Mailman 3 April 2017 - Kwant-discuss

Query regarding Carbon Nano Tube (CNT) simulation using Kwant 1.2.2
by Biswadeep Upadhyay 20 Apr '17

20 Apr '17

Hello Sir, It's Biswadeep from India. I'm an electronics engineering student currently persuading my study on electronic transport in nano devices. My recent area of focus is to simulate the system and lead structure of Carbon nano tube. So far my approach has not given me the edge . As i'm new with this specific tube structure , I'm failing to come up with the python code needed to simulate CNT. So I need a guidance on how to do it. If possible any source link to find code for such tube structures would definitely help in this regard. So I request you to help me with your guidance. Regards, Biswadeep Upadhyay

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installing kwant 1.1.2 for python 2.7 on the enthought canopy environment
by Peotta Sebastiano 19 Apr '17

19 Apr '17

Hi all, I am a new user of kwant and I have some troubles in installing kwant on the Enthought Canopy environment. I have always used this environment for my numerical work with Python and it is annoying to change it. The downside is that it only supports Python 2.7 and not the latest Python 3.x. This is the source of problems. I follow the instructions in this webpage <https://support.enthought.com/hc/en-us/articles/204469690-Installing-packag…>. Essentially they ask to run the command 'pip install kwant' within their Canopy Terminal. If I try to do this I get the following error message """"" (Canopy 64bit) peottas1@penrose:/m/home/home5/50/peottas1/unix$ pip install kwant Collecting kwant Downloading kwant-1.2.2.tar.gz (820kB) 100% |████████████████████████████████| 829kB 996kB/s Complete output from command python setup.py egg_info: This version of Kwant requires Python 3.4 or above. Kwant 1.1 is the last version to support Python 2. ---------------------------------------- Command "python setup.py egg_info" failed with error code 1 in /tmp/pip-build-VoZTHV/kwant/ """ According to the install instructions in the Kwant homepage the pip command should automatically install the Python 2 version, while the command pip3 should install the Python 3 version. However this does not seem the case. Moreover if I try to do a search with pip I get the following """ (Canopy 64bit) peottas1@penrose:/m/home/home5/50/peottas1/unix$ pip search kwant kwant (1.2.2) - Package for numerical quantum transport calculations (Python 3 version) """ so it seems that only kwant-1.2.2 is available for download. On the contrary in the webpage "https://pypi.python.org/pypi/kwant" both version 1.1.2 and 1.2.2 are available. Can you give me some hint on how to solve the problem? Best, Sebastiano

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Wrapping leads around a channel
by LaGasse, Samuel 13 Apr '17

13 Apr '17

Hi all, I am working on trying to create a system where there is a lead which wraps around the edge of a sample. For example, I have included a code which generates a square scattering region with two contacts, one extending vertically and the other extending horizontally. What I would like to do is combine these two contacts into a single contact, so this might represent a system where there is a contact that has been etched surrounding a central region. I think that the way to approach this would be to generate one wide contact, whose width is equal to the combined with of the contacts in my code (40 in my example). This contact would then be attached to the channel and its interface would be wrapped around the channel. Unfortunately, I am at a bit of a loss as to how I would code this. My main reason for wanting to do this rather than treat the problem with two separate leads is that the modes are going to be different between the two. Ultimately, it would be nice to be able to do this in a way that wraps the lead completely around the outer edge of the channel to simulate some sort of open boundary condition. Additionally, it would be nice to do something similar with a hole in the middle of the scattering region, to construct a Corbino disk. If anyone as any advice, I would love to hear it! Thanks, Sam LaGasse Here is the code: import kwant import matplotlib.pyplot as plt lat = kwant.lattice.square(a=1) s = 20.0 t = 1 def square(pos): x, y = pos return x <= s/2 and x >= -s/2 and y <= s/2 and y >= -s/2 y_min = -10. y_max = +10. x_min = -10. x_max = +10. def y_contact(pos): x, y = pos return x >= y_min and x <= y_max def x_contact(pos): x, y = pos return y >= y_min and y <= y_max sys = kwant.Builder() sys[lat.shape(square, (0,0))] = 4*t sys[lat.neighbors()] = -t sym0 = kwant.TranslationalSymmetry((-1,0)) sym_down = kwant.TranslationalSymmetry((0,1)) lead0 = kwant.Builder(sym_down) lead0[lat.shape(y_contact, (0, 0))] = 0 lead0[lat.neighbors()] = -t lead1 = kwant.Builder(sym0) lead1[lat.shape(x_contact, (0, 0))] = 0 lead1[lat.neighbors()] = -t sys.attach_lead(lead0) # top sys.attach_lead(lead1) # left kwant.plot(sys) plt.show()

