[Pw_forum] Confused on nat definition

[Huiqun Zhou]

Hmm ..., because there are 4 lattice points, that are (0,0,0) and 3 end 
points of the
basis vectors as stated, in a diamond structure, and there are two primitive 
elements
which are located at (0,0,0) and (1/4,1/4,1/4) for this structure, so there 
should be
4 x 2 = 8 atoms in a conventional unit cell of diamond structure.

But because of periodicity, theoretically you can choose either only the 
primitive
elements (2 atoms) in an unit cell with appropriatelly selected large number 
of k points
for calculation, or the whole member (8 atoms) in an unit cell with less k 
points for
calculation. I may be wrong, please correct me.

Huiqun Zhou

----- Original Message ----- 
From: "Eyvaz Isaev" <eyvaz_isaev at yahoo.com>
To: <pw_forum at pwscf.org>
Sent: Thursday, May 11, 2006 2:51 AM
Subject: Re: [Pw_forum] Confused on nat definition


> Hi,
>
>> This may seem silly, but I'm confused as to what
>> exactly the number of
>> atoms in a unit cell is (nat).  For example, the
>> Silicon example says
>> that nat is only two, yet a diamond structure such
>> as this should have
>> much more than two atoms per unit cell.
>
> Let us consider the diamond case. If you choose as
> basis vectors next 3 vectors (which are the standard
> choice)
>
> 1/2, 1/2, 0
> 1/2, 0  , 1/2
> 0  , 1/2, 1/2
>
> you have only 2 atoms in the unit cell
> (parallelepiped) spanned by these vectors:
> 0,     0, 0
> 1/4, 1/4, 1/4
>
> If you decide to choose as basis vectors next 3 ones
>
> 1 0 0
> 0 1 0
> 0 0 1
>
> you have 8 atoms in the unit cell which is now a cub.
> If your choice is the latter  for CaF2 structure you
> will have 12 atoms, but using the former - only 3.
>
> So, number of atoms (nat) in a unit cell depends on
> your unit cell choice defined by 3 basis vectors.
>
>> Is the definition of nat the  number of basis
>> vectors?
> To me it is not so clear, but see above.
>
> Bests,
> Eyvaz.
>
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