Author Topic: Perturbed MoS2 monolayer  (Read 4226 times)

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Offline ams_nanolab

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Perturbed MoS2 monolayer
« on: November 3, 2012, 07:00 »
I am looking to find out the effect of perturbations like ripples, twists etc. of a monolayer MoS2 sheet on the system Hamiltonian. If ATK treats such rippled twisted sheets like a supercell, then does the symmetry points of the system changes from that of an unit cell? If so then how do we know the new symmetry points and their relation to the original(unperturbed) system?

Offline zh

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Re: Perturbed MoS2 monolayer
« Reply #1 on: November 5, 2012, 01:17 »
In principle, it should change as the perturbations.  For example, the flat graphene nanoribbon is bent into a nanotube. There are many studies on the rippled and twisted graphene sheet. You can refer to them.
« Last Edit: November 8, 2012, 01:19 by zh »

Offline Anders Blom

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Re: Perturbed MoS2 monolayer
« Reply #2 on: November 6, 2012, 17:53 »
About the symmetry points, these are not really related to the basis (the positions of the atoms inside the cell) but rather the shape of the cell, i.e. the symmetries of the Bravais lattice. So the ripples etc will not change the location of the high symmetry points, although some degeneracies may of course break as a result of the distortions.

It can be difficult to keep track of symmetries when making super cells. For instance, you can easily convert a hexagonal cell to orthorhombic with a simple transformation (A'=2A+B, B'=B, C'=C), but since it's a super cell there will be zone folding, and the symmetry points attain new labels and new coordinates. If a,c are the original hexagonal lattice constants, the orthorhombic lattice vectors are sqrt(3)*a*x, a*y, c*z.

I will leave it as an exercise to prove that the symmetry points of the hexagonal lattice are mapped onto the symmetry points of the simple orthorhombic lattice as follows:

M -> S
L -> R
A -> Z
K -> a point between G and Y
H -> a point between T and Z