Four Bosons vertex related to gauge symmetry

[Andrea_G]

Hi,
is correct to say that there is no interaction between four photons because the gauge group of QED is U (1) while there are interactions of four gluons or four W's because the gauge group of QCD is SU (3) and EW's one is SU (2) xU (1)?

I know that the interaction between four photons is not allowed for the parity conservation but I was wondering if the gauge symmetry is connected with this thing.

Thank you

 

[Vanadium 50]

I know that the interaction between four photons is not allowed for the parity conservation

That's not true. One photon has the same paity as three.

is correct to say that there is no interaction between four photons because the gauge group of QED is U (1)

A little. The U(1) nature means the photon is uncharged. However, there is a $$(F_{\mu\nu})^4$$ interaction in QED once you allow an electron field. This is what causes the light-by-light scattering discussed in another thread.

 

[nrqed]

That's not true. One photon has the same paity as three.
A little. The U(1) nature means the photon is uncharged. However, there is a $$(F_{\mu\nu})^4$$ interaction in QED once you allow an electron field. This is what causes the light-by-light scattering discussed in another thread.

I don't want to sound like I am nitpicking, but strictly speaking this interaction is not present in what is typically considered to be the QED Lagrangian. I know that it appears in the Euler-Heisenberg lagrangian and it must be present of we think of QED as an effective field theory (as it must be) but if we talk about QED being the theory described in most textbooks, there is no tree level four photon interaction.

To the OP: there is no tree level four photon interaction in QED but of course, there are higher order (loop) diagrams that do produce four photon interactions. the reason there is no tree level four photon interactions but there is a tree level four gluon vertex is indeed because of the difference between U(1) and SU(N), the former is abelian while the later are non abelian.
The difference appears through the definition of $F_{\mu\nu}$. The key point is that it contains the commutator $[A_\mu, A_\nu ]$ For an abelian group like U(1), this vanishes. For SU(N), it does not. Because of this the term $F_{\mu \nu} F^{\mu \nu}$ will contain an explicit term of the form $A_\mu A_\nu A^\mu A^\nu$ for SU(N).