Help understanding square root notation of supercell

Question

I often see supercells and slabs explained like this:

• $2\sqrt{2} \times 6$ slab of $\ce{MoS2}$ (001)

• $\sqrt{2} \times \sqrt{2} \times 2$ supercell of $\ce{Y Ba(1−x)Sr(x)CuFeO5}$ (see here).

Take $\ce{MoS2}$, for example. Can someone please help me understand how we can have $\sqrt{2}$ of a unit cell for this case? Thank you.

Answer 1

The main idea here is to preserve the periodic boundary conditions when creating a fractional supercell. Therefore, not all combinations are possible, such as $\sqrt{2}\times\sqrt{2}$ or$\sqrt{5}\times\sqrt{5}$ for the case of MoS$_2$.

Basically, you need to define a rotation matrix for creating such supercells. e.g., B=RA (in matrix form), where A is the original lattice vectors (in matrix form), R is the rotation matrix and B is the result.

For example, one can define a rotation matrix R to create $\sqrt{3}\times\sqrt{3}R30^{0}$ supercell as following:\begin{bmatrix}2 & 1 & 0\\-1 & 1& 0 \\0 & 0& 1\end{bmatrix}

I showed here three pictures for comparison of a fractional supercell of Sqrt3xSqrt3R30 to normal cells, see how the supercell looks like and how you define matrix. So, to make a cell of size Sqrt3xSqrt3R30, you traverse 2 time x and then 1 time y, making [2 1 0] the first line of matrix, similarly, 1 times in -x and 1 time y makes [-1 1 0] the second line of matrix. You can't have any combination giving Sqrt2xSqrt2 with periodic boundary conditions. Hope it will be clear now.

Answer 2

If you want to achieve this using VESTA you can follow what Shahid said in his answer but you have to keep in mind how lattice vectors are defined in VESTA which in turn will change the rotation matrix. For example I show here an unit cell of TaS2, notice how a and b vectors are defined on the lower left corner

Then look at the left side of the transformation matrix option in VESTA, se how they are defined, which matrix element are getting multiplied with which lattice vector.

So the correct matrix for VESTA would be:

\begin{bmatrix}1&1&0\\-1&2&0\\0&0&1\end{bmatrix}

Which will give you the correct root3 x root3 x 1 supercell