University Quantum Field Theory I · QED at Tree Level: Moller, Bhabha and Compton Scattering and Pair Annihilation · 10.8
Pair annihilation e⁺e⁻ → γγ by crossing Compton scattering
Pair annihilation is a cross-section you get without a new trace. Turn Compton's incoming photon and outgoing electron around, pay one minus sign for the crossed fermion, and topic 10.05's result becomes Dirac's 1930 formula: isotropic and growing as 1/v at threshold, crowded along the beam pipe at high energy.
Build the model
Connect the measurement to the mechanism.
Crossing says that one analytic function describes every process its diagrams can draw. Compton scattering γe⁻ → γe⁻ and annihilation e⁺e⁻ → γγ share the same two diagrams, an electron line with two photons attached, read with different legs in and out: reverse the incoming photon, k → −k₁, and the outgoing electron, p′ → −p₂, and Compton's s- and u-channel diagrams become the t- and u-channel exchanges in which the electron emits one photon and meets the positron, which emits the other. The spin-summed |M|² carries over by substitution with one change: the crossed fermion's spin sum p̸′ + m becomes −(p̸₂ − m), a factor −1 that a direct trace confirms and positivity demands.
After that the kinematics does the physics. Both exchanged electrons are spacelike, and one of them comes close to its mass shell whenever a photon leaves along the beam, so at high energy the photons crowd into cones of half-angle 2m/√s and the total, (2πα²/s)[ln(s/m²) − 1], carries the logarithm of that crowding. At threshold each photon still leaves with 511 keV while the incoming flux vanishes, so σ grows as 1/v and σv tends to πα²/m².
The model stops where its assumptions do: below v ≈ α the pair's Coulomb attraction, absent at tree level, enhances the rate and ends in positronium, and at relative order α radiative corrections and three-photon final states enter.
- Simple definition
- Pair annihilation e⁺e⁻ → γγ is the tree-level QED process in which an electron and a positron turn into two photons by t- and u-channel electron exchange; its spin-summed |M|² is Compton's at crossed momenta, times −1 for the one crossed fermion.
- Example
- At threshold ⟨|M|²⟩ = 4e⁴ in every direction and σv → πrₑ²c = 7.48 × 10⁻¹⁵ cm³ s⁻¹. A positron moving at v = 0.30 through electrons at rest has σ = 0.830 b, within 0.13% of the threshold law πrₑ²/v = 0.832 b.
Σu(p′)ū(p′) = p̸′ + m at p′ = −p₂ is −(p̸₂ − m), minus the positron's Σvv̄. The photon crosses free, since −gμν is even in k. Compton's s-channel becomes the t-channel.
Σ: all spins and polarisations; Compton's p, k come in and p′, k′ go out; momenta in GeV. The −1 is the crossed fermion. Derived.
The substitution p⋅k → −p₁⋅k₁ and the −1 turn Compton's m² term positive and its m⁴ term negative. At threshold, p₁⋅k₁ = p₁⋅k₂ = m², it is 2e⁴(2 + 4 − 4) = 4e⁴ in every direction.
Average over 2 × 2 spins in, sum over polarisations; dimensionless; p₁⋅k₁ = (m² − t)/2, p₁⋅k₂ = (m² − u)/2 in GeV². Derived.
Isotropic at threshold, where D = 1 and the bracket is 1. At high energy it is (1 + cos²θ*)/sin²θ* off the beam and peaks at (1 + β²)/(1 − β²) ≈ s/2m² along it, with half-width 2m/√s.
β: each beam's CM speed; θ*: angle from e⁻ to a photon; s in GeV², σ in GeV⁻². Counts each photon direction: no 1/2 yet. Derived.
Integrate the angular distribution over −1 ≤ cos θ* ≤ 1 and halve. At β = 0.5 it is 1.108πrₑ² = 0.276 b, as Dirac's laboratory form with γ = 5/3 also gives.
Includes the 1/2 for identical photons; σ in GeV⁻², times 0.3894 mb GeV² for barns. Dirac's 1930 result in CM variables. Derived.
Falls as 1/s, like e⁺e⁻ → μ⁺μ⁻, but with a logarithm from the nearly on-shell exchanged electron. At √s = 1.00 GeV it is 1845 nb, within 2 parts in 10⁶ of the exact form.
s in GeV², σ in GeV⁻²; m is the electron mass. The logarithm is collinear: equal shares per decade of angle from θ* ≈ 2m/√s out to about 1 rad. Derived.
