University Quantum Field Theory I · QED at Tree Level: the Lagrangian, Feynman Rules and Electron-Muon Processes · 9.8

Crossing symmetry: from e⁺e⁻ → μ⁺μ⁻ to e⁻μ⁻ → e⁻μ⁻

Topic 9.05 spent a page of traces on e⁺e⁻ → μ⁺μ⁻. Here that page becomes electron-muon scattering by reversing two momenta, and we pin down the one thing fermions add to topic 4.07's crossing rule: a minus sign for each fermion moved across, carried by the spin sum rather than by the amplitude.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Crossing is LSZ read in both directions: the field that absorbs an incoming particle also emits the outgoing antiparticle, so one correlation function holds every process made by moving a leg across the reaction with its four-momentum reversed, and topic 4.07 turned that into one function ℳ(s, t, u) for scalars. Fermions add a twist. The amplitude does not continue cleanly, because a spinor such as u(p) carries √(E + m) and is not a polynomial in p, but the spin-summed |ℳ|² does: Casimir's trick builds it from the spin sums p̸ + m and p̸ − m, and reversing the momentum in an antifermion's p̸ − m gives −(p̸ + m), minus the fermion's own sum.

So each crossed fermion multiplies Σ|ℳ|² by −1. Crossing the e⁺ and μ⁺ of e⁺e⁻ → μ⁺μ⁻ moves two fermions, the signs cancel, and the s-channel photon becomes a t-channel one: a bounded 1 + cos²θ distribution turns into a forward pole, and a direct trace agrees term by term. Crossing does not hand over a number, though.

The two processes read one rational function in regions that never touch, joined only through an unphysical gap. At tree level the continuation between them is trivial because ⟨|ℳ|²⟩ is rational; once loops add branch cuts it is a theorem quoted rather than proved here, and identical fermions or external photons bring extra diagrams and new averaging factors that must be recounted in Unit 10.

Simple definition: Crossing for fermions: the spin-summed |ℳ|² of one process gives that of another, in which an incoming antifermion becomes an outgoing fermion or the reverse, by sending that leg's momentum p to −p and multiplying by −1 for every fermion so moved.
Example: Reversing the e⁺ and μ⁺ momenta in 2e⁴(t² + u²)/s² costs (−1)² = 1 and gives 2e⁴(s² + u²)/t² for e⁻μ⁻ → e⁻μ⁻: at θ = 90°, where annihilation has ⟨|ℳ|²⟩ = e⁴, scattering has 10e⁴.

A crossed spin sumΣₛ v(k, s)v̄(k, s) = k̸ − m, which at k = −p is −(p̸ + m) = −Σₛ u(p, s)ū(p, s)

Spin sums are polynomials in the momentum, so they continue to −p where spinors cannot. Each crossed fermion multiplies Σ|ℳ|² by −1; photon and scalar legs carry no such sign.

p, k four-momenta and m the fermion mass, in GeV; spinors normalised to ūu = 2m (UAQM-04.06); sums from UAQM-04.08

The crossing rule for spin-summed |ℳ|²Σ|ℳ′|² = (−1)ⁿ Σ|ℳ|², with each crossed leg's momentum p replaced by −p

Topic 4.07's scalar rule with one sign per fermion. The spin average belongs to the process, so it is recounted whenever crossing changes which particles arrive.

n is the number of fermions and antifermions moved across; ℳ and ℳ′ dimensionless for 2 → 2; derived for tree amplitudes

The dictionary for this crossingp = p₁, p′ = −p₃, k = p₄, k′ = −p₂, so (s, t, u) of e⁺e⁻ → μ⁺μ⁻ = (t, u, s) of e⁻μ⁻ → e⁻μ⁻

The photon's p + p′ becomes p₁ − p₃, so 1/s² turns into 1/t². The annihilation result is symmetric under t ↔ u, so in effect s and t trade places.

