University Quantum Field Theory I · QED at Tree Level: Moller, Bhabha and Compton Scattering and Pair Annihilation · 10.3

Bhabha scattering: s and t channels interfering

Bhabha scattering is where you must fix the sign between two diagrams with no identical particles to blame, and a wrong sign moves the 90° rate by 44 per cent. Get the result twice, by crossing Møller and by a trace, and see why every collider counts small-angle Bhabha pairs to know its luminosity.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Elastic e⁺e⁻ scattering has two tree-level diagrams: the pair can exchange a spacelike photon in the t channel, or annihilate into a timelike virtual photon of invariant mass √s that rematerialises as a new pair in the s channel. Both end in the same final state, so their amplitudes add, and the sign between them is not a convention. The field ψ̄ both creates the outgoing electron and destroys the incoming positron, and the two diagrams differ only in which of two anticommuting ψ̄ fields does which job: one transposition, worth −1.

Crossing Møller scattering by s ↔ u turns its u channel into Bhabha's annihilation and carries that sign along, giving at high energy dσ/dΩ = (α²/(2s))[(s² + u²)/t² + (t² + u²)/s² + 2u²/(st)]; a direct trace confirms it, with the interference coming out as one trace threaded through both fermion lines. The last term is negative throughout the physical region, where s is positive and t negative, a sign fixed by the propagators as much as by the minus, for Møller's interference, built with the same minus, is positive. The exchange pole makes forward scattering large, pure QED and calculable, which is why luminosity is counted with it.

The model stops where its assumptions do: masses matter only for |t| ≲ m², radiative corrections and the running of α move it by per cents, and near √s = 91 GeV the Z boson, not the photon, rules the s channel.

Simple definition: Bhabha scattering is elastic electron-positron scattering, e⁺e⁻ → e⁺e⁻, whose tree-level amplitude is the t-channel photon-exchange diagram minus the s-channel annihilation diagram.
Example: At √s = 10 GeV and θ = 90°, the exchange, annihilation and interference terms are 5, 0.5 and −1 in units of α²/(2s) = 0.1037 nb, so dσ/dΩ = 0.467 nb sr⁻¹, nine times the μ-pair rate at that angle.

The amplitude with its relative signM = e²[(ū₃γ^μu₁)(v̄₂γ_μv₄)/t − (v̄₂γ^μu₁)(ū₃γ_μv₄)/s]

The minus is a swap of two anticommuting ψ̄ fields, not a convention. Only the relative sign is physical: an overall phase drops out of |M|², which is dimensionless for 2 → 2.

e⁻(p₁)e⁺(p₂) → e⁻(p₃)e⁺(p₄); s = (p₁ + p₂)², t = (p₁ − p₃)² in GeV²; spinors GeV¹⁄²; e² = 4πα. Derived.

Crossing from MøllerMøller with p₂ → −p₄, p₄ → −p₂: s ↔ u, t unchanged, overall sign (−1)² = +1

A relabelling turns a known result into a new one. Each crossed fermion turns Σuū = p̸ + m into −Σvv̄, and two such signs cancel.

Topic 10.01's t- and u-channel diagrams become Bhabha's exchange and annihilation diagrams, keeping their relative sign. Derived.

The interference is one traceΣ Mₜ Mₛ* = (e⁴/(st)) Tr[p̸₃γ^μp̸₁γ^νp̸₂γ_μp̸₄γ_ν] = −8e⁴u²/(st)

The four spinor chains close into a single loop through both lines, and contractions cut the eight-γ trace to −32(p₁⋅p₄)(p₂⋅p₃) = −8u².

Mₜ, Mₛ: the two terms of M without the minus; massless sums Σuū = p̸, Σvv̄ = p̸; u = (p₁ − p₄)² in GeV². Derived.

Bhabha cross-section at high energydσ/dΩ = (α²/(2s))[(s² + u²)/t² + (t² + u²)/s² + 2u²/(st)]

Exchange, annihilation, interference. With s positive and t negative the last term always subtracts, and it vanishes only at θ = 180°, where u = 0.

Centre-of-mass frame, s ≫ m²; α = 1/137.036; dσ/dΩ in GeV⁻² sr⁻¹, times 0.3894 mb GeV² for millibarns. Derived.

