New pseudos and coulombian cut

[Flex]

Hello,

I have two questions about the latest Yambo :

- Does it support the latest ONCVPSP pseudos ? These are norm conserving Vanderbilt with several VKB projectors. Those I am using come from the pseudodojo database. I attached one of them in UPF format as an example.

- I study 2D materials and I want to use a vertical coulombian cut (i.e. a XY plane each side of the layer). Is this the way to do it ?

CUTGeo= "box Z" # [CUT] Coulomb Cutoff geometry: box/cylinder/sphere X/Y/Z/XY..
% CUTBox
0.00 | 0.00 | 20.00 | # [CUT] [au] Box sides

Thanks in advance

[Daniele Varsano]

Dear Thierry,

- Does it support the latest ONCVPSP pseudos ?

Yes, they are supported.

- I study 2D materials and I want to use a vertical coulombian cut (i.e. a XY plane each side of the layer). Is this the way to do it ?

Correct, but you need to:
1) Activate also the RIM, something like:
RandQpts= 1000000 # [RIM] Number of random q-points in the BZ
RandGvec= 111 RL # [RIM] Coulomb interaction RS components

as it is needed to build up the cutoff potential.

2) The z box side should be just a bit smaller of the dimension of your cell in that direction, so your definition it is correct if you have a supercell side of 21 Bohr, otherwise increase that accordingly.

Best,

Daniele

[Flex]

Thanks for the answers,

I still have a question. As I want to compute GW corrections on a monolayer material, I need a unit cell with a lot of empty space to converge. The goal of the coulombian cut here is to truncate the coulomb term quite a lot, about half of the unit height. It should help to converge faster.
The method I want to use is detailed in this article : https://journals.aps.org/prb/abstract/1 ... .93.235435
I use this method in Abinit as well.

Can I use the coulombian cut as it is implemented in Yambo for this kind of truncation ?

Is the random integration method significantly slower or faster than the traditional integration ? I saw in the documentation that the RIM is quite better for confined materials, like monolayers. In this case, do I need a cut to converge faster or is using RIM only fast enough ?

Thanks in advance

[Daniele Varsano]

Dear Thierry,
The method implemented in Yambo is very similar to the one you mention and the goal is exactly the same.

The goal of the coulombian cut here is to truncate the coulomb term quite a lot, about half of the unit height.

Indeed, half of the unit cell is needed to avoid cell-cell interaction.

Is the random integration method significantly slower or faster than the traditional integration?

The RIM is not time-consuming, you calculate the coulomb integral at the beginning and they are stored in a database to be used in all other steps of the calculation. It is needed to remove divergences of the coulomb potential that arise from the fact you do not have a 3D sampling.
The RIM alone does not speed up the convergences with respect the vacuum, so I suggest you to use the cutoff.


Best,

Daniele

[martinspenke]

Dear Daniele,

According to some older posts of you, i thought that setting
RandGvec= 1 RL # [RIM] Coulomb interaction RS components
should be enough, however i see now you are setting RandGvec to 111 RL.

Should one check the results for different RandGvec values ?


Bests
Martin

[Daniele Varsano]

Dear Martin,

yes usually considering the RIM for the first Brillouin zone is enough, but it depends on the geometry of the BZ itself.
Beside the divergence, it happens that for non-3D sampling the \int_Vi 1/q^2 d^3q in each small portion you divide the BZ
is different from 1/q^2_i x V_i (V_i volume of the fraction of the Bz, where q_i is contained), in particular for the q_i close to the origin.
Out of the first BZ, if it is large enough then the two expressions are very similar, so no RIM is needed for the integral beyond the first Bz. Note that in the report when calculating the RIM you can look at:

Code: Select all

Summary of Coulomb integrals for non-metallic bands

Here there is indeed the ratio between these two expressions for each q point, which will be very different from 1 for the smallest q, while becomes to be close to 1 for large q at the border to the Bz. If you see that also the q at the border differs from 1 (let's more than 10^-3) then you can add shells of G vectors where the RIM is considered (RandGvec). Of course, you can do a check of you final results by changing the number of the variables (e.g the Sigma_x which is fast and very sensible to the RIM).
Hope it is clear enough,

Best,

Daniele

[martinspenke]

Dear Daniele,

Thanks a lot for your response.
I have in my report file the following :

Code: Select all

 Summary of Coulomb integrals for non-metallic bands |Q|[au] RIM/Bare:

 Q [1]:0.1000E-40.9797     * Q [7]: 0.136714 1.085033
 Q [2]: 0.136714 1.085731 * Q [9]: 0.193343 1.066889
 Q [4]: 0.193343 1.066650 * Q [3]: 0.273428 1.026921
 Q [8]: 0.273428 1.027096 * Q [10]: 0.305702 1.023585
 Q [5]: 0.305702 1.023358 * Q [6]: 0.386686 1.015361

So, my RIM values seem to be okay and non-divergent, and hence RandGvec = 1 RL should be enough.

Is there any physical reason, why the number of empty bands for convergence in GW has to be increased, if you increase the vacuum between the 2 mono-layers ?

Best wishes
Martin

[Daniele Varsano]

Dear Martin,

So, my RIM values seem to be okay and non-divergent, and hence RandGvec = 1 RL should be enough.

RIM is not divergent, by definition, as the Coulomb integral in 3D do converge. What is important here is that at the border of the Brillouin zone the ratio between the Monte Carlo integral and the simple 1/q_i^2dV are nearly one.

Is there any physical reason, why the number of empty bands for convergence in GW has to be increased if you increase the vacuum between the 2 mono-layers ?

Yes, particle in a box, increasing the vacuum, ie the box in the same energy window, you will find more and more unbound states.
In Yambo you can also define the number of states as energy range, so the number is defined consequently, but me personally do not find it very practical

Best,

Daniele

[martinspenke]

Dear Daniele,

many thanks.

What is important here is that at the border of the Brillouin zone the ratio between the Monte Carlo integral and the simple 1/q_i^2dV are nearly one.

I totally understand but I do not see this from the data I presented. What I see is that the numbers produced by RIM for a 2D system do not explode and hence the coulomb integral converges.
Where can I see that at borders Monte Carlo integral and the simple 1/q_i^2dV are nearly one ?

Best wishes
Martin

[Daniele Varsano]

Hi Martin,

Code: Select all

Q [6]: 0.386686 1.015361

This is the larger q vector and the ratio is nearly 1.

The q->0 limit

Code: Select all

 Q [1]:0.1000E-4 0.9797 

do never explode as also without RIM the divergence is cured with an analytic integral.

Other small q, can have problems and explode when small q are considered, i.e. large sampling.
In you case, considering your sampling, you do not have problems: all components ratio are close to 1, the larger q the better,
as it should be.

Best,

Daniele