University Quantum Field Theory I · The One-Loop Vertex: the Anomalous Magnetic Moment and Infrared Divergences · 14.9
The coherent soft-photon cloud and exponentiation
Topic 14.08 cancelled the photon mass but left a logarithm that order α cannot hold: make the detector sharp enough and the corrected rate turns negative. Here the soft photons are treated as what they are, the coherent radiation of a classical kicked charge, and summing them turns that logarithm into an exponential.
Build the model
Connect the measurement to the mechanism.
Topic 14.08 cancelled the photon mass at order α but left a logarithm, (α/π) fIR ln(−q²/Eₗ²), that grows without limit as the resolution Eₗ shrinks. What is it the first term of? A soft photon cannot resolve the hard collision: to it the electron is a classical point charge whose velocity jumps from p/m to p′/m, and the eikonal vertex is that charge's current.
A c-number current drives the photon field into a coherent state (topic 1.09, UAQM-11.07), so photons are emitted independently, their number is Poisson-distributed, and the mean between k₁ and k₂ is (α/π) fIR ln(k₂²/k₁²): the same in every decade, infinite as k₁ → μγ → 0, while the energy they carry stays finite. Virtual soft photons obey the same independence, and their sum is the cloud's no-photon probability e(−N̄) up to √(−q²). Multiplied by the real-emission sum below Eₗ, e(+N̄), it loses μγ at every order and leaves dσ₀ exp[−(α/π) fIR ln(−q²/Eₗ²)], the probability that no photon lies above Eₗ.
It vanishes as Eₗ → 0, since a deflected charge always radiates, yet the Poisson probabilities still sum to one and the inclusive rate is untouched. The argument is leading-logarithmic: it fixes the cloud's upper edge only up to a constant, ignores recoil and photons hard enough to see the collision, and leaves the all-orders proof to Yennie, Frautschi and Suura (1961), quoted.
- Simple definition
- Exponentiation is the statement that soft photons are emitted independently, like the radiation of a classical kicked charge, so their number is Poisson-distributed and the chance that none exceeds Eₗ is e(−N̄), N̄ being the mean number above Eₗ.
- Example
- For 2.0 GeV electrons turned through 60° by a heavy nucleus, −q² = 4.00 GeV² and fIR = 15.54, so N̄ above Eₗ = 1 MeV is 0.549: a fraction e(−0.549) = 0.578 of dσ₀ has no such photon, where order α says 0.451.
Derived: the Fourier transform of a straight worldline that bends at x = 0. It is topic 14.07's eikonal factor, conserved (k⋅j̃ = 0), and it falls as 1/k.
p, p′ the electron's momenta before and after, k the photon's, in GeV; e = √(4πα) = 0.303; j̃μ in GeV⁻¹.
Derived from Σλ |ε*⋅j̃|²: each decade of photon energy holds (2α/π) fIR ln 10 photons, so N̄ diverges as k₁ → μγ → 0.
k₁ ≤ k₂ photon energies, any one unit; fIR = θ coth θ − 1 with cosh θ = p⋅p′/m², dimensionless.
Derived: a c-number current leaves the field in a coherent state (topic 1.09, UAQM-11.07), and P(0) = e(−N̄) is its squared overlap with the photon vacuum.
n photons in a band of mean N̄; dimensionless; Σₙ P(n) = 1 whatever N̄ is.
Derived: finite as k₁ → 0 though N̄ is not. At −q² = 4.00 GeV² the cloud below 10 MeV carries 0.72 MeV however many photons it holds.
Ē in the unit of k; the mean energy per photon is (k₂ − k₁)/ln(k₂/k₁).
Shown: n independent soft loops carry 1/n! and exponentiate topic 14.06's one-loop term. It is the cloud's P(0) between μγ and √(−q²).
μγ the photon mass in GeV; for −q² ≫ m² (m² replaces −q² when −q² ≪ m²); leading infrared logarithm only.
Shown here from independent emission; its first order is topic 14.08. The all-orders proof, with infrared-finite remainders, is Yennie, Frautschi and Suura (1961), quoted.
dσ₀ the tree-level cross-section; Eₗ ≪ √(−q²), in GeV; μγ has cancelled; equivalently dσ ∝ Eₗ(2α fIR/π).
A soft photon sees a classical current
Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout. Topic 14.07 found that a photon with k ≪ E couples to an external electron line through the eikonal factor e p⋅ε*/(p⋅k), blind to the electron's spin and to whatever the hard collision did. Read that factor as a source: it is exactly ε*⋅j̃(k) for the current of a point charge moving with p/m until x⁰ = 0 and with p′/m afterwards. The current is a c-number because the electron does not notice the photon: emitting 1 MeV changes a 2 GeV electron's momentum by 5 × 10⁻⁴ of itself. Topic 1.09, extending UAQM-11.07 to every mode, says what a c-number source does to the free photon field: it leaves it in a coherent state, displaced in mode (k, λ) by an amplitude proportional to ε*λ⋅j̃(k)/√(2k). So the scattered electron emerges dressed in a coherent cloud. Current conservation, k⋅j̃ = 0, lets Unit 8's −gμν polarisation sum be used, and the 1/k fall of j̃ is the whole infrared problem in one line.
