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The Hamiltonian for a helium atom with two electrons is given by $$\hat H=-\frac{\hbar^2}{2m_e}(\nabla_1^2+\nabla_2^2)-\frac{e^2}{4\pi \epsilon_0}\left(\frac{2}{|\vec r_1|}+\frac{2}{|\vec r_2|}-\frac{1}{|\vec r_2-\vec r_1|}\right)$$ This makes sense to me: the terms from left to right are the kinetic energy, the potential energy due to the nucleus and the potential energy due the electron-electron repulsion. What I don't understand is that the repulsion term has a one as numerator. When you write down the Hamiltonian for the first electron you get $$\hat H_1=-\frac{\hbar^2}{2m_e}\nabla_1^2-\frac{e^2}{4\pi \epsilon_0}\left(\frac{2}{|\vec r_1|}-\frac{1}{|\vec r_2-\vec r_1|}\right)$$ and vice versa for the second electron.

Because both the Hamiltonian have a $\frac{1}{|\vec r_2-\vec r_1|}$ term you would expect that the total Hamiltonian $\hat H=\hat H_1+\hat H_2$ would have two such terms, resulting in $\frac{2}{|\vec r_2-\vec r_1|}$. Why is the numerator just one?

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    $\begingroup$ Hint: Why do you try to split the Hamiltonian in 2 parts when the He atom consists of 3 particles? $\endgroup$ – Qmechanic Mar 5 '18 at 10:16
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This just comes from the classical potential energy, which intuitively is the work done to get to a certain configuration starting in a reference configuration. The reference (0 potential energy) is taken to be the nucleus and the electrons all at infinite separations. When bringing the first electron to position $r_1$ in a coordinate system centered at the nucleus, your work is aided by the attraction of the nucleus (no interaction with the second electron). It is only when you move the second electron there, that the work is influenced by the presence of the other electron.

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  • $\begingroup$ Ah that makes sense. The repulsion term only appears when one of the electrons is already at $r_1$, so it can't appear in both $H_1$ and $H_2$ only in $H_2$. $\endgroup$ – user3502079 Mar 5 '18 at 15:12

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