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  How to apply the Faddeev-Popov method to a simple integral

+ 5 like - 0 dislike

Some time ago I was reviewing my knowledge on QFT and I came across the question of Faddeev-Popov ghosts. At the time I was studying thеse matters, I used the book of Faddeev and Slavnov, but the explanation there is on very transparent, specially not for someone like me who was just starting to learn QFT. Therefore, I never understood fully what was meant. To clear my doubts how the mehtod works and what are the Gauge orbits I decided to think how the method will work on a simple toy problem.

The local Gauge transformation in the non-Abelian case acts non-linearly i.e.

$$ F[\mathscr{A}_{\mu}] = g\mathscr{A}_{\mu}g^{-1} + g_{\mu}g_{\mu}^{-1} $$

Since in the generating functional we integrate over the fields,

$$ Z=\int \mathcal{D}\mathscr{A}_{\mu}e^{iS[\mathscr{A}_{\mu}]} $$

double counting is introduced, due to the integration over many equivalent fields generated by the local Gauge transformation. To fix this L.D. Faddeev and V. Popov proposed to introduce the constraint of the Gouge transformation in the form:

$$ \Delta_L(\mathscr{A}) \int \delta(F[\mathscr{A}_{\mu}^{\omega}])d\omega=1 $$ There are different methods how to get $\Delta_L(\mathscr{A})$, but I think that the simplest one is to use just the definition of the delta function. Of course using the properties of the Haar measure, the above expression is Gauge invariant. Say $U(1)$ with $\omega=e^{i\phi}$ this can be checked by using fixed $U'$

$$ \mathcal{D}\omega\omega' = \mathcal{D}\omega, $$

which in my opinion is just the product rule.

Plugging into the generating functional we get

$$ Z=\iint \mathcal{D}\mathscr{A}_{\mu} d \omega \Delta_L(\mathscr{A}) \delta(F[\mathscr{A}_{\mu}^{\omega}])e^{iS[\mathscr{A}_{\mu}]}, $$

which produces a multiplicative volume factor.

Now comes my question, how do we use this on a toy problem. Suppose we were integrating

$$ I=\iint e^{-(x^2+y^2)}dxdy $$

The integration is redundant and by going to cylindrical coordinates $(r,\phi)$ we can easily factor out the $\int d\phi$ part. Let's do this with the Faddeev-Popov method.

Our integral is rotational invariant and the only real contribution comes form moving in the direction $r \to \infty$. I visualise our Gauge transformation as rotation around the origin and I have the feeling that the Gauge orbits are concentric circles. Since we would like to use only non-equivalent orbits we fix the $y$ variable. To do so use the value $y_{\phi} = x\sin\phi+y\cos\phi$

For our unity integral we have $$ 1=\int d\phi\delta(x\sin\phi+y\cos\phi)|\frac{\partial(x\sin\phi+y\cos\phi)}{\partial \phi}| $$ Since we have rotational-invariant integral, let's pick $\phi=0$ this gives

$$ I=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(x^2+y^2)}\delta(y_{\phi}) |x| dx dy d \phi $$ $$ I=\int_0^{\infty} e^{-x^2} x dx \times \int_0^{2\pi}d\phi = \pi $$

What we have done above is just rotating, so that the integral is taken along the positive real axis $y_{\phi}=0$. This looks like a complicated way of doing change of variables or introducing constraints.

If the above is correct, what are the gauge orbits in the general case. According to Faddeev himself, his intuition was purely geometrical and the non-Abelian case produces lines that intersect the Gauge orbits at different angles.

Coming back to my example instead of circles $F[\mathscr{A}_{\mu}]$ defines a manifold and the Gauge condition $\partial_{\mu}\mathscr{A}^{\mu}$ gives a cut trough this manifold equivalent to the intersection $y_{\phi}=0$.

I would appreciate your critical review of my question.

This post imported from StackExchange Physics at 2015-12-02 17:57 (UTC), posted by SE-user Alexander Cska
asked Dec 2, 2015 in Theoretical Physics by Alexander Cska (25 points) [ no revision ]
All of this is absolutely correct and is basically how FP gauge fixing is explained in modern field theory/string theory books. Good going if you figured out this picture on your own!

This post imported from StackExchange Physics at 2015-12-02 17:57 (UTC), posted by SE-user childofsaturn

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