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I have a questions regarding Feynman diagrams when it involves hadrons. I think the easiest way is considering this reactions:


1) $\Lambda_{c}^{+} \rightarrow p+\bar{K}^{0}$

I need to draw the feynman diagram for this reaction. I know that the quark contents of the particles are the following: $\Lambda_{c}^{+}=u d c,\; p=u u d, \;\bar{K}^{0}=s \bar{d}$.
I know that it is a weak interaction and the c-quark decay directly to the s-quark or the d-quark + any other particles allowed by the appropriate laws. So after analyzing the possible decays (quark mixing and so on), I made the following feynman diagram:
enter image description here

However I looked up the solution, and it was very similar but still different:
enter image description here
The solution doesn't group together the quarks in the correct way, so that it looks like the quarks form the hadrons given in the interaction. Is that because it's not important that you group them together, as long the correct quark constituents are drawn on the right side? I mean if the reaction created the $\Lambda=uds$, $\pi^{+}= u \bar{d}$, would they have the excact same feynman diagram?

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A Feynman diagram shouldn't show spectator particles. It shouldn't care about anything composite. The fact that the incoming and outgoing quarks remain trapped in a certain hadron isn't relevant to the interaction the diagram shows. You just want the classic $2$-in,-$2$-out$^\dagger$ setup per diagram, so each vertex/edge is elementary, and you can explain the hadronic implications afterwards.

$^\dagger$ Where anti-in=out, anti-out=in.

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