 Let's start, so this is part of a Professor Kim talk on topology and thrombium. Okay, so thank you very much. This is the second part of this discussion about the correlation and topology in quantum materials. So this was my outline and I think I covered most of the important topics that I wanted to say last time. So now I'm gonna switch the gear. So in the previous example of particle reading, I was mostly talking about the band topology, essentially thinking about things like topological insulators, bicep numerals. Those are essentially property of the band topology, topology of the band structure. Now for, and those pages were the examples of some kind of symmetry protected topological phase. So on the other hand, the quantum spin liquid state I'm gonna talk about today is an example of intrinsic topological order, topological phase, just like a quantum whole state. So you can take that as an example of intrinsic topological phase or meta. And I mean, you guys will have a more extensive discussion about the spin liquid from a professor Hideo Takagi's lectures tomorrow. So I think my job is just to give you some enough theoretical background so that you can enjoy Professor Takagi's lecture tomorrow. So that's basically what I'm going to aim at. Okay, so there are many classes of spin liquid materials, but today I just wanna focus on only one class of system and one of the most famous one is this honeycomb iridate. So this class of material has this chemical formula. So A is some alkali ions like sodium or lithium. And in this material, the iridium four plus iron that makes up this famous pseudo spin half degree of freedom. They are sitting on a honeycomb lattice and alkali ions are actually sitting at the center of each hexagon. An important thing in this class of material is that the oxygen octahedra around the iridium ions are sharing the edge. So under this condition, these gentlemen, George Jekli and Kenya Kallugin, they show that if you start from this pseudo spin half degree of freedom, remember that that was obtained as a consequence of strong spin over coupling in the iridium ions, then you can end up with the interesting spin model. And here, the spin model is essentially a bone dependent ising interaction. So for example, depending on the bone direction, either red, blue, or green direction, for example, along the blue direction, you only have an ising interaction along say x component of spin. But in the other bone direction, you get a different kinds of ising interaction. And the reason why this model is interesting is that if you look at this model, then try to solve this model classically. Let's don't think about the quantum mechanical model yet. If you try to solve this model classically, then you will end up with exponentially large number of equally unhappy classical spin configuration. What that means is that the system is extremely frustrated. There's a macroscopic number of degenerate classical spin configuration. So in this situation, the question is what happens if you include the quantum fluctuation? And that problem was solved beautifully by Kitai. So that's why this type of bone dependent interaction is called Kitai interaction. So this model is bone dependent ising interaction. It's Hamiltonian. This model can be solved exactly. And this is really a very small number of models that we can actually solve exactly. And this is actually the dom model that can be solved analytically. So Kitai solved this model as follows. And he showed that the ground state of this model is a quantum spin liquid state. So he decided that it's useful to represent spin operator as a product of myoniformean. So for example, for x, y, z component of spin, you use one of these myoniformeans, then you multiply there by another myoniformean. So if you're not familiar with the myoniformean, what that means is that these are the fermionic operators in the sense that they satisfy anti-commutational relation. But if you take the Hermitian conjugate of a c-degga, then that's essentially the same as itself. So in this sense, you can think about that as some kind of real fermion in contrast to the ordinary fermion that's sometimes called complex fermion. And it's just the c-degga equals to c. So in that sense, it's a real fermion. And those fermions are called myoniformean. So that's all you need to know. And but in order to satisfy spin-commutational relation with this kind of strange form, you actually have to impose a constraint that the product of all these operators actually has to be one. So there's some kind of constraint under this myoniformean degree of freedom. But if you do so, you can solve this problem exactly. And the way you solve this is very simple. Basically, notice that if I put this representation here, then each spin operates a bilinear in myoniformeans. So usually this kind of spin-spin interaction becomes full fermion interaction. And normally we cannot solve that exactly. But here one can show that this combination of myoniformean operator actually commute with the Hamiltonian. So what that means is that I can take that as a number, a c number, a complex number. And by the time you decide that this is just a number, not an operator, then this Hamiltonian is quadratic. And so it becomes a free fermion problem. And essentially we are all trained to solve a free particle problem, right? So we can solve this model just because it's quadratic. The only certainty is that not every choice of a number for this combination of operator, now it's a c number, is distinct choices. So it turns out that a better quantity to look at is the product of these operators along the hexagon. So it basically has six spawns. You think about a loop product of this operator. And it turns out that this one also commute with the Hamiltonian. And what can show that the eigenvalue of this operator has to be either plus or minus one. So there are a number of choices of Uij that will give you the same eigenvalue of Wp. And those are all equivalent configurations. So they don't represent different physical state. They actually represent the same state. So you can think of that as some kind of gauge degree of freedom. But nonetheless, by the time you can make a choice here, I think I can solve a free particle problem. And it turns out that for the honeycomb lattice, the choice for the ground is very simple. You just take all Uij to be plus one. It's as simple as that. So if you do that, then you can solve this quadratic problem. Then essentially, your C-miniformion becomes your cosine particle. And just like the case in graphene, two-dimensional graphene, you get a direct dispersion at the k point in the brilliant zone. So this is the, the ground is a quantum spin liquid. It