Re: Not enough states to converge the Fermi Level
Posted: 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.
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.