Chapter 3 — Summary and Problems

Part I ★ Summary & Problems Bettini pp. 125–129 · ~13 min read

  • chapter summary
  • conservation-law checklist
  • isospin ratios

Two kinds of question hide in these thirty-one: which law forbids this reaction, and — when none of them does — what ratio do the same laws predict?

🎯 Why this matters

From here on the book will assert that a process is weak, or that it cannot happen at all, without showing the working. It expects you to run the check yourself in one line, which is what thirty-one variations are for.

Chapter 2’s problems were experiments redone with numbers. Chapter 3’s are different again: almost all of them are the same question asked thirty-one waysis this allowed, and if not, which law stops it? Nine of the thirty-one are lists of reactions to adjudicate, nine are isospin ratios, and the rest are the two measurements of §3.5 and the kinematics that supports them.

That repetition is the point. By the end you should be running the checklist below without thinking, because every chapter after this one assumes you can.

What the chapter established

The chapter summary (p. 129), with where each item was built
You should now haveWhat that means in practice§
the different types of symmetryGauge symmetries force conservation as a theorem; dynamical ones sort particles into multiplets; discrete ones give multiplicative quantum numbers. And breaking comes in two kinds — explicit (the interaction does not respect it) and spontaneous (the ground state does not).3.1
P, C and T, and when they give a quantum numberP and C are unitary involutions, so their eigenvalues are ±1 and they label particles. T is antiunitary, so it labels nothing — its "eigenvalue" moves when you rephase the state.3.2–3.4
CPT invarianceA theorem, not a hypothesis: Lorentz invariance plus locality force it. So a CPT test is a test of special relativity, and the antiproton's charge-to-mass ratio matches the proton's to 7 parts in 10¹¹.3.4
the parity of the pionsπ⁻d → nn, where two identical fermions with J = 1 have exactly one available state; and the π⁰ from the angle between two e⁺e⁻ planes — the method reused for the Higgs in §9.15.3.5
the quark and lepton flavour numbersAdditive counts, conserved by the strong and electromagnetic interactions and violated by the weak. Baryon and lepton number have no gauge symmetry behind them, which is why 22 500 t of water sits under a mountain watching for proton decay.3.6–3.7
SU(2) and the rules for summing isospinsA multiplet is a degenerate eigenspace; Clebsch–Gordan is a change of basis; and the ratios of cross-sections follow with no dynamical input at all — 9 : 1 : 2 against a measured 195 : 22 : 45.3.8–3.10

One line of that summary is doing more work than the others: <strong>“when they correspond to quantum numbers and when not”</strong>. Half of this chapter's problems are decided by knowing which law exists at all.

The checklist, in the order that saves the most work

Nine problems — 3.4, 3.5, 3.6, 3.7, 3.8, 3.22, 3.23, 3.27, 3.28 — are lists of reactions to adjudicate, and the fastest route through them is always the same. Test the cheap, absolute laws first; only reach for the discrete symmetries once the reaction has survived.

They carry five different topic chips below — allowed or forbidden, strangeness, lepton number, flavour numbers, bookkeeping — and the chip is not a category of question but the law that ends up doing the work. Filter by one and you are reading a set of reactions that all fail the same way.

absolute — a failure here means the reaction simply does not happen1 · charge Qgauge — a theorem2 · B and Land each lepton flavour3 · J and J_zhalf-integer ≠ integer4 · energyΣm_final ≤ √s?selective — a failure here only tells you WHICH interaction is responsible5 · flavour: S, C, B̃, Tfails → it must be weak6 · I but not I_zfails → electromagnetic7 · P and Cfails → weakboth I and I_zfail → weak, evenwith only hadronsWhy this orderSteps 1–4 are cheap arithmetic and they kill most of the reactions in these problem sets outright.Steps 5–7 never forbid a reaction — they only decide how slowly it goes. Λ → pπ⁻ fails 5, 6 and 7 and happens anyway.

Problem 3.1 asks for that pattern as a grid, and the book never prints the answer. Here it is, with the reason and the measurement behind every cell:

What each interaction conserves — the answer to Problem 3.1

quantum numberstrongelectromagneticweak
Iisospin
I_zthird component of isospin
Sstrangeness
flavourthe other quark flavours (C, B̃, T)
Bbaryon number
Llepton number (total)
L_e, L_μ, L_τlepton flavours
Pparity
Cparticle–antiparticle conjugation
Ttime reversal
Jangular momentum
J_zthird component of angular momentum
Qelectric charge

parity · weak · violated

Violated maximally, not slightly. This is Chapter 7, and it is the reason a neutrino parity cannot even be defined: neutrinos have only weak interactions.

