University Quantum Field Theory I · The One-Loop Vertex: the Anomalous Magnetic Moment and Infrared Divergences · 14.8
Bloch-Nordsieck: virtual and real divergences cancel
Two lessons have each left a logarithm of a photon mass nobody can measure: one drags the elastic rate below zero, the other makes emission infinitely likely. Add them the way a detector does and the mass disappears. What survives, a correction of tens of percent fixed by the energy resolution, is what experiments really correct for.
Build the model
Connect the measurement to the mechanism.
In natural units ħ = c = 1 with metric (+, −, −, −), topic 14.06's one-loop vertex gives the elastic rate dσ₀[1 − (α/π) fIR(q²) ln(−q²/μγ²)], which drops below zero as the photon mass μγ → 0, and topic 14.07 gives the probability of radiating a photon softer than Eₗ as (α/π) fIR(q²) ln(Eₗ²/μγ²), which diverges as fast the other way. Neither is a measurement. A detector of energy resolution Eₗ cannot tell an elastic event from one that shed a photon below Eₗ, so the rate it records is the sum of the two, added as probabilities because the final states differ.
In that sum the photon mass cancels identically, because soft virtual and soft real photons couple through one eikonal current, and what is left is dσ₀[1 − (α/π) fIR(q²) ln(−q²/Eₗ²)]: finite, free of the regulator, and dependent on the apparatus, as a measured rate must be. That is Bloch and Nordsieck's 1937 result, derived here at order α. The model is the soft limit at leading logarithm.
It sets aside F₂, vacuum polarisation, photons too hard for the eikonal limit and non-logarithmic constants, and it keeps the electron mass, whose logarithm the Kinoshita-Lee-Nauenberg theorem must handle when m → 0. It fails as Eₗ → 0: at √(−q²) = 500 MeV the correction passes −100% near Eₗ = 24 eV, the sign that no charge scatters without radiating and that the logarithms must be summed to all orders, topic 14.09.
- Simple definition
- The Bloch-Nordsieck cancellation: the infrared divergence of the elastic vertex correction and that of emitting photons too soft to detect are equal and opposite, so the rate recorded by a detector of energy resolution Eₗ is finite and depends on Eₗ, not on the photon mass.
- Example
- At √(−q²) = 500 MeV, with μγ = 1 eV and Eₗ = 5 MeV, the virtual correction is −118.8% of dσ₀ and soft emission +91.5%. Their sum, −27.3%, is the same for any photon mass.
A definition, not a result. A photon below Eₗ leaves no trace, so both final states are counted, as probabilities: they differ by a photon and cannot interfere.
Eₗ the energy resolution and k the photon energy, in MeV; each dσ in GeV⁻² sr⁻¹, or mb sr⁻¹ via 1 GeV⁻² = 0.3894 mb
Leading log, from |F₁(q²)|². It falls without limit as μγ → 0, so the exclusive elastic rate is not an observable on its own.
dσ₀ the tree-level rate; −q² in MeV², with −q² ≫ m²; μγ the photon mass in MeV; α and fIR pure numbers
Every factor of e in photon energy adds the same probability, 2(α/π) fIR, so the integral diverges as μγ → 0, with the sign opposite to the virtual piece.
photon energies k from μγ to Eₗ, in MeV; soft means Eₗ ≪ √(−q²) and Eₗ ≪ the beam energy
Derived here, with v′ = tanh χ. Emission and the soft virtual loop are built from one eikonal current, so μγ cancels at every q², not only at leading log.
I the angular integral of the eikonal factor, a pure number; v′ the outgoing speed in the rest frame of p, in units of c
Follows here from the angular integral, I = 2(χ coth χ − 1), and topic 14.07's I = 2fIR; topic 14.06's Feynman-parameter integral gives the same. It is 12.77 at √(−q²) = 500 MeV; its small-q² limit is the dipole spectrum of a kicked classical charge.
m = 0.511 MeV; χ the rapidity of p′ seen from p; fIR → −q²/(3 m²) for −q² ≪ m², → ln(−q²/m²) − 1 for −q² ≫ m²
The sum of the two pieces: μγ cancels and Eₗ stays. At √(−q²) = 500 MeV and Eₗ = 5 MeV it is −27.3% of dσ₀, and halving Eₗ costs a further 4.1%.
