University Quantum Field Theory I · Covariant Quantisation of the Electromagnetic Field · 8.6

The Ward identity for external photons: k_μM^μ = 0

Topic 8.05 demanded k_μM^μ = 0 of every external photon; here QED delivers it. You will reduce a photon with LSZ, watch the amputated leg become the current, find the contact terms conservation leaves inside time-ordered products, and see why they vanish on shell. The payoff is licence to choose your gauge.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Topic 8.05 showed that Lorentz invariance needs every amplitude with an external photon to vanish when ε is replaced by k; this topic shows that QED supplies it, and says exactly when. Repeat topic 4.04's LSZ reduction for the photon: the field is massless, so □ replaces □ + m², the physical polarisation is projected out with a minus sign because ε⋅ε* = −1, and the residue √Z₃ is 1 at tree level and Unit 12's business beyond. In Feynman gauge the field equation is □A^μ = j^μ, so the □ that amputates the photon leg turns A^μ into the current: every amplitude takes the form ε_μM^μ, with M^μ a matrix element or correlator of j^μ.

Conservation then does the work, but not naively. Between exact momentum eigenstates ∂_μ becomes ik_μ and k_μM^μ = 0 at once; in the time-ordered correlators that Feynman diagrams compute, ∂₀ also hits the step functions and leaves equal-time commutators, contact terms that push the photon's momentum into one charged leg. Each lacks that leg's pole, so LSZ sends it to zero once the charged particles are on shell.

Two dividends follow: ε → ε + ck changes no physical amplitude, and the ξ-dependent k_μk_ν part of an internal photon dies where it meets a complete, on-shell sum. The identity fails for single diagrams, for off-shell Green's functions and wherever a charged line is left off shell.

Simple definition: The Ward identity for an external photon says that the amplitude ε_μM^μ obeys k_μM^μ = 0 whenever every charged external particle is on its mass shell, so adding any multiple of k to the polarisation changes nothing.
Example: In X → φ⁺φ⁻γ the two diagrams contract with k to −eg and +eg. Replacing ε by ε + (5 GeV⁻¹)k moves them from 15.89 to 10.89 and from 3.56 to 8.56, in units of eg GeV⁻¹, while their sum stays at 19.45.

LSZ for an external photonincoming: −i ε_μ ∫ d⁴x e^(−ik⋅x) □ₓ ⟨Ω|T A^μ(x) ⋯|Ω⟩outgoing: ε_μ → ε*_μ, e^(−ik⋅x) → e^(+ik⋅x)

Topic 4.04's reduction with m = 0: □ cancels the pole at k² = 0, and contracting with −ε picks one physical photon out of A's four Feynman-gauge components.

k² = 0; ε⋅ε* = −1 gives the minus sign; [A] = 1 in mass units; √Z₃ = 1 at tree level, Unit 12 beyond. Derived.

The amputated photon leg is the current⟨f|S|ik, ε⟩ = −i ε_μ ∫ d⁴x e^(−ik⋅x) ⟨f|j^μ(x)|i⟩, so M = ε_μM^μ with M^μ = −⟨f|j^μ(0)|i⟩

Exact to first order in this photon and to all orders in everything else. Lowest-order perturbation theory with −j⋅A gives the same, which fixes the sign.

Feynman gauge, □A^μ = j^μ; j^μ the full Heisenberg current, [j] = 3; ⟨f|, |i⟩ the other particles, on shell. Derived.

Conservation inside T: Ward-Takahashi∂_μ⟨T j^μ(x) Φ₁(x₁)⋯Φₙ(xₙ)⟩ = −e Σᵢ qᵢ δ⁴(x − xᵢ) ⟨T Φ₁(x₁)⋯Φₙ(xₙ)⟩

∂₀ also hits the step functions of T and leaves contact terms. In momentum space each feeds k into one charged leg, which then has no pole at pᵢ² = m².

qᵢ the charge, in units of e, that Φᵢ removes: +1 for φ, −1 for φ†; e dimensionless. Derived in outline.

