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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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  • $\begingroup$ Welcome to our new community! Why don't you register your account? Please take a look at the edits that I made to your question and keep them in mind for next time! $\endgroup$ Aug 1 at 3:04
  • $\begingroup$ Thanks, I registered :) $\endgroup$ Aug 1 at 18:54
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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
    Aug 4 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? $\endgroup$ Aug 4 at 16:29
  • $\begingroup$ This you can decide now! $\endgroup$ Aug 5 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. $\endgroup$ Aug 5 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? $\endgroup$ Aug 13 at 14:41

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