University Quantum Field Theory I · Renormalising QED: the Electron Self-Energy and the Ward-Takahashi Identity · 12.6

The Ward-Takahashi identity derived to all orders

Topic 9.04 checked current conservation one tree diagram at a time. Here it becomes an exact relation between the full vertex and the full electron propagator, proved for every diagram at once, and you will see what it costs, a regulator that lets loop momenta shift, and what it buys: δ₁ = δ₂.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Gauge invariance is a property of the action; the Ward-Takahashi identity is what it becomes for Green's functions. Insert a photon of momentum q anywhere in a diagram and contract its index with q_μ. On a free electron line the contracted vertex is a difference, S₀(p + q)(−ieq̸)S₀(p) = e[S₀(p) − S₀(p + q)], so along an open charged line successive insertions cancel in pairs and only the two ends survive: the diagram without the photon, minus the same diagram with q carried through it.

Around a closed loop there are no ends, and the insertions cancel outright once the loop momentum is shifted by q, a step dimensional regularisation allows and a momentum cutoff does not. Summed over every diagram, the bookkeeping gives −iq_μΓ^μ(p + q, p) = S⁻¹(p + q) − S⁻¹(p) for the 1PI vertex and the full propagator, off shell, in every covariant gauge and at every order, and the divergence of ⟨0|T j^μ ψ ψ̄|0⟩ gives it again without diagrams. Its weight lies in what it constrains: the vertex's ultraviolet divergence must mirror the divergence in the p̸ part of Σ, so Z₁/Z₂ is finite, δ₁ and δ₂ cancel one and the same divergence, and δₘ is left free.

It holds only as far as the regulator respects shifts: with a cutoff, closed loops leave surface terms, growing as Λ² for the two-vertex loop and finite for the four-vertex one, and a cutoff routed through the electron rather than the photon spoils it already at one loop.

Simple definition: The Ward-Takahashi identity, −iq_μΓ^μ(p + q, p) = S⁻¹(p + q) − S⁻¹(p), says that the full electron-photon vertex contracted with the photon's momentum equals the difference of the full inverse electron propagators at its two ends, exactly and to all orders.
Example: At tree level S⁻¹(p) = −i(p̸ − m), and both sides are −iq̸. At one loop in Feynman gauge, the −(α/4π)(2/ε)p̸ in Σ₂'s pole forces a vertex pole +(α/4π)(2/ε)γ^μ, so δ₁ and δ₂ share the pole −(α/4π)(2/ε) = −1.16 × 10⁻³/ε.

Contracting one vertex on a free lineS₀(p + q)(−ieq̸)S₀(p) = e[S₀(p) − S₀(p + q)], with S₀(p) = i/(p̸ − m)

Derived from q̸ = (p̸ + q̸ − m) − (p̸ − m). It holds for integrands, off shell and in any gauge, and it is the only algebra the all-orders proof needs.

p, q, m in GeV and S₀ in GeV⁻¹; e = √(4πα) = 0.303, dimensionless; ħ = c = 1, metric (+, −, −, −)

Telescoping along an open linesum over j = 0 … V of e(Dⱼ₊₁ − Dⱼ) = e(D_(V+1) − D₀) = e[G(p) − G(p + q)]

Derived. Interior terms cancel in pairs. With no loop momentum routed along the line itself, D₀ is G(p + q) point by point, so no shift is needed.

V vertices and V + 1 propagators on the line; Dⱼ carries q on propagators j to V; G a two-point diagram, in GeV⁻¹

Closed loops need a shift of loop momentumsum round a closed loop = e∫d⁴l [F(l) − F(l + q)], zero if l → l + q is allowed

Derived. A loop has no ends, so telescoping leaves only a shifted integral, which vanishes in any translation-invariant regularisation.

F the loop's integrand without the photon, l its loop momentum in GeV; the shift is exact in dimensional regularisation

What a Euclidean cutoff leaves behind∫_{(|l| ≤ Λ)} d⁴l [F(l + a) − F(l)]^ν = a^ν(π²/2)Λ⁴/(Λ² + m²)ⁿ + O(a³), F^ν = l^ν/(l² + m²)ⁿ

Derived by Gauss's theorem. It grows as Λ² for n = 1, tends to π²/2 for n = 2 and dies for n = 3: a cutoff spoils the two- and four-vertex loops.

