Chapter 4 — Summary and Problems · Elementary Particle Physics

Thirty-four problems, and the instructive ones are where the obvious calculation gives the wrong answer: a mean instead of a tail, a threshold formula missing two terms, a ratio that cannot exist.

🎯 Why this matters

Each of those shortcuts is plausible, which is why they survive into published work. A mean is the right statistic until you are asked about a tail, and a dropped term stays invisible until the masses stop being small.

Chapter 3 taught the rules. Chapter 4 spent fifty pages applying them to a hundred particles, and its thirty-four problems are where you find out whether the application stuck.

They fall into five families, and the family tells you which tool to reach for before you have finished reading the question:

nine ask is this allowed, and what stops it — 4.1, 4.2, 4.3, 4.6, 4.7, 4.8, 4.12, 4.28, 4.29. This is the checklist of §3.6, now with charm and beauty in it;
seven are quark-content bookkeeping — 4.15 to 4.21, a solid run. Read the flavour numbers, count the slots, check the charge;
eight are numerical — 4.4, 4.22 to 4.25, 4.32 to 4.34: two flight lengths, a vertex-resolution design, a collider design, three thresholds and one integral;
six are isospin ratios — 4.5, 4.9, 4.10, 4.11, 4.26, 4.27, which are Clebsch–Gordan coefficients and nothing else;
four ask which quantum numbers a composite state can have at all — 4.13, 4.14, 4.30, 4.31.

What the chapter established

The chapter summary (p. 182), with where each item was built

You should now have	What that means in practice	§
many hadrons decay strongly and are seen as resonances	A width IS a lifetime, and a lifetime of 10⁻²⁴ s is a flight of a fraction of a femtometre. You find such a state as a peak in a cross-section (formation) or in an invariant mass (production) — never as a track. Problem 4.4 is this fact in one line.	4.1–4.2
three families of quarks, each a doublet of charge −1/3 and +2/3	Two of the three were found rather than predicted; charm was predicted by GIM and beauty arrived before anyone had asked for it. The structure repeats exactly and the masses do not repeat at all.	4.9–4.11
baryons are qqq, mesons are qq̄	Which is the whole content of eight of the problems below: given the quantum numbers, there is exactly one way to fill three slots or two, and the charge arithmetic closes it.	4.6–4.8
quarks are never free	With one exception. The top decays in 5 × 10⁻²⁵ s, faster than the ~3 × 10⁻²⁴ s hadronization takes, so it is the only quark whose mass is measured as if it were a free particle.	4.10
u and d are very light, s is light-ish, the rest are heavy	A proton weighs 100 times its quarks. The mass of ordinary matter is colour-field energy, not constituent mass — which is why Chapter 6 exists.	4.11
isospin SU(2) and flavour SU(3) are consequences of those small masses	Both are accidental. Isospin holds to about 1 % because m_d − m_u is a few MeV against a GeV; SU(3)_f is far worse because 93 MeV is not negligible, and the decuplet rungs are 145 MeV apart in consequence.	4.6–4.8
only some SU(3) × spin combinations exist — those with an antisymmetric colour wave function	The chapter's deepest result, and the only one that is dynamics rather than classification. Twelve pairings are conceivable; Eq. (4.56) leaves the 3/2⁺ decuplet and the 1/2⁺ octet, which is what nature shows.	4.8

One line of that summary does more work than the rest: “only some multiplets are realized”. Everything before it is bookkeeping that a sufficiently patient cataloguer could have found. That line is QCD, three chapters early.

The checklist, in the order that saves the most work

Problems 4.28 and 4.29 hand you fourteen reactions to adjudicate, and several more are the same question in disguise. Run the tests in this order and stop at the first failure — most reactions die on the third line.