9.05's e⁻(p)e⁺(p′) → μ⁻(k)μ⁺(k′) against e⁻(p₁)μ⁻(p₂) → e⁻(p₃)μ⁻(p₄); invariants in GeV²; n = 2

e⁻μ⁻ → e⁻μ⁻ in dot products⟨|ℳ|²⟩ = (8e⁴/t²)[(p₁⋅p₂)(p₃⋅p₄) + (p₁⋅p₄)(p₂⋅p₃) − m²(p₁⋅p₃)]

Derived twice: by crossing 9.05's bracket, whose m²(p⋅p′) turns into −m²(p₁⋅p₃), and by a direct trace. Topic 9.09 takes it to the muon's rest frame.

⟨|ℳ|²⟩ = (1/4)Σ|ℳ|², dimensionless; e² = 4πα; m = 0.10566 GeV, the muon mass; mₑ neglected, safe while |t| ≫ mₑ²

The same function in s, t and u⟨|ℳ|²⟩ = (2e⁴/t²)[(s − m²)² + (u − m²)² + 2m²t], with s + t + u = 2m²

One rational function read in two regions. Its mass term, +2m²s in annihilation, is 2m²t ≤ 0 here: it flipped through p⋅p′ → −p₁⋅p₃, not through a fermion sign.

e⁺e⁻ → μ⁺μ⁻ is (2e⁴/s²)[(t − m²)² + (u − m²)² + 2m²s]; s, t, u and m² in GeV²; derived

The high-energy limit⟨|ℳ|²⟩ = 2e⁴(s² + u²)/t² = 2e⁴[4 + (1 + cos θ)²]/(1 − cos θ)²

Against annihilation's bounded 1 + cos²θ, the t-channel photon gives a forward pole: 10e⁴ at 90°, rising as 1/sin⁴(θ/2) toward θ = 0.

m² ≪ s and |t|; θ the centre-of-mass angle between incoming and outgoing electron; dσ/dΩ = ⟨|ℳ|²⟩/(64π²s)

Why a crossed fermion brings a minus sign

Work in ħ = c = 1 with metric (+, −, −, −). Topic 4.04's LSZ formula builds every S-matrix element from one correlation function, and the field that absorbs an incoming particle also creates the outgoing antiparticle. For the Dirac field, an incoming e⁻ enters as u(p)e^(−ip⋅x) and an outgoing e⁺ of momentum k as v(k)e^(+ik⋅x): one function, evaluated at p = −k. At the level of the amplitude this gives no usable formula, because u(p) carries factors such as √(E + m) that are not polynomials in p and have no clean continuation to negative energy. The spin-summed squared amplitude does give one. Casimir's trick (topic 7.06) writes it as traces of the spin sums Σu ū = p̸ + m and Σv v̄ = k̸ − m (UAQM-04.08), polynomials that can be evaluated at any momentum. Put k = −p in the antifermion's sum and k̸ − m = −(p̸ + m), minus the fermion's. Every crossed fermion therefore multiplies Σ|ℳ|² by −1, derived. A photon's sum −g_μν does not depend on k and a scalar has none, so bosons cross without a sign.

Two crossed legs, two signs, and a spacelike photon

Topic 9.05 wrote e⁻(p)e⁺(p′) → μ⁻(k)μ⁺(k′) as ℳ = (e²/s) v̄(p′)γ^μ u(p) ū(k)γ_μ v(k′) and averaged it into (e⁴/4s²) Tr[(p̸′ − mₑ)γ^μ(p̸ + mₑ)γ^ν] Tr[(k̸ + m)γ_μ(k̸′ − m)γ_ν]. To reach e⁻(p₁)μ⁻(p₂) → e⁻(p₃)μ⁻(p₄), keep p = p₁ and k = p₄, turn the incoming e⁺ into the outgoing e⁻ with p′ = −p₃, and the outgoing μ⁺ into the incoming μ⁻ with k′ = −p₂. The electron trace's p̸′ − mₑ becomes −(p̸₃ + mₑ) and the muon trace's k̸′ − m becomes −(p̸₂ + m): two signs, product +1. The photon's momentum p + p′ becomes p₁ − p₃, so s = (p + p′)² in the propagator becomes t = (p₁ − p₃)². That is the physics of this crossing. In annihilation the photon is timelike, q² = s ≥ 4m², and stays far from its pole at every angle, so the angular distribution is bounded; in scattering it is spacelike, q² = t ≤ 0, and it reaches its pole q² = 0 as the electron goes forward.

