[Pw_forum] Confused on nat definition

Thu May 11 07:29:45 CEST 2006

OK. Let me try to shed some light on this unexpectedly obscure point.

On May 11, 2006, at 6:41 AM, Huiqun Zhou wrote:

> 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.

I am sorry, Huiqun, but I am afraid that this argument is wrong and  
confusing. What do you mean by "4 lattice points"? Of course, every  
triplet of vectors originating at (0,0,0) has one common point (the  
origin), and 3 end points. AND SO WHAT?

> 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.

Although it is certainly true that, for a same system, doing a  
calculation with a larger conventional unit cell would require a  
smaller number of k points, I feel that it is confusing to try to  
explain a property of the real-space lattice with something that has  
only to do with the way calculations are technically done (the number  
of k points).

Let me try to clarify the matter as much as I can.

1) The Bravais lattice of the diamond structure is face-centered  
cubic (FCC). A basis for this Bravais lattice is, e.g., (1/2,1/2,0),  
(1/2,0,1/2), (0,1/2,1/2).

2) Each elementary cell of this FCC lattice contains two equivalent  
atoms, located at (000)+R(l,m,n) and (1/4,1/4,1/4)+R(l,m,n), where R 
(l,m,n) = l*(1/2,1/2,0) + m*(1/2,0,1/2) + n*(0,1/2,1/2) is a generic  
point of the Bravais lattice (l,m,n being integer numbers).

3) The above basis for the Bravais lattice defines a unit cell of  
*minimum volume*. Of ourse, one is free to choose as a basis any  
triplet of integer linear (and linearly independent) combinations of  
the minimum basis. For instance, one could choose: (1,0,0) =  
(1/2,1/2,0)+(1/2,0,1/2)-(0,1/2,1/2), (0,1,0)= [guess what?], and  
(0,0,1)=[guess what?]. These three vectors form a basis for the  
simple cubic (SC) Bravais lattice. Calculate the volume of the  
original FCC unit cell (defined at point 1 above) and compare it with  
the volume of the SC unit cell considered here (hint: the volume of a  
parallelepiped is the absolute value of the determinant of the 3x3  
matrix whose columns are the coordinates of the three vectors which  
form the edges of the parallelepiped). You will find that the the  
volume of the SC cell is 4 times larger than the volume of the FCC  
cell. This means that if you choose to describe the diamond lattice  
as SC instead of FCC (which you are free to do), the unit cell will  
contain 4x2 instead of 2 atoms. You see where the factor 4 comes  
from? Nothing to do with the "number of lattice points" (whatever  
these lattice points may be).

4) Now, the volume of the Brillouin zone (BZ) is (2\pi)^3 divided by  
the volume of the unit cell. Hence, the volume of the SC BZ is 1/4  
the volume of the FCC BZ. That's why, in order to sample a SC BZ with  
the same accuracy of an FCC cell you will need 1/4 as many k points.  
But this is another story which has little to do with the main point  
of this thread.

Hope this clarifies a little bit the muddy waters

Stefano Baroni

>
> 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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---
Stefano Baroni - SISSA  &  DEMOCRITOS National Simulation Center -  
Trieste
[+39] 040 3787 406 (tel) -528 (fax) / stefanobaroni (skype)

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