Dear WIEN-users,
I am trying to calculate the plasma frequency (in order to obtain the
electronic conductivity) for a magnetic system, so I put 'SO' and 'Hubbard U'
For a test system I calculated Ni, what I found is that I obtained the same
value for -up and for -dn values, is this
I don't know what I was doing wrong, now I am getting correct results
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de delamora
delam...@unam.mx
Enviado: viernes, 02 de mayo de 2014 12:52 p.m.
Para: wien
For the plasma frequency you do not need to use -so, if you do not use then you
will have *in2 and if you use -so then you will have *in2 and *in2c and the one
that you have to modify is *in2c
if you do not find *in2 then you are doing something wrong elsewere
What I do is to edit spaghetti_ps, You can put titles and move them...
For example, a Ni-2 band;
0.0 16.0 M
/Times-Roman findfont 20.0 scalefont setfont
(Ni-2) show
/Times-Roman findfont 20.0 scalefont setfont
(
What are the Heusler compounds, what are their crystal structure, otherwise it
is difficult to help you...
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de Peram sreenivasa reddy
peramsreeni...@gmail.com
Enviado:
: [Wien] Reg: AFM structure in 111, 110 and 100
On Sat, May 31, 2014 at 03:05:15PM +, delamora wrote:
What are the Heusler compounds, what are their crystal structure, otherwise
it is difficult to help you...
https://en.wikipedia.org/wiki/Heusler_alloy
wikipedia is not the most credible source
The Li crystal has no bonds because the atoms are quite far, while in Fe the
bonds are much stronger and the distances are quite close. In graphite the
inlayer bonds are quite short, while between layers it is quite large, too
large.
De:
Dear WIEN2k users,
I am trying to calculate the plasma frequency for a magnetic system.
I find that if the calculation is with spin-orbit then the 'up' and 'dn'
values are the same, so I tried the CrO2 system;
a=b=4.36A
c=2.897
angles=90
group: 136 (P 42/m n m)
This is a half
Dear WIEN2k users,
I want to make some corrections;
I was not calculating the optic properties correctly, I was using
x optic -up
x optic -dn
for the calculations with spin-orbit coupling
but when I use the correct options;
x optic -c -so -up
x optic -c -so -dn
I get the same
--
delamora delam...@unam.mx wrote:
I am trying to calculate the band structure of a magnetic system and I
having trouble, so I used Ni for a simple system.
I calculate Ni with spin-orbit coupling and Hubbard U:
runsp -so -orb
then I calculate the band
you need to open the connection without password;
man ssh-keygen -N -t dsa
this creates in the directory
~/.ssh
the id_dsa and id_dsa.pub
now copy id_dsa.pub at the end of
authorized_keys
and the permisions of the files in '.ssh' should be 700, that is:
chmod 700 *
that is, only
you need to open the connection without password;
ssh-keygen -N -t dsa
***
not;
man ssh-keygen -N -t dsa
***
this creates in the directory
~/.ssh
the id_dsa and id_dsa.pub
now copy id_dsa.pub at the end of
authorized_keys
and the permisions of the files in '.ssh' should be
Now that the Mexico and Brazil game ended I can continue...
You need to open the connection without password;
ssh-keygen -N -t dsa
this creates in the directory
~/.ssh
also the
id_dsa
id_dsa.pub
files
now copy id_dsa.pub at
Yes, in the WIEN2k you find a, b and c optimization all set for you, but ti
include the angles the number of structures would increase exponentially;
1 dimensionn
2 dimensionsn**2
3 dimensionsn**3
4 dimensionsn**4
5 dimensionsn**5
6 dimensionsn**6
4**6=4096
So
the changes in the cell parameters change little.).
According to user guide at page #66 (listed from 1-6 ) in volume optimization
case.
Please guide me according to list in user guide step wise.
I would be very thankful to you.
On Tue, Jun 24, 2014 at 12:30 AM, delamora
delam
In the option 6 you can put one of the mentioned numbers
27, 64,125 in which the parameters are varied by 3x3x3 (=27 different cases)
'a' is varied by -3 0 3, the same for 'b and 'c', then you have to fit a curve
to these results.
De:
It is relatively easy but there are many little details and you have to do it
yourself;
the file
case.int
has the atom or orbital/atom that you want to plot, the DOS of these are put in
case.dos1ev, case.dos2ev (for non sp-cases)
the 'case.dos1ev' has the first 7 cases of the
Experimental results are the real thing, but usually the optimized parameters
are close to these values, so you can use the experimental results.
