Bethe against Welton, and the ground-state shift

Units are SI, with energies also quoted as frequencies ΔE/h; this non-relativistic topic needs no four-vectors, but the course's metric (+, −, −, −) stands. Welton's route, quoted from Quantum Mechanics II: the zero-point field jiggles a free electron by ⟨(δr)²⟩ = (2α/π)(ħ/mc)² L, and averaging the Coulomb potential over the jiggle adds (1/6)⟨(δr)²⟩∇²V. Bethe's route, derived earlier in this unit: after the mass subtraction, ΔEₙ = (2α/(3πm²c²)) Σₘ |pₘₙ|²(Eₘ − Eₙ) ∫₀^K dE/(E + Eₘ − Eₙ). If every state shared one logarithm L, the sum rule Σₘ |pₘₙ|²(Eₘ − Eₙ) = (ħ²/2)⟨∇²V⟩ would give (α/(3π))(ħ/mc)²⟨∇²V⟩ L, and Welton's (1/6)(2α/π) is the same α/(3π). Dimensions: (ħ/mc)² is m² and ∇²V is J m⁻². With ∇²V = 4παħc δ³(r) and |ψₙ₀₀(0)|² = 1/(πn³a³), an ns level shifts by (4α⁵mc²/(3πn³)) L. Limits: a free electron, ∇²V = 0, keeps no shift, its whole self-energy having gone into the mass; and α → 0 removes it as α⁵. The coefficient is settled; the dispute is what L is.

Each intermediate state carries the factor (Eₘ − Eₙ)/(E + Eₘ − Eₙ). Far above the interval it becomes (Eₘ − Eₙ)/E, and summed with the sum rule the integrand is (α/(3π))(ħ/mc)²⟨∇²V⟩ dE/E, Welton's free-electron jiggle term for term: a photon well above the atom's frequencies drives the electron as if it were free. Below the interval the factor tends to 1 and stops growing. A field slower than the atom's own transitions meets the bound electron's stiff response, eE/(mωₘₙ²), not the free eE/(mω²) that rises without limit as ω falls. Integrating gives (Eₘ − Eₙ) ln(1 + K/(Eₘ − Eₙ)). For 1s → 2p, Eₘ − Eₙ = 10.20 eV and K/(Eₘ − Eₙ) = 50 077, so ln 50 078 and ln 50 077 differ by 2 × 10⁻⁵. Per unit ln E, and divided by Eₘ − Eₙ, the integrand is a step, E/(E + Eₘ − Eₙ): photons below the knee add ln 2, exactly what the step withholds above it, so the smooth knee integrates to a hard cutoff at Eₘ − Eₙ itself. Welton chose ωₘᵢₙ; the spectrum supplies it, state by state.

Summing the per-state logarithms with weights wₘ = |pₘₙ|²(Eₘ − Eₙ) turns L into ln(K/K₀), with ln K₀ the weighted mean of ln(Eₘ − Eₙ). Since |pₘₙ|² = m²ωₘₙ²|rₘₙ|², wₘ = (3/2)m fₘₙ(Eₘ − Eₙ)²: a state counts as its oscillator strength times its interval squared. For 1s, Σ fₘ(Eₘ − E₁)² = 16/3 Ry² by the sum rule. The 2p state, f = 0.4162 at 0.75 Ry, carries 0.234 Ry², only 4.4%; all bound p states together carry 6.8%, though they hold 56.5% of the oscillator strength. The continuum carries 93.2%, and its weight per unit energy falls only as E^(−3/2), so half of all the weight lies above 164 eV and 22% above 1.36 keV. These shares are computed here from the closed-form 1s oscillator strengths, quoted, and the same sum returns ln(K₀/Ry) = 2.984, the quoted Bethe logarithm. So K₀ = 19.8 Ry = 269 eV, twenty times the binding energy. The weights measure ∇²V at the nucleus, and only high-momentum continuum states resolve so small a region.

With K = mc², assumed, and ln(K₀/Ry) = 2.984, the logarithm is ln(mc²/Ry) − 2.984 = 10.534 − 2.984 = 7.550. For n = 1 the coefficient is 4α⁵mc²/(3π) = 4.488 μeV, h × 1085.2 MHz, so ΔE₁ₛ = 33.9 μeV, or 8.19 GHz. The measured ground-state Lamb shift, extracted from Doppler-free two-photon 1S–2S spectroscopy, is 8.173 GHz (quoted), 0.24% lower. Welton's natural lower cutoffs fail here: the 1 Ry ionisation energy gives 11.43 GHz, the 10.2 eV Lyman-α energy 11.74 GHz, the atomic unit α²mc² = 2 Ry 10.68 GHz, all 31% to 44% high, because each sits an order of magnitude below the true K₀. The upper cutoff is the softer worry: halving or doubling K moves the answer by 1085.2 MHz × ln 2 = 752 MHz, 9%, to 7.44 or 8.94 GHz. So the 0.24% is better than the method can promise. The covariant corrections quoted in the next topic shift the pieces by hundreds of MHz and nearly cancel, netting about −65 MHz at this order; the near-coincidence is partly luck, and 9% is the honest error bar.

Welton's argument, with one lower cutoff for every level, predicts ΔE₁ₛ/ΔE₂ₛ = 8 exactly, the ratio of the densities at the nucleus. Bethe's predicts 8 × ln(mc²/K₀(1s))/ln(mc²/K₀(2s)) = 8 × 7.5495/7.7219 = 7.821, using ln(K₀/Ry) = 2.812 for 2s, quoted, which with K = mc² puts 2s at 1047 MHz. The 2s mean excitation energy, 16.6 Ry = 226 eV, is 67 times its own binding energy and close to the 1s value, because both are set by the cusp at r = 0, not by the size of the orbit. The measured ratio needs the 2S₁/₂ level's own shift: the measured 1057.8 MHz interval includes the 12.8 MHz by which QED, quoted, lowers 2P₁/₂, so 2S₁/₂ itself rises 1057.8 − 12.8 = 1045.0 MHz, and 8172.8/1045.0 = 7.821. The test is sharp because the upper cutoff nearly cancels: halving K gives 7.804 and doubling it 7.836, while the departure from 8 is 0.18. That 2% failure of the 1/n³ law is the lower cutoff's signature, the one feature of the ground-state shift that Welton's picture cannot produce.

Give the nucleus charge Ze. ∇²V gains one factor of Z and |ψ₁₀₀(0)|² = Z³/(πa³) three more, so the coefficient grows as Z⁴. Every momentum matrix element scales as Z and every interval as Z², so the weights scale together and the ion's mean excitation energy is Z² × 19.77 Ry. The upper cutoff does not move, so the logarithm is ln(mc²/(Z²K₀)) = ln(1/(Zα)²) − 2.291, down 2 ln Z; Z = 1 returns 7.550. The product Z⁴(7.550 − 2 ln Z) peaks near Z = 34, where the logarithm is 0.50, and vanishes at Z = √1899.8 = 43.6, where Zα = 0.32 and Z²K₀ has reached mc². Past it the formula turns negative: for hydrogen-like uranium it gives −480 eV, against a measured 1s Lamb shift of +460.2 ± 4.6 eV and an all-orders theory of 464.0 eV, about 265 eV of it QED and 199 eV the nucleus's finite size (quoted). The zero is not physics. Bethe's result keeps only the large logarithm of an expansion in Zα; once that logarithm is no larger than the constants beside it and the electron moves at Zαc, only calculations to all orders in Zα, quoted, describe heavy ions.