Hydrogen fine-structure corrections

For a point nucleus of charge $+Ze$ and mass $M$, separating center-of-mass motion gives a relative-coordinate Coulomb Hamiltonian with reduced mass $\mu=m_eM/(m_e+M)$. Its negative energies scale as $\mu Z^2/n^2$. This mass replacement is exact for the nonrelativistic two-body model. It explains a leading isotope effect: a heavier nucleus gives a slightly larger reduced mass, hence more tightly bound levels. It does not incorporate nuclear size or relativistic recoil.

The expansion parameter for low-lying hydrogenic states is $Z\alpha$, with $\alpha$ the dimensionless fine-structure constant. For small nuclear charge this makes relativistic energy corrections much smaller than Coulomb binding. To state an unambiguous leading fine-structure formula below, take an infinitely heavy point nucleus and electron mass $m_e$. Replacing that mass everywhere by the reduced mass is a useful leading estimate for some purposes, but is not an exact finite-mass relativistic solution.

Expanding relativistic kinetic energy produces the negative operator $-p^4/(8m_e^3c^2)$. Spin-orbit coupling adds an interaction between orbital and spin angular momenta, including the Thomas-precession factor one half. The Darwin term is a contact correction for a point Coulomb source; its expectation is nonzero for Schrödinger s states. These are terms in an effective Hamiltonian obtained from relativistic electron theory. The contact term should not be pictured as an electron following a tiny classical orbit through the nucleus.

Although individual corrections depend on orbital angular momentum, their sum depends only on $n$ and total electronic angular momentum $j$ at this order. The spin-orbit term vanishes for orbital angular momentum zero, but the total fine-structure correction does not. Omitting the Darwin term therefore gives a particularly misleading answer for s states. At fixed $n$, the surviving degeneracy between states with the same $j$ is a special property of the point-Coulomb Dirac problem, rather than a general consequence of rotational symmetry.

For the two p levels at $n=2$, the smaller $j$ has the more negative correction. Their energy difference is positive when defined as the $j=3/2$ energy minus the $j=1/2$ energy: specify upper minus lower before calculating. Holding the mass approximation fixed, this interval scales as the fourth power of nuclear charge. Gross Coulomb transition energies instead scale as its square. Confusing the two scalings makes a helium-ion comparison wrong by a factor of four.

A laboratory line also includes differences of radiative shifts, nuclear recoil, finite-size effects, and possibly hyperfine contributions. The fine-structure approximation predicts coincidence of the $2s_{1/2}$ and $2p_{1/2}$ levels, but their measured separation is not zero. Calling that discrepancy an experimental failure would confuse a controlled approximation with the full theory. The attached parabola lab explores square-law scaling only; its position variable is a dimensionless comparison parameter, not a continuously adjustable physical nuclear charge.