doesn't break, it doesn't break trans-relation symmetry. And there's no magnetic order. But you have an interesting quadri-particle, quadri-particle agitation in your system. Those are the Dirac-miniformion, okay. So by this time, you may wonder what's the connection between, say, this mineral-miniformion construction and the spin-liquid construction that I told you about yesterday. Do you still remember that? I told you that spin-liquid is essentially a projected version of the superconductor. I started with a superconductor and I explicitly constructed the spin-liquid wave function. Now here, it looks like there is no connection to superconductor, right? The question is, what's the connection between this picture and the picture that I presented yesterday? So here is the connection. So it turns out that you can actually make a following unitary transformation. So what you do, you take a linear combination of two of the mineral-miniformion, another combination of mineral-miniformion, then call that some-formion with the up spin, some-formion with the down spin. Just rewriting the operators, right? So instead of using this four-miniformion, I'm using two complex-formion, two ordinary-formion. And if you do that, then you can show that this representation of spin, namely, in mineral-formion representation, spin operator looks like this with this constraint that you can rewrite this like that. So now a spin operator can be written in terms of this f-formion operator like that. And the constraint on the mineral-formion now becomes this constraint, namely that the number of-formion per site is exactly one. Now what happens to Kittai Hamiltonian is something like this, one can show that that Hamiltonian looks exactly like a DCS superconductor Hamiltonian. So he's hoping term, he's a pairing term. It's just that it's not a spin-singular pairing, it's actually a spin-triple pairing. And also because of the fact that original Kittai model has a bond-dependent interaction, the pairing amplitude or the structure of the pairing, triply pairing in x, y, z bond, they are actually different. So in terms of say, I think, you know, you learned about t-vector in triply superconductor, right? From previous lectures. If you like the t-vector for a different bond direction, it's different in this type of spin-triply superconductor. So notice that this is essentially a BCS form, but I have this strong constraint, the number of particles exactly one. That's why the ground state of this model is not a superconductor. The ground state of this model is actually insulator because I have this exactly one particle beside constraint. Remember that? Yeah. So that connection exists here. So even though it's written differently, you see that one can still think about this Kittai spin-triple ground state as a projected triply superconductor. So often in the literature, you encounter terminology like you want spin liquid and G2 spin liquid. So the idea is as follows. Imagine that I had the Hamiltonian like this, but imagine that there was no pairing term in your Hamiltonian. It just then it becomes just free particle like problem with this constraint. And in that case, you can easily convince yourself that I can do the following phase transformation for formula operator. And for the Hoping amplitude, I can make some kind of gauge transformation like this. If you assume that this matrix element is a magnitude and phase. So and this Hamiltonian is invariant under this type of phase transformation and this type of phase transformation for the phase. And this is exactly like U1 gauge transformation in electromagnetism. This is essentially A plus gradient set up, right? So in that sense, this Hamiltonian will be invariant under U1 gauge transformation like this. So whenever you encounter a situation like that, then this type of spin liquid is called U1 spin liquid, which there is no pairing term. On the other hand, in the case of Kittai spin liquid, pairing is always present. So now the Hamiltonian is not invariant under this type of U1 transformation. In fact, it's invariant only under some discrete transformation, namely that you can change the sign of this formula operator. You can think about the corresponding sign changes for Hoping and pairing. And the Hamiltonian is invariant under that sign transformation. So that's why it's called Z2 spin liquid, but sometimes called the Eijing gauge theory. Yeah, yes sir? Yeah, at this point, this is just an index, yeah. But if you look at this representation, then you can see that you can clearly identify this as some kind of spin projection. But this representation is not unique. Another question? Yeah, would you say it? Where does the, yeah, yeah. No, I mean for the Kittai model, for the Kittai model, maybe I didn't emphasize this clearly, that for the Kittai spin liquid, this Uij is some number, not an operator. That's why I could solve this model exactly, yeah? Okay, now imagine that. Imagine that I start from this model and instead of using my other representation, I just decided to use this representation, yeah? A lot of people will do that, right? Then you end up with this model like that. But you see the beauty of the Kittai spin liquid is that in the Kittai model, both pairing and hoping amplitude, those guys again becomes a C number so that it becomes a quadratic problem. If I take an arbitrary spin model, imagine that you cook up some model, give it to me, and you ask me to solve it, then I can do some kind of mean field theory like this, but I know that this combination of operator is not a C number, and I have to make an approximation. But in Kittai spin liquid, your approximation becomes exact, if you like. So if you like, in Kittai spin liquid, your mean field theory becomes exact, and that's why it's solvable, yeah? Okay, so it's by construction. That's all right, so the question was, the question was in the original Kittai model, there was a gapless, there's a gapless dilacformion, and the question is whether we can still see that here, yet the answer is yes, if you diagonalize this, if you diagonalize this Hamiltonian using your portable transformation, then that portable quadriple particles essentially have the same dispersion, essentially the same particle, yeah. Okay, so there is the connection, okay. So now that's just a model, that's just a model, and you could add other interactions to the Kittai model, like here basically they were adding a Heisenberg interaction, and asking how stable the Kittai spin liquid is, right? So if Kittai spin liquid is so fragile, say if I add an arbitrary additional interaction, if this immediate is destroyed, then we have no chance to see it in the real material. So you want to see some stability. So