University Quantum Field Theory I · Renormalising QED: the Electron Self-Energy and the Ward-Takahashi Identity · 12.9

A transverse self-energy keeps the photon massless

Every loop on a photon line is a chance to give the photon a mass, and in QED none of them takes it. Here you derive why: the Ward identity forces a factor q² onto the self-energy, so summing the insertions rescales the propagator by Z₃ and leaves its pole at zero, a protection the scalar of Unit 6 never had.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A photon line in QED is dressed by a chain of one-particle-irreducible blobs iΠ^μν(q), and the danger is the one topic 6.05 met for the scalar: a self-energy that does not vanish on the mass shell moves the pole, and with a cutoff it moves it by an amount set by the highest scale in the theory. The photon escapes because two facts constrain its self-energy. Topic 12.06's Ward-Takahashi identity, applied to a function whose external legs are all photons, telescopes around closed fermion loops and leaves nothing behind, so q_μΠ^μν = 0 at every order and every q; Lorentz covariance, allowing only g^μν and q^μq^ν, then forces Π^μν = (q²g^μν − q^μq^ν)Π(q²).

Summing the chain gives −ig_μν/(q²[1 − Π(q²)]) once the q_μq_ν pieces, which conserved currents never feel, are dropped. The factor q² in front of Π is the whole story: so long as Π is regular at q² = 0 the pole cannot leave zero, and Π(0), even divergent, only rescales the residue to Z₃ = 1/(1 − Π(0)), which becomes charge renormalisation. Each condition can fail.

A momentum cutoff spoils the shift of loop momentum the cancellation needs and leaves a photon mass behind; a massless state that the current can create may put a 1/q² pole into Π, which is how a superconductor, or QED in one space dimension, gives the photon a mass while its self-energy stays transverse.

Simple definition: The photon self-energy Π^μν(q) obeys q_μΠ^μν = 0, so it equals (q²g^μν − q^μq^ν)Π(q²); the factor q² lets it rescale the photon propagator by Z₃ = 1/(1 − Π(0)) but never move its pole from q² = 0.
Example: An electron loop regulated at the Planck scale gives Π(0) = −0.080, so Z₃ = 0.926: the residue moves by 7% and the pole not at all. Experiment agrees, with m_γ ≤ 1 × 10⁻¹⁸ eV, a reduced Compton wavelength ħc/m_γ of at least 1.97 × 10¹¹ m.

The insertion identity at one loopS_F(k) q̸ S_F(k + q) = i[S_F(k) − S_F(k + q)], with S_F(p) = i/(p̸ − m)

Derived. Contracting q into one vertex of the electron loop splits it into two tadpoles that differ only by k → k + q, so shifting the loop momentum cancels them.

S_F in GeV⁻¹, q̸ = γ^μ q_μ and m in GeV; from q̸ = (k̸ + q̸ − m) − (k̸ − m), each bracket cancelling one propagator

The photon self-energy is transverseq_μ Π^μν(q) = 0 for every q, on or off shell

Derived to all orders from topic 12.06's telescoping: a function whose legs are all photons has no open charged line to leave end terms. Needs a shift-invariant regulator.

Π^μν in GeV², where iΠ^μν is the sum of 1PI diagrams with two amputated photon legs; q in GeV

Covariance leaves one scalar functionΠ^μν(q) = (q² g^μν − q^μ q^ν) Π(q²)

Derived. A = q²Π, so the photon-mass coefficient A(0) vanishes unless Π has a 1/q² pole, which needs a massless state coupled to the current.

Π(q²) dimensionless; from Π^μν = A g^μν + B q^μ q^ν with A + q²B = 0, so A = q²Π and B = −Π

The resummed propagator in any covariant gaugeD_μν(q) = −i(g_μν − q_μ q_ν/q²)/(q²[1 − Π(q²)]) − iξ q_μ q_ν/q⁴

Derived as a geometric series on the projector. The ξ term is exactly the tree-level one; between conserved currents only −ig_μν/(q²[1 − Π(q²)]) survives.

D_μν in GeV⁻²; ξ the gauge parameter of topic 8.04; g_μν − q_μ q_ν/q² is a projector, equal to its own square

The pole stays at zero and the residue movesD_μν(q) → −i Z₃ g_μν/q² as q² → 0, with Z₃ = 1/(1 − Π(0))

Derived. With Z₁ = Z₂ from topic 12.07 this residue is the whole of charge renormalisation, e² = Z₃e₀², and it moves no pole.

