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. >>> >>>> _______________________________________________ >>>> Pw_forum mailing list >>>> Pw_forum at pwscf.org <mailto:Pw_forum at pwscf.org> >>>> http://www.democritos.it/mailman/listinfo/pw_forum >>>> >>> >>> >>> __________________________________________________ >>> Do You Yahoo!? >>> Tired of spam? Yahoo! Mail has the best spam protection around >>> http://mail.yahoo.com >>> _______________________________________________ >>> Pw_forum mailing list >>> Pw_forum at pwscf.org <mailto:Pw_forum at pwscf.org> >>> http://www.democritos.it/mailman/listinfo/pw_forum >>> >> >> _______________________________________________ >> Pw_forum mailing list >> Pw_forum at pwscf.org <mailto:Pw_forum at pwscf.org> >> http://www.democritos.it/mailman/listinfo/pw_forum > > > --- > Stefano Baroni - SISSA & DEMOCRITOS National Simulation Center - Trieste > [+39] 040 3787 406 (tel) -528 (fax) / stefanobaroni (skype) > > Please, if possible, don't send me MS Word or PowerPoint attachments > Why? See: http://www.gnu.org/philosophy/no-word-attachments.html > > > > = -------------- next part -------------- An HTML attachment was scrubbed... URL: /pipermail/attachments/20060511/27f53ab1/attachment.htm
