University Quantum Field Theory I · Cross-Sections, Decay Rates and Unitarity · 5.3

Decay rates, and the scalar's decay to ψψ̄ in seconds

This is where an amplitude first becomes a clock. We take the one-particle case of the rate formula, watch time dilation fall out of a normalisation factor instead of being put in, and carry the scalar Yukawa decay from ℳ = −g to a width in MeV, a lifetime in seconds and a decay length in femtometres.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

A decay is the simplest rate quantum field theory computes, because only one particle comes in. Put topic 5.01's box around it: the squared momentum delta becomes one delta times VT, the initial norm 2EV absorbs the V, each final particle's norm cancels against its count of states, and what is left per unit time is dΓ = |ℳ|² dΠₙ/(2E). Both |ℳ|² and dΠₙ are Lorentz invariant, so a decay's whole frame dependence sits in that 1/(2E): a particle of energy E = γm decays at Γ/γ, time dilation obtained from a state normalisation.

At rest, two-body phase space gives Γ = |ℳ|²|p_f|/(8πm²), and with topic 2.09's ℳ = −g the scalar Yukawa decay is Γ(φ → ψψ̄) = g²β/(16πm), where β = √(1 − 4M²/m²) is the daughters' speed and M the ψ mass. It vanishes at threshold with the final momentum, carries no 1/2 because ψ and ψ̄ are different particles, falls as 1/m for a heavy parent because g carries a power of mass, and becomes a lifetime through ħ = 6.582 × 10⁻²⁵ GeV s. The model stops where its assumptions do: Γ is a lowest-order rate for a particle treated as if it were a stable in-state, trustworthy only while Γ ≪ m, and the exponential survival law that makes ħ/Γ a mean life is quoted from Weisskopf-Wigner theory (UAQM-12.06), not derived here.

Simple definition: A particle's decay rate Γ is its probability per unit time of turning into the final states open to it, Γ = (1/(2E))∫|ℳ|² dΠₙ summed over channels, and at rest its mean life is τ = ħ/Γ.
Example: With g = 0.50 GeV, m = 2.0 GeV and M = 0.80 GeV the daughters leave at β = √(1 − 0.64) = 0.60, so Γ(φ → ψψ̄) = 0.25 × 0.60/(16π × 2.0) GeV = 1.49 MeV and τ = 6.582 × 10⁻²² MeV s/1.49 MeV = 4.4 × 10⁻²² s.

Decay rate of one particledΓ = |ℳ|² dΠₙ/(2E)

Derived from topic 5.01's box: the initial norm 2EV leaves 1/(2E) once every V and T has cancelled. Integrate over all final momenta, with 1/n! for n identical daughters; at rest E = m.

E the parent's energy and Γ, both in GeV; ℳ of mass dimension 3 − n; dΠₙ topic 5.02's invariant phase space

Time dilation from the normalisationΓ_lab = (m/E) Γ = Γ/γ

Derived, not assumed: |ℳ|² and dΠₙ are invariant, so 1/(2E) is the only frame-dependent factor in a decay rate.

Γ the rest-frame rate in GeV, E = γm the lab energy in GeV; mean decay length (|p|/m) ħc/Γ, in fm for Γ in MeV

Two-body decay at restΓ = S |ℳ|² |p_f|/(8πm²)

Derived from ∫dΠ₂ = |p_f|/(4πm), quoted from topic 5.02. A spin-0 parent decaying to two spin-0 daughters always has a constant ℳ.

|p_f| the daughter momentum and m the parent mass, in GeV; S = 1/2 for identical daughters, 1 otherwise; |ℳ|² independent of angle

The scalar Yukawa decay φ → ψψ̄Γ = (g²/(16πm)) √(1 − 4M²/m²)

Derived with ℳ = −g. The root is the daughters' speed β, so Γ vanishes at threshold as √(m − 2M) and falls as 1/m once m ≫ M.

g the φψ*ψ coupling, m the φ mass, M the ψ mass, all in GeV; S = 1, because ψ and ψ̄ differ

Width to lifetimeτ = ħ/Γₜₒₜ

Γ = 1 MeV is τ = 6.58 × 10⁻²² s and cτ = 197 fm. The total width sets the single lifetime that every channel shares.

ħ = 6.582 × 10⁻²⁵ GeV s, so Γ in GeV gives τ in s; cτ = ħc/Γₜₒₜ with ħc = 197.327 MeV fm

Branching fractionBᵢ = Γᵢ/Γₜₒₜ

Widths add because different channels are orthogonal final states. Opening a channel shortens τ and lowers every other Bᵢ, leaving the other Γᵢ untouched.