I did an optimization of Li2O2 to discern between 2 possible structures:
Féher et al (1953) Chem. Ber. 86, 1429
Föppl (1957) Z. Anorg. Allg. Chem
For totally symmetric structures forces are zero, for example NaCl, Li2O, so
the forces criterion does not apply!
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de Parker, David S.
parke...@ornl.gov
Enviado:
Sometimes you have to increase the emax in case.in1
K-VECTORS FROM UNIT:4 -7.0 ***2.5***13 red emin/emax/nband
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de ding dingming...@qq.com
Enviado: jueves,
You should look at the case.struct, there should be two atoms at the same
place, for example;
Fe has two atoms in the conventional cell, 000 .5.5.5, but if you put these two
atoms and also put the cell as body center then the atom at 000 will be
repeated at .5.5.5 and viceversa
1. Fermi Energy results are listed in case.scf? yes
2. Fermi Energy is same for a borate mineral (insulator), or it changes
according to the type of atoms in the structure (say O, B, Ca, K, Na)? it
changes
3. Fermi Energy is usually around 0.0 Ry for an insulator mineral? Energy is
relative
You could also try rotating the cell to have a different axis, x=y, y=z, z=x
or x=y, etc., and maybe supercell will accept it.
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de Tomas Kana
k...@seznam.cz
Enviado:
Are you calculating Density of States (DOS) or bands.
In DOS the k-points are widened to give a smooth curve, so Ef is surpassed by a
little DOS tail.
In the bands Ef should be touching the upper edge of the band.
Saludos
Pablo
De:
Temuujin and WIEN2k users,
I want to add that this also happens with Fedora 19 and up.
XCrySDen works fine for Fedora 18 and down.
Pablo de la Mora
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de bayarr
Dear WIEN2k users,
I want to include the DMFT into the calculations and apparently
wannier90 does this type of calculations, but I cannot see how to do them. The
wannier90 user-guide does not even mention DMFT.
Is there a guide for these type of calculations?
: lunes, 13 de octubre de 2014 03:23 a.m.
Para: A Mailing list for WIEN2k users
Asunto: Re: [Wien] DMFT
Hi Pablo,
On 10/13/2014 03:55 AM, delamora wrote:
I want to include the DMFT into the calculations and apparently
wannier90 does this type of calculations, but I cannot see how to do
.
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de Elias Assmann
elias.assm...@gmail.com
Enviado: martes, 14 de octubre de 2014 07:56 a.m.
Para: A Mailing list for WIEN2k users
Asunto: Re: [Wien] DMFT
On 10/13/2014 02:46 PM, delamora wrote:
I
O A Yassin
You do not describe the structure...
There is a cubic Sr2MnTiO6 in the ICDD there the Mn and Ti are occupying
the same positions with 0.5 each.
What you can do is to make a supercel and put these two atoms in alternated
positions.
This is what I get for the cubic
Dear WIEN2k users;
I am using the 14.2 version and I am doing spin polarized calculation and
when I want to plot
'simple' DOS (without 'both spins at once')
I get a plot with the energy axis limits normal, but the DOS axis is [0:1]
both spins DOS (without neg spin-dn DOS)
it is
I forgot to say that I used the intel 11.1 compiler
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de delamora
delam...@unam.mx
Enviado: domingo, 19 de octubre de 2014 10:00 a.m.
Para: A Mailing list for WIEN2k
I solved this problem for RED HAT;
In a terminal I tried to run
xcrysden
and there was a program missing, so I installed it, then another one, again I
installed it.
With these two programs and their dependences XCrySDen worked.
Pablo
De:
The crystal field splitting depends on the crystal symmetry!!, but for a
compound with metallic characteristics the bandwidth will be larger that the
field splitting.
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en
?the forces are in .scf, so you can look at the end of that file or
grepline :for .scf 10
where .scf = real name of scf
10 = # of lines you want to see. grepline lists the last '#' lines of the
.scf? file that have ':for' or :FOR', so you will see the results of the
last iteration, but
...@ymail.com
From: ?delamora? ?delam...@unam.mx?
To: ?A Mailing list for WIEN2k users? ?wien@zeus.theochem.tuwien.ac.at?
Sent: Saturday, 8 November 2014, 6:56:18
Subject: Re: [Wien] relaxation (was:_nb in dscgst.F 256 128)
?the forces are in .scf, so you can look
This problem is quite interesting;
You have Ni in a FCC arrangement, so you have Ni tetrahedra, so how can you
place them in an antiferromangnetic order???
