University Quantum Field Theory I · The One-Loop Vertex: the Anomalous Magnetic Moment and Infrared Divergences · 14.2

The one-loop vertex: Feynman parameters and the numerator

This is the calculation the rest of the unit lives on: three propagators merged over a triangle of Feynman parameters, a Dirac string collapsed onto γ^μ and σ^(μν)q_ν, and a count of powers that says which piece can blow up and where. Do it carefully once, and Schwinger's g − 2, Z₁ = Z₂ and the infrared problem all start from its output.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Exchanging a photon between the incoming and outgoing electron is the only one-particle-irreducible order-α correction to the vertex, and Unit 9's rules write it at once: two electron propagators carrying k and k + q, a Feynman-gauge photon carrying k − p with topic 8.08's mass μ_γ, and the Dirac string γ^ν(k̸′ + m)γ^μ(k̸ + m)γ_ν. Topic 6.02's Feynman formula, applied twice, merges the three denominators into (xA + yB + zC)⁻³ over the triangle x + y + z = 1, and the shift ℓ = k + yq − zp, with the on-shell relation p⋅q = −q²/2, leaves (ℓ² − Δ)⁻³ with Δ = (1 − z)²m² + zμ_γ² − xyq². The rest is algebra between on-shell spinors: odd powers of ℓ vanish, topic 12.03's d-dimensional contractions remove γ^ν and γ_ν, the Dirac equation removes p̸ and p̸′, a q^μ term cancels under x ↔ y as current conservation demands, and the Gordon identity splits (p + p′)^μ; what remains is γ^μ δF₁ + (iσ^(μν)q_ν/(2m))F₂ with explicit integrands.

Power counting sorts the trouble. Only the ℓ² term diverges in the ultraviolet, and it multiplies γ^μ alone; the corner z → 1, where the photon goes soft and both electrons go on shell, makes the rest of the γ^μ part grow as ln(m/μ_γ); the σ^(μν)q_ν part has no ℓ² term, and its factor 1 − z kills the logarithm, so it escapes both. The result is exact at order α, valid as written below q² = (2m)², and still a set of integrals; their values are the business of topics 14.03 to 14.06.

Simple definition: The one-loop vertex correction is the diagram with a photon exchanged between the incoming and outgoing electron, rewritten with Feynman parameters x + y + z = 1 as one denominator (ℓ² − Δ)³ and a numerator reduced to a γ^μ part and a σ^(μν)q_ν part.
Example: At q² = 0 and μ_γ = 0, Δ = (1 − z)²m². At z = 0.9 the σ^(μν)q_ν integrand 2z(1 − z)m²/Δ is 18 and the γ^μ integrand (1 − 4z + z²)m²/Δ is −179; times the width 1 − z of the triangle there they are 1.8, finite at the corner, and −17.9, heading for −2/(1 − z).

The vertex diagram from Unit 9's rulesū′δΓ^μ u = −ie² ∫d⁴k/(2π)⁴ ū′γ^ν(k̸′ + m)γ^μ(k̸ + m)γ_ν u/(ABC)

Two −ieγ vertices, two electron propagators and the Feynman-gauge photon give −ie²; six powers of k up and six down make it dimensionless, topic 12.01's D = 0.

A = k² − m², B = (k + q)² − m², C = (k − p)² − μ_γ², each + iε; k′ = k + q, q = p′ − p; momenta and masses in GeV; derived

Feynman parameters for three denominators1/(ABC) = 2 ∫₀¹ dx dy dz δ(x + y + z − 1) (xA + yB + zC)⁻³

Three propagators cost an integral over a triangle of area ½ and buy one denominator. Setting A = B = C returns A⁻³, which checks the 2.

x, y, z dimensionless weights; A, B, C in GeV²; topic 6.02's formula applied twice; derived

The shift and Δℓ = k + yq − zp, Δ = (1 − z)²m² + zμ_γ² − xyq²

Below q² = (2m)² Δ is positive on the whole triangle. It falls to μ_γ² at the corner z = 1, where the photon is soft and both electrons are on shell.

