Welcome everyone. So today we have the pleasure to have with us Andrei Bernovik from Princeton. So Andrei is a world leader in everything which concerns the political properties of electrons and solids, be that insulators, superconductors, whale semimedals, and his work also is characterized by close attention to experiments. In fact, some many of his theoretical predictions have been confirmed experimentally. And recently he's been doing some intriguing work on some special quantum SCAR states in various condensed matter systems, which also have been searched for experimentally. And that's what you're going to be talking about today. So please, Andrei. Many thanks. So yeah, thanks for having me. And this is the first seminar online in Paris, I wish I was a different person, but such is life. And what I'll be talking about today is work that was done in collaboration with my long-term collaborator, Nicolas Renaud, and a student who's now postdoc Sanjay Mulgalia, and more recently Frank Schindler. I'll be presenting some results, some new results from work that Nicolas and Frank Schindler and I have done together. And some results from the collaboration with the second row of people, in particular, Stefan Rachelle, Rahul Makishar and Paul Fendley. These are the references. And let's get into it. So basically, to motivate the talk experimentally, if needed, it turns out that, you know, while we didn't really know how to study non-equilibrium dynamics in, let's say, 20 or 30 years ago to the extent of controllability that we do today, cold atoms basically have solved this problem in the sense that they have enabled the study of non-equilibrium dynamics of interacting quantum systems. And of course, this brings about realization of phenomena or states of matter that can only be realized out of equilibrium. For example, one of these is time crystals. And hence, there's been a lot of effort on strong theoretical foundations of interacting systems. And this work is, you know, sort of a very small piece of that in the sense that it tries to review whether the eigenstate thermalization hypothesis in its strong sense is violated or not. And this is the outline of the talk. I'll basically be presenting the first exact model where the strongly T H violation was observed and then connected to what's later known as scars. And then I represented a lot of other systems where scars exist. So it turns out that this phenomenon is quite, it's not, of course, it's not entirely generic in the sense that it won't appear in any disordered non-translation environment. But it's quite generic for some set of Hamiltonians. And I'll make some connections. We still don't know the entire theory, but I'll make some connections to what we know. So I'll give a quick review of ergodicity. And I'm not an expert in this, but the one thing that I'll be focusing on is entanglement entropy. And basically, the property of ergodic systems that I'll be using is entanglement. Entanglement of states in the middle of the spectrum. And these are some properties here of ergodic systems. But really, what I'll be using is that the fact that the eigenstate termination hypothesis says that in the middle of the spectrum, so in this region of energy where there's a finite density of states of levels, where it's basically exponential density of states of levels, then the eigenstates are thermal and their entanglement entropy obeys a volume law. So you take one of these eigenstates, a generic eigenstate in the middle of the spectrum, and the strong ETH hypothesis says that any state you take here, you cut half of the system, you compute the reduced density matrix. Once you compute the reduced density matrix, you can compute the entanglement entropy. And this obeys a volume law, which basically means that any matrix per state representation of this state would have a bond dimension that is basically exponential in the volume of the system that's traced out. And this is very, very different from the ground states of or highest excited states, because the highest excited state is the ground state of minus a Hamiltonian. And the ground state of the system and the low lying states, and by this I refer to, I'm not sure if you can see my cursor, my mouse, but for this I refer to the ground state and the ground state like states, which means the low energy excitations before the middle of the spectrum or before the density of the level spacing between the states exponentially small. So before that, before you reach that, just the ground state and take like excitation or the upper, the highest state and excitations are around below the highest state. These ones obey an area law. And this is, of course, people misogynists know this is the area law entanglement. Now the strong ETH says that whatever ion state I pick here in the middle of the spectrum, they will satisfy ETH and they'll have the one thing that will be focusing on the one property that will show it's violated in these new states is that their entanglement entropy would be proportional to the volume if they respected ETH. Now there's many Hamiltonians that were thought to obey ETH and only a few examples that were known not to obey ETH and one of these examples are, of course, integrable systems. And another example is many body localized systems. Many body localized systems can be sort of thought as, this was the work of Babanin as integrable systems with stable integrability. So in many body localized systems, of course, you need disorder. But once you have disorder, you can write quasi local or local constants of motion. And then if you perturb the Hamiltonian a little bit, they still remain integrable. So you can still write integrals of motion. So that's why we call stable integrability. And these ETH violations, ETH violation was known and thought was to occur only in these two types of systems. So now the question we wanted to pose is can ETH be violated in the absence of an extensive number of conserved quantities in the absence of an integrable systems and in the absence of disorder. So no many body localization also. So in a purely translational invariant system interacting, of course, without disorder and without a number of, without integrability. And we solved this question analytically. And this was later known by the independent work, kind of these two works basically didn't know about each other. This was later given the name of many body scars by Serbin Papin and Papich and Abanin and Turner at all. And these, in this paper, and they were coming at this question from trying to model an experiment by Misha looking. And these quantum antibody scars basically, the name was given because of similarity with classical systems trajectories of the billiard ball that are non-chaotic. So the questions, is there a paradigm of ETH breaking beyond integrability and many body localized states? And what are the issues? Yes, so when it's beyond integrability, how would you test non-integrrability? Yeah, so the one way we test it is, so of course, as you know, as you're thinking, it might be, maybe it's not visible, it's not clear, integrability, maybe there's something hidden. But the one thing that's done in these is large scale level statistics. And there, now I know there's some issues with, there's some very special cases in the level statistics where you have Wigner-Dyson, but it might still be, I think, Shaman was pointing them out to still be integral, but these are not those special cases. So they're very simple models that have perfect, even on small size numerics, not small size, but, you know, 10 to 5. So by testing the distribution of eigenvalue, the statistics. Exactly, exactly. And then you can do more. Yeah, I'll show you the plots in two slides. Thank you. But indeed, you know, that's tested numerically because, as you pointed out, it's, you know, there might be, you might be missing analytically the integrability, but all these have perfect Wigner-Dyson distribution. So I'll show you the plots. Thank you very much. So then there's, so what can you do to find a paradigm, to find the ETH breaking? Well, the certain issues, strong issues arise in the fact that we don't know how to solve for eigenstates in the middle of the spectrum. Obviously, these are very complicated in many body systems. The Hilbert space curve is exponentially, and you don't have control over the eigenstates of spectrum, even for models where you do have control over the, over the ground state, for example. So hence, what most people in the field, or almost all the papers in the field of many body localization do, is basically just very smart conjecture supported by numerical calculations. And these numerical calculations are as big as, as one can, one can, one can do. It really just depends on your numerical skills. Roughly, there are 20 sites of the order of that order roughly. So, and they're mostly one D because in one in two D, you immediately get swamped by the Hilbert space dimension. So then what I want to claim is that we found a simple set of states that span from the ground state all the way to the highest excited states. This, the kind of peers through the spectrum of a non-integral systems, a non-integral system, and we're able to give their expression analytically, and then we're able to calculate the entanglement entropy analytically, and to show that they don't, that they, you know, it doesn't, that the entanglement entropy does not follow a volume log, even though these states kind of interpolate between the ground state and the highest excited state, not interpolate, but they appear at energies that form a tower. And since then there's been many examples of systems, of systems where this happened, and these are the so-called exact scar states. They should be kind of sort of distinguished from the, from the approximate scar states, which are the ones in the experiment, which, you know, they have a volume component of the entanglement entropy, but it's very small. They're very small, and these ones have no volume component of the entanglement entropy. Okay, so what's the model that we used? Well, the model that you use is the AKLT spin chain, and this is a famous model, where you have a sum of nearest neighbor projections, which, for example, cannot have, two nearest neighbors cannot be in a spin two state. This is spin one model, and two nearest neighbors cannot be in a spin two state, and the Hamiltonian can be written more, more familiarly as this, basically, as this bi-quadratic Hamiltonian, and this is, of course, the model where you, so the camera being spin two state in the ground state. This can be, it's a projector Hamiltonian, and this has been used to prove the holding gaps by, you know, these, of course, the AKLT paper. It's also, this ground state serves the prototype for symmetry, technological states, matrix product state wave function. It's got a matrix product state expression, the ground state with a bond dimension two, etc. So now, how do you find the ground state in the highest excited state? Well, the ground state is actually very simple, in the sense that you form, you take every spin one on one side, and you basically say that, oh, I'm going to form the spin one out of two, spin one half degrees of freedom is symmetrized, and then I'm going to dimerize the spin one half degrees of freedom in between sites, so that they form a singlet, and this is the, you know, the pictorial representation of