# Solid State: Crystals

1. Sep 1, 2009

### Niles

Hi

When talking about crystals, then what is a "conventional cell"? I know what a primitive cell is, but not what a conventional cell is.

E.g. see this: http://www.scribd.com/doc/91701/Crystal-Lattices [Broken]

Here they do not describe a conventional cell, but just mention the name.

Last edited by a moderator: May 4, 2017
2. Sep 2, 2009

### olgranpappy

The cubes shown in the picture in the section on convetional cells are conventional cells. The shaded regions are primitive cells. The conventional cells are larger than the primitive cell, and they contain more than one atom (unlike the primitive cell, which in the picture contains one cell).

Basically, a conventional cell is any cell which, when repeated along "conventional lattice vectors", makes up the lattice, but is not the primitive cell---i.e., contains more atoms than the primitive cell.

People usually choose to discuss a lattice in terms of a conventional cell rather than a primitive cell because the conventional cell is usually easier to describe (e.g., a simple cube)... of course, the downside is that now the conventional basis (set of atoms in the cell) is harder to describe.

For example, we can either use a simple cubic conventional cell (conventional lattice vectors (a,0,0),(0,a,0),(0,0,a)) and a four atom basis ((0,0,0),(a/2,a/2,0),(0,a/2,a/2),(a/2,0,a/2)) to describe an fcc lattice. Or, we can use the primitive cell (with primitive lattice vectors (0,a/2,a/2),(a/2,0,a/2),(a/2,a/2,0)) and a single atom basis.

Last edited by a moderator: May 4, 2017
3. Sep 9, 2009

### Niles

I am sorry to respond so late.

First, thank you.

Second, does this mean that I can calculate various things using the conventional lattice-vectors (since they are easy to work with), and get the same result I would have gotten using primitive vectors?

An example: I want to find the length of the reciprocal lattice vector G for diamond. If I use the primitive vectors for diamond-structure I get the same answer as when I use the x-axis, y-axis and z-axis as my vectors. Is this a coincidence or does it confirm my statement above?

4. Sep 9, 2009

### olgranpappy

That doesn't sound right. Maybe you should show what you did. In general different real space lattices will lead to different reciprocal space lattices. E.g., the reciprocal of the primitive lattice will be the primative Brillouin zone, the reciprocal of the conventional lattice will be the conventional Brillouin zone.

5. Sep 9, 2009

### Niles

Thank you for that explanation. The reciprocal lattice is just a change of basis. We look at two scenarios for bcc:

1) The reciprocal lattice space is constructed using x, y and z unitvectors as a1, a2 and a3.
2) The reciprocal lattice space is constructed using the primitive vectors.

Will the length of a unit vector G in R.L. (short for reciprocal lattice) #1 be the same as the length of a unit vector G in RL #2?

Regarding my approach: Using the conventional cell with lattice vectors x, y and z, then the length of b1, b2 and b3 is 2π, and thus the length of G is 2π√3.

Now the primitive vectors for diamond (fcc) are given by (below x, y and z are unit vectors, and a is the length of one side in the cube of fcc)

a1 = ½a(x + y)
a2 = ½a(y + z)
a3 = ½a(z + x)

Finding the reciprocal vectors from these in Maple and then finding the length of G I get 2π√3 as well.

Last edited: Sep 10, 2009
6. Sep 13, 2009

### Niles

Ok, just disregard my post #5, because it is confusing. I mailed the admin in order to get the text changed, but no luck.

So my question is: We know that a reciprocal lattice is something that every real, physical lattice has. Do I construct the same reciprocal lattice using primitive vectors as when using non-primitive vectors? They must span the same space (R3), but is the length of a unit vector the same for both reciprocal spaces?

Last edited: Sep 13, 2009
7. Sep 13, 2009

### olgranpappy

No.

Given the vectors.
$${\bf a_1}\qquad{\bf a_2}\qquad{\bf a_3}\;,$$
the reciprocal lattice vectors are given by
$${\bf b_1}=2\pi \frac{{\bf a_2}\times{\bf a_3}}{{\bf a_1}\cdot{\bf a_2}\times{\bf a_3}}\;,$$
and
$${\bf b_2}=2\pi \frac{{\bf a_3}\times{\bf a_1}}{{\bf a_1}\cdot{\bf a_2}\times{\bf a_3}}\;,$$
and
$${\bf b_3}=2\pi \frac{{\bf a_1}\times{\bf a_2}}{{\bf a_1}\cdot{\bf a_2}\times{\bf a_3}}\;.$$

The a_1, a_2, a_3 are different for the primitive lattice as for the conventional lattice, so the reciprocal lattice vectors will also be different.

