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I am trying to do a tree level calculation of cross section in a process that involves sbottom exchange. There is a $ \widetilde{b} q \bar{q} $ vertex, where $q$ and $\bar q$ are quark and anti-quark of the same flavor. I have following questions:

  1. How is the color conserved in such vertices? The problem is that (s)quarks carry only one unit of color.

  2. What will be the color factor from this vertex if written explicitly? (Just like Griffiths does in his book)

PS: My knowledge of QFT, QCD, and SUSY is very limited, almost none. I got the vertex factors and propagators from the superpotential for MSSM without R-parity using the method outlined in Griffiths' book. My answer is correct upto a color factor, hence I know THAT is the source of the problem.

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It seems that there are problems with charge conservation. There could be something mixed up about particles and antiparticles, so I'm disregarding signs, but b-squark has charge 1/3 and two quarks/anti-quarks of the same flavour have 0 or 2/3 (down) or 4/3 (up). Probably they are not of the same flavour, maybe they are from the same family.

For example red b-squark can decay into anti-blue u-antiquark and anti-green d-antiquark. This can be also interpreted as blue u-quark receiving red b-squark and becoming anti-green d-antiquark and so on — component $\tilde b u d + H.c.$ in the Lagrangian. For chosen colour of squark and flavours of up-type and down-type quarks there are two possible combinations of colours of the quarks.

I don't know anything about colour factors, so in best case this is one half of an answer, but It's to long for a comment so I send it as an answer.

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