University Quantum Field Theory I · Cross-Sections, Decay Rates and Unitarity · 5.1
From |M|² to a rate: squared deltas and the flux factor
This is where amplitudes become seconds and barns. Meet δ⁴(0), see why it measures a volume rather than signalling a sickness, and leave with the two formulae every later rate in the course uses: the decay rate of one particle, and a cross-section whose flux factor is the same at a collider and on a fixed target.
Build the model
Connect the measurement to the mechanism.
An S-matrix element is not yet a number anyone measures. Unit 4 delivered ⟨f|S − 1|i⟩ = i(2π)⁴δ⁴(pf − pᵢ)M between plane waves, and plane waves fill all of space for all of time, so the probability, its square, contains δ⁴(0). That infinity is bookkeeping, not a disease: in a box of volume V watched for a time T, (2π)⁴δ⁴(0) is VT, so the probability grows with the spacetime volume in which the particles can meet, and the rate per unit volume is finite.
Divide by the norm 2EV of every state, count V d³p/(2π)³ final states per particle, and for one initial particle every V cancels, leaving dΓ = |M|²dΠₙ/(2EA): Quantum Mechanics II's golden rule in relativistic dress. Two initial particles leave one factor 1/V, the density of beam particles in the box, and dividing the rate by the flux |v₁ − v₂|/V gives a cross-section. Wave packets narrow in momentum reach the same result without a box and show where the 1/|v₁ − v₂| comes from: the time the packets take to cross.
For collinear beams the combination 4E₁E₂|v₁ − v₂| equals 4√((p₁⋅p₂)² − m₁²m₂²), so σ is one number on a fixed target and at a collider. The construction fails where M changes across the packets' momentum spread, as on a resonance narrower than the beams' energy spread, and exactly forward, where the scattered wave interferes with the unscattered one.
- Simple definition
- A cross-section is the transition rate per target particle divided by the incident flux; in relativistic normalisation it is dσ = |M|² dΠₙ/(4E₁E₂|v₁ − v₂|), the flux factor 4E₁E₂|v₁ − v₂| undoing the 2E-per-volume normalisation of the two incoming states.
- Example
- A 10.0 GeV massless beam on a proton at rest has F = 4mₚE = 4 × 0.938 × 10.0 = 37.5 GeV². A process with ∫|M|² dΠ = 1 then has σ = 1/37.5 GeV⁻² = 0.0267 × 0.3894 mb = 10.4 μb, in the lab or in any frame boosted along the beam.
(2π)⁴δ⁴(0) = ∫d⁴x = VT. A transition that can happen anywhere and at any time has a probability growing as VT, so per unit volume and time it is finite.
V the box volume, T the time the interaction acts: m³ and s in SI, GeV⁻³ and GeV⁻¹ here, with 1 fm = 5.068 GeV⁻¹ (derived)
Divide |⟨f|S|i⟩|² by the norm of every initial and final state. That is why each external particle brings a 1/(2E) into a rate.
Eₚ = √(p² + m²) in GeV and V in GeV⁻³; ⟨p|q⟩ = 2Eₚ(2π)³δ³(p − q) is topic 1.03's normalisation, quoted
Every V cancels for a single initial particle, so the rate owes nothing to the box. Its rest-frame value and its 1/γ in flight are topic 5.03's business.
dΠₙ = Πf d³pf/((2π)³2Ef) × (2π)⁴δ⁴(pA − Σpf); Γ in GeV, and 1 GeV/ħ = 1.519 × 10²⁴ s⁻¹ (derived)
Rate per target over incident flux. A box holding one of each gives the rate |M|²dΠₙ/(4E₁E₂V) and the flux |v₁ − v₂|/V, and the V cancels in the ratio.
