How would one go about drawing the Feynman diagram for the following reaction?

$$\Xi^{-} \rightarrow n\space+\pi^{-}$$

In terms of quarks:

$$dss\rightarrow udd\space + d\overline{u} $$

Could the reaction be mediated by two $W^{\pm}$ bosons similar to the $K^{0}\rightarrow \overline{K^{0}}$ reaction? In the $\Xi^{-}$ case, could the $s$ quarks change into two $d$ quarks while one of them also emits a gluon which gives the $u\overline{u}$ quarks ?

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Image from B.R. Martin's Nuclear and Particle Physics Ch.6 - Electroweak reactions - page 202.


1 Answer 1


Indeed, this must be a doubly weak decay, as you can see from the BR in PDG, and must involve two virtual Ws to eliminate two s quarks. The basic reaction is $$s\rightarrow u W^{-} \to u d \overline{u}, $$ and you need two of those.

So, overall $$ dss \to d ~~ u d \overline{u} ~~ u d \overline{u}\leadsto udd+ d\overline{u}, $$ with a $\overline{u} u$ pair annihilating to a gluon you attach anywhere you like.


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