Y* marks a law that holds in every collision and decay yet fails somewhere else — lepton flavour fails for neutrinos in flight, and T fails only in the neutral-kaon system. Click any cell for the reason and the measurement behind it.

⚙️ Engineer’s bridge — validate cheap and absolute first

The ordering above is not physics taste; it is the same discipline as any validation pipeline.

Cheap before expensive. Charge and baryon number are integer sums you can do in your head. Isospin decomposition needs Clebsch–Gordan coefficients. Do the arithmetic that costs nothing first, and most candidates never reach the expensive test.

Absolute before conditional. A failure in steps 1–4 is a hard rejection — the process does not occur, at any rate, ever. A failure in steps 5–7 is a routing decision: the process still happens, just through a slower interaction. Mixing the two categories is the single commonest error in these problems, and it shows up as answers like “forbidden because parity is violated”, which is never a reason for anything.

And the residue is informative. When every absolute law passes and several selective ones fail, you have not found a contradiction — you have identified the interaction. Λ → pπ⁻ violates II, IzI_z, SS and PP, contains nothing but hadrons, and is simply weak. That inference is worth more than the verdict.

Where it breaks: “validate cheap and absolute first” works because the cheap checks here are exact conservation laws, and it fails the moment a rule is only approximate. Isospin and parity are not absolute — they are strong-interaction rules, and a violated one signals “weak process”, not “impossible”. So the fast path returns a classification rather than a verdict, and reading a failed isospin check as “forbidden” would reject most of the decays in this book.

The nine ratio problems, in one widget

Problems 3.9, 3.10, 3.11, 3.12, 3.13, 3.14, 3.15, 3.16 and 3.25 are all the same calculation: decompose both states over total isospin, multiply the coefficients, square — with the two isospin amplitudes as the only unknowns. The widget from §3.9 does exactly that, from lib/cg.ts rather than from a table — set A1/2=0|A_{1/2}| = 0 for the Δ\Delta-region problems and read the rows.

πN cross-section ratios from isospin alone

reactionamplitudeσ (arb.)÷ σ(π⁻p elastic)
π⁺pπ⁺p(3.42)1 A₃⁄₂1.00009.00
π⁻pπ⁻p(3.44)1/3 A₃⁄₂ + 2/3 A₁⁄₂0.11111.00
π⁻pπ⁰n(3.43)√2/3 A₃⁄₂ √2/3 A₁⁄₂0.22222.00
π⁻nπ⁻n(3.45)1 A₃⁄₂1.00009.00
π⁺nπ⁺n1/3 A₃⁄₂ + 2/3 A₁⁄₂0.11111.00
π⁺nπ⁰p√2/3 A₃⁄₂ √2/3 A₁⁄₂0.22222.00
prediction (scaled to σ(π⁺p) = 195 mb) against the measured values at the Δ(1236)π⁺p → π⁺p195.0 predicted195 mb measuredπ⁻p → π⁻p21.7 predicted22 mb measuredπ⁻p → π⁰n43.3 predicted45 mb measured

This is Eq. (3.46). With the I = 1/2 amplitude switched off, the three cross-sections stand in the ratio 9 : 1 : 2, and the measured 195 : 22 : 45 gives 21.7 and 43.3 against 22 and 45. Nothing about the strong interaction was used — only how the states decompose.

💡 What this really says — why so many of the ratios are 1/2 or 2

Work through 3.11 to 3.14 and the same two numbers keep appearing. That is not a coincidence: all four have a singlet plus a doublet in one state and a triplet times a doublet in the other, so the only Clebsch–Gordan coefficients in play are 1/3\sqrt{1/3} and 2/3\sqrt{2/3} — and their squares are in the ratio 1:21 : 2.

Once you recognise the shape, the answer arrives before the arithmetic. A 0120 \otimes \frac12 initial state is pure, so it selects exactly one total isospin, and the two final channels then split that one amplitude in the fixed proportion 1:21 : 2. Which of the two is the “1” depends only on which member of the triplet appears.

The two-amplitude problems (3.10, 3.15, 3.16) are the interesting ones, because there the relative sign survives into the answer — and a sign is exactly what a measurement of two channels can determine.