It is frustratng!!! Well this is known as geometric frustration, which is a
very interesting field!!!
On the opposite stance immagine a
the hexagonal structure where you see
the Ni-1 planes and Ni-2 planes while with F3 you see the primitive cell.
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de delamora
delam...@unam.mx
Enviado: domingo, 30 de noviembre de
...@zeus.theochem.tuwien.ac.at en nombre de Abed Reg
jazai...@gmail.com
Enviado: domingo, 30 de noviembre de 2014 05:53 p.m.
Para: wien@zeus.theochem.tuwien.ac.at
Asunto: Re: [Wien] What's the crystal structure of NaCl in antiferromagnetic
calculation
Thank you Mr Delamora for your reply
First, we can
Answering your question of NiO with a P lattice;
If you take a P structure with 4 Ni and 4 O then you can flip two Ni atoms and
you have an antiferro in the 100 direction, which is not the one that you sent
in the figure
In my earlier message you can see the antiferro ordering in the 111.
Both
Both structures are almost the same, the rhombohedral one is shortened or
elongated in the 111 direction
If the cell is shortened then the Ni-Ni distance in the 111 planes is larger
than the distance between planes, then there is an antiferromagnetic ordering,
with Ni up in one plane and Ni dn
?There is a discussion about this in the WIEN in December 1:
What's the crystal structure of NaCl in antiferromagnetic calculation
http://www.mail-archive.com/wien%40zeus.theochem.tuwien.ac.at/msg11525.html?
De: wien-boun...@zeus.theochem.tuwien.ac.at
Mr Delamora for your reply
But I want to know how to do a paramagnetic calculation
A paramagnetic calculation is a non magnetic one or not?
Thanks
--
Mr: A.Reggad
Laboratoire de Génie Physique
Université Ibn Khaldoun - Tiaret
Algerie
___
Wien
If someone has a previous Intel ifort compiler you can put it into your
computer. In my computer it is in /opt/intel
so you copy into the new computer and it will work.
Pablo
De: wien-boun...@zeus.theochem.tuwien.ac.at
Dear WIEN2k colleagues,
The ghost bands are a problem, but I have not found how to get rid of
them.
One simple example, cubic BaRuO3
Running with
runsp -so -orb
When r*k=7 or 8 (k-points=100) it finishes without problem
but when it is increased to r*k=9
large effective RKmax (in your case for Sr: 9 * 2.5/1.74) lead to linear
dependencies
and ghost bands.
PPS: With Oxygen as smallest atom, RKmax=8 should be more than enough.
Am 16.02.2015 um 06:55 schrieb delamora:
Dear WIEN2k colleagues,
The ghost bands are a problem, but I have
...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de delamora
delam...@unam.mx
Enviado: miércoles, 18 de marzo de 2015 05:48 p. m.
Para: A Mailing list for WIEN2k users
Asunto: Re: [Wien] DOS spin polarized in WIEN2k-14.2
Dear Professor Blaha,
I did not answer your mail
there is adosplot.ini file which is used by default
if it is present (dosplot2 -h)
On 10/19/2014 05:00 PM, delamora wrote:
Dear WIEN2k users;
I am using the 14.2 version and I am doing spin polarized
calculation and when I want to plot
'simple' DOS (without 'both spins at once
Dear WIEN2k community,
I am using WIEN2k 14.2
When I am plotting the DOS and I do not chose the y range and I choose
Do you want to plot both spins at once: 'n' with neg spin-dn
DOS: 'n'
then the 'y' range is:0:1
Do you want to plot both
/27/2015 10:56 AM, Lyudmila Dobysheva wrote:
On 26.02.2015 00:28, delamora wrote:
Indeed there is no fit for the EOS2, Murnaghan, but for the
Birch-Murnaghan there is a fit.
V0,B(GPa),BP,E0NaNNaNNaN
Equation of state: Murnaghaninfo
If you want no symmetry, or in other words lowest symmetry just put
SG = 1 (P1)
This will have only the identity symmetry.
In this case you can run
x kgen
Pablo
De: wien-boun...@zeus.theochem.tuwien.ac.at
Dear Wien comunity,
I calculated BaRuO3
Cubic P (SG 221)
a=b=c=4A
Ba .5,.5,.5
Ru 0,0,0
O .5,0,0
I do an optimization with low quality RxK=7 Kpoints=100
Vol; 0,-5,-10,-15,-20
I try to plot Energy-Vol and I get a flat line above 0 with NaN,NaN...
