University Quantum Field Theory I · The One-Loop Vertex: the Anomalous Magnetic Moment and Infrared Divergences · 14.1

The vertex function: form factors F₁(q²) and F₂(q²)

Before topic 14.02 evaluates a single loop, this lesson settles what any answer is allowed to look like. Symmetry leaves two functions of q², and the work here is learning what each one measures, so that when a calculation hands you a σ^μν q_ν term you read it at once as g − 2.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The vertex function is the electron's complete answer to a photon: every diagram that absorbs one, sandwiched between on-shell spinors. Its form is decided before any of it is computed. On shell the only invariant is q², so the answer is a few functions of one variable; the Dirac equation strips the matrix structure down to γ^(μ), (p + p′)^μ and q^μ; parity forbids γ⁵; the Gordon identity trades (p + p′)^μ for γ^μ and σ^μν q_ν; and current conservation removes q^μ.

What survives, γ^μ F₁(q²) + (iσ^μν q_ν/2m)F₂(q²), is exact to every order in α and holds for any spin-½ particle, the proton included. The two functions then acquire meanings. A static potential at long wavelength couples to eF₁(0), so F₁(0) is the charge; a slowly varying magnetic field meets the spin with strength F₁(0) + F₂(0), so g = 2(F₁(0) + F₂(0)), and the anomaly a is F₂(0), the low-energy value of the Pauli term in UAQM-06.04.

That term has dimension five, so renormalisable QED holds no counterterm for it, and F₂ is a number the theory must predict rather than absorb into a parameter. The frame stops where its assumptions do: off shell the vertex has more structures, a P- and T-violating dipole would need another term, above q² = 4m² the functions turn complex, and symmetry says nothing about their values. Those need the loop of topic 14.02.

Simple definition: The form factors F₁(q²) and F₂(q²) are the two real functions of the momentum transfer that, by Lorentz invariance, parity and current conservation, fix the whole coupling of an on-shell spin-½ particle to a photon: ū(p′)[γ^μ F₁ + (iσ^μν q_ν/2m)F₂]u(p).
Example: For the electron F₁(0) = 1 and F₂(0) = aₑ = 0.00115965218, so g = 2(1 + 0.00115965218) = 2.00231930436; for the proton F₁(0) = 1 and F₂(0) = 1.79285, so g = 5.5857. Both are measured.

General on-shell vertex, parity evenū(p′)Γ^μ u(p) = ū(p′)[γ^μ A(q²) + (p + p′)^μ B(q²) + q^μ C(q²)]u(p)

Lorentz invariance leaves q² as the only variable, the Dirac equation strips products of gammas, and parity bans γ⁵ and ε^μνρσ.

q = p′ − p; p² = p′² = m²; A dimensionless, B and C of mass dimension −1 (GeV⁻¹). Derived in outline.

Form-factor decomposition (derived)Γ^μ = γ^μ F₁(q²) + (iσ^μν q_ν/2m) F₂(q²)

Gordon (UAQM-05.07) turns (p + p′)^μ into 2mγ^μ − iσ^μν q_ν; the Ward identity q_μū(p′)Γ^μ u(p) = 0 forces C = 0, because q²C cannot vanish.

F₁ = A + 2mB and F₂ = −2mB, both dimensionless; σ^μν = (i/2)(γ^μ γ^ν − γ^ν γ^μ); m in GeV. Tree level: F₁ = 1, F₂ = 0.

Charge seen in the Breit frame (derived)ū(p′)Γ⁰u(p) = 2m G_E(q²) ξ′†ξ, with G_E = F₁ + (q²/4m²) F₂

No spin-dependent piece survives, so G_E is the charge form factor a probe reads; as q² → 0 it tends to F₁(0), the charge in units of e.

Breit frame p = −q/2, p′ = q/2, so q⁰ = 0; a static field gives it as exact backscattering. ξ, ξ′ unit two-component spinors.

The g-factor (derived)g = 2[F₁(0) + F₂(0)] = 2G_M(0), with G_M = F₁ + F₂

Gordon splits Γ^μ into convection F₁(p + p′)^μ/2m and magnetisation (F₁ + F₂)iσ^μν q_ν/2m. Only the spin term sees F₂; orbital g_L stays F₁(0) = 1.

g dimensionless; moment μ = g(e/2m)S. Measured: electron g = 2.00231930436, proton 5.58569.

