University Quantum Field Theory I · Covariant Quantisation of the Electromagnetic Field · 8.9

Scalar electrodynamics: vertex, seagull and Ward identity

The charged scalar is where gauge invariance does its bookkeeping in plain sight. Minimal coupling hands you a momentum-dependent vertex and a two-photon contact term; Compton scattering shows that the pole diagrams alone fail the Ward identity, and that the contact term is Thomson scattering.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Scalar electrodynamics is what you get by taking UAQM-09.03's complex field and replacing ∂_μ with D_μ = ∂_μ + ieA_μ, exactly as UAQM-03.04 coupled the Klein-Gordon equation. Because the kinetic term is quadratic in D_μ, the coupling comes in two pieces: a three-point vertex −ie(p + p′)^μ which, unlike the Dirac vertex, depends on the momenta it joins, and a four-point seagull 2ie²g^μν with no momentum at all. The seagull is neither decoration nor a higher-order correction.

In Compton scattering it enters at the same order, e², as the s- and u-channel pole diagrams, and it is the only diagram that can cancel the −2e²k^ν those two leave when the photon's polarisation is replaced by its momentum: the Ward identity, and with it the right to sum polarisations with −g_μν, needs all three. Go to the scalar's rest frame with polarisations that have no time component and the roles reverse: both pole diagrams vanish identically, the seagull alone is the tree amplitude 2e²ε⋅ε′*, and as ω/m → 0 it gives Thomson's σ_T = 8πα²/(3m²), the field-theoretic form of UAQM-11.08's A² term. The model is a point particle.

A real π⁺ is polarisable, which corrects the amplitude at order ω², and beyond tree level the theory needs a (φ†φ)² coupling to stay renormalisable, a result quoted here.

Simple definition: Scalar electrodynamics couples a charged spin-0 field to photons through D_μ = ∂_μ + ieA_μ, which yields a vertex −ie(p + p′)^μ and a seagull 2ie²g^μν; gauge invariance needs both, and the seagull's coefficient is exactly what the Ward identity demands.
Example: For a π⁺ at rest the seagull alone is the whole tree amplitude, and as the photon energy falls far below 139.57 MeV the cross-section tends to 8πα²/(3m_π²) = 8.92 μb: the electron's 0.665 b scaled by (mₑ/m_π)² = 1.34 × 10⁻⁵.

Minimal coupling of the charged scalarℒ = (D_μφ)†D^μφ − m²φ†φ − (1/4)F_μνF^μν, with D_μ = ∂_μ + ieA_μ

Expanding gives ℒᵢₙₜ = −ieA^μ(φ†∂_μφ − φ∂_μφ†) + e²A_μA^μφ†φ, derived. Gauge invariance fixes both terms; neither is a modelling choice.

φ and A_μ of mass dimension 1 (GeV), e dimensionless with α = e²/(4π) = 1/137.036; ħ = c = 1, metric (+, −, −, −)

The three-point vertex−ie(p + p′)^μ, with p in and p′ out along the arrow of charge

Derived by topic 3.07's rule: ∂_μφ brings −ip_μ, ∂_μφ† brings +ip′_μ. For a slow scalar its time component is −2iem, the charge coupling in relativistic normalisation.

p, p′ four-momenta in GeV, so the vertex has mass dimension 1; an antiparticle's physical momenta enter with a minus sign

The seagull2ie²g^μν, from the term e²A_μA^μφ†φ

Derived. It carries no momentum and the same order e² as two vertices joined by a propagator, so it is never subleading in Compton scattering.

g^μν the metric; dimensionless, as a four-point coupling must be; the 2 counts the two ways to attach two identical photons

Compton amplitude on a scalarM^μν = −e²(2p + k)^μ(2p′ + k′)^ν/(2p⋅k) + e²(2p′ − k)^μ(2p − k′)^ν/(2p⋅k′) + 2e²g^μν

Derived: s-channel pole, u-channel pole and seagull. The first two swap under k ↔ −k′ with μ ↔ ν, as crossing requires.

M = ε_μ ε′*_ν M^μν, photon k in and k′ out; 2p⋅k = s − m² and 2p⋅k′ = m² − u in GeV²; M dimensionless

The Ward identity, diagram by diagramk_μM^μν = −e²(2p′ + k′)^ν + e²(2p − k′)^ν + 2e²k^ν = 0

Shown. The pole diagrams alone leave −2e²k^ν, which only the seagull cancels; that is what licenses replacing a polarisation sum by −g_μν.