ℓ the shifted loop momentum; Δ in GeV²; uses p² = p′² = m², so p⋅q = −q²/2; derived

The γ^μ coefficient: carrier of both divergencesN₁ = −((d − 2)²/(2d))ℓ² + (1 − x)(1 − y)q² + (1 − 4z + z²)m²

Its ℓ² term is the only ultraviolet divergence. At z → 1 the rest tends to q² − 2m², not zero, which is the source of the logarithm in μ_γ.

GeV²; multiplies γ^μ under 2ie²μ^ε ∫d^dℓ/(2π)^d ∫dF 2(ℓ² − Δ)⁻³; (d − 2)²/(2d) = ½ at d = 4; derived

The σ^(μν)q_ν coefficient: F₂ as an integralF₂(q²) = (α/2π) ∫dF 2z(1 − z)m²/Δ

No ℓ² term, and a factor 1 − z that vanishes at the soft corner: F₂ converges at both ends, so topic 14.03 may set d = 4 and μ_γ = 0 before integrating.

∫dF = ∫₀¹ dx dy dz δ(x + y + z − 1); F₂ dimensionless; α = e²/(4π); derived

The ℓ² piece of δF₁ in d = 4 − εℓ² piece of δF₁ = (α/2π) ∫dF [2/ε − γ_E + ln(4πμ²/Δ) − 2]

The −2 is (1 − ε/2)² times 2/ε, the d-dimensional Dirac algebra at work. Set (d − 2)²/(2d) = ½ before integrating and δF₁(0) moves by 3α/(8π).

μ the scale from e → eμ^(ε/2), in GeV; γ_E = 0.5772; ∫dF 1 = ½, so the pole is α/(2πε); derived

The diagram and its prefactor

Work in natural units, ħ = c = 1, with the metric (+, −, −, −). An electron of momentum p absorbs a photon of momentum q and leaves with p′ = p + q, both on shell. At order α the only one-particle-irreducible correction has the electron emit a virtual photon before the absorption and take it back after: the photon carries p − k, and the electron carries k into the external vertex and k′ = k + q out of it. Unit 9's rules supply vertices −ieγ^ν and −ieγ^ρ, two propagators i(k̸ + m)/(k² − m² + iε), and the Feynman-gauge photon −ig_νρ/((k − p)² − μ_γ² + iε) with topic 8.08's mass μ_γ as the infrared regulator. Writing the full vertex as −ieΓ^μ with Γ^μ = γ^μ + δΓ^μ at this order, the factors are (−ie)² × i² × (−i) = −ie², so ū(p′)δΓ^μ u(p) = −ie² ∫d⁴k/(2π)⁴ ū(p′)γ^ν(k̸′ + m)γ^μ(k̸ + m)γ_ν u(p)/(ABC), derived. Four powers of k from the measure and two from the numerator against six from the denominators make δΓ^μ dimensionless like γ^μ and superficially logarithmic, topic 12.01's D = 0. In d = 4 − ε, e → eμ^(ε/2) keeps e dimensionless, as topic 12.03 did for Σ₂.

Three denominators, one triangle

Topic 6.02 wrote 1/(AB) = ∫₀¹ du E⁻² with E = uA + (1 − u)B. The same formula for 1/(EC), differentiated with respect to E, gives 1/(E²C) = ∫₀¹ dv 2v[vE + (1 − v)C]⁻³. Multiply the two, rename x = uv, y = (1 − u)v and z = 1 − v, whose Jacobian is v, and 1/(ABC) = 2∫dx dy dz δ(x + y + z − 1)(xA + yB + zC)⁻³, derived. The triangle has area ½, so A = B = C returns A⁻³. With A = k² − m², B = (k + q)² − m² and C = (k − p)² − μ_γ², the terms linear in k are 2k⋅(yq − zp), so ℓ = k + yq − zp completes the square. With p² = m² and p⋅q = −q²/2, which is p′² = m² rewritten, the denominator is ℓ² − Δ with Δ = (1 − z)²m² + zμ_γ² − xyq², derived. The corners read the diagram back: Δ = m² at x = 1 or y = 1, where one electron line dominates, and Δ = μ_γ² at z = 1, where the photon does. On the edge z = 0, Δ = m² − x(1 − x)q² first vanishes at x = ½ when q² = (2m)², the threshold for a real e⁺e⁻ pair.

