University Quantum Field Theory I · QED at Tree Level: Moller, Bhabha and Compton Scattering and Pair Annihilation · 10.5

The Compton trace: the spin- and polarisation-averaged |M|²

Here the Compton amplitude becomes a number. Average over what comes in, sum over what goes out, and two diagrams turn into four Dirac traces of up to eight gamma matrices. Reduce them with contraction identities and on-shell kinematics and one invariant bracket covers Thomson scattering, the collinear pole and everything between.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Topic 10.04 left the Compton amplitude as two terms, an s-channel diagram with denominator s − m² = 2p⋅k and a u-channel one with u − m² = −2p⋅k′, whose sum, and only whose sum, obeys k_μM^μν = 0. This topic spends that licence. An unpolarised measurement averages over the incoming spin and polarisation and sums over the outgoing ones: the spin sums close the fermion line with p̸ + m, the Ward identity lets both photon sums become −g_μν, and |M|² becomes four Dirac traces, two squared diagrams and two equal interference terms.

Contraction identities cut their eight gamma matrices to four, and the on-shell relations p⋅k − p⋅k′ = k⋅k′ and s + t + u = 2m² do the rest. Out comes one Lorentz-invariant bracket in p⋅k and p⋅k′, symmetric under s ↔ u. Its m² and m⁴ terms are not small corrections: in the laboratory frame they add to −sin²θ at every energy, and at threshold they turn 2 into Thomson's 1 + cos²θ.

Its massless limit, −2e⁴(u/s + s/u), has a pole where a massless electron emits its photon collinearly; the mass keeps u at or below m⁴/s and caps the backward peak at s/m² + m²/s. The result stops at tree level, with corrections of order α/π; at atomic binding, for electrons that are not free; and at polarised beams, which need spin projectors in place of p̸ + m and explicit polarisation vectors in place of −g_μν.

Simple definition: The Compton trace turns the sum over electron spins and photon polarisations of |M|² for γe⁻ → γe⁻ into Dirac traces, giving the unpolarised squared amplitude as a function of the two invariants p⋅k and p⋅k′ alone.
Example: A 1.00 MeV photon scattered through 90° by a free electron at rest leaves with 0.338 MeV, so p⋅k = 0.511 MeV² and p⋅k′ = 0.173 MeV². The bracket is 0.338 + 2.957 − 2 + 1 = 2.295, and ⟨|M|²⟩ = 4.59e⁴ = 0.0386.

The average as four traces⟨|M|²⟩ = (e⁴/4)[I/(2p⋅k)² + (II + III)/((2p⋅k)(−2p⋅k′)) + IV/(2p⋅k′)²]

Σu ū = p̸ + m closes the electron line; Σεε* → −g_μν is legal because the summed amplitude obeys k_μM^μν = 0. The u-channel denominator is negative, and that sign rides on II + III.

⟨ ⟩ averages 2 spins × 2 polarisations in and sums those out; I to IV are traces in GeV⁴; p⋅k, p⋅k′ in GeV²; e² = 4πα. Derived.

The squared s-channel traceI = 32[(p⋅k)(p⋅k′) + m²(p⋅k) + m⁴]

γ^νp̸γ_ν = −2p̸ and γ_μp̸′γ^μ = −2p̸′ cut eight gammas to four. The swap k ↔ −k′ turns it into the u-channel square, IV = 32[(p⋅k)(p⋅k′) − m²(p⋅k′) + m⁴].

I = Tr[(p̸′ + m)γ^μ(p̸ + k̸ + m)γ^ν(p̸ + m)γ_ν(p̸ + k̸ + m)γ_μ], in GeV⁴; m the electron mass in GeV. Derived.

The interference traceII = III = 16[2m⁴ + m²(p⋅k) − m²(p⋅k′)]

γ^μa̸γ^νb̸γ_μ = −2b̸γ^νa̸ leaves (p + k)⋅(p − k′) = (s + t + u)/2 = m², so every term carries m²: with m = 0 the diagrams do not interfere in this bookkeeping.

II = Tr[(p̸′ + m)γ^μ(p̸ + k̸ + m)γ^ν(p̸ + m)γ_μ(p̸ − k̸′ + m)γ_ν], in GeV⁴; III is II read backwards. Derived.