the singlet, of the singlet is denoted by a line here, and that means that in the ground state, for example, since these two spin one halves, the right one on this side and the left one on this side have formed a spin singlet, I remain with two spin one halves, they can only be in spin zero or spin one state, so they cannot be in spin two state, and hence the projector term on sites N and N plus one, out of which the Aklity-Cannuptonium is formed, which basically projects onto the spin two, gives you zero on this state, okay, because this state cannot be on spin, this is the state that two spins one on one on two ADS in size, they cannot be in spin in a spin two state, and of course the highest state is the ferromagnetic state, and this is, you know, this, this is the expression because it's, it projects onto spin two, and the ground state of this Hamiltonian has energy zero with this, with this, of course you can shift it, but the one that I wrote before, which is the sum of these projectors says energy zero, and the ferromagnetic state is the highest excited state, so you can write the ground state exactly, and despite this fact the model is non-integral, so you can see very clear GOE level repulsion being done, so which is the rate to the, to the question before, that's, you know, you can solve, even this is the thing for 10 sides or something like this, and it satisfies very neatly the Dyson level repulsion. Now the other projectors Hamiltonians that we know, there's many of them, these are, these are sort of called frustration-free Hamiltonians, where you just have a Hamiltonian being a sum of projectors, each of which annihilates the ground state, these projectors do not commute, so it's not kind of a stabilizer code, like the other, like, like the exactly soluble Hamiltonians, so they don't commute, so it's not exactly soluble, and there's many examples of these in any dimension really, for example, the Halding-Südder potentials are one example for the fractional quantum whole state, the Schastri-Sullin model, the Madrugnagosh model, and all of them that I know of are non-integrable, these, and then how do, so then what we said, well, was let's look at the spectrum of this model in finite size, and you look at the spectrum of this model in finite size, and this is for helical 16 particles, okay, and this is for helical 16 sites, so it's, you know, quite a large system, you've got, you know, 3 to the 16 by 3 to the 16 matrix basically cut into symmetry sectors, and what you see is that, you know, you've got a bunch of energies, which is, of course, a huge amount of energies, but then what you observe is you observe some remarkably, some remarkable energies which are integers, and this is kind of like the way we, Nikolai and Sanjim, myself, initially thought about this, you know, the Hamiltonian basically has an integer, it's an integer matrix, so really if you're going to get out of a 3 by 3 to the 16 times 3 to the 16 integer matrix, you're going to get some integer energies, and those energies, the specific polynomial is certainly not going to give you, you know, integer numbers as roots generically, and if you get integer energies basically out of a huge integer matrix, those basically define some very special Krilov subspaces, and out of these integer energies, I want to point out that or they're not all integer, but they're some rational, okay, and you can get this out of high-scale numerics by doing some, by basically, you know, converting numbers to ratios, and then, and then, and then, you know, you can see within machine precision that they're these numbers, and what I'll be focusing on in this talk are these integer energies denoted here in red, which as you can see, formal ladder, so you can see zero as the ground state, 16 is the highest excited state, but then you've got two, four, six, eight, ten, twelve, fourteen. So after numerical observations, has anyone proved algebraically or analytically that they are integers, true integers? I mean, yeah, yeah, I'll give you the expression in the next slide. Okay, thanks. Yeah, yeah, yeah, very good, yes, so this is how you basically, this is how you hunt for them, right, because given a model, you're not gonna, just by looking at it, it's not entirely, I mean, in retrospect, it's easy to see that these are the excited states, and you'll see them immediately, but it's not, it's not, given a generic model, it's not easy to find them, and this is how we hunt for them, and you know, this procedure has been kind of now used by some people, but it hasn't been used to the extent that I had hoped for, so even now we find using the same procedure, other models that have exact eigenstates, and I'll show you in a little bit, but yeah, this is how you hunt for them, so now let me give you their expression, so their expression is the following, let's define the basis state mn, which is a S, a spin on site n, raising operator, applied twice, so it's a spin to magnon on site n, and this what it does is it breaks the bonds on site n minus 1 and n in the ground state, acting on the ground state, based on the bonds between site n minus 1 and n, in between site n and n plus 1, and it has to create the ground state of spin 0, it has to create the spin 2, so it just polarizes those spins and you can see it here, so now you can immediately see that the beauty of these projector Hamiltonians, of these frustration free Hamiltonians, that you know, the only thing that will matter in the scattering problem is what happens around this site n, where I hit, where I flip the spin twice, and this is because you know, the projector Hamiltonian will kill everything that's, that's to the left of site n minus 1, is to the right of site n plus 1, so that one you don't care about, and you find out that the Hamiltonian scatters this mn configuration to bond inversion symmetric configuration, so it scatters them to, for example, this configuration plus the left configuration plus the right configuration, okay, and now you can immediately see that, so you have the scattering, a scattering equation which is, which is h times mn is equal to some eigenvalue times mn plus this n, n states are the left one and the right one, okay, but you can see that the left one here, I'm not sure if you see my cursor, but the left one in the second row is, is a translation or the right one is the translation of the left one, okay, so now if you make a momentum pi excitation of this, then this scattering cancels and you're left with an eigenstate, okay, and this is an eigenstate that energy 2, because of this 2, spin 2, and it happens for all even l, by the way these states happen for even, even l because of this, this, this issue, so, so okay, so good momentum pi you have to hit, you have to have even l, so this is the first eigenstate of energy 2 is minus 1 to the ns n plus squared on the ground state, which we'll call this an operator p, so you act with p on the ground state and the energy is, is, is 2, now, you know, I've, we've thought and a lot of people thought so for a long time that, that only momentum pi scars, it can exist or, you know, momentum zero or pi, but recently there's been a paper by Nusha Chandran and collaborators at Boston University basically found scars, they were able to come, kind of reverse engineer scars at every momentum, so okay, so now, so now you take this operator p, which is again, is this momentum pi spin 2 magnon, you take this operator p and you act it n times on the ground state, okay, and this is the tower of states that we had found numerically and you can kind of see that, you know, if you acted once it gives you one spin 2 magnon and momentum pi, if you acted twice it gives you two spin magnons which scatter a little bit, but only very close to each other at total momentum zero, so they're, each of them is a momentum pi, etc, and if you acted l over two times, you get, you get to the ferromagnetic state, so these are the exact expressions, so now having the exact expressions, we ask what are the implications for the eigenstate thermalization hypothesis, well the implications are immediate because you can, having these exact expressions and having this operator and having this very simple form of these states, you can immediately estimate their entanglement entropy, so what you can do is you can write this s2n as a matrix product state with a small bond dimension, and the bond dimension, so this n here is the number of times you hit you apply this operator p, okay, and basically you can write this operator p as a matrix product operator, etc, but this, but the fact that it's a local operator that you apply, you know, n times, so some local operators that you apply n times, basically gives you a matrix product state with a small bond dimension, you can, you can find exactly this bond, an upper bound on the bond dimension, and this bond dimension is two times n plus one, okay, and now you can basically see that the entanglement entropy, because by bond dimension you can, you can bound the entanglement entropy is, has to be less than log of this, which is two times n plus one, log of two times, two times n plus one, so it's got log bounded, log of n, where n now is the number of times you hit, you know, the, the, the, the ground state with this projector, if I'm in the thermodynamic limit and one states in the middle of the spectrum, n over l is finite, is constant, is some fraction, n over l, so that means that s is bounded by log l, so these are log l states as opposed to s being volume law as predicted by the age, so these states analytically violate the eigenstate organization hypothesis, you can see them in a spectrum where you, where you compute the entanglement entropy of all the states in every, of all the states of, for example, this is 16 site AKLT model, so the entanglement spectrum of all the states has this kind of nice semi-circle law, but then you have these states, these cars, these x's, which have much lower entanglement entropy, they kind of, they kind of, you know, they, they, they, they, they're, they're much, much lower than, than the ones that satisfy DTH, so all of these states that sit on this, on this semi-circle is, I mean, semi-circle is parabola, some, some curve is not, is, are, are not, are volume law and then you get these cars, which are area law, which are log l entanglement entropy. Okay, so if there's questions now, I'll take some questions, if not I'll go to the connection between these and the matrix rock states and try to get to the spectrum-generating algebras that describe these states. Okay, I can ask a question, I mean, if nobody else has to, but it's just too similar to the coordinate beta ansatz to me, so how do you explain this or, or maybe it's the recognition, maybe like the sub part, subsystem is actually integrable or like? Actually, related question, maybe, maybe could, so yeah, on the landscape of spin one, spin chains, there is an integrable system, right? There is a way to write the Hamiltonian using gamma functions probably, which will be integrable, which will be solvable by beta ansatz. So how far is the AKLT model from that integrable point? Yeah, so I'll, so I'll answer both of these questions in the matrix cross-state because I'll construct the pathway to integrability. So, so the very good, I mean, it's quite the model, the AKLT model by itself is quite far from integrable, you can see it from the vignetization