
Nuclear, particle & quantum-information sequence · Interactive lesson
Decay widths, phase space and lineshapes
Every unstable particle hands you one number — a lifetime, or equivalently a width — and this topic is how to take that number apart. How much of it is the matrix element, how much is the room the final state has to move in, and what is left over that names the interaction responsible.
UPV-17.08 · Lesson 8 of 9 in this unit. Position is not a completion score.Nuclear, particle & quantum-information sequence. These units continue into nuclear and particle physics and quantum information. Follow your university’s prerequisites, mathematical depth and assessment; this independent pathway is not a universal programme specification.
Build the model
At similar energy release, phase space shifts a width by factors of a few; the interaction shifts it by 10¹³.
The key idea
The total decay width Γ of an unstable state is the sum of its partial widths; it equals ħ divided by the mean lifetime, and it is the full width at half maximum of the Breit-Wigner lineshape that the state's own exponential decay produces.
Example. The Δ(1232) has Γ = 117 MeV, so τ = ħ/Γ = (6.582 × 10⁻²² MeV s)/(117 MeV) = 5.6 × 10⁻²⁴ s and cτ = ħc/Γ = (197.3 MeV fm)/(117 MeV) = 1.7 fm — in that time light itself barely crosses a nucleon.
The complete physical model and assumptions
A decay rate is not one thing but a product of two, and the whole skill here is separating them. Fermi's golden rule gives the rate out of a state as (2π/ħ)|⟨f|H′|i⟩|²ρ(Ef): a dynamical factor, the matrix element, and a kinematic factor, the density of final states. Made covariant, the second becomes the n-body Lorentz-invariant phase space dΦₙ = (2π)⁴δ⁴(P − Σpᵢ) Πᵢ d³pᵢ/[(2π)³2Eᵢ], and for a two-body decay it collapses to one number, the centre-of-mass momentum p*, fixed by the three masses alone. What survives is dΓ = |M|² dΦₙ/(2M), and every partial width in the Particle Data Group tables is one integral of that form. Summing the partial widths gives the total width Γ, whose reciprocal is the lifetime — because the Fourier transform of an amplitude falling as e(−Γt/2ħ) has a squared modulus that is a Lorentzian of full width Γ, so Γτ = ħ is the width of a transform pair and not a separate law. The cost of the split is that phase space is cheap to compute and the matrix element is not, and the two live on wildly different scales. At comparable energy release, kinematics moves a width by factors of a few, occasionally a few hundred; the matrix element separates the strong, electromagnetic and weak interactions by thirteen orders of magnitude. That is why a measured lifetime, once the kinematics is divided out, names the interaction that drove the decay.
1From the golden rule to an invariant width
Fermi's golden rule reads Γ = (2π/ħ)|⟨f|H′|i⟩|²ρ(Ef), and every relativistic decay formula is that line with both factors made covariant. Relativistic states are normalised as ⟨p′|p⟩ = 2E(2π)³δ³(p − p′), which puts a 1/2E with every particle and a 1/2M in front of the parent. The density of final states becomes the n-body Lorentz-invariant phase space dΦₙ = (2π)⁴δ⁴(P − Σpᵢ) Πᵢ d³pᵢ/[(2π)³2Eᵢ], and what survives is dΓ = |M|² dΦₙ/(2M), with |M|² the invariant amplitude squared, summed over final spins and averaged over initial ones. Two consequences are worth saying out loud. Each d³p/2E is invariant, so the integral may be done in whatever frame is convenient; and Γ is defined in the rest frame, which is exactly why a moving particle decays at Γ/γ in the laboratory. Everything else in this topic is the arithmetic of that one line.
2Two bodies: three masses fix everything but an angle
For 1 → 2 the delta function eats all six momentum integrals but one solid angle, and the momentum magnitude is fixed by the masses alone: p* = √([M²−(m₁+m₂)²][M²−(m₁−m₂)²])/(2M), the square root of the Källén function λ(M², m₁², m₂²) divided by 2M. The invariant volume is Φ₂ = p*/(4π√s), so an unpolarised spin-0 parent decays isotropically and Γ = |M|²p*/(8πM²). Numbers: Δ⁺⁺(1232) → pπ⁺ gives p* = 227 MeV/c, Λ(1116) → pπ⁻ gives 101 MeV/c, π⁺ → μ⁺ν gives 29.8 MeV/c. All the interesting behaviour is at threshold. As M falls towards m₁ + m₂, p* → 0 like √(M − m₁ − m₂), so a channel switches off smoothly rather than abruptly; an orbital barrier sharpens that, since an L-wave decay carries an extra p*(2L). The Δ → Nπ is P-wave, so its width runs as p*³ across the resonance — which is why a constant-width fit to it is already an approximation.
