Commutator of massless scalar field Hello I'm trying to calculate $\langle 0|[\phi(x),\phi(0)]|0\rangle$ where $\phi (x)$ is a free massless scalar field.
I've computed $$\langle 0|\phi (x) \phi (0)|0\rangle = \frac{1}{4\pi^2}\frac{1}{(x_0 - i\epsilon)^2 - (\vec x)^2}$$ where $x = (x_0, \vec x)$ and I am using the (+, -, -, -) signature.
I'm confused about computing the commutator. If I try to compute it I get something like $$\frac{i\epsilon}{4\pi^2}\frac{1}{((x_0 - i\epsilon)^2 - (\vec x)^2)((-x_0 - i\epsilon)^2 - (\vec x)^2) },$$ where I have discarded terms of $O(\epsilon^2)$. 
I cannot see that this will simplify to anything meaningful or simple and how I this will be 0 when $x$ is spacelike. Can anyone offer any help?
 A: \begin{align}\langle 0|\phi (x) \phi (0)|0\rangle &= \frac{1}{4\pi^2}\frac{1}{(x_0 - i0_+)^2 - (\vec x)^2} \\ &= \frac{1}{8\pi^2 |\vec x|}\left(\frac{1}{(x_0 - i0_+) - |\vec x|}
- \frac{1}{(x_0 - i0_+) + |\vec x|}\right)\:,\end{align}
that is 
\begin{align} \langle 0|\phi (x) \phi (0)|0\rangle  &= \frac{1}{8\pi^2 |\vec x|}\left(\frac{1}{x_0 - |\vec x| - i0_+ }
- \frac{1}{x_0 + |\vec x|- i0_+ }\right)\\ & = \frac{1}{8\pi^2 |\vec x|}\left(PV \frac{1}{x_0 - |\vec x| }-
PV \frac{1}{x_0 + |\vec x| }\right) +  \frac{\pi i}{8\pi^2 |\vec x|}
\left(\delta(x_0 - |\vec x| )-
\delta(x_0 + |\vec x| )\right)\:.\end{align}
Similarly
\begin{align} \langle 0|\phi (0) \phi (x)|0\rangle &=  \frac{1}{8\pi^2 |\vec x|}\left(PV \frac{1}{-x_0 - |\vec x| }-
PV \frac{1}{-x_0 + |\vec x| }\right) +  \frac{\pi i}{8\pi^2 |\vec x|}
\left(\delta(-x_0 - |\vec x| )-
\delta(-x_0 + |\vec x| )\right)\:.\end{align}
Taking the difference, using $PV1/(-z) = -PV 1/z$ and $\delta(z)=\delta(-z)$, we have an expression for the causal propagator like this
$$\langle 0|[\phi (x), \phi (0)]|0\rangle  =   \frac{2\pi i}{8\pi^2 |\vec x|}
\left(\delta(x_0 + |\vec x| )-
\delta(x_0 - |\vec x| )\right)\:.\tag{1}$$
Using $$\delta(f(x)) = \sum_{i} \frac{\delta(x-x_i)}{\left|\frac{\mathrm df}{\mathrm dx}\bigg |_{x_i}\right|}\tag{2}$$
where $x_i$ are distinct simple zeros of $f$, the found identity can be recast into the form
$$\langle 0|[\phi (x), \phi (0)]|0\rangle =   -\frac{4\pi i}{8\pi^2}
\mbox{sgn}(x_0)\delta(x^2_0 - (\vec x)^2 )\:.$$
Barring wrong coefficients (please check everything!), the final result should be
$$\langle 0|[\phi (x), \phi (0)]|0\rangle =   \frac{1}{2\pi i }
\mbox{sgn}(x_0)\delta(x^2_0 - (\vec x)^2 )\:.$$
It is evident that the right-hand side vanishes for spacelike related arguments of the causal propagator. However it also vanishes for timelike related arguments! This is a feature of free massless theories in flat spacetime.