Problems

📝 All 31 problems, with worked solutions

0/31 solved
  1. 3.1conservation lawstheory
    For each interaction — strong (S), electromagnetic (EM) and weak (W) — mark whether each of II, IzI_z, SS, B\mathcal{B}, L\mathcal{L}, TT, CC, PP, JJ, JzJ_z is conserved.
  2. 3.2thresholdstheory
    A π\pi^- beam hits a liquid hydrogen target. Find the threshold energy for KK^- production.
    • mπ=139.57m_\pi = 139.57 MeV, mp=938.27m_p = 938.27 MeV, mK±=493.68m_{K^\pm} = 493.68 MeV, mK0=497.61m_{K^0} = 497.61 MeV, mn=939.57m_n = 939.57 MeV
  3. 3.3thresholdstheory
    The anti-lambda was discovered by Baldo-Ceolin and Prowse in 1958 with a π\pi^- beam of Eπ=4.6E_\pi = 4.6 GeV on an emulsion stack. (a) What is the lightest final state containing a Λˉ\bar\Lambda from πp\pi^- p? (b) Find the threshold for free protons, and approximately for protons bound with Fermi momentum pf=150p_f = 150 MeV. (c) If N0=106N_0 = 10^6 cm2^{-2} pions were produced l=8l = 8 m upstream, how many reached the emulsion?
    • mΛ=1115.68m_\Lambda = 1115.68 MeV, mn=939.57m_n = 939.57 MeV, mp=938.27m_p = 938.27 MeV, mπ=139.57m_\pi = 139.57 MeV, cτπ=7.80c\tau_\pi = 7.80 m
  4. 3.4allowed or forbiddentheory
    For each reaction: is it allowed? If not, why (there may be several reasons)? If yes, by which interaction? (1) πpπ0n\pi^- p \to \pi^0 n; (2) π+μ+νμ\pi^+ \to \mu^+\nu_\mu; (3) π+μ+νˉμ\pi^+ \to \mu^+\bar\nu_\mu; (4) π02γ\pi^0 \to 2\gamma; (5) π03γ\pi^0 \to 3\gamma; (6) e+eγe^+e^- \to \gamma; (7) ppˉΛΛp\bar p \to \Lambda\Lambda; (8) ppΣ+π+pp \to \Sigma^+\pi^+; (9) npen \to p e^-; (10) npπn \to p\pi^-.
  5. 3.5allowed or forbiddentheory
    Allowed or not, and why? (1) μ+e+γ\mu^+ \to e^+\gamma; (2) eνeγe^- \to \nu_e\gamma; (3) ppΣ+K+pp \to \Sigma^+K^+; (4) pppΣ+Kpp \to p\Sigma^+K^-; (5) pe+νep \to e^+\nu_e; (6) ppΛΣ+pp \to \Lambda\Sigma^+; (7) pnΛΣ+pn \to \Lambda\Sigma^+; (8) pnΞ0ppn \to \Xi^0 p; (9) pne+νep \to n e^+\nu_e; (10) npeνˉen \to p e^-\bar\nu_e.
  6. 3.6allowed or forbiddentheory
    Give the reasons forbidding each decay: (a) npen \to p e^-; (b) nπ+en \to \pi^+ e^-; (c) npπn \to p\pi^-; (d) npγn \to p\gamma.
  7. 3.7strangenesstheory
    Which of these are allowed and which are forbidden by strangeness conservation? (a) πpKp\pi^- p \to K^- p; (b) πpK+Σ\pi^- p \to K^+\Sigma^-; (c) KpK+Ξ0πK^- p \to K^+\Xi^0\pi^-; (d) K+pKΞ0πK^+ p \to K^-\Xi^0\pi^-.
  8. 3.8lepton numbertheory
    Allowed or not, and why? (a) pne+p \to n e^+; (b) μ+νμe+\mu^+ \to \nu_\mu e^+; (c) e+eνμνˉμe^+e^- \to \nu_\mu\bar\nu_\mu; (d) νμpμ+n\nu_\mu p \to \mu^+ n; (e) νμnμp\nu_\mu n \to \mu^- p; (f) νμnep\nu_\mu n \to e^- p; (g) e+npνee^+ n \to p\nu_e; (h) epnνee^- p \to n\nu_e.
  9. 3.9isospin ratiostheory
    Assuming (unrealistically) that they proceed only through the I=3/2I = 3/2 channel, evaluate the ratios of the cross-sections πpK0Σ0\pi^- p \to K^0\Sigma^0, πpK+Σ\pi^- p \to K^+\Sigma^- and π+pK+Σ+\pi^+ p \to K^+\Sigma^+ at the same energy.
