Lattice basis changes#

Overview#

A change of basis is used when converting between crystallographic cells, including symmetry settings, conventional and primitive descriptions, or representations motivated by computational convenience. The basis change in direct space induces a corresponding change in the reciprocal basis, fractional coordinates, and Miller indices.

Direct lattice#

Let the original direct-lattice basis vectors be \(\mathbf{a}, \mathbf{b}, \mathbf{c}\), arranged as columns of

\[\begin{split}\mathbf{A} = \begin{bmatrix} | & | & | \\ \mathbf{a} & \mathbf{b} & \mathbf{c} \\ | & | & | \end{bmatrix}.\end{split}\]

Let the new basis vectors be expressed in terms of the original basis through a 3×3 matrix \(\mathbf{P}\)

\[\mathbf{A}' = \mathbf{A}\,\mathbf{P}.\]

The columns of \(\mathbf{P}\) are the components of the new basis vectors in the original basis, e.g.

\[\begin{split}\mathbf{a}' = p_{11}\mathbf{a} + p_{21}\mathbf{b} + p_{31}\mathbf{c}, \\ \mathbf{b}' = p_{12}\mathbf{a} + p_{22}\mathbf{b} + p_{32}\mathbf{c}, \\ \mathbf{c}' = p_{13}\mathbf{a} + p_{23}\mathbf{b} + p_{33}\mathbf{c}.\end{split}\]

The unit-cell volume transforms as

\[V' = (\det \mathbf{P})\,V.\]

Fractional coordinates#

Let \(\mathbf{x} = (x, y, z)^T\) and \(\mathbf{x}' = (x', y', z')^T\) denote fractional coordinates in the original and new basis. Since

\[\mathbf{r} = \mathbf{A}\mathbf{x} = \mathbf{A}'\mathbf{x}' = \mathbf{A}\mathbf{P}\mathbf{x}',\]

fractional coordinates transform as

\[\mathbf{x}' = \mathbf{P}^{-1}\mathbf{x}.\]

Reciprocal lattice#

Let the original reciprocal basis vectors be collected as columns of

\[\begin{split}\mathbf{B} = \begin{bmatrix} | & | & | \\ \mathbf{a}^* & \mathbf{b}^* & \mathbf{c}^* \\ | & | & | \end{bmatrix} = \mathbf{A}^{-T}.\end{split}\]

If the direct basis transforms as \(\mathbf{A}' = \mathbf{A}\mathbf{P}\), then

\[\mathbf{B}' = \mathbf{P}^{-T}\,\mathbf{B}.\]

Thus, the reciprocal basis vectors transform with the inverse-transpose of \(\mathbf{P}\).

Miller indices#

Let \(\mathbf{h}\) and \(\mathbf{h}'\) denote Miller index column vectors in the original and new bases. Since

\[\mathbf{Q} = 2\pi\,\mathbf{B}\mathbf{h} = 2\pi\,\mathbf{B}'\mathbf{h}' ,\]

and \(\mathbf{B}' = \mathbf{P}^{-T}\mathbf{B}\), the transformation satisfies

\[\mathbf{B}\mathbf{h} = \mathbf{P}^{-T}\mathbf{B}\mathbf{h}' \quad \Rightarrow \quad \mathbf{h} = \mathbf{P}^{-T}\mathbf{h}'.\]

Equivalently,

\[\mathbf{h}' = \mathbf{P}^{T}\mathbf{h}.\]

Transformations summary#

Given the direct-space transformation \(\mathbf{A}' = \mathbf{A}\mathbf{P}\):

  • Direct lattice basis:

    \[\mathbf{A}' = \mathbf{A}\mathbf{P}.\]

  • Reciprocal lattice basis:

    \[\mathbf{B}' = \mathbf{P}^{-T}\mathbf{B}.\]

  • Fractional coordinates:

    \[\mathbf{x}' = \mathbf{P}^{-1}\mathbf{x}.\]

  • Miller indices:

    \[\mathbf{h}' = \mathbf{P}^{T}\mathbf{h}.\]

Hexagonal to orthorhombic transformation (example)#

Consider a hexagonal cell with basis vectors \(\mathbf{a},\mathbf{b},\mathbf{c}\) (\(120^\circ\) in the basal plane). An orthorhombic cell aligned with a Cartesian frame can be defined by

\[\mathbf{a}' = \mathbf{a}, \qquad \mathbf{b}' = 2\mathbf{b} - \mathbf{a}, \qquad \mathbf{c}' = \mathbf{c}.\]

The corresponding direct transformation matrix is

\[\begin{split}\mathbf{P} = \begin{bmatrix} 1 & -1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{bmatrix}.\end{split}\]

Fractional coordinates transform as

\[\mathbf{x}' = \mathbf{P}^{-1}\mathbf{x}.\]

Miller indices transform as

\[\mathbf{h}' = \mathbf{P}^{T}\mathbf{h}.\]

Centered lattices#

Centered conventional cells (A, B, C, I, F, R) can be expressed in terms of a primitive basis through the matrix \(\mathbf{P}\). Fractional coordinates and Miller indices transform with \(\mathbf{P}^{-1}\) and \(\mathbf{P}^{T}\), respectively.

Primitive lattice#

\[\begin{split}\mathbf{P}_P = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}\end{split}\]

Base-centered lattices#

A-centered#

\[\begin{split}\mathbf{P}_A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \tfrac{1}{2} & \tfrac{1}{2} \\ 0 & -\tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix}\end{split}\]

B-centered#

\[\begin{split}\mathbf{P}_B = \begin{bmatrix} \tfrac{1}{2} & 0 & \tfrac{1}{2} \\ 1 & 0 & 0 \\ -\tfrac{1}{2} & 0 & \tfrac{1}{2} \end{bmatrix}\end{split}\]

C-centered#

\[\begin{split}\mathbf{P}_C = \begin{bmatrix} \tfrac{1}{2} & \tfrac{1}{2} & 0 \\ -\tfrac{1}{2} & \tfrac{1}{2} & 0 \\ 0 & 0 & 1 \end{bmatrix}\end{split}\]

Body-centered lattice#

\[\begin{split}\mathbf{P}_I = \begin{bmatrix} \tfrac{1}{2} & \tfrac{1}{2} & -\tfrac{1}{2} \\ -\tfrac{1}{2} & \tfrac{1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & -\tfrac{1}{2} & \tfrac{1}{2} \end{bmatrix}\end{split}\]

Face-centered lattice#

\[\begin{split}\mathbf{P}_F = \begin{bmatrix} 0 & \tfrac{1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & 0 & \tfrac{1}{2} \\ \tfrac{1}{2} & \tfrac{1}{2} & 0 \end{bmatrix}\end{split}\]

Rhombohedral lattice#

Hexagonal axes#

\[\begin{split}\mathbf{P}_R = \begin{bmatrix} \tfrac{2}{3} & \tfrac{1}{3} & \tfrac{1}{3} \\ \tfrac{1}{3} & \tfrac{2}{3} & \tfrac{1}{3} \\ \tfrac{1}{3} & \tfrac{1}{3} & \tfrac{2}{3} \end{bmatrix}\end{split}\]