University Quantum Field Theory I · Cross-Sections, Decay Rates and Unitarity · 5.2

Lorentz-invariant phase space and two-body final states

Every decay rate and cross-section from here on is |M|² times one phase-space integral, so we do that integral once and carefully: why it needs no frame, how four delta functions leave two bodies only a direction to choose, and why |p_f| dΩ/(16π² E_cm) is the factor you will reuse all term.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Every rate in this course is |M|² summed over the final states a process can reach, and dΠₙ is the measure that counts them. It comes out of topic 5.01's box: a sum over final states gives V d³p/(2π)³ per particle, each state's relativistic norm 2EV divides it, and the squared S-matrix element leaves one (2π)⁴δ⁴ of four-momentum conservation. What survives, a product of d³p/((2π)³ 2E) with that delta function, is Lorentz invariant by topic 1.03's identity, so it can be evaluated in whatever frame is easiest and a rate's frame dependence lives only in its 1/(2E) or flux prefactor.

For two bodies the four delta functions fix everything except the direction of the decay axis: both products carry one momentum magnitude |p_f|, set by √s and the two masses through the Källén function, and the measure collapses to |p_f| dΩ/(16π² E_cm). Integrated over angles it vanishes at threshold as the square root of the energy release when both products are massive, where it reproduces the non-relativistic density of states with one 1/(2m) per particle, and saturates at 1/(8π) once the masses are negligible; identical products add 1/n! when every angle is integrated. The measure stops there: it knows nothing of the dynamics in |M|², it assumes stable products with sharp masses, which a final particle with a width violates (topic 5.09), and for three or more bodies the delta functions no longer fix the momenta (topic 5.05).

Simple definition: Lorentz-invariant phase space dΠₙ is the measure over the momenta of n final particles, d³p/((2π)³ 2E) for each one, constrained by (2π)⁴δ⁴ of total four-momentum, that counts final states with the same weight in every inertial frame.
Example: A 1.000 GeV scalar decaying at rest to two distinct massless particles has |p_f| = 0.500 GeV and ∫dΠ₂ = |p_f|/(4π E_cm) = 1/(8π) = 0.0398; if each product has mass 0.300 GeV, |p_f| = 0.400 GeV and ∫dΠ₂ falls to 0.0318.

n-body Lorentz-invariant phase spacedΠₙ = ∏_f d³p_f/((2π)³ 2E_f) × (2π)⁴ δ⁴(P − Σ_f p_f)

Every factor is a Lorentz scalar, so a rate's frame dependence sits wholly in its 1/(2E) or flux prefactor. Derived from the box of topic 5.01.

P the total initial four-momentum, p_f and E_f each product's momentum and energy in GeV; [dΠₙ] = GeV⁽²ⁿ⁻⁴⁾

Invariant one-particle measure (topic 1.03, quoted)d³p/((2π)³ 2Eₚ) = d⁴p δ(p² − m²) θ(p⁰)/(2π)³

d⁴p, p² and the sign of p⁰ survive any proper orthochronous Λ, and δ⁴(Λq) = δ⁴(q); together they make dΠₙ invariant.

p⁰ integrated against the mass shell, θ(p⁰) keeping the positive-energy sheet; m the particle's mass in GeV

Two bodies in the centre-of-mass framedΠ₂ = |p_f| dΩ/(16π² E_cm)

Derived: δ³ removes one momentum, the energy delta's Jacobian p E_cm/(E₁E₂) cancels the 4E₁E₂, and only the decay axis is left free.

|p_f| either product's momentum and E_cm = √s, both in GeV; dΩ in sr about one product's direction; dΠ₂ is dimensionless

Final momentum from the Källén function|p_f| = λ¹⁄²(s, m₁², m₂²)/(2√s), λ(a, b, c) = a² + b² + c² − 2ab − 2bc − 2ca

Symmetric in the two masses, zero at threshold √s = m₁ + m₂, and tending to s² for massless products, so |p_f| → √s/2. Derived.

s = E_cm² in GeV², m₁ and m₂ the product masses in GeV; λ = [s − (m₁ + m₂)²][s − (m₁ − m₂)²]

Energy of each productE₁ = (s + m₁² − m₂²)/(2√s)

The heavier product takes more of the energy while both carry the same |p_f|; E₁² − m₁² = |p_f|² is the one-line check.