here they demonstrated that if they add a Heisenberg interaction like that to a Kittai interaction, then indeed there's a finite region in the phase diagram where the Kittai spin liquid is stable. So it doesn't immediately become unstable. Eventually for Heisenberg interactions very large, then you can get into a magnetical order state, for example if Heisenberg interaction, anti-ferromagnetic Heisenberg interaction on the honeycomb lattice is a domino interaction, then obviously you will just end up with a Nell anti-ferromagnetic, just up-down, up-down spin configuration. It turns out that if the strengths of two terms are comparable, then you have some kind of compromise state, that's what people call a stripy anti-ferromagnetic. So in principle, but main message here is that Kittai spin liquid can be robust to a small Heisenberg interaction. And in fact, you don't even have to do this calculation. If you remember that there were actually gavel-less elementary excitations in the Kittai spin liquid, then you realize that what it means is that the densest of state of this quasi-particles goes to zero, the zero energy limit. And because of that, any short-range interaction between my and our family will be an irrelevant perturbation in the real-migration group set. So we know from the real-migration group argument that this spin liquid cannot be immediately unstable. And this computation clearly shows that. So at the beginning, people thought that this may be a good model for the sodium medi-day or the lithium medi-day, the material that I talked about, the honeycomb medi-day. So maybe you may get excited to see either spin liquid or if you find the stripy anti-ferromagnetic, you could have claimed that you're actually pretty close to the spin liquid ground state. Even if you don't get into this space, if you discover this space, you may have claimed that, oh, okay, I could have been there, for example, things like that. But people did experiment, and people found that none of this thing actually happens. This is one of those things that theorists predict something but experimented to discover something else, right? So there was a scattering experiment on sodium medi-day and they found that magnetic order is not any one of these. In fact, they found the zigzag magnetic order. So this zigzag magnetic order, if you look at it very closely, it's different from stripy anti-ferromagnetic. So quite frankly, I have never understood why this is called striped, this is called zigzag. This one looks also striped to me, but it depends on how you look at it, right? But they are different magnetic order, okay? So now the question is, why this happens? Why is such a prediction doesn't agree with the experiment? So it turns out that the real spin model is much more complicated, and I'm gonna explain why. So it'll be really nice if Kittab interaction is the only spin-spin interaction I have in this material. It turns out that if you try to write down the most general exchange model on the honeycomb lattice, but imagine that you only want to take into account nearest neighbor interaction, then this is the most general form, okay? So in principle, all of these terms are allowed. So this is the usual Heisenberg interaction, the Kittab interaction. This is what we call a symmetric or an isotropic interaction. So to be more explicit, the structure of this interaction looks like this. So for example, if you look at the X-bone, remember that Kittab interaction had this Heisenberg interaction along the X-direction. This additional interaction has an off-diagrammatic segment, for the X-bone, I have YZ, ZY. For the Y-bone, I have ZX and XZ, et cetera, et cetera. The index is basically staggering. It has to be off-diagram. And if this sign was negative, then this interaction could have been a Jaloczynski-Moria interaction that you are familiar with. It turns out that on this lattice, the Jaloczynski-Moria is actually not allowed because there's an inversion center at the bone. But you allow this symmetric combination. You don't allow anti-symmetric combination. Okay, good, so okay, right. Okay, so I'm gonna sort of sketch the derivation of this model because it doesn't really exist in the literature, so I'd like to show you how this general model may be derived. So you don't have to think about spin exchange interaction between, say, two iridium local moments sitting at two nearby sites. So I'm gonna pick one of the bone. Here I'm picking the red bone. Red bone. So the orbital configuration is slightly different for different bone directions, so I have to pick a one-direction like this. So here I'm thinking about a T2G orbital at this site, another T2G orbital at site two. Then there could be all kinds of hopping processes between the T orbital is here, T orbital is there. And one way to characterize this is to think about some kind of wave function overlap or orbital overlap. For example, I could have direct iridium-iridium orbital overlap. That's the way that the electrons may hop or I can go through, say, the oxygen orbital here. So I first go to the oxygen orbital and go to another D orbital, et cetera, et cetera. So this process, going from D orbital to P orbital of the oxygen and going to another D orbital, so this would be what you may call a super exchange process, and those guys corresponds to the direct process, meaning it comes from a direct overlap of D orbital. So the point is, if you have, say, 3D transition metal oxide, if you're dealing with a 3D orbital, and then typically this is the dominating process, meaning oxygen mediated hopping is usually dominating. The reason is that the D orbital, the size of the D orbital is so much smaller, but by the time you go to a 4D or 5D element, the D orbital size is big. So it turns out that the direct overlap is as strong as the indirect overlap like this. So you actually have to take into account all of these processes. So, for example, using all these processes, we can construct the hopping matrix element, and all the hopping matrix element consists of either a super exchange-like amplitude or it comes from a direct overlap. And notice that if I only had a super exchange-like process, say, hopping through the oxygen, then I only have this type of hopping amplitude, that those guys will be absent. Good. So, this is an inherently multi-orbital system. So even though I'm thinking about an effective J-half system, remember that there were actually J-effective 3-half state below, they are filled out, but when I'm deriving a speed exchange model, then I have to second-order perturbation theory. Remember that when you do the second-order perturbation theory, you have to take into account all the excited state as an intermediate state. So, you