The Dirac string in d dimensions

Topic 12.03's contractions, γ^ν a̸ γ_ν = −(2 − ε)a̸, γ^ν a̸ b̸ γ_ν = 4a⋅b − ε a̸ b̸ and the five-matrix rule γ^ν a̸ γ^μ b̸ γ_ν = −2b̸ γ^μ a̸ + ε a̸ γ^μ b̸, collapse the string: γ^ν(k̸′ + m)γ^μ(k̸ + m)γ_ν = −2k̸ γ^μ k̸′ + ε k̸′γ^μ k̸ + 4m(k + k′)^μ − εm(k̸′γ^μ + γ^μ k̸) − (2 − ε)m²γ^μ. This is derived; now substitute k = ℓ − yq + zp and k′ = ℓ + (1 − y)q + zp. Terms odd in ℓ integrate to zero, and ℓ^α ℓ^β → g^(αβ)ℓ²/d turns the two ℓ̸ γ^μ ℓ̸ terms into ((d − 2)²/d)ℓ²γ^μ. That ε-dependence must be kept: the ℓ² integral has a pole 2/ε, and ε × 2/ε is a finite number. Every other ε term multiplies ∫d^dℓ (ℓ² − Δ)⁻³, finite in the ultraviolet and, with μ_γ inside Δ, in the infrared too, so those terms vanish as ε → 0 and the rest of the algebra may be done in four dimensions, where the string is −2[k̸ γ^μ k̸′ + m²γ^μ − 2m(k + k′)^μ] and the diagram is 2ie² times that bracket. This clean split is why the course keeps a photon mass rather than letting d regulate the infrared as well.

Dirac equation, a q^μ that cancels, and Gordon

Set ℓ aside. After the contraction k̸ stands next to ū(p′) and k̸′ next to u(p), so write k = zp′ − (1 − x)q and k′ = zp + (1 − y)q: ū(p′)p̸′ = mū(p′) and p̸u(p) = m u(p) remove the momenta, and three sandwich rules finish the job, ū′q̸ u = 0, ū′q̸ γ^μ u = ū′(2mγ^μ − 2p^μ)u and ū′γ^μ q̸ u = ū′(2p′^μ − 2mγ^μ)u. The bracket becomes γ^μ[(1 − x)(1 − y)q² + (1 − 2z − z²)m²] + mz(z − 1)(p + p′)^μ + m(z − 2)(x − y)q^μ. This is derived, and checked against explicit Dirac matrices. Δ and the measure are symmetric under x ↔ y, so the q^μ term integrates to zero, as it must: contracted with q_μ it would leave q² times a spinor product and break topic 12.06's Ward identity. The Gordon identity, (p + p′)^μ = 2mγ^μ − iσ^(μν)q_ν between spinors (UAQM-05.07, quoted), moves −2z(1 − z)m² into the γ^μ coefficient, making it (1 − 4z + z²)m², and leaves 2z(1 − z)m² × iσ^(μν)q_ν/(2m). With ∫d⁴ℓ/(2π)⁴ (ℓ² − Δ)⁻³ = −i/(32π²Δ), the prefactor 2ie² × 2 becomes (α/2π)/Δ, so F₂(q²) = (α/2π)∫dF 2z(1 − z)m²/Δ, derived.

Power counting: where each divergence lives

Only the ℓ² term can diverge in the ultraviolet: ∫d⁴ℓ ℓ²/(ℓ² − Δ)³ is logarithmic, while ∫d⁴ℓ (ℓ² − Δ)⁻³ converges. Topic 6.03's master integral with an ℓ² numerator is (i/(4π)^(d/2))(d/4)Γ(ε/2)Δ^(−ε/2); with the coefficient −(d − 2)²/(2d) the ℓ² piece of δF₁ is (α/2π)∫dF [2/ε − γ_E + ln(4πμ²/Δ) − 2], derived, the −2 being (1 − ε/2)² times 2/ε. Since ∫dF 1 = ½ the pole is α/(2πε), spread evenly over the triangle, and it sits in γ^μ alone, which is why the γ^μ counterterm δ₁ can absorb it (topic 14.05). The infrared needs a different count. Near z = 1 put w = 1 − z, x = wξ and y = w(1 − ξ), so the measure becomes w dw dξ and Δ → w²[m² − ξ(1 − ξ)q²] + μ_γ². The γ^μ numerator tends to q² − 2m², so at q² = 0 the integrand behaves as −2w/(w² + μ_γ²/m²) and the w integral is −2 ln(m/μ_γ): finite while μ_γ ≠ 0, divergent as μ_γ → 0, derived. The σ^(μν)q_ν numerator 2z(1 − z)m² brings one more power of w, and ∫w² dw/w² converges: F₂ is finite at both ends, as the figure shows.