Terms from the s-channel, the u-channel and the seagull in that order; each in GeV. Uses only p² = p′² = m² and p + k = p′ + k′

Rest-frame cross-section from the seagull alonedσ/dΩ = (α/m)²(ω′/ω)²|ε⃗⋅ε⃗′*|², with ω′ = ω/[1 + (ω/m)(1 − cos θ)]

Computed from M = −2e²ε⃗⋅ε⃗′*. Averaged over polarisations it is (α²/2m²)(ω′/ω)²(1 + cos²θ), and as ω/m → 0 the total is Thomson's 8πα²/(3m²).

ω, ω′ photon energies and m the scalar's mass in GeV; θ the scattering angle; αħc/(mc²) is 2.818 fm for mₑ, 0.01032 fm for m_π

Minimal coupling gives two vertices, not one

Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Take UAQM-09.03's complex scalar and replace ∂_μ by D_μ = ∂_μ + ieA_μ, as UAQM-03.04 did for the Klein-Gordon equation: under φ → e^(−ieΛ)φ and A_μ → A_μ + ∂_μΛ, D_μφ transforms like φ, so |D_μφ|² is invariant. Expanding it gives ℒᵢₙₜ = −ieA^μ(φ†∂_μφ − φ∂_μφ†) + e²A_μA^μφ†φ, derived. Topic 3.07's rule reads off the vertices. In the first term φ annihilates the incoming particle, so ∂_μφ brings −ip_μ, and φ† creates the outgoing one, so ∂_μφ† brings +ip′_μ; i times the term gives −ie(p + p′)^μ. The second term holds two identical A fields, and attaching two photons in either order doubles it: 2ie²g^μν. Dimensions agree: [φ] = [A] = 1 and [e] = 0, so the vertex has dimension 1, like the momentum in it, and the seagull 0. For a slow scalar the time component, −ie(E + E′) ≈ −2iem, is the coupling to A⁰; the spatial part, from −ie(p⃗ + p⃗′), is the p⃗⋅A⃗ term of UAQM-11.08's Hamiltonian, and the seagull is its A⃗² term, each multiplied by the factor of twice the mass that relativistic normalisation brings.

A derivative coupling is read off with the T* product

The vertex carries momenta because ℒᵢₙₜ contains ∂₀φ, and that changes the canonical structure. The momentum conjugate to φ becomes π = (D₀φ)†, and a short Legendre transform gives Hᵢₙₜ = −ℒᵢₙₜ + e²(A⁰)²φ†φ: an extra term with no Lorentz covariance. At the same time ⟨0|T ∂₀φ(x) ∂₀φ†(y)|0⟩ is not ∂/∂x⁰ ∂/∂y⁰ acting on ⟨0|T φ(x)φ†(y)|0⟩, because ∂₀ also acts on the step functions of the time ordering and leaves a contact term proportional to δ⁴(x − y). The two non-covariant pieces cancel order by order, so the Dyson series may be built from −ℒᵢₙₜ with the covariant T* product, in which derivatives act outside the time ordering. That is topic 2.04's remark, stated here and not proved, and it is why the vertex and the seagull can be read straight off ℒ as though nothing had a derivative. With it the internal scalar carries topic 1.05's i/(p² − m² + iε), and Unit 4's LSZ formula puts a factor 1 on each external scalar and ε_μ or ε′*_ν on each external photon.

Three diagrams at order e²

For γ(k, ε) + φ(p) → γ(k′, ε′) + φ(p′), with p + k = p′ + k′, three diagrams contribute. In the s-channel the scalar absorbs k first and propagates with p + k, so (p + k)² − m² = 2p⋅k = s − m². In the u-channel it emits k′ first and propagates with p − k′, giving −2p⋅k′ = u − m². The seagull joins both photons at one point. Writing M = ε_μ ε′*_ν M^μν, the rules give M^μν = −e²(2p + k)^μ(2p′ + k′)^ν/(2p⋅k) + e²(2p′ − k)^μ(2p − k′)^ν/(2p⋅k′) + 2e²g^μν, derived. Squaring p′ = p + k − k′ with p′² = m² gives p⋅k − p⋅k′ = k⋅k′, hence p′⋅k = p⋅k′ and p′⋅k′ = p⋅k: the on-shell identities that do the later work. Each term is dimensionless, as a 2 → 2 amplitude must be, either two momenta over a momentum squared or the bare e². The first two terms exchange under k ↔ −k′ with μ ↔ ν, which is crossing symmetry, and the seagull is symmetric on its own. Count the couplings: each pole diagram has two vertices of order e, the seagull one of order e², so all three are tree level and none corrects the others.

