University Quantum Field Theory I · QED at Tree Level: the Lagrangian, Feynman Rules and Electron-Muon Processes · 9.7

Helicity amplitudes and chirality conservation

The trace of topics 9.05 and 9.06 hands you 1 + cos²θ and hides where it comes from. Here e⁺e⁻ → μ⁺μ⁻ is taken apart one helicity at a time: chirality leaves four amplitudes, each a spin-one state seen along a rotated axis, and the muon mass adds the flips that take over near threshold.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

The trace of topics 9.05 and 9.06 gives (1/4)Σ|M|² = e⁴(1 + cos²θ) for e⁺e⁻ → μ⁺μ⁻ at high energy and says nothing about why. Helicity amplitudes do. Evaluate M with every lepton in a definite helicity and the vector coupling does most of the work: by UAQM-05.08's chirality rule, quoted, it joins an e⁻ only to an e⁺ of opposite helicity, so the annihilating pair carries J_z = +1 or −1 along the beam, never 0, and the muon pair leaves with J = ±1 along its own axis.

Of sixteen helicity combinations four survive, and explicit massless Weyl spinors give them as e²(1 + cos θ) when the muons carry the same J along their axis as the beam pair along the beam, and e²(1 − cos θ) when opposite, derived. Each is 2e² times a rotation matrix element d¹(θ), so the zeros at θ = 0 and π are angular-momentum conservation, not algebra. Different final helicities never interfere, so the squares add, and (1 + cos θ)² + (1 − cos θ)² = 2(1 + cos²θ) is the trace result.

The picture is exact only for massless leptons. The muon mass lets the pair flip into J = 0 along its axis, with amplitude e²(m_μ/E) sin θ, which makes the distribution isotropic at threshold. The electron's own flips, of size mₑ/E, are negligible above a few MeV, and a coupling that weighted the two channels unequally, as the Z does, would add a forward-backward asymmetry that tree-level QED lacks.

Simple definition: A helicity amplitude is M evaluated with every incoming and outgoing particle in a definite helicity state; squaring and summing over the helicities nobody measured gives back the spin-averaged |M|² that traces compute.
Example: At high energy e⁻_R e⁺_L → μ⁻_R μ⁺_L has M = e²(1 + cos θ). At θ = 45°, with e² = 4πα = 0.0917, that is 0.157, against 0.0269 for μ⁻_L μ⁺_R: right-handed μ⁻ outnumber left-handed 34 to 1 at that angle.

Chirality selection at the annihilation vertexv̄(p₂, h₂) γ^μ u(p₁, h₁) = 0 unless h₂ = −h₁, up to terms of order mₑ/E

Quoted from UAQM-05.08: γ^μ joins L to L and R to R, and an e⁺ of helicity −½ is R-chiral. The pair's spins line up: J_z = ±1 along the beam, never 0.

h₁, h₂: e⁻ and e⁺ helicities, ±½; mₑ = 0.511 MeV; E: beam energy in GeV; at √s = 10 GeV, mₑ/E = 1.0 × 10⁻⁴

The beam current is a spin-one polarisationv̄(p₂)γ^μ u(p₁) = 2√2 E ε^μ_±(ẑ), with ε^μ_± = ∓(0, 1, ±i, 0)/√2

Derived from massless Weyl spinors. The time component is zero, as q_μ j^μ = 0 with q = (2E, 0, 0, 0) requires: the virtual photon is J = 1 with J_z = ±1.

E: beam energy (GeV); ẑ: e⁻ direction; upper signs e⁻_R e⁺_L, lower e⁻_L e⁺_R; u†u = 2E; ε dimensionless

Each helicity amplitude is a rotation overlap|M(J_z = λ → Jₙ = λ′)| = 2e² |d¹_λλ′(θ)|, with d¹₁₁ = (1 + cos θ)/2 and d¹₁,₋₁ = (1 − cos θ)/2

Derived: M = 2e² ε⃗_λ′(n̂)*⋅ε⃗_λ(ẑ). Same J along both axes gives e²(1 + cos θ), opposite J gives e²(1 − cos θ); signs and phases are conventions.

