Mailman 3 June 2024 - Kwant-discuss

To to separate spins, electron and hole
by Khani Hosein Oct. 11, 2024

Oct. 11, 2024

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

4 5

Unexpected Landau Level Discrepancies in Multilayer Graphene Compared to Published Results
by yl June 22, 2024

June 22, 2024

Dear Kwant Group, I attempted to replicate the Landau levels of ABA-trilayer graphene as described in the paper "Landau level spectra and the quantum Hall effect of multilayer graphene," Physical Review B 83, 165443 (2011), Figure 2, using Kwant. The results obtained with Kwant match the paper's results approximately well; however, there is an issue. The number of Landau levels (LLs) under each magnetic field does not match the results presented in the paper. For example (see the attached figure): At a magnetic field strength of ( B = 10 ) Tesla: The left figure shows a total of 46 LLs (18 + 28). The right figure shows a total of 34 LLs (16 + 18). For the specific cases labeled as 'a' to 'd', each of these cases has 2 more LLs compared to a reference. For the case labeled 'e', this case has 4 more LLs compared to a reference. I believe the reference is correct as I was able to reproduce the completely identical LLs results using WannierTools(an open source code which can calculate LLs by Peierls substitution method). I am currently unsure where the problem lies or if it is a bug. I have attached the code (test.py) and a small modification to the spectrum function in plotter.py. Thank you very much to the Kwant development team for your efforts. I hope to get an answer, as it would be very helpful to me. Versions used: 1.4.3 + landau_levels.py Best regards, ac76888 ac76888(a)sjtu.edu.cn

4 4

[kwant] How to make a finite lead?
by 김봉수 June 12, 2024

June 12, 2024

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

2 2

Error in calculating conductance of 3D system with periodic boundaries
by Naveen Yadav June 7, 2024

June 7, 2024

Dear Kwant Developers, I am trying to calculate the conductance of a 3D system with dimension L × W × H, by imposing periodic boundary conditions in all three directions. I am able to calculate the band structure but unable to calculate the corresponding conductance using kwant.smatrix. I am not sure whether kwant is able to do transport calculations on systems with periodic boundaries using smatrix. Please advise. The code is following import kwant import scipy.sparse.linalg as sla import matplotlib.pyplot as plt import tinyarray import numpy as np from numpy import cos, sin, pi import cmath from cmath import exp from matplotlib import rc, rcParams rc('text', usetex=True) rc('axes', linewidth=2) rc('font', weight='bold') #rcParams['text.latex.preamble'] = [r'\usepackage{sfmath} \boldmath'] #rcParams['text.latex.preamble'] = [r'\usepackage{sfmath} \boldmath'] 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]]) def make_sys(a=1, m_0 = 2, L=1, W=100, H=1, t=1.0, t_z=0.0, phi=0.01): #L=10; W=80; H=60; def onsite(site): return 2 * m_0 * sigma_x def hoppingx(site0, site1): y = site1.pos[1] return (t * sigma_x) def hoppingy(site0, site1): return -1j * t * sigma_y - (m_0/2) * sigma_x def hoppingz(site0, site1): y = site1.pos[1] return (-1j * t * sigma_z - (m_0/2) * sigma_x - 1j * t_z * sigma_0) * exp(2 * pi * 1j * (y) * phi) lat =kwant.lattice.cubic(norbs=2) #sys = kwant.Builder(kwant.TranslationalSymmetry(lat.vec((0, 0, H)))) #sys[(lat(x, y, z) for x in range(L) for y in range(W) for z in range(H))] = onsite #sys = kwant.Builder(kwant.TranslationalSymmetry((L, 0, 0),(0, 0, H))) #sys[(lat(0, y, 0) for y in range(W))] = onsite sys = kwant.Builder(kwant.TranslationalSymmetry((L, 0, 0),(0, W, 0),(0, 0, H))) sys[lat.shape(lambda p: True, (0,0,0))] = onsite sys[kwant.builder.HoppingKind((1, 0, 0), lat, lat)] = hoppingx sys[kwant.builder.HoppingKind((0, 1, 0), lat, lat)] = hoppingy sys[kwant.builder.HoppingKind((0, 0, 1), lat, lat)] = hoppingz sys = kwant.wraparound.wraparound(sys, keep=None) leadL = kwant.Builder(kwant.TranslationalSymmetry((-a, 0, 0),lat.vec((0, W, 0)),lat.vec((0, 0, H)))) leadL[(lat(0, y, z) for y in range(W) for z in range(H))] = onsite leadL[kwant.builder.HoppingKind((1, 0, 0), lat, lat)] = hoppingx leadL[kwant.builder.HoppingKind((0, 1, 0), lat, lat)] = hoppingy leadL[kwant.builder.HoppingKind((0, 0, 1), lat, lat)] = hoppingz leadR = kwant.Builder(kwant.TranslationalSymmetry((a, 0, 0),lat.vec((0, W, 0)),lat.vec((0, 0, H)))) leadR[(lat(0, y, z) for y in range(W) for z in range(H))] = onsite leadR[kwant.builder.HoppingKind((1, 0, 0), lat, lat)] = hoppingx leadR[kwant.builder.HoppingKind((0, 1, 0), lat, lat)] = hoppingy leadR[kwant.builder.HoppingKind((0, 0, 1), lat, lat)] = hoppingz leadL = kwant.wraparound.wraparound(leadL, keep=0) leadR = kwant.wraparound.wraparound(leadR, keep=0) sys.attach_lead(leadL) sys.attach_lead(leadR) sys = sys.finalized() return sys syst=make_sys() def plot_bands_2d(syst, momenta=(101, 101, 101), ky=0, kz=0): if not isinstance(syst, kwant.system.FiniteSystem): raise TypeError("Need a system without symmetries.") fig = plt.figure(figsize=(5, 5)) #kwant.plotter.bands(syst.leads[0], momenta=np.linspace(-np.pi, np.pi, 101), show=False) kxs = np.linspace(-np.pi, np.pi, momenta[0]) def get_energies(ky=ky, kz=kz): energies = [np.linalg.eigvalsh(syst.hamiltonian_submatrix ((kx, ky, kz))) for kx in kxs] return energies energies = np.array(get_energies(ky=ky, kz=kz)) # Plot the band structure for band in np.array(energies).T: plt.plot(kxs, band) plt.ylim(-2.5, 2.5) plt.xlim(-pi, pi) plt.xlabel("kx") plt.ylabel("Energy") plot_bands_2d(syst) kzs = np.linspace(-pi,pi,101) kys = np.linspace(-pi,pi,101) energies = np.linspace(0.0, 0.5, 30) data=[] for energy in energies: Tkx = np.zeros(101) for i in np.arange(len(kzs)): kz=kzs[i] ky=kys[i] smatrix = kwant.smatrix(syst, energy, [kz], [ky]) Tkx[i]=smatrix.transmission(1, 0) data.append(sum(Tkx)) plt.plot(energies, data) plt.show() Also as a separate question, I am trying to run the notebook Kubo_Bastin_conductivity_with_periodic_boundary_conditions.ipynb from https://gitlab.com/kpm-tools/bloch/-/tree/master/examples?ref_type=heads to understand the kubo conductivity calcualton on a system with periodic boundaries but various errors are there in importing bloch.py. I am using kwant.__version__ = '1.4.4'. Please help in this regard. Thanks in advance. -- Best Regards, Naveen Yadav Department of Physics & Astrophysics University Of Delhi New Delhi-110007

