University Quantum Field Theory I · Renormalising QED: the Electron Self-Energy and the Ward-Takahashi Identity · 12.5

The electron's Z₂ and its infrared divergence

Topic 12.04 took the value of the electron's self-energy at its pole; this lesson takes the slope. The slope fixes Z₂, and with it come two things a first course meets here: a logarithm of the photon mass that no ultraviolet regulator touches, and a dependence on the gauge. Neither breaks QED, and seeing exactly why is the point.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Renormalised perturbation theory asks the electron propagator for residue 1 at the physical mass, and the counterterm that enforces it is δ₂ = Z₂ − 1, at one loop the slope dΣ₂/dp̸ at p̸ = m. Differentiating topic 12.03's Σ₂ splits that slope in two. The p̸ standing openly in the numerator carries the whole ultraviolet pole, −(α/4π)(2/ε), spread across the Feynman-parameter range.

The p² hidden in Δ carries no pole but brings 1/Δ, and at x → 1, where the photon goes soft, Δ shrinks to μ_γ². The value Σ₂(m) survives that corner because a logarithm is integrable; the slope does not, and returns 2 ln(m²/μ_γ²). The cause is kinematic: with a massless photon the eγ threshold sits exactly on the pole, so near the pole Σ₂ − Σ₂(m) goes like (p² − m²) ln(m² − p²), finite with an infinite slope.

The result, δ₂ = −(α/4π)[2/ε − γ_E + ln(4πμ²/m²) + 4 + 2 ln(μ_γ²/m²)], is not an observable. Topic 12.02's rescaling strips topic 7.03's √Z₂ from every external leg, so δ₂ lives only in counterterms, where Z₁ = Z₂ sets its infrared logarithm against the vertex's; and it depends on the gauge, its pole on ξ and its infrared logarithm on 3 − ξ, while the pole mass does not. The calculation stops at one loop and at a fictitious photon mass, whose honest replacement is soft emission, topic 14.08.

Simple definition: Z₂ is the electron's field-strength renormalisation, ψ₀ = √Z₂ ψ, chosen so the renormalised propagator has residue 1 at the physical mass; at one loop δ₂ = Z₂ − 1 is the slope dΣ₂/dp̸ of the self-energy at p̸ = m.
Example: In Feynman gauge, lowering the photon mass from 1 keV to 1 eV raises δ₂ by (α/π) ln 1000 = 0.0160, while the self-mass m − m₀ moves by only (α/2)(999 eV) = 3.6 eV, a term that vanishes as μ_γ → 0.

The residue conditionδ₂ = Z₂ − 1 = dΣ₂/dp̸ at p̸ = m

Derived from the on-shell slope condition of topic 12.02, and equal to topic 7.03's 1/Z₂ = 1 − dΣ/dp̸ at order α. The value Σ₂(p̸ = m) is the self-mass of topic 12.04.

Σ₂ is the one-loop 1PI self-energy, mass dimension 1, a function of p̸ with p² = p̸²; m is the pole mass; δ₂ is a pure number

The slope as a Feynman-parameter integralδ₂ = (α/4π) ∫₀¹ dx [−2x(2/ε − γ_E + ln(4πμ²/Δ₀)) + 2x + 4x(1 − x)(2 − x)m²/Δ₀]

Derived by differentiating topic 12.03's Σ₂: the first two terms come from its explicit p̸, the last from the p² inside Δ through d(p²)/dp̸ = 2m.

Δ₀ = (1 − x)²m² + xμ_γ², mass dimension 2; x weights the photon line; μ is the regularisation scale, μ_γ the photon mass

One-loop δ₂ in Feynman gaugeδ₂ = −(α/4π)[2/ε − γ_E + ln(4πμ²/m²) + 4 + 2 ln(μ_γ²/m²)]

Derived. ∫2x dx = 1 fixes the pole; the finite parts −3 and +1 come from the explicit p̸, and the p² inside Δ gives −2 − 2 ln(μ_γ²/m²), the logarithm from the soft corner.

d = 4 − ε; α/4π = 5.807 × 10⁻⁴; valid for μ_γ ≪ m, with corrections of order μ_γ/m

The soft-photon logarithm∫₀¹ 4y dy/(y² + μ_γ²/m²) = 2 ln(m²/μ_γ²) + O(μ_γ²/m²)