The flux vanishes, not the final state: each photon still leaves with about 511 keV. It is 3/8 of Thomson's (8π/3)rₑ² times c, and below v ≈ α the Coulomb attraction outgrows it.
v: positron speed in the electron's rest frame; rₑ = α/m = 2.818 fm. The first correction is −0.15v⁴. Derived.
Crossing the Compton legs
Natural units ħ = c = 1 and the metric (+, −, −, −) hold throughout. Quoted from topic 10.05: for γ(k)e⁻(p) → γ(k′)e⁻(p′), ⟨|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]. Crossing (topic 9.08) moves a leg to the other side of the reaction by reversing its four-momentum. The incoming photon becomes an outgoing one, k = −k₁, and the outgoing electron an incoming positron, p′ = −p₂, while p = p₁ and k′ = k₂ stay. So p⋅k → −p₁⋅k₁ and p⋅k′ → p₁⋅k₂, and Compton's s = (p + k)² becomes (p₁ − k₁)² = t, its t becomes the annihilation s, and u stays u. The s- and u-channel Compton diagrams are now the t- and u-channel exchanges of e⁺e⁻ → γγ. Both exchanged electrons are spacelike: t = m² − 2p₁⋅k₁ and u = m² − 2p₁⋅k₂ never exceed −m²(1 − β)/(1 + β), with β each beam's centre-of-mass speed, so both are negative across the physical region. Both also lie below their mass shell, t − m² = −2p₁⋅k₁ and u − m² = −2p₁⋅k₂, where Compton's s − m² = 2p⋅k was positive. No new diagram, and no new trace, is needed.
One crossed fermion, one factor −1
The substitution acts on the spin-summed |M|², and the spin sums are where the sign hides. Compton's outgoing electron supplies Σu(p′)ū(p′) = p̸′ + m; annihilation's incoming positron supplies Σv(p₂)v̄(p₂) = p̸₂ − m (UAQM-04). At p′ = −p₂ the first is −p̸₂ + m = −(p̸₂ − m). Vertices, propagators and the −gμν polarisation sums carry over by substitution alone, and a crossed photon brings no sign, because −gμν does not care which way k points. Both processes average over four initial states, 2 spins × 2 polarisations against 2 × 2 spins, so the average inherits the −1 directly, giving the annihilation ⟨|M|²⟩ of the formula cards, derived. The two ratio terms change sign twice and survive; the m² term turns positive and the m⁴ term negative. At threshold, p₁⋅k₁ = p₁⋅k₂ = m², the result is 2e⁴(2 + 4 − 4) = 4e⁴; without the −1 it would be −4e⁴, a negative sum of squared moduli. Møller to Bhabha crosses two fermions, and their two signs cancel.
The direct trace agrees
Unit 9's rules give the annihilation amplitude directly: M = −e²ε*μ(k₁)ε*ν(k₂)v̄(p₂)[γν(p̸₁ − k̸₁ + m)γμ/(t − m²) + γμ(p̸₁ − k̸₂ + m)γν/(u − m²)]u(p₁). Summing both photons' polarisations with −gμν, which the Ward identity licenses for the two diagrams together (topic 10.04), the squared t-channel numerator is T = Tr[(p̸₂ − m)γν(p̸₁ − k̸₁ + m)γμ(p̸₁ + m)γμ(p̸₁ − k̸₁ + m)γν]. The inner contraction gives γμ(p̸₁ + m)γμ = −2p̸₁ + 4m; the outer one, by γνa̸b̸c̸γν = −2c̸b̸a̸, γνa̸b̸γν = 4a⋅b and γνa̸γν = −2a̸, leaves traces of at most four γ matrices. With p₂⋅k₂ = p₁⋅k₁ and p₁⋅p₂ = p₁⋅k₁ + p₁⋅k₂ − m², T = 32[(p₁⋅k₁)(p₁⋅k₂) + m²(p₁⋅k₁) − m⁴], derived. Compton's s-channel square, I = 32[(p⋅k)(p⋅k′) + m²(p⋅k) + m⁴], crossed, is −32[(p₁⋅k₁)(p₁⋅k₂) + m²(p₁⋅k₁) − m⁴], and the factor −1 turns it into T. At β = 0.6, θ* = 60°, where p₁⋅k₁ = 1.09375m² and p₁⋅k₂ = 2.03125m², T = 74.09m⁴. The interference and u-channel traces go the same way, and the full sum, evaluated with explicit 4 × 4 Dirac matrices, matches the crossed formula to rounding.
