 All right, it's time to get started with this session. So we have two speakers in this session talking about Dirac materials. The first is Oscar Wafak, who will tell us about twisted biographies. And I believe the next talk is closely related. Jet Pixley will give the second talk about some closely related physics. Thank you very much. And thank you very much to the organizers for the opportunity to speak here. This work was done at the National High Magnetic Field Lab in Tallahassee and Florida State University and was supported by the US National Science Foundation. Jian Kang is my collaborator. He's a postdoc at the Magnet Lab. And this talk will be mostly based on this preprint, which was recently accepted for peer-reviewed publication. So let me start by describing the system. Imagine two monolayer graphene sitting on top of each other. And then you twist one of them by an angle theta. If this angle is small and commensurate, you will generate a new periodic pattern on top of the underlying atomic periodicity of each individual monolayer. And this new periodic potential will then reconstruct the plane waves where the valence or conduction electrons of the individual graphene, the famous Dirac cones. So because of this reconstruction, there will be new band openings. There will be new smaller Brillouin zone, because in real space, this is a larger unit cell. And now we can talk about filling these mini bands in these so-called Moiré Brillouin zone. By the way, this pattern is called the Moiré pattern. So as you can imagine, if this angle gets smaller and smaller, this pattern gets larger and larger wavelength. And in principle, we are now in a qualitatively new regime from the original monolayer graphene physics. Of course, this field has seen remarkable development since the announcement at the APS March meeting by Pablo Herrero-Harrero group from MIT of rather remarkable effects. So let me just walk you through the results and then try to put it in a perspective of what I just discussed. So on the top graph, the x-axis is the carrier concentration measured relative to the neutrality point of graphene. At that point, the conductance shows a dip, but it doesn't go to zero. So we are dealing with possibly a semi-metal there. As we increase the carrier concentration, at first the conductance goes down, but then in a small window of carrier concentration, there is a sharp suppression of conductance and actually an insulating behavior unless you look at very, very low temperatures, where in some of the samples, it can become superconductive in some others. It's not, and it basically becomes insulating. And then if you dope near this point, superconductivity comes in. And if you dope too much, you empty some band because you enter into an insulator. Now, similar effects happen on both the electron and whole-doped side of this system. So experimentalists argue that the insulating state that happens between the neutrality point and this point, where, by the way, we would have four electrons in the unit cell that span this doping range, that this insulating state happens at commensurate filling. If it happens at commensurate filling, presumably this is happening because of some strong correlation effects. And the natural way to think about this problem is then to start thinking about the interaction energy as being the dominant term in the Hamiltonian, ask what kind of an insulating state that would stabilize and then turn on the hopping as a perturbation to see which one of the multitude of insulating states, many body insulating states, get selected. So the natural first step in a theoretical analysis would be to understand what kind of Vanier states we can write down for these four bands and then start thinking about what kind of insulating many body states we could generate. So this talk is about the first part of it. It turns out that the Vanier states are highly non-trivial and there is some debate about exactly how to construct them. So I'd like to walk you through those arguments. This talk will, at this point, have nothing to do with strong correlations, although it is motivated by trying to understand the strongly correlated behavior. OK, so how should we think about this problem? Imagine that the blue is the large, brillant zone of the top layer. The red is the large, brillant zone of the bottom layer. They're twisted relative to each other and therefore the Dirac cones are not exactly on top of each other. Now, if the twist angle is commensurate, we will have a new brillant zone, the more a brillant zone. And these two Dirac points will fold into the corners of this new brillant zone. And similarly, the opposite, so this is coming from one valley and similarly from the opposite valley, the Dirac cones will fold into the same point. OK, now, because we have some interlayer coupling, this will introduce band hybridizations and therefore we'll reconstruct the band structure and for some special values of these angles for the experimentally, for the realistic coupling between the layers, it was predicted theoretically some time ago before this was discovered that some of these bands that you generate can become very, very thin in energy. And there are so-called special magic angles where some of these bands get very, very thin and I will try to make this precise. And so once you have a thin band, you have these narrow bands, you would expect that the correlation effects, the many body effects play some important role. OK, so that's sort of the basic playground in which I would like to discuss the physics. OK, and so in terms of those bands, well, there are, not including spin, four of them. But this would be the top of that four band structure and this would be the bottom. So these two are presumably the trivial band insulators and everything in between, if you have an insulator, should be interesting. It should be due to many body effects. So this in particular would be happening for two electrons or two holes per Moray unit cell. OK, so strictly speaking, that's actually quarter-filling. OK, so to build these vanes days, what we consider is the following structure. We started from a monolayer, let's