Resonances · Elementary Particle Physics

A resonance never leaves a track — a yoctosecond at γ = 300 is still nuclear distances — so everything known about one is read off the shape of a curve.

🎯 Why this matters

A lifetime here is quoted by an experiment that never ran a clock. The number comes out of a fitted lineshape, which is why resonance lifetimes and flight-distance lifetimes are measured by entirely different apparatus.

Every particle in Chapters 2 and 3 decayed weakly or electromagnetically, and lived long enough to leave a track. This chapter is about the ones that do not. Once a hadron is heavy enough that a strong decay is open — no selection rule in the way — its lifetime collapses to about a yoctosecond, 10⁻²⁴ s, and it decays essentially where it was born. Nothing in any detector will ever see one directly.

So the entire hadron spectrum of the next fifty pages had to be found indirectly, as a resonance : a peak in a cross-section, or a bump in an invariant mass. This section builds that tool, and it begins somewhere unexpected — with a classical oscillator.

Just how invisible is a yoctosecond?

🔢 Worked example — the flight distance, and what would have to see it

Take a resonance with $\tau = 1$ ys $= 10^{-24}$ s and produce it with a Lorentz factor as large as $\gamma = 300$ — a violent collision by 1960s standards. Its flight distance is

\beta\gamma\, c\tau \approx 300 \times (3\times10^8\ \text{m/s}) \times 10^{-24}\ \text{s} = 9\times10^{-14}\ \text{m} \approx 90\ \text{fm}.

Ninety femtometres. A bubble chamber (§1.13b) resolves about 100 μm, so the production and decay points are separated by one part in a billion of what it can measure. Compare the metastable particles of Chapter 2, which fly centimetres.

Reproduce it

hbar, c = 6.582119569e-22, 2.99792458e8      # MeV s, m/s
d = 300 * c * 1e-24
print(f"a 1 ys resonance at gamma = 300 travels {d*1e15:.1f} fm")
print(f"   for scale: proton radius 0.84 fm, bubble chamber resolution {1e-4*1e15:.1e} fm")
print(f"   so the two vertices are {d/1e-4:.1e} of a resolution element apart")
G = 120.0                                     # MeV, Example 4.1
print(f"Gamma = {G:.0f} MeV (Example 4.1) -> tau = hbar/Gamma = {hbar/G:.3e} s")
print(f"   the book's arithmetic: 1/(1.52e21 x {G:.0f}) = {1/(1.52e21*G):.3e} s")
print(f"   c*tau = {hbar/G*c*1e15:.2f} fm — the resonance decays inside its own size")

prints

a 1 ys resonance at gamma = 300 travels 89.9 fm
 for scale: proton radius 0.84 fm, bubble chamber resolution 1.0e+11 fm
 so the two vertices are 9.0e-10 of a resolution element apart
Gamma = 120 MeV (Example 4.1) -> tau = hbar/Gamma = 5.485e-24 s
 the book's arithmetic: 1/(1.52e21 x 120) = 5.482e-24 s
 c*tau = 1.64 fm — the resonance decays inside its own size

The last line is the one to keep: $c\tau = 1.6$ fm for a 120 MeV width, which is smaller than a proton. A resonance does not travel between two points; it is a transient in a collision.

4.1 The classical resonance, which is not an analogy

Bettini starts with a naïve atom: a heavy central charge with a cloud of opposite charges around it, treated classically. It has normal modes; pick one, of angular frequency $\omega_0$ and width $\Gamma$ , excite it and let it go.

\htmlClass{t-psi}{\Psi(t)} = \Psi_0 \exp\left(-\frac{t}{\htmlClass{t-tau}{2\tau}}\right)\cos \htmlClass{t-w}{\omega_0} t, \qquad \htmlClass{t-tau}{\tau} = 1/\htmlClass{t-G}{\Gamma}

(4.1)

Bettini p. 131 — the free decay. Note the 2 in the exponent: it makes τ the time constant of the intensity, not of the amplitude.

