Mailman 3 December 2019 - Kwant-discuss

Bound States in a JJ
by Denise Puglia 10 Dec '19

10 Dec '19

Dear All, Has anyone tried to calculate bound states in a JJ? I used the package by Benoit Rossignol available at: https://gitlab.kwant-project.org/kwant/boundstate . I am sending attached the code I used to simulate it but I am having some difficulties interpreting the results. First, the algorithm could not find bound states, i.e., psi_tot returned by the solver is an empty array, for N<5. Second, it returns a total wavefunction for larger N, however the wavefunction oscillates between each site and decays very slowly in the leads. In the original proposal of this method, Istas et al (arXiv:1711.08250) solves the wavefunction of quantum billiard (Fig 5) in which this does not seem to happen. Has anyone found something similar? I do know if it a mistake in the code, an attribute of discretization or something else. Scanning over N=[1,2,3,5,10,20] and ,u=[2, Delta, 0, -2] the algorithm returns the following non-zero wavefunctions: SNS junction with N=10, S=20 and mu=2 [image: N_10_S_20_mu_2.png] SNS junction with N=20, S=20 and mu=2 [image: N_20_S_20_mu_2.png] SNS junction with N=5, S=20 and mu=-2 (in this case the amplitude of the bound states decays to zero in the lead) [image: N_5_S_20_mu_-2.png] SNSNS junction with 20/10/10/10/20 sites and mu=2. The wavefunction in the N region is similar to the SNS junction but there seems to be no decay in the middle S region. [image: N_10_S_20_mu_2.png] and of course it's mirrored version: [image: NSN_10_mu_2.png] I appreciate any comments on this subject. Best regards, Denise

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Calculating the total energy in a superconducting lead
by Jannis Neuhaus-Steinmetz 09 Dec '19

09 Dec '19

Dear all, I am trying to calculate the total energy of the electrons in an s-wave superconducting lead with spins. But I am not sure, if my method is correct. Currently I separate the particle and hole Hamiltonians with a projector H = syst.hamiltonian_submatrix(params=params) P_electron = np.kron(np.eye(L+1), np.array([[1, 0, 0, 0], [0, 1, 0, 0], [0, 0,0,0], [0, 0,0,0]])) H_electron = P_electron.conjugate().transpose() @ H @ P_electron where L is the number of sites. L+1 because I get a dimension mismatch otherwise. My guess here is, that kwant generates an extra site because of the translational symmetry. For the conservation laws I used particle_hole=pauli.sxs0, conservation_law=-pauli.szs0 where pauli.sisj is the kroneker product of the i-th and the j-th pauli matrix. For my Hamiltonian I used the base (e_up, e_down, h_down, h_up). Then I calculate the eigenvalues of H_electron and multiply them with the fermi distribution to get the total energy. spectrum= np.linalg.eigvalsh(H_electron) fermi_dist=kwant.kpm.fermi_distribution(spectrum, params["mu"], T) total_energy=np.dot(fermi_dist,spectrum) Now to the part, why I suspect a mistake in my method. I use a spin-dependent hopping term and scan over a parameter /b/, that changes the spin dependency. When doing so the qualitative behavior of the total energy /E_total(b)/ changes with the size of my unit cell. But all sites in my unit cell are identical. So I would suspect, that the results should only differ by a factor. For large systems the curvature of /E_total(b)/ converges, but for small systems the results vary quit a lot. I'd really appreciate some help on this. Best regards, Jannis

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matrix
by Nafise Nouri 08 Dec '19

08 Dec '19

Dear all, Is there any one to know how we can use one element of matrix? For example we have matrix A as follows: A=[2 4;6,0] We have A matrix with dimension 2*2. If we want to use one element of matrix such as A(2,1), (that is number 6 in the matrix), How we can write in kwant? Any help appreciate. Best, Nafise

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Lead
by tavakkolidjawad 08 Dec '19

08 Dec '19

Hello everyone I want to attach a lead on top of lattice but it doesn't work. Please check the following code and tell me the error code in the lead construction section Thank you import kwant from matplotlib import pyplot lat = kwant.lattice.general([(0, 0.5, 0.5), (0.5, 0, 0.5), (0.5, 0.5, 0)], [(0, 0, 0), (0.25, 0.25, 0.25)], name=['a1', 'a2']) a1, a2 = lat.sublattices def make_cuboid(a=15, b=15, c=15): def cuboid_shape(pos): x, y, z = pos return 0 <= x < a and 0 <= y < b and 0 <= z < c syst = kwant.Builder() syst[a1.shape(cuboid_shape, (0, 0, 0))] = 1 ## the same on-site foar a1 and b1 syst[a2.shape(cuboid_shape, (0, 0, 0))] = 1 syst[lat.neighbors()] = 1 sym_lead = kwant.TranslationalSymmetry((-1,0)) top_lead = kwant.Builder(sym_lead) top_lead[(lat(0,0,j) for j in range(c))] = 1 top_lead[lat.neighbors()]=-1 syst.attach_lead(top_lead) return syst def main(): syst = make_cuboid() kwant.plot(syst) syst = syst.finalized() syst = make_cuboid(a=1.1, b=1.1, c=1.1) def family_colors(site): return 'r' if site.family == a1 else 'g' kwant.plot(syst, site_size=0.18, site_lw=0.01, hop_lw=0.05, site_color=family_colors) ### Conventional unit cell "FCC" #pyplot.show() if __name__ == '__main__': main()