8. Sep 13, 2009

### Niles

To be absolutely clear, here is the exact problem. In a crystal we have that λ = 1.54 Å (angstrom), and the lattice spacing is 5.43 Å. I want to find the Bragg-angle θ for the plane with Miller indices (111), and we are talking about a diamond structure.

I find |G| (using my above approach), and since I know that

$$2d\sin \theta = \lambda$$

where

$$d = \frac{2\pi}{|G|},$$

I find θ from this. My problem is that I found |G| using the conventional vectors of a diamond structure (i.e. the usual x, y and z-coordinates), and I get the same result when I use the primitive vectors.

Now, if the two reciprocal spaces are not the space, then why do I get the same answer in both cases?

Last edited: Sep 13, 2009
9. Sep 13, 2009

### olgranpappy

But this is the part I don't understand. How can you get the same result for |G|? Could you show your work?

10. Sep 13, 2009

### Niles

For the conventional unit vectors (which have length a in direction x, y and z), I get the trivial result (2π√3)/a (the calculations are straighforward, so I will not write them up).

The primitive vectors of silicon (diamond, i.e. fcc) are (in the following x,y and z are unit vectors)

a1 = ½a(x + y)
a2 = ½a(y + z)
a3 = ½a(z + x),

and using your formulas for b1, b2 and b3 I get (I am using Maple in order to avoid mistakes!)

b1 = (2π/a) (1, 1, -1)
b2 = (2π/a) (-1, 1, 1)
b3 = (2π/a) (1, -1, 1)

Thus the size of a unit vector G=b1+b2+b3 is (2π/a)2(12+12+12)) = (2π√3)/a.

Hence, I get the same result. I believe this is very strange.

Basically, the question is: Then we have the equation d = 2π/G, then is the unit vector G in the space spanned by primitive reciprocal vectors or conventional reciprocal vectors?

Last edited: Sep 14, 2009
11. Sep 14, 2009

### olgranpappy

No. No, you don't. You get G=2pi/a.

12. Sep 14, 2009

### Niles

?

In the "conventional vector"-case we have

a1 = ax
a2 = ay
a3 = az

Thus

b1 = (2π/a)x
b2 = (2π/a)y
b3 = (2π/a)z

The length of G = b1+b2+b3 is (2π√3)/a.

13. Sep 14, 2009

### olgranpappy

ah yes, you are right. Sorry, I was thinking of the length of a single reciprocal lattice vector, not the sum of the three reciprocal lattice vectors.

So, you can see that these magnitudes should be the same by looking at the real space crystal too; the crystal plane which has normal $\hat x+\hat y+\hat z$ and goes through the points
$(a,0,0)$
and
$(0,a,0)$
and
$(0,0,a)$
(the simple cubic nearest neighbors).

Also goes through the points
$(0,a/2,a/2)$
and
$(a/2,0,a/2)$
and
$(a/2,a/2,0)$
(the fcc nearest neighbors to the origin).

The equation for this plane in real space is
$$x+y+z=a$$
and so the distance of this plane from a parallel the plane through the origin is given
by setting x=y=z in the above equation to find
$$x=y=z=a/3\;.$$

And the distance of this point from the origin is
$$d=a/\sqrt{3}\;.$$

14. Sep 14, 2009

### Niles

Ahh, I see.

Is the lesson from this that I am allowed to calculate these things with whatever geometry is easiest? Or am I only allowed to work with the primitive vectors?

15. Sep 14, 2009

### olgranpappy

You can use whatever is easiest, but it will probably be easiest to work with the primitive vectors, I think, because to determine if that Bragg reflections really shows up you still have to calculate the form factor due to the basis (there's a two atom basis in diamond, for example, using the primitive fcc vectors).

E.g. consider an fcc lattice with no basis. I.e., one atom per primitive fcc cell. I could factor this lattice as a simple cubic lattice with 4 atoms per cell instead. But,
a simple cubic lattice and a fcc lattice will not give the same bragg spots. You *can* factor your fcc lattice into a simple cubic lattice plus a basis, but the extra Bragg spots that show up due to the (larger in real-space) simple cubic lattice must vanish due to the form factor arising from the basis of four atoms.

Cheers,