University Quantum Field Theory I · QED at Tree Level: Moller, Bhabha and Compton Scattering and Pair Annihilation · 10.4

The Compton amplitude: two diagrams, one Ward identity

Compton scattering is the first QED process with a photon at each end, and the place where gauge invariance stops being a slogan and becomes two lines of Dirac algebra. Write both diagrams with the photon indices open, then watch the Dirac equation turn each one's gauge violation into the other's with the sign reversed.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

At order e², γ(k) e⁻(p) → γ(k′) e⁻(p′) has two diagrams on one open electron line. In the s-channel diagram the electron absorbs k first and carries p + k, off shell by 2p⋅k; in the u-channel diagram it emits k′ first and carries p − k′, off shell by −2p⋅k′. Both reach the external electrons through the same ψ̄ and ψ and differ only in where the commuting photon fields land, so they add with no relative sign.

Strip off the polarisations and the amplitude becomes a tensor, M = ε_μ ε′*_ν M^μν, and gauge invariance becomes a statement about it: ε and ε + λk describe the same photon, so k_μ M^μν must vanish. One identity does the work. k̸ = (p̸ + k̸ − m) − (p̸ − m) writes the photon's momentum as a difference of inverse propagators; in each diagram one piece cancels the internal propagator and the other dies on an external spinor, leaving ∓e² ū(p′)γ^ν u(p). Neither diagram is gauge invariant; their sum is, for every ν, and that free index licenses −g_μν for both photons in the next topic's trace.

The electron's coupling is linear in A, so unlike a charged scalar it needs no seagull: the Thomson amplitude a scalar draws from its A² term, the electron draws from the γ⁰ part of its propagator. The argument needs both electrons on shell, not the photons; put an electron line inside a larger diagram and k_μ M^μν keeps the inverse-propagator remainders of the Ward-Takahashi identity.

Simple definition: The Compton amplitude is the sum of QED's two order-e² diagrams for γe⁻ → γe⁻, written as a tensor M^μν in the two photon indices; gauge invariance demands k_μ M^μν = 0 and k′_ν M^μν = 0, which the sum obeys and each diagram alone violates.
Example: A 511 keV photon scattered through 90° by an electron at rest, polarisations in the scattering plane, spin kept: radiation gauge gives Mₛ = −0.224e² and Mᵤ = +0.447e²; shifting ε by k/ω makes them −1.118e² and +1.342e², and the sum stays 0.224e².

Two diagrams, photon indices freeM^μν = −e² ū(p′)[γ^ν(p̸ + k̸ + m)γ^μ/(2p⋅k) − γ^μ(p̸ − k̸′ + m)γ^ν/(2p⋅k′)]u(p)

Derived from Unit 9's rules. Both terms sit on one fermion line and reach ū(p′) and u(p) through the same operators, so they add with no relative sign.

μ for the incoming photon k, ν for the outgoing k′; M = ε_μ ε′*_ν M^μν; momenta in MeV; ū u = 2m; e² = 4πα; M dimensionless

Photon momentum as two inverse propagatorsk̸ = (p̸ + k̸ − m) − (p̸ − m)

An identity, not an approximation. One bracket cancels the internal propagator, the other kills an external spinor, and nowhere is k² = 0 needed.

p̸ − m annihilates u(p); (p̸ + k̸ − m)(p̸ + k̸ + m) = (p + k)² − m², in MeV²

Each diagram alone fails the Ward identityk_μ Mₛ^μν = −e² ū(p′)γ^ν u(p) = −k_μ Mᵤ^μν

Derived. Under ε → ε + λk/ω one diagram moves by ∓λe² ū(p′)ε̸′*u(p)/ω; for λ = 1 at 511 keV and 90° that is four times the physical amplitude.

Mₛ, Mᵤ the s- and u-channel terms of M^μν; ū(p′)γ^ν u(p) in MeV; both electrons on shell

The sum obeys it, the other index freek_μ M^μν = 0 for all four values of ν

Derived. The free index is what lets −g_μν replace both photons' polarisation sums at once in topic 10.05's trace.

