Quantum Chemistry and ground-state energy

[Blamo_slamo]

Homework Statement

If we were to ignore the interelectronic repulsion in helium, what would be it's ground state energy and wave function?

Homework Equations

I have created my ground state wave function \psi for 1s:

\psi = (1/\sqrt{}\pi)(z/a)^3/2(e^-zr/a)

The operator is the laplacian, in spherical polar coordinates.

The Attempt at a Solution

So the energy of the two particles is the hamiltonian operating on \psi,
and I should get an eigen function out which would be the energy for one of the two particles.
Using the laplacian operator I got:

E = [(-\hbar ²/2m)(1/\sqrt{}\pi)(z/a)^3/2](z²/a² e^-zr/a - 2z/ar e^-zr/a) + V(r)\psi

For the energy of the one particle. My problem is,
that this isn't an eigen function of the laplacian, and I've managed to hit a brick wall.
I'm completely stumped on what I could do, any help would be greatly appreciated!

 

[TeethWhitener]

If we were to ignore the interelectronic repulsion in helium, what would be it's ground state energy and wave function?

It might help to start by writing out the Hamiltonian. Ignoring the electron-electron repulsion term gives:
$$H = -\frac{\hbar^2}{2m}(\nabla^2_{\mathbf{r}_1}+\nabla^2_{\mathbf{r}_2}) -\frac{Ze^2}{4\pi \varepsilon_0 |\mathbf{r}_1-\mathbf{R}|}-\frac{Ze^2}{4\pi \varepsilon_0 |\mathbf{r}_2-\mathbf{R}|}$$
But hopefully you'll see this is just a Hamiltonian for two isolated hydrogen atoms. Couple that with the fact that $Z_{He}=2Z_{H}$ and you get that the ground state energy for each electron is $E_{He} = 4E_H$. So the total ground state energy (for 2 electrons) is $E_{He} = 8E_H$ and the wavefunction is just $\Psi_{He}(\mathbf{r}_1,\mathbf{r}_2) = \Psi_H(\mathbf{r}_1)\Psi_H(\mathbf{r}_2)$, where $\Psi_H$ is the ground state wavefunction for the hydrogen atom.

EDIT: one quick note, since the electrons are fermions, the total wavefunction must be antisymmetric with respect to exchange of $\mathbf{r}_1$ and $\mathbf{r}_2$. Obviously, our ground state helium wavefunction above is not. However, if we attach spin labels to the electrons, then we can get back antisymmetry as long as the electrons have opposite spins.