level statistics, but there is a path, and this is another standing question, whether, whether these cars are remnants of what was, you know, like a one of the beta ansatz states which remains. So the answer is probably yes, you know, but they're remnants very far away from integrable. So, so, so, so I'll answer both of these questions right now. You can, you're just preempted basically my next five slides. Thank you. So, so, so yeah, so if I can take 10 minutes to basically, or five minutes to, to point out this matrix cross-state connection, then construct the pathway to integrability. So basically, the ground state AKLT is a matrix, can be written as a matrix cross-state. And this matrix, these A matrices are two by two matrices. And then you can create quasi-particles and mental k, kind of excited states are on top of this, by this operator, you act with some operator on side j, and, and this is the simplest quasi-particle that you can create. And, and you put a momentum k. And this basically acting with this operator basically modifies the MPS of this A matrix in the AKLT case, its translation invariant, so all the A's are the same. It modifies the A matrix by basically putting a matrix block operator on them. And, and you create this new B matrix, which is this operator O on the MPS matrix. Okay. So now the question is, how do you construct a, how do you, you know, backtrack to find B from a given A? Okay. And the reason is, if I, if, if we manage to find this, we can have a pathway so you can tune the A matrices, the MPS matrices, to find a pathway between the AKLT and the integrable model. And then if you find a prescription to find a tower of quasi-particles, kind of scars in this matrix cross-state prescription, you can kind of follow it all the way to integrability. Okay. And I won't bore you with the details, but this is very similar to how you construct exact Hamiltonians from MPS ground state. So it turns out that if you have a ground state or eigenstate, which has an MPS description with a bond dimension that's finite or that doesn't grow, that doesn't grow as the volume. So let's take it to a bond dimension that's finite. Then you can find a frustration-free Hamiltonian that, that, that has this as a ground state. And the prescription is simple, is simpler. This is a simpler way of doing it. Then there was first done by Verstrasse and Sirach and other people. I think there were two or three groups that did it at the same time. But basically, what you do is you want to find an operator H that annihilates some parts of this matrix contraction. And then given that you have, that you have a finite bond dimension, all you need to do is count how many, how many times the Hilbert space onsite, how fast it grows. And once it grows, once it's become larger than the span of the bond dimension matrices, then you can write the Hamiltonian. So for example, for the AKLT, these matrices are two by two. And then the image of the, so since they're two by two, I can't write a one-body Hamiltonian because a one-body Hamiltonian, the physical dimension would be three. So three is less than two. But if I have two sides, now the physical dimension is nine. Whereas, whereas, whereas actually the bond dimension is, I have two here and two here, so it's four. So I can project into the complement of, of, of, of the image of this image of the two-side MPS. And this projection Hamiltonian will basically annihilate the, will annihilate the ground state if this projector here is Ej matrix is positive semi-definite. So very easy construction that you can build it actually, and you can generalize it to excited states. So this is some sort of, you know, MPS generalization to excited states. This B operator is the operator of the, of the excited states. So it's the, this operator on the matrix A. Okay. And you can find out the conditions for which, for which this excited state is now the ground state of the Hamiltonian that, that annihilated the way function with matrix products, state A. Okay. So, Bogdan, can I ask a question? So are you saying that for these parent Hamiltonians, you can construct course particles, eigenstates analytically? So, so I, so there's, yeah, so there's several, so I'm saying there's, there's, you can find the conditions, and then you have to solve the conditions. So, so given, so given a matrix product state A, if the matrix product state A has a finite bond dimension, then you, then you can find the projector Hamiltonian. You just, you basically just count how many times, how many sites you have to, to, to get to the space of those sites is larger than the bond dimension squared, for example. Yeah. Here, and then you project into that, and then you found, you found, you found Hamiltonian that annihilates that state, so you can find the Hamiltonian that, this is, this is, this is the frustration. That's, that's simple, but now you're in the second slide, you were saying how you can find quite, quite a particle situation. Yes, and now there's a more complicated condition, and the condition is the following. The condition, the condition is, is, is, is this one. You have this, this, this one here, so now you have to solve it, and you have to find out if this, if this exists or not. So, so it's not, so it's not immediate that it will exist for all the frustration free Hamiltonians even. But it doesn't exist, where's the mistake? So we know that the quasi-paric, for example, let's take ATLT, as far as I, I thought that for ATLT, quasi-paric excitations are not exactly solvable. So something is well, but the Hamiltonian is of this parent Hamiltonian form. Ah, right, right, right. This, this is not, this is, okay. So first of all, this quasi-particle, so this, I'm claiming that here you can find this, you can find a solution that's solved for this tower of states, not for every quasi-particle, right? So this is, so for example, you can immediately see that this is a very simple solution, just takes an operator, this, the, the, the ansatz for this solution is an operator, local operator on site J, times a momentum, right? This is not the most generic form of all, right? I could have operated that act on two sites, right? So this is how, so we try to kind of backtrack from, so we know the form of these tower of states in the ATLT, and we try to backtrack from there. Okay, so you're saying these ansatz would not be sufficient state to describe the, the first quasi-particle, but it might be sufficient for some states in the middle of the spectrum. That's right. Well, it's certainly sufficient for this tower of states that I, that integer energies that I, that I, that I showed. So this is, so yeah, so it will certainly want to describe all the, all the, well, it will certainly want to describe both of the states, right? It will just describe these, these, these ones that are local operator, times a momentum, and then, and then you want to apply this operator in times, okay? So, so for example, right, so, so, so you find these conditions, and then you know the matrix A, and you can kind of do a numerical search in this matrix A is two by two, you can kind of do a numerical search in the space of operator satisfying this condition for both the momentum and the operator, and the AKLT reproduces, you find the AKLT tower of states. But what's more interesting here, and this coming to your question, is that, so for example, this, the way it works for the AKLT is you have this AKLT matrices, this is A for spin, 1 for spin, 0 for spin minus, then you find the subspaces that I was talking about that are spanned by, by the, by the bond MPS, and you find three subspaces. This is the subspace of, of, of, if you project into the complement of this of A, you find the ground state, but if you project in the complement of, so this is the B is the quasi-particle space, because we chose the operator to be S, S, S plus squared, so it's spanned by S plus squared on, on, on the ground state, and this is another complement space, which is the complement space of A minus B, and then you find the most general nearest neighbor Hamiltonian that has this, so the tower of states that was, it's the exact eigenstates that I presented is a tower of states not only for the AKLT, but for this more general Hamiltonian, okay? So for example, this more general Hamiltonian would be AKLT if this, I guess, if these ZJs and if these, if Epson was two, and if these ZJs were one, for example. So this is kind of a more generic Hamiltonian that has the exact same tower of stars. So now going to answer the question that was posed before, you can start, you can change these matrices, right? You don't need to start from the AKLT, from the AKLT matrices which are these ones. You can change these MPS matrices and you can change them to be, for example, these constants, C plus, C0, C minus, okay? And you can tune these C's around, okay? And you can ask, when is this operator P, which is the momentum pi spin 2 magnon, when are P to the n times the ground state, exact states of a Hamiltonian that also has this as the ground state, okay? And you can find from this, by tuning the C plus and C minus, you can find the continuous family of spin 1 Hamiltonians, which have these states times the ground state. Now notice that as I change C plus, C0, and C minus, the ground state changes, but the scars are just P to the n powers times the ground state, whatever the ground state changes, the scars will change, but there's still P to the n times the ground state. And then there's a continuous family of spin 1 scarred Hamiltonians connecting the AKLT point to the integrable pure by quadratic point, okay? And then this raising operator becomes a symmetry at the integrable point. It's not a symmetry, it doesn't commute to the Hamiltonian away from the integrable point. It's, even though, you know, applying it on the ground state gives you exact excitations. There doesn't mean commute to the Hamiltonian. I'll tell you in like the next five minutes, the last time I talk what it actually means, it's a restricted spectrum generating algebra, but at the integrable point, this operator does transform into the one of one of the symmetries of the pure by quadratic chain. So I think that this answers the two questions that were posed before. So, you know, even though the ground state changes over this interpolation, to this, you know, you find interpolation with an angle, the ground state changes over this interpolation, the P to the n times the ground state are still exact eigenstates. And they're eigenstates that interpolate between the ground state and the highest excited states. And as I go to the integrable point, this operator P that was used to create the states becomes a symmetry. Okay, and there's more involved scars you can do scars in the perturbed POTS model that involve more more you can sort of generalize this is Slava question you can generalize this this construction to more complicated operators, you can't get all the excited states, but in the case that when there are scars, it works for more complicated operators. For example, for the perturbed POTS model, you find this operator that creates scar states. Okay, so now I want to basically for the last part of for the last five minutes or three minutes of the talk, I want to basically sort of try to give a more algebraic description of these of these of these scars. And we do this by looking at the Haber model in any dimension