  10. 3.10isospin ratiostheory
    Evaluate the same three cross-sections with both isospin amplitudes A1/2A_{1/2} and A3/2A_{3/2} contributing.
  11. 3.11isospin ratiostheory
    Evaluate the ratio of the cross-sections of πpΛK0\pi^- p \to \Lambda K^0 and π+nΛK+\pi^+ n \to \Lambda K^+ at the same energy.
  12. 3.12isospin ratiostheory
    Evaluate the ratio of the cross-sections of p+d3He+π0p + d \to {}^3\mathrm{He} + \pi^0 and p+d3H+π+p + d \to {}^3\mathrm{H} + \pi^+ at the same s\sqrt s. (3^3He and 3^3H form an isospin doublet.)
  13. 3.13isospin ratiostheory
    Evaluate the ratio σ(ppdπ+)/σ(pndπ0)\sigma(pp \to d\pi^+)/\sigma(pn \to d\pi^0) at the same energy.
  14. 3.14isospin ratiostheory
    Evaluate σ(K4HeΣ03H)/σ(K4HeΣ3He)\sigma(K^- {}^4\mathrm{He} \to \Sigma^0 {}^3\mathrm{H}) / \sigma(K^- {}^4\mathrm{He} \to \Sigma^- {}^3\mathrm{He}) at the same energy.
  15. 3.15isospin ratiostheory
    Express the ratios of σ(Kpπ+Σ)\sigma(K^- p \to \pi^+\Sigma^-), σ(Kpπ0Σ0)\sigma(K^- p \to \pi^0\Sigma^0) and σ(KpπΣ+)\sigma(K^- p \to \pi^-\Sigma^+) in terms of the isospin amplitudes A0A_0, A1A_1 and A2A_2.
  16. 3.16isospin ratiostheory
    Express the ratio of the elastic πpπp\pi^- p \to \pi^- p and charge-exchange πpπ0n\pi^- p \to \pi^0 n cross-sections in terms of A1/2A_{1/2} and A3/2A_{3/2}.
  17. 3.17paritytheory
    A π\pi^- is captured by a deuteron (JP=1+J^P = 1^+) giving πdnn\pi^- d \to nn. (a) If capture is from an SS wave, what are the total spin and orbital momentum of the two neutrons? (b) Show that if capture is from a PP state, the neutrons are in a spin singlet.
  18. 3.18C-paritytheory
    Positronium is an e+ee^+e^- atom. (1) What relation does CC conservation impose between ll, ss and CC? (2) What relation between ll, ss and nn allows e+enγe^+e^- \to n\gamma? (3) What is the minimum number of photons for ortho-positronium (3S1^3S_1) and para-positronium (1S0^1S_0)?
  19. 3.19parity and Ctheory
    From which of the pˉp\bar p p states 1S0^1S_0, 3S1^3S_1, 1P1^1P_1, 3P0^3P_0, 3P1^3P_1, 3P2^3P_2, 1D2^1D_2, 3D1^3D_1, 3D2^3D_2, 3D3^3D_3 can pˉpnπ0\bar p p \to n\pi^0 proceed with parity conservation? (1) for any nn; (2) for n=2n = 2.
  20. 3.20isospin and statisticstheory
    Consider the strong process KˉKπ+π\bar K K \to \pi^+\pi^- (where KˉK\bar K K means both K+KK^+K^- and Kˉ0K0\bar K^0 K^0). What angular momenta are possible (1) if the initial total isospin is I=0I = 0, and (2) if I=1I = 1?
  21. 3.21parity and Ctheory
    From which of the pˉp\bar p p states 1S0^1S_0, 3S1^3S_1, 1P1^1P_1, 3P0^3P_0, 3P1^3P_1, 3P2^3P_2, 1D2^1D_2, 3D1^3D_1, 3D2^3D_2, 3D3^3D_3 can pˉpπ+π\bar p p \to \pi^+\pi^- proceed if the two pions are (1) in an SS wave, (2) in a PP wave, (3) in a DD wave?
  22. 3.22flavour numberstheory
    With Λc=udc\Lambda_c = udc, D+=cdˉD^+ = c\bar d and D=cˉdD^- = \bar c d, which are allowed? (a) π+pD+p\pi^+ p \to D^+ p; (b) π+pDΛcπ+π+\pi^+ p \to D^-\Lambda_c\pi^+\pi^+; (c) π+pD+Λc\pi^+ p \to D^+\Lambda_c; (d) π+pDΛc\pi^+ p \to D^-\Lambda_c.
