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Hello,

I'm wondering if it is possible or of physical meaning to employ the Coulomb cutoff technique of bulk systems for BSE calculations. The situation is that my system is a layered bulk semiconductor. The exciton in real space (solved by BSE and plot by ypp) is confined only in one layer, i.e. the exciton is 2D. So the bright exciton in this case should have a linear dispersion due to the 2D 1/q Coulomb potential. However, if without Coulomb cutoff, one would always end up with parabolic dispersion. With Coulomb cutoff as I tried, there are some strange behaviors of the dispersion, say not smooth any more. So any comments or suggestions for my situation?

Best,
Xiaoming

Dear Xiaoming,

in my opinion you should not cut the potential, even if you have a 2D confined exciton the e-h interaction should be 1/q^2 and I'm not sure that the dispersion should be linear.
But I can be wrong and maybe others can have different opinion.

With Coulomb cutoff as I tried, there are some strange behaviors of the dispersion, say not smooth any more.

This is probably due because in order to make the coulomb cutoff potential meaningful you need enough vacuum between your layers.

Best,

Daniele

Dear Daniele,

Thanks for your comments.

Daniele Varsano wrote: ↑Tue Aug 03, 2021 7:55 am even if you have a 2D confined exciton the e-h interaction should be 1/q^2

Why is that? The exciton is 2D means the e-h pair is also 2D, right?
From a different point of view, if one take the exciton wave function as basis and make a k dot p expansion for finite Q, the Coulomb interaction between exciton states in the exchange part is obviously 1/q. This would lead to a linear dispersion. Not sure if this argument is correct.

Best,
Xiaoming

Dear Xiamonig,

I would say because the 2D exciton you get is anyway the solution of a BSE in 3D, and you get it considering a potential in 3D.
So, I see it as an electron and an hole confined in a layer interacting with a 3D coulomb potential. As you say the linear dispersion comes from the exchange term only if an 1/q interaction is considered.

Note: this is just my personal interpretation.

Best,
Daniele