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Path Integrals and Non-Abelian Gauge Theories

  1. Apr 28, 2007 #1
    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 [tex]\mathcal{G}[/tex] is:

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

    where

    [tex]\mathcal{D}U = \Pi d\theta[/tex],

    and

    [tex]U \in \mathcal{G}[/tex].

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

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

    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
  2. jcsd
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