Obtain optical spectrum from SBE using BSE kernel
Posted: Fri Dec 09, 2022 5:19 am
Dear users and developers,
I'm trying to solve Semiconductor Bloch Equation (SBE) to obtain the optical spectrum for TMD materials.
My scheme is to use Yambo to generate Coulomb matrixes input for the SBE and Wannier90 for velocity and other matrixes required for solving SBE in velocity gauge.
I've got a good agreement for the independent particle absorption spectrum (without including Coulomb matrixes), compared with results from Wannier90.
But when I include Coulomb matrixes in my calculation, I can't get it right compared with Yambo BSE result.
I output the Coulomb potential follow from the forum link: viewtopic.php?p=6693&sid=44332c65502e15 ... b688#p6693
Yambo grid differs from Wannier90, so I modified subroutine k_ibz2bz.F to get the grid. The Coulomb potential from Yambo seems to exclude the symmetrical part (k<->k'), so I expand it by W_{vcv'c'kk'}=conjg(W_{v'c'vck'k}).
Wannier90 uses different gauges for the wave function, so I take the Wannier definition matrix and multiply it with the diagonalization matrix to get the additional phase, and then multiply it with the states in the Coulomb matrixes.
Yambo-BSE results in a sharp peak with a binding energy of about 0.5 eV, my result also shows a peak at this position but is really small, and other peaks are in the wrong positions. I've checked to rule out the causes of the broadening factor, the number of kpoints sampling dependence and the number of bands to include the excitonic effect.
I don't know if I handle the Coulomb matrixes correctly, especially in the unit of the potential is Ha (I see from the thread) and the division by volume in the potential, which causes me to believe I only have to sum up all kpoints to get Coulomb contribution to the SBE and don't have to divide it by any factor. I also use Yambo grid as an input for Wannier90, which should result in the same spectrum as Wannier90 kmesh.pl because I think the wave function is the same for kpoints different by a K unit vector, but the spectrum is somewhat different.
I hope you can help me.
Best regards,
Minh
I'm trying to solve Semiconductor Bloch Equation (SBE) to obtain the optical spectrum for TMD materials.
My scheme is to use Yambo to generate Coulomb matrixes input for the SBE and Wannier90 for velocity and other matrixes required for solving SBE in velocity gauge.
I've got a good agreement for the independent particle absorption spectrum (without including Coulomb matrixes), compared with results from Wannier90.
But when I include Coulomb matrixes in my calculation, I can't get it right compared with Yambo BSE result.
I output the Coulomb potential follow from the forum link: viewtopic.php?p=6693&sid=44332c65502e15 ... b688#p6693
Yambo grid differs from Wannier90, so I modified subroutine k_ibz2bz.F to get the grid. The Coulomb potential from Yambo seems to exclude the symmetrical part (k<->k'), so I expand it by W_{vcv'c'kk'}=conjg(W_{v'c'vck'k}).
Wannier90 uses different gauges for the wave function, so I take the Wannier definition matrix and multiply it with the diagonalization matrix to get the additional phase, and then multiply it with the states in the Coulomb matrixes.
Yambo-BSE results in a sharp peak with a binding energy of about 0.5 eV, my result also shows a peak at this position but is really small, and other peaks are in the wrong positions. I've checked to rule out the causes of the broadening factor, the number of kpoints sampling dependence and the number of bands to include the excitonic effect.
I don't know if I handle the Coulomb matrixes correctly, especially in the unit of the potential is Ha (I see from the thread) and the division by volume in the potential, which causes me to believe I only have to sum up all kpoints to get Coulomb contribution to the SBE and don't have to divide it by any factor. I also use Yambo grid as an input for Wannier90, which should result in the same spectrum as Wannier90 kmesh.pl because I think the wave function is the same for kpoints different by a K unit vector, but the spectrum is somewhat different.
I hope you can help me.
Best regards,
Minh