Counting the cloud: the same number in every decade
Sum |ε*λ⋅j̃|² over both polarisations and integrate over d³k/((2π)³ 2k): the mean photon number is dN̄ = (α/π) I dk/k, where I is the angular average of 2p⋅p′/((k̂⋅p)(k̂⋅p′)) − m²/(k̂⋅p)² − m²/(k̂⋅p′)² with k̂ = (1, k/|k|). A Feynman parameter combines the two denominators and gives I = 2fIR(q²), the function F₁ carried in topic 14.06; doing the ξ integral gives fIR = θ coth θ − 1, θ the rapidity between p and p′. Since dN̄ goes as dk/k, every decade of photon energy holds the same number: for 2.0 GeV electrons at 60°, −q² = 4.00 GeV², fIR = 15.54 and each decade holds 0.166 photons. A coherent state is a product of single-mode coherent states, each Poissonian, so the count in any band is Poisson with mean N̄. Between μγ = 1 eV and 2 GeV that mean is 1.55; lower μγ to 10⁻⁶ eV and it is 2.54, growing without limit as μγ → 0.
Infinitely many photons, a finite energy
Weight the spectrum by photon energy and the logarithm disappears: Ē = (2α/π) fIR (k₂ − k₁), set by the top of the band and blind to its bottom. In the 2 GeV, 60° example the cloud below 10 MeV holds ever more photons as μγ → 0 yet carries 0.72 MeV, and below 1 MeV only 0.072 MeV. That is the classical answer. For −q² ≪ m², θ ≈ |Δv| and fIR → θ²/3 ≈ |Δv|²/3, and dĒ/dk = (2α/3π)|Δv|² is the flat low-frequency spectrum of a charge whose velocity changes by Δv in a collision too brief to resolve; a 20 keV electron turned straight back has fIR = 0.102 against |Δv|²/3 = 0.099, a gap of order v². What diverges is only a count of photons too soft for any calorimeter. Its consequence is sharp all the same: the probability that the cloud is empty, e(−N̄), goes to zero, so 'electron deflected, no photons' is a final state of probability zero. The divergence of F₁ in topic 14.06 is this fact seen from the virtual side.
Independent emission: the orderings sum to a product
On a Feynman diagram n soft photons leave a line in a definite order, so why are they independent? Take two from the outgoing leg. With k₂ emitted nearer the external end the propagators give 1/(2p′⋅K) × 1/(2p′⋅k₂), K = k₁ + k₂, each vertex supplying 2p′μ; the other order swaps k₂ for k₁ in the second factor. Adding, 1/(2p′⋅K) × [1/(2p′⋅k₁) + 1/(2p′⋅k₂)] = 1/((2p′⋅k₁)(2p′⋅k₂)): the sum over orderings is the product of two one-photon factors. Induction handles n photons, and emission from both legs adds the two currents, so the n-photon amplitude is M₀ Πᵢ εᵢ*⋅j̃(kᵢ), exactly what the coherent state encodes. Squared, summed over polarisations and integrated with 1/n! for identical photons, it gives dσₙ = dσ₀ N̄ⁿ/n!, and summing every n with all photons below Eₗ multiplies dσ₀ by e(+N̄(μγ, Eₗ)). The identity drops the kᵢ⋅kⱼ terms in inner propagators and the electron's recoil, both of relative size k/E.
Virtual photons exponentiate too, and μγ cancels
Soft virtual photons obey the same identity: each exchanged photon contributes the eikonal factors of both its ends, and n of them, with the 1/n! of identical exchanges, exponentiate topic 14.06's one-loop logarithm into |M|² = |M₀|² exp[−(α/π) fIR ln(−q²/μγ²)]. That is the cloud's P(0) = e(−N̄) for the band from μγ to √(−q²), the scale above which a loop photon resolves the collision and stops being soft. Multiply by the real sum e(+N̄(μγ, Eₗ)) and the bottom of the band drops out: dσ = dσ₀ e(−N̄(Eₗ, √(−q²))) = dσ₀ exp[−(α/π) fIR ln(−q²/Eₗ²)]. The photon mass cancels at every order, not only at order α, and the exponent is the mean number of photons above Eₗ, so the rate goes as a power of Eₗ. For 2.0 GeV at 60° and Eₗ = 1 MeV it gives 0.578 of dσ₀ where order α gives 0.451; at 1 keV, 0.351 where order α gives −0.048. Summed over every photon number, Σ P(n) = 1 and the inclusive rate stays at dσ₀.
