[Wien] Problem with DOS for Fe-Pd system

[Robert Laskowski]

Hi,
usually to get an elegant DOS  more k-points are required then in the scf.
Assuming your charge and potential are converged vs number of k-points.
For dos doubling this number in each direction, usually is fine. And, yes, you 
can reuse your potential, you do not need to run the scf cycle, just run kgen, 
lapw1 and lapw2 -qtl.

regards

Robert

On Friday 17 September 2010 11:37:31 Maxim Rakitin wrote:
>   Dear Prof. Blaha and WIEN2k community,
> 
> I'm sorry that I'm posting the same problem twice, but I didn't received
> any reply on my previous email and didn't have time to sort the problem
> out a half year ago. Actually I met the same problem both for Fe53Pd and
> Fe53Ti systems.
> 
> What I need is to know how many k-points I need to have to draw DOS for
> these systems correctly. As I wrote before, I need initial struct file
> with 54 atoms (without sgroup changes).
> 
> As far as I get this number, is it possible to perform x lapw2 -qtl
> -up/-dn to get correct DOS without running of SCF cycle? I suspect that
> I can't use already existing files from the calculations with 24
> k-points. Any ideas?
> 
> Thank you in advance,
> Maxim
> 
> 30.04.2010 10:48, Maxim Rakitin пишет:
> > Dear Prof. Blaha,
> > 
> > Thank you for your explanation. When I use x sgroup during
> > initialization, I do have 48 symmetry operations, but all atoms are
> > splitted in 8 groups. I need to have all inequivalent atoms, but in
> > this case I have only 1 operation. There is the following information
> > in case.outputs file:
> > ...........................................
> > pointgroup is 1 (neg. iatnr!!)
> > ...........................................
> > ****** IATNR IN STRUCT_ST CHANGED TO A NEGATIVE NUMBER ******
> > 
> > I don't know what this means. Maybe this is the reason of the problem.
> > Any thoughts?
> > 
> > Thanks for your help.
> > 
> > Best regards,
> > 
> >    Maxim Rakitin
> >    Postgraduate student
> >    South Ural State University,
> >    76 Lenin av., Chelyabinsk, Russia, 454080
> >    Email: rms85 at mail.ru
> >    Web: http://www.susu.ac.ru
> > 
> > 28.04.2010 14:34, Peter Blaha пишет:
> >> Your k-mesh is wrong.
> >> All 4 k-points are "the same" and thus of course the eigenvalues are
> >> identical.
> >> 
> >> >     K=   0.25000   0.25000   0.25000            1
> >> >     K=   0.25000   0.25000   0.75000            2
> >> 
> >> You have a simple cubic lattice with inversion, rotations and "mirror
> >> planes".
> >> Hence you can subtract (0,0,1) from the second k-point and after
> >> mirroring z
> >> you come to the coordinates of the first k-point.
> >> 
> >> Make sure you case.struct file is correct and contains the correct
> >> symmetry operations, and
> >> no warnings in case.outputs. You should still have 48 operations!
> >> Rerun   x kgen.
> >> with 24 k-points as input it gives only 1 (shifted) k-point.
> > 
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> 
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-- 
Dr Robert Laskowski
Vienna University of Technology, Institute of Materials Chemistry, 
Getreidemarkt 9/165-TC, A-1060 Vienna, Austria
tel. +43 1 58801 15675               Fax  +43 1 58801 15698