Test in this order; each line is cheaper than the one below it

#	test	why it comes here
1	electric charge	Exact, gauge-protected, and one addition. Two of the fourteen reactions die here (4.28c, 4.29e) and cost nothing to spot.
2	baryon number	Equally cheap. Also the fastest way to notice that a printed final state is not a possible decay at all — see the misprint in 4.10.
3	strangeness (and charm, beauty)	The workhorse. Seven of the fourteen die here. Note the two distinct verdicts: \|ΔS\| = 2 is forbidden to EVERY interaction in one step, while \|ΔS\| = 1 is merely not strong — it is exactly what the weak interaction does.
4	energy	Only now, because it needs numbers. Two reactions are forbidden by mass alone (4.28d, 4.29f), and one of those is ALSO forbidden by isospin — a reaction can fail more than one test and the problem asks for all the reasons.
5	isospin and its third component	Costs a Clebsch–Gordan lookup, so it goes last. I_z is additive and cheap; I itself needs the coupling.
6	P, C, G	Only for decays and only when the state has them. G-parity in particular is undefined the moment strangeness is non-zero — problem 4.2 is precisely that trap.

“Forbidden” always needs a qualifier. Every one of these reactions is forbidden to some interaction, and half of the fourteen happen constantly — as weak decays. Writing “forbidden” without saying by what is the commonest wrong answer in this set.

The one zero that appears six times

Six of the thirty-four problems — 4.3, 4.7, 4.11, 4.12, 4.26 and 4.30 — are decided by a single Clebsch–Gordan coefficient being zero:

\langle 1,0 \mid 1,0;1,0 \rangle = 0

Two isovectors with $I_z = 0$ each have no $I = 1$ component. Every consequence in this chapter follows from that one number:

One vanishing coefficient, six problems

problem	what is observed	what it establishes
4.3, 4.7	ρ⁰ → π⁰π⁰ never happens	the ρ has I = 1 — no other isospin would forbid it
4.12	—	a 2π⁰ system can only have I = 0 or 2
4.11(b)	p̄p(I = 1) → ρ⁰π⁰ is absent	the ratios come out 1 : 0 : 1, and the zero measures the I = 0 fraction of the initial state
4.26	a baryon decays to Σ⁺π⁻ and Σ⁻π⁺ but never Σ⁰π⁰	its isospin is 1, and the two observed widths must then be equal
4.30, 4.31	—	ρ⁰ρ⁰ and ρ⁰π⁰ states cannot have I = 1
4.27(2)	K⁻p → π⁰Σ⁰ has no I = 1 amplitude	that channel measures A₀ on its own — a rare clean handle

Worth internalising rather than looking up. The symmetry statement behind it: for two identical isovectors the I = 1 coupling is antisymmetric, so Bose statistics kills it — and the coefficient is zero whether or not the particles are identical, which is why 4.27 and 4.31 get the same result without any exchange argument.

Why ⟨1,0 | 1,0 ; 1,0⟩ = 0 — the zero is forced by a symmetry, not a coincidence

⟨1, M | 1,m₁ ; 1,m₂⟩ for the I = 1 outputm₂ = +10−1m₁ = +10−1—+1/√2+1/√2−1/√20+1/√2−1/√2−1/√2—the diagonal m₁ = m₂Read the grid across the diagonal:every entry is MINUS its mirror image.The I = 1 output is antisymmetric underexchanging the two isovectors.An antisymmetric function must vanishwhere its two arguments are equal.Six problems — 4.3, 4.7, 4.11, 4.12, 4.26, 4.30 — are decided by that one box, and none of them needs the others.Two identical π⁰s, two identical ρ⁰s, anything with I_z = 0 twice over: no I = 1 component exists to decay through.

Supplied — the section states the zero and lists its six consequences; the grid shows why it cannot be otherwise. Coefficients from Appendix 4, for 1 ⊗ 1 → 1. The dashes are the entries with no I = 1 partner at that M. Nothing about pions or isospin enters the argument: combining two identical representations into the antisymmetric product always kills the diagonal, which is the same statement as Pauli’s for two identical fermions in the same state, and the same statement as “the cross product of a vector with itself is zero” — because 1 ⊗ 1 → 1 is the cross product, in disguise.

Thresholds: three problems, one formula, and a lesson about beams

Problems 4.32, 4.33 and 4.34 all ask the same thing, and the arithmetic is the easy half. The hard half is working out what else has to be produced.

