# 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, 2019 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.

• Is there a way to get the same answer without adding the mass term? Jul 18, 2019 at 8:00
• 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.
– Jon
Jul 18, 2019 at 11:25

There is already an accepted answer, but the OP is asking whether one can get the same answer without adding a mass term, so let me show how this is done.

An alternative to using mass regularization is to use regularization in the scaling dimension. Let us assume for a minute that we are not dealing with a free field, but with an interacting field in a conformal field theory. Then the 2-point function would be $$\int d^2p \left[ (p^0)^2 - (p^1)^2 + i \epsilon \right]^{\Delta - 1} e^{-i (p^0 t - p^1 x)}$$ where $$\Delta$$ is the scaling dimension of the interacting field. The free field theory limit is $$\Delta \to 0$$. The only subtlety here is that we need to define what it means to take a non-integer power of a quantity that can be negative. This is why I have added an $$i \epsilon$$ prescription in the propagator. This was implicit before, but somehow it was already needed to specify how to deal with the singular points where $$(p^0)^2 = (p^1)^2$$.

Now we can apply the same the steps as in the other answer by Jon:

1. Wick rotation $$p^0 \to i p^0$$ (with $$t \to i t$$ as well): $$i \int d^2p \left[ -(p^0)^2 - (p^1)^2 + i \epsilon \right]^{\Delta - 1} e^{i (p^0 t + p^1 x)}$$ The $$i \epsilon$$ prescription tells us that this is equal to $$i e^{i \pi (\Delta - 1)} \int d^2p \left[ (p^0)^2 + (p^1)^2 \right]^{\Delta - 1} e^{i (p^0 t + p^1 x)}$$
2. Spherical coordinates: $$i e^{i \pi (\Delta - 1)} \int_0^\infty dp \int_0^{2\pi} d\theta p^{2\Delta - 1} e^{i p r \cos\theta}$$
3. Integral over $$\theta$$: $$i e^{i \pi (\Delta - 1)} \int_0^\infty dp ~ p^{2\Delta - 1} J^0(p r)$$
4. Integral over $$p$$: first rescale $$p \to p/r$$ to get $$i e^{i \pi (\Delta - 1)} r^{-2 \Delta} \int_0^\infty dp ~ p^{2\Delta - 1} J^0(p)$$ When $$\Delta$$ is positive but small enough this is a convergent integral that gives $$i e^{i \pi (\Delta - 1)} r^{-2 \Delta} \frac{2^{2\Delta - 1} \Gamma(\Delta)}{\Gamma(\Delta - 1)}$$

The result is divergent in the limit $$\Delta \to 0$$: the $$\Gamma$$ function has a pole there. But if we expand in powers of $$\Delta$$, we get $$i \left[ -\frac{1}{2\Delta} + \log(r) + \ldots \right]$$ This is exactly the same result as with a massive field: an infinite constant, plus the log term (the constant can be absorbed in the definition of what is the "physical" scale).