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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?

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I also needed the Fourier transform of the propagator

$$\tilde{G}^{\mu\nu}(p)=\frac{-ig^{\mu\nu}}{p^2}+\xi\frac{ip^{\mu}p^{nu}}{p^4}.$$

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

$$p_\mu\tilde{G}^{\mu\nu}(p)=0$$

This means that in position space

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

should hold. Solving this condition, we find

$$b=-\frac{1}{2(4-D)}$$

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

$$G^{\mu\nu}(x)=\frac{-i}{4\pi^2}\frac{g^{\mu\nu}}{|x|^{D-2}}+\frac{i\xi}{8\pi^2}\left[\frac{g^{\mu\nu}}{|x|^{D-2}}-(D-2)\frac{x^{\mu}x^{\nu}}{|x|^D}\right].$$

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  • $\begingroup$ How do you reproduce the delta function in OPs eq. (2)? $\endgroup$
    – dennis
    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. $\endgroup$ Nov 9, 2023 at 15:43
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    $\begingroup$ 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. $\endgroup$ Nov 13, 2023 at 10:40

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