λ, λ′ = ±1: pair spin along the beam ẑ and along the μ⁻ direction n̂; θ: μ⁻ angle from e⁻; e² = 4πα; M dimensionless

Squares added: the unpolarised distribution(1/4) Σ|M|² = (e⁴/4) × 2[(1 + cos θ)² + (1 − cos θ)²] = e⁴(1 + cos²θ)

Derived with no trace, and equal to topic 9.06's result. Different final helicities never interfere, so 1 + cos²θ is two incoherent channels of equal total rate.

1/4 averages the e⁻ and e⁺ helicities; the 2 counts e⁻_R e⁺_L and e⁻_L e⁺_R; massless leptons; dimensionless

Helicity flips at finite muon mass|M(e⁻_R e⁺_L → μ⁻_R μ⁺_R)| = |M(e⁻_R e⁺_L → μ⁻_L μ⁺_L)| = e² (m_μ/E) sin θ

Derived: the equal-helicity current −2m_μ n̂ is Jₙ = 0, reached with d¹₁₀ ∝ sin θ. It adds (4m_μ²/s) sin²θ to 1 + cos²θ and makes threshold isotropic.

m_μ = 105.66 MeV; E: beam energy, at least m_μ; the opposite-helicity amplitudes stay e²(1 ± cos θ) at every muon speed

Forward-backward asymmetry from the two channelsA_FB = (3/4)(wₛₐₘₑ − wₒₚₚ)/(wₛₐₘₑ + wₒₚₚ)

Derived: 7/8 of a (1 + cos θ)² channel lies forward. QED's equal weights give A_FB = 0; PETRA's measured ≈ −0.1 near √s = 35 GeV is the Z, quoted.

wₛₐₘₑ, wₒₚₚ: total weights of the (1 + cos θ)² and (1 − cos θ)² channels; A_FB dimensionless; massless leptons

Chirality leaves the beam pair only J_z = ±1

Units are natural, ħ = c = 1, with the metric (+, −, −, −), and e² = 4πα = 0.0917, so a 2 → 2 amplitude M is dimensionless. Label e⁻(p₁) e⁺(p₂) → μ⁻(p₃) μ⁺(p₄) in the centre-of-momentum frame, with beam energy E, s = 4E², the e⁻ along ẑ and the μ⁻ along n̂ at angle θ; the subscripts R and L name helicity. UAQM-05.08's rule, quoted, is that ψ̄γ^μψ = ψ̄_Lγ^μψ_L + ψ̄_Rγ^μψ_R never joins opposite chiralities. For a massless particle chirality is helicity, but a v spinor of R chirality describes an antiparticle of helicity −½. So v̄(p₂)γ^μu(p₁) is non-zero only for e⁻_R e⁺_L and e⁻_L e⁺_R: opposite helicities, which for particles moving in opposite directions means parallel spins, J_z = +1 or −1. Of the 16 helicity choices for four leptons, 4 survive. What is left over is of order mₑ/E, 1.0 × 10⁻⁴ at √s = 10 GeV, and it enters rates squared. A scalar vertex would do the reverse, joining L to R and allowing only J_z = 0, so the selection belongs to the photon's vector coupling, not to annihilation as such.

Explicit spinors: the beam current is a polarisation vector

In the Weyl representation, with u†u = 2E and one common choice of phases, a massless e⁻_R moving along +ẑ has u(p₁) = √(2E)(0, 0, 1, 0), and an e⁺_L moving along −ẑ has v(p₂) = √(2E)(0, 0, 0, −1). Since γ⁰γ^μ = diag(σ̄^μ, σ^μ), only the lower, right-chiral halves meet, and v̄(p₂)γ^μu(p₁) = −2E(0, 1, i, 0), derived. Three features carry the rest of the topic. The time component vanishes, as current conservation with q = (2E, 0, 0, 0) demands. The spatial part is proportional to ε⃗₊(ẑ) = −(1, i, 0)/√2, the circular polarisation of a spin-one state with J_z = +1, so the current is 2√2E ε^μ₊(ẑ): the virtual photon is born with J = 1 and J_z = +1. And it carries one power of energy, from two spinors of √(2E) each, so that two currents over the propagator's s leave M dimensionless and independent of energy. For e⁻_L e⁺_R the same steps give 2E(0, 1, −i, 0) = 2√2E ε^μ₋(ẑ), with J_z = −1.