know, taking into account all the excited state, I cannot just restrict myself to the J-effective-half manifold. I have to worry about what happens to those J-effective-3-half manifold. So I have to actually worry about all five orbital. That's why I have to write down some kind of multi-orbital Hobart model. So here, the simplified version of that, this is called Kanamori Hamiltonian. So there's Hobart U, screen kilometer action. J-H is the Hons coupling. And then S is the total spin quantum number. L is the total angular momentum quantum number of each atom. So this is the Kanamori Hamiltonian for two atom. And then you use the hopping process of perturbation to this type of interaction term. So this is the interaction term. This is the hopping term. So you do the second-order perturbation theory. And you wanna derive a model in terms of J-effective-half spin, J-effective-half degree of freedom. But as an intermediate step, so that's what I mean by end state, these are excited state. And in order to take into account, you have to actually worry about configuration, what happens to the electrons in the J-equal-three-half manifold. For example, I can dig a hole in the J-effective-half, three-half manifold and move that particle to the J-effective-half-half band. It's half-level, it's things like that. So those are all possible excited state. So when you do all that, then you arrived at the model. But now imagine that initially, I didn't worry about direct overlap between the orbital. And imagine that I only had a super-exchange process like that. So this just represents the hopping between P-overl and D-overl square. The delta PD is the energy level difference between P-overl and D-overl. If you do that, then you only get a guitar interaction. So if you only have oxygen mediated hopping, then you only have a guitar atom. So that will be an ideal situation. But unfortunately, other processes are as important as this process. So if you include all the direct overlap between D-overl, so this dd pi, dd delta, you don't have to know exactly what they are. They represent the direct overlap of D-overl. And those guys basically arise. And because of that, you're not just getting a guitar interaction, you also get other kinds of interaction term. So I mean, in principle, this tells us how we may get a pure guitar interaction. Basically, one way to get there is to suppress all the direct overlap. If the super-exchange like process is dominating, then in principle, I can go to the regime where the K is dominating. That's essentially what you'd want. But practically, it turns out that there's hard. The reason why it's hard is as follows. So from the first place, in order to write down a model like this, I have to use this J-effective half degree of freedom. And that was obtained by taking a large spin-over hopping level. And but typically, large spin-over hopping is obtained when your local moment element is heavy, right? The heavy element usually has a large spin-over coupling. But usually, the heavy element has a larger orbital. So if you try to achieve a large spin-over coupling, typically, the outer shell, the orbital is big, so that it's very difficult to avoid this relatively large direct overlap. So that's why it's so difficult to achieve a pure guitar model. So that's one thing that you should have in mind. Okay, good. Okay, so you will hear more about this from Hideo Takagi tomorrow, but here is the sort of summary of where we are in terms of the discovery of guitar and quantum spin-over. So this is the example of 2D honeycomb system. So these are all layered materials. The iridium basically makes the honeycomb lattice. This is quite special. This is pre-recent. The Hideo Takagi's group took this material and they managed to replace some of the lithium by hydrogen. And it turns out that this material, without doing anything, does not show any magnetic order. And this is also another interesting material called alpharotenium chloride. Here, ruteinium is actually a 4D element, not a 5D element. But nonetheless, this material shows some characteristics of this J-fective half-degree freedom. There are also three-dimensional versions of this, but the important thing is that most of this material actually order magnetically. So this material, all three order magnetically, only this new material doesn't order. And these three-dimensional material also order magnetically with the very complex, incommensurate spiral order. So now the game is to find out how you may suppress the magnetic order so that you may be able to reach some kind of putative quantum spin liquid state. For example, for this material, this hydrogen substitution was used for ruteinium chloride III. People apply some kind of in-plane magnetic field. It turns out that if you apply a magnetic field, you can suppress the magnetic order. For those materials, there are some pressure experiments. When you apply a pressure on the system, it turns out that you could suppress the magnetic order. So the question is, what happens after that? So you lose your magnetic order, but you still have to characterize what the ground state or elementary excitations are. So I'm gonna tell you a little bit about mostly theoretical ideas about these three-dimensional honeycomb lattice. So it's called hyperhoneycomb lattice. So there are two kinds of three-dimensional system here. Again, I'm showing you the lattice of the iridium ions. So each iridium ion is connected to three other iridium ions. But in this structure, this is called beta polymorph, gamma polymorph. In the beta polymorph, you don't find any hexagon here. So maybe calling that honeycomb is a misnomer, but that's why it's called hyperhoneycomb. If you are beyond the honeycomb structure. But local connectivity is the same. And for the gamma polymorph, you do have a hexagon, but notice that this is not a layered honeycomb structure. This honeycomb plane is alternating and it's arranged in the three-dimensional structure. And they also, both of them automatically. And in both cases, the iridium ions are surrounded by edgyshering octahedral, just like a two-dimensional system. Okay, so you may wonder what happens to the guitar model on these lattices. So I told you the guitar absorbed this model in two-dimensional honeycomb lattice, right? So you can do the same. Everything will be the same. Everything will be the same. The only difference is that now these c-myanophomians live on a three-dimensional lattice. And it turns out that if you solve this model, remember that for two-dimensional honeycomb lattice, you get a Dirac spectrum, just like a graphene structure. For three-dimensional system, it turns out that this myanophomian has a nodal line spectrum, nodal