/2}$ fine-structure splitting.

quantum electrodynamics changes the electron's interaction with the Coulomb field. electron self-energy and vacuum polarization give state-dependent radiative corrections; a complete prediction also keeps specified recoil and nuclear-size contributions. An $S$ wave samples the nucleus differently from a $P$ wave, so equal $n,j$ no longer ensures equal energy. A leading-log estimate for an $nS$ radiative contribution scales as $\alpha (Z\alpha {)}^4{m}_e{c}^2\ln [(Z\alpha {)}^{-2}]/n^3$, with a numerical prefactor and further state-dependent terms. This is a scale estimate, not a replacement for renormalized bound-state QED. The measurable shift is a difference relative to a specified reference Hamiltonian, rather than an unregulated absolute vacuum energy.^[ref]

The original microwave experiment used a tungsten oven to dissociate hydrogen, an electron beam to populate metastable $2S$ atoms, and electron ejection from a metal target to measure surviving metastable atoms. Microwaves transferred population to rapidly decaying $2P$, reducing detector current. At a fixed microwave frequency, interruption of the radiation and a scan of static magnetic field revealed Zeeman-tuned resonances. Comparing resonance locations with the Dirac-level prediction showed the $S$ level shifted upward. Field calibration and homogeneity, excessive microwave intensity, unresolved hyperfine structure and stray-electric-field quenching limit the measurement; the signal is not direct absorption by a large equilibrium excited population.^[ref]