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 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
> Date: Monday, October 17, 2016 at 11:58 AM
> To: "Leon Maurer (lmaurer)"
> Cc: "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 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 righ
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 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
Re: [Kwant] a step tripping up Kwant
Hi Abbout,
Thanks for the help.
-Leon
From: Abbout Adel mailto:abbout.a...@gmail.com>>
Date: Monday, October 17, 2016 at 1:59 PM
To: "Leon Maurer (lmaurer)" mailto:lmau...@sandia.gov>>
Cc: "kwant-discuss@kwant-project.org<mailto: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
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 mailto:abbout.a...@gmail.com>>
Date: Monday, October 17, 2016 at 11:58 AM
To: "Leon Maurer (lmaurer)" mailto:lmau...@sandia.gov>>
Cc: "kwant-discuss@kwant-project.org<mailto: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
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.t
Re: [Kwant] a step tripping up Kwant
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 mailto:abbout.a...@gmail.com>>
Date: Monday, October 17, 2016 at 11:58 AM
To: "Leon Maurer (lmaurer)" mailto:lmau...@sandia.gov>>
Cc: "kwant-discuss@kwant-project.org<mailto: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
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
