[Wien] problem with fdd2 symmetry group

[antia sanchez]

Thanks for your reply.

Best regards,

Antia

2009/10/23 <wmppemam at lg.ehu.es>

>
> 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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