Chapter 8 — Summary and Problems · Elementary Particle Physics

Twelve problems for a chapter containing four particle systems, because the four are one problem with different numbers in it.

🎯 Why this matters

That is the chapter’s real claim, stated as a workload rather than as a sentence. Learn the kaon and the B mesons cost almost nothing extra — the equation is the same one and only the constants move.

Twelve problems, and they are unusually well chosen — three of them are the chapter’s own derivations handed back to the reader, and several of the rest quietly re-derive results the text asserts.

six are kaon kinematics and conservation laws — 8.1, 8.3, 8.4, 8.5, 8.6, 8.12. Between them they rebuild the whole experimental basis of §8.2–8.5: purification by flight, associated production, and why an absorber regenerates;
two are group theory — 8.2 and 8.12, on which initial and final states can connect;
three are the algebra of ε′ — 8.9, 8.10, 8.11, which the text explicitly skips as “not difficult, but long”;
and 8.7 and 8.8 are the beauty factory, read one event at a time and then one exposure at a time.

The book supplies no solutions. All twelve are worked here.

What the chapter established

Every entry traces back to one number: δ₁₃ = 65.6°, the irremovable CKM phase of §7.11.

what↕	how↕	the number↕
Neutral mesons mix	the propagating states are not the flavour states	four systems, $T/\tau$ = 13, 1600, 8, 0.24
The superposition is coherent	regeneration — an absorber brings the $K_S$ back	$1 \to \tfrac12 \to \tfrac14 \to \tfrac18$
CP is violated in the mixing	$K_L \to \pi^+\pi^-$ , which CP forbids	$\|\varepsilon\| = 2.232\times10^{-3}$
…and in the decay	a double ratio of four rates	$\mathrm{Re}(\varepsilon'/\varepsilon) = 1.66\times10^{-3}$
…and in the interference	$B^0\to J/\psi K_S$ , tagged and time-resolved	$\sin2\beta = 0.675 \pm 0.020$
It all fits one triangle	two sides from moduli, two angles from asymmetries	$(\bar\rho, \bar\eta) = (0.159, 0.348)$

One number decides what each of the four systems can be measured at all: T / τ

0.1110100100010⁴

oscillation period ÷ lifetime (log scale) — how much of one cycle happens before the meson is gone

you SEE it oscillatenothing oscillates before everything decaysB_s⁰0.24 — 12.5 cyclesΔm to 0.03 %B⁰8.2 — a third of a cycleneeds a ±6 ps fitK⁰13.3 — a quarter cyclebut τ_S ≠ τ_L rescues itD⁰1584 — 0.002 of a cyclestill never seen directlyNote the ordering is not the mass ordering, and not the lifetime ordering.It is Δm × τ — a mixing rate against a decay rate — and the two are set by unrelated physics.The B_s is the hardest of the four to produce and yields the best-measured Δm in the chapter.

Supplied — the book gives the four systems four separate sections and never puts their organising number side by side. Everything about the experimental difficulty of a neutral-meson system is on this axis. Left of 1 you watch the oscillation happen and precision is bought with cycles; right of ~100 the meson has decayed before the phases separate and you are reduced to indirect extraction, which is why charm mixing was the last to be established and its oscillation has still never been observed directly. The K⁰ sits in the awkward middle and is measurable only because its two eigenstates have wildly different lifetimes — a second lever the other three do not have.

The numbers, all at once

every numeric answer in the chapter's problems

import numpy as np
mphi, mKc, mK0, mp, mn, mL, mpic = 1019.461, 493.677, 497.611, 938.272, 939.5654, 1115.683, 139.57039
c = 0.299792458                                  # m/ns, and um/ps

pc = np.sqrt((mphi/2)**2 - mKc**2); p0 = np.sqrt((mphi/2)**2 - mK0**2)
print(f"8.1   phi -> K+K- / K0K0bar = {(pc/p0)**3:.3f}  (measured 1.44), NOT 1")
print(f"        a {(mK0-mKc)/mK0*100:.2f}% mass difference gives a {((pc/p0)**3-1)*100:.0f}% rate difference, because")
print( "        L = 1 puts p*^3 in front and both pairs are near threshold")
print( "      the neutral pair is PURELY K1 K2 -- frequencies 0, 1, 0")

