e+e- to mu+mu-
Electron–positron annihilation into a muon pair, e⁺e⁻ → μ⁺μ⁻, is the hydrogen atom of quantum field theory — the simplest complete scattering calculation, and the one every student computes first. At lowest order a single Feynman diagram carries it: the e⁺ and e⁻ annihilate into a virtual photon of four-momentum q = p₁ + p₂, so q² = s, the square of the centre-of-mass energy, and that photon rematerialises as μ⁻μ⁺. Because the muon is just a heavier electron — same charge, spin, and coupling — the final state differs from the initial one only by a mass. No hadronic mess, no strong force, no substructure: two pointlike QED vertices joined by one propagator, which is why this is the cleanest test of electrodynamics ever devised and the standard "candle" for collider luminosities and quark counts. The amplitude factorises with beautiful economy into an electron current and a muon current contracted through the photon propagator, ℳ = −(e²/s) jₑ·j_μ. Squaring and spin-averaging — by the trace machinery of Casimir's trick, or more transparently by tracking helicities one configuration at a time — collapses to a single compact prediction: dσ/dΩ = (α²/4s)(1 + cos²θ), integrating to σ = 4πα²/3s. The (1 + cos²θ) shape is no accident of algebra; it is the fingerprint of a spin-1 mediator coupling to fermions whose chirality the interaction conserves. Only four of the sixteen helicity combinations survive, and their d-function shapes add up to that curve. Measured at SPEAR, at PETRA's JADE detector, and at every e⁺e⁻ collider since, it is QED shown working to the decimal point.
- A single tree-level QED diagram: e⁺e⁻ annihilate into a virtual photon with q² = s that converts to μ⁺μ⁻ — no strong force, no substructure.
- The amplitude splits into two fermion currents joined by the photon propagator, ℳ = −(e²/s) jₑ·j_μ.
- Spin-averaging gives ⟨|ℳ|²⟩ = e⁴(1 + cos²θ) in the ultra-relativistic limit, so dσ/dΩ = (α²/4s)(1 + cos²θ).
- Integrating gives the point cross section σ = 4πα²/3s = 86.8 nb ÷ s[GeV²] — 0.87 nb at √s = 10 GeV.
- Of 16 helicity combinations only four are non-zero, from chirality conservation at the vertex; their (1 ± cosθ)² shapes build the (1 + cos²θ) distribution.
- The reaction is forward–backward symmetric in pure QED (A_FB = 0); the measured asymmetry is γ–Z interference, and R = σ_had/σ_μμ = 3Σ Q_q² counts quark colours.
- Key idea: the annihilation amplitude is a product of two independent fermion currents joined by the photon propagator, and because QED conserves chirality only specific helicity pairings couple — which is what shapes the (1 + cos²θ) distribution.
The annihilation amplitude is a product of two independent fermion currents joined by the photon propagator, and because QED conserves chirality, only specific helicity pairings of the incoming and outgoing fermions actually couple — this is what shapes the (1 + cos²θ) distribution.
Where this lives today
Muon-pair production is a live tool, not a historical exercise. Because its cross section is calculable to sub-percent precision, e⁺e⁻ → μ⁺μ⁻ is the standard candle for luminosity: count the muon pairs and you know how many collisions the machine delivered, which normalises every other measurement — at the B-factories BaBar, Belle, and Belle II (√s = 10.58 GeV) it and the Bhabha process pin down the luminosity behind rare-decay branching fractions. Once the Z is added, the asymmetry that vanishes in pure QED becomes one of the sharpest electroweak observables: at LEP, A_FB in e⁺e⁻ → μ⁺μ⁻ and → bb̄ measured sin²θ_W to four significant figures and bounded the top and Higgs masses before either was seen. The muon-pair-normalised R ratio remains the primary spectroscopy of quark thresholds and the running of α_s, and its low-energy hadronic cousin feeds the hadronic vacuum-polarisation term of the muon g − 2 — the largest theoretical uncertainty in that landmark test. Proposed machines like FCC-ee and a linear collider would return to muon pairs as a precision benchmark. The one-diagram process a student computes first is, decades on, still doing frontier work.
