University Quantum Field Theory I · The One-Loop Vertex: the Anomalous Magnetic Moment and Infrared Divergences · 14.5
F₁'s ultraviolet pole and the one-loop check of Z₁ = Z₂
Topic 12.07 proved Z₁ = Z₂ to all orders in one stroke; here you watch it happen at order α. Take the γμ part of the one-loop vertex to q² = 0, line its pole, its constant and its photon-mass logarithm up against δ₂, and see why vacuum polarisation alone renormalises the electron's charge.
Build the model
Connect the measurement to the mechanism.
Every divergence of the one-loop vertex lives in F₁. Topic 14.03 found F₂ finite at both ends, so what remains is the γμ coefficient, and in d = 4 − ε dimensions its ultraviolet part comes from one term, the l² piece of the numerator, which gives (α/2π)∫dF [2/ε − γE + ln(4πμ²/Δ) − 2]. The pole multiplies the area of the Feynman-parameter simplex and nothing else, so it is the same constant, (α/4π)(2/ε), at every momentum transfer, and one local counterterm, −ieγμ δ₁, removes it everywhere; the on-shell condition F₁(0) = 1 fixes δ₁ = −δF₁(0).
At q² = 0 the simplex collapses onto the photon's parameter z, and the corner z → 1 supplies a logarithm of the photon mass. The result, derived here, is (α/4π)[2/ε − γE + ln(4πμ²/m²) + 4 + 2 ln(μγ²/m²)]: term for term minus topic 12.05's δ₂, so δ₁ = δ₂ and Z₁ = Z₂, Ward's identity checked at one loop. Physically, at zero momentum transfer the vertex correction cancels the half of δ₂ carried by each external leg, so whatever charged particle you scatter, the charge it shows at long distance is renormalised by Z₃ alone.
The check needs the same gauge, the same photon mass and d-dimensional Dirac algebra on both sides, and it fixes F₁ at one point only: away from q² = 0 the renormalised F₁(q²) is finite and free of μ but still depends on μγ, which topic 14.06 takes up.
- Simple definition
- The one-loop check of Z₁ = Z₂ is the calculation showing that the γμ part of the vertex correction at zero momentum transfer, δF₁(0), equals −δ₂ term by term, so the vertex counterterm δ₁ fixed by F₁(0) = 1 equals δ₂.
- Example
- With μ = m and μγ = 10⁻³ m, δF₁(0) = (α/4π)[2/ε − γE + ln 4π − 23.63]: beyond 2/ε − γE + ln 4π it is −0.01372, while topic 12.05's δ₂ has +0.01372 there and the opposite pole, so the sum is zero and δ₁ = δ₂.
The first four terms are the l² piece of the numerator, the only one that diverges; the −2 is O(ε) Dirac algebra times the pole. F₂ receives none of it.
∫dF = ∫dx dy dz δ(x+y+z−1); Δ = (1−z)²m² + zμγ² − xyq² in MeV²; m, μ, μγ in MeV; d = 4 − ε
No q², m or μγ multiplies it, so a single local counterterm −ieγμ δ₁ cancels it at every momentum transfer.
∫dF = ½, the simplex's area; α/4π = 5.807 × 10⁻⁴; ε dimensionless
Every term is a pure number. The photon-mass logarithm comes from the corner z → 1, where 1 − 4z + z² → −2 and Δ₀ → (1 − z)²m² + μγ².
μγ ≪ m the photon mass and μ the regularisation scale, in MeV; the 4 is −2 + 1 + 5 from three z-integrals
With Z₁ = 1 + δ₁ and Z₂ = 1 + δ₂ this is Z₁ = Z₂ at order α, topic 12.07's identity checked in pole, constant and infrared logarithm.
δ₁ fixed by F₁(0) = 1; δ₂ quoted from topic 12.05, same Feynman gauge and photon mass; all dimensionless
At q = 0 the vertex correction cancels the ½δ₂ each external leg carries, so a muon's charge shifts exactly as an electron's does.
e₀ the bare and e the physical charge, both dimensionless in d = 4; Z₃ from vacuum polarisation, Unit 13
ε, γE and μ have cancelled and F₁(0) = 1 exactly; μγ survives at q² ≠ 0, the subject of topic 14.06, and the slope is topic 15.02's.