really. So the Haber model or translation symmetry, this is my graph, for example, but let's for right now consider translational symmetry. So the Haber model is nearest neighbor hopping plus Hubbard and it was known very since, you know, early work by Sien Yang that it this model besides having, you know, st2 symmetry spin symmetry also has an eta bearing symmetry, which is a pseudo spin symmetry. And it's a it's it's an operator that doesn't commute with the Hamiltonian, but it commutes up to itself. Okay. And this is kind of the spectrum spectrum generating algebra. It's a very powerful right. I'm sure this audience knows very well, very powerful construction in the sense that once you found the state of the Hamiltonian, you can hit it with eta dagger and and find a number of times and find eigenstates at equally spaced energies. Now, this of course reminds you of scars, except that in this case, eta is an exact symmetry. Okay. And it's an it actually gives you another recipe to the tree. And, you know, of course, as I said, at some spacing starting from any any state. Now we know that the AKLT model or the models that I presented before do not have this symmetry because the spectrum does not come, you can find only one tower of states, even, you know, if it had the symmetry, the spectrum for any state that you that you start with for me for any eigenstate, you could find this tower of states. You'd find many towers, but you know, you find only one. Okay. So, so the, of course, the difference is that in the Haber model, eta is a symmetry, whereas in the AKLT, this p operator that I used to that we used to define the states is not a symmetry. So what is it? Okay. Well, it turns out that it's not a spectrum generating algebra, but it's a restricted what we call restricted spectrum generating algebra, which means the following thing. It basically means that you have spectrum generating algebra property, but only when acting on one on one state. So for example, you focus on some tower of eigenstates and you ask, which is eta or p in the AKLT state to some power n acting on some on some state be the ground state or another state, and you can prove that the tower of eigenstates exists when the following things happen. Psi zero is an eigenstate of the Hamiltonian. The commutator of h and eta gives you epsilon, some energy times eta, but only when acting on this psi zero. And this is the case of the AKLT state. And, and the second commutator of eta vanishes. So you can see this is kind of like, it's kind of like, right, kind of like a symmetry, but the symmetry that's cut when in, in, in the expansion of commutator. Right. So if I tried to transform a Hamiltonian by a symmetry, I would have, you know, a lot of commutators, but at some point this commutator cuts. Okay. And this we call restricted spectrum generating algebra of order one. And I'll show you why. And you can actually modify the Haber model by adding extra terms to the Haber model so that this eta pairing is not a symmetry anymore, but you can still find the tower of the exact states. And for example, one of the interactions that you can put on is nearest neighbor spin-spin interaction, which would kill the eta pairing as a symmetry, but would remain what would still have scar states because all these conditions would be satisfied. Okay. And you can generalize this, of course. Schematically, what does the commutator of h and eta dagger looks like? It's, so you're saying it should be something which commutes with eta dagger. So it's, it looks like what like schematically? It's zero. It's not just epsilon eta zero.  That's a good question. So, so I didn't think about exactly how h eta dagger is not eta dagger, because if it was, you'd have a spectrum generator. Yeah, it's not eta dagger plus some extra terms. Exactly. It's eta dagger plus something that commutes with, plus something that commutes with eta dagger. Right. So, yeah. So it's a good question. I don't, I don't know a priority answer to that. But there are other many things which commute with eta dagger, which could appear in that side. For example, for this simple model that you present here by adding this spin-spin interaction, probably you know what in this case the commutator is. Yeah. Now I know this, it would be easy to write it down. I don't know in my heart what the commutator is, but it's easy to write it down. Yeah, let me, yeah, I'll get back to you on that. That's a very good question. Yeah, it's a very good question. But okay, anyway, it's clear to me that it's a more general condition. Yeah. Yeah. And I mean, you can generalize this condition to the following condition, which is, which is, you know, you act, you act with, you have a, you have a psi zero state, which is, which is an eigenstate of Hamiltonian. You have a commutator with eta, which is, which gives you a spectrum generating algebra, but only when acting on psi zero, then the commutator n times, small n times of eta with h is zero only when you acted on the, on the, on the state psi zero, but then from n plus one, it's, it's, it's, it's non-zero. It's, it's for any state that you act on. So I don't put a state here. And for example, the AKLT, the AKLT state has, so this is, this is called the restricted spectrum generator of order m. And the AKLT state is when m is equal to two in this case. So only when I, when I do the commutator three times, does it kill the, the, it's, it's commuting with the Hamiltonian in the AKLT case. Okay. So there's many scars in other things such as non-linear messenger liquid, but I will basically stop here because I've taken, I think enough of your time with this, but the main message of this is that the scars are, I think generic, at least in these simple Hamiltonians that we look at, certainly they exist in the Haber model, they exist, you know, in, in, in, in many other models, they exist in this projector, frustration-free models, and the relationship with integrability is not, not, not yet known. So that's, that's what I wanted to say. Thank you. Thanks, Andrei. Thank you. Are there any questions, including from guests from outside the YGS? I actually have a question. Andrei said this, you know, of course I'm not an expert, but I know I heard about this parent Hamiltonians and like AKLT and so on. And I know that people usually say that for many aspects, for understanding the properties of the ground state and maybe the first excited states, this parent Hamiltonians are as good as more realistic Hamiltonians, which don't have this property of being made of projectors all annihilating the ground state. But usually they say that, okay, well, for, for the excited states, it's clear not the case. These guys are very different. In the sense, your work is showing, is giving some very dramatic example of how these Hamiltonians made of pieces annihilating the ground states are different for the excited spectrum. But in your, in your abstract, you were mentioning that these scar states were actually observed in experiments. So how is this possible that in experiments that they were able to reproduce this very special Hamiltonians, how did this work out for them? Yes, they reproduce, they didn't reproduce. They're not exact scars in the, they're, they're these things called the PXP models. They're not, they're not actually exact scars. They're basically, they're basically Hamiltonians that where you, where you have the kind of, I don't know if you know this friendly types Hamiltonians or whatever, say a Sigma X and some Sigma Z operators on, on, on the ADS and sites. So basically this is kind of like, this is projector one minus Sigma Z on sites, one Sigma X on site two and one minus Sigma Z on sites, on site three. So these things, they can, they can, they can kind of get them in these Rydberg atom experiments where, okay, so, so basically where you, where you kind of tune to some resonance between, between two different levels and kind of forbid, forbid some, some spin flips, depending on whether the spins here or here are up or down. So, so these are, so these are basically these Rydberg atom experiments. I don't know exactly how they, you know, the details, but they're basically, they basically realize this PXP model. Now in this PXP model, what you can show, so this, this is, you know, this is the model where this, this P is the, you know, one minus Sigma Z, for example, and, and they don't commute between themselves, some of operators don't commute. Now this model, you don't know the exact ground state, but what you do know, and, and the scars are not exact, but what you, what you do know is that you can find some, you know, almost exact expression of the scars, they don't, they still scatter, scatter very weakly, and this is what Serbin, Pappic and Habanind did. And these ones have a volume law entanglement part, but it's very small. So normally the ETH would predict there, there's, you know, basically some, some coefficient to the volume law of the entanglement entropy. And these ones, because they're not exact scars, they have that, but it's much smaller than the ETH would, would, would, would predict.  So this type of model where I have, where I have, so this is, this is a simple model that, for example, you also don't know the ground state where you take two particles on site one and two and then hop them on sites zero and three. Okay. So this is the process, the pair hopping. And these appear in like, you know, you can realize them, for example, in strong, in strong electric fields, like this is with interactions in the strong electric field, you realize exactly this Hamiltonian plus some, some, plus some electrostatic terms, or you can realize in quantum whole on the thin torus plus some electrostatic terms. Okay. And this model in, in some, so this model is actually very special because, because it's Hilbert space breaks up into Krulloff subspaces that are, that are not generic in the sense that, you know, Krulloff subspace, if I start with a state with a random state and hit the Hamiltonian many times, hit the state of Hamilton, I mean, does I'll go into and cover the whole Hilbert space. For this Krulloff subspace, I started with a special with the product state, I hit, I hit it with the Hamiltonian and the Hilbert space does not cover the whole, the Hilbert space, such form by hitting the Hamiltonian does not spend the full Hilbert space. So if you do this, you'll find out that these Hamiltonians have, you know, what's called, have, now there's a fancy name for this in this, in quantum evidence, it's called Hilbert space fragmentation, but it's really just Krulloff subspaces that are not, that don't spend the full, the full Hilbert space. So what you find out is that this, this pair hopping Hamiltonian some integrable, has some integrable subspaces and some finite dimensional subspaces and has some non integrable spaces. And this PXP Hamiltonian, this Hamiltonian for the Rydbeck atoms belongs to one of these non integrable subspaces. And here you don't, you know, there's nothing exact here in these things, there's only approximate, you can only approximately argue that there's cars and you can only find them numerically. So it's not, so it's not a, you know, but, but so, no, so they're not, so my point is cars are not, and this was, you know, a bun in, in Papich and Serb and Turner that showed this are not particularized to like these exact models, they exist even in models where you can't find the exact ground state, when you can't find almost anything exactly, but you still see some, some, some scar states, and you can see them numerically, like what they did is, you know, you can see them numerically here. So you can see, you know, this, these are the entanglement entropy with the one, two, three, four, five, six, seven are the scar states, very low entanglement entropy.  