The physical picture
Fire an electron and a positron head-on. At high energy they do not scatter through a slow exchange of momentum — they annihilate outright, their charges cancelling into a single quantum of the electromagnetic field. That quantum is a virtual photon carrying the full centre-of-mass energy and momentum, q = p₁ + p₂, and therefore off its mass shell, q² = s ≫ 0 (a real photon has q² = 0). It lives only the instant the uncertainty principle allows, and — being a spin-1 object that couples identically to every charged fermion — it can convert into any charged pair light enough to make. Look at the μ⁺μ⁻ channel and you are watching electrodynamics with initial and final states differing by nothing but a mass. The first figure is that story drawn as the one diagram that carries it.
The angular distribution is where the physics becomes visible. Each external fermion can be right- or left-handed, giving sixteen helicity combinations in principle — but the vector coupling ū γ^μ u links only same-chirality states, so the electron and positron must arrive with opposite helicities to feed a spin-1 photon, and likewise the muon pair. That kills twelve and leaves four. Each survivor is the overlap of an initial spin-1 state (along the beam) with a final spin-1 state (along the muon direction), which quantum mechanics fixes to be a Wigner d-function of the angle: the amplitudes go as (1 ± cosθ). Squaring and averaging turns the four (1 ± cosθ)² shapes into the smooth (1 + cos²θ) curve — a parameter-free prediction. The second figure plots it, with the two helicity families that build it.
The core equations
The starting point is the Feynman amplitude. Reading the single diagram with the QED rules — a spinor per external leg, a factor −ieγ^μ at each vertex, the propagator −i g_{μν}/q² between — gives an amplitude that factorises into an electron current and a muon current contracted through the propagator. All else follows.
where: e — the positron charge (α = e²/4π ≈ 1/137); s = q² = (p₁ + p₂)² — the CoM energy squared carried by the virtual photon; u, v — Dirac spinors for particles and antiparticles; γ^μ — the vertex gamma matrices. The bracket [v̄(p₂)γ^μu(p₁)] is the electron current jₑ^μ, [ū(p₃)γ^νv(p₄)] the muon current j_μ^ν, and 1/s the propagator — the amplitude is literally jₑ·j_μ divided by s. For an unpolarised cross section we square the amplitude, sum over final spins, and average over the four initial spins (a factor ¼). Casimir's trick turns each spin sum into a trace of gamma matrices; evaluating and contracting the two traces gives a result built purely from four-momentum dot products.
where: m_μ — the muon mass, kept in the final-state trace (the electron mass is dropped, since √s ≫ m_e); the three dot-product terms are the muon-tensor–electron-tensor contraction, and only m_μ² appears because m_e is negligible. Inserting the centre-of-mass kinematics — beam energy E = √s/2, muon momentum p = βE with β = √(1 − 4m_μ²/s), muon angle θ to the e⁻ beam — turns the dot products into the angular form, which for β → 1 reduces to the clean (1 + cos²θ).
where: θ — angle between the outgoing μ⁻ and incoming e⁻ in the CoM frame; the mass term (4m_μ²/s)sin²θ fills the θ = 90° region near threshold and vanishes at high energy. At √s = 10 GeV, 4m_μ²/s ≈ 4.5×10⁻⁴ is negligible. Folding this into the two-body master formula dσ/dΩ = (1/64π²s)(p_f/p_i)⟨|ℳ|²⟩, with flux ratio p_f/p_i = β and e⁴ = (4πα)² = 16π²α², gives the differential cross section.
where: β = √(1 − 4m_μ²/s) — the muon velocity, a threshold factor switching the process on at √s = 2m_μ; α²/4s sets the scale, falling as 1/s. At √s = 10 GeV the θ = 90° value is 0.052 nb/sr, doubling at the poles. Integrating (1 + cos²θ) over solid angle gives 16π/3, and the total takes its famous compact form — the "point cross section" benchmarking every e⁺e⁻ machine. The mass-dependent version carries the threshold factor β(3 − β²)/2.
where: the numerical form uses (ħc)² = 3.894×10⁵ nb·GeV²; at √s = 10 GeV, σ = 0.87 nb. Near threshold σ ∝ β (S-wave onset of a vector current); far above it σ = 4πα²/3s exactly. Tracking helicities instead of traces gives the same total more transparently: only four of the sixteen combinations survive, each a squared Wigner d¹-function of θ — the amber and green curves of the second figure.
where: the labels give (e⁻ e⁺ → μ⁻ μ⁺) helicities, R = right, L = left; the twelve unlisted combinations vanish by chirality conservation. Averaging, ¼[2(1+cosθ)² + 2(1−cosθ)²] = 1 + cos²θ, recovers equation (3). Finally, why this reaction is a workhorse: normalising the hadronic rate to the muon-pair rate gives the ratio R, counting quark charges and colours directly.
where: N_c = 3 is the number of colours, the sum running over quark flavours accessible at that energy. Below the b threshold (u, d, s, c), R = 3(4/9 + 1/9 + 1/9 + 4/9) = 10/3 ≈ 3.33 — the factor of 3 was decisive evidence for colour.
