University Quantum Field Theory I · Renormalising QED: the Electron Self-Energy and the Ward-Takahashi Identity · 12.8

Unphysical photons decouple and ξ drops out, to all orders

Feynman gauge buys a covariant photon with two debts: four polarisations, two of them unphysical, and a parameter ξ that no experiment fixes. This topic pays both at every order with one identity, and marks where the protection ends: at the mass shell, so Σ₂ and Z₂ may depend on ξ while no cross-section can.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A covariant photon is harmless only if no measurable amplitude ever sees its timelike and longitudinal states or its gauge parameter ξ. Topic 8.06 settled that at tree level; here one tool settles it at every order, topic 12.06's Ward-Takahashi identity in its current form. Contract a Green's function with the momentum k of one of its photons and the identity returns contact terms only, each the same function with one charged leg's momentum moved by k.

Those terms have lost the pole at p² = m² that LSZ extracts, so once every charged particle is on shell they vanish and k_μM^μ = 0 exactly, derived. Photon legs give no contact terms because the photon carries no charge, so the result holds whatever the other photons are doing, and that clause does the rest of the work: timelike and longitudinal photons cancel in every polarisation sum, S carries topic 8.03's physical subspace into itself, and ξ, which enters only through q_μq_ν terms that end on conserved currents, leaves every S-matrix element, all shown in outline. The protection stops at the mass shell, where LSZ acts: Σ₂, Z₂ and the off-shell vertex keep their ξ-dependence.

It also needs a regulator that respects the identity, dimensional regularisation and not a momentum cutoff, and gauge bosons without charge. Gluons carry colour, and for them the argument fails without ghost fields, beyond this course.

Simple definition: To every order in QED, an S-matrix element vanishes when one external photon's polarisation is replaced by its momentum, k_μM^μ = 0, and is unchanged when the gauge parameter ξ of the internal photons is changed.
Example: For k = (ω, 0, 0, ω) the identity forces M⁰ = M³. With M^μ = (0.4, 0.6, 0.3i, 0.4), −g_μν M^μ M^ν* = −0.16 + 0.36 + 0.09 + 0.16 = 0.45 = |M¹|² + |M²|²: the timelike and longitudinal terms cancel.

Ward-Takahashi identity with one photon contractedk_μG^μ(kpᵢqᵢ) = −e Σᵢ [G(…, qᵢ − k, …) − G(…, pᵢ + k, …)]

Topic 12.06's identity in the form an amplitude needs: every term has one charged leg pushed off its momentum by k, and photon legs contribute no term at all.

G^μ: Green's function with j^μ inserted, momentum k flowing in; pᵢ in, qᵢ out, in GeV. Overall factor set by j^μ's normalisation.

LSZ factor on a shifted charged leg(p² − m²) × i/((p + k)² − m²) → 0 as p² → m², at fixed k ≠ 0

LSZ keeps only terms with a pole at p² = m² on every charged leg; a contact term has its pole at (p + k)² = m² and dies. Send k → 0 first and it survives: topic 12.07.

p, m: the leg's four-momentum and mass, in GeV. For k² = 0 the denominator is p² − m² + 2p⋅k, finite on shell.

An external photon decouples, all ordersk_μM^μ = 0

So ε → ε + ck leaves every S-matrix element unchanged at every order, not only at tree level, and the photon's √Z₃, a common factor, cannot spoil the zero.

M^μ: LSZ-amputated amplitude with one photon index free. Every charged leg on shell; other photons' polarisations arbitrary.

Timelike cancels longitudinal−g_μν M^μ M^ν* = |M¹|² + |M²|²

The negative-norm timelike term −|M⁰|² cancels the longitudinal +|M³|², so topic 8.07's −g_μν sum holds in loop amplitudes too, for every photon at once.

k = ω(1, 0, 0, 1), so k_μM^μ = ω(M⁰ − M³) = 0. M carries GeV⁽⁴⁻ⁿ⁾ for n external particles.

ξ enters only through q_μq_ν∂D_μν(q)/∂ξ = −iq_μq_ν/(q² + iε)²

Both ends of the q_μq_ν term meet conserved currents. With charged legs on shell all that survives rescales √Z₂, which LSZ divides out, so ∂S/∂ξ = 0.

D_μν = −i[g_μν − (1 − ξ)q_μq_ν/q²]/(q² + iε), Unit 8's propagator. q in GeV, D_μν in GeV⁻².

