University Quantum Field Theory I · Cross-Sections, Decay Rates and Unitarity · 5.9
Unstable particles: the width and the Breit-Wigner peak
This is the lesson where a pole learns to decay. We take the ψ loop, already complex above threshold, let its imaginary part replace the iε in the φ propagator, and watch a tree cross-section that blew up at √s = m turn into a finite peak whose height no coupling can change.
Build the model
Connect the measurement to the mechanism.
The tree amplitude for ψψ̄ → φ → ψψ̄ is −g²/(s − m²), infinite at s = m², and the cross-section built from it breaks the s-wave unitarity limit wherever |s − m²| is below about mΓ. The cure needs no new physics, only diagrams already present. Every self-energy insertion on the s-channel φ is the ψ loop, and above the ψψ̄ threshold that loop is complex: unitarity of S, applied to one φ, fixes Im Σ(m²) = −mΓ, with Γ exactly topic 5.03's decay rate.
Summing the chain as in topic 3.08 and absorbing Re Σ into the physical mass turns the propagator into i/(p² − m² + imΓ). The iε of a stable particle has become a finite +imΓ of the same sign, the pole has moved off the real axis to √s ≈ m − iΓ/2, and in time that pole is the exponential decay e(−Γt). The cross-section now peaks with full width Γ, and at the top the coupling cancels: σ = 4π/p², the s-wave limit, reached because all of Γ returns to ψψ̄.
Near the peak s − m² ≈ 2m(E − m) and the line is the non-relativistic Breit-Wigner. The construction holds to leading order in Γ/m and near the peak; far from it the energy dependence of Σ, the t-channel exchange and the residue Z return, and φ itself, never reaching t → ±∞, has no LSZ formula and lives only as an internal line.
- Simple definition
- An unstable particle's width enters its propagator through the imaginary part of its self-energy, i/(p² − m² + imΓ), so a cross-section through it peaks at √s = m with full width Γ instead of diverging.
- Example
- With m = 0.80 GeV, M = 0.20 GeV and g = 0.50 GeV the ψ loop gives Im Σ(m²) = −4.31 × 10⁻³ GeV², so Γ = 5.38 MeV; ψψ̄ → ψψ̄ then peaks at √s = 0.800 GeV, 5.38 MeV wide at half height, at 4π/p² = 105 GeV⁻² = 40.8 mb.
Derived from unitarity: −Σ is the forward φ → φ amplitude, and 2 Im ℳ = ∫|ℳ(φ → f)|² dΠf = 2mΓ. Below the ψψ̄ threshold Im Σ = 0 and φ is stable.
Σ and mΓ in GeV²; m the φ pole mass and Γ its total decay rate, both in GeV; at order g², Σ is the ψ loop
Derived from topic 3.08's chain. The +iε of a stable particle becomes a finite +imΓ of the same sign, and the pole moves to √p² ≈ m − iΓ/2.
p² in GeV², D in GeV⁻²; m the pole mass with Re Σ absorbed; valid while |p² − m²| is of order mΓ and Γ ≪ m
Derived from ℳ = −g²/(s − m² + imΓ) and topic 5.04's dσ/dΩ = |ℳ|²/(64π²s). Far from the peak it returns the tree result g⁴/(16πs(s − m²)²).
g in GeV, s in GeV², σ in GeV⁻² (× 0.3894 mb GeV²); isotropic, and no 1/2 because ψ and ψ̄ differ
Derived: the g⁴ of |ℳ|² cancels the g⁴ in m²Γ². It is the s-wave unitarity limit, reached because all of Γ returns to ψψ̄; g sets only the width and area.
p = (m/2)√(1 − 4M²/m²), the centre-of-mass momentum at √s = m, in GeV; σ in GeV⁻²
UQMII-15.07's resonance and UAQM-12.07's Lorentzian, quoted, are this limit. Integrated over E the peak has area (π/2)Γσₚₑₐₖ, which grows as g².
E = √s in GeV; from s − m² ≈ 2m(E − m), good to order (E − m)/m; full width at half maximum Γ
Derived as for one channel: every channel feeds Im Σ, so Γ is the total. Spin adds (2J + 1)/((2s₁ + 1)(2s₂ + 1)), quoted; the Z peak gives 41.5 nb.
