Solution of Poisson's equation

Consider the solution of Poisson's equation:

$$\frac{d^2\phi(x)}{dx^2} = \rho(x),$$ (303)

where $\rho(x) = n(x)/n_0-1$. Note that $n_0 = N/L$ in normalized units. Let $\phi_j\equiv\phi(x_j)$ and $\rho_j\equiv \rho(x_j)$. We can write

$$\phi_j = \sum_{j'=0,J-1}\hat{\phi}_{j'} \,{\rm e}^{\,{\rm i}\,j\,j'\,2\,\pi/J},$$ (304)

$$\rho_j = \sum_{j'=0,J-1}\hat{\rho}_{j'} \,{\rm e}^{\,{\rm i}\,j\,j'\,2\,\pi/J},$$ (305)

which automatically satisfies the periodic boundary conditions $\phi_J=\phi_0$ and $\rho_J=\rho_0$. Note that $\hat{\rho}_0=0$, since $\int_0^L n(x)\,dx = n_0$. The other $\hat{\rho}_j$ are obtained from

$$\hat{\rho}_j = \frac{1}{J}\sum_{j'=0,J-1}\rho_{j'} \,{\rm e}^{-{\rm i}\,j\,j'\,2\,\pi/J},$$ (306)

for $j=1,J-1$. The Fourier transformed version of Poisson's equation yields

$$\hat{\phi}_0 = 0$$ (307)

and

$$\hat{\phi}_j = - \frac{\hat{\rho}_j}{j^2\,\kappa^2}$$ (308)

for $j=1,J/2$, where $\kappa = 2\pi / L$. Finally,

$$\hat{\phi}_j = \hat{\phi}^\ast_{J-j}$$ (309)

for $j=J/2+1$ to $J-1$, which ensures that the $\phi_j$ remain real. The discretized version of Eq. (297) is

$$E_j = \frac{\phi_{j-1}-\phi_{j+1}}{2\,\delta x}.$$ (310)

Of course, $j=0$ and $j=J-1$ are special cases which can be resolved using the periodic boundary conditions. Finally, suppose that the coordinate of the $i$th electron lies between the $j$th and $(j+1)$th grid-points: i.e., $x_j < r_i < x_{j+1}$. We can then use linear interpolation to evaluate the electric field seen by the $i$th electron:

$$E(r_i) = \left(\frac{x_{j+1}-r_i}{x_{j+1}-x_j}\right)E_j+\left(\frac{r_i - x_j}{x_{j+1}-x_j}\right) E_{j+1}.$$ (311)