The ABINIT Users Mailing List ( CLOSED )

[PGanesh <pganesh@ciw.edu>]

Dear Matthieu and Takeshi and all,

Thank you very much.  Replies to comments by Matthieu:

> > If I choose ratsph=2.1 a.u. on the Cr sites, what I get for the 
> > integrated LDOS is the following:
> >
> > l =                         0         1          2         3        4 
> > Cr1: spin up               0.09      3.05       2.54      0.03     0.01
> >    spin down               0.01      0.10       0.03      0.00     0.00
> > Cr2: spin-up               0.09       3.05      0.89      0.03     0.01
> >    spin-down               0.01       0.10      0.07      0.00     0.00
> This is indeed strange: the two Cr should be equivalent through some 
> (AFM) symmetry operation, and the other channels are identical, but 
> only for the spin up, instead of swapping spin up/down.
>
> Could you try to output all the atomic LDOS, to see if the atoms are 
> being selected wrong? And if any other atoms are incorrectly 
> inequivalent?
I am quite sure that is not the case, but I will do a run printing LDOS 
for all the 10 atoms and let you know if I see any difference. (I have 
done it before for an intial Ferromagnetic arrangement, which actually 
ends up being non-magnetic, and had similar problems interpreting the 
LDOS on all the 5 atoms in the unit cell).

> > If I compute the moment on Cr1 and Cr2 by integrating the density 
> > within a cube...
> which density are you integrating? The spin up/down in the full unit 
> cell?
I am using the python gz2.py which calls the 'cut3d' program .  The 
integration is within a cube around each atom.  The python program 
performs a crude integration of the difference in the density between 
spin up and spin down within this cube. (tutorial on spin in Abinit)

> > Thank you very much again.  (please look at the input file variables 
> > as pasted below).  Note I am not using nsppol=1, nspden=2 here. Do 
> > you think that might be part of the problem? I would expect that it 
> > only speeds up the calculation and the final result wouldn't change.
> Should be correct. Using nsppol=1 would definitely give the same 
> result for up/down, as the wavefunctions are the same, and no 
> reference to the density is made in the projection on Ylm. In your 
> case the d channel is different, so it's not that.
If I use nsppol=1, nspden=2 then this is what I get for the two Cr atoms:

Cr1:
No.           Ef            l=0           l=1        l=2           
l=3       l=4                    l=0     l=1   l=2     l=3       l=4
33463   -0.12685   0.0069   0.0208 165.5058   0.0051   
0.0013              0.05    5.68    4.26    0.01    0.00

Cr2:
33463   -0.12685   0.1604   0.2169  34.4107   0.0143   
0.0023              0.04    5.67    1.37    0.01    0.00

The last 5 columns are the integrated DOS.  There is now mainly 
difference in the l=2 value, which is good.  But does this difference 
indicate that the system is in an anti-ferromagnetic state?  Since the 
values are not spin projected anymore I am assuming that the DOS is the 
sum of both up and dn on each Cr site.  That would indicate that the 
moments on Cr1 and Cr2 are not equal in magnitude!! 


Reply to comments by Takeshi:
> This week, I will concentrate on this problem. Or you may help me.
Thank you for working on this problem.  Being a new user to Abinit, I am 
afraid I will be of little help to you, if any at all.  I can still be 
helpful in testing newer versions once you have removed the use of 
symmetries to get the m-decomposed PDOS.

General question:

How to get LDOS if I use PAW potentials?

Thanks,
-Ganesh