Γᵢ the partial width into channel i, Γₜₒₜ = Σ Γᵢ, both in GeV; Bᵢ dimensionless, summing to 1

One particle in: what the box leaves behind

Work in natural units, ħ = c = 1, with the metric (+, −, −, −). Put topic 5.01's box of volume V, switched on for a time T, around one particle of momentum p decaying to n particles. The amplitude is ⟨f|iT|p⟩ = (2π)⁴δ⁴(p − Σpⱼ) iℳ, with ℳ topic 4.01's invariant amplitude, written in script because M is the ψ mass below, and its square carries (2π)⁴δ⁴ × VT. Divide by the norms, ⟨p|p⟩ = 2EV for the parent and 2EⱼV for each daughter; multiply by the number of final states, V d³pⱼ/(2π)³ per daughter; divide by T. The V from the delta function cancels the V in the parent's norm, each daughter's V cancels against its own count, and T goes, leaving dΓ = |ℳ|² dΠₙ/(2E), derived. Check the dimensions: ℳ for 1 → n has mass dimension 3 − n and dΠₙ has 2n − 4, so Γ carries −1 + 2(3 − n) + (2n − 4) = 1, a mass, which in natural units is a rate.

In flight, the 1/(2E) is time dilation

Nothing in |ℳ|² or dΠₙ knows the frame: the amplitude is built from invariants, and dΠₙ is invariant by topic 5.02. So the number of decays per unit volume per unit time, |ℳ|² dΠₙ in this normalisation, is invariant, and it is shared among 2E particles per unit volume, a density that grows with E because lengths contract along the motion. Hence a particle of energy E = γm decays at Γ_lab = (m/E)Γ = Γ/γ. That is time dilation, derived and not put in. Between production and decay it travels v × γτ = (|p|/m)cτ, the mean decay length. Measured check, quoted: CERN's muon storage ring held muons at γ = 29.33 and measured a lab lifetime of 64.4 μs, against γτ₀ = 29.33 × 2.197 μs = 64.44 μs, agreement to about 0.1%. The muon's decay is weak and three-body, which does not matter, because the argument never used ℳ or n.

Two bodies at rest: Γ(φ → ψψ̄) = g²β/(16πm)

In the scalar Yukawa model a real φ of mass m couples through −gψ*ψφ to a complex ψ of mass M, and topic 2.09 found ℳ(φ → ψψ̄) = −g, a constant. At rest E = m, and topic 5.02's two-body phase space, quoted, gives ∫dΠ₂ = |p_f|/(4πm) over the full 4π of solid angle. So Γ = (1/(2m)) g² |p_f|/(4πm) = g²|p_f|/(8πm²). Energy conservation, m = 2√(|p_f|² + M²), fixes |p_f| = (m/2)√(1 − 4M²/m²), and since each daughter has energy m/2, that root is its speed β = |p_f|/E_f. Hence Γ(φ → ψψ̄) = g²β/(16πm), derived. There is no 1/2: ψ and ψ̄ carry opposite ψ-number, so ψ at p with ψ̄ at −p is a different state from ψ at −p with ψ̄ at p, and the angular integral must count both.

Four checks: dimensions, threshold, heavy parent, narrowness

Dimensions: a cubic scalar coupling in four dimensions has mass dimension 1 (topic 2.01), so g²/m is a mass and Γ a rate. Threshold: write Q = m − 2M for the energy released. For Q ≪ M each daughter carries kinetic energy Q/2 = Mβ²/2, so β ≈ √(Q/M) and Γ ∝ √Q, the s-wave threshold law of a constant ℳ times phase space growing as |p_f|. Massless limit: for M ≪ m, β → 1 and Γ → g²/(16πm), which falls as the parent gets heavier, because the only rate g and m can build is g²/m. At fixed M the two trends meet at m = 2√2 M, where β = 1/√2 and Γ is largest. Narrowness: Γ/m = (g/m)²β/(16π), so even g = m gives Γ/m ≤ 1/(16π) = 2.0%. The 16π of two-body phase space is why tree-level widths come out narrow, and why treating φ as a particle at all is self-consistent here.

From GeV to seconds and femtometres

A width in GeV becomes a time through ħ = 6.582 × 10⁻²⁵ GeV s and a length through ħc = 197.327 MeV fm: Γ = 1 MeV means τ = ħ/Γ = 6.58 × 10⁻²² s and cτ = 197 fm. Take the figure's default, g = 1 GeV, m = 1.2 GeV and M = 0.30 GeV. Then 4M²/m² = 0.25, β = 0.866 and Γ = 0.866/(16π × 1.2) GeV = 14.4 MeV, so τ = 4.6 × 10⁻²³ s and cτ = 13.7 fm. Made with |p| = 2 GeV it travels (|p|/m)cτ = 1.67 × 13.7 fm = 22.9 fm before decaying. No tracker resolves that, so such a φ is seen only as a peak in the ψψ̄ mass spectrum, topic 5.09's Breit-Wigner; the measured ρ(770), quoted, has a 149 MeV width and cτ = 1.32 fm. At the other extreme a width of 10⁻¹⁴ GeV gives cτ = 2.0 cm, which is why weak decays leave visible gaps between vertices.

