[Pw_forum] Confused on nat definition

Luke Thulin lukethulin at netscape.net
Thu May 11 13:08:12 CEST 2006


Thank you for the clarification, it makes fine sense to me.  The 
original problem I am having is that I'm trying to input the titania 
antase structure.  It is fairly complicated, but I found a source, 
http://cst-www.nrl.navy.mil/lattice/struk/c5.html,  that spells it out 
clearly.  It says it's a body-centered tetragonal and lists six basis 
vectors, presumably to describe the peroidicity of the space group that 
it is in.  So I figured that ibrav = 7 for bct, nat = 6, and use the 
basis vectors to enter the atomic coordinates.  Yet when I view my 
structure with XCrysDen, it looks nothing like anatase.  I then started 
looking at the examples and how they are input and starting thinking 
that my choice for nat was wrong (based on all the discussion about 
Si).  I did see that the internet source and PWscf do not use the same 
vectors to describe the bct lattice, I thought that they were equivalent 
and resulted in the same volume of the unit cell, though I will double 
check.  Sorry about all this, but I wouldn't have asked if I hadn't 
spend so much time trying to enter this structure already.

Thanks for your help,
Luke
.
baroni at sissa.it wrote:

> 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 <mailto:eyvaz_isaev at yahoo.com>>
>> To: <pw_forum at pwscf.org <mailto: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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>>>
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>>
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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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>
>
>
> = 

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