# Explicit derivation that the Faddeev-Popov functional determinant is gauge invariant

I am trying to show that the Faddeev-Popov functional determinant used in the quantisation of non-Abelian gauge theory is indeed gauge invariant. As shown in my previous question when we follow the Faddeev-Popov procedure we make use of the fact that the Faddeev-Popov functional determinant is gauge invariant.

First let's write down the Faddeev-Popov determinant \begin{aligned}\det\left(\frac{\delta G((A^\alpha(x))^a)}{\delta\alpha(y)^b}\right)&=\int{\mathcal{D}c\mathcal{D}\bar{c}\exp\left(i\int{d^4x d^4y}\bar{c}(x)\frac{\delta G((A^\alpha(x))^a)}{\delta\alpha(y)^b}c(y)\right)} \end{aligned} My problem here is the following: We choose $$G(A)=\partial^\mu A^a_\mu(x)-\omega^a(x)$$ so the gauge transformed functional will be $$G(A^\alpha)=\partial^\mu[(A^\alpha)^a_\mu]-\omega^\alpha(x)\\ =\partial^\mu\left(A^a_\mu+\frac{1}{g}D_\mu\alpha^a\right)-\omega^\alpha(x)\\ =\partial^\mu A^a_\mu+\frac{1}{g}\partial^\mu D_\mu\alpha^a-\omega^a(x)$$

which means that we can calculate \begin{aligned}\frac{\delta G((A^\alpha(x))^a)}{\delta\alpha(y)^b}&=\frac{\delta}{\delta\alpha(y)^b}\left(\partial^\mu A^a_\mu+\frac{1}{g}\partial^\mu D_\mu\alpha^a-\omega^a(x)\right)\\ &=\frac{1}{g}\partial^\mu D_\mu\frac{\delta\alpha(x)^a}{\delta\alpha(y)^b}\\ &=\frac{1}{g}\partial^\mu D_\mu\delta^{(4)}(x-y)\delta^{ab} \end{aligned} If we try and calculate $$\frac{\delta G(A)}{\delta\alpha}$$ will we not just get zero? How is then the determinant gauge invariant?

• The functional determinant $\det(d\star D)$ is not gauge invariant. Under a gauge transformation, the covariant differential $D$ transforms as $D\rightarrow UDU^{-1}$. The real problem is that you missed the definition of the Faddeev-Popov determinant. The Faddeev-Popov determinant is evaluated at $\alpha=0$, i.e at the identity element of the (infinite dimensional) Lie group of gauge transformations. i.e $\det(\delta G)/\delta\alpha)|_{\alpha=0}$, and gauge redundancies are already removed. Peskin's writing is not clear. May 30, 2022 at 4:54

You must be careful with the definition of the Faddeev-Popov determinant you inserted in the path-integral.

The Faddeev-Popov determinant in the path-integral is evaluated at the identity element of the Lie group of gauge transformations. i.e, what you should insert into the path-integral is $$\mathrm{Det}\left(\frac{\delta\mathcal{F}(A[U])}{\delta U}\right)\Bigg|_{U=\mathrm{id}}.$$

In your notations, it corresponds to evaluating the functional determinant at $$\alpha^{a}(x)=0$$.

Then, assume $$U(x)=\exp\left(iT^{a}\Lambda_{a}(x)\right)$$, the Faddeev-Popov determinant is \begin{align} &\,\,\,\,\,\,\,\mathrm{Det}\left(\frac{\delta\mathcal{F}(A[U])}{\delta U}\right)\Bigg|_{U=\mathrm{id}} \\ &=\mathrm{Det}\left(\int d^{4}z\frac{\delta\mathcal{F}(A[U])}{\delta\Lambda(z)}\Bigg|_{\Lambda(z)=0}\cdot\frac{\delta\Lambda(z)}{\delta U}\Bigg|_{\Lambda(z)=0}\right) \\ &=\mathrm{Det}\left(\frac{\delta\mathcal{F}(A[U])}{\delta\Lambda}\right)\Bigg|_{\Lambda=0}. \end{align}

It does not make any sense to consider a gauge transformation on the determinant since it is already evaluated at $$\Lambda=0$$!

In my previous answer, I said that there's a more fundamental definition of the Faddeev-Popov determinant, which is the functional integral $$\mathrm{Det}\mathcal{M}^{-1}=\int\mathcal{D}[U]\,\delta[\mathcal{F}(A[U])].$$

Using the formula $$\int d\theta\,\delta(f(\theta))=\int df\det\Bigg|\frac{d\theta}{df}\Bigg|\delta(f(\theta))=\det\Bigg|\frac{d\theta}{df}\Bigg|_{f=0}$$

you can immediately realize that the determinant is evaluated $$\mathcal{F}=0$$, which is equivalent to $$U=\mathrm{id}$$, or $$\Lambda=0$$.

It is precisely because of the above functional integral definition of the Faddeev-Popov determinant that then one must make sure $$\mathrm{Det}\mathcal{M}$$ is a gauge-invariant quantity so that it can be inserted into the path-integral.

To sum up, the functional determinant $$\mathrm{Det}\left(\frac{\delta\mathcal{F}(A[U])}{\delta U}\right)$$

is of course NOT gauge invariant. In fact, under a gauge transformation, one has $$D\rightarrow UDU^{-1},\quad\mathrm{hence}\quad\mathrm{Det}(\partial^{\mu}D_{\mu})\rightarrow\mathrm{Det}(\partial^{\mu}UD_{\mu}U^{-1}).$$

But this is not what you are inserting in the path-integral. What you are actuall inserting is $$\mathrm{Det}\left(\frac{\delta\mathcal{F}(A[U])}{\delta U}\right)\Bigg|_{U=\mathrm{id}}=\mathrm{Det}(\partial^{\mu}D_{\mu})|_{\Lambda=0},$$

which is of course gauge invariant because it is evaluated at $$U=\mathrm{id}$$. This is exactly the reason why one can factor out the $$\int\mathcal{D}\Lambda$$ integral, which gives the volume of the (infinite dimensional) Lie group of gauge transformations. Everything inside the integrand should be independent of $$\Lambda$$.

I didn't read Peskin's book, but I believe he assumes readers already know a lot about quantum field theory. It's like asking why the function $$f(x)|_{x=0}$$ is a constant under a shift $$x\rightarrow x+x^{\prime}$$.

I hope it makes sense to you now.