E₁, E₂ in GeV; v₁, v₂ the incoming velocities, in units of c, along the beam; σ in GeV⁻², 1 GeV⁻² = 0.3894 mb (derived)
E₁E₂ and |v₁ − v₂| each change under a boost along the beam; their product does not. F = 4m₂|plab| on a fixed target, 4|pcm|√s in the CM frame, 2s for massless beams.
p₁⋅p₂ the four-vector product in GeV², m₁ and m₂ the masses in GeV; valid for beams along one line (shown)
A boost along the beam adds one constant to every rapidity, so y₁ − y₂, and F with it, cannot change. With the target at rest, m₁ sinh y₁ = |plab| gives F = 4m₂|plab|.
y = ½ ln((E + pz)/(E − pz)) the rapidity along the beam, dimensionless; E = m cosh y, v = tanh y (derived)
The square of a delta function is a spacetime volume
Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Unit 4 gave ⟨f|S|i⟩ = ⟨f|i⟩ + i(2π)⁴δ⁴(pf − pᵢ)M, quoted; for f ≠ i only the second term survives. A probability needs its modulus squared, and in [δ⁴(pf − pᵢ)]² the second factor sits where the first has already set pf = pᵢ, so it is δ⁴(0). Plane waves are at fault, not the theory: they fill all of space for all time, so the collision can occur anywhere and at any moment, and its probability must grow with the spacetime volume on offer. Put the system in a periodic box of volume V and let the interaction act for a time T. Since (2π)⁴δ⁴(q) = ∫d⁴x e(iq⋅x), at q = 0 the integral is ∫d⁴x = VT, so [(2π)⁴δ⁴(pf − pᵢ)]² = (2π)⁴δ⁴(pf − pᵢ) VT, derived. The time half is QM-II's step from sin²(ΔE T/2)/(ΔE/2)² to 2πT δ(ΔE) (UQMII-12.06, quoted); the space half is its twin for momentum. A probability proportional to VT is a rate per unit volume, which is what a detector counts.
Divide by the norms and count the final states
In topic 1.03's normalisation a state's norm is of order EV: ⟨p|p⟩ = 2Eₚ(2π)³δ³(0) = 2EₚV. So P = |⟨f|S|i⟩|²/(⟨i|i⟩⟨f|f⟩), and a final particle has V d³p/(2π)³ box states in d³p. For A decaying to n particles, P = VT(2π)⁴δ⁴|M|²/(2EAV) × Πf V d³pf/((2π)³2EfV). Each final V cancels its norm, VT cancels the initial V and leaves T, and P/T = |M|²dΠₙ/(2EA), with dΠₙ = Πf d³pf/((2π)³2Ef) × (2π)⁴δ⁴(pA − Σpf): derived, with no trace of the box. Dimensions: M for 1 → n has mass dimension 3 − n, so |M|² has 6 − 2n, dΠₙ has 2n − 4 and 1/(2EA) has −1, leaving 1: a rate in GeV, and 1/ħ turns 1 GeV into 1.519 × 10²⁴ s⁻¹. In box-normalised states the golden rule's matrix element is, up to sign, M V/Πⱼ √(2EⱼV) for momentum-conserving pairs, and QM-II's rule, quoted, returns this same rate; what is derived here is its relativistic normalisation.
Two incoming particles: a rate per volume and a flux
With two particles in the box the counting leaves one norm uncancelled: P/T = |M|²dΠₙ/(2E₁ 2E₂ V). That 1/V is physical. The box holds one particle of each kind, and the bigger the box, the less often they meet. A cross-section removes exactly this: σ is the number of events per unit time per target particle divided by the incident flux, the number of beam particles crossing unit area per unit time. One beam particle in V, closing on the target at |v₁ − v₂| in the chosen frame, is a flux |v₁ − v₂|/V. Dividing gives dσ = |M|²dΠₙ/(4E₁E₂|v₁ − v₂|), derived. For 2 → n, |M|² has mass dimension 4 − 2n, dΠₙ has 2n − 4 and the flux factor 2, so σ has −2: an area, with 1 GeV⁻² = (0.197327 fm)² = 0.3894 mb. Read backwards, the formula is how an experiment counts: events per second = σ × beam flux × number of targets in the beam, which is σ times the luminosity.