Tree-level vertex identityk_μ(2p + k)^μ = [(p + k)² − m²] − [p² − m²]

A vertex contracted with k cancels one of its two propagators or the other: the diagram-level source of every contact term. Topic 9.04 repeats it with k̸.

p and p + k the charged scalar's momenta at the vertex −ie(2p + k)^μ, in GeV; true for k² ≠ 0 as well. Derived.

The Ward identity and the freedom it buysk_μM^μ = 0 ⟹ M(ε + c k) = M(ε)

Only the sum over all insertions is invariant: in X → φ⁺φ⁻γ each diagram shifts by ∓c eg. Choose c to make a diagram vanish outright.

Every charged external particle on shell; other photons' polarisations arbitrary; c any number, of dimension GeV⁻¹. Shown.

Where ξ meets a sub-amplitudeM_μ D^μν N_ν = −i M⋅N/(k² + iε) + i(1 − ξ)(k⋅M)(k⋅N)/(k² + iε)²

The ξ term dies once either end is a complete sum of attachments with its charged lines on shell; a single diagram or an off-shell leg keeps it.

M^μ, N^μ: sub-amplitudes at the two ends of an internal photon of momentum k, in GeV; ξ topic 8.04's gauge parameter. Shown.

LSZ for a massless vector: □ in place of □ + m²

Natural units, ħ = c = 1, and the metric (+, −, −, −) throughout. Topic 4.04 reduced a scalar leg with i∫d⁴x e^(−ip⋅x)(□ + m²). The photon differs twice. Its mass is zero, so □ alone multiplies the Fourier transform by −k² and cancels the pole at k² = 0. And A^μ has four components: in Feynman gauge, topic 8.02's commutator −g_μν makes a physical photon of polarisation ε, with k⋅ε = 0 and ε⋅ε* = −1, come out of A^μ when it is contracted with −ε for an incoming photon or −ε* for an outgoing one, the minus undoing that negative norm. So an incoming photon contributes −i ε_μ∫d⁴x e^(−ik⋅x) □ₓ⟨Ω|T A^μ(x)⋯|Ω⟩, and an outgoing one the same with ε* and e^(+ik⋅x), derived. The residue of the photon pole, Z₃, is 1 at tree level; its loop value, and why the pole stays at k² = 0, are Unit 12's. For k = (1, 0, 0, 1) GeV and ε = (0, 1, 0, 0), ε⋅ε = −1 and k⋅ε = 0: the reduction picks one transverse photon and nothing timelike or longitudinal.

□A^μ = j^μ turns the photon leg into the current

With L = −¼F_μνF^μν − ½(∂⋅A)² − j_μA^μ, Feynman gauge's field equation is ∂_μF^μν + ∂^ν(∂⋅A) = j^ν, and the two ∂^ν(∂⋅A) terms cancel to leave □A^ν = j^ν, derived. Inside the reduction formula the □ therefore acts as a substitution: up to equal-time terms with no photon pole, □⟨T A^μ(x)⋯⟩ = ⟨T j^μ(x)⋯⟩. Reduce the other legs too and only j^μ is left between the in and out states. Since ⟨f|j^μ(x)|i⟩ = e^{(i(p_f − pᵢ)⋅x)}⟨f|j^μ(0)|i⟩, the x integral gives (2π)⁴δ⁴(p_f − pᵢ − k), and M = ε_μM^μ with M^μ = −⟨f|j^μ(0)|i⟩, derived. It is exact to first order in this photon and to all orders in every other interaction, because j^μ is the full Heisenberg current; lowest-order perturbation theory with −j⋅A gives the same, fixing the sign. Dimensions agree: [j] = 3 and a one-particle state has dimension −1, so a charged scalar's ⟨p′|j^μ(0)|p⟩ has dimension 1, as its tree value e(p + p′)^μ, quoted from UAQM-09.03, does.

