University Quantum Field Theory I · QED at Tree Level: Moller, Bhabha and Compton Scattering and Pair Annihilation · 10.1
Møller scattering: t and u channels with opposite signs
The first QED process here with identical fermions in and out, and so the first where the sign between two Feynman diagrams stops being bookkeeping and starts setting a cross-section. Fix it from the operator ordering, square with a trace that runs through both electron lines, and watch it reshape the angular distribution and a counting rate.
Build the model
Connect the measurement to the mechanism.
At tree level two electrons scatter by exchanging one photon, but the outgoing electrons are identical, so there is no fact about which one left which vertex. Wick's theorem therefore delivers two diagrams: the t channel, in which p₃ leaves the vertex that absorbed p₁, and the u channel, in which p₄ does. Bringing the second contraction into the operator order of the first costs one transposition of anticommuting fields, so ℳ = ℳₜ − ℳᵤ, an amplitude odd under exchange of the final electrons: the Pauli principle written as a scattering amplitude. Squaring gives two direct terms, each a product of two traces, and an interference term that cannot factorise, because the conjugated u-channel spinors close both fermion lines into a single trace.
That trace is negative at high energy, so the minus sign makes the interference add. The result is Møller's 1932 formula, valid with the mass kept, which at s ≫ m² becomes (α²/(2s))[(s² + u²)/t² + (s² + t²)/u² + 2s²/(tu)]. It stops working in four places. Its poles at θ = 0 and π make the total cross-section infinite, so every measurement needs a cut.
It needs α ≪ v, because the exchange term's Coulomb phase is invisible at tree level. At order α, vacuum polarisation, the vertex and soft photon emission enter, and they wait for Unit 12. And at large s, Z⁰ exchange adds a parity-violating piece, of order 10⁻⁷ at fixed-target energies.
- Simple definition
- Møller scattering is elastic e⁻e⁻ → e⁻e⁻; at tree level its amplitude is the t-channel photon exchange minus the u-channel exchange with the two outgoing electrons swapped, the minus sign being Fermi statistics.
- Example
- At √s = 1.00 GeV and θ = 90° in the CM frame, t = u = −s/2 and the bracket is 5 + 5 + 8 = 18, so dσ/dΩ = 9α²/s = 4.79 × 10⁻⁴ GeV⁻² sr⁻¹ = 0.187 μb sr⁻¹, 44% of it interference.
Swapping p₃ ↔ p₄ swaps t ↔ u and the two terms, so ℳ → −ℳ: the amplitude is odd under exchange of the outgoing electrons, as Fermi statistics demands.
ħ = c = 1, metric (+, −, −, −); uᵢ = u(pᵢ, sᵢ) with ūu = 2m; t = (p₁ − p₃)², u = (p₁ − p₄)² in GeV²; e = √(4πα) = 0.303; ℳ dimensionless
Massless it is −8s². The Fermi sign makes the cross term −2Re(ℳₜℳᵤ*), so the interference is positive above s = 6m² and negative below it.
the spin sum of ℳₜℳᵤ* is e⁴/(tu) times this trace; s and m² in GeV², the trace in GeV⁴
Møller's 1932 result at tree level: the direct traces 8(s² + u² + 8m²t − 8m⁴) and its t ↔ u mirror, plus the single interference trace, each over 4 spin states.
s + t + u = 4m²; CM frame: s = 4(p² + m²), t = −2p²(1 − cos θ), u = −2p²(1 + cos θ), all in GeV²
In CM angles it is α²(3 + cos²θ)²/(s sin⁴θ): 9α²/s at 90°, even under θ → π − θ, with 1/θ⁴ poles at 0 and π.
s ≫ m²; α = 1/137.036; dσ/dΩ in GeV⁻² sr⁻¹, times 0.3894 mb GeV² for mb sr⁻¹
For a small cut it grows as 16πα²/(sθₘᵢₙ²), so the uncut total is infinite. Integrating both hemispheres without a factor ½ counts every event twice.
s ≫ m², CM angles; one electron counted over half the sphere, so each event counts once; σ in GeV⁻²
Runs from −1/2 at threshold, the destructive Born exchange term behind Mott's formula, to +4/5 as s/m² → ∞, where it supplies 8 of the 18 units at 90°.