E₂ follows by swapping 1 and 2, so E₁ + E₂ = √s; energies in GeV

Integrated two-body phase space∫dΠ₂ = |p_f|/(4π S √s)

Times |M|²/(2m) it is a decay rate whenever |M| is independent of angle. Near the threshold of two massive products |p_f| ≈ √(2μQ), so the phase space opens as √Q.

S = 1 for distinct products, 2 for an identical pair (n! for n identical); dimensionless and never above 1/(8πS) = 0.0398/S

Where dΠₙ comes from: states counted per unit volume

Work in natural units, ħ = c = 1, with the metric (+, −, −, −), so p² = E² − |p|² = m². Topic 5.01 put the process in a box of volume V for a time T. There a sum over final states becomes V d³p/(2π)³ for each final particle, and each state carries the relativistic norm 2EV, so one V cancels per particle and d³p_f/((2π)³ 2E_f) is left. Of the squared (2π)⁴δ⁴ one factor became VT and one survives. Their product is dΠₙ, and a rate is a prefactor times ∫|M|² dΠₙ. Nothing in dΠₙ remembers the box, the beams or the initial state beyond its total four-momentum P, which is why one measure serves decays and collisions alike. Count dimensions: each d³p/(2E) is GeV² and the delta function GeV⁻⁴, so [dΠₙ] = GeV⁽²ⁿ⁻⁴⁾. Two-body phase space is a pure number. An amplitude with N external legs has dimension GeV^(4−N), so a 1 → n decay rate (1/(2m))∫|M|² dΠₙ carries GeV⁽⁶⁻²ⁿ⁾ × GeV⁽²ⁿ⁻⁴⁾/GeV = GeV, and ħ = 6.582 × 10⁻²⁵ GeV s turns it into s⁻¹.

Why no frame has to be chosen

Topic 1.03 showed, and we quote, that d³p/((2π)³ 2Eₚ) = d⁴p δ(p² − m²) θ(p⁰)/(2π)³. Every piece on the right is a Lorentz scalar: d⁴p because every Lorentz transformation has |det Λ| = 1, p² by construction, and θ(p⁰) because an orthochronous Λ never turns a future-pointing timelike vector into a past-pointing one. The four-dimensional delta function is invariant for the same determinant reason, δ⁴(Λq) = δ⁴(q)/|det Λ| = δ⁴(q). So dΠₙ is invariant, derived, and ∫|M|² dΠₙ is a Lorentz scalar whenever |M|² is. A rate's frame dependence therefore sits entirely in its prefactor: 1/(2E) for a decay, which makes a moving particle's width Γ/γ, and the flux factor for a collision, unchanged by boosts along the beam (topics 5.03 and 5.01). The payoff is a licence to evaluate dΠₙ wherever its delta functions are easiest. For two bodies that is the frame in which the total three-momentum vanishes: the rest frame of a decaying particle, or the centre-of-mass frame of a collision.

Spending the delta functions for two bodies

In the centre-of-mass frame P = (E_cm, 0) with E_cm = √s. Split δ⁴ into δ³(p₁ + p₂) δ(E_cm − E₁ − E₂) and let δ³ do the d³p₂ integral: p₂ = −p₁, so both products carry the same magnitude p, and E₂ = √(p² + m₂²) now changes with p as well. With d³p₁ = p² dp dΩ, the powers of 2π collapse to (2π)⁴/(2π)⁶ = 1/(4π²) and dΠ₂ = p² dp dΩ δ(E_cm − E₁ − E₂)/(16π² E₁E₂). The energy delta function fixes p at its root |p_f|, and integrating it divides by the derivative of its argument, d(E₁ + E₂)/dp = p/E₁ + p/E₂ = p E_cm/(E₁E₂). The product E₁E₂ cancels, one power of p survives, and dΠ₂ = |p_f| dΩ/(16π² E_cm), derived. Six momentum components less four constraints leave two variables, the angles of the decay axis; three bodies leave five, which is why topic 5.05 needs a Dalitz plot. The recoil term p/E₂ is the step most often lost: drop it and you get |p_f| dΩ/(16π² E₂), too large by the factor E_cm/E₂ for any masses, and so twice too large when the two masses are equal.