Checks, and what the representation leaves open

Dimensions: Δ, both numerators and m² are GeV², and every integral over the triangle is a pure number, so F₁ and F₂ are dimensionless, as form factors must be. The free limit e → 0 returns F₁ = 1 and F₂ = 0, topic 14.01's tree-level vertex. At q² = 0 and μ_γ = 0, Δ depends on z alone and the σ^(μν)q_ν weight integrated over x is the polynomial 2z, so F₂(0) is a finite pure number with no logarithm of anything; topic 14.03 evaluates it as α/(2π) = 0.0011614, to be set against the measured aₑ = 0.00115965218 of UAQM-06.04, both quoted here. The representation is exact at order α but valid as written only below threshold: above q² = (2m)², Δ ≤ 0 on part of the triangle, the iε of the propagators decides how the zero is passed, and F₁ and F₂ acquire imaginary parts from real pair production. Nor is it yet a prediction. δF₁ still carries the 2/ε, which topic 14.05 cancels against δ₂, and the ln μ_γ, which δ₁ removes only at q² = 0 and real soft photons remove in general, in topics 14.06 to 14.08.

Change one variable at a time

Make the relationship visible.

Interactive model

Momentum transfer −q²/m²1.00

Photon mass μ_γ = m × 10⁻ⁿ, n3.0

Slide n from 1 to 4: μ_γ falls a thousandfold, the dashed γ^μ plateau stretches toward z → 1 and its area keeps growing, while the solid σ^(μν)q_ν hump is already near zero there and barely moves, its area settling on F₂/(α/2π). Raise −q²/m² and the plateau deepens as the solid area falls.

F₂/(α/2π) AT μ_γ → 00.861

γ^μ PLATEAU HEIGHT-2.582 α/2π

PLATEAU LENGTH ln(m/μ_γ)6.91

γ^μ IR LOG, HEIGHT × LENGTH-17.8 α/2π

Live interpretationF₂/(α/2π) AT μ_γ → 0: 0.861. γ^μ PLATEAU HEIGHT: −2.582 α/2π. PLATEAU LENGTH ln(m/μ_γ): 6.91. γ^μ IR LOG, HEIGHT × LENGTH: −17.8 α/2π

Catch the common trap

Explain before calculating.

Misconception check

“ε goes to zero at the end, so do the Dirac algebra in four dimensions and keep only the integrals in d.” That is safe for every term except the one that diverges. The ℓ² coefficient (d − 2)²/(2d), times the integral's d/4, is ½(1 − ε/2)²; against Γ(ε/2) = 2/ε − γ_E it leaves a finite −2 in the bracket, where the four-dimensional ½, times d/4 = 1 − ε/4, leaves −½. The difference, 3/2 at every point of the triangle, raises δF₁(0) by 3α/(8π) = 8.7 × 10⁻⁴, three quarters of Schwinger's α/(2π). Topic 12.05's δ₂ was computed with the d-dimensional identities, so the mixed calculation finds a spurious δ₁ − δ₂ = −8.7 × 10⁻⁴ and seems to break Z₁ = Z₂. Elsewhere ε multiplies finite integrals and may be dropped, but only because μ_γ, not d, regulates the infrared.

