University Quantum Field Theory I · The Lamb Shift Completed at Order α (Z α)⁴ · 15.7
Matching across the overlap: ln(m/μγ) → ln(m/(2K)) + 5/6
Two correct calculations, each carrying its own fictitious parameter, do not add up to a right answer until you can say exactly which photons each one counted. Here that bookkeeping is done once, on one soft-photon integral, and it turns Unit 14's photon mass into Bethe's cutoff so that both can disappear.
Build the model
Connect the measurement to the mechanism.
The order α(Zα)⁴ self-energy of an s state comes in two halves that cannot be added as they stand. The free-electron vertex is right for virtual photons far above atomic energies and carries a photon mass μγ; Bethe's bound-state dipole sum is right for photons far below m and carries a cutoff K on photon energy. Both regulators are fictions, but different fictions: K deletes every photon below one energy, while a mass suppresses soft photons smoothly, makes each transverse photon cost an energy ω above κ, and adds a longitudinal polarisation.
Matching uses the window (Zα)²m ≪ k ≪ m, where binding no longer matters and the photon is still soft, so both halves reduce to one dipole integral for a free electron kicked by the potential, worth (α/3π)(q²/m²) per unit ln k. Evaluating that integral once with the mass and once with the cutoff recomputes the high part with exactly the photons below K removed, and gives ln(m/μγ) → ln(m/(2K)) + 5/6, derived. Bethe's ln(K/K₀) then cancels K: the ns self-energy is (4α(Zα)⁴m/(3πn³))[ln(m/(2K₀)) + 11/24], 1015.2 MHz on hydrogen's 2s.
The construction stops where the window closes. Its dropped terms are of relative order K/m and (Zα)²m/K, at best Zα, which is why it is an α(Zα)⁴ result and why it has nothing to stand on in hydrogen-like uranium, where (Zα)²m is already 0.45m.
- Simple definition
- Matching rewrites the photon-mass high part of the Lamb shift as if the virtual photons below Bethe's energy cutoff K had been removed instead, by replacing ln(m/μγ) with ln(m/(2K)) + 5/6, so that it can be added to Bethe's low part and K cancels.
- Example
- Cut Bethe's part at K = 5.00 keV and the high part's bracket becomes ln(511.0/10.00) + 11/24 = 4.392, exactly what a photon mass of 0.869K = 4.35 keV gives; on hydrogen's 2s level that is 4.392 × 135.6 MHz = 595.6 MHz.
Bethe's integrand above atomic energies and the vertex's soft corner both reduce to this, so only the weight ρ, the regulator, differs between them.
⊃ marks the soft part; κ = |k| photon momentum (MeV); split Λ with K, μγ ≪ Λ ≪ m; ρ in MeV⁻¹; q² = −|q|²
Transverse modes give ln(2Λ/μγ) − 1, since a massive photon of momentum κ costs an energy ω above κ; the longitudinal mode adds 1/6.
ω = √(κ² + μγ²) the massive photon's energy; first term the two transverse modes, second the longitudinal one
Read backwards, a photon mass μγ = 2e(−5/6)K = 0.869K removes as much of the soft dipole integral as a sharp cutoff at K does, though not the same photons.
from ∫₀Λ θ(κ − K) dκ/κ = ln(Λ/K) minus the card above; K Bethe's cutoff on photon energy (MeV)
At K = αm = 3.73 keV the bracket is 4.685, which is 635.3 MHz on hydrogen's 2s level.
11/24 = 5/6 − 3/8; prefactor ÷ h = 135.6 MHz for Z = 1, n = 2; m = 0.511 MeV; zero for l ≥ 1
Bethe's ln(K/K₀) plus the card above: 7.487 units, 1015.2 MHz on 2s. The anomalous moment and the Uehling term are added in topic 15.08.
K₀ Bethe's mean excitation energy, quoted: ln(K₀/Ry) = 2.8118 for 2s, so K₀ = 226.4 eV; Ry = 13.606 eV
At the best K each term is Zα ≈ 0.73%, the relative size of the α(Zα)⁵ corrections; as Zα → 1 the window closes.
hydrogen: (Zα)²m = 27.2 eV, Zαm = 3.73 keV, m = 511.0 keV; both ratios are pure numbers
Two halves, two regulators, one double count
Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout; a static Coulomb kick has q⁰ = 0 and q² = −|q|². Topic 15.03 turned the slope of F₁ into an ns shift (4α(Zα)⁴m/(3πn³))[ln(m/μγ) − 3/8], derived in this course; topic 15.06 quoted Bethe's low part (4α(Zα)⁴m/(3πn³))ln(K/K₀) from UAQM-13.04 to 13.07. Added as they stand they leave ln(mK/(μγK₀)) − 3/8, which moves with two unphysical parameters at once. The fault is bookkeeping. The free vertex counts virtual photons down to about μγ and Bethe counts them up to K, so the photons between are counted twice, or not at all if μγ exceeds K. Setting μγ = K does not cure that, because one edge is cut smoothly and the other sharply. The cure is to recompute the high part with exactly the photons below K removed, the ones Bethe keeps. Because K ≪ m only soft photons are touched; the hard region is common to both. Hydrogen leaves room: (Zα)²m = 27.2 eV, Zαm = 3.73 keV, m = 511 keV.