Every threshold in the chapter

mp, mpi, mK = 0.93827, 0.13957, 0.49368

def threshold(products, m_beam):
    S = sum(products)
    exact = (S**2 - m_beam**2 - mp**2) / (2*mp)   # s = m_b^2 + m_p^2 + 2 m_p E
    book = S**2 / (2*mp)                          # the high-energy approximation
    return S, exact, book

cases = [("pi-", "Lambda_b0 B0",          [5.6196, 5.2796],                 mpi),
         ("pi-", "Sigma_c++ D- pi-",      [2.4540, 1.8697, 0.1396],         mpi),
         ("pi-", "Omega_c0 D- K+ K0",     [2.698, 1.869, 0.494, 0.498],     mpi),
         ("K-",  "Omega_c0 D- K+",        [2.698, 1.869, 0.494],            mK),
         ("K+",  "Omega_c0 D- K+ K+ K+",  [2.698, 1.869, 0.494, 0.494, 0.494], mK)]

print("beam   final state                        sum m    E(exact)   E(book)")
for beam, name, prods, mb in cases:
    S, ex, bk = threshold(prods, mb)
    print(f"{beam:5s}  {name:32s} {S:6.3f}   {ex:6.1f} GeV   {bk:6.1f} GeV")
print("\nthe K- beam is cheapest and the K+ dearest, by 5.8 GeV, and nothing")
print("about the produced particles changed -- only whether the beam's own")
print("strangeness helped or hindered")

prints

beam   final state                        sum m    E(exact)   E(book)
pi-    Lambda_b0 B0                     10.899     62.8 GeV     63.3 GeV
pi-    Sigma_c++ D- pi-                  4.463     10.1 GeV     10.6 GeV
pi-    Omega_c0 D- K+ K0                 5.559     16.0 GeV     16.5 GeV
K-     Omega_c0 D- K+                    5.061     13.1 GeV     13.6 GeV
K+     Omega_c0 D- K+ K+ K+              6.049     18.9 GeV     19.5 GeV

the K- beam is cheapest and the K+ dearest, by 5.8 GeV, and nothing
about the produced particles changed -- only whether the beam's own
strangeness helped or hindered

📏 The book’s threshold formula drops two terms

Bettini’s printed solution to 4.34 uses

E \simeq \frac{\left(\sum m_f\right)^2}{2m_p}

which is the exact expression $E = [(\sum m_f)^2 - m_{\text{beam}}^2 - m_p^2]/(2m_p)$ with the two initial masses dropped. That is a good approximation when the produced masses are large: the error is 3 % for the charm thresholds and under 1 % for beauty, and it never changes a conclusion.

It is worth knowing which one you are using, though, because the difference is 0.5 GeV — larger than the mass of the pion doing the colliding.

💡 What this really says — choose the beam that already carries the quantum number

The three answers to 4.34 differ by nearly 6 GeV, and the produced particles are identical in all three. The whole spread comes from the beam.

A $K^-$ brings $S = -1$ into the collision, so one of the two units of strangeness the $\Omega_c^0$ needs is already paid for. A $K^+$ brings $S = +1$ , of the wrong sign, so three kaons have to be made instead of two — and each one costs 494 MeV inside a squared sum.

This is not a trick of the problem; it is how experiments are designed. It is why Barnes hunted the Ω⁻ in a $K^-$ beam (§4.8), why Ting used protons and Richter used $e^+e^-$ for the same particle (§4.9), and why a beauty factory sits exactly on the ϒ(4S) rather than anywhere else (§4.10). Conservation laws are constraints when you are checking a reaction and design parameters when you are building one.

The kinematics problems, on one axis

Four problems ask how far something travels before it decays, and the answers span twenty orders of magnitude. Putting them on one axis makes the chapter’s central experimental distinction visible at a glance.

resonances — no track possible
J/ψ at 5 GeV/c — narrow, still invisible
weak decays — vertex detectors reach these
Ω⁻ — visible in a bubble chamber

Row 4: the K* of problem 4.4 at 90 GeV/c — 400 fm. Row 3: the J/ψ of 4.22 — 3.5 pm, ten thousand times longer and still hopeless. Row 2: the D⁰ of 4.23 at 20 GeV (1.3 mm) and the B of 4.25 at a beauty factory (257 μm), both weak decays and both resolvable with silicon. Row 1: the Ω⁻ of §4.8 at 24.6 mm. The dashed lines are the two technologies. Everything to the left of them is found as a mass peak; everything to the right is found as a track.