Rotating to the muon axis: e²(1 + cos θ) and e²(1 − cos θ)

The muon line is the same calculation for momenta along ±n̂, with n̂ = (sin θ, 0, cos θ), each spinor's phase fixed, as the beam's were, by the polar angles of its own momentum: θ for the μ⁻, and π − θ at azimuth π for the μ⁺. For μ⁻_R along n̂ and μ⁺_L along −n̂ that gives ū(p₃)γ^μv(p₄) = 2E(0, cos θ, −i, −sin θ), derived; spinors turned rigidly through θ from the beam's would flip the sign of every muon current. The photon propagator −ig_μν/s joins the two currents, M = (e²/s)[v̄(p₂)γ^μu(p₁)][ū(p₃)γ_μv(p₄)]. Both time components vanish, so the contraction is minus a three-vector product: −(−2E)(2E)[cos θ + i(−i) + 0] = 4E²(1 + cos θ). With s = 4E², M(e⁻_R e⁺_L → μ⁻_R μ⁺_L) = e²(1 + cos θ), derived. Reversing both muon helicities gives the current −2E(0, cos θ, i, −sin θ) and M(e⁻_R e⁺_L → μ⁻_L μ⁺_R) = e²(1 − cos θ). From e⁻_L e⁺_R the same steps give e²(1 + cos θ) for μ⁻_L μ⁺_R and e²(1 − cos θ) for μ⁻_R μ⁺_L. Overall signs and phases depend on the spinor conventions, the magnitudes do not, and no rate sees the phases. Check at θ = 90°: both amplitudes are e² = 0.0917, so a fast μ⁻ emitted sideways is equally likely to have either helicity.

Each amplitude is a spin-one overlap, d¹(θ)

The muon current is a polarisation vector too: 2E(0, cos θ, −i, −sin θ) = −2√2E ε^μ₊(n̂)*, where ε⃗₊(n̂) = −(cos θ, i, −sin θ)/√2 is ε⃗₊(ẑ) rotated through θ about y. Substituting both currents, M = 2e² ε⃗₊(n̂)*⋅ε⃗₊(ẑ), and that dot product is the overlap ⟨1, +1 along n̂ | 1, +1 along ẑ⟩ = d¹₁₁(θ) = (1 + cos θ)/2, derived; likewise ε⃗₋(n̂)*⋅ε⃗₊(ẑ) = d¹₁,₋₁(θ) = (1 − cos θ)/2. So annihilation makes a J = 1 state with J_z = ±1 along the beam, the muon pair leaves in J = 1 with Jₙ = ±1 along its own axis, and each amplitude is 2e² times the amplitude to find the one state in the other. The zeros need no algebra. At θ = 0, Jₙ = −1 would mean J_z = −1, so e²(1 − cos θ) vanishes; at θ = π, n̂ = −ẑ and Jₙ = +1 would again mean J_z = −1, so e²(1 + cos θ) vanishes. A consistency check: d¹₁₁ = cos²(θ/2) is the product of two spin-½ overlaps, one per muon, as it must be for a state of two parallel spins.

