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I have a question regarding the Hartree potential in spherical symmetry. Specifically, The Hartree potential reads: $$V_H(\mathbf{r})=\int\frac{n(\mathbf{r}^\prime)}{|\mathbf{r}-\mathbf{r}^\prime|}d\mathbf{r}^\prime$$ where $n(\mathbf{r})$ is the electron density. In spherical symmetry, the density is only function of radius $r: n(\mathbf{r})=n(r)$. I found in the spherical symmetry, the above Hartree potential can be rewritten as: $$V_H(r)=\frac{1}{4\pi}\int V_H(\mathbf{r})\ d\Omega=\int\frac{4\pi n(r^\prime)r^{\prime 2}}{r_>}dr^\prime$$ where $r_>=\max(r,r^\prime)$.

There is a website about this derivation: Hatree_Potential_In_Spherical_Symmetry, but it is a little complicate that I failed to follow it.

Alternately, I would like to prove this result by using the following generating function: $$\frac{1}{|\mathbf{r}-\mathbf{r}^\prime|}=\sum_{lm}\frac{r_<^l}{r_>^{l+1}}\left(\frac{4\pi}{2l+1}\right)Y_l^m(\theta,\phi)Y_l^{m\ast}(\theta^\prime,\phi^\prime)$$ where $r_<=\min(r,r^\prime),r_>=\max(r,r^\prime)$. By using the above generating function, I wrote Hartree potential as follows: \begin{align*} &V_H(r)=\frac{1}{4\pi}\int V_H(\mathbf{r})\ d\Omega=\frac{1}{4\pi}\int \frac{n(\mathbf{r}^\prime)}{|\mathbf{r}-\mathbf{r}^\prime|}d\mathbf{r}^\prime d\Omega\\ &=\frac{1}{4\pi}\int n(r^\prime)\sum_{lm}\frac{r_<^l}{r_>^{l+1}}\left(\frac{4\pi}{2l+1}\right)Y_l^m(\theta,\phi)Y_l^{m\ast}(\theta^\prime,\phi^\prime)r^{\prime 2}\ dr^\prime\ d\Omega\ d\Omega^\prime\\ &=\int dr^\prime\ \frac{4\pi n(r^\prime)r^{\prime 2}}{r_>}\int d\cos\theta d\cos\theta^\prime\ \frac{1}{4}\sum_{l}\frac{r_<^l}{r_>^l} P_l(\cos\theta)P_l(\cos\theta^\prime) \end{align*} where in the last step, the integral over $\phi$ has been done.

Next, I failed to simplify the above equation to the final result. Can anyone give some help? Thank you so much for any help.

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  • $\begingroup$ The minus sign seems very unfortunate. If you're going to use the same symbol $V_H$ both for the multivariate function and for the univariate function (which is already somewhat confusing), one would expect them to have the same sign. $\endgroup$ – joriki Apr 30 '13 at 17:48
  • $\begingroup$ There's still one minus sign left, in the first line of the last equation. That same expression is also missing an integral sign. $\endgroup$ – joriki Apr 30 '13 at 17:51
  • $\begingroup$ To joriki: Thanks a lot for your pointing out the minus sign errors. $\endgroup$ – Hui Zhang Apr 30 '13 at 20:50
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$\int P_l(\cos\theta)d\cos\theta=0$ unless $l=0$ in which case the integral is 2 (You know $\int P_l(\cos(\theta))P_{l'}(\cos(\theta))d\cos(\theta)=\frac{2}{2l'+1}\delta_{ll'}$ and $P_0=1$.That basically kills the sum except for $l=0$, from which your result follows.

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  • $\begingroup$ Thank you so much, now I know how to obtain my results. $\endgroup$ – Hui Zhang Apr 30 '13 at 20:49

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