say, described by the red, which is the bottom layer, and then the blue, which is the top layer. And then we twist the top one by some angle theta over 2, the bottom one by minus theta over 2, and then we introduce interlayer coupling between them. Now, we start from a situation in which the atoms here were in registry and we twist about this position. And we ask, what kind of point group symmetries does this new crystal have? Well, it's periodic because the angle was chosen to be commensurate. It has a three-fold rotational symmetry around the stacked site, the carbon registered site. It also has a two-fold rotational symmetry, which is perpendicular to this main axis. So imagine rotating in the plane of the two monolayers by 180 degrees. It will bring the structure back into itself. And this generates the rest of the group, which in this case, the point group is so-called D3. OK, so as I said, we start with an end with a situation where the so-called AA sites are in registry. Now, we can get to the end of this unit cell in two different ways. We can either track the blue lattice and the triangular unit cell vectors of the top monolayer as well as the bottom monolayer. We have to come to the same point. And so this commensurate structure is therefore described by how many times you go along each of these legs. And so a pair of integers m and n defines this commensurate structure and the twist angle. So we will be referring then to this pair of integers. So if these are some small numbers, well, then your unit cell will be small. These are large numbers, a little bit offset from each other. Then your unit cell can be actually quite large. So many of these band structures have been calculated before the discovery of the correlated effects. And the band structure that we calculate from the model that was published, for example, in Moon and Koshino's paper. And it is essentially a tie-binding model for this large unit cell where the hopping amplitudes, which can be within a layer, either top or bottom, or between the layers, are described by a two-parameter family of functions. These are fixed experimentally, as well as the rest of the coefficients, to match the LDA calculation for the large twist angle, so small unit cell, where the LDA calculation can actually be done. So this tie-binding model has been calibrated. Unfortunately, the LDA calculation for the realistic magic angle is too costly. And so the best that one has to actually compute the band structure microscopically is to fit these tie-binding coefficients for larger angles to LDA. They work there. And then you just extrapolate to small twist angle. So that's what we did. We used these to compute the band structure. And as you can see, the progression, for example, is shown in their paper of these angles. And the set of integers that they show is shown here. So once you have these larger integers, you can see there's a new pattern, clearly new pattern that emerges. OK. Now, it turns out that the experimentally observed, yes, please. Not be able to extrapolate this thing, right? Or maybe at least? Well, they're apparently there. So experimentally. OK. So now, experimentally, it turns out that the magic angle is close to 1.06 or so as measured. And if you use this model to calculate the band structure at that angle, the closest commensurate integers, simple integers, to that would be 30 and 31. It turns out that for 30 and 31, this model gives band structure, which indeed gives four bands. But these four bands are not separated from the rest of the spectrum, I think, from the bottom. And so as a result, one doesn't actually get an insulator, as one should, at the bottom of these bands. Now, people who add the relaxation to this claim that, in fact, that does produce the gaps from the top and the bottom, but we were just simply taking the un-relaxed model. Now, if you want to understand the qualitative effects, like the shapes of the vanille orbitals, et cetera, well, I think you are justified in taking a pair of integers, which is large enough, although not necessarily exactly at the magic angle, where this kind of model does give you bands that are separated. And that turns out to be true for 25 and 26. So this unit cell contains now something like almost 8,000 sites, not quite the 11,000 sites that are there for 30 and 31, but close enough. But qualitatively now, you can reproduce these gaps. So the band structure that we get from here looks something like this. And I will discuss its details in a moment. Now, if we were to zoom in at the gamma point, which is the center of the Moray-Brillant zone, we would find that the two top energy states are degenerate. And if we check how they transform under the three-fold rotation, they form a doublet. They transform as an E representation of the D3 group. The same is true at the bottom of this band structure. On the other hand, if we zoom in to the corners of the Brillant zone, the Moray k and k there, we see one doublet. And this indeed transforms as a doublet under rotations. But then we see split pair of bands. And these are one-dimensional representations, A2 and A1. Now we have a setup that we can try to build our Vanier states for. We have a gap that separates our four bands from the top and from the bottom. And so in principle, we should just be able to construct such exponentially localized Vanier states. Now, before you computed the Vanier states, you looked at the block states. And from the block states, you computed the local density of states, let's say for positive energy. You would discover that this local density of states is peaked at a series of triangle lattice sites, the Moray triangle lattice sites. So this really suggests that the Vanier states should be centered around the triangle lattice sites. So perhaps you have some type of a triangle lattice Hubbard model or something like that. And there are people who try to argue for this kind of reasoning. I will argue to that this is, in fact, incorrect. This was also pointed out by others. Let me present the argument for that. So let us assume that we indeed have four Vanier states associated with the triangle lattice sites. Now let's say that these