University Quantum Field Theory I · The Lamb Shift Completed at Order α (Z α)⁴ · 15.2
The slope of F₁ and its infrared logarithm
Unit 14 left F₁ finite in the ultraviolet and still holding a photon mass. Here we take its first derivative at q² = 0, the part of the vertex an atom actually feels, and find the photon mass sitting in it as a logarithm: a debt Unit 15 must pay by matching onto Bethe's low-energy calculation.
Build the model
Connect the measurement to the mechanism.
Charge conservation pins F₁(0) = 1, so the first thing the one-loop vertex says about how the electron's charge is spread is the slope: how fast F₁ falls as the momentum transfer grows. Expanding Unit 14's renormalised Feynman-parameter integral to first order in q² gives F₁(q²) = 1 + (α/3π)(q²/m²)[ln(m/μγ) − 3/8], and the calculation shows where each piece comes from. The hard region w = 1 − z ∼ 1 supplies a constant, while the corner z → 1, where the virtual photon is soft and both electron lines are nearly on shell, contributes the same amount per decade from w ≈ μγ/m up to w ≈ 1 and so builds a logarithm of the photon mass.
The ultraviolet is gone, the infrared is not: read as a charge radius, the slope gives ⟨r²⟩ = (2α/(πm²))[ln(m/μγ) − 3/8], which grows without limit as μγ → 0. The model is a free electron scattered once by a slowly varying field, and it stops working in two places: at |q²| comparable with m², where higher powers of q² matter, and at virtual-photon energies so low that the electron is no longer free, which in hydrogen means below about (Zα)²m. The second boundary is why the logarithm is a debt rather than an answer.
Unit 15 pays it by matching this slope onto Bethe's low-energy calculation, which carries the same coefficient, so that the photon mass and Bethe's cutoff cancel.
- Simple definition
- The slope of F₁ is the coefficient of q² in the electron's renormalised charge form factor near q² = 0; at one loop with a photon mass μγ it is (α/(3πm²))[ln(m/μγ) − 3/8], which depends logarithmically on that infrared regulator.
- Example
- At −q²/m² = 10⁻³ with μγ/m = 10⁻⁶ (μγ = 0.51 eV) the bracket is 13.82 − 0.375 = 13.44 and F₁ − 1 = −1.04 × 10⁻⁵; halving μγ adds ln 2 = 0.69 to the bracket and makes F₁ − 1 = −1.09 × 10⁻⁵.
Topics 14.02 and 14.05 in one line: no ε and no scale μ remain, only the photon mass μγ. Everything in this lesson is a Taylor expansion of this integral in q².
w = 1 − z, x = wξ, y = w(1 − ξ); Δ₀ = m²w² + μγ²(1 − w), Δ = Δ₀ − q²xy, N₀ = m²(w² + 2w − 2); masses in MeV
F₁(0) = 1 is a normalisation, so this coefficient is the first thing the vertex says about the charge. It is positive for μγ ≪ m, so F₁ dips below 1 at spacelike q².
q² = (p′ − p)², negative for scattering; m = 0.511 MeV; photon mass μγ ≪ m; α/(3π) = 7.743 × 10⁻⁴; F₁ is dimensionless
h sits near 1 on every decade from μγ/m to about 0.1, so each decade adds ln 10 = 2.30. The hard region w ∼ 1 gives −5/8 and the photon mass's rounded edge gives +1/4; the O(μγ/m) remainder is under 3 parts in 10⁴ once μγ/m ≤ 10⁻³.
h = (3w²/2)[1/6 + (w² + 2w − 2)w²/(6Δ₀²) + (1 − w)/Δ₀ + w²/(6Δ₀)] with m = 1; h → (3/4)w² − w + 1 for w ≫ μγ/m
Soft-photon reasoning reproduces the logarithm's coefficient but not the −3/8, which only the full Feynman-parameter integral supplies.
fIR(q²) = ∫₀¹ dξ (m² − q²/2)/(m² − ξ(1 − ξ)q²) − 1, from topic 14.06; ≈ −q²/(3m²) − q⁴/(20m⁴)
Each factor of 10 by which μγ falls adds 1595 fm² to ⟨r²⟩ and nothing stops it, so a free electron has no charge radius until the soft photons are treated.
