# Can a photon see ghosts?

Does it make sense to introduce Faddeev–Popov ghost fields for abelian gauge field theories?

Wikipedia says the coupling term in the Lagrangian "doesn't have any effect", but I don't really know what that means. If it doesn't work at all (probably because structure constants are zero?) then why doesn't it work, physically? I mean you still have to do things like getting rid of unphysical degrees of freedom/gauge fixing.

Since it isn't done in QED afaik, I guess its not a reasonable way of doing things like computation of self energy/computing corrections of propagators. I wondered, because the gauge field-ghost-ghost vertices for Yang-Mills basically look just like the gauge field-fermion-fermion vertex in electrodynamics.

If you look at the Faddeev-Popov ghosts' Lagrangian, you will see that for Abelian groups, the structure constants $f_{abc}$ vanish and we're left with $${\mathcal L}_{\rm ghost} = \partial_\mu \bar c^a \partial^\mu c^a$$ which means that the ghosts are completely decoupled. They don't interact with the gauge fields (photons). You may still use the Faddeev-Popov machinery and the BRST formalism based upon it to identify the physical states as the cohomologies of $Q$, the BRST operator.
But what this BRST machinery tells you is something you may easily describe without any Faddeev-Popov ghosts, too. It just tells you that the excitations of $\bar c,c$ are unphysical much like the excitations of time-like and longitudinal photons. That's why the BRST problem in the case of Abelinan gauge groups is "solvable" in such a way that you may simply eliminate the ghosts completely, together with 2 unphysical polarizations of the photon. And that's why QED may be taught without any Faddeev-Popov ghosts and one may still construct nice Feynman rules for any multiloop diagrams.