It's much smaller than it should be based on, based on the TH. So my point is that, you know, an experiment can create something that's close to, you know, you don't need these fine tuned models to get scars, that's my point, or to get, to get the same phenomenon, of course, you know, as I said, they're not exact scars, but, but, you know, an experiment over some time scale doesn't matter. So they, they see, what they see is they prepare a state, a special, a special state, and then they evolve it, and they see revivals of this state, which basically tells you that this state has a large component on some thing that's almost an eigenstate of the system. Okay. Thanks. Thanks. Any other questions? People are not leaving, so presumably somebody has. Yeah, I have a question. So that's one way, whenever you hear the ETH hypothesis, you normally you hear that you take a generic state from the middle of the spectrum, and you have certain properties. And then to me, not, not, not being an expert, that the word generic encodes maybe some complicated tests that you should take. So in one sense, do these scar states fail some of these tests? Or is it just that you can't compute their energy? Is this the main reason why then everything fails? Yeah, I mean, so basically, so the main thing that you can do is you can, you know, prove that their entanglement entropy is not volume law. So that's one thing, but there's many other things that you can prove. So you can, for example, prove that the matrix elements of these states to ADS and states, okay, are of local operators are zero, you know. So these states are basically don't basically kind of don't mix with states, even though the other states are, right, exponentially close, they sit in a background of those exponentially close to them, right. So this, they sit there, they're like, right, there's states here, just there's an explanation, the system has a finite bandwidth, there's an exponential number of states here. So there's going to be some states very, very close to this to these scar states, but they still won't mix, even though they're exponentially close, they still won't mix with these states by local operators. So that's kind of one way that, you know, they don't satisfy ETH, but the main thing that you can compute here, so that's a hard thing to compute the matrix elements, because you don't know the expression of these other states, but the one thing you can compute is their entanglement properties, which are completely different. So basically, the volume law would tell you that if I cut a state in half in the middle of the spectrum, I have connections between every volume on this side, every part on this side, and across the cut, right, whereas these states have log L, right. So they look kind of like gapless ground states, right, this log L. Maybe I can rephrase a bit. So even before you found these states, let's say you take some spin chain from Newtonian, you diagonalize it numerically, maybe for finite size, you compute the entanglement entropy, and you will certainly find some non-generic states that will have low entanglement entropy. Not in the middle of the spectrum, you shouldn't. Well, you could have done this before finding the scar states in this model, for example. Right, right. And you can imagine in some other model, you could find the same. If you find the same, you probably found scars. So you're saying that it is the only way to break KTH for now with scars, so you should be looking at scars and maybe trying to prove this other property you just mentioned. Yeah, the only way to break KTH if you don't have many body localization or integrability is this way. That's the statement. So yeah, so you're certainly right. If I can, you can compute, and this is what, you know, for non-exact scars, this is what people do, right. So like, so this is a plot from the original paper of the PXP model, where is the non-exact scars and, you know, they compute the entanglement entropy and they saw this low line, right. Most states are here. There's a huge amount of states here at large entanglement entropy volume, but there's tons of them. There's this low set of scar states. So that's how they basically found. So if you find some of some low ones, like that look like this, those are scars. And they might not be exact scars, like in this PXP model, this is the puppet serving up on plot. In this PXP model, they're not exact, but that's, you know, that's how they found them. So yeah, so that's exactly what you pointed out. You can just compute and if you find that low, then you find scars. Thanks. We have some younger people, students and so maybe they have some questions. So they shouldn't be afraid to speak up. Let me ask something because probably I didn't fully get the talks. So what are the scars in the continuum limit? What are what? What are the scars in the continuum limit? Yeah, so they're basically spin to momentum pi magnons. That's the first state. I mean, right, so let's not talk about lattice, but there, you know, you can still build a spin. Okay, so it's a spin to momentum pi magnon. Okay, this state in particular is a low lying state. Okay, because it only has energy to not energy. Right. So the bandwidth of the model is order L, okay, the bandwidth of the model. This L grows in the continuum limit. It's L is the number of sites. Okay. And this is a low state, but then all the other states, you just take this spin to magnon and momentum pi and you hit the ground state with it. Right. So it's some it's some combination of spin to magnons with some scattering between them. Okay, so for example, this one schematically looks like this. You create two spin magnons when they're far away from each other. They don't interact. They interact when they're close to close to each other. Okay, so this so each of these p operators contains a spin pi magnon in real space, you kind of act one and then you act the other one. There's going to be, you know, they're going to act differently on the in the real space when they're far away. They're not interact when they're close together interact now, as you get to P to the N where N is a fraction of L, those are in the middle of the spectrum when it is a fraction of L, then you get a lot of them. Right. So their interaction is not particularly easy to to to express like this in in in real space when there's a thermodynamic limit, thermodynamic number of them, but you can express in matrix product state form. But but yeah, so this is, you know, this is their their their expression in the momentum space, you just in the in the sorry in the in the in the real space in the continuum just spin to spin to meant to pi magnons on top of each other. And I propose that that we thank Andre and and then people who have more questions like myself can stay for a little longer so that we can say it says some people who want to leave they should they shouldn't be you. Okay, so thanks Andre again for thank you. Great job. Thank you.


Hello, let's talk about the GDP and are we in a bad situation or what? The GDP numbers for the April-June quarter which is the first quarter for the current financial year have been released. It is contracted by nearly 24% when compared to the last quarter and by nearly 23% when compared to the same quarter in the last year. Mosque sectors of the economy, sands agriculture, have suffered huge blows with manufacturing seeing a decline of 39%, construction a decline of 50% and so on. All of this is not entirely new. We knew that the pandemic and the lockdown would have an impact. But the extent of the blow indicates that there are deeper structural issues in the economy. We talked to journalist Paranjay Guwathakurtha on some of these issues. Thank you Paranjay for talking to us. So we have taken a look at the numbers of course and they are much worse than many experts, at least many of the mainstream experts predicted they had said there would be a 20% decline maybe but nobody was or even that was considered much more than what was possible. So before we go into some of the details, could you take a look at the larger picture and see was this something that was writing on the walls or were we all a lot of experts living in denial or was this something out of the blue? Government economists and supporters of the government and the ruling party were obviously in a state of denial because until very recently, until maybe a few months ago they were claiming none of the Indian economy is not going to be badly impacted by the hardy manner in which the lockdown was imposed and the suddenness with which the harshest arguably one of the harshest ever lockdown was imposed by Prime Minister Narendra Modi on the midnight of the 24th of March. Now we have official data. The government of India's own data shows that there's been a dramatic fall in the country's national income across domestic product by almost 23% about 22.8% in the period in the April to June period if you compare it with the April to June period of 2019. And if you want to compare the figures for April, May, June to the figures for January, February and March 2020, then the fall is almost 24%, 23.9%. This is the steepest fall in the last 40 years. This is the sharpest quarterly fall since these figures, since these quarterly statistics were published by the government that is from 1996 onwards. And what is very important to realize is that this fall has come after a steady decline in the rate of growth of India's GDP over the last two years. It has successfully fallen every quarter for the last eight quarters. Right. So, Paranjai, like you said, this is not a new phenomenon, it's part of a larger trend. So an key argument that has been raised, of course, is that this is completely due to the pandemic. I mean, it was an emergency situation, nothing much could have been done. So is that argument, does that, is any credit to that argument at all? Well, the lockdown is really what was responsible. The very, very harsh lockdown, I think, was primarily responsible. And I think these figures that have come out in the public domain, Paul, you know, there may be actually the actual figures, maybe worse than what you see, right? And I'll tell you why. Because the data for the unorganized sector or the informal sector, that data is based on extrapolations. Now, this is a big part of India's economy. The informal or unorganized sector accounts for about half of India's gross domestic product. More importantly, it accounts for over 90% of our total employment. Now, these figures are based on surveys that have conducted every five years. And the rest of the time, you're extrapolating on the basis of data. So I suspect that the fall may be even steeper in what the government has been is claiming. The other point that is often made that no, India has done not so badly compared to other countries. No, that's not so. If you look at the same quarter, we compare it. And if you look at the data that has been compiled by the OECD, the Organization of Economic Cooperation and