The final momentum and the Källén triangle function

Energy conservation and the shared momentum fix everything else. Since E₁² − m₁² = E₂² − m₂² = |p_f|², E₁² − E₂² = m₁² − m₂²; divide by E₁ + E₂ = √s to get E₁ − E₂ = (m₁² − m₂²)/√s, so E₁ = (s + m₁² − m₂²)/(2√s) and |p_f|² = E₁² − m₁² = λ(s, m₁², m₂²)/(4s), derived. Factorised as [s − (m₁ + m₂)²][s − (m₁ − m₂)²], λ vanishes at threshold, is symmetric in the two masses, and tends to s² for massless products, giving |p_f| = √s/2. A measured check: for π⁺ → μ⁺ν_μ at rest, with m_π = 139.570 MeV, m_μ = 105.658 MeV and a massless neutrino, |p_f| = (m_π² − m_μ²)/(2m_π) = 29.79 MeV/c, and the muon's kinetic energy is (m_π − m_μ)²/(2m_π) = 4.12 MeV. That single sharp energy, about 4.1 MeV, is what the emulsion tracks of 1947 showed: every muon from a stopped pion travelled the same range before stopping, the signature of a two-body decay. A three-body decay would have spread the ranges out.

Threshold, high energy and the golden rule's density of states

Integrated over all directions, ∫dΠ₂ = |p_f|/(4π√s) = λ¹⁄²/(8πs) for distinct products, and since λ¹⁄² ≤ s it never exceeds the massless value 1/(8π) = 0.0398, where it saturates at high energy. Near threshold write √s = m₁ + m₂ + Q with Q small next to both masses: then s − (m₁ + m₂)² ≈ 2(m₁ + m₂)Q and s − (m₁ − m₂)² ≈ 4m₁m₂, so |p_f| ≈ √(2μQ) with μ = m₁m₂/(m₁ + m₂) the reduced mass. Phase space opens as √Q, the s-wave threshold law; higher partial waves bring further powers of |p_f| through |M|². The limit is also a consistency check with Quantum Mechanics II: near threshold dΠ₂ ≈ [μ|p_f| dΩ/(2π)³] × 2π/(2m₁ × 2m₂), the golden rule's density of relative-motion states (UQMII-12.06, quoted) times one 1/(2m) per particle from relativistic normalisation and the 2π of the energy delta. For two 0.500 GeV products with Q = 10 MeV, √(2μQ) = 70.7 MeV against the exact 70.9 MeV.

Identical particles: each configuration counted n! times

For two identical bosons |p₁, p₂⟩ and |p₂, p₁⟩ are the same vector. The overlap ⟨p₁, p₂|q₁, q₂⟩ has two delta function terms, one per pairing, so letting d³p₁ d³p₂ range over all values visits every final state twice, and the completeness relation for the sector needs 1/2!, shown; n identical particles need 1/n!, and several repeated species a product of 1/nₖ!. In the centre-of-mass frame the double counting is plain: a product at (θ, φ) with its partner at (π − θ, φ + π) is the same state as the one with the labels swapped. Integrate over a hemisphere, or over the full 4π with a factor 1/2, never both. The factor belongs to the total: a counter at angle θ that records any product sees particles from both labellings, so its count follows the full |M|² dΠ₂/dΩ with no 1/2. It also belongs only to genuinely identical particles. In φ → ψψ̄ the particle and antiparticle are distinct, so topic 5.03's rate has no 1/2. Identical fermions take the same 1/n!; their antisymmetry lives in the sign between amplitudes, not in the count.

Change one variable at a time

Make the relationship visible.