Pauli's term is F₂ at q² = 0 (derived)−(μ′/2)ψ̄σ^μν ψF_μν yields the F₂ vertex with μ′ = eF₂(0)/2m, so a = F₂(0)

UAQM-06.04's constant μ′ is F₂ frozen at q² = 0. The operator has dimension 5, so renormalisable QED has no counterterm for it: F₂ must come out finite.

μ′ in GeV⁻¹ in natural units. Electron, measured: μ′ = −aₑ μ_B = −1.0755 × 10⁻²⁶ J T⁻¹, or −3.436 × 10⁻⁴ MeV⁻¹, negative because e = −|e|.

Static-field rate with both form factors (derived)Σ|ū(p′)Γ⁰u(p)|² = 8E²[F₁² + τF₂² − β² sin²(θ/2)(F₁ + F₂)²]

Forward it is F₁², the charge; F₂ = 0 gives Mott's 1 − β² sin²(θ/2). The τF₂² term grows as |p|²/m², so F₂ must fall at large −q².

Σ over both spins; recoil-free target, E′ = E; τ = −q²/4m² = (|p|/m)² sin²(θ/2); β = |p|/E; F's at q² = −4|p|² sin²(θ/2).

What the vertex function is, and why only q² enters

Natural units ħ = c = 1 and the metric (+, −, −, −) throughout, with Unit 9's vertex −ieγ^(μ), e = −|e| the electron's charge and |e| = √(4πα) = 0.3028. Sum every amputated diagram with an incoming electron p, an outgoing electron p′ and one photon bringing q = p′ − p, and call the result −ieΓ^μ(p′, p): at tree level Γ^μ = γ^(μ), and topic 14.02 adds the one-loop triangle. Inside a bigger loop Γ^μ can be anything, but with external electrons we only ever need ū(p′)Γ^μ u(p), and on shell that is tightly constrained. Its coefficients are scalars built from p and p′, and p² = p′² = m² are fixed while p⋅p′ = m² − q²/2, so every coefficient is a function of q² alone. The photon need not be on shell: q² = −4|p|² sin²(θ/2) is negative for scattering off a static source, and above 4m² in e⁺e⁻ annihilation. Hermiticity of the current makes the coefficients real for every spacelike q², quoted.

The Dirac equation and parity leave three structures

Any 4 × 4 matrix is a combination of 1, γ⁵, γ^(μ), γ^μ γ⁵ and σ^μν, multiplied by vectors made from p and p′. Between on-shell spinors p̸u(p) = m u(p) and ū(p′)p̸′ = mū(p′), so anticommuting each p̸ to the right and each p̸′ to the left turns products such as p̸′γ^μ p̸ into numbers times γ^(μ), p^μ and p′^μ; σ^μν p_ν acting on u(p) is i(mγ^μ − p^μ), and σ^μν p′_ν reduces the same way on the left. Electromagnetism conserves parity and J^μ is a true vector, so neither γ⁵ nor ε^μνρσ may appear. What survives is ū(p′)[γ^μ A + (p + p′)^μ B + q^μ C]u(p), derived in outline, with A dimensionless and B and C of mass dimension −1. Dropping parity would admit two more terms: an anapole, odd under P, and σ^μν q_ν γ⁵, whose value at q² = 0 is an electric dipole moment, odd under both P and T. Both are named only; the JILA experiment bounds the electron's at |dₑ| below 4.1 × 10⁻³⁰ e cm, quoted.

Gordon removes (p + p′), and the Ward identity removes q

The Gordon identity of UAQM-05.07, quoted, says that between these spinors (p + p′)^μ acts as 2mγ^μ − iσ^μν q_ν. Substituting, γ^μ A + (p + p′)^μ B becomes γ^μ(A + 2mB) + (iσ^μν q_ν/2m)(−2mB). Now contract with q_μ. Current conservation gives q_μ⟨p′|J^μ(0)|p⟩ = 0, and Unit 12's Ward-Takahashi identity makes q_μΓ^μ a difference of inverse propagators, S⁻¹(p′) − S⁻¹(p), which vanishes between on-shell spinors. Term by term: ū(p′)q̸u(p) = (m − m)ū(p′)u(p) = 0, q⋅(p + p′) = p′² − p² = 0, and σ^μν q_μq_ν = 0 by antisymmetry, but q_μq^μ C = q²C survives, so C = 0. The result, derived: Γ^μ = γ^μ F₁(q²) + (iσ^μν q_ν/2m)F₂(q²), with F₁ = A + 2mB and F₂ = −2mB. The 2m is a convention that makes F₂ a pure number. Tree level, Γ^μ = γ^(μ), is F₁ = 1 and F₂ = 0. Two real functions of one variable hold everything an on-shell electron can show a photon.

