University Quantum Field Theory I · Cross-Sections, Decay Rates and Unitarity · 5.4

2 → 2 cross-sections and φ⁴ in microbarns

This is the formula you will reach for more than any other in the course: divide |M|² by 64π² s, correct for the momenta out and in, and an amplitude becomes an area. We derive it once, recast it in t so it holds in any frame along the beam, and carry φ⁴ through to microbarns and events per second.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A 2 → 2 cross-section is the dynamics, |M|², multiplied by a count of final states and divided by the flux that brings the pair together, and in the centre-of-mass frame every kinematic factor collapses onto three numbers: s, the incoming momentum |pᵢ| and the outgoing |p_f|. That is why dσ/dΩ = |M|²|p_f|/(64π² s|pᵢ|) separates so cleanly what a theory predicts from what kinematics imposes, and why trading the scattering angle for the invariant t gives a dσ/dt that every frame boosted along the beam agrees on. φ⁴ theory is the cleanest test: its tree amplitude is the single number −λ, so the angular distribution is flat, all of the physics is λ², and the only subtlety is the ½ that stops two identical φ being counted twice. The result, σ = λ²/(32π s), falls as 1/s because a dimensionless coupling leaves s as the only scale, and at λ = 1 and s = 1 GeV² it is 3.87 μb.

The model stops in three places. It is a tree amplitude: M = −λ is real, yet unitarity demands an imaginary part of order λ², which the one-loop bubble supplies in topic 5.08, while its real part, the rest of the order-λ² correction to M, waits for Unit 6. It assumes LSZ's stable, well-separated external particles.

And it has two bodies out; a third needs topic 5.05's Dalitz variables.

Simple definition: A 2 → 2 differential cross-section is the rate of scatters into a solid angle per target particle per unit incident flux; in the centre-of-mass frame it equals |M|² times |p_f|/|pᵢ|, divided by 64π² s.
Example: For φ⁴ with λ = 1 at s = 1 GeV², dσ/dΩ = 1/(64π²) GeV⁻² sr⁻¹ = 0.616 μb sr⁻¹ in every direction; half its integral over 4π is σ = 3.87 μb, about 3.9 × 10³ events per second at L = 10³³ cm⁻² s⁻¹.

Centre-of-mass differential cross-sectiondσ/dΩ = |M|² |p_f| / (64π² s |pᵢ|)

Derived: flux 4|pᵢ|√s divided into two-body phase space |p_f|dΩ/(16π²√s). All dynamics sits in |M|²; the prefactor is kinematics fixed by s and the masses.

|pᵢ|, |p_f| the CM momenta in and out (GeV), s = E_cm² (GeV²), dΩ in sr; dσ/dΩ in GeV⁻² sr⁻¹

Lorentz-invariant form in tdσ/dt = |M|² / (64π s |pᵢ|²)

Derived from dt = 2|pᵢ||p_f| d cos θ. It holds unchanged on a fixed target; for massless beams it is |M|²/(16π s²).

t = (p₁ − p₃)² in GeV²; 4s|pᵢ|² = [s − (m₁ + m₂)²][s − (m₁ − m₂)²], so dσ/dt needs only s and t

Total, for an angle-independent |M|σ = |M|² |p_f| / (16π S s |pᵢ|)

4π times dσ/dΩ, then 1/S because the configurations at n̂ and −n̂ are one state. It covers φ⁴ and any contact amplitude.

S = 2 when the two final particles are identical, 1 otherwise; σ in GeV⁻²

φ⁴ at tree levelσ(φφ → φφ) = λ²/(32π s)

Derived from M = −λ (topic 4.05), isotropic and a pure s wave. λ = 1 at s = 1 GeV² gives 1/(32π) GeV⁻² = 3.87 μb.

λ dimensionless, s ≥ 4m² in GeV²; exact at tree level at every energy, since |p_f| = |pᵢ| for equal masses

From GeV⁻² to barnsσ/mb = 0.3894 × σ/GeV⁻²

Restoring ħ and c turns an inverse energy squared into an area a measurement can be set against.

(ħc)² = 0.3894 GeV² mb; 1 b = 10⁻²⁴ cm² = 100 fm², so 1 GeV⁻² = 389.4 μb

Event rate at a colliderR = L σ

At L = 10³³ cm⁻² s⁻¹ every nanobarn is one event per second, so φ⁴'s 3.87 μb at s = 1 GeV² would give about 3.9 × 10³ s⁻¹.

