Coulomb cutoff of hexagonal cell.

Concerns any physical issues arising during the setup step (-i option). This includes problems with symmetries, k/q-point sets, and so on. For technical problems (running in parallel, etc), refer to the Technical forum.

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vvu
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Coulomb cutoff of hexagonal cell.

Post by vvu » Thu Aug 31, 2023 5:00 pm

Dear developers,

I want to study a 0D system (a molecule of significant size, around 15 A with only a gamma point) of the point group D6h. So, I chose ibrav=4 and placed the molecule's center at (0,0,0) to detect all system symmetries instead of putting it in the center of the cell. Please, I have questions about a truncated Coulomb potential.
As I found there (viewtopic.php?t=45), the truncated Coulomb potential depends only on |r-r'| so the box or sphere or whatever is not centered at any point of the cell. If I understand, for the 0D system with a gamma point, I do not need to calculate any Random Integration, which means there is no need to mention the rim_cut option in GW and BSE inputs, do I?

If not, I'm misunderstood, so I must perform the Random Integrations. In my case, my primitive cell is hexagonal. Does it cause any problems If I choose a box or a sphere?

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Daniele Varsano
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Re: Coulomb cutoff of hexagonal cell.

Post by Daniele Varsano » Thu Sep 07, 2023 5:19 pm

Dear vvu,
please conform with the rule of the forum:
1) Sign your post with your name and affiliation, you can do once for all by filling the signature in your user profile
2) Post your question in the dedicated forum: this question is clearly not related to "Announcements and Job offer".

Your understanding is correct:
You can use both sphere or box (Sphere here is suggested). If you use the sphere you do not need the Random Integration, while for the box it is needed as the potential is not analytical, it requires integration over the BZ to be calculated, so the RIM is necessary. If you use the sphere choose a radius about half of the size of the cell while if you want to use the box choose the sides of the box slightly smaller than its size.

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/

vvu
Posts: 3
Joined: Tue Jul 11, 2023 3:25 pm
Location: Paris-Saclay
Contact:

Re: Coulomb cutoff of hexagonal cell.

Post by vvu » Mon Sep 11, 2023 4:50 pm

Daniele Varsano wrote: Thu Sep 07, 2023 5:19 pm Dear vvu,
please conform with the rule of the forum:
1) Sign your post with your name and affiliation, you can do once for all by filling the signature in your user profile
2) Post your question in the dedicated forum: this question is clearly not related to "Announcements and Job offer".

- I'm sorry. I'm not familiar with Yambo Forum. I will ask my colleagues for help then.

Your understanding is correct:
You can use both sphere or box (Sphere here is suggested). If you use the sphere you do not need the Random Integration, while for the box it is needed as the potential is not analytical, it requires integration over the BZ to be calculated, so the RIM is necessary. If you use the sphere choose a radius about half of the size of the cell while if you want to use the box choose the sides of the box slightly smaller than its size.

- If I use the box, and I have only gamma point, do I need to perform RIM?. And In my case, my primitive cell is hexagonal. Does it cause any problems If I choose a box or a sphere?

Best,
Daniele

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