Mailman 3 September 2014 - Kwant-discuss

How the spin is positioned in the Smatrix?
by Qingtian Zhang March 6, 2022

March 6, 2022

Dear all, I am trying square lattice with spin-orbit coupling and Zeeman splititng just like the example of Tutorial 2.3.1. Now I want to get the spin-dependent conductance：Guu,Gud,Gdu,and Gdd, where "u" indicates up spin, d indicates down spin. We can have the scattering matrix through Smatrix=kwant.smatrix(sys, energy), but how the spin is positioned in the scattering matrix? I have checked some simple cases, it seems the Smatrix is organized like this: | r t | S= | t' r' | but how the spin is included? Thanks in advance. Qingtiian Zhang The code is: import kwant from matplotlib import pyplot from numpy import matrix import tinyarray sigma_0 = tinyarray.array([[1, 0], [0, 1]]) sigma_x = tinyarray.array([[0, 1], [1, 0]]) sigma_y = tinyarray.array([[0, -1j], [1j, 0]]) sigma_z = tinyarray.array([[1, 0], [0, -1]]) sys=kwant.Builder() lat=kwant.lattice.square() a=1 t=1.0 alpha=0.5 e_z=0.08 W=2 L=2 sys[(lat(x, y) for x in range(L) for y in range(W))] = \ 4 * t * sigma_0 + e_z * sigma_z # hoppings in x-direction sys[kwant.builder.HoppingKind((1, 0), lat, lat)] = \ -t * sigma_0 - 1j * alpha * sigma_y # hoppings in y-directions sys[kwant.builder.HoppingKind((0, 1), lat, lat)] = \ -t * sigma_0 + 1j * alpha * sigma_x #### Define the left lead. #### lead = kwant.Builder(kwant.TranslationalSymmetry((-a, 0))) lead[(lat(0, j) for j in xrange(W))] = 4 * t * sigma_0 + e_z * sigma_z # hoppings in x-direction lead[kwant.builder.HoppingKind((1, 0), lat, lat)] = \ -t * sigma_0 - 1j * alpha * sigma_y # hoppings in y-directions lead[kwant.builder.HoppingKind((0, 1), lat, lat)] = \ -t * sigma_0 + 1j * alpha * sigma_x #### Attach the leads and return the finalized system. #### sys.attach_lead(lead) sys.attach_lead(lead.reversed()) kwant.plot(sys) sys=sys.finalized() print "Hamiltonian=" print sys.hamiltonian_submatrix() Smatrix=kwant.smatrix(sys, energy=3.) print "The scattering matrix=" print matrix.round((Smatrix.data),2) print "T=",(Smatrix.transmission(1, 0))

3 3

How to caculate hall resistance
by ZHANG Bing Sept. 12, 2021

Sept. 12, 2021

Dear Sir, I am a PhD student of Hong Kong University of Science and Technology. I want to use KWANT to caculate Hall resistance of a Hall bar structure.We can get the conductance between 6 electrodes, but how to get hall resistance? Can you give me some help? Thank you very much. Best Regards, Zhang Bing