Z₃ dimensionless, 0 ≤ Z₃ ≤ 1 by topic 4.03; q_μq_ν terms dropped, as between conserved currents; Unit 13's Π(0) ≈ −(α/3π) ln(Λ²/m²) gives Z₃ = 0.926 at Λ = 1.22 × 10¹⁹ GeV

What a non-transverse piece would costΠ_T(q²) = q²Π(q²) + a ⇒ m_γ² = Z₃ a + O(a²)

Derived as a contrast, from the zero of q²[1 − Π(q²)] − a. The measured m_γ ≤ 1 × 10⁻¹⁸ eV needs a below 1.1 × 10⁻⁵⁴ GeV²; QED supplies a = 0 exactly.

a in GeV², a g^μν term surviving at q² = 0; Π_T the coefficient of the transverse projector; m_γ in GeV

One loop: contract with q and the loop splits in two

Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Write iΠ^μν(q) for the sum of one-particle-irreducible diagrams with two amputated photon legs, so Feynman gauge's −ig_μν/(q² + iε) is corrected by (−ig_μρ/q²)(iΠ^ρσ)(−ig_σν/q²) and its repeats. At order e², with topic 9.02's vertex −ieγ^μ and topic 7.04's minus sign for a closed loop, iΠ₂^μν(q) = −(−ie)² ∫ d^dk/(2π)^d tr[γ^μ S_F(k) γ^ν S_F(k + q)]. Contract with q_ν. Writing q̸ = (k̸ + q̸ − m) − (k̸ − m) lets each bracket cancel one propagator, so S_F(k) q̸ S_F(k + q) = i[S_F(k) − S_F(k + q)], derived. Then q_ν iΠ₂^μν = ie² ∫ d^dk/(2π)^d (tr[γ^μ S_F(k)] − tr[γ^μ S_F(k + q)]), and k → k − q in the second term makes the two equal: q_ν Π₂^μν = 0, derived. Each term alone is a divergent tadpole, ∫ 4ik^μ/(k² − m²), so the result rests on shifting the variable of a divergent integral: legitimate in dimensional regularisation, illegitimate under a momentum cutoff, whose price topic 13.01 counts.

All orders: a function with only photon legs telescopes away

Topic 12.06 showed that contracting q_μ into a photon on a charged line, summed over every place the photon can attach, telescopes: each insertion −ieq̸ between S_F(p + q) and S_F(p) becomes e[S_F(p) − S_F(p + q)], and neighbouring terms cancel. On an open line two end terms survive, the inhomogeneous terms of the Ward-Takahashi identity. On a closed loop there are no ends: what survives is the loop without the photon minus the same loop with every momentum shifted by q, and the two cancel once the loop momentum is shifted. Every diagram in Π^μν has photons for both external legs, so the photon labelled μ attaches to a closed fermion loop, and the diagrams differing only in where on that loop it attaches all appear at the same order with the same factors. Summing them gives q_μ Π^μν(q) = 0 at every order, derived. Because no charged line is external, nothing has to be on shell: unlike topic 12.08's amplitudes, this holds at every q, timelike, spacelike or lightlike, and whatever gauge the internal photons are in, since they never enter the telescoping.

One function survives, and only a 1/q² pole could give a mass

Π^μν(q) is built from the single vector q in an invariant vacuum, so it can only be A(q²)g^μν + B(q²)q^μq^ν; the one other candidate, ε^μνρσ q_ρ q_σ, vanishes by antisymmetry. Contracting with q_μ gives (A + q²B)q^ν = 0, so A = −q²B, and naming Π(q²) = −B gives Π^μν = (q²g^μν − q^μq^ν)Π(q²), derived. Dimensions: iΠ^μν adds to an inverse propagator of size q², so Π^μν has mass dimension 2 and Π is a pure number, as e² ∫ d⁴k S_F S_F confirms with e dimensionless and each propagator of dimension −1. Now the subtle step. Transversality fixes A = −q²B but not B, and A(0) is precisely a photon-mass term: it vanishes unless B = −Π has a 1/q² pole. A pole at q² = 0 in a two-point function needs a massless single-particle state that the current creates. Π is one-particle irreducible, so the photon itself is excluded, and every charged line in QED carries the electron's mass, so the e⁺e⁻ threshold sits at twice the electron mass, far from zero. Π is therefore regular at q² = 0 and A(0) = 0, derived for QED; a massless single-particle state coupled to the current would undo this step, and only this step.