Vanier states are chosen to have a corresponding site symmetry of the D3 group. So if there is a point group operation that now acts on these Vanier states, there will be some unitary matrix which transforms between our set of four states that depends on which symmetry we picked. Now if we construct block states out of these Vanier states in the usual way and then act with this symmetry operation, we discover that if we pick triangle lattice sites, then the block states would have to transform the same way, this matrix U has to be the same, as the Vanier states at high symmetry points, the points which get mapped onto themselves due to these point group operations. That turns out to be a problem. Because once you fix your Vanier states, you fix this matrix. And what this is telling us is that at high symmetry points which map onto themselves, let's say under three-fold rotation, such as the gamma point and the corners of the Brillouin zone, this matrix has to be the same. But I just showed you that at the gamma point, we have two E representations, two doublets. And at the k point, we have one doublet and two separated states. So this is, therefore, inconsistent. So as I said, the representations at gamma and k, according to this, would have to be identical, but they are not. So this suggests, oh, and the similar argument was presented by Liangfu's group, and then Centel and Ashton Vishwanak's group before, although it was a slightly different argument. And so this suggests that we should not think of this as a triangle lattice, despite the fact that the local density of states is peaked on the Moray triangle lattice, but rather as a honeycomb lattice, the dual lattice of the triangle lattice. Now it turns out that this argument doesn't quite go through the same way if we pick those sides, because now under some of the point group operations, let's say rotation around this point, you see some of these Vanier states, they change position. You're no longer rotating around the triangular side, and they go into different units. That's why under this operation, there's an additional shift. The shift depends on which atom you're considering and which point group operation. That's why R prime depends on G and I. So now if you follow the logic I presented on previous slide and ask how do the block states transform, you discover that they don't actually quite transform the same way. There's an additional phase factor that you pick up. And this phase factor depends on this shift R prime. Now notice that this does not matter at gamma, because at gamma K is equal to zero and this factor is just one, and so at gamma, the block states should transform the same way as the Vanier states. And I told you that the block states are two E's, and therefore we should construct our Vanier states in such a way that they transform as E under the point group operation, the site operations. But because at the K point, this factor does no longer disappear, it's present, we have the desired results that the representations at gamma and K need not be identical anymore. Okay, so to construct our Vanier states, we're gonna focus on the gamma point where this factor is not present. And because the threefold rotation around the Z axis at the AA registered point does not change the sub lattice of the honeycomb lattice, the Vanier states must transform the same way as the block states. And so this means that U must be diagonal for G that is C3, so for the threefold rotation. Okay, and so now let's imagine that we apply C3 here and then ask what happens. Well, let us pick the one one element of our diagonal matrix to be a phase factor with two pi over three rotation. So that would fix the transformation properties of our first Vanier state, W1. But once we have W1, you see, we can then apply time and resource symmetry and that doesn't change the site, it will generate a new state, let's call this W2. And because under time and resource symmetry this phase factor will change a sign that gives us the second element of our diagonal matrix. Now to get W3, this is generated by a rotation around this dashed line. Well, we're just gonna be taken from one honeycomb sub lattice, more a honeycomb sub lattice to the other. It also tells us exactly how that new Vanier states, W3, should transform under threefold rotation. That's fixed for us, this phase is now fixed for us. And similarly, or lastly to get the fourth Vanier state we just apply time and resource symmetry to W3 and that forces this phase factor to be this. Okay, so now what do our block states look like at the gamma point? I'm plotting the absolute value squared for the top energy state at gamma which transforms under rotation such that it picks up a phase two pi over three. This is separated into the top layer and sub lattice A and B and the bottom layer and sub lattice A and B. So it has this peculiar, if you wish, propeller-like structure. Similarly, but it's smooth other than that. If you similarly look at the block state below in the doublet below which transforms as epsilon, also has rotation eigenvalue with the phase two pi over three then this side, I'm sorry, yes. This side is the A-A side, the corner here. So that would be the triangle lattice. Same, yeah, this is a triangle lattice. Yes, thank you for your question. Okay, so this is what the block states look like. Now we're gonna use these block states, this is just to give you a feel for what they look like. We're gonna use these block states to construct our Vanier states. Now the method for constructing Vanier states was developed by Marsari and Vanderbilt and to optimize their spread and we just simply follow that method. Perhaps I can walk you very quickly through how this works. Remember the Vanier states are related to block states by Fourier transformation. Now numerically, when you compute the block states, there is nothing that fixes the phase of the block state at one k point and the nearby k point. And so as a result of that, numerically or in practice, this block state will have a random phase in k space or at least that phase will not necessarily be smooth. And that presents a problem because when you Fourier transform that