## Green’s function in dimension two

Any hint to compute the Green's function:

If $\Delta_z G(z,z') = 2\pi \delta^2(z-z')$, then

$$G(z,z')=-2\pi \int \frac{d^2q}{4\pi^2}\frac{e^{iq(z-z')}}{q^2} = ln|\mu(z-z')|$$ where $\mu$ is some infrared cutoff at $q=0$.

I can see the first step is Fourier transform and inverse Fourier transform but I don't know how to figure out the second step. Thank you.

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How is $mu$ defined? – Helge Sep 27 2010 at 15:21
This is really not a research level mathematics question. Please see mathoverflow.net/faq for some other places you might get help from. – jc Oct 15 2010 at 0:11

Defining $\frac{\partial}{\partial \bar z}=\frac12(\partial_x+i\partial_y)$, $\frac{\partial}{\partial z}=\frac12(\partial_x-i\partial_y)$, we have $$\frac{\partial}{\partial \bar z}(\frac{1}{\pi z})=\delta_0$$ since $\frac{\partial}{\partial \bar z}(\frac{1}{\pi z})=0$ on $\mathbb R^2$ \ $0$ so that $support \frac{\partial}{\partial \bar z}(\frac{1}{\pi z})\subset{0}$ and $\frac{\partial}{\partial \bar z}(\frac{1}{\pi z})$ is homogeneous with degree $-2$. As a result, $\frac{\partial}{\partial \bar z}(\frac{1}{\pi z})=c\delta$. To check $c=1$, we test $\frac{\partial}{\partial \bar z}(\frac{1}{\pi z})$ against $e^{-\pi z\bar z}$. In two dimensions we have $\Delta= 4\frac{\partial}{\partial \bar z}\frac{\partial}{\partial z}$ so that $$\Delta(\frac{1}{2\pi}\ln \vert x\vert)=\frac{1}{\pi}\frac{\partial}{\partial \bar z}\frac{\partial}{\partial z}\ln (z\bar z)= \frac{\partial}{\partial \bar z}(\frac{1}{\pi z})=\delta.$$