Qustion about the appearance of $\Delta_{FP}[A_{\mu}]$ in the path integral of gauge field

Question

Why is the Faddeev-Popov quantization of a $U(1)$ gauge field not the naive solution $$\int {\cal D}A \, \, \delta\left[F(A_\mu) \right]\exp \left\{ -\frac{i}{4}\int \mathrm{d}^4 x \, F_{\mu\nu}F^{\mu\nu}\right\} $$

rather than $$\int {\cal D}A \, \, \delta\left[F(A_\mu) \right]\, \Delta_{FP} \exp \left\{ -\frac{i}{4}\int \mathrm{d}^4 x \, F_{\mu\nu}F^{\mu\nu}\right\} $$

I know how to derive the correct one, but I want to know the significance/reason of the appearance of the term $\Delta_{FP}[A_{\mu}]$.

Answer 1

Recall the delta function identity,

$$\int \delta [f(x)]g(x) \mathrm{d} x = \sum_i \frac{g(x_i)}{|f'(x_i)|}$$

where $x_i$ are the solutions to $f(x)=0$. The identity arises from the general formula,

$$\underbrace{\delta[f(x)] \, \, \frac{\mathrm{d}f}{dx}}_{\text{evaluated at all }a_i}= \sum_i \delta[x-a_i]$$

For the case of a multi-variable function, the derivative generalizes to a Jacobian factor. The Faddeev-Popov determinant is a functional generalization or analogue of this 'Jacobian.'

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In addition, the formula for the functional determinant of an operator, which is used to introduce the ghost fields, follows naturally by generalizing the Berezin/Grassmann integral,

$$\prod_i \int \mathrm{d}\theta_i \int \mathrm{d}\bar{\theta}_i \, e^{-\bar{\theta} A\theta} = \det(A)$$

to a functional integral, i.e.

$$\int \mathcal{D}c \mathcal{D}\bar{c} \, \, e^{-\bar{c}\Delta c} \sim \det(\Delta)$$