We can write this in terms of the ( ) Hydrogen Hamiltonian for each electron plus a perturbation,

where . Note that is about the same size as the the rest of the Hamiltonian so first order perturbation theory is unlikely to be accurate.

For our zeroth order energy eigenstates, we will use
**product states of Hydrogen wavefunctions**.

These are not eigenfunctions of H because of , the electron coulomb repulsion term. Ignoring , the problem separates into the energy for electron 1 and the energy for electron 2 and we can solve the problem exactly.

We can write these

Note that is above ionization energy, so the state can decay rapidly by ejecting an electron.

Now let's look at the (anti) **symmetry of the states** of two identical electrons.
For the ground state, the spatial state is symmetric,
so the spin state must be antisymmetric
.

For excited states, we can make either symmetric or antisymmetric space states.

The first state is or spin

We label the states according to the spin quantum numbers, singlet or triplet. We will treat V as a perturbation. It is very large, so first order perturbation theory will be quite inaccurate.

Jim Branson 2013-04-22