# Path Integrals and Non-Abelian Gauge Theories

1. Apr 28, 2007

### BenTheMan

Ok, I have a question about this Fade'ev Popov procedure of teasing out the ghosts when one quantizes a non-Abelian gauge theory with path integrals.

The factor of 1 that people insert, for some gauge fixing function f, and some non-Abelian symmetry $$\mathcal{G}$$ is:

$$1=\int \mathcal{D}U \delta[f(\mathbf{A})] \Delta[\mathbf{A}]$$,

where

$$\mathcal{D}U = \Pi d\theta$$,

and

$$U \in \mathcal{G}$$.

This is probably a stupid question, but the function $$\Delta$$ works out just to be a Jacobian of some sort over the manifold $$\mathcal{G}$$, right?

$$\Delta[\mathbf{A}] = det \left(\frac{\delta f}{\delta \theta}\right)$$

I am confused because no one actually says this. Am I completely off base?

Thanks in advance for helping me, and tolerating a (possibly) stupid question!

Last edited: Apr 28, 2007
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