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
Dealing with Coulomb divergencies
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 Posts: 99
 Joined: Fri Mar 26, 2021 11:27 am
Re: Dealing with Coulomb divergencies
Quxiao in BIT,calculate the exciton
 Daniele Varsano
 Posts: 3949
 Joined: Tue Mar 17, 2009 2:23 pm
 Contact:
Re: Dealing with Coulomb divergencies
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
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
S3CNR Institute of Nanoscience and MaX Center, Italy
MaX  Materials design at the Exascale
http://www.nano.cnr.it
http://www.maxcentre.eu/
S3CNR Institute of Nanoscience and MaX Center, Italy
MaX  Materials design at the Exascale
http://www.nano.cnr.it
http://www.maxcentre.eu/

 Posts: 99
 Joined: Fri Mar 26, 2021 11:27 am
Re: Dealing with Coulomb divergencies
Dear Daniele：
I want to confirm what the twodimensional 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
I want to confirm what the twodimensional 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
 Daniele Varsano
 Posts: 3949
 Joined: Tue Mar 17, 2009 2:23 pm
 Contact:
Re: Dealing with Coulomb divergencies
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/s41524023009897
How to use it is explained here:
https://www.yambocode.eu/wiki/index.ph ... 2D_systems
Best,
Daniele
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/s41524023009897
How to use it is explained here:
https://www.yambocode.eu/wiki/index.ph ... 2D_systems
Best,
Daniele
Dr. Daniele Varsano
S3CNR Institute of Nanoscience and MaX Center, Italy
MaX  Materials design at the Exascale
http://www.nano.cnr.it
http://www.maxcentre.eu/
S3CNR Institute of Nanoscience and MaX Center, Italy
MaX  Materials design at the Exascale
http://www.nano.cnr.it
http://www.maxcentre.eu/