In the overlap both halves are one dipole integral
Above atomic energies, k ≫ |Eₘ − Eₙ|, Bethe's integrand (2α/(3πm²))Σₘ|pₘₙ|²(Eₘ − Eₙ)/(Eₘ − Eₙ + k) tends to (2α/(3πm²))Σₘ|pₘₙ|²(Eₘ − Eₙ)/k, and the sum rule Σₘ|pₘₙ|²(Eₘ − Eₙ) = ½⟨∇²V⟩, quoted, leaves (α/(3πm²))⟨∇²V⟩ per unit ln k with no atomic level in it. For a free electron kicked from p to p + q, ⟨p + q|∇²V|p⟩ = q²Ṽ(q), so this is a correction (α/3π)(q²/m²) to F₁ per unit ln k: the coefficient of ln(m/μγ) in the slope, built in the vertex's soft corner. For (Zα)²m ≪ k ≪ m both halves are the soft-photon factor of a slow electron whose velocity jumps by δ = q/m. The photon couples to the conserved current Jμ = p′μ/(p′⋅k) − pμ/(p⋅k), whose space part is δ/ω and time part δ⋅k/ω² here, and the virtual correction is minus half the probability of emitting that photon, the identity behind topic 14.08's infrared cancellation, quoted. Hence F₁ − 1 ⊃ (α/3π)(q²/m²)∫₀Λ ρ(κ)dκ, with ρ = 1/κ for massless photons, derived. Only ρ knows the regulator.
Evaluate it with a photon mass
Unit 14's vertex uses the Feynman-gauge propagator −igμν/(k² − μγ² + iε). Because k⋅J = 0, adding kμkν/μγ² changes nothing, so the photon acts as a massive vector with three polarisations and energy ω = √(κ² + μγ²). The two transverse modes give |J⊥|², which averages over directions to (2/3)δ²/ω². The longitudinal mode, εL = (κ, ωk̂)/μγ, gives εL⋅J = −μγ(δ⋅k̂)/ω², averaging to (1/3)δ²μγ²/ω⁴. Normalised so that a massless photon gives 1/κ, the weight is ρμ = κ²/ω³ + μγ²κ²/(2ω⁵). Put κ = μγ sinh t: the transverse integral is t − tanh t and the longitudinal one (1/6)tanh³t, so ∫₀Λ ρμ dκ = ln(2Λ/μγ) − 5/6 + μγ²/(2Λ²) + O(μγ⁴/Λ⁴), derived. Each piece has a reading. The 2 is arcsinh x → ln 2x. The −1 comes from transverse photons near κ ∼ μγ, which cost an energy ω above κ and so count for less than 1/κ. The +1/6 is the longitudinal mode a real photon does not have. At Λ = 100μγ the exact integral, 4.46503, exceeds the asymptote by 5.0 × 10⁻⁵.
Swap the regulators and watch K and μγ leave
Applied to the high part, Bethe's cutoff keeps only massless transverse photons above K, so ρK = θ(κ − K)/κ and ∫₀Λ ρK dκ = ln(Λ/K). Take the split Λ with K, μγ ≪ Λ ≪ m. Above Λ both calculations are the same free vertex; below it they differ only through ρ. So the high part with photons below K removed is the photon-mass result, minus its soft integral, plus the cutoff one: ln(m/μγ) − 3/8 − [ln(2Λ/μγ) − 5/6] + ln(Λ/K) = ln(m/(2K)) + 11/24. Λ and μγ have gone, and this is the replacement ln(m/μγ) → ln(m/(2K)) + 5/6, derived. Adding Bethe's ln(K/K₀) gives the ns self-energy (4α(Zα)⁴m/(3πn³))[ln(m/(2K₀)) + 11/24], with K gone, derived here with K₀ quoted. For 2s, ln(K₀/Ry) = 2.8118 gives K₀ = 226.4 eV and ln(m/(2K₀)) = ln(1/α²) − 2.8118 = 7.0287, so the bracket is 7.487 and the shift 7.487 × 135.6 MHz = 1015.2 MHz. At K = αm = 3.73 keV it divides into 635.3 MHz above K and 379.9 MHz below; at K = 25 keV, into 377.3 and 637.9 MHz.