θ = 90° in the CM frame; p the CM momentum of each electron; zero at p = m/√2, i.e. s = 6m² and v = 1/√3
Two ways to connect the electrons, one relative sign
Between ⟨p₃p₄| and |p₁p₂⟩ the second-order Dyson term holds two interaction vertices ψ̄γμψAμ, at x and y, each bringing −ieγμ. Contract the photon fields into one Feynman propagator; each remaining ψ must annihilate an incoming electron and each ψ̄ create an outgoing one. Swapping x and y only cancels the 1/2!, so two assignments are left: p₃ leaves the vertex that absorbed p₁, with momentum transfer p₁ − p₃ (the t channel), or p₄ does, with p₁ − p₄ (the u channel). Bringing the second assignment's operators into the order of the first takes one extra transposition of anticommuting fields, so it enters with −1: topic 7.05's rule, used here, not re-derived. The result, derived, is ℳ = ℳₜ − ℳᵤ. The overall sign depends on whether |p₁p₂⟩ was built as b†(p₁)b†(p₂)|0⟩ or the reverse, and it drops out of every rate; the relative sign does not, and it makes ℳ odd under p₃ ↔ p₄.
Where the poles sit
In the CM frame each electron has momentum p and energy E = √(p² + m²), so s = 4E², t = −2p²(1 − cos θ) and u = −2p²(1 + cos θ), with s + t + u = 4m². The t-channel photon carries the momentum transfer of the scattering: −t vanishes as θ → 0, the propagator's 1/t blows up, and the long range of the Coulomb force shows as a forward peak. The u channel does the same as θ → π. Swapping the outgoing electrons sends θ → π − θ and t ↔ u, and because they are identical an electron at θ and one at π − θ are the same event, so every rate must be even in cos θ. Away from 90° one channel dominates. At √s = 200 MeV and θ = 20°, −t = 1.21 × 10⁻³ GeV² while −u = 3.88 × 10⁻² GeV²: the t-channel photon is 32 times nearer its pole, and its direct term outweighs the u-channel one by a factor of about 2000.
Squaring: two traces, then one trace through both lines
Average over the four initial spin states and sum over the final ones with Casimir's trick. |ℳₜ|² factorises into one trace per fermion line, Tr[(p̸₃ + m)γμ(p̸₁ + m)γν] Tr[(p̸₄ + m)γμ(p̸₂ + m)γν] = 8(s² + u² + 8m²t − 8m⁴), derived from topic 7.06's four-γ trace with p₁⋅p₂ = (s − 2m²)/2, p₁⋅p₃ = (2m² − t)/2 and p₁⋅p₄ = (2m² − u)/2. The cross term cannot factorise. Conjugating ℳᵤ turns ū₄γνu₁ into ū₁γνu₄, and the spinors chain as ū₃…u₁ ū₁…u₄ ū₄…u₂ ū₂…u₃: one closed loop through both lines, a single trace of eight gamma matrices. Massless, γμp̸₁γνp̸₄γμ = −2p̸₄γνp̸₁ and then γνp̸₁p̸₂γν = 4p₁⋅p₂ reduce it to −32(p₁⋅p₂)(p₃⋅p₄) = −8s², derived. With the mass kept the same reductions give −8(s − 2m²)(s − 6m²), quoted here; evaluating all sixteen spin amplitudes with explicit Dirac spinors reproduces both.
The high-energy cross-section, and why the interference adds
In the CM frame the final momentum equals the initial one in size, so dσ/dΩ = ¼Σ|ℳ|²/(64π²s), and e⁴ = 16π²α² turns 2e⁴/(64π²s) into α²/(2s). Dropping m gives dσ/dΩ = (α²/(2s))[(s² + u²)/t² + (s² + t²)/u² + 2s²/(tu)], derived. The interference collects two minus signs, the Fermi sign and the trace's −8s², and tu is positive, so it adds. At θ = 60° the bracket is 25 + 1.89 + 10.67 = 37.6; at 90° it is 5 + 5 + 8 = 18, so dσ/dΩ = 9α²/s. In CM angles the whole bracket is 2(3 + cos²θ)²/sin⁴θ. A helicity count, with the massless amplitudes quoted, confirms the sign: equal helicities give ℳ ∝ s(1/t + 1/u), where both diagrams contribute and t and u share a sign, while opposite helicities keep one diagram each, ∝ u/t or t/u. Summing the six non-zero |ℳ|² and dividing by 4 returns the bracket term by term. Dimensions: α²/s is GeV⁻², an area per steradian.
