# What is the lattice vector description of diamonds structure for a conventional fcc unit cell?

I understand how diamond has a conventional fcc unit cell with 4 extra atoms in it's basis. I also know that we can construct a primitive lattice cell using the lattice vectors:

$$\pmatrix{\frac{1}{2} \\ \frac{1}{2} \\ 0}$$,$$\pmatrix{\frac{1}{2} \\ 0 \\ \frac{1}{2}}$$ and $$\pmatrix{0 \\ \frac{1}{2} \\ \frac{1}{2}}$$ given that the conventional fcc unit cell has length 1 edges.

However, I need to find the lattice vectors that describe the conventional unit cell and I don't understand what makes them different from the primitive unit cell vectors. Must they be the same vectors to span the whole lattice? If so, doesn't this mean that the reciprocal space lattice vectors has to be the same for both unit cells?

• Well, diamond cubic is an fcc lattice with two atoms as a basis. The conventional fcc unit cell (the cube with an atom at each face center) has 4 atoms in it, so the '4 extra atoms' are one for each fcc site. There are an infinite number of primitive unit cells, with the Wigner-Seitz being my personal favorite. – Jon Custer May 28 at 13:47