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?

  • $\begingroup$ No it is not the Born-Oppenheimer approximation: You separate the movement of the center of mass from the relative movement (just as in classical mechanics). This leaves an equation with the reduced mass for the relative coordinates. $\endgroup$ Commented Sep 16, 2015 at 15:05
  • $\begingroup$ Which basically means we take the proton as a rigid point in the coordinate system and only consider the movement of the electron relative to it. And why can we do that? Because the proton has much more mass compared the the electron, which then basically is the B.O. Appr, isn't it? $\endgroup$ Commented Sep 17, 2015 at 9:41
  • 1
    $\begingroup$ No, that is not what we do: We separate centre of mass and relative coordinates, that is the total wave function is something like $\Psi(\vec r_1, \vec r_2) = \phi\left(\frac {m_1 \vec r_1 + m_2 \vec r_2} M\right)\psi(\vec r_1 - \vec r_2)$. Then you will see, that $\psi$ follows a Schrödinger equation where the kinetic energy term is $-\frac{\hbar^2\Delta}{2\mu}$, where $\mu = \frac{m_1m_2}{m_1 + m_2}$ is the reduced mass and $\vec r := \vec r_1 - \vec r_2$ is the relative coordinate. (Side note: This is done explicitly in any elementary quantum mechanics textbook that is worth its money). $\endgroup$ Commented Sep 17, 2015 at 10:42
  • $\begingroup$ Thank you, I got it. I tried this coordinate transformation right now, but somehow don´t get the right results. I posted a question about it. Could you have a look at it? $\endgroup$ Commented Sep 17, 2015 at 13:17

2 Answers 2


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:

  1. 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.

  2. 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.

  • $\begingroup$ What if the forces for 1) and 2) were not the same? (See my answer) I understand your point, though. $\endgroup$ Commented Sep 17, 2015 at 9:39
  • $\begingroup$ @EpsilonDelta They are the same due to the fact, that the electric field is conservative (i.e. has a potential). $\endgroup$ Commented Sep 17, 2015 at 10:43
  • $\begingroup$ However, if you want to disassemble the system you would need twice the force of the assembling, because you just assumed the first particle fixed in place, when in reality it would be repelled (if charges are opposite). You are right per definition of the potential energy but don´t consider both fields. $\endgroup$ Commented Sep 17, 2015 at 11:56

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. :(


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