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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.

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    $\begingroup$ 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). $\endgroup$
    – Jon Custer
    Commented Oct 2, 2018 at 12:51

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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.

Things are basically the same in solids. An electron is a "dressed" electron, not the free naked electron of particle physics. It has an effective mass, which can be smaller or hundreds of times larger ("heavy fermion"). Same for holes: not a vacuum, but a quasiparticle, describing the behaviour of the rest of the electrons.

I always compare with a bubble in water. It goes up in a gravitational field, so one could say that it has negative mass. One does not really see the water flowing down around it. The bubble is a much more economical and much more intuitive description.

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  • $\begingroup$ Sorry for commenting so late, but does this understanding arise due to incomplete/improper modeling of quantum systems? Or do we fully understand what is going on? $\endgroup$
    – Sagar Patil
    Commented Dec 22, 2020 at 1:38
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    $\begingroup$ @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. $\endgroup$
    – user137289
    Commented Dec 22, 2020 at 9:14

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