The Ward identity needs all three diagrams

Replace ε_μ by k_μ, topic 8.06's test. For the s-channel, k⋅(2p + k) = (p + k)² − p², which is the propagator's denominator once p² = m², so k_μMₛ^μν = −e²(2p′ + k′)^ν. For the u-channel, k⋅(2p′ − k) = p′² − (p′ − k)² = −[(p − k′)² − m²], which cancels its denominator with a sign: k_μMᵤ^μν = e²(2p − k′)^ν. The seagull gives 2e²k^ν. The two pole diagrams leave 2e²(p − p′ − k′)^ν = −2e²k^ν, and the seagull cancels it exactly: k_μM^μν = 0, shown. No single diagram and no pair satisfies the identity. Note what was used: p² = p′² = m² and momentum conservation, but not k² = 0, k′² = 0 or ε′*⋅k′ = 0, so the identity holds for each ν separately, and by crossing k′_νM^μν = 0 as well. Read backwards, the identity fixes the seagull: a contact term λ × 2ie²g^μν passes only at λ = 1. The Dirac vertex γ^μ carries no momentum, and in topic 10.04's electron Compton amplitude the s- and u-channel diagrams cancel between themselves, so no contact term is needed there.

In the rest frame the seagull is the whole amplitude

Evaluate in the scalar's rest frame, p = (m, 0⃗), with polarisations that have no time component, ε⁰ = ε′⁰ = 0, the radiation gauge Advanced QM used for real photons. Then ε⋅p = 0 and ε⋅k = 0, so ε⋅(2p + k) = 0 and the s-channel diagram vanishes; likewise ε′*⋅p = 0 and ε′*⋅k′ = 0 kill the u-channel's factor ε′*⋅(2p − k′). What remains is the seagull, M = 2e²ε⋅ε′* = −2e²ε⃗⋅ε⃗′*, shown. This is exact at tree level and at every ω, not a low-energy approximation: the energy dependence enters only through phase space. It is also a statement about this gauge, not about the diagrams. Shift ε → ε + ck and the s-channel reappears with c k_μMₛ^μν ε′*_ν while the seagull moves by the opposite amount, as the Medium example shows. In the centre-of-mass frame, for ω/m = 0.7 and θ = 60° in the rest frame, a numerical evaluation gives the u-channel 0.78 of the in-plane amplitude and the seagull only 0.22. The rest-frame result is UAQM-11.08's structure: the A⃗² term gave Thomson scattering at first order, while the p⃗⋅A⃗ term at second order vanished for a free charge at rest.

Thomson's limit, recoil and where the point scalar fails

Unit 5's cross-section for a target at rest, dσ/dΩ = |M|²(ω′/ω)²/(64π²m²), quoted, turns M = −2e²ε⃗⋅ε⃗′* into (α/m)²(ω′/ω)²|ε⃗⋅ε⃗′*|², with ω′ = ω/[1 + (ω/m)(1 − cos θ)] from kinematics. In ordinary units α/m is the length αħc/(mc²). Averaging ε⃗ and summing ε⃗′ gives (1 + cos²θ)/2, and as ω/m → 0 the total tends to (8π/3)(α/m)² = 8πα²/(3m²), Thomson's cross-section, derived, as UAQM-12.09 found it from the A² term. At the electron's mass that is 0.66525 b, the CODATA value, although the electron has spin ½: the low-energy limit depends only on charge and mass, a low-energy theorem quoted here. Spin first shows at order (ω/m)². Integrating the recoil factor gives σ/σ_T = 0.835 at ω/m = 0.1 against Klein-Nishina's 0.841, the two agreeing at first order, σ ≈ σ_T(1 − 2ω/m). For π⁺ the point-scalar value is 8.92 μb at low energy. Its polarisability, measured by COMPASS as α_π = (2.0 ± 0.6 ± 0.7) × 10⁻⁴ fm³ with α_π + β_π = 0 assumed, quoted, adds an amplitude about 0.5% of the seagull's at ω = 100 MeV, growing as ω².

Change one variable at a time

Make the relationship visible.

Interactive model

Photon energy ω/m0.7

Seagull coefficient λ1.00 × 2ie²g^μν

Marker angle θ60 °

At λ = 1 the dashed curve lies on the solid one for every ω and θ: the Ward identity at work. Move λ off 1 and they split, the −g curve going negative at λ = 0, so without the exact seagull there is no cross-section. Then drag ω/m to 0 to land on Thomson.

