Feynman diagram for a strong interaction decay

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SUMMARY

The discussion focuses on the lowest order Feynman diagram for the strong interaction decay process \(\Sigma^{++}_c \rightarrow \Lambda^+_c \pi^+\). The quark compositions are identified as \(\Sigma^{++}_c = (c u u)\), \(\Lambda^+_c = (c u d)\), and \(\pi^+ = (\bar{d} u)\). The main question raised is why the charm quark (C) cannot emit the gluon, with initial considerations pointing towards isospin conservation. However, the conclusion suggests that any quark can emit a gluon, leading to the same final state, indicating that the process is not strictly limited by isospin constraints.

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Gal Winer
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hey,

i want to find the lowest order strong interaction feynman diagram for the following process:
\Sigma^{++}_c \rightarrow \Lambda^+_c \pi^+

the quark composition is:

\Sigma^{++}_c = (c u u) \ \Lambda^+_c = (c u d). \pi^+ =( \bar{d} u )
an image of a correct process is attached .

my question is: why can't the C quark be the one emitting the gluon?
my initial thought was Isospin conservation, as C doesn't carry isospin and the d \bar{d} pair may have a I = 1. but the pair may also have a I=0.

thanks for any help
 

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I don't think it's Isospin, cause in the case of u, it's isospin is 1/2, we look for conservation of isospin only in total of all the process, i.e isospin in LHS should be the same as the RHS, not part of RHS with part of LHS.

But who knows...

I think that the case of c transmitting gluon would have different products, as in

(u u d)+ (\bar{d} c)
which is a proton and some Kaon I believe.
 
Any of the quarks can emit a gluon and result in the same final state.
 

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