# No "i" in the Triple gluon vertex Feynman Rule

• I
Gold Member
This is probably a simple question but puzzled me a little when trying to explain something to somebody. In all resources online and in most books I've seen, the triple gluon vertex has no overall 'i' factor while e.g. the four gluon vertex always does. The photon and gluon propagators as well as the 3 point quark and gluon/photon vertices also have an 'i'.

Is there a conceptual reason for this?

Looking in Yndurain's book, 'Relativistic QM and intro. to field theory', however, there is an overall 'i' factor in the triple gluon vertex, while the other Feynman rules continue to also have an overall 'i'. This means in some computation, depending on the Feynman rules used, one may get a relative 'i' factor between diagrams contributing to an amplitude (instead of e.g. a relative sign). Of course, an overall 'i' for the amplitude is irrelevant but not a relative 'i' between diagrams.

So, how would one reconcile the two approaches?

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Staff Emeritus
2019 Award
There are a lot of conventions, but there must be a relative i. gg -> gg by two 3g couplings interferes destructively with via a 4g coupling. It needs to be there. I am pretty sure if it's not it won't even by gauge invariant.

Gold Member
There are a lot of conventions, but there must be a relative i. gg -> gg by two 3g couplings interferes destructively with via a 4g coupling. It needs to be there. I am pretty sure if it's not it won't even by gauge invariant.
Thanks. Why must the two 3g couplings contribution interfere destructively (due to relative i) with the 4g one? I suppose you mean we must have ##A = A_{3g} - i A_{4g}## and not ## A = A_{3g} + A_{4g}## or ## A = A_{3g} - A_{4g}##?

Staff Emeritus
2019 Award
Why must the two 3g couplings contribution interfere destructively (due to relative i) with the 4g one?
A reasonable answer would be "I'm not saying they have to. I am saying they do." However, calculating gg -> gg using each piece separately is infinite. So if you want a finite cross-section, the two pieces have to have opposite sign.

I suppose you mean
No, I meant what I said. gg -> gg via the 3g vertex has two 3g vertices in it. Each one has a relative i phase w.r.t. 4g, so together it's a minus sign.

Gold Member
No, I meant what I said. gg -> gg via the 3g vertex has two 3g vertices in it. Each one has a relative i phase w.r.t. 4g, so together it's a minus sign.
I’ve seen the 3 point vertex written more without an i and the 4 point vertex with an i. If both have an i, then the amplitude for this particular process is ##A = -A_{3g} \pm i A_{4g}##, while if the 4 point only has an i, then ##A = A_{3g} \pm i A_{4g}##. Both give the same cross section. However, if both do not have an overall i, then the cross section differs. Why is that?

Moreover, if another process diagram has only a single contributing 3 point gluon vertex, then defining the 3 point vertex with an i and the 4 point with an i will make its relative contribution to an amplitude different if one has only an i in the 4 point vertex.

Staff Emeritus
2019 Award
First, I answered your question. You seem to want to argue with me. Sorry, not interested.

Second, I said quite specifically that you have two 3g vertices to one 4g. That means the relative phase is -1, not i. You keep writing i's.

Gold Member
First, I answered your question. You seem to want to argue with me. Sorry, not interested.
No, you said something and I wanted to understand it.
Second, I said quite specifically that you have two 3g vertices to one 4g. That means the relative phase is -1, not i. You keep writing i's.
The relative phase of two 3g vertices to a 4g vertex is -1 only if the 4g vertex doesn’t have an i and the 3g vertices do. If the 4g vertex has an i too, then there can be a relative phase of i too (or if the 4g vertex does and the 3g vertex doesn’t). So this ambiguity is what I’d like to understand/clear up - depending on the Feynman rule convention, one seems to get different relative contributions of diagrams.

Staff Emeritus