Suppression of kaon decay compared to pion decay due to cabibbo angle

Question

I'm a bit confused by something I've read around and also heard in class.

Let's look at these decay channels for kaons and pions:

$$K^- \rightarrow \mu^- \ \overline{\nu}_\mu, \quad \pi^- \rightarrow \mu^- \ \overline{\nu}_\mu \ .$$ The argument I've heard is that the kaon decay is suppressed compared to the pion one because of quark mixing and CKM matrix.

$$\frac{\Gamma_{K}}{\Gamma_{\pi}} \propto \frac{|V_{us}|^2}{|V_{ud}|^2} = \tan{\theta_c}^2 \simeq 0.05$$

Which is fine and makes a lot of sense, however I wanted to try to back this up with the actual experimental data showing such a suppression. I went to the particle data group website and looked for the branching ratios of the Kaon and the pion. However these branching ratios are not as different as I expected them to be (their ratio is $63.56/99.98 \sim 0.635$). My guess is that we have to compare the actual decay widths and not the branching ratio, but still my calculations did not lead to the expected suppression.

My guess is that I also have to take into account the mass of the decaying particle but I'm still not able to make anything useful out of this. I hope that someone may clear things up and tell me how exactly the experimental data should suggest a suppression of the Kaon decay compared to the pion decay. Thanks in advance

Answer 1

The decay width of a $P_{ \ell 2}$ decay is given by

$\Gamma_{P_{ \ell 2}} =\frac{G_F^2 |V_P|^2 F_P^2}{4 \pi} M_P m_\ell^2 (1-m_\ell^2/M_P^2)^2$,

where $P$ denotes the pseudoscalar meson ($\pi^\pm$ or $K^\pm$), $\ell$ denotes the charged lepton ($e$ or $\mu$) and $V_{\pi^\pm}= V_{ud}$, $V_{K^\pm}=V_{us}$ are the relevant Cabibbo-Kobayashi-Maskawa matrix elements. $F_{\pi^\pm}$ is the decay constant of the charged pion and $F_{K^\pm}$ the one of the charged kaon. Taking the ratio $\Gamma_{K_{\ell 2}}/ \Gamma_{\pi_{\ell 2}}$ answers your question.