Many channels, one lifetime, and what Γ is not

When φ can decay several ways, each channel i has its own partial width Γᵢ, computed as above with its own ℳ and phase space. Final states of different channels are orthogonal, so the probabilities per unit time add: Γₜₒₜ = Σ Γᵢ, the branching fraction is Bᵢ = Γᵢ/Γₜₒₜ, and there is one mean life, ħ/Γₜₒₜ, for every channel. A φ carries no separate clock for each fate. Opening a channel shortens τ and lowers every other Bᵢ while leaving the other Γᵢ unchanged, and a boost divides every Γᵢ by the same γ, so branching fractions are frame-independent. Two limits on what Γ means: it is a lowest-order rate, computed as if φ were a stable in-state, which topic 4.09's LSZ does not strictly allow; and that survival then falls as e^(−Γt) is UAQM-12.06's Weisskopf-Wigner result, quoted, not derived. Topic 5.09 meets the same Γ in the φ self-energy at its pole, Im Σ = −mΓ.

Change one variable at a time

Make the relationship visible.

Interactive model

ψ mass M0.30 GeV

Energy released Q = m − 2M0.60 GeV

Lab momentum of φ, p2.00 GeV

Drag Q down to 0.02 GeV and β and Γ collapse together as √Q: the rate dies with the daughters' momentum. Push Q past the peak at m = 2√2 M and Γ falls as 1/m, since g carries a mass. Raise p and the dashed in-flight curve sinks by m/E, most for light parents.

DAUGHTER SPEED β0.866

REST WIDTH Γ14.36 MeV

MEAN LIFE τ = ħ/Γ4.58 10⁻²³ s

LAB DECAY LENGTH22.9 fm

Live interpretationDAUGHTER SPEED β: 0.866. REST WIDTH Γ: 14.36 MeV. MEAN LIFE τ = ħ/Γ: 4.58 10⁻²³ s. LAB DECAY LENGTH: 22.9 fm

Catch the common trap

Explain before calculating.

Misconception check

The tempting error is a 1/2 in Γ(φ → ψψ̄), because the daughters share the mass M and fly apart back to back. The 1/n! corrects an overcount of identical particles: integrating each daughter over every direction counts |p, −p⟩ and |−p, p⟩ separately, and for two identical bosons those are one and the same state. ψ and ψ̄ are not identical. They carry opposite ψ-number, the conserved charge of the complex field, so ψ going up with ψ̄ going down is a different final state from the reverse; a detector that tells the charges apart records two distinguishable events, and the angular integral must count both. Equal masses are irrelevant, identical quantum numbers are what matter, and the spurious 1/2 halves Γ and doubles τ. The 1/2 does belong to φ → χχ for a real scalar χ coupled through −(μ/2)φχ², where the two contractions give ℳ = −μ and the rate keeps the 1/2.

In the scalar Yukawa model the coupling g and the ψ mass M are held fixed while the φ mass is doubled, from m = 4M to m = 8M. By what factor does the width Γ(φ → ψψ̄) change?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasyIn the scalar Yukawa model take a φ of mass m = 2.5 GeV, a ψ of mass M = 0.70 GeV and g = 0.60 GeV. Compute Γ(φ → ψψ̄) in MeV, the mean life in seconds and cτ in femtometres, and check that the width is narrow enough for a lowest-order rate to make sense.

Daughter speed: 4M²/m² = 4(0.49)/6.25 = 0.3136, so β = √0.6864 = 0.8285 and |p_f| = (m/2)β = 1.036 GeV.
Rate, with no 1/2 since ψ and ψ̄ differ: Γ = g²β/(16πm) = (0.36 GeV²)(0.8285)/(125.66 GeV) = 2.373 × 10⁻³ GeV = 2.37 MeV.
Seconds and femtometres: τ = ħ/Γ = 6.582 × 10⁻²⁵ GeV s/2.373 × 10⁻³ GeV = 2.77 × 10⁻²² s, and cτ = 197.327 MeV fm/2.373 MeV = 83.1 fm.
Checks: g²/m is a mass, so Γ has the dimension of a rate. With M → 0 the width would be 2.86 MeV, so the threshold factor costs 17%. Γ/m = 9.5 × 10⁻⁴, narrow, so φ behaves as a particle and the lowest-order rate is self-consistent.