  23. 3.23flavour numberstheory
    With Λb=dub\Lambda_b = dub, D0=cuˉD^0 = c\bar u, B+=ubˉB^+ = u\bar b, B=uˉbB^- = \bar u b and B0=dbˉB^0 = d\bar b, which are allowed? (a) πpD0Λb\pi^- p \to D^0\Lambda_b; (b) πpB0Λb\pi^- p \to B^0\Lambda_b; (c) πpB+Λbπ\pi^- p \to B^+\Lambda_b\pi^-; (d) πpBΛbπ+\pi^- p \to B^-\Lambda_b\pi^+; (e) πpBB+\pi^- p \to B^-B^+.
  24. 3.24kinematicstheory
    An η\eta meson of energy Eη=5E_\eta = 5 GeV, moving along xx, decays to 2γ2\gamma. (1) If the photons go along +x+x and x-x, what are their energies? (2) If they are emitted at equal and opposite angles ±θ\pm\theta to xx, what is the angle between them?
    • mη=547.86m_\eta = 547.86 MeV, Eη=5000E_\eta = 5000 MeV
  25. 3.25isospin ratiostheory
    The Δ(1232)\Delta(1232) has I=3/2I = 3/2. (1) What is the ratio between the decay rates Δ0pπ\Delta^0 \to p\pi^- and Δ0nπ0\Delta^0 \to n\pi^0? (2) What would it have been if I=1/2I = 1/2?
  26. 3.26detectorstheory
    A water Cherenkov detector has n=1.33n = 1.33. (1) Find the minimum velocity for radiation. (2) Find the minimum kinetic energy for an electron and for a KK meson. (3) Is the charged particle above threshold in pe+π0p \to e^+\pi^0 and in pK+νp \to K^+\nu?
    • n=1.33n = 1.33, me=0.511m_e = 0.511 MeV, mK+=493.68m_{K^+} = 493.68 MeV, mp=938.27m_p = 938.27 MeV, mπ0=134.98m_{\pi^0} = 134.98 MeV
  27. 3.27allowed or forbiddentheory
    Forbidden or allowed, with justification? (a) μeγ\mu^- \to e^-\gamma; (b) π+μ+νμνˉμ\pi^+ \to \mu^+\nu_\mu\bar\nu_\mu; (c) Σ0Λγ\Sigma^0 \to \Lambda\gamma; (d) ηγγγ\eta \to \gamma\gamma\gamma; (e) γpπ0p\gamma p \to \pi^0 p; (f) pπ0e+p \to \pi^0 e^+; (g) πμγ\pi^- \to \mu^-\gamma.
  28. 3.28quantum-number bookkeepingtheory
    Identify the particle XX: (a) πpΣ0X\pi^- p \to \Sigma^0 X; (b) e+npXe^+ n \to p X; (c) Ξ0ΛX\Xi^0 \to \Lambda X.
  29. 3.29isospintheory
    The Ξ0\Xi^0 is produced in π+pK+K+Ξ0\pi^+ p \to K^+ K^+ \Xi^0. Knowing the isospin and third component of every other particle, establish II and IzI_z of the Ξ0\Xi^0.
  30. 3.30kinematicstheory
    Consider π\pi^- capture at rest giving πpnγ\pi^- p \to n\gamma. Calculate the energy of the photon and the kinetic energy of the neutron.
    • mπ=139.57m_\pi = 139.57 MeV, mp=938.27m_p = 938.27 MeV, mn=939.57m_n = 939.57 MeV
  31. 3.31invariant masstheory
    A 12 GeV π+\pi^+ beam in a bubble chamber gives an event with two positive tracks from the primary vertex and two V0V^0s pointing back to it. The first V0V^0 has p+=0.4p_+ = 0.4 GeV, p=1.9p_- = 1.9 GeV, θ1=24.5°\theta_1 = 24.5°; the second has p+=0.75p_+ = 0.75 GeV, p=0.25p_- = 0.25 GeV, θ2=22°\theta_2 = 22°. With 5 % accuracy, identify the two neutral particles and guess the reaction.
    • M2=m12+m22+2(E1E2p1p2cosθ)M^2 = m_1^2 + m_2^2 + 2(E_1E_2 - p_1p_2\cos\theta); mΛ=1.116m_\Lambda = 1.116 GeV, mK0=0.498m_{K^0} = 0.498 GeV