The spin- and polarisation-averaged |M|²⟨|M|²⟩ = 2e⁴[p⋅k′/p⋅k + p⋅k/p⋅k′ + 2m²(1/p⋅k − 1/p⋅k′) + m⁴(1/p⋅k − 1/p⋅k′)²]

1/p⋅k − 1/p⋅k′ = −k⋅k′/((p⋅k)(p⋅k′)) ≤ 0, so the m² term lowers the bracket and the m⁴ term raises it. In the lab frame the two add to −sin²θ at every energy.

Dimensionless, as any 2 → 2 amplitude squared; p⋅k = (s − m²)/2 and p⋅k′ = (m² − u)/2, in GeV². Derived.

The massless limit⟨|M|²⟩ → −2e⁴(u/s + s/u) as m → 0

The interference and the mass terms drop out. s/u diverges as θ* → π, where the u-channel electron p − k′ goes on shell because a massless electron emits its photon collinearly.

s is positive and u ≤ 0, in GeV², so both terms are positive; in the centre-of-mass frame −u/s = cos²(θ*/2). Derived.

The mass caps the u-channel poleu ≤ m⁴/s, and at θ* = π: ⟨|M|²⟩ = 2e⁴(s/m² + m²/s)

At θ* = π the m² and m⁴ terms are −4 and +4 and cancel; the cap is kinematic. The massless form fails only inside the cone cos²(θ*/2) ≲ m²/s, of half-angle about 2m/√s.

θ*: centre-of-mass angle between k and k′; the bound is p⋅k′ ≥ m²(s − m²)/(2s), in GeV². Derived.

Turning the sums into traces

Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout. Quoted from topic 10.04: M = −e²ε*_μ(k′)ε_ν(k)ū(p′)Γ^μν u(p), with Γ^μν = γ^μ(p̸ + k̸ + m)γ^ν/(2p⋅k) + γ^ν(p̸ − k̸′ + m)γ^μ/(−2p⋅k′). An unpolarised photon beam on unpolarised electrons asks for the average over 2 spins × 2 polarisations in and the sum over both out, a factor 1/4. The spin sums Σu ū = p̸ + m (UAQM-04) close the line into Tr[(p̸′ + m)Γ^μν(p̸ + m)Γ̄^ρσ], where Γ̄ = γ⁰Γ†γ⁰ reverses the order of the γ matrices. Each photon sum Σε_με*_ρ becomes −g_μρ, legal only because the summed amplitude obeys k_μM^μν = 0, and the two minus signs multiply to +1. Expanding Γ gives four traces: I and IV from each diagram squared, II and III from the cross terms. III is II read backwards, and a trace of gamma matrices equals the trace of the reversed product, so II = III, both real.

Eight gamma matrices down to four

Odd numbers of gamma matrices trace to zero, so I splits into pieces with eight, six and four. In the eight-γ piece the inner contraction γ^νp̸γ_ν = −2p̸ and, by cyclicity, the outer one γ_μp̸′γ^μ = −2p̸′ leave 4Tr[p̸′(p̸ + k̸)p̸(p̸ + k̸)] = 16[2(p′⋅(p + k))(p⋅(p + k)) − (p⋅p′)s], by topic 7.06's four-γ trace. Now spend the kinematics: k² = k′² = 0 and p² = p′² = m², and squaring p′ = p + k − k′ gives p⋅k − p⋅k′ = k⋅k′, hence p′⋅k′ = p⋅k, p′⋅k = p⋅k′ and p⋅p′ = m² + p⋅k − p⋅k′. With the six- and four-γ pieces added, I = 32[(p⋅k)(p⋅k′) + m²(p⋅k) + m⁴], derived. The u-channel diagram is the s-channel one with k ↔ −k′, so IV = 32[(p⋅k)(p⋅k′) − m²(p⋅k′) + m⁴] needs no new trace. At p⋅k = 1.5m² and p⋅k′ = 0.5m² these are 104m⁴ and 40m⁴.

The interference trace starts at order m²

II = Tr[(p̸′ + m)γ^μ(p̸ + k̸ + m)γ^ν(p̸ + m)γ_μ(p̸ − k̸′ + m)γ_ν] interleaves its indices, so the short contractions do not apply. The longer identity does: γ^μa̸γ^νb̸γ_μ = −2b̸γ^νa̸ turns the eight-γ piece into −2Tr[p̸′p̸γ^ν(p̸ + k̸)(p̸ − k̸′)γ_ν], and γ^νa̸b̸γ_ν = 4a⋅b collapses that to −32[(p + k)⋅(p − k′)](p⋅p′). Now (p + k)⋅(p − k′) = (s + u + t)/2, and the Mandelstam sum s + t + u = 2m² makes it m². So even the piece with no explicit mass is proportional to m², and with the rest, II = 16[2m⁴ + m²(p⋅k) − m²(p⋅k′)], derived. At p⋅k = 1.5m², p⋅k′ = 0.5m² the eight-γ piece is −64m⁴ and the whole trace 48m⁴. As m → 0 the cross terms vanish, a fact about this bookkeeping rather than a licence: the −g_μν sums hold only for the sum of diagrams, so I, II and IV are not separately physical.

