# Why is the electronic wavefunction of helium a product of the two 1s wavefunctions when ignoring electron-electron repulsion?

This question has also been asked on physics stack exchange:
Why is the helium atom wavefunction a product of the two 1s wavefunctions?

From [1, p. 116]:

In seeking an approximation to the ground state, we might first work out the solution in the absence of the $$1/r_{12}$$ term. In the Schrodinger equation thus simplified, we can separate the variables $$r_{1}$$ and $$r_{2}$$ to reduce the equation to two independent hydrogen like problems. The ground state wavefunction (not normalized) for this hypothetical helium atom would be:

$$\psi(r_1, r_2) = \psi_{1s}(r_1)\psi_{1s}(r_2) = e^{−Z(r_1 + r_2)}$$

Why is it only the product and not some linear combination of the two wavefunctions? I heard somewhere that it has something to do with "tensor product". Can someone provide a detailed explanation about this?

### Reference:

1. Blinder, S. M. Introduction to Quantum Mechanics: in Chemistry, Materials Science, and Biology; Elsevier, 2012,. ISBN 978-0-08-048928-5.
• I did a quick search and found the source for the quoted text and I edited it in. In the future, please do cite the source you are quoting. – andselisk Dec 12 '17 at 8:18
• The exact same question was posed by the same OP, here: physics.stackexchange.com/questions/373985/… Quite different answers too! – Gert Dec 12 '17 at 15:54
• @Gert Exactly! The answers here didn't quite give any information about the tensor product which is why I had to go to physics SE. – Pauling0304 Dec 13 '17 at 2:02
• @qforquantum Cross-posting is usually frowned upon, especially if undisclosed. You could have commented on the answers here for the missing information, or flagged a moderator for migration to physics.SE instead – Tobias Kienzler Dec 13 '17 at 7:10
• @qforquantum You should at least mention the cross-post, otherwise users may waste their time writing an answer almost identical to one on the other site. It also helps others getting all answers. See meta.stackexchange.com/q/4708/146482. But actually I agree with you, there's a nice feature request: Build and strengthen the Stack Exchange community with “crossover questions” between sites – Tobias Kienzler Dec 13 '17 at 11:51

The same mechanics as for the separation of angular and radial parts of the hydrogen electronic wavefunction are in place here: You have a Schrödinger equation that is separable into two (i.e. one for each electron). You solve them and in order to combine them, they must be multiplied.

It so happens in this case that the solution for each equation is known and recognizable: the electronic wavefunction of $\ce{He+}$. The fact that the two wavefunctions are multiplied is not related at all to electronic structure theory per se (no involvement of LCAO etc.), but merely a result of the separability of the Schrödinger equation.

• If an atom is represented by a wavefunction does this mean that they are also waves ? – ado sar May 2 '19 at 14:57
• @adosar: This question goes deeper than you may think. Some would say that atom (i.e. nucleus and electrons) are represented by one wave-function. Some would separete nucleus and electrons into two wave-functions to make the problem easier and justify that by the large mass difference between the particles. Some go further than that in their handling. (For more discussion, consider a chat) – TAR86 May 2 '19 at 15:39

As TAR86 answered, this is the separation of variables. This is only possible due to the neglected $1/r_{12}$ term, which would otherwise make separation impossible. Therefore, this multiplication is actually merely an approximation.

For completenesses sake, just like with the hydrogen orbitals (or any solution to a linear PDE obtained via separation of variables) the full solution to this approximated $He$ is a superposition of all multiplicative orbitals, i.e.

$$\Psi(r_1,r_2) = \sum_{n,l,n',l'}c_{nln'l'}\psi_{nl}(r_1)\psi_{n'l'}(r_2)$$