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?


1 Answer 1


${\cal D}$ 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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.