Σ₂'s ultraviolet pole in a general covariant gaugeΣ₂(p)|pole = (α/4π)(2/ε)[(3 + ξ)m − ξp̸]

At p̸ = m it is 3(α/4π)(2/ε)m for every ξ, so δm is gauge-independent; the slope −ξ(α/4π)(2/ε) is δ₂'s pole, absent in Landau gauge, ξ = 0.

d = 4 − ε. Σ₂ and m in GeV, α = 1/137.036. At ξ = 1 it is topic 12.03's Feynman-gauge pole.

Contract the photon and only contact terms are left

Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Topic 8.06 established k_μM^μ = 0 for an external photon at tree level, checked in topic 9.04, and with it the −g_μν polarisation sum; this topic carries both to every order. By topic 8.06's LSZ step the amplitude is ε_μM^μ with M^μ taken from a Green's function G^μ in which the current j^μ is inserted at momentum k. Topic 12.06's identity, in its current form, evaluates k_μG^μ exactly: a sum of contact terms, one for each external charged leg, each the Green's function without the photon and with that leg's momentum moved by k, pᵢ → pᵢ + k for an incoming electron and qᵢ → qᵢ − k for an outgoing one. Nothing else appears. A contact term comes from the equal-time commutator of j⁰ with a charged field, and A^ν commutes with j⁰, so photon legs contribute none: the other photons may carry any polarisation, physical or not, and need not be contracted with anything.

LSZ keeps poles, and the contact terms have lost theirs

LSZ multiplies by p² − m², or by p̸ − m for a fermion, on every external leg and keeps what survives as the leg goes on shell. G^μ has a pole at pᵢ² = mᵢ², so its residue survives. A contact term with leg i moved has its pole at (pᵢ + k)² = mᵢ² instead, and on shell (pᵢ + k)² − mᵢ² = 2pᵢ⋅k, which is not zero for an on-shell photon of nonzero momentum. Zero times something finite is zero, so every contact term dies and k_μM^μ = 0 to all orders in e, derived, whenever every charged external particle is on shell. The photon's √Z₃, left to this unit by topic 8.06, multiplies every term alike and cannot spoil a zero. The approach is linear: for a 0.511 MeV photon absorbed by an electron 10⁻⁴ off shell in p²/m², the surviving fraction is 5.0 × 10⁻⁵; for a 10 eV photon at the same off-shellness it is 0.72, because the shifted pole then sits only 3.9 × 10⁻⁵ away. Send k → 0 first and the two poles merge: that limit is Ward's identity of topic 12.07, not decoupling.

Timelike and longitudinal photons cancel beyond tree level

For k = ω(1, 0, 0, 1) the identity reads ω(M⁰ − M³) = 0, so M⁰ = M³ at every order. The covariant sum −g_μν M^μ M^ν* = −|M⁰|² + |M¹|² + |M²|² + |M³|² then equals |M¹|² + |M²|²: the negative-norm timelike term cancels the longitudinal one exactly, and topic 8.07's −g_μν replacement, proved there at tree level, holds for loop amplitudes too, shown. Because the identity holds with the other photons' indices free, it applies to each photon while the rest are timelike, longitudinal or transverse, so in e⁺e⁻ → γγ at one loop both sums are replaced at once; the Medium example counts the twelve entries that cancel. The same cancellation runs inside loops. When the optical theorem cuts a Feynman-gauge photon line, the cut numerator −g_μν is a sum over all four polarisations, and the identity on each side of the cut reduces it to the two transverse ones, so unitarity closes on physical states alone, shown in outline.

S keeps physical states physical

Topic 8.03 calls a state physical when L(k)|ψ⟩ = 0, with L(k) = a(k, 0) − a(k, 3) for k along z. Start from a physical |i⟩ and ask whether S|i⟩ is still physical. Its overlap with any state, ⟨f|L(k)S|i⟩, is the amplitude for i to go to f plus the null photon that L(k)† creates, whose wavefunction points along k^μ; the overlap is therefore proportional to k_μM^μ and vanishes. Here ⟨f| runs over the whole indefinite-metric Fock space, unphysical photons included, so the argument needs precisely the clause that the other photons' polarisations are arbitrary; with it, L(k)S|i⟩ = 0 for every k. S therefore maps topic 8.03's physical subspace into itself, shown in outline. Whatever null admixtures S|i⟩ carries are orthogonal to every physical state and add nothing to any probability, so probabilities summed over transverse photons alone add to one, order by order.

