# Can we show that the ground state of the He atom is a spin singlet rather than triplet?

The ground state of Helium atom is a state in which the space part of the wavefunction is symmetric and the spin part is antisymmetric under the exchange of the electrons. Therefore, the ground state is a spin singlet state. Theoretically, can we show that the singlet state has a lower energy than the triplet state? I think to derive it from the perturbation theory calculation, we need to show that the exchange integral $$J= \int d^3 x_1 \int d^3x_2 \psi_{100}(\vec{x}_1)\psi_{nlm}(\vec{x}_2)\frac{e^2}{r_{12}}\psi^*_{100}(\vec{x}_2)\psi^*_{nlm}(\vec{x}_1)$$ is positive. This is because the energy associated with the spin singlet and triplet states are respectively $$I-J$$ and $$I+J$$ where $$I$$ is the direct integral which is always positive.

• – Ruslan Nov 4 '18 at 16:49
• @Ruslan I think (although I'm not sure) that the no-node theorem for ground states only applies for single-particle problems. – tparker Nov 4 '18 at 17:28
• @tparker as I've said in the comment to your answer, it works whenever Pauli principle doesn't forbid it from working. In the case of two particles a spatially-symmetric state is not forbidden, so it is the ground state. – Ruslan Nov 4 '18 at 17:29
• @Ruslan Do you have a reference for that? It frankly doesn't seem right to me. – tparker Nov 4 '18 at 17:30
• @tparker no, no reference. But it simply follows from the following. If you ignore spin, then the theorem obviously works (it doesn't depend on number of dimensions). As you include the spin (without spin-orbit coupling etc.), you just have to classify spatial states according to their symmetry and associate with corresponding spinor part. But as you didn't introduce any coupling, energies remain the same, so ground state still obeys the theorem. – Ruslan Nov 4 '18 at 17:34

First of all, note that the simple description of helium atom, where spin-orbital coupling is neglected, allows to solve for spinor and spatial part of the wavefunction separately. Moreover, for each symmetric and asymmetric spatial eigenfunction there exist corresponding opposite-type-symmetry spinor values, which means that none of the spatial eigenfunctions are forbidden by Pauli principle.

Now, what remains to be shown is that lowest energy spatial wavefunction is symmetric. This is true by Courant nodal domain theorem$$^\dagger$$: lowest energy eigenfunction must have no nodes. Since an antisymmetric in electron positions function must have a node at the locus of $$e^--e^-$$ collision, lowest energy eigenfunction must indeed be symmetric.

$$^\dagger$$ Courant, Hilbert, "Methods of Mathematical Physics", vol. 1, §VI.6 "Nodes of eigenfunctions"

• Sakurai's book mentions that we can show $J$ to be positive. How do we show that? By brute force calculation? – mithusengupta123 Mar 8 at 16:28
• @mithusengupta123 it's a different question, and it has been discussed at Chemistry.SE. – Ruslan Mar 8 at 16:58
• Actually that was the real question. See that I asked "we need to show that the exchange integral is positive." – mithusengupta123 Mar 8 at 18:00

To make the spin triplet state while satisfying the exclusion principle, you would have to change one of the spatial wavefunctions to $$n=2$$. That requires an amount of energy equal to $$E_2-E_1$$.

• It's worth mentioning that this explanation is only heuristic and assumes the Hartree-Fock approximation. The two electrons in a helium atom are highly entangled because of their Coulomb repulsion and can't be described as having individual single-particle wave functions with individual quantum numbers. – tparker Nov 4 '18 at 17:23
• @tparker: What you say is true, but this approximation is plenty good enough to explain the very basic fact that the OP wants to know the reason for. – Ben Crowell Nov 4 '18 at 18:41
• Agreed (I wasn't the downvote btw) – tparker Nov 4 '18 at 18:45