Development, which is really a grouping of some of the richest countries in the world, it shows that only three countries really grew in this field. And this was like Australia, Russia, China. These were the only three major economies that grew. Other than that, the slump in the Indian economy has been steeper than the slump in other major economies. And these would include countries like the United Kingdom, France, the 27 countries in the European Union, Germany, Japan, the United States, South Africa, Brazil. So the point is, in my opinion, and it's not just my opinion, a lot of other economists would share this opinion, the hurry manner and the hasty manner in which the lockdown was imposed from midnight of the 24th or should not have been done in the way it was. It should have been done in a far more calibrated manner. Please remember that the time the lockdown was imposed, there were barely 500 odd corona positive cases in the country. Today that number is over 3.6 million. These are the confirmed COVID cases. Already we've seen over 64,000 deaths, including the former president of India, Pranabukkaji. So the situation was, in a sense, it was almost inevitable. And I really am amazed how some government economists and some supporters of the ruling dispensation are perpetually in a state of denial. I mean, the fact is, India is going through a very, very deep recession. We can debate as to how long it is going to last, how steep the decline would be. But to deny that you're in a recession would be like living in a fool's paradise. Let's get some of the statistics which have come out. Private spending, which accounts for over half of our GDP, about 55, 56%, it's come down by almost 27%. Government spending has gone up by about 24%, but government spending only accounted for 18% of our GDP. Our investments have gone down, gross fixed capital formation have been down by almost half, 48%. Unemployment has been on the rise. The Center for Monitoring Indian Economy said it's sold to 23% in April. That is one out of four Indians. Almost one out of four Indians was unemployed that month in April. And then this number came down to 11% in June, by which time the harvesting had begun and about 8.3% at the end of August. The government clearly failed to put enough money in the hands of people. Despite some transfers that were made through the Jandhan accounts to women, the fact is we've seen a terrible situation and this situation is continuing. The decline in the manufacturing sector has been almost 40%, 39%. The decline in construction activities, it's halved. Trade, hospitality, commerce has been down by almost half, 47%. Mining and quarrying down by 20%. The only saving grace is agriculture, which is grown by about 3.4%. But agriculture is a relatively small portion of our Indian GDP. So what we are looking at is a huge shock that the Indian economy has received on account of the very harsh manner in which the lockdown was imposed. And I really don't know how long the impact is going to last. And the government's own statistics, however flawed or inadequate they may be, confirms that this indeed is the situation. Absolutely. And Paranjee, so we have a series of factors here. One, as you've put it, there's been the lockdown induced shock to the economy. Two, again, like you said, considering that the number of cases is continuing to rise, considering that right now we are leaders in terms of the number of COVID-19 cases. The impact is not likely to suddenly cease anytime soon. So we're going to see a continuing impact even in the coming months. And above all, there's a larger aspect that this is not a one-time decline, but something that has been happening for quite some time. So the GDP has been declining over various quarters over the past two years. So it's not a sudden dip after a series of growth or whatever. And like you said, key sectors like manufacturing also have been facing a decline. So all of this seems to indicate a larger bankruptcy in economic thinking as far as the government is concerned and planning as well. So could you maybe talk a bit about that aspect in terms of why is the government without answers to this kind of a situation? And not just the pandemic, but something a longer time period as well. The Narendra Modi government has a track record of using sudden, shock and awe tactics all ostensibly meant to improve the state of the economy. I mean, look what happened in November 2016, demonetization. It was a huge shock to the economy. The government refused to acknowledge that. And the impact was felt for many, many years. The hurried and botched up manner in which the goods and services tax was implemented is another example of how this government is completely apathetic to consultations with experts and others. In the lockdown, we've seen the states, the governments of the states that are supposed to implement all the relief packages, all the major rehabilitation packages meant they are the ones who are supposed to rehabilitate all the migrant workers. They have been so badly squeezed for funds. The union government has not been able to give the revenues that were promised for the goods and services tax. And instead they're telling them borrow, borrow, borrow more. The so-called 20 lakh crore project announced in five phases by finance minister Nirmala Sitaraman. I mean, out of the 20 lakh crores, 18 lakh crores were old schemes. They were recycled and made to appear as if there were new schemes. No, they were not. And most of them were schemes that were already in under implementation or telling people to borrow. And then the additionality was just about 2 lakh crores, which in a country which had a GDP of 200 lakh crores, that was 1% of GDP. Now, the story doesn't end there. Now, you are completely in a state of denial as to what should be done. And you seem to have not, I mean, the government does not seem to have the focused idea. It should be printing, it should be spending money. Even the Reserve Bank of India actually has been seeing all this. Instead, the government in its desperation to raise resources has been privatizing with the vengeance, selling assets that belong to the people of this country and painting it out as if this is the way forward in this kind of a deep recessionary condition that the country is going through. The corporate conglomerates headed by the two richest men in India, Mr. Mukesh Ambani and Mr. Gautam Atani, they seem to be doing rather well. Everybody else is in a mess. All of it indicates how bankrupt, how ideologically biased this government's economic policy regime is. And I regret to say this, all of which has very, very negative consequences for the country. It's going to sharpen and widen an already extremely unequal society. And despite paying lip service to the conditions of the poor and the migrant workers, the government seems to have done little or nothing to alleviate their miseries. I mean, in the months of April, May and June, we saw the kind of internal migration that's never been seen in the history of this country, perhaps never in the history of humankind. Even in the mid 40s, when the subcontinent was partitioned, you had about one and a half crores of about 15 million people possibly displaced. We've had many more people who have migrated from different parts of the country in the months of April, May and June. But what do you, when the government just refuses to look at the reality on the ground, refuses to consult not just its ideological opponents, but experts, believing it sort of knows it all. And it's unfortunate, but true, that I do not see the economic situation improving significantly in the near future. Thank you so much, Konrad, for talking to us. Thank you, Prashant. That's all we have time for today. We'll be back tomorrow with major news developments from the country. Until then, keep watching NewsClick.


Hi, I'm Zor. Welcome to InDesert Education. Well, after listening the lecture about what trigonometric equations actually are, well, it's time to solve few problems in this. So we will solve some trigonometric equations. Now, we do know how to solve the basic trigonometric equation, something like this one. We already discussed that during the lecture. So any kind of a trivial manipulation with this particular equation, something like two cosine of 3x plus seven equals to one. I'm not going to do this type of thing because it's really trivial and it's very easily reduced to the basic one. So my purpose right now is to talk about equations which are not trivial. Don't forget that the whole purpose of this course is not to give you certain skills. Okay, this is the case and if you need a situation like this in the future, you have to do that. No, the purpose is to prepare you to unknown, to challenge your mind with something which you just don't know how to do it and you have to think about how to, in this case, solve trigonometric equation. There are no methodology to solve any particular situation to help you to do anything else in the future. No, every problem is separate and you have to really go through many of them to develop the whole repertoire of different approaches, different methodologies, etc. It all accumulates into your mind. So my purpose right now is to present you with a few trigonometric equations which are not that trivial as I was just trying to exemplify. Okay, so I have three different equations for this particular lecture, so let's do it one by one. Okay, the first one is simple. Sine of x plus cosine of x equals a. Well, obviously, a cannot be anything. As we know, the sine and cosine are limited. They're from minus one to one, so a cannot be 25, for instance, obviously. Well, actually, it's much lower the boundaries of the a, which you can derive with. But anyway, how to solve this particular equation? Well, it goes to some property of sine plus cosine and in the corresponding lecture about sum of two angles, I have a lecture dedicated specifically to linear combination of sine and cosine of the same argument x in this particular case. So I will use exactly the same approach, but instead of general approach as in that lecture, I'll use a very concrete example. So what I'm going to do is the following. I will multiply the equation by square root of two over two. That does not change the solutions, the domain of the functions, et cetera. It's a completely equivalent transformation. Now, let's recall that square root of two over two is a cosine of 45 degrees and a sine as well. So sine 45 degrees equal cosine of 45 degrees equals square root of two over two. Remember, the Pythagorean theorem, if this is one, this is one, this is 45 degree, this is 45 degree, and hypotenuse by the Pythagorean theorem is one square plus one square is two, so it's square root of two. So the cosine of 45 degree of this angle, for instance, is equal to one over square root of two, which is the same as square root of two over two, right? And same is sine. It's exactly the same. It's also one opposite calculus over hypotenuse. So we know this. So that's what I'm going to use. And instead of square root of two, in this case, I will put sine of 45 degrees or pi over four. I prefer to do the regions if you're talking about equations. Sine x plus, and this square root over two, I will put as a cosine of p over four. Cosine x is equal to this. Now, what is this? Remember formula for cosine of the difference between two angles. It's cosine