## Converging the Coulomb potential Random Integration

Run-time issues concerning Yambo that are not covered in the above forums.

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### Converging the Coulomb potential Random Integration

Dear forum,

I am trying to calculate the quasiparticle corrections for a 1d system (periodic along x) using a coulomb cutoff "box yz" and the random integration method for the Coulomb potential.
Using the standard parameters RandQpts = 1000000 and RandGvec = 1, I obtain the following report

Code: Select all
 Coloumb potential Random Integration (RIM) =============================================== [RD./SAVE//ndb.RIM]-----------------------------------------  Brillouin Zone Q/K grids (IBZ/BZ):  33   64   33   64  Coulombian RL components        : 1  Coulombian diagonal components  :yes  RIM random points               : 1000000  RIM  RL volume             [a.u.]:  0.02669  Real RL volume             [a.u.]:  0.02660  Eps^-1 reference component       :0  Eps^-1 components                : 0.00      0.00      0.00  RIM anysotropy factor            : 0.000000 - S/N 002743 --------------------------- v.03.04.01 r.3187 - Summary of Coulomb integrals for non-metallic bands |Q|[au] RIM/Bare: Q :0.1000E-40.8187 * Q :  0.02132  0.19701 Q :  0.04264  0.42336 * Q :  0.06395  0.59031 Q :  0.08527  0.70378 * Q : 0.106590 0.780348 Q : 0.127908 0.832871 * Q : 0.149226 0.869775 Q : 0.170544 0.896378 * Q : 0.191862 0.916030 Q : 0.213180 0.930879 * Q : 0.234497 0.942328 Q : 0.255815 0.951317 * Q : 0.277133 0.958490 Q : 0.298451 0.964296 * Q : 0.319769 0.969056 Q : 0.341087 0.973005 * Q : 0.362405 0.976313 Q : 0.383723 0.979111 * Q : 0.405041 0.981498 Q : 0.426359 0.983550 * Q : 0.447677 0.985326 Q : 0.468995 0.986873 * Q : 0.490313 0.988228 Q : 0.511631 0.989423 * Q : 0.532949 0.990480 Q : 0.554267 0.991421 * Q : 0.575585 0.992261 Q : 0.596903 0.993015 * Q : 0.618221 0.993693 Q : 0.639539 0.994306 * Q : 0.660857 0.994862 Q : 0.682175 0.995382

i.e. the ratio RIM/bare differs quite significantly from 1 for low q. In the forum Daniele suggested that for convergence it only matters that the ratio approaches 1 at the Brillouin zone boundary(?). Can this value of 0.995382 then be considered converged?

As far as the understanding of the method is concerned, I am still struggling a lot and I would be very grateful for a few hints.
I have read through the formulae in http://www.yambo-code.org/theory/docs/doc_RIM.php . At the end there is a comparison between "RIM" and "without RIM" for a chain of H2 molecules.
What does "without RIM" mean? Does it mean that the integral over dq has been replaced by a sum over q_i for *both* F(q+G) and the potential (as in the formula after "The integral over the BZ is usually numerically evaluated as ...")?
Wouldn't this mean that the potential is evaluated at q+G=0, leading immediately to a divergence?

And in connection to this: what is the difference between the "Bare" and "RIM" Coulomb integrals?

Best,
Leopold
Leopold Talirz
Swiss Federal Laboratories for Materials Science and Technology, Dübendorf, Switzerland
http://www.surfaces.ch
leoteo

Posts: 30
Joined: Tue Apr 09, 2013 5:40 pm

### Re: Converging the Coulomb potential Random Integration

Dear Leopolod,
i.e. the ratio RIM/bare differs quite significantly from 1 for low q. In the forum Daniele suggested that for convergence it only matters that the ratio approaches 1 at the Brillouin zone boundary(?). Can this value of 0.995382 then be considered converged?

Indeed, as the code will use the available integral evaluated with RIM (1st Brillouin zone in case of G=1), and then sum_qi for the rest. If at the border they are similar it means that the sum_qi for bigger |q+G| is reasonable. Anyway you can test it performing RIM for G>1, for instances converging the exchange part of the self energy (which is extremely sensible to this issue) with respect the G included int he RIM.

Wouldn't this mean that the potential is evaluated at q+G=0, leading immediately to a divergence?