4 3

Density of States and Fermi-Dirac
by Daniel Rodgers-Pryor Oct. 24, 2014

Oct. 24, 2014

Hi, Firstly, thanks for creating Kwant - it's so nice to use physics code written by people who understand software-engineering as well as physics :) I've got a few questions about units and density-of-states in Kwant, please respond if you know anything about any of them; don't feel the need to respond to them all at once. I'm trying to add self-consistent electrostatics to my Kwant system (using FiPy as a finite-element/volume Poisson-solver). Obviously, I need to calculate the electron-carrier-density of the system by integrating over the Fermi-Dirac occupation, then feed that (via real-space basis functions) to my Poisson solver. I'm not quite sure how to handle the density of states in the context of Kwant: I assume that just summing over the LDOS (integrating over all space) will give the density of states as a function of energy, D(E). Plotting it seems to produce reasonable bands, but I'm not quite sure about the units, or how it scales with system size. In the system I'm modelling (low-temperature p-donors in silicon), every lattice site adds an electron to the system (the temperatures are low enough that the silicon is frozen-out as a conductor and can just be treated as a background dielectric constant). I should be able to integrate over the density-of-states until the total equals the (known) number of electrons, but the density of states obtained by summing over the LDOS calculated by Kwant does not scale properly with the number of sites in the system; larger systems always need higher Fermi energies, which isn't physical at all. What am I missing here? Are the units of the LDOS Kwant calculates somehow normalised? How can I get a density-of-states which scales appropriately with the total number of electrons/sites in my system? The lead unit-cell of my system will need to be solved self-consistently too; how can I calculate the local density of states (and thus, via Fermi-Dirac, the electron-density) of a lead? Am I correct in assuming that the LDOS produced by Kwant is equivalent to summing over the state-density-weighted scattering-wavefunctions from the modes in all leads (and thus that integrating it over the occupied-energies will produce a sensible total electron-density)? Finally, and slightly unrelated, do my chosen energy-units need to be accounted-for anywhere in Kwant's Schroedinger-solutions? I'm writing my Hamiltonian terms in meV; will bands, LDOS etc. all naturally scale to make this choice transparent? Similarly, does effective electron-mass need to be accounted for at all? Thanks so much for your help, Daniel R-P

2 1

Fwd: About graphen in KWANT
by ANANT VIJAY Sept. 26, 2014

Sept. 26, 2014

Dear authors, I am trying to simulate graphene using KWANT.But I am not sure about on site energies and hopping paramater.I am sending you the code Please tell me how to impregnate these all. ANANT VIJAY VARMA IISER KOLKATA INDIA

2 1

Infinite Anti-ferromagnetic leads
by Collins Akosa Sept. 14, 2014

Sept. 14, 2014

Hello, I have been trying to build an antiferromagnetic g-type lead with the code attached below. It seems i have a problem with the band-structure because i was expecting to have a ferromagnet if if set m1 = m2 = 0.0 in the code below. (But apparently i got a different band structure with that of a ferromagnet) I will appreciate your help. best Regards Colins :::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::: import kwant # For plotting. from matplotlib import pyplot from math import * from numpy import * from mpl_toolkits.mplot3d.axes3d import Axes3D # imports 3D plotting from matplotlib import cm # module for color pattern from random import seed from random import uniform import tinyarray import time s_0 = tinyarray.array([[1, 0], [0, 1]]) s_x = tinyarray.array([[0, 1], [1, 0]]) s_y = tinyarray.array([[0, -1j], [1j, 0]]) s_z = tinyarray.array([[1, 0], [0, -1]]) def make_lead(a=1, t=1.0, Vat= 0.0, EF= 7.0, Jex=-1.0, W=40): # Start with an empty lead with a single square lattice lat = kwant.lattice.square(a) sym_lead = kwant.TranslationalSymmetry((-2*a, 0)) lead = kwant.Builder(sym_lead) # build up one unit cell of the lead, and add the hoppings # to the next unit cell for j in xrange(W): mz1 = cos((1.0 + j) * pi) mz2 = cos(j * pi) lead[lat(0, j)] = (4 * t + Vat)* s_0 + Jex * mz1 * s_z lead[lat(1, j)] = (4 * t + Vat)* s_0 + Jex * mz2 * s_z if j > 0: lead[lat(0, j), lat(0, j - 1)] = -t * s_0 lead[lat(1, j), lat(1, j - 1)] = -t * s_0 lead[lat(1, j), lat(0, j)] = -t * s_0 lead[lat(2, j), lat(1, j)] = -t * s_0 return lead def main(): lead = make_lead().finalized() kwant.plotter.bands(lead, show=False) pyplot.xlabel("momentum [(lattice constant)^-1]") pyplot.ylabel("energy [t]") pyplot.title("Band structure G-type AF") pyplot.show() if __name__ == '__main__': main() ::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::