line spectrum. And interestingly, in all of these cases, the capacity coming from those myanophomians should be given by temperature square, t-square, if you discover this, yeah, you discover this. So in fact, you can think about all sorts of related structures. So you start from a hyperhoneycomb. There's no hexagon. For stripy honeycomb, there's a one-hexagon layers. And you can have two of those guys and three of those guys, and eventually you may reach the completely layered structure like a 2D honeycomb system. So in all of these cases, you can solve guitar model exactly. The exact solution exists for all of these lattices, okay? All of these lattices. And in all of these cases, the Hamiltonian for C-phomium has this structure. So diagonal matrix element is zero, but only the octagonal matrix element is finite, okay? And it turns out that this structure of the Hamiltonian is intimately related to the solution of the guitar model. In fact, there's a symmetry. So I'm not gonna go into details, so just some people call that BD1 class, maybe it doesn't mean to you. So it doesn't mean anything to you. So it doesn't really matter how you call it. There's certain symmetries coming from the Hamiltonian. So important thing is that if you have a structure like this, it's pretty clear that in order to get a zero energy spectrum, I have to require that both real and imaginary part of determinant of D has to be zero. Only then I get a zero energy eigenvalue of such an Hamiltonian. So if it's not clear to you, you can go home and do it in two seconds. So what happens is as follows. So in two-dimension, one of these conditions will give you a circle. That's the circle in brilliant zone will give you zero energy, satisfy this condition. The imaginary part will give you another circle in order to have both of them to be zero. You have basically whenever these two circles meet, that's the solution, right? So that's why you get a direct point. In three-dimension, each condition will give you surface of the sphere. There's another surface of sphere. Two surface of sphere intersect, you get a nodal line. That's why solution is a nodal line spectrum, yeah? And again, the density of state goes to zero in the low energy limit. So by the time you get to there, get there, such a spin liquid is actually quite robust. It's robust against a small perturbation. So it's a good thing, okay? And this kind of system is also characterized by some kind of topological invariant. So here I'm showing you the nodal line spectrum. This is actually the zero energy contour in the three-dimension in brilliant zone. And if you take a loop around this zero energy contour in the momentum space and take this combination and do the integral over all this line, then you can show that this becomes some kind of integer topological invariant. Actually, that's one way to characterize such a face. So you can think of that as some kind of spin-on semi-metal. So there is an ordinary formula version of this. It's called topological semi-metal. This is like a spin-liquid version of that, spin-liquid version of that. So it's, yeah. So, and another interesting aspects of this, the Githam spin-liquid solution is that now I can deform this contour in momentum space around the ring to a one-dimensional contour. Think about a closed contour, but going from minus infinity to plus infinity, going all the way and come back. I can deform that. I can deform that line. So it turns out that if you evaluate this integral along this line, say within the ring, then you can show that this one-dimensional system, you just take the momentum position along this line. Now think about that as a one-dimensional system. And that one-dimensional system is non-trivial, but the other one-dimensional system is trivial in the sense that if you evaluate this invariant, this is one, this is zero. So practically what that means is that for this trajectory, since this system is non-trivial, there is an interesting boundary state if you have an open boundary. But for this direction, even if I have an open boundary, there is no interesting edge state, just like a topological insulator. So in fact, because of this, if you go to a surface, if you go to the surface or the open boundary of the system, then one can show that depending on the direction, so if you think about the projection of this ring to the surface brilliant zone, then within this circle, that's the projection of this ring to the surface brilliant zone, then you get a large number of zero energy mode. And you get actually flat spectrum. The spectrum is flat, but you have a large number of zero mode within the circle. And depending on the open boundary direction, it may look slightly different. So you get an interesting boundary state. So here, of course bulk is interesting, but boundary is also interesting. So this is three-dimensional Kittai solution has this interesting property. I don't know whether we can use that ever, but it's good to have in mind. So I told you that for this three-dimensional system, there was a magnetic order. So experimentally it's found the very complex in convention magnetic order. And we can explain this by using this model that we wrote down. Yeah, so for the surface, you do have a huge density of state. Huge density of state. So in fact, interesting possibility is that bulk is stable to the short, the small perturbation, but the surface may not be stable because of that. So it's possible that you may get an interesting instability at the boundary if you add some other interactions. So experimentally the material order magnetically, you don't find the Kittai spin look here. So okay, so then you may say, I forget it, but the life is not as simple, right? Life is supposed to be more interesting. So first of all, we can actually explain this just by looking at the classical model. It turns out that the material actually has a very large Kittai of interaction. That's a good thing, but bad thing was this additional interaction is equally large. And the magnitude of these two couplings are actually comparable. And because of that, you just get a magnetic load state. So, but good news is that Hideo Takagi's group applied a large pressure, hydrostatic pressure, and they managed to make this magnet of the disappear. Yeah, he's a magician, you know? So, so you will hear about that. You will hear about that. So after hearing about the experiment, we went back to our steam model and tried to see how those Kittai and gamma interaction may change on the hydrostatic pressure. So you can investigate this using say some kind of electronic structure calculation. Then what we found is that your Kittai of interaction, this K is the Kittai of interaction. Kittai of interaction