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How 4 atoms with the positions (0,0,0), (0,0,a), (0,a,0), and (a,0,0)? 
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#1
Jun814, 12:32 AM

P: 11

We could have defined the supercell as
a cube with side length 2a containing four atoms located at (0,0,0), (0,0,a), (0,a,0), and (a,0,0). Kindly explain how can a cell with side length 2a is possible with the following postions (0,0,0), (0,0,a), (0,a,0), and (a,0,0) ? 


#3
Jun814, 03:09 AM

P: 11

I am unable to understand the positions of the atoms if we change the cubic lattice from 'a' to '2a'. The cell with lattice constant 'a' is primitive while with '2a' the cell is just a unit cell. Kindly explain the positions in case of cell with lattice constant 2a ? The original text (We could have defined the supercell as
a cube with side length 2a containing four atoms located at (0,0,0), (0,0,a), (0,a,0), and (a,0,0) ) is taken from a book. 


#4
Jun814, 03:47 AM

Admin
P: 23,400

How 4 atoms with the positions (0,0,0), (0,0,a), (0,a,0), and (a,0,0)?
Would treating it as a cell with constant 'a' and positions (0,0,0), (0,0,a/2), (0,a/2,0) and (a/2,0,0) help?
I must admit I haven't dealt with these things for many years, but it looks like 'a' vs '2a' is just a matter of convention. 


#5
Jun814, 07:45 AM

P: 660

Borek is right. The new primitive lattice parameter of the supercell could be, say, A=2a or a'=2a, and
within this cell the positions are (a'/2,0,0) or (A/2,0,0), etc. When you deal with this all the time you get tired of defining new lattice parameters and just use 2a directly. Defining a supercell is just a first step. In the following step you make the atoms at (a'/2,0,0) different from the ones at (0,0,0), e.g. by attaching opposite magnetic moments, displacing them a little bit, etc. You cannot do that in the original cell because (in this case) there is only one atom that cannot be different from itself. 


#6
Jun814, 10:23 AM

P: 11

But why there are only four positions mentioned for atoms ? I think in case of cubic lattice constant of side "a" there is one atom per unit cell (each atom at corner is shared by 8 adjcent unit cells). But in case of cubic lattice constant of side "2a" there might be four atoms per unit cell (I am not sure). My question is that why there are four atoms per unit cell in case of "2a" and how the author described their positions ? Kindly help



#7
Jun814, 10:42 AM

Sci Advisor
P: 3,567

I don't understand very well what you are asking or what the original statement of the book you are referring to are.
However, if you want to describe a primitive lattice with a cell of double side lenght, I would expect it to contain 8 atoms not four. Especially, there should be also atoms at (a,a,0), (a,0,a), (0,a,a) and (a,a,a), in addition to the ones you mentioned. 


#8
Jun814, 11:31 AM

P: 11

Kindly have a look at the orignal text "For example, we could have defined the supercell as
a cube with side length 2a containing four atoms located at (0,0,0), (0,0,a), (0,a,0), and (a,0,0). Repeating this larger volume in space defines a simple cubic structure just as well as the smaller volume we looked at above. There is clearly something special about our first choice, however, since it contains the minimum number of atoms that can be used to fully define the structure (in this case, 1). The supercell with this conceptually appealing property is called the primitive cell." First choice refers to cube with length "a". Reference is DENSITY FUNCTIONAL THEORY A Practical Introduction by DAVID S. SHOLL (Chapter 2 page number 36) Book can be downloaded from http://en.bookfi.org/s/?q=DENSITY+FU...troduction&t=0 Thank you for your attention 


#9
Jun814, 02:04 PM

P: 660

DrDu is right. If you double the side of the cube, you increase the volume by a factor of 2*2*2=8. So if in
the original primitive cubic cell there was 1 atom/a^3, then in the new cell there have to be 8 atoms/(8 a^3) to get the same density. 


#10
Jun814, 11:28 PM

P: 11

Dear sir So what we conclude the text in the book is incorrect ?



#11
Jun914, 01:48 AM

P: 660

Yes.



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