In the tutorial zerothi/ts-tbt-sisl-tutorial/ts_02, it is mentioned that "applying a bias to a bulk system is wrong" and "if you can't immediately figure this out, try and create a longer system by replacing device = elec.tile(3, axis=0) with, say: device = elec.tile(6, axis=0) and redo the calculation for a given bias. then compare the potential profiles."

I know that applying a bias to a pristine system has no physical meaning, but why is it also wrong to apply a bias to a bulk system? From the hint given, it seems that the author simply increased the size of the device model by a factor of two along the current direction. Is this error related to ballistic transport, i.e., when the device size is no longer much smaller than the electron's mean free path? Is my understanding correct?

Here are their potential profiles(the potential in the figures has been amplified by a factor of ten for better visualization). I cannot discern the meaning the author is trying to convey.

device = elec.tile(3, axis=0) device = elec.tile(6, axis=0)


1 Answer 1


The thing is that it will be experimentally impossible (at least to my knowledge) to attach electrodes of the same pristine device structure without disrupting the electronic structure of the electrodes.

Imagine a gold wire where you want to alter the potentials at some fixed positions on the wire. It would require you to attach something to the wire at the positions of the electrodes. This would effectively change the electronic structure at those given positions, meaning it is not the system you'll be simulating.

Lastly, the key point about increasing the length of the system is that the potential drop changes depending on the length of the system since there is no scattering. So it becomes a bit dubious to create a device with bias for a pristine system since it requires 1) non-interaction attachment of a battery and 2) the exact placement of the batteries at the distances you simulate since other placements would result in a different potential profile.

  • $\begingroup$ Thank you for your patient explanation, I understand now. $\endgroup$
    – Ming
    Apr 13 at 13:52

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