The Diamond Lattice (e.g. here) is an fcc-lattice. For an fcc-lattice, one set of primitive translation vectors could be

$\left\{ \frac{a}{2}\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}, \frac{a}{2}\begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \frac{a}{2}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \right\}. $

However, this set of vectors must be slightly modified for the Diamond Lattice, and I'm not quite sure how. Any hints would be appreciated!

One question I am asking myself is: If I translated the crystal by the vector $$\frac{a}{4}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix},$$ would it be invariant under this translation (I cannot imagine it right now)?

  • $\begingroup$ Diamond cubic is an fcc lattice with a two atom basis. $\endgroup$
    – Jon Custer
    Jul 4, 2020 at 16:05

1 Answer 1


The diamond structure is fcc with two atoms in the basis. If you pick the lattice vectors as you chose:

$$ \begin{eqnarray} \mathbf{a}_1=\frac{a}{2}(1,1,0), \\ \mathbf{a}_2=\frac{a}{2}(1,0,1), \\ \mathbf{a}_3=\frac{a}{2}(0,1,1), \end{eqnarray} $$ then you can place the atoms at the following positions to obtain the diamond structure:

$$ \begin{eqnarray} (0,0,0), \\ \frac{a}{4}(1,1,1). \\ \end{eqnarray} $$

The vector $\mathbf{v}=\frac{a}{4}(1,1,1)$ is the vector connecting the two atoms in the basis, but is not a translation vector of the lattice. For the latter, you need a linear combination of the lattice vectors.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy