most recent 30 from http://mathoverflow.net 2013-05-20T07:26:57Z http://mathoverflow.net/feeds/question/40085 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40085/greens-function-in-dimension-two Dan 2010-09-27T05:44:01Z 2012-03-14T13:20:30Z

<p>Any hint to compute the Green's function:</p> <p>If $\Delta_z G(z,z') = 2\pi \delta^2(z-z')$, then</p> <p>$$ G(z,z')=-2\pi \int \frac{d^2q}{4\pi^2}\frac{e^{iq(z-z')}}{q^2} = ln|\mu(z-z')| $$ where $\mu$ is some infrared cutoff at $q=0$.</p> <p>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.</p>

http://mathoverflow.net/questions/40085/greens-function-in-dimension-two/91173#91173 Bazin 2012-03-14T13:20:30Z 2012-03-14T13:20:30Z

<p>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 $$ \Delta(\frac{1}{2\pi}\ln \vert x\vert)=\frac{1}{\pi}\frac{\partial}{\partial \bar z}\frac{\partial}{\partial z}\ln (z\bar z)= \frac{\partial}{\partial \bar z}(\frac{1}{\pi z})=\delta. $$</p>