University Quantum Field Theory I · Renormalising QED: the Electron Self-Energy and the Ward-Takahashi Identity · 12.7

Z₁ = Z₂ and the universality of charge

Charge renormalisation looks as if it needs a separate correction for every particle the photon touches, set by that particle's mass, its other forces, the gauge and the photon mass. This topic shows the vertex and the field rescaling always cancel, so one number, the photon's Z₃, renormalises every charge alike.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Topic 12.02 wrote the measured charge as e = e₀Z₂√Z₃/Z₁: a factor √Z₂ from each charged leg, √Z₃ from the photon and 1/Z₁ from the vertex at zero momentum transfer. Only Z₃ belongs to the photon. Z₁ and Z₂ belong to the charged particle, and at one loop Z₂ by itself depends on the gauge, diverges as the photon mass goes to zero, and differs between muon and electron by 1.9 per cent.

Charge would be neither universal nor measurable unless the vertex undid the field rescaling exactly, and it does. Let the photon momentum q go to zero in topic 12.06's Ward-Takahashi identity and the vertex becomes the slope of the inverse propagator, Γ^μ(p, p) = γ^μ − ∂Σ/∂p_μ, because a zero-momentum photon is a constant, pure-gauge potential that only relabels p. On shell that slope is 1/Z₂ and the vertex is 1/Z₁, so Z₁ = Z₂ to all orders and e = √Z₃ e₀.

The identity needs nothing but a conserved current, so the same holds for a muon, a charged scalar, or a proton whose vertex is dressed by strong interactions no one can compute: charges equal in the Lagrangian stay equal in the laboratory, and matter is neutral to 10⁻²¹ e. The argument stops in three places. It needs a regulator that respects gauge invariance; it fixes only the γ^μ coefficient at q = 0, leaving magnetic moments and charge radii free; and it cannot say why the bare charges were equal in the first place.

Simple definition: Z₁ = Z₂ is the all-orders statement, from Ward's identity, that a charged field's vertex renormalisation equals its field-strength renormalisation, so the physical charge is e = √Z₃ e₀ whichever field the photon couples to.
Example: With the electron loop and a Pauli-Villars cutoff at the Planck mass, 2.39 × 10²² mₑ, Z₃ = 0.9202, so the bare charge is e₀ = 1.042e for the electron, the muon and the proton alike, although their Z₂ all differ.

Ward's identity at zero momentum transferΓ^μ(p, p) = γ^μ − ∂Σ(p)/∂p_μ

A zero-momentum photon is pure gauge and only relabels p, so its vertex is the slope of the inverse propagator, off shell and in every gauge ξ.

Γ^μ: full amputated vertex with −ie₀ removed, dimensionless; Σ: 1PI self-energy, mass dimension 1. Derived from topic 12.06.

The vertex constant is the propagator residueZ₁⁻¹ = Z₂⁻¹ = 1 − dΣ/dp̸, evaluated at p̸ = m

Between on-shell spinors ūp^μu = m ūγ^μu turns ∂Σ/∂p_μ into (dΣ/dp̸)γ^μ, so every gauge-dependent and infrared piece lands in both constants alike.

Z₁ from ūΓ^μ(p, p)u = Z₁⁻¹ūγ^μu on shell; Z₂ from S ≈ iZ₂/(p̸ − m); both dimensionless. Derived, all orders.

The photon alone renormalises chargee = √Z₃ e₀

Z₃ depends on neither ξ nor the photon mass, so the charge fixed by α = 1/137.036 at q² = 0 is gauge-independent and infrared-finite.

From topic 12.02's e₀Z₂√Z₃ = Z₁e with Z₁ = Z₂; e and e₀ dimensionless in d = 4; Z₃ the photon's residue. Derived.

Charge ratios are not renormalisedq_f / q_g = q_f⁽⁰⁾ / q_g⁽⁰⁾ for any two charged fields f and g

Each field's own Z₁ = Z₂ cancels its own dressing and leaves the shared √Z₃: proton and positron agree to about 10⁻²¹ e (quoted) despite the strong force.

q: physical charge in units of e; q⁽⁰⁾: the bare coupling in units of e₀. Derived for elementary fields, argued for hadrons.

