No "i" in the Triple gluon vertex Feynman Rule

In summary, there are different conventions for the inclusion of an 'i' factor in the Feynman rules for the triple and four gluon vertices. While most sources have an overall 'i' factor for the four gluon vertex and no 'i' factor for the triple gluon vertex, Yndurain's book includes an 'i' factor for both vertices. This can lead to different relative contributions of diagrams in amplitude calculations. However, it is important to note that the inclusion of an 'i' factor is relative and does not affect the final result. Any source suggesting otherwise is incorrect.
  • #1
CAF123
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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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  • #2
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.
 
  • #3
Vanadium 50 said:
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}##?
 
  • #4
CAF123 said:
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.

CAF123 said:
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.
 
  • #5
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.
 
  • #6
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.
 
  • #7
Vanadium 50 said:
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.
 
  • #8
It doesn't matter where the i goes. It's relative. If you find a book suggesting otherwise, it's wrong.
 
  • #9
Vanadium 50 said:
It doesn't matter where the i goes. It's relative. If you find a book suggesting otherwise, it's wrong.
Yes in most sources indeed I find there to be a relative i between the three point and four point vertex. However in Yndurain he gives them both with an i. Could just be a typo after all.
 

Related to No "i" in the Triple gluon vertex Feynman Rule

1. What is the "i" in the Triple gluon vertex Feynman Rule?

The "i" in the Triple gluon vertex Feynman Rule represents the imaginary unit, which is a mathematical concept used in quantum field theory to describe the behavior of particles and their interactions.

2. Why is the "i" important in the Triple gluon vertex Feynman Rule?

The "i" is important because it helps to calculate the amplitude of the interaction between three gluons, which is a fundamental part of the Standard Model of particle physics.

3. How does the "i" affect the Triple gluon vertex Feynman Rule?

The "i" affects the Triple gluon vertex Feynman Rule by introducing a phase factor into the calculation, which accounts for the spin and charge of the gluons involved in the interaction.

4. Can the "i" be removed from the Triple gluon vertex Feynman Rule?

No, the "i" cannot be removed from the Triple gluon vertex Feynman Rule as it is a necessary component of the mathematical formula used to describe the interaction between gluons.

5. What is the significance of "No "i" in the Triple gluon vertex Feynman Rule"?

The phrase "No "i" in the Triple gluon vertex Feynman Rule" is used to emphasize the fact that the "i" is not included in the final calculation of the amplitude, but rather is used as a mathematical tool to simplify the calculation process.

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