Check it: dimensions, limits and the measured splitting
Dimensions: every logarithm holds a ratio of energies, m/K, K/K₀ or Λ/μγ, and the prefactor 4α(Zα)⁴m/(3πn³) is an energy, 135.6 MHz times h for Z = 1 and n = 2. Limits: α → 0 removes the shift, as the tree level must; d/dK of ln(m/(2K)) + ln(K/K₀) is −1/K + 1/K = 0 across the window; and for l ≥ 1 the contact term vanishes with |ψ(0)|² while the sum rule removes Bethe's K-dependence, so matching touches s states only and 2P₁/₂ keeps its K-free Bethe logarithm. The decisive check is the measurement. With the conversion, the splitting assembled in topic 15.08, quoted ahead, is about 1051.9 MHz against the measured 1057.8 MHz. Identify μγ with K instead and 2s drops by (5/6 − ln 2) × 135.6 MHz = 19.0 MHz, to 1032.8 MHz. Keep the 2 but forget the longitudinal mode, using ln(m/(2K)) + 1, and it rises by 135.6/6 = 22.6 MHz, to 1074.5 MHz. Each wrong join misses by 17 to 25 MHz, three to four times the 5.9 MHz the right one leaves.
What the matching assumes, and who got it right first
Gauge: Coulomb below K, Feynman above. In the overlap only the conserved current J enters, so any propagator term carrying kμ or kν drops against k⋅J = 0 and the gauges agree there; that is a check at this order, not a theorem about the split. Regulator: K cuts the photon energy in the atom's rest frame, which the atom selects, and μγ must sit far below Λ, the μγ²/(2Λ²) term being dropped. Orders: the high part ignores binding near K, an error of relative order (Zα)²m/K; the dipole integral ignores retardation, magnetic coupling and recoil near K, relative order K/m. Their sum is least at K ≈ Zαm, where each is Zα ≈ 0.7%, the relative size of the α(Zα)⁵ terms a complete treatment supplies; 0.7% of 1015 MHz is about 7 MHz, the scale of the 5.9 MHz still missing (topic 15.09). History, quoted: Feynman's and Schwinger's first relativistic results mismatched the two regulators, while French and Weisskopf, and Kroll and Lamb, matched them consistently in 1949 and found about 1051 MHz. Feynman later acknowledged the slip in print.
Change one variable at a time
Make the relationship visible.
Start at μγ = K, the naive join: the smooth curve deletes more than the step and 2s comes out 19.0 MHz low. Slide μγ/K to 0.87 and the gap closes. Then drag K across the window: the high and low readouts trade hundreds of megahertz while their sum stays at 1015.2 MHz.
HIGH PART ABOVE K, 2s623.8 MHz
BETHE LOW PART, 2s391.4 MHz
SUM, K CANCELS1015.2 MHz
GAP IF μγ IS READ AS K-19.0 MHz
Live interpretationHIGH PART ABOVE K, 2s: 623.8 MHz. BETHE LOW PART, 2s: 391.4 MHz. SUM, K CANCELS: 1015.2 MHz. GAP IF μγ IS READ AS K: −19.0 MHz
Catch the common trap
Explain before calculating.
Hydrogen's high part is computed with a photon mass μγ and joined to Bethe's low part, cut at K, by setting μγ = K. Compared with the correct matching, how does this move the predicted 2S₁/₂–2P₁/₂ splitting? Take 4α(Zα)⁴m/(3πn³) ÷ h = 135.6 MHz for n = 2.
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyBethe's low part for hydrogen is cut at K = 5.00 keV. Find the photon mass μγ that removes as much of the soft dipole integral from the high part as the cutoff does, evaluate the high-part bracket both as ln(m/μγ) − 3/8 and as ln(m/(2K)) + 11/24, and convert it to megahertz on 2s with the unit 135.6 MHz.
- The replacement ln(m/μγ) = ln(m/(2K)) + 5/6 fixes μγ = 2e(−5/6)K = 2 × 0.43460 × 5.00 keV = 4.346 keV.
- Photon-mass form: ln(510.999 keV/4.346 keV) − 0.375 = ln(117.58) − 0.375 = 4.7671 − 0.375 = 4.3921.
- Cutoff form: ln(510.999 keV/10.00 keV) + 11/24 = 3.9338 + 0.4583 = 4.3921, the same, as it must be.
- On 2s: 4.3921 × 135.6 MHz = 595.6 MHz. Dimensions: each logarithm holds a ratio of energies, and the unit carries the energy.
