We're working through the Griffith's introduction to elementary particle physics and are wondering about the quark antiquark annihilation process. There are actually three first order processes, but my question is about the two feynman diagrams I sketched in paint.

If we look at a QED annihilation process, like $e^-$ and $e^+$ we get two photons. The diagrams eventually look the same, if you replace the gluon lines with photon lines. I understand, that we can not differentiate the two photons, that's why we have to draw both diagrams. However, the gluons have different color charge, so I'd say I can differentiate them and therefore they are not interchangeable, thus I only get the left diagram.

Can someone shed some light on this?

enter image description here


1 Answer 1


Experimentally speaking, it is found we detect only $\mathrm{SU}(3)$ colour singlets in nature so the gluons we draw in our feynman diagrams will ultimately hadronise to form such states at time scales longer than the hard scattering with which they were involved. So at the end of the day the resultant states will carry $\mathrm{SU}(3)$ colour quantum numbers with a net red/green/blue configuration. Colour combinations other than this, as carried by the gluons in your picture, are not discernible to detectors so from this viewpoint we expect to consider the two topologically distinct diagrams as drawn.

BTW those two diagrams are also equivalent to enter image description here


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