University Quantum Field Theory I · QED at Tree Level: the Lagrangian, Feynman Rules and Electron-Muon Processes · 9.5

e⁺e⁻ → μ⁺μ⁻: the amplitude and its trace

Every first course in QED computes this process first, and it earns the place: one diagram, two spinor chains, and a trace that turns sixteen spin amplitudes into three dot products. Do it once with the muon mass kept and you own the technique that every later rate in this course reuses.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Electrons and muons interact in QED only through the photon, and the photon never turns one into the other, so at order e² the annihilation e⁻(p) e⁺(p′) → μ⁻(k) μ⁺(k′) has exactly one diagram: the pair becomes a virtual photon of momentum q = p + p′, far off its mass shell with q² = s, which rematerialises as muons. Topic 9.02's rules turn that diagram into two spinor chains divided by s, and topic 9.04 lets the gauge-dependent q_μq_ν part of the propagator go, because q contracted into either conserved current vanishes. No experiment here resolves the individual spin states, so what is measured is |M|² averaged over the four beam-spin states and summed over the four muon ones.

Casimir's trick replaces each spinor pair by p̸ ± m, each chain closes into a trace, and the identity for a trace of four gamma matrices reduces them to dot products. The result, (8e⁴/s²)[(p⋅k)(p′⋅k′) + (p⋅k′)(p′⋅k) + m_μ²(p⋅p′)], is frame-free, fixes the angular distribution, and is what topic 9.06 turns into a cross-section. It is honest only to lowest order.

It drops the electron mass, at a cost of order mₑ²/s that never exceeds about 10⁻⁵; it ignores loops and real radiation, of order α/π but swollen by logarithms of s/mₑ²; and it leaves out the Z boson, whose own s-channel diagram makes |M|² lopsided in cos θ, an asymmetry the PETRA experiments measured near √s = 35 GeV.

Simple definition: The spin-averaged squared amplitude ¼Σ|M|² for e⁺e⁻ → μ⁺μ⁻ averages |M|² over the four beam-spin states and sums it over the four muon-spin states; Casimir's trick writes it as an electron trace times a muon trace.
Example: At √s = 0.30 GeV and θ = 90°, the terms (p⋅k)(p′⋅k′) and (p⋅k′)(p′⋅k) each give 0.500 e⁴ and m_μ²(p⋅p′) adds 0.496 e⁴, so ¼Σ|M|² = 1.496 e⁴ = 0.0126, a third of it from the muon's mass.

The one-photon amplitudeM = (e²/s) [v̄(p′)γ^μ u(p)] [ū(k)γ_μ v(k′)]

One diagram, so its overall sign is invisible in |M|². Here q² = s ≥ 4m_μ², so +iε never acts, and q_μ on either current gives zero, so g_μν alone suffices.

e² = 4πα = 0.0917; s = (p + p′)² in GeV²; each chain has mass dimension 1, so M is dimensionless. Derived from topic 9.02.

Casimir's trick: a product of two traces¼Σ|M|² = (e⁴/4s²) Tr[(p̸′ − mₑ)γ^μ(p̸ + mₑ)γ^ν] Tr[(k̸ + m_μ)γ_μ(k̸′ − m_μ)γ_ν]

Derived. Conjugating a chain reverses it, because γ⁰γ^μ†γ⁰ = γ^μ. So each chain meets its mirror image and closes into a trace with no spinor left.

Σuū = p̸ + m and Σvv̄ = p̸ − m (UAQM-04.08, quoted); ¼ averages the 2 × 2 beam spins; muon spins summed.

The lepton tensorTr[(p̸′ − m)γ^μ(p̸ + m)γ^ν] = 4[p′^μ p^ν + p^μ p′^ν − g^μν (p⋅p′ + m²)]

Derived. Real, symmetric and conserved: contracted with q = p + p′ it gives zero, the Ward identity surviving the square. A sign slip in a spin sum fails that test.

From UAQM-04.03's Tr(a̸γ^μ b̸γ^ν) = 4(a^μ b^ν + a^ν b^μ − g^μν a⋅b), quoted; odd traces vanish. In GeV².

The trace result, electron mass dropped¼Σ|M|² = (8e⁴/s²)[(p⋅k)(p′⋅k′) + (p⋅k′)(p′⋅k) + m_μ²(p⋅p′)]

The k⋅k′ pieces of the two tensors cancel, so the muon mass enters once. Exchanging k and k′ swaps the first two terms, so the result is even under μ⁻ ↔ μ⁺.