F₁′ = dF₁/dq²; from F₁ = 1 − |q|²⟨r²⟩/6 at q⁰ = 0; 2α/(πm²) = 692.8 fm² with ħ/(mc) = 386.16 fm
Start from Unit 14's renormalised integral
Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout, so q² = (p′ − p)² is negative when an electron scatters. Topic 14.02 combined the vertex's denominators into Δ = (1 − z)²m² + zμγ² − xyq², and topic 14.05 fixed δ₁ = δ₂ by F₁(0) = 1, removing ε and the scale μ; both are quoted. What remains is the subtracted integral of the first formula card, with w = 1 − z, x = wξ and y = w(1 − ξ), so the simplex becomes ∫₀¹ w dw ∫₀¹ dξ. Hydrogen needs only small transfers, |q| ∼ Zαm/n, so −q²/m² ∼ (Zα)² ≈ 5 × 10⁻⁵ and first order in q² is ample. Expanding under the integral is safe even as μγ → 0, because the expansion parameter is bounded: xy = w²ξ(1 − ξ) ≤ w²/4 while Δ₀ ≥ m²w², so |q²|xy/Δ₀ ≤ |q²|/(4m²) everywhere, the soft corner included.
Expand the integrand to first order in q²
Set m = 1 for the algebra and write λ = μγ². The logarithm gives ln(Δ₀/Δ) ≈ q²xy/Δ₀ = q²ξ(1 − ξ), a finite piece from which the photon mass can be dropped. The fraction gives q²[N₀xy/Δ₀² + (1 − x)(1 − y)/Δ₀]. Insert N₀ = w² + 2w − 2 and (1 − x)(1 − y) = 1 − w + w²ξ(1 − ξ), average over ξ with ∫₀¹ ξ(1 − ξ) dξ = 1/6, and define S by F₁ − 1 = (α/2π)q²S. Then S = ∫₀¹ w dw [1/6 + (w² + 2w − 2)w²/(6Δ₀²) + (1 − w)/Δ₀ + w²/(6Δ₀)], with Δ₀ = w² + λ(1 − w). Four parts are finite as λ → 0 and can be done at once: 1/12 from the logarithm, 1/12 from the last term, 5/12 from the 2w + w² part of the second term, and −1 from the −w in 1 − w. The two left over, the −2 in N₀ and the 1 in 1 − w, both behave as 1/w near w = 0.
The soft corner builds the logarithm
Small w means z → 1: the photon's Feynman parameter takes almost everything, the virtual photon is soft and both electron lines sit nearly on shell. There the two remaining integrands fall as 1/w until w drops to about μγ/m, where Δ₀ stops shrinking. With u = w², ∫₀¹ w³ dw/(w² + λ)² = ½[ln(1/λ) − 1] and ∫₀¹ w dw/(w² + λ) = ½ ln(1/λ), up to terms that vanish with λ. The first enters S with weight −1/3 and the second with weight 1, so the logarithms add to (1/3)ln(1/λ) = (2/3)ln(m/μγ), and the constants, the +1/6 from the first integral with the four finite parts, add to −1/4. Hence S = (2/3)[ln(m/μγ) − 3/8] and F₁(q²) = 1 + (α/3π)(q²/m²)[ln(m/μγ) − 3/8], derived. The dropped terms are proportional to μγ/m: the exact first-order integral exceeds the formula by 3 parts in 10⁴ at μγ/m = 10⁻³ and by 2 parts in 10⁵ at 10⁻⁴. Integrating the unexpanded one-loop F₁ at μγ/m = 10⁻⁵ and −q²/m² = 10⁻³ reproduces the slope to 1.4 parts in 10⁴, and the gap shrinks tenfold with q², as a q⁴ term should.
Check dimensions, limits and the coefficient
Dimensions: q²/m² and m/μγ are pure numbers, so F₁ stays dimensionless and the slope dF₁/dq² has mass dimension −2, as 1/m² does. Limits: α → 0 returns the tree-level F₁ = 1, and q² → 0 returns F₁(0) = 1, the condition that fixed δ₁ = δ₂. The logarithm's coefficient has an independent check. Topic 14.06 found F₁ − 1 → −(α/2π)fIR(q²)ln(m²/μγ²) at small −q², and fIR(q²) ≈ −q²/(3m²), so the logarithm is (α/3π)(q²/m²)ln(m/μγ): the same corner reached through soft photons. The −3/8 has no such check, because fIR knows only the logarithm. For contrast, the same expansion of topic 14.03's F₂ gives (α/2π)(1 + q²/(6m²)), stated, a slope with no logarithm. No experiment measures this slope alone, since none has a photon mass; its first meeting with data is topic 15.08's 2s splitting, quoted ahead as about 1052 MHz against the measured 1057.8 MHz.
