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$$H_{tot}=\sum \dfrac{p_i^2}{2m}+\sum\dfrac{p_I^2}{2M_I}+\sum V_{nucl}(r_i)+\dfrac{1}{2}\sum_{i\ne j} \dfrac{e^2}{|r_i-r_j|}+\dfrac{1}{2}\sum_{I\ne J}\dfrac{z_Iz_Je^2}{|R_I-R_J|} $$

with: $$V_{nuc}(r)=-\sum_I\dfrac{z_Ie^2}{|r-R_I|}$$

This equation is for the total non-relativistic Hamiltonian of a system of electron and nucleus in mutual interaction via Coulomb forces. Here first term is for K.E. Of electron, second term is K.E for nucleus. My problem is on third term where negative sign is used. Is that potential energy of electron and nucleus? and in fourth and fifth term 1/2 is used. My question is why did 1/2 used in potential energy equation for both electron and nucleus?

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1) The minus sign shows the fact, that the interaction between electrons and protons is attractive, not repulsive.

2) The coefficient $\frac{1}{2}$ is used in the cases, where you count the interaction between the same particles not to sum the same potential twice. I think you have slight mistake in the range of summation, let me try to explain your problem in form of a equation.

$$ \frac{1}{2}\sum_{i \neq j}\frac{e^2}{|r_i - r_j|} = \sum_{i < j}\frac{e^2}{|r_i - r_j|}$$

Edit: Corrected my own mistakes as pointed out by Peter. Thank you!

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  • $\begingroup$ Jan, you probably should write $i\neq j$ on the left and $i<j$ on the right. $\endgroup$ Jun 3 '13 at 8:27
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Negative sign just indicates the attractive potential. The factor of $1/2$ is to average the double counting of the same term. It is the average of two terms for every value of $i$ and $j$.

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