The goal is to turn the compact amplitude of equation (1) into the spin-averaged square of equations (2) and (3), using Casimir's trick to trade explicit spinors for traces of gamma matrices, and to show the result in the centre-of-mass frame is exactly e⁴(1 + cos²θ) once the muon mass is neglected. The method is mechanical, which is its virtue: it never requires choosing a spin basis, and the same steps recur in every QED and electroweak cross-section, so the bookkeeping pays off everywhere.
Step 1. Form |ℳ|² = ℳℳ. Because the amplitude factorises, so does the square: |ℳ|² = (e⁴/s²) L_e^{μν} L^{(μ)}_{μν}, where each L is a spin sum over one current times its conjugate. Why: the electron and muon currents are contracted only through the metric, so their spin sums separate into two tensors computed one at a time.*
Step 2. Apply Casimir's trick to each tensor. Summing a bilinear [ū Γ u][ū Γ u] over spins, using Σ u ū = (p̸ + m) and Σ v v̄ = (p̸ − m), collapses the spinor indices into a trace. The electron tensor (massless) is L_e^{μν} = Tr[p̸₁γ^μ p̸₂γ^ν], the muon tensor is L^{(μ)}_{μν} = Tr[(p̸₃ + m_μ)γ_μ(p̸₄ − m_μ)γ_ν]. Why: the spinor completeness relations convert an explicit spin sum into a basis-free trace, which is why no helicity choice is needed.*
Step 3. Evaluate the traces with Tr[γ^αγ^βγ^γγ^δ] = 4(g^{αβ}g^{γδ} − g^{αγ}g^{βδ} + g^{αδ}g^{βγ}) and Tr[odd γ] = 0. The electron tensor gives 4(p₁^μp₂^ν + p₁^νp₂^μ − g^{μν} p₁·p₂); the muon tensor the same with 3,4 and an extra −g_{μν}m_μ². Contracting the two gives 8[(p₁·p₃)(p₂·p₄) + (p₁·p₄)(p₂·p₃) + m_μ²(p₁·p₂)]. Why: every dot product in equation (2) is a direct output of contracting the two trace results, with the m_μ² term arising solely from the mass insertion in the muon tensor.
Step 4. Insert CoM kinematics. With p₁·p₂ = s/2 and (massless muons) p₁·p₃ = p₂·p₄ = (s/4)(1 − cosθ), p₁·p₄ = p₂·p₃ = (s/4)(1 + cosθ), the bracket becomes (s²/8)(1 + cos²θ). Multiplying by 8e⁴/s² and averaging over the four initial spins (÷4) yields ⟨|ℳ|²⟩ = e⁴(1 + cos²θ). Why: the abstract dot products become an angle, and the (1 + cos²θ) drops out with no free parameters, confirming equation (3). *Note: the massless spin average e⁴(1 + cos²θ) reproduces equation (6) exactly, and retaining the muon mass yields the threshold factor β(3 − β²)/2 in the total cross section.*
Go deeper · Derivation B · 2 minThe four surviving helicity amplitudes and where the shape comes from
This derivation reaches the same ⟨|ℳ|²⟩ by a more physical road: computing the amplitude for each definite helicity configuration, finding that only four of sixteen are non-zero, and seeing the survivors' angular shapes dictated by angular-momentum conservation alone. The aim is to expose why the (1 + cos²θ) curve looks as it does — that it is the overlap of a spin-1 state along the beam with a spin-1 state along the muon direction — which the trace method hides.
Step 1. Impose chirality conservation at each vertex. In the massless limit the vector coupling ū γ^μ u connects only equal-chirality spinors, and for a massless fermion chirality equals helicity. The e⁻ and e⁺ annihilating into a spin-1 photon must carry opposite helicities (spins adding to ±1 along the beam), and likewise the μ⁻μ⁺. Why: a spin-1 mediator cannot couple to a J = 0 fermion pair, so the twelve same-helicity combinations give an identically vanishing current — the origin of "16 → 4", dynamical not kinematic.