Keeping the mass: the interference changes sign
The factor (s − 6m²) means the Fermi sign alone does not decide whether the diagrams reinforce. At 90°, with P = p²/m², the interference divided by the direct terms is (4P² − 1)/(5P² + 4P + 2), derived from the mass-kept |ℳ|²: +4/5 for s ≫ m², zero at p = m/√2 (s = 6m², CM speed 1/√3 = 0.577), and −1/2 at threshold. Near threshold the bracket becomes 8m⁴[1/t² + 1/u² − 1/(tu)], the Born-level exchange pattern of two spin-½ particles, which the next topic turns into Mott's formula. For a beam on electrons at rest s = 4m² + 2mT, so the flip comes at kinetic energy T = m = 511 keV. The slow end has a floor: the exchange term of Mott's formula carries a Coulomb phase proportional to α/v that tree level sets to zero, so this result needs α ≪ v.
Counting events: a cut and half a sphere
The poles go as 1/θ⁴ and 1/(π − θ)⁴, so ∫(dσ/dΩ)dΩ diverges and a rate exists only inside an angular cut. The electrons leave back to back, at θ and π − θ, so every event has exactly one electron in the forward hemisphere: count events by integrating one electron over θₘᵢₙ ≤ θ ≤ 90°, the same as the symmetric window with a factor ½. Writing 3 + cos²θ = 4 − sin²θ makes the integral elementary, and its logarithms cancel: σ = (2πα²/s) c(9 − c²)/(1 − c²), c = cos θₘᵢₙ, derived, which grows as 16πα²/(sθₘᵢₙ²) for a small cut. At √s = 200 MeV with θₘᵢₙ = 30°, σ = 0.239 GeV⁻² = 93.1 μb. The formula has been tested head-on: the 1966 Princeton–Stanford colliding-beam experiment, two 300 MeV electron beams, found wide-angle e⁻e⁻ scattering in agreement with it, quoted; a plus sign would have put the 90° rate nine times lower.
Change one variable at a time
Make the relationship visible.
Lower v from 0.99 to below 0.58: the faint interference curve crosses zero and the 90° dot drops below 1, because the interference flips sign at s = 6m². Back at v = 0.99, set η = +1 to see the wrong sign cut the 90° value about eightfold.
√s (CM ENERGY)7.245 MeV
INTERFERENCE ÷ DIRECT AT 90°0.79
dσ/dΩ AT 90°3.59 mb sr⁻¹
σ, θₘᵢₙ TO 90°72.1 mb
Live interpretation√s (CM ENERGY): 7.245 MeV. INTERFERENCE ÷ DIRECT AT 90°: 0.79. dσ/dΩ AT 90°: 3.59 mb sr⁻¹. σ, θₘᵢₙ TO 90°: 72.1 mb
Catch the common trap
Explain before calculating.
At s ≫ m² and θ = 90° in the CM frame, the t-channel and u-channel direct terms each contribute 5 to the bracket in dσ/dΩ = (α²/(2s))[(s² + u²)/t² + (s² + t²)/u² + 2s²/(tu)]. What is the whole bracket, and why?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA 6.0 GeV electron beam strikes electrons at rest in a thin foil, treated as free. Find √s and the Møller dσ/dΩ at 90° in the CM frame in μb sr⁻¹, say how much of it the interference supplies, and find what a plus sign between the two diagrams would have predicted.
- Fixed target: s = 2m² + 2mE = 2m(E + m) = 2(5.110 × 10⁻⁴ GeV)(6.0005 GeV) = 6.133 × 10⁻³ GeV², so √s = 78.3 MeV. With s/m² = 2.35 × 10⁴ the mass corrections at 90° are below 10⁻⁴, and the high-energy formula is enough.
- At θ = 90°, t = u = −s/2. The bracket's three terms are (s² + u²)/t² = (1 + ¼)/¼ = 5, its mirror 5, and 2s²/(tu) = 2/(¼) = 8, a total of 18.
- dσ/dΩ = (α²/(2s)) × 18 = 9α²/s = 9 × 5.325 × 10⁻⁵/(6.133 × 10⁻³ GeV²) = 7.815 × 10⁻² GeV⁻² sr⁻¹. Times 0.3894 mb GeV², that is 3.043 × 10⁻² mb sr⁻¹ = 30.4 μb sr⁻¹.
- The interference supplies 8 of the 18 units, 44%. A plus sign would flip it to 5 + 5 − 8 = 2, giving 30.4 × 2/18 = 3.38 μb sr⁻¹, nine times less.
- Checks: α²/s is GeV⁻², an area per steradian; t ↔ u leaves 90° unchanged, as identical electrons require.