Ok, this may be misleading in the documentation. q+G=0 also in case of 3D sampling is regularized. For the case 3D using an integration over a small sphere around q=0, or other smarter surface. The example is there for showing that the problem for non 3D sampling arise not only for the q+G=0 case, but also for the rest of the q when they are small, ie for larger k point sampling. The approximation \Sum_qi breakdown and the smaller q_i (or bigger 1/q_2) the worst is the approximation and tend to explode when the first non zero q became too small.
For larger q+G the function 1/|q+g|^2 become smoother and the discretization is not that bad.

And in connection to this: what is the difference between the "Bare" and "RIM" Coulomb integrals?

I do not know exactly in which context you read "Bare" and 'RIM" Coulomb integrals, but I suppose that it referes:
Bare \int d^3q 1/|q+G|^2 ~ \sum_q_i 1/|q+G|^2 Vol_qi
RIM \int d^3q 1/|q+G|^2 evaluated using Monte Carlo sampling

Hope it is clear enough,

Best,
Daniele
Dr. Daniele Varsano
S3-CNR Institute of Nanoscience and MaX Center, Italy
MaX - Materials design at the Exascale
http://www.nano.cnr.it
http://www.max-centre.eu/ Daniele Varsano

Posts: 1974
Joined: Tue Mar 17, 2009 2:23 pm

### Re: Converging the Coulomb potential Random Integration

Dear Daniele,

thanks a lot for your kind help! Things already look much clearer.

Anyway you can test it performing RIM for G>1, for instances converging the exchange part of the self energy (which is extremely sensible to this issue) with respect the G included int he RIM.

Thank you for the suggestion, I will check that.

So, if I understand correctly, the ratios given in yambo's report can only be used to judge the convergence with respect to G (namely, if the ratios are very nearly 1 at the boundary of the Brillouin zone).

The convergence of the Monte-Carlo integral itself (i.e. with respect to RandQpts) can not be judged from one calculation, correct? In particular, it is not "a bad sign" if the ratio RIM/Bare is different from 1 for small q.
One should increase RandQpts and make sure that the ratios do not change. I guess the idea is that 1000000 are much more than enough points for almost all cases(?).

Ok, this may be misleading in the documentation. q+G=0 also in case of 3D sampling is regularized. For the case 3D using an integration over a small sphere around q=0, or other smarter surface.

Ok, this makes sense. But then I have a question: In my report ratio of RIM/Bare for q+G=0 is given as 0.8187. Does this indicate that the RIM is off by 18% from the correct value? Or is the analytic integration for the "Bare" potential wrong?

I do not know exactly in which context you read "Bare" and 'RIM" Coulomb integrals, but I suppose that it referes:
Bare \int d^3q 1/|q+G|^2 ~ \sum_q_i 1/|q+G|^2 Vol_qi
RIM \int d^3q 1/|q+G|^2 evaluated using Monte Carlo sampling

I read it in the report: "Summary of Coulomb integrals for non-metallic bands |Q|[au] RIM/Bare"
Concerning the ratios reported there, do they belong to the formulae you wrote?

If this is the case, I agree that "Bare" is probably not the best word here, since it is already used in other contexts (e.g. bare vs screened).
Also the description "without RIM" was not so clear to me in the beginning. When I read it, I expect that instead of performing a Monte Carlo integration, the problem would be solved somehow exactly (without randomness).
But in reality, the alternative to "RIM" is "no integration", right?

Thanks,
Leo

EDIT: This terminology is also used in the yambo paper. This is from figure 5
Screen Shot 2014-09-18 at 18.40.43 .png

So, if I understand correctly what is being called "bare integrals" here is really just v(q) = 1/q^2 (no integration)?
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Leopold Talirz
Swiss Federal Laboratories for Materials Science and Technology, Dübendorf, Switzerland
http://www.surfaces.ch
leoteo

Posts: 30
Joined: Tue Apr 09, 2013 5:40 pm

### Re: Converging the Coulomb potential Random Integration

Dear Leopolod,
So, if I understand correctly, the ratios given in yambo's report can only be used to judge the convergence with respect to G (namely, if the ratios are very nearly 1 at the boundary of the Brillouin zone).

Right

The convergence of the Monte-Carlo integral itself (i.e. with respect to RandQpts) can not be judged from one calculation, correct? In particular, it is not "a bad sign" if the ratio RIM/Bare is different from 1 for small q.

Right, of RIM/Bare is different form 1 for small q , just means that it has been a good idea to use the RIM.

One should increase RandQpts and make sure that the ratios do not change. I guess the idea is that 1000000 are much more than enough points for almost all cases(?).

