Strange quark decay into two down quarks and an anti-down quark I saw that a $\Sigma^+$ can decay into $n+\pi^+$, which means that the $s$-quark must decay into $dd\bar{d}$. However, is there a Feynman diagram to represent this? I cannot find one for either the $\Sigma^+$ decay or the $s$-quark decay. I have only just started learning so I am not sure. Thank you.
 A: One possibility for this decay is as follows. Note that the initial quark content is $uus$ while the final quark content is $udd+u\bar{d}$. 
The $s$-quark can emit a $W^-$ and so turn into a $u$-quark, and this $W^-$ can then be absorbed by one of the initial $u$ quarks turning it into a $d$-quark. We then have quark content $uud$. A gluon (or a photon or a $Z$-boson) can then be emitted (from any of these quarks) and subsequently turn into a $d\bar{d}$ pair. We then have the correct final quark content, and these five quarks can then form into an $n+\pi^+$ final state.
A: This is a leading order Feynman diagram (which may be referred to as a 'gluonic penguin') for a $s \to dd\bar{d}$ transition. It also holds for any other kind of $q \to q'q'\bar{q'}$ flavour-changing neutral current transition (i.e. when $q$ and $q'$ have the same charge). FCNC decays are pretty interesting because the leading order diagrams contain a loop, which makes them more sensitive to contributions from BSM particles entering at loop-level.