Every symbol, one at a time

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💡 What this really says — a resonance is a second-order system rung once and left alone

This is a damped sinusoid — a second-order system rung once and left alone. The only two numbers in it are where the pole sits ( $\omega_0$ ) and how fast the ringing dies ( $\Gamma$ ).

Everything in this section, and the reason resonances are useful at all, is that those same two numbers govern what happens when you drive the system instead. Nothing new enters; the same pole is simply being looked at from the frequency side.

\htmlClass{t-R}{R(\omega)} = \frac{\Gamma^2\omega^2}{\left(\htmlClass{t-det}{\omega_0^2 - \omega^2}\right)^2 + \omega^2\Gamma^2}

(4.2)

Bettini p. 132 — the response of the same oscillator to a periodic drive of angular frequency ω, once the transient has died away.

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💡 What this really says — two experiments, one number

Compare (4.1) and (4.2) and the conclusion is immediate: the width of the resonance curve equals the reciprocal of the lifetime of the free oscillations.

The book adds the sharper version of the same statement, which is worth keeping because it is a theorem rather than a coincidence: the squared Fourier transform of the decaying oscillation (4.1) is proportional to the response function (4.2). Ring the system and watch it die, or drive it and map its response — you are measuring the same pole through two different windows.

That is why a hadron nobody can time can still have its lifetime measured. You cannot watch a 10⁻²⁴ s particle decay. You can scan a beam energy across it and read the width off a peak.

🪜 From the full response (4.2) to the Breit–Wigner (4.3)

Step 1 of 4 — Factor the detuning

\left(\omega_0^2 - \omega^2\right)^2 + \omega^2\Gamma^2 = (\omega_0-\omega)^2(\omega_0+\omega)^2 + \omega^2\Gamma^2

Why you may do this: A difference of squares, nothing more. It separates the one factor that varies fast near the peak, (ω₀ − ω), from the ones that do not.

Bettini p. 132. Three lines of algebra, and the only physics input is Γ ≪ ω₀ — the resonance is narrow compared with where it sits.

\htmlClass{t-L}{L(\omega)} = \frac{(\htmlClass{t-G}{\Gamma}/2)^2}{\left(\htmlClass{t-w0}{\omega_0}-\omega\right)^2 + (\htmlClass{t-G}{\Gamma}/2)^2}

(4.3)

Bettini p. 132 — the Breit–Wigner shape function, and the single most-fitted curve in particle physics.

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the resonance line shape · one pole, two domains

mass m = 1232 MeVwidth Γ = 117 MeV

lifetime τ = ħ/Γ

5.63 × 10⁻²⁴ s

decay length cτ

1.687 fm

quality factor m/Γ

10.53

oscillations in a lifetime

1.676

Breit–Wigner, Eq. (4.3)

The energy domain. Γ is the full width at half maximum by construction — the two dashed verticals are exactly where L falls to 1/2.

free decay Ψ(t) = e^(−t/2τ) cos ω₀t
envelope e^(−t/2τ)

The time domain, on the same two numbers. The envelope decays with time constant 2τ, so the INTENSITY — the square — decays with τ. The number of oscillations before it dies is m/2πΓ, which is exactly the quality factor above.

The first resonance ever found (Fermi 1952). Q ≈ 10 — barely a resonance at all by engineering standards, and it still dominates the πp cross-section.

Both plots move together because they are the same pole. Widening Γ broadens the peak and shortens the ringing in exactly the same proportion — that is Γτ = ħ, the time–bandwidth relation, and it is why a resonance's width is a lifetime measurement.

⚙️ Engineer’s bridge — this is a second-order system, and you can read its Q

Nothing here is an analogy. (4.1) and (4.2) are the impulse response and the magnitude-squared frequency response of a resonant second-order system, and the particle physics differs only in what the pole is called.