The charge: what a static potential sees

Let the electron scatter from a static potential φ(x), so q⁰ = 0 and q² = −|q|². The amplitude is iM = −ieū(p′)Γ⁰u(p)φ̃(q), and the Born approximation of Quantum Mechanics II, with relativistic normalisation, reads iM = −iṼ(q)(2m)ξ′†ξ. For a slow electron ū(p′)γ⁰u(p) = 2mξ′†ξ, while the F₂ term is second order in the momenta, so Ṽ = eF₁(q²)φ̃, and a slowly varying potential, which probes only q → 0, sees V = eF₁(0)φ: F₁(0) is the charge in units of e, derived. To make this exact, use the Breit frame, p = −q/2 and p′ = q/2, where the photon carries no energy; a static field realises it as exact backscattering. With u = √(E + m)(ξ, σ⋅p ξ/(E + m)), ū(p′)γ⁰u(p) = 2mξ′†ξ and ū(p′)u(p) = 2Eξ′†ξ; Gordon gives Γ⁰ = (F₁ + F₂)γ⁰ − (E/m)F₂, and E² = m² − q²/4 sums these to 2mG_E(q²)ξ′†ξ, with G_E = F₁ + (q²/4m²)F₂, derived. That loops keep F₁(0) = 1 is Z₁ = Z₂, quoted from Unit 12 and checked in topic 14.05; atoms are neutral to about 10⁻²¹ e, quoted, so the proton's structure leaves its F₁(0) alone too. The slope of F₁ waits for Unit 15.

The moment: g = 2(F₁(0) + F₂(0)), and Pauli's term

Run Gordon the other way: Γ^μ = F₁(p + p′)^μ/2m + (F₁ + F₂)iσ^μν q_ν/2m, a charge F₁e moving at the average velocity plus a magnetisation of strength F₁ + F₂. Put the slow electron in a static vector potential A(x). Then iM = +ieū(p′)Γⁱ u(p)Ã⁽ⁱ⁾, and at small momenta ū(p′)σ^ij u(p) → 2mε^ijk ξ′†σ^k ξ, so the spin part is −i(F₁ + F₂)ε^ijk q^j Ãⁱ ξ′†σ^k ξ = (F₁ + F₂)ξ′†σ⋅B̃ξ, with B̃ = iq × Ã. The Born potential is V = −(eF₁(0)/2m)(p⋅A + A⋅p) − (e/2m)(F₁(0) + F₂(0))σ⋅B, derived. The orbital term keeps g_L = F₁(0) = 1; the spin term is −μ⋅B with μ = g(e/2m)S and g = 2(F₁(0) + F₂(0)). So F₂(0) = (g − 2)/2 = a. Read the Pauli term of UAQM-06.04, −(μ′/2)ψ̄σ^μν ψF_μν, as a vertex: ∂_μ becomes −iq_μ on the absorbed photon, giving μ′σ^μν q_ν ε_μ, which matches −ie(iσ^μν q_ν/2m)F₂ε_μ when μ′ = eF₂(0)/2m. Pauli's constant is F₂ frozen at q² = 0.

Why F₂ is a prediction, and where the frame stops

Count dimensions: ψ has mass dimension 3/2 and F_μν has 2, so ψ̄σ^μν ψF_μν has dimension 5 and μ′ has dimension −1. A renormalisable Lagrangian cannot contain it, so QED has no counterterm of the form σ^μν q_ν: whatever loops put into F₂ must be finite by itself, a prediction rather than a parameter, and topic 14.03 finds α/2π. Nothing in the decomposition used perturbation theory, only Lorentz invariance, parity and current conservation, so it holds for any spin-½ particle. The proton has F₁(0) = 1 and F₂(0) = 1.793; the neutron has F₁(0) = 0 and its entire moment, −1.913 nuclear magnetons, in F₂, both measured. It stops in three places. Off shell, inside a larger loop, the vertex has twelve independent structures, not two, quoted. Above q² = 4m² a real e⁺e⁻ pair can be made, and F₁ and F₂ acquire imaginary parts. And a constant F₂, which is what Pauli's term is, holds only while −q² is small beside the scale on which F₂ varies: m² for the electron, about 0.7 GeV² for the proton.

Change one variable at a time

Make the relationship visible.

