Diamond structure post

In class on Thursday Zack and I seemed to have different opinions on the distinguishability of sites in the Si crysal structure (Zincblende, or diamond). This structure has bravais lattice vectors $a(\frac{1}{2},\frac{1}{2},0)$,$a(\frac{1}{2},0,\frac{1}{2})$,$a(0,\frac{1}{2},\frac{1}{2})$. The basis vectors for this structure are $(0,0,0)$ and $a(\frac{1}{4},\frac{1}{4},\frac{1}{4})$.

For those unfamiliar with crystal structures, the basis of a crystal is the unit that has translational symmetry, that is, the part that repeats in space. So, this structure has two types of atoms that repeat, one for each basis vector, and the two atoms are connected by the vector $a(\frac{1}{4}-0 , \frac{1}{4}-0 , \frac{1}{4}-0)$, namely the difference vector between the two basis vectors.

Now that we have our basis, we put a copy of the basis at each point generated by the bravais lattice vectors. In other words, we take the bravais lattice vectors and form all possible linear combinations with integer (...,-2,-1,0,1,2,...) coefficients. (For example, $5a(\frac{1}{2},\frac{1}{2},0)+3a(\frac{1}{2},0,\frac{1}{2})-2a(0,\frac{1}{2},\frac{1}{2}) = (4a,\frac{3a}{2},\frac{a}{2})$ is one such point generated by a linear combination of the bravais lattice vectors with integer coefficients 5,3,-2.) Then, we put a copy of our basis on each of these points. Since our basis is $\{(0,0,0),(\frac{a}{4},\frac{a}{4},\frac{a}{4})\}$, we add these vectors to our generated points to find out where all the atoms are. For our example of the point $(4a,\frac{3a}{2},\frac{a}{2})$, we have atoms at $(4a,\frac{3a}{2},\frac{a}{2})+(0,0,0)$ = $(4a,\frac{3a}{2},\frac{a}{2})$ and $(4a,\frac{3a}{2},\frac{a}{2})+(\frac{a}{4},\frac{a}{4},\frac{a}{4})$ = $(\frac{17a}{4},\frac{7a}{4},\frac{3a}{4})$.

With a picture of the structure (somewhat) firmly established, let us now come to the apparent disagreement between me and Zack.

Zack commented that the two sites in our basis ($(0,0,0)$ and $(\frac{a}{4},\frac{a}{4},\frac{a}{4})$) are identical by symmetry. From this conclusion, he mused that it was puzzling two basis vectors were required to describe the structure.

On the other hand, I argued that the problem was more subtle. The sites were identical and not at the same time. While this seems an obvious contradiction, this is not the case.

I agree with Zack in saying that the two points of the basis are indistinguishable in one sense. Let atom A be at $(0,0,0)$ and atom B at $(\frac{a}{4},\frac{a}{4},\frac{a}{4})$.

If we translate the whole crystal by $(-\frac{a}{4},-\frac{a}{4},-\frac{a}{4})$ and then rotate by 90 degrees about the z-axis, atom B now looks like it is at $(0,0,0)$. (Atom A is off in a different place than $(\frac{a}{4},\frac{a}{4},\frac{a}{4})$, but that doesn't matter--we just needed to show that both atom A and atom B could be seen as at point $(0,0,0)$.)

So, this seems to show that atom A and atom B are identical, since either can be at $(0,0,0)$. However, this is not the whole story. The difference between the two lies in their positions relative to each other. Say atom A is at $(0,0,0)$ and atom B is at $(\frac{a}{4},\frac{a}{4},\frac{a}{4})$.

There are 8 possible locations for atom A and atom B to have nearest neighbors. The eight locations relative to either atom A or atom B comprise all choices for $(\pm\frac{a}{4},\pm\frac{a}{4},\pm\frac{a}{4})$. The distinction between atom A and atom B lies in that atom A has exactly 4 of these neighbors, and atom B has the other 4--there is no overlap. No matter how you rotate or translate the crystal, atom A will have 4 neighbors in 4 different directions, and atom B has 4 neighbors in 4 different directions, but the directions never coincide.

If we choose atom A to be at $(0,0,0)$, make our origin the bottom front left, make the front axis going to the right the x-axis, the line going into the board the y-axis, and the vertical the z-axis, here are the directions of the nearest neighbors for atoms A and B:

(NOTE: the directions are actually multiplied by $\frac{a}{4}$, but this seemed cumbersome to repeat over and over.)

atom A: (+,+,+);(-,+,-);(-,-,+);(+,-,-)

atom B: (-,-,-);(+,+,-);(-,+,+);(+,-,+)

So, while either atom A or atom B can be at $(0,0,0)$, the two sites are distinct from each other. In fact, there are actually two classes of atoms here--those with neighbors in the same direction as atom A's neighbors and those with atoms in the same direction as atom B's neighbors. This is why we need to have a basis of two atoms--if atoms A and B were truly identical, then only one basis vector would be required, namely $(0,0,0)$.

This problem shows that it is critical that we look at the neighbors of an atom to judge whether it is identical to another atom, for this is often how we can tell different sites apart. Moreover, in order to find the smallest building block for a crystal (our basis), we need to know how many unique atom sites there are.

If any of this is incorrect or confusing, please comment!