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combining two s matrices
by Prof. CHAN Kwok Sum 13 Apr '17

13 Apr '17

Dear Kwant developer, Is it possible to combine the s matrices for two structures to obtain the s matrix of a large system? The evanescent modes in a lead can be excited, does the s matrix or some other object contain the evanescent mode information so that it can be used for combining two s matrices? KS Chan Disclaimer: This email (including any attachments) is for the use of the intended recipient only and may contain confidential information and/or copyright material. If you are not the intended recipient, please notify the sender immediately and delete this email and all copies from your system. Any unauthorized use, disclosure, reproduction, copying, distribution, or other form of unauthorized dissemination of the contents is expressly prohibited.

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Making 3D leads in a monoatomic lattice
by John Eaves 13 Apr '17

13 Apr '17

Good day I am trying to make a simple 3D geometry; a cylinder. Now, the only tutorial available for 3D is https://kwant-project.org/doc/1.0/tutorial/tutorial6#d-example-zincblende-s… so there is a chance my problems come from that. The thing is that I cannot seem to attach any leads. The scattering region code works just fine, but when I try to attach the leads I get the error Site family <unnamed Monatomic lattice, vectors [0.0 1.0 1.0], [1.0 0.0 1.0], [1.0 1.0 0.0], origin [0.0 0.0 0.0]> does not have commensurate periods with symmetry <kwant.lattice.TranslationalSymmetry object at 0x00000000093B17F0>. The error is not completely opague of course; something seems to be problematic about the lattice vectors and the proposed symmetry. For that it is perhaps good to include the full code: def make_system(a=1, t=1.0, W=10, L=5, r2=20): lat = kwant.lattice.general([(0, a, a), (a, 0, a), (a, a, 0)]) sys = kwant.Builder() def ring(pos): (x, y,z) = pos rsq = x ** 2 + y ** 2 return rsq < r2 ** 2 and 0 <= z < L sys[lat.shape(ring, (1, 1,1))] = 4 * t sys[lat.neighbors()] = -t sym_lead = kwant.TranslationalSymmetry((0, 0,-a)) lead = kwant.Builder(sym_lead) def lead_shape(pos): (x, y,z) = pos rsq = x ** 2 + y ** 2 return rsq < r2 ** 2 lead[lat.shape(lead_shape, (0, 0,0))] = 4 * t lead[lat.neighbors()] = -t sys.attach_lead(lead) sys.attach_lead(lead.reversed()) return sys I personally do not see what is wrong with this, especially as the scattering region looks fine without the leads. According to the error there is something wrong with the lattice vectors themselves? Should I change them in some way? I don't care about any particular structure here really, I am just looking for the 3D equivalent of the square lattice and according to the source code that is defined as def square(a=1, name=''): """Make a square lattice.""" return Monatomic(((a, 0), (0, a)), name=name) so I figured my definition should work for 3D. Could someone point out the obvious mistake I am making?

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transverse spin currents
by puromate 11 Apr '17

11 Apr '17

HI, I'm interested in computing transverse spin currents and/or something related to spin-Hall conductivity in spin-orbit coupled wires. Any example out there? Thanks in advance. Best, Jose

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Finite temperature, finite bias conductance, and units of energy
by John Eaves 11 Apr '17

