University Quantum Field Theory I · The One-Loop Vertex: the Anomalous Magnetic Moment and Infrared Divergences · 14.6
F₁'s infrared divergence and the vanishing elastic rate
This is where QED first returns a negative probability, and the job is to see why it is the right answer to the wrong question. Topic 14.05 cleared F₁ of ε and μ but not of the photon mass; here that leftover is traced to soft virtual photons, and its coefficient fIR(q²) is computed in closed form.
Build the model
Connect the measurement to the mechanism.
Renormalisation fixes F₁(0) = 1, and with δ₁ = δ₂ that removes every trace of the ultraviolet regulator. It cannot remove the photon mass. The logarithm of μγ that the vertex shares with δ₂ at q² = 0 has a coefficient that changes with q², so a subtraction made at one point leaves −(α/2π) fIR(q²) ln(1/μγ²) at every other.
The source is the corner z → 1 of the Feynman-parameter simplex, where both electron propagators sit on shell and the virtual photon carries almost no momentum: there the integral over 1 − z is logarithmic and only μγ stops it. The coefficient depends on nothing but the rapidity χ between the incoming and outgoing electron, fIR = χ coth χ − 1, zero when the velocity does not change. Squared into a rate, the strictly elastic cross-section is dσ₀[1 − (α/π) fIR ln(−q²/μγ²)], and it falls through zero once μγ is small enough: below about 27 eV for 1 GeV electrons at 30°.
That is where the model stops. At order α the rate turns negative, at all orders (topic 14.09) it tends to zero, and at large −q² the full integral trades the leading logarithm for Sudakov's double logarithm. The cure is not a counterterm but a better question: no detector can separate elastic scattering from scattering with a photon below its resolution, and that sum is finite (topics 14.07 and 14.08).
- Simple definition
- F₁'s infrared divergence is the logarithm of the photon mass left in the renormalised one-loop vertex, −(α/2π) fIR(q²) ln(1/μγ²): it comes from virtual photons of vanishing momentum and is absent only at q² = 0.
- Example
- For 1 GeV electrons scattered through 30°, −q² = 0.268 GeV² and fIR = 12.8; with μγ = 1 eV the order-α elastic factor is 1 − 0.02983 × 40.13 = −0.197, a negative probability.
Derived from the corner z → 1. The coefficient of ln μγ² is exact; choosing M² is the leading-log step, and its error stays finite as μγ → 0.
α = 1/137.036, dimensionless; q², m² and μγ² in GeV² with ħ = c = 1; μγ ≪ m is the photon mass; fIR dimensionless
Derived: the numerator −2(m² − q²/2) and Δ at z = 1, with the −1 from subtracting at q² = 0. fIR(0) = 0, so the forward point keeps F₁ = 1.
ξ = x/(x + y) splits the weight between the two electron propagators, 0 ≤ ξ ≤ 1; q² and m² in GeV², q² ≤ 0 in scattering
Derived by doing the ξ integral. It tends to −q²/(3m²) for −q² ≪ m² and to ln(−q²/m²) − 1 for −q² ≫ m²; soft bremsstrahlung returns the same function.
χ is the rapidity of p′ seen from the rest frame of p, dimensionless; p and p′ are the electron four-momenta in GeV
Shown: |F₁|² = 1 + 2(F₁ − 1) at order α. Negative once μγ falls below √(−q²) exp[−π/(2α fIR)], about 27 eV for 1 GeV electrons at 30°.
dσ₀ is topic 9.09's Mott cross-section in GeV⁻² sr⁻¹, with 1 GeV⁻² = 0.3894 mb; −q² ≫ m², else m² replaces −q² in the log
Shown by keeping ln A(ξ) under the ξ integral: the ends ξ ≲ m²/(−q²) give ln²(−q²/m²)/2, half what the leading-log form assigns.
For −q² ≫ m² ≫ μγ²; single logarithms and constants dropped; each logarithm is of a ratio of GeV² quantities
Derived from Γ⁰ = (F₁ + F₂)γ⁰ − F₂E/m and the Mott trace. No μγ: F₂'s numerator vanishes at z = 1. It dies as β → 0, where only the charge acts.