Step 2. The survivors are e⁻_R e⁺_L and e⁻_L e⁺_R initially, μ⁻_R μ⁺_L and μ⁻_L μ⁺_R finally — four combinations. Each initial state is a J = 1 object with J_z = ±1 along the beam; each final state a J = 1 object with J_z′ = ±1 along the muon direction. Why: with massless leptons the process is the transition of one unit of angular momentum from the beam axis to the muon axis, so the amplitude is fixed by rotation.
Step 3. Read off the amplitudes as Wigner d-functions. The overlap of J_z = +1 (beam) with J_z′ = +1 (muons at θ) is d¹_{1,1}(θ) = ½(1 + cosθ); with J_z′ = −1 it is d¹_{1,−1}(θ) = ½(1 − cosθ). Hence |ℳ_{RL→RL}|² ∝ (1 + cosθ)² and |ℳ_{RL→LR}|² ∝ (1 − cosθ)², the L-initial states giving the mirror pair. Why: the (1 ± cosθ) factors are literally rotation matrix elements — the forward-peaked survivor keeps the spin projection aligned, the backward-peaked one flips it, which is why the distribution is symmetric.
Step 4. Average over the four initial spin states (two of which — RL and LR — are non-zero, but the ¼ runs over all four): ⟨|ℳ|²⟩ = ¼ e⁴[(1 + cosθ)² + (1 − cosθ)² + (1 − cosθ)² + (1 + cosθ)²] = e⁴(1 + cos²θ). Why: 2(1 + cosθ)² + 2(1 − cosθ)² = 4(1 + cos²θ), and the ¼ delivers exactly equation (3) — the same answer as the traces, now with the shape explained. *Note: ¼[2(1+cosθ)² + 2(1−cosθ)²] = 1 + cos²θ identically — for example at θ = 37° both sides equal 1.638 — and the Wigner d¹-function values d¹_{1,1} = ½(1+cosθ), d¹_{1,−1} = ½(1−cosθ) are standard results.*
Go deeper · Derivation C · 1 minFrom the amplitude to the total cross section
The purpose is to complete the calculation: take the spin-averaged amplitude and produce the differential and total cross sections of equations (4) and (5), threshold factor included, by folding ⟨|ℳ|²⟩ through two-body phase space. This is where the abstract amplitude becomes a number a detector normalises against, and the integral that yields 4πα²/3s is worth doing by hand once, because that combination is the reference cross section of the entire e⁺e⁻ programme.
Step 1. Write the CoM master formula dσ/dΩ = (1/64π²s)(p_f/p_i)⟨|ℳ|²⟩. With massless electrons p_i = √s/2 and muons p_f = βE = (√s/2)β, the phase-space ratio is p_f/p_i = β. Why: flux and phase space contribute only the kinematic prefactor and the velocity ratio β; all the dynamics sit in ⟨|ℳ|²⟩, cleanly separating "how many final states" from "how strongly they couple".
Step 2. Substitute ⟨|ℳ|²⟩ and use e⁴ = 16π²α²: the prefactor 16π²α²/64π²s = α²/4s, giving equation (4). Why: converting e⁴ to α² turns a formula in the bare coupling into one in the measured fine-structure constant, ready for numbers.
Step 3. Integrate: using ∫(1 + cos²θ)dΩ = 16π/3 and ∫sin²θ dΩ = 8π/3, σ = (α²/4s)β[16π/3 + (4m_μ²/s)(8π/3)]. Writing 4m_μ²/s = 1 − β² gives σ = (2πα²/3s)β(3 − β²) = (4πα²/3s)·β(3 − β²)/2. Why: the two angular integrals are the only calculus here, and 2 + (1 − β²) = 3 − β² packages the mass correction into the single threshold factor β(3 − β²)/2.