11 Apr '17

Hello, My question is based on the reply to https://www.mail-archive.com/kwant-discuss@kwant-project.org/msg00983.html. Here it is written that one can extend the KWANT zero temperature infinitesimal source drain bias conductance to finite bias, finite temperature by numerical integration. The answer refers the user to the Datta book to see how exactly this is done. However, I have looked at the Datta book and I cannot figure out how to implement it in KWANT. My problem has a few parts, so below I will build it up step by step so that one can see if perhaps I make a mistake somewhere. However, I can start with a very simple question: In order to account for finite temp, you need the fermi dirac distribution (or its derivative). Inside the distribution we have k_B*T; how do you rescale k_B*T to agree with the units of the kinetic term? The kinetic term is t = hbar^2/(2m a^2). If I set t = 1, what do I do to kB*T? I don’t understand this as a (the lattice constant) enters the expression, but changing a and keeping t fixed does not change any of my results. So somehow I need to relate hbar^2/2m with kB*T as these are the fixed constants? It is the fact that a enters into t but not into kb*T that makes it difficult to understand I think. Now, for my main question, lets start from the basics. If I am not mistaken, in the default zero T zero bias case KWANT essentially calculates G = 2e^2/h * T(E_fermi), where T(E_fermi) is the quantity calculated from smatrix = kwant.smatrix(sys, energy) T = smatrix.transmission(1, 0) This makes sense in the context of linear response. Now if we want to go to finite temp, finite bias we take the expression from Datta I = 2e/h \int{dE T(E) (f_L(E) - f_R(E))} where L,R is the left and right leads at potentials \mu_L, \mu_R abd f_L is then the fermi-dirac distribution of the left lead. Now this is the current; we don’t want the current, we want the conductance. So let us set \mu_L = E_f – eV_sd/2 and \mu_R = E_f + eV_sd/2 (which I suppose one should be able to do without loss of generality, but correct me if I am wrong). If we then want to find the conductance we simply take the derivative with respect to V_sd giving us G = 2e^2/h \int{dE T(E) d(f_L(E) - f_R(E))/dVsd} One simple limit here is to take T = 0, in which case the fermi-dirac distribution becomes a step function, its derivative a delta function, and we (not forgetting to carry the minus sign and the ½) get that G = e^2/h (T(E_f+V_sd/2) + T(E_f-V_sd/2) So if I am not mistaken, one can calculate the finite-bias zero-temperature conductance without numerical integration; you simply have to evaluate smatrix.transmission(1,0) at two different energies.This seems correct as it recovers the standard KWANT expression for zero bias. Now for finite temperature and zero bias things are different of course. One can taylor expand (f_L(E) - f_R(E)) as df_L/dV_sd *eV_sd = -df_L/dE *eV_sd so that G = e^2/h \int{dE T(E) -df_L(E) /dE} Now here things get a bit tricky: the interval is over all energies. Of course the fermi-dirac derivative is strongly peaked around E_fermi, but numerically this will still be annoying, will it not? If I were to numerically calculate the above current I’d need to evaluate smatrix.transmission(1, 0) for a huge range of energies, right? Could you tell me something about how one would go about this in a smart way? Do you use the trapezoid method? How do you keep it from taking ages? What I would come up with myself is something like data = [] integ = [] for energy in energies: for erange in np.linspace(0.5*energy,1.5*energy,num=100): smatrix = kwant.smatrix(sys, erange,args=[params]) integ.append(smatrix.transmission(1,0)*-dfd(erange,energy,T)) data.append(1.5*energy/100*(integ[0]+integ[100-1]+2*sum(integ[1:100-2]))) integ = [] where in the above dfd(energy,mu,temp) is the derivative of the fermi-dirac distribution. But this seems horribly inefficient, and I don't see how this would recover the KWANT limit (as we get some delta-like peak around E_f), but perhaps that is too much to ask. Finally we return to the full expression for finite bias and finite T: G = 2e^2/h \int{dE T(E) d(f_L(E) - f_R(E))/dVsd} I suppose evaluating this is equally difficult/easy as finite T with zero bias; all that changes is that one now has to calculate two derivatives of the fermi-dirac distribution which will form some sort of window around E_fermi. Here my question this repeats itself; how does one compute such an integral efficiently? Kind regards, John

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Lead self-energy in Green Function method
by 胡勇亮 10 Apr '17

10 Apr '17

Dear Sir/Madam， Hello, I am interested in Green Function, so I read related Kwant codes (kwant.solver.greens_function, make_linear_system, solve_linear_system, etc.). I found a lot of optimizations in Kwant, such as LU Decomposition and Matrix Partition, which make Kwant more efficient. To calculate Green Function, we need to calculate the self-energy of leads. Though I know how to get self-energy by using command "lead.selfenergy(energy)". I want to know how this command works, the codes in "leads.py" are too diffcult to me. Sincerely, Hu Yongliang

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Impurity Average - salt
by Camilla Espedal 07 Apr '17

07 Apr '17

Dear Kwant-team, I have a question regarding impurities. I have a system where I want to add a random impurity potential between -w/2 and w/2. I do this using w_imp*uniform(repr(s), salt)*s0 - w_imp/2 * s0 which seems to be working fine. Then, I want the impurity averaged conductance. So I make a list of salts using range(0,200) and loop through the salts, and find the smatrix for all of them: smatrix= kwant.smatrix(syst, energy=energy, args=(str(salts[n])) and then calculate the transmission for each salt and take the average. I guess I just want to know if this is the correct way to do it? Have I understood correctly how salt and this uniform-function works? Best regards, Camilla

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kwant.builder.BuilderLead
by Татьяна Григоренко 05 Apr '17

05 Apr '17

Hi, sorry to bother you again. How interface parameter of class kwant.builder.BuilderLead(builder, interface) is specified? How to define a sequence of sites which were attached to the lead? С уважением, Татьяна Григоренко

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