Schrödinger eq. for H-Atom The stationary Schrödinger equation for the hydrogen atom is given by
$
\left(-\frac{\hbar^2}{2m_e}\Delta_e -\frac{\hbar^2}{2m_k}\Delta_k - \frac{Z e^2}{4\pi \varepsilon_0 |\vec r_e - \vec r_k|} \right) \Psi(\vec r_e, \vec r_k) = \mathrm E \Psi(\vec r_e, \vec r_k)
$
where the subindex $e$ stands for electron and likewise $k$ for the core.
I don´t understand the Coulomb potential. The Coulomb potential of the total system is the electron in the field of the core and the other way around, the core in the field of the electron. So why is there a factor 2 missing in the Coulomb potential?
I know that in further analysis we can neglect the kinetic energy of the core and just consider a relative motion $\vec r=\vec r_e - \vec r_k$. Is this already the Born-Oppenheimer Approximation in disguise?
 A: Consider two charges $q_1$ and $q_2$ kept at some separation.
Suppose we want to calculate the potential energy of the system.
By definition, potential energy is the work done to assemble such distribution.
We can assemble the system in two ways:


*

*Bring $q_1$ to its place; no work done during this as there is no field present. Then bring $q_2$ to its place; now you are moving $q_2$ in the field of $q_1$ so work is done.

*Bring $q_2$ first and then move $q_1$ to its place in the presence of the electric field of $q_2$.
In either case, the work done is the same, but you have done it only one way. Therefore, no $2$ factor.
A: You seem to be thinking that the electron has a Coulomb potential energy  and the nucleus has a separate Coulomb potential energy. That's not how you should think.
Potential energy belongs to the whole system interacting with other parts of the system. Here we have an electron interacting with a proton.  That system has potential energy.  If we were to add another charged particle, we would add potential energy to the system.
Sometimes when dealing with objects falling to Earth we assign a potential energy to an object ($mgh$), but technically that's incorrect. The energy belongs to the system of the Earth and the object.  The Earth's mass shows up when we consider the $g$ term.  Because the Earth has so much mass compared to the traditional falling object, and because of conservation of momentum, most of the potential energy change goes to the change of the object's kinetic energy, so we (justifiably)  ignore what happens to the planet in that situation.  But, unfortunately, ignoring that without explanation leads to a misunderstanding of potential energy. :(
