I have a charge density $\rho(\vec{r})$ which is given as an expansion of spherical harmonics $Y_{l}^{m}(\hat{r})$.

$$\rho(\vec{r}) = \sum_l^{l_{max}} \sum_{m = -l}^{l} \rho_{l}(r) Y_{l}^{m}(\hat{r}) $$

Now I want to solve the Poisson equation $\nabla^2V(\vec{r}) = -4 \pi \rho(\vec{r})$. One way of solving this equation would be to use a Green's function. However, I want to explicitly solve the differential equation for each ${l,m}$.

For that, I am expanding the Poisson operator into spherical harmonics.

$$\nabla^2V(\vec{r}) = \sum_l^{l_{max}} \sum_{m = -l}^{l} \nabla^2 \left( V_l(r) \, Y_{l}^{m}(\hat{r}) \right) = \sum_l^{l_{max}} \sum_{m = -l}^{l} V_l(r) \, \nabla^2 Y_{l}^{m}(\hat{r}) + \nabla^2 V_l(r) \, Y_{l}^{m}(\hat{r}) $$

$$ \nabla^2 Y_{l}^{m}(\hat{r}) = - \frac{l (l + 1)}{r^2} \, Y_{l}^{m}(\hat{r}) $$

After some algebra, I am left with:

$$\nabla^2V_l(r) = \frac{d^2V}{dr^2} + \frac{2}{r} \frac{dV}{dr} = V_l(r) \, \frac{l (l + 1)}{r^2} - 4 \pi \rho(r)$$

This Euler-Cauchy ODE equation can than be solved. My question is now how to I get the correct initial values for solving this ODEs?

  • $\begingroup$ Requiring $V(0)$ to be nonsingular will give you one appropriate restriction. Keep in mind also that any dynamics arising from a potential $V$ are invariant under shifts (i.e $V\mapsto V+V_0$) so this one condition should already be enough. If you like you can require $V(0)=0$ which certainly appeals to physical intuition. $\endgroup$
    – K.defaoite
    Commented Sep 29, 2022 at 20:50
  • $\begingroup$ In practice this is more difficult because the ODE is singular at $r = 0$. $\endgroup$
    – pmu2022
    Commented Oct 4, 2022 at 14:15
  • $\begingroup$ Are you certain there is no typo in the last equation? Also, when you expand $\rho$ in spherical harmonics, the coefficients $\rho_l(r)$ have no $m$ dependence- is this deliberate? In any case, you find eventually an ODE for the $V_l$. The appropriate boundary conditions for the $V_l$ are determined by your choice of boundary conditions for $V$, which you haven't specified $\endgroup$
    – Sal
    Commented Oct 4, 2022 at 17:11


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