Summing the chain on the transverse projector

Write Δ_μν = g_μν − q_μq_ν/q², which obeys Δ_μρ Δ^ρ_ν = Δ_μν and q^μ Δ_μν = 0, so one insertion is iΠ^ρσ = i q² Δ^ρσ Π(q²). Topic 8.04's free propagator in a general covariant gauge is −i(Δ_μν + ξ q_μq_ν/q²)/q². One insertion between two free propagators gives (−i)(i)(−i)(q²Π/q⁴)(Δ + ξqq/q²)Δ(Δ + ξqq/q²) = −iΔ_μν Π/q², because Δ annihilates the longitudinal pieces. n insertions give −iΔ_μν Πⁿ/q², and the geometric series sums to D_μν(q) = −iΔ_μν/(q²[1 − Π(q²)]) − iξ q_μq_ν/q⁴, derived. Two lessons sit in that line. The longitudinal part, the only place ξ appears, receives no correction at any order: loops never touch the gauge-fixing sector. And in any amplitude the propagator ends on charged lines whose currents are conserved, where topic 12.08's k_μM^μ = 0 removes every q_μ or q_ν, so it acts as −ig_μν/(q²[1 − Π(q²)]), the form usually quoted. The free-field limit Π → 0 returns −ig_μν/q².

The pole stays put and only its residue moves

Near q² = 0 the propagator behaves as −ig_μν/(q²[1 − Π(0)]): a pole at q² = 0 with residue Z₃ = 1/(1 − Π(0)), derived, provided only that Π(0) is finite and not equal to 1. Compare topic 6.05's scalar, whose self-energy adds to the inverse propagator, p² − m₀² − Σ, so m² = m₀² + Σ moves with every loop, by about λΛ²/(32π²) under a cutoff; keeping the scalar light means choosing m₀² to cancel that. The electron sits in between: chiral symmetry makes its δm proportional to m and only logarithmic, topic 12.04. The photon needs no choice at all, because the zero is supplied by the factor q² and Π(0) only sets the slope there. In numbers: Unit 13's electron loop, cut off gauge-invariantly at a mass Λ, gives Π(0) ≈ −(α/3π) ln(Λ²/m²), quoted, which is −0.0798 at Λ = 1.22 × 10¹⁹ GeV, so Z₃ = 0.926 while the pole stays at exactly zero. With Z₁ = Z₂, topic 12.07, that residue is all of charge renormalisation, e² = Z₃e₀², for every charged species alike.

Where the protection ends, and what experiment says

Each step has a condition, and each can fail. A momentum cutoff is not invariant under k → k − q, so the tadpoles of the first section stop cancelling and a g^μν term of order (α/π)Λ² survives, a regulator-made photon mass that topic 13.01 exhibits. A massless state coupled to the current is the other way out: if it gives Π a 1/q² pole, Π^μν stays transverse and the zero of q²[1 − Π] moves off q² = 0. In QED in one space dimension with a massless electron, Π = e²/(πq²), with e there of mass dimension 1, and the photon acquires the mass e/√π, Schwinger's result, quoted; in a superconductor the Cooper-pair condensate plays that role, and a London penetration depth of 50 nm corresponds to ħc/λ ≈ 4 eV inside the metal. Neither loophole exists in the vacuum of QED with a massive electron, and experiment agrees: the Particle Data Group bounds m_γ ≤ 1 × 10⁻¹⁸ eV, so any Yukawa factor e^{(−m_γ r)} on the Coulomb potential has a range ħc/m_γ of at least 1.97 × 10¹¹ m, about 1.3 AU, the non-relativistic face of a pole at exactly q² = 0.

Change one variable at a time

Make the relationship visible.

Interactive model

Π(0), the self-energy at q² = 0−0.08

Slope s, in Π(q²) = Π(0) + s q²/mₑ²−0.10

Non-transverse piece a0.00 mₑ²

Keep a at 0 and sweep Π(0) and s end to end: the solid curve tilts and bends, yet its zero never leaves q² = 0; only the slope there, 1/Z₃, changes. Then raise a, the g^μν term a momentum cutoff would leave, and the zero walks to q² ≈ Z₃a: a photon mass.

