Question about sign convention in electron-hole crystal momentum

Question

I was trying to understand a toy model Raman scattering diagram from a paper on pressure tuned moire phonons, when I realized the standard electron hole pair creation diagram confuses me. By conservation of crystal momentum, assuming the absorbed photon has negligible $k$,

$k_e = -k_h$

however, drawings are always drawn with a direct vertical transition. Look at any direct bandgap semiconductor, the e/h pair end up occupying the same $k$, but with opposite sign on energy. The sign on $k$ for both the electron and hole are the same. Ignore the other things going on in this figure, just look at how the electron and hole states have same $k$.

To remedy this, you can think of the conservation of crystal momentum from start to end. Both the e and h occupy such that initially the crystal momentum was 2k and then after absorption they still both have crystal momentum 2k. This conflicts with the initial assertation that they have negative signs.

I understand that the negative effective mass of electrons and holes makes it so that under external electric field they will respond oppositely, but conservation of crystal momentum has nothing to do with that.

Have I just confused myself or am I missing something?

Answer 1

It is good to keep in mind that, ultimately, a hole is just a vacant electron state in a valence band. It merely provides a convenient (but optional) language for describing conduction due to valence electrons (or lack thereof) in an intuitive way.

The absorption of a photon involves two electron states, with roughly the same value of $\bf k$. Initially, the state in the valence band is occupied and that in the conduction band is vacant, so the sum of the crystal momenta of the electrons in these states is $\hbar\bf k$. Following the absorption, the valence band state is vacant and conduction band state is occupied: the total electron crystal momentum is still $\hbar\bf k$. Optical recombination can be described similarly.