ω′/ω AT THE MARKER0.741

TRANSVERSE SUM0.343 (α/m)² sr⁻¹

−g REPLACEMENT0.343 (α/m)² sr⁻¹

THOMSON (1 + cos²θ)/20.625 (α/m)² sr⁻¹

Live interpretationω′/ω AT THE MARKER: 0.741. TRANSVERSE SUM: 0.343 (α/m)² sr⁻¹. −g REPLACEMENT: 0.343 (α/m)² sr⁻¹. THOMSON (1 + cos²θ)/2: 0.625 (α/m)² sr⁻¹

Catch the common trap

Explain before calculating.

Misconception check

The seagull is often treated as a small correction: it carries e² where the vertex carries e, spinor QED has no such diagram, and the pole diagrams look like the real physics of absorption and re-emission. All three reasons fail. In Compton scattering each pole diagram has two vertices, so all three diagrams are order e² and tree level. Without the seagull k_μM^μν = −2e²k^ν, so the amplitude changes when ε is shifted along k, and the −g_μν polarisation sum stops agreeing with the sum over physical photons: for a 150 MeV photon scattered through 120° by a π⁺ at rest it even returns a negative Σ|M|². And in the rest frame with ε⁰ = 0 both pole diagrams vanish, so dropping the seagull would leave no scattering at all. The electron needs no seagull because its vertex γ^μ has no momentum dependence; that is a property of the Dirac vertex, not evidence that contact terms are optional.

In Compton scattering on a charged scalar, the incoming photon's polarisation is replaced by its momentum k. What do the s- and u-channel diagrams give together, k_μ(Mₛ + Mᵤ)^μν?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA π⁺ (m = 139.57 MeV) at rest scatters a 30.0 MeV photon through θ = 90°, with both photons polarised perpendicular to the scattering plane. Using polarisations with no time component, find the amplitude and dσ/dΩ in μb sr⁻¹, and compare with the Thomson value.

In the rest frame with ε⁰ = ε′⁰ = 0, ε⋅p = ε⋅k = 0 kills the s-channel and ε′*⋅p = ε′*⋅k′ = 0 kills the u-channel, so only the seagull survives: M = 2e²ε⋅ε′* = −2e²ε⃗⋅ε⃗′*. Both vectors are ŷ, so ε⃗⋅ε⃗′* = 1 and M = −2e² = −8πα = −0.1834, dimensionless as a 2 → 2 amplitude must be.
Recoil: ω′ = ω/[1 + (ω/m)(1 − cos θ)] = 30.0 MeV/(1 + 0.2149) = 24.69 MeV, so ω′/ω = 0.8231 and (ω′/ω)² = 0.6775.
Then dσ/dΩ = |M|²(ω′/ω)²/(64π²m²) = (α/m)²(ω′/ω)², an area per steradian. In ordinary units αħc/(mc²) = 1.43996 MeV fm ÷ 139.57 MeV = 0.010317 fm, whose square is 1.0644 × 10⁻⁴ fm² = 1.0644 μb, since 1 fm² = 10 mb.
So dσ/dΩ = 1.0644 μb × 0.6775 = 0.721 μb sr⁻¹. As ω → 0 the recoil factor goes to 1 and this tends to Thomson's 1.064 μb sr⁻¹. For in-plane polarisations ε⃗⋅ε⃗′* = cos θ = 0 at 90°, so a photon scattered through 90° is polarised perpendicular to the plane, as in Thomson scattering.

AnswerM = −2e² = −0.183, all of it from the seagull; dσ/dΩ = 0.721 μb sr⁻¹, which is 0.68 of the low-energy Thomson value 1.064 μb sr⁻¹.

MediumThe same π⁺ at rest scatters a 100 MeV photon k = (ω, 0, 0, ω) through θ = 60° in the xz-plane, with in-plane polarisations ε = (0, 1, 0, 0) and ε′ = (0, cos θ, 0, −sin θ). Evaluate each diagram in this gauge and again after ε → η = ε + k/ω, then find dσ/dΩ for this polarisation pair.

