 Okay so our next speaker is Xieok Xiong Liu from the University of Zürich and he will talk about the symmetrization of very curvature in the magnetic moment. Okay, hello everyone, I'm very happy to be able to talk today. In this talk I will talk about how to symmetrize the very curvature and the magnetic moment with one-year interpolation. At first I will talk about why I need the symmetry. The recently I'm doing, I'm trying to calculate the conductivity as a response function of different order of electric field and magnetic field and all the conductivities can be solved by a Barry Boseman equation and each of the conductivity should be a function of energy and very curvature and magnetic moment. For example, looks like this integral for something over all the brain zone and here is a nonlinear anonymous Hall effect conductivity. So to make the conductivity converge we need a very dense k-grade and we need a lot of time to calculate. A easy way to reduce the calculation time is only to calculate the symmetry reduce the k-grade. But usually the if we use one-year interpolation, the one-year function slightly break the symmetry a little bit like this one. Here is a plot of a very curvature z component of one band from Togono-Talurium. The Talurium have C3 rotation. We can see is roughly half the C3 rotation but for the details not it's clearly here. It break a little bit C3 rotation and the details around here as well. If we only calculate the reduced k-grade it's like this. Now we do have some symmetry but the first problem that's the two-figure equivalent which is other. I think it's no because you only choose part of the k-point from the non-symmetric data site. You will lose some information and it showed in the result here. Here is a very curvature dipole result based on different energy and some energy use reduce the k-grade converge to a different result with the full k-grade. The second problem is we can clearly see with only part of the k-point the data is not smooth. The non-smooth data will give us a lot of noise when we evaluate the integral. That means we need more k-point to converge. It also showed in the in the result here with full k-grade is have a very good convergence here but with part of the k-grade is not. In this part because there is a well point we should have the bad convergence it's understandable. If we focus on the normal part here without well point and the band intersection the convergence is bad with reduced k-grade. So that's why I am thinking about if I have a symmetric data site can we improve it? The most cheaper way for me to have some symmetry is directly symmetrized the output from 1 year 90 and have some symmetry. Okay the basic idea to symmetrize the Hamiltonian is from Tommy who is the postdoc in University of Freiburg. The basic idea is following. Let's think about two dimensional grade. The yellow dot means original point and each orange dot means the position of autumn. That means at each unit cell we have only one autumn located at the zero point. Let's assume if each autumn only have one S orbital how to symmetrize this hopping? The idea is find all the oh from here only consider the C3 rotation only find all the symmetry in equivalent hopping and sum together calculate the average we can get a symmetric hopping map. It's only for the S orbital but for the P orbital or D orbital we need to rotate the orbital as well like here. If we want to rotate the orbital from beta prime to beta we can build a local coordinate system at beta and another local coordinate system at beta prime which after C3 rotation we can find a projector to the E prime system to E system. The projector it should be the rotation matrix of the orbitals I mean the one-year functions. Okay if our Hamiltonian is respect to the symmetry we can have this relation of the hopings. If we break the symmetry a little bit we can use this function to symmetrize the hopings and this symmetrization method is implemented in one-year tools only for the Hamiltonian. But for me only Hamiltonian is not enough I need to symmetrize a barrier curvature and magnetic moment. And actually if we consider a well point at a high symmetry point the barrier curvature is more sensitive with the symmetry than the energy eigenvalue. If we break the symmetry a little bit the well point will have to move a little bit away and around the high symmetry point. If we consider the high symmetry point the energy eigenvalue will give us a little gap. But the barrier curvature is changed very fast around the band intersection that's the number if we only consider the the high symmetry point the value of the barrier curvature will be definitely different with it on the right on the well point. That's one point and another point is energy eigenvalue is just the number but the barrier curvature is a vector they can break the symmetry in three-dimensional space. That's why I say it's more sensitive. Okay to try to symmetrize it we need to know how to interpolate a barrier curvature. To interpolate the barrier curvature we need to first interpolate the barrier connection A. With Vanier function they have two terms one is A terms and one is D term. The A term we can use a position element to calculate and D terms is used. Hamiltonian is enough. Hamiltonian we already know how to symmetrize it and how to symmetrize the position element. Actually it's the same way. The only difference we need a additional S is the rotation matrix because it's a vector. We need to rotate in a vector again. There is some special event when r equals to zero because when r equals to zero the diagonal element is of the A matrix is Vanier-Centers. If we rotated Vanier-Centers by S the Vanier-Centers sometimes from the home unit cell changed to another unit cell. We need an additional term to push the Vanier-Centers back to the home unit cell. For the magnetic moment the spin part is very simple. We can use the same formula. For the orbital part first we need to find the interpolate function as well like this. The omega we already know and the new term is the G term. The G term is the details is a lot of another matrix which is new for us is the B and C matrix. We need to know how to symmetrize these two matrix. For the B matrix we can use the totally the same way with the A matrix to symmetrize it. For the C matrix which is the tensor matrix so we need two rotation matrix to rotate the vector each vector column. So up to here we can finish the symmetrize the energy eigenvalue and bary curvature and magnetic moment all I need. Let's move to the spin orbital copy. Before I only talk about how to rotate the axis of the orbital if we have spin orbital coupling the simple idea is to add the spin rotation matrix to the orbital rotation matrix P to make it working for a spin orbital coupling system. And for a magnetic system we need an additional input which is the magnetic moment of each autumn. We can use that to to improve the group from space group into magnetic group to find the which symmetry is respect to our system. Okay from the result part we made a same result as before after symmetrization. Here we have a totally perfect symmetry now. From the result side if we only use a reduced k-grade after symmetrization it can converge to very close to the full k-grade calculation before symmetrization and keep a very good convertency this part and this part except the venue point part. Okay that's all I want to talk today. Thank you very much. Okay so there is time for a few questions if you are joining on Zoom please raise your hand virtually with the button or write your message in the chat and we will read your question otherwise fair in presence here just raise your hand and I'll give you the microphone. So let's you have like a px orbital you do a threefold rotation then it becomes a combination of px and py. Yeah and then can you still do this averaging or not? Actually it is to change by group of orbitals just like they're based on the complex axis just like x, y, d. I mean px, y, d we will fix together. Oh so it still works. And then just like like this minimum is maximum authorization. Is there is there a simple answer to why does it break the symmetry or? I'm not sure. I don't know okay maybe somebody can comment on that I'm kind of confused. Okay any other questions? I've seen none so if not we can thank again our speaker and move to the next stop that is actually connected to yours right it's like a sort of part one and part two right the talk of me. Okay so it will be a sort of a continuation of this work. Oh okay so you're giving part one. Yeah the prequel. Oh you can use this one I think you're louder.