Euclidean l, a, m, Λ in GeV; the result has mass dimension 5 − 2n; at n = 1 the O(a³) term is −(π²/6)a²a^ν

The Ward-Takahashi identity−iq_μΓ^μ(p + q, p) = S⁻¹(p + q) − S⁻¹(p)

Derived to all orders, off shell, for every ξ. Equivalently q_μδΓ^μ = −[Σ(p + q) − Σ(p)]; between on-shell spinors it gives topic 9.04's zero.

Γ^μ = γ^μ + δΓ^μ the 1PI vertex, dimensionless; S(p) = i/(p̸ − m₀ − Σ(p)) in GeV⁻¹; q enters along the photon

The renormalised identity and the counterterms−iq_μΓ_R^μ(p + q, p) = (Z₁/Z₂)[S_R⁻¹(p + q) − S_R⁻¹(p)]

Derived. Finite Γ_R and S_R make Z₁/Z₂ finite, so δ₁ and δ₂ share their poles, −(α/4π)(2/ε) each in Feynman gauge; on shell, Z₁ = Z₂ is topic 12.07's.

S_R = S/Z₂ and Γ_R^μ = Z₁Γ^μ from topic 12.02's rescaling; Z₁ and Z₂ dimensionless; δₘ cancels from the difference

From one tree diagram to every diagram

Work in natural units, ħ = c = 1, with the metric (+, −, −, −), and write ultraviolet poles for dimensional regularisation in d = 4 − ε, as topic 12.03 does. Topic 9.04 proved k_μM^μ = 0 for on-shell external electrons by telescoping insertions along an open line; here the ends go off shell and loops appear. With S₀(p) = i/(p̸ − m) and the vertex −ieγ^μ, the split q̸ = (p̸ + q̸ − m) − (p̸ − m) gives (p̸ + q̸ − m)⁻¹q̸(p̸ − m)⁻¹ = (p̸ − m)⁻¹ − (p̸ + q̸ − m)⁻¹, and with the factors of i restored S₀(p + q)(−ieq̸)S₀(p) = e[S₀(p) − S₀(p + q)], derived. Nothing here needs p or p + q on shell, and nothing needs a gauge: the internal photons, whose propagators carry all the dependence on ξ, are never touched. With the propagators stripped, this line is the identity at lowest order, −iq̸ = −i(p̸ + q̸ − m) + i(p̸ − m). Dimensions agree: S₀ is GeV⁻¹ and e is a pure number, 0.303, so both sides are GeV⁻¹.

Open lines telescope, with no shift needed

Take any diagram G of the electron two-point function. Its open line has V vertices and V + 1 free propagators, numbered 0 to V from the incoming end. Inserting the photon on propagator j adds q to it and to every propagator after it, and contracting gives e(Dⱼ₊₁ − Dⱼ), where Dⱼ is the diagram with q carried by propagators j to V, D_(V+1) the diagram with no q at all. Summed over the V + 1 insertion points, each interior Dⱼ appears once with each sign, leaving e(D_(V+1) − D₀). D₀ has q on every propagator of the line, which is G evaluated at p + q, so the sum is e[G(p) − G(p + q)], derived. If no loop momentum is routed along the line itself, each running through a photon or a closed loop, D₀ and G(p + q) are the same integrand point by point: no change of variable, so the step survives a cutoff on the photon momenta. For the one-loop Σ₂ diagram, V = 2: e(D₃ − D₂) + e(D₂ − D₁) + e(D₁ − D₀) = e(D₃ − D₀).

Closed loops cancel only if the loop momentum may shift

A closed electron loop with N vertices has N propagators and no ends. Inserting the photon on each in turn and contracting telescopes all the way round, leaving e∫d⁴l [F(l) − F(l + q)], with F the loop's integrand without the photon. That vanishes only if l → l + q is a legitimate change of variable. Dimensional regularisation is translation invariant by construction, so the closed-loop terms are zero exactly. A cutoff is not. With |l| ≤ Λ in Euclidean space, Gauss's theorem turns the first-order difference into a surface integral: for a shift a and F^ν = l^ν/(l² + m²)ⁿ it is a^ν(π²/2)Λ⁴/(Λ² + m²)ⁿ, derived. At n = 1, the integrand left by contracting a two-vertex loop, it grows as Λ²: 488.6 in units of m² per unit shift at Λ/m = 10. Furry's theorem disposes of odd loops in pairs, so the loops a cutoff spoils are those with two and four vertices; that the photon stays massless, topic 12.09, therefore needs a gauge-invariant regulator.