R in s⁻¹, L the instantaneous luminosity in cm⁻² s⁻¹, σ in cm²; 1 nb = 10⁻³³ cm²

Flux over phase space, in the centre-of-mass frame

With ħ = c = 1 and the metric (+, −, −, −) throughout, topic 5.01 gave dσ = |M|² dΠ₂/F with flux factor F = 4E₁E₂|v₁ − v₂|, and topic 5.02 reduced two-body phase space in the centre-of-mass frame to dΠ₂ = |p_f| dΩ/(16π²√s); both are quoted. In that frame the beams carry ±pᵢ, so |v₁ − v₂| = |pᵢ|/E₁ + |pᵢ|/E₂ and E₁E₂|v₁ − v₂| = |pᵢ|(E₁ + E₂) = |pᵢ|√s, giving F = 4|pᵢ|√s. Dividing, dσ/dΩ = |M|²|p_f|/(64π² s|pᵢ|), derived. Dimensions check it: an amplitude with n external legs has mass dimension 4 − n, so a 2 → 2 |M|² is a pure number and dσ/dΩ is GeV⁻² per steradian, an area. The ratio |p_f|/|pᵢ| is kinematics alone: exactly 1 for elastic scattering of equal masses, small just above the threshold of a heavier final pair, and large when slow particles turn into lighter ones.

Trading the angle for t

A scattering angle belongs to one frame; t = (p₁ − p₃)² belongs to none. In the centre-of-mass frame t = m₁² + m₃² − 2E₁E₃ + 2|pᵢ||p_f| cos θ, and at fixed s the energies are fixed, so dt = 2|pᵢ||p_f| d cos θ. Nothing depends on the azimuth, so dΩ = 2π d cos θ and dσ/dt = [2π/(2|pᵢ||p_f|)] dσ/dΩ = |M|²/(64π s|pᵢ|²), derived; |p_f| has cancelled. Because 4s|pᵢ|² = [s − (m₁ + m₂)²][s − (m₁ − m₂)²] is built from s and the masses alone, dσ/dt is a function of invariants, and a fixed-target experiment can use it in its own laboratory frame without converting angles. For massless beams |pᵢ|² = s/4 and dσ/dt = |M|²/(16π s²). The physical region is a segment of length 4|pᵢ||p_f| in t; for elastic equal masses it runs from −4|pᵢ|² (backward) to 0 (forward).

φ⁴: one number for an amplitude, and a factor ½

Topic 4.05 found the tree amplitude for φφ → φφ with Lᵢₙₜ = −(λ/4!)φ⁴: the 4! ways of attaching four external lines to one vertex cancel the 1/4!, so M = −λ, quoted. It depends on neither s nor t, so dσ/dΩ = λ²/(64π² s) is isotropic, a pure s wave, since a constant has only the P₀ term in a partial-wave expansion; and for equal masses |p_f| = |pᵢ| at every energy. Integrating over the whole sphere would now count each final state twice, because two identical φ along n̂ and −n̂ form one state whichever is called particle 3. So σ = ½ × 4π × λ²/(64π² s) = λ²/(32π s), derived. The ½ belongs to the total, not to a counter: every event puts one φ in each hemisphere, so a small detector covering ΔΩ counts L(dσ/dΩ)ΔΩ particles per second with the unhalved dσ/dΩ, while the number of events is half the number of particles seen over 4π. The initial pair needs no factor: it is one prepared state.

From GeV⁻² to microbarns and events per second

At λ = 1 and s = 1 GeV², σ = 1/(32π) GeV⁻² = 9.95 × 10⁻³ GeV⁻². Restoring ħ and c, an inverse energy squared becomes an area through (ħc)² = 0.3894 GeV² mb, so σ = 3.87 × 10⁻³ mb = 3.87 μb, computed; with 1 fm² = 10 mb that is 3.9 × 10⁻⁴ fm². To turn an area into counts, multiply by the luminosity, R = Lσ. Since 1 nb = 10⁻³³ cm², a machine delivering L = 10³³ cm⁻² s⁻¹ records one event per second per nanobarn, and 3.87 μb would give about 3.9 × 10³ events per second. φ⁴ describes no particle we have measured, so for scale take QED: e⁺e⁻ → μ⁺μ⁻ has σ = 4πα²/(3s) = 86.9 nb/(s/GeV²), quoted here and derived in topic 9.06, a 1/s law that e⁺e⁻ colliders have confirmed across a wide range of energies. At s = 1 GeV² it is 45 times smaller than φ⁴ at λ = 1: α² = 5.3 × 10⁻⁵ against λ² = 1, partly offset by QED's larger numerical factor, 4π/3 against 1/(32π).

