[Wien] DOS of cubic structures

jan kunes kunes at yammer.ucdavis.edu
Wed Jan 28 16:17:09 CET 2004


if you have caculated the p-PDOS using lapw2,  only the 
irreducible wedge of BZ was used without any symmetrization. In this way 
you can obtain correctly only the total p-PDOS=px+py+pz. However, because 
of the cubic symmetry px=py=pz=p_tot/3.


On Wed, 28 Jan 2004, Dominik Legut wrote:

> Dear Wien2k users,
> recently, I have calculated DOS of Pb in its ground state fcc cubic
> structure and decomposed it into the s, p, px, py, pz components according
> to the line corresponding line of the *.qtl file. As in the cubic
> structure the x, y and z directions are equivalent, I supposed that
> the DOSs for px, py and pz states will also be the same. However, this
> is not the case!
> Why do the DOS p-contributions for px, py and pz states differ if one
> has a cubic crystal? What files should I check  ?
> I should appreciate any comment on that. Thanks!
> Dominik Legut
> ---------------------------------------------------------------------
>  Dominik Legut                               tel.+420/5/32290461
>  Institute of Physics of Materials           fax.+420/5/41218657
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>  CZ-616 62
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