Kinematics: ω′ = 100 MeV/(1 + 0.7165 × 0.5) = 73.62 MeV and k′ = ω′(1, sin θ, 0, cos θ), so p′ = p + k − k′ = (165.95, −63.76, 0, 63.19) MeV, with p′² = m² to the precision kept. Both ε and ε′ are transverse to their photons and orthogonal to p = (m, 0⃗).
In this gauge ε⋅(2p + k) = 0 and ε′⋅(2p − k′) = 0, so Mₛ = Mᵤ = 0, and the seagull gives M = 2e²ε⋅ε′ = −2e² cos θ = −e².
Shift to η = (1, 1, 0, 1). It is still transverse, η⋅k = 0, with the same norm, η⋅η = −1, but now η⋅(2p + k) = 2m_π = 279.14 MeV. With ε′⋅(2p′ + k′) = 2ε′⋅k = 2ω sin θ = 173.21 MeV and 2p⋅k = 2m_πω = 2.7914 × 10⁴ MeV², Mₛ = −e² × 279.14 × 173.21/(2.7914 × 10⁴) = −1.732e² = −2e² sin θ.
The u-channel still vanishes, since ε′ is unchanged and ε′⋅(2p − k′) = 0. The seagull becomes 2e²η⋅ε′ = 2e²(sin θ − cos θ) = 0.732e². The sum, −1.732e² + 0.732e² = −e², is unchanged. Drop the seagull and the same shift would move the amplitude from 0 to −1.732e²: a gauge-dependent answer.
With M = −e² = −4πα = −0.0917, dσ/dΩ = (α/m)²(ω′/ω)² cos²θ = 1.0644 μb × 0.5421 × 0.25 = 0.144 μb sr⁻¹. The perpendicular pair gives 0.577 μb sr⁻¹, four times as much, as cos²60° = 1/4 requires.

AnswerIn the rest-frame gauge Mₛ = Mᵤ = 0 and the seagull gives −e²; after ε → ε + k/ω, Mₛ = −1.732e² and the seagull 0.732e², with the same total. dσ/dΩ = 0.144 μb sr⁻¹ for the in-plane pair.

HardAn unpolarised 150 MeV photon scatters through θ = 120° off a π⁺ at rest. Using the invariants P = p⋅k and Q = p⋅k′, evaluate Σ|M|² with −g_μν for both photons, with and without the seagull, and compare with the transverse sum 4e⁴(1 + cos²θ). Then find dσ/dΩ in nb sr⁻¹.

ω′ = 150 MeV/(1 + 1.0747 × 1.5) = 57.43 MeV, so P = mω = 20936 MeV², Q = mω′ = 8015 MeV² and k⋅k′ = P − Q = 12921 MeV². Write M^μν = A^μν + 2e²g^μν, with A the two pole terms built from a = 2p + k, b = 2p′ + k′, c = 2p′ − k and d = 2p − k′. With g_μνg^μν = 4, Σ|M|² = A_μνA^μν + 4e²A^μ_μ + 16e⁴.
Trace: A^μ_μ = −e²(a⋅b)/(2P) + e²(c⋅d)/(2Q). The on-shell relations give a⋅b = 4(m² + P) + P − Q = 174582 MeV² and c⋅d = 4(m² − Q) + P − Q = 58780 MeV², so A^μ_μ = e²(−4.1695 + 3.6670) = −0.5025e².
Square: a⋅a = b⋅b = 4(m² + P) = 161661 MeV², c⋅c = d⋅d = 4(m² − Q) = 45860 MeV², and a⋅c = b⋅d = 4(m² + P) − 2(P + Q) = 103760 MeV². Then A_μνA^μν = e⁴[(a⋅a)(b⋅b)/(4P²) + (c⋅c)(d⋅d)/(4Q²) − (a⋅c)(b⋅d)/(2PQ)] = e⁴(14.907 + 8.185 − 32.082) = −8.990e⁴.
With the seagull, Σ|M|² = e⁴(−8.990 − 2.010 + 16) = 5.000e⁴ = 4e⁴(1 + cos²120°), the transverse sum exactly, as k_μM^μν = 0 guarantees. Without it the −g sum is −8.990e⁴, negative as no sum of squared moduli can be, while the transverse sum in this frame is 0: no cross-section at all.
dσ/dΩ = (1/2)Σ|M|²(ω′/ω)²/(64π²m²) = (α/m)²(ω′/ω)²(1 + cos²θ)/2 = 1.0644 μb × 0.14656 × 0.625 = 0.0975 μb sr⁻¹ = 97.5 nb sr⁻¹. Thomson at this angle would give 665 nb sr⁻¹; the recoil factor (ω′/ω)² = 0.147 cuts it almost sevenfold.

AnswerΣ|M|² = 5.000e⁴ with the seagull, equal to the transverse 4e⁴(1 + cos²θ); without it the −g sum is −8.99e⁴ and the transverse sum 0. dσ/dΩ = 97.5 nb sr⁻¹.