`lead.cell_hamiltonian()` gives 32000. The thickness of my slab is 50. Is this something not possible to calculate?

Ok, now it certainly makes sense that the memory blows up; the kwant paper[1] indicates
that the memory requirement for solving this sort of size should be ~ 100GB.

Anton reminded me of a workaround written by Christoph for dealing with periodic
systems. You can find the code over here[2] (if you don't want to download it
using git you can get the raw python module here[3]). You should be able to
use this instead of replicating your unit cell in the y direction, which should drastically
reduce your memory/runtime requirements.



[1]: http://downloads.kwant-project.org/doc/kwant-paper.pdf
[2]: https://gitlab.kwant-project.org/cwg/wraparound
[3]: https://gitlab.kwant-project.org/cwg/wraparound/raw/master/wraparound.py