generally becomes weaker. So you may think that there's a bad sign because I'm making Kittai of interaction smaller and smaller. But you are actually making gamma interaction larger and larger. Yeah, larger and larger. So what is interesting is that if you take a steam model now I have only gamma interaction, only gamma interaction, it turns out that that model is also highly frustrating. So classically, again, you end up with experiencing a large number of equally unhappy classical steam configuration. So it is possible that even that limit may produce an interesting spin of Kittai. We don't know. This model, we cannot serve exactly. So, and I'm not as smart as Kittai, so I cannot serve it. So what happens is that, remember how we derived the spin model? And in order to derive that, I used all the hopping processes and hopping processes were expressed as an overlap integral of orbital. That's how you change the hopping amplitude. Then using the same exercise, I can derive the spin model. Then all the exchange parameter will be changed under pressure. That's the way that it is estimate. Yeah. Okay, so let me see. How much time do I have? About five minutes. Yeah, yeah. Question? So I don't know. I don't know the answer. I mean, he's asking whether I expect to see a new spin locate when Kittai interactions almost disappearing or somewhere in the middle. I wish I know that answer. I don't know the answer. Yeah, question? Three-dimensional lattice. Yeah, that's right. But classical spin lick really means you end up with an exponentially large number equally unhappy spin configuration. And it doesn't order into any particular magnetic loader state even if you decrease your temperature. Now, what happens for the quantum model? We don't know. Nobody was able to solve that. You can solve a classical model and confirm that there is no magnetic order down to very low temperature. But we don't know what happens to the quantum model. That's essentially what you want to know. Okay, so since I don't have a lot of time, I just want to mention, I want to briefly mention connection to, I have more time? More than 10 minutes. 10 minutes, okay, okay, yeah. But, how much time do I have? Oh, okay, I have a lot of time there, okay, good. No, I want to use it. So, another interesting material is this Alpalusinium chloride tree. So, I don't know how much Hideo is going to talk about this. Are you going to talk about these two? Only these two slides, okay. So, this material is also an interesting candidate system for the spin rocket. And the point is that Lutinium III plus in this system is also D5 configuration, even though it's a 4D not 5D. So, you can in principle use the same kind of picture. You may say that this Lutinium III plus iron may also carry this effective spin hop. In that case, since Lutinium III plus are sitting on a honeycomb lattice, you may expect to see a key time spin rocket physics. But again, this material also order magnetically with the zigzag order, just like any other honeycomb everyday. But, there was this very interesting set of experiment claiming that they can see something more than the magnum in the magnetically order state. They somehow see continuum of excitations here. And the idea is something like this. So, this is the neutron scattering intensity as a function of energy. And these two peaks are supposed to represent the magnum excitations, order the magnetic state, the zigzag order. But on top of that, they somehow see a broad continuum of excitations. And when they increase the temperature above the narrow-growing temperature, so narrow-growing temperature like a seven Kelvin, then magnum disappears, of course, but this continuum still stays. So, the interesting idea was, well, imagine that this continuum comes from some kind of underlying spin rocket. So, if you had some spin rocket, then in the neutron scattering, if you are measuring a spin flip process, right? Neutron measures spin-on excitations. So, magnons are spin-on excitations, for example. But if spin-ons are usually, they usually carry spin-hop quantum number. So, in order to create a spin-on object, you have to excite two of them. So, you have to excite spin-on and this spin-on pair excitations to generate a spin-on excitations. And these pair excitations, just like a particle hole continuum in metals, they will give you some kind of continuum. And the dispersion of the spin-on essentially determines what the dispersion of the bottom of the continuum looks like. So, this will be a generic expectation. There's some kind of threshold energy for pair excitation, but that threshold energy has some two-dimensional dispersion. So, that's what you may want to see. So, there was a suggestion that this continuum may be something like that. So, there's one idea. But then, the question is, why then in this case continuum coexist with the magnetic order? Doesn't mean that the spin-on coexist with the magnetic order. This sounds weird, right? So, some people propose that, perhaps there is some unknown coupling constant here that you are in the zigzag order state, but this zigzag order is kind of weak and maybe it's about to be destroyed, what we destroy. And then, the finite temperature property here may look like the property of the spin-on rocket. In fact, it turns out that for the quitaib spin-on rocket, there is no finite temperature transition. So, the quitaib spin-on rocket grounds there. If you heat up the system, there is no finite temperature transition. So, it's only a crossover. So, you may expect to see that, well, if this spin-on rocket state is very powerful, then you may see some signatures of that, yeah? Question? Good question. There is no theory. It's a speculation at this point, right? Yeah. So, I support your theorist. Yeah, so, you're welcome to work it out. Good. Another interesting discovery was that when they did look at the momentum dependence of this continuum, they see this interesting star shape, this person in a two-dimensional brilliant zone, in the low energy regime. And they don't quite see that in the high energy regime, but they see this. And the reason why this may be important is that if you solve pure quitaib model, you don't see this structure in the dynamical structure factor. So, it's clear that if this is the signature spin-on rocket, we know, we know the underlying Hamiltonian is not a pure quitaib model, something else. So, that much we know, okay? So, but then very recently, people managed to suppress the magnetic order by applying an in-plane magnetic field like a eight tesla, then the zigzag magnetic order disappear. So, then obviously you will get into some kind of paramagnetic state. And most boring scenario will be, you immediately go to a fully