The substitution, term by term

With mₑ neglected, 9.05's average is (8e⁴/s²)[(p⋅k)(p′⋅k′) + (p⋅k′)(p′⋅k) + m²(p⋅p′)], m the muon mass. Substitute p⋅k = p₁⋅p₄, p′⋅k′ = p₂⋅p₃, p⋅k′ = −p₁⋅p₂, p′⋅k = −p₃⋅p₄ and p⋅p′ = −p₁⋅p₃. The two minus signs in (p⋅k′)(p′⋅k) cancel, p′⋅k′ = (−p₃)⋅(−p₂) carries none, and the result is (8e⁴/t²)[(p₁⋅p₂)(p₃⋅p₄) + (p₁⋅p₄)(p₂⋅p₃) − m²(p₁⋅p₃)], derived. Notice what flipped the mass term: one dot product changing sign, not a fermion sign, since the two trace signs were spent cancelling each other. In invariants, with s + t + u = 2m², p₁⋅p₂ = p₃⋅p₄ = (s − m²)/2, p₁⋅p₄ = p₂⋅p₃ = (m² − u)/2 and p₁⋅p₃ = −t/2, so ⟨|ℳ|²⟩ = (2e⁴/t²)[(s − m²)² + (u − m²)² + 2m²t]. Dimensions check: e⁴ is a pure number and the bracket over t² is GeV⁴ over GeV⁴, as a 2 → 2 amplitude must be. For s ≫ m² it becomes 2e⁴(s² + u²)/t², the annihilation's 2e⁴(t² + u²)/s² with s and t exchanged.

A direct trace confirms it

Compute e⁻μ⁻ → e⁻μ⁻ from scratch to be sure no sign went astray. Topic 9.02's rules give one t-channel diagram, ℳ = (e²/t) ū(p₃)γ^μ u(p₁) ū(p₄)γ_μ u(p₂), up to an overall phase. Averaging over both incoming spins and summing over both outgoing ones, ⟨|ℳ|²⟩ = (e⁴/4t²) L^μν W_μν with L^μν = Tr[p̸₃γ^μ p̸₁γ^ν] = 4[p₃^μ p₁^ν + p₃^ν p₁^μ − g^μν(p₁⋅p₃)] and W^μν = Tr[(p̸₄ + m)γ^μ(p̸₂ + m)γ^ν] = 4[p₄^μ p₂^ν + p₄^ν p₂^μ − g^μν(p₂⋅p₄ − m²)], every spin sum now p̸ + m. In the contraction the momentum terms give 2(p₁⋅p₂)(p₃⋅p₄) + 2(p₁⋅p₄)(p₂⋅p₃) and the terms carrying g give 2(p₁⋅p₃)(p₂⋅p₄ − m²) − 2(p₁⋅p₃)(p₂⋅p₄) = −2m²(p₁⋅p₃). So L^μν W_μν = 32[(p₁⋅p₂)(p₃⋅p₄) + (p₁⋅p₄)(p₂⋅p₃) − m²(p₁⋅p₃)] and ⟨|ℳ|²⟩ = 8e⁴/t² times that bracket, the crossed result exactly. The 1/4 is common to both because each process starts with two spin-½ particles.