Step 4. In the high-energy limit β → 1 the factor β(3 − β²)/2 → 1 and σ → 4πα²/3s. Putting in (ħc)² = 3.894×10⁵ nb·GeV² gives σ = 86.8 nb ÷ s[GeV²]. Why: the point cross section, whose 1/s fall-off is the defining signature of pointlike production — the baseline every measured e⁺e⁻ cross section is quoted against. *Note: evaluating explicitly, 4π/3·α²·(ħc)² = 86.86 nb·GeV², giving σ(√s = 10 GeV) = 0.869 nb; the differential cross section at 90° is dσ/dΩ = α²/4s·(ħc)² = 0.052 nb/sr, rising to 0.104 nb/sr at the poles.*
Limiting cases
| Case | Result | Meaning |
|---|---|---|
| Ultra-relativistic, √s ≫ m_μ | dσ/dΩ = (α²/4s)(1 + cos²θ), σ = 4πα²/3s | the headline QED result; symmetric, A_FB = 0, falls as 1/s |
| Near threshold, √s → 2m_μ, β → 0 | σ → (2πα²/s)·β; angular term picks up (4m_μ²/s)sin²θ | S-wave switch-on of a vector current; slow muons, the 90° dip fills in |
| Keep muon mass exactly | σ = (4πα²/3s)·β(3 − β²)/2 | β(3 − β²)/2 interpolates from 0 at threshold to 1 at high energy |
| Add Z exchange, √s → M_Z | γ–Z interference: A_FB ≠ 0, resonant peak at √s = 91.2 GeV | electroweak beyond QED; the asymmetry measures sin²θ_W |
| Hadronic final state | R = σ_had/σ_μμ = 3Σ_q Q_q² (= 10/3 for u,d,s,c) | the factor 3 counts colour; step jumps at each quark threshold |
| Bhabha channel e⁺e⁻ → e⁺e⁻ | extra t-channel diagram, forward divergence ∝ 1/θ⁴ | identical final state adds Coulomb scattering; not pure annihilation |
Worked examples
1 · The muon-pair cross section and event rate at a B-factory (fully worked)
A symmetric e⁺e⁻ collider runs at √s = 10.0 GeV with luminosity L = 1.0×10³⁴ cm⁻²s⁻¹. Find the total μ⁺μ⁻ cross section, the differential cross section at θ = 90° and the poles, and the muon-pair rate per second and per day.
Step 1. The muon mass is negligible: 4m_μ²/s = 4(0.1057)²/100 = 4.5×10⁻⁴, so β = 0.99978 ≈ 1 and β(3 − β²)/2 = 1.0000 to four figures. Use the massless formulas. The pure-QED limit applies.
Step 2. Total from equation (5): σ = 4πα²/3s = 86.8/100 = 0.87 nb. (Long form: (4π/3)(1/137.036)²(3.894×10⁵ nb·GeV²)/100 GeV² = 0.869 nb.)
Step 3. Differential from equation (4): α²/4s·(ħc)² = 0.052 nb/sr at θ = 90° (where 1 + cos²θ = 1), rising to 0.104 nb/sr at the poles — a factor-of-two swing, the 1 → 2 range of (1 + cos²θ).
Step 4. Event rate. Convert luminosity: L = 1.0×10³⁴ cm⁻²s⁻¹ = 10 nb⁻¹s⁻¹ (1 nb⁻¹ = 10³³ cm⁻²). Then dN/dt = σL = (0.869 nb)(10 nb⁻¹s⁻¹) = 8.7 muon pairs per second, or 7.5×10⁵ per day — a clean, precisely calculable rate ticking at a few Hz, which is exactly why μ⁺μ⁻ serves as a luminosity monitor. σ = 0.87 nb, dσ/dΩ = 0.052–0.104 nb/sr, rate ≈ 8.7 s⁻¹.
2 · Reading the angular distribution: the helicity decomposition at 60°
At θ = 60° in the CoM frame, evaluate the total (1 + cos²θ) and its two helicity families (1 + cosθ)²/2 and (1 − cosθ)²/2, confirm they sum, and find the forward–backward asymmetry in pure QED.
Step 1. Total: at θ = 60°, cosθ = 0.5, so 1 + cos²θ = 1.25 (units of α²/4s) — a quarter of the way up from the 90° minimum of 1 toward the pole value of 2.
Step 2. Forward family: (1 + cosθ)²/2 = (1.5)²/2 = 1.125. Backward family: (1 − cosθ)²/2 = (0.5)²/2 = 0.125. Their sum 1.125 + 0.125 = 1.25 matches Step 1 — the two families always add to the total, as equation (6) requires.
Step 3. This is why the distribution looks forward-heavy at 60°: the forward family dominates there by 1.125/0.125 = 9 to 1. But over the whole sphere the forward excess of one family exactly cancels the backward excess of the other.
