How to solve the electric potential due to a charge distribution using fast Fourier transforms I'm writing a toy program to simulate 2D wave functions. I'm using a split-operator method to solve the Schrödinger equation and have no problems with arbitrary potentials. However, I'd now like to have the wave functions react to the electric potential due to a 2D charge distribution.
The Poisson equation relates the potential with a charge distribution (I'm ignoring possible constants here since I'm not interested in actual units)
$\nabla^2 V = -\rho$.
If we take the Fourier transform we get
$-k^2 \hat{V} = -\hat{\rho} \Longleftrightarrow \hat{V} = \hat{\rho}/k^2$.
I'm well aware that at the origin $k^2 = 0$ and that the zero mode corresponds to the average value of $V$. Normally I'd just set $\hat{\rho}/k^2$ to zero there since the absolute value of the potential doesn't matter in most cases - it's its derivatives that appear in the equations.
Now, however, as far as I understand, in the Schrödinger equation we have
$\left(-\nabla^2 + V\right)\psi$,
meaning that it is actually the potential, not its derivatives, that shows up in the equation. How do I work out the proper value for $\hat{\rho}/k^2$ at the origin? Is it somehow related to the total charge in the system? My plan B is to solve $V$ using Jacobi over-relaxation, but I'd really like to stick with FFTs that are very efficient and noniterative.
I tried searching for previous questions and their answers here and elsewhere but couldn't find the answer I was looking for (or possibly identify one as such). Thanks a lot in advance!
 A: $\newcommand\ket[1]{|#1\rangle}$The absolute part of $V$ is still irrelevant in the Schrödinger equation. If you have an initial state $\ket\psi,$ then its time evolution is $$\ket{\psi(t)}=\exp(-iHt/\hbar)\ket\psi=\exp(-i(-\nabla^2+V)t/\hbar)\ket\psi.$$ If you add a constant to $V,$ so $V\mapsto V+E$ (equivalently, add a constant to $H$ for $H\mapsto H+E$), then the new time evolution is just (remember $\exp(A+B)=\exp A\exp B$ when $A,B$ commute) $$\ket{\psi'(t)}=\exp(-iEt/\hbar)\ket{\psi(t)}.$$ Now remember that quantum states are only physical up to global phases: $\ket\psi$ and $e^{i\theta}\ket\psi$ always represent the same physical situation. In particular, all observables (Hermitian operators) have the same probability distributions. So at every time $t,$ the evolved state with the shifted potential $\ket{\psi'(t)}$ is the same physical state as the state resulting from the unshifted potential $\ket{\psi(t)}$: the absolute values of $V$ do not matter, even for the Schrödinger equation.
So the reason you didn't find many answers to your question is because there's no question to answer. You can do whatever is most convenient for handling the 0-frequency component of $V$ and you will still get all the right physics.
