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I am studying the Faddeev-Popov procedure for quantizing Gauge fields. I am stuck in the step where it says that the measure is gauge invariant for the $U(1)$ case.

I came across this question on stackexchange: How to apply the Faddeev-Popov method to a simple integral

Here OP says in the question that ${\cal D}\omega\omega' = {\cal D}\omega$, for fixed $\omega'$, which follows from the product rule, but I don't see how. I figured:

$D\omega\omega' = \omega'{\cal D}\omega + \omega {\cal D}\omega'$, where the second term goes to zero as $\omega'$ is just a fixed gauge transformation. But then ${\cal D}\omega\omega' = \omega'{\cal D}\omega$.

So, what am I missing here?

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1 Answer 1

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${\cal D}$ is not a differentiation; ${\cal D}\omega=\prod_x d\omega(x)$ is a path integral measure, and the right invariance of the Haar measure is being used.

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  • $\begingroup$ I get that it is not a differentiation, but still how can the invariance be shown explicitly? A derivation would be really helpful. $\endgroup$
    – Razor
    Nov 6, 2016 at 22:32
  • $\begingroup$ You mean an explicit formula for the Haar measure for the $SU(N)$ group? $\endgroup$
    – Qmechanic
    Nov 6, 2016 at 22:50
  • $\begingroup$ Not just the formula. A proof that it really is invariant under a fixed U(1) transformation. $\endgroup$
    – Razor
    Nov 7, 2016 at 5:20
  • $\begingroup$ Well, the path integral measure is not a well-defined mathematical object. The proof is only formal. $\endgroup$
    – Qmechanic
    Nov 7, 2016 at 6:54
  • $\begingroup$ There obviously has to be some sort of proof! The entire Faddeev-Popov method stands on this very fact. Introduction to Gauge Field Theory by Bailin & Love page 120 for reference. $\endgroup$
    – Razor
    Nov 7, 2016 at 7:06

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