One counterterm in two placesδ₁ = δ₂ = −(α/4π)[2/ε − γ_E + ln(4πμ²/m²) + 4 + 2 ln(μ_γ²/m²)]

Pole, constant and infrared logarithm agree term by term, so the vertex counterterm needs no condition of its own and the charge counterterm is δ₃ alone.

Feynman gauge, d = 4 − ε, scale μ, photon mass μ_γ, all masses in GeV; δ₂ computed in topic 12.05, δ₁ checked in topic 14.05.

What the vertex must cancel: muon against electronZ₂(muon)/Z₂(electron) − 1 = (3α/2π) ln(m_μ/mₑ) = 0.0186

Dropping Z₁ would make the muon's charge 1.9 per cent larger than the electron's; muonium finds them equal within a few parts in 10⁹ (quoted).

One loop, Feynman gauge; independent of the cutoff and of any photon mass μ_γ ≪ mₑ. Derived from topic 12.05's δ₂.

Three factors, and two of them belong to the particle

Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. Topic 12.02's rescaling gave e₀Z₂√Z₃ = Z₁e, and it reads as a recipe for what Thomson scattering measures at q → 0 (topic 10.07): LSZ puts √Z₂ on each of the two charged legs and √Z₃ on the photon, and the bare vertex at zero momentum transfer contributes 1/Z₁, so e = e₀Z₂√Z₃/Z₁. Z₃ belongs to the photon. Z₁ and Z₂ belong to the charged particle and depend on its mass, on every other force it feels, on the gauge parameter ξ and on the photon mass μ_γ. Topic 12.05's δ₂ makes this concrete: its mass dependence, (3α/4π) ln m², survives any cutoff and photon mass, so at one loop the muon's Z₂ exceeds the electron's by (3α/2π) ln 206.77 = 0.0186. Unless the vertex compensated, the two charges would differ by 1.9 per cent and each would diverge as μ_γ → 0; Z₂ even exceeds 1 once μ_γ is small enough, which no probability could. Muonium spectroscopy finds the muon's and electron's charges equal within a few parts in 10⁹, quoted. Something must cancel, exactly and at every order.

Let the photon's momentum go to zero

Topic 12.06 derived −iq_μΓ^μ(p + q, p) = S⁻¹(p + q) − S⁻¹(p), with S(p) = i/(p̸ − m₀ − Σ(p)) the full propagator, Σ the sum of one-particle-irreducible self-energy graphs and Γ^μ the full amputated vertex with −ie₀ stripped off, all built from bare fields. Expand the right side to first order: S⁻¹(p + q) − S⁻¹(p) = −iq_μ(γ^μ − ∂Σ/∂p_μ) + O(q²). The identity fixes only q_μΓ^μ, so a transverse piece T^μ with q_μT^μ = 0 could hide; but if the vertex is regular at q = 0, as it is off shell or with a photon mass, q_μT^μ(p, p) = 0 for every direction of q forces T^μ(p, p) = 0. Hence Γ^μ(p, p) = γ^μ − ∂Σ/∂p_μ, Ward's 1950 identity, derived. Checks: with Σ = 0 it returns the tree vertex γ^μ, and Σ and p_μ both carry mass dimension 1, so ∂Σ/∂p_μ is dimensionless like γ^μ. The reason is physical. A zero-momentum photon is a constant A_μ, a pure gauge that only shifts p by a multiple of e₀A, so the amplitude to absorb it is the derivative of the inverse propagator. No internal photon propagator entered, so the result holds in every gauge ξ.

On shell, the vertex constant is the residue

Lorentz covariance and parity give Σ(p) = A(p²)p̸ + B(p²)m, so ∂Σ/∂p_μ = Aγ^μ + 2p^μ(A′p̸ + B′m), a prime meaning d/dp². Sandwich it between on-shell spinors: p̸u = μ, and the Gordon identity at zero momentum transfer (UAQM-05.07, quoted) gives ūp^μu = m ūγ^μu. The p^μ terms become γ^μ terms, and ū(∂Σ/∂p_μ)u = Σ′(m) ūγ^μu with Σ′(m) = A + 2m²(A′ + B′), which is exactly dΣ/dp̸ at p̸ = m once p² is written as p̸². Ward's identity becomes ūΓ^μ(p, p)u = [1 − Σ′(m)]ūγ^μu. The left side defines 1/Z₁ and topic 12.05's residue condition defines 1/Z₂ = 1 − Σ′(m), so Z₁ = Z₂, derived, to all orders; with Z = 1 + δ, δ₁ = δ₂ order by order. A model shows the bookkeeping. Σ = ap̸ + bm + κ(p² − m²)p̸/m² with a = −0.012 and κ = −0.0035 has Σ′(m) = a + 2κ = −0.019, the κ term entering only through 2p^μ, so Z₁ = Z₂ = 1/1.019 = 0.9814.

