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!

  • $\begingroup$ Which page? Which equation? $\endgroup$ Nov 26, 2020 at 4:22
  • $\begingroup$ Chapter 9.3, pg 295. Also, chapter 9.2 has a small section of functional derivatives but I wasn't able to use it. $\endgroup$
    – Ivan
    Nov 26, 2020 at 4:24

2 Answers 2


$$\begin{align}{\rm Det} \left(\frac{\delta G}{\delta \alpha}\right) ~=~&\int {\cal D}c{\cal D}\bar{c}\exp\left(\int \!d^4x \int \!d^4y ~\bar{c}(x)\frac{\delta G(x)}{\delta \alpha(y)}c(y) \right) \cr ~=~&\int {\cal D}c{\cal D}\bar{c}\exp\left(\int \!d^4x \int \!d^4y ~\bar{c}(x) \frac{1}{e}\partial_x^2\delta(x-y) c(y) \right)\cr ~=~& {\rm Det} \left(\frac{1}{e}\partial^2\right).\end{align}$$


This should be roughly familiar from obtaining the usual Feynman rules from the path integral formulation, though perhaps it usually isn't phrased this way.

The long story short is that the functional determinant of an operator is defined by the trace log formula, $$ \det A=e^{\text{tr}\log A} $$ which can be proven by thinking about eigenvalues. For these operations to make sense on a differential operator, however, we usually move to the Fourier basis where all the derivatives are just momenta.

In fact, this is what we do when writing down the Feynman rules. Remember that computing the saddle point approximation of the path integral (with source $J\phi$) results in $$ Z[J]\sim e^{J^T\Delta J}, $$ appropriate integrations implied. From here we employ the usual tricks of taking $J$ derivatives to obtain correlators, this $\Delta$ being the Feynman propagator.

The way this approximation works is by writing the action Taylor expanded to second order around the saddle point (which is usually taken to be $\phi=0$ in Peskin and Schroeder, but need not be and indeed this fact has important implications related to instantons) so we write $$ S[\phi]=S_0+\frac{1}{2}\phi \frac{\delta^2 S}{\delta \phi\delta\phi}\phi+J\phi + \mathcal{O}(\phi^3), $$ integrations again implied as necessary. In the typical case of a scalar field, $$ \frac{\delta^2 S}{\delta \phi\delta\phi}=\partial^2-m^2 $$ up to signs.

This is where it all comes together, we of course know that the Feynman propagator should be the inverse of this operator $\partial^2-m^2$, but the trick we usually employ is moving to Fourier basis where this operator is $-k^2-m^2$ so its inverse is given by $$ \frac{-1}{k^2+m^2}, $$ which is the propagator up to a factor of $i$.

So asking about the determinant of $\partial^2$ shouldn't be so terrible in light of the fact that we invert the operator $\partial^2-m^2$ all the time (though admittedly it is objectively worse to calculate).

  • 3
    $\begingroup$ I think OP's asking about how $\frac{\delta (\partial^2\alpha(x))}{\delta\alpha} = \partial^2$, not necessarily about functional determinants. $\endgroup$ Nov 26, 2020 at 6:13
  • $\begingroup$ Is it possible that $\frac{\delta(\partial^2 \alpha(x))}{\delta \alpha} \neq \partial^2$ while $\det\left( \frac{\delta(\partial^2 \alpha(x))}{\delta \alpha} \right) = \det ( \partial^2) $ $\endgroup$
    – Ivan
    Nov 27, 2020 at 6:31
  • 2
    $\begingroup$ @Ivan $\frac{\delta \partial^2\alpha(x)}{\delta\alpha(y)}=\partial^2\delta(x-y)$. Integrating by parts gives $\partial^2$. This of this as the moral equivalent of taking an $x_j$ derivative of $A^i_jx^j$. The result of the computation is, in fact, the elements of the operator $A$. $\endgroup$ Nov 27, 2020 at 20:56
  • $\begingroup$ @RichardMyers you know any reference where I can get more familiar with this type of computations? I need to compute $ \frac{\delta_\mu \alpha(x)}{\delta \alpha}$ and other similar terms. $\endgroup$
    – Ivan
    Nov 28, 2020 at 1:00
  • 1
    $\begingroup$ @Ivan There are many QFT references, all of which cover functional derivatives and determinants to varying degrees of depth. But I believe David Tong's gauge theory notes contain an explicit calculation of a functional determinant. Even if they don't they are very nice notes in general. He also has a set of QFT notes (which I haven't looked at) that might contain useful information and calculations. $\endgroup$ Nov 28, 2020 at 1:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.