The same against the angledσ/dΩ = (α²/(2s))[(4 + (1 + c)²)/(1 − c)² + (1 + c²)/2 − (1 + c)²/(1 − c)], c = cos θ

At 90° the terms are 5, 1/2 and −1. The middle one is topic 9.06's (α²/(4s))(1 + cos²θ), and forward of 102° the interference outweighs it.

From t = −s(1 − c)/2 and u = −s(1 + c)/2, with θ the angle between outgoing and incoming electron. Derived.

Small-angle acceptance of a luminometerσ_acc = (16πα²/s)(1/θₘᵢₙ² − 1/θₘₐₓ²)

Counting N = σ_acc ∫L dt events measures the luminosity. The inner edge dominates: for 30 to 60 mrad, δσ/σ = −2.67 δθₘᵢₙ/θₘᵢₙ.

From the forward limit dσ/dΩ → 16α²/(sθ⁴); θ in rad; σ_acc in GeV⁻², times 0.3894 mb GeV². Derived; its use for luminosity quoted.

Two diagrams from one field, and the sign between them

Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. Label e⁻(p₁)e⁺(p₂) → e⁻(p₃)e⁺(p₄), with s = (p₁ + p₂)², t = (p₁ − p₃)² and u = (p₁ − p₄)². At order e² the vertex ψ̄γ^μψA_μ acts at two points x and y. ψ destroys electrons and creates positrons; ψ̄ creates electrons and destroys positrons. After the x ↔ y relabelling that cancels the 1/2! of the Dyson series, ψ(x) absorbs the incoming electron and ψ(y) emits the outgoing positron, and the two ψ̄ fields share the remaining jobs. Let ψ̄(x) emit the outgoing electron and the chains are (ū₃γ^μu₁)(v̄₂γ_μv₄): exchange, with photon momentum p₁ − p₃. Let it absorb the incoming positron and they are (v̄₂γ^μu₁)(ū₃γ_μv₄): annihilation, with photon momentum p₁ + p₂. The two pairings differ by swapping the partners of two anticommuting fields, so Wick's theorem gives them opposite signs. No identical particles are involved: the minus belongs to the Fermi field, not to the final state.

Crossing Møller by s ↔ u

Topic 10.01 found Møller's massless result ⟨|M|²⟩ = 2e⁴[(s² + u²)/t² + (s² + t²)/u² + 2s²/(tu)]. Bhabha is the same pair of fields at crossed momenta. Turn Møller's incoming electron p₂ into an outgoing positron and its outgoing electron p₄ into an incoming positron: p₂ → −p₄, p₄ → −p₂. Møller's s = (p₁ + p₂)² becomes (p₁ − p₄)², Bhabha's u; its u becomes Bhabha's s; t = (p₁ − p₃)² is untouched. Topic 9.08's rule gives −1 per crossed fermion on the summed |M|², because Σuū = p̸ + m becomes −(p̸ − m) = −Σvv̄ when p → −p, so two crossed fermions give (−1)² = +1, and both processes average over the same four initial spin states. Diagram by diagram, Møller's t channel becomes the exchange diagram and its u channel the annihilation diagram, so their relative minus sign is inherited, not re-derived. The result is ⟨|M|²⟩ = 2e⁴[(s² + u²)/t² + (t² + u²)/s² + 2u²/(st)], and dσ/dΩ = ⟨|M|²⟩/(64π²s) with e² = 4πα gives the cross-section on the card.