The translation is exact:

oscillator	resonance
natural frequency $\omega_0$	mass $m$
bandwidth (FWHM) $\Gamma$	width $\Gamma$
time constant of the energy, $\tau = 1/\Gamma$	lifetime $\tau = \hbar/\Gamma$
quality factor $Q = \omega_0/\Gamma$	$m/\Gamma$
impulse response ↔ frequency response	free decay ↔ line shape

And the Q factor is worth computing, because it says how good a resonator a particle is. Press through the widget’s presets: the $\rho$ has $Q = 5$ — it rings less than once before it dies, which is why “particle” and “bump” are the same statement for it. The Δ manages $Q = 10$ . The J/ψ has $Q = 33\,000$ and the ϒ $175\,000$ , which by the standards of a quartz crystal is respectable. Chapter 4 spends its second half explaining why those two are so different from the first two, and the answer will be that a decay channel is closed.

Where the analogy breaks: a driven oscillator’s response is a classical amplitude you could watch on a scope, whereas $L(\omega)$ is a probability. And a hadron has many open channels at once, which the next few equations handle with a quantity — the elasticity — that has no place in the single-pole picture.

Where it breaks: and the elasticity is the tip of it. A second-order system is linear and its pole is a property of the system alone; a hadronic resonance sits in a channel where the “system” and the “input” are the same particles, so the pole position shifts with what else is open. Once a second channel opens the single pole is no longer even the right model — the width acquires energy dependence, thresholds put cusps in the line shape, and the Breit–Wigner parameters become convention-dependent. The two-pole intuition is exact for a narrow, isolated resonance and degrades continuously from there.

⚠️ Natural units — where ħ enters and leaves

Equation (4.1) sets $\tau = 1/\Gamma$ , with $\Gamma$ a rate in s⁻¹. Equation (4.3) is quoted with $\Gamma$ in MeV. Both are right, because in $\hbar = c = 1$ units a rate and an energy are the same thing — and to get back to seconds you put $\hbar = 6.582\times10^{-22}$ MeV s in:

\tau = \frac{\hbar}{\Gamma} = \frac{6.582\times10^{-22}\ \text{MeV s}}{120\ \text{MeV}} = 5.5\times10^{-24}\ \text{s}.

This is the time–bandwidth relation yet again — the third appearance on this site after §1.6–1.7 and §3.3 — and here it is doing real work: it converts a quantity you can measure (the width of a peak) into one you cannot (the lifetime of something that lives 10⁻²⁴ s).

Two ways to find one: formation and production

The two methods, and when each is the only option

	formation	production
what you vary	the beam energy — the resonance is the whole intermediate state	nothing; you take one beam energy and sort events afterwards
what you plot	a cross-section against √s	the invariant mass of a chosen subset of the final state
the equation	a + b → c + d + … + f, Eq. (4.4)	a + b → R + e → c + d + e, Eqs. (4.8)–(4.9)
the background under it	other partial waves and non-resonant scattering
the hard constraint		none — any resonance produced in any reaction can be looked for
what it gives you directly	mass, width, and — through partial wave analysis — J, P and I	mass and width; the quantum numbers need an angular or Dalitz analysis (§4.4)

<strong>Neither is optional.</strong> §4.2 finds the Δ(1232) in formation because a π beam on hydrogen has exactly its quantum numbers, and finds the Σ(1385) in production because no beam of Λ exists. The Ξ(1530) and every meson in §4.5 are production-only.

Formation: what a scan actually measures

\htmlClass{t-i}{a + b} \;\longrightarrow\; \htmlClass{t-f}{c + d + \ldots + f}

(4.4)

Bettini p. 133, Fig. 4.1. The resonance is not written down — it is the intermediate state whose formation makes the cross-section jump.

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Measuring the angular distribution, channel by channel and energy by energy, lets the amplitude be decomposed into terms of definite orbital momentum $L$ , total angular momentum $J$ , parity and isospin. That is partial wave analysis , and its terms are labelled $L_{2I,2J}$ — so $S_{11}$ , $S_{31}$ , $P_{11}$ , $P_{13}$ , $P_{31}$ , $P_{33}$ , and the Δ(1232) turns out to be $P_{33}$ .

\htmlClass{t-L}{L_{2I,2J}} = \frac{1}{2i\htmlClass{t-q}{q}}\left(\htmlClass{t-eta}{\eta}\,\exp\left(2i\htmlClass{t-d}{\delta}\right) - 1\right)

(4.6)

Bettini p. 134. One complex number per partial wave, per energy — and both of its parameters are physically named.

