[Wien] DOS of cubic structures

[jan kunes]

Dominik,

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.

Jan

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
> 
> 
> 
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>  Institute of Physics of Materials           fax.+420/5/41218657
>  Academy of Sciences of the Czech Republic   email: legut at ipm.cz
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