I’m interested in the form of the photon propagator in position space when expressed in a general $R_{\xi}$ gauge. The integral representation of this propagator is usually written as the sum of two Fourier transforms:

$$G_F^{\mu{}\nu{}}\left(x-y,\xi{}\right)=-i{\ g}^{\mu{}\nu{}}\int\frac{d^4k}{{\left(2\pi{}\right)}^4}\frac{e^{-ik\bullet{}\left(x-y\right)}}{k^2+i\epsilon{}}+i\ \xi{}\int\frac{d^4k}{{\left(2\pi{}\right)}^4}\frac{e^{-ik\bullet{}\left(x-y\right)}k^{\mu{}}k^{\nu{}}}{{(k^2+i\epsilon{})}^2}.\tag{1}$$

The value of the gauge parameter $\xi$ is arbitrary in general, but often is selected to impose a specific gauge condition on the propagator (e.g., $\xi$=0 for Feynman or $\xi$=1 for Landau). However, I want to leave $\xi$ unspecified.

Now the first integral in (1) just gives the massless scalar Feynman propagator, which has a well-known form in position space, so the term in (1) proportional to the metric is explicitly:

$$i\ g^{\mu{}\nu{}}G_F(x-y)=g^{\mu{}\nu{}}\left[\frac{1}{{4\pi{}}^2}PV\frac{1}{{(x-y)}^2}+\frac{i}{4\pi{}}\delta{}\left({(x-y)}^2\right)\right].\tag{2}$$

But what is the explicit form of the second photon propagator term proportional to $\xi$? The second integral involves higher-order poles and a more complicated tensor structure. Can anyone point me to a reference which demonstrates how to evaluate this second integral and then displays the result in position space?


1 Answer 1


I also needed the Fourier transform of the propagator


I tried computing it, but couldn't do it, so here's a different solution.

From the action we know that the gluon field has length dimension $[A]=\frac{-D+2}{2}$ and so the propagator has dimension $[<AA>]=-D+2$. Since the vectors in the numerator of the second term become derivatives in the Fourier transform, the $D$-dimensional position space propagator has the general form

\begin{align} G^{\mu\nu}(x)&=a\left(\frac{g^{\mu\nu}}{|x|^{D-2}}+b\xi\partial_x^{\mu}\partial_x^{\nu}\frac{1}{|x|^{D-4}}\right)\\ &=a\left(\frac{g^{\mu\nu}}{|x|^{D-2}}+b\xi(4-D)\left[\frac{g^{\mu\nu}}{|x|^{D-2}}-(D-2)\frac{x^{\mu}x^{\nu}}{|x|^D}\right]\right) \end{align}

where $a$ and $b$ are constants. Now it seems that in four dimensions the gauge dependent term vanishes, which is not ideal. We will ignore that for now and compute the relative constant $b$. Setting $\xi=1$ the propagator becomes transversal and in momentum space we have the identity


This means that in position space

$$\partial_\mu G^{\mu\nu}(x)=0$$

should hold. Solving this condition, we find


which cancels the problematic factor in the gauge term of the propagator.

(I imagine something similar happens when performing the Fourier transform. After taking derivatives there will probably be a $D-4$ factor that has to be cancelled by the pole of a gamma function from the spherical integration.)

Using OPs result for the first term to fix the overall constant ($a=\frac{-i}{4\pi^2}$) we finally have


  • $\begingroup$ How do you reproduce the delta function in OPs eq. (2)? $\endgroup$
    – dennis
    Commented Nov 7, 2023 at 21:02
  • $\begingroup$ ah, yes. good point. the delta function and the principal value come from the epsilon prescription and the (distributional) identity $\frac{1}{x+\mathrm{i}\epsilon}=\mathrm{PV}\frac{1}{x}-\mathrm{i}\pi\delta(x)$ (see e.g. here). also, my prefactor is only correct for dimension 4. in general there are probably some gamma functions involved. i'll update the answer when a have a minute. $\endgroup$ Commented Nov 9, 2023 at 15:43
  • 1
    $\begingroup$ yes, but $\partial\partial G_F$ does not give the fourier transform of the $\xi$-dependent term. the $x$-derivatives appear when you use the $e^{-ikx}$-term in the fourier transform to get rid of the $k$-vectors in the numerator. after that youre left with the fourier transform of $1/k^4$, which will not give you $G_F$. you can also see in your second comment, that $g$ and $\xi\partial\partial$ have incompatible dimensions (0 and -2, respectively), so you cant add them. $\endgroup$ Commented Nov 13, 2023 at 10:40

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