How could I fold a specific high symmetry point to the Gamma point?

Question

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?

Answer 1

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 $$\tag{6} \begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{pmatrix}. $$