by cosine plus sine by sine. So that's exactly where it is. Now, and this is already a basic trigonometric equation which we have considered before during the lecture. So x minus pi over four is equal to plus minus arc cosine of square root of two over two a, from which x is equal to pi over four plus minus arc cosine. So that's the general solution. Actually, to be exact, you have to add 2 pi n in this case and 2 pi n because the cosine is a periodic function. So all solutions are expressed in this particular formula. Well, that's it. That's a relatively simple thing. Now, what kind of a lesson I would like to actually for you to learn from this particular thing? Just looking at that particular equation from the very beginning, it's not obvious, of course, what's the approach how to solve it. But however, this is a guess, if you wish. You really have to be prepared that solving certain trigonometric equations, well, actually solving any good mathematical problem, requires you to find a good approach. Because most likely you never dealt with this particular problem before. So every time you solve another problem which you don't know how to solve, it's very important for you to come up with new methods. That's what develops your creativity. And that's the whole purpose of this course, by the way. So I am trying to present to you as many things which you just don't know what to do. And you have to really think about what to do. And that's why I'm always asking you to try to solve all these problems before you listen to the lecture where I'm explaining the solutions. So by the way, if you didn't do it for this particular lecture, I do encourage you to pause the video and try to solve these two remaining problems just by yourself first, and then listen to the lecture. All right, next one. Next one is sine square x plus sine square 2x equals sine square 3x plus sine square 4x. Again, I basically claim that this is not your typical equation, which you know beforehand, or you are taught beforehand how to solve. You have to invent a new methodology to solve this particular equation. All right. So first of all, if you are thinking about methodology, how to solve this equation, first which I personally think is natural, you should think about these being square of something. Nobody likes squares. Everybody likes the first degree. So the lower degree, the easier it is. Can v represent square sine square x in terms of lower degree of some trigonometric function? Well, yes, of course. Remember this. Cosine 2 phi is equal to its cosine of phi plus phi. So it's a sum of two angles, which means it's cosine phi times cosine phi, which is cosine square minus, this is plus phi plus phi. So this is minus sine phi times sine phi, which is sine square phi. Or since I need sines, I can replace cosine square with 1 minus sine square, because cosine square plus cosine square is equal to 1. So that's 1 minus sine square and another sine square. So it's 2 sine square phi. OK, this is the key which I can use to convert sine square with a clean cosine. So sine square phi equals 1 minus cosine 2 phi over 2, from this. 1 minus cosine sine square goes here and divided by 2. So that's what I'm going to use, where phi can be either x or 2x or 3x or 4x, and all squares will be turned into the first degree of the cosine. All right, fine, let's do it.  That's the equation which I have. Now, obviously, I can multiply it by 2. And 1 will be reduced. It's 2 on this side and 2 on this side. And then minus, minus, minus, minus, I will change to plus, plus, plus, plus. I'll just multiply by minus 1. And the final equation which I have, which seems to be easier than the original, is this one. Now, these are completely equivalent transformations. So it's invariant transformations. So let me wipe out this. Actually, I don't need this one either. So I have reduced my original trigonometric equation to this form using invariant transformations. And this looks easier, right? Now, what's next? So first, I converted the second degree to the first degree. Now, I have pluses and minuses of different cosines. Now, I have it on the left. I have it on the right. And if you remember, any sum of two cosines, or difference between the cosines, can be represented as a product. And that's exactly what I'm going to do. And I would like, if I will convert it into product, something, if I do it smartly, something might actually be reduced. So let me try. Now, instead of 2x, now you remember actually I did exactly the same whenever I was talking about trigonometric series. If you have a cosine of the same of the angles which are separated by the same difference, I divided the difference in half. And I converted every member as half minus or half plus, this particular. So if I have 2x and then 4x, then I can have 3x. And this is 3x minus x. And this is 3x plus x. 2x is 3x minus x and 4x is 3x plus x. So that's what I'm going to do here. And then, if I will do it correctly, I hope it will be, something will be reduced. So this is this. And here I will do exactly the same thing. In between, you have 7x. So it's 7x minus x plus 7x plus. So that's the same equation. I just replaced 2x with 3x minus x and 4x with 3x plus x. And the same there. Why is it easier? Well, think about this way. This is cosine times cosine plus sine times sine. This is cosine times cosine minus sine times sine. And sines will be reduced. So I will have a product of cosine 3x times cosine x. That's it, right?  And this is minus. Actually, there is a multiplier, too. One from here and one from there. And here is the same thing. Cosine 7x times cosine x, then plus sine times sine. And this will be minus sine times sine. So that's exactly what will be left. And obviously, I can reduce by 2. Now, is it easier? Of course it is easier. Because number one, we have common multiplier. So immediately, we have cosine x equals to 0 as potential solutions to our equation. Because if it's equal to 0, then left part is 0, and the right part is 0. Now, this one is x is equal to pi over 2 plus pi n. OK, so we found a few solutions. Now, we want to find other solutions when the cosine of x is not equal to 0. But if it's not equal to 0, I can reduce it. And have only cosine 3x equals to cosine 7x. And then I will do exactly the same. It's cosine 7x minus cosine 3x is equal to 0. And that will convert the difference between these two cosines into a product. So what I will do, the same thing. It's cosine 5x. 5 is in between the 7 and 3, minus, sorry, plus 2x, minus cosine 5x minus 2x, right? Now, this is cosine times cosine minus sine sine. This is cosine cosine plus sine sine, right? So the cosine will be reduced, and the sines will remain. So we'll have from here minus sine 5x sine 2x. And from here, this will be the plus, right? So this minus will be minus again, the same thing. Now, they are identical. So basically, the solution is sine 5x sine 2x is equal to 0. So that's another thing, which has, in turn, two different solutions. Either sine 2x is equal to 0. And the result is 2x is equal to pi n, where n is any integer, and x is equal to pi over 2n. Or sine of 5x is equal to 0, which means 5x is equal to pi n, and x is equal to pi over 5n. So we have three different cases, and this one. But if you look at this, this one includes these points. That's very easy to determine, right? Because this is pi over 2, pi, 3 pi over 2, 2 pi, et cetera. And this one, just every other one from these points. So basically, we can disregard this solution, and these two constitute the solution to a trigonometric equation, which was in the beginning. Which looked kind of scary initially, but after a few transformations, and what we did actually, we always used exactly the same approach. We transformed a difference between trigonometric functions into their product, and then something actually was reduced. That's it. Next one. Next one is the game. It looks a little unusual, but it's still kind of solvable, more or less the same methodology. Sine of pi cosine x is equal to cosine of pi sine x. Yeah, looks, again, a little bit unusual. Scary, if you wish. So what can we do? Well, let's think about what we were using before. We were converging a difference or a sum of cosines or sines into the product. Well, let's try to do exactly the same thing, except we have sine and cosine. They are different functions, and we don't really know what sine minus cosine is. I mean, how to convert it. But I can always convert cosine into a sine, right? Remember, cosine of pi is equal to sine of pi over 2 minus pi. This is one of the classic identities of trigonometry. So I will replace the cosine on the right with the corresponding sine, and then I will bring everything to the left. So we'll have sine of pi cosine x minus sine of pi over 2 minus pi sine of x equals to 0, right? So cosine is equal to sine of pi over 2 minus angle, and then I transferred it to the left. OK, great. Now, let's think about how to convert difference between the sines into their product. And that's very useful, because when it will be the product and it's equal to 0, which means every multiplier is equal to 0 by itself, right? So it's easy. I'll use exactly the same approach as in the previous problem. Remember, if you have sine of alpha, in this case minus sine of beta, how to convert it into product and a product of different trigonometric functions? Well, alpha can be represented as sine of alpha plus beta over 2 minus beta minus alpha over 2, right? Minus beta, yes. And this can be represented as alpha plus beta over 2 plus beta minus alpha over 2. Alpha will be reduced, beta will be 2, yes. And this is the difference between the sine, the difference between two angles is what? Sine cosine minus cosine sine, right?  But this is minus in front, so I will have everything with a minus sign.  So what's reduced? This one. So I will have minus 2 cosine alpha plus beta over 2 sine of beta minus alpha over 2. That's what I have, right? So that's what sine minus sine alpha minus sine beta is. This is how to convert it into a product. OK, so let's use it in this particular case. So this is alpha. This is beta, right? And since it's equal to 0, minus and 2 are insignificant. I will retain only cosine and sine. So it would be cosine of alpha plus beta over 2. So it's pi cosine x over 2 plus pi over 4 minus pi sine x over 2. Right? This is sum of two angles alpha plus beta divided by 2. So it's half of this, half of this, and half of this times sine of the difference beta minus alpha, which is pi over 4 minus pi over 2 sine x minus pi cosine x over 2. And this is equal to 0, which means this can be equal to 0, and that's the solutions. Or this can be equals to 0, which is the solution. Well, let's just consider two cases separately. So when the cosine of something is equal to 0, when it's equal to pi over 2 plus pi n, right? So pi cosine x over 2 plus pi over 4 minus pi sine x over 2 equals to pi over 2 plus pi n. So that's one set of solutions. And as you see, pi is conveniently reduced, right? So if we will multiply everything by what? By 2, right? So we reduce pi, and we multiply by 2. I will have cosine x minus sine x equals 2, pi over 2 minus pi over 4, it's pi over 4. But you multiply by 2, so it's pi over 2 plus 2 pi n. Am I right? I'm sorry. 