Step 4. Asymmetry: A_FB = (σ_F − σ_B)/(σ_F + σ_B). Since (1 + cos²θ) is symmetric under θ → π − θ, the forward (0–90°) and backward (90–180°) hemispheres carry equal rate, so A_FB = 0 exactly in pure QED. A non-zero A_FB directly signals physics beyond one-photon exchange — γ–Z interference. Total 1.25 = 1.125 + 0.125; A_FB = 0.
3 · Counting quark colours with the R ratio
At √s = 10 GeV (below the open-beauty threshold), predict R = σ_had/σ_μμ and the hadronic cross section, and say what a measured R ≈ 3.6 implies.
Step 1. Accessible quarks at 10 GeV are u, d, s, c with charges +2/3, −1/3, −1/3, +2/3, so Σ Q_q² = 4/9 + 1/9 + 1/9 + 4/9 = 10/9.
Step 2. With colour, equation (7) gives R = N_c Σ Q_q² = 3 × 10/9 = 10/3 ≈ 3.33. Without colour (N_c = 1) it would be only 10/9 ≈ 1.11 — a factor of three smaller.
Step 3. Hadronic cross section: σ_had = R × σ_μμ = 3.33 × 0.87 nb = 2.9 nb, measured by counting multi-hadron final states against clean muon pairs.
Step 4. A measured R ≈ 3.6 sits just above 3.33; the excess is the leading QCD correction R = 3Σ Q_q²(1 + α_s/π), which pins α_s ≈ 0.2 here. The gross value near 3⅓, not 1, was historically the cleanest proof of three quark colours. R = 10/3, σ_had ≈ 2.9 nb, and R ≈ 3.6 measures α_s.
Physics of the portrait
The theory of e⁺e⁻ → μ⁺μ⁻ was pure paper until someone built a machine that could collide matter with antimatter head-on. That someone was Bruno Touschek, who turned the annihilation diagram into an experimental programme.
Origins · 1 minBruno Touschek and how this was discovered
Bruno Touschek was an Austrian-born physicist whose life mirrored the century's turbulence — barred from university under the Nazi race laws, he studied physics half-clandestinely, worked on the German betatron project, and was imprisoned by the Gestapo in 1945, surviving to finish his degree in Göttingen and settle in Rome. His gift to particle physics was conceptual and audacious. In 1960, at Frascati, he argued that the way to reach high energy cleanly was not to smash a beam into a fixed target — where most of the energy is wasted on recoil — but to store electrons and positrons in a single ring, circulating in opposite directions in the same magnetic field, and let them annihilate head-on. Because a positron is an electron running backward, one field steers both, and every joule is available in the centre of mass. He built the proof himself: AdA (Anello di Accumulazione), a ring just over a metre across and the world's first electron–positron storage ring, which stored counter-rotating beams in 1961 and, moved to Orsay for a brighter source, produced the first observed e⁺e⁻ collisions in 1964. Every collider that followed — ADONE, SPEAR, PETRA, LEP, the B-factories, the proposed FCC-ee — descends from Touschek's ring, and e⁺e⁻ → μ⁺μ⁻ is the reaction they were built to exploit: the calculable QED yardstick that calibrates their luminosity and measures their new physics. Touschek gave the annihilation amplitude a laboratory.
Going further
Texts. Thomson, Modern Particle Physics, §6.2 (electron–positron annihilation), Eqs. 6.3–6.23 · Peskin & Schroeder, An Introduction to QFT, §5.1 (e⁺e⁻ → μ⁺μ⁻), Eqs. 5.1–5.13 · Halzen & Martin, Quarks and Leptons, §6.2, Eq. 6.29 · Griffiths, Introduction to Elementary Particles, 2nd ed., §7.7 · Schwartz, QFT and the Standard Model, §13.1. The key results of this article are: the spin-averaged amplitude ⟨|ℳ|²⟩ = e⁴(1 + cos²θ), obtained consistently from both the trace contraction and the four helicity amplitudes; the differential cross section dσ/dΩ = (α²/4s)(1 + cos²θ) and total σ = 4πα²/3s = 86.8 nb ÷ s[GeV²]; the numerical values σ = 0.87 nb, dσ/dΩ = 0.052–0.104 nb/sr, and rate 8.7 s⁻¹ at √s = 10 GeV with L = 10³⁴ cm⁻²s⁻¹; R = 10/3 for u, d, s, c; and the threshold factor β(3 − β²)/2.
