QuantumATK Forum
QuantumATK => General Questions and Answers => Topic started by: yang su on November 29, 2012, 07:42
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I am kind of confused...The boundary conditions are set by determining the k points sampling right? What does the k-point sampling mean in both electrode and transmission spectrum part? My understanding is that if I set the k-point sampling to be 1 in a direction, in that direction the calculation will be infinite...
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"The boundary conditions" are defined by the lattice vectors of unit cell. The k-point sampling is done within the Brillouin zone. The unit cell and the Brillouin zone have an exact correspondence.
The number of k-point sampling and the boundary conditions do not have an exact correspondence, because they are two different concepts.
" if I set the k-point sampling to be 1 in a direction, in that direction the calculation will be infinite."
This is not true.
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The opposite is however true - if the system is infinite in some direction, you only need one k-point in that direction.
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Thank you but I have searched through this forum and that I got this answer to some post
"In X and Y, you can basically have two situations: the structure can be finite (like a nanotube, or a ribbon). Since the model is anyway periodic, you add some vacuum so the periodic copies don't interact. In this case, 1x1x100 would be the appropriate k-point sampling.
If you have periodicity in X and Y, like if the electrodes are formed by a cleaved crystal surface. Then you need NAxNBx100 k-points, and how many NB and NB depends on the length of the period - the shorter the period length, the more k-points. So 1 might be still ok if the period is really long, which usually means you make several repetition of the system.
You can also have a mixture in X and Y, like finite in X and periodic in Y. This would be the same of a graphene sheet in the YZ plane. Then 1xNBx100 would be an appropriate sampling, with the same rules applying for NB as just above.
So in short: for a graphene sheet - 1x9x100 (or 12, or 15, etc). For a ribbon, 1x1x100."
Does that mean if I have periodicity and infinity in some direction I need to use multiple k-points in that direction and if it is finite in some direction I can just use 1 k-point in that direction?
And do I need to use neumann boundary conditions in the finite direction? Thank you
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First question yes.
About Neumann - yes, in principle you could use Neumann for a finite system. However, the multigrid Poisson solver is much slower than FFT, and as long as your simulation cell is large enough, the difference should not be noticeable. So, we don't recommend using the Neumann (or Dirchlet) boundary conditions except if you have gates in the system simply from a performance perspective.
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Ok, but in my previous simulation where I did not use a neumann boundary condition in the direction I have gates, the simulation did not converge. However, after the neumann boundary conditions added and multigrid solver used it converged better. So can I say that if I have gates in some direction it is better to use neumann and multigrid and if there are no spatial regions I can simply add vacuum at the edge of the cell and use single k-point? Thank you
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You are still slightly mixing the concepts. If your system is finite in one direction, you can use a single k-point (it has nothing to do with gates or not). But yes, if you have gates, you need multigrid.
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So how can I tell if a direction is finite or infinite? By boundary conditions or the vacuum added? Thank you
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??? If you know what your system looks like, it should be rather obvious... A hint: a nanowire is finite in 2 dimensions and periodic in 1.
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Sorry but I mean how can I "mark" in the script that a direction is infinite?
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ATK doesn't contain an special object classes for 1D or 2D objects - in fact all calculations are carried out in 3D. So, if you want the system to be effectively finite in any direction, you add some vacuum around it to make sure it doesn't have any basis set overlap with the repeated copy in that direction, and then a bit extra so that the electrostatic influence is minimized.
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I think I got it..Thank you very much
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Can anyone please help me to find the reciprocal coordinates of the K and M points in a simple hexagonal Brillouin zone? I have the lattice vectors and the reciprocal vectors, but I can't get the right coordinates for the K point (the corner of the hexagon) and the M point (half way between 2 K points).Is there any formula for it?
Thanks.
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Here are the co-ordinates of High-symmetry points in the BZ (in reciprocal lattice vectors)
Gamma = 0 0 0
M = 0 1/2 0
K = 1/3 1/3 0
H= 1/3 1/3 1/2
A= 0 0 1/2
L= 0 1/2 1/2