3Three bodies: a Dalitz plot and a power of Q
Three products leave two independent invariants after the delta function, conventionally m₁₂² and m₂₃², and the physical region is the Dalitz plot. A constant |M|² fills that region uniformly, so any structure on a Dalitz plot is dynamics, not kinematics — that is the whole reason the plot is drawn. Rates near threshold follow a power law: for n non-relativistic products the phase-space volume goes as Q((3n−5)/2), which returns the two-body √Q, while for beta decay, with a relativistic electron and neutrino, it becomes Sargent's rule Γ ∝ Q⁵. Muon decay is the clean test. The V−A calculation gives Γ = GF²mμ⁵/192π³; with GF = 1.1664 × 10⁻¹¹ MeV⁻² and mμ = 105.658 MeV that is 3.009 × 10⁻¹⁶ MeV, so τ = ħ/Γ = 2.187 μs against a measured 2.197 μs, the 0.4% gap being radiative corrections and the neglected electron mass. That fifth power is phase space and the mass dimension of GF together. It gives the τ lepton, 16.8 times heavier, an eνν width (mτ/mμ)⁵ = 1.35 × 10⁶ times the muon's; its other open channels, μνν and the hadronic modes, multiply that by 1/B(eνν) = 5.6, so the τ lives 7.6 × 10⁶ times less long.
4Partial widths add; lifetimes do not
With several channels open, each contributes a partial width and the total is their sum, Γ = ΣᵢΓᵢ. Rates add because probabilities per unit time add; lifetimes do not, so 1/τ = Σᵢ1/τᵢ and the observed lifetime is shorter than every partial lifetime — opening a second door can only speed the exit. The branching ratio is a rate fraction, Bᵢ = Γᵢ/Γ, and that is what the tables quote. Read a real one. The π⁰ lives 8.43 × 10⁻¹⁷ s, so Γ = ħ/τ = 7.81 eV; with B(γγ) = 98.82% and B(e⁺e⁻γ) = 1.17% the partial widths are 7.72 eV and 0.0914 eV. The partial lifetime of the Dalitz channel is ħ/(0.0914 eV) = 7.2 × 10⁻¹⁵ s, some 85 times the pion's actual life — the right number if you want to know how fast that one process is, and the wrong number to call a lifetime. Confusing the two is the commonest arithmetic slip in the subject.
5The lineshape is the Fourier transform of the decay
An unstable state has no sharp energy, and that is the same statement as its finite lifetime. Write the rest-frame amplitude as a(t) = e(−iM₀t/ħ)e(−Γt/2ħ) for t > 0, so |a|² = e(−Γt/ħ) and the mean life is τ = ħ/Γ. Its Fourier transform is ã(E) ∝ 1/(E − M₀ + iΓ/2), and the measured lineshape is |ã|² ∝ 1/[(E − M₀)² + Γ²/4] — a Lorentzian whose full width at half maximum is Γ. Watch the factor of two: the amplitude decays at Γ/2ħ, the probability at Γ/ħ. So Γτ = ħ is the width of a transform pair, with ħ = 6.582 × 10⁻²² MeV s and ħc = 197.327 MeV fm doing the arithmetic. Which measurement is even available depends on Γ. Above about an MeV you fit the lineshape; below about 10⁻³ eV you measure a flight distance, since cτ = ħc/Γ is 7.80 m for the π⁺, 2.68 cm for the KS, 455 μm for the B⁰ and 87 μm for the τ. In between, neither method reaches, and the lifetime must be got at indirectly.
6Divide out the kinematics and read off the interaction
Once phase space is divided out, a lifetime is a statement about |M|², and the three interactions occupy separate bands: strong decays near 10⁻²³ s, electromagnetic ones from 10⁻²¹ to 10⁻¹⁶ s, weak ones from 10⁻¹³ s upward. Δ(1232) → Nπ sits at 5.6 × 10⁻²⁴ s, π⁰ → γγ at 8.4 × 10⁻¹⁷ s, Λ → pπ⁻ at 2.6 × 10⁻¹⁰ s. The band names the selection rules that were in force, because only the weak interaction changes strangeness, violates parity or violates C — so a strange baryon decaying to non-strange products has no faster option. The Σ⁰ shows the logic at its sharpest. It could in principle go strongly to Λπ⁰, but it outweighs the Λ by only 76.96 MeV, well under mπ⁰ = 134.98 MeV, so the strong channel is closed by kinematics and Σ⁰ → Λγ runs electromagnetically at 7.4 × 10⁻²⁰ s. Two cautions. The bands are broad — within the weak class the muon lives 2.2 μs and the B⁰ 1.5 ps, six orders apart on m⁵ and CKM factors alone, and the top quark, with some 90 GeV to release, decays weakly in 5 × 10⁻²⁵ s — and a decay squeezed against threshold is slow in any interaction.
