= Solution
The long-lived, electrically neutral <kaon> often traverses the <particle detector> without decaying into a reconstructable charged final state. Identifying it through a hadronic interaction or missing <four-momentum> is much less efficient than reconstructing the short-lived <kaon> through its nearby two-<pion> <decay vertex>. The efficiencies must be applied to the observed counts before taking their asymmetry:
$$
N_S=\frac{1107}{0.93}=1190.32,\qquad N_L=\frac{49}{0.05}=980,
\qquad \mathcal R=\frac{N_S-N_L}{N_S+N_L}=0.09691.
$$
Using the leading aligned-amplitude approximation from <2/c>,
$$
\boxed{\theta_C=\arctan\sqrt{\mathcal R/2}=0.21667\ \mathrm{rad}=12.41^\circ.}
$$
The exact aligned formula instead gives $12.43^\circ$, an immaterial correction here. Treating the observed counts as independent <Poisson random variables> and the efficiencies as known, <propagation of uncertainty> gives
$$
\sigma_S=\frac{\sqrt{1107}}{0.93}=35.78,\qquad \sigma_L=\frac{\sqrt{49}}{0.05}=140,
$$
$$
\sigma_{\mathcal R}^2=\left[\frac{2N_L}{(N_S+N_L)^2}\right]^2\sigma_S^2+
\left[\frac{2N_S}{(N_S+N_L)^2}\right]^2\sigma_L^2,
\qquad \boxed{\sigma_{\mathcal R}=0.0723.}
$$
The separate contributions to $\sigma_{\mathcal R}$ are $0.0149$ and $0.0708$; the small $K_L$ sample dominates. Linear <propagation of uncertainty> yields $\sigma(\theta_C)\simeq4.49^\circ$, although the nonlinear extraction and physical boundary make a symmetric <confidence interval> a rough summary. Unknown efficiency uncertainties, especially that of $K_L$, would be additional systematic contributions.
The supplied <CKM matrix> gives $\sin\theta_C\simeq0.224$, hence $\theta_C\simeq12.94^\circ$ and $\mathcal R\simeq2(0.224/0.975)^2=0.1056$. The asymmetry difference is only about $0.12$ statistical <standard deviations>, so the measurement is compatible. Without the phase assumption, these data measure $\tan^2\theta_C\cos\delta$ to leading order and do not uniquely measure the <Cabibbo angle>.
Back to article page
