# Question about excited Helium states

• I
Most of the books I've seen they say that the first excited state of Helium (with two electrons, one in orbital 1s and other in 2s) can have the two electrons with parallel spin (orthohelium) or anti-parallel spin (parahelium).
If ##\operatorname{X_{↑}}{\left (n \right )}## represent the state of particle n with spin up, there are the following possible states:

Anti-parallel spin:

##
A \left(\operatorname{X_{↑}}{\left (1 \right )} \operatorname{X_{↓}}{\left (2 \right )} - \operatorname{X_{↑}}{\left (2 \right )} \operatorname{X_{↓}}{\left (1 \right )}\right) \left(\operatorname{Ψ_{1s}}{\left (r_{1} \right )} \operatorname{Ψ_{2s}}{\left (r_{2} \right )} + \operatorname{Ψ_{1s}}{\left (r_{2} \right )} \operatorname{Ψ_{2s}}{\left (r_{1} \right )}\right)
##

Parallel spin:

##
A \left(\operatorname{Ψ_{1s}}{\left (r_{1} \right )} \operatorname{Ψ_{2s}}{\left (r_{2} \right )} - \operatorname{Ψ_{1s}}{\left (r_{2} \right )} \operatorname{Ψ_{2s}}{\left (r_{1} \right )}\right) \operatorname{X_{↑}}{\left (1 \right )} \operatorname{X_{↑}}{\left (2 \right )}
##

##
A \left(\operatorname{Ψ_{1s}}{\left (r_{1} \right )} \operatorname{Ψ_{2s}}{\left (r_{2} \right )} - \operatorname{Ψ_{1s}}{\left (r_{2} \right )} \operatorname{Ψ_{2s}}{\left (r_{1} \right )}\right) \operatorname{X_{↓}}{\left (1 \right )} \operatorname{X_{↓}}{\left (2 \right )}
##

##
A \left(\operatorname{X_{↑}}{\left (1 \right )} \operatorname{X_{↓}}{\left (2 \right )} + \operatorname{X_{↑}}{\left (2 \right )} \operatorname{X_{↓}}{\left (1 \right )}\right) \left(\operatorname{Ψ_{1s}}{\left (r_{1} \right )} \operatorname{Ψ_{2s}}{\left (r_{2} \right )} - \operatorname{Ψ_{1s}}{\left (r_{2} \right )} \operatorname{Ψ_{2s}}{\left (r_{1} \right )}\right)
##

From what I understand, any linear combinations of the above functions also represent two electrons in 1s and 2s orbitals, but if for example i add the first and the four function I get:

##
A \left(\operatorname{X_{↑}}{\left (1 \right )} \operatorname{X_{↓}}{\left (2 \right )} \operatorname{Ψ_{1s}}{\left (r_{1} \right )} \operatorname{Ψ_{2s}}{\left (r_{2} \right )} - \operatorname{X_{↑}}{\left (2 \right )} \operatorname{X_{↓}}{\left (1 \right )} \operatorname{Ψ_{1s}}{\left (r_{2} \right )} \operatorname{Ψ_{2s}}{\left (r_{1} \right )}\right)
##

Becasue I add the Anti-parallel spin function and one of the Parallel spin functions, now I get a function that also represent two electrons in 1s and 2s orbitals, but it is neither parallel or antiparallel.(it's a superposition of both possibilities)

Is this correct or I'm missing something?

Last edited:

## Answers and Replies

king vitamin
Science Advisor
Gold Member
Yes, that is correct - any linear combination of the four eigenstates will also be an eigenstate with the same energy.

But the particular choice you've made in your post is a rather special one. You are ignoring electron-electron interactions and fine structure (effects due to electron spin), but if you were to take those into account, you would in principle need to use degenerate perturbation theory, which involves diagonalizing your degenerate subspace with the perturbed Hamiltonian. But the perturbing Hamiltonian conserves total angular momentum (the sum of the electron spins in this case, since these are $/ell=0$ states), so you know that the correct basis is the one which are eigenstates of $(\mathbf{S}_1 + \mathbf{S}_2)^2$ and $S^z_1 + S^z_2$. The is precisely the basis you've written down, so to leading order we expect the actual Helium atom with electron-electron interactions will actually split these four states into a single nondegenerate state (the first one you've written down) and a threefold degenerate state.

• Rafael
Yes, that is correct - any linear combination of the four eigenstates will also be an eigenstate with the same energy.

But the particular choice you've made in your post is a rather special one. You are ignoring electron-electron interactions and fine structure (effects due to electron spin), but if you were to take those into account, you would in principle need to use degenerate perturbation theory, which involves diagonalizing your degenerate subspace with the perturbed Hamiltonian. But the perturbing Hamiltonian conserves total angular momentum (the sum of the electron spins in this case, since these are $/ell=0$ states), so you know that the correct basis is the one which are eigenstates of $(\mathbf{S}_1 + \mathbf{S}_2)^2$ and $S^z_1 + S^z_2$. The is precisely the basis you've written down, so to leading order we expect the actual Helium atom with electron-electron interactions will actually split these four states into a single nondegenerate state (the first one you've written down) and a threefold degenerate state.

I'm a bit confused because I have't studied perturbation theory yet. So if we take account of electron repulsion and fine structure, what happens with the linear combination I have writted? Is it impossible for helium? Does it have more energy than the others?
Thanks for the answer.

king vitamin
Science Advisor
Gold Member
I'm a bit confused because I have't studied perturbation theory yet.

I'm sorry! I shouldn't have confused you by going beyond what you know. In some courses it is common to study the Helium atom along with perturbation theory, but I can see how the problem you're dealing with could be suited to a first semester quantum course.

So if we take account of electron repulsion and fine structure, what happens with the linear combination I have writted?

If you take those into account, the linear combination you wrote down won't be an eigenstate anymore, it'll be a superposition of two different eigenstates (its average energy will lie between them). Basically, the first state you wrote down will be an eigenstate with some energy E1, and the other three will all be eigenstates with degenerate energy E2, and you've just taken some superposition of different energy eigenstates.

But anyways, if you have not seen perturbation theory yet perhaps you shouldn't worry about this right now. You can come back to this thread later on and see if what's I've said makes sense once you've worked a little with degenerate perturbation theory.

• Rafael