Each e-fold by which μ_γ falls adds α/π = 2.32 × 10⁻³ to δ₂. The self-mass has ln Δ₀ where the slope has 1/Δ₀, and stays finite as μ_γ → 0.

y = 1 − x is dimensionless; the integrand is the x → 1 form of 4x(1 − x)(2 − x)m²/Δ₀

Gauge-dependent part of the self-energyΣ₂(ξ) − Σ₂(1) = (1 − ξ)(p̸ − m)[(α/4π)(2/ε − γ_E + ln(4πμ²/μ_γ²)) + ie²J(p̸)(p̸ − m)]

Derived from k̸ = (p̸ − m) − (p̸ − k̸ − m). It vanishes at p̸ = m, so the self-mass and the pole mass do not depend on ξ; its slope does not vanish.

ξ = 1 Feynman, 0 Landau, 3 Yennie; J(p̸) = ∫ dᵈk/(2π)ᵈ (p̸ − k̸ − m)⁻¹/(k² − μ_γ²)² is finite at p̸ = m while μ_γ ≠ 0

δ₂ in a general covariant gaugeδ₂(ξ) = −(α/4π)[ξ(2/ε − γ_E + ln(4πμ²/m²)) + (3 − ξ) ln(μ_γ²/m²) + c(ξ)]

Derived from the two cards above. The pole vanishes in Landau gauge and the infrared logarithm in Yennie's gauge ξ = 3; no gauge removes both.

c(ξ) = 4 with μ_γ in both factors of the gauge term, as here; other placements change c(ξ) for ξ ≠ 1, not the two coefficients

Value and slope: two conditions at one pole

Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout, with d = 4 − ε and Feynman gauge unless ξ is named. Topic 12.02 writes the bare field as ψ₀ = √Z₂ ψ and asks the renormalised propagator i/(p̸ − m − Σ_R(p̸)) for a pole at the physical mass m with residue 1. Expanding Σ_R(p̸) ≈ Σ_R(m) + (p̸ − m)Σ_R′(m) shows this is two conditions, Σ_R(m) = 0 and Σ_R′(m) = 0. The counterterm iδ₂p̸ enters Σ_R as −δ₂p̸ and the mass counterterms are constants, so the slope condition reads δ₂ = dΣ₂/dp̸ at p̸ = m, however the mass counterterm is defined; the value condition gave topic 12.04's self-mass m − m₀ = Σ₂(p̸ = m). Topic 7.03 reached the same δ₂ from the bare side, 1/Z₂ = 1 − dΣ/dp̸, so Z₂ = 1 + δ₂ at order α. The value fixes where the pole sits and the slope how much of it there is, and a function can have a finite value and an infinite slope at the same point.

The explicit p̸ carries the ultraviolet pole

Quoted from topic 12.03: Σ₂(p) = (α/4π) ∫₀¹ dx [(4m − 2xp̸)(2/ε − γ_E + ln(4πμ²/Δ)) − 2m + 2xp̸], with Δ = −x(1 − x)p² + (1 − x)m² + xμ_γ², x weighting the photon propagator; the last two terms are the O(ε) parts of γ^μγ_μ = d and γ^μk̸γ_μ = −(d − 2)k̸ meeting the pole. Σ₂ depends on p̸ twice, openly and through p² = p̸², so d/dp̸ = ∂/∂p̸ + 2p̸ ∂/∂p². The open p̸ gives −2x[2/ε − γ_E + ln(4πμ²/Δ₀)] + 2x, with Δ₀ = (1 − x)²m² + xμ_γ² the on-shell Δ. Since ∫₀¹ 2x dx = 1, the entire ultraviolet pole of δ₂ is −(α/4π)(2/ε), a third the size of δm/m's and opposite in sign. Writing ln(4πμ²/Δ₀) = ln(4πμ²/m²) − ln(Δ₀/m²) leaves ∫2x ln(1 − x)² dx = −3 as μ_γ → 0, and the 2x from the O(ε) algebra adds +1. A four-dimensional Dirac algebra would lose that +1 and turn the final constant 4 into 5: the constant belongs to the scheme, and the vertex must follow the same rule for Z₁ = Z₂ to hold term by term.

