The sum rule that confines the shift to s states

Units are SI throughout, with energies in eV and shifts quoted as frequencies δE/h; the electron stays non-relativistic, so the course metric (+, −, −, −) never enters. Topic 13.04's mass subtraction, quoted here, left δEₙ = (2α/(3πm²c²)) Σₘ |pₘₙ|²(Eₘ − Eₙ) ln(K/|Eₘ − Eₙ|), for a cutoff K far above the energies that matter. Split each logarithm as ln(K/K₀) + ln(K₀/|Eₘ − Eₙ|) and choose K₀, as topic 13.06 does, so that the second pieces cancel once weighted. What is left is δEₙ = (2α/(3πm²c²)) Sₙ ln(K/K₀), with Sₙ = Σₘ |pₘₙ|²(Eₘ − Eₙ). Sₙ alone says how the level answers the cutoff: doubling K adds (2α/(3πm²c²)) Sₙ ln 2, whatever K₀ is. It is a sum over every bound and continuum state with weights of both signs, positive for states above n and negative for states below, and the rest of this topic evaluates it.

Energy differences are commutators in disguise: (Eₘ − Eₙ)⟨m|p|n⟩ = ⟨m|[H, p]|n⟩, and because p commutes with p²/2m, [H, p] = [V, p] = iħ∇V. Used on the ket side with closure over bound and continuum states, this gives Sₙ = ⟨n|p⋅[H, p]|n⟩; used on the bra side, Sₙ = ⟨n|[p, H]⋅p|n⟩. Their average is half a double commutator, and [pᵢ, iħ∂ᵢV] = ħ²∂ᵢ²V turns it into Sₙ = (ħ²/2)⟨n|∇²V|n⟩, derived. This is the Thomas-Reiche-Kuhn move of Quantum Mechanics II with p in place of x. In oscillator strengths it is the third rung of a ladder: Σₘ fₘₙ = 1; Σₘ fₘₙ(Eₘ − Eₙ) = (2/(3m))⟨p²⟩, which fed topic 13.02's linear divergence; and Σₘ fₘₙ(Eₘ − Eₙ)² = (ħ²/(3m))⟨∇²V⟩, this sum. Each rung leans harder on the high-energy continuum, and the next, (2ħ²/(3m))⟨|∇V|²⟩ ∝ ⟨r⁻⁴⟩, diverges for s states.

The hydrogen-like V = −Ze²/(4πε₀r) is harmonic everywhere except the origin: ∇²(1/r) = −4πδ³(r), so ∇²V = (Ze²/ε₀)δ³(r). The expectation value collapses onto the density at the nucleus, and with e²/ε₀ = 4παħc the sum is Sₙ = 2πZαħ³c|ψₙ(0)|², derived. An ns state has |ψₙ₀₀(0)|² = Z³/(πn³a₀³) with a₀ = ħ/(mcα), which makes Sₙ = 2Z⁴mEₕ²/n³, where Eₕ = α²mc² = 27.211 eV. Dimensions agree: |pₘₙ|² times an energy is kg² m² s⁻² × J = kg J², and so is mEₕ². Multiplying by 2α/(3πm²c²) gives δE(ns) = (4Z⁴α⁵mc²/(3πn³)) ln(K/K₀). The prefactor is h × 1085 MHz for 1s and h × 135.6 MHz for 2s, falling as n⁻³. With K = mc² and topic 13.06's quoted K₀ = 16.64 Ry = 226.4 eV, ln(K/K₀) = 7.72 and hydrogen's 2s rises 1047 MHz, against the measured 1057.8 MHz of the 2S₁/₂–2P₁/₂ interval.

Every state with l ≥ 1 vanishes at the nucleus, so Sₙₗ = 0 exactly; its individual terms do not vanish. Take 2p. Its one partner below is 1s, with radial integral ∫R₁₀R₂₁r³ dr = (128√6/243)a₀ = 1.290 a₀, so |⟨1s|r|2p⟩|² = (1.290 a₀)²/3 = 0.5549 a₀² for each magnetic sublevel. With |pₘₙ|² = (m/ħ)²(Eₘ − Eₙ)²|rₘₙ|² and Eₘ − Eₙ = −(3/8)Eₕ, its weight is −(27/512) × 0.5549 = −0.02926 in units of mEₕ². That weight is the decay: the rate Γ = (4α/(3m²c²ħ)) times the size of the downward weights, quoted from topic 13.01, gives 6.27 × 10⁸ s⁻¹, the measured 1.60 ns lifetime of 2p. The sum rule then demands that 3s, 3d, 4s, 4d and the continuum supply +0.02926 mEₕ² between them. For 2s nothing below needs cancelling: 1s is dipole-forbidden and 2p is degenerate, and S₂ₛ = 0.25 mEₕ² is 8.5 times the 2p cancellation.

Because Sₙ is an expectation value, the whole s-level shift is the expectation of one operator, δH = (αħ²/(3πm²c²)) ln(K/K₀) ∇²V, with each level's own K₀. Unit 6's Darwin term is (ħ²/(8m²c²))∇²V: the same operator with a different coefficient. Their ratio, (8α/(3π)) ln(K/K₀) = 0.0478 for 2s, puts Bethe's shift at 4.8 per cent of the 21.9 GHz Darwin energy of 2s, which reproduces the 1047 MHz. In the Dirac energies the Darwin term does for s levels what spin-orbit coupling does for l ≥ 1, which is why 2S₁/₂ and 2P₁/₂ stay together; Bethe's contact term has no such partner, so it parts them. The confinement to s states belongs to Coulomb's point source, not to the sum rule. In an isotropic oscillator ∇²V = 3mω² everywhere, every level has Sₙ = (3/2)mħ²ω², and no interval moves.

Sₙₗ = 0 removes ln K from a p level's shift, not the shift itself. The cancelling weights sit at different energies, and their logarithms do not cancel: measuring each energy in units of Z² Ry, δE(nl) = −(4Z⁴α⁵mc²/(3πn³)) ln k₀(n, l), where ln k₀(n, l) is Σₘ |pₘₙ|²(Eₘ − Eₙ) ln(|Eₘ − Eₙ|/(Z² Ry)) divided by 2Z⁴mEₕ²/n³, the s-level value of Sₙ, as topic 13.06 computes it. The quoted ln k₀(2, 1) = −0.0300 gives +4.07 MHz for 2p, with no cutoff in it and 0.4 per cent of the 2s shift. The same zero is why K₀ cannot be defined for a p level as it is for an s level: it would mean dividing by Sₙ = 0. In full QED, 2P₁/₂ ends near −13 MHz, because the anomalous moment that topic 13.09 quotes lowers it by 17.0 MHz. And the s-level formula needs ln(K/K₀) to be large; for hydrogen-like ions K₀ grows as Z², so the logarithm shrinks, which the hard example follows to its failure.