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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?

Community
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Razor
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1 Answers1

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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.

Qmechanic
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  • I get that it is not a differentiation, but still how can the invariance be shown explicitly? A derivation would be really helpful. – Razor Nov 06 '16 at 22:32
  • You mean an explicit formula for the Haar measure for the $SU(N)$ group? – Qmechanic Nov 06 '16 at 22:50
  • Not just the formula. A proof that it really is invariant under a fixed U(1) transformation. – Razor Nov 07 '16 at 05:20
  • Well, the path integral measure is not a well-defined mathematical object. The proof is only formal. – Qmechanic Nov 07 '16 at 06:54
  • 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. – Razor Nov 07 '16 at 07:06
  • The formal proof typically discretizes spacetime. Within each lattice site $x$, the gauge transformation $\omega^{\prime}$ is given by a group element $\omega^{\prime}(x)$. Next apply right invariance property of the Haar measure wrt. that group element. End formal proof. – Qmechanic Nov 07 '16 at 07:16
  • @Qmechanic I believe OP is asking for the proof of invariance of $d\omega(x)$. It is indeed trivial: the Haar measure is invariant by construction (an explicit formula for the measure on $SU(n)$ in group coordinates would be nice, but I am too lazy to look it up). Anyways, for $U(1)$ the measure is $d\omega = d\varphi / (2\pi)$ with $\varphi \in [0..2\pi)$. On the other hand, $\mathcal{D} \omega$ is invariant under the $SU(n)$ gauge group (which is a product of $SU(n)$ at all points of spacetime). This is of course highly formal and is far from being well-defined, as you've mentioned. – Prof. Legolasov Nov 08 '16 at 18:45
  • @Solenodon Paradoxus: Agree. That is what I was trying to say in my previous comment. – Qmechanic Nov 08 '16 at 18:50