Hi Abbout,

Thanks for the help.

-Leon

From: Abbout Adel <abbout.a...@gmail.com<mailto:abbout.a...@gmail.com>>
Date: Monday, October 17, 2016 at 1:59 PM
To: "Leon Maurer (lmaurer)" <lmau...@sandia.gov<mailto:lmau...@sandia.gov>>
Cc: "kwant-discuss@kwant-project.org<mailto:kwant-discuss@kwant-project.org>" 
<kwant-discuss@kwant-project.org<mailto:kwant-discuss@kwant-project.org>>
Subject: [EXTERNAL] Re: [Kwant] a step tripping up Kwant


Hi Leon,

The upper and bottom limits of the conduction band are obtained from the 
relation of dispersion.

E=V-2t *cos(k)                   (#  in your case V=2t for left lead and V0 for 
the right lead ).

For more details, you can look for example to :
http://www-personal.umich.edu/~sunkai/teaching/Fall_2014/Chapter6.pdf

Regards,
Adel

On Mon, Oct 17, 2016 at 10:34 PM, Maurer, Leon 
<lmau...@sandia.gov<mailto:lmau...@sandia.gov>> wrote:
Hi Abbout,

Is the upper bound on the energies in the leads documented somewhere? I guess 
it’s implicit in Sec. 2.4 of the tutorial, and now that you mention it, it 
makes sense given the periodic lattice.

-Leon

From: Abbout Adel <abbout.a...@gmail.com<mailto:abbout.a...@gmail.com>>
Date: Monday, October 17, 2016 at 11:58 AM
To: "Leon Maurer (lmaurer)" <lmau...@sandia.gov<mailto:lmau...@sandia.gov>>
Cc: "kwant-discuss@kwant-project.org<mailto:kwant-discuss@kwant-project.org>" 
<kwant-discuss@kwant-project.org<mailto:kwant-discuss@kwant-project.org>>
Subject: [EXTERNAL] Re: [Kwant] a step tripping up Kwant

Dear Leon,

the value of the parameter 't' in your program is  around 16. this means that 
the conduction band for the left lead is
band_l=[0, 4 t ]=[0,64] and the conduction band for the right lead is 
band_r=[V0, V0+4 t]=[100, 164]

as you can notice there is no energy which conducts in both leads. In order to 
have a non zero transmission you need to use a value of V0<4 t.

The result you are calling "exact" is valid for a continuous model: on a 
lattice, the dispersion relation is not quadratic.

To compare, your result with the continuous limit, you need to choose V0<< 2t.

For a "non uniform finite differences", you can look at the article [1]



Hope that this helps.
Adel

[1]: http://scitation.aip.org/content/aip/journal/jap/68/8/10.1063/1.346245

On Mon, Oct 17, 2016 at 7:10 PM, Maurer, Leon 
<lmau...@sandia.gov<mailto:lmau...@sandia.gov>> wrote:
Hello everyone,

I’ve been playing around with Kwant and come across some situations where the 
transmission between two leads is identically equal to zero when I wouldn’t 
expect that result.

I’ve come up with a simple working example: a 1D step function with a step 
height that’s large relative to the lattice spacing (code below), altho this 
problem seems to sometimes crop up in other, somewhat less extreme situations.

I understand that the numerical result should become less accurate as the 
step-height-to-lattice-spacing ratio increases, but why does the transmission 
become identically equal to zero at some point? Are there well-defined 
conditions for when this happens? Is there some way to know that the 
transmission is zero because of numerical issues rather than the underlying 
physics?

(Ultimately, I’m interested in modeling some systems where the potential mostly 
varies gradually but has a few small regions with abrupt changes in potential. 
Moving to a finer mesh (smaller lattice constant) everywhere is 
cost-prohibitive. Having some tool to easily refine the mesh in a region would 
be very useful.)

Thanks.

-Leon


(Below code taken from jupyter notebook.)


# In[1]:

get_ipython().magic('load_ext autoreload')
get_ipython().magic('autoreload 2')
from numpy import *
import matplotlib.pyplot as plt
get_ipython().magic('matplotlib inline')
import tqdm
import kwant


# In[2]:

m0 = 9.10938215e-31 # Electron mass, [kg]
hbar = 1.054571726e-34 # hbar in [J] [s]
q = 1.602176565e-19 # Elementary charge, [C]
mt = 0.19
ml = 0.92

m = mt*m0

# In[3]:

V0 = 100 # step height
x = linspace(0,100,30) # thirty grid points
U = zeros_like(x)
U[len(x)//2:] = V0
plt.plot(x,U)


# In[4]:

a = x[1]-x[0] # grid spacing [nm]
t = hbar**2/(2.*m*(a*1e-9)**2)/q*1e3 #hopping parameter [meV]

lat = kwant.lattice.chain(a) # Set up the transport simulation on a 1D latice
sys = kwant.Builder() # initialize the transport simulation
for i in range(len(U)): # populate based on the potential landscape
    sys[lat(i)]=U[i]+2*t

sys[lat.neighbors()] = -t # set the finite-difference hopping parameters
leftLead = kwant.Builder(kwant.TranslationalSymmetry((-a,))) # the lead to the 
left
leftLead[lat(0)] = 2*t + U[0]
leftLead[lat.neighbors()] = -t
sys.attach_lead(leftLead) # attach it

rightLead = kwant.Builder(kwant.TranslationalSymmetry((a,))) # the lead to the 
right
rightLead[lat(0)] = 2*t + U[-1]
rightLead[lat.neighbors()] = -t
sys.attach_lead(rightLead) # attach it
sys = sys.finalized()


# In[5]:

def plot_conductance(sys, energies):
    # Compute transmission numerically
    data = []
    for energy in tqdm.tqdm(energies,leave=True):
        smatrix = kwant.smatrix(sys, energy)
        data.append(smatrix.transmission(1, 0))

    # Compute exact conductance
    k1 = sqrt(2*m*energies/hbar**2)
    k2 = sqrt(2*m*(energies - V0)/hbar**2)
    T = 4*k1*k2/(k1+k2)**2
    T[energies <= V0] = 0

    plt.figure()
    plt.plot(energies, data, energies, T)
    plt.legend(('numerical','exact'), loc=4)
    plt.xlabel("energy [V0]")
    plt.ylabel("Transmission")
    plt.show()
    return data, T


# In[6]:

stepNumerical, stepExact = plot_conductance(sys,linspace(1e-9,2*V0,201))


# In[7]:

print(stepNumerical)




--
Abbout Adel



--
Abbout Adel

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