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Access to eigenvalue, eigenvector and number of points in each unit cell
by Saj.ZiaBorujeni 02 Dec '19

02 Dec '19

Dear all, I need to access to the number of atoms of my unit cell, the eigenvalue and eigenvectors for each eigenvalue of my Hamiltonian. Is there any one to help me and let me know if it is possible in kwan to access them. Best wishes, Sajad

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Re: [Kwant] armchair nanoribbon with defect
by Nafise Nouri 02 Dec '19

02 Dec '19

Nafise Nouri Tue, Nov 5, 10:37 PM Dear Adel, Best regards, Nafise On Wed, Nov 6, 2019 at 1:27 AM Adel Belayadi <adelphys(a)gmail.com> wrote: > Dear nafise. I have seen you have asked the question again. Please state > that your query was solved so other might know and benifit. Please sharing > is the purpose of mailing list. > Know back to your problem. It is better to avoid attach.reversed.Instead > create for instance lead2 with -Num. > Please state that your problem was solved in mail listing as Josef advised > u. > Best, Adel > > Le mar. 5 nov. 2019 à 21:07, Adel Belayadi <adelphys(a)gmail.com> a écrit : > >> You are welcome. >> >> Le mar. 5 nov. 2019 à 20:08, Nafise Nouri <nafise.nour(a)gmail.com> a >> écrit : >> >>> Dear Adel, >>> >>> Thank you for your reply and help. It is so useful for me. >>> >>> Best regards, >>> >>> Nafise >>> >>> On Tue, Nov 5, 2019 at 12:03 AM Adel Belayadi <adelphys(a)gmail.com> >>> wrote: >>> >>>> Dear nafise, >>>> Good day, it seem that you have problem with symmetry. >>>> To some up you have to fix the following issues: >>>> 1. define the lattice.general correctly >>>> 2. provide larger limit along x for the rectangle, so we might avoid >>>> isolated atoms(unconnected) >>>> 3. chose the symmetry as Num = 8*(a1*sqrt(3)/2) and then use sym = >>>> kwant.TranslationalSymmetry([0, Num]) >>>> 4. shift the origin of y direction in your delet function by >>>> (y-a1*sqrt(3))**2 intead of y**2 >>>> 5. the boundary in lead_shape is for x not y. If we use y the code will >>>> bug (infinity system) >>>> Please, you have one thing left to do. kindly make sure that the lead >>>> is 100% armchair, perhaps Mrs Josefe or About would be for a help. >>>> I have plotted the bands for the lead, it seems armchair but I still do >>>> not muster 100% the source code of kwant.plotter.bands. Please check this >>>> pont. >>>> See the code with explanation >>>> Happy kwanting >>>> >>>> Best Adel >>>> >>>> import kwant >>>> from math import sqrt >>>> import matplotlib.pyplot as plt >>>> import tinyarray >>>> import numpy as np >>>> import math >>>> import cmath >>>> #import scipy.linalg as la >>>> import matplotlib >>>> d=1.42; >>>> a1=d*sqrt(3); >>>> >>>> #on-site energy................................................... >>>> t=-2.7; >>>> >>>> latt = kwant.lattice.general([[a1, 0], [a1/2, a1*sqrt(3)/2]], # >>>> lattice vectors >>>> [[0, 0], [0, a1/sqrt(3)]]) # >>>> Coordinates of the sites >>>> a, b = latt.sublattices >>>> >>>> syst= kwant.Builder() >>>> >>>> >>>> ## you have to provide quite big boundaries to avoid isolated atoms so >>>> abs(x)=4*a1 >>>> def rectangle(pos): >>>> x, y = pos >>>> z=x**2+y**2 >>>> return -4*a1<x<4*a1 and -6*d<y<7*d >>>> >>>> syst[latt.shape(rectangle,(1,1))]=0 >>>> >>>> ## you have to shift the origine of your diffect by a1*sqrt(3) >>>> def delet(pos, shift = a1*sqrt(3)): >>>> x, y = pos >>>> z=x**2+(y-shift)**2 >>>> return z<(2*a1)**2 >>>> >>>> del syst[latt.shape(delet, (1,1))] >>>> >>>> >>>> #nearestneighbors............................................................. >>>> syst[[kwant.builder.HoppingKind((0,0),a,b)]]=t >>>> syst[[kwant.builder.HoppingKind((0,1),a,b)]]=t >>>> syst[[kwant.builder.HoppingKind((-1,1),a,b)]]=t >>>> >>>> ## using [0, Num] as True >>>> Num = 8*(a1*sqrt(3)/2) >>>> sym = kwant.TranslationalSymmetry([0, Num]) >>>> >>>> lead = kwant.Builder(sym) >>>> >>>> ##the lead shape must be lilited in x not y >>>> def lead_shape(pos): >>>> x, y = pos >>>> return -4*a1<x<4*a1 >>>> >>>> lead[latt.shape(lead_shape, (0,0))]=0 ##or use the folowing lambda form >>>> >>>> ##lead[latt.shape((lambda pos: abs(pos[0]) < 4*a1 ), (0, 0))] = 0 >>>> >>>> def delet_lead(pos, shift = +a1*sqrt(3)): >>>> x, y = pos >>>> z=x**2+(y-shift)**2 >>>> return z<(2*a1)**2 >>>> >>>> del lead[latt.shape(delet_lead, (1,1))] >>>> >>>> lead[[kwant.builder.HoppingKind((0,0),a,b)]]=t >>>> lead[[kwant.builder.HoppingKind((0,1),a,b)]]=t >>>> lead[[kwant.builder.HoppingKind((-1,1),a,b)]]=t >>>> >>>> syst.attach_lead(lead,add_cells=0) >>>> kwant.plot(syst); >>>> syst.attach_lead(lead.reversed(),add_cells=0) >>>> kwant.plot(syst); >>>> >>>> ### for armchair -pi/3*d<ky<pi/3*d >>>> from numpy import pi, linspace >>>> momenta = list(linspace(-pi/3, pi/3, 50)) >>>> kwant.plotter.bands(lead.finalized(), momenta); >>>> >>>> Le dim. 3 nov. 2019 à 10:55, Nafise Nouri <nafise.nour(a)gmail.com> a >>>> écrit : >>>> >>>>> Dear Joseph >>>>> >>>>> At first thank you for kwant team. My problem for zigzag nanoribbon >>>>> with defect (hole) was solved with your help. Now I have a bit problem with >>>>> armchair nanoribbon. I have changed zigzag nanoribbon according to making >>>>> the armchair nanoribbon. It is OK when we have no defect on the system but >>>>> when I try to make hole on the scattering region and leads I have a >>>>> problem. The shape of holes on the scattering region and leads are not >>>>> same. My code is as the following. That is very king of you if you help me >>>>> where is the problem. >>>>> >>>>> Thanks In advance for your help. >>>>> >>>>> Beat wishes, >>>>> Nafise >>>>> >>>>> import kwant >>>>> from math import sqrt >>>>> import matplotlib.pyplot as plt >>>>> import tinyarray >>>>> import numpy as np >>>>> import math >>>>> import cmath >>>>> #import scipy.linalg as la >>>>> import matplotlib >>>>> d=1.42; >>>>> a1=d*math.sqrt(3); >>>>> >>>>> #on-site energy................................................... >>>>> t=-2.7; >>>>> >>>>> latt = kwant.lattice.general([(0,a1),(a1*math.sqrt(3)/2,a1*0.5)], >>>>> [(-d/2,a1/2),(d/2,a1/2)]) >>>>> a,b = latt.sublattices >>>>> syst= kwant.Builder() >>>>> >>>>> def rectangle(pos): >>>>> x, y = pos >>>>> z=x**2+y**2 >>>>> return -2.4*a1<x<2.4*a1 and -5*d<y<5*d >>>>> >>>>> syst[latt.shape(rectangle,(1,1))]=0 >>>>> >>>>> def delet(pos): >>>>> x, y = pos >>>>> z=x**2+y**2 >>>>> return z<(2*a1)**2 >>>>> >>>>> del syst[latt.shape(delet, (1,1))] >>>>> >>>>> #nearest >>>>> neighbors............................................................. >>>>> syst[[kwant.builder.HoppingKind((0,0),a,b)]] =t >>>>> syst[[kwant.builder.HoppingKind((0,1),a,b)]] =t >>>>> syst[[kwant.builder.HoppingKind((-1,1),a,b)]] =t >>>>> >>>>> ax=kwant.plot(syst); >>>>> >>>>> sym = kwant.TranslationalSymmetry(latt.vec((-2,4))) >>>>> >>>>> lead = kwant.Builder(sym) >>>>> >>>>> def lead_shape(pos): >>>>> x, y = pos >>>>> return -5*d<y<5*d >>>>> >>>>> lead[latt.shape(lead_shape, (1,1))]=0 >>>>> >>>>> def delet_lead(pos): >>>>> x, y = pos >>>>> z=x**2+y**2 >>>>> return z<(2*a1)**2 >>>>> >>>>> del lead[latt.shape(delet_lead, (1,1))] >>>>> >>>>> lead[[kwant.builder.HoppingKind((0,0),a,b)]]=t >>>>> lead[[kwant.builder.HoppingKind((0,1),a,b)]]=t >>>>> lead[[kwant.builder.HoppingKind((-1,1),a,b)]]=t >>>>> >>>>> syst.attach_lead(lead,add_cells=0) >>>>> syst.attach_lead(lead.reversed(),add_cells=0) >>>>> ax=kwant.plot(syst); >>>>> >>>>

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Kwant-discuss December 2019