Interactive model

Mass m₁ of product 10.60 GeV

Mass m₂ of product 20.35 GeV

Energy release Q = √s − m₁ − m₂0.50 GeV

Set m₂ = 0 and the dashed curve collapses onto the axis, since a massless product has no non-relativistic limit, while the solid curve now opens linearly in Q. Then put both masses at 0.60 GeV and slide Q up from 0.05 GeV: the two momenta agree near threshold and part as Q nears the masses, while the filled marker climbs the solid curve towards 1/(8π).

|p_f| EXACT (KÄLLÉN)0.540 GeV

|p_f| NON-REL. √(2μQ)0.470 GeV

ENERGY E₁0.807 GeV

∫dΠ₂ DISTINCT PAIR0.0296

Live interpretation|p_f| EXACT (KÄLLÉN): 0.540 GeV. |p_f| NON-REL. √(2μQ): 0.470 GeV. ENERGY E₁: 0.807 GeV. ∫dΠ₂ DISTINCT PAIR: 0.0296

Catch the common trap

Explain before calculating.

Misconception check

Because dΠ₂ is Lorentz invariant, it is tempting to use |p_f| dΩ/(16π² E_cm) in any frame, with a laboratory momentum and solid angle inserted, and to conclude that a decay isotropic at rest is isotropic in the laboratory. Invariance belongs to the whole measure with its delta functions, and to integrals over it, not to the variables left once the delta functions are spent. The two-body formula spent δ³ where the total momentum vanishes, so its |p_f| and dΩ are centre-of-mass quantities. In the laboratory, with p₁ and P the momentum magnitudes and E the parent's energy, the same measure reads p₁² dΩ_lab/(16π² |p₁E − E₁P cos θ|), an angle-dependent Jacobian: a 0.500 GeV scalar of energy 2.50 GeV decaying to massless products has 98 times the rest-frame density per steradian straight ahead and 25 times less at 90°, yet the same total, 1/(8π). Work in the centre-of-mass frame, then boost the momenta.

A spinless particle of mass 2.00 GeV decays at rest into two distinct spinless particles of masses 1.00 GeV and 0.50 GeV. What is the integrated two-body phase space ∫dΠ₂, the number that multiplies |M|²/(2m) in the decay rate when |M| is constant?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyA spinless particle of mass 750 MeV decays at rest into two spinless particles of 140 MeV each. Find |p_f|, the speed of each product and ∫dΠ₂ (a) when the two are a distinct particle-antiparticle pair and (b) when they are identical.

Equal masses make the energy formula give E₁ = E₂ = √s/2 = 375 MeV, so |p_f| = √(E₁² − m²) = √(375² − 140²) MeV = √121025 MeV = 347.9 MeV.
Each product moves at β = |p_f|/E₁ = 347.9/375 = 0.928, which is √(1 − 4m²/s), the form every equal-mass pair decay takes.
(a) Distinct products: ∫dΠ₂ = |p_f|/(4π√s) = 347.9 MeV/(4π × 750 MeV) = 0.0369. The MeV cancel, as they must for a dimensionless measure, and the result is β/(8π), 92.8% of the massless value 0.0398.
(b) Identical products: the configurations (θ, φ) and (π − θ, φ + π) are one state, so the full sphere counts each twice; divide by 2! to get 0.0185.

Answer|p_f| = 348 MeV and β = 0.928; ∫dΠ₂ = 0.0369 for a distinct pair and 0.0185 for an identical pair.

MediumA spinless particle of mass 1.000 GeV decays at rest into two distinct spinless particles of masses 0.600 GeV and 0.350 GeV, releasing Q = 50 MeV. Find |p_f| from the Källén function and the energy and speed of each product, compare |p_f| with the non-relativistic √(2μQ), and evaluate ∫dΠ₂.