2 1

Leads with larger unit cell
by Michal Papaj Sept. 14, 2014

Sept. 14, 2014

Hello, I have a question regarding the construction of leads with periods greater than one lattice constant and having longer hoppings in those leads. I have constructed such a lead, but when I plot its band structure I am getting a different result than when plotting band structure of the exactly same system in PythTB (from David Vanderbilt's group at Rutgers University). My minimal working examples showing the issue are included below the message (I uploaded the resulting images to imgur: http://imgur.com/a/5uzyo , first image being the kwant's output). The band structure calculated by kwant seems to be folded in half - the energies in both programs are exactly the same, but they correspond to different crystal momentum. Am I defining the lead period in a wrong way? Best regards, Michal Papaj Kwant code: import kwant import matplotlib.pyplot as plt E = 0.0 t1 = 1.0 t2 = -1.0 width = 50 lat = kwant.lattice.square(1) sym = kwant.TranslationalSymmetry((-2, 0)) lead = kwant.Builder(sym) lead[(lat(0, y) for y in range(width))] = E lead[(lat(-1, y) for y in range(width))] = E lead[kwant.builder.HoppingKind((1, 0), lat, lat)] = t1 lead[kwant.builder.HoppingKind((0, 1), lat, lat)] = t1 lead[kwant.builder.HoppingKind((2, 0), lat, lat)] = t2 lead[kwant.builder.HoppingKind((0, 2), lat, lat)] = t2 kwant.plotter.bands(lead.finalized(), momenta=100) PythTB code: from pythtb import * import pylab as pl E = 0.0 t1 = 1.0 t2 = -1.0 width = 50 lat=[[1.0, 0.0],[0.0, 1.0]] orb=[[0.0,0.0]] test_model = tb_model(2,2,lat, orb) test_model.set_onsite([E]) test_model.set_hop(t1,0,0,[1.0,0.0]) test_model.set_hop(t1,0,0,[0.0,1.0]) test_model.set_hop(t2,0,0,[2.0,0.0]) test_model.set_hop(t2,0,0,[0.0,2.0]) test_model_cut=test_model.cut_piece(width,1,glue_edgs=False) path=[[-0.5],[0.0],[0.5]] kpts=k_path(path,50) evals_cut=test_model_cut.solve_all(kpts) fig=pl.figure() for i in range(width): pl.plot(evals_cut[i]) pl.show()

2 1

A bug with kwant and matplotlib 1.4
by Michel Fruchart Sept. 10, 2014

Sept. 10, 2014

Hello, When trying to use kwant with matplotlib 1.4.0 (which is the last released version), I had a problem when trying to draw a system. For exemple, the following code import kwant import matplotlib as mpl import matplotlib.pyplot as plt %matplotlib inline # to be used in ipython notebook lat = kwant.lattice.square(1.) sys = kwant.Builder() sys[(lat(x, y) for x in xrange(20) for y in xrange(10))] = 1. sys[lat.neighbors(1)] = 1. lead = kwant.Builder(kwant.TranslationalSymmetry((-1, 0))) lead[(lat(0, y) for y in xrange(10))] = 1. sys.attach_lead(lead) sys.attach_lead(lead.reversed()) p = kwant.plot(sys, show=False) plt.show() produces a huge traceback ending with ValueError: Transforms must be a Nx3x3 numpy array The exact result depends on the matplotlib backend, but when an image is produced, the sites of the leads are not drawn. The bug seems to come from an update in matplotlib API, and is (in this case) fixed by changing line 94 of plotter.py (http://git.kwant-project.org/kwant/tree/kwant/plotter.py#n94) from self._transforms = [matplotlib.transforms.Affine2D().scale(x) for x in sizes] to self._transforms = [matplotlib.transforms.Affine2D().scale(x).get_matrix() for x in sizes] There may be other instances of this in kwant, but fixing this one is enough to make the previous code work. A similar issue in 3D is described in https://www.mail-archive.com/kwant-discuss@kwant-project.org/msg00111.html (I did not look further into this one.) Best, Michel

2 1