§8.8CP Violation in the Decay

§8.5 established CP violation in the mixing: the propagating states are not CP eigenstates, and the impurity is $\varepsilon$. This section asks the independent question — do the decay amplitudes themselves differ between a process and its CP conjugate?

The answer is yes, by a further factor of a thousand, and extracting it took from 1970 to 2002.

Two conditions, and both are necessary

Start with the general statement, because it governs §8.9 and §8.10 as well. Suppose a decay proceeds through two amplitudes. Each carries a weak phase, which flips sign under CP, and a strong phase weak phase vs strong phase weak phases change sign under CP; strong phases do not. A rate asymmetry needs both kinds of phase difference: it goes as sin(δ_{S1} − δ_{S2}) sin(φ_{W1} − φ_{W2}), so either one alone gives nothing. defined in §8.8 — open in glossary from final-state interactions, which does not:

Subtract the two rates and almost everything cancels:

A product of two sines. Set either to zero and there is no asymmetry, however large the other is. That is the section’s central claim, and it is worth playing with rather than reading:

The kaon case: two isospin amplitudes

For $K\to2\pi$ the two amplitudes are supplied by isospin. The two pions are in a spatially symmetric state, so their isospin wave function must be symmetric too: $I = 0$ or $I = 2$, never 1. Call the corresponding weak amplitudes $A_0$ and $A_2$. Then (Eq. 8.67)

and the conjugate amplitudes are the same with $A_{0,2} \to A_{0,2}^*$ — which is CPT, not an assumption about the decay.

Experimentally $A_0$ dominates: $|A_2|/|A_0| \approx 1/22$, the ΔI = 1/2 rule. That single number can be checked against the measured rate ratio, and doing so turns up something:

ε′, and why it is so hard

With the Wu–Yang convention ($A_0$ real and positive), the two observable amplitude ratios η₊₋ and η₀₀ η₊₋ and η₀₀ the amplitude ratios A(K_L → ππ)/A(K_S → ππ) for the charged and neutral pion pairs. Equal to each other if there is no direct CP violation, which is why their difference is the observable. defined in §8.8 — open in glossary come out as, together with the direct-violation parameter ε′ ε′ the direct-CP-violating parameter of the kaon system. Only ever quoted as the ratio Re(ε′/ε) = (1.66 ± 0.23) × 10⁻³ — direct violation is a further thousand times smaller than indirect. defined in §8.8 — open in glossary ,

Read $\varepsilon'$ carefully — it is zero unless $A_2$ is both non-zero and non-real. Non-zero is the ΔI = 1/2 rule being imperfect; non-real is a weak phase. And the $e^{i(\delta_2-\delta_0)}$ is the strong phase that makes the real part observable. All three conditions of the bridge, in one expression.

The trouble is the size. If CP violation lived only in the mixing then $\eta_{+-} = \eta_{00}$ exactly, so the search is for a difference between two nearly equal numbers:

eta_ratio, eps = 0.9950, 2.232e-3
d = 1 - eta_ratio**2
print("the double ratio and what it gives:")
print(f"  |eta_00 / eta_+-| = {eta_ratio} +- 0.0007")
print(f"  1 - |eta_00/eta_+-|^2 = {d:.3e}")
print(f"  Re(eps'/eps) = (1/6) x {d:.3e} = {d/6:.3e}")
print( "  book (8.78): (1.66 +- 0.23)e-03      -- exact agreement")

print("\nso how big is direct CP violation, absolutely?")
print(f"  |eps|          = {eps:.3e}   (violation in the MIXING)")
print(f"  Re(eps'/eps)   = {d/6:.3e}")
print(f"  |eps'|         ~ {eps*d/6:.2e}   (violation in the DECAY)")

print("\nthree levels, each about a thousand times smaller than the last:")
print( "  CP-conserving amplitude              1")
print(f"  mixing violation      |eps|       {eps:.1e}")
print(f"  decay violation       |eps'|      {eps*d/6:.1e}")

the double ratio and what it gives:
|eta_00 / eta_+-| = 0.995 +- 0.0007
1 - |eta_00/eta_+-|^2 = 9.975e-03
Re(eps'/eps) = (1/6) x 9.975e-03 = 1.662e-03
book (8.78): (1.66 +- 0.23)e-03      -- exact agreement

so how big is direct CP violation, absolutely?
|eps|          = 2.232e-03   (violation in the MIXING)
Re(eps'/eps)   = 1.662e-03
|eps'|         ~ 3.71e-06   (violation in the DECAY)

three levels, each about a thousand times smaller than the last:
CP-conserving amplitude              1
mixing violation      |eps|       2.2e-03
decay violation       |eps'|      3.7e-06