e = √Z₃ e₀ for every charged field

Put Z₁ = Z₂ into e₀Z₂√Z₃ = Z₁e and the particle's own factors leave: e = √Z₃ e₀, derived. The argument used only that field's propagator and vertex, so a muon, a τ or a charged scalar, each coupled with its own bare charge q⁽⁰⁾e₀, ends with physical charge q⁽⁰⁾√Z₃ e₀, and ratios of charges are not renormalised. Z₃ is one number for the photon, summing the vacuum polarisation of every charged species (Unit 13). With the electron loop and a Pauli-Villars cutoff at the Planck mass, 1.22 × 10¹⁹ GeV = 2.39 × 10²² mₑ, Z₃ = 1 − (α/3π) ln(Λ²/mₑ²) = 0.9202 (quoted from topic 13.03), so the bare charge is e/√Z₃ = 1.042e for every particle. The gauge dependence and infrared logarithms sat in Z₁ and Z₂ identically and cancelled, and Z₃ has neither, so e is gauge-independent and infrared-finite, as a measured number must be. In counterterm language δ₁ = δ₂ and the charge counterterm is δ₃ alone, which is all topic 13.06 needs for QED's beta function.

The proton: dressed by strong forces, charged like a positron

The diagrams of topic 12.06 assumed perturbative QED, but its operator form needs only two facts: ∂_μJ^μ = 0, and an equal-time commutator saying that the charge Q = ∫J⁰ d³x generates the field's phase rotation. Strong interactions conserve electric charge, so the identity holds for the proton's full propagator and vertex with every pion cloud and gluon included, though none of it can be computed in perturbation theory. Hence Z₁ = Z₂ for the proton too, argued, and its charge is q⁽⁰⁾√Z₃ e₀. QED cannot say why the proton's bare charge is minus the electron's; it says only that renormalisation cannot pull them apart. Experiment agrees: |qₚ + qₑ| and the neutron's charge both lie below about 10⁻²¹ e, quoted. What the identity protects is narrow. It fixes the γ^μ coefficient at q = 0 and nothing else: the strong dressing raises the proton's magnetic moment from Dirac's 1 to 2.793 nuclear magnetons, F₂(0) = 1.793 (quoted), and spreads its charge over a radius near 0.84 fm, while leaving F₁(0) = 1 exactly.

A regulator that breaks gauge invariance breaks the equality

Both steps assumed that the regulated Green's functions still obey the Ward-Takahashi identity. Dimensional regularisation and Pauli-Villars do; so does a photon mass, because the external photon's insertions telescope along charged lines and never touch an internal photon propagator, which is why δ₁ and δ₂ can carry the same ln μ_γ. A sharp cutoff on the Euclidean loop momentum does not, stated: Σ is superficially linearly divergent (D = 1, topic 12.01), so shifting its loop momentum by a finite amount changes it by a finite surface term proportional to αp̸, which moves Z₂ while the logarithmically divergent vertex stays put. Z₁ − Z₂ becomes a finite, regulator-dependent number with no reason to be the same for every species, and keeping hydrogen neutral would need a separately tuned counterterm for every particle at every order. An untuned mismatch of one-loop size, α/4π = 5.8 × 10⁻⁴, would exceed the neutrality bound by a factor near 6 × 10¹⁷. Unit 13 meets the same rule when a cutoff hands the photon a mass.

Change one variable at a time

Make the relationship visible.

