I'm following The oxford solid state basics By S. Simon. I'm sort of confused, How he has written the basis in different places?

The definition that is given goes like this:

The description of objects in the unit cell with respect to the reference lattice point in the unit cell known as a basis.

He has given the following example:

enter image description here

Basis for crystal

Large Light Gray Atom have Position : $[a/2,a/2]$

Large Light Gray Atom have Position $[a/4,a/4]$, $[a/4,3a/4]$ , $[3a/4,a/4]$ and $[3a/4,3a/4]$. I have no problem with this.

The reference points (the small black dots in figure) forming the square lattice have positions $$\mathbf{R}_{[n_1,n_2]}=[an_1,an_2]=an_1\hat{x}+an_2\hat{y}$$ so that the large light gray atoms have positions $$\mathbf{R}^\mathbf{light-gray}_{[n_1,n_2]}=[an_1,an_2]+[a/2,a/2]$$ whereas the small dark gray atoms have positions

\begin{align*} \mathbf{R}^\mathbf{dark-gray^1}_{[n_1,n_2]} &= [an_1,an_2]+[a/4,a/4] \\ \mathbf{R}^\mathbf{dark-gray^2}_{[n_1,n_2]} &= [an_1,an_2]+[a/4,3a/4] \\ \mathbf{R}^\mathbf{dark-gray^3}_{[n_1,n_2]} &= [an_1,an_2]+[3a/4,a/4] \\ \mathbf{R}^\mathbf{dark-gray^4}_{[n_1,n_2]} &= [an_1,an_2]+[3a/4,3a/4] \end{align*} And I have no problem till this.

For the BCC structure, he has written the basis to be $[0,0,0]$ and $[1/2,1/2,1/2]$. By which one can write, \begin{align*} \mathbf{R}_\mathrm{corner} &= [n_1,n_2,n_3] \\ \mathbf{R}_\text{center} &= [n_1,n_2,n_3] + \left[\frac{1}{2},\frac{1}{2},\frac{1}{2}\right] \end{align*}

The sodium has a BCC structure and thus it should follow this. But in his examples, he has written that the basis $=$ Na at $[0,0,0]$. I do not understand, How he did that?

Further for the diamond which has an fcc structure, He has given basis to be C at $[0,0,0]$ and C at $[1/4,1/4,1/4]$ and a similar problem is with Cu, again an fcc, he has given basis to be Cu at $[0,0,0]$? I don't understand these examples.

Can anybody make me understand what's going on here?


1 Answer 1


I presume that the reason for the confusion is due to the possibility of describing highly symmetric crystal structures in an alternative way using different Bravais lattices of symmetry translations. Depending on the choice of the underlying Bravais lattice, a bcc or a diamond lattice may be described with different Bravais lattices, then different unit cells, and consequently, different atomic bases.

In the case of the bcc lattice, it can be described as $1)$ a crystal structure characterized by a bcc Bravais lattice with a non-cubic primitive cell, therefore a one-atom basis, or $2)$ as a simple cubic Bravais lattice with a cubic unit cell containing a two-atom basis.

In the former case, a possible choice of the basis vectors of the bcc-Bravais lattice is (using an orthogonal set of unit vectors) $$ \begin{align} {\bf a}_1&=\frac{a}{2}(-{\bf \hat x}+{\bf \hat y}+{\bf \hat z})\\ {\bf a}_2&=\frac{a}{2}( {\bf \hat x}-{\bf \hat y}+{\bf \hat z})\\ {\bf a}_3&=\frac{a}{2}( {\bf \hat x}+{\bf \hat y}-{\bf \hat z}), \end{align} $$ the Bravais lattice is described by $$ {\bf R}_{n_1,n_2n_3}=n_1{\bf a}_1+n_2{\bf a}_2+n_3{\bf a}_3,\tag{1} $$ with $n_i \in {\mathbb Z}$, and a possible basis has one atom at $(0,0,0)$.

In the latter case, the simple cubic base can be described as $$ \begin{align} {\bf a'}_1&=a{\bf \hat x}\\ {\bf a'}_2&=a{\bf \hat y}\\ {\bf a'}_3&=a{\bf \hat z} \end{align} $$ the Bravais lattice as in equation $(1)$ with ${\bf a'}_i$ in place of the ${\bf a}_i$, and a possible two-atom basis with atoms in $(0,0,0)$ and $a\left(\frac12,\frac12,\frac12\right)$.

The case of the diamond lattice is analogous. It can be described as an fcc Bravais lattice with a two-atom basis, for instance at $(0,0,0)$ and $a\left(\frac14,\frac14,\frac14\right)$. Alternatively, it can be described as a simple cubic Bravais lattice with an eight-atom unit cell.

  • $\begingroup$ Yes, the ambiguity arises as Simon doesn't describe the non-cubic unit cell (even though he mentions the existence of the alternatives). I have a question though, is it always possible to find a primitive cell enclosing 1 atom for any lattice made of single type of atom? For instance, is it possible to find non-cubic primitive cell for diamond such that we will need only 1 basis or single-atom basis? $\endgroup$ Feb 9, 2021 at 9:09
  • $\begingroup$ @AbhishekAnand, no, it is not always possible to have always a one-atom basis even for one-component systems. The case of Silicon or Carbon are illustrations of such a point. It is evident that the vector joining the positions of the two atoms of the minimal basis does not represent a symmetry translation of the lattice. Therefore, using (at least) a two-atom basis is unavoidable. $\endgroup$ Feb 9, 2021 at 12:04

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