What is shown, what is quoted, and where it stops
All of the above is leading-logarithmic and shown here; that the infrared-divergent factor exponentiates exactly, leaving a remainder finite order by order, is Yennie, Frautschi and Suura (1961), quoted. Three limits matter. The cloud's upper edge is fixed only up to a constant, so non-logarithmic terms need F₁'s finite part, F₂, vacuum polarisation and hard photons. A detector that bounds the total radiated energy rather than each photon gets the same logarithm times e(−γE a)/Γ(1 + a), a = (2α/π) fIR, quoted: 0.996 at −q² = 4.00 GeV². And fIR ≈ ln(−q²/m²) − 1 grows without bound as m → 0, a collinear logarithm soft photons do not cure. A measured test: at the Z peak in e⁺e⁻ → hadrons the width plays Eₗ's part, β = (2α/π)(ln(MZ²/m²) − 1) = 0.108 and (ΓZ/MZ)β = 0.68, or 0.73 with the quoted factor 1 + 3β/4, a 27% cut against the roughly 26% the LEP fits unfold from the measured peak (quoted). Order α would cut 31%.
Change one variable at a time
Make the relationship visible.
Lower Eₗ from 10 MeV: the dashed order-α line falls straight through zero, near 2 keV at 2 GeV and 60°, while the solid curve bends and stays positive. The filled bar P(0) matches the dot; the open bars are events with photons above Eₗ, which an inclusive count keeps.
fIR(q²)15.54
PHOTONS ABOVE Eₗ, x0.549
EXPONENTIATED e(−x)0.578 of dσ₀
ORDER α, 1 − x0.451 of dσ₀
Live interpretationfIR(q²): 15.54. PHOTONS ABOVE Eₗ, x: 0.549. EXPONENTIATED e(−x): 0.578 of dσ₀. ORDER α, 1 − x: 0.451 of dσ₀
Catch the common trap
Explain before calculating.
Electrons scatter from a heavy nucleus at a momentum transfer where (α/π) fIR(q²) = 0.030, and the detector rejects any event with a photon above Eₗ, with ln(−q²/Eₗ²) = 20. To leading logarithmic accuracy, what fraction of dσ₀ does it record, and what happens to the rate summed over all photon numbers as μγ → 0?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA 200 MeV electron is turned through 60° by a heavy nucleus that does not recoil. Treating its soft radiation as the coherent cloud, find the mean number of photons with energies between 1 MeV and 10 MeV, the probabilities of finding 0, 1 and 2 of them, and the mean energy they carry; then compare the cloud between 1 eV and 1 MeV.
- Momentum transfer: −q² = 4E² sin²(θ/2) = 4(0.200 GeV)²(0.250) = 0.0400 GeV², so −q²/m² = 0.0400/(2.611 × 10⁻⁷) = 1.53 × 10⁵ ≫ 1 and fIR ≈ ln(−q²/m²) − 1 = 11.94 − 1 = 10.94; the exact θ coth θ − 1 agrees to within 2 × 10⁻⁵.
- Mean number: N̄ = (α/π) fIR ln(k₂²/k₁²) = (0.0023228)(10.94) ln 100 = 0.02541 × 4.605 = 0.117, dimensionless as a count must be.
- Poisson: P(0) = e(−0.117) = 0.890, P(1) = 0.117 × 0.890 = 0.104, P(2) = 0.117² × 0.890/2 = 0.0061; three or more photons, 2.4 × 10⁻⁴.
- Energy: Ē = (2α/π) fIR (k₂ − k₁) = 0.05082 × 9 MeV = 0.457 MeV, or 0.457/0.117 = 3.9 MeV per photon, which is (k₂ − k₁)/ln(k₂/k₁) = 9 MeV/ln 10.
- Six decades lower, from 1 eV to 1 MeV: N̄ = 0.02541 × ln 10¹² = 0.702, six times as many photons, but Ē = 0.05082 × 1 MeV = 0.051 MeV, a ninth of the energy. Extending the band towards μγ → 0 adds photons without limit and energy hardly at all.
AnswerN̄ = 0.117 between 1 and 10 MeV, with P(0) = 0.890, P(1) = 0.104 and P(2) = 0.0061, carrying 0.46 MeV on average; the 1 eV to 1 MeV band holds 0.70 photons but only 0.051 MeV.
MediumAn electron leaves a hard collision with energy 100 MeV along z and emits two soft photons: k₁ of 2.0 MeV at 10° to z and k₂ of 5.0 MeV at 90° to z, both in the xz-plane. Evaluate the eikonal denominators for both orders of emission, show that their sum equals the product of the single-photon factors, and estimate the error made by dropping k₁⋅k₂ in the inner propagator.