Squares add: 1 + cos²θ, and what an asymmetry would mean

The surviving final states differ in helicity, so they are orthogonal and their amplitudes never interfere: rates add. Averaging over the four beam-helicity choices, two of which give zero, (1/4)Σ|M|² = (e⁴/4) × 2[(1 + cos θ)² + (1 − cos θ)²] = e⁴(1 + cos²θ), derived without a trace and equal to topic 9.06's result. Each square integrates to 16π/3 over the sphere, so the two channels carry equal shares of σ = 4πα²/(3s), but not of each hemisphere: (1 + cos θ)² puts 7/8 of its events forward of the beam's transverse plane, (1 − cos θ)² puts 7/8 behind it. With channel weights wₛₐₘₑ and wₒₚₚ the forward-backward asymmetry is A_FB = (3/4)(wₛₐₘₑ − wₒₚₚ)/(wₛₐₘₑ + wₒₚₚ), derived, and QED's vector coupling, which treats R and L alike, makes the weights equal: A_FB = 0 at tree level. The PETRA experiments measured about −0.1 for μ⁺μ⁻ near √s = 35 GeV, quoted. The Z couples unequally to R and L, an effect outside QED, and the helicity split shows exactly where it enters.

Muon mass: helicity flips that fill in the threshold

With m_μ kept, each spinor's two chiral halves carry √(E + p) and √(E − p), where p = βE. For opposite muon helicities the current collects (E + p) + (E − p) = 2E, so e²(1 ± cos θ) hold at any muon speed. Equal helicities, forbidden at m = 0, now survive through the cross terms √(E + p)√(E − p) = m_μ: for μ⁻_R μ⁺_R and for μ⁻_L μ⁺_L the current is −2m_μ(0, sin θ, 0, cos θ), longitudinal along n̂, a Jₙ = 0 state reached with d¹₁₀(θ) = −sin θ/√2. The amplitude is e²(m_μ/E) sin θ for each, derived, so the spin average becomes e⁴[1 + cos²θ + (m_μ/E)² sin²θ], the trace's mass term with (m_μ/E)² = 4m_μ²/s. At threshold m_μ/E = 1 and (1 + cos θ)² + (1 − cos θ)² + 2 sin²θ = 4 at every angle: isotropic, the S-wave, because the sum of |d¹₁λ′(θ)|² over λ′ = 1, 0, −1 is 1 at every angle. The flips' share of all pairs is (1 − β²)/(3 − β²): 2.2 × 10⁻⁴ at √s = 10 GeV, 26% at 0.25 GeV, and a third at threshold.

Change one variable at a time

Make the relationship visible.

Interactive model

Muon speed β0.60

μ⁻ angle θ60 °

Set β = 0.99 and drag θ to 0°, then 180°: the opposite-J readout dies at 0° and the same-J one at 180°, since J_z = +1 cannot become J = −1 along one line. Then drop β to 0: the flips fill in sin²θ and the solid sum flattens at 4, the isotropic threshold.

SAME J |M|/e²1.500

OPPOSITE J |M|/e²0.500

EACH FLIP |M|/e²0.693

FLIP SHARE OF ALL PAIRS0.242

Live interpretationSAME J |M|/e²: 1.500. OPPOSITE J |M|/e²: 0.500. EACH FLIP |M|/e²: 0.693. FLIP SHARE OF ALL PAIRS: 0.242

Catch the common trap

Explain before calculating.

Misconception check

The tempting reading of helicity conservation is that each lepton keeps its handedness through the process, so a right-handed e⁻ beam makes right-handed μ⁻. The chirality rule says something narrower. It acts at each vertex separately, forcing the e⁻ and e⁺ that meet at one vertex to have opposite helicities, and likewise the μ⁻ and μ⁺ that leave the other; it says nothing linking the e⁻ to the μ⁻, which do not share a fermion line. What links them is the angular momentum carried by the spin-one photon: J_z = +1 along the beam is re-expressed along the muon axis, where both Jₙ = +1 and Jₙ = −1 appear, with amplitudes e²(1 + cos θ) and e²(1 − cos θ). At θ = 90° a pure e⁻_R beam gives right- and left-handed μ⁻ equally often, and at θ = 120° the μ⁻ is left-handed 90% of the time. Near threshold even the muons' opposite pairing fails, as equal-helicity flips approach a third of all pairs.