ν not contracted with any ε′; the same steps give k′_ν M^μν = 0 for all μ; needs p² = p′² = m², not k² = 0

A charged scalar needs its seagullscalar: k_μ(Mₛ + Mᵤ)^μν = −2e² k^ν = −k_μ M_c^μν

Derived. For a scalar the A² term is forced by gauge invariance; ψ̄(iD̸ − m)ψ is linear in A, so the electron's two diagrams close the identity alone.

M_c^μν = 2e² g^μν, the contact term of topic 8.09's scalar QED; vertices −ie(p + p′)^μ; k^ν in MeV

Soft limit: the seagull's value without a seagullM → −2e² ε⃗⋅ε⃗′* δₛ′s as ω → 0, electron at rest

Derived from the γ⁰ part of k̸. It equals a scalar's seagull amplitude and gives dσ/dΩ = rₑ²|ε⃗⋅ε⃗′*|² with rₑ² = 79.4 mb: Thomson's limit.

ε⃗, ε⃗′ the photons' 3-vector polarisations, ε⁰ = ε′⁰ = 0; δₛ′s keeps the spin; e² = 4πα = 0.0917

Two diagrams on one line, added with a plus sign

Work in natural units, ħ = c = 1, with the metric (+, −, −, −). The e² term of the Dyson series, (1/2!)(−ie)²∫d⁴x d⁴y T[ψ̄γ^μψA_μ(x) ψ̄γ^νψA_ν(y)], must annihilate the incoming electron with a ψ, create the outgoing one with a ψ̄, absorb k with one A and emit k′ with the other, and contract the leftover ψ and ψ̄ into a propagator. The four ways to do this pair up under x ↔ y, cancelling the 1/2!, and leave two diagrams. In the s-channel diagram k is absorbed where the incoming electron arrives, so the internal line carries p + k, with (p + k)² − m² = 2p⋅k ≥ 0; in the u-channel diagram k′ is emitted there first, so the line carries p − k′, with (p − k′)² − m² = −2p⋅k′ ≤ 0. Neither vanishes for a photon of non-zero energy. The terms differ only in which vertex each A field contracts with, and A fields commute, while ψ̄ and ψ reach ū(p′) and u(p) in the same order in both: a plus sign, derived. Møller and Bhabha carry a minus because their diagrams swap which external fermion meets which operator (topic 7.05).

The photon momentum is two inverse propagators

Replace the incoming polarisation ε_μ by k_μ and write k̸ = (p̸ + k̸ − m) − (p̸ − m). In the s-channel term k̸ stands between the propagator and u(p). The piece p̸ − m meets u(p) and gives zero by the Dirac equation; the piece p̸ + k̸ − m turns the numerator p̸ + k̸ + m into (p + k)² − m², which cancels the denominator exactly. The term collapses to −e² ū(p′)γ^ν u(p). In the u-channel term k̸ stands next to ū(p′), and momentum conservation makes the internal momentum p − k′ = p′ − k, so split k̸ = (p̸′ − m) − (p̸′ − k̸ − m) instead. Now ū(p′)(p̸′ − m) = 0, and (p̸′ − k̸ − m)(p̸′ − k̸ + m) = (p′ − k)² − m² cancels this denominator with the opposite sign: +e² ū(p′)γ^ν u(p). The residues are equal and opposite, so k_μ M^μν = 0 for every ν, derived, and the same steps with ε′* → k′ give k′_ν M^μν = 0. What was used: both Dirac equations and momentum conservation, never k² = 0, so virtual photons obey it too. Dimensions agree: each residue, e² times ū(p′)γ^ν u(p), carries MeV, as k_μ times a dimensionless M^μν must. It is topic 9.04's open-line identity, closing over two vertices.