N = (1−x)(1−y)q² + (1−4z+z²)m²; Δ₀ = Δ at q² = 0, both in MeV²; q² ≤ 0 for scattering
Only the l² term diverges, and it is pure γμ
Working in ħ = c = 1 with the metric (+, −, −, −), topic 14.02, quoted, left the vertex as 2ie²με ∫dF ∫ddl/(2π)d 2Nμ/(l² − Δ)³, with Δ = (1 − z)²m² + zμγ² − xyq². Power counting gives D = 4 − B − (3/2)F = 4 − 1 − 3 = 0 (topic 12.01), so only the part of Nμ quadratic in l can diverge, and only logarithmically. That part is −½ γν l̸ γμ l̸ γν; with lρ lσ → g(ρσ) l²/d and topic 12.03's contractions it becomes −((2 − ε)²/(2d)) l² γμ, a multiple of γμ with no σ(μν) qν. That is why F₂ is ultraviolet finite and every ultraviolet divergence belongs to F₁. The master integral ∫ddl/(2π)d l²/(l² − Δ)³ = i(d/4)Γ(ε/2)/((4π)(d/2) Δ(ε/2)) (topic 6.03) brings a d/4 that cancels the 1/d, leaving (α/4π)((2 − ε)²/2)Γ(ε/2)(4πμ²/Δ)(ε/2) = (α/2π)[2/ε − γE + ln(4πμ²/Δ) − 2]. The −2 is the O(ε) part of (2 − ε)²/4 = 1 − ε + ε²/4 multiplied by 2/ε. The other terms of Nμ converge and give (α/2π)N/Δ.
The pole does not know q², so one counterterm suffices
The pole is (α/2π)(2/ε) times ∫dF = ∫₀¹dz (1 − z) = ½, the area of the simplex: (α/4π)(2/ε) = 1.161 × 10⁻³/ε, multiplied by no q², m or μγ. Δ enters only through (μ²/Δ)(ε/2) = 1 + (ε/2)ln(μ²/Δ), which against the pole is a finite logarithm. So the divergent part of the vertex is a constant times γμ, exactly the shape of topic 12.02's counterterm vertex −ieγμ δ₁, and one number removes it at every momentum transfer. This is locality at work: a divergence of degree D = 0 is a polynomial of degree zero in the external momenta. Had the pole come with q²γμ, a counterterm with two derivatives would be needed, the mark of a non-renormalisable coupling. The renormalisation condition, vertex equal to −ieγμ at q = 0, fixes δ₁ = −δF₁(0), and subtracting at q² = 0 turns ln(μ²/Δ) into ln(Δ₀/Δ): the logarithm of the regulator scale becomes a logarithm of the momentum transfer.
At q² = 0 the simplex collapses to one parameter
With q² = 0, Δ₀ = (1 − z)²m² + zμγ² depends on z alone, and integrating x from 0 to 1 − z at fixed z gives ∫dF f(z) = ∫₀¹dz (1 − z) f(z). Three integrals remain. The constant bracket gives (α/4π)(2/ε − γE + ln 4π − 2). The logarithm gives (α/2π)∫(1 − z)ln(μ²/Δ₀)dz = (α/4π)[ln(μ²/m²) + 1], using ∫₀¹u ln u du = −¼ with u = 1 − z; μγ can be set to zero here because ln u² is integrable. The finite term, (α/2π)∫(1 − z)(1 − 4z + z²)m²/Δ₀ dz, is where the corner bites. Write 1 − 4z + z² = (1 − z)² − 2z: the first piece gives ½, and the second leaves −2∫₀¹u(1 − u)du/(u² + μγ²/m²) = ln(μγ²/m²) + 2, up to corrections of order μγ/m. So this term is (α/4π)[5 + 2 ln(μγ²/m²)], and adding the three, δF₁(0) = (α/4π)[2/ε − γE + ln(4πμ²/m²) + 4 + 2 ln(μγ²/m²)], derived.
The ledger against δ₂, and what it needs
Topic 12.05 computed δ₂, the slope dΣ₂/dp̸ at p̸ = m, as −(α/4π)[2/ε − γE + ln(4πμ²/m²) + 4 + 2 ln(μγ²/m²)], quoted. Set the two side by side. Pole: +2/ε against −2/ε. Scale: ln(4πμ²/m²) − γE in both. Constant: the vertex's 4 is −2 + 1 + 5, from the O(ε) Dirac term, the logarithm and the corner, and δ₂'s is 4. Photon mass: 2 ln(μγ²/m²) in both. So δF₁(0) + δ₂ = 0 identically, not just its poles. That needs the same gauge, since δ₂'s pole is proportional to the gauge parameter, 1 in Feynman gauge, and the vertex's moves with it, both vanishing in Landau gauge (stated, not derived here); the same μγ, since a vertex computed with μγ′ would leave (α/2π)ln(μγ′²/μγ²); and d-dimensional algebra in both. Writing 4 for (2 − ε)² in the vertex alone drops the −2 and leaves δF₁(0) + δ₂ = α/(2π) = 1.16 × 10⁻³, a slip, not physics. It had to work: differentiating the electron propagator inside Σ inserts a zero-momentum photon, Ward's identity (topic 12.07) holding integrand by integrand.
