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 ways — is 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
| You should now have↕ | What that means in practice↕ | § |
|---|---|---|
| the different types of symmetry | Gauge 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 number | P 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 invariance | A 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 pions | π⁻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 numbers | Additive 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 isospins | A 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.
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 number | strong | electromagnetic | weak |
|---|---|---|---|
| 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 , , and , 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 isospin amplitude one complex amplitude per total-isospin channel (A_{1/2}, A_{3/2}); every cross-section in a related family is a Clebsch–Gordan combination of the same few, so ratios follow with no dynamical input at all.
defined in §3.8-3.10 — open in glossary
as the only unknowns. The widget from §3.9 does exactly that, from
lib/cg.ts rather than from a table — set for the -region
problems and read the rows.
πN cross-section ratios from isospin alone
| reaction | amplitude | σ (arb.) | ÷ σ(π⁻p elastic) |
|---|---|---|---|
| π⁺p → π⁺p(3.42) | 1 A₃⁄₂ | 1.0000 | 9.00 |
| π⁻p → π⁻p(3.44) | 1/3 A₃⁄₂ + 2/3 A₁⁄₂ | 0.1111 | 1.00 |
| π⁻p → π⁰n(3.43) | √2/3 A₃⁄₂ − √2/3 A₁⁄₂ | 0.2222 | 2.00 |
| π⁻n → π⁻n(3.45) | 1 A₃⁄₂ | 1.0000 | 9.00 |
| π⁺n → π⁺n | 1/3 A₃⁄₂ + 2/3 A₁⁄₂ | 0.1111 | 1.00 |
| π⁺n → π⁰p | √2/3 A₃⁄₂ − √2/3 A₁⁄₂ | 0.2222 | 2.00 |
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 and — and their squares are in the ratio .
Once you recognise the shape, the answer arrives before the arithmetic. A 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 . 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- For each interaction — strong (S), electromagnetic (EM) and weak (W) — mark whether each of , , , , , , , , , is conserved.
- A beam hits a liquid hydrogen target. Find the threshold energy for production.
- MeV, MeV, MeV, MeV, MeV
- The anti-lambda was discovered by Baldo-Ceolin and Prowse in 1958 with a beam of GeV on an emulsion stack. (a) What is the lightest final state containing a from ? (b) Find the threshold for free protons, and approximately for protons bound with Fermi momentum MeV. (c) If cm pions were produced m upstream, how many reached the emulsion?
- MeV, MeV, MeV, MeV, m
- For each reaction: is it allowed? If not, why (there may be several reasons)? If yes, by which interaction? (1) ; (2) ; (3) ; (4) ; (5) ; (6) ; (7) ; (8) ; (9) ; (10) .
- Allowed or not, and why? (1) ; (2) ; (3) ; (4) ; (5) ; (6) ; (7) ; (8) ; (9) ; (10) .
- Give the reasons forbidding each decay: (a) ; (b) ; (c) ; (d) .
- Which of these are allowed and which are forbidden by strangeness conservation? (a) ; (b) ; (c) ; (d) .
- Allowed or not, and why? (a) ; (b) ; (c) ; (d) ; (e) ; (f) ; (g) ; (h) .
- Assuming (unrealistically) that they proceed only through the channel, evaluate the ratios of the cross-sections , and at the same energy.
- Evaluate the same three cross-sections with both isospin amplitudes and contributing.
- Evaluate the ratio of the cross-sections of and at the same energy.
- Evaluate the ratio of the cross-sections of and at the same . (He and H form an isospin doublet.)
- Evaluate the ratio at the same energy.
- Evaluate at the same energy.
- Express the ratios of , and in terms of the isospin amplitudes , and .
- Express the ratio of the elastic and charge-exchange cross-sections in terms of and .
- A is captured by a deuteron () giving . (a) If capture is from an wave, what are the total spin and orbital momentum of the two neutrons? (b) Show that if capture is from a state, the neutrons are in a spin singlet.
- Positronium is an atom. (1) What relation does conservation impose between , and ? (2) What relation between , and allows ? (3) What is the minimum number of photons for ortho-positronium () and para-positronium ()?
- From which of the states , , , , , , , , , can proceed with parity conservation? (1) for any ; (2) for .
- Consider the strong process (where means both and ). What angular momenta are possible (1) if the initial total isospin is , and (2) if ?
- From which of the states , , , , , , , , , can proceed if the two pions are (1) in an wave, (2) in a wave, (3) in a wave?
- With , and , which are allowed? (a) ; (b) ; (c) ; (d) .
- With , , , and , which are allowed? (a) ; (b) ; (c) ; (d) ; (e) .
- An meson of energy GeV, moving along , decays to . (1) If the photons go along and , what are their energies? (2) If they are emitted at equal and opposite angles to , what is the angle between them?
- MeV, MeV
- The has . (1) What is the ratio between the decay rates and ? (2) What would it have been if ?
- A water Cherenkov detector has . (1) Find the minimum velocity for radiation. (2) Find the minimum kinetic energy for an electron and for a meson. (3) Is the charged particle above threshold in and in ?
- , MeV, MeV, MeV, MeV
- Forbidden or allowed, with justification? (a) ; (b) ; (c) ; (d) ; (e) ; (f) ; (g) .
- Identify the particle : (a) ; (b) ; (c) .
- The is produced in . Knowing the isospin and third component of every other particle, establish and of the .
- Consider capture at rest giving . Calculate the energy of the photon and the kinetic energy of the neutron.
- MeV, MeV, MeV
- A 12 GeV beam in a bubble chamber gives an event with two positive tracks from the primary vertex and two s pointing back to it. The first has GeV, GeV, ; the second has GeV, GeV, . With 5 % accuracy, identify the two neutral particles and guess the reaction.
- ; GeV, 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'}") 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 states as ”, , , and ”. does not exist — with gives — and is missing.
- 3.19(2) then concludes “only ”, which follows from the dropped . Restore it and it passes every test, so the answer is and .
- 3.21 concludes “(1) ” while its own table puts the Y under . is and an -wave is .
- 3.26 quotes 469 MeV for the positron in , which is the formula with the term dropped; the value is 459 MeV. It also uses the neutral kaon mass for a .
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 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. violates , , and — and happens, with a lifetime of s. What does that tell you?
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.In problem 3.3 the free-proton threshold for production is 4.88 GeV, and the beam was 4.6 GeV. How was the anti-lambda seen at all?
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.Why does Super-Kamiokande quote its best limit on rather than on (problem 3.26)?