pᵢ the initial centre-of-mass momentum in GeV; Bᵢ, Bf the branching fractions Γᵢ/Γ and Γf/Γ; spin 0 throughout
Above threshold the ψ loop has an imaginary part
Work in natural units, ħ = c = 1, with the metric (+, −, −, −). In the scalar Yukawa model, −gψ*ψφ, the only order-g² 1PI insertion on the φ line is the ψ loop, which topic 3.08 called −iΣ(p²). Read as an amplitude, −iΣ = iℳ(φ → φ), so ℳ(φ → φ) = −Σ. Topic 5.07's unitarity argument never used the number of incoming particles: for one φ at rest, 2 Im ℳ(φ → φ) = Σf ∫|ℳ(φ → f)|² dΠf, which topic 5.03 named 2mΓ. Hence Im Σ(m²) = −mΓ, derived. At this order f is ψψ̄ with ℳ = −g and ∫dΠ₂ = β/(8π), so Im Σ(p²) = −(g²/(16π))√(1 − 4M²/p²) above p² = 4M² and zero below; topic 5.08's cut gives the same, without the 1/2 of φ⁴'s identical pair. Both sides are GeV². With g = m = 1 GeV and M = 0.25 GeV, Im Σ = −0.0172 GeV² and Γ = 17.2 MeV. The real part diverges; the imaginary part is finite.
Resum the chain: the iε becomes +imΓ
Topic 3.08 summed the insertions into D(p²) = i/(p² − m₀² − Σ(p²) + iε). Split Σ into Re Σ + i Im Σ. The real part, divergent and slowly varying, is absorbed by defining the pole mass through m² = m₀² + Re Σ(m²), which Unit 6 makes the on-shell renormalisation condition; its slope at the pole changes the residue by Z = 1 + O(g²), a change to ℳ at order g⁴, dropped here. Near the pole what is left is p² − m² − i Im Σ(m²) = p² − m² + imΓ, so D(p²) ≈ i/(p² − m² + imΓ), derived. The sign is not a convention. The Feynman iε, put in by hand in topic 1.06, has become a finite imaginary part of the same sign: iε is the Γ → 0⁺ limit of a width, and the opposite sign would describe a state that grows in time. For a φ lighter than 2M, Im Σ(m²) = 0 and only the infinitesimal iε survives.
The pole leaves the real axis, and that is decay
The denominator vanishes at p² = m² − imΓ, so √p² = m√(1 − iΓ/m) ≈ m − iΓ/2; strictly that pole lies on the second sheet, reached by continuing down through the ψψ̄ cut. Fourier transform the propagator at rest over p⁰: for positive t the contour closes below and picks up p⁰ = m − iΓ/2, giving e(−imt)e(−Γt/2)/(2m). The amplitude for the φ to remain falls as e(−Γt/2) and the probability as e(−Γt), UAQM-12.06's exponential law recovered as a pole position, with small non-exponential pieces from the cut at very early and very late times, quoted. At momentum p the pole sits at p⁰ ≈ Eₚ − imΓ/(2Eₚ), a rate Γm/Eₚ: topic 5.03's time dilation again. For Γ = 17.2 MeV, τ = 3.82 × 10⁻²³ s and cτ = 11.5 fm. Such a φ never reaches t → ±∞, so it has no in or out states and no LSZ formula, topic 4.09's gap: it appears only as an internal line between stable ψ, which carry the conserved ψ-number.
At the peak g cancels and σ reaches 4π/p²
For ψψ̄ → ψψ̄ the s-channel φ now gives ℳ = −g²/(s − m² + imΓ). The t-channel exchange remains, but |ℳₜ| ≤ g²/m² while at the peak |ℳₛ| = g²/(mΓ), so it is smaller by at least Γ/m and is dropped. The pair ψψ̄ is not identical and ℳ is isotropic, so topic 5.04 gives σ = |ℳ|²/(16πs) = g⁴/(16πs[(s − m²)² + m²Γ²]), derived; it is half its top where |s − m²| = mΓ, a full width Γ in √s. At s = m² insert Γ = g²β/(16πm): σ = g⁴/(16πm⁴Γ²) = 16π/(m²β²), and with p = mβ/2 there, σₚₑₐₖ = 4π/p², derived, the coupling gone. With M = 0.25 GeV and m = 1 GeV, p² = 0.1875 GeV² and σₚₑₐₖ = 67.0 GeV⁻² = 26.1 mb, by (ħc)² = 0.3894 GeV² mb. Limits: for |s − m²| ≫ mΓ the tree's g⁴/(16πs(s − m²)²) returns, and as g → 0 the peak keeps its height and narrows, its area (π/2)Γσₚₑₐₖ vanishing as g².
Why 4π/p²: the s wave on the unitarity circle
A spin-0 φ in the s channel makes the amplitude isotropic, so f = ℳ/(8π√s) is a pure s wave, and partial-wave unitarity (UQMII-15, quoted) writes an elastic s wave as pf = e(iδ)sin δ: |f| ≤ 1/p and σ ≤ 4π/p². With s − m² ≈ 2m(E − m) the resummed amplitude becomes pf = (Γ/2)/(m − E − iΓ/2), derived. As E rises through m that point runs round the circle of radius 1/2 centred on i/2 and reaches pf = i, δ = π/2, at E = m, the top. Topic 5.07's optical theorem agrees: (4π/p)Im f(0) = 4π/p², the σ found directly. The tree amplitude is real and breaks |f| ≤ 1/p wherever |s − m²| falls below g²p/(8π√s) ≈ mΓ, inside the half-height band: resummation repairs perturbation theory exactly where it failed. Keeping Im Σ(s) = −(g²/(16π))β(s) at every s, as the figure does, puts the amplitude on the circle at all energies, because any real term in the denominator leaves it there.