Wave packets: where 1/|v₁ − v₂| comes from
A box is a device; real beams are packets. Build each incoming particle as ∫d³k/(2π)³ φ(k)|k⟩/√(2Eₖ) with ∫d³k/(2π)³ |φ(k)|² = 1, shift the second by an impact parameter b transverse to the beam, and define σ = ∫d²b P(b), the area over which collisions happen. With pf the total final momentum, the amplitude carries δ⁴(k₁ + k₂ − pf) inside the packets' integrals, and its complex conjugate a second δ⁴(k̄₁ + k̄₂ − pf) with integration variables of its own. The b integral gives (2π)²δ²(k̄₂⊥ − k₂⊥), the transverse part of the second δ⁴ fixes k̄₁⊥, and what is left, δ(k̄₁ᶻ + k̄₂ᶻ − k₁ᶻ − k₂ᶻ) δ(Ē₁ + Ē₂ − E₁ − E₂), is integrated over k̄₁ᶻ and k̄₂ᶻ. The first sets k̄₂ᶻ; the second's argument then changes with k̄₁ᶻ at the rate v₁ − v₂, vanishing at k̄ = k, so it yields 1/|v₁ − v₂|, derived: the time the packets take to pass through each other. If M, E and v barely change across each packet's momentum spread they leave the integrals, ∫|φ|² = 1 does the rest, and the box result returns without a box.
One flux factor for a fixed target and a collider
For beams along z, v = p/E with p the z component, so E₁E₂|v₁ − v₂| = |E₂p₁ − E₁p₂|. Expand (p₁⋅p₂)² − m₁²m₂² with p₁⋅p₂ = E₁E₂ − p₁p₂ and m² = E² − p²: the E₁²E₂² and p₁²p₂² terms cancel, leaving (E₁p₂ − E₂p₁)², so F = 4E₁E₂|v₁ − v₂| = 4√((p₁⋅p₂)² − m₁²m₂²), shown. E₁p₂ − E₂p₁ is the determinant of the two (E, p) columns, and a boost along z has determinant γ²(1 − β²) = 1. On a fixed target F = 4m₂|plab|; in the CM frame F = 4|pcm|√s; for massless beams F = 2s. Two 6.8 TeV proton beams meeting head-on have F = 4 × 6800² × 2 = 3.70 × 10⁸ GeV²; a cosmic-ray proton of 9.86 × 10¹⁶ eV striking a proton at rest reaches the same s, and 4mₚ|plab| = 4 × 0.938 × 9.86 × 10⁷ = 3.70 × 10⁸ GeV² too. |M|² is a scalar and dΠₙ is invariant by topic 1.03's identity for d³p/(2E), as topic 5.02 spells out, so σ is the same in all such frames: an area across the beam, which a boost along it does not contract.
Checks, and where the formula stops
Non-relativistic limit: E → m makes F → 4m₁m₂|v₁ − v₂|, so if |M|² and the final momenta stay fixed as the speed falls, as in an exothermic capture of a neutral particle, σ ∝ 1/v. ³He(n, p)³H, about 5330 b at 2200 m/s (quoted), follows that 1/v law across the thermal range, which is why a ³He counter's rate measures neutron density rather than flux. High-energy limit: F → 2s, so a dimensionless |M|² gives σ ∝ 1/s. Three conditions bound the derivation. M must vary little across the packets' momentum spread: the J/ψ, 93 keV wide, showed up as a peak as wide as the colliding beams' energy spread (quoted), so what was measured was |M|² averaged over the beams. A final momentum equal to an initial one, exactly forward, keeps the cross term between the 1 and iT in |⟨f|S|i⟩|²: scattered and unscattered waves interfere there, and topic 5.07 makes that the optical theorem. And 4E₁E₂|v₁ − v₂| needs collinear beams; for crossing beams only the invariant form survives.
Change one variable at a time
Make the relationship visible.
Slide the frame from 0 to 1: 4E₁E₂ dips as |v₁ − v₂| climbs toward 2, and the open dot's offset from the 1 line always matches the flat line's offset from the filled dot. Raise m₁ at fixed p: the curves reshape, F stays put, and the CM ring moves onto the U's floor, the equal-speed frame.
