BSE kmesh

[Longjun Xiang]

Dear developers:

Why we need a denser K mesh for BSE calculation? What's the physical reason behind this? Can I understand this from solving the BSE equation?

Thanks in advance.

[Daniele Varsano]

Dear Longjun,

in general, this is something which is system dependent, in particular, from the band structure and the nature of the excitons.
If you have rapidly varying bands with many transitions contributing to the excitons, you need a dense k point grid to resolve them.
Best,
Daniele

[Longjun Xiang]

If I use a less dense K mesh, I will just miss some transitions in the absorption spectrum. How does that missing affect the final results, such as absorption peak and position? I mean I can always obtain the G0W0 bandstructure with a moderate K mesh and find where the minimal direct gap is (taking that as K1). Assume the band dispersion around K1 is rapid and but K1 has contained the mesh of BSE calculation, whereas the K points around K1 is not fully considered, what will happen in the final absorption spectrum?

[Daniele Varsano]

Dear Longjun,
it really depends on the system, what can happen is that the calculation of the excitation is not fully converged (both position and intensity).
You can also see this in term of localization of the exciton: if your exciton is very delocalized in real space (let's call Ro its bohr radius), it will be very localized in reciprocal space and you need a k space grid such that 1/delta_k is large enough (~Ro) to represent the exciton localization. If this is not satisfied, the binding energy of the exciton will be not converged. Clearly, the denser the k grid the more accurate is your calculation, the degree of accuracy you have with a less dense grid is something you need to check.

Best,
Daniele

[Longjun Xiang]

Thanks for your sharings.