- Contrast the naive μγ = K = 5.00 keV: ln(102.20) − 0.375 = 4.6269 − 0.375 = 4.2519, short by 0.1402, which is 19.0 MHz.
Answerμγ = 0.869K = 4.35 keV; both forms give 4.392, which is 595.6 MHz on 2s. Setting μγ = K instead gives 4.252, 19.0 MHz too low.
MediumTest the photon-mass dipole integral. With ρμ = κ²/ω³ + μγ²κ²/(2ω⁵) and ω = √(κ² + μγ²), evaluate ∫₀Λ ρμ dκ exactly for Λ = 10μγ and Λ = 100μγ, split it into transverse and longitudinal parts, compare with ln(2Λ/μγ) − 5/6, and say what the residual would cost on 2s if Λ = 10μγ were used as the split point.
- Put κ = μγ sinh t, so ω = μγ cosh t and dκ = μγ cosh t dt. The transverse part is ∫ tanh²t dt = t − tanh t; the longitudinal part is ½∫ tanh²t sech²t dt = (1/6)tanh³t. Both are pure numbers, since ρ carries MeV⁻¹ and dκ carries MeV.
- Λ = 10μγ: t = arcsinh 10 = 2.99822 and tanh t = 10/√101 = 0.99504, so the transverse part is 2.00319 and the longitudinal part (1/6)(0.98519) = 0.16420; the total is 2.16738.
- Asymptote: ln 20 − 5/6 = 2.99573 − 0.83333 = 2.16240. The residual, 0.00498, matches μγ²/(2Λ²) = 0.00500.
- Λ = 100μγ: transverse 5.29834 − 0.99995 = 4.29839, longitudinal 0.16664, total 4.46503, against ln 200 − 5/6 = 4.46498; the residual is 5.0 × 10⁻⁵.
- The longitudinal part tends to 1/6 and the transverse part to ln(2Λ/μγ) − 1, so −5/6 = −1 + 1/6. On 2s the residual costs 0.00498 × 135.6 MHz = 0.68 MHz at Λ = 10μγ and 0.0068 MHz at Λ = 100μγ.
AnswerExact 2.1674 against 2.1624 at Λ = 10μγ, and 4.46503 against 4.46498 at Λ = 100μγ; the residual is close to μγ²/(2Λ²), 0.68 MHz on 2s at Λ = 10μγ. The longitudinal mode supplies the +1/6 in −5/6 = −1 + 1/6.
HardFor hydrogen's 2s level, with K₀ = 226.4 eV quoted and the unit 135.6 MHz, evaluate the high part ln(m/(2K)) + 11/24 and Bethe's ln(K/K₀) in megahertz at K = 2.00 keV and at K = 25.0 keV, and show that their sum does not depend on K. Then use K/m + (Zα)²m/K to judge each choice, find the best K, and say whether the window survives for hydrogen-like neon, Z = 10, and uranium, Z = 92.
- K = 2.00 keV: the high part is ln(127.75) + 0.4583 = 4.8501 + 0.4583 = 5.3084, which is 719.8 MHz; the low part is ln(2000 eV/226.4 eV) = ln(8.834) = 2.1786, which is 295.4 MHz; the sum, 7.4870, is 1015.2 MHz.
- K = 25.0 keV: the high part is ln(10.220) + 0.4583 = 2.7827, which is 377.3 MHz; the low part is ln(110.42) = 4.7043, which is 637.9 MHz; the sum is again 1015.2 MHz. Some 342 MHz moved from one half to the other while the total stood still, because the sum's derivative in K is −1/K + 1/K = 0.
- Dropped terms at K = 2.00 keV: K/m = 0.0039 and (Zα)²m/K = 27.21 eV/2000 eV = 0.0136, about 1.8% in all, mostly binding. At 25.0 keV: 0.0489 + 0.0011, about 5.0%, now mostly the dipole approximation.
- The sum K/m + (Zα)²m/K is least where the two are equal, K = Zαm = 3.73 keV, each then Zα = 0.0073, about 1.5% in all. That is an order-of-magnitude estimate, not a computed error; a complete treatment replaces it by definite α(Zα)⁵ terms of this relative order.
- Neon, Z = 10: Zα = 0.0730, (Zα)²m = 2.72 keV and Zαm = 37.3 keV, so a window remains but the dropped terms grow to about 15%. Uranium, Z = 92: Zα = 0.671 and (Zα)²m = 230 keV, almost half of m, so no K can sit both far above the binding scale and far below m.
Answer2s self-energy 1015.2 MHz at both K (719.8 + 295.4 and 377.3 + 637.9 MHz). Dropped terms about 1.8% and 5.0%, least, about 2Zα = 1.5%, at K = Zαm = 3.73 keV. Neon keeps a window at about 15%; uranium has none.