Past exam of the particle physics course of the University of Oxford / 2022 / 2 / d

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:Using the leading aligned-amplitude approximation from c,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 givesThe separate contributions to ${\sigma }_{\mathcal{R}}$ are $0.0149$ and $0.0708$; the small $K_L$ sample dominates. Linear propagation of uncertainty yields $\sigma ({\theta }_C)\simeq 4.49^{\circ }$, 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 $\sin {\theta }_C\simeq 0.224$, hence ${\theta }_C\simeq 12.94^{\circ }$ and $\mathcal{R}\simeq 2(0.224/0.975{)}^2=0.1056$. 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.