Summing every diagram: the full vertex and propagator

Let G^μ(p + q, p) be the sum of all diagrams with the photon inserted somewhere in the electron's two-point function. Each diagram is a chain of free propagators and 1PI self-energy blobs, so the insertion lands on a link of the chain, giving γ^μ, or inside a blob, giving a vertex correction: G^μ = S(p + q)(−ieΓ^μ)S(p), with S(p) = i/(p̸ − m₀ − Σ(p)) and Γ^μ = γ^μ + δΓ^μ. The two previous steps give q_μG^μ = e[S(p) − S(p + q)]. Multiplying by S⁻¹(p + q) on the left and S⁻¹(p) on the right, −iq_μΓ^μ(p + q, p) = S⁻¹(p + q) − S⁻¹(p), derived to all orders, off shell and for every ξ; written out, q_μδΓ^μ = −[Σ(p + q) − Σ(p)]. At the physical mass ū(p′)S⁻¹(p′) = 0 = S⁻¹(p)u(p), so ū(p′)q_μΓ^μu(p) = 0: topic 9.04's statement to all orders, and the reason the on-shell vertex has form factors F₁ and F₂ but none multiplying q^μ, which would leave q²F₃ ≠ 0.

The same identity from the conserved current

Current conservation gives a second route. j^μ = ψ̄γ^μψ obeys ∂_μj^μ = 0 by the full equations of motion, and the equal-time anticommutator (ψ(x), ψ†(y)) = δ³(x − y), quoted from Advanced QM, gives [j⁰(z), ψ(x)] = −δ³(z − x)ψ(x) and [j⁰(z), ψ̄(y)] = +δ³(z − y)ψ̄(y). Differentiate ⟨0|T j^μ(z)ψ(x)ψ̄(y)|0⟩ with respect to z^μ. On j^μ the derivative gives zero; on the step functions of the time ordering it gives δ(z⁰ − x⁰) and δ(z⁰ − y⁰) times those commutators. So ∂_μ⟨0|T j^μ(z)ψ(x)ψ̄(y)|0⟩ = [δ⁴(z − y) − δ⁴(z − x)]⟨0|Tψ(x)ψ̄(y)|0⟩: contact terms only, outlined. Fourier transforming with q entering at z turns ∂_μ into iq_μ and the contact terms into S(p + q) − S(p), and amputating both legs gives the identity again. This route needs no perturbation theory, but it needs what the diagrams needed: a regulator that keeps the current conserved.

What the identity says about the counterterms

Topic 12.02's rescaling gives S_R = S/Z₂ and Γ_R^μ = Z₁Γ^μ, so −iq_μΓ_R^μ = (Z₁/Z₂)[S_R⁻¹(p + q) − S_R⁻¹(p)]. Both sides are finite, so Z₁/Z₂ is finite to all orders; in minimal subtraction, where counterterms are pure poles, Z₁ = Z₂ exactly. Order by order, the counterterm vertex adds −iq̸δ₁ to the left side and i(p̸δ₂ − δₘ) adds −iq̸δ₂ to the right, its mass term cancelling between p + q and p: the renormalised identity keeps its bare form only if δ₁ = δ₂, and it says nothing about δₘ, nor about δ₃, which the photon's own identity governs. At one loop the vertex must mirror the p̸ part of topic 12.03's Σ₂ pole, (α/4π)(2/ε)(4m − p̸): the identity predicts δΓ^μ's pole, +(α/4π)(2/ε)γ^μ, before Unit 14 computes it, and gives δ₁ and δ₂ the same pole, −(α/4π)(2/ε) in Feynman gauge and none in Landau gauge. The free-field limit e → 0 sends every δ to zero. That the on-shell constants agree too, and so charge renormalises alike for every particle, measured to about 10⁻²¹ e, is topic 12.07's.

Change one variable at a time

Make the relationship visible.

Interactive model

Power n in l^ν/(l² + m²)ⁿ1

Cutoff Λ/m10

Set n = 1 and drag Λ/m from 1 to 30: the leftover grows a hundredfold per decade of cutoff, the two-vertex loop's problem. At n = 2 it settles onto π²/2, finite but never zero; at n = 3 it dies as Λ⁻², the only case a cutoff gets right.

