[Wien] AFM calculations for YBCO6

[delamora]

Hi John,
        You have the famous 1236 derived from the more famous 1237
In here you have one plane of Cu1 and two of Cu2
What order would you expect?
Cu1 are quite far away so there would be no strong correlation between them.
On the other hand Cu2 have O connecting them in the plane, so you would expect an inplane antiferromagnetic ordering.
>From your question it seems that you already knew this, but you did not make it clear.
So you need two kind of Cu in the same plane in alternating magnetic directions, like 001 plane in NaCl, Na alternates with Cl
So you make a 2 2 1 supercell
Mark Cu in 000 and 1/2 1/2 0 as Cu1 
            and 1/2 0 0 and 0 1/2 0 as Cu2
and erase the number in all other atoms
Then the program choose a simpler cell with the same symmetry, and you put with "instgen" alternating spins, up and dn, for Cu and n for all others
The afm operation is
100 .5
010 .5
001 0
I did the calculation and MM converges to 0, so you need the Hubbard U
________________________________________
De: wien-bounces at zeus.theochem.tuwien.ac.at <wien-bounces at zeus.theochem.tuwien.ac.at> en nombre de delamora <delamora at unam.mx>
Enviado: viernes, 15 de mayo de 2015 04:03 p. m.
Para: A Mailing list for WIEN2k users
Asunto: Re: [Wien] AFM calculations for YBCO6

What magnetic ordering do you have?
If Cu1 is up and Cu2 is dn then you have to calculate it with this ordering!
For an afm calculation you need a symmetry operation that moves Cu1 to Cu2.
In that case you do an afm calculation with runafm, otherwise you assign Cu1 as up and Cu2 as dn and do a runsp calculation.
There is another issue, if you have a symmetry operation that moves Cu1 to Cu2 then the primitive cell probably has only one Cu (I mean, for a simple system)
Cr is the example, Cr has B symmetry, and has one Cr. To make the afm calculation then you change to P symmetry with two Cr and assign one as up and the other as dn.
One interesting exception is the rutile symmetry, where there is an M atom at 0,0,0 and another at 1/2,1/2,1/2, but to go from one to the other you need;
0 -1 0  0.5
1 0 0   0.5
0 0 1   0.5
that is, a rotation by 90 degrees in the 001 axis.

        Pablo
________________________________________
De: wien-bounces at zeus.theochem.tuwien.ac.at <wien-bounces at zeus.theochem.tuwien.ac.at> en nombre de pieper <pieper at ifp.tuwien.ac.at>
Enviado: viernes, 15 de mayo de 2015 08:28 a. m.
Para: A Mailing list for WIEN2k users
Asunto: Re: [Wien] AFM calculations for YBCO6

Good eveneing, Ioannis Madesis

As you observed yourself: The (primitve) unit cell of your structure
does not have enough Cu1 to support AFM on that sublattice. For that you
need at least two Cu1 atoms. You achieve that by doubling (at least) the
unit cell (perhaps using supercell) and assigning two different numbers
to the Cu-sites you want to assign opposite spins to. As far as I recall
you should double the unit cell along the a- AND the b-axis to be able
to assign the correct spin directions. If you want to describe AF
stacking sequences along the c-axis you will have to double the unit
cell along that axis. Once you assigned numbers to Cu according to their
two spin directions in your supercell (say, leave Cu1 for up, rename the
Cu1 you want to point down into Cu3) you can let nn and symmetry do the
job of finding the corresponding space group - which will necessarily be
different from 123 (consult the UG for this procedure).

Good luck

Martin Pieper


---
Dr. Martin Pieper
Karl-Franzens University
Institute of Physics
Universitätsplatz 5
A-8010 Graz
Austria
Tel.: +43-(0)316-380-8564


Am 15.05.2015 14:32, schrieb Madesis Ioannis(John):
> Good evening everyone
>
> I am having trouble with the struct file for these calculations. To be
> more specific: This material has 2 nonequivalent Cu atoms, Cu1 and
> Cu2. Each one of these sits at the four corners of the unit cell, Cu1
> at z=0 and Cu2 at approximately z=1/3, and the space group is 123.
> In order to achieve AFM ordering, I need to have 2 types of Cu1, and
> the spacegroup doesn't let me do that. Once I place Cu1(up), all 4
> corners have Cu1(up). What can I do to solve this?
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