When applying Faddeev and Popov method (am using Peskin and Schroeder as reference), we use the identity: $$1=\int \mathcal{D}\alpha(x)\delta(G(A^\alpha)) \det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right) \tag{9.53}$$ to write $$ \int \mathcal{D}Ae^{iS[A]}=\det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right)\int\mathcal{D}\alpha(x)\int\mathcal{D}Ae^{iS[A]}\delta (G(A^\alpha))\tag{9.54}$$ When we use the Lorentz gauge, we obtain: $$G(A)=\partial^\mu A_\mu+\frac{1}{e}\partial^2\alpha(x)$$

My questions is: how do we obtain the following: $$ \det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right) = \det\left(\frac{1}{e}\partial^2\right) $$

I am also confused about how can we obtain an operator by taking a functional derivative, if someone could give an intuitive explanation of this it would be highly appreciated!

When applying Faddeev and Popov method (am using Peskin and Schroeder as reference), we use the identity: $$1=\int \mathcal{D}\alpha(x)\delta(G(A^\alpha)) \det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right) \tag{9.53}$$ to write $$ \int \mathcal{D}Ae^{iS[A]}=\det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right)\int\mathcal{D}\alpha(x)\int\mathcal{D}Ae^{iS[A]}\delta (G(A^\alpha))\tag{9.54}$$ When we use the Lorenz gauge, we obtain: $$G(A)=\partial^\mu A_\mu+\frac{1}{e}\partial^2\alpha(x)$$

My questions is: how do we obtain the following: $$ \det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right) = \det\left(\frac{1}{e}\partial^2\right) $$

I am also confused about how can we obtain an operator by taking a functional derivative, if someone could give an intuitive explanation of this it would be highly appreciated!

When applying Faddeev and Popov method (am using Pesking and Shcroeder as reference), we use the identity: $$1=\int \mathcal{D}\alpha(x)\delta(G(A^\alpha)) \det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right)$$ to write $$ \int \mathcal{D}Ae^{iS[A]}=\det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right)\int\mathcal{D}\alpha(x)\int\mathcal{D}Ae^{iS[A]}\delta (G(A^\alpha))$$ When we use the Lorentz gauge, we obtain: $$G(A)=\partial^\mu A_\mu+\frac{1}{e}\partial^2\alpha(x)$$

My questions is: how do we obtain the following: $$ \det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right) = \det\left(\frac{1}{e}\partial^2\right) $$

I am also confused about how can we obtain an operator by taking a functional derivative, if someone could give an intuitive explanation of this it would be highly appreciated!

When applying Faddeev and Popov method (am using Peskin and Schroeder as reference), we use the identity: $$1=\int \mathcal{D}\alpha(x)\delta(G(A^\alpha)) \det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right) \tag{9.53}$$ to write $$ \int \mathcal{D}Ae^{iS[A]}=\det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right)\int\mathcal{D}\alpha(x)\int\mathcal{D}Ae^{iS[A]}\delta (G(A^\alpha))\tag{9.54}$$ When we use the Lorentz gauge, we obtain: $$G(A)=\partial^\mu A_\mu+\frac{1}{e}\partial^2\alpha(x)$$

My questions is: how do we obtain the following: $$ \det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right) = \det\left(\frac{1}{e}\partial^2\right) $$

I am also confused about how can we obtain an operator by taking a functional derivative, if someone could give an intuitive explanation of this it would be highly appreciated!