non-smooth function in k, you will not necessarily generate something that is nice and localized in real space. So you would like to make this block state as smooth as possible in k because that would then produce as localized over Vanier state as possible. Okay, so the way this is done is that you start with some trial functions. In the case of an isolated band, just a single trial function. That's usually something that has some nice smooth peak and but it should have the right symmetry of the band if you'd like to reproduce that. And then you project that onto the Hilbert space at a particular k point, spent by your isolated band. This is done this way. This defines your new basis which lives in the correct subspace of the block state you are interested in and has an additional advantage that it is smooth in k because this projector is smooth in k. So if you have a random phase here, well, the random phase here exactly cancels that. That defines our, at this point, complex k dependent term and to normalize it is trivial if you have a single band. Okay, now in the case, this is not our case because we have four bands. These four bands are sort of intermingled but they're separated from the rest of the bands by a gap. So instead of just taking one of them, we're gonna consider all four block states. We're going to find a linear combination of them and we try to make this linear combination as smooth as possible in k space. Okay, so there's a procedure for this. You have to find the single value decomposition of the overlap matrix of these states and from there find the square root of an inverse of this but this is a well documented procedure, okay? So as long as this inverse exists, in other words, none of the single value decomposition eigenvalues vanish anywhere in Brillouin's zone, then this procedure can be done and there is no obstruction to building these binary states. Okay, so this is what we follow. To get a good trial state, we construct them as following. We just take these block states and then we imagine them centered around the dual, the honeycomb lattice sites and we're just simply going to take a wave function that lives in here on the top layer in sublattice A. So from the top energy state and from the bottom in sublattice B and then mix in a state that lives in the different energy, the bottom doublet, but they're both coming from the same C3 eigenvalues. So that now our block state, even though it's mixed in energy, sorry, now our Vanie state, even though it's mixed in energy, transforms under rotations as it should. It should pick up an epsilon and should be an eigenstate. And so then we mix in these two together that gives us our trial state. We fit into this machinery, go through the single value decomposition and outcome the Vanie state after the so-called projection method. Now one can do a little bit better than that. One can try to then optimize these unitary matrices that you still have a freedom of acting with the unitary matrix and try to get the spread of these Vanie functions to be as small as possible. This was developed by Marzari and Vanderbilt, the so-called maximally localized Vanie functions. And you can do that, you will achieve about 20% improvement in the spread. Now, what do these functions look like? So remember, they're all centered on the honeycomb lattice. So the dual lattice of the triangle lattice. But interestingly enough, they all have three peaks which are centered approximately on the triangle lattice. So this immediately explained how you're going to recover the local density of states from something that is centered at the honeycomb lattice, but while the density of states is peaked on the triangle lattice. Okay, and I can tell you more about these if you are interested, but once you build these Vanie functions, they live in the correct subspace of interest in these four flat or nearly flat bands. And then you can sandwich the original Hamiltonian in this basis and compute your type binding Hamiltonian from that, yeah. Yes, this shows the triangle lattice. Just consider the middle of these triangles. That is the honeycomb lattice. That is the more honeycomb lattice. Pretty close to that, yes. Yeah, triangle lattice. The block state. But this is just a gamma point. Remember, you sum over all Brillouin's zone. So once you project it and then do the integrals with the projectors throughout the Brillouin's zone, this is the structure you pick up, see? So it naturally explains the peaks in local density of states and it doesn't suffer from this issue of the symmetry mismatch. Okay, so the story here, therefore, is once you pick W1 to be the eigenstate of C3, you can generate W2 by time reversal, W3 by C2 prime, which is the twofold rotation I discussed, W4, and by time reversal of this one. Okay, so all the symmetries that I discussed, the exact symmetries of the D3 structure are therefore represented locally in these Vanier states. And so how well can you do with this? Well, this is how well it reproduces the original band structure, just from the tie-binding model of these Vanier states. The blue is computed from the Vanier-based tie-binding and the red is the original tie-binding problem with almost 8,000 sites. Okay, okay, now I hope I have another hour to discuss the relation to the obstruction that has been discussed in this field. And so let me just tell you how this is related to this. So let me start with the claim. The claim is that if the system has a twofold rotational symmetry around the z-axis followed by time reversal symmetry, and then in addition to that, it has a valid conservation symmetry. Then there's an obstruction to prevent, to constructing exponentially localized states, exponentially localized Vanier functions which are symmetric in areas which transform locally under the operations of the crystal. Okay, so by symmetric I mean represent the crystalline time reversal and valley symmetries locally. So the question is how are we avoiding this obstruction? So this argument has been quite insightful argument has been pointed out by Senkel and Vishwanak collaboration. So where does this come from? How does one understand this? So imagine for just