In $H_2^+$, what is the Hamiltonian of the movement of the electron? An electron is orbiting two protons. With the Born-Oppenheimer approximation that the protons do not move, I'd write the Hamiltonian of the electron's movement as:
$$ \mathbf{H} = -\frac{\hbar^2}{2m}\nabla^2 + E_p$$
with
$$E_p = -\frac{e^2}{4\pi \epsilon_0}\left(\frac{1}{r_1}+\frac{1}{r_2}\right) $$
as the potential caused by the protons, with $r_1$ and $r_2$ denoting distances to the protons. Apparently it should be
$$E_p = -\frac{e^2}{4\pi \epsilon_0}\left(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{r_0}\right), $$
where $r_0$ denotes the distance between the protons. How can that term be explained?
 A: Is'nt
$$\frac{e^2}{4\pi\epsilon_0 \, r_0}$$
the (constant) electrostatic energy of repulsion between the two stationary protons at a separation $r_0$?
EDIT:
The full problem of $H_2^+$ molecule $\mathcal{H} \Psi = \mathcal{E} \Psi$ is given by the Hamiltonian
$$\mathcal{H} = \frac{\mathbf{p}^2}{2m} 
+ \frac{\mathbf{P}_1^2}{2M} + \frac{\mathbf{P}_2^2}{2M}
- \frac{e^2}{4\pi\epsilon_0} \Big(\frac{1}{|\mathbf{r}_1|} + \frac{1}{|\mathbf{r}_2|} - \frac{1}{|\mathbf{R_1}-\mathbf{R_2}|} \Big)$$
where $\mathbf{r}$ is the position of the electron, $\mathbf{R}_i$ the nuclei location, and $\mathbf{r}_i = \mathbf{r} - \mathbf{R}_i$.
In the Born-Oppenheimer approximation for massive nuclei, we take
$$\Psi(\mathbf{r}, \mathbf{R}_1, \mathbf{R}_2) = \phi(\mathbf{R}_1, \mathbf{R}_2) \, \psi(\mathbf{r}, \mathbf{R}_1, \mathbf{R}_2)$$
where it is assumed that the $\mathbf{R}_i$ are treated as constant parameters in $\psi$. The nuclear and electronic motions then decouple.
One can write the decoupled equations in two ways
(1) including the electrostatic energy of the protons in the electronic equation
$$\Big[ \frac{\mathbf{p}^2}{2m} -
\frac{e^2}{4\pi\epsilon_0} \Big(\frac{1}{|\mathbf{r}_1|} + \frac{1}{|\mathbf{r}_2|} - \frac{1}{|\mathbf{R_1}-\mathbf{R_2}|} \Big) \Big] \psi = E \psi$$
$$\Big[ \frac{\mathbf{P}_1^2}{2M} + \frac{\mathbf{P}_2^2}{2M} + E \Big] \phi = \mathcal{E} \phi$$
(2) including the electrostatic energy of the protons in the equations for the nuclei
$$\Big[ \frac{\mathbf{p}^2}{2m} -
\frac{e^2}{4\pi\epsilon_0} \Big(\frac{1}{|\mathbf{r}_1|} + \frac{1}{|\mathbf{r}_2|}  \Big) \Big] \psi = E' \psi$$
$$\Big[ \frac{\mathbf{P}_1^2}{2M} + \frac{\mathbf{P}_2^2}{2M} + \frac{e^2}{4\pi\epsilon_0} \frac{1}{|\mathbf{R_1}-\mathbf{R_2}|} + E' \Big] \phi = \mathcal{E} \phi$$
where $E$ and $E'$ are the electronic energy contributions, and are functions of the $\mathbf{R}_i$.
In both the cases above, you solve the equation for $\psi$ assuming constant $\mathbf{R}_i$. The electronic energy $V_1 = E(\mathbf{R}_1, \mathbf{R}_2)$ then acts as an effective potential for the motion of the nuclei in case (1) above while in case (2) the nuclei experience an effective potential energy $V_2 = \frac{e^2}{4\pi\epsilon_0} \frac{1}{|\mathbf{R_1}-\mathbf{R_2}|} + E'(\mathbf{R}_1, \mathbf{R}_2)$.
It looks like your course material refers to case (1) above, and this is standard in many places, for e.g., Chap 7 of Atkins. I also found places where case (2) is considered.
In either case, the effective potential experienced by the nuclei (either $V_1$ or $V_2$) seems to have the same structure. So probably either case is fine though I would like to know if there is any subtle difference between them.
A: The $\frac{e^2}{4\pi\epsilon_0}\frac{1}{r_0}$ term appears in the potential for the electron motion, as Luboš and Vijay point out, to keep the whole energy accounting in place so that the nuclear motion can be properly quantized. The key point is that this potential does not involve the electron coordinates, so that as far as the electronic wavefunction is concerned it acts like a constant and therefore does not affect the solution to the electronic eigenvalue equation.
If you do include that term, and write the potential energy of the molecule as
$$E_p=-\frac{e^2}{4\pi\epsilon_0}\left(\frac{1}{r_1}+\frac{1}{r_2}-\frac{1}{r_0}\right)$$
then you can write the potential for the nuclear coordinates directly as
$$E_n=\langle\psi(\mathbf{R}_1,\mathbf{R}_2)|E_p|\psi(\mathbf{R}_1,\mathbf{R}_2)\rangle$$
where the total wavefunction is split into electronic and nuclear parts as $$\langle \mathbf{r},\mathbf{R}_1,\mathbf{R}_2|\Psi\rangle=\langle\mathbf{r}|\psi(\mathbf{R}_1,\mathbf{R}_2)\rangle\langle\mathbf{R}_1,\mathbf{R}_2|\phi\rangle.$$
The Schrödinger equation for the nuclear coordinates is then
$$\left(\frac{\mathbf{p}_1^2}{2M}+\frac{\mathbf{p}_2^2}{2M}+E_n\right)|\phi\rangle=E|\phi\rangle.$$