The hidden p² finds the soft corner

The p² inside Δ brings no pole, because ∂Δ/∂p² = −x(1 − x) only differentiates a logarithm. It contributes (4m − 2xm)(2m)x(1 − x)/Δ₀ = 4x(1 − x)(2 − x)m²/Δ₀, and with y = 1 − x the numerator is (4y − 4y³)m². The 4y³ piece over y²m² is harmless and gives −2. The 4y piece is not: near y = 0, where the photon line takes all the Feynman weight and the photon is soft, Δ₀ → y²m² + μ_γ², and ∫₀¹ 4y dy/(y² + μ_γ²/m²) = 2 ln(m²/μ_γ²). Collecting, δ₂ = −(α/4π)[2/ε − γ_E + ln(4πμ²/m²) + 4 + 2 ln(μ_γ²/m²)], derived. Quadrature of the finite integrand at μ_γ = 10⁻³m returns −23.640 for the bracket's 4 + 2 ln(μ_γ²/m²), against −23.631 from the formula, the gap falling in proportion to μ_γ/m. The value escapes: the self-mass integrand holds ln Δ₀, which near y = 0 is ln y² and integrable, whereas the slope's 1/Δ₀ times the y from ∂Δ/∂p² leaves dy/y. Checks: δ₂ is a pure number with a ratio of squared masses in every logarithm, and it vanishes with α, the free field's Z₂ = 1.

Why the slope diverges: the threshold sits on the pole

With μ_γ = 0, Δ = (1 − x)(m² − xp²), which at p² = m² vanishes at x = 1: the eγ continuum starts at (m + μ_γ)² = m², on top of the pole. Two-body phase space for one massive and one massless particle opens linearly, as 1 − m²/p², so the absorptive part of Σ₂ grows like p² − m² above threshold and Σ₂ − Σ₂(m) behaves like (p² − m²) ln(m² − p²), continuous with a logarithmically infinite slope. This is the failure topic 7.03 flagged: LSZ wants an isolated pole, and QED's electron has none. A photon mass opens a gap of about 2mμ_γ in p² between pole and cut, which makes Z₂ finite and leaves it depending on μ_γ. Plotted against ln(1/(1 − x)), as in the figure, the soft corner becomes a plateau: every e-fold of scale between μ_γ and m adds the same α/π to δ₂, the dω/ω of soft radiation. At μ_γ = 1 eV the infrared part alone is (α/π) ln(m/μ_γ) = 0.0305, thirteen times α/π.

Legs carry 1, and δ₂ moves into the counterterms

Topic 7.03 put √Z₂ on every external electron, Z₂ being the residue of the bare propagator. Topic 12.02's rescaling ψ₀ = √Z₂ ψ divides that residue out: the renormalised two-point function is Z₂⁻¹ times the bare one and has residue 1 exactly when δ₂ = dΣ₂/dp̸, so LSZ for renormalised fields puts 1 on each leg, and a self-energy on an external line, amputated with the rest of the full propagator, is never drawn. δ₂ has not disappeared. It sits in the kinetic counterterm and, through Z₁ = Z₂ (topic 12.07, quoted here), in the vertex counterterm −ieγ^μδ₁. At zero momentum transfer the Ward-Takahashi identity of topic 12.06, quoted, makes the one-loop vertex correction −δ₂γ^μ between on-shell spinors, infrared logarithm included, so the counterterm cancels it and the charge stays the Thomson-limit e, as topic 14.05 checks. At q² ≠ 0 the vertex keeps an infrared logarithm of its own, and that one cancels only against the emission of soft photons, topic 14.08.