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\htmlClass{t-f}{f} = \frac{\htmlClass{t-Gel}{\Gamma_{el}}/2}{\htmlClass{t-m}{m} - \htmlClass{t-E}{E_{cm}} - i\htmlClass{t-G}{\Gamma}/2}

(4.7)

Bettini p. 134 — the Breit–Wigner amplitude. Its squared modulus is the shape function (4.3); its phase is the new information.

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the partial-wave amplitude in the complex plane

energy, in half-widths from the peak: (E − m)/(Γ/2) = 0.00elasticity η = 1.00

phase shift δ	90.0°
\|f\|, in units of 1/q	1.000
\|f\|², relative to the peak	1.000

|f|², the cross-section — what a scan shows you
δ in degrees ÷ 180 — what the Argand loop shows you

The same resonance, twice. The peak (solid) is what a cross-section scan measures and it shrinks as η² — drop η to 0.25 and it nearly vanishes into a background. The phase (dashed) sweeps 45° → 90° → 135° across one width whatever η is, which is why the Argand plane finds resonances the cross-section hides.

Fig. 4.2(a). The only open channel is the one you came in through, so the trajectory rides the unitarity circle itself — the scattering probability reaches its maximum allowed value at the peak.

💡 What this really says — why the loop matters more than the peak

Write (4.7) in units of half-widths, $x = 2(m - E_{cm})/\Gamma$ , and it becomes $f = \eta\,(x + i)/(x^2+1)$ . Two facts follow immediately:

the trajectory is a circle of radius $\eta/2$ centred at $i\eta/2$ , traced anticlockwise as the energy rises;
the phase is independent of $\eta$ — only the size of the circle shrinks.

That second point is the whole reason the Argand diagram exists. The cross-section goes as $\eta^2$ , so a resonance that couples weakly back to the entrance channel is quadratically suppressed in a scan and can vanish into the background entirely. The phase sweep does not care. Drag the elasticity down to 0.25 in the widget: the peak all but disappears while the loop stays perfectly legible.

§4.2 makes this concrete — the Δ(1232) is the one resonance in the πp cross-section you can see by eye, and “the vast majority cannot be seen by eye; rather, they are found in the Argand diagram.”

⚙️ Engineer’s bridge — an Argand diagram is a Nyquist plot

You have drawn this figure before, with different axis labels.

Plot the complex response of a system parametrically as the frequency is swept, and a single resonant pole traces a circle in the complex plane. That is a Nyquist plot, and every feature of Fig. 4.2 is a feature you already read off one:

the loop is a pole; its diameter is the coupling strength (here $\eta$ );
the phase passing through 90° at the top is the pole’s phase response, the same 90° a second-order system shows at its natural frequency;
the speed of the sweep is the bandwidth: one width takes you from 45° to 135°;
the unitarity circle is a bound with no control-theory analogue — it says the scattering probability cannot exceed 1, so the amplitude physically cannot leave that disc.

The engineering habit that transfers directly: when a magnitude plot is ambiguous, look at the phase. A small resonance buried in a large background is hard to see in $|H|$ and obvious in $\arg H$ , because the background contributes a slowly varying phase and the pole contributes a fast one. Hadron spectroscopy in the 1960s ran on exactly that instinct.

Where it breaks: a Nyquist plot is drawn for a transfer function you chose to measure, while $f$ here is a scattering amplitude constrained by unitarity — the disc is not a convention but a law, and an amplitude that appears to leave it means the analysis is wrong.

Where it breaks: a Nyquist plot is drawn from a measured transfer function you can sweep at will. Here the “frequency sweep” is a beam energy scan costing months, the phase is never measured directly — only through interference with something else — and the unitarity disc constrains the total amplitude, so a single partial wave leaving it means the decomposition is wrong rather than the physics. The disc is a law about the sum; reading it as a bound on each curve individually is the standard misuse.

