Dear Kristjan,

>>> " And the conductance is the sum of the transmission probabilities of the conduction channels, right? "

Yes this is correct.

The number of conducting channels goes to zeros at higher energies because you work in a lattice: the dispersion relation is not anymore E=k^2 .

In a lattice, the dispersion relation for a mode 'm' is E_m=4t-e_m-2*cos(k) (where e_m=-2*cos(m*pi/(W+1)))

in order to recover the results of the continuum, you need to work at low energies.

in your case, you need to change the theoretical relation by putting a shift 2-2*cos(pi/(W+1)) for the energies and this will work for the first mode. you need to do the same thing (shift 2-e_m) for the other modes and make a sum of the transmissions to obtain the results for higher modes.

a script is included below to show how we recover the exact result.

I hope that this helps

Adel

import kwant

from numpy import sqrt,sin,cos,pi,sinh

# For plotting

from matplotlib import pyplot

def make_system(a=1, t=1.0, W=10, L=30,V0=0):

lat = kwant.lattice.square(a)

sys = kwant.Builder()

sys[(lat(x, y) for x in range(L) for y in range(W))] = 4*t+V0

sys[lat.neighbors()] = -t

#### Define and attach the leads. ####

lead = kwant.Builder(kwant.TranslationalSymmetry((-a, 0)))

lead[(lat(0, j) for j in range(W))] = 4 * t

lead[lat.neighbors()] = -t

sys.attach_lead(lead)

sys.attach_lead(lead.reversed())

return sys

def plot_conductance(sys, energies):

# Compute conductance

data = []

for energy in energies:

smatrix = kwant.smatrix(sys, energy)

data.append(smatrix.transmission(1, 0))

# pyplot.figure()

pyplot.plot(energies, data,lw=2,label="numerics")

pyplot.xlabel("well depth [t]")

pyplot.ylabel("conductance [e^2/h]")

pyplot.legend()

pyplot.show()

def main():

V0=0.1

sys = make_system(V0=V0)

# Check that the system looks as intended.

# kwant.plot(sys)

# Finalize the system.

sys = sys.finalized()

e1=2-2*cos(pi/11) #energy of the first transverse mode

def Cond(E,V0,L=30):

if E-e1-V0<=0:

k=sqrt(-E+e1+V0)

return 1/(1+((V0*sinh(k*L))**2 )/(4*(E-e1)*(-E+e1+V0)))

else:

k=sqrt(E-e1-V0)

return 1/(1+((V0*sin(k*L))**2 )/(4*(E-e1)*(E-e1-V0)))

energies=[0.1+0.001 * i for i in range(300)]

Cond_analytic=[Cond(E,V0) for E in energies]

pyplot.plot(energies,Cond_analytic,"ro",label="analytic")

# We should see conductance steps.

plot_conductance(sys, energies)

# Call the main function if the script gets executed (as opposed to imported).

if __name__ == '__main__':

main()

On Mon, Dec 19, 2016 at 4:25 PM, Kristjan Eimre <kristjaneimre@gmail.com> wrote:

Hello again,Thanks for the ideas. I have fiddled around with Kwant a bit now, and I'm trying to recreate quantum tunneling through rectangular potential barrier (see [1] for analytical solution). I have also added a figure of the analytical solution (rect_trans.png), which shows the transmission probability dependence on electron energy.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?I probably should delve deep into books on quantum transport to understand some of these things, but perhaps you could give me some answers for a good start. 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.Is it even possible to recreate the analytical quantum tunneling result for the rectangle barrier in Kwant?Best regards,

Kristjan

Abbout Adel