Key equations
Read the units, assumptions and limits before substituting.
The golden rule, made covariant
dΓ = |M|² dΦₙ / (2M)
Splits every rate into dynamics you must compute and kinematics the masses already fix.
M the parent mass in MeV; |M|² spin-summed and −averaged, of mass dimension 2 for a 1 → 2 decayLorentz-invariant phase space
dΦₙ = (2π)⁴ δ⁴(P − Σpᵢ) Πᵢ d³pᵢ/[(2π)³ 2Eᵢ]
The same integral in any frame, so Γ is unambiguous once you say it is the rest-frame width.
The 1/2Eᵢ per particle is what makes each measure invariant; δ⁴ removes four of the 3n integralsTwo-body momentum and width
p* = √([M²−(m₁+m₂)²][M²−(m₁−m₂)²])/2M , Γ = |M|² p*/(8πM²)
Three masses fix the entire kinematics: p* = 227 MeV/c for Δ → Nπ, 101 MeV/c for Λ → pπ⁻.
Masses in MeV, p* in MeV/c, Γ in MeV; the isotropic dΩ integral is already donePartial widths and branching ratios
Γ = Σᵢ Γᵢ , Bᵢ = Γᵢ/Γ , 1/τ = Σᵢ 1/τᵢ
Channels add as rates, never as lifetimes, so the total lifetime is shorter than every partial one.
Γᵢ in MeV or eV; Bᵢ dimensionless and summing to 1 over every open channelBreit-Wigner lineshape and Γτ = ħ
|A(E)|² ∝ 1/[(E − M₀)² + Γ²/4] , Γτ = ħ
One transform pair: an amplitude decaying at Γ/2ħ gives a Lorentzian of full width Γ.
ħ = 6.582 × 10⁻²² MeV s, ħc = 197.327 MeV fm; Γ is the FWHM in energyThe interaction hierarchy in one line
τ ~ 10⁻²³ s (strong) · 10⁻²¹–10⁻¹⁶ s (EM) · ≥ 10⁻¹³ s (weak)
Divide the phase space out and the residue names the interaction that drove the decay.
Δ(1232) 5.6 × 10⁻²⁴ s; Σ⁰ → Λγ 7.4 × 10⁻²⁰ s; Λ → pπ⁻ 2.6 × 10⁻¹⁰ sInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Set M₀ = 1232 MeV and Γ = 117 MeV — the Δ — and read τ = 5.63 × 10⁻²⁴ s off the reciprocal, with 11.5% of the Lorentzian already sitting below threshold. Now walk M₀ down towards 1080 MeV: that fraction climbs to 49%, and the constant-width Breit-Wigner has stopped being physical.
Live results
- LIFETIME τ = ħ/Γ
- 5.63 ×10⁻²⁴ s
- DECAY LENGTH cτ
- 1.69 fm
- TWO-BODY p* AT M₀
- 227.2 MeV/c
- BW AREA BELOW THRESHOLD
- 11.5 %
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Breit-Wigner width Γ: 117 MeV; Resonance mass M₀: 1232 MeV
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The usual error is to explain the lifetime hierarchy with phase space: a decay is slow, the reasoning goes, because there is little energy to share out. Test it. Δ⁺⁺ → pπ⁺ releases Q = 154 MeV and leaves p* = 227 MeV/c; Λ → pπ⁻ releases Q = 37.8 MeV and leaves p* = 101 MeV/c. Two-body phase space goes as p*, so it differs by a factor 2.26 — and even charging the P-wave Δ its full p*³ barrier gives only 11.5. The measured widths differ by 117 MeV against 2.5 × 10⁻¹² MeV, a factor of 4.7 × 10¹³. Kinematics covers 11.5 of that; the other 4 × 10¹² is |M|², small for the Λ because the decay must change strangeness, which no strong or electromagnetic vertex can do. Phase space rescales a width; the interaction chooses its order of magnitude.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
The Δ(1232) appears in the πN invariant-mass spectrum as a bump of full width at half maximum Γ = 117 MeV. Find its mean lifetime and its cτ, and say whether a vertex detector resolving 5 μm could ever see it decay away from the production point.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- The FWHM of a Breit-Wigner is the total width itself, so Γ = 117 MeV is read straight off the fit — no extra constant is involved.
- τ = ħ/Γ = (6.582 × 10⁻²² MeV s)/(117 MeV) = 5.63 × 10⁻²⁴ s.
- cτ = ħc/Γ = (197.327 MeV fm)/(117 MeV) = 1.687 fm, the same result reached without ever converting to seconds.
