Solving Schrödinger's Equation for the electronic energies of the Molecular Ion Hydrogen H2+ in the Elliptic coordinate system Electronic Energies of Molecular Ion Hydrogen $H_2^{+}$
$r_1$ is the distance between the proton $1$ and the electron.
$r_2$ is the distance between the proton $2$ and the electron.
$R$ is the distance between the two protons, fixed parameter in Born Oppenheimer approximation.
Schrödinger's Equation:
$$
\hat{H} \Psi = E_{el} \Psi
$$
$$
-\frac{{\hbar}^2}{2 m_e} \Delta \Psi -\frac{e^2}{4 \pi \varepsilon_0} \left(\frac{1}{r_1}+\frac{1}{r_2}\right)\Psi = E_{el} \Psi
$$
Atomic Units:
$$
a_0=\frac{4 \pi \varepsilon_0 \hbar^2}{m_e e^2} = 0.5291 *10^{-10} m
$$
$$
E_h=\frac{\hbar^2}{m_e a_0^2}=27.21 eV
$$
$$
\Delta \Psi + \frac{2}{a_0}\left(\frac{1}{r_1}+\frac{1}{r_2}\right)\Psi=-2 \frac{E_{el}}{E_h a_0^2} \Psi
$$
We introduce the adimentional parameters $ r= \frac{R}{a_0} $ and $ \varepsilon = \frac{E_{el}}{E_h} $
Elliptic coordinate system:
$ \xi = \frac{r_1 + r_2}{R} $ , $ \eta=\frac{r_1 - r_2}{R} $ and $ \phi $ is the angle between the position vector of the electron and the $xy$ plane.
The Laplacian operator will become:
$$ \Delta =\frac{4}{R^2 (\xi^2 - \eta^2) } \left[\frac{\partial }{\partial \xi} (\xi^2 -1)\frac{\partial }{\partial \xi}+\frac{\partial }{\partial \eta} (1-\eta^2)\frac{\partial }{\partial \eta}+\frac{\xi^2 - \eta^2}{(\xi^2-1)(1-\eta^2)}\frac{\partial^2 }{\partial \phi^2}\right] $$
then we obtain:
$ \left[\frac{\partial }{\partial \xi} (\xi^2 -1)\frac{\partial }{\partial \xi}+\frac{\partial }{\partial \eta} (1-\eta^2)\frac{\partial }{\partial \eta}+\frac{\xi^2 - \eta^2}{(\xi^2-1)(1-\eta^2)}\frac{\partial^2 }{\partial \phi^2}\right] \Psi + 2r\xi \Psi = - \frac{1}{2} r^2\varepsilon (\xi^2 - \eta^2)\Psi $
We can separate the variables using $ \Psi = \Xi_{\xi} H_{\eta} \Phi_{\phi} $ so we can obtain three indipendent differential equations:
$$ \frac{\partial }{\partial \xi} (\xi^2 -1)\frac{\partial \Xi }{\partial \xi} + \left(2r\xi+A+\frac{1}{2}r^2 \varepsilon \xi^2 - \frac{\Lambda^2}{\xi^2-1}\right)\Xi=0 $$
$$ \frac{\partial }{\partial \eta} (1-\eta^2)\frac{\partial H }{\partial \eta} + \left(-A-\frac{1}{2}r^2 \varepsilon \eta^2 - \frac{\Lambda^2}{1-\eta^2}\right)H=0 $$
$$ \frac{\partial^2 \Phi }{\partial \phi^2} = - \Lambda^2 \Phi $$
where A is a separation parameter and $ \Lambda \in \mathbb{N} $
focusing on the first one we can divide both bembers for $ \Xi $ and do this substitution: $ f_{\xi} =- \frac{1}{\Xi} \frac{\partial \Xi}{\partial \xi} $ obtaining this new differential equation:
$$ 2 \xi f + (\xi^2-1)(f'+f^2)-\left(2r\xi+A+\frac{1}{2}r^2 \varepsilon \xi^2 - \frac{\Lambda^2}{\xi^2-1}\right)=0 $$
and from now on I don't know how to move. Is there someone who knows how to solve numerically this equation?
Thank you very much
 A: As you have discovered, the H$_2^+$ ion is separable but it is not exactly solvable. Spheroidal coordinates allow you to separate the three-dimensional time-independent Schrödinger equation into three separate one-dimensional Schrödinger problems, one of which is trivial, but that's as far as you can go.
On the analytical side, you can do some simple approximations, such as Hund-Mulliken and Guillemin-Zener wavefunctions, but these are fairly crude. You can make more sophisticated approximations and stay on the analytical side, but you're stuck with (arguably) blind attempts at what the solution ought to look like.
Numerically that's a complicated problem, because the equations are not quite as separated as you'd like: the eigenvalue $\varepsilon$ and the separation constant $A$ appear in both the $\xi$ and the $\eta$ equations, so they make a 'bispectral' problem and must be solved in tandem.
For a good modern update on current numerical methods, try

New Approach for the Electronic Energies of the Hydrogen Molecular Ion. T.C. Scott et al. Chem. Phys. 324 no. 2-3, 323 (2006), arXiv:physics/0607081.

If you want more accessible, low-powered numerical methods to get a feel for the thing, you're somewhat more likely to find an answer at Computational Science, though.