bg = 10000/mK0; lS, lL = bg*c*89.54e-3, bg*c*51.16
print(f"8.3   10 GeV beam: beta.gamma = {bg:.1f}, lambda_S = {lS:.3f} m, lambda_L = {lL:.0f} m")
print(f"        N_S/N_L = 1 at production, {np.exp(-10/lS)/np.exp(-10/lL):.1e} at 10 m")
print(f"        2pi fraction with CP conserved = {0.5*0.9989:.4f}, all in the first metre")

lam = (2000/mKc)*c*12.38
print(f"8.4   2 GeV K+ beam: lambda = {lam:.2f} m, 10% survival at {lam*np.log(10):.1f} m")

th = lambda M: (M**2 - mpic**2 - mp**2)/(2*mp)
print(f"8.5   pi- p -> K0 Lambda    threshold {th(mK0+mL):6.1f} MeV")
print(f"      pi- p -> K0bar n K0   threshold {th(mK0+mn+mK0):6.1f} MeV   -- {th(mK0+mn+mK0)/th(mK0+mL):.1f}x higher,")
print( "        because NO baryon carries S = +1")

EK = np.sqrt(400**2+mK0**2); rs = np.sqrt(mK0**2+mp**2+2*mp*EK)
print(f"8.6   at p_K = 400 MeV, sqrt(s) = {rs:.0f} MeV:")
print( "        K0 p  -> K+ n only (elastic + charge exchange)")
print( "        K0bar p -> Lambda pi+, Sigma+ pi0, Sigma0 pi+ all OPEN")
print( "        -> the K0bar cross-section is far larger.  that IS regeneration.")

step = 0.56*299.792458*1.520
dt = (500-120)/step*1.520; ph = 0.5065*dt
print(f"8.7   one lifetime of flight = {step:.1f} um")
print(f"        120 um = {120/step:.2f} lifetimes;  500 um = {500/step:.2f} lifetimes")
print(f"        between the two decays: dt = {dt:.3f} ps, dm x dt = {ph:.3f} rad")
print(f"        P(mu+) = {(1+np.cos(ph))/2:.3f},  P(mu-) = {(1-np.cos(ph))/2:.3f}  -- both signs occur")

smu = 86.8/10.58**2
print(f"8.8   100 fb^-1 at 1e34 = {1e41/1e34:.1e} s = {1e41/1e34/86400:.0f} days")
print(f"        sigma_had = 3 x {smu:.3f} = {3*smu:.2f} nb -> {1e41*3*smu*1e-33:.1e} pairs")
print(f"        average production-to-decay separation = {step:.0f} um")

prints

8.1   phi -> K+K- / K0K0bar = 1.516  (measured 1.44), NOT 1
      a 0.79% mass difference gives a 52% rate difference, because
      L = 1 puts p*^3 in front and both pairs are near threshold
    the neutral pair is PURELY K1 K2 -- frequencies 0, 1, 0
8.3   10 GeV beam: beta.gamma = 20.1, lambda_S = 0.539 m, lambda_L = 308 m
      N_S/N_L = 1 at production, 9.2e-09 at 10 m
      2pi fraction with CP conserved = 0.4995, all in the first metre
8.4   2 GeV K+ beam: lambda = 15.04 m, 10% survival at 34.6 m
8.5   pi- p -> K0 Lambda    threshold  907.5 MeV
    pi- p -> K0bar n K0   threshold 1515.3 MeV   -- 1.7x higher,
      because NO baryon carries S = +1
8.6   at p_K = 400 MeV, sqrt(s) = 1525 MeV:
      K0 p  -> K+ n only (elastic + charge exchange)
      K0bar p -> Lambda pi+, Sigma+ pi0, Sigma0 pi+ all OPEN
      -> the K0bar cross-section is far larger.  that IS regeneration.
8.7   one lifetime of flight = 255.2 um
      120 um = 0.47 lifetimes;  500 um = 1.96 lifetimes
      between the two decays: dt = 2.263 ps, dm x dt = 1.146 rad
      P(mu+) = 0.706,  P(mu-) = 0.294  -- both signs occur
8.8   100 fb^-1 at 1e34 = 1.0e+07 s = 116 days
      sigma_had = 3 x 0.775 = 2.33 nb -> 2.3e+08 pairs
      average production-to-decay separation = 255 um

Two of those repay a second look.