a second that we are going to conserve the number of particles in a particular valley. So now you twist the two layers and so naively because the two direct cones in the Morel-Brillant zone come from nearby valleys at small angle. In the one valley problem, their winding numbers should be the same because the winding number of this is presumably the same as the winding number of this on the other layer. And so in the band structure if you now get the two direct cones for each valley separately, then they should have an equal winding number. This is not the situation for each monolayer. In each monolayer the winding number at the corner of the Brillant zone k and minus k are opposite. But here the claim is that they are the same. I just made it clear that this is just a very naive argument but it is a lot more subtle but it's actually correct conclusion. So now remember for this to work you need to have a twofold rotational symmetry around the z-axis. That is true if you combine the twofold rotation axis that we have present in our structure and then an additional twofold rotation axis that's perpendicular to that but still in the plane of the bilayer. So it's only if we have this additional C to double prime combined with C to prime that the twofold rotation is present. Now to establish that the winding number is the same is not still enough to establish obstruction. For that one as they have done argued that in some open region in the single valley band structure that surrounds the corners of the Brillant zone one can define a K smooth basis on the two narrow bands and that in this basis the structure of the effective Hamiltonian has to look like this in other words it doesn't contain sigma three matrix. And then what one can ask is because it's now a smooth field how do the components of this field wind as you go around these two point. And in the presence of the additional twofold rotational symmetry the point in the middle between K and K prime so the endpoint turns out to have block states which are eigenstates of this twofold rotation symmetry with opposite eigenvalues. With that it is enough to show that N one under a mirror sorry under a twofold rotation operation that goes through the end points and therefore takes one to the other one direct point to the other under that operation it turns out that N one does not change sign so the X component of this phase field will stay but the Y component changes sign. So if you were to plot the zeros of these two functions under this symmetry it turns out that they would have to have the same winding number. Now so these two components of the argument not only force the phase field to lie in the XY plane but it also shows that the winding numbers are the same and there are but there are no periodic functions N one and N two that you could now extend throughout the entire Brillouin zone which would have two same winding numbers in the entire Brillouin zone but only two. In other words you would have to accompany that with opposite winding numbers. And this is the basis of the essentially a proof that there has to be an obstruction to building exponentially localized vignette states provided that the value symmetry which we needed to define this and C to T are good symmetries. Okay now it turns out that in the D three structure the C to the prime is not an exact symmetry and because of that the C to T neither is C to T and so therefore our if we did have valley number conservation then our effective Hamiltonian would have to be two by two but it could contain sorry about this typo but it could contain the sigma three matrix and that immediately tips this vector out of plane and there's no longer any obstruction. The obstruction only works if this is in plane. Moreover because we are on the lattice the U1 valley symmetry is not exact so there is some mixing between the two different valleys. So neither the valley symmetry is exact in the microscopic construction that I showed you nor the two fold rotation symmetry followed by time reversal along the Z axis are exact and therefore there's actually no obstruction. And so in practice when we look at our Vanya states after the projection but before the but before Vanya 90 which is essentially the maximum localization we find that a significant portion of the wave numbers that compose our Vanya states are actually coming from one valley but there is some spillage into the other valley in this case it's something like nine to one and if we then maximally localize it then it's something like five to one never do you have a perfect valley polarization. I like to connect this to the so-called obstruction issue that has been discussed by Soryanov and David Vanderbilt in the case of 2D topological insulators. So it is known that there is an obstruction to building exponential Vanya states if you have a separated band but that band has a finite churn number. Now Z two topological insulators do not break time in symmetry so they don't have churn numbers. So the question was is there any obstruction in this case? The immediate answer is that there shouldn't be because you don't have churn numbers but it turns out that if you try to build exponentially localized Vanya states in the 2D topological insulator with a Z two index that's odd and you require your Vanya functions to respect the time of symmetry locally then there is an obstruction. So you have to relax this condition that the two Vanya states come in Kramer's pairs. And that's because loosely speaking the basis that composes your Hilbert space at the two different K points is completely orthogonal. And so it's actually quite related but there is a way around it that they proposed which is that you simply pick up something that does not, you pick up Vanya states which don't go into each other locally they're not Kramer's pairs under time overs. Okay so I think I don't have enough time to talk about what happens to this band structure as you start twisting the angle closer and closer to the precise magic angle. They're actually quadratic band touchings that in itself has many body instabilities but let me just stop here and flash my conclusions so thank you for your attention.