leading log, m² ≪ −q² and Eₗ ≪ √(−q²); F₂, vacuum polarisation, hard photons and constants set aside
What a detector of resolution Eₗ actually counts
A magnetic spectrometer measures the scattered electron's energy to a resolution of perhaps 0.2%, which at 1 GeV is Eₗ = 2 MeV. An electron that also emitted a 0.5 MeV photon lands inside the elastic peak: the photon escapes unseen, and the electron's energy is short by less than the spectrometer resolves. So no experiment measures dσ(p → p′). It measures dσₘₑₐₛ(Eₗ), the elastic rate plus the rate for p → p′ + γ with photon energy below Eₗ, and at higher orders any number of such photons. At order α relative to dσ₀ exactly two contributions carry the photon mass: the interference 2 Re(M₀*M₁) of the tree amplitude with the one-loop vertex, of order e⁶, and |Mγ|² for one real photon, also e⁶. Vacuum polarisation is of order α too, but it is infrared-finite and is set aside below. The two add as probabilities, not amplitudes, because their final states differ by a photon. Counting powers of e is the first check that the sum is consistent: each term carries exactly one extra α.
A ledger of three regulators
Three cut-offs have appeared in this unit, and each must end in a different place. The ultraviolet pole 1/ε of the vertex was cancelled by δ₁ = δ₂ in topic 14.05 and plays no part here. The photon mass μγ was put in by hand to make the infrared finite; nothing measures it, so it must cancel. The resolution Eₗ belongs to the apparatus, so it must survive. Quote the two pieces at leading logarithm, −q² ≫ m²: the virtual one, from |F₁(q²)|², is −(α/π) fIR(q²) ln(−q²/μγ²) times dσ₀, and the real one, over photon energies μγ to Eₗ, is +(α/π) fIR(q²) ln(Eₗ²/μγ²) times dσ₀. At √(−q²) = 500 MeV, where fIR = 12.77, with μγ = 1 eV and Eₗ = 5 MeV, they are −118.8% and +91.5%. Each alone is large and set by a number nobody can measure, and the first makes the elastic rate negative. Their sum is −27.3%, and lowering μγ to 1 meV adds −41.0% to the first and +41.0% to the second, leaving the sum untouched.
Why both pieces carry the same function
The cancellation is not a coincidence of leading logarithms: it holds for the whole coefficient of ln μγ at every q². A photon much softer than every other scale couples to the external lines only through the eikonal current e(p′μ/p′⋅k − pμ/p⋅k) of topic 14.07. Real emission squares that current and integrates it over d³k/2k; the infrared part of the vertex joins the same current to itself through a virtual photon, so both inherit one angular integral, I(v, v′). In the rest frame of p, with v′ the outgoing speed, it needs only ∫dΩ/4π (1 − v′cos θ)⁻¹ = artanh(v′)/v′ and ∫dΩ/4π (1 − v′cos θ)⁻² = 1/(1 − v′²), giving I = 2[artanh(v′)/v′ − 1]. With v′ = tanh χ and cosh χ = p⋅p′/m², that is 2(χ coth χ − 1) = 2fIR(q²), the function topic 14.06 found in F₁; at v′ = 0.6 both sides are 0.3105. Unitarity fixes the sign: probability that goes into soft photons must leave the no-photon channel.
The measured rate: finite, and set by the resolution
Since ln(−q²/μγ²) − ln(Eₗ²/μγ²) = ln(−q²/Eₗ²), the sum is dσₘₑₐₛ = dσ₀[1 − (α/π) fIR(q²) ln(−q²/Eₗ²)], derived. Read it as an uncancelled band: virtual photons up to about √(−q²) deplete the elastic rate, real ones up to Eₗ restore it, and those between Eₗ and √(−q²) are counted in the loss but vetoed in the gain. Dimensions: each logarithm is a ratio of squared energies and fIR is a pure number, so dσₘₑₐₛ carries the units of dσ₀. Limits: as Eₗ rises toward √(−q²) the logarithm vanishes; for −q² ≪ m², fIR → −q²/(3 m²) = |Δv|²/3, the photon spectrum becomes (2α/3π)|Δv|² dk/k, the low-frequency radiation of a classically kicked charge, and m² replaces −q² inside the logarithm. The slope d(dσₘₑₐₛ)/d ln Eₗ = 2(α/π) fIR dσ₀ is exact at order α in the soft limit: halving Eₗ at 500 MeV costs 4.1% of dσ₀. Elastic electron-scattering experiments apply corrections of this size, typically 10 to 30%, from the full order-α formula of Mo and Tsai, quoted.
