# Free, massless boson propagaor

1. Jun 4, 2012

### physicus

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
$\langle X(z,\overline{z})X(w,\overline{w})\rangle = \frac{\int_X exp(-S[X])X(z,\overline{z})X(w,\overline{w})}{\int_X exp(-S[X])}$

3. 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.

2. Jun 5, 2012

### sgd37

I think you will find the following attachment useful

File size:
163.1 KB
Views:
70
3. Jun 5, 2012

### 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:
$\frac{\delta S}{\delta X(z,\overline{z})} = \frac{1}{2\pi}\frac{\delta}{\delta X(z,\overline{z})}\int dzd\overline{z} \partial X \overline{\partial}X$
$= \frac{1}{2\pi} \int dzd\overline{z}\frac{\delta} {\delta X(z,\overline{z})}(\partial X)\overline{\partial}X+\partial X\frac{\delta}{\delta X(z,\overline{z})}(\overline{\partial}X)$
$=-\frac{1}{\pi}\partial\overline{\partial} X(z,\overline{z})$
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?