- #1

dm4b

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Conceptually, I got what's going on - badly divergent integrals from redundantly integrating over physically equivalent field configurations and a gauge-fixing trick to isolate the interesting part of the path integral counting each physical configuration once and only once, yada, yada.

However, I didn't quite get the math. I absorbed enough to move on, but it's coming back to bite me now in Chapter 16 on Quantization of Non-Abelian Gauge Fields.

So, here is the initial equation that threw me off. It's eq. 9.53 and I don't quite get why it is equal to 1.

1 = ∫ D [itex]\alpha[/itex] (x) [itex]\delta[/itex] (G(A[itex]^{\alpha}[/itex])) det([itex]\delta[/itex]G(A[itex]^{\alpha}[/itex])/[itex]\delta\alpha[/itex])

I've been assuming the determinant is the Jacobian of the transformation. The delta is the gauge-fixing condition G(A) = 0, but can somebody fill in any extra steps that shows why this is equal to 1? Any extra insight on the what's going on even conceptually would be nice too.

Also, I suck at using Latex, but an equation two down from this one is a slightly simpler form if that helps elucidate things.

Anyhow, if I can clear that up, I'm sure the rest will fall into place.

Thanks!