# Ground state of ortho-helium?

1. Apr 7, 2010

### Oddbio

I want to construct the ground state wave function of orthohelium.
Here is my reasoning, please let me know if I am correct or not.

[EDIT]
I forgot to add that for this post I am ignoring the electron-electron interactions.
[/EDIT]

Orthohelium has the triplet spin state which is symmetric. Because the total wavefunction must be antisymmetric that means that orthohelium must have an antisymmetric spatial wave function to go with the symmetric spin part.

However, I read that the ground state of helium in general MUST have a symmetric spatial wave function. Therefore that cannot also be the ground state of orthohelium.

So we must go to the first excited state, which can be written as:
$$u(r1,r2) = \frac{1}{\sqrt{2}}[\psi_{100}(r_{1}) \psi_{2lm}(r_{2}) \pm \psi_{100}(r_{2}) \psi_{2lm}(r_{1})] X$$

where X is the spin wave function. Either singlet or triplet.

the 100 means (n=1 l=0 m=0) which must be true for the electron in the ground state:
the 2lm subscript is (n=2, l=0 or 1, m=-1 or 0 or 1)

Now, because orthohelium must have the X as the triplet spin which is symmetric, then we must use the minus sign in the equation to get the antisymmetric spatial state.

Therefore I get that the ground state wave function of orthohelium is:
$$u(r1,r2) = \frac{1}{\sqrt{2}}[\psi_{100}(r_{1}) \psi_{2lm}(r_{2}) - \psi_{100}(r_{2}) \psi_{2lm}(r_{1})] X_{triplet}$$

Then also I am curious as to the degeneracy of the ground state of orthohelium.
IF I am correct, then would I be correct in saying that this ground state of orthohelium it quadruply degenerate?
Because we can have 4 different wave functions for the (l,m) values: (0,0) (1,-1) (1,0) (1,1)
and because the energy only depends on the n quantum number then all four have the same energy right?

Any advice will be GREATLY appreciated.
Thanks.