Gauge dependence: the pole mass is measured, Z₂ is not

In a covariant gauge the photon propagator gains i(1 − ξ)k_μk_ν/(k² − μ_γ²)². On the electron line it gives k̸(p̸ − k̸ − m)⁻¹k̸, and k̸ = (p̸ − m) − (p̸ − k̸ − m) reduces this to (p̸ − m)(p̸ − k̸ − m)⁻¹(p̸ − m) − (p̸ − m) − k̸, whose last term integrates to zero. So Σ₂(ξ) − Σ₂(1) carries an overall factor (1 − ξ)(p̸ − m), derived. It vanishes at p̸ = m: the self-mass, and the pole mass with it, do not depend on ξ at one loop. Its slope survives and adds (α/4π)(1 − ξ)[2/ε − γ_E + ln(4πμ²/μ_γ²)] to δ₂, giving a pole −(α/4π)ξ(2/ε) and an infrared logarithm with coefficient 3 − ξ in place of 2. The pole vanishes in Landau gauge, ξ = 0, the infrared logarithm in Yennie's, ξ = 3, and no gauge removes both. That is acceptable because ψ is not gauge invariant, so its residue is bookkeeping. What is measured is the pole mass, 0.510999 MeV, and rates in which Z₂ never appears alone.

Change one variable at a time

Make the relationship visible.

Interactive model

Photon mass log₁₀(μ_γ/m)−3.0

Pole term L = 2/ε − γ_E + ln(4πμ²/m²)4.0

Lower μ_γ/m from 10⁻² to 10⁻⁶: the plateau stretches by ln 10 per decade, adding (α/π) ln 10 to δ₂ each time, while the dashed ultraviolet hump near x = ½ stays put. Raise L and only the hump deepens; the δm readout never feels μ_γ.

IR LOG 2 ln(m²/μ_γ²)27.63

δ₂ IN UNITS OF α/4π19.63

Z₂ = 1 + δ₂1.0114

δm/m IN UNITS OF α/4π16.0

Live interpretationIR LOG 2 ln(m²/μ_γ²): 27.63. δ₂ IN UNITS OF α/4π: 19.63. Z₂ = 1 + δ₂: 1.0114. δm/m IN UNITS OF α/4π: 16.0

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading is that Z₂ is a physical property of the electron, the probability of finding the dressed electron as a bare one, so that an infrared-divergent Z₂ is a disaster and the cure is to find the gauge that makes it finite: Landau's for the pole, Yennie's for the logarithm. Both halves are wrong. ψ is not gauge invariant, so the residue of its propagator may depend on ξ, and it does, through ξ(2/ε) and (3 − ξ) ln(μ_γ²/m²); no single gauge removes both, and none needs to. The √Z₂ that LSZ asks for is cancelled by the field rescaling, and what is left of δ₂ sits in counterterms, where Z₁ = Z₂ cancels it against the vertex. No measured number contains Z₂ alone, so a calculation is judged by whether its pole mass and its infrared-safe rates come out independent of ξ and μ_γ, not by the size of Z₂.

In Feynman gauge with a photon mass, δ₂ = −(α/4π)[2/ε − γ_E + ln(4πμ²/m²) + 4 + 2 ln(μ_γ²/m²)]. A student lowers μ_γ by a factor of 10 and redoes the one-loop e⁻e⁻ → e⁻e⁻ amplitude in renormalised perturbation theory. Which statement is right?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasySuppose the photon really had a mass at today's experimental bound, μ_γ = 1 × 10⁻¹⁸ eV. In Feynman gauge, how much does each decade of photon mass contribute to δ₂, how many decades separate μ_γ from the electron's 0.511 MeV, and how large is the infrared part of δ₂? Compare the self-mass's sensitivity to μ_γ.

The infrared term of δ₂ is −(α/4π) × 2 ln(μ_γ²/m²) = (α/π) ln(m/μ_γ): the plateau of height 4, in units of α/4π, running out to ln(m/μ_γ).
One decade adds (α/π) ln 10 = 2.3228 × 10⁻³ × 2.3026 = 5.348 × 10⁻³.
m/μ_γ = 5.110 × 10⁵ eV ÷ 10⁻¹⁸ eV = 5.110 × 10²³, which is 23.71 decades, or ln(m/μ_γ) = 54.59.
The infrared part is (α/π)(54.59) = 0.1268, a pure number as δ₂ must be. Limit: as μ_γ rises towards m the logarithm, and this term, fall to zero.
The self-mass feels μ_γ only near x = 1, where its integrand shifts by −2 ln(1 + μ_γ²/(m²y²)), and 2∫₀^∞ ln(1 + μ_γ²/(m²y²)) dy = 2πμ_γ/m: a shift −(α/4π)(2π)μ_γ = −(α/2)μ_γ = −3.6 × 10⁻²¹ eV, against 0.511 MeV.