Isotropic at threshold, crowded along the beam at high energy
In the centre-of-mass frame each beam has energy E = √s/2 and speed β, each photon has energy E, and θ* is the angle from the electron to photon k₁. Then p₁⋅k₁ = E²(1 − β cos θ*), p₁⋅k₂ = E²(1 + β cos θ*) and m² = E²(1 − β²), so with D = 1 − β²cos²θ*, ⟨|M|²⟩ = 4e⁴[(2 − D)/D + 2(1 − β²)/D − 2(1 − β²)²/D²]. Topic 5.04's dσ/dΩ = ⟨|M|²⟩|pf|/(64π²s|pᵢ|), with |pf|/|pᵢ| = 1/β, turns it into the angular distribution of the formula cards, derived. At threshold D = 1 and the bracket is 1 + 2 − 2 = 1: isotropic, two back-to-back 511 keV photons in any direction. At high energy, for sin θ* ≫ 2m/√s, the bracket tends to (1 + cos²θ*)/sin²θ*, because the exchanged electron, with t − m² = −2E²(1 − β cos θ*), is nearly on shell when a photon leaves along the electron. At θ* = 0 the two mass terms, +2 and −2, cancel, and the peak is (1 + β²)/(1 − β²) ≈ s/2m², falling to about half height at cos θ* = β, where sin θ* = √(1 − β²) = 2m/√s. At √s = 1.00 GeV the peak stands 1.9 × 10⁶ times the 90° value, with half-width 1.02 mrad.
Identical photons and Dirac's total cross-section
The two photons are identical bosons, so photon k₁ along n̂ with k₂ along −n̂ is the same final state as the swap, and integrating dσ/dΩ over the whole sphere counts every state twice: σ = (1/2)∫dΩ dσ/dΩ. With c = cos θ* from −1 to 1 and L = ln((1 + β)/(1 − β)), the integrals ∫dc/D = L/β and ∫dc/D² = 1/(1 − β²) + L/(2β) give Dirac's formula of the cards, derived. Every term carries α²/s, of mass dimension −2, an area, as a cross-section must. With the electron at rest and the positron's Lorentz factor γ = (1 + β²)/(1 − β²), the same function is Dirac's 1930 laboratory form, σ = (πrₑ²/(γ + 1))[((γ² + 4γ + 1)/(γ² − 1)) ln(γ + √(γ² − 1)) − (γ + 3)/√(γ² − 1)], quoted; at β = 0.5, γ = 5/3, both give 1.108πrₑ² = 0.276 b. As s ≫ m², L → ln(s/m²) and σ → (2πα²/s)[ln(s/m²) − 1], derived: 1845 nb at √s = 1.00 GeV. The logarithm is collinear: near the beam (1 + c²)/(1 − c²) dc becomes 2dθ*/θ*, so every decade of angle down to the cutoff at 2m/√s adds the same amount. The LEP experiments measured e⁺e⁻ → γγ at √s up to about 209 GeV and found this QED prediction, with its order-α corrections, within errors of a few per cent (quoted).
Threshold: σ as 1/v, σv as πrₑ², and where tree level stops
As β → 0, L/β → 2, Dirac's bracket tends to 2 and σ → πα²/(2m²β). The positron's speed in the electron's rest frame is v = 2β/(1 + β²), so σv → πα²/m² = πrₑ², derived, approached as πrₑ²(1 − 0.15v⁴ + …). With c restored, πrₑ²c = 7.48 × 10⁻¹⁵ cm³ s⁻¹, three-eighths of Thomson's cross-section times c. The 1/v is the flux: |M|² stays 4e⁴ and the photons leave with 511 keV however slowly the pair moves, whereas in e⁺e⁻ → μ⁺μ⁻ it is the final phase space that closes, and σ vanishes. Treating water's 3.34 × 10²³ electrons cm⁻³ as free, the rate per positron is nπrₑ²c = 2.50 × 10⁹ s⁻¹ at any slow speed. Tree level fails as v nears α: the Coulomb attraction it omits multiplies the rate by Sakharov's factor (2πα/v)/(1 − e(−2πα/v)), quoted, 1.08 at v = 0.30 and 6.3 at v = α, and ends in positronium, topic 10.09. Crossing once more gives γγ → e⁺e⁻, named: head-on, a 1 TeV γ-ray needs a partner above 0.26 eV.
