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.

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2 Answers 2


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. sqrt3 supercell diagrams 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.

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    $\begingroup$ It would probably be better to consolidate this into your existing answer rather than having two separate answers. $\endgroup$
    – Tyberius
    Commented Aug 4, 2021 at 14:05
  • $\begingroup$ I'll accept this as the correct answer for now. Thank you so much for your clarification! Just one quick question - it looks like you rotate 30 degrees to the left here (counter-clockwise), is this correct? $\endgroup$ Commented Aug 4, 2021 at 16:29
  • $\begingroup$ This you can decide now! $\endgroup$ Commented Aug 5, 2021 at 13:18
  • $\begingroup$ Ya, it looks like it, but I think the point of this website is to facilitate dual (often redundant) agreement on a result, for the sake of others' clarity. $\endgroup$ Commented Aug 5, 2021 at 15:43
  • $\begingroup$ When I apply the above transformation matrix to a 1x1 cell of MoS2, I do not get the sqrt(3) super cell you show. Can you please try it with VESTA? $\endgroup$ Commented Aug 13, 2021 at 14:41

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

TaS2 unit cell

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.

VESTA transformation

So the correct matrix for VESTA would be:


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

TaS2 root3 x root3 x 1 supercell

  • $\begingroup$ Just saw this .. thanks! $\endgroup$ Commented Jun 29, 2022 at 0:50

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