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How could I get the Brillouin Zone of 2*2 supercell? I tried to use see k-path website, but when I upload a supercell, it still generates the Brillouin Zone of the unit cell.

What I want to do is to know when building a supercell, how could I fold a specific high symmetry point to the Gamma point.
As shown below, how could I fold M point back to Gamma point?

Band structure of the MoS2/WS2 heterostructure.

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  • $\begingroup$ What is the position of the M point in terms of primitive cell reciprocal lattice vectors? $\endgroup$
    – ProfM
    Jan 11, 2022 at 20:18
  • $\begingroup$ @ProfM the coordinate of M point is (0.5,0,0). $\endgroup$
    – Jack
    Jan 11, 2022 at 20:21
  • $\begingroup$ Then a supercell of size $2\times1\times1$ will map the M point to the Gamma point. $\endgroup$
    – ProfM
    Jan 11, 2022 at 20:23
  • $\begingroup$ @ProfM is there a general formula for this kind of manipulation? $\endgroup$
    – Jack
    Jan 11, 2022 at 20:27
  • $\begingroup$ Just provided a full answer. $\endgroup$
    – ProfM
    Jan 11, 2022 at 20:45

1 Answer 1

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Setup. Consider a primitive cell with lattice vectors $\mathbf{a}_{p_1}$, $\mathbf{a}_{p_2}$, and $\mathbf{a}_{p_3}$. You can construct a supercell with lattice vectors $\mathbf{a}_{s_1}$, $\mathbf{a}_{s_2}$, and $\mathbf{a}_{s_3}$ by making linear combinations of the primitive cell lattice vectors: $$\tag{1} \begin{pmatrix} \mathbf{a}_{s_1} \\ \mathbf{a}_{s_2} \\ \mathbf{a}_{s_3} \end{pmatrix} = \begin{pmatrix} S_{11} & S_{12} & S_{13} \\ S_{21} & S_{22} & S_{23} \\ S_{31} & S_{32} & S_{33} \end{pmatrix} \begin{pmatrix} \mathbf{a}_{p_1} \\ \mathbf{a}_{p_2} \\ \mathbf{a}_{p_3} \end{pmatrix}, $$ where $S_{ij}\in\mathbb{Z}$ are the elements of the so-called supercell matrix.

The corresponding reciprocal space lattice vectors between the primitive cell and the supercell are related by: $$\tag{2} \begin{pmatrix} \mathbf{b}_{s_1} \\ \mathbf{b}_{s_2} \\ \mathbf{b}_{s_3} \end{pmatrix} = \begin{pmatrix} \overline{S}_{11} & \overline{S}_{12} & \overline{S}_{13} \\ \overline{S}_{21} & \overline{S}_{22} & \overline{S}_{23} \\ \overline{S}_{31} & \overline{S}_{32} & \overline{S}_{33} \end{pmatrix} \begin{pmatrix} \mathbf{b}_{p_1} \\ \mathbf{b}_{p_2} \\ \mathbf{b}_{p_3} \end{pmatrix}, $$ where $\overline{S}_{ij}=(S^{-1})_{ji}$.

Reciprocal space points. A $\mathbf{k}$-point in reciprocal space can be written in terms of the reciprocal space primitive lattice basis vectors or in terms of the reciprocal space superlattice basis vectors. The corresponding fractional coordinates are related by: $$\tag{3} \begin{pmatrix} k_{s_1} \\ k_{s_2} \\ k_{s_3} \end{pmatrix} = \begin{pmatrix} S_{11} & S_{12} & S_{13} \\ S_{21} & S_{22} & S_{23} \\ S_{31} & S_{32} & S_{33} \end{pmatrix} \begin{pmatrix} k_{p_1} \\ k_{p_2} \\ k_{p_3} \end{pmatrix}. $$ If the reciprocal space superlattice fractional coordinates $(k_{s_1},k_{s_2},k_{s_3})$ are all integers, then this $\mathbf{k}$-point is mapped to the $\Gamma$-point in the corresponding supercell.

There are an infinite number of supercells that map a given $\mathbf{k}$-point to the $\Gamma$-point, and one is usually interested in the smallest possible supercell. A general discussion of how to find this smallest possible supercell is presented in this paper [disclaimer: I am a co-author].

Your question. As an example, let's consider your question, where the $\mathbf{k}$-point of interest is the M point with reciprocal space primitive lattice fractional coordinates $(k_{p_1},k_{p_2},k_{p_3})=(\frac{1}{2},0,0)$. We need to find a supercell matrix $S$ such that this $\mathbf{k}$-point written in terms of the reciprocal space supercell lattice gives integer fractional coordinates $(k_{s_1},k_{s_2},k_{s_3})$. By inspection, we see that picking this supercell matrix will do the trick: $$\tag{4} \begin{pmatrix} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}. $$ This also happens to be the smallest such supercell. Other supercells that will also map M to the $\Gamma$-point (but that are not the smallest possible) include: $$\tag{5} \begin{pmatrix} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ or $$\tag{6} \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}. $$

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  • $\begingroup$ +1 as always! I just added equation labels in case anyone in the future wants to cite this answer and say refer to a specific equation numerically! $\endgroup$ Jan 11, 2022 at 21:44
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    $\begingroup$ @NikeDattani thanks! $\endgroup$
    – ProfM
    Jan 13, 2022 at 8:08

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