β = |p|/E of the electron, θ the scattering angle; F₂(q²) = (α/2π) χ/sinh χ, dimensionless and never above α/2π
Where the photon mass hides after renormalisation
Units throughout are ħ = c = 1 with the metric (+, −, −, −), so q² ≤ 0 in scattering. Topic 14.05 left F₁(q²) = 1 + δF₁(q²) − δF₁(0), free of the pole in ε and of the scale μ. It still holds μγ, for a structural reason: δF₁(0) and −δ₂ share one infrared logarithm, so subtracting at q² = 0 cancels it there, but the logarithm's coefficient changes with q² and a subtraction at one point cannot follow it. Find the region by power counting in topic 14.02's Δ = (1 − z)²m² + zμγ² − xyq². Near the corner z → 1 put w = 1 − z, x = ξw and y = (1 − ξ)w, so that dx dy dz δ(x + y + z − 1) = w dw dξ and Δ ≈ w²(m² − ξ(1 − ξ)q²) + μγ². The 1/Δ terms then go as ∫ w dw/w², a logarithm stopped only where w²m² ≈ μγ². The ln Δ term is integrable there, and so is F₂, whose numerator 2m²z(1 − z) brings an extra w. In momentum space this corner is a loop photon with ℓ → 0 and both electrons on shell: long distances, not short.
The corner integral produces fIR
At z = 1 the numerator (1 − x)(1 − y)q² + (1 − 4z + z²)m² of the γμ part becomes q² − 2m² = −2(m² − q²/2), and the w integral is elementary: ∫₀¹ w dw/(w²A + μγ²) = ln[(A + μγ²)/μγ²]/(2A), with A(ξ) = m² − ξ(1 − ξ)q². So the corner adds −(α/2π) ∫₀¹ dξ [(m² − q²/2)/A] ln(A/μγ²) to δF₁(q²); at q² = 0, where A = m², this is −(α/2π) ln(m²/μγ²). Subtract, and the coefficient of ln(1/μγ²) in F₁ − 1 is −(α/2π)[∫ dξ (m² − q²/2)/A − 1] = −(α/2π) fIR(q²). That coefficient is exact. Only the next step, setting ln A ≈ ln M² with M² the larger of −q² and m², is the leading-logarithm approximation. A numerical check made for this lesson: the full two-parameter integral at −q² = 100 m², evaluated at μγ = 10⁻³ m and 10⁻⁵ m and differenced in ln μγ², gives 3.625 in units of α/2π, against fIR = 3.626.
What fIR measures: how sharply the velocity turns
Define cosh χ = p⋅p′/m² = 1 − q²/(2m²). Then ∫₀¹ dξ/A(ξ) = χ/(m² sinh χ) and m² − q²/2 = m² cosh χ, so fIR = χ coth χ − 1, derived. Here χ is the rapidity of the outgoing electron seen from the rest frame of the incoming one. The divergence therefore knows nothing about spin or about what did the scattering, only about the change of velocity. For a gentle kick χ² ≈ −q²/m² and fIR ≈ χ²/3 = −q²/(3m²), which reproduces the (α/3π)(q²/m²) ln(m/μγ) that topic 15.02 will find in the slope of F₁. For a hard kick χ ≈ ln(−q²/m²) and fIR ≈ ln(−q²/m²) − 1. In numbers, fIR = 0.291 at −q² = m², 3.63 at 100 m² and 12.8 at 10⁶ m². And fIR(0) = 0: an undeflected charge carries the same Coulomb field before and after, and no soft photon can tell the difference. That is why the forward point, and only the forward point, is infrared safe.
The strictly elastic rate turns negative
Scattering off topic 9.09's recoil-free heavy target with no photon emitted carries |F₁|², which at order α is 1 + 2(F₁ − 1); setting aside the finite terms of the last section, dσₑₗ = dσ₀[1 − (α/π) fIR ln(−q²/μγ²)], shown. Take 1 GeV electrons at 30°: −q² = 4E² sin²15° = 0.268 GeV², −q²/m² = 1.03 × 10⁶, fIR = 12.84, and the rate falls by (α/π) fIR = 0.0298 per unit of logarithm. With μγ = 1 keV the bracket is 1 − 0.02983 × 26.31 = 0.215; with 1 eV it is 1 − 0.02983 × 40.13 = −0.197. It crosses zero at μγ* = √(−q²) exp[−π/(2α fIR)] ≈ 27 eV. Checks: the bracket is dimensionless, it returns to 1 and Mott's formula as q² → 0, and to tree level as α → 0. Keeping the whole one-loop integral rather than the leading logarithm moves the zero to about 0.5 eV but cannot remove it, because the coefficient of ln μγ² is exact. The negative number is the theory saying the question is wrong: no detector can select a final state with no photon at all.
