Mailman 3 February 2024 - Kwant-discuss

To to separate spins, electron and hole
by Khani Hosein 11 Oct '24

11 Oct '24

Dear all, I am using Kwant to study a system with both spins and electron and hole, so the hopping matrix is 4X4. Now I want to know how to separate the spins, electron and hole. I study this through the "2.6. Superconductors: orbital degrees of freedom, conservation laws and symmetries" smatrix = kwant.smatrix(syst, energy) data.append(smatrix.submatrix((0, 0), (0, 0)).shape[0] -smatrix.transmission((0, 0), (0, 0)) +smatrix.transmission((0, 1), (0, 0))) I do not find more information about this for kwant. For my system, if I have 3 leads, and the hopping and onsite energy is set in this order: (e↑,0,0,0)- spin up electron,first row of the 4X4matrix (0,e↓,0,0)- spin down electron,second row of the 4X4matrix (0,0, h↑ ,0)- spin up hole,third row of the 4X4matrix (0,0, 0, h ↓ )- spin down hole,fourth row of the 4X4matrix I want to obtain the transmissions from 0 →2. smatrix = kwant.smatrix(syst, energy) The transmission from h↑ to e↑ is: smatrix.transmission((2, 1), (0, 2)) The transmission from h ↓ to e↑ is: smatrix.transmission((2, 1), (0, 3)) Is my understanding correct? Thanks very much in advance! Hosein Khani

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[kwant] How to make a finite lead?
by 김봉수 12 Jun '24

12 Jun '24

Dear kwant users, I am trying to reproduce the paper following. https://arxiv.org/abs/1110.4280 (in this paper, fig5, 6 will help to understand my questions.) What I’m struggling with now is to build a finite(not infinite, and not having any translational symmetry) lead as below. I wonder 1. I think "kwant.builder.SelfEnergyLead()” function might be proper thing to build a finite lead without any translational symmetry. Is it correct? 2. If it's right, when I run this code using SelfEnergyLead(), the following error occurs. How can I solve it? ------------------------------------------------------------------------------------------------ lat = kwant.lattice.chain(norbs=1) syst = kwant.Builder() syst[(lat(i) for i in range(4))] = 4 syst[lat.neighbors()] = -1 def self_energy(energy, args=(), *, params=None): return -1j left_lead = kwant.builder.SelfEnergyLead(self_energy, interface=[lat(0)], parameters = []) right_lead = kwant.builder.SelfEnergyLead(self_energy, interface=[lat(3)], parameters = []) syst.leads.append(left_lead) syst.leads.append(right_lead) syst = syst.finalized() H = syst.leads[0].kwant.hamiltonian_submatrix kwant.plot(syst) ------------------------------------------------------------------------------------------------ AttributeError: 'SelfEnergyLead' object has no attribute ‘kwant' Any help is appreciated! Best, Bongsu Kim

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fill command continuum Hamiltonian
by jorge huamani 19 Mar '24

19 Mar '24

Dears, Is it possible to obtain a finite system with three different regions using a discretize continuum Hamiltonian approach? I am using the fill command but I have issues obtaining a finite system, my idea is to have a finite system and fill each part of the system with different Hamiltonian talking about those three regions. I would appreciate if you have any idea to make it, thank in advance Best regards, -- Dr. Jorge Huamaní Correa Professor at Universidad Tecnológica del Perú CTI Vitae : https://ctivitae.concytec.gob.pe/appDirectorioCTI/VerDatosInvestigador.do?i…

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Conductivity bilayer graphene
by Arturo 22 Feb '24

22 Feb '24

Hi everyone, I would like to reproduce figure 2.c of the manuscript http://doi.org/10.1103/PhysRevB.88.125409 in that they calculate the conductivity of a bilayer zigzag graphene. Even though I get similar transmission values, the steps do not occur at the same values. I'm not sure what my mistake would be. I would appreciate any suggestions. The code is the following: import numpy as np import kwant import matplotlib.pyplot as plt a = 0.142 c0 = 0.5 # to try t = 3.16 t1 = 0.39/3.16 widthh= 23.7 graphene = kwant.lattice.general([(a*np.sqrt(3), 0,0), (a*np.sqrt(3) / 2, 3 / 2 * a,0)], [(0, 0,0), (0, -a,0), (0, a,c0), (0, 0,c0) ]) a1, b1, a2, b2 = graphene.sublattices syst = kwant.Builder() def make_system(length=20, width=widthh): def rectangle(pos): x, y, z = pos return -length / 2 < x < length / 2 and -width / 2 < y < width / 2 and -5 < z < 5 syst[graphene.shape(rectangle, (0, 0,0))] = 0 syst[graphene.neighbors()] = t syst[kwant.builder.HoppingKind((0, 0), b2, a1)] = t1 make_system() symmetry_left = kwant.TranslationalSymmetry((-a*np.sqrt(3), 0, 0)) symmetry_right = kwant.TranslationalSymmetry((a*np.sqrt(3), 0, 0)) lead_left = kwant.Builder(symmetry_left) lead_right = kwant.Builder(symmetry_right) def lead_shape(pos): x, y, z = pos return -widthh / 2 < y < widthh / 2 for lead in [lead_left, lead_right]: lead[graphene.shape(lead_shape, (0, 0, 0))] = 0 lead[graphene.neighbors()] = t lead[kwant.builder.HoppingKind((0, 0), b2, a1)] = t1 syst.attach_lead(lead_left); syst.attach_lead(lead_right); kwant.plot(syst) syst_finalized = syst.finalized() lead_index = 0 lead = syst_finalized.leads[lead_index] # Conductivity calculation data = [] for i in range(51): energias = 0+0.07*3.16*i/50.0+0.000001 smatrix_bilayer = kwant.smatrix(syst_finalized, energy=energias) data.append([energias, smatrix_bilayer.transmission(1, 0)]) data = np.array(data) plt.plot(data[:,0]/3.16,data[:,1]) In the case of the monolayer (figure 2.a) I obtain the same results. thanks

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LDOS and Wave function for a specific kpoint
by Gusthavo Miranda 14 Feb '24

14 Feb '24

Dear Kwant users, I am trying to compute localized density of states and the wavefunction for a specific energy and kpoint. Looking at the tutorials, I understand that the LDOS can be computed as ldos = kwant.ldos(syst, energy=0.8) However, when I try to compute for a specific kpoint in the following manner ldos = kwant.ldos(syst, energy=0.8, params=dict(k_x = 1.7)) the plot of the LDOS looks the same. Am I missing something? Is this the correct way of plotting for a specific kpoint? Thanks in advance Gusthavo

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