Conservation leaves contact terms, and LSZ removes them

Between exact momentum eigenstates the proof takes one line: ∂_μ⟨f|j^μ(x)|i⟩ = i(p_f − pᵢ)_μ⟨f|j^μ(x)|i⟩ with p_f − pᵢ = k, so ∂_μj^μ = 0 gives k_μM^μ = 0. The work hides in the word eigenstates. Feynman diagrams compute time-ordered correlators with every leg off shell, and there ∂₀ also differentiates the step functions of T, producing δ(x⁰ − xᵢ⁰)[j⁰(x), Φᵢ(xᵢ)]. The local form of charge conservation, [j⁰(x), Φᵢ(y)] = −e qᵢ δ³(x − y)Φᵢ(y) at equal times, turns these into the Ward-Takahashi identity, derived in outline, with topic 2.04's T* wherever couplings carry derivatives. In momentum space each contact term is the photonless Green's function with k fed into leg i, so that leg's propagator has its pole at (pᵢ ± k)² = m², not at pᵢ² = m². LSZ multiplies by pᵢ² − m² and sets it to zero, and the term dies. The photon carries no charge, so [j⁰, A^ν] = 0 and other photons add no contact terms: the identity holds whatever their polarisations. Topic 9.04 checks it along a fermion line and topic 12.08 to all orders.

Anatomy at tree level: two diagrams, two contact terms

Let a neutral spin-zero X couple as gXφ†φ to a scalar of charge e and mass m. Then X → φ⁺φ⁻γ has two diagrams at order eg, the photon leaving the φ⁺ (A) or the φ⁻ (B), and no seagull. Each uses the vertex −ie(p + p′)^μ along the charge arrow, the matrix element of UAQM-09.03's Noether current, quoted here and derived in topic 8.09. Contract with k: the vertex identity cancels one propagator or the other, so k⋅G_A = e[G₀(p₁, p₂) − G₀(p₁ + k, p₂)] and k⋅G_B = −e[G₀(p₁, p₂) − G₀(p₁, p₂ + k)], with G₀ the photonless Green's function, derived. The G₀(p₁, p₂) pieces cancel in the sum because the charges +1 and −1 do, leaving k⋅G = e[G₀(p₁, p₂ + k) − G₀(p₁ + k, p₂)]: one contact term per charged leg. Amputated, k⋅M = eg[δ₁/((p₁ + k)² − m²) − δ₂/((p₂ + k)² − m²)] with δᵢ = pᵢ² − m², zero on shell. Yet on shell each diagram alone gives k⋅M_A = −eg and k⋅M_B = +eg, at any momenta and photon energy: no single diagram obeys the identity.

Choosing the gauge: external ε and internal ξ

With k_μM^μ = 0, M(ε + ck) = M(ε) + c k_μM^μ = M(ε) for any c, shown, so any polarisation equal to a physical one up to a multiple of k gives the same amplitude. Use the freedom: c = −ε⋅p₁/(p₁⋅k) makes ε′⋅p₁ = 0 and deletes diagram A of X → φ⁺φ⁻γ outright, moving its whole contribution into B, as the second example shows. Internal photons inherit the protection. Topic 8.04's propagator joins two ends, and M_μD^μνN_ν carries i(1 − ξ)(k⋅M)(k⋅N)/(k² + iε)², shown. The vertex identity never used k² = 0: for a virtual photon, k⋅(2p₁ + k) still equals (p₁ + k)² − m² with p₁ on shell, so k⋅M_A = −eg and k⋅M_B = +eg exactly, and the ξ term dies at that end even if the far end were not conserved. Where it stops: a single diagram keeps its ξ dependence, and so does any end whose charged lines are internal and off shell, as in loops, where only the full set of diagrams at that order, with the external legs on shell, is ξ independent (topic 12.08).