Interactive model

Dirac form factor F₁1.00

Pauli form factor F₂1.79

Momentum p/m0.60

Set F₂ to 0 and the curve lies on Mott's dashed one. Raise F₂ to the proton's 1.79 and backward scattering collapses as G_E = F₁ − (p/m)²F₂ nears zero; set F₁ to 0 and F₂ to −1.91, a neutron, and the forward rate vanishes. At p/m = 0.1 every curve is flat: magnetism costs v².

g = 2(F₁ + F₂)5.58

G_E AT 180°0.356

N AT 180°0.093

MOTT N AT 180°0.735

Live interpretationg = 2(F₁ + F₂): 5.58. G_E AT 180°: 0.356. N AT 180°: 0.093. MOTT N AT 180°: 0.735

Catch the common trap

Explain before calculating.

After renormalisation, a calculation returns the on-shell vertex at q → 0 as ū(p′)[1.0012γ^μ − 0.0006(p + p′)^μ/m]u(p), with no σ^μν term anywhere. What are F₁(0), F₂(0) and g?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyThe electron's measured g is 2.00231930436 and the proton's measured moment is 2.79284734 nuclear magnetons, both quoted. Find F₁(0), F₂(0) and g for each, then the Pauli coefficient μ′ = eF₂(0)/2m in MeV⁻¹ and in J T⁻¹, and compare the two anomalous moments.

F₁(0) is the charge in units of the particle's own e, so F₁(0) = 1 for both, the electron's charge being e = −|e| and the proton's |e|. Then g = 2(F₁(0) + F₂(0)) gives F₂(0) = g/2 − 1.
Electron: F₂(0) = 1.00115965218 − 1 = 0.00115965218 = aₑ. Proton: a moment in nuclear magnetons |e|/2mₚ is g/2, so g = 5.58569468 and F₂(0) = 2.79284734 − 1 = 1.79284734 = κₚ.
Natural units, |e| = √(4πα) = 0.302822: μ′ₑ = −0.302822 × 0.00115965218/(2 × 0.51099895 MeV) = −3.4361 × 10⁻⁴ MeV⁻¹, negative because e = −|e|, and μ′ₚ = 0.302822 × 1.79284734/(2 × 938.27209 MeV) = 2.8932 × 10⁻⁴ MeV⁻¹.
SI: μ′ₑ = −aₑ μ_B = −0.00115965218 × 9.2740101 × 10⁻²⁴ J T⁻¹ = −1.07546 × 10⁻²⁶ J T⁻¹ and μ′ₚ = κₚ μ_N = 1.79284734 × 5.0507837 × 10⁻²⁷ J T⁻¹ = 9.0553 × 10⁻²⁷ J T⁻¹. The ratio μ′ₚ/μ′ₑ = −0.842 in either system: each anomalous moment points along its own Dirac moment, which for the electron is opposite to its spin.
Checks: μ′ has mass dimension −1, as the coefficient of a dimension-5 operator must, and F₂(0) → 0 returns g = 2. Relative to its own Dirac moment the proton's anomaly is 1546 times the electron's, yet in absolute terms it is smaller, because |e|/2mₚ is 1836 times smaller than |e|/2mₑ.

AnswerElectron: F₁(0) = 1, F₂(0) = 0.00115965218, g = 2.00231930436, μ′ = −3.436 × 10⁻⁴ MeV⁻¹ = −1.0755 × 10⁻²⁶ J T⁻¹. Proton: F₁(0) = 1, F₂(0) = 1.79285, g = 5.58569, μ′ = 2.893 × 10⁻⁴ MeV⁻¹ = 9.055 × 10⁻²⁷ J T⁻¹.

MediumThe neutron has F₁ⁿ(0) = 0 and a measured moment of −1.91304273 nuclear magnetons, quoted. Writing its vertex with 2mₙ, find F₂ⁿ(0); then suppose F₁ⁿ vanishes at every q² and use the Breit-frame G_E to find the mean-square charge radius ⟨r²⟩ = −6 dG_E/dQ² at Q² = −q² = 0. Compare with the measured −0.1155 fm².