The lab in numbers: each diagram moves, the sum does not

Put the electron at rest, the photon along z with ε = (0, 1, 0, 0), and the scattered photon at θ in the x-z plane with ε′ = (0, cos θ, 0, −sin θ). Then p⋅ε = p⋅ε′ = 0, so (p̸ + m)ε̸u(p) = 2(p⋅ε)u(p) = 0 and M = −e² ū(p′)[ε̸′k̸ε̸/(2mω) + ε̸k̸′ε̸′/(2mω′)]u(p). Multiplying out the 2 × 2 blocks with Dirac-basis spinors, spin up along z kept, x = ω/m, x′ = ω′/m and n = 2 + x − x′, gives Mₛ = −e²[(2 − x′)cos θ + x′]/√(2n) and Mᵤ = −e²[(2 + x)cos θ − x]/√(2n), derived. Shift ε → ε + λk/ω and they move by ∓2λe² sin θ/√(2n), the residue ∓λe² ū(p′)ε̸′u(p)/ω evaluated; the sum, −e²[4 cos θ − (x − x′)(1 − cos θ)]/√(2n), does not. At θ = 0 each term is −e² at any energy. At 511 keV and 90°, x′ = 0.5 and n = 2.5: Mₛ = −0.224e², Mᵤ = +0.447e², and λ = 1 makes them −1.118e² and +1.342e², a move four times the physical amplitude, while the sum stays 0.224e².

Why the free index matters: −g_μν for both photons

Topic 8.07's physical polarisation sum, quoted, is Σ ε_μ ε*_ρ = −g_μρ + (k_μ n_ρ + n_μ k_ρ)/(k⋅n) − k_μ k_ρ/(k⋅n)², with n = (1, 0, 0, 0) in the frame where the polarisations are transverse. Every extra term carries k_μ or k_ρ. In Σ|M|² = M^μν (M^ρσ)* (Σ ε_μ ε*_ρ)(Σ ε′*_ν ε′_σ) those terms die against k_μ M^μν = 0, but only because the identity holds for all four values of ν: once the outgoing photon's sum is also replaced by −g_νσ, ν runs over timelike and longitudinal values that no physical ε′ has. The second section proved k_μ M^μν = 0 with ν free and k′_ν M^μν = 0 with μ free, so both sums can be replaced at once, which is what topic 10.05's trace does. Applied to |Mₛ|² alone, the same replacement keeps the extra terms, because k_μ Mₛ^μν = −e² ū(p′)γ^ν u(p) is not zero, and returns a gauge-dependent number with no meaning. −g_μν is licensed for a gauge-invariant set of diagrams, never diagram by diagram.

No seagull: where the electron keeps its A² term

A charged scalar (topic 8.09) has |D_μφ|² in its Lagrangian, which contains e²A_μA^μ φ*φ, so its Compton amplitude has a third, contact diagram with vertex 2ie²g^μν. Run the Ward test on the scalar's pole diagrams, vertices −ie(p + p′)^μ: k⋅(2p + k) = (p + k)² − m² cancels one propagator, k⋅(2p′ − k) = −[(p′ − k)² − m²] the other, and they leave −2e²k^ν, which the contact term's 2e²k^ν cancels. The Dirac Lagrangian ψ̄(iD̸ − m)ψ is linear in A, so there is no contact vertex and two diagrams suffice. The soft limit agrees anyway. For the scalar in the lab with ε⁰ = ε′⁰ = 0, (2p + k)⋅ε = 0 and (2p − k′)⋅ε′* = 0 kill both pole terms, leaving −2e²ε⃗⋅ε⃗′*. For the electron as ω → 0 only γ⁰ survives in k̸/ω between rest spinors, and γ⁰ε̸ = −ε̸γ⁰ turns the lab bracket into −2(ε⋅ε′*)γ⁰/(2m), so M → −2e²ε⃗⋅ε⃗′* δₛ′s, derived: the same number, so dσ/dΩ = rₑ²|ε⃗⋅ε⃗′*|² with rₑ² = 79.4 mb, and Thomson's 0.665 b (topic 10.07). In old-fashioned perturbation theory this γ⁰ piece is the state with an extra pair, quoted.

Change one variable at a time

Make the relationship visible.

Interactive model

Gauge shift λ in ε → ε + λk/ω1.00

Photon energy ω ÷ m (m = 511 keV)1.0

Marker angle θ90 °

Drag λ from −1 to 1: the dashed curves swing apart by ∓2λ sin θ/√(2n), with n = 2 + (ω − ω′)/m, while the solid sum never moves: the Ward identity at work. Lower ω to 51 keV and the sum settles onto Thomson's −2 cos θ, with no seagull diagram anywhere.

s-CHANNEL Mₛ ÷ e²-1.118

u-CHANNEL Mᵤ ÷ e²1.342

SUM M ÷ e²0.224

THOMSON −2 cos θ0.000

Live interpretations-CHANNEL Mₛ ÷ e²: −1.118. u-CHANNEL Mᵤ ÷ e²: 1.342. SUM M ÷ e²: 0.224. THOMSON −2 cos θ: 0.000