1 0

ValueError: Value functions must not take *args or **kwargs
by wilson2048＠outlook.com June 4, 2024

June 4, 2024

Dear kwant community: I have encountered this error in my code. Basically I don't want to redefine the hop in the y direction, so I use the existing function `hop`. However, the code throw me an error which I don't know how to fix. Below is a simplest code example, the real situation is more complex but the spirit is the same. ``` lat = kwant.lattice.square() def make_system(r=10, t=1): def circle(pos): x, y = pos return x**2+y**2<r**2 syst = kwant.Builder() def onsite(site): return 0 syst[lat.shape(circle, (0, 0))] = onsite def hop(site1, site2, some_param): # some complex calculations not shown. return t def modified_hop(alpha, hop): # I don't want to define a hopy function, just use the exist hop return lambda *a, **kw: alpha*hop(*a, **kw) alpha = 0.5 syst[kwant.builder.HoppingKind(lat.prim_vecs[0], lat, lat)] = hop syst[kwant.builder.HoppingKind(lat.prim_vecs[1], lat, lat)] = modified_hop(alpha, hop) # alpha*hop would be very nice, but it won't work! syst.eradicate_dangling() return syst syst = make_system() fsyst = syst.finalized() H = fsyst.hamiltonian_submatrix(params=dict(some_param=1)) ``` Best, Wilson

1 0

Leads self-energy matrix elements not ordered as system's hamiltonian_submatrix?
by wilson2048＠outlook.com June 4, 2024

June 4, 2024

Dear kwant community: I am investigating the `kwant.greens_function`。 I would like to get the lead self-energies and add to the original system Hamiltonian and then inverse it, thus obtaining the retarded green's function. I tested, but the resulted green's function is not identical to the one kwant calculated. I want to know the reason. Is it because what I describe in the title? Below is a minimal code example: ``` lat = kwant.lattice.square() def make_system(r=20, t=-1): def circle(pos): x, y = pos return x**2 + y**2 < r**2 syst = kwant.Builder() syst[lat.shape(circle, (0, 0))] = 0 syst[lat.neighbors()] = t syst.eradicate_dangling() def lead_shape(pos): return -12 < pos[0] < 12 lead = kwant.Builder( kwant.TranslationalSymmetry((0, 1))) lead[lat.shape(lead_shape, (0, 0))] = 0 lead[lat.neighbors()] = -t syst.attach_lead(lead) return syst syst = make_system() fsyst = syst.finalized() kwant.plot(fsyst) greens_function=kwant.greens_function(fsyst) Hc = fsyst.hamiltonian_submatrix(sparse=False) index = fsyst.lead_interfaces[0] Hc[np.ix_(index, index)] += greens_function.lead_info[0] Gr = np.linalg.inv(-Hc) Gr[np.ix_(index, index)]-greens_function.submatrix(0, 0) # should be a zero matrix ``` Bests, Wilson

2 6

Kwant-compatible Hartree-Fock Package
by Johanna June 3, 2024

June 3, 2024

Hi all, We're excited to share a new Python package we've developed at the Quantum Tinkerer group: MeanFi. MeanFi is a self-consistent Hartree-Fock solver for tight-binding models designed to work smoothly with Kwant. Our algorithm is optimized to easily and extensibly perform all your mean-field calculations. We made it such that you can: 1) define the initial Hamiltonian and the density-density interaction as Kwant models, 2) turn it into a tight-binding format, 3) find the groundstate solution, and 4) finally turn it back to a Kwant system to do all the analysis you want! You can find the documentation, tutorials, and installation instructions here: <https://meanfi.readthedocs.io/en/latest/>. We welcome your feedback and contributions. If you have any questions or run into issues, feel free to reach out or open an issue on GitHub here: <https://github.com/quantum-tinkerer/MeanFi>. All the best, MeanFi Developers Kostas, Johanna, Anton and Antonio

1 0