[Wien] Magnetic dipole term and Orbital polarization

Pavel Novak novakp at fzu.cz
Thu Apr 14 07:40:53 CEST 2011


you need the radius in denominator. For this you have to put the negative 
integer in RINDEX.

Regards Pavel Novak

On Wed, 13 Apr 2011, gimbert at cemes.fr wrote:

> Dear Pavel Novak, dear Peter Blaha,
>
> I thank you for your replies to my message.
>
> Concerning the difference between spin moments in scf and scfdmup files:
> indeed, the difference is due to the contribution of orbitals s and p, thank
> you for this information.
>
> Concerning the magnetic dipole term Tz: I have always a problem with that. 
> For
> the radial part, I used RINDEX = 1 (which corresponds to a constant radial
> part) and LSINDEX = 8 as you have told me, but the result which I obtained
> for bulk hcp Co is high, while it should be nearly zero: I obtained 3.3
> instead of 0.014 (Wu, Freeman, J. Magn. M. Mat. 200, 498 (1999)) or instead
> of -0.003 (with Wien97, Komelj, PRB 66, 140407, 2002). I tried with other
> values of RINDEX but the results are worse (and I think that the magnetic
> dipole term Tz does not depend on a radial function, doesn't it?)
> I think I have used the correct succession of commands to obtain this too 
> high
> value of Tz:
>
> runsp_lapw -p -so -i 100 -cc 0.00001 -ec 0.0000001
> x lapw0
> x lapw1  -up
> x lapw1  -dn
> x lapwso -up
> x lapw2 -c -up -so
> x lapw2 -c -dn -so
> x lapwdm -up  -so -c
> x lcore -up
> x lcore -dn
> x mixer
>
> and I have checked that this succession of commands (with RINDEX = 1 and
> LSINDEX = 3) give values for the spin and orbital moments of hcp Co close to
> litterature.
>
> The  file Co.indmc which I used is the following:
>
> -10.                      Emin cutoff energy
> 1                       number of atoms for which density matrix is
> calculated
> 1  1  2      index of 1st atom, number of L's, L1
> 1 8           r-index, (l,s)index
>
> Do you have an idea why the value of Tz which I obtain is so bad, and how I
> can process to make a correct calculation?
>
> Thank you for your help,
>
> Florian Gimbert
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