Dealing with Coulomb divergencies

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Quxiao
Posts: 96
Joined: Fri Mar 26, 2021 11:27 am

Re: Dealing with Coulomb divergencies

Post by Quxiao » Tue Dec 05, 2023 2:20 am

Dear Daniele:
Thanks again for your help!
In fact, what I want is W (q=0), and I can get v from yambo.py,the value of {G=0, G '=0} (q=0) should only be 4 * pi/q ^ 2.
Then, I can obtain vx to construct \ epsilon ^ {-1}=1+vx,
and use W (q=0)=\ epsilon ^ {-1} * v_ {G=0, G '=0} (q=0) yields W (q=0), is that correct?
Looking forward to your reply!
Best wishes!
Quxiao
Quxiao in BIT,calculate the exciton

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Daniele Varsano
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Re: Dealing with Coulomb divergencies

Post by Daniele Varsano » Thu Dec 07, 2023 9:01 am

Dear Quxiao,
yes, that's correct, just pay attention to the coulomb potential that it is 4*pi/q^2 in 3D but it has different shape if you have lower diumension and you used a coulomb cutoff technique.

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/

Quxiao
Posts: 96
Joined: Fri Mar 26, 2021 11:27 am

Re: Dealing with Coulomb divergencies

Post by Quxiao » Fri Dec 08, 2023 2:16 am

Dear Daniele:
I want to confirm what the two-dimensional Coulomb equation would look like in yambo without Coulomb truncation,the W_00(q) is just in the form of \frac{2\pi}{|q|(1+2\pi\alpha_{2D}|q|)S},S is the area of the unit?(W_{00}(q)=\varspsilon^{-1}*v(q),v_{00}(q)=\frac{2\pi}{q})

If Coulomb truncation is used, is it only in the form of v_00(q)=\frac{2\pi*L_z}{q})when q tends towards 0,L_z is the cutoff distance? If I use Monte Carlo to integrate and solve the W_{00}(q=0) formed by frac{2\pi}{|q|(1+2\pi\alpha_{2D}|q|)S} near q=0, is the given scatter range or scatter boundary similar to Coulomb truncation technique?

This equation v_00(q)=\frac{2\pi*L_z}{q}) is what I saw below equation (5) in this article :PHYSICAL REVIEW B 94, 155406 (2016)https://journals.aps.org/prb/abstract/1 ... .94.155406

But usually, I also often see the form of equation v_{00}(q)=\frac{2\pi}{q}) , and when q tends to 0, is there only one L_z difference between them? This confuses me.
Thanks for your help!
Best wishes!
Quxiao
Quxiao in BIT,calculate the exciton

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Daniele Varsano
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Joined: Tue Mar 17, 2009 2:23 pm
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Re: Dealing with Coulomb divergencies

Post by Daniele Varsano » Mon Dec 11, 2023 11:21 am

Dear Quxiao,

W is calculated as W=v+vXv where v is the 3D Coulomb potential even if your system is 2D.
If you use the truncated Coulomb potential, than the Vcut is used. Vcut has the shape of a box or a slab (suggested).
The paper you indicate describe an acceleration for the integration of W in the BZ.
In Yambo this technique has been recently improved and it is described here:
https://www.nature.com/articles/s41524-023-00989-7

How to use it is explained here:
https://www.yambo-code.eu/wiki/index.ph ... 2D_systems

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/

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