ξ reaches the S-matrix only through conserved currents

Changing ξ changes each internal photon propagator by ∂D_μν/∂ξ = −iq_μq_ν/q⁴, so ∂S/∂ξ is a sum of diagrams in which one internal photon line carries q_μq_ν. Open that line: each end is a current insertion contracted with q, summed over every point where it can attach, closed loops included. The identity slides the q_μ end onto an external charged leg i, shifting its momentum by q, and the q_ν end onto a leg j, shifting it by −q. If i ≠ j, two legs have lost their poles and LSZ kills the term. If i = j, the shifts cancel, the pole survives, and the term only rescales leg i by a ξ-dependent constant, the change in √Z₂ that LSZ divides out. So ∂S/∂ξ = 0 at every order, shown in outline, provided all diagrams of the order are summed, the infrared is regulated by a small photon mass, which a conserved current tolerates, and the regulator permits topic 12.06's shift of loop momentum, as dimensional regularisation does and a momentum cutoff does not.

Off shell, ξ stays: Σ₂ and Z₂

Nothing protects a Green's function off shell. With S(p) = 1/(p̸ − m), writing q̸ = (p̸ − m) − (p̸ − q̸ − m) on both sides gives q̸S(p − q)q̸ = (p̸ − m)S(p − q)(p̸ − m) − (p̸ − m) − q̸, so the ξ-dependent part of the one-loop self-energy is (1 − ξ)(p̸ − m) times a function regular at the shell, the lone q̸ integrating to zero; its pole is (1 − ξ)(α/4π)(2/ε)(p̸ − m). Added to topic 12.03's Feynman-gauge pole, (α/4π)(2/ε)m(4 − p̸/m), it gives (α/4π)(2/ε)[(3 + ξ)m − ξp̸], derived. At p̸ = m that is 3(α/4π)(2/ε)m in every gauge: δm, and the pole mass mₑ = 0.51099895 MeV that experiment measures, cannot depend on ξ. The slope can. δ₂'s pole is −ξ(α/4π)(2/ε), absent in Landau gauge, ξ = 0, while its infrared logarithm carries a factor 3 − ξ and vanishes in the Fried-Yennie gauge, ξ = 3, quoted. No gauge removes both, and none needs to: Z₂ is not measured, and the vertex carries the opposite ξ-dependence, as the Hard example shows.

Change one variable at a time

Make the relationship visible.

Interactive model

Gauge parameter ξ2.00

Off-shell point p̸/m1.6

Slide ξ to 0, Landau gauge: the line goes flat, δ₂ loses its pole and both bars vanish, yet the line still crosses the shell at 3. Then park p̸/m at 1 and sweep ξ: the marker never moves, because the gauge-dependent part of Σ₂ carries a factor p̸ − m.

Σ₂ POLE AT CHOSEN p̸/m1.80 (α/4π)(2/ε)m

GAUGE-DEPENDENT PART-0.60 (α/4π)(2/ε)m

δ₂ POLE, THE SLOPE-2.00 (α/4π)(2/ε)

δm POLE, VALUE AT p̸ = m3.00 (α/4π)(2/ε)m

Live interpretationΣ₂ POLE AT CHOSEN p̸/m: 1.80 (α/4π)(2/ε)m. GAUGE-DEPENDENT PART: −0.60 (α/4π)(2/ε)m. δ₂ POLE, THE SLOPE: −2.00 (α/4π)(2/ε). δm POLE, VALUE AT p̸ = m: 3.00 (α/4π)(2/ε)m

Catch the common trap

Explain before calculating.

Misconception check

The tempting shortcut is to use the identity graph by graph: an internal photon's q_μq_ν term meets a conserved current, so drop it from each diagram. It does not vanish diagram by diagram. On shell the one-loop vertex alone keeps a ξ-dependent part, −(1 − ξ)(α/4π)L γ^μ with L = 2/ε − γ_E + ln(4πμ²/λ²), and the self-energy keeps (1 − ξ)(α/4π)L(p̸ − m) plus terms quadratic in p̸ − m; only the sum over every place the q_μ end can attach, the legs' self-energies included through their √Z₂ factors, telescopes to zero. The mirror-image error is to expect every quantity to be gauge-independent and to hunt for a mistake when δ₂ differs between Feynman and Landau gauge. Off-shell Green's functions, Z₂ and the vertex function depend on ξ, and should. Only the S-matrix, with every charged leg on shell, and what is read from it, the pole mass, the charge, F₂ and cross-sections, is protected.