8.1 is a trap with a payoff. Asked to neglect the kaon mass difference, you get a ratio of 1; asked whether that is legitimate, the answer is emphatically no. A 0.79 % mass difference produces a 52 % rate difference, because the $\phi$ is a vector so the pair comes out in a P wave, the rate carries $p^{*3}$ , and both pairs sit barely above threshold. Then the second half of the problem delivers the KLOE tag of §7.12 for free: $L = 1$ makes the pair antisymmetric, every $K_1K_1$ and $K_2K_2$ term cancels identically, and what is left is purely $K_1K_2$ .

8.5 explains why a kaon beam is pure. Producing a $K^0$ costs a $\Lambda$ and 908 MeV. Producing a $\bar K^0$ requires a partner with $S = +1$ , $B = 1$ , $Q = 0$ — and no such baryon exists, so you must make an entire extra kaon, and the threshold jumps to 1515 MeV. That asymmetry is what lets §8.3 start from a state of definite strangeness at all.

0/12 solved

8.1why a 0.8 % mass difference matterstheory
DAΦNE is an $e^+e^-$ collider at $\sqrt s = m_\phi$ . Calculate the ratio of the annihilation rates into $K^+K^-$ and $K^0\bar K^0$ , neglecting the charged–neutral kaon mass difference. Is that a good approximation? For the $K^0\bar K^0$ case, find the relative frequency of $K_1^0K_1^0$ , $K_1^0K_2^0$ and $K_2^0K_2^0$ .
- $m_\phi = 1019.46$ , $m_{K^\pm} = 493.68$ , $m_{K^0} = 497.61$ MeV
- the $\phi$ is $J^{PC} = 1^{--}$ , so the two pseudoscalars come out in a P wave
- a decay to two spinless particles with orbital $L$ has rate $\propto p^{*\,2L+1}$
8.2which p̄p states can reach which kaon pairtheory
From which of the $\bar pp$ initial states $^1S_0$ , $^3S_1$ , $^1P_1$ , $^3P_0$ , $^3P_1$ , $^3P_2$ can each of $\bar pp \to K^+K^-$ , $\bar pp \to K_1^0K_1^0$ and $\bar pp \to K_1^0K_2^0$ proceed?
- for $\bar pp$ in $^{2S+1}L_J$ : $P = (-1)^{L+1}$ , $C = (-1)^{L+S}$
- for two pseudoscalars with orbital $\ell$ : $J = \ell$ , $P = (-1)^\ell$
- $CP$ of a kaon pair $= \eta_a\eta_b(-1)^\ell$ , with $\eta(K_1) = +1$ , $\eta(K_2) = -1$
8.3purifying a kaon beamtheory
A $\pi^-$ hits a target producing neutral $K$ mesons and $\Lambda$ hyperons. For the component with $p = 10$ GeV: what is the $K_S/K_L$ ratio at production? At $l = 10$ m? What fraction of decays into $2\pi$ would be observed if CP were conserved?
- $\tau_S = 89.54$ ps, $\tau_L = 51.16$ ns, $m_{K^0} = 497.61$ MeV
- $\mathrm{BR}(K_S\to\pi\pi) = 99.89\,\%$
8.4how long a K⁺ beam can betheory
An experiment needs an almost monochromatic $K^+$ beam of $p = 2$ GeV, selected with a magnetic spectrometer and slits. The usable length is limited by the $K^+$ lifetime. At what distance is the intensity reduced to 10 % of its initial value?
- $\tau_{K^+} = 12.38$ ns, $m_{K^+} = 493.68$ MeV
8.5why an anti-strange particle costs so much moretheory
For $\pi^-p \to K^0 X$ and $\pi^-p \to \bar K^0 Y$ , find the minimum masses of $X$ and $Y$ compatible with the conservation laws, and the two threshold energies.
- strong interactions conserve $S$ , $B$ and $Q$ separately
- $m_\Lambda = 1115.68$ , $m_n = 939.57$ , $m_{K^0} = 497.61$ , $m_p = 938.27$ , $m_{\pi} = 139.57$ MeV
8.6why an absorber regeneratestheory