I do the same witn Na and I get a nice
: [Wien] Vol optimization for BaRuO3
It might be useful to provide more information from Optimize volume,
c/a-ratio, …”
Just include the whole page (list of scf files read, EOS fitting tables).
Oleg
On Feb 25, 2015, at 14:22, delamora delam...@unam.mx wrote:
Dear Wien comunity,
I calculated
for the fitting program to determine the minimum and the proper
curvature.
A non-linear fit is not always converging. It depends on the starting values
and on the details of the input points.
Am 26.02.2015 um 00:55 schrieb delamora:
Dear Wien comunity,
I did a very rough calculation for two
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de delamora
delam...@unam.mx
Enviado: miércoles, 25 de febrero de 2015 05:55 p. m.
Para: A Mailing list for WIEN2k users
Asunto: Re: [Wien] Vol optimization for BaRuO3
them
to the Murnagham formulas for BaRuO3, but for Na it does well for all the
formulas!!!
Saludos
Pablo
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de delamora
delam...@unam.mx
Dear Wien comunity,
I did a very rough calculation for two systems
Na
BaRuO3
I did them in the lowest of quality
Now, when I tried to plot the energy .vs. volume I got that the fit was only
done for the Birch-Murnagham formula for the BaRuO3, while for the Na system I
got a nice fit for
The simplest perivskite is
a=b=c
alpha=beta=gamma=90
A: 1/2,1/2,1/2
B: 0,0,0
O: 1/2,0,0
0,1/2,0
0,0,1/2
space group 221
so, if you have alpha=89.999 you can adjust to 90.
Now, if you have alpha=89.9 you can adjust to 90 for learning purposes, the
calculation will be quite fast, but the
As far as I know, the zero of the Fermi Energy is not important, but once it is
taken then all the energies of the calculation are relative to this value.
The Ecut value is relative to the ionization energy of the individual atoms.
De:
...@zeus.theochem.tuwien.ac.at] im Auftrag von delamora
[delam...@unam.mx]
Gesendet: Montag, 20. April 2015 15:48
An: A Mailing list for WIEN2k users
Betreff: Re: [Wien] Fermi level
As far as I know, the zero of the Fermi Energy is not important, but once it is
taken then all the energies
Here I am sending the structure from the ICSD (Find it)
*.struct-orig
with F symmetry
After using supercell and you get a P symmetry, then you differentiate two Dy
as Dy1 and two as Dy2;
*.struct-P
with sgroup it will change to tetragonal;
*.struct
this structure crystallographically
Dear Gerhard,
Sorry for my rudeness, I did not want to offend you.
What I see in the ICSD is that there is only DyPdBi reported (see below), and
it is
Dysprosium Palladium Bismuthide (1/1/1)
There there is only one Dy in this structure, so we are talking about two
different compounds.
Dear P Blaha and F Tran,
I would like to ask if with the
modified Becke Johnson potential
one should use the Hubbard U?
Saludos
Pablo
___
Wien mailing list
Wien@zeus.theochem.tuwien.ac.at
Sorry for this question, it has been answered before;
- (2/22/2012)
So most likely mBJ + U gives the better solution, but probably the U should be
smaller than in LDA+U, because mBJ already shifts the f-states a bit.
but I am still confused, in GW and DMFT calculations the
Dear WIEN2k community,
I have a discussion with my Professor, John B Goodenough, and I would
like to share with you.
Is the U or the effective U (U-J) the same for up and down spin?
We are working in ruthenates and in some cases they are half metals, so
the mobility of
Gerhard,
No, Murugan is right, it is DyPdBi
Although I found in the structure base find it
a=b=c=6.643
Dy: 0,0,0
Pd: 3/4,3/4,3/4
Bi: 1/2,1/2,1/2 and Murugan has 1/4,1/4,1/4
In terms of frustration this difference does not make a difference, since Bi is
magnetically neutral.
Saludos
Dear Murugan,
The Cr example has BCC symmetry, your compound is FCC.
You have to be more explicit in your letter.
I looked at the structure;
DyPd make a ZnS structure, that is, it is diamond-like. Bi is
non-magnetic.