Checks: soft and slow limits, and a kaon decay

Dimensions first: [g] = 1 for a gXφ†φ coupling and e is pure, so a three-body amplitude, of dimension 0, scales as eg/ω, and k⋅M = ∓eg carries the one power of mass that k adds. With ε⋅k = 0 the exact amplitude is ε⋅M = eg[ε⋅p₂/(p₂⋅k) − ε⋅p₁/(p₁⋅k)], the form of topic 8.05's soft factor, here exact at every ω because X's coupling carries no momentum. For slow pions p⋅k ≈ mω, and ε⋅M → (eg/ω) ε⃗⋅(v⃗₁ − v⃗₂), derived: the classical dipole radiation of a suddenly created current e(v⃗₁ − v⃗₂). Now a measurement. K_S → π⁺π⁻, a spin-zero particle going to two, has a single constant amplitude g, so this is the full inner-bremsstrahlung prediction for K_S → π⁺π⁻γ, with g cancelling against the two-body width. Summing the two transverse polarisations and integrating numerically over the Dalitz plot gives Γ(π⁺π⁻γ)/Γ(π⁺π⁻) = 6.99 × 10⁻³ for photons above 20 MeV in the kaon's rest frame, computed here; Fermilab's E731 measured (7.10 ± 0.22) × 10⁻³, quoted.

Change one variable at a time

Make the relationship visible.

Interactive model

φ⁺ angle to the photon θ60 °

Gauge shift cω0.5

φ⁻ off-shellness δ/(2p₂⋅k)0.00

Keep δ at 0 and drag cω: the open markers slide down diagram A's line and up diagram B's by equal amounts while the filled marker on the solid sum keeps its height, because k⋅M = 0. Then take the φ⁻ off shell with δ: diagram B's slope leaves +1 and the sum tilts by the contact term. Changing θ moves the amplitude but never the flatness.

ε′⋅M, DIAGRAM A0.655 eg/ω

ε′⋅M, DIAGRAM B0.995 eg/ω

ε′⋅M, A + B1.650 eg/ω

k⋅M OF THE SUM0.000 eg

Live interpretationε′⋅M, DIAGRAM A: 0.655 eg/ω. ε′⋅M, DIAGRAM B: 0.995 eg/ω. ε′⋅M, A + B: 1.650 eg/ω. k⋅M OF THE SUM: 0.000 eg

Catch the common trap

Explain before calculating.

Misconception check

The tempting shortcut: since ∂_μj^μ = 0, every vertex, and so every Feynman diagram, satisfies k_μM^μ = 0 on its own. Contract a vertex with k and you get a difference of inverse propagators, k⋅(2p + k) = [(p + k)² − m²] − [p² − m²], not zero; in X → φ⁺φ⁻γ the φ⁺ diagram alone gives k⋅M_A = −eg at every momentum. Conservation does not pass through time ordering unchanged: it leaves contact terms, which cancel only in the sum over every place the photon can attach to every charged line, and only once those lines are on shell. Two practical errors follow. Dropping a diagram, even a small one, breaks gauge invariance: the remainder then depends on the arbitrary c in ε + ck, by 5.0 eg GeV⁻¹, a third of diagram A, under the first example's shift. And k_μM^μ = 0 is not the transversality ε⋅k = 0 of a polarisation, which every physical ε obeys whether or not the amplitude is gauge invariant.

In X → φ⁺φ⁻γ the amputated tree amplitude obeys k⋅M = eg[δ₁/((p₁ + k)² − m²) − δ₂/((p₂ + k)² − m²)], with δᵢ = pᵢ² − m². A student keeps the φ⁺ on shell and takes the φ⁻ off shell by δ₂ = 0.030 GeV², where (p₂ + k)² − m² = 0.150 GeV². What does she find, and what does it mean?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA neutral X at rest decays to φ⁺φ⁻γ through gXφ†φ, with m = 0.140 GeV. The photon has k = (0.100, 0, 0, 0.100) GeV and the φ⁺ has p₁ = (E₁, 0.240, 0, 0.180) GeV. With ε = (0, 1, 0, 0), find each diagram's k⋅M and ε⋅M, then repeat with ε′ = ε + (5.0 GeV⁻¹)k.