AnswerΓ ≈ 2.37 MeV, τ ≈ 2.8 × 10⁻²² s and cτ ≈ 83 fm, with Γ/m ≈ 10⁻³.

MediumA feebly coupled φ of mass 0.80 GeV decays only to ψψ̄, with M = 0.25 GeV and g = 3.0 × 10⁻⁷ GeV, and is produced with lab momentum 12 GeV. Using the 1/(2E) in the rate, find its lab decay rate and mean decay length, then the fraction that decay inside a detector shell from 1.0 m to 3.0 m from the production point, taking the survival law e^(−Γt) as quoted.

At rest: 4M²/m² = 0.3906, so β = 0.7806 and Γ = g²β/(16πm) = (9.0 × 10⁻¹⁴ GeV²)(0.7806)/(40.21 GeV) = 1.747 × 10⁻¹⁵ GeV. Then τ = ħ/Γ = 3.77 × 10⁻¹⁰ s and cτ = ħc/Γ = 0.113 m.
In the lab |ℳ|² = g² and dΠ₂ are unchanged; only 1/(2E) changes, with E = √(12² + 0.80²) GeV = 12.03 GeV. So Γ_lab = Γ m/E = 1.747 × 10⁻¹⁵ GeV × 0.0665 = 1.16 × 10⁻¹⁶ GeV, a lab mean life of 5.66 × 10⁻⁹ s, γ = 15.0 times the rest value.
Mean decay length: L = v × (lab mean life) = (|p|/m)cτ = 15.0 × 0.113 m = 1.69 m.
With the quoted exponential law the probability of surviving a distance d is e^(−d/L), so the fraction decaying between 1.0 m and 3.0 m is e^(−0.590) − e^(−1.771) = 0.554 − 0.170 = 0.384.
Checks: Γ/m = 2 × 10⁻¹⁵, so lowest order is safe. Without time dilation the mean path would be only about cτ = 0.113 m, and e^(−8.85) = 1.4 × 10⁻⁴ of the φ would reach 1 m: the detector would see almost nothing.

AnswerΓ_lab ≈ 1.16 × 10⁻¹⁶ GeV and L ≈ 1.69 m, so about 38% of the φ decay between 1.0 m and 3.0 m.

HardThe K_S meson (spin 0, m = 497.61 MeV) has mean life 8.954 × 10⁻¹¹ s and branching fractions 69.20% to π⁺π⁻ and 30.69% to π⁰π⁰, all measured and quoted; m_π± = 139.57 MeV, m_π⁰ = 134.98 MeV. Extract |ℳ| for each channel in eV, and test the quoted isospin expectation that a ΔI = 1/2 decay has |ℳ(π⁰π⁰)| = |ℳ(π⁺π⁻)|.

Total width from the lifetime: Γₜₒₜ = ħ/τ = 6.582 × 10⁻¹⁶ eV s/8.954 × 10⁻¹¹ s = 7.351 × 10⁻⁶ eV. Partial widths: Γ₊₋ = 0.6920 Γₜₒₜ = 5.087 × 10⁻⁶ eV and Γ₀₀ = 0.3069 Γₜₒₜ = 2.256 × 10⁻⁶ eV.
Daughter momenta from |p_f| = (m/2)√(1 − 4m_π²/m²): 4m_π²/m² = 0.3147 for π⁺π⁻ and 0.2943 for π⁰π⁰, so p₊₋ = 206.0 MeV and p₀₀ = 209.0 MeV.
Spin 0 to two spin 0 allows only an s-wave, so each ℳ is a constant and Γ = S|ℳ|²|p_f|/(8πm²). For π⁺π⁻, S = 1: |ℳ₊₋|² = 8πm²Γ₊₋/p₊₋ = 1.537 × 10⁵ eV², so |ℳ₊₋| = 392 eV.
For π⁰π⁰ the daughters are identical, S = 1/2: |ℳ₀₀|² = 2 × 8πm²Γ₀₀/p₀₀ = 1.343 × 10⁵ eV², so |ℳ₀₀| = 367 eV. Dropping the 1/2 would give 259 eV.
Compare: with the 1/2 the ratio |ℳ₀₀|/|ℳ₊₋| = 0.935, within 7% of isospin's 1; without it the ratio is 0.661, close to 1/√2, a failure by √2. The remaining 7% is the small ΔI = 3/2 amplitude and isospin breaking, quoted. Dimensions: a 1 → 2 amplitude carries one power of mass, and |ℳ|/m ≈ 8 × 10⁻⁷ measures how weak the decay is.

Answer|ℳ(π⁺π⁻)| ≈ 392 eV and |ℳ(π⁰π⁰)| ≈ 367 eV, equal to 7% as isospin predicts once the 1/2 for identical pions is kept; without it the π⁰π⁰ amplitude would be 259 eV.