One function, two regions, a pole between them

Call F(s, t, u) = (2e⁴/s²)[(t − m²)² + (u − m²)² + 2m²s] the annihilation average, physical for s ≥ 4m² with t and u negative. e⁻μ⁻ → e⁻μ⁻ reads the same F at (t, s, u), its own invariants, so F's first argument, the photon's q², is t ≤ 0 there. The regions never touch, and between them sits the double pole at q² = 0 that annihilation never approaches and scattering reaches only as θ → 0. That is why the shapes differ so much: masses neglected, at θ = 90° annihilation has ⟨|ℳ|²⟩ = e⁴ and scattering 10e⁴, and at 30° it is 1.75e⁴ against 834e⁴. Because F is symmetric in its last two arguments, its third region, e⁻μ⁺ → e⁻μ⁺, has the same ⟨|ℳ|²⟩ as e⁻μ⁻ → e⁻μ⁻: at lowest order a μ⁺ and a μ⁻ scatter electrons identically, the sign of the muon's charge entering only at order α³, through the interference of one- and two-photon exchange together with that of photons radiated by the electron and by the muon, whose infrared divergences cancel between them, quoted. So topic 4.07's rule holds for these tree amplitudes, fermions included, once each crossed fermion pays its −1.

Where crossing stops being a substitution

Three limits. First, the crossing sign is not the relative minus between diagrams. Bhabha and Møller scattering have two diagrams whose relative sign (topic 7.05) survives as interference, so each diagram must be crossed with its sign attached, which is Unit 10's work; Compton scattering crossed into pair annihilation moves a single fermion and needs its −1 written in. Second, the spin average belongs to the process, not to the function: Yukawa theory's ψφ → ψφ averages by 1/2 where its crossed ψψ̄ → φφ averages by 1/4. Third, loops. A one-loop amplitude has branch cuts: vacuum polarisation (Unit 13) is real at spacelike q² but turns complex once q² passes 4mₑ², so the continuation must go around the cut with the Feynman iε, and that it always succeeds is stated, not proved, in this course. The tree-level shape derived here is also a measuring instrument: CERN's MUonE experiment plans to scatter a muon beam off atomic electrons and, with radiative corrections added, read the hadronic part of α's running at spacelike q² from it, quoted.

Change one variable at a time

Make the relationship visible.

Interactive model

Centre-of-mass angle θ60 °

Collision energy √s5.0 GeV

Lower θ toward 30°: the e⁻μ⁻ dot climbs the photon pole at q² = 0 while the e⁺e⁻ dot stays between 1 and 2, since a timelike photon never nears its pole. Then drag √s: the dots slide and the gap stretches, but their heights hold, because without masses ⟨|ℳ|²⟩ depends on θ alone.

ANNIHILATION ⟨|ℳ|²⟩1.25 e⁴

SCATTERING ⟨|ℳ|²⟩50.0 e⁴

ANNIHILATION, PER SOLID ANGLE0.259 nb sr⁻¹

SCATTERING, PER SOLID ANGLE10.37 nb sr⁻¹

Live interpretationANNIHILATION ⟨|ℳ|²⟩: 1.25 e⁴. SCATTERING ⟨|ℳ|²⟩: 50.0 e⁴. ANNIHILATION, PER SOLID ANGLE: 0.259 nb sr⁻¹. SCATTERING, PER SOLID ANGLE: 10.37 nb sr⁻¹

Catch the common trap

Explain before calculating.

Misconception check

Many students file the crossing sign with the Fermi-statistics signs of Unit 7, as a −1 multiplying the amplitude, and conclude that it can never matter in a rate. It is not a phase of ℳ. It appears in the spin-summed |ℳ|², where an antifermion's spin sum k̸ − m continued to k = −p becomes −(p̸ + m), so it multiplies the probability itself. In crossing e⁺e⁻ → μ⁺μ⁻ into e⁻μ⁻ → e⁻μ⁻ two fermions move and the signs cancel, which is why dropping them goes unnoticed here. Cross a single fermion, as Compton scattering crossed into pair annihilation does, and ignoring the sign makes ⟨|ℳ|²⟩ negative across the whole physical region. The genuine amplitude-level sign, the relative minus between t- and u-channel diagrams of identical fermions, is a different object: it lives inside ℳ, survives as interference, and is carried along with the diagrams when such a process is crossed.

Masses neglected, e⁺e⁻ → μ⁺μ⁻ has ⟨|ℳ|²⟩ = 2e⁴(t² + u²)/s², which is 1.25e⁴ at θ = 60° in its centre-of-mass frame. Cross the e⁺ and μ⁺ to obtain e⁻μ⁻ → e⁻μ⁻ and evaluate its ⟨|ℳ|²⟩ at the same √s = 10 GeV, with the electron scattered through θ = 60° in the centre-of-mass frame.