Electron scattering from a heavy nucleus is computed at one loop in Feynman gauge, ξ = 1, and again in Landau gauge, ξ = 0. The vertex correction and δ₂ both come out different in the two gauges. What should be concluded?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA contact term from topic 12.06's identity carries an incoming electron's line at p + k instead of p, k an absorbed on-shell photon of energy ω. Before the LSZ limit take p = (E, 0, 0, 0) with p² = m²(1 + δ), m = 0.511 MeV. LSZ multiplies the term by p² − m², so it arrives scaled by R = (p² − m²)/((p + k)² − m²). Find R for ω = 0.511 MeV at δ = 10⁻² and 10⁻⁴, for ω = 10 eV at δ = 10⁻⁴, and on shell.

With k² = 0 and p at rest, (p + k)² − m² = (p² − m²) + 2p⋅k = m²δ + 2Eω, so R = m²δ/(m²δ + 2Eω). Both terms are squared masses, so R is dimensionless, as a factor multiplying an amplitude must be.
ω = 0.511 MeV, δ = 10⁻²: E = m√1.01 = 0.5135 MeV, m²δ = 2.611 × 10⁻³ MeV² and 2Eω = 0.5248 MeV², so R = 2.611 × 10⁻³/0.5275 = 4.95 × 10⁻³.
δ = 10⁻⁴: m²δ = 2.611 × 10⁻⁵ MeV² and 2Eω = 0.5223 MeV², so R = 5.00 × 10⁻⁵, a hundred times smaller. R falls linearly, R ≈ δm/(2ω) = δ/2 here, because the shifted line's pole sits a fixed distance 2ω/m = 2 away in δ.
ω = 10 eV = 1.0 × 10⁻⁵ MeV, δ = 10⁻⁴: 2Eω = 1.022 × 10⁻⁵ MeV², now smaller than m²δ, so R = 2.611/(2.611 + 1.022) = 0.719. The shifted pole is only 2ω/m = 3.9 × 10⁻⁵ away in δ, and this point is not yet close enough to the shell for the term to fade.
On shell, δ → 0 at fixed ω: m²δ → 0 while 2Eω → 2ωm ≠ 0, so R → 0 for every nonzero ω and the term is gone from the S-matrix element. Reverse the limits, ω → 0 first, and R → 1: the contact term survives, which is why topic 12.07's q → 0 limit yields Ward's identity rather than zero.

AnswerR = 4.95 × 10⁻³ and 5.00 × 10⁻⁵ for the 0.511 MeV photon, 0.719 for the 10 eV photon at δ = 10⁻⁴, and exactly 0 on shell for any ω above zero.

MediumA one-loop code for e⁺e⁻ → γγ, photon 1 along +z with k₁ = ω(1, 0, 0, 1) and photon 2 along −z, returns through one loop, for one spin configuration and in units of e², M⁰⁰ = 0.20, M⁰¹ = 0.30, M⁰² = −0.10, M¹⁰ = 0.10i, M²⁰ = 0.25, M¹¹ = 0.80, M¹² = M²¹ = 0.10i and M²² = −0.60. Supply the seven components with an index 3 from the identity, then compare the doubly covariant polarisation sum with the transverse one.

Photon 1: k₁ lowers to ω(1, 0, 0, −1), so k₁_μM^μν = 0 says that the row μ = 3 of M^μν equals the row μ = 0 for every ν, whatever photon 2 is doing. Photon 2: k₂ = ω(1, 0, 0, −1) lowers to ω(1, 0, 0, 1), so k₂_νM^μν = 0 makes the column ν = 3 minus the column ν = 0.
Column ν = 3: M⁰³ = −0.20, M¹³ = −0.10i, M²³ = −0.25. Row μ = 3: M³⁰ = 0.20, M³¹ = 0.30, M³² = −0.10, and M³³ = M⁰³ = −0.20, which agrees with −M³⁰: the two identities are consistent.
The double sum (−g_μρ)(−g_νσ)M^μν M^ρσ* weights |M^μν|² by +1 when both indices are timelike or both spatial and by −1 when exactly one is timelike. Row by row: μ = 0 gives 0.04 − 0.09 − 0.01 − 0.04 = −0.10; μ = 1 gives −0.01 + 0.64 + 0.01 + 0.01 = 0.65; μ = 2 gives −0.0625 + 0.01 + 0.36 + 0.0625 = 0.37; μ = 3 gives −0.04 + 0.09 + 0.01 + 0.04 = 0.10. Total 1.02.
Transverse sum over the two physical polarisations of each photon: |M¹¹|² + |M¹²|² + |M²¹|² + |M²²|² = 0.64 + 0.01 + 0.01 + 0.36 = 1.02, the same.
The twelve unphysical entries cancel in pairs: the four entries built from M⁰⁰ give +0.04 − 0.04 − 0.04 + 0.04 = 0, which needs photon 1's identity with photon 2 timelike or longitudinal, exactly the clause that photons carry no charge supplies; the photon-1-unphysical entries give −0.10 + 0.10 and the photon-2-unphysical ones −0.0725 + 0.0725. M^μν is dimensionless, as a 2 → 2 amplitude must be.