Answer√s = 78.3 MeV and dσ/dΩ(90°) = 30.4 μb sr⁻¹, of which 44% is interference; with a plus sign it would be 3.38 μb sr⁻¹.
MediumKeep the electron mass. Show that the Møller interference term vanishes at every angle when s = 6m², and find the beam kinetic energy T at which that happens for a beam on electrons at rest. Then, for T = 2.00 MeV, find the interference at 90° in the CM frame as a fraction of the direct terms, and dσ/dΩ at 90° compared with the massless formula.
- The interference in ¼Σ|ℳ|² is 2e⁴ × 2(s − 2m²)(s − 6m²)/(tu). Since s ≥ 4m², the factor s − 2m² never vanishes, so the whole term vanishes at every angle only at s = 6m², and with tu positive it is negative below that.
- On a fixed target s = 2m² + 2m(T + m) = 4m² + 2mT, so s = 6m² gives T = m = 0.511 MeV. In the CM frame s = 4(p² + m²) gives p = m/√2, a speed v = 1/√3 = 0.577.
- For T = 2.00 MeV: s = 4m² + 2mT = 1.044 + 2.044 = 3.088 MeV², and P = p²/m² = T/(2m) = 1.957.
- At 90°, t = u = −2Pm². Each direct term is (5P² + 4P + 2)/P² = 7.566 and the interference is 2(4P² − 1)/P² = 7.478, so the ratio is 7.478/15.13 = 0.494, against 0.800 at high energy.
- The bracket is 22.61, so dσ/dΩ = α² × 22.61/(2s) = 5.325 × 10⁻⁵ × 22.61/(6.177 MeV²) = 1.949 × 10⁻⁴ MeV⁻² = 194.9 GeV⁻², or 75.9 mb sr⁻¹. The massless 9α²/s gives 60.4 mb sr⁻¹: keeping m raises the 90° value by 26% here.
AnswerThe interference vanishes at s = 6m², i.e. T = m = 0.511 MeV (v = 0.577 in the CM frame). At T = 2.00 MeV it is 0.494 of the direct terms, and dσ/dΩ(90°) = 75.9 mb sr⁻¹, 26% above the massless 60.4 mb sr⁻¹.
HardAn 8.0 GeV electron beam on a liquid-hydrogen target makes Møller pairs, and a detector records a pair when its CM angles lie between 50° and 130°. With the high-energy formula, derive the counted cross-section for the window θₘᵢₙ ≤ θ ≤ π − θₘᵢₙ, evaluate it in μb, find the event rate at a luminosity of 1.0 × 10³⁵ cm⁻² s⁻¹, and show what happens as θₘᵢₙ → 0.
- In CM angles dσ/dΩ = (α²/s)(3 + x²)²/(1 − x²)² with x = cos θ; the poles at x = ±1 make the uncut integral diverge.
- The electrons leave back to back, at θ and π − θ, so each recorded event has exactly one electron between 50° and 90°. Counting events means integrating one electron over that half: σ = 2π∫₀c (dσ/dΩ) dx with c = cos 50° = 0.6428. Integrating over 50° to 130° would count every event twice.
- Write 3 + x² = 4 − (1 − x²): the integrand becomes 16/(1 − x²)² − 8/(1 − x²) + 1, whose antiderivative is 8x/(1 − x²) + x, the logarithms cancelling. So σ = (2πα²/s) c(9 − c²)/(1 − c²).
- s = 2m(E + m) = 8.177 × 10⁻³ GeV² (√s = 90.4 MeV). Then c(9 − c²)/(1 − c²) = 0.6428 × 8.587/0.5868 = 9.406 and 2πα²/s = 4.092 × 10⁻² GeV⁻², so σ = 0.3849 GeV⁻² = 0.1499 mb = 150 μb.
- Rate = σL = 1.499 × 10⁻²⁸ cm² × 1.0 × 10³⁵ cm⁻² s⁻¹ = 1.5 × 10⁷ s⁻¹.
- Checks: α²/s is an area. For a small cut c(9 − c²)/(1 − c²) → 8/θₘᵢₙ², so σ → 16πα²/(sθₘᵢₙ²): a 10° cut gives 262.3 in place of 9.406, 27.9 times the count, and θₘᵢₙ → 0 gives infinity, the Coulomb pole.
Answerσ = (2πα²/s) c(9 − c²)/(1 − c²) = 150 μb for 50° to 130° at E = 8.0 GeV, a rate of 1.5 × 10⁷ s⁻¹; the count grows as 1/θₘᵢₙ² and has no finite limit without a cut.