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

[saqib.javaid at ipcms.u-strasbg.fr]

Dear Gabrielle,

I was bit concerned because if you are right, the order to PDOS for Spin-orbit,
non-collinear case would be different from collinear, scaler relativistic case
where order is sorted out according to magnitude of Ml i.e. 0,+-1, +-2 ..
thanks a lot for the clarification.
best regard
saqib javaid
University of strasbourg.

Selon Gabriele Sclauzero <sclauzer at sissa.it>:

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




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