How do electron holes have spin, and how do they interact with "real" particles / vice versa?

So, I thought I had an uneasy truce with quantum mechanics but I read something today which made spotted combat flare up in my mind again. In solid state semiconductor free electrons certainly have spin, and in order to "annihilate" meeting a hole, it also must certainly also have a spin. All this is perfectly acceptable to me, it is just this idea of a quasiparticle that is in essence a vacuum, an absence, having something as tangible and measurable as spin sits uneasy with me - for some reason. And I know that it is likely me, not the mass of knowledge on QM that is to blame. I suspect I am clinging to a far too "visual" imagination, most things quantum seem to be unsuited for mental visualisation.

My question is basically - how does hole spin interact with the electron spin? Does it work similarly as between any pair of electrons?

I read that in an exciton there are exchange interactions between hole and electron - and they can be both singlet and triplet. I can understand this, if they both have spin - certainly the bound state can be both singlet and triplet. I can also understand exchange interactions between electron and electron - but hole and electron? How? Imagining a sea of electrons with aligned spin, a hole's spin is thus related to the entire sea. The hole is an empty state, and it's state is defined by the state of all electrons in the stea. So this interaction is it an exchange between a single electron and the entire sea? But it is a short range range interaction... This does not compute...

PS: As must be obvious, I am not really a physicist. I am just an intrigued chemist.

• The allowed energy levels (more broadly $E$ vs $k$) are constructed including spin. So, yes, an empty state in the valence band is an empty state with a specific spin included (we just don't really think about it much mostly). Oct 2 '18 at 12:51

It is easier to start thinking about atoms, for example the transition atoms. A closed shell has no spin. But a $$d^9$$ shell has spin 1/2. And it is much easier to deal with the spectroscopy when one talks about a $$d$$-hole coupling with the angular momentum of the outer $$s$$ and $$p$$ electrons.
• @SagarPatil The description is completely equivalent in many cases, it is just easier to understand and easier to describe. I am just an experimentalist, but calculations for copper atoms are done with all the electrons, also the $3d$ shell which is mostly $d^9.$ A computer has no problems with such a model.