Why the charge is left to Z₃
Scatter an electron off a weak static potential and let q → 0. In bare perturbation theory the amplitude carries e₀ from the vertex, 1 + δF₁(0) from the loop, a factor √Z₂ = 1 + ½δ₂ for each external electron (topic 7.03's LSZ factor) and √Z₃ from the photon line. To order α the electron factors combine to 1 + δF₁(0) + δ₂ = 1, so the charge seen at long distance is e₀√Z₃: the cloud of virtual photons around the electron, which is what makes δ₂ depend on m, changes nothing a distant observer measures. In renormalised language this is e₀Z₂√Z₃ = Z₁e with Z₁ = Z₂, so e = √Z₃ e₀ (topic 12.02). The mass never entered the cancellation, so a muon or a tau shows the same charge for the same e₀, and topic 12.07 extends the argument to the proton; matter is neutral to about 10⁻²¹ e, quoted. δ₁ also multiplies γμ only, so F₂ needs no counterterm and Schwinger's α/(2π) = 0.0011614 stands as a prediction, set against the measured aₑ = 0.00115965218 in topic 14.04.
What survives in F₁(q²): finite, μ-free, photon mass kept
With δ₁ = −δF₁(0), the renormalised form factor F₁(q²) = 1 + δF₁(q²) − δF₁(0) is 1 + (α/2π)∫dF [ln(Δ₀/Δ) + N/Δ − (1 − 4z + z²)m²/Δ₀], derived. The pole, γE, ln 4π, μ and the −2 are gone, because each was the same at every q². Three checks. Dimensions: Δ, Δ₀ and N are all squared masses, so every term is a pure number, as a form factor must be. Free-field limit: α → 0 gives F₁ = 1, topic 14.01's tree-level value. Long-distance limit: q² → 0 gives Δ → Δ₀ and F₁(0) = 1 exactly, the charge of Thomson scattering, α = 1/137.036 (topic 13.03). What has not gone is μγ. With x = (1 − z)ξ, Δ = (1 − z)²(m² − ξ(1 − ξ)q²) + zμγ², and at the corner z → 1 with q² ≠ 0 the terms N/Δ and (1 − 4z + z²)m²/Δ₀ no longer cancel, so a photon-mass logarithm with a q²-dependent coefficient, zero at q² = 0, survives. Renormalisation removes ultraviolet divergences; this infrared one goes only in a rate measured with finite resolution, topics 14.06 to 14.08. The slope at q² = 0 is topic 15.02's.
Change one variable at a time
Make the relationship visible.
Move ε and the photon mass: the solid curve always lands on the −δ₂ line, the pole arriving in the first decade of 1 − z and the photon-mass logarithm along the straight fall that stops at 1 − z = μγ/m. Then set the (2 − ε)² switch to 0: a false gap of 2, in units of α/4π, opens.
VERTEX δF₁(0)-17.68 α/4π
−δ₂ FROM 12.05-17.68 α/4π
δ₁ − δ₂0.00 α/4π
IR LOG 2 ln(μγ²/m²)-27.63 α/4π
Live interpretationVERTEX δF₁(0): −17.68 α/4π. −δ₂ FROM 12.05: −17.68 α/4π. δ₁ − δ₂: 0.00 α/4π. IR LOG 2 ln(μγ²/m²): −27.63 α/4π
Catch the common trap
Explain before calculating.
In Feynman gauge with a photon mass μγ, the γμ part of the one-loop vertex at q² = 0 is δF₁(0) = (α/4π)[2/ε − γE + ln(4πμ²/m²) + 4 + 2 ln(μγ²/m²)], and topic 12.05's δ₂ is exactly minus this. What follows?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyUsing the one-loop vertex in d = 4 − ε, show that the pole of δF₁(q²) is the same at every q² and find its coefficient with α = 1/137.036. Then find how δF₁(0), δ₁ and the renormalised F₁(q²) change when the regularisation scale μ is doubled.
- The pole comes only from the l² term, (α/2π)[2/ε − γE + ln(4πμ²/Δ) − 2] under ∫dF. Δ appears only inside the logarithm, so the pole's coefficient is (α/2π)(2/ε) times ∫dF, whatever q² is.
- ∫dF = ∫₀¹dz ∫₀(1−z)dy = ∫₀¹(1 − z)dz = ½, so the pole is (α/4π)(2/ε).
- α/(4π) = 1/(4π × 137.036) = 5.807 × 10⁻⁴, so the pole is 1.161 × 10⁻³/ε at every momentum transfer, equal and opposite to δ₂'s.
- Doubling μ raises ln(4πμ²/Δ) by ln 4 = 1.386 at every point of the simplex, so δF₁(q²) rises by (α/2π)(½)(1.386) = (α/4π)(1.386) = 8.05 × 10⁻⁴ at every q², q² = 0 included.