Why 1/s: what a dimensionless coupling allows

A cross-section has mass dimension −2 in every theory. In φ⁴ theory [φ] = 1 and [λ] = 0, so once s ≫ m² nothing but s is left to carry the dimension: σ must be λ² c/s with c a pure number, and the tree calculation fixes c = 1/(32π). The fall is set by dimensions before any dynamics, and the same 1/s runs through QED, where α is also dimensionless. A coupling with a dimension changes the power. Fermi's constant has [G_F] = −2, so σ ∝ G_F² s grows with energy, as the quoted inverse-muon-decay result G_F² s/π does, and cannot continue for ever without breaking unitarity. The scalar Yukawa coupling has [g] = 1, so at fixed angle dσ/dΩ ∝ g⁴/s³ falls faster. In the massless limit λ²/(32π s) = λ²/(128π|pᵢ|²): φ⁴ scatters into a fixed fraction of the area 1/|pᵢ|² set by the reduced de Broglie wavelength, which is also the form the s-wave unitarity limit takes.

Threshold, the non-relativistic limit, and unitarity

At threshold s = 4m² the φ⁴ result becomes λ²/(128π m²), a finite constant: the behaviour of low-energy s-wave scattering. Matching dσ/dΩ = |M|²/(64π² s) to |f|², with the usual phase, gives f = M/(8π√s). M already contains every pairing of the four legs, so it plays the part of the symmetrised f(θ) + f(π − θ), which is −2a at zero energy; then −λ/(16π m) = −2a gives the scattering length a = λ/(32π m), positive as a repulsive contact force requires, and the identical-boson 8πa² returns λ²/(128π m²), derived. For m = 0.5 GeV and λ = 1, a = 0.0199 GeV⁻¹ = 3.9 × 10⁻³ fm, and since 4m² = 1 GeV², 8πa² is the same 3.87 μb. The second test fails, instructively. The optical theorem of topic 5.07 demands Im M(forward) = 2√s|pᵢ|σ = λ²|pᵢ|/(16π√s), yet the tree M = −λ is real. The order-λ² imaginary part comes from the one-loop bubble, which is topic 5.08's subject.

Change one variable at a time

Make the relationship visible.

Interactive model

Coupling strength1.0

Final-pair mass0.25 GeV

Centre-of-mass energy1.00 GeV

At m′ = 0.25 GeV the solid curve lies on the dashed line: elastic φ⁴. Drop m′ to 0.05 GeV and σ climbs as 1/|pᵢ| near threshold; raise it to 0.50 GeV and σ opens from zero at √s = 1 GeV. Every curve joins the 1/s line at high s.

CROSS-SECTION AT MARKER3.873 μb

MOMENTUM RATIO OUT/IN1.000

DASHED-LINE VALUE3.873 μb

EVENT RATE AT L = 10³³3873 s⁻¹

Live interpretationCROSS-SECTION AT MARKER: 3.873 μb. MOMENTUM RATIO OUT/IN: 1.000. DASHED-LINE VALUE: 3.873 μb. EVENT RATE AT L = 10³³: 3873 s⁻¹

Catch the common trap

Explain before calculating.

In φ⁴ theory with λ = 0.60, two φ collide head-on at √s = 3.0 GeV. At tree level, how many φφ → φφ events per second does a collider with L = 1.0 × 10³³ cm⁻² s⁻¹ record?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyTwo φ of equal energy collide head-on at √s = 2.5 GeV in φ⁴ theory with λ = 0.40. At tree level find dσ/dΩ in nb sr⁻¹ and σ in nb, the event rate at L = 2.0 × 10³² cm⁻² s⁻¹, and the counting rate of a small detector covering 0.010 sr at 90°.

M = −λ (topic 4.05) and |p_f| = |pᵢ| for equal masses, so dσ/dΩ = λ²/(64π² s) with s = 6.25 GeV²: 0.16/(64π² × 6.25) = 4.05 × 10⁻⁵ GeV⁻² sr⁻¹.
Convert with 1 GeV⁻² = 0.3894 mb = 3.894 × 10⁵ nb: dσ/dΩ = 15.8 nb sr⁻¹, the same in every direction.
Total: σ = ½ × 4π sr × 15.78 nb sr⁻¹ = 99.2 nb, the ½ because two identical φ would otherwise fill the sphere twice.
Events: R = Lσ = 2.0 × 10³² cm⁻² s⁻¹ × 99.2 × 10⁻³³ cm² = 19.8 s⁻¹.
Counter: it registers particles, not events, and every event sends one φ into each hemisphere, so it uses the unhalved dσ/dΩ: 2.0 × 10³² × 15.78 × 10⁻³³ × 0.010 = 0.0316 s⁻¹, about 114 per hour.