Interactive model

Lepton mass log₁₀(M/mₑ)2.30

Photon mass log₁₀(mₑ/μ_γ)6.0

Cutoff log₁₀(Λ/mₑ)19.0

Move the lepton from the electron (0) to the muon (2.3) or the τ (3.55), then drag the photon-mass slider right to lower μ_γ: Z₁ = Z₂ shifts by several per cent, yet the dashed charge curve never moves, because each lepton's vertex cancels its own field rescaling and leaves only the photon's Z₃.

Z₁ = Z₂, THIS LEPTON0.9971

Z₂ ÷ Z₂ OF THE ELECTRON1.0189

e/e₀ = √Z₃, EVERY SPECIES0.9655

e/e₀ IF Z₁ WERE SET TO 10.9628

Live interpretationZ₁ = Z₂, THIS LEPTON: 0.9971. Z₂ ÷ Z₂ OF THE ELECTRON: 1.0189. e/e₀ = √Z₃, EVERY SPECIES: 0.9655. e/e₀ IF Z₁ WERE SET TO 1: 0.9628

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading is that Z₁ = Z₂ is a coincidence found by computing: the one-loop vertex and self-energy happen to share a pole, so the equality might fail at two loops and must be re-examined for a proton, whose vertex is non-perturbative. It is wrong because the equality is never computed here. It is the q → 0 limit of an identity whose only input is a conserved current, so it holds at every order, and beyond perturbation theory, for every field whose charge the photon measures, provided the regulator respects gauge invariance; topic 14.05's one-loop calculation is a check, not the proof. The opposite slip is to read Z₁ = Z₂ as saying a particle's other interactions leave its electromagnetic vertex alone. They reshape it everywhere but one point, giving the proton F₂(0) = 1.793 and a 0.84 fm radius; only the γ^μ coefficient at zero momentum transfer, the charge, is protected.

In a gauge-invariant regularisation at some cutoff, the muon's field-strength constant is Z₂ = 0.97 and the photon's is Z₃ = 0.92. The proton's Z₂ is unknown, being set by strong interactions. What is the ratio e/e₀ of physical to bare charge for the muon, and for the proton?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyCharge universality makes a hydrogen atom exactly neutral if the bare proton and electron charges cancel, and experiment bounds |qₚ + qₑ| below 1 × 10⁻²¹ e (quoted). A 1.00 L flask holds hydrogen gas at 0 °C and 101.325 kPa. Find the largest net charge the bound allows, and compare it with the charge if qₚ + qₑ were α/4π in units of e, the size of an uncancelled one-loop mismatch.

Ideal gas: n = pV/RT = (101325 Pa × 1.00 × 10⁻³ m³)/(8.3145 J mol⁻¹ K⁻¹ × 273.15 K) = 0.04462 mol.
Molecules: N = nN_A = 0.04462 × 6.022 × 10²³ = 2.687 × 10²² H₂, each with two protons and two electrons, so 5.374 × 10²² atoms.
Bound: |Q| ≤ 5.374 × 10²² × 1 × 10⁻²¹ e = 53.7 e, and 53.7 × 1.602 × 10⁻¹⁹ C = 8.6 × 10⁻¹⁸ C.
Mismatch of α/4π = 5.807 × 10⁻⁴: Q = 5.374 × 10²² × 5.807 × 10⁻⁴ e = 3.12 × 10¹⁹ e = 5.0 C. A 1.00 L sphere has radius 6.20 cm, so its surface would sit at kQ/r = 8.99 × 10⁹ × 5.0/0.0620 = 7.2 × 10¹¹ V.
Check: both answers scale with the number of atoms, so their ratio is just (5.807 × 10⁻⁴)/(1 × 10⁻²¹) = 5.8 × 10¹⁷, the margin by which measured neutrality beats a one-loop accident.

AnswerAt most 54 e, about 8.6 × 10⁻¹⁸ C, among 5.4 × 10²² atoms; a mismatch of α/4π would give 5.0 C and about 7 × 10¹¹ V at the flask's surface. Universality has to be exact.

MediumIn Feynman gauge with a Pauli-Villars cutoff Λ = 1.0 × 10¹⁶ GeV and photon mass μ_γ = 1.0 eV, a lepton of mass M has Z₂ = 1 − (α/4π)[ln(Λ²/M²) + 9/2 + 2 ln(μ_γ²/M²)], the Pauli-Villars form of topic 12.05's δ₂ (quoted). Find Z₂ and Z₁ for the electron and the muon, their charges relative to e₀ given the photon's Z₃ = 0.9312, and the charges if the vertex were ignored.