Read the slope as a charge radius
For a static, spherically symmetric charge density ρ(r) of unit total charge, probed with q⁰ = 0, the Fourier transform is ∫d³r ρ(r)e(iq⋅r) = 1 − |q|²⟨r²⟩/6 + …. Matching F₁(−|q|²) to that form defines the Dirac radius ⟨r²⟩ = 6 dF₁/dq² at q² = 0, so ⟨r²⟩ = (2α/(πm²))[ln(m/μγ) − 3/8]: a reading of the derived slope, stated, not a new calculation. The Sachs charge radius, read from GE = F₁ + (q²/(4m²))F₂, adds a finite 3α/(4πm²) = 260 fm² and leaves the logarithm untouched. Restoring ħ and c, (2α/π)(ħ/mc)² = 0.0046456 × (386.16 fm)² = 692.8 fm². A photon mass of 10 eV gives ⟨r²⟩ = 7.25 × 10³ fm² and √⟨r²⟩ = 85 fm, about a fifth of ħ/(mc), and the radius keeps growing as μγ → 0, though only as the square root of a logarithm. Virtual photons of every softness, which probe ever larger distances, contribute equally per decade, so nothing in the free vertex gives the electron's charge an edge. A free electron's radius is not defined until the soft photons are treated, and there is no measured value to compare with.
The debt Unit 15 must pay
The flat stretch of the slope's integrand runs from w ≈ μγ/m upward, so the photon mass decides how many decades of soft virtual photons the free vertex counts. A bound electron is not free for photon energies below about (Zα)²m (topic 15.05), so for hydrogen the decades counted below there are wrong and must be replaced, not merely regulated. Two features of this result make the replacement possible. Topic 15.03 turns the slope into an s-state shift (4α(Zα)⁴m/(3πn³))[ln(m/μγ) − 3/8], quoted ahead, and the coefficient in front is exactly the one that multiplies ln K in Bethe's low-energy shift, quoted from UAQM-13.04 to 13.07, so a cutoff can cancel between the two. And the −3/8 is specific to the photon-mass regulator, not a property of the electron: a sharp cutoff on w moves it by 1/4, and topic 15.07's swap of ln(m/μγ) for ln(m/(2K)) + 5/6, quoted ahead, is the same bookkeeping for Bethe's cutoff K.
Change one variable at a time
Make the relationship visible.
Drag log₁₀(μγ/m) from −2 to −8: the plateau stretches left and the bracket grows by 2.30 per decade without ever settling. Then raise Z: the (Zα)² marker moves right, and by Z = 92 hardly any plateau is left between it and w = 1.
BRACKET ln(m/μγ) − 3/813.44
RMS RADIUS √⟨r²⟩96.5 fm
F₁ − 1 AT |q| = Zαm/2-0.1385 ppm
PLATEAU BELOW (Zα)²1.7 decades
Live interpretationBRACKET ln(m/μγ) − 3/8: 13.44. RMS RADIUS √⟨r²⟩: 96.5 fm. F₁ − 1 AT |q| = Zαm/2: −0.1385 ppm. PLATEAU BELOW (Zα)²: 1.7 decades
Catch the common trap
Explain before calculating.
The renormalised one-loop F₁ is evaluated with a photon mass μγ, and a student lowers μγ from 1 keV to 1 eV. What happens to the mean square charge radius ⟨r²⟩ = 6 dF₁/dq² at q² = 0?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyIn hydrogen's 2s state a typical momentum transfer is |q| ≈ αm/2. Evaluate F₁(q²) − 1 there from the slope formula with a photon mass μγ = 10 eV, then with μγ = 1 eV, and say what the comparison shows.
- Static transfer, q⁰ = 0: q² = −|q|² = −(α/2)²m², so q²/m² = −(0.0072974/2)² = −1.3313 × 10⁻⁵, with |q| = 1.86 keV. First order is ample: the expansion parameter is at most |q²|/(4m²) ≈ 3 × 10⁻⁶.
- Bracket at μγ = 10 eV: ln(510 999 eV/10 eV) − 3/8 = 10.8415 − 0.375 = 10.4665.
- F₁ − 1 = (α/3π)(q²/m²)[ln(m/μγ) − 3/8] = 7.7427 × 10⁻⁴ × (−1.3313 × 10⁻⁵) × 10.4665 = −1.079 × 10⁻⁷.