The direct trace: two squares and one loop

Check it head-on, with M = Mₜ − Mₛ and massless spin sums. The exchange square is a product of two traces, (e⁴/t²)Tr[p̸₃γ^μp̸₁γ^ν]Tr[p̸₂γ_μp̸₄γ_ν] = 32e⁴[(p₁⋅p₂)(p₃⋅p₄) + (p₁⋅p₄)(p₂⋅p₃)]/t², which is 8e⁴(s² + u²)/t² once p₁⋅p₂ = p₃⋅p₄ = s/2 and p₁⋅p₄ = p₂⋅p₃ = −u/2. The annihilation square is topic 9.06's, 8e⁴(t² + u²)/s². The cross term is different in kind. In Σ Mₜ Mₛ*, the chains ū₃…u₁, ū₁…v₂, v̄₂…v₄ and v̄₄…u₃ close into one loop, Tr[p̸₃γ^μp̸₁γ^νp̸₂γ_μp̸₄γ_ν]. Contract γ^μ(p̸₁γ^νp̸₂)γ_μ = −2p̸₂γ^νp̸₁, then γ^ν(p̸₁p̸₄)γ_ν = 4p₁⋅p₄, leaving −8(p₁⋅p₄)Tr[p̸₃p̸₂] = −32(p₁⋅p₄)(p₂⋅p₃) = −8u². So −2Re Σ Mₜ Mₛ* = +16e⁴u²/(st), and dividing all three pieces by the four initial spin states reproduces the crossed result term by term.

Reading the angle: a pole, a familiar term, a subtraction

In the centre-of-mass frame t = −s(1 − cos θ)/2 and u = −s(1 + cos θ)/2. The exchange term (4 + (1 + c)²)/(1 − c)² grows without limit as θ → 0, where the photon's virtuality t goes to zero. The annihilation term (1 + c²)/2 alone is topic 9.06's μ-pair distribution (α²/(4s))(1 + cos²θ), the limit in which the exchange diagram is switched off. The interference, −(1 + c)²/(1 − c), is negative at every angle, vanishes at 180° where u = 0, and reaches −25 per cent of the total near 76°. At √s = 10 GeV, α²/(2s) = 0.1037 nb, and at 90° the three terms are 5, 0.5 and −1, so dσ/dΩ = 0.467 nb sr⁻¹, nine times the μ-pair rate; dropping the interference would give 0.570 nb sr⁻¹, 22 per cent high. Forward of 102° the interference outweighs the whole annihilation term, and forward of 112° it already outweighs half of it, so there keeping annihilation while dropping the interference is worse than keeping exchange alone.

Why the forward pole measures luminosity

At small angles t ≈ −sθ²/4 and u ≈ −s, so the exchange term is 2s²/t² = 32/θ⁴ and dσ/dΩ → 16α²/(sθ⁴): Rutherford's θ⁻⁴ for two unit charges, the same as e⁻μ⁺ scattering, because the annihilation term is of order 1 and the interference, with only a single pole in t, is a fraction θ²/4 of the exchange term. With dΩ ≈ 2πθ dθ, the band θₘᵢₙ to θₘₐₓ holds σ_acc = (16πα²/s)(1/θₘᵢₙ² − 1/θₘₐₓ²). At √s = 91.19 GeV and 30 to 50 mrad that is 89 nb, twice the 41.5 nb hadronic peak cross-section measured at the Z, and even there the t-channel photon at |t| of a few GeV² dwarfs the Z's s channel. Large, pure photon exchange and calculable: that is why colliders count small-angle Bhabha pairs, N = σ_acc∫L dt, to measure luminosity, and LEP's luminometers at a few tens of milliradians reached a theory error below 0.1 per cent (quoted). The price is geometry: the inner edge dominates σ_acc, so θₘᵢₙ must be known to a few parts in 10⁴.

Where tree-level Bhabha stops

Four limits bound the result. Masses: the massless formula needs |t| ≫ m², which fails only within θ ≈ 2m/√s of the beam, 0.1 mrad at 10 GeV; the photon pole itself survives. Loops, from Unit 12: initial-state radiation brings factors of (2α/π) ln(s/m²) ≈ 0.09 at 10 GeV, multiplied by logarithms of the energy resolution, and the running of α raises the s-channel rate by roughly 8 per cent at 10 GeV (quoted). The Z: near √s = 91 GeV a resonant Z boson coupled to weak charges overwhelms the photon in the s channel at large angles, a later course's subject. Low energy: as v → 0 the exchange diagram becomes the Coulomb attraction of a distinguishable pair, Born-level only for α ≪ v, while annihilation shrinks to a contact term of relative size about v², felt only by the spin triplet. In positronium it supplies 3/7 of the lowest-order hyperfine splitting (7/12)mα⁴ = 8.45 × 10⁻⁴ eV, or 204.4 GHz, against 203.39 GHz measured (quoted).