Kinematics: E₁ = √(0.140² + 0.240² + 0.180²) = √0.1096 = 0.3311 GeV. X at rest fixes p⃗₂ = −(p⃗₁ + k⃗) = (−0.240, 0, −0.280) GeV, so E₂ = √(0.0196 + 0.0576 + 0.0784) = 0.3945 GeV and m_X = 0.3311 + 0.3945 + 0.100 = 0.8255 GeV.
Invariants: p₁⋅k = 0.100 × (0.3311 − 0.180) = 0.01511 GeV² and p₂⋅k = 0.100 × (0.3945 + 0.280) = 0.06745 GeV². On shell, M_A^μ = −eg(2p₁ + k)^μ/(2p₁⋅k) and M_B^μ = +eg(2p₂ + k)^μ/(2p₂⋅k).
Contract with k: k⋅(2p₁ + k) = 2p₁⋅k because k² = 0, so k⋅M_A = −eg, and likewise k⋅M_B = +eg, at these or any on-shell momenta. Neither diagram satisfies the identity; their sum does.
Polarisation ε = (0, 1, 0, 0): ε⋅k = 0, ε⋅p₁ = −0.240 GeV and ε⋅p₂ = +0.240 GeV. Then ε⋅M_A = eg × 0.240/0.01511 = 15.89 eg GeV⁻¹ and ε⋅M_B = eg × 0.240/0.06745 = 3.56 eg GeV⁻¹, a total of 19.45 eg GeV⁻¹.
Shift: ε′ = (0.500, 1, 0, 0.500) still has ε′⋅ε′ = −1 and ε′⋅k = 0. Each diagram moves by c k⋅M: ε′⋅M_A = 15.89 − 5.00 = 10.89 and ε′⋅M_B = 3.56 + 5.00 = 8.56 eg GeV⁻¹, and the sum stays 19.45 eg GeV⁻¹.

Answerk⋅M_A = −eg and k⋅M_B = +eg, summing to zero. ε⋅M = 15.89 + 3.56 = 19.45 eg GeV⁻¹; with ε′ the diagrams become 10.89 and 8.56 eg GeV⁻¹ and the total is still 19.45 eg GeV⁻¹.

MediumFor the decay of the Easy example, show that the photon is completely polarised in the decay plane. Then compute the amplitude with the in-plane polarisation shifted along k so that ε′⋅p₁ = 0, and again so that ε″⋅p₂ = 0. Which diagram carries the amplitude in each case, and what is the polarisation-summed |M|²?

Out of the plane: ε₂ = (0, 0, 1, 0) has ε₂⋅p₁ = ε₂⋅p₂ = ε₂⋅k = 0, so ε₂⋅M_A = ε₂⋅M_B = 0, and adding any multiple of k keeps it zero since k⋅M = 0. Only ε₁ = (0, 1, 0, 0) radiates: the photon is 100% linearly polarised in the decay plane.
Reference p₁: ε′ = ε₁ + ck with ε′⋅p₁ = ε₁⋅p₁ + c p₁⋅k = 0 needs c = 0.240/0.01511 = 15.89 GeV⁻¹, so ε′ = (1.589, 1, 0, 1.589). Check: ε′⋅ε′ = 1.589² − 1 − 1.589² = −1 and ε′⋅k = 0.1589 − 0.1589 = 0.
Then ε′⋅M_A = −eg ε′⋅(2p₁ + k)/(2p₁⋅k) = 0, since ε′⋅p₁ = ε′⋅k = 0: diagram A has vanished. ε′⋅M_B = 3.56 + 15.89 = 19.45 eg GeV⁻¹ carries everything.
Reference p₂: c = −ε₁⋅p₂/(p₂⋅k) = −0.240/0.06745 = −3.56 GeV⁻¹ deletes diagram B instead, and ε″⋅M_A = 15.89 + 3.56 = 19.45 eg GeV⁻¹.
Summing polarisations: |ε₁⋅M|² + |ε₂⋅M|² = 19.45² + 0 = 378 (eg)² GeV⁻², the same in all three gauges. A diagram's share of the amplitude is a gauge artefact: diagram A carries 82%, 0% or 100% of it, as you choose.

