[Pw_forum] question about order of PDOS in fully relativistic case

[Gabriele Sclauzero]

Il giorno 07/feb/2011, alle ore 13.48, saqib.javaid at ipcms.u-strasbg.fr ha scritto:

> 
> Dear PWSCF users,
> 
> I have a question regarding the order in which PDOS is written in fully
> relativistic case. As per documentation, it is
> 
>  E LDOS(E) PDOS_1(E) ... PDOS_2j+1(E)
> 
> Does PDOS_1(E) corresponds to -mj and '2j+1' to +mj. e.g. in case of P orbital,
> the order would corresponds to mj='-3/2,-1/2,1/2,3/2' for 4 columns printed or

If I remember well it should be as you state here above. Moreover it sounds like the most reasonable choice.

For the p orbital (if the corresponding pseudopotential has the SO coupling) you should have two files, 
one for j=1/2 (which will contain columns corresponding to m_j=-1/2 and +1/2)
another for j=3/2 with the values of m_j that you've already listed.


HTH

GS



> something else???
> I would appreciate a clarification
> Saqib Javaid
> University of strasbourg.
> 
> ----------------------------------------------------------------
> This message was sent using IMP, the Internet Messaging Program.
> _______________________________________________
> Pw_forum mailing list
> Pw_forum at pwscf.org
> http://www.democritos.it/mailman/listinfo/pw_forum


§ Gabriele Sclauzero, EPFL SB ITP CSEA
   PH H2 462, Station 3, CH-1015 Lausanne

-------------- next part --------------
An HTML attachment was scrubbed...
URL: http://www.democritos.it/pipermail/pw_forum/attachments/20110207/e2a0c00a/attachment-0001.htm