- δ₁ = −δF₁(0) falls by 8.05 × 10⁻⁴, and δ₂, through its −(α/4π)ln(4πμ²/m²), falls by the same, so δ₁ = δ₂ survives. F₁(q²) = 1 + δF₁(q²) − δF₁(0) does not move: a shift that is the same at every q² cancels in the difference.
AnswerPole (α/4π)(2/ε) = 1.161 × 10⁻³/ε at every q²; doubling μ raises δF₁(0) by 8.05 × 10⁻⁴, lowers δ₁ and δ₂ by 8.05 × 10⁻⁴ each, and leaves F₁(q²) unchanged.
MediumTake μ = m and a photon mass μγ = 10⁻³ m. Evaluate the corner integral I = ∫₀¹dz (1 − z)(1 − 4z + z²)m²/Δ₀, with Δ₀ = (1 − z)²m² + zμγ², first at leading order in μγ/m and then numerically. Assemble δF₁(0) and compare its finite part with topic 12.05's δ₂.
- Split 1 − 4z + z² = (1 − z)² − 2z. The first piece gives ∫(1 − z)³m²/Δ₀ dz → ∫₀¹(1 − z)dz = ½, since μγ matters only where (1 − z)² is already tiny.
- With u = 1 − z and ρ = μγ²/m² = 10⁻⁶, the second piece is −2∫₀¹u(1 − u)du/(u² + ρ) = −2[½ ln(1/ρ) − 1] = ln ρ + 2 = −13.816 + 2 = −11.816 at leading order.
- So I = 5/2 + ln ρ = −11.316. Numerical integration gives −11.320; the difference, −0.0047, matches the leading correction −(3π/2)(μγ/m), which vanishes in the limit δ₂ is quoted in.
- Assemble at μ = m, where the logarithm integral gives (α/4π)(1): δF₁(0) = (α/4π)[2/ε − γE + ln 4π − 2 + 1 + 2I] = (α/4π)[2/ε − γE + ln 4π + 4 + 2 ln ρ] = (α/4π)[2/ε − γE + ln 4π − 23.63].
- Beyond 2/ε − γE + ln 4π: 5.807 × 10⁻⁴ × (−23.63) = −0.01372. Topic 12.05's δ₂ at the same μ and μγ is −(α/4π)[2/ε − γE + ln 4π − 23.63], the same part +0.01372: the two cancel, pole against pole and constant against constant.
AnswerI = 5/2 + ln(μγ²/m²) = −11.32 (numerically −11.320); δF₁(0) = (α/4π)[2/ε − γE + ln 4π − 23.63], whose −23.63 gives −0.0137, the exact negative of δ₂'s.
HardAn electron and a muon scatter off the same weak, slowly varying electrostatic potential. Using δ₂ from topic 12.05, find δ₂(mμ) − δ₂(mₑ) and say which regulators it depends on. Then find the ratio of the charges the two leptons show at q → 0 if the vertex correction were left out, and with it, and compare with experiment. Take mμ/mₑ = 206.768 and α = 1/137.036.
- δ₂ = −(α/4π)[2/ε − γE + ln 4π + ln μ² − ln m² + 4 + 2 ln μγ² − 2 ln m²] depends on the lepton mass only through +(3α/4π)ln m². The pole, γE, μ and μγ are common to both leptons.
- So δ₂(mμ) − δ₂(mₑ) = (3α/4π)ln(mμ²/mₑ²) = (3α/2π)ln(mμ/mₑ): finite and independent of ε, μ and μγ, although each δ₂ depends on all three.
- Numbers: 3α/(2π) = 3 × 1.1614 × 10⁻³ = 3.484 × 10⁻³ and ln 206.768 = 5.3316, so the difference is 0.01858.
- Without the vertex correction each lepton's coupling at q → 0 would be e₀√Z₃ × (√Z₂)² = e₀√Z₃(1 + δ₂), and the ratio eμ/eₑ = 1 + δ₂(mμ) − δ₂(mₑ) = 1.0186, a 1.9 per cent difference.
- With it, the factor is 1 + δF₁(0) + δ₂ = 1 for each lepton, because the derivation of δF₁(0) = −δ₂ never fixed m. Both charges are e₀√Z₃ and the ratio is 1 at order α: Z₁ = Z₂ for each species separately.
- Muonium 1S–2S spectroscopy finds the μ⁺ and e⁻ charges equal and opposite to a few parts in 10⁹, quoted, so the vertex-free answer is excluded by a factor of about 10⁷.
Answerδ₂(mμ) − δ₂(mₑ) = (3α/2π)ln(mμ/mₑ) = 0.0186, free of every regulator; without the vertex eμ/eₑ = 1.019, with it exactly 1 at order α, as muonium confirms to a few parts in 10⁹.