Dy will certainly have a
May 2015, delamora wrote:
Sorry for this question, it has been answered before;
- (2/22/2012)
So most likely mBJ + U gives the better solution, but probably the U should
be smaller than in LDA+U, because mBJ already shifts the f-states a bit.
but I am still confused
...@zeus.theochem.tuwien.ac.at en nombre de Madesis Ioannis(John)
imade...@physics.uoc.gr
Enviado: sábado, 16 de mayo de 2015 01:09 p. m.
Para: Wien
Asunto: Re: [Wien] AFM calculations for YBCO6
Mr. Delamora, first of all thank you for your dedication, and thorough
examination of my problem.
I haven't fully tested your
001 0
I did the calculation and MM converges to 0, so you need the Hubbard U
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de delamora
delam...@unam.mx
Enviado: viernes, 15 de mayo de 2015 04:03 p. m.
Para
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
Dear WIEN2k community,
I am interested in evaluating the Hubbard U and sometime ago I tried;
WIEN2k-Textbooks:
Notes about constraint LDA calculations to determine U
by G. Madsen and P. Novak (not updated)
but I did not get very good results, U was
Depending on the symmetry you can put 2 cells in a, b or c direction. The
direction would be in which the replaced atom is furthest apart.
From the formula it seems to be tetragonal, so you can put 2x2x1 and replace
in cells in the diagonal and you will have after group a rotated cell by 45.
Institute for Chemical Physics of Solids
01187 Dresden
Von: wien-boun...@zeus.theochem.tuwien.ac.at
[wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von delamora
[delam...@unam.mx]
Gesendet: Montag, 13. April 2015 16:32
An: A Mailing list for WIEN2k users
I do not know what happened with my reply, but here it is again;
Yes, you are right; if after symmetry you have N k-points, lets say for example
53 k-points, and you use 5 cores, then the machine will run 10 k-points in each
core and after this it will 1 k-point in 3 cores.
But as Laurence
Yes, so, if you have 53 points and 5 cores, there will be 10k-points in each
core and at the end it runs the 3 remaining points in separate processors. In
all you will have 8 '.machine*' files.
De: wien-boun...@zeus.theochem.tuwien.ac.at
One k point are for large cells, such as surfaces, and I have done calculations
and I did not have a problem.
Pablo
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de Laurence Marks
lo que pasa es que mandaste en doble página y yo solo veía la izquierda.
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de delamora
delam...@unam.mx
Enviado: domingo, 14 de junio de 2015 10:57 a. m.
Para: A Mailing
mBJ is a potential and not a functional, so it is very clumsy; Ef has to be in
the gap, and you cannot make any changes;
If you run without mBJ
runsp -so -orb
then in the mBJ run you have to run the same
runsp -so -orb
and for example you cannot change the Hubbard U value
You have to change the symmetry to tetragonal, change the c parameter by 0.001
and the program will recognize that it is not cubic.
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de Seyyed Amir Abbas Emami
This is the same that I simplified and symmetized. If you are going to add H
then you need set the space group as P1 (#1) so the H can move freely and not
in a symmetrical path.
Now, if you use my proposed structure then you can put inversion (SG P-1, #2)
and add two H, one on each surface,
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de delamora
delam...@unam.mx
Enviado: domingo, 26 de julio de 2015 02:15 p. m.
Para: A Mailing list for WIEN2k users
Asunto: Re: [Wien] Reconstructed Si 100 surface
Muhamad
As I said, the first structure is F (press f3 in xcrysden and you will see the
primitive cell) the second structure is P (if you press f3 nothing will happen
in xcrysden).
So the P structure is 4x larger.
Sorry, I was wrong with the counting in the last mail, in the F cell you have 4
atoms
The reciprocal cell is also reciprocal in volume, so you need to reduce the #
of k-points accordingly.
But I would start the run with very small quality, that is, # of k-point for
the non-doped I would use 100, and then grow it to 1000, etc. that is a quicker
way to converge systems. So for a
-boun...@zeus.theochem.tuwien.ac.at en nombre de Seyyed Amir Abbas Emami
a.a.em...@birjand.ac.ir
Enviado: sábado, 25 de julio de 2015 01:20 p. m.
Para: wien@zeus.theochem.tuwien.ac.at
Asunto: Re: [Wien] k-points of doped material
thank you dear delamora.
I did not grow the cell.actually supercell
for one Si atom. Note that the H-atoms are
oriented to keep tetrahedral coordination of Si.