Assembling the invariant bracket

Put each trace over its denominator. The s-channel square gives 2e⁴[p⋅k′/p⋅k + m²/p⋅k + m⁴/(p⋅k)²]; the u-channel square gives 2e⁴[p⋅k/p⋅k′ − m²/p⋅k′ + m⁴/(p⋅k′)²]; and the interference, entering as −2II/(4(p⋅k)(p⋅k′)) because the u-channel denominator is negative, gives 2e⁴[m²(1/p⋅k − 1/p⋅k′) − 2m⁴/((p⋅k)(p⋅k′))]. The m² pieces add to 2m²(1/p⋅k − 1/p⋅k′) and the m⁴ pieces complete a square, giving the averaged |M|² of the formula cards. Three checks follow at once. It is dimensionless, as |M|² for 2 → 2 must be. It is symmetric under p⋅k ↔ −p⋅k′, which is s ↔ u, the exchange of the two diagrams. And since 1/p⋅k − 1/p⋅k′ = −k⋅k′/((p⋅k)(p⋅k′)) ≤ 0, the m² term is never positive. Flip the sign of the interference and, at p⋅k = 1.5m², p⋅k′ = 0.5m², the result jumps from 4.89e⁴ to 20.9e⁴: the sign of u − m² is not a detail.

The massless limit and the collinear pole

Set m = 0 and only the ratios survive: ⟨|M|²⟩ = −2e⁴(u/s + s/u), with s positive and u ≤ 0. In the centre-of-mass frame −u/s = cos²(θ*/2), so the bracket is cos²(θ*/2) + 1/cos²(θ*/2): 2 forward, and unbounded as θ* → π. The culprit is the u-channel propagator 1/(u − m²): for a massless electron the internal line p − k′ goes on shell when the photon leaves along the incoming electron's direction. The s-channel pole is out of reach, since s − m² = 2p⋅k is positive. Restore the mass and p⋅k′ ≥ m²(s − m²)/(2s), so u ≤ m⁴/s, short of the pole at u = m². At θ* = π the ratio p⋅k/p⋅k′ is exactly s/m², the m² and m⁴ terms are −4 and +4, and the bracket is s/m² + m²/s. Well outside a backward cone, cos²(θ*/2) ≲ m²/s or π − θ* ≲ 2m/√s, the massless form is accurate. Integrated over angle, the cap turns its logarithmic divergence into ln(s/m²), the logarithm behind topic 10.07's high-energy fall.

Threshold, a measured number, and the limits

At threshold p⋅k′ → p⋅k, yet the mass terms do not vanish. In the lab p⋅k = mω and p⋅k′ = mω′, and the Compton relation 1/ω′ − 1/ω = (1 − cos θ)/m (UQMI-01.04) makes 1/p⋅k − 1/p⋅k′ = −(1 − cos θ)/m² exactly, so the mass terms total −2(1 − cos θ) + (1 − cos θ)² = −sin²θ at every energy and ⟨|M|²⟩ = 2e⁴[ω′/ω + ω/ω′ − sin²θ]. As ω → 0 this is 2e⁴(1 + cos²θ), Thomson's angular factor (UAQM-12.09). Topic 10.06 turns it into the Klein-Nishina cross-section, 0.2112 b per electron at 1 MeV; water's attenuation coefficient there, 0.0707 cm² g⁻¹ (quoted), is 0.2115 b per electron, agreement to 0.2%. The result is for a free electron at tree level. Binding enters when ω nears atomic binding energies, 88.0 keV for lead's K shell (quoted), and radiative corrections, double Compton scattering among them, at order α/π ≈ 0.23%.

Change one variable at a time

Make the relationship visible.