The double logarithm at large −q²
The leading logarithm replaced ln(A/μγ²) by ln(−q²/μγ²), but A = m² − ξ(1 − ξ)q² runs from m² at the ends of the ξ interval to m² − q²/4 in the middle, while the weight (m² − q²/2)/A piles up at the ends, within ξ ≲ m²/(−q²). Keep the logarithm under the integral. Near ξ = 0, with u = −ξq²/m², the end contributes (1/2)∫ du ln(1 + u)/(1 + u) ≈ (1/4) ln²(−q²/m²), so the two ends give ln²(−q²/m²)/2 where the leading-log form put ln²(−q²/m²). The result, shown, is Sudakov's double logarithm: F₁ − 1 ≈ −(α/4π)[ln²(−q²/μγ²) − ln²(m²/μγ²)] plus single logarithms. At −q² = 10⁶ m² and μγ = 10⁻⁴ m, in units of α/2π, the leading-log form gives −413.1, the double-log form −349.9, and a direct numerical evaluation of the full integral −311.1. The last gap, 38.8, is ln(m²/μγ²) + (3/2) ln(−q²/m²) + π²/6 − 2 to 0.01, a form identified numerically here and linear in the logarithms. Only the leading-log form and the full integral carry the exact coefficient of ln(1/μγ²), −fIR in these units; the double-log form alone has −ln(−q²/m²), one unit too large, and the ln(m²/μγ²) in the gap restores it. That exact coefficient is all the vanishing elastic rate needs.
Finite companions: F₂ and vacuum polarisation
Two other order-α effects enter the same rate, and neither knows about μγ. With the Gordon identity (UAQM-05.07) the vertex is Γμ = (F₁ + F₂)γμ − F₂(p + p′)μ/(2m); a static field selects Γ⁰ = (F₁ + F₂)γ⁰ − F₂E/m, and the spin-summed trace becomes 4(2E² + q²/2)(F₁ + F₂)² − 16F₂E² to first order in F₂, derived, which gives the correction −2F₂ β² sin²(θ/2)/(1 − β² sin²(θ/2)). The same ξ integral that built fIR gives F₂(q²) = (α/2π) χ/sinh χ, at most α/2π and falling as (α/π)(m²/(−q²)) ln(−q²/m²): at 1 GeV and 30° it is 3 × 10⁻⁸, utterly negligible. Vacuum polarisation multiplies the rate by 1/(1 − Π̂₂(q²))², and Unit 13's Π̂₂ ≈ (α/3π)[ln(−q²/m²) − 5/3], quoted, makes that +1.9% there. F₂ is infrared finite because its numerator vanishes at z = 1, and the photon self-energy's loop holds only massive electrons. The photon mass therefore enters the elastic rate through F₁ alone, and only real soft photons, topic 14.07's subject, can take it out.
Change one variable at a time
Make the relationship visible.
The default is 1 GeV electrons at 30° (−q² ≈ 10⁶ m²) with μγ ≈ 0.5 eV: the solid dot sits below zero. Lower k to 4 and it climbs back; move the marker to small −q² and both dots return to 1, because fIR(0) = 0. The dashed curve never crosses zero.
fIR AT THE MARKER12.816
ORDER α dσₑₗ/dσ₀-0.234
ALL ORDERS (QUOTED)0.291
ZERO CROSSING k*4.3 decades
Live interpretationfIR AT THE MARKER: 12.816. ORDER α dσₑₗ/dσ₀: −0.234. ALL ORDERS (QUOTED): 0.291. ZERO CROSSING k*: 4.3 decades
Catch the common trap
Explain before calculating.
Renormalised so that F₁(0) = 1, the one-loop F₁(q²) still depends on the photon mass μγ. For 1 GeV electrons scattered through 30° by a recoil-free nucleus, −q² = 0.268 GeV² and fIR = 12.8. What does the order-α strictly elastic cross-section do as μγ → 0?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron of total energy 2.0 MeV scatters through 60° off a heavy nucleus treated as recoil-free. Find −q²/m², the rapidity χ and fIR, test both limiting forms of fIR, and give the coefficient of ln(1/μγ²) in the renormalised F₁ and in the elastic rate.
- Momentum: |p| = √(E² − m²) = √(4.000 − 0.2611) MeV = 1.934 MeV. No recoil keeps |p′| = |p|, so −q² = 4|p|² sin²30° = |p|² = 3.739 MeV², and −q²/m² = 3.739/0.2611 = 14.32.
- Rapidity: cosh χ = 1 − q²/(2m²) = 1 + 7.159 = 8.159, so χ = ln(8.159 + √(8.159² − 1)) = 2.789. Check: p⋅p′/m² = (E² − |p|² cos 60°)/m² = (4.0000 − 1.8694)/0.26112 = 8.159.
- fIR = χ coth χ − 1 = 2.789 × 1.0076 − 1 = 1.810, dimensionless as it must be.
- Limits: −q²/(3m²) = 4.77 overshoots by a factor 2.6, and ln(−q²/m²) − 1 = 1.66 is 8% low. At −q² = 14 m² the hard-kick form is already the better guide, but neither replaces the closed form.
- Coefficients: in F₁ − 1 it is −(α/2π) fIR = −2.10 × 10⁻³ per unit of ln(1/μγ²); in the rate, twice that, −4.20 × 10⁻³. With μγ = 1 eV, ln(−q²/μγ²) = 28.95 and the elastic factor is 1 − 0.004204 × 28.95 = 0.878.