I hope it will help
Oleg
On Jul 19, 2015, at 3:00 PM, delamora delam...@unam.mx wrote:
Sorry, I wanted to say symmetrize
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun
If you systematize your system you will have a much simpler problem;
Here in the SiH100.struct I moved the central plane to 0 0 1/2, then I averaged
the positions of the atoms above the plane and the equivalent atom below this
plane.
After this sgroup changed the structure to a simpler and
Sorry, I wanted to say symmetrize
De: wien-boun...@zeus.theochem.tuwien.ac.at
wien-boun...@zeus.theochem.tuwien.ac.at en nombre de delamora
delam...@unam.mx
Enviado: domingo, 19 de julio de 2015 02:56 p. m.
Para: A Mailing list for WIEN2k users
Asunto: Re
to
passivate dangling bonds. I enclosed a sketch (I hope it will come through)
where the proposed changes are shown for one Si atom. Note that the H-atoms are
oriented to keep tetrahedral coordination of Si.
I hope it will help
Oleg
On Jul 19, 2015, at 3:00 PM, delamora
delam...@unam.mxmailto:delam
You can enlarge the cell with "supercell" and then you will have more atoms of
the same kind so you can replace half for the other atom.
The problem is that it will be replaced in an ordered way
De: wien-boun...@zeus.theochem.tuwien.ac.at
You can enlarge the cell with "supercell" and then you will have more atoms o
De: wien-boun...@zeus.theochem.tuwien.ac.at
en nombre de wasim raja Mondal
Enviado: domingo, 1 de noviembre de 2015
Yes, "runafm" cannot be ran with SOC, so the calculation could slightly shift
to a magnetic state, but if you put more strict conditions the magnetic moment
should tend to cero
Pablo
De: wien-boun...@zeus.theochem.tuwien.ac.at
d Analytical Chemistry
Johannes Gutenberg - University
55099 Mainz
and
Max Planck Institute for Chemical Physics of Solids
01187 Dresden
Von: wien-boun...@zeus.theochem.tuwien.ac.at
[wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von delamora
[dela
Have you changed Mn2 by Fe? so you would see different color for Mn1 and Mn2,
is this what you are asking?
De: wien-boun...@zeus.theochem.tuwien.ac.at
en nombre de Manish Jain
Just a comment;
The structure that you put as MnO is the usual rock salt structure, where the
Mn form a FCC structure, you can arrange an antiferromagnetic structure by
alternating Mg up and Mg dn in the 111 direction, which is what you put in the
structure.
But the FCC structure with
"start SCF cycle" button, then click back on
"Bandstructure" in the left menu. Then, you should see "x lapwso -up
-p" (with a checked "orb" box next to it) on the Band structure page.
On 10/5/2015 10:45 AM, delamora wrote:
> Yes, you are right, but the
The volume of the reciprocal cell is inverse of the unit cell, therefore the
number of k points is proportional to the volume... in other words, if the unit
cell is increased then the reciprocal cell is reduced and the # of k- points
should be reduced proportionally to give the same k-point
Dear WIEN2k Users,
This seems to be an old problem, see the Mail Archive 2013-02-22
"Incomplete DOS", but I do not know how to solve.
I am doing Pd;
FCC a=b=c=3.89A
But when I plot DOS it does not reach Ef.
Following the suggestions of the <2013-02-22 "Incomplete DOS"> discussion
<wien-boun...@zeus.theochem.tuwien.ac.at> en nombre de delamora
<delam...@unam.mx>
Enviado: lunes, 14 de septiembre de 2015 10:01 p. m.
Para: A Mailing list for WIEN2k users
Asunto: [Wien] Incomplete DOS
Dear WIEN2k Users,
This seems to be an old problem, see the Mail Archive 2013-02-22
&quo
>
Enviado: martes, 15 de septiembre de 2015 08:43 a. m.
Para: A Mailing list for WIEN2k users
Asunto: Re: [Wien] Incomplete DOS
You never told us that you do SO !!
There is another Emax in case.inso, which you need to increase.
On 09/15/2015 03:39 PM, delamora wrote:
> Gerhard,
> Thanks
&g
Physics of Solids
01187 Dresden
Von: wien-boun...@zeus.theochem.tuwien.ac.at
[wien-boun...@zeus.theochem.tuwien.ac.at] im Auftrag von delamora
[delam...@unam.mx]
Gesendet: Dienstag, 15. September 2015 06:16
An: A Mailing list for WIEN2k users
Betreff: Re: [Wien] Incomplete DOS
I w
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