⚙️ Engineer’s bridge — the resolution question in 4.23 is a tail probability, not a mean

Problem 4.23 is the one most often got wrong, and the error is instructive. The mean decay length of a 20 GeV D⁰ is 1.3 mm, so it is tempting to answer “about a millimetre”.

But decay lengths are exponentially distributed, and the problem asks for 90 % of events, not the typical one. $P(\ell > x) = e^{-x/L}$ , so demanding $P = 0.9$ gives $x = L\ln(1/0.9) = 0.105\,L$ — a factor of ten below the mean. The answer is 138 μm.

Any engineer who has sized a buffer against a latency requirement has met this: the mean tells you almost nothing about the tail, and a specification quoted at the 90th percentile of an exponential sits an order of magnitude away from the average. Here the factor of ten is the difference between a technology that existed in 1976 and one that did not.

Where it breaks: “an order of magnitude in resolution is the difference between existing and not” is true of this measurement and is not a general law of instrumentation. The step mattered here because the quantity being resolved — a decay length of a few hundred micrometres — sat just past what 1976 emulsion and bubble chambers could reach, so a factor of ten crossed a threshold. Where no threshold sits nearby, a factor of ten in resolution buys a factor of ten and nothing more. The dramatic version of the claim needs a discontinuity to be dramatic about.

Problems

Thirty-four, in the book’s order. Statement first, then a hint, then the worked solution, then the answer — nothing appears until you ask for it, and the topic chips let you take one family at a time.