Answerdσ/dΩ = 15.8 nb sr⁻¹ and σ = 99.2 nb; 19.8 events per second, and 0.032 s⁻¹ (about 114 per hour) in the 0.010 sr counter.

MediumA φ beam of laboratory energy 3.0 GeV strikes φ at rest in φ⁴ theory with m = 0.40 GeV and λ = 1.5. Find s and the centre-of-mass |pᵢ|, show that the flux factor is the same in the laboratory and centre-of-mass frames, find dσ/dt in μb GeV⁻² and the range of t, and integrate to σ.

s = 2m² + 2mE_lab = 0.32 + 2.40 = 2.72 GeV², so √s = 1.649 GeV, and |pᵢ|² = s/4 − m² = 0.680 − 0.160 = 0.520 GeV², |pᵢ| = 0.721 GeV.
Flux: on the fixed target F = 4m|p_lab| with |p_lab| = √(3.0² − 0.40²) = 2.973 GeV, giving 4.757 GeV²; in the centre-of-mass frame 4|pᵢ|√s = 4 × 0.7211 × 1.6492 = 4.757 GeV². The same number, as topic 5.01 promised.
dσ/dt = λ²/(64π s|pᵢ|²) = 2.25/(64π × 2.72 × 0.520) = 2.25/284.4 = 7.91 × 10⁻³ GeV⁻⁴; times 0.3894 mb GeV² that is 3.08 μb GeV⁻², flat in t. It depends on s and t only, so it holds as written in the laboratory.
Range: elastic scattering of equal masses gives −4|pᵢ|² ≤ t ≤ 0, a span of 4 × 0.520 = 2.08 GeV².
σ = ½ × 2.08 GeV² × 3.081 μb GeV⁻² = 3.20 μb, the ½ again for identical φ. Check: λ²/(32π s) = 2.25/(32π × 2.72) = 8.23 × 10⁻³ GeV⁻² = 3.20 μb.

Answers = 2.72 GeV² and |pᵢ| = 0.721 GeV; F = 4.76 GeV² in both frames; dσ/dt = 3.08 μb GeV⁻² over −2.08 GeV² ≤ t ≤ 0; σ = 3.20 μb.

HardAdd a scalar χ of mass m′ = 0.30 GeV coupled by Lᵢₙₜ = −(κ/4)φ²χ² with κ = 0.80, and take m = 1.0 GeV for φ. At tree level find σ(φφ → χχ) for centre-of-mass momenta |pᵢ| = 0.20 GeV and 0.10 GeV, show that σ|v₁ − v₂| is nearly constant, and find σ at √s = 20 GeV.

Amplitude: the vertex has 2 ways to annihilate the φ pair and 2 to create the χ pair, cancelling the ¼, so M = −κ, isotropic. With an identical χ pair, σ = ½ × 4π × κ²|p_f|/(64π² s|pᵢ|) = κ²|p_f|/(32π s|pᵢ|).
|pᵢ| = 0.20 GeV: s = 4(m² + |pᵢ|²) = 4.16 GeV² and |p_f| = √(s/4 − m′²) = √(1.04 − 0.09) = 0.9747 GeV, so |p_f|/|pᵢ| = 4.87 and σ = 0.64 × 0.9747/(32π × 4.16 × 0.20) = 7.46 × 10⁻³ GeV⁻² = 2.90 μb.
|pᵢ| = 0.10 GeV: s = 4.04 GeV², |p_f| = 0.9592 GeV, ratio 9.59, σ = 1.51 × 10⁻² GeV⁻² = 5.89 μb. Halving the incoming momentum has doubled σ (× 2.03): the reaction releases energy, and σ ∝ 1/|pᵢ| near threshold.
In the centre-of-mass frame |v₁ − v₂| = 4|pᵢ|/√s = 0.392 and 0.199, so σ|v₁ − v₂| = 1.14 μb and 1.17 μb (velocities in units of c); as |pᵢ| → 0 it tends to κ²√(1 − m′²/m²)/(64π m²) = 3.04 × 10⁻³ GeV⁻² = 1.18 μb. The 1/v rise comes from the flux, not from a growing amplitude.
At √s = 20 GeV, |p_f|/|pᵢ| = √(99.91/99.00) = 1.0046 and σ = 1.0046 × 0.64/(32π × 400) GeV⁻² = 6.23 nb, within 0.5% of the massless κ²/(32π s) = 6.20 nb.

Answerσ = 2.90 μb at |pᵢ| = 0.20 GeV and 5.89 μb at 0.10 GeV; σ|v₁ − v₂| = 1.14 and 1.17 μb, tending to 1.18 μb at threshold; σ = 6.23 nb at √s = 20 GeV.