I W boson decay

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Hello! In Modern Particle Physics by Mark Thomson, in the Electroweak Unification chapter, pg. 412 he talks about the branching ration of the W decay to quarks. And for this he includes both the ##W\to q \bar{q'}## and ##W\to q \bar{q'}g## i.e. the state with a gluon and 2 quarks in the final state. I am not sure I understand this. Isn't the decay defined just at the W vertex? The gluon is produced later by the quark and it has nothing to do with the W properties. Also, one can have a photon, too coming out of the quark and even in the leptonic case, one can have a photon emission by one of the leptons resulting from the decay. Why is it just this gluon case considered? Thank you!
 
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"Later" is not a well-defined concept if it happens in the same Feynman diagram. If you calculate it your integral will also run over an "earlier" gluon emission.
The quarks will hadronize in some way afterwards anyway, but you can also have the emission of a gluon with a high energy - in that case you get three jets instead of two. Photon emission is possible as well, it is relatively rare as the electromagnetic interaction is much weaker.
 
"Later" is not a well-defined concept if it happens in the same Feynman diagram. If you calculate it your integral will also run over an "earlier" gluon emission.
The quarks will hadronize in some way afterwards anyway, but you can also have the emission of a gluon with a high energy - in that case you get three jets instead of two. Photon emission is possible as well, it is relatively rare as the electromagnetic interaction is much weaker.
I am sorry, I am still confused. The diagram he shows, is not "1 particle irreducible" (1PI). I remember from my QFT class that, when trying to understand a process, one looks only at diagrams that't can't be split into 2 or more diagrams (and this one can). For example, one can add lots of self interactions on the quark propagator, but that doesn't contribute to the Wqq vertex. Of course Wqq is just a first order approximation, but but higher order must still be 1PI, while W->qqg is not.
 

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I am sorry, I am still confused. The diagram he shows, is not "1 particle irreducible" (1PI). I remember from my QFT class that, when trying to understand a process, one looks only at diagrams that't can't be split into 2 or more diagrams (and this one can). For example, one can add lots of self interactions on the quark propagator, but that doesn't contribute to the Wqq vertex. Of course Wqq is just a first order approximation, but but higher order must still be 1PI, while W->qqg is not.
I do not think you have understood what 1PI means and when it is used. It is used in connection to propagators, not interaction diagrams. What you want is the interaction diagram with no 1PIs on the external legs. The leg with the extra gluon is not a 1PI because it has three legs. A 1PI only has two legs, the ”incoming” and ”outgoing” particle.
 
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That depends on the process you want to understand. Do you want to understand vertices with W or do you want to understand W decays?
 

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