After Feynman parameters and the numerator reduction the one-loop vertex is γ^μ δF₁ + (iσ^(μν)q_ν/(2m))F₂, the ℓ² term sitting only in δF₁ and F₂ = (α/2π)∫dF 2z(1 − z)m²/Δ with Δ = (1 − z)²m² + zμ_γ² − xyq². Which statement about its divergences is right?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTake m = 0.511 MeV and a photon mass μ_γ = 1.0 keV. For a spacelike q² = −2.0 MeV², evaluate Δ = (1 − z)²m² + zμ_γ² − xyq² at the three corners of the Feynman triangle and at its centre, and turn each corner into a distance ħc/√Δ. Then, for a timelike q² = (1.5 MeV)², find where on the edge z = 0 Δ turns negative, and say what that means.

m² = 0.2611 MeV². At x = 1 or y = 1 only one electron propagator survives and Δ = m² = 0.2611 MeV², a distance ħc/m = 197.327 MeV fm / 0.511 MeV = 386 fm.
At z = 1 only the photon survives: Δ = μ_γ² = 1.0 × 10⁻⁶ MeV², 2.6 × 10⁵ times smaller, a distance ħc/μ_γ = 1.97 × 10⁵ fm = 0.197 nm. The soft corner reaches atomic distances, which is why it carries the infrared problem.
Centre, x = y = z = ⅓: Δ = (4/9)(0.2611) + (⅓)(1.0 × 10⁻⁶) + (1/9)(2.0) = 0.1161 + 0.0000003 + 0.2222 = 0.3383 MeV². For any spacelike q², −xyq² ≥ 0, so every term is non-negative and Δ is positive on the whole triangle: the loop functions are real.
Timelike, edge z = 0: Δ = m² − x(1 − x)q² is negative when x(1 − x) exceeds m²/q² = 0.2611/2.25 = 0.1161, that is for x between (1 − √(1 − 0.4642))/2 = 0.134 and (1 + 0.732)/2 = 0.866. At x = ½, Δ = 0.2611 − 0.5625 = −0.301 MeV².
Meaning: q² = 2.25 MeV² exceeds (2m)² = 1.044 MeV², so the virtual photon can become a real e⁺e⁻ pair, and there F₁ and F₂ gain imaginary parts. Check: since xy ≤ (1 − z)²/4, Δ ≥ (1 − z)²(m² − q²/4) + zμ_γ², so as q² falls to (2m)² the negative segment shrinks to the single point x = ½ on z = 0.

AnswerΔ = 0.261 MeV² (386 fm) at the electron corners, 1.0 × 10⁻⁶ MeV² (0.197 nm) at the photon corner and 0.338 MeV² at the centre, positive everywhere for spacelike q². For q² = (1.5 MeV)², Δ is negative for x between 0.134 and 0.866 on the edge z = 0.

MediumEvaluate the ℓ² piece of the one-loop vertex in d = 4 − ε. Show that it adds (α/2π)∫dF [2/ε − γ_E + ln(4πμ²/Δ) − 2] to δF₁ and find its pole. Then find how much δF₁(0) changes if the coefficient (d − 2)²/(2d) is replaced by its four-dimensional value ½ before integrating, and compare that shift with α/(2π).

The ℓ² term of the bracket is −((d − 2)²/(2d))ℓ²γ^μ: it comes from ℓ^α ℓ^β → g^(αβ)ℓ²/d and γ^ν γ^α γ^μ γ_α γ_ν = (d − 2)²γ^μ. Topic 6.03 gives ∫d^dℓ/(2π)^d ℓ²/(ℓ² − Δ)³ = (i/(4π)^(d/2))(d/4)Γ(ε/2)Δ^(−ε/2).
Multiply by 2ie²μ^ε × 2. The coefficient becomes −((d − 2)²/(2d))(d/4) = −½(1 − ε/2)², and 2ie² × 2 × (−½) × i/(16π²) = α/(2π). So the piece is (α/2π)(1 − ε/2)²Γ(ε/2)(4πμ²/Δ)^(ε/2) = (α/2π)[2/ε − 2 − γ_E + ln(4πμ²/Δ)] + O(ε), since (1 − ε)(2/ε) = 2/ε − 2.
Pole: ∫dF 1 = ½, so the pole in δF₁ is (α/2π)(½)(2/ε) = α/(2πε), the same at every q². Check: a pure number, as F₁ must be, and the same size as the pole of −δ₂ in topic 12.05, (α/4π)(2/ε).
With ½ in place of (d − 2)²/(2d) the coefficient is −½(d/4) = −½(1 − ε/4), and (1 − ε/4)(2/ε) = 2/ε − ½: the constant −2 becomes −½. The bracket rises by 3/2 at every point of the triangle, so δF₁(0) rises by (α/2π)(3/2)(½) = 3α/(8π).
Numbers: 3α/(8π) = 3/(8π × 137.036) = 8.71 × 10⁻⁴, against α/(2π) = 1.161 × 10⁻³, a shift of 75% of Schwinger's term. It is the same at every q², so F₁(q²) − F₁(0) and F₂ are untouched; only the comparison with δ₂, computed with the d-dimensional identities, goes wrong, by exactly this amount.

