Past exam of the sub-atomic physics course of the University of Oxford / 2016 / 2 / b / Solution

The charged meson decays proceed through a virtual $W^{+}$:The first quark vertex has CKM factor $V_{ud}$, while the second has $V_{us}$. In the simple Cabibbo description these are approximately $\cos {\theta }_C$ and $\sin {\theta }_C$, so the strange amplitude carries a suppression $\tan {\theta }_C$ relative to the nonstrange one, or ${\tan }^2{\theta }_C\simeq 0.053$ in their corresponding squared weak-coupling factors.

For a charged pseudoscalar and a negligibly massive neutrino, the leading leptonic energy width has the formin natural units. The helicity factor $m_{\mu }^2$ is common to these two muon channels, while the threshold factor is much less favorable for the pion. Using the usual masses, its value is about $0.182$ for the pion and $0.910$ for the charged kaon: the pion's threshold factor alone is only about one fifth as large. Including the explicit $m_P$ factor gives a kinematic ratio about $0.0564$. The pseudoscalar decay constants $f_P$ also matter, so Lorentz-invariant phase space and CKM factors alone do not determine the full lifetime ratio.^[ref]

For $K^{+}\to {\pi }^{+}{\pi }^0$, one tree topology has the spectator $u$ unchanged and $\bar{s}\to \bar{u}+W^{+\ast }$ followed by $W^{+\ast }\to u\bar{d}$. The spectator $u$ can combine with $\bar{d}$ to form ${\pi }^{+}$, and the $u\bar{u}$ component supplies ${\pi }^0$. This carries the strangeness-changing weak factor $V_{us}^{\ast }{V}_{ud}$. Hadronization, other quark rearrangements and strong final-state interactions determine the hadronic amplitude; the diagram is not a complete precision width calculation. The small pion–muon energy release suppresses the pion leptonic channel kinematically, whereas the strange weak vertex suppresses the kaon channels through Cabibbo suppression.