More channels, the Z line, and where it stops
If φ has other channels, each adds to Im Σ, so the propagator carries the total Γ while the vertices carry the couplings of the channels in and out. The same algebra gives σₚₑₐₖ(i → f) = (4π/pᵢ²)BᵢBf, derived for spin 0: the elastic peak drops to Bᵢ² of the limit and the total to Bᵢ. With spin, a factor (2J + 1)/((2s₁ + 1)(2s₂ + 1)) is quoted, 3/4 for a spin-1 Z formed by spin-½ e⁺e⁻. Then σₚₑₐₖ(e⁺e⁻ → hadrons) = (12π/MZ²)Bₑₑ Bhad, and MZ = 91.1876 GeV, Bₑₑ = 3.363% and Bhad = 69.91% give 41.5 nb against LEP's measured 41.541 nb, a consistency check since LEP's fit uses this relation; the measured line is ΓZ = 2.4952 GeV wide, Γ/M = 2.7%. The construction stops where its approximations do: it needs Γ ≪ m, √s near the peak, one isolated resonance, and no threshold close by, where Im Σ(s) changes quickly and the width is no longer a constant.
Change one variable at a time
Make the relationship visible.
Drag g from 0.5 to 2.5 GeV: Γ grows 25-fold but the top stays pinned to 4π/p², so the coupling buys width, not height; near 2.5 GeV the line skews as Γ/m reaches 10%. Raise M to lift the peak. Put the probe at m ± Γ/2: half the peak, where the dashed tree crosses the limit.
WIDTH Γ17.2 MeV
PEAK 4π/p² AT √s = m26.1 mb
σ AT THE PROBE26.10 mb
σ ÷ 4π/p² AT THE PROBE1.000
Live interpretationWIDTH Γ: 17.2 MeV. PEAK 4π/p² AT √s = m: 26.1 mb. σ AT THE PROBE: 26.10 mb. σ ÷ 4π/p² AT THE PROBE: 1.000
Catch the common trap
Explain before calculating.
In the scalar Yukawa model, with the masses m and M held fixed, the coupling g is tripled. What happens to the peak that ψψ̄ → φ → ψψ̄ shows in the cross-section near √s = m?
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 = 1.4 GeV, a ψ of mass M = 0.50 GeV and g = 1.2 GeV. Use the ψ loop's imaginary part to find Γ, locate the pole of the resummed propagator in √s, and give the peak cross-section for ψψ̄ → ψψ̄ in millibarns.
- Threshold factor: 4M²/m² = 1.00/1.96 = 0.5102, so β = √0.4898 = 0.6999.
- Self-energy: Im Σ(m²) = −g²β/(16π) = −(1.44 GeV²)(0.6999)/50.27 = −2.005 × 10⁻² GeV². Then Γ = −Im Σ(m²)/m = 1.432 × 10⁻² GeV = 14.3 MeV, topic 5.03's g²β/(16πm) as it must be.
- Pole: the denominator vanishes at s = m² − imΓ = (1.960 − 0.0200i) GeV², so √s ≈ m − iΓ/2 = (1.4000 − 0.0072i) GeV, and the mean life is τ = ħ/Γ = 6.582 × 10⁻²⁵ GeV s/0.01432 GeV = 4.60 × 10⁻²³ s.
- Peak: p² = m²/4 − M² = 0.49 − 0.25 = 0.24 GeV², so σₚₑₐₖ = 4π/p² = 52.4 GeV⁻², and with 0.3894 mb GeV² that is 20.4 mb. The coupling never entered.
- Checks: 4π/p² has the dimension of an area, GeV⁻²; Γ/m = 1.0%, so the narrow-width expansion is safe; a massless ψ would give p² = 0.49 GeV² and a lower peak, 10.0 mb.
AnswerΓ ≈ 14.3 MeV, pole at √s ≈ (1.4000 − 0.0072i) GeV with τ ≈ 4.6 × 10⁻²³ s, and σₚₑₐₖ = 4π/p² ≈ 20.4 mb whatever g is.
MediumA ψψ̄ collider studies a φ with m = 3.0 GeV, M = 1.0 GeV and g = 0.40 GeV. Its centre-of-mass energy is spread uniformly over Δ = 20 MeV centred on m, and its luminosity is 2.0 × 10³² cm⁻² s⁻¹. Find Γ, the peak cross-section, the cross-section averaged over the spread, and the event rate.