Interactive model

Lab photon energy, log₁₀(ω/MeV)0.5

Centre-of-mass angle θ*170 °

Start at log₁₀(ω/MeV) = −2, where the solid curve lies near Thomson's 1 + cos²θ* on the floor. Raise the energy: well right of the vertical line the solid curve follows the dashed one, and left of it the solid curve levels off at s/m². Then slide θ* to 178° and compare the readouts.

s/m²13.4

⟨|M|²⟩/2e⁴ AT θ*11.97

MASSLESS −(u/s + s/u) AT θ*131.65

BACKWARD PEAK s/m² + m²/s13.5

Live interpretations/m²: 13.4. ⟨|M|²⟩/2e⁴ AT θ*: 11.97. MASSLESS −(u/s + s/u) AT θ*: 131.65. BACKWARD PEAK s/m² + m²/s: 13.5

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading is that the m² and m⁴ terms are mass corrections: small at low energy, where p⋅k′ ≈ p⋅k, so that 2e⁴[p⋅k′/p⋅k + p⋅k/p⋅k′] ≈ 4e⁴ is a fair first answer. It is wrong because each mass term is a difference of large pieces. At 20 keV, 2m²/p⋅k and 2m²/p⋅k′ are 51.1 and 54.1, yet they combine to −3.00 at 120°; in the laboratory frame the two mass terms total −sin²θ at every energy. At low energy they are as large as the leading terms, they turn 2 into 1 + cos²θ, and dropping them makes the Thomson cross-section 3/2 times too large. The opposite slip at high energy is to set m = 0 everywhere. Dropping the explicit mass terms is then harmless, at most −1 against a bracket that reaches s/m², but dropping the mass from the kinematics is not: it moves the pole at u = m² onto the edge of the physical region and makes the cross-section diverge.

Compton scattering at √s = 5.11 MeV, so s = 100m². What is ⟨|M|²⟩/(2e⁴) for exact backscattering, θ* = 180° in the centre-of-mass frame, and why?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA 20.0 keV photon scatters through θ = 120° off a free electron at rest. Find p⋅k and p⋅k′, evaluate each of the four terms in the bracket of ⟨|M|²⟩ = 2e⁴[…], and compare the total with Thomson's angular factor 1 + cos²θ.

Compton relation (UQMI-01.04), with m = 511.0 keV: ω′ = ω/(1 + (ω/m)(1 − cos θ)) = 20.0/(1 + 0.03914 × 1.5) = 20.0/1.05871 = 18.891 keV.
In the lab p = (m, 0), so p⋅k = mω = 1.0220 × 10⁴ keV² and p⋅k′ = mω′ = 9.653 × 10³ keV².
Ratio terms: p⋅k′/p⋅k = ω′/ω = 0.9445 and p⋅k/p⋅k′ = 1.0587, together 2.0033.
Mass terms: 2m²/p⋅k = 51.10 and 2m²/p⋅k′ = 54.10, so 2m²(1/p⋅k − 1/p⋅k′) = −3.00, and m²(1/p⋅k − 1/p⋅k′) = −1.50 squares to the m⁴ term, +2.25. These are −2(1 − cos θ) and (1 − cos θ)², whatever ω is.
Bracket: 2.0033 − 3.00 + 2.25 = 1.2533, against 1 + cos²θ = 1.2500: 0.26% above Thomson's factor, the first trace of recoil. With e² = 4πα = 0.09170, ⟨|M|²⟩ = 2e⁴ × 1.2533 = 0.0211.

AnswerBracket = 0.9445 + 1.0587 − 3.00 + 2.25 = 1.2533, 0.26% above 1 + cos²θ = 1.25, so ⟨|M|²⟩ = 2.507e⁴ = 0.0211. The mass terms, −0.75 in all, are as large as the angular dependence itself.

MediumAt one point of γe⁻ → γe⁻, p⋅k = 1.5m² and p⋅k′ = 0.5m². Show the point is physical and find its lab and centre-of-mass angles, evaluate the traces I, II and IV, assemble ⟨|M|²⟩ and check it against the closed bracket. What does writing the u-channel denominator as +2p⋅k′ do?

