# [Pw_forum] Re: Crystallographic question. (sir_puding at tut.by)

Eduardo Ariel Menendez P emenendez at macul.ciencias.uchile.cl
Fri Mar 10 13:29:38 CET 2006

```Hi,
One way would be use the Brillouin zone of the 8-atoms cell for both
cases. This is straightforward, but it may be not very pretty. The
8-atomos Brilloiun cell has the same information as the 2 atoms cell, it has 1/4 of
the volume, and 4 times the number of bands.

Other way would to represent the bands in the extended zone scheme (see
Ashcroft and Mermin). This is straightforward in a
1-dimensional case (just adding 2N pi/a to the k-vector of the Nth-band).
In the cubic cell, you can just plot the dispersion law for the
x,y, or z direction, and to each Nth band add 2NPi/a to the wave vector,
where a is the
cubic lattice constant. Then you will have only one band that is discontinuous
at the borders of the Brillouin zones. If the structure has additional
peridiocity, such as the perfect lattice diamond, some discontinuities
will vanish. I do not know hoy to do it in a general direction. I would be
interested if you find an algorithm like that.

Other straightforward possiblity is to forget the bands, and only
compare the density of states.
Eduardo

>
> Hi, All
>
> Maybe my question is simple to answer, but I'm little confused with it.
>
> I want to calculate dispersion curves for an electron energy in diamond.
>
> So we have the FCC lattice with two atoms as the primitive basis. On
> the other hand this lattice is equivalent to the Simple Cubic lattice with 8
> atoms as the primitive basis. (Let assume that we have a periodical defect
> such as very very tiny displacement of one of 8 atoms).
>
> The crystals are almost identical, but BZ are different. In the case
> of SP lattice BZ would be cubic with different volume than in the case
> of FCC. How one can transform one type of BZ to another when we place
> displaced atom at the right place in SC lattice. How do i need to
> sample this two BZs tio obtain equivalent dispersion curves?
>
> Thanx.
>
> Sergey/
>

```