Change one variable at a time

Make the relationship visible.

Interactive model

Scattering angle θ90 °

Relative sign η of the s diagram−1

Sweep θ forward from 180°: the interference bar is zero at 180°, where u = 0, sinks to −25 per cent near 75° and shrinks again as the exchange pole takes over. Then set η to +1 or 0 and watch the 90° total move from 4.5 to 6.5 or 5.5.

EXCHANGE (s²+u²)/t²5.00 α²/(2s)

ANNIHILATION (t²+u²)/s²0.500 α²/(2s)

INTERFERENCE −2ηu²/(st)-1.000 α²/(2s)

dσ/dΩ AT √s = 10 GeV0.467 nb sr⁻¹

Live interpretationEXCHANGE (s²+u²)/t²: 5.00 α²/(2s). ANNIHILATION (t²+u²)/s²: 0.500 α²/(2s). INTERFERENCE −2ηu²/(st): −1.000 α²/(2s). dσ/dΩ AT √s = 10 GeV: 0.467 nb sr⁻¹

Catch the common trap

Explain before calculating.

Misconception check

Because e⁺ and e⁻ are distinguishable, students often add the two diagrams' cross-sections, Rutherford-like exchange plus the μ-pair result, or give the diagrams a relative plus sign since there is no exchange symmetry to impose. Both fail. The diagrams share a final state, so amplitudes add, and the minus has nothing to do with identical particles: ψ̄ creates the outgoing electron and destroys the incoming positron, and the two diagrams swap those jobs between two anticommuting fields. A plus sign turns the 90° interference from −1 to +1, giving 6.5 instead of 4.5 in units of α²/(2s); the incoherent sum gives 5.5. Nor does the minus alone make the interference destructive: Møller has the same minus and constructive interference at high energy, because its 2s²/(tu) is positive, t and u both negative, while Bhabha's 2u²/(st) has s positive and t negative.

At √s = 10 GeV, where α²/(2s) = 0.1037 nb, a detector at θ = 60° in the centre-of-mass frame records Bhabha events. Which tree-level dσ/dΩ should its rate be compared with?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAt √s = 5.0 GeV, find s, t and u for Bhabha scattering at θ = 120° in the centre-of-mass frame, evaluate the three terms of the tree-level cross-section, and give dσ/dΩ in nb sr⁻¹ and its ratio to e⁺e⁻ → μ⁺μ⁻ at the same angle.

Massless kinematics: s = 25.0 GeV², t = −s(1 − cos θ)/2 = −25.0 × 1.5/2 = −18.75 GeV² and u = −s(1 + cos θ)/2 = −6.25 GeV². Check: s + t + u = 0, as it must be for four massless particles.
Exchange: (s² + u²)/t² = (625 + 39.06)/351.56 = 1.889. Annihilation: (t² + u²)/s² = (351.56 + 39.06)/625 = 0.625. Interference: 2u²/(st) = 78.13/(−468.75) = −0.167. Each is a ratio of GeV⁴, so dimensionless.
Their sum is 2.347. With α² = 5.325 × 10⁻⁵, α²/(2s) = 1.065 × 10⁻⁶ GeV⁻², and times 0.3894 mb GeV² that is 4.147 × 10⁻⁷ mb = 0.4147 nb.
dσ/dΩ = 2.347 × 0.4147 nb = 0.973 nb sr⁻¹. Limit check: the annihilation term alone, 0.625 × 0.4147 = 0.259 nb sr⁻¹, equals topic 9.06's μ-pair rate (α²/(4s))(1 + cos²θ) = 0.2074 × 1.25 = 0.259 nb sr⁻¹, as it must.
Ratio to μ pairs: 0.973/0.259 = 3.76. The interference is −0.167/2.347 = −7.1 per cent of the total, small this far back because u is small.

Answerdσ/dΩ = 0.973 nb sr⁻¹ (terms 1.889, 0.625 and −0.167 in units of α²/(2s) = 0.415 nb), 3.76 times the μ-pair rate at 120°.