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading is that the s-channel diagram is the history in which the electron absorbs the photon and then emits one, and the u-channel diagram the history in which it emits first, each a physical amplitude in its own right, so that one may keep whichever diagram dominates, or square each with −g_μν and add. Neither diagram is an observable. Under ε → ε + λk, which changes nothing about the photon, each moves by ∓λe² ū(p′)ε̸′*u(p); at 511 keV and 90° a shift of k/ω moves each by four times the whole amplitude, and even as ω → 0 it does not shrink, staying comparable to the Thomson amplitude itself. Nor are diagrams time orderings: each covariant propagator holds both orders, including the one with an extra electron-positron pair, which is where the electron's Thomson amplitude lives. Only the gauge-invariant sum has a meaning, and only the sum may be squared with −g_μν.

In the tree-level Compton amplitude for on-shell electrons, the incoming photon's polarisation ε is replaced by ε + λk. What happens to the s-channel diagram, the u-channel diagram and their sum?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA ¹³⁷Cs photon of 661.7 keV is scattered straight back, θ = 180°, by a free electron at rest. Find the scattered energy, the virtuality of the internal electron in each Compton diagram and t, check s + t + u = 2m², and compare with the backscatter peak and Compton edge that γ-ray spectrometers record.

Four-momentum conservation gives Compton's relation, quoted from UQMI-01.04: ω′ = ω/[1 + (ω/m)(1 − cos θ)] = 661.7/(1 + 2 × 1.2949) = 661.7/3.5898 = 184.3 keV. The electron recoils with T = ω − ω′ = 477.4 keV.
s-channel line: (p + k)² − m² = 2p⋅k, which in the lab is 2mω = 2 × 0.5110 × 0.6617 = 0.676 MeV², positive. u-channel line: (p − k′)² − m² = −2p⋅k′ = −2mω′ = −2 × 0.5110 × 0.1843 = −0.188 MeV², negative. Both carry MeV², the dimension of an inverse propagator.
t = (k − k′)² = −2ωω′(1 − cos θ) = −4 × 0.6617 × 0.1843 = −0.488 MeV². Then (s − m²) + t + (u − m²) = 0.676 − 0.488 − 0.188 = 0.000, so s + t + u = 2m², the sum of the four squared masses, as it must be.
Neither denominator can vanish: 2p⋅k is positive and −2p⋅k′ negative for any photon of non-zero energy, so the internal electron is never on shell. Here the u-channel line is nearer its shell by ω/ω′ = 3.59, so its propagator is the larger; the two still add with the same sign, because both sit on one fermion line.
Measured ¹³⁷Cs spectra show a backscatter peak near 184 keV, photons Compton-scattered back into the detector by its surroundings, and a Compton edge at 477 keV, the most energy one scatter can leave behind. Both are this kinematics.

Answerω′ = 184 keV and T = 477 keV, matching the measured backscatter peak and Compton edge; s − m² = 0.676 MeV², u − m² = −0.188 MeV², t = −0.488 MeV², and s + t + u = 2m².

MediumA 1022 keV photon (ω = 2m) scatters through 60° off an electron at rest, both polarisations in the scattering plane and the electron's spin along the beam kept. Using the lab-frame formulas of this lesson, find Mₛ and Mᵤ in radiation gauge and after ε → ε − 0.5k/ω, then find dσ/dΩ for this channel in mb sr⁻¹ and compare it with the Thomson value.

