Lambda -> n + pi^0 decay

1. Jun 26, 2011

Sebastian

The $\Lambda$ baryon (quark content $uds$) decays into $n + \pi^0$ or $p + \pi^-$. In the case $\Lambda \to p + \pi^-$, the $s$ quark decays into a $u$ quark, releasing a $W^-$ in the process (which subsequently decays into a $\pi^-$ meson). What happens in the $\Lambda \to n + \pi^0$ case? (I tried Google, but couldn't find anything about this specific decay.)

Thanks!

2. Jun 26, 2011

phyzguy

At the quark level, these two decays are the same. The s quark decays into a u quark plus a W-, which decays into and up-bar antiquark and a down quark. In the first case, the u quark from the s decay ends up with the u and d from the original lambda, and the upbar and down pair together, so we end up with a proton and a pi-. In the second case, the d quark from the W- decay ends up with the u and d from the original lambda, and the upbar and up quark pair together, so we end up with a neutron and a pi-0.