λ = [s − (m₁ + m₂)²][s − (m₁ − m₂)²] = (1 − 0.9025)(1 − 0.0625) GeV⁴ = 0.0975 × 0.9375 GeV⁴ = 0.091406 GeV⁴, so |p_f| = √(0.091406 GeV⁴)/(2 × 1.000 GeV) = 0.15117 GeV.
E₁ = (s + m₁² − m₂²)/(2√s) = (1 + 0.36 − 0.1225)/2 GeV = 0.61875 GeV and E₂ = (1 + 0.1225 − 0.36)/2 GeV = 0.38125 GeV. They sum to 1.000 GeV, and E₁² − m₁² = 0.022852 GeV² = |p_f|², as the check requires.
Equal momenta, unequal speeds: v₁ = |p_f|/E₁ = 0.244 and v₂ = |p_f|/E₂ = 0.397, and the lighter product carries 31.25 MeV of the 50 MeV of kinetic energy.
The reduced mass is μ = 0.600 × 0.350/0.950 GeV = 0.22105 GeV, so √(2μQ) = √(2 × 0.22105 × 0.050) GeV = 0.14868 GeV, 1.6% below the exact value: a relativistic correction already visible with the lighter product at 0.4c.
∫dΠ₂ = |p_f|/(4π√s) = 0.15117/(4π × 1.000) = 0.01203, only 30% of the massless value 1/(8π), because the products are barely above threshold.

Answer|p_f| = 0.1512 GeV (√(2μQ) gives 0.1487 GeV, 1.6% low); E₁ = 0.6188 GeV, E₂ = 0.3813 GeV, v₁ = 0.244, v₂ = 0.397; ∫dΠ₂ = 0.01203.

HardA spinless particle of mass m = 0.500 GeV flies along z with energy E = 2.50 GeV and decays to two distinct massless spinless particles 1 and 2. Find dΠ₂/dΩ in its rest frame; then, in the laboratory, the momentum of particle 1 at angle θ, the density dΠ₂/dΩ_lab at θ = 0 and θ = 90°, and show that the laboratory density integrates to the rest-frame ∫dΠ₂.

Rest frame: |p_f| = m/2 = 0.250 GeV, so dΠ₂/dΩᵣₑₛₜ = |p_f|/(16π² m) = 1/(32π²) = 3.17 × 10⁻³ sr⁻¹ in every direction, and ∫dΠ₂ = 1/(8π) = 0.0398.
Laboratory: |P| = √(E² − m²) = √(6.00 GeV²) = 2.449 GeV and γ = E/m = 5.00. With particle 2 massless, (P − p₁)² = 0 gives m² = 2P⋅p₁ = 2|p₁|(E − |P| cos θ), so |p₁| = m²/[2(E − |P| cos θ)]: 2.475 GeV at θ = 0 and 0.0500 GeV at θ = 90°.
Spend δ³ on p₂ and hold the direction of particle 1 fixed. The energy delta's Jacobian is d(E₁ + E₂)/d|p₁| = 1 + (|p₁| − |P| cos θ)/E₂ = (E − |P| cos θ)/E₂, so dΠ₂/dΩ_lab = |p₁|²/(16π² E₁E₂) × E₂/(E − |P| cos θ) = m²/[32π²(E − |P| cos θ)²].
At θ = 0, E − |P| = 0.05051 GeV and the density is 0.250/(32π² × 0.002551) sr⁻¹ = 0.310 sr⁻¹, 98.0 times the rest value, which is (E + |P|)²/m². At 90° it is 0.250/(32π² × 6.25) sr⁻¹ = 1.27 × 10⁻⁴ sr⁻¹, 1/γ² = 1/25 of the rest value.
Integrate: ∫dΩ_lab gives 2π × m²/(32π²) × ∫ d(cos θ)/(E − |P| cos θ)² over [−1, 1] = m²/(16π) × 2/(E² − |P|²) = m²/(16π) × 2/m² = 1/(8π). The laboratory density is wildly anisotropic, yet its integral is the invariant 0.0398.
So the rate in flight differs from the rest-frame rate only through its 1/(2E) prefactor: Γ_lab = Γ m/E = Γ/5, topic 5.03's time dilation with nothing borrowed from the phase space.

AnswerRest frame 3.17 × 10⁻³ sr⁻¹, isotropic; laboratory 0.310 sr⁻¹ forward and 1.27 × 10⁻⁴ sr⁻¹ at 90°; both integrate to 1/(8π) = 0.0398, and Γ_lab = Γ/5.