📝 Chapter 4 — all 34 problems

0/34 solved

4.1G-paritytheory
Consider the three states $\pi^0$ , $\pi^+\pi^+\pi^-$ and $\rho^+$ . Which of them is a G-parity eigenstate, and in that case what is the eigenvalue?
- $G = C\,(-1)^I$ , evaluated on the neutral member of the multiplet
- G-parity is multiplicative
4.2G-paritytheory
Consider the particles $\omega$ , $\phi$ , $K$ and $\eta$ . Which of them is a G-parity eigenstate, and in that case what is the eigenvalue?
4.3quantum numberstheory
From the observation that the strong decay $\rho^0 \to \pi^+\pi^-$ exists but $\rho^0 \to \pi^0\pi^0$ does not, what can be extracted about the $\rho$ quantum numbers $J, P, C, G, I$ ?
4.4resonancestheory
Find the distance travelled by a $K^*$ with momentum $p = 90$ GeV in one lifetime.
- $m(K^*) = 892$ MeV
- $\Gamma(K^*) = 50$ MeV
- $\hbar c = 197.33$ MeV fm
4.5isospintheory
In a bubble-chamber experiment on a $K^-$ beam, events of $K^- + p \to \Lambda^0 + \pi^+ + \pi^-$ are selected. A resonance appears in both the $\Lambda^0\pi^+$ and $\Lambda^0\pi^-$ mass distributions, at $M = 1385$ MeV with $\Gamma = 50$ MeV; it is called $\Sigma(1385)$ . (a) What are the strangeness, hypercharge, isospin and third component of the resonance seen in $\Lambda^0\pi^+$ ? (b) If the angular distributions establish that the $\Lambda^0\pi$ orbital angular momentum is $L = 1$ , what are the possible $J^P$ ?
- $\Lambda$ : $J^P = 1/2^+$ , $I = 0$ , $S = -1$
- $\pi$ : $J^P = 0^-$ , $I = 1$
4.6strangenesstheory
The $\Sigma(1385)$ hyperon is produced in $K^- + p \to \pi^- + \Sigma^+(1385)$ but is not observed in $K^+ + p \to \pi^+ + \Sigma^+(1385)$ . Its width is $\Gamma = 50$ MeV and its main decay channel is $\pi^+\Lambda$ . (a) Is the decay strong or weak? (b) What are the strangeness and the isospin of the hyperon?
4.7selection rulestheory
State the three reasons forbidding the decay $\rho^0 \to \pi^0\pi^0$ .
4.8selection rulestheory
The $\rho^0$ has spin 1; the $f$ meson has spin 2. Both decay into $\pi^+\pi^-$ . Is the $\pi^0\gamma$ decay forbidden for one of them, for both, or for neither?
4.9isospin ratiostheory
Calculate the branching ratio $\Gamma(K^{*+} \to K^0 + \pi^+)\,/\,\Gamma(K^{*+} \to K^+ + \pi^0)$ , assuming in turn that the isospin of the $K^*$ is $I = 1/2$ and $I = 3/2$ .
- $K^+ = |1/2, +1/2\rangle$ , $K^0 = |1/2, -1/2\rangle$
- $\pi^+ = |1,+1\rangle$ , $\pi^0 = |1,0\rangle$
4.10isospin ratiostheory
Calculate the ratios $\Gamma(K^-p)\,/\,\Gamma(\bar K^0 n)$ and $\Gamma(\pi^-\pi^+)\,/\,\Gamma(\bar K^0 n)$ for the $\Sigma(1915)$ , which has $I = 1$ .
- $K^- = |1/2,-1/2\rangle$ , $\bar K^0 = |1/2,+1/2\rangle$
- $p = |1/2,+1/2\rangle$ , $n = |1/2,-1/2\rangle$
4.11isospin ratiostheory
Low-energy antiprotons are stopped in a bubble chamber, filled with hydrogen in one exposure and deuterium in another. A stopped antiproton is captured into an atom and annihilates from an S wave; the $\bar p p$ and $\bar p n$ S-wave states are the triplet ${}^3S_1$ and the singlet ${}^1S_0$ . List the possible $J^P$ and $I$ , say which are eigenstates of $C$ and of $G$ with their eigenvalues, identify the possible initial states of $\bar p n \to \pi^-\pi^-\pi^+$ , and compute the ratios within each group: (a) $\bar p n \to \rho^0\pi^-$ vs $\rho^-\pi^0$ ; (b) $\bar p p (I{=}1) \to \rho^+\pi^-$ , $\rho^0\pi^0$ , $\rho^-\pi^+$ ; (c) the same three from $\bar p p (I{=}0)$ .
4.12isospintheory
Establish the possible total isospin values of the $2\pi^0$ system.
4.13Dalitz plotstheory
Find the Dalitz-plot zeros for the $3\pi^0$ states with $I = 0$ and $J^P = 0^-$ , $1^-$ and $1^+$ .
4.14spectroscopic notationtheory
Knowing that the spin and parity of the deuteron are $J^P = 1^+$ , give its possible states in spectroscopic notation.
- The deuteron is a proton and a neutron, each $J^P = 1/2^+$
- $P = P_1 P_2 (-1)^L = (-1)^L$
4.15quark contenttheory
What are the possible charm values $C$ of a baryon in general? What are they if the charge is $Q = +1$ , and what if $Q = 0$ ?
4.16quark contenttheory
A particle has $\mathcal{B} = 1$ , $Q = +1$ , $C = 1$ , $S = 0$ , $B = 0$ , $T = 0$ . Give its valence quark content.
4.17quark contenttheory
For each of the following, with $\mathcal{B} = 1$ and $T = 0$ throughout, give the valence quark content: $(Q, C, S, B) = (-1, 0, -3, 0)$ ; $(2, 1, 0, 0)$ ; $(1, 1, -1, 0)$ ; $(0, 1, -2, 0)$ ; $(0, 0, 0, -1)$ .
4.18quark contenttheory
For each of the following, with $\mathcal{B} = 0$ and $T = 0$ throughout, give the valence quark content: $(Q, S, C, B) = (1, 0, 1, 0)$ ; $(0, 0, -1, 0)$ ; $(1, 0, 0, 1)$ ; $(1, 0, 1, 1)$ .
4.19quark modeltheory