Change one variable at a time
Make the relationship visible.
Start at β = 0.010: the curve lies flat on the axis, isotropic at 4e⁴, and σv/πrₑ² reads 1.000 while σ grows as 1/v; much lower and v/α nears 1, where tree level fails. Then drag towards 0.999: the beam peaks rise to about s/2m² and σ closes on the high-energy logarithm.
σ, DIRAC (TREE LEVEL)0.0696 b
σv / πrₑ² (v: e⁺ lab speed)0.278
σ / (2πα²/s)(ln(s/m²) − 1)1.435
e⁺ LAB SPEED v/α136.3
Live interpretationσ, DIRAC (TREE LEVEL): 0.0696 b. σv / πrₑ² (v: e⁺ lab speed): 0.278. σ / (2πα²/s)(ln(s/m²) − 1): 1.435. e⁺ LAB SPEED v/α: 136.3
Catch the common trap
Explain before calculating.
Treat a slow positron in a gas of free electrons at rest, at tree level. Its speed falls from v = 0.20 to v = 0.10, both well above α and well below 1. What happens to the annihilation cross-section σ and to the annihilation rate per positron, nσv?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyConsider e⁺e⁻ → γγ in the centre-of-mass frame, each beam at β = 0.600, with photon k₁ at θ* = 60.0° to the electron. Find p₁⋅k₁ and p₁⋅k₂, evaluate Compton's averaged bracket at the crossed invariants, apply the crossing sign to get ⟨|M|²⟩, check it against the centre-of-mass form, and find dσ/dΩ in b sr⁻¹.
- Each beam has E = m/√(1 − β²) = m/√(1 − 0.36) = 1.25m, so E² = 1.5625m². Then p₁⋅k₁ = E²(1 − β cos θ*) = 1.5625m² × 0.700 = 1.09375m² and p₁⋅k₂ = 1.5625m² × 1.300 = 2.03125m². Check: they sum to 2E² = s/2 = 3.125m².
- Crossed Compton invariants: p⋅k = −1.09375m² and p⋅k′ = +2.03125m². Compton's bracket term by term: p⋅k′/p⋅k = −1.85714, p⋅k/p⋅k′ = −0.53846, 2m²(1/p⋅k − 1/p⋅k′) = 2(−0.91429 − 0.49231) = −2.81319, and m⁴(1/p⋅k − 1/p⋅k′)² = +1.97851. Sum: −3.23029.
- One crossed fermion, one −1: ⟨|M|²⟩ = −2e⁴ × (−3.23029) = 6.4606e⁴. Without it the result is −6.46e⁴, negative and so impossible.
- Centre-of-mass check: D = 1 − 0.36 × 0.25 = 0.910, and (2 − D)/D + 2(0.64)/D − 2(0.64)²/D² = 1.19780 + 1.40659 − 0.98925 = 1.61514, so ⟨|M|²⟩ = 4e⁴ × 1.61514 = 6.4606e⁴, as crossed. With e⁴ = (4πα)² = 8.409 × 10⁻³, ⟨|M|²⟩ = 0.05433.
- dσ/dΩ = ⟨|M|²⟩/(64π²sβ) with s = 6.25m² = 1.6320 MeV²: 0.05433/(64π² × 1.6320 MeV² × 0.600) = 8.784 × 10⁻⁵ MeV⁻², and 1 MeV⁻² = 389.4 b, so 0.0342 b sr⁻¹. That is per photon direction; the 1/2 for identical photons enters only when integrating over the sphere.
Answerp₁⋅k₁ = 1.094m², p₁⋅k₂ = 2.031m²; ⟨|M|²⟩ = 6.461e⁴ = 0.0543, against −6.46e⁴ without the crossing sign; dσ/dΩ = 0.0342 b sr⁻¹.
MediumAn e⁺e⁻ collider runs at √s = 1.00 GeV and counts γγ events in a barrel detector covering |cos θ*| ≤ 0.80. Find the total tree-level cross-section in nb, the differential cross-section dσ/dcos θ* at 90°, the cross-section inside the barrel, the fraction of all γγ pairs the barrel sees, and the angular half-width of the beam peaks.
- s/m² = (1000 MeV/0.5110 MeV)² = 3.830 × 10⁶, so 1 − β² = 4m²/s = 1.04 × 10⁻⁶: the massless form (1 + c²)/(1 − c²), with c = cos θ*, holds everywhere except within about 2m/√s of the beam. ln(s/m²) = 15.158.