🔢 Worked example — every numeric answer on this page, checked

The thresholds, ratios and kinematics above are all short enough to get wrong by hand, so here they are computed together. The Clebsch–Gordan coefficients come from the same Racah formula as src/lib/cg.ts.

Reproduce it

import numpy as np
from math import sqrt, factorial as f
def cg(j1, m1, j2, m2, J, M):                       # Condon-Shortley, = src/lib/cg.ts
    if m1 + m2 != M or abs(M) > J or J > j1 + j2 or J < abs(j1 - j2): return 0.0
    pre = sqrt((2*J+1)*f(int(j1+j2-J))*f(int(j1-j2+J))*f(int(-j1+j2+J))/f(int(j1+j2+J+1)))
    pre *= sqrt(f(int(j1+m1))*f(int(j1-m1))*f(int(j2+m2))*f(int(j2-m2))*f(int(J+M))*f(int(J-M)))
    s = 0.0
    for k in range(40):
        d = [j1+j2-J-k, j1-m1-k, j2+m2-k, J-j2+m1+k, J-j1-m2+k]
        if any(x < 0 or x != int(x) for x in d): continue
        s += (-1)**k/(f(k)*f(int(d[0]))*f(int(d[1]))*f(int(d[2]))*f(int(d[3]))*f(int(d[4])))
    return pre * s

mp, mn, mpi, mpi0 = 938.27209, 939.56542, 139.57039, 134.9768
mK, mK0, mL, me, meta = 493.677, 497.611, 1115.683, 0.51099895, 547.862
thr = lambda M, mb, mt: (sum(M)**2 - mb**2 - mt**2)/(2*mt)

print(f"3.2  threshold E_pi(K- K+ n)  = {thr([mK,mK,mn],mpi,mp):.0f} MeV = "
      f"{thr([mK,mK,mn],mpi,mp)/1000:.2f} GeV")
print(f"     the two channels:  m_n + m_K+ = {mn+mK:.1f}  vs  m_p + m_K0 = {mp+mK0:.1f} MeV")
print(f"3.3  lightest final state Lbar L n, sum m = {2*mL+mn:.1f} MeV")
print(f"     free-proton threshold  = {thr([mL,mL,mn],mpi,mp)/1000:.2f} GeV   -> ABOVE the 4.6 GeV beam")
EN, need = np.hypot(150., mp), (2*mL+mn)**2
sof = lambda E: mpi**2 + mp**2 + 2*(E*EN + np.sqrt(E**2-mpi**2)*150.)
lo, hi = mpi, 20000.
for _ in range(200):
    mid = (lo+hi)/2; lo, hi = (mid, hi) if sof(mid) < need else (lo, mid)
print(f"     with p_f = 150 MeV     = {(lo+hi)/2000:.2f} GeV   -> below it, so the reaction runs on bound protons")
bg = np.sqrt((4600./mpi)**2 - 1)
print(f"     pion survival over 8 m: beta*gamma = {bg:.1f}, decay length {bg*7.8045:.0f} m, "
      f"N/N0 = {np.exp(-8/(bg*7.8045)):.4f}")

r = lambda a, b: (cg(1,a[0],.5,a[1],.5,sum(a))*cg(1,b[0],.5,b[1],.5,sum(b)))**2
print(f"3.11 sigma(pi-p -> L K0) / sigma(pi+n -> L K+) = "
      f"{cg(1,-1,.5,.5,.5,-.5)**2/cg(1,1,.5,-.5,.5,.5)**2:.3f}")
print(f"3.12 sigma(3He pi0)      / sigma(3H pi+)       = "
      f"{cg(1,0,.5,.5,.5,.5)**2/cg(1,1,.5,-.5,.5,.5)**2:.3f}")
print(f"3.13 sigma(pp -> d pi+)  / sigma(pn -> d pi0)  = {1/0.5:.3f}")
print(f"3.14 sigma(Sig0 3H)      / sigma(Sig- 3He)     = "
      f"{cg(1,0,.5,-.5,.5,-.5)**2/cg(1,-1,.5,.5,.5,-.5)**2:.3f}")
amp = lambda m1, m2, I: (-1 if I == 0 else 1)/np.sqrt(2)*cg(1,m1,1,m2,I,0)
print("3.15 K-p amplitudes:  " + " | ".join(
    f"{n} = {amp(a,b,0):+.4f} A0" + (f" {amp(a,b,1):+.4f} A1" if abs(amp(a,b,1)) > 1e-9 else "")
    for n, a, b in (("pi+Sig-",1,-1), ("pi0Sig0",0,0), ("pi-Sig+",-1,1))))
print(f"     1/sqrt(6) = {1/np.sqrt(6):.4f}, and A2 never appears: <K-p|2,0> = 0")