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTopic 9.06 gives e⁺e⁻ → μ⁺μ⁻ at high energy as ⟨|ℳ|²⟩ = 2e⁴(t² + u²)/s². Cross it into e⁻μ⁻ → e⁻μ⁻ and find dσ/dΩ in the centre-of-mass frame at √s = 10.0 GeV and θ = 90°, masses neglected, in nb sr⁻¹. Compare it with annihilation at the same √s and θ.

Crossing the e⁺ and μ⁺ moves two fermions, so the signs cancel and s and t exchange: ⟨|ℳ|²⟩ = 2e⁴(s² + u²)/t² for e⁻μ⁻ → e⁻μ⁻.
At θ = 90°, t = −(s/2)(1 − cos θ) = −s/2 and u = −(s/2)(1 + cos θ) = −s/2. Scattering: 2e⁴(1 + 1/4)/(1/4) = 10e⁴. Annihilation: 2e⁴(1/4 + 1/4) = e⁴.
The scattering is elastic, so in the centre-of-mass frame the final momentum equals the initial one and dσ/dΩ = ⟨|ℳ|²⟩/(64π²s). With e² = 4πα this is 10 × 16π²α²/(64π²s) = 2.5α²/s.
With α = 1/137.036 and s = 100 GeV²: 2.5 × 5.325 × 10⁻⁵/100 = 1.331 × 10⁻⁶ GeV⁻² per steradian, and × 0.3894 mb GeV² gives 5.18 × 10⁻⁷ mb = 0.518 nb sr⁻¹. Annihilation gives α²/(4s), ten times less: 0.0518 nb sr⁻¹.
Checks: GeV⁻² is an area, as a cross-section per steradian must be; the ratio 10 does not depend on s, since without masses ⟨|ℳ|²⟩ depends on θ alone; and as θ → 0, t → 0 and the scattering result grows as 1/sin⁴(θ/2), Rutherford's forward rise.

Answerdσ/dΩ = 0.518 nb sr⁻¹ for e⁻μ⁻ → e⁻μ⁻, ten times the 0.0518 nb sr⁻¹ of e⁺e⁻ → μ⁺μ⁻ at the same √s = 10.0 GeV and θ = 90°.

MediumElectron-muon scattering's muon line gives W^μν = Σ[ū(p₄)γ^μ u(p₂)][ū(p₄)γ^ν u(p₂)]* = 4[p₄^μ p₂^ν + p₄^ν p₂^μ − g^μν(p₂⋅p₄ − m²)]. Cross only the incoming μ⁻ into an outgoing μ⁺, p₂ = −k′ and p₄ = k, to get the pair tensor T^μν of a virtual photon at rest with √s = 1.00 GeV, the μ⁻ along +z. Evaluate T⁰⁰, T¹¹ and T³³ with and without the crossing sign, and decide which is right.

Kinematics with m = 0.10566 GeV: E = √s/2 = 0.500 GeV, |k| = √(E² − m²) = 0.4887 GeV, k = (E, 0, 0, |k|), k′ = (E, 0, 0, −|k|), so k⋅k′ = E² + |k|² = 0.4888 GeV² = s/2 − m².
Substituting: W^μν becomes 4[−k^μ k′^ν − k^ν k′^μ − g^μν(−k⋅k′ − m²)] = −4B^μν, with B^μν = k^μ k′^ν + k^ν k′^μ − g^μν(k⋅k′ + m²) and k⋅k′ + m² = s/2 = 0.500 GeV².
Components, with g⁰⁰ = 1 and g¹¹ = g³³ = −1: B⁰⁰ = 2E² − 0.500 = 0; B¹¹ = 0.500 GeV²; B³³ = −2|k|² + 0.500 = 0.0223 GeV², which is 2m².
Without a sign, T¹¹ = −4B¹¹ = −2.00 GeV². But T¹¹ = Σ|ū(k)γ¹v(k′)|² is a sum of squared moduli and cannot be negative. One crossed fermion, one −1: T^μν = +4B^μν, so T⁰⁰ = 0, T¹¹ = T²² = 2.00 GeV², T³³ = 0.0893 GeV².
Checks: the direct trace Tr[(k̸ + m)γ^μ(k̸′ − m)γ^ν] gives 4B^μν; current conservation q_μT^μν = 0 with q = (√s, 0, 0, 0) demands T⁰⁰ = 0; T³³/T¹¹ = 4m²/s = 0.0447 vanishes at high energy, the helicity suppression of topic 9.07; and −g_μνT^μν = 4(s + 2m²) = 4.09 GeV².

