What effect causes energy difference between $2 \:^1 S$ and $2\: ^1 P$ electron state in a helium atom

Question

I'm wondering where the energy difference between the two electron states $2 \:^1 S$ and $2\: ^1 P$ comes from in a helium atom. I thought it is the fine structure, but the fact that $S = 0$ is confusing me because there shouldn't be any fine structure influence for the two states.

Answer 1

I imagine that you're referring to the second and fourth excited states, $^1S$ with configuration $1s2s$ and $^1P$ with configuration $1s2p$.

Frankly, I'm surprised that the difference between those two states isn't larger: the $^1S$ state sits $20.62 \:\rm eV$ above the ground state, and the $^1P$ state is $21.22 \:\rm eV$ above the ground state, and the relative difference at $0.6\:\rm eV$ is frankly pretty minor.

This difference can mostly be attributed to differences in electrostatic shielding produced by the $1s$ electron. It's a relatively standard fact that $s$ electrons wander further inwards towards the core than $p$ electrons on the same shell, because of the absence of the centrifugal barrier. This means that the $2s$ electron spends more 'time' at low-$r$ regions where it experiences a higher effective nuclear charge, and that brings its energy down.

In addition to that, there probably are indeed fine-structure effects that meddle with both of those energies at a level that's not too far below the $0.6\:\rm eV$ energy difference, including both a spin-orbit coupling on the $^1P$ state as well as kinetic-energy and Darwin terms on both states. But you don't need to bring those out to explain the difference.

Simply put, multi-electron atoms are messy places. The nice symmetries of the hydrogen atom, where e.g. equal-$\ell$ subshells were magically degenerate (so long as you're happy to ignore all the ways in which they're not degenerate) fly straight out of the window starting with helium upwards.