p = np.sqrt(5000.**2 - meta**2)
print(f"3.24 eta at 5 GeV: collinear photons {(5000+p)/2:.1f} and {(5000-p)/2:.1f} MeV; "
      f"symmetric opening {2*np.degrees(np.arcsin(meta/5000)):.2f} deg")
print(f"3.25 Delta0 -> p pi- / n pi0 :  I=3/2 gives "
      f"{cg(1,-1,.5,.5,1.5,-.5)**2/cg(1,0,.5,-.5,1.5,-.5)**2:.3f},  I=1/2 would give "
      f"{cg(1,-1,.5,.5,.5,-.5)**2/cg(1,0,.5,-.5,.5,-.5)**2:.3f}")

n = 1.333; gm = 1/np.sqrt(1 - 1/n**2)
print(f"3.26 beta_min = {1/n:.4f}, gamma-1 = {gm-1:.4f} -> T_min(e) = {me*(gm-1):.2f} MeV, "
      f"T_min(K+) = {mK*(gm-1):.0f} MeV")
print(f"     p -> e+ pi0: T(e+) = {(mp**2+me**2-mpi0**2)/(2*mp)-me:.1f} MeV, above threshold")
print(f"     p -> K+ nu : T(K+) = {(mp**2+mK**2)/(2*mp)-mK:.1f} MeV, below threshold")
M = mpi + mp; Eg = (M**2 - mn**2)/(2*M)
print(f"3.30 pi- p -> n gamma at rest: E_gamma = {Eg:.2f} MeV, T_n = {M-mn-Eg:.3f} MeV")
def minv(p1, p2, th, m1, m2):
    return np.sqrt(m1**2 + m2**2 + 2*(np.hypot(p1,m1)*np.hypot(p2,m2)
                                      - p1*p2*np.cos(np.radians(th))))
for tag, pp, pm, th in (("V0_1", 0.4, 1.9, 24.5), ("V0_2", 0.75, 0.25, 22.)):
    L = minv(pp, pm, th, mp/1000, mpi/1000); K = minv(pp, pm, th, mpi/1000, mpi/1000)
    print(f"3.31 {tag} as Lambda = {L:.3f} GeV, as K0 = {K:.3f} GeV  -> it is a "
          f"{'K0' if abs(K-0.4976) < abs(L-1.1157) else 'Lambda'}")
prints
3.2  threshold E_pi(K- K+ n)  = 1499 MeV = 1.50 GeV
   the two channels:  m_n + m_K+ = 1433.2  vs  m_p + m_K0 = 1435.9 MeV
3.3  lightest final state Lbar L n, sum m = 3170.9 MeV
   free-proton threshold  = 4.88 GeV   -> ABOVE the 4.6 GeV beam
   with p_f = 150 MeV     = 4.16 GeV   -> below it, so the reaction runs on bound protons
   pion survival over 8 m: beta*gamma = 32.9, decay length 257 m, N/N0 = 0.9694
3.11 sigma(pi-p -> L K0) / sigma(pi+n -> L K+) = 1.000
3.12 sigma(3He pi0)      / sigma(3H pi+)       = 0.500
3.13 sigma(pp -> d pi+)  / sigma(pn -> d pi0)  = 2.000
3.14 sigma(Sig0 3H)      / sigma(Sig- 3He)     = 0.500
3.15 K-p amplitudes:  pi+Sig- = -0.4082 A0 +0.5000 A1 | pi0Sig0 = +0.4082 A0 | pi-Sig+ = -0.4082 A0 -0.5000 A1
   1/sqrt(6) = 0.4082, and A2 never appears: <K-p|2,0> = 0
3.24 eta at 5 GeV: collinear photons 4984.9 and 15.1 MeV; symmetric opening 12.58 deg
3.25 Delta0 -> p pi- / n pi0 :  I=3/2 gives 0.500,  I=1/2 would give 2.000
3.26 beta_min = 0.7502, gamma-1 = 0.5123 -> T_min(e) = 0.26 MeV, T_min(K+) = 253 MeV
   p -> e+ pi0: T(e+) = 458.9 MeV, above threshold
   p -> K+ nu : T(K+) = 105.3 MeV, below threshold
3.30 pi- p -> n gamma at rest: E_gamma = 129.41 MeV, T_n = 8.870 MeV
3.31 V0_1 as Lambda = 1.845 GeV, as K0 = 0.520 GeV  -> it is a K0
3.31 V0_2 as Lambda = 1.114 GeV, as K0 = 0.358 GeV  -> it is a Lambda

