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

saqib.javaid at ipcms.u-strasbg.fr saqib.javaid at ipcms.u-strasbg.fr
Mon Feb 7 13:48:07 CET 2011

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
something else???
I would appreciate a clarification
Saqib Javaid
University of strasbourg.

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