SURFACE TERM S488.5943 m⁴⁻²ⁿ

GROWTH FROM Λ TO 10Λ100.990 ×

LARGE-Λ SLOPE 4 − 2n2

Live interpretationSURFACE TERM S: 488.5943 m⁴⁻²ⁿ. GROWTH FROM Λ TO 10Λ: 100.990 ×. LARGE-Λ SLOPE 4 − 2n: 2

Catch the common trap

Explain before calculating.

Misconception check

A common slip is to read the identity as q_μΓ^μ = 0, the full vertex transverse to the photon's momentum, as current conservation seems to suggest. That is the on-shell statement only. Off shell the contracted vertex is a difference of inverse propagators, not zero: at tree level q_μγ^μ = q̸, and at one loop q_μδΓ^μ = −[Σ(p + q) − Σ(p)], whose pole is (α/4π)(2/ε)q̸ in Feynman gauge. If q_μΓ^μ vanished at all momenta, the identity would force S⁻¹(p + q) = S⁻¹(p) for every q, an electron propagator with no momentum dependence at all. The zero appears only between spinors at the physical mass, where ū(p′)S⁻¹(p′) and S⁻¹(p)u(p) vanish, and it is that on-shell zero, not an off-shell one, that topic 12.08 turns into the decoupling of unphysical photons.

Topic 12.02's renormalised Lagrangian carries δ₁, δ₂ and δₘ for the electron and δ₃ for the photon. With a regulator that respects shifts of loop momentum, what does the Ward-Takahashi identity for the electron vertex demand of these counterterms?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyIn a general covariant gauge the ultraviolet pole of the one-loop self-energy is (α/4π)(2/ε)[(3 + ξ)m − ξp̸], quoted. Use the Ward-Takahashi identity to find the pole of the one-loop vertex correction and the minimal-subtraction δ₁ for ξ = 0, 1 and 3, and compare each with δ₂.

At one loop the identity reads q_μδΓ^μ(p + q, p) = −[Σ₂(p + q) − Σ₂(p)]. The mass term (3 + ξ)m does not depend on momentum and cancels in the difference, leaving the pole −(α/4π)(2/ε)(−ξq̸) = +ξ(α/4π)(2/ε)q̸.
The vertex has superficial degree 0, so its pole is momentum independent, and by Lorentz covariance and parity it is a constant times γ^μ. Matching q_μ times that constant γ^μ to ξ(α/4π)(2/ε)q̸ gives the pole ξ(α/4π)(2/ε)γ^μ.
The counterterm −ieγ^μδ₁ cancels it, so δ₁ = −ξ(α/4π)(2/ε). The p̸ pole of Σ₂ is cancelled by the ip̸δ₂ in the propagator counterterm, so δ₂ = −ξ(α/4π)(2/ε) too. The mass pole fixes δₘ alone, which the identity never sees.
With α/4π = 1/(4π × 137.036) = 5.807 × 10⁻⁴: ξ = 0 (Landau) gives δ₁ = δ₂ = 0; ξ = 1 (Feynman) gives −5.807 × 10⁻⁴ × (2/ε); ξ = 3 gives −1.742 × 10⁻³ × (2/ε).
Checks: the δ's are dimensionless, as renormalisation constants must be; they vanish in the free limit e → 0; and ξ = 1 reproduces topic 12.05's pole in δ₂.

AnswerVertex pole ξ(α/4π)(2/ε)γ^μ and δ₁ = δ₂ = −ξ(α/4π)(2/ε): 0 for ξ = 0, −5.81 × 10⁻⁴ × (2/ε) for ξ = 1 and −1.74 × 10⁻³ × (2/ε) for ξ = 3.

MediumWith a Euclidean cutoff |l| ≤ Λ, the closed-loop step leaves ∫d⁴l [F(l + a) − F(l)]. For the power-counting stand-ins F^ν = l^ν/(l² + m²)ⁿ with n = 1, 2 and 3, find the first-order surface term Sₙ at Λ/m = 10 and 100, its behaviour as Λ → ∞, and the QED closed loops each n stands for.