AnswerThe photon is fully polarised in the decay plane. Shifting ε along k so that ε′⋅p₁ = 0 (c = 15.9 GeV⁻¹) puts all 19.4 eg GeV⁻¹ in diagram B; ε″⋅p₂ = 0 (c = −3.56 GeV⁻¹) puts it all in A. Σ|M|² = 378 (eg)² GeV⁻² either way.

HardKeep the Easy example's momenta but let the charged legs leave their mass shells, pᵢ² = m² + δᵢ, as they would inside a larger diagram. Derive k⋅M for the amputated sum of the two diagrams, evaluate it for δ₁ = 0 and δ₂ = 0.0100 GeV² (raising E₂ at fixed p⃗₂), then for δ₂ = 0.0010 GeV², and find how much ε′ = ε + (5.0 GeV⁻¹)k changes ε⋅M at δ₂ = 0.0100 GeV².

With D(p) = i/(p² − m²), the identity k⋅(2p₁ + k) = [(p₁ + k)² − m²] − [p₁² − m²] cancels one propagator at a time: k⋅G_A = e[G₀(p₁, p₂) − G₀(p₁ + k, p₂)] and k⋅G_B = −e[G₀(p₁, p₂) − G₀(p₁, p₂ + k)], where G₀(q₁, q₂) = −ig D(q₁)D(q₂).
Adding them, the G₀(p₁, p₂) terms cancel because the two charges sum to zero: k⋅G = e[G₀(p₁, p₂ + k) − G₀(p₁ + k, p₂)], the Ward-Takahashi identity with one contact term per charged leg.
Amputate by multiplying by D(p₁)⁻¹D(p₂)⁻¹ = −δ₁δ₂: k⋅M = eg[δ₁/((p₁ + k)² − m²) − δ₂/((p₂ + k)² − m²)]. Each contact term has lost the pole of its own leg, so it keeps a factor δᵢ.
Numbers: E₂ = √(0.1556 + 0.0100) = 0.4069 GeV, 2p₂⋅k = 0.200 × (0.4069 + 0.280) = 0.1374 GeV² and (p₂ + k)² − m² = 0.1474 GeV², so k⋅M = −(0.0100/0.1474)eg = −0.0678eg. At δ₂ = 0.0010 GeV², E₂ = 0.3957 GeV, the denominator is 0.1361 GeV² and k⋅M = −0.00735eg: ten times nearer the shell, 9.2 times smaller.
Gauge shift at δ₂ = 0.0100 GeV²: ε⋅M = eg(0.240/0.015106 + 0.480/0.14739) GeV⁻¹ = (15.888 + 3.257) eg GeV⁻¹ = 19.145 eg GeV⁻¹, and ε′ changes it by c k⋅M = 5.0 × (−0.0678) = −0.339 eg GeV⁻¹, or 1.8%. On shell the same shift changed nothing.
Meaning: an off-shell Green's function need not be gauge independent. Were the φ⁻ an internal line, the photon could also attach beyond it, and those diagrams' contact terms cancel this one; only the complete amplitude with on-shell external legs obeys k⋅M = 0.

Answerk⋅M = eg[δ₁/((p₁ + k)² − m²) − δ₂/((p₂ + k)² − m²)]: −0.068eg at δ₂ = 0.0100 GeV², −0.0073eg at 0.0010 GeV², zero on shell. Off shell the 5.0 GeV⁻¹ shift moves ε⋅M by −0.34 eg GeV⁻¹, 1.8%.