Not enough states to converge the Fermi Level

Concerns any physical issues arising during the setup step (-i option). This includes problems with symmetries, k/q-point sets, and so on. For technical problems (running in parallel, etc), refer to the Technical forum.

Moderators: Davide Sangalli, andrea.ferretti, myrta gruning, Daniele Varsano

User avatar
Davide Sangalli
Posts: 614
Joined: Tue May 29, 2012 4:49 pm
Location: Via Salaria Km 29.3, CP 10, 00016, Monterotondo Stazione, Italy
Contact:

Re: Not enough states to converge the Fermi Level

Post by Davide Sangalli » Fri Feb 02, 2024 10:31 am

The code diagonalizes the full Hamiltonian, also including the kinetic term. The 1/Nq factor is in the code for the interaction terms.

This is my interpretation about what is happening. Suppose you have an exciton which is very localized in k space around k=k0.
In a coarse k-grid, only the diagonal interaction term k=k'=k0 matters, while all other k/=k0 or k'/=k0 terms are zero.
However, by refining the grid sampling, the diagonal term will go to zero, as observed. Then the off-diagonal terms must become non zero for some k.
This would imply that, no matter how much an exciton can be localized in k-space, the wave-function cannot be just a delta(k-k0). It must be some peaked and smooth function, e.g. a very peaked gaussian. Its eigenvalue will be stable against different k-grids, despite the single matrix elements going to zero.

Best,
D.
Davide Sangalli, PhD
CNR-ISM, Division of Ultrafast Processes in Materials (FLASHit) and MaX Centre
https://sites.google.com/view/davidesangalli
http://www.max-centre.eu/

Post Reply