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I have been trying to calculate the massless scalar field propagator in position space by directly Fourier transforming the momentum space propagator. $$\int{d^2p\frac{1}{(p^0)^2-(p^1)^2}e^{-i(p^0t-p^1x)}}$$

Upon referring to multiple sources (linked below), I realize that the answer is actually proportional to $ln|x|$ but I don't see how this integral will result in that answer. All of these sources obtain that answer by finding the massive propagator and then taking the $m\rightarrow 0$ limit. I don't see what I am missing by directly doing doing the above integral.

To see how the above integral does not give $ln|x|$:

Evaluate the $dp^0$ integral using the Feynman prescription for avoiding the poles and this will give: $$\int\frac{i}{2\pi p^1}e^{-ip^1 (t-x)}dp^1$$This integral is actually a constant multipled by a step function.

I also head into a similar problem in the (1+3)-D case where a direct Fourier transform gives a different answer from the known propagator and from the answer got by taking the limit on the massive case. So, what am I missing by directly Fourier transforming the propagator from momentum space?

Sources:

http://max2.physics.sunysb.edu/~rastelli/HW4Solutions.pdf

https://arxiv.org/pdf/0811.1261.pdf

Two-point function of massless scalar theory in 2d CFT

Massless limit of the Klein-Gordon propagator

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  • $\begingroup$ Just Wick-rotate the integral and then observe that $d^2p=pdpd\theta$. $\endgroup$ – Jon Jul 8 at 15:13
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The idea for this kind of computation is the following. Firstly, add a mass term to the propagator. This will yield $$ \int \frac{d^2p}{(2\pi)^2}\frac{1}{p^2-m^2}e^{ip\cdot x}. $$ This integral can be evaluated provided we make the rotation $p_0\rightarrow ip_0$ that yields $$ i\int \frac{d^2p}{(2\pi)^2}\frac{1}{p^2+m^2}e^{ip\cdot x}. $$ Now, one us $d^2p=pdpd\theta$ and $p\cdot x = pr\cos\theta$ and one has to evaluate the integral $$ \frac{1}{4\pi^2}\int_0^\infty dp\int_0^{2\pi}d\theta\frac{p}{p^2+m^2}e^{ipr\cos\theta}. $$ Firtstly, we integrate on $\theta$. This can be done remembering that $$ e^{ia\cos\theta}=\sum_{n=0}^\infty i^nJ_n(a)e^{in\theta} $$ being $J_n$ the Bessel functions of the first kind of integer order. Integration in $\theta$ leaves just $J_0$ and so, our integral becomes $$ -\frac{1}{4\pi^2i}\int_0^\infty dp\frac{p}{p^2+m^2}J_0(pr). $$ This integral can be evaluated with techniques in complex integration, with a proper choice of the integration path, yielding $$ G(r)=-\frac{1}{2\pi}K_0(mr) $$ being $K_0$ the modified Bessel function of 0 order. This is the point where the references you cite bring you. The next step is to note that, for $m\rightarrow 0$, the massless limit, $$ K_0(mr)\sim -\ln r $$ and you are done. Note the presence of an infinite constant, $\ln m$, that is generally omitted taking the massless limit. The reason is that, in the massless limit, one can always add an arbitrary constant to the propagator.

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  • $\begingroup$ Is there a way to get the same answer without adding the mass term? $\endgroup$ – adithya Jul 18 at 8:00
  • $\begingroup$ That is an interesting question. I guess so but note that, without the mass term, after integrating on the angle you are left with a divergent integral. This is reflected by the infinite constant $\ln m$ that one gets with the mass term. $\endgroup$ – Jon Jul 18 at 11:25

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