Dot products in GeV² and s² in GeV⁴, so the result is dimensionless. Derived; the error from mₑ = 0 is of order mₑ²/s.

The same result in Mandelstam variables¼Σ|M|² = (2e⁴/s²)[(t − m_μ²)² + (u − m_μ²)² + 2m_μ²s]

Frame-free, and the form topic 9.08 will cross into other channels. Far above threshold it reduces to 2e⁴(t² + u²)/s².

t = (p − k)², u = (p − k′)², s + t + u = 2m_μ² with mₑ = 0; all in GeV². Derived using p⋅k = (m_μ² − t)/2.

In the centre-of-mass frame¼Σ|M|² = e⁴[1 + 4m_μ²/s + (1 − 4m_μ²/s) cos²θ]

Even in cos θ. It tends to e⁴(1 + cos²θ) at high energy and to 2e⁴, isotropic, at threshold, where the mass term supplies half; at cos θ = ±1 it is 2e⁴ always.

θ: angle from the e⁻ beam to the μ⁻; 1 − 4m_μ²/s = β², the muon speed squared. Derived from p⋅k = (s/4)(1 − β cos θ).

One diagram, read against the fermion arrows

Natural units ħ = c = 1 and the metric (+, −, −, −) throughout. Label the process e⁻(p) e⁺(p′) → μ⁻(k) μ⁺(k′). The photon couples each lepton to itself and never turns an electron into a muon, so at order e² only the annihilation diagram exists. Read each fermion line against its arrow, as topic 9.02 prescribes: the electron line starts at the incoming positron, v̄(p′), passes the vertex −ieγ^μ and ends on the incoming electron, u(p); the muon line gives ū(k)(−ieγ^ν)v(k′). Join them with −ig_μν/(q² + iε) at q² = s. Then iM = (−ie)²(−i/s) times the two chains, so M = (e²/s)[v̄(p′)γ^μ u(p)][ū(k)γ_μ v(k′)], derived. The gauge-dependent q_μq_ν piece of the propagator drops because v̄(p′)q̸u(p) = 0 by the Dirac equation (topic 9.04), and since s ≥ 4m_μ² = 0.0447 GeV² the photon is never on shell, so iε is idle. Dimensions: each chain carries one power of mass, spinors scaling as √E, so two chains over s leave M dimensionless, as a 2 → 2 amplitude must be.

Squaring: sixteen spin amplitudes, one contraction

The beams of most colliders are unpolarised and the detectors do not see muon spins, so the observable is ¼Σ|M|²: averaged over the 2 × 2 beam-spin states and summed over the 2 × 2 muon ones, sixteen amplitudes in all. Each chain is a complex number whose conjugate is the reversed chain: [v̄(p′)γ^ν u(p)]* = u†γ^ν†γ⁰v = ū(p)γ^ν v(p′), because γ⁰γ^ν†γ⁰ = γ^ν. So |M|² = (e⁴/s²)[v̄(p′)γ^μ u(p) ū(p)γ^ν v(p′)][ū(k)γ_μ v(k′) v̄(k′)γ_ν u(k)]. Now apply topic 7.06's Casimir trick with the spin sums Σuū = p̸ + m and Σvv̄ = p̸ − m, quoted from UAQM-04.08: a chain that begins and ends on the same spinor becomes a trace. The result is ¼Σ|M|² = (e⁴/4s²) L^μν H_μν, with the electron tensor L^μν = Tr[(p̸′ − mₑ)γ^μ(p̸ + mₑ)γ^ν] and the muon tensor H_μν = Tr[(k̸ + m_μ)γ_μ(k̸′ − m_μ)γ_ν], derived. A polarised beam would replace its spin sum by a helicity projector.

Two traces from one identity

Expand L^μν. The terms linear in mₑ hold three gamma matrices and vanish, because the trace of an odd number of them is zero. The mₑ² term is −mₑ² Tr(γ^μ γ^ν) = −4mₑ² g^μν. The rest is UAQM-04.03's identity Tr(a̸γ^μ b̸γ^ν) = 4(a^μ b^ν + a^ν b^μ − g^μν a⋅b), quoted. So L^μν = 4[p′^μ p^ν + p^μ p′^ν − g^μν (p⋅p′ + mₑ²)], and in the same way H_μν = 4[k_μ k′_ν + k′_μ k_ν − g_μν (k⋅k′ + m_μ²)], both derived. Two properties check them. Each is real and symmetric, because a pure vector coupling brings no γ⁵ and so no ε tensor; the Z's axial coupling would add an antisymmetric piece. And each is conserved: with q = p + p′, q_μ L^μν = 4[(q⋅p′)p^ν + (q⋅p)p′^ν − q^ν (p⋅p′ + mₑ²)] = 0, since q⋅p = q⋅p′ = p⋅p′ + mₑ². That is topic 9.04's Ward identity surviving the square, and it makes a sharp test of the spin sums before any contraction is done.