- On the outgoing line the propagator next to the hard vertex carries p′ + K, K = k₁ + k₂, and (p′ + K)² − m² = 2p′⋅K + K²; the next carries p′ + kb, kb the photon nearer the external end. With numerators 2p′μ, the two orders differ only in 1/(2p′⋅K) × 1/(2p′⋅kb).
- Dot products, p′⋅k = E′k(1 − β cos ψ) with β = 0.999987: p′⋅k₁ = 200 MeV² × 0.015205 = 3.041 MeV², p′⋅k₂ = 500 MeV² × 1 = 500.0 MeV², so p′⋅K = 503.04 MeV².
- Orders: k₁ outermost gives 1/(1006.08 × 6.082) = 1.634 × 10⁻⁴ MeV⁻⁴; k₂ outermost gives 1/(1006.08 × 1000.0) = 9.94 × 10⁻⁷ MeV⁻⁴. Their sum is 1.644 × 10⁻⁴ MeV⁻⁴.
- Product of single-photon factors: 1/((2 × 3.041)(2 × 500.0)) = 1/6082 MeV⁻⁴ = 1.644 × 10⁻⁴ MeV⁻⁴, equal to every digit, because 1/(a + b) × (1/a + 1/b) = 1/(ab) is an identity.
- Dropped term: K² = 2k₁⋅k₂ = 2k₁k₂(1 − cos 80°) = 2 × 10 × 0.8264 = 16.53 MeV², against 2p′⋅K = 1006.1 MeV², a 1.6% shift; recomputed with it, the sum falls by 1.6%. That is the k/E′ accuracy of the eikonal picture.
AnswerThe orders give 1.634 × 10⁻⁴ and 9.94 × 10⁻⁷ MeV⁻⁴; their sum, 1.644 × 10⁻⁴ MeV⁻⁴, equals 1/((2p′⋅k₁)(2p′⋅k₂)). Keeping k₁⋅k₂ changes it by 1.6%, the size of the neglected recoil.
HardElectrons of 3.0 GeV scatter through 40° from a recoil-free nucleus, and the detector rejects events with any photon above Eₗ = 1.0 MeV. Show, with photon masses of 1 eV and 10⁻³ eV, that the virtual and real soft factors change but their product does not; compare the exponentiated rate with order α; and predict both ways how much the accepted rate grows when Eₗ is raised to 5.0 MeV.
- −q² = 4E² sin²20° = 36 GeV² × 0.11698 = 4.211 GeV², −q²/m² = 1.613 × 10⁷, fIR = ln(−q²/m²) − 1 = 15.60, so (α/π) fIR = 0.03623.
- μγ = 1 eV = 10⁻⁹ GeV. Virtual: exp[−0.03623 ln(4.211 × 10¹⁸)] = exp(−0.03623 × 42.88) = e(−1.554) = 0.211. Real photons below Eₗ, summed over number: exp[+0.03623 ln(10⁻⁶/10⁻¹⁸)] = e(+1.001) = 2.72. Product: 0.575.
- μγ = 10⁻³ eV: the virtual exponent grows to 0.03623 × 56.70 = 2.054, factor 0.128, and the real one to 0.03623 × 41.45 = 1.501, factor 4.49. Each moves by 0.03623 × ln 10⁶ = 0.500, in opposite directions, and the product is 0.575 again.
- So dσ = dσ₀ e(−x) with x = 0.03623 ln(4.211/10⁻⁶) = 0.553, giving 0.575 dσ₀. Order α gives 1 − 0.553 = 0.447 dσ₀, 22% low; the gap is x²/2 − x³/6 + …, the multi-photon terms order α leaves out.
- Raising Eₗ to 5.0 MeV: exponentiation gives the power law (5.0/1.0)(2 × 0.03623) = 50.0725 = 1.124, a 12% rise. Order α gives (1 − 0.436)/(1 − 0.553) = 0.564/0.447 = 1.26, a 26% rise. The ratio is free of dσ₀ and of μγ, so it tests the exponent cleanly.
- Checks: x is a count and dimensionless; as Eₗ → √(−q²) = 2.05 GeV, x → 0 and the leading-log rate returns to dσ₀; the rejected 0.425 of dσ₀ is not lost but is the events with photons above Eₗ, P(n ≥ 1) = 1 − e(−x).
AnswerVirtual and real factors are 0.211 and 2.72 at μγ = 1 eV and 0.128 and 4.49 at 10⁻³ eV, product 0.575 both times; order α gives 0.447. Raising Eₗ to 5.0 MeV lifts the rate 12% exponentiated, against 26% at order α.