Can Kwant calculate density of states of quasi one-dimensional systems?
Dear Kwant users, In Kwant manual there is a section entitled "Calculating spectral density with the kernel polynomial method". I find that examples in this section are all talking about finite-size systems. I wonder whether Kwant can calculate density of states of a quasi one-dimensional system that is finite along y-direction but is translationally invariant along x-direction. For a minimal test, I have run the following codes but errors occurred. #-------------------------------------------------------------------------------------- import kwant import numpy a=1 W=30 t=1 lat = kwant.lattice.square(a) syst = kwant.Builder(kwant.TranslationalSymmetry((-a, 0))) syst[(lat(0, j) for j in range(W))] = 4 * t syst[lat.neighbors()] = -t fsyst = syst.finalized() energies = numpy.linspace(-0.1,0.1, 10) spectrum = kwant.kpm.SpectralDensity(fsyst) spectrum.add_moments(10) spectrum.add_vectors(5) densities = spectrum(energies) #-------------------------------------------------------------------------------------- What's wrong in above codes? Currently, are we able to employ Kwant to calculate density of states of 3D systems being translationally invariant along one or two directions? Regards, Zhan
Hi Zhan, SpectralDensity indeed doesn't support translationally invariant systems. You have two options instead. If you want to obtain the DOS at a specific energy, you should use the output of syst.modes—specifically the wave functions of the propagating modes. If you need the DOS of the full spectrum, you should diagonalize the Hamitlonian and perform a summation over all momenta on your own. Best, Anton On Sat, 21 Dec 2019 at 13:51, Cao Zhan <caozhan@baqis.ac.cn> wrote:
Dear Kwant users,
In Kwant manual there is a section entitled "Calculating spectral density with the kernel polynomial method". I find that examples in this section are all talking about finite-size systems. I wonder whether Kwant can calculate density of states of a quasi one-dimensional system that is finite along y-direction but is translationally invariant along x-direction. For a minimal test, I have run the following codes but errors occurred. #-------------------------------------------------------------------------------------- import kwant import numpy
a=1 W=30 t=1 lat = kwant.lattice.square(a) syst = kwant.Builder(kwant.TranslationalSymmetry((-a, 0))) syst[(lat(0, j) for j in range(W))] = 4 * t syst[lat.neighbors()] = -t fsyst = syst.finalized() energies = numpy.linspace(-0.1,0.1, 10)
spectrum = kwant.kpm.SpectralDensity(fsyst) spectrum.add_moments(10) spectrum.add_vectors(5) densities = spectrum(energies) #-------------------------------------------------------------------------------------- What's wrong in above codes? Currently, are we able to employ Kwant to calculate density of states of 3D systems being translationally invariant along one or two directions?
Regards, Zhan
Hi Anton, Thanks for your two suggestions. I am able to perform the second method but I don't know how to call syst.modes. Would you please show me the codes using syst.modes to calculate DOS(energy=0) of my minimal example? #------------------------------------------------------------------- import kwant import numpy a=1 W=30 t=1 lat = kwant.lattice.square(a) syst = kwant.Builder(kwant.TranslationalSymmetry((-a, 0))) syst[(lat(0, j) for j in range(W))] = 4 * t syst[lat.neighbors()] = -t fsyst = syst.finalized() # Then how to use fsyst.modes to calculate DOS(energy=0) ? #------------------------------------------------------------------- Thanks in advance! Regards, Zhan 发件人：Anton Akhmerov <anton.akhmerov+kd@gmail.com> 发送日期：2019-12-21 21:55:34 收件人：Cao Zhan <caozhan@baqis.ac.cn> 抄送人：kwant-discuss <kwant-discuss@kwant-project.org> 主题：Re: [Kwant] Can Kwant calculate density of states of quasi one-dimensional systems?>Hi Zhan,
SpectralDensity indeed doesn't support translationally invariant systems. You have two options instead. If you want to obtain the DOS at a specific energy, you should use the output of syst.modes—specifically the wave functions of the propagating modes. If you need the DOS of the full spectrum, you should diagonalize the Hamitlonian and perform a summation over all momenta on your own.
Best, Anton
On Sat, 21 Dec 2019 at 13:51, Cao Zhan <caozhan@baqis.ac.cn> wrote:
Dear Kwant users,
In Kwant manual there is a section entitled "Calculating spectral density with the kernel polynomial method". I find that examples in this section are all talking about finite-size systems. I wonder whether Kwant can calculate density of states of a quasi one-dimensional system that is finite along y-direction but is translationally invariant along x-direction. For a minimal test, I have run the following codes but errors occurred. #-------------------------------------------------------------------------------------- import kwant import numpy
a=1 W=30 t=1 lat = kwant.lattice.square(a) syst = kwant.Builder(kwant.TranslationalSymmetry((-a, 0))) syst[(lat(0, j) for j in range(W))] = 4 * t syst[lat.neighbors()] = -t fsyst = syst.finalized() energies = numpy.linspace(-0.1,0.1, 10)
spectrum = kwant.kpm.SpectralDensity(fsyst) spectrum.add_moments(10) spectrum.add_vectors(5) densities = spectrum(energies) #-------------------------------------------------------------------------------------- What's wrong in above codes? Currently, are we able to employ Kwant to calculate density of states of 3D systems being translationally invariant along one or two directions?
Regards, Zhan
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Anton Akhmerov -
Cao Zhan