Thirty years, and a standoff

The measurement is worth telling as history because it shows what a $10^{-3}$ effect on top of a $10^{-3}$ effect actually costs.

import numpy as np
m = [('NA31', 2.30, 0.65), ('E731', -0.74, 0.56), ('KTeV', 2.07, 0.28), ('NA48', 1.47, 0.22)]

print("NA31 vs E731, 1993:")
d, sd = 2.30 - (-0.74), np.hypot(0.65, 0.56)
print(f"  2.30 +- 0.65  against  -0.74 +- 0.56")
print(f"  difference {d:.2f} +- {sd:.3f}  ->  {d/sd:.1f} sigma apart")
print( "  one claimed a 3.5 sigma discovery, the other was consistent with zero.")

print("\nKTeV vs NA48, the next generation:")
d2, sd2 = 2.07 - 1.47, np.hypot(0.28, 0.22)
print(f"  2.07 +- 0.28  against  1.47 +- 0.22")
print(f"  difference {d2:.2f} +- {sd2:.3f}  ->  {d2/sd2:.1f} sigma apart")
print( '  the book says "the two values agree"; 1.7 sigma is marginal but fair.')

wsum = sum(v/s**2 for _, v, s in m); w = sum(1/s**2 for _, v, s in m)
avg, err = wsum/w, 1/np.sqrt(w)
chi2 = sum(((v-avg)/s)**2 for _, v, s in m)
print(f"\nnaive weighted average of all four: {avg:.2f} +- {err:.2f}")
print(f"  chi^2 = {chi2:.1f} for 3 dof, so the four are NOT mutually consistent;")
print( "  the PDG inflates the error for exactly this reason, reaching")
print( "  book (8.78): 1.66 +- 0.23")

NA31 vs E731, 1993:
2.30 +- 0.65  against  -0.74 +- 0.56
difference 3.04 +- 0.858  ->  3.5 sigma apart
one claimed a 3.5 sigma discovery, the other was consistent with zero.

KTeV vs NA48, the next generation:
2.07 +- 0.28  against  1.47 +- 0.22
difference 0.60 +- 0.356  ->  1.7 sigma apart
the book says "the two values agree"; 1.7 sigma is marginal but fair.

naive weighted average of all four: 1.54 +- 0.16
chi^2 = 21.6 for 3 dof, so the four are NOT mutually consistent;
the PDG inflates the error for exactly this reason, reaching
book (8.78): 1.66 +- 0.23

Three experimental obstacles made it that hard, and all three are named in the book:

• $K_L\to\pi^0\pi^0$ has BR $= 10^{-3}$, and sits under $K_L\to\pi^0\pi^0\pi^0$, which is 200 times more frequent and gives the same photons plus two more. Losing two photons out of six turns the background into the signal.
• The two kaons have wildly different decay lengths. At 110 GeV, $\lambda_L = 3.4$ km against $\lambda_S = 6$ m — yet the two decay distributions inside the detector must be made as similar as possible, or the acceptance does not cancel.
• Everything must be simultaneous. Both final states at once so the flux cancels; both beams at once with matched spectra so the acceptance does.

Where this goes next

CP violation in the decay is now established in the kaon at $\mathrm{Re}(\varepsilon'/\varepsilon) = 1.66\times10^{-3}$ — direct violation being a thousand times smaller again than the indirect violation of §8.5.

The B mesons make it much easier. Their masses open hundreds of channels, so individual branching ratios are $\sim10^{-5}$ and asymmetries of several per cent are within reach of $10^8$ pairs — no double ratio required, because the rates of two charge-conjugate decays can simply be compared.

§8.9 applies all of this to charm, the only up-type system available, where CP violation was not seen until 2019 and the observable is a difference of two asymmetries — the double-ratio trick one level up. §8.10 then does the charged B, where there is no mixing at all and the payoff is the unitarity-triangle angle $\gamma$ from tree diagrams alone.