- At μγ = 1 eV the bracket gains ln 10 = 2.3026 and becomes 12.7691, so F₁ − 1 = −1.316 × 10⁻⁷: 22 per cent larger in size, from a regulator that is meant to be unphysical.
- Checks: the result is a pure number, and it vanishes as α → 0 or q² → 0, as a correction to F₁(0) = 1 must.
AnswerF₁ − 1 ≈ −1.08 × 10⁻⁷ at μγ = 10 eV and −1.32 × 10⁻⁷ at μγ = 1 eV: tiny, yet 22% dependent on a photon mass that no measurement fixes.
MediumRead the slope as a mean square charge radius. Find √⟨r²⟩ in femtometres for photon masses of 1 keV and 1 eV, then estimate how small μγ would have to be for √⟨r²⟩ to reach the Bohr radius, 52 918 fm.
- ⟨r²⟩ = 6 dF₁/dq² = (2α/(πm²))[ln(m/μγ) − 3/8]. Restore units with 1/m → ħ/(mc) = 197.327 MeV fm/0.510999 MeV = 386.16 fm, so 2α/(πm²) → 0.0046456 × (386.16 fm)² = 692.75 fm².
- μγ = 1 keV: bracket = ln(511.00) − 0.375 = 6.2364 − 0.375 = 5.8614, so ⟨r²⟩ = 4060 fm² and √⟨r²⟩ = 63.7 fm.
- μγ = 1 eV: bracket = 13.1441 − 0.375 = 12.7691, so ⟨r²⟩ = 8846 fm² and √⟨r²⟩ = 94.1 fm. A thousandfold change of regulator moved the radius by 30 fm.
- Bohr radius: the bracket must reach (52 918 fm)²/(692.75 fm²) = 2.800 × 10⁹ fm²/692.75 fm² = 4.04 × 10⁶, so ln(m/μγ) ≈ 4.04 × 10⁶ and μγ/m ≈ 10(−1.76 × 10⁶).
- The growth is slow but never stops: the radius is a logarithm of the regulator, not a property of the electron, so no choice of μγ is the right one.
Answer√⟨r²⟩ = 63.7 fm at μγ = 1 keV and 94.1 fm at 1 eV; reaching the Bohr radius would need μγ/m ≈ 10(−1.76 × 10⁶). The radius never converges as μγ → 0.
HardSwap the photon mass for a sharp cutoff: keep the massless integrand h₀(w) = (3/4)w² − w + 1 per unit ln w and integrate it only from wc to 1. Find the bracket this gives, choose wc to reproduce ln(m/μγ) − 3/8 for μγ = 100 eV, and say what the careless choice wc = μγ/m would cost on hydrogen's 2s level.
- The integral of h₀ dw/w from wc to 1 is the integral of (3/4)w − 1 + 1/w, which is (3/8)(1 − wc²) − (1 − wc) + ln(1/wc), or ln(1/wc) − 5/8 once wc ≪ 1.
- Match it to the photon-mass result: ln(1/wc) − 5/8 = ln(m/μγ) − 3/8, so wc = e(−1/4) μγ/m = 0.7788 μγ/m.
- For μγ = 100 eV, μγ/m = 100/510 999 = 1.9570 × 10⁻⁴ and wc = 1.524 × 10⁻⁴. Check: ln(1/wc) − 0.625 = 8.7890 − 0.625 = 8.164, and ln(5110.0) − 0.375 = 8.539 − 0.375 = 8.164.
- The plateau is the same in both schemes; only the rounding of its lower edge differs, and the photon mass's soft edge is worth +1/4 against a sharp edge at the same w. The careless wc = μγ/m gives 8.539 − 0.625 = 7.914, low by 1/4.
- Topic 15.03, quoted ahead, multiplies the bracket by 4α(Zα)⁴m/(3πn³), which is 135.6 MHz for Z = 1 and n = 2 (topic 15.01), so the careless choice shifts 2s by 135.6/4 = 33.9 MHz: nearly six times the 5.8 MHz by which Unit 15's final answer misses 1057.8 MHz.
AnswerThe sharp cutoff gives ln(1/wc) − 5/8, so wc = e(−1/4) μγ/m = 1.52 × 10⁻⁴ for μγ = 100 eV. The logarithm's coefficient survives any regulator; the constant does not, and wc = μγ/m errs by 1/4, about 34 MHz on 2s.