x = 2 and cos θ = 0.5, so x′ = 2/(1 + 2 × 0.5) = 1, ω′ = 511 keV, n = 2 + x − x′ = 3 and √(2n) = √6 = 2.449.
Radiation gauge: Mₛ = −e²[(2 − 1)(0.5) + 1]/2.449 = −0.612e², and Mᵤ = −e²[(2 + 2)(0.5) − 2]/2.449 = 0. In this gauge the whole amplitude sits in the s-channel diagram.
With λ = −0.5 each term moves by ∓2λe² sin θ/√(2n) = ±2 × 0.5 × 0.866e²/2.449 = ±0.354e²: Mₛ = −0.612e² + 0.354e² = −0.259e² and Mᵤ = 0 − 0.354e² = −0.354e². Now the u-channel diagram is the larger. The sum is −0.612e² both times, as the closed form −e²[4 × 0.5 − 1 × 0.5]/2.449 confirms.
M is dimensionless, as a 2 → 2 amplitude must be. With e² = 4πα, |M|² = 0.375e⁴ = 0.375 × 16π²α². The lab-frame phase space of topic 10.06, quoted, gives dσ/dΩ = (ω′/ω)²|M|²/(64π²m²) = 0.25 × 0.375α²/(4m²) = 0.0234 rₑ², with rₑ = α/m.
Restoring ħ and c, rₑ = αħc/(mc²) = 2.818 fm and rₑ² = 79.4 mb, so dσ/dΩ = 0.0234 × 79.4 = 1.86 mb sr⁻¹. The soft-limit amplitude −2e² cos θ would give rₑ² cos²θ = 19.9 mb sr⁻¹ in this channel: at 1 MeV, recoil and the electron's spin have cut it more than tenfold.

AnswerMₛ = −0.612e² and Mᵤ = 0 in radiation gauge, −0.259e² and −0.354e² after the shift; the sum is −0.612e² both ways, so dσ/dΩ = 1.86 mb sr⁻¹ against the Thomson value of 19.9 mb sr⁻¹.

HardUnder ε → ε + λk/ω each Compton diagram moves by ∓λe² ū(p′)ε̸′u(p)/ω. For an electron at rest and a real outgoing polarisation with ε′⁰ = 0, find the spin-summed square of that shift by a trace, evaluate it for a ⁶⁰Co photon of 1.3325 MeV scattered through 60° with λ = 1, compare it with the physical spin-summed |M|² for in-plane polarisations, and find its limits as θ → 0 and as ω → 0.

Casimir's trick with a real ε′: the sum over both spins of |ū(p′)ε̸′u(p)|² is Tr[(p̸′ + m)ε̸′(p̸ + m)ε̸′]. Odd traces vanish, and topic 7.06's traces of two and four γ matrices give 4[2(p⋅ε′)(p′⋅ε′) − (p⋅p′ − m²)(ε′⋅ε′)].
In the lab p⋅ε′ = mε′⁰ = 0 and ε′⋅ε′ = −1, so the trace is 4(p⋅p′ − m²) = 4m(E′ − m) = 4mT, with T = ω − ω′ the recoil energy. So Σ|ΔMₛ|² = Σ|ΔMᵤ|² = 4λ²e⁴ m(ω − ω′)/ω², dimensionless as it must be.
Numbers: ω/m = 2.6076, so ω′ = 1.3325/(1 + 0.5 × 2.6076) = 1.3325/2.3038 = 0.5784 MeV and T = 0.7541 MeV. Then 4m(ω − ω′)/ω² = 4 × 0.511 × 0.7541/1.7756 = 0.868, so Σ|ΔMₛ|² = 0.868e⁴ at λ = 1.
The physical value for in-plane polarisations is the polarised Klein-Nishina bracket, Σ|M|² = 2e⁴[ω/ω′ + ω′/ω − 2 + 4cos²θ], quoted here (topic 10.06 derives its polarisation-summed form): 2e⁴[2.3038 + 0.4341 − 2 + 1] = 3.48e⁴. One diagram's gauge artefact is a quarter of the whole.
Forward, θ → 0: ω′ → ω and p′ → p, so the trace becomes 8(p⋅ε′)² = 0 and the artefact vanishes. Soft, ω → 0: ω − ω′ → ω²(1 − cos θ)/m, so Σ|ΔMₛ|² → 4λ²e⁴(1 − cos θ) = 2λ²e⁴ at 60°, exactly the Thomson value 2 × 4e⁴cos²θ = 2e⁴. At low energy a single diagram's gauge dependence is as large as the answer.

AnswerΣ|ΔMₛ|² = 4λ²e⁴m(ω − ω′)/ω² = 0.868e⁴ at λ = 1, a quarter of the physical 3.48e⁴; it vanishes forward and tends to 4λ²e⁴(1 − cos θ), equal to Thomson's 2e⁴ at 60°, as ω → 0.