Explain why each of the following cannot exist according to the quark model: a meson with positive strangeness and negative charm; a spin-0 baryon; an antibaryon with charge $+2$ ; a positive meson with strangeness $-1$ .
4.20Gell-Mann–Nishijimatheory
Suppose you do not know the electric charges of the quarks. Find them using the other columns of Table 4.5.
4.21quark modeltheory
What are the possible electric charges in the quark model of (a) a meson and (b) a baryon?
4.22kinematicstheory
The J/ψ has $m_J = 3.097$ GeV and $\Gamma = 91$ keV. What is its lifetime? If it is produced with $p_J = 5$ GeV in the laboratory, how far does it travel in one lifetime? For a symmetric $J/\psi \to e^+e^-$ decay — electron and positron at equal and opposite angles $\pm\theta_e$ to the J/ψ direction — find $\theta_e$ and the electron energy in the laboratory. Then find $\theta_e$ for $p_J = 50$ GeV.
- $\hbar = 6.582 \times 10^{-25}$ GeV s
- $\hbar c = 197.33$ MeV fm
- $m_e$ is negligible at these energies
4.23detectorstheory
A $D^0$ meson is produced with energy $E = 20$ GeV. We wish to resolve its production and decay vertices in at least 90 % of cases. What spatial resolution is needed? Mention adequate detectors.
- $m(D^0) = 1.865$ GeV
- $\tau(D^0) = 0.41$ ps, so $c\tau = 123$ μm
4.24resonancestheory
For $e^+e^- \to f\bar f$ near a resonance of mass $M_R$ and total width $\Gamma$ , assume the Breit–Wigner line shape and calculate the integral of the cross-section over energy — the 'peak area'. Assume $\Gamma/2 \ll M_R$ .
- $\sigma(E) = \dfrac{3\pi}{E^2}\dfrac{\Gamma_e\Gamma_f}{(E-M_R)^2 + \Gamma^2/4}$
4.25kinematicstheory
A beauty factory is a high-luminosity $e^+e^-$ collider running at the $\Upsilon(4S)$ , 10 580 MeV — only 20 MeV above $2m_B$ . At equal beam energies the two B mesons are nearly at rest and travel unmeasurably short distances, so beauty factories are asymmetric. For PEP-II at SLAC, with $p_{e^-} = 9$ GeV and $p_{e^+} = 3$ GeV, and taking the case where both B mesons have the same energy, find the distance they travel in one lifetime and the angles of their directions to the beams.
- $m(B^0) = 5279.6$ MeV, $\tau(B^0) = 1.5$ ps so $c\tau = 450$ μm
- $\sqrt s = 10\,580$ MeV at the resonance
4.26isospin ratiostheory
A baryon decays strongly into $\Sigma^+\pi^-$ and $\Sigma^-\pi^+$ , but not into $\Sigma^0\pi^0$ nor into $\Sigma^+\pi^+$ , even though all are energetically possible. (1) What can you say about its isospin? (2) Check the conclusion against the ratio of the two observed widths. Neglecting phase-space differences, what value do you expect?
4.27isospintheory
Write the scattering amplitudes of the following processes in terms of the total isospin amplitudes: (1) $K^-p \to \pi^-\Sigma^+$ , (2) $K^-p \to \pi^0\Sigma^0$ , (3) $K^-p \to \pi^+\Sigma^-$ , (4) $\bar K^0 p \to \pi^0\Sigma^+$ , (5) $\bar K^0 p \to \pi^+\Sigma^0$ .
- $\bar K N$ has $I = 0$ or 1 only (two isodoublets)
- Amplitudes $A_0$ and $A_1$ for the two total isospins
4.28allowed or forbiddentheory
Which of these is allowed or forbidden by the strong interaction? Give the reason in each case. (a) $\pi^-p \to K^-\Sigma^+$ · (b) $\pi^-p \to K^0\Lambda$ · (c) $\pi^+p \to K^0\Sigma^+$ · (d) $\Lambda \to \Sigma^-\pi^+$ · (e) $K^-p \to K^0 n$ · (f) $\Xi^- \to \Lambda\pi^-$ · (g) $\Omega^- \to \Xi^-\pi^0$ .
4.29allowed or forbiddentheory
Which of these is allowed or forbidden by the strong interaction? Give the reason. (a) $\pi^-p \to \bar K^0\Sigma^0$ · (b) $\pi^-p \to \Omega^-K^+K^0\pi^0$ · (c) $\pi^+p \to \Lambda K^+\pi^+$ · (d) $\pi^-p \to \Xi^-K^+K^0$ · (e) $\pi^-p \to \Lambda\pi^-$ · (f) $\Xi^0 \to \Sigma^+\pi^-$ · (g) $\Xi^- \to p\pi^-\pi^-$ .
4.30quantum numberstheory
Find the possible values of isospin, parity, charge conjugation, G-parity and total angular momentum $J$ , up to $J = 2$ , for a $\rho^0\rho^0$ state. For each $J$ , give also the orbital momentum $L$ and total spin $S$ .
4.31quantum numberstheory
Find the possible values of isospin, parity, charge conjugation, G-parity and spin, up to $J = 2$ , for a $\rho^0\pi^0$ state.
4.32thresholdstheory
Find the threshold energy needed to produce a $\Lambda_b(udb)$ with a $\pi^-$ beam on a hydrogen target.
- $m(\Lambda_b^0) = 5619.6$ MeV, $m(B^0) = 5279.6$ MeV
- $m_\pi = 139.6$ MeV, $m_p = 938.3$ MeV
4.33thresholdstheory
Find the threshold energy needed to produce a $\Sigma_c^{++}(uuc)$ with a $\pi^-$ beam on a hydrogen target.
- $m(\Sigma_c^{++}) = 2454.0$ MeV, $m(D^-) = 1869.7$ MeV, $m_\pi = 139.6$ MeV, $m_p = 938.3$ MeV
4.34thresholdstheory
Find the threshold energy needed to produce an $\Omega_c^0(ssc)$ with a $\pi^-$ , a $K^-$ or a $K^+$ beam on a hydrogen target.
- $m(\Omega_c^0) = 2698$ MeV, $m(D^-) = 1869$ MeV, $m(K^\pm) = 494$ MeV, $m(K^0) = 498$ MeV, $m_p = 938$ MeV

