We usually plot electronic bands with the help of high symmetry points of the irreducible zone of primitive cell of particular material. But if we want to plot bands with conventional cell, we have to map the high symmetry points from primitive cell to conventional cell. so how can we map the high symmetry points from primitive cell to conventional cell ?? basically how can we transform the points ?


1 Answer 1


Note. This answer is adapted from an answer to a related question: Translating fractional coordinates resulting from conventional cell and primitive cell

Let $(\mathbf{a}_{p_1},\mathbf{a}_{p_2},\mathbf{a}_{p_3})$ be your primitive cell lattice vectors, and let $(\mathbf{a}_{s_1},\mathbf{a}_{s_2},\mathbf{a}_{s_3})$ be your supercell lattice vectors. The key ingredient you need to answer your question is the supercell matrix $\mathbf{S}$ relating the two as: \begin{equation} \begin{pmatrix} \mathbf{a}_{\text{s}_1}\\ \mathbf{a}_{\text{s}_2}\\ \mathbf{a}_{\text{s}_3} \end{pmatrix} = \begin{pmatrix} S_{11}&S_{12}&S_{13}\\ S_{21}&S_{22}&S_{23}\\ S_{31}&S_{32}&S_{33} \end{pmatrix} \begin{pmatrix} \mathbf{a}_{\text{p}_1}\\ \mathbf{a}_{\text{p}_2}\\ \mathbf{a}_{\text{p}_3} \end{pmatrix}. \end{equation} Note that $S_{ij}\in\mathbb{Z}$, and the size of the supercell is given by the determinant of $\mathbf{S}$.

The reciprocal primitive lattice has basis vectors: \begin{equation} \begin{pmatrix} \mathbf{b}_{\text{p}_1}\\ \mathbf{b}_{\text{p}_2}\\ \mathbf{b}_{\text{p}_3} \end{pmatrix} =2\pi \begin{pmatrix} \mathbf{a}_{\text{p}_1}\\ \mathbf{a}_{\text{p}_2}\\ \mathbf{a}_{\text{p}_3} \end{pmatrix}^{-\text{T}}, \end{equation} and we can obtain the basis vectors of the reciprocal superlattice as: \begin{equation} \begin{pmatrix} \mathbf{b}_{\text{s}_1}\\ \mathbf{b}_{\text{s}_2}\\ \mathbf{b}_{\text{s}_3} \end{pmatrix} = \begin{pmatrix} \bar{S}_{11}&\bar{S}_{12}&\bar{S}_{13}\\ \bar{S}_{21}&\bar{S}_{22}&\bar{S}_{23}\\ \bar{S}_{31}&\bar{S}_{32}&\bar{S}_{33} \end{pmatrix} \begin{pmatrix} \mathbf{b}_{\text{p}_1}\\ \mathbf{b}_{\text{p}_2}\\ \mathbf{b}_{\text{p}_3} \end{pmatrix}, \end{equation} where $\bar{S}_{ij}=(S^{-1})_{ji}$.

Any $\mathbf{k}$ point, for example one corresponding to a high-symmetry point in the irreducible Brillouin zone, can be written in fractional coordinates in terms of either the reciprocal primitive basis vectors, $(k_{\text{p}_1},k_{\text{p}_2},k_{\text{p}_3})$, or in terms of the reciprocal supercell basis vectors, $(k_{\text{s}_1},k_{\text{s}_2},k_{\text{s}_3})$. You can relate the two with: \begin{equation} \begin{pmatrix} k_{\text{s}_1}\\ k_{\text{s}_2}\\ k_{\text{s}_3} \end{pmatrix} = \begin{pmatrix} S_{11}&S_{12}&S_{13}\\ S_{21}&S_{22}&S_{23}\\ S_{31}&S_{32}&S_{33} \end{pmatrix} \begin{pmatrix} k_{\text{p}_1}\\ k_{\text{p}_2}\\ k_{\text{p}_3} \end{pmatrix}. \end{equation}

Example. Imagine you have an orthogonal primitive cell, and let the supercell be simply twice as large along the first direction compared to the primitive cell. In this example: \begin{equation} \mathbf{S}= \begin{pmatrix} 2&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix}. \end{equation} Now imagine that you consider a $\mathbf{k}$-point at the boundary of the irreducible Brillouin zone of the primitive cell, such that $(k_{\text{p}_1},k_{\text{p}_2},k_{\text{p}_3})=(\frac{1}{2},0,0)$. The same point in terms of the reciprocal supercell basis vectors becomes: \begin{equation} \begin{pmatrix} k_{\text{s}_1}\\ k_{\text{s}_2}\\ k_{\text{s}_3} \end{pmatrix} = \begin{pmatrix} 2&0&0\\ 0&1&0\\ 0&0&1 \end{pmatrix} \begin{pmatrix} \frac{1}{2}\\ 0\\ 0 \end{pmatrix} = \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}. \end{equation} Conceptually, as we have doubled the size of the cell in real space along the first direction, the size of the supercell in reciprocal space halves along the same direction. Thus we need to go twice as far to get to the same point in reciprocal space.

As an additional comment, note that when you plot a band structure in a supercell, you will get band folding. In the example above where we double the size of the primitive cell, you will get twice as many bands at every $\mathbf{k}$ point.

  • 1
    $\begingroup$ Thank you dear Prof M. . I have also tried this before but I was in doubt that my way of conversion may be wrong perhaps. But thank you for your clarification. It helped me a lot. $\endgroup$
    – Alpha_Roy
    Jun 13, 2022 at 13:26
  • $\begingroup$ Dear Prof.M, Here you have taken (kp1,kp2,kp3)=(1/2,0,0). is this in fractional coordinate or in basis coordinate ? I mean here (kp1,kp2,kp3) = fractional coordinate * (b1,b2,b3) or only fractional coordinate. $\endgroup$
    – Alpha_Roy
    Jun 15, 2022 at 7:09
  • $\begingroup$ @Alpha_Roy the choice is the fractional coordinate only. To be absolutely clear, this means that for the first BZ of any compound, each $k_{p_i}$ will only run between $-1/2$ and $1/2$. $\endgroup$
    – ProfM
    Jun 15, 2022 at 8:04

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