Contracting the tensors: where k⋅k′ goes

Set mₑ = 0 first and write K = k⋅k′ + m_μ². In L^μν H_μν/16, the outer products pair up into 2(p⋅k)(p′⋅k′) + 2(p⋅k′)(p′⋅k). Each tensor's outer products against the other's g term give −2(p⋅p′)K and −2(p⋅p′)(k⋅k′), and the two g terms meet with g^μν g_μν = 4 to give +4(p⋅p′)K. Collect: 2(p⋅p′)(K − k⋅k′) = 2m_μ²(p⋅p′). The muon pair's own invariant k⋅k′ cancels and the muon mass survives as a single term. With the 16 restored and e⁴/4s² in front, ¼Σ|M|² = (8e⁴/s²)[(p⋅k)(p′⋅k′) + (p⋅k′)(p′⋅k) + m_μ²(p⋅p′)], derived. Keeping mₑ, the same steps add mₑ²(k⋅k′) + 2mₑ²m_μ² inside the bracket. The full bracket is unchanged under (p, p′, mₑ) ↔ (k, k′, m_μ), as it must be, since μ⁺μ⁻ → e⁺e⁻ is the same diagram read backwards, and the new terms are of relative order mₑ²/s: 4.2 × 10⁻⁶ at √s = 0.25 GeV, falling as 1/s.

Reading the result in the centre-of-mass frame

In the centre-of-mass frame each beam has energy E = √s/2, with p = E(1, 0, 0, 1) and p′ = E(1, 0, 0, −1), and the μ⁻ leaves at angle θ to the e⁻ with speed β = √(1 − 4m_μ²/s). Then p⋅k = p′⋅k′ = (s/4)(1 − β cos θ), p⋅k′ = p′⋅k = (s/4)(1 + β cos θ), p⋅p′ = s/2 and m_μ² = (s/4)(1 − β²). The bracket becomes (s²/16)[(1 − β cos θ)² + (1 + β cos θ)² + 2(1 − β²)], so ¼Σ|M|² = e⁴[1 + 4m_μ²/s + β² cos²θ], derived. Three limits read off at once. As β → 1 it becomes e⁴(1 + cos²θ), the shape topic 9.07 explains by helicity. At threshold, β → 0, it is 2e⁴ at every angle: the pair is made at rest in an s wave, and the mass term supplies half. And at cos θ = ±1 it is 2e⁴ at any energy. Away from the poles the exact result exceeds e⁴(1 + cos²θ) by e⁴(4m_μ²/s) sin²θ: at √s = 1.00 GeV that is at most 4.5%, at θ = 90°. Topic 9.06 multiplies this by phase space to make a cross-section.

What this amplitude leaves out

The result is exact at order e⁴ in QED once mₑ is dropped, and the omitted pieces have sizes worth knowing. Loops and real photon emission enter at relative order α/π = 2.3 × 10⁻³ (Unit 12), but photons radiated by the light beams bring ln(s/mₑ²), about 20 at √s = 10 GeV, so the electron mass, harmless here, returns there and makes initial-state radiation a correction of several per cent that depends on how hard a photon the analysis accepts (quoted). Near threshold the slow muons attract each other, raising the rate by about πα/(2β), 4% at √s = 0.22 GeV (quoted). The largest omission lies outside QED. The Z boson has its own s-channel diagram with vector and axial couplings, and its interference with the photon adds terms odd in cos θ that the pure-vector traces above cannot produce. Its amplitude grows relative to the photon's as s/(M_Z² − s), from under 1% at √s = 10 GeV to the size that let the PETRA experiments measure a forward-backward asymmetry of about −10% near √s = 35 GeV (quoted).

Change one variable at a time

Make the relationship visible.

Interactive model

Collision energy √s0.30 GeV

Angle θ from e⁻ beam to μ⁻60 °

Drag √s down to 0.22 GeV: the flat mass term climbs to 0.92 and the dashed terms flatten, so the sum nears 2e⁴ at every angle, an s wave. Push √s to 3 GeV and the mass term fades, leaving 1 + cos²θ. Sweep θ to watch the three markers trade places.