Erratum — problem 4.10 asks for an impossible ratio

Problem 4.10 asks for $\Gamma(\pi^-\pi^+)/\Gamma(\bar K^0 n)$ for the $\Sigma(1915)$ . A $\pi^-\pi^+$ final state has baryon number zero and cannot be a decay channel of a Σ at all; the intended channel is presumably $\Sigma^-\pi^+$ .

Even repaired, the ratio is not calculable: isospin relates the charge states within one channel, and $\Sigma\pi$ and $\bar K N$ are different channels with independent reduced matrix elements. The solution below computes what is well posed — $\Gamma(K^-p)/\Gamma(\bar K^0 n) = 1$ , and $\Gamma(\Sigma^+\pi^-)/ \Gamma(\Sigma^-\pi^+) = 1$ with $\Gamma(\Sigma^0\pi^0) = 0$ — and says why the cross-channel ratio is not.

The general point is worth more than this problem: a Clebsch–Gordan coefficient never tells you the relative strength of two different final-state channels. That is the commonest way these ratio calculations get misused.

🔑 If you remember only three things

The same zero turns up six times. One forbidden quantity, six different disguises — recognising the repeat is worth more than solving any one of them.
A threshold problem is two decisions. What you must produce fixes the number; which beam you choose fixes how much of the energy is wasted reaching it.
A printed problem can be wrong. 4.10 asks for a ratio between final states that cannot both exist, and spotting that is the same skill the chapter spent fifty pages teaching.

Where this goes next

Chapter 5 leaves spectroscopy for dynamics. The problems change character completely: from is this allowed to how big is it.
Chapter 6 is the colour force whose antisymmetric wave function decided which multiplets exist in §4.8 — and where the other 99 % of a proton’s mass lives.
§7.9 explains the flavour hierarchy that problems 4.32–4.34 take as given: why c → s is favoured, why b → c costs a family, and why that makes B mesons longer-lived than D mesons.
The isospin machinery of the seven ratio problems is used unchanged in Chapter 8, where the two states being mixed are a neutral meson and its antiparticle.

✅ Check yourself — working the chapter's problems

0/6 answered · 0 correct

1.You are handed fourteen reactions and asked which are allowed by the strong interaction. What do you test first, and why?
2.Several problems turn on the single fact that $\langle 1,0|1,0;1,0\rangle = 0$ . What does that coefficient being zero actually say?
3.Problem 4.23 asks for the resolution needed to separate a D⁰'s production and decay vertices in 90 % of cases. The mean decay length is 1.3 mm. What is the answer?
4.Problem 4.34 asks for the threshold to make an $\Omega_c^0$ with a π⁻, a K⁻ or a K⁺ beam. The three answers differ by nearly 6 GeV. Why?
5.Problem 4.18 asks for the quark content of a meson with $B = +1$ and $Q = +1$ . What is it, and what is the trap?
6.Problem 4.10 asks for a ratio between a $\Sigma\pi$ width and a $\bar K N$ width. Why is that not calculable from isospin?