4E₁E₂ IN THIS FRAME18.77 GeV²
|v₁ − v₂| IN THIS FRAME1.000
F = 4m₂p, EVERY FRAME18.76 GeV²
√s3.207 GeV
Live interpretation4E₁E₂ IN THIS FRAME: 18.77 GeV². |v₁ − v₂| IN THIS FRAME: 1.000. F = 4m₂p, EVERY FRAME: 18.76 GeV². √s: 3.207 GeV
Catch the common trap
Explain before calculating.
A massless particle of energy 50.0 GeV strikes a particle of mass 1.00 GeV at rest, so s = 101 GeV². What is the flux factor F = 4E₁E₂|v₁ − v₂| evaluated in the centre-of-mass frame?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn asymmetric collider brings a 7.0 GeV electron beam head-on into a 4.0 GeV positron beam. Treating both as massless, find s and the flux factor F = 4E₁E₂|v₁ − v₂| in the laboratory and in the CM frame. Then find the electron energy a fixed-target machine, firing electrons at electrons at rest, needs for the same s, and its flux factor.
- Laboratory: each speed is 1 to within mₑ²/(2E²) ≈ 8 × 10⁻⁹, and the beams point opposite ways, so |v₁ − v₂| = 2 and F = 4 × 7.0 × 4.0 × 2 = 224 GeV².
- Invariant mass: s = (E₁ + E₂)² − (p₁ + p₂)², with the momenta +7.0 and −4.0 GeV along the beam, so s = 11.0² − 3.0² = 112 GeV² and √s = 10.58 GeV.
- CM frame: each beam carries √s/2 = 5.29 GeV and the closing speed is still 2, so F = 4 × 5.29² × 2 = 4 × 28.0 × 2 = 224 GeV². The energies changed, 7.0 and 4.0 to 5.29 and 5.29, but their product, 28.0 GeV², did not: the boost multiplies one by e(−y) and the other by e(y), with y = ½ ln(7.0/4.0) = 0.280.
- Invariant check: for massless beams 4√((p₁⋅p₂)² − 0) = 4p₁⋅p₂ = 2s = 224 GeV².
- Fixed target: s = 2mₑ² + 2mₑElab, so Elab = (112 − 2mₑ²)/(2 × 0.000511 GeV) = 1.10 × 10⁵ GeV, about 110 TeV. Then F = 4mₑ|plab| = 4 × 0.000511 × 1.096 × 10⁵ = 224 GeV²: a beam 1.6 × 10⁴ times more energetic than the electron beam, the same flux factor, and so the same σ.
Answers = 112 GeV² (√s = 10.58 GeV), and F = 2s = 224 GeV² in the laboratory, in the CM frame and for a 110 TeV electron on a fixed target: one flux factor, so one cross-section.
MediumNeutrons at 2200 m/s are absorbed by ¹⁰B through ¹⁰B(n, α)⁷Li with a measured cross-section of 3840 b. Use dσ = |M|²dΠ₂/F to predict the cross-section at a neutron kinetic energy of 1.00 eV, and find the absorption rate per ¹⁰B nucleus in a neutron gas of density 5.0 × 10⁷ cm⁻³ at either energy. Take mₙ = 0.9396 GeV and a ¹⁰B nucleus of mass 9.324 GeV at rest.
- Target at rest and v ≪ 1, so E₁ → mₙ, E₂ = mB and F = 4mₙ mB v. At v = 2200 m/s = 7.338 × 10⁻⁶ c, F = 4 × 0.9396 × 9.324 × 7.338 × 10⁻⁶ = 2.5715 × 10⁻⁴ GeV².
- The reaction releases 2.31 or 2.79 MeV, so the α and ⁷Li momenta, and ∫dΠ₂ with them, are set by the Q-value, not by a neutron energy of order an eV; and a neutral neutron's s-wave |M|² tends to a constant as v → 0. So σF = ∫|M|²dΠ₂ is the same at both energies. From the data, σ = 3.840 × 10⁶ mb ÷ 0.3894 = 9.861 × 10⁶ GeV⁻², and σF = 2.536 × 10³, dimensionless as a 2 → 2 |M|² must be.