AnswerThe ℓ² piece is (α/2π)∫dF [2/ε − γ_E + ln(4πμ²/Δ) − 2], with pole α/(2πε). Four-dimensional algebra in its coefficient shifts δF₁(0) by 3α/(8π) = 8.7 × 10⁻⁴, 75% of α/(2π), a spurious breach of δ₁ = δ₂; F₂ and F₁(q²) − F₁(0) are unchanged.

HardYukawa theory with Dirac fermions (topic 7.04), −gφψ̄ψ with φ of mass M, adds a second one-loop vertex: φ exchanged between the incoming and outgoing electron. Repeat this topic's steps for it, show that its F₂(0) = (g²/(8π²))∫₀¹ dz (1 − z)²(1 + z)/((1 − z)² + zr²) with r = M/m, and evaluate it for r → 0, r = 1 and r = 10 with g²/(4π) = α. Compare with QED's α/(2π).

Rules: two Yukawa vertices −ig, a scalar propagator i/((k − p)² − M² + iε) and the same electron propagators give (−ig)² × i³ = ig² and the numerator (k̸′ + m)γ^μ(k̸ + m), with no γ^ν and γ_ν to contract. The denominators are QED's with μ_γ → M, so the same parameters give Δ = (1 − z)²m² + zM² − xyq².
Shift: k′ = ℓ + zp′ + xq and k = ℓ + zp − yq. The ℓ² term multiplies γ^μ only. The rest, between spinors, is ū′[(1 + z)m + xq̸]γ^μ[(1 + z)m − yq̸]u.
Expand with ū′q̸ γ^μ u = ū′(2mγ^μ − 2p^μ)u, ū′γ^μ q̸ u = ū′(2p′^μ − 2mγ^μ)u and ū′q̸ γ^μ q̸ u = −q²ū′γ^μ u. The vector part is −2(1 + z)m(xp + yp′)^μ = −(1 − z²)m(p + p′)^μ plus a q^μ term odd under x ↔ y, which integrates to zero.
Gordon turns −(1 − z²)m(p + p′)^μ into −2(1 − z²)m²γ^μ + 2(1 − z²)m² × iσ^(μν)q_ν/(2m). With ig² × 2 × (−i/(32π²Δ)) = g²/(16π²Δ), F₂(q²) = (g²/(16π²))∫dF 2(1 − z²)m²/Δ, and at q² = 0 the x integral supplies 1 − z, giving the stated F₂(0).
With g²/(8π²) = α/(2π): r → 0 gives ∫(1 + z)dz = 3/2, so F₂(0) = 3α/(4π) = 1.742 × 10⁻³. At r = 1 the integrand is z + (1 − 2z)/(z² − z + 1), whose second part integrates to zero, so the integral is ½ and F₂(0) = α/(4π) = 5.81 × 10⁻⁴. At r = 10 it is 0.0355 numerically, and F₂(0) = 4.12 × 10⁻⁵.
Checks: dimensionless, and positive, so scalar exchange raises g as the photon does. For r ≫ 1 the integral approaches (ln r² − 7/6)/r², checked numerically (8.04 × 10⁻⁴ against 8.05 × 10⁻⁴ at r = 100), so heavy scalars decouple. A massless scalar gives 1.5 times QED's α/(2π) = 1.161 × 10⁻³: QED's γ^ν…γ_ν turns 2(1 − z²) into 2z(1 − z) under a prefactor twice as large.

AnswerF₂(0) = (g²/(8π²))∫₀¹ dz (1 − z)²(1 + z)/((1 − z)² + zr²): 3α/(4π) = 1.74 × 10⁻³ for r → 0, 1.5 times QED's α/(2π); α/(4π) = 5.81 × 10⁻⁴ at r = 1; and 4.12 × 10⁻⁵ at r = 10.