No charge scatters without radiating
Let Eₗ → 0 and the order-α rate does not settle on an elastic value: it runs to −∞, crossing zero where (α/π) fIR ln(−q²/Eₗ²) = 1. At √(−q²) = 500 MeV that is Eₗ = 500 MeV × e(−16.85) ≈ 2.4 × 10⁻⁵ MeV, about 24 eV. Nobody builds a 24 eV electron spectrometer, but the breakdown states physics. The expected number of photons per octave of energy, 2(α/π) fIR ln 2 = 0.041 at this q², is the same in every octave, and there are infinitely many octaves below any Eₗ. A charge whose velocity changes radiates infinitely many arbitrarily soft photons that carry a finite total energy, so the probability of scattering with no photon above Eₗ goes to zero as Eₗ → 0, while the rate summed over all photon energies stays finite. First-order perturbation theory cannot express a probability that tends to zero; summing the logarithms to all orders, topic 14.09, can.
What is set aside, and the theorems behind it
The leading-log result omits, by choice: F₂'s share of the rate, finite and free of both μγ and Eₗ; vacuum polarisation, which raises the rate by (2α/3π)[ln(−q²/m²) − 5/3] ≈ +1.9% at √(−q²) = 500 MeV, quoted from Unit 13; photons too hard for the eikonal approximation, corrections of order Eₗ/E; and every single logarithm and constant the complete calculation adds. None depends on μγ, so none disturbs the cancellation. Bloch and Nordsieck showed in 1937, treating the electron as a classical current, that the rate with soft photons included is finite to all orders, and Yennie, Frautschi and Suura proved it in full QED in 1961; we derive it at order α and quote the rest. One logarithm stays: fIR ≈ ln(−q²/m²) − 1 grows without bound as m → 0, a collinear singularity that summing over final soft photons cannot remove. The Kinoshita-Lee-Nauenberg theorem, named and quoted, removes it by summing over every degenerate initial and final state, an electron accompanied by a collinear photon included.
Change one variable at a time
Make the relationship visible.
Drag the photon mass across seven decades: the virtual and real readouts swing by about a hundred percent in opposite directions while the measured sum never moves. Then tighten Eₗ from 10 to 0.5 MeV and watch the sum fall: the measured rate remembers the detector, never the regulator.
VIRTUAL, ELASTIC-118.8 %
REAL, k BELOW Eₗ91.5 %
MEASURED SUM-27.3 %
fIR(q²)12.77
Live interpretationVIRTUAL, ELASTIC: −118.8 %. REAL, k BELOW Eₗ: 91.5 %. MEASURED SUM: −27.3 %. fIR(q²): 12.77
Catch the common trap
Explain before calculating.
An order-α prediction for electron scattering at √(−q²) = 400 MeV, where fIR = 12.3, is redone twice: once with the photon mass halved, and once with the detector resolution improved from Eₗ = 10 MeV to 5 MeV. What happens to the predicted measured cross-section?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyElectrons scatter with √(−q²) = 300 MeV into a spectrometer of resolution Eₗ = 3 MeV. Using the leading-log pieces and fIR = χ coth χ − 1, evaluate the virtual and real order-α corrections with a photon mass of 1 eV, repeat with 1 μeV, and show that the measured rate is the same both times.
- fIR first: √(−q²)/(2 m) = 300/1.022 = 293.5, so χ = 2 asinh(293.5) = 12.750 and coth χ = 1 to ten figures, giving fIR = 11.750. The large-q² form ln(−q²/m²) − 1 = 11.750 agrees, and (α/π) fIR = 0.0023228 × 11.750 = 0.027294.
- μγ = 1 eV = 10⁻⁶ MeV. Virtual: ln(−q²/μγ²) = 2 ln(3 × 10⁸) = 39.039, so δV = −0.027294 × 39.039 = −1.0655. Real: ln(Eₗ²/μγ²) = 2 ln(3 × 10⁶) = 29.828, so δR = +0.027294 × 29.828 = +0.8141.
- Sum: δ = −1.0655 + 0.8141 = −0.2514. Directly, ln(−q²/Eₗ²) = 2 ln 100 = 9.2103, and −0.027294 × 9.2103 = −0.2514, the same.
- μγ = 1 μeV = 10⁻¹² MeV: each logarithm grows by 2 ln 10⁶ = 27.631, worth 0.027294 × 27.631 = 0.7542. So δV = −1.8197 and δR = +1.5683, and their sum is still −0.2514.
- Checks: δV alone is below −1, an elastic rate less than zero, so it cannot be a measurement; every logarithm is a ratio of energies, so δ is a pure number; and the sum depends on Eₗ and q² only.
