[QE-users] lda_plus_u_kind=2

[Timrov Iurii]

Dear Andrew,


> From looking into the code a bit it appears to me that it generates a 3x3x3 supercell with (I am guessing) ±R in each direction of the lattice vectors.


Right


> I see that for the orginal unit cell the indices go 1-nat


Right


> ...but I am unsure on how the indices in the adjacent unit cells are chosen.


See the routine q-e/PW/src/intersite_V.f90, lines 88-113.


When sc_size=1 you have a supercell 3x3x3, when sc_size=2 you have 5x5x5, etc. In most cases 3x3x3 is enough.


> I would like to specify a hubbard V between atom1 at the edge of the unit cell R=[0,0,0] and a nearest neighbor, atom2, in an adjacent unit cell R=[0,0,1].


It can be quite painful to do all this "by hand". I suggest to use the HP code, and you will get all possible pairs of neighbors automatically, and then you choose what you want. We will try to prepare some tutorial about DFT+U+V in the near future.


Greetings,

Iurii


--
Dr. Iurii Timrov
Postdoctoral Researcher
STI - IMX - THEOS and NCCR - MARVEL
Swiss Federal Institute of Technology Lausanne (EPFL)
CH-1015 Lausanne, Switzerland
+41 21 69 34 881
http://people.epfl.ch/265334
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From: users <users-bounces at lists.quantum-espresso.org> on behalf of Andrew Supka <supkaand at gmail.com>
Sent: Sunday, July 19, 2020 1:27:09 AM
To: Quantum ESPRESSO users Forum
Subject: Re: [QE-users] lda_plus_u_kind=2

Thank you for the response. From looking into the code a bit it appears to me that it generates a 3x3x3 supercell with (I am guessing) ±R in each direction of the lattice vectors. I see that for the orginal unit cell the indices go 1-nat but I am unsure on how the indices in the adjacent unit cells are chosen.

Is there a part of that code that you'd suggest I check for information on that?

Andrew Supka
Postdoc Physics
Central Michigan University


On Sat, Jul 18, 2020, 16:54 Matteo Cococcioni <matteo.cococcioni at unipv.it<mailto:matteo.cococcioni at unipv.it>> wrote:
This part of the code is still not so user-friendly (in fact still under improvement).
yes the code recognizes all the neighbors at the same distance and puts the same V between them. You just need to find one of them in the unit cell and make sure that that atom (or one of its periodic replicas) belongs to the proper star of neighbors you are treating with V.
In order to do that it might be a good idea to shift the reference atom to the center of the cell, so you increase the chances to find one of its neighbors in the unit cell.

HTH

Matteo

Il giorno sab 18 lug 2020 alle ore 17:38 Andrew Supka <supkaand at gmail.com<mailto:supkaand at gmail.com>> ha scritto:
I have been experimenting with the developers version cloned from gitlab. I would like to specify a hubbard V between atom1 at the edge of the unit cell R=[0,0,0] and a nearest neighbor, atom2, in an adjacent unit cell R=[0,0,1]. It appears that the code automatically finds the equivalent atoms to the site you put a hubbard U and applies the V to those also.

I have attempted to put the V on the equivalent to atom2 in R=[0,0,0] but it assumes I want the V between the two atoms in R=[0,0,0].

I have also tried adding the V using an index corresponding to another unit cell. I'm not sure how the indices for the atoms are chosen when the algorithm finds equivalent sites.

Is there a simple way to do what I detailed above? Also if I'm way off please let me know.

Thank you for any information you can provide.

Andrew Supka
Postdoc Physics
Central Michigan University
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Matteo Cococcioni
Department of Physics
University of Pavia
Via Bassi 6, I-27100 Pavia, Italy
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