polarized paramagnetic. You destroy the magnetic order, all this means a polarized, because you apply a magnetic field after all. They all polarized, then you have a transition from boring magnetic order state to boring paramagnetic. That's it, right? They're the boring scenario. But it turns out that the experiment that is in Japan, Kyoto University, is led by Yuji Matsuda. They measure the thermal hole conductivity around this area before you reach the fully polarized paramagnetic. And they found that for a range of magnetic field, which is, window is actually quite small, but they found this almost constant quantized, they call quantized value of the thermal hole conductivity. And the reason why this is interesting is that, yeah, so somewhere before you reach the fully polarized state, the reason why it's interesting is that, according to Gitae, if you take a pure Gitae model and apply a strong magnetic field, that you can get into a different kind of spin-lucky state. There's Tanya Mussel-Smithy breaking spin-lucky state, that the bulkiest gap, your Mylana excitation acquire a mass. But you get an interesting boundary state, just like integer quantum state. And this boundary state is some kind of Mylana chiral mode. And that will give you quantized kappa xy. And it's quantized in the sense that, the unit of kappa xy to this kb square by h bar. So if you use what people call convenient unit, meaning every constant equals to one, then it should be given by pi by 12. So Gitae emphasized that this coefficient here has to be half, because for the integer quantum state, it actually has to be pi by six, because the edge current is k by usual fermion. But here, because of fact that boundary fermions and Mylana's, the coefficient actually has to be half. It's just that number of digital freedom basically is half. Remember that I can make one ordinary fermion by combining two Mylana fermions. So, you know, in terms of Mylana fermion, it's half of the ordinary fermion. So that's why the coefficient half. And you can also see that at a finite temperature, a finite temperature, if you go to a low temperature, you actually reach this quantized value. So this is theoretical calculation of the pure Gitae model. So you should reach the pi by 12. So the claim in the experiment is that they see that. Claim in the experiment is that they see pi by 12. That's interesting, isn't it? I think some people don't believe it. But that's basically the data. So then the question is, okay, then what's really the model for this material? So there was a serious amnesia computations from various different groups. And I think that most of them converge to a model like this. They don't precisely agree, but roughly this is kind of the model that people end up with. So you do have a Gitae interaction. You also have this gamma interaction that I told you about. There's also a Heisenberg interaction, but it turns out that the third neighbor Heisenberg interaction is the largest instead of the nearest neighbor. And you may think that this is weird. Usually nearest neighbor has to be the biggest. But the reason why third neighbor, in this case, it turns out to be big, is number of exchange paths for the third neighbor is actually larger. So this is a competition, right? Of course, for the individual exchange path, nearest neighbor is, of course, dominating. It has to be biggest. But you have to multiply that by the number of exchange paths. So when you work out all the details, it turns out that third neighbor, somehow is the biggest for this system. But you could keep both third neighbor and nearest neighbor. Actually it turns out it doesn't make a huge difference. But important thing is the largest coupling is not this J. Largest coupling is K and gamma. But you cannot even know a gamma term. Both of them are very strong. Both of them are very strong. And we know the sign, the ferro-like type and anti-ferro-like gamma interaction. So, okay. So from now on, so I'm gonna be a little bit biased because I have to present my own research result. So far, I have been trying to be very objective, but now I'm becoming a little more subjective. Yeah. There is a neutron scattering data in the field. And around that critical field, they do see a continuum, some kind of continuum showing up around the gamma point. But normally, if you really reach the spin-like state, you would expect to see that the continuum exists in different parts of the momentum space. But in their case, it's more or less concentrated near the gamma point. Exactly, yeah. As, yeah, just. We don't know. We don't know, yeah. I mean, there's, there's. No, no, no, something changes in between. But by the time you suppress the magnetic order, looks like that continuum say, you know, shows up again. It's something like that, yeah. Okay. Very, very high field, yeah. All polarized. Should be there, yeah. Yeah. Yeah, I think so. I think that some kind of multi-magnon bounds that should condense below the single-magnon. Yeah, yeah, yeah. But then there are so many possibilities. You can have like dimetic order and, yeah. So one thing we try to do was, since K and gamma is the largest energy scale in this problem, why don't we just start from a K-gamma model? It's a natural thing to do, isn't it? So I take a K-gamma model. Just, you just keep K and gamma. But you know what to make the story a little more interesting? We make K-type coupling along X and, X, Y and G bond to be slightly different. I give out some, an isotropic here. So when they are completely isotropic, along this line, we found that there's a force of a transition from K-type spin circuit. So this is the K-type limit. This is the gamma limit. So the pure K-type, pure gamma. We found that we didn't find any obvious magnetic order in the entire line, but there was a force of a transition from one state to the other. Yeah. We know that this guy has to be a K-type spin circuit because this one is continuously connected to the pure K-type limit. We don't know what this guy is. And it's separate by force of transition. But what is interesting is that when we introduce an isotropy, this force of transition line actually terminates somewhere. It looks as if I could go around this point from here to here. That's the numerical result. Okay. So, you know, and this was done by some kind of exact diagonalization and, you know, then symmetric diagonalization group computation. So the question is, what's the nature of this phase here? You know, is it really distinct from K-type spin circuit? Or maybe there's a possibility that this, some kind of paramagnetic state exists here and that has some character of K-type spin circuit. And another interesting thing is that if you take this model, simple