[Wien] problem with fdd2 symmetry group

wmppemam at lg.ehu.es wmppemam at lg.ehu.es
Fri Oct 23 15:26:34 CEST 2009


To change to the standard setting Fd2d to Fdd2, the only important thing 
is to put the binary axis (b-direction in the Fd2d setting) along the c 
direction, and to keep a right handed a,b,c set of vectors. So it is a 
matter of choice to decide which is a and which is b in the standard 
setting among the the remaining a and c parameters of the Fd2d cell.

But the problem with the transformation that you say sgroup does is that 
it does not keep the right-handedness of the  three vectors (the 
determinant of the transformation is -1). Probably you have overlooked a 
minus sign in the transformation. I do not think that sgroup does such 
mistakes.

Regards,

Manuel

*************************************************
J. Manuel Perez-Mato
Fac. Ciencia y Tecnologia,
Universidad del Pais Vasco, UPV
48080 BILBAO,
Spain

tel. +34 946012473
fax. +34 946013500
*************************************************

On Fri, 23 Oct 2009, antia sanchez wrote:

> Dear WIEN2k users,
>
> According to the Bilbao crystallographic server data, the rotational matrix
> which transforms the coordinates (a,b,c) given in the fd2d setting to the
> fdd2 (standard) one is:
> (0  0  1)(1  0  0)
> (0  1  0)
>
> Hence, (a, b, c)(fd2d) transforms to --> (c, a, b)(fdd2).
>
> I've introduced an atomic structure given in the fd2d setting using the
> StructGen and I've initialized the calculation. When sgroup and symmetry are
> run, the setting that the code appears to want is the standard one. The
> point is that, observing the initial and final lattice parameters, a
> different matrix seems to be used for the symmetry operation as: (a, b,
> c)(fd2d) is transformed to --> (a, c, b)(fdd2).
>
> Does anybody know why WIEN2k does this?.
>
> Regards,
>
> P.S. I am using the WIEN2k 8.3 version.
>


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