Erratum — the phase shifts at m ∓ Γ/2 are 45° and 135°, not 0 and 180°

Page 134 states: “In the elastic case, δ = 0 when the centre of mass energy is half a width below the resonance, that is, δ = m − Γ/2, and δ = 180° when δ = m + Γ/2.”

Two things are wrong, one typographic and one physical.

The typo: δ = m ∓ Γ/2 should be E_cm = m ∓ Γ/2. A phase shift is being equated to an energy, twice in one sentence.
The physics: at those energies the phase is 45° and 135°, not 0 and 180°. Put $x = \pm1$ into $f = (x+i)/(x^2+1)$ and you get $(\pm1 + i)/2$ , whose arguments are exactly 45° and 135°. Those are the two points where the trajectory is widest — the left and right of the circle, which is precisely where Fig. 4.2(a) puts its own $m \mp \Gamma/2$ labels.

δ = 0 and 180° are the asymptotic limits far below and far above the peak, where the amplitude has returned to the origin. Set the energy slider to ±1 in the widget above and read δ off directly; the figure and the formula agree with each other, and not with the sentence.

Production: what an invariant mass actually measures

When no beam of the right kind exists — and for every meson and every hyperon resonance, none does — you produce the particle instead and reconstruct it from its daughters.

a + b \to \htmlClass{t-R}{R} + e \to \htmlClass{t-cd}{c + d} + e, \qquad \htmlClass{t-M}{M_{cd}} = \sqrt{\left(E_c + E_d\right)^2 - \left(\mathbf p_c + \mathbf p_d\right)^2}

(4.9, 4.10)

Bettini p. 135, Fig. 4.3. The resonance is reconstructed after the fact, from particles the detector did see.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

💡 What this really says — an invariant-mass distribution is a histogram of a hypothesis

An invariant mass distribution is a histogram of a hypothesis. For every event you assume two particular final-state particles came from a common parent, compute what that parent’s mass would have been, and fill a bin.

If the hypothesis is right for some fraction of events, those events all give the same answer and pile up into a peak at $M_R$ , smeared by the Breit–Wigner width and the detector resolution. If it is wrong — the two particles were produced independently — the answer is whatever kinematics allows, and those events spread into a smooth phase-space background.

The book’s warning is worth taking seriously: “the simple observation of a peak is not enough to establish a resonance.” A bump can come from a reflection of some other resonance, from a threshold, or from an acceptance edge. Establishing one takes the quantum numbers, and that is what §4.3–4.4 is for.

🔑 If you remember only three things

The particle and the curve are the same object. There is no further fact about a resonance waiting behind the shape it makes.
A yoctosecond is not a short time; it is no time at all. Even at γ = 300 the flight distance is smaller than a nucleus, so no detector will ever see one directly.
Mass and width come out of one fit, not two measurements. They are quoted as separate numbers in every table and they were never separately observed.

Where this goes next

§4.2 The 3/2⁺ baryons applies both methods: the Δ(1232) in formation with the Argand analysis that fixes $J^P = 3/2^+$ , and the Σ(1385) and Ξ(1530) in production because no beam can form them.
§4.3–4.4 is the missing half of production — how to get spin, parity and isospin out of a three-body final state with the Dalitz plot.
§4.5 finds the η and ω as bumps in a 3π mass distribution, and §4.9 finds the J/ψ as the narrowest peak in this book.
§6.1 returns to e⁺e⁻ resonances with the R ratio, where the same Breit–Wigner is read for a completely different purpose.

✅ Check yourself — resonances

0/5 answered · 0 correct

1.Why is a resonance's width a measurement of its lifetime, when no clock is involved anywhere?
2.In the BreitWigner widget, switch from the ρ to the J/ψ. The quality factor $m/\Gamma$ jumps from about 5 to about 33 000. What does the time-domain plot do?
Hint: How many oscillations fit inside a lifetime?
3.Set the elasticity to $\eta = 0.25$ in the Argand widget and compare the two plots. What survives, and why does it matter?
4.Why can the Σ(1385) not be found by a formation experiment, when the Δ(1232) can?
5.The book warns that "the simple observation of a peak is not enough to establish a resonance". What is the strongest reason?