📏 Four slips in the book’s printed solutions

Ten of the thirty-one problems have solutions printed at the back (pp. 513–515). Four of them do not survive checking, and all four are flagged in the relevant problem’s note above:

  • 3.19(1) lists the surviving pˉp\bar pp states as ”1S0^1S_0, 3P1^3P_1, 3P2^3P_2, 3P3^3P_3 and 1D2^1D_2”. 3P3^3P_3 does not exist — l=1l = 1 with s=1s = 1 gives J2J \le 2 — and 3P0^3P_0 is missing.
  • 3.19(2) then concludes “only 3P2^3P_2”, which follows from the dropped 3P0^3P_0. Restore it and it passes every test, so the answer is 3P0^3P_0 and 3P2^3P_2.
  • 3.21 concludes “(1) 1S0^1S_0” while its own table puts the Y under 3P0^3P_0. 1S0^1S_0 is 0+0^{-+} and an SS-wave π+π\pi^+\pi^- is 0++0^{++}.
  • 3.26 quotes 469 MeV for the positron in pe+π0p \to e^+\pi^0, which is the formula with the mπ02m_{\pi^0}^2 term dropped; the value is 459 MeV. It also uses the neutral kaon mass for a K+K^+.

None of them changes a physical conclusion, and the last two are arithmetic. But 3.19 and 3.21 change the answer, which is why the solutions above derive everything rather than reproducing what is printed.

🔑 If you remember only three things

  • A symmetry can fix a ratio with no dynamics at all. Nine of these get a number out of isospin alone, without a coupling constant appearing anywhere in the calculation.

  • The first law that says no ends the calculation. Nothing further needs computing once a reaction violates charge or baryon number, which is what makes the order of the checks worth having.

  • Thirty-one variations is the point. A checklist only becomes useful once running it takes no thought, and the rest of the book assumes you can.

Where this goes next

  • Chapter 4 is where the machinery gets used: the Δ(1232) of problems 3.9–3.10 and 3.25 becomes a particle with a width, and SU(2) becomes SU(3).
  • §4.2 is the resonance itself — the reason A3/2|A_{3/2}| dominates and 9 : 1 : 2 works at all.
  • §7.2–7.3 removes P and C from the checklist for weak processes, and §8.5 removes their product too.
  • §9.15 is problem 3.18’s argument at 125 GeV: a two-photon final state, an angular distribution, and a spin–parity assignment.

Check yourself — the chapter, end to end

0/5 answered · 0 correct

  1. 1.Λpπ\Lambda \to p\pi^- violates II, IzI_z, SS and PP — and happens, with a lifetime of 2.6×10102.6\times10^{-10} s. What does that tell you?

  2. 2.Problems 3.11, 3.12 and 3.14 all come out as ratios of exactly 1 or 1/2. What structural feature causes that?

    Hint: Look at the isospins of the initial state in each.

  3. 3.In problem 3.3 the free-proton threshold for Λˉ\bar\Lambda production is 4.88 GeV, and the beam was 4.6 GeV. How was the anti-lambda seen at all?

  4. 4.The solutions above disagree with the book's printed answer for problems 3.19 and 3.21. What went wrong in the book?

  5. 5.Why does Super-Kamiokande quote its best limit on pe+π0p \to e^+\pi^0 rather than on pK+νp \to K^+\nu (problem 3.26)?

Study aid derived from A. Bettini, Introduction to Elementary Particle Physics, 3rd ed., Cambridge University Press 2024 — published Open Access under CC-BY-NC 4.0, DOI 10.1017/9781009440745. Not the book: an independently written interactive companion, figures redrawn.