Electron, M = 0.511 MeV: ln(Λ²/M²) = 88.841 and 2 ln(μ_γ²/M²) = −52.577, so the bracket is 88.841 + 4.500 − 52.577 = 40.764 and Z₂ = 1 − 5.807 × 10⁻⁴ × 40.764 = 0.9763.
Muon, M = 105.66 MeV: ln(Λ²/M²) = 78.178 and 2 ln(μ_γ²/M²) = −73.903, so the bracket is 8.775 and Z₂ = 0.9949. The difference, 0.0186, equals (3α/2π) ln 206.77: Λ and μ_γ cancel out of it.
Ward's identity at q → 0 gives each lepton its own Z₁ = Z₂: 0.9763 for the electron, 0.9949 for the muon. Lowering μ_γ to 1 meV raises both by (α/4π) × 4 ln 1000 = 0.0160 and leaves their difference alone.
Charges: e/e₀ = Z₂√Z₃/Z₁ = √0.9312 = 0.9650 for both leptons, although their Z₂ differ by 1.9 per cent.
Ignoring the vertex would give Z₂√Z₃ = 0.9763 × 0.9650 = 0.942 for the electron and 0.9949 × 0.9650 = 0.960 for the muon: 1.9 per cent apart, and both drifting as μ_γ → 0. Muonium finds the two charges equal within a few parts in 10⁹ (quoted).

AnswerZ₁ = Z₂ = 0.9763 for the electron and 0.9949 for the muon; e/e₀ = 0.9650 for both. Without the vertex the charges would be 0.942e₀ and 0.960e₀, 1.9 per cent apart.

HardA charged spin-0 particle of physical mass M couples through |D_μφ|², with tree vertex −ie₀(p + p′)^μ and full propagator D(p) = i/(p² − m₀² − Π(p²)). Its other interactions give Π(p²) = −0.040M² − 0.030(p² − M²) + 0.008(p² − M²)²/M². Derive the q → 0 Ward identity for its vertex, find m₀, Z₂ and Z₁, and compare its e/e₀ with the electron's, taking the photon's Z₃ = 0.9312.

The scalar's conserved current gives an identity of topic 12.06's form, −iq_μΓ^μ(p + q, p) = D⁻¹(p + q) − D⁻¹(p), with D⁻¹ = −i(p² − m₀² − Π). Tree check: q_μ(2p + q)^μ = (p + q)² − p² holds identically.
To first order in q: q_μΓ^μ(p, p) = q_μ∂(p² − Π)/∂p_μ = q_μ2p^μ[1 − Π′(p²)], and regularity at q = 0 removes any transverse piece, so Γ^μ(p, p) = 2p^μ[1 − Π′(p²)], derived. Dimensions: Π is a mass squared and Π′ a pure number, so Γ^μ is a momentum, like the tree vertex 2p^μ.
Pole at p² = M²: m₀² = M² − Π(M²) = 1.040M², so m₀ = 1.020M. Residue: D ≈ iZ₂/(p² − M²) with Z₂ = 1/(1 − Π′(M²)) = 1/1.030 = 0.9709.
On shell the vertex is 2p^μ × 1.030 = 2p^μ/Z₁, so Z₁ = 0.9709 = Z₂. The (p² − M²)² term changes the vertex only off shell: at p² = 1.25M², Π′ = −0.030 + 0.016 × 0.25 = −0.026 and the factor is 1.026.
Two √Z₂ legs, √Z₃ and 1/Z₁ give e/e₀ = √Z₃ = 0.9650 per unit bare charge, the electron's ratio too; dropping the vertex would give 0.9709 × 0.9650 = 0.9369, 2.9 per cent low. Limit check: with Π = 0, Z₁ = Z₂ = 1 and still e = √Z₃ e₀.

AnswerΓ^μ(p, p) = 2p^μ[1 − Π′(p²)]; m₀ = 1.020M; Z₁ = Z₂ = 0.9709; e/e₀ = 0.9650 per unit bare charge, the same as the spin-½ electron's.