POLE AT q²0.000 mₑ²

PHOTON MASS0.000 mₑ

Z₃ = 1/(1 − Π(0))0.926

RESIDUE AT THE POLE0.926

Live interpretationPOLE AT q²: 0.000 mₑ². PHOTON MASS: 0.000 mₑ. Z₃ = 1/(1 − Π(0)): 0.926. RESIDUE AT THE POLE: 0.926

Catch the common trap

Explain before calculating.

Misconception check

Fresh from the electron's mass shift in topic 12.04, students see a nonzero, even divergent, Π(0) and read it as a photon mass that a counterterm must cancel, or that the on-shell condition Π̂(0) = 0 tunes away. Neither is right. The electron's Σ enters its inverse propagator additively, p̸ − m₀ − Σ(p̸), so its value at the pole moves the pole. The photon's Π enters multiplied by q², as q²[1 − Π(q²)], so its value at q² = 0 cannot move a zero that the factor q² already supplies; it changes only the slope there, 1 − Π(0) = 1/Z₃, and so the residue and the charge, e² = Z₃e₀². The renormalised Lagrangian of topic 12.02 has no photon-mass counterterm at all, and its δ₃ vertex, −iδ₃(q²g^μν − q^μq^ν), is itself transverse. The photon is massless because gauge invariance puts the q² there, not because anything was subtracted.

Once its q_μq_ν terms are dropped, the photon's full propagator is −ig_μν/(q²[1 − Π(q²)]). Which pair of facts guarantees that its pole stays at exactly q² = 0 to all orders in QED?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA spacelike photon carries q^μ = (2, 0, 0, 3) GeV. A one-loop calculation returns Π⁰⁰ = −0.108 GeV², Π⁰³ = Π³⁰ = −0.072 GeV², Π¹¹ = Π²² = 0.060 GeV² and Π³³ = −0.048 GeV², all others zero. Check that q_μΠ^μν = 0, extract Π(q²) and the coefficients A and B of g^μν and q^μq^ν, and say what a stray term 0.010 GeV² × g^μν would do to the check.

Lower the index with the metric (+, −, −, −): q_μ = (2, 0, 0, −3) GeV and q² = 4 − 9 = −5 GeV², spacelike, as for a photon exchanged in scattering. Every component of Π^μν carries GeV², the dimension of q².
Contract: q_μΠ^μ0 = 2(−0.108) + (−3)(−0.072) = −0.216 + 0.216 = 0, and q_μΠ^μ3 = 2(−0.072) + (−3)(−0.048) = −0.144 + 0.144 = 0, in GeV³. The ν = 1 and ν = 2 columns have no entries along q, so all four contractions vanish: the result is transverse.
Read Π off a component transverse to q: Π¹¹ = q²g¹¹ Π = −q²Π = 5Π, so Π = 0.060/5 = 0.012, a pure number, close to the 0.0117 that Unit 13's renormalised electron loop gives at this q². Check another: Π⁰⁰ = (q² − q⁰q⁰)Π = (−5 − 4)(0.012) = −0.108 GeV².
So A = q²Π = (−5)(0.012) = −0.060 GeV² and B = −Π = −0.012, with A + q²B = −0.060 + 0.060 = 0 as transversality requires. The trace g_μνΠ^μν = 3q²Π = −0.180 GeV² matches −0.108 − 0.060 − 0.060 + 0.048 from the components.
A stray δ g^μν with δ = 0.010 GeV², the kind of term a momentum cutoff leaves, adds q_μ δ g^μν = δ q^ν to the contraction: 0.010 × 2 = 0.020 GeV³ for ν = 0 and 0.010 × 3 = 0.030 GeV³ for ν = 3. The check fails, and A(0) would no longer vanish.

AnswerTransverse: q_μΠ^μν = 0 for every ν. Π(−5 GeV²) = 0.012, A = −0.060 GeV², B = −0.012. A 0.010 GeV² g^μν term would leave q_μΠ^μν = (0.020, 0, 0, 0.030) GeV³.

MediumRegulate QED gauge-invariantly at a mass Λ = 1.22 × 10¹⁹ GeV, the Planck scale, so that Unit 13's electron loop gives Π(0) = −(α/3π) ln(Λ²/m²), quoted, with m = 0.511 MeV and α = 1/137.036. Find Π(0), say where the photon's pole sits, and find Z₃, its one-loop truncation 1 + Π(0), and the bare coupling 1/α₀.