MediumA detector at √s = 10.0 GeV accepts Bhabha events with |cos θ| ≤ 0.8. Integrate the tree-level cross-section over this acceptance, split it into exchange, annihilation and interference, and compare it with e⁺e⁻ → μ⁺μ⁻ in the same acceptance.

With c = cos θ and dΩ = 2π dc, σ = 2π(α²/(2s))∫F dc over −0.8 ≤ c ≤ 0.8, F being the bracket of the angle form, and α²/(2s) = 0.1037 nb at s = 100 GeV². Put w = 1 − c, which runs from 0.2 to 1.8, and 1 + c = 2 − w.
Exchange: (4 + (2 − w)²)/w² = 8/w² − 4/w + 1, and [−8/w − 4 ln w + w] from 0.2 to 1.8 is (−4.444 − 2.351 + 1.800) − (−40.000 + 6.438 + 0.200) = 28.367.
Annihilation: ∫(1 + c²)/2 dc = [c + c³/3]/2 from −0.8 to 0.8 = (1.6 + 0.3413)/2 = 0.9707. Interference: −∫(2 − w)²/w dw = −[4 ln w − 4w + w²/2] from 0.2 to 1.8 = −(−3.229 + 7.218) = −3.989.
Sum 25.348, so σ = 2π × 25.348 × 0.1037 nb = 16.5 nb: exchange 18.48 nb, annihilation 0.632 nb, interference −2.60 nb, which is −15.7 per cent of the total.
μ pairs: σ = 2π(α²/(4s))∫(1 + c²)dc = 2π × 0.05184 nb × 1.9413 = 0.632 nb, identical to the annihilation piece, as the s-channel limit demands. Bhabha events outnumber μ pairs 26 to 1 even in this central barrel, and the interference alone is four times the whole μ-pair rate.

Answerσ = 16.5 nb (exchange 18.5 nb, annihilation 0.632 nb, interference −2.60 nb, or −15.7 per cent), 26 times the 0.632 nb of μ pairs in the same acceptance.

HardA collider at √s = 10.0 GeV counts Bhabha pairs in forward calorimeters 2.50 m from the collision point, accepting the electron at 30 ≤ θ ≤ 60 mrad with the positron opposite. Find the accepted cross-section from the small-angle limit and check it against the full tree-level formula, find the integrated luminosity if a fill yields 5.0 × 10⁶ events, and find how precisely the inner edge must be placed for a 0.1 per cent systematic.

At small θ, t ≈ −sθ²/4 and u ≈ −s, so the exchange term is 2s²/t² = 32/θ⁴ and dσ/dΩ = 16α²/(sθ⁴). Here |t| runs from 0.0225 to 0.090 GeV², far above m² = 2.6 × 10⁻⁷ GeV², so dropping the mass is safe.
With dΩ = 2πθ dθ, σ_acc = (16πα²/s)(1/θₘᵢₙ² − 1/θₘₐₓ²). 16πα²/s = 2.677 × 10⁻⁵ GeV⁻² = 10.42 nb, and 1/θₘᵢₙ² − 1/θₘₐₓ² = 1111.1 − 277.8 = 833.3, so σ_acc = 8686 nb = 8.69 μb.
The full formula, integrated numerically over the same band, gives 8679 nb, 0.083 per cent lower: −0.042 per cent from the exact exchange term with sin θ in dΩ, −0.042 per cent from the interference, and 1 × 10⁻⁷ of the total from annihilation. The luminometer sees pure t-channel photon exchange.
Luminosity: ∫L dt = N/σ_acc = 5.0 × 10⁶/8679 nb = 576 nb⁻¹ = 0.576 pb⁻¹, with a statistical uncertainty 1/√N = 0.045 per cent.
Edge: ∂ln σ_acc/∂ln θₘᵢₙ = −(2/θₘᵢₙ²)/833.3 = −2.67. A 0.1 per cent systematic allows δθₘᵢₙ/θₘᵢₙ = 3.75 × 10⁻⁴, that is 11.3 μrad, which at 2.50 m is 28 μm of radius.

Answerσ_acc = 8.69 μb (8.68 μb from the full formula, 0.083 per cent lower); ∫L dt = 0.576 pb⁻¹ with 0.045 per cent statistical error; the inner edge must be known to 11 μrad, 28 μm at 2.50 m.