Free, Massless Boson Propagator in Two Dimensions?

[physicus]

Homework Statement

Cosider a single, free, massless boson with action S=\int\mathcal{L}=\frac{1}{2\pi}\int\partial X \overline{\partial}X in two dimensions \overline{\partial}X(z,\overline{z}) = \partial_{\overline{z}} X(z,\overline{z})
Show, that the propagator \langle X(z,\overline{z})X(w,\overline{w})\rangle=-\frac{1}{2}log|z-w|.
Use z=\sigma^{1}+i\sigma^{0} and the integration measure 2i\, dz\wedge d\overline{z}=d\sigma^{1}\wedge d\sigma^{0}.
\sigma^{0}, \sigma^{1} are the real coordiates.

Homework Equations

\langle X(z,\overline{z})X(w,\overline{w})\rangle = \frac{\int_X exp(-S[X])X(z,\overline{z})X(w,\overline{w})}{\int_X exp(-S[X])}<br />

The Attempt at a Solution

Unfortuntely, I don't really know how to start. I don't even know why the integration measure is 2i\, dz\wedge d\overline{z}=d\sigma^{1}\wedge d\sigma^{0}.
It would be very nice if someone could just give me an ansatz.

 

[sgd37]

I think you will find the following attachment useful

 

[physicus]

Thanks, that is helpful, indeed. But I still have some trouble with the formulation in complex coordinates.

That's how I far I got:
The path integral of a total derivative vanishes. Therefore, I obtain:
0=\int\mathcal{D}X\frac{\delta}{\delta X(z,\overline{z})}(e^{-S}X(w,\overline{w})) = \int\mathcal{D}X\, e^{-S}\left[-\frac{\delta S}{\delta X(z,\overline{z})}X(w,\overline{w})+\delta(w-\overline{w})\delta(z-\overline{z})\right]

Now, I have to evaluate the functional derivative of the action:
\frac{\delta S}{\delta X(z,\overline{z})}<br /> = \frac{1}{2\pi}\frac{\delta}{\delta X(z,\overline{z})}\int dzd\overline{z} \partial X \overline{\partial}X
= \frac{1}{2\pi} \int dzd\overline{z}\frac{\delta} {\delta X(z,\overline{z})}(\partial X)\overline{\partial}X+\partial X\frac{\delta}{\delta X(z,\overline{z})}(\overline{\partial}X)
=-\frac{1}{\pi}\partial\overline{\partial} X(z,\overline{z})
In the last step I have used integration by parts with vanishing boundary terms.

Plugging this into the equation above I find:
0=\int\mathcal{D}X\, e^{-S} \left[\frac{1}{\pi} \partial \overline{\partial} X(z,\overline{z})X(w,\overline{w})+\delta(w-\overline{w})\delta(z-\overline{z})\right]
\Rightarrow \partial\overline{\partial} X(z,\overline{z})X(w,\overline{w}) = -\pi\delta(w-\overline{w})\delta(z-\overline{z})

Next, I must show that X(z,\overline{z})X(w,\overline{w})=-\frac{1}{2}ln|z-w| solves this equation, but I am unable to do so:
It would be suffiecient to show: -\frac{1}{2}\int dzd\overline{z} \,\partial\overline{\partial}ln|z-w| = \pi
But here I fail:
\int dzd\overline{z}\, \partial\overline{\partial}ln|z-w|=\int dzd\overline{z}\,\overline{\partial}\frac{1}{z-w} = \ldots

Can someone tell me how to proceed?