The moment is G_M(0)|e|/2mₙ, while a nuclear magneton is |e|/2mₚ, so F₂ⁿ(0) = G_Mⁿ(0) = −1.91304273 × mₙ/mₚ = −1.91304273 × 939.56542/938.27209 = −1.91568.
Breit frame: G_E = F₁ + (q²/4mₙ²)F₂ = F₁ − (Q²/4mₙ²)F₂. With F₁ⁿ ≡ 0, G_Eⁿ(Q²) = −(Q²/4mₙ²)F₂ⁿ(0) + O(Q⁴): a neutral particle still shows a charge form factor, because F₂ enters G_E away from q² = 0.
Comparing with G_E ≈ G_E(0) − Q²⟨r²⟩/6 gives ⟨r²⟩_F = 6F₂ⁿ(0)/4mₙ² = (3/2)F₂ⁿ(0)/mₙ². Restoring ħc: (ħc/mₙc²)² = (197.327 MeV fm/939.565 MeV)² = 0.044108 fm².
⟨r²⟩_F = 1.5 × (−1.91568) × 0.044108 fm² = −0.12675 fm², Foldy's term. Dimensions: F₂ is a pure number and (ħc/mₙc²)² is an area.
Measured, quoted: −0.1155 fm². The Foldy term alone is 110% of it, and the remainder, +0.0112 fm², is the slope of F₁ⁿ, the neutron's own charge distribution. Limit: mₙ → ∞ removes the term, which is set by the Compton wavelength ħ/mₙc, the same contact interaction UAQM-06.04 met as a scattering length.

AnswerF₂ⁿ(0) = −1.9157; ⟨r²⟩_F = (3/2)F₂ⁿ(0)(ħc/mₙc²)² = −0.1268 fm², 110% of the measured −0.1155 fm², the rest being F₁ⁿ's own slope.

HardElectrons of 300 MeV scatter through 60° from a heavy, recoil-free nucleus. Using the form-factor decomposition, derive the spin-summed factor Σ|ū(p′)Γ⁰u(p)|² in terms of F₁, F₂, β and θ; find the fractional change in the rate to first order in F₂ with F₁ = 1; and evaluate what the τF₂² term would give if F₂ kept its value aₑ = 0.00115965 at q² = 0.

A static field means E′ = E, so (p + p′)⁰ = 2E and Gordon gives Γ⁰ = G_Mγ⁰ − (E/m)F₂, with G_M = F₁ + F₂. Spacelike F's are real, so the spin sum is Tr[(p̸′ + m)Γ⁰(p̸ + m)Γ⁰].
Three traces, with s = sin(θ/2): Tr[(p̸′ + m)γ⁰(p̸ + m)γ⁰] = 4(2E² − p⋅p′ + m²) = 8E²(1 − β²s²); each cross trace Tr[(p̸′ + m)γ⁰(p̸ + m)] = 4m(E + E′) = 8mE; and Tr[(p̸′ + m)(p̸ + m)] = 4(p⋅p′ + m²) = 8m²(1 + τ), with τ = |p|²s²/m².
Collect: 8E²[G_M²(1 − β²s²) − 2G_MF₂ + F₂²(1 + τ)] = 8E²[F₁² + τF₂² − β²s²G_M²], since G_M² − 2G_MF₂ + F₂² = F₁². Checks: F₂ = 0 gives Mott's 1 − β²s²; θ → 0 gives F₁², the charge; and it equals (G_E² + τG_M²)/(1 + τ) − β²s²G_M², the Rosenbluth form.
Kinematics: |p| = 299.9996 MeV, β² = 0.999997, s² = 0.25, −q² = 4|p|²s² = 9.000 × 10⁴ MeV², τ = 8.617 × 10⁴. With F₁ = 1 the first-order change is −2β²s²F₂/(1 − β²s²) = −0.6667F₂, or −7.73 × 10⁻⁴ if F₂ = aₑ.
But τaₑ²/(1 − β²s²) = 8.617 × 10⁴ × 1.3448 × 10⁻⁶/0.75 = +0.1545: a constant Pauli coupling would raise the rate by 15%, because σ^μν q_ν/2m grows as |q|/m. A constant F₂ cannot survive −q² ≫ m²; F₂(q²) must fall.
Topic 14.03's one-loop F₂, quoted, evaluates to 8.59 × 10⁻⁸ at Q² = −q² = 0.0900 GeV², matching (α/π)(m²/Q²)ln(Q²/m²). F₂'s true share is then −5.7 × 10⁻⁸ from the linear term and 8 × 10⁻¹⁰ from τF₂².

AnswerΣ|ū′Γ⁰u|² = 8E²[F₁² + τF₂² − β² sin²(θ/2)(F₁ + F₂)²]; δN/N = −0.667F₂. A constant F₂ = aₑ would add +15% through τF₂²; the falling one-loop F₂ leaves a share of −5.7 × 10⁻⁸.