Hi,

I modified the 1st tutorial code to include the rectangular barrier (see my code in [2]). I have attached the figures i got from the code (conductance.png and rect_trans.png). If i have understood correctly, then Kwant outputs the conductance depending on excitation energy. And the conductance is the sum of the transmission probabilities of the conduction channels, right?

Yes, this is correct. You may also get transmission probablities between individual channels directly from the scattering matrix.

I have also calculated the number of conduction channels, but this goes to 0 at higher energies. Why does this occur? In the simple analytical case, there is no upper bound on electron energies.

This is because the tight-binding model corresponds to a discretization of the continuum model. If you do the analytical calculation for the actual (discretized) case that you have, you will find the dispersion relation `E(k) = 2 * t (1 - cos(k))` (t the hopping), which reduces to E(k) = k**2 (the continuum result) in the limit k->0 as it should. This is natural, as at large enough k, the wavelength of your solution will be of comparable size to the lattice spacing, and will start to "see" the discretization.

In addition, as you have created a quasi-1D system, as opposed to a purely 1D (chain) system, there will be an energy offset to the above dispersion relation for each band.

Is it even possible to recreate the analytical quantum tunneling result for the rectangle barrier in Kwant?

Of course. You only need to make sure that your discretization is fine enough for the range of energies (viz. wavelengths) that you wish to look at, if you wish to approximate well the continuum result.

Hope that helps,

Joe