[QE-users] nscf with Hubbard

[Paolo Giannozzi]

I do not think Hubbard corrections are the problem here. A scf 
calculation contains several self-consistency iterations, each of which 
is less expensive than a similar non-scf calculations (either because 
strict convergence is not needed at the beginning of the scf procedure, 
or because the previous step is a good starting point for iterative 
diagonalization). So in the end a non-scf calculation may turn out take 
ano more time than a scf one, unless one computes many more bands in the 
former than in the latter

Paolo

On 08/11/2022 12:22, Aleksandra Oranskaia wrote:
> Dear users and developers,
> 
> Are there any specific tricks for nscf with Hubbard corrections?
> 
> Strangely it takes a comparable amount of resources as scf. E.g. scf for 
> 4 4 4 grid (for a huge cell) takes just a bit less time as scf for 2 2 2 
> -> nscf 4 4 4. I am using the simplest correction (lda_plus_u_kind = 1) 
> in qe-6.4 (because I want to finish project with one version, not the 
> most new).
> 
> Thank you!
> -- 
> Best wishes,
> Al., phd candidate in chemical sciences
> 'I.. a universe of atoms, an atom in the universe' (Richard P. Feynman)
> https://cpms.kaust.edu.sa/ 
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-- 
Paolo Giannozzi, Dip. Scienze Matematiche Informatiche e Fisiche,
Univ. Udine, via delle Scienze 206, 33100 Udine Italy, +39-0432-558216