# Why is the interaction energy of the electrons in an atom positive?

Consider a simple Hamiltonian for the Helium atom (where $e'^2 = e^2/4\pi \epsilon_0)$:

$H=\frac{P_1^2}{2\mu}+\frac{P_2^2}{2\mu}-\frac{Ze'^2}{R_1}-\frac{Ze'^2}{R_2}+\frac{e'^2}{|\vec{R}_1-\vec{R}_2|}$

I understand the first two kinetic terms are positive because they contribute to the ionization; the second two are negative because they correspond to the attractive potential of the nucleus. But is the third term positive? If I calculate the ground state energy ($-\frac{\mu Z^2e'^4}{2\hbar^2}\sim-Z^2$Ry) without electron interactions, I obtain $\sim-4$ Ry, whereas the experimental value is $\sim-5,8$ Ry, which leads me to conclude that to correct this value the interaction term must be attractive (in order to "attract" the system towards its center), so it should have a minus sign.

However, intuitively, I know the interaction between electrons should be repulsive, and have a positive sign!

## 1 Answer

Your intuition is correct. The error is in your calculation of the ground state energy without interactions. There are two electrons, each with energy $-4\,{\rm Ry}$, for a total of $-8\,{\rm Ry}$. The repulsion raises the energy to the experimental value.

• Thank you for your answer, you're absolutely right. I see you're proposing edits over the units of the electron charge and the sign of the ground state energy. I think they should have the opposite sign of the ionization energy, right? – Sebgr May 12 '16 at 1:41