- At 1.00 eV, v = √(2T/mₙ) = √(2 × 1.00 × 10⁻⁹/0.9396) = 4.614 × 10⁻⁵ c = 13 830 m/s, so F = 1.6168 × 10⁻³ GeV².
- σ(1.00 eV) = 2536/(1.6168 × 10⁻³ GeV²) = 1.569 × 10⁶ GeV⁻² = 6.11 × 10⁵ mb = 611 b: the 1/v law, 3840 b × 2200/13 830.
- Rate per nucleus = σ × flux = σvn. σv = 3.840 × 10⁻²¹ cm² × 2.20 × 10⁵ cm s⁻¹ = 8.45 × 10⁻¹⁶ cm³ s⁻¹ at both energies, so the rate is 8.45 × 10⁻¹⁶ × 5.0 × 10⁷ = 4.2 × 10⁻⁸ s⁻¹. Faster neutrons sweep past the nucleus more often, and each pass is less likely to be absorbed by the same factor.
Answerσ(1.00 eV) ≈ 611 b, and the absorption rate per ¹⁰B nucleus is 4.2 × 10⁻⁸ s⁻¹ at both energies, because F ∝ v carries the only speed dependence and σv is fixed.
HardA massless particle of energy 10.0 GeV moving along +z strikes a proton (0.93827 GeV) at rest. Find F in the laboratory. Then view the collision from a frame moving at β = 0.600 along +x, transverse to the beam, and evaluate 4E₁E₂|v₁ − v₂| there. Find the combination of velocities that does stay invariant, and use the particle densities to decide which flux gives a frame-independent event rate.
- Laboratory: F = 4m₂|plab| = 4 × 0.93827 × 10.0 = 37.53 GeV², which is 4p₁⋅p₂ with p₁⋅p₂ = m₂E₁ = 9.383 GeV².
- Boost with γ = 1.25 using E′ = γ(E − βpₓ) and p′ₓ = γ(pₓ − βE). The beam becomes E₁′ = 12.50 GeV with momentum (−7.50, 0, 10.0) GeV and velocity v₁ = (−0.600, 0, 0.800); the proton becomes E₂′ = 1.17284 GeV with momentum (−0.70370, 0, 0) GeV and velocity v₂ = (−0.600, 0, 0).
- Collinear formula: |v₁ − v₂| = 0.800, so 4E₁′E₂′|v₁ − v₂| = 4 × 12.50 × 1.17284 × 0.800 = 46.91 GeV², larger than the laboratory value by exactly γ = 1.25. It does not survive a transverse boost.
- Invariant: p₁⋅p₂ = 12.50 × 1.17284 − (−7.50)(−0.70370) = 14.6605 − 5.2778 = 9.383 GeV², unchanged, so 4√((p₁⋅p₂)² − m₁²m₂²) = 37.53 GeV². With p = Ev and m² = E²(1 − v²), (p₁⋅p₂)² − m₁²m₂² = E₁²E₂²[(1 − v₁⋅v₂)² − (1 − v₁²)(1 − v₂²)] = E₁²E₂²[(v₁ − v₂)² − |v₁ × v₂|²].
- Here (v₁ − v₂)² = 0.640 and v₁ × v₂ = (0, −0.480, 0), so √(0.640 − 0.230) = 0.640 and 4 × 12.50 × 1.17284 × 0.640 = 37.53 GeV². The cross-product term corrects for velocities that are not parallel.
- The event rate per unit volume, σn₁n₂w, must be invariant, since both the event count and d⁴x are. Densities grow with energy, n′ = n E′/E, so n₁n₂ is 1.25² = 1.5625 times larger here; w = 0.640 restores the laboratory n₁n₂ × 1 (1.5625 × 0.640 = 1.000), while the collinear w = 0.800 would give 1.25.
AnswerF = 37.53 GeV² in the laboratory and, from 4√((p₁⋅p₂)² − m₁²m₂²), in every frame; 4E₁E₂|v₁ − v₂| gives 46.91 GeV² after the transverse boost, too large by γ = 1.25. Off-axis, the closing speed must become √((v₁ − v₂)² − |v₁ × v₂|²).