MUON SPEED β0.710

¼Σ|M|² / e⁴1.622

MASS-TERM SHARE30.6 %

¼Σ|M|² × 10³, α = 1/137.03613.64

Live interpretationMUON SPEED β: 0.710. ¼Σ|M|² / e⁴: 1.622. MASS-TERM SHARE: 30.6 %. ¼Σ|M|² × 10³, α = 1/137.036: 13.64

Catch the common trap

Explain before calculating.

Misconception check

The tempting slip is to give the antiparticle the particle's spin sum, Σvv̄ = p̸ + m. It survives because its answer looks physical: the muon tensor's mass term flips sign, the bracket's last term becomes −m_μ²(p⋅p′), and the centre-of-mass result turns into e⁴β²(1 + cos²θ), which vanishes at threshold as though slow pairs were hard to make. It is wrong three ways. v(p) solves (p̸ + m)v = 0, so the sum of vv̄ must be annihilated by p̸ + m, and (p̸ + m)(p̸ − m) = p² − m² = 0 while (p̸ + m)² = 2m(p̸ + m) is not zero. The wrong tensor breaks current conservation, giving q^μ H_μν = 8m_μ² q_ν instead of zero. And a photon, J^PC = 1⁻⁻, has the quantum numbers of a ³S₁ fermion pair at rest, so nothing forces the amplitude to vanish at threshold: the true ¼Σ|M|² there is 2e⁴, and the cross-section falls to zero only through phase space.

At √s = 0.25 GeV a student evaluates the spin-averaged |M|² for e⁺e⁻ → μ⁺μ⁻ at θ = 60° between the e⁻ beam and the μ⁻, with mₑ neglected and m_μ = 105.66 MeV. Which value of ¼Σ|M|²/e⁴ is right?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAt √s = 0.500 GeV a μ⁻ emerges at θ = 120° to the e⁻ beam. With mₑ neglected, find p⋅k, p⋅k′ and p⋅p′ in the centre-of-mass frame, evaluate ¼Σ|M|² from the trace result, and check it against the centre-of-mass formula. Take m_μ = 0.10566 GeV and α = 1/137.036.

Kinematics: E = √s/2 = 0.2500 GeV and |k| = √(E² − m_μ²) = √(0.06250 − 0.01116) GeV = 0.22658 GeV, so β = |k|/E = 0.9063, while cos θ = −0.5.
Dot products: p⋅k = p′⋅k′ = E² − E|k| cos θ = 0.06250 + 0.02832 = 0.09082 GeV²; p⋅k′ = p′⋅k = 0.06250 − 0.02832 = 0.03418 GeV²; p⋅p′ = 2E² = 0.1250 GeV².
Bracket: 0.09082² + 0.03418² + 0.011164 × 0.1250 = 0.008249 + 0.001168 + 0.001395 = 0.010812 GeV⁴. The prefactor is 8/s² = 8/0.0625 GeV⁻⁴ = 128 GeV⁻⁴, so ¼Σ|M|² = 1.3840 e⁴, a pure number as it must be.
Check: 1 + 4m_μ²/s + β² cos²θ = 1 + 0.1786 + 0.8214 × 0.25 = 1.3840. The two routes agree to every digit carried.
With e² = 4πα = 0.09170, e⁴ = 0.008409 and ¼Σ|M|² = 0.01164. The backward-going μ⁻ makes (p⋅k)(p′⋅k′) the largest term, 76% of the bracket.

Answerp⋅k = 0.0908 GeV², p⋅k′ = 0.0342 GeV², p⋅p′ = 0.125 GeV²; ¼Σ|M|² = 1.384 e⁴ = 0.0116, matching the centre-of-mass formula.

MediumA τ-charm factory collides e⁺e⁻ at √s = 4.00 GeV, above the τ⁺τ⁻ threshold. The τ amplitude is the muon's with m_μ replaced by m_τ = 1.7769 GeV. Find β_τ, the ratio of ¼Σ|M|² at θ = 0 to its value at θ = 90°, the share of the angle-averaged ¼Σ|M|² carried by the mass term, and how the angle-averaged ¼Σ|M|² for τ pairs compares with that for μ pairs at the same √s.

