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 $\Delta_{FP}[A_{\mu}]$.


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.'

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.

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


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