Answer−q²/m² = 14.3, χ = 2.79, fIR = 1.81; F₁ − 1 carries −2.10 × 10⁻³ ln(1/μγ²) and the rate −4.20 × 10⁻³ ln(1/μγ²), a 12% deficit at μγ = 1 eV.
MediumA 500 MeV electron beam strikes a heavy, recoil-free target. At leading logarithm, find the photon mass below which the order-α strictly elastic cross-section turns negative at θ = 20° and at θ = 120°. Then compare the order-α and the exponentiated elastic factors, the latter quoted from topic 14.09, at μγ = 10 eV.
- With |p| = 500 MeV to the precision needed, −q² = 4|p|² sin²(θ/2) = 3.015 × 10⁴ MeV² at 20° and 7.500 × 10⁵ MeV² at 120°, so −q²/m² = 1.155 × 10⁵ and 2.872 × 10⁶.
- Both are hard kicks: fIR = χ coth χ − 1 = 10.66 and 13.87, equal to ln(−q²/m²) − 1 within 10⁻⁴. Per unit of logarithm the rate falls by (α/π) fIR = 0.02475 and 0.03222.
- The bracket vanishes when ln(−q²/μγ²) = π/(α fIR): 40.40 at 20° and 31.04 at 120°. So μγ* = √(−q²) e(−20.20) = 173.6 MeV × 1.69 × 10⁻⁹ = 0.29 eV at 20°, and √(−q²) e(−15.52) = 866.0 MeV × 1.82 × 10⁻⁷ = 158 eV at 120°.
- At μγ = 10 eV the logarithms are 33.34 and 36.55, so the order-α factors are 1 − 0.02475 × 33.34 = 0.175 and 1 − 0.03222 × 36.55 = −0.178. The exponentiated forms, e(−0.825) = 0.438 and e(−1.178) = 0.308, stay positive and fall towards zero as μγ → 0.
- The backward angle fails first because the velocity turns further: larger χ, larger fIR, steeper fall. Check: as θ → 0, q² → 0, fIR → 0 and both factors return to 1, the Mott limit.
Answerμγ* ≈ 0.29 eV at 20° and ≈ 160 eV at 120°. At μγ = 10 eV the order-α factors are 0.175 and −0.178; the exponentiated ones are 0.438 and 0.308.
HardA 2.0 MeV electron scatters through 120° off a heavy nucleus treated as recoil-free, with higher orders in Zα ignored. Build the order-α ledger for dσₑₗ/dσ₀ at μγ = 1 eV: the leading-log infrared term from F₁, F₂'s interference term, and vacuum polarisation. Then say which entries move when μγ is lowered to 1 meV.
- Kinematics: |p|² = 3.739 MeV² and sin²60° = 0.75, so −q² = 4 × 3.739 × 0.75 = 11.22 MeV² and −q²/m² = 42.96. cosh χ = 1 + 21.48 = 22.48 gives χ = 3.805, and fIR = χ coth χ − 1 = 3.805 × 1.00099 − 1 = 2.809.
- Infrared term: ln(−q²/μγ²) = ln(1.122 × 10¹³) = 30.05, so −(α/π) fIR × 30.05 = −0.006525 × 30.05 = −0.1961.
- F₂: the ξ integral gives F₂ = (α/2π) χ/sinh χ = 1.1614 × 10⁻³ × 3.805/22.46 = 1.968 × 10⁻⁴. With β² = 3.739/4.000 = 0.9347, β² sin²60° = 0.7010, so the term is −2 × 1.968 × 10⁻⁴ × 0.7010/0.2990 = −9.23 × 10⁻⁴.
- Vacuum polarisation, Unit 13's Π̂₂ quoted: (2α/π)∫₀¹ dx x(1 − x) ln[1 + 42.96 x(1 − x)] = 1.719 × 10⁻³ numerically, so 1/(1 − Π̂₂)² adds +3.44 × 10⁻³. The large-log form (α/3π)[ln 42.96 − 5/3] = 1.62 × 10⁻³ is 6% low at this modest −q².
- Ledger at μγ = 1 eV: 1 − 0.1961 − 0.0009 + 0.0034 = 0.806. At 1 meV the logarithm is 43.86, the first entry −0.2862 and the factor 0.716; F₂ and Π̂₂ do not move. The full one-loop F₁ in place of its leading log shifts the first entry by +0.020 at both photon masses: the slope in ln μγ is exact.
Answerdσₑₗ/dσ₀ = 1 − 0.196 − 0.0009 + 0.0034 = 0.806 at μγ = 1 eV and 0.716 at 1 meV; only the F₁ infrared entry depends on μγ.