4m_τ²/s = 4 × 3.1574 GeV²/16.00 GeV² = 0.7893, so β_τ² = 0.2107 and β_τ = 0.4590: the taus are slow.
At θ = 0: 1 + 0.7893 + 0.2107 = 2.000, in units of e⁴, as at every energy. At θ = 90°: 1.7893. The ratio is 2.000/1.789 = 1.118, so the distribution is nearly isotropic.
Over the sphere the average of cos²θ is 1/3, so the angle average is e⁴(1 + 4m_τ²/s + β_τ²/3) = e⁴(1.7893 + 0.0702) = 1.8596 e⁴, of which the mass term supplies 0.7893, a share of 42%.
Muons at the same energy: 4m_μ²/s = 0.04465/16.00 = 0.00279 and β_μ² = 0.99721, so the angle average is (1 + 0.00279 + 0.33240) e⁴ = 1.3352 e⁴.
Ratio: 1.8596/1.3352 = 1.393. The heavier pair has the larger |M|², because the mass term gains more than the β² cos²θ term loses. Only Unit 5's phase-space factor |p_f|/|pᵢ| in dσ/dΩ (quoted), here |p_τ|/|p_μ| = 0.460, makes the τ rate the smaller, at 0.640 of the μ rate.

Answerβ_τ = 0.459; ¼Σ|M|²(0°)/¼Σ|M|²(90°) = 1.118; mass-term share 42%; the angle-averaged ¼Σ|M|² is 1.39 times the μ-pair value, though phase space leaves the τ rate at 0.64 of the μ rate.

HardKeep the electron mass. (a) Show that the contraction adds mₑ²(k⋅k′) + 2mₑ²m_μ² to the bracket and that the result is unchanged under (p, p′, mₑ) ↔ (k, k′, m_μ). (b) Reduce it to the centre-of-mass frame. (c) Find the largest fractional error made by dropping mₑ at √s = 0.250 GeV, and compare it with the one-loop scale α/π.

(a) Divide L^μν H_μν by 16 and write P = p⋅p′ + mₑ², K = k⋅k′ + m_μ². The g terms now give −2(p⋅p′)K − 2P(k⋅k′) + 4PK = 2m_μ²(p⋅p′) + 2mₑ²(k⋅k′) + 4mₑ²m_μ². Since L^μν H_μν/16 is twice the bracket, the bracket gains mₑ²(k⋅k′) + 2mₑ²m_μ².
The swap (p, p′, mₑ) ↔ (k, k′, m_μ) exchanges m_μ²(p⋅p′) with mₑ²(k⋅k′) and leaves (p⋅k)(p′⋅k′), (p⋅k′)(p′⋅k) and 2mₑ²m_μ² alone: the bracket is symmetric, as it must be for μ⁺μ⁻ → e⁺e⁻, the same diagram read backwards.
(b) Now |p| = √(E² − mₑ²) and |k| = √(E² − m_μ²), so p⋅k = E² − |p||k| cos θ, p⋅p′ = 2E² − mₑ² and k⋅k′ = 2E² − m_μ². The ±2E²|p||k| cos θ cross terms cancel, as do the mₑ²m_μ² terms, and the bracket is 2E⁴ + 2|p|²|k|² cos²θ + 2E²(mₑ² + m_μ²).
Multiplying by 8/s² = 1/(2E⁴): ¼Σ|M|² = e⁴[1 + 4(mₑ² + m_μ²)/s + βₑ²β_μ² cos²θ], symmetric in the two masses; βₑ = 1 and mₑ = 0 return the lesson's formula.
(c) The fractional change is (4mₑ²/s)(1 − β_μ² cos²θ)/(1 + 4m_μ²/s + β_μ² cos²θ), largest at θ = 90°. At √s = 0.250 GeV, 4mₑ²/s = 1.671 × 10⁻⁵ and 4m_μ²/s = 0.7145, giving 1.671 × 10⁻⁵/1.7145 = 9.7 × 10⁻⁶; at θ = 0 it falls to 8mₑ²m_μ²/s² = 6.0 × 10⁻⁶.
Compare α/π = 2.32 × 10⁻³, about 240 times larger: the loop corrections of Unit 12 matter long before the electron mass does in this tree amplitude.

AnswerThe bracket gains mₑ²(k⋅k′) + 2mₑ²m_μ²; ¼Σ|M|² = e⁴[1 + 4(mₑ² + m_μ²)/s + βₑ²β_μ² cos²θ]; dropping mₑ errs by at most 9.7 × 10⁻⁶ at √s = 0.250 GeV, about 240 times below α/π.