At √s = 20 GeV, where both lepton masses are negligible, an e⁻_R e⁺_L pair (R and L naming helicity) annihilates into μ⁻μ⁺ with the μ⁻ at θ = 60° from the e⁻. What is the ratio of the rate into μ⁻_R μ⁺_L to the rate into μ⁻_L μ⁺_R?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyAt √s = 20.0 GeV, where both lepton masses can be neglected, an e⁻_L e⁺_R pair annihilates into μ⁻μ⁺ with the μ⁻ at θ = 120° from the e⁻. Find the magnitudes of the two surviving amplitudes from the d¹ functions and the probability that the μ⁻ is left-handed, check against e⁴(1 + cos²θ), and estimate the neglected muon-mass term.

Beam pair: the e⁻_L moving along +ẑ has spin −½ along ẑ; the e⁺_R moving along −ẑ has spin +½ along its own motion, so −½ along ẑ. The total is J_z = −1.
Muon pair: μ⁻_L μ⁺_R has Jₙ = −1 along the μ⁻ direction, the same as the beam pair; μ⁻_R μ⁺_L has Jₙ = +1, the opposite. So |M| = 2e²|d¹₋₁,₋₁(120°)| = e²(1 + cos 120°) = 0.500e² for μ⁻_L μ⁺_R, and 2e²|d¹₋₁,₁(120°)| = e²(1 − cos 120°) = 1.500e² for μ⁻_R μ⁺_L. With e² = 0.0917 these are 0.0459 and 0.138.
Rates go as squares, 0.250 against 2.250, so P(μ⁻ left-handed) = 0.250/2.500 = 0.100. At this backward angle the muon axis leans against the beam, and J_z = −1 along the beam looks mostly like Jₙ = +1 along the muons.
Check: the e⁻_L e⁺_R squares sum to 2.50e⁴, e⁻_R e⁺_L gives the same sum, and averaging over the four beam-helicity choices gives (2 × 2.50/4)e⁴ = 1.25e⁴, equal to e⁴(1 + cos²120°) = 1.25e⁴. Dimensionless, as a 2 → 2 amplitude squared must be.
Mass term: with E = 10.0 GeV, (m_μ/E)² sin²θ = (0.010566)² × 0.750 = 8.37 × 10⁻⁵, a fraction 6.7 × 10⁻⁵ of 1.25. Negligible.

Answer|M| = 0.500e² = 0.0459 for μ⁻_L μ⁺_R and 1.500e² = 0.138 for μ⁻_R μ⁺_L; the μ⁻ is left-handed with probability 0.100 although the e⁻ was; the mass term is below 10⁻⁴ of the rate.

MediumAn electron beam with longitudinal polarisation P = +0.80, that is 90% right-handed, meets unpolarised positrons at √s = 10.0 GeV. For e⁺e⁻ → μ⁺μ⁻, find the μ⁻ polarisation (N_R − N_L)/(N_R + N_L) in the forward hemisphere and in the backward one, the forward-backward asymmetry, and the cross-section for forward right-handed μ⁻, taking σ = 4πα²/(3s) from topic 9.06.