To first order in a the difference is a^μ∫d⁴l ∂_μF^ν, which Gauss's theorem turns into a^μΛ³∮dΩ n_μn^ν Λ/(Λ² + m²)ⁿ over the sphere |l| = Λ. With ∮dΩ = 2π² and n_μn^ν averaging to δ_μν/4, this is a^ν Sₙ with Sₙ = (π²/2)Λ⁴/(Λ² + m²)ⁿ, of mass dimension 4 − 2n.
n = 1: S₁ = 4.9348 × 10⁴/101 = 488.6 at Λ/m = 10 and 4.9348 × 10⁸/10001 = 4.934 × 10⁴ at Λ/m = 100, in units of m²: it grows as Λ². A direct 4D integration at a = m gives 486.9 in units of m³, the difference being the Λ-independent −(π²/6)a³ = −1.645.
n = 2: S₂ = 4.838 at Λ/m = 10 and 4.934 at 100, dimensionless, approaching π²/2 = 4.935: finite, but never zero however large Λ is.
n = 3: S₃ = 4.790 × 10⁻² at Λ/m = 10 and 4.933 × 10⁻⁴ at 100, in units of m⁻²: it falls as Λ⁻², so the shift is safe.
F^ν has superficial degree 5 − 2n, and a loop with N propagators has 4 − N. So n = 1 matches the one-propagator integrand left by contracting a two-vertex loop, n = 2 the three-propagator integrands left by contracting a four-vertex loop, and n = 3 loops of six vertices or more. Dimensional regularisation gives zero in all three cases.

AnswerS₁ = 488.6 and 4.934 × 10⁴ in units of m², growing as Λ²; S₂ = 4.838 and 4.934, tending to π²/2 = 4.935; S₃ = 4.79 × 10⁻² and 4.93 × 10⁻⁴ in units of m⁻², falling as Λ⁻². Only n = 3 is safe with a cutoff.

HardCut off the one-loop electron self-energy in Feynman gauge at |l_E| ≤ Λ on the electron's loop momentum l = p − k instead of on the photon's k. Find how Σ₂ changes as Λ → ∞, how far the Ward-Takahashi identity then fails, and what happens to δ₁ − δ₂.

In four dimensions, Σ₂(p) = −ie²∫d⁴k/(2π)⁴ G(k), with G = [2k̸ − 2p̸ + 4m]/[((p − k)² − m²)(k² − m_γ²)], k the photon's momentum. Its superficial degree is 1: at large k, G → 2k̸/(k²)².
The electron cutoff integrates k over a ball centred on p, so the change is ∫d⁴k [G(k + p) − G(k)] = p^μ∫d⁴k ∂_μG plus higher terms. Those involve ∂G, which falls as k⁻⁴ against a surface growing only as Λ³, so they vanish as Λ → ∞.
Only the leading term survives, 2γ^νk_ν/(k² − M²)² with M any mass. By covariance ∫d⁴k ∂_μ[k_ν/(k² − M²)²] = Cg_μν; its divergence is −4M²/(k² − M²)³, which the Wick rotation integrates to 2iπ², so 4C = 2iπ² and C = iπ²/2, whatever M is.
So the change in Σ₂ is −ie²/(16π⁴) × 2γ^νp^μ(iπ²/2)g_μν = e²p̸/(16π²) = (α/4π)p̸ = 5.81 × 10⁻⁴ p̸, independent of Λ, m and m_γ. A direct 4D integration of the Euclidean k̸ part of G, with m = |p| and m_γ = 0.3|p|, at Λ = 100|p| gives 0.99989 of this.
The photon-routed Σ₂ obeys the identity exactly, by integrand-level telescoping, and the vertex, only logarithmically divergent, is the same in either routing. With the electron cutoff, q_μδΓ^μ + Σ₂(p + q) − Σ₂(p) = (α/4π)q̸ instead of 0, δ₂ grows by α/4π and δ₁ − δ₂ = −5.81 × 10⁻⁴.
Checks: the defect has Σ's mass dimension, 1, and vanishes as e → 0; in dimensional regularisation the two routings are one integral, so it is exactly zero. At α/4π it is half the entire one-loop anomalous moment α/2π = 1.16 × 10⁻³, itself 0.15% above the measured aₑ = 1.15965 × 10⁻³.

AnswerΣ₂ rises by (α/4π)p̸ = 5.81 × 10⁻⁴ p̸ as Λ → ∞, a shift that never goes away; the identity fails by (α/4π)q̸ and δ₁ − δ₂ = −5.81 × 10⁻⁴. Routing the cutoff through the photon, or using dimensional regularisation, removes the defect.