Does the Z boson couple to photons?

[maani]

Hi I can't find a Feynman rule for the coupling of two Z bosons to a photon. Does the Z boson couple to a photon at all?

[tom.stoer]

I guess that there is a four-boson vertex with W+, W-, Z0 and A. But to be sure one would have to look at the el.-weak Lagrangian:

eq. 2.47 in http://arxiv.org/PS_cache/hep-ph/pdf/0001/0001283v1.pdf
fig. 11. in http://arxiv.org/PS_cache/arxiv/pdf/0901/0901.0241v1.pdf

Anyway: even w/o such a vertex in higher orders an interaction would be created an via fermion loops.

[genneth]

You can get Z+Z -> gamma + gamma.

[ansgar]

zz to fermion to 2gamma

zzA is not in the SM lagrangian

[tom.stoer]

That's what I say: the only direct vertex is

W+ + W- => Z0 + Photon

but via intermediate states you can get (nearly) anything you like.

[Vanadium 50]

There are no all-neutral gauge boson couplings in the SM.

It's easiest to see this by working in the w1, w2, w3, B basis rather than the W+/-, Z, A basis.

Tom is right, these can appear at loop level, but these will be second-order weak and GIM surpressed. However, you still have to comply with conservation laws. For example, you can't get any electric dipole radiation because the Z is neutral. I need to think more about magnetic dipole and electric quadrupole.

[fortuneteller]

The coupling of gauge bosons self interaction in non-abelian gauge theory is proportional to f_{abc}, i.e. only diffrent bosons can interact. So, no Z-Z-A and no Z-Z-Z interactions. You can also think that photon only couple to charged particles directly.

[tom.stoer]

I do not see how this applies directly. In the el.-weak theory we have U(1)*SU(N); in addition the photon is "mixed" according to the original basis.