Each e⁻ helicity finds a positron of the opposite helicity half the time, so e⁻_R e⁺_L and e⁻_L e⁺_R occur in the ratio 0.90 : 0.10, and the total rate does not depend on P.
Hemisphere integrals over c = cos θ: ∫₀¹(1 + c)² dc = 7/3 and ∫₀¹(1 − c)² dc = 1/3; in the backward hemisphere the two values swap, so each channel holds 8/3 in all.
Forward: e⁻_R e⁺_L gives μ⁻_R (same J) with weight 0.90 × 7/3 and μ⁻_L with 0.90 × 1/3; e⁻_L e⁺_R gives μ⁻_L (same J, now −1) with 0.10 × 7/3 and μ⁻_R with 0.10 × 1/3. So N_R = 6.4/3 and N_L = 1.6/3, and the forward μ⁻ polarisation is 4.8/8.0 = +0.60.
Backward the roles swap: N_R = 1.6/3 and N_L = 6.4/3, so the polarisation is −0.60; in general the forward value is (3/4)P. Both hemispheres hold 8/3 of weight, so A_FB = 0: polarising the beam moves the muon helicities but not the angular distribution, because QED weights the two channels equally.
Rate: 4πα²/(3s) = 2.2306 × 10⁻⁴/(100 GeV²) = 2.2306 × 10⁻⁶ GeV⁻², and times 0.3894 mb GeV² that is 0.8686 nb; the muon-mass factor β(3 − β²)/2 differs from 1 by under 10⁻⁷. Forward right-handed μ⁻ are (6.4/3)/(16/3) = 0.400 of all events, so σ = 0.347 nb.

Answerμ⁻ polarisation +0.60 forward and −0.60 backward, A_FB = 0, and σ(forward, μ⁻_R) = 0.347 nb out of a total 0.869 nb.

HardAt √s = 3.80 GeV, just above the τ⁺τ⁻ threshold, take m_τ = 1.77686 GeV and massless electrons. Find (a) the helicity amplitudes from an e⁻_R e⁺_L pair at θ = 90°, (b) the unpolarised angular distribution in the form A + B cos²θ, (c) the fraction of τ pairs produced with equal helicities, and (d) σ in nb, by integrating the helicity sum and using Unit 5's dσ/dΩ = |M|²|p_f|/(64π²s|pᵢ|).

Kinematics: E = 1.900 GeV, m_τ/E = 0.93519 and (m_τ/E)² = 0.87458, so β² = 0.12542 and β = 0.35415.
(a) At θ = 90° the opposite-helicity pairs τ⁻_R τ⁺_L and τ⁻_L τ⁺_R each have |M| = e²(1 ± 0) = e², and the equal-helicity pairs τ⁻_R τ⁺_R and τ⁻_L τ⁺_L each have e²(m_τ/E) sin 90° = 0.935e². Near threshold the flips are almost as large as the rest.
(b) With c = cos θ, (1/4)Σ|M|² = (e⁴/2)[(1 + c)² + (1 − c)² + 2(m_τ/E)²(1 − c²)] = e⁴[1 + (m_τ/E)² + (1 − (m_τ/E)²)c²] = e⁴(1.8746 + 0.1254 cos²θ). The forward rate exceeds the sideways one by only 2/1.8746 = 1.067.
(c) Over the sphere ∫(1 + c²) dΩ = 16π/3 and ∫ sin²θ dΩ = 8π/3, so the equal-helicity share is (m_τ/E)²(8π/3)/[16π/3 + (m_τ/E)²(8π/3)] = (1 − β²)/(3 − β²) = 0.304.
(d) With |p_f|/|pᵢ| = β and e⁴ = 16π²α², σ = βe⁴(16π/3)[1 + (m_τ/E)²/2]/(64π²s) = [4πα²/(3s)] β(3 − β²)/2. Here 4πα²/(3s) = 2.2306 × 10⁻⁴/(14.44 GeV²) = 1.5447 × 10⁻⁵ GeV⁻² = 6.015 nb, and β(3 − β²)/2 = 0.5090, so σ = 3.062 nb.
Checks: GeV⁻² becomes nb through 0.3894 mb GeV²; as β → 1 the factor tends to 1 and the massless result returns; as β → 0, σ falls as (3/2)β times 4πα²/(3s), the S-wave threshold. Left out: the Coulomb attraction of the slow τ pair, a higher-order rise of about πα/(2β) = 3% quoted from topic 9.06, and radiation from the beams.

Answer(a) e², e² and 0.935e² twice; (b) dσ/dΩ ∝ 1.875 + 0.125 cos²θ; (c) 30.4% of τ pairs have equal helicities; (d) σ = 3.06 nb at tree level.

Chirality leaves the beam pair only Jz = ±1