- Even at γ = 100 the flight distance is βγcτ ≈ 169 fm = 1.7 × 10⁻¹³ m, about 3 × 10⁷ times below a 5 μm vertex resolution.
- Compare with the light-crossing time of the hadron itself, 1 fm/c = 3.3 × 10⁻²⁴ s: the Δ survives about 1.7 of those, so it is a pole in the amplitude rather than a particle that flies.
Answer. τ = 5.63 × 10⁻²⁴ s and cτ = 1.69 fm. Only the lineshape can measure this width; no detector will ever resolve the flight.
MediumExample 2
The π⁰ has mean life 8.43 × 10⁻¹⁷ s and decays 98.82% to γγ and 1.17% to e⁺e⁻γ. Find its total width, both partial widths, and the partial lifetime of the Dalitz channel. Then name the interaction responsible and say why the strong interaction is not.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Γ = ħ/τ = (6.582 × 10⁻²² MeV s)/(8.43 × 10⁻¹⁷ s) = 7.81 × 10⁻⁶ MeV = 7.81 eV.
- Partial widths are branching fractions of the total: Γ(γγ) = 0.9882 × 7.81 eV = 7.72 eV and Γ(e⁺e⁻γ) = 0.0117 × 7.81 eV = 0.0914 eV. They add back to 7.81 eV, as partial widths must.
- Partial lifetime of the Dalitz mode: τᵢ = ħ/Γᵢ = (6.582 × 10⁻¹⁶ eV s)/(0.0914 eV) = 7.2 × 10⁻¹⁵ s, some 85 times the pion's actual life — what it would live if that channel were the only one open.
- Both final states are photons and charged leptons, and 8.43 × 10⁻¹⁷ s sits squarely in the electromagnetic band, so the decay is electromagnetic.
- The strong interaction is shut out by kinematics and by the final state together: it has no vertex turning a quark-antiquark pair into photons, and m(π⁰) = 134.98 MeV cannot produce even one more pion, so every hadronic channel is below threshold.
Answer. Γ = 7.81 eV; Γ(γγ) = 7.72 eV and Γ(e⁺e⁻γ) = 0.0914 eV; the Dalitz partial lifetime is 7.2 × 10⁻¹⁵ s. The decay is electromagnetic.
HardExample 3
The π⁺ decays to μ⁺νμ with branching essentially 1 and to e⁺νₑ with 1.230 × 10⁻⁴. Taking mπ = 139.570, mμ = 105.658 and mₑ = 0.511 MeV, find the two-body momentum in each mode, show that phase space alone favours the electron mode, extract the ratio of squared matrix elements, and compare with the tree-level V−A result Γ ∝ m_ℓ²(1 − m_ℓ²/mπ²)².
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Neglecting the sub-eV neutrino mass, p* = (mπ² − m_ℓ²)/(2mπ). Muon mode: (19479.8 − 11163.6)/279.14 = 8316.2/279.14 = 29.79 MeV/c. Electron mode: (19479.8 − 0.26)/279.14 = 69.78 MeV/c.
- Two-body phase space goes as p*, so kinematics alone favours the electron mode by 69.78/29.79 = 2.342. It is instead observed 1/(1.230 × 10⁻⁴) = 8130 times slower.
- Since Γ = |M|²p*/(8πmπ²), the dynamics carries the whole discrepancy: |Mₑ|²/|Mμ|² = (Γₑ/Γμ)(p*μ/p*ₑ) = 1.230 × 10⁻⁴ × 0.4269 = 5.25 × 10⁻⁵, a suppression of 1.9 × 10⁴.
- Tree level predicts Γₑ/Γμ = (mₑ/mμ)²[(1 − mₑ²/mπ²)/(1 − mμ²/mπ²)]² = 2.339 × 10⁻⁵ × [0.999987/0.42691]² = 2.339 × 10⁻⁵ × 5.487 = 1.283 × 10⁻⁴.
- Measured is 1.230 × 10⁻⁴, so the tree-level number is 4.3% high; radiative corrections close the gap. The m_ℓ² is helicity suppression: a spin-0 parent forces the antilepton into the neutrino's helicity, which the V−A current supplies only at order m_ℓ/E.
Answer. p* = 29.79 and 69.78 MeV/c; phase space favours the electron mode 2.34 to 1, while |Mₑ|²/|Mμ|² = 5.25 × 10⁻⁵. Tree level gives 1.283 × 10⁻⁴ against 1.230 × 10⁻⁴ measured.
Quick check
Select an answer and explain it before checking.
Review & continue
The total decay width Γ of an unstable state is the sum of its partial widths; it equals ħ divided by the mean lifetime, and it is the full width at half maximum of the Breit-Wigner lineshape that the state's own exponential decay produces.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.