AnswerδV = −1.07 then −1.82 and δR = +0.81 then +1.57, but the measured correction is −0.251 both times: dσₘₑₐₛ = 0.749 dσ₀ whatever the photon mass.
MediumAt √(−q²) = 450 MeV an experiment records the elastic peak with resolution Eₗ = 9 MeV and then, after an upgrade, with Eₗ = 3 MeV. Predict both measured rates as fractions of dσ₀ at order α, find the change, and say which events the upgrade removes.
- fIR at 450 MeV: √(−q²)/(2 m) = 450/1.022 = 440.3, χ = 2 asinh(440.3) = 13.561, so fIR = χ coth χ − 1 = 12.561 and (α/π) fIR = 0.0023228 × 12.561 = 0.029178.
- Eₗ = 9 MeV: ln(−q²/Eₗ²) = 2 ln 50 = 7.824, so δ₉ = −0.029178 × 7.824 = −0.2283 and dσₘₑₐₛ = 0.772 dσ₀.
- Eₗ = 3 MeV: ln(−q²/Eₗ²) = 2 ln 150 = 10.021, so δ₃ = −0.029178 × 10.021 = −0.2924 and dσₘₑₐₛ = 0.708 dσ₀.
- Change: δ₃ − δ₉ = −(α/π) fIR × 2 ln 3 = −0.029178 × 2.197 = −0.0641. Neither μγ nor the virtual correction appears, because the virtual piece is identical at both resolutions and drops out of the difference.
- Meaning: 0.0641 is the expected number of photons per scattering with energies from 3 to 9 MeV, (α/π) I ln 3 with I = 2fIR = 25.12. An event that radiated one of them sat inside the 9 MeV window and falls outside the 3 MeV one.
Answerdσₘₑₐₛ = 0.772 dσ₀ at Eₗ = 9 MeV and 0.708 dσ₀ at 3 MeV: the upgrade lowers the rate by 0.064 dσ₀, the events that radiated a photon of 3 to 9 MeV.
HardElectrons of 750 MeV scatter through 25° from a heavy nucleus treated as recoil-free, and a spectrometer with Eₗ = 1.5 MeV records dσ/dΩ = 2.6 μb sr⁻¹, a hypothetical reading. At leading log and order α, what tree-level cross-section does it imply? Find the resolution at which the order-α correction would reach −100%, and compare the correction with vacuum polarisation.
- Kinematics: recoil-free means |p′| = |p|, and m/E = 6.8 × 10⁻⁴ lets |p| = E, so −q² = 4E² sin²(θ/2) and √(−q²) = 2 × 750 MeV × sin 12.5° = 324.7 MeV, or −q² = 0.1054 GeV². Eₗ/√(−q²) = 0.0046, safely soft.
- fIR: √(−q²)/(2 m) = 317.7, χ = 2 asinh(317.7) = 12.908, fIR = 11.908, equal to ln(−q²/m²) − 1 to six figures. So (α/π) fIR = 0.027661.
- Correction: ln(−q²/Eₗ²) = 2 ln(324.7/1.5) = 10.755, so δ = −0.027661 × 10.755 = −0.2975. The spectrometer sees 70.3% of the tree-level rate; the rest radiated more than 1.5 MeV.
- Unfold: dσ₀/dΩ = 2.6/(1 − 0.2975) = 3.70 μb sr⁻¹. No nuclear model is needed, since the soft factor multiplies whatever the hard cross-section is, form factor included. Multiplying by 1 + 0.2975 instead gives 3.37 μb sr⁻¹: the two agree at order α but differ by 9.7%, an order-α² ambiguity that topic 14.09 settles.
- Breakdown: δ = −1 when ln(−q²/Eₗ²) = 1/0.027661 = 36.15, so Eₗ = 324.7 MeV × e(−18.08) = 4.6 × 10⁻⁶ MeV, about 4.6 eV. Below that the order-α rate is negative, the sign that no charge scatters without radiating.
- Set aside, stated: vacuum polarisation, quoted from Unit 13, raises the rate by (2α/3π)[ln(−q²/m²) − 5/3] = +1.7%, a seventeenth of the soft correction. F₂, single logarithms and constants, photons too hard for the eikonal limit, nuclear recoil and higher orders in Zα are omitted too.
Answerdσ₀/dΩ ≈ 3.7 μb sr⁻¹ (3.70 dividing by 1 + δ with δ = −0.297, 3.37 at strict order α); the order-α correction reaches −100% only at Eₗ ≈ 4.6 eV; vacuum polarisation adds just +1.7%.