3

What is the correct series expansion for the $U(1)$ Faddeev-Popov ghosts?

I know that the $U(1)$ ghosts are only a phase such that they can be neglected in most cases but it turns out that this is not true in curved spaces even for $U(1)$ theories so please don't answer this...

In this thread Faddeev-Popov ghost propagator in canonical quantization I found that $c$ is hermitian and $\bar{c}$ anti-hermitian which makes sense since $\bar{c} = c^\dagger \gamma_0$.

But in the $U(1)$ case the ghost are Grassmann variables such that $\bar{c} = c^\dagger \gamma_0$ doesn't make sense does it?

For those willing to help me even more. I think that the source of my problem is a poor understanding of the Faddeev-Popov mechanism. More precisely, what happens when $\det(\square)$ is written as a path integral? What exactly do the $c$ and $\bar{c}$ fields mean? Why is it said that one is a ghost and the other an anti ghost?

When quantizing them I obtain $\{ c_k , \bar{c_{k'}}\} = -\delta(k-k')$ how does this tell us anything regarding the norm of these ghosts?

I read Peskin and Schroeder but they do not answer this question (or I missed it).

Finally, my sincere aplogies for this "all over the place" type question. I fail to pinpoint the exact sources of my confusion that's why my question is rather broad. I hope that someone more experiences can pinpoint it with the above information.

Community
  • 1
gertian
  • 1,333

1 Answers1

4

As discussed in Kugo and Ojima 1979, "ghost is Hermitian, anti-ghost is anti-Hermitian" is just a convention, another being that both fields are Hermitian, which results in a factor of $i$ in the FP-ghost term so that the Lagrangian is still Hermitian. In their notation $c,\,\overline{c}$ are both Hermitian while $C:=c,\,\overline{C}:=i\overline{c}$ provide a half-Hermitian convention. Then $$\mathcal{L}_{FP}=-i\partial_\mu\overline{c}D^\mu c=-\partial_\mu\overline{C}D^\mu C.$$

J.G.
  • 25,615