A neutral $K$ beam of $p_K = 400$ MeV hits a liquid hydrogen target. Determine the reaction channels open to each of the $K^0$ and $\bar K^0$ components, and estimate which has the larger cross-section.
- strangeness is conserved by the strong interaction
- $m_\Lambda = 1115.68$ , $m_{\Sigma^+} = 1189.37$ , $m_{\Sigma^0} = 1192.64$ MeV
8.7reading one beauty-factory eventtheory
An asymmetric beauty factory at $\sqrt s = 10\,580$ MeV studies $e^+e^-\to B^0\bar B^0$ , with $\langle\beta\gamma\rangle = 0.56$ . One $B$ decays 120 μm from the primary vertex with a $\mu^-$ among its products. How many lifetimes did it live, and what can be said about the two flavours? The second decays at 0.5 mm, again with a muon. How many lifetimes? Can that muon be positive, and why?
- $\tau_B = 1.520$ ps, $\Delta m_B = 0.5065$ ps⁻¹
- $\bar b \to \bar c\ell^+\nu$ so $B^0 \to \ell^+$ , and $b \to c\ell^-\bar\nu$ so $\bar B^0 \to \ell^-$
8.8what 100 fb⁻¹ buystheory
For a BaBar sample of integrated luminosity 100 fb⁻¹ with $\langle\beta\gamma\rangle = 0.56$ : how many seconds at $L = 10^{34}$ cm⁻²s⁻¹ are needed? Taking $R = 3$ at the $\Upsilon(4S)$ , how many $B^0\bar B^0$ pairs are collected? What is the average separation between production and decay vertices?
- 1 fb⁻¹ $= 10^{39}$ cm⁻²
- $\sigma(e^+e^-\to\mu^+\mu^-) = 86.8\;\text{nb}/s[\text{GeV}^2]$
- $\tau_B = 1.520$ ps
8.9prove η₊₋ = ε + ε′ and η₀₀ = ε − 2ε′theory
Prove expression (8.71), neglecting terms of order higher than the first in $\varepsilon$ and $\varepsilon'$ .
- $|K_S\rangle \propto |K_1^0\rangle + \varepsilon|K_2^0\rangle$ and $|K_L\rangle \propto \varepsilon|K_1^0\rangle + |K_2^0\rangle$ , Eq. (8.27)
- $|K_{1,2}^0\rangle = (|K^0\rangle \pm |\bar K^0\rangle)/\sqrt2$
- Eqs. (8.74) and (8.75): $|\bar A/A| = 1 - 2\mathrm{Re}\,\varepsilon'$ for $\pi^+\pi^-$ and $1 + 4\mathrm{Re}\,\varepsilon'$ for $\pi^0\pi^0$
8.10prove the expression for ε′theory
Prove expression (8.72), $\varepsilon' = \dfrac{i}{\sqrt2}\dfrac{\mathrm{Im}\,A_2}{A_0}e^{i(\delta_2-\delta_0)}$ , to first order.
- Eqs. (8.67a) and (8.67b) for the four amplitudes
- the Wu–Yang convention: choose the phase so that $A_0$ is real and positive
- $\delta_f \equiv (A-\bar A)/(A+\bar A)$ from problem 8.9
8.11prove the amplitude ratiotheory
Prove expression (8.74), $\left|A(\bar K^0\to\pi^+\pi^-)/A(K^0\to\pi^+\pi^-)\right| \simeq 1 - 2\,\mathrm{Re}\,\varepsilon'$ , to first order.
- the two amplitudes of problem 8.10, differing only in the sign of $\phi_2$
- $r \equiv |A_2|/(\sqrt2 A_0) \approx 1/31$ is small, and so is $\phi_2$
8.12isospin of the three-pion final statetheory
For $K_1^0 \to \pi^+\pi^-\pi^0$ and $K_2^0 \to \pi^+\pi^-\pi^0$ , let $\ell$ be the angular momentum of the $\pi^+\pi^-$ pair, $I_{\pi\pi}$ its isospin, and $L$ that of the $\pi^0$ relative to it. Supposing the kaon were much heavier so that higher $\ell$ and $L$ were not suppressed, establish the possible values of $I_{\pi\pi}$ and of the total $I$ , assuming CP is conserved.
- $CP(\pi^+\pi^-\pi^0) = (-1)^{\ell+1}$ , from §8.2
- two pions are bosons: the state must be symmetric under their exchange
- $I_{\pi\pi} = 0$ and 2 are symmetric under exchange; $I_{\pi\pi} = 1$ is antisymmetric