- Total: σ = (2πα²/s)[ln(s/m²) − 1] = 3.346 × 10⁻⁴ GeV⁻² × 14.158 = 4.737 × 10⁻³ GeV⁻², and × 0.3894 mb GeV² this is 1.845 × 10⁻³ mb = 1845 nb. Dirac's exact form agrees to 2 parts in 10⁶.
- At 90°: dσ/dcos θ* = (2πα²/(βs)) × 1 = 3.346 × 10⁻⁴ GeV⁻² = 130.3 nb per unit cos θ*.
- Inside the barrel, with the 1/2 for identical photons: σbarrel = (1/2)(2πα²/s)∫(1 + c²)/(1 − c²) dc over |c| ≤ 0.80 = (πα²/s)[2 ln 9 − 1.6] = 65.14 nb × 2.7944 = 182.0 nb.
- Fraction: 182.0/1845 = 9.9%. The other 90% lies within 36.9° of the beam, spread almost evenly in ln θ*, since near the beam (1 + c²)/(1 − c²) dc becomes 2dθ*/θ*: 65% of all pairs fall within 0.1 rad of the beam, 32% within 10 mrad, and only 5% inside the beam peaks proper, which stand s/2m² = 1.9 × 10⁶ times the 90° value with half-width 2m/√s = 1.02 mrad.
Answerσ = 1845 nb in all; dσ/dcos θ* = 130.3 nb at 90°; 182.0 nb, or 9.9% of the total, inside |cos θ*| ≤ 0.80, the rest spread almost evenly in ln θ* towards the beam; beam peaks of half-width 1.02 mrad, holding only 5% of the total.
HardA positron of kinetic energy 1.00 MeV annihilates in flight on an electron treated as free and at rest. Find s, √s and the centre-of-mass speed β; the tree-level σ from both the laboratory and centre-of-mass forms; how far the threshold law πrₑ²/v is off; the ratio of the beam-axis to the 90° angular density in the centre-of-mass frame; and the laboratory photon energies at those angles.
- γ = 1 + T/m = 1 + 1.00/0.5110 = 2.9570 and v = √(1 − 1/γ²) = 0.9411. s = 2m²(1 + γ) = 7.9139m², so √s = 2.8131m = 1.4375 MeV, and β = √(1 − 4m²/s) = 0.70325. Check: (1 + β²)/(1 − β²) = 2.9570 = γ.
- Laboratory form: ((γ² + 4γ + 1)/(γ² − 1)) ln(γ + √(γ² − 1)) = 2.78572 × 1.74740 = 4.86777, and (γ + 3)/√(γ² − 1) = 2.14069; their difference over γ + 1 = 3.95695 is 0.68919. In units of πrₑ² the centre-of-mass form is ((1 − β²)/(4β))[(3 − β⁴)L/β − 2(2 − β²)], and L = ln((1 + β)/(1 − β)) = 1.74740 is the same logarithm, so it gives 0.68919 too. So σ = 0.68919 × 0.24947 b = 0.1719 b.
- Threshold law: πrₑ²/v = 0.24947 b/0.9411 = 0.2651 b, 1.54 times too large; σv/πrₑ² = 0.649 here rather than 1. The law assumed √s ≈ 2m, but s = 7.91m² and σ is already turning over towards its 1/s high-energy form.
- Angular density: at θ* = 0 the bracket is (1 + β²)/(1 − β²) = 2.957, the positron's γ, and at 90° it is 1 + 2(1 − β²)β² = 1.4999, so photons along the beam axis are 1.97 times as likely per unit cos θ* as photons at 90°.
- Laboratory energies: Eₜₒₜ = m + γm = 2.0220 MeV and p = m√(γ² − 1) = 1.4220 MeV. Along the axis the photons carry (Eₜₒₜ ± p)/2 = 1.7220 MeV forward and 0.3000 MeV backward; at θ* = 90° each carries Eₜₒₜ/2 = 1.0110 MeV. In-flight annihilation spreads the photons from 0.300 to 1.722 MeV, weighted towards the ends, instead of making the 511 keV line.
Answers = 7.914m², √s = 1.4375 MeV, β = 0.7033; σ = 0.689πrₑ² = 0.172 b by both forms, against πrₑ²/v = 0.265 b; beam axis 1.97 times the 90° density; photons of 1.722 and 0.300 MeV along the axis, 1.011 MeV each at 90°.