Answer5.35 × 10⁻³ per decade over 23.7 decades, so the infrared part of δ₂ is 0.127; the self-mass shifts by 3.6 × 10⁻²¹ eV.

MediumA muon and an electron are renormalised in one Feynman-gauge calculation, with a common ultraviolet regulator and a common photon mass μ_γ = 1 eV. Find δ₂ for the muon minus δ₂ for the electron, show which parts of the formula drop out, and split the difference into its ultraviolet and infrared shares. Take m_μ/mₑ = 206.768.

In δ₂ = −(α/4π)[2/ε − γ_E + ln(4πμ²/m²) + 4 + 2 ln(μ_γ²/m²)] the mass enters through the ultraviolet logarithm once and through the infrared one twice over, so ln m² carries a total weight of three.
Subtracting, the pole, γ_E, 4π, μ, the constant 4 and μ_γ all cancel: δ₂(muon) − δ₂(electron) = −(α/4π)[ln(mₑ²/m_μ²) + 2 ln(mₑ²/m_μ²)] = (3α/4π) ln(m_μ²/mₑ²), finite and regulator-free.
ln(m_μ²/mₑ²) = 2 ln 206.768 = 10.663, and 3α/4π = 1.7421 × 10⁻³, so the difference is 0.01858.
The ultraviolet share is (α/4π)(10.663) = 0.00619 and the infrared share twice that, 0.01238: the heavier muon gives soft photons a longer run of scales, ln(m_μ/μ_γ) against ln(mₑ/μ_γ).
Checks: a ratio of masses, so dimensionless; and the difference vanishes as m_μ → mₑ. Through Z₁ = Z₂ (topic 12.07) the two vertex counterterms differ by the same 0.0186, which is why the two charges still renormalise alike.

Answerδ₂(muon) − δ₂(electron) = (3α/4π) ln(m_μ²/mₑ²) = 0.0186, independent of ε, μ and μ_γ: 0.0062 from the ultraviolet logarithm and 0.0124 from the infrared one.

HardKeep only the leading logarithm of a cutoff, replacing 2/ε − γ_E + ln(4πμ²/m²) by ln(Λ²/m²) with Λ = 1 TeV. In Feynman gauge with μ_γ = 1 eV find δ₂ and Z₂, find the photon mass below which Z₂ exceeds 1, explain why topic 7.03's 0 ≤ Z₂ ≤ 1 is not violated in any meaningful sense, and say which logarithm survives in Landau gauge and how large it is at μ_γ = 1 eV.

With m = 0.511 MeV: ln(Λ²/m²) = 2 ln(10⁶ MeV ÷ 0.511 MeV) = 2 ln(1.957 × 10⁶) = 28.974, and ln(μ_γ²/m²) = 2 ln(1 eV ÷ 510 999 eV) = −26.288.
δ₂ = −(α/4π)[28.974 + 4 + 2(−26.288)] = −(5.807 × 10⁻⁴)(−19.603) = +0.01138, so Z₂ = 1.0114.
Z₂ exceeds 1 once 2 ln(μ_γ²/m²) = 4 ln(μ_γ/m) falls below −32.974, that is ln(μ_γ/m) below −8.243: μ_γ/m below 2.630 × 10⁻⁴, or μ_γ below 134 eV.
The bound came from ∫ρ₁ dμ² = 1 with ρ₁ ≥ 0, and ρ₁ ≥ 0 needs every intermediate state to have positive norm. Feynman gauge keeps all four photon polarisations under Gupta-Bleuler's indefinite metric (Unit 8), whose timelike photons have negative norm, so the bound need not hold and Z₂ is no probability here.
In Landau gauge, ξ = 0, the coefficient ξ removes ln(Λ²/m²) altogether and the infrared coefficient grows from 2 to 3: −(α/4π)(3)(−26.288) = +0.0458, before the regulator-dependent constant c(0).
Checks: every term is dimensionless; the pole mass 0.511 MeV is the same in both gauges; and α → 0 returns Z₂ = 1.

Answerδ₂ = +0.0114 and Z₂ = 1.011; Z₂ exceeds 1 for μ_γ below about 134 eV. In Landau gauge only the infrared logarithm remains, +0.046 at μ_γ = 1 eV.