Faddeev-Popov procedure

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When we naïvely evaluate the path integral for the vector potential,

Z = \int\!\mathcal{D}A_\mu\,e^{iS}\,,

we find that by integrating over all field configurations within a particular class of configurations, i.e., all configurations A_\mu\, related to one another by a gauge transformation A_\mu \to A^g_\mu = A_\mu + \partial_\mu g\,, we grossly overcount equivalent field configurations, and since S\, is gauge invariant, Z\, is necessarily divergent. Although we always deal with a form of Z\, that is normalized, it is necessary to quantify this divergence. Naively, we may simply integrate over a representative from each class of field configurations using a gauge condition F[A_\mu] = 0\, by inserting a delta functional of the form \delta[F[A_\mu]]\, into the integrand:

\delta[F[A_\mu]] \equiv \prod_x \delta(F[A_\mu(x)])\,,
Z \to \int\!\mathcal{D}A_\mu\,\delta[F[A_\mu]] e^{iS}\,.

However, this is problematic since, under a change of variables A_\mu \to A'_\mu\,, the delta functional picks up a Jacobian factor. Instead, consider the integral,

\Delta^{-1}_F[A_\mu] \equiv \int\!\mathcal{D}g\, \delta[F[A^g_\mu]]\,.

Using the invariance of the Haar measure for compact groups, we see that this object is gauge invariant:

\Delta^{-1}_F[A^{g'}_\mu] = \int\!\mathcal{D}g\, \delta[F[A^{g'g}_\mu]] = \int\!\mathcal{D}(g'g)\, \delta[F[A^{(g'g)}_\mu]] = \Delta^{-1}_F[A_\mu]\,.

We can rewrite the integral as

1 = \Delta_F[A_\mu] \int\!\mathcal{D}g\, \delta[F[A^g_\mu]]\,.

For ordinary functions, a property of the delta function gives

\delta(x-x_0) = \left|\frac{df(x)}{dx}\right|_{x=x_0}\delta(f(x))\,

assuming f(x)\, has only one zero at x=x_0\, and is differentiable there. Integrating both sides gives

1 = \left|\frac{df(x)}{dx}\right|_{x=x_0}\int\!dx\,\delta(f(x))\,.

Extending over n variables, suppose f(x^i) = 0\, for some x^i_0\,. Then, replacing \delta(x-x_0)\, with \prod_i^n \delta^i(x^i-x^i_0)\,

1 = \left(\prod_i \left|\frac{\partial f(x^i)}{\partial x^i}\right|\right) \int\!\left(\prod_i dx^i\right)\,\delta(f(x^i))\,.

Recognizing the first factor as the determinant of the diagonal matrix \frac{\partial f(x^i)}{\partial x^i}\delta^{ij}\, (no summation implied), we can generalize to the functional version of the identity:

1 = \det\left|\frac{\delta F}{\delta g}\right|_{F=0} \int\!\mathcal{D}g\,\delta[F[A^g_\mu]]\,,

where \Delta_F[A_\mu] \equiv \det\left|\frac{\delta F}{\delta g}\right|_{F=0}\, is the Faddeev-Popov determinant.

We can insert this identity into the original form of Z\,, and understand that the determinant is evaluated where F[A^g_\mu]\, is zero. Then,

Z = \int\!\mathcal{D}g\int\!\mathcal{D}A_\mu \det \left|\frac{\delta F}{\delta g} \right| \delta[F[A^g_\mu]]e^{i \int dx \mathcal{L}}\,.

Supposing F[A_\mu]\, to be linear in A_\mu\,, we note that the determinant is independent of g\,, and since \mathcal{D}A^g_\mu = \mathcal{D}A_\mu\,, and since A^g_\mu\, is merely a dummy variable, we obtain

Z = \left(\int\!\mathcal{D}g\right)\int\!\mathcal{D}A_\mu \det \left|\frac{\delta F}{\delta g} \right| \delta[F[A_\mu]]e^{i \int dx \mathcal{L}}\,.

Here we have isolated the divergent part of Z\,, i.e., \int\!\mathcal{D}g\,, which we recognize as v[G]^V\,, where v[G]\, is the volume of the gauge group and V\, is the volume of the spacetime.

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