k⋅k′ = p⋅k − p⋅k′ = m² ≥ 0, as two real photons require. In the lab, with m = 0.5110 MeV, ω = p⋅k/m = 0.7665 MeV and ω′ = p⋅k′/m = 0.2555 MeV, so cos θ = 1 − m(1/ω′ − 1/ω) = 1 − (2 − 2/3) = −1/3: θ = 109.5°.
s = m² + 2p⋅k = 4m². In the centre-of-mass frame p⋅k′/p⋅k = [(s + m²) + (s − m²)cos θ*]/(2s) = (5 + 3cos θ*)/8 = 1/3, so cos θ* = −7/9 and θ* = 141.1°; p⋅k′ lies inside its range, 0.375m² to 1.5m².
Traces in units of m⁴: I = 32(0.75 + 1.5 + 1) = 104, IV = 32(0.75 − 0.5 + 1) = 40, II = III = 16(2 + 1.5 − 0.5) = 48. Of II, the eight-γ piece is −32m²(p⋅p′) = −64, since p⋅p′ = m² + p⋅k − p⋅k′ = 2m².
Assemble: ⟨|M|²⟩/e⁴ = (1/4)[104/(4 × 2.25) − 2 × 48/(4 × 0.75) + 40/(4 × 0.25)] = (1/4)(11.556 − 32 + 40) = 4.889.
Closed bracket: 1/3 + 3 + 2(1/1.5 − 1/0.5) + (1/1.5 − 1/0.5)² = 0.3333 + 3 − 2.6667 + 1.7778 = 2.4444, and 2 × 2.4444 = 4.889, as assembled. With e⁴ = (4πα)² = 8.409 × 10⁻³, ⟨|M|²⟩ = 0.0411.
With +2p⋅k′ the interference changes sign: (1/4)(11.556 + 32 + 40) = 20.89, 4.3 times too large, and at low energy the error in ⟨|M|²⟩/e⁴ would grow as 8m²/(ωω′) instead of leaving 1 + cos²θ.

Answerθ = 109.5° in the lab and θ* = 141.1°; I = 104m⁴, II = III = 48m⁴, IV = 40m⁴; ⟨|M|²⟩ = 44e⁴/9 = 4.889e⁴ = 0.0411. The wrong sign gives 20.9e⁴.

HardCompton scattering at √s = 10.0 MeV, in the centre-of-mass frame. Find the smallest p⋅k′ and the largest u; evaluate ⟨|M|²⟩ at θ* = 180°, showing what the mass terms do; compare the massive and massless brackets at θ* = 175°; and find the backward cone inside which the massless formula overstates ⟨|M|²⟩ by more than a factor of 2, comparing its half-angle with 2m/√s.

s/m² = (10.0/0.5110)² = 383.0. The photon has ω* = (s − m²)/(2√s) = 4.987 MeV and the electron E* = (s + m²)/(2√s) = 5.013 MeV, so p⋅k = ω*(E* + ω*) = (s − m²)/2 = 49.87 MeV².
p⋅k′ = ω*(E* + ω* cos θ*) is smallest at θ* = 180°: ω*(E* − ω*) = m²(s − m²)/(2s) = 0.1302 MeV². So u = m² − 2p⋅k′ ≤ m⁴/s = 6.82 × 10⁻⁴ MeV², almost a full m² = 0.2611 MeV² below the pole at u = m².
At 180°: p⋅k/p⋅k′ = 383.0 and its inverse 0.0026; m²(1/p⋅k − 1/p⋅k′) = 0.2611 × (0.02005 − 7.679) = −2.000, so the m² term is −4.000 and the m⁴ term +4.000. They cancel: the bracket is s/m² + m²/s = 383.0 and ⟨|M|²⟩ = 2e⁴ × 383.0 = 6.44.
At 175°: p⋅k′ = 0.2249 MeV², giving 0.0045 + 221.78 − 2.312 + 1.336 = 220.8. The massless bracket is cos²(87.5°) + 1/cos²(87.5°) = 0.0019 + 525.6 = 525.6, 2.38 times larger.
Near the back the massive bracket is about (s/m²)/(1 + (s/m²)cos²(θ*/2)) and the massless one 1/cos²(θ*/2), so their ratio is about 1 + m²/(s cos²(θ*/2)), which is 2 at cos²(θ*/2) = m²/s: π − θ* ≈ 2m/√s = 0.1022 rad = 5.86°. Solving the full brackets numerically puts the factor 2 at θ* = 174.12°, a half-angle of 5.88°.

Answerp⋅k′ ≥ 0.1302 MeV² and u ≤ 6.82 × 10⁻⁴ MeV²; at 180° the mass terms cancel and ⟨|M|²⟩ = 2e⁴ × 383.0 = 6.44; at 175°, 220.8 against 525.6; the factor-2 cone has half-angle 5.88°, against 2m/√s = 5.86°.