Don't Understand Screened Potential Expression Can someone explain to me please why this screened potential has the following expression?
$$ V(r)=\frac{e^{2}}{\epsilon}\sum_{n=-\infty}^{\infty}\frac{(-1)^{n}}{\sqrt{r^{2}+\xi^{2}n^{2}}} $$
It is relative to a graphene sheet between two gates $ \xi $ apart. It is equation (C1) of the paper https://arxiv.org/abs/2009.12376.
I particularly don't understand the role of $ n $ and why the enumerator is $ (-1)^n $.
If I'm not mentioning something you think is important, please say!
Thanks in advance!
 A: It is the solution for the potential on a charge in between two conducting planes (position $x=\xi/2$). In order to agree with the boundary conditions (potential plus the potential of the charge must be zero at the boundaries), the only solution is the infinite sum you wrote. The only way (that I know) to reach this results is to construct the potential using the image method and the God-given ansatz that the solution is given by a series of equidistant charges with alternating signs but same absolute charge. The properties of Poisson's equation guarantee that the solution is unique. You can check which boundary conditions are satisfied every time that you add a charge.
As for the alternating sign: Suppose that you start with one charge in between the plates, the left and right plates have a non-zero potential, so you have to add an image charge with the opposite sign to the right and to the left. However, this does not solve the isuee, as you added these image charges, you now need new image charges with the original sign farther away to cancel the effect of the added charges on the opposite plates. And so on...
A: I'd use a mode expansion to write
$$
G({\bf r}, {\bf r}', z,z') = \frac 1 {2\pi} \sum_{m=-\infty}^{\infty} \int \frac{d^2k}{(2\pi)^2}
\frac{e^{i{\bf k}\cdot ({\bf r}- {\bf r}')}}{k^2 + m^2\pi^2/(2\xi)^2} 
\sin( m\pi z/2\xi) \sin( m\pi z'/2\xi)
$$
and then
$$
\frac 1{2\pi} \sum_{m=-\infty}^\infty \frac{e^{im\tau}}{m^2+k^2}=
   \sum_{n=-\infty}^\infty   \frac 1{2|k|} e^{-|k||\tau+2\pi n|}, \quad \hbox{(Poisson Summation)}
$$
to do the sum over $m$. This should give the image charge pattern without having to visualize them
