This document https://arxiv.org/abs/1008.4884

presents in Tables 2 and 3 the mathematical expression of many dimension-6 operators,

for example (just an example)

enter image description here

The mathematical expression does not help so much to visualize the corresponding expression (in particular those who have a derivative).

Are there Feynman diagrams for dimension 6 operator ? or is drawing Feynman diagram for that does not make sense ?

  • 1
    $\begingroup$ What about the generic method of drawing Feynman diagrams makes you think it might not apply if the Lagrangian has these particular operators? $\endgroup$
    – ACuriousMind
    Sep 7, 2020 at 20:40
  • $\begingroup$ I thought that it was applying only when there were no operators. But now, I have understood with the comment of Cosmas Zachos $\endgroup$ Sep 7, 2020 at 20:41

1 Answer 1


Of course drawing Feynman diagrams for such vertices makes sense. (I'm leaving the meaningfulness of loop corrections for nonrenormalizable interactions aside; these are effective field theory terms.)

For example, a term $(\phi^\dagger \phi)^3$ amounts to a plain vertex with six incoming scalar field lines.

The fGGG term trilinear in gauge field strengths, includes diagrams with six gauge fields, or lesser numbers, supplanting momenta for gauge fields, etc.

The term you wrote has several Feynman diagrams corresponding to its pieces: the easiest one has four scalars and two gauge fields coming in; there is one with four scalar fields, and two suitable momenta with the right symmetries for the "kinetic-like" term, and mixed terms with four scalars, one momentum and one gauge field. And so on...

The textbook Feynman rules are identical in this case. It's what you would do with the answers that bears further discussion, judiciously excluded here.

  • $\begingroup$ Thank you. Ok, you give examples for the basic ones with field without operator on them. But how would Feynman diagrams for terms that make appear a derivation (or any operator) would look like ? How would you represent that in the diagram ? If there is no way to represent it, then it would look like the same as ones without operators. I'm not sure what you mean about GGG : maybe you mean GGGGGG ? Tiny point : for $\phi$ with dag, you miss a antislash at the beginning. $\endgroup$ Sep 7, 2020 at 19:19
  • $\begingroup$ remark : aboutu your sentence "what you would do with the answers" : the goal of Feynman diagram is to visualize graphically what the lagrangian does, nothing more : I'm interested in how would appear the derivative terms and all non basic terms. (I'm not hidding anything for my "goal", which is to understand such terms graphically) $\endgroup$ Sep 7, 2020 at 19:23
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    $\begingroup$ G is the fields strengths, as depicted in your reference. No, actually, Feynman diagrams are precise shorthand for writing down precise terms in covariant perturbation theory. To indicate a derivative/momentum at a vertex, you write the momentum next to the relevant line in the diagram: this is what QFT books specify exactly in their elaborate rules. This has nothing to do with dimension 6, and you already see it in the kinetic term $(D\phi)^\dagger (D\phi)$ of dimension 4: the piece with two scalar fields, a gauge field, and a momentum on one of the two scalar lines. $\endgroup$ Sep 7, 2020 at 20:30
  • $\begingroup$ I understand thanks to your explanation. Thank you $\endgroup$ Sep 7, 2020 at 20:40
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    $\begingroup$ Here is a dimension 5 term with two momenta, two photons, and a scalar ... $\endgroup$ Sep 7, 2020 at 20:44

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