model, K-gamma model, and compute a dynamic structure factor, and you do find the star-shaped dynamic structure factor, as seen in the experiment. So there must be some truth in that model. Now if you add a J3, if you add a J3, indeed you go into a zigzag magnetic order. So the model actually has some correct characteristics that you could go into a zigzag magnetic order state by adding J3. And if you don't have it, then it looks like you get some kind of quantum paramagnetic. Yeah, okay. So here is the speculative picture of many people, not just me. So imagine that alpha-routine cloratory somewhere here. And there's some control parameter. In this case, I'm using third neighbor J. Yeah? So here, then I'm applying a magnetic field. Then perhaps what happens is that my magnetic order is suppressed. So I, you know, in an ideal world, I may get into a Kyler spin circuit state. Then eventually I go to polarized paramagnetic. Okay? So there's no computation here. It's just a speculative phase diagram. So this is one possibility. If you truly believe using Matilda's thermal conductivity data, then this could be a one possibility. So in Kyler spin circuit, when you apply a magnetic field, you immediately become a Kyler spin circuit. Yeah. Here, of course, we don't know. Yeah, I think I'm gonna skip this. So maybe in the last five minutes, I just wanna briefly mention that now, one interesting work that we have done recently is that we ask this question, what happens to a Kyler spin circuit when you couple that to an itinerary electron? So think about a condo lattice like problem. So imagine that I have some kind of a formula system, local moments are sitting on a hard income lattice that interact each other in terms of Kyler interaction so that if I only have a local moment, they form a Kyler spin circuit. Now I add a conduction electron using the usual condo coupling. What happens then? In the ordinary condo problem, when I add a conduction electron, there's some kind of condo screening, I get into a heavy for me liquid state. That's the usual scenario of, that's what you find a textbook. So now here what happens is as follows. So just, I'd like to remind you that we could rewrite Kyler model in this form, the BCS form. So basically claim was that Kyler spin liquid is a failed superconductor, almost superconductor, but failed to do so. So now I take this, so some kind of project is spin triple superconductor. Now I couple this guy to a conduction electron and I use the textbook slave particle representation of the spin operator and I basically repeat the same exercise. So it's just that one technically mark is that since this model doesn't have a spin rotation symmetry, it turns out that this is to the expert. When you decouple this condo interaction in many different hybridization channel, essentially you have to include all possible singlet and triplet channels. So this is, we have done that. So this is the result. So what is interesting is that when the condo coupling is very, very small, the small condo coupling is an irrelevant perturbation again to the spin liquid. Many because the density of state of my anachromions in the Kyler spin liquid state goes to zero in the zero energy limit. Because of that, any short range interaction added to that system is an irrelevant perturbation. Essentially for small guitar coupling, the guitar spin liquid and your itinerary electron, they happily coexist. They don't want to mix. So that state persists for a while. So you may call that using central nomenclature, you may say the formal liquid star phase. But eventually something happens. But what is interesting is that you get a superconductor. And this is not an ordinary superconductor. We found that we actually end up with a topological superconductor. And it spontaneously breaks a tiny vasosymmetry. It becomes a pharomagnetic topological superconductor. And in this superconductor, the bulk is capped, but the boundary has a Kyler myelophorium. And you can sort of understand why you obtain such a strange superconducting state here. If you remember, or if you know, that if I apply a magnetic field to the guitar spin liquid, I told you that it becomes a Kyler spin liquid. And this state, I can do the exact same internet transformation from myelon representation to complex formula representation. Then again, I end up with a bichet like Hamiltonian. That Hamiltonian is actually some kind of topological superconductor Hamiltonian. It's just that I have to impose the constraint the number of particles per side exactly one. Now, by putting it in an electron, now a charged degree of freedom is now active, if you like. And because of that, the first thing the copper system wants to do is to take the advantage of that. To take the advantage of the fact that if you break Tary muscle in the guitar spin liquid, it was almost topological superconductor. So by the time you couple this thing to a conduction electron, it immediately wants to be there. Now, difference between the Kyler spin liquid and this guy is that this guy now is a true superconductor. It's not a spin liquid. It's not an insulator. Now it's a superconductor. Your superconductivity is in a way liberated by coupling this thing to a conduction electron. So if you have this, then you have a boundary Kyler monophenium. For this state, indeed, copper XY has to be quantized. And interestingly, if you increase the conduct coupling further, then you suppress this ferromagnetic order. It becomes a paramagnetic superconductor. It turns out that this state is also topological. It just belongs to different class. In this case, there are two counter-propagating edge modes, not just one. It's not Kyler anymore because you have a tiny muscle symmetry back. So one of them moving in a circularly direction. The other guy moving in an anti-circularly, clockwise and anti-clockwise direction. So this is essentially the solution we obtain. So if you're interested in, you can probably see the different reference. But what I'm trying to emphasize here is that this is perhaps a new route to the topological superconductor. So this route has not been explored. Basically using the counter coupling, try to get a topological superconductor. Question? So yeah, I'm basically done. So I just wanted to show you this slide because I think it's interesting. And this may be a new route to get into a topological superconductor. This is the plain slave particle mean field theory that you will do with any condolence model. I mean, you cannot solve it using like quantum decal because sign problem. So at the moment, the only thing that we have is this slave particle mean field theory. Yeah. Yeah, so I think I like to end my lecture with that. Thank you very much.