α/(3π) = 7.743 × 10⁻⁴, and Λ/m = 1.22 × 10¹⁹ GeV ÷ 5.11 × 10⁻⁴ GeV = 2.39 × 10²², so ln(Λ²/m²) = 2 × 51.53 = 103.05 and Π(0) = −0.0798: dimensionless, as it must be, and negative.
The transverse inverse propagator is q²[1 − Π(q²)]. At q² = 0 it vanishes because of the factor q², whatever Π(0) is, so the pole stays at exactly q² = 0 and the photon stays massless, with no counterterm involved.
The residue is Z₃ = 1/(1 − Π(0)) = 1/1.0798 = 0.9261. The one-loop truncation 1 + Π(0) = 0.9202 differs by 0.0059, of the size of Π(0)² = 0.0064: a two-loop-sized difference, which the resummed form contains without being more accurate than one loop.
With Z₁ = Z₂ from topic 12.07, e² = Z₃e₀², so α₀ = α/Z₃ and 1/α₀ = 137.036 × 0.9261 = 126.9. The bare charge exceeds the measured one: virtual pairs in the vacuum screen it.
Checks: Z₃ lies between 0 and 1, as topic 4.03 requires of any residue; Λ → m sends Π(0) → 0 and Z₃ → 1, the free field; doubling Λ changes Π(0) by only −(2α/3π) ln 2 = −0.0011, so even a Planck-scale cutoff makes a mere 7% change in the residue.

AnswerΠ(0) = −0.0798; the pole stays at q² = 0; Z₃ = 0.926, against 0.920 truncated, and 1/α₀ = 126.9 for an electron loop regulated at the Planck scale.

HardSuppose a regulator broke transversality and left the photon a piece a g^μν, so its transverse self-energy is Π_T(q²) = q²Π(q²) + a. With Z₃ = 0.926, find the photon mass this gives, the largest a allowed by the measured m_γ ≤ 1 × 10⁻¹⁸ eV, and that a as a fraction of Λ² at Λ = 1.22 × 10¹⁹ GeV. Contrast a φ⁴ scalar with λ = 0.1 held at 125 GeV against the same cutoff.

The transverse propagator is −iΔ_μν/(q² − Π_T(q²)), the geometric series of the fourth section with Π_T in place of q²Π. Its pole solves q²[1 − Π(q²)] = a; to first order in a, q² = a/(1 − Π(0)) = Z₃a. So m_γ² = Z₃a: GeV² on both sides, and a → 0 returns the massless photon.
m_γ ≤ 1 × 10⁻¹⁸ eV = 1 × 10⁻²⁷ GeV means m_γ² ≤ 1 × 10⁻⁵⁴ GeV², so a ≤ 1 × 10⁻⁵⁴/0.926 = 1.08 × 10⁻⁵⁴ GeV².
Λ² = (1.22 × 10¹⁹ GeV)² = 1.488 × 10³⁸ GeV², so a/Λ² ≤ 7.3 × 10⁻⁹³. A one-loop coefficient is naturally of size α/(4π) = 5.8 × 10⁻⁴, so the allowed a sits about 89 orders of magnitude below a typical loop term.
Topic 6.05's cutoff tadpole gives the scalar Σ ≈ λΛ²/(32π²) = 0.1 × 1.488 × 10³⁸/315.8 = 4.71 × 10³⁴ GeV², a shift √Σ = 2.2 × 10¹⁷ GeV. Holding m = 125 GeV needs m₀² = m² − Σ with m²/Σ = 1.56 × 10⁴/(4.71 × 10³⁴) = 3.3 × 10⁻³¹: a cancellation to one part in 3 × 10³⁰.
QED needs no such cancellation. Transversality and the absence of a 1/q² pole make Π_T(q²) = q²Π(q²) identically, so a = 0 with no counterterm; topic 12.02's Lagrangian has no photon-mass term to tune, and its δ₃ counterterm is itself transverse.
Measured check, restoring ħc: ħc/m_γ ≥ 1.973 × 10⁻⁷ eV m ÷ 1 × 10⁻¹⁸ eV = 1.97 × 10¹¹ m, about 1.3 AU. Coulomb's 1/r carries no Yukawa cutoff out to at least the Earth's orbit, the non-relativistic face of a pole at q² = 0.

Answerm_γ² = Z₃a; the bound needs a ≤ 1.1 × 10⁻⁵⁴ GeV², or 7 × 10⁻⁹³ of Λ², 89 orders below α/(4π). A φ⁴ scalar would need its bare mass tuned to one part in 3 × 10³⁰; QED's transversality sets a = 0 exactly.