Right, 1000000 should be ok, but you can easily increase it as it is a part of the code which is not heavy at all.

But then I have a question: In my report ratio of RIM/Bare for q+G=0 is given as 0.8187. Does this indicate that the RIM is off by 18% from the correct value? Or is the analytic integration for the "Bare" potential wrong?

It means that analytic integration of the "Bare" potential is wrong as it is not done on the correct volume around gamma, but on an approximate volume (a sphere containing it). The Bz is divided in pieces accroding to the q-sampling BZ=U_i Bz_i , where Bz_i a region of the Bz around the q_i point. The accuracy of this approximation strongly depend on the shape of te Bz or better the shape of the region around gamma Bz_i=o.

I read it in the report: "Summary of Coulomb integrals for non-metallic bands |Q|[au] RIM/Bare"
Concerning the ratios reported there, do they belong to the formulae you wrote?

Yes, RIM integrate the 1/|q^2| on the volumes Bz_i around each point q_i(always 3D sampling), Bare the integral is calculated as \sum_i 1/|q_i|^2 Vol(Bz_i). This is called "Rectangle method" integration if I'm not wrong. For a 3d sampling it reduces to a convergence in k-point, as the denser the grid the more accurate should be the integrals. If the q-grid is not 3d this is not true anymore and the "Rectangle method" cannot be applied.

f this is the case, I agree that "Bare" is probably not the best word here, since it is already used in other contexts (e.g. bare vs screened).
...
EDIT: This terminology is also used in the yambo paper. This is from figure 5

Sorry, short memory, I did that figure and I used that terminology But in reality, the alternative to "RIM" is "no integration", right?
...
So, if I understand correctly what is being called "bare integrals" here is really just v(q) = 1/q^2 (no integration)?

Well, not exactly, the alternative to RIM is Rectangle method integration, which works in 3D while cannot work in less than 3D.
Note that in the Figure I did not plot the q=0 Bare integrals as it is divergent or regolarized by analytical integration that would have given a non-monotonic behaviour.

Best,
Daniele
Dr. Daniele Varsano
S3-CNR Institute of Nanoscience and MaX Center, Italy
MaX - Materials design at the Exascale
http://www.nano.cnr.it
http://www.max-centre.eu/ Daniele Varsano

Posts: 1974
Joined: Tue Mar 17, 2009 2:23 pm

### Re: Converging the Coulomb potential Random Integration

Dear Daniele,

thank you very much for the detailed answer. Almost everything has been clarified, I am just wondering about the very last point:

So, if I understand correctly what is being called "bare integrals" here is really just v(q) = 1/q^2 (no integration)?

Well, not exactly, the alternative to RIM is Rectangle method integration, which works in 3D while cannot work in less than 3D.
Note that in the Figure I did not plot the q=0 Bare integrals as it is divergent or regolarized by analytical integration that would have given a non-monotonic behaviour.

Just to make sure: Is the "Bare integral" shown in the plot I_q(G=0) = v(q) or is it not?
If yes, then I would say I_q(G=0) isn't an integral (v is just evaluated at one point). If not, what is it?

Best,
Leo
Leopold Talirz
Swiss Federal Laboratories for Materials Science and Technology, Dübendorf, Switzerland
http://www.surfaces.ch
leoteo

Posts: 30
Joined: Tue Apr 09, 2013 5:40 pm

### Re: Converging the Coulomb potential Random Integration

Dear Leopold,
Just to make sure: Is the "Bare integral" shown in the plot I_q(G=0) = v(q) or is it not?
If yes, then I would say I_q(G=0) isn't an integral (v is just evaluated at one point). If not, what is it?

I_q(G=0)=v(q)*Vol_q

so evaluated at one point, assumed constant in that volume and hence multiplied for the volume.
Best,
Daniele
Dr. Daniele Varsano
S3-CNR Institute of Nanoscience and MaX Center, Italy
MaX - Materials design at the Exascale
http://www.nano.cnr.it
http://www.max-centre.eu/ Daniele Varsano

Posts: 1974
Joined: Tue Mar 17, 2009 2:23 pm

### Re: Converging the Coulomb potential Random Integration

Thanks Leopold Talirz
Swiss Federal Laboratories for Materials Science and Technology, Dübendorf, Switzerland
http://www.surfaces.ch
leoteo

Posts: 30
Joined: Tue Apr 09, 2013 5:40 pm