# 2 dimensional massless scalar field propagator in position space

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:

• Just Wick-rotate the integral and then observe that $d^2p=pdpd\theta$.
– Jon
Jul 8 '19 at 15:13

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.
• 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.