- Width: β = √(1 − 4M²/m²) = √(5/9) = 0.7454, so Γ = g²β/(16πm) = (0.16 GeV²)(0.7454)/(150.8 GeV) = 7.91 × 10⁻⁴ GeV = 0.791 MeV, far narrower than Δ.
- Peak: p² = m²/4 − M² = 1.25 GeV², so σₚₑₐₖ = 4π/p² = 10.05 GeV⁻² = 3.91 mb.
- Line shape: Γ/m = 2.6 × 10⁻⁴, so s − m² ≈ 2m(E − m) and σ(E) = σₚₑₐₖ(Γ²/4)/[(E − m)² + Γ²/4]. Its integral over ±Δ/2 is Γσₚₑₐₖ arctan(Δ/Γ), so the average is σ̄ = σₚₑₐₖ(Γ/Δ)arctan(Δ/Γ) = 3.91 mb × 0.03954 × 1.5313 = 0.237 mb. The narrow-width value (π/2)Γσₚₑₐₖ/Δ = 0.243 mb is 2.6% higher: the tails beyond ±10 MeV.
- Rate: Lσ̄ = 2.0 × 10³² cm⁻² s⁻¹ × 2.37 × 10⁻²⁸ cm² = 4.7 × 10⁴ s⁻¹. Every collision at the peak would give 7.8 × 10⁵ s⁻¹, 16.5 times more: the energy spread, not the coupling, limits the rate.
- Checks: at the window's edge |ℳₛ| ≈ g²/(2m × 0.010 GeV) = 2.7 while |ℳₜ| ≤ g²/m² = 0.018, so the t channel is below 1%. Doubling g would quadruple Γ but raise the rate only 3.7-fold, as the wider tails spill past ±10 MeV, while σₚₑₐₖ stays put.
AnswerΓ ≈ 0.79 MeV, σₚₑₐₖ ≈ 3.9 mb, σ̄ ≈ 0.24 mb over the 20 MeV spread, and about 4.7 × 10⁴ events per second.
HardGive the φ of mass m = 2.0 GeV a second channel: besides ψ (M = 0.60 GeV, g = 2.0 GeV) it couples through −g′χ*χφ to a complex scalar χ of mass 0.20 GeV, with g′ = 1.5 GeV. At the peak in ψψ̄ collisions find σ(ψψ̄ → χχ̄) directly from |ℳ|² and from branching fractions, then the elastic and total cross-sections, and verify the optical theorem.
- Widths: βψ = √(1 − 1.44/4) = 0.800 and βχ = √0.96 = 0.9798, so Γψ = g²βψ/(16πm) = 31.83 MeV and Γχ = g′²βχ/(16πm) = 21.93 MeV. Both loops feed Im Σ, so the propagator carries Γ = 53.76 MeV, with Bψ = 0.5921 and Bχ = 0.4079.
- Directly: at s = m², ℳ = −gg′/(imΓ), so |ℳ|² = g²g′²/(m²Γ²) = 9.00/(4 × 2.890 × 10⁻³) = 778.5. With pχ = 0.9798 GeV out and pψ = 0.800 GeV in, σ = |ℳ|²pχ/(16πs pψ) = 778.5 × 1.2247/201.1 GeV⁻² = 4.742 GeV⁻² = 1.847 mb.
- By branching: 4π/pψ² = 19.63 GeV⁻², and 19.63 × 0.5921 × 0.4079 = 4.742 GeV⁻², the same number with g and g′ entering only through the B. Elastic: 19.63 × 0.5921² = 6.884 GeV⁻² = 2.681 mb.
- Optical theorem: σₜₒₜ = 6.884 + 4.742 = 11.63 GeV⁻² = 4.527 mb = (4π/pψ²)Bψ. The forward elastic amplitude is ℳ = ig²/(mΓ) = 37.20i, and 2Ecm pcm σₜₒₜ = 2 × 2.0 × 0.800 × 11.63 = 37.20, so Im ℳ(forward) = 2Ecm pcm σₜₒₜ holds.
- Argand: pf = pℳ/(8π√s) = 0.800 × 37.20i/50.27 = 0.592i = iBψ, 0.092 from the circle's centre i/2 and so inside it: the open χχ̄ channel pulls the s wave off the unitarity circle, and the elastic peak falls from 4π/p² to Bψ² of it.
Answerσ(ψψ̄ → χχ̄) ≈ 1.85 mb at the peak, σₑₗ ≈ 2.68 mb and σₜₒₜ ≈ 4.53 mb = (4π/p²)Bψ, with Im ℳ(0) = 37.2 = 2Ecm pcm σₜₒₜ as the optical theorem requires.