Prerequisites. Feynman Rules of QED · Dirac Spinor Solutions · Spin and Helicity States · The Photon Propagator · Differential Cross Sections. Leads to. Trace Techniques · Chirality in QED · Bhabha and Møller Scattering · Z Resonance at LEP · QCD in e⁺e⁻ Annihilation.
Ask the tutor: Ask me to redo the spin sum keeping the full muon mass and recover the β(3 − β²)/2 threshold factor, or to add Z exchange and derive the forward–backward asymmetry A_FB that LEP measured.
Why is e⁺e⁻ → μ⁺μ⁻ called the "hydrogen atom" of QFT?
Because it is the simplest complete calculation that still contains all the essential machinery. A single tree diagram means no summing over topologies; two pointlike leptonic vertices mean no strong force, no form factors, no substructure; and identical electron and muon couplings mean the states differ only by a mass. A student can carry the whole thing from Feynman rules to a measured number in a few pages — meeting Casimir's trick, the traces, and helicity methods along the way — and every harder QED and electroweak process reuses exactly those steps. It is the universal first entry in the cross-section chapter.
The photon here has q² = s > 0 but a real photon has q² = 0. Is that a contradiction?
No — the photon in this diagram is virtual, an internal line that is never observed, and internal lines need not satisfy the mass-shell condition q² = 0. It is a stand-in for the field configuration produced when the e⁺e⁻ currents overlap, and it carries whatever four-momentum conservation forces on it, here q = p₁ + p₂ with q² = s. The propagator factor 1/q² = 1/s encodes this off-shell-ness and is exactly why the cross section falls as 1/s. Only external, detected photons must have q² = 0; the virtual one lives and dies inside the diagram.
Why does the cross section fall as 1/s, and what does that tell us?
A cross section is an area, so in natural units it scales as 1/energy² = 1/s once there is no other scale in the problem — and for pointlike particles well above threshold the lepton masses are negligible, so there is none. The 1/s fall-off is thus the direct signature of structureless, pointlike production. Any deviation would flag new structure: a resonance (the Z bump), a form factor (if the leptons were composite), or a contact interaction. Confirming that the muon-pair cross section really tracks 4πα²/3s across a wide energy range is a stringent proof that electrons and muons are pointlike down to the shortest distances probed.
What is the physical origin of the (1 + cos²θ) shape rather than isotropy?
It comes entirely from angular momentum. The ultra-relativistic e⁺e⁻ pair, coupling through a vector current, forms a spin-1 state aligned along the beam; the outgoing muons form a spin-1 state aligned along their own direction at angle θ. The amplitude is the overlap of these two spin-1 states, which rotational invariance fixes to be a Wigner d¹-function ∝ (1 ± cosθ). Summing the four allowed helicity configurations turns those into (1 + cosθ)² and (1 − cosθ)² terms whose sum is 1 + cos²θ. An isotropic distribution would instead signal a spin-0 mediator — so the shape of the curve measures the photon's spin.
If pure QED gives A_FB = 0, how does the measured asymmetry arise?
It comes from adding the Z boson to the photon. Both the γ and the Z are exchanged in the s-channel and their amplitudes interfere. The Z couples with both vector and axial-vector pieces, and the parity-violating axial coupling introduces a term in the angular distribution proportional to cosθ (not cos²θ). That linear term is antisymmetric under θ → π − θ, so the forward and backward hemispheres differ and A_FB ≠ 0. Well below the Z pole it is a small negative asymmetry (a few percent, as JADE saw at 34 GeV); on the resonance at LEP it is large and was measured to extract sin²θ_W. A non-zero A_FB is a clean window on electroweak physics beyond electrodynamics.
Why keep the muon mass in the trace but drop the electron mass?
Because the two enter at very different scales. At any collider that makes muon pairs √s is at least a few GeV, so the electron mass (0.511 MeV) gives corrections of order m_e²/s ≲ 10⁻⁸ — negligible, and dropping it simplifies the electron tensor to one clean trace. The muon mass (105.7 MeV) sets the production threshold √s = 2m_μ, so near threshold m_μ²/s is not small and the (4m_μ²/s)sin²θ term and the velocity factor β genuinely matter. Far above threshold both reduce to the massless (1 + cos²θ) result. Keeping m_μ symbolically lets one formula cover both regimes.