the total number of k point in the full BZ

[sky-zhou]

I am new to quantum espresso and yambo, but has some experience with abinit and DP codes. Recently, I find something that is wired to me:

I start from LDA+U, giving the k point grid as 4 4 4 0 0 0 (Gamma centered 4x4x4 grid) in QE that should give 64 k points in the full BZ (12 k points in the Irreducible part BZ), however, yambo finds only 58 k points in the FBZ. When I changed to LDA calculation of Si, giving the same k grid, yambo finds the correct number 64 for the FBZ....

I attached " r_setup" file from the LDA+U calculation to make my question clear.


Can someone tell me the reason? Maybe it is better to tell me how does yambo find the kpoints in FBZ from the k points in IBZ using the symmetry, e.g., tell me the routine where this is done.

Many thanks!

Cheers,

Sky

[Daniele Varsano]

Dear Sky,
-please note the r_setup is not attached.
-yambo at the moment does not support DFT+U calculation as starting point: we are working on that but at the moment it is not supported
-yambo reads the K points in the IBZ and symmetries from QE, from that reconstruct the points in the BZ. Please note that non-symmorphic symmetries are not supported.

You can compare if all the symmetries of PW are correctly reported in the yambo report.

Best,

Daniele

[sky-zhou]

Dear Daniele,

1. I am sorry for the missing file, but now I realize that the problem is in QE instead of yambo... So it does not matter anymore...

2. I understand your concern about DFT+U but as far as I know, it only matters at q=0 limit because yamboo needs the commutator to the Hamiltonian which contains U. All the rest q values should be safe, right? I hope you can confirm that so I will be able to calculate some chi at q!= 0 .

3. The problem now is that the symmetries report by QE is not complete.... I am working on that these days, and find that QE report 12 symmetries but what it actually uses to get the inequivalent k poitns are 23 symmetries....maybe because of +k and -k problem.... I do not know the reason yet but trying to figure out this point.

4. I put force_symmorphic in QE calculation so that the non-symmorphic symmetries are not taken into account in QE.

Many thanks!

Sky

Dear Sky,
-please note the r_setup is not attached.
-yambo at the moment does not support DFT+U calculation as starting point: we are working on that but at the moment it is not supported
-yambo reads the K points in the IBZ and symmetries from QE, from that reconstruct the points in the BZ. Please note that non-symmorphic symmetries are not supported.

You can compare if all the symmetries of PW are correctly reported in the yambo report.

Best,

Daniele

[Daniele Varsano]

Dear Sky,

Code: Select all

2. I understand your concern about DFT+U but as far as I know, it only matters at q=0 limit because yamboo needs the commutator to the Hamiltonian which contains U. All the rest q values should be safe, right? I hope you can confirm that so I will be able to calculate some chi at q!= 0 .

It depends on what you need to calculate. If you are interested in linear response it should be sage, but if you need to do a GW calculation then you need to subctract from the self energy the U contribution to each KS state and this part needs coding we are planning to do in the next days/months.

About symmetry problem you can check in the r_setup if yambo recognize that there is the time reversal symmetry.

Best,

Daniele

[sky-zhou]

Dear Daniele,

Thank you very much for your explanation. At this moment I think I will only do a linear response RPA calculation. If I want to do GW, I will start from LDA but not LDA+U...

I have checked carefully the time reversal symmetry in QE code, it does not make any difference on this problem.... Actually, I will use shifted grid or double grid such that symmetry will not be used, so at this moment, it is fine to run calculations. But I will also try to figure out what is happening when a non-shifted grid is used.

Cheers,

Sky

Dear Sky,

Code: Select all

2. I understand your concern about DFT+U but as far as I know, it only matters at q=0 limit because yamboo needs the commutator to the Hamiltonian which contains U. All the rest q values should be safe, right? I hope you can confirm that so I will be able to calculate some chi at q!= 0 .

It depends on what you need to calculate. If you are interested in linear response it should be sage, but if you need to do a GW calculation then you need to subctract from the self energy the U contribution to each KS state and this part needs coding we are planning to do in the next days/months.

About symmetry problem you can check in the r_setup if yambo recognize that there is the time reversal symmetry.

Best,

Daniele