Three kinds of CP violation, and why only the third one is big

producedas fin the MIXINGthe two paths are M and M̄|q/p| ≠ 1 — the long-lived stateis not a CP eigenstateK_L → π⁺π⁻, 1964|ε| = 2.2 × 10⁻³in the DECAYtwo amplitudes to one final stateneeds a strong phase AND a weakone — the only kind B± can showε′/ε, 1999 · charm, 2019|ε′| = 3.7 × 10⁻⁶in the INTERFERENCEdecay directly, or mix firstthen decay — both reach J/ψ K_Sno small amplitude anywhereB⁰ → J/ψ K_S, 2001sin 2β = 0.675The third is 300 times the first and 180 000 times the second — and it is NOT a bigger CKM phase.It is the same δ₁₃, in the one configuration where the two interfering paths have comparable amplitudes.

Supplied — the book introduces the three kinds across four sections and never draws the comparison. The sizes come from the block above (|ε| = 2.232×10⁻³, |ε′| = 3.71×10⁻⁶, sin2β = 0.675), and they are different kinds of quantity — an amplitude ratio, an amplitude ratio, and an asymmetry amplitude — so the ratios between them are indicative rather than exact. The lesson survives that caveat: every one of these numbers is driven by the same single phase δ₁₃ = 65.6°, and what separates them by five orders of magnitude is not the phase but whether the two interfering paths are comparable in size. Eq. (8.63)‘s prefactor is 4|A₁||A₂|, and nature made A₂/A₁ ≈ 1/22 for the kaon and ≈ 1 for the B.

🔑 If you remember only three things

Three of these hand back the chapter’s own derivations. Knowing in advance that an answer is reachable changes what you are willing to attempt.
What separates the four systems is one dimensionless ratio. How far a state oscillates before it decays is the whole of the difference between them.
Several problems quietly re-derive what the text asserted. The chapter states more than it proves, and this page is where the balance is settled.

Where this goes next

Chapter 9 supplies what this chapter kept assuming: the electroweak theory behind the W and the Z, why $M_W$ has the value that makes $G_F$ what it is, and the $g_L$ , $g_R$ that §7.13 named without explaining.

Chapter 10 runs this entire chapter again for neutrinos. The formalism is identical — OscillationLab gets mounted a fifth time — but the mixing angles are large rather than small, and the phenomenon requires neutrino mass, which nothing in Chapters 7–9 provides.

And one thread leaves the Standard Model. Every CP violation established here, added together, is far too small to explain why the universe contains matter and not antimatter. Sakharov’s conditions require CP violation and the CKM phase supplies some — short of what is needed by many orders of magnitude. That gap is among the clearest evidence that the Standard Model is incomplete, and Chapter 12 returns to it.

✅ Check yourself — Chapter 8 in review

0/6 answered · 0 correct

1.Problem 8.1 asks for the φ → K⁺K⁻ / K⁰K̄⁰ rate ratio neglecting the kaon mass difference, then whether that is a good approximation. What is the answer?
2.In problem 8.1's second part, what is the K₁K₂ content of the neutral kaon pair from φ decay?
3.Problem 8.5: why is the threshold for π⁻p → K̄⁰Y so much higher than for π⁻p → K⁰X?
4.Problem 8.6 asks which of K⁰p and K̄⁰p has the larger cross-section at p_K = 400 MeV. What is the answer, and why does it matter?
5.In problem 8.7, the first B decays with a μ⁻ at 120 μm and the second decays at 500 μm. Can the second muon be positive?
6.Problems 8.9–8.11 ask you to derive η₊₋ = ε + ε′, η₀₀ = ε − 2ε′ and |Ā/A| = 1 − 2 Re ε′. What is the significance of the relative factor of −2 between η₊₋ and η₀₀?