Reading the spectrum: n, j and hydrogen-like uranium

Work in SI units, with α = e²/4πε₀ħc = 1/137.036 and mc² = 510.999 keV; where four-vectors appear the metric is (+, −, −, −), though nothing in this topic needs one. Topic 7.06's termination condition gave E = mc²[1 + (Zα)²/(nᵣ + γ)²]^(−1/2) with γ = √(κ² − (Zα)²). Look at where κ enters: inside γ as κ², and in n = nᵣ + |κ| through its size. Reverse the sign of κ at fixed nᵣ and nothing in the formula changes, so the states κ and −κ of one shell are degenerate, and since nothing was expanded the statement holds to all orders in Zα. With this course's convention the pair is l = j − ½, negative κ, and l = j + ½, positive κ: 2S₁/₂ with 2P₁/₂, 3S₁/₂ with 3P₁/₂, 3P₃/₂ with 3D₃/₂. The exception is the top level j = n − ½, which needs nᵣ = 0, and nᵣ = 0 terminates only for negative κ, so 1s₁/₂, 2P₃/₂ and 3D₅/₂ stand alone. Bohr's degeneracy is lifted only part of the way: the 18 states of n = 3 fall into three levels holding 4, 8 and 6 states, not five.

The partners are not one state written twice. 2S₁/₂ has an s-wave upper component and a p-wave lower one, 2P₁/₂ the reverse; they have opposite parity and opposite eigenvalues of K, and topic 7.04's radial equations differ between them through the ∓κ/r terms. Near the origin both G and F go as r^γ. Put that into dG/dr = −(κ/r)G + (E + mc² − V)F/ħc, where −V = Zαħc/r dominates, and γ = −κ + Zα F/G, so F/G = (κ + γ)/(Zα). For κ = −1 that is −(1 − γ)/(Zα) = −0.0036 in hydrogen; for κ = +1 it is (1 + γ)/(Zα) = 274. Close to the nucleus a 2P₁/₂ electron is almost all ‘small’ component, an s-wave. In uranium the two ratios are −0.386 and 2.59. So no rotation or reflection forces the coincidence: it is dynamical, a property of 1/r like Bohr's l-degeneracy, and an operator found by Johnson and Lippmann in 1950, a Dirac cousin of the Runge–Lenz vector, commutes with H and carries κ into −κ (quoted). Any departure from a point charge's 1/r splits the pair.

Expand the exact energy in (Zα)² and it gives Bohr's −mc²(Zα)²/2n² and then −mc²(Zα)⁴(n/(j + ½) − ¾)/2n⁴, the order-α⁴ formula Quantum Mechanics II quoted and Unit 6 assembled from three Foldy–Wouthuysen terms. Each of those terms depends on l; their sum cannot, because the exact answer does not. Check it for n = 2 in units of mc²(Zα)⁴. For 2S₁/₂ the kinetic term gives −13/128 and the Darwin term +8/128, and spin–orbit is zero at l = 0. For 2P₁/₂ the kinetic term gives −7/384 and spin–orbit −8/384, and the Darwin term, a contact term, is zero. Both sum to −5/128: the Darwin term supplies at l = 0 exactly what spin–orbit cannot. For 2P₃/₂, −7/384 + 4/384 = −1/128, so the doublet is mc²(Zα)⁴/32, which in hydrogen is 4.53 × 10⁻⁵ eV or 10.95 GHz, against 10.969 GHz measured. The 0.2% left over is mostly the electron's anomalous magnetic moment, quoted, which the Dirac equation does not contain.

Hydrogen cannot tell the exact formula from its first correction: the next term, mc²α⁶/16, is 4.8 × 10⁻⁹ eV. Hydrogen-like uranium can. With Z = 92, Zα = 0.6714, (Zα)² = 0.4507 and γ₁ = √(1 − (Zα)²) = 0.7411, the exact 1s energy is mc²γ₁, a binding of mc²(1 − γ₁) = 132.28 keV. Bohr's (Zα)²mc²/2 gives 115.16 keV, and adding Unit 6's (Zα)⁴mc²/8 = 12.98 keV gives 128.13 keV, still 4.15 keV short. The (Zα)⁶ term supplies 2.92 keV of that and the (Zα)⁸ term 0.82 keV: the series converges, but slowly, and Unit 6's term carries only 76% of the 17.12 keV relativistic correction. The checks hold. The binding is mc² times a pure number, so the dimensions are right; as Z → 0, mc²(1 − √(1 − (Zα)²)) → (Zα)²mc²/2, Bohr's value; and only at Zα = 1 does the binding reach mc², where topic 7.09 takes over.

For n = 2 the formula collapses into closed forms. The j = ½ pair has nᵣ = 1 and γ₁, and (1 + γ₁)² + (Zα)² = 2(1 + γ₁), so E = mc²√((1 + γ₁)/2) for both 2S₁/₂ and 2P₁/₂, a binding of 34.22 keV in uranium. 2P₃/₂ has nᵣ = 0 and γ₂ = √(4 − (Zα)²) = 1.8840, so E = mc²γ₂/2 = mc²√(1 − (Zα)²/4), a binding of 29.65 keV. Bohr puts all eight n = 2 states at 28.79 keV. The doublet is 4.57 keV, against Unit 6's mc²(Zα)⁴/32 = 3.24 keV. Expanding both closed forms one order further gives the splitting as mc²[(Zα)⁴/32 + 5(Zα)⁶/256 + …], so the exact splitting exceeds the α⁴ value by about (5/8)(Zα)² at first, and by more as Zα grows: 2% in iron, 10% by Z = 53, and 41% in uranium, where (5/8)(Zα)² alone gives 28%. The quantum defect δ = (Zα)²/(|κ| + γ) shows why the shift is lopsided. It is 0.259 for j = ½ and 0.116 for j = 3/2, so the j = ½ pair, whose spinors reach the nucleus as r^{(γ₁ − 1)}, is pulled 18.8% beyond Bohr while 2P₃/₂ moves only 3.0%.

X-ray measurements on U⁹¹⁺ in the ESR storage ring at GSI found the ground state 460 ± 5 eV less bound than the point-nucleus Dirac value (Gumberidze and co-workers, 2005, quoted). Theory accounts for it in two comparable pieces, both quoted. The nucleus's finite size, with a 5.86 fm rms charge radius, takes about 0.20 keV, because a 1s density that grows toward the nucleus as r^{(2γ₁ − 2)} feels the flattened potential inside it. The radiative corrections take about 0.27 keV: the self-energy raises the level by about 0.36 keV and vacuum polarisation lowers it by about 0.09 keV. So the measured binding is about 131.82 keV. The exact point-nucleus value is right to 0.35%, while the α⁴ formula is 2.8% short and Bohr 12.6%. Both pieces lie outside the Dirac–Coulomb Hamiltonian: the finite nucleus and vacuum polarisation change the point charge's 1/r potential, and the self-energy couples the electron to the quantised radiation field. So they also split every κ, −κ pair. In hydrogen the same physics separates 2S₁/₂ from 2P₁/₂ by the measured 1057.8 MHz Lamb shift, which Unit 13 computes; the Dirac formula predicts zero there at every order.