AnswerRow μ = 3 equals row μ = 0 and column ν = 3 is minus column ν = 0; both the covariant and the transverse sums give 1.02 e⁴, the twelve timelike and longitudinal entries cancelling in pairs.

HardFor electron scattering from a heavy nucleus at one loop, in a covariant gauge with photon mass λ, find the ξ-dependent parts of the on-shell vertex and of δ₂ using q̸ = (p̸ − m) − (p̸ − q̸ − m), and show that they cancel in the amplitude. Evaluate each for Landau against Feynman gauge with λ = 1.0 eV and the pole removed by MS-bar at μ = m, and say what this means for aₑ.

Relative to Feynman gauge the photon propagator gains i(1 − ξ)q_μq_ν f(q²), here f = 1/(q² − λ²)², though any f serves the algebra, so each ξ-dependent piece carries q̸ … q̸ along the electron line. With S(p) = 1/(p̸ − m), on shell ū(p′)q̸S(p′ − q) = ū(p′)[(p̸′ − m) − (p̸′ − q̸ − m)]S(p′ − q) = −ū(p′), and likewise S(p − q)q̸u(p) = −u(p).
Vertex: (−ie)³, two electron propagators' factors of i and the i of the propagator term multiply to e³, so −ie ū δΓ^μ u = e³(1 − ξ) ū γ^μ u ∫d⁴q/(2π)⁴ f. With that integral equal to iL/16π², L = 2/ε − γ_E + ln(4πμ²/λ²), the ξ-dependent vertex is −(1 − ξ)(α/4π)L γ^μ and nothing else.
Self-energy: −iΣ_ξ = e²(1 − ξ)∫d⁴q/(2π)⁴ f q̸S(p − q)q̸, and q̸S(p − q)q̸ = (p̸ − m)S(p − q)(p̸ − m) − (p̸ − m) − q̸. The q̸ integrates to zero and the first term vanishes, with its slope, at p̸ = m, so Σ_ξ = (1 − ξ)(α/4π)L(p̸ − m) + … and the ξ-dependent part of δ₂ = dΣ/dp̸ is +(1 − ξ)(α/4π)L.
The amplitude carries √Z₂ per electron leg, together 1 + δ₂, plus the vertex, so its ξ-dependent part is ū γ^μ u [(1 − ξ) − (1 − ξ)](α/4π)L = 0: pole, μ and λ dependence alike. α and L are pure numbers, as a correction to a dimensionless form factor must be, and at ξ = 1 both pieces vanish, the Feynman-gauge limit.
Numbers, ξ = 0 against ξ = 1: MS-bar at μ = m leaves L = ln(m²/λ²) = 2 ln(5.11 × 10⁵) = 26.29, so (α/4π)L = 5.807 × 10⁻⁴ × 26.29 = 0.0153. Moving to Landau gauge raises δ₂ by 0.0153 and lowers the vertex by 0.0153, each 13 times the anomalous moment itself.
The ξ-dependent part is proportional to γ^μ alone and touches F₁ only: F₂(0) = α/2π = 1.1614 × 10⁻³, Unit 14's Feynman-gauge result, quoted, holds in every covariant gauge, derived here at one loop. The measured aₑ = 1.15965 × 10⁻³, quoted, lies 0.15% below it, the size of the α² term, not of any gauge effect.

AnswerThe vertex's ξ-dependent part −(1 − ξ)(α/4π)L γ^μ cancels δ₂'s +(1 − ξ)(α/4π)L; from Feynman to Landau gauge each moves by 0.0153, yet F₂(0) = α/2π = 1.161 × 10⁻³ in both.