a puzzle about "Why are so many k-points needed in the transport direction"

[njuxyh]

Hi all:

i saw the notes "Why are so many k-points needed in the transport direction in a device calculation?"

and there is a sentence  "Now, clearly you will only get perfect transmission in this system if HL=HR=HCR. To obtain this, the same k-point sampling should be used in the A and B directions in the electrode and the open boundary condition calculation",

i do not very understand why the K-Point sampling in the A and B directions?

for example, in my device, the electrode cell lattice is 15 angstrom in the x direction, 45 angstrom in the y direction, the lattice in the x and y direction are different,  so when i  relaxed the central region geometry as a device,  how to set the the k-sample,  also should set the same k-sampling in the x and y direction? if so, i think it is some strange!

or? only calculate the transmission spectrum, i should set the same k-sampling in the x and y direction?

i need your help!

thanks very much!! 

[Anders Blom]

The point is just that you must use the same sampling (kA,kB) (so, for instance kA=6 and kB=3) in the left electrode, right electrode, and in the central region.

As for the transmission calculation, you should generally use more k-points than for the selfconsistent calculation, since the Green's functions (which determine the transmission) have a stronger variation with k than the Fourier components of the electron density do.

[njuxyh]

OK, I GOT IT,

THANKS VERY MUCH!