AnswerWith the sign: T⁰⁰ = 0, T¹¹ = T²² = 2.00 GeV², T³³ = 0.0893 GeV². Without it T¹¹ = −2.00 GeV², a negative sum of squares, so one crossed fermion demands one −1.

HardA 150 GeV μ⁻ beam strikes electrons at rest. Find √s and the largest momentum transfer −t. Then, keeping m = 0.10566 GeV but neglecting mₑ in ⟨|ℳ|²⟩, cross the massive e⁺e⁻ → μ⁺μ⁻ result into e⁻μ⁻ → e⁻μ⁻ and evaluate ⟨|ℳ|²⟩/e⁴ and dσ/dΩ in the centre-of-mass frame at θ = 90°. Compare with the massless formula and locate the crossed point in the annihilation's variables.

s = m² + mₑ² + 2mₑE = (0.011164 + 0.00000026 + 2 × 0.000511 × 150) GeV² = 0.16446 GeV², √s = 0.4055 GeV. The centre-of-mass momentum √[(s − (m + mₑ)²)(s − (m − mₑ)²)]/(2√s) = 0.1890 GeV, so −tₘₐₓ = 4|p|² = 0.1429 GeV², reached when the electron reverses direction in the centre-of-mass frame.
Crossing (2e⁴/s²)[(t − m²)² + (u − m²)² + 2m²s] with s and t exchanged gives ⟨|ℳ|²⟩ = (2e⁴/t²)[(s − m²)² + (u − m²)² + 2m²t]. The terms −mₑ²(p₂⋅p₄) + 2mₑ²m² that mₑ would add to the dot-product bracket total −6 × 10⁻⁹ GeV⁴ here, against 7.2 × 10⁻³ GeV⁴.
At θ = 90°, t = −2|p|² = −0.07145 GeV² and u = 2m² − s − t = −0.07069 GeV². Then (s − m²)² = 0.02350 GeV⁴, (u − m²)² = 0.006700 GeV⁴ and 2m²t = −0.001595 GeV⁴, summing to 0.02861 GeV⁴; over t² = 0.005105 GeV⁴ and doubled, ⟨|ℳ|²⟩ = 11.21e⁴.
dσ/dΩ = ⟨|ℳ|²⟩/(64π²s) = 11.21α²/(4s) = 9.07 × 10⁻⁴ GeV⁻², which × 0.3894 mb GeV² is 3.53 × 10⁻⁴ mb = 353 nb sr⁻¹. The massless 2e⁴(s² + u²)/t² at the same s and θ gives 10e⁴, so the mass raises ⟨|ℳ|²⟩ by 12%.
Read as an annihilation, this point puts the photon at q² = t = −0.0714 GeV², spacelike, while e⁺e⁻ → μ⁺μ⁻ needs q² = s ≥ 4m² = 0.0447 GeV². The same function is read where no annihilation can reach, which is why crossing transfers a formula, never a number.

Answer√s = 0.4055 GeV and −tₘₐₓ = 0.143 GeV²; at 90°, ⟨|ℳ|²⟩ = 11.2e⁴ and dσ/dΩ = 353 nb sr⁻¹, 12% above the massless 10e⁴; its photon, q² = −0.071 GeV², lies outside annihilation's q² ≥ 0.0447 GeV².