1 half. There is no pi here. It's 1 half minus 1 quarter, we reduce pi and plus n. OK, so that's one thing. Now, the second thing is when this particular expression is equal to 0, and when sine is equal to 0, when the angle equals to pi n, right? So pi over 4 minus pi sine x over 2 minus pi cosine x over 2 equals to pi n. Again, pi is conveniently reduced. So what do we have instead? We have sine and cosine. Let's put it to the right with a plus sine. So we'll have cosine x plus sine x equals 2. So we multiply by 2, so it's 1 pi over 4, so it's 1 half. Now, pi n would be on this side. So it's n, actually, because pi will be reduced. And we multiply by 2, so it's minus 2n. Now, since n is any integer number, it doesn't really matter whether we put n, minus 2n or plus 2n. So it's exactly the same thing. So we have two different equations, actually. So our one equation, whatever it was in the beginning, is split into two different equations. Well, actually, not two, infinite number of equations, because n can be any integer number, but can it actually not? Let's consider this is cosine and sine. Each one of them is not greater than 1 and not less than minus 1. So the sum of them or the difference between them, just by definition, cannot exceed 2 or be less than minus 2. So obviously, n cannot be anything. n can be 0. That's true. But if n is equal to 1, as you see, in both cases, we have 2 and 1 half. So it's completely out. It's greater than 2. And this cannot exceed 2, because each one cannot exceed 1. Now, on the negative side, minus 1 is still possible, because minus 1 would be what? Minus 3 seconds, right? So we have either 1 half or minus 3 halves. And same here. So that's the only available ends. n is equal to 0 or n is equal to minus 1. When these equations at least make sense. But let's go even further. Now, you remember that the first equation which I did today in the lecture was cosine x plus sine x is equal to a. Remember this? And I solved it. So let's solve it again. So what I did the first time, I multiplied by square root of 2 over 2 and having cosine x cosine pi over 4, that's square root of 2 divided by 2. Plus, well, let's start with the top one, with a minus. Minus sine x sine pi over 2 equals 2. Well, let's do 1 half first. So I multiplied by square root of 2 over 2, right? So that's square root of 2 over 4. That's my equation with 1 half, which is actually what? A cosine of x plus pi over 4 is equal to square root of 2 over 4. From this, we can say that x plus pi over 4 is equal to plus minus arc cosine square root of 2 over 4 plus 2 pi. I'll use another number, k. k is signifying any integer number. So x is equal to minus pi over 4 plus minus arc cosine square root of 2 over 4 plus 2 pi k. So that's what I have if I take 1 half. OK, fine, this is done. Now, what if I have minus 3 halves? If I have minus 3 halves, I will have minus 3 halves times square root of 2 over 2, right? Which is minus 3 square root of 2 over 4. Now, let's think about it. Square root of 2 is approximately 1.4, et cetera. Times 3 would be 4.2, et cetera. Divided by 4 would be greater than 1 by absolute value. Actually, in this case, it's less than minus 1. And the cosine cannot be less than minus 1. So the whole case does not have any solutions, because sine cannot be less than minus 1 or greater than 1. So this is no good. Now, let's go to this one. So it's plus. And again, let's start with 1 half. So I have here square root of 2 over 4. Now, since this is a plus, this is a minus. Cosine of x minus pi over 4 is this. And it's equal to square root of 2 over 4. And now I have minus here. So the x is equal to plus pi over 4. Now, to combine both solutions, the old one was minus, and the new one was plus, I'll put plus minus here. So now I have basically four different combinations of pi over 4 and arc cosine of square root of 2 over 4. Plus plus, plus minus, minus plus, or minus minus. And now, so this is done. Now, for minus 3 half, I have exactly the same consideration. I will have minus 3 quarters of square root of 2, which is by absolute value greater than 1. So there is no solution. So basically, that's it. We finally got all the solutions. And that's the result, where k is any integer number. Well, that's it for today. These three problems are important in some way, because I was trying to put something non-trivial. But let's just think about what kind of methodology I was using. Primarily, it's just converting minus sine or sine, plus sine into product. And then the solution is easier, because if the product of two variables is equal to 0, then either one variable or another is equal to 0. So just remember, this is approach. This is the methodology. And whenever something else comes up, this is one of your instruments in your toolbox. And that's the purpose, to put not the exact solutions to all the problems that you can just see in the future, but to equip you with sufficient number of tools that you can use to achieve the goal. Well, that's it for today. Thank you very much. And good luck with new problems, which I am definitely going to put on my website. Thank you.


The story of Dead Bart is a long and confusing one. To summarize, Dead Bart was a production by Matt Groening, and one of the first episodes in the Simpsons series. Today, not much footage from Dead Bart exists, since most copies have been destroyed. Including those made by the studio. The production has been considered to be a mistake, and those who have seen the episode, or worked on it, have said that the creation of it should never have happened. Fox labels all of its episodes according to a smart and efficient system. But the creation of Dead Bart has caused an inconsistency in the amount of episodes reported. The original production label given to the episode was 7G06, but searching for 7G06 brings up another episode, entitled Moning Lisa. People believe that this is to hide the fact that Dead Bart was ever created. Many people know the plot of the story quite well, having read about it in the endless creepypasta. But not many know the actual origin of the production. During the early part of 1989, Matt Groening was experiencing quite a few health problems. He had been experiencing serious delusion and had visions of his death from time to time, many of them eerie and vomit inducing. He was under a lot of pressure to produce quality content for Fox, who were forcing him to work nights in order to get the first season of The Simpsons finalized and ready for broadcast. Having made several mistakes in his life so far, he had decided to make one more, and that mistake was Dead Bart. He wrote the episode in hopes that it would allow him to sleep easy. Like many people write their feelings down on paper, Matt felt that doing the same would take his mind off things. Except for instead of just writing them down, he put them into a storyline, and made his characters play out the scenario. The episode starts with four characters, Bart, Homer, Lisa and Marge. Maddie hadn't made an appearance in the episode, as the overall idea for The Simpsons was a lot different when he had started writing Dead Bart. This includes Maddie being absent. There are several fan theories regarding continuity as to when and how Maddie was born. The Simpsons show was originally going to make stylistic use of death, and the ramifications of someone dear dying. The characters had stronger feelings, due to the effects of death on them in earlier years. Bart had more hatred for his elders, Marge was subject to depression, Lisa was more anxious and Homer was a lot enrear. This concept wasn't scrapped completely in newer episodes of The Simpsons though. Homer still has a strong sense of anger, as reminiscent of him strangling Bart all the time. One of them is related to the characters' feelings, Homer being angry, and Marge being depressed may drive them away from reproduction, plus Marge's depression may have been caused by previous experiences. Maybe she had an abortion. The episode Madd had wrote wasn't supposed to be an actual production though, it was simply a log of his feelings turned into a script. But somewhere along the line, one of the head animators had picked up Madd's work, and took the episode seriously, animating the entire thing, and had voice tracks produced to go along. The voice actors became suspicious as to why they had been asked to read such deluded lines, but they never brought it up with Madd, as they didn't want to mess with an individual who was in such a state. Or somebody in more authority than them for that matter. The production eventually came to the rendering stage. Madd was brought into a conference room, where the head animator had the final render on tape. Madd was dumbfounded as to why he was asked in, seeing that he was screening another episode that week, and official production hadn't been announced yet. The tape started, with the family, as described above. As Madd watched he became pale and looked flushed. The first scene eventually came to a close, and he was already on the verge of tears. I'm not going to summarize the rest here, if you need to, go look at the original creepypasta, entitled Bed Bart. As time ran out, and the credits were fading out, a black and white fox logo appeared on the top of the screen, and the production label, 7G06 faded in just below that, along with the words, final sales copy. That's where the tape ended. After that, Madd left and went home. The team didn't see him for another two days, and when he finally turned up, he completely ignored everyone around him. After being asked about his absence by the receptionist, he quoted, we shall never talk about it again. Since that time, the entire team has uttered not one word about the day. The tape was thrown out. Not burnt, not destroyed, just thrown out. This allowed a needier fan to pick up the tape from the alleyway outside. He watched it several times. His 12-year-old son found out about the tape by accident. He then attempted to set the tape light. Though not a successful destruction, this caused the footage to become shaky and filled with static. An anonymous man recovered the videocassette and uploaded it to a web server under the name of deadbart.complete and exploited.net slash pwtrue slash 7G06, which when opened, would bring up a blank black page with a yellow download link. The page doesn't exist today, and Google failed to get a cached version of it. Not many people can recollect memories of the page while it existed, but some can, including Matt Groening himself. After many weeks of moping over the incident, Matt had received an email on his personal address, containing the link, deadbart.complete and exploited.net slash pwtrue slash 7G06, which was masked as the SimpsonsOfficial.com. Seeing no harm and assuming it was their rebooted site, he went ahead and clicked it. This being in the days of 128 kilobytes per second internet, the page slowly loaded, and then internet explorer packed up under the pressure. Matt decided to take a look later on his home PC. After finally getting onto the site, he downloaded the 20 millibar file entitled just4u.flv and loaded it into his Flash Player. To his dismay, the file downloaded was a ripped file of deadbart. He closed down his computer and attempted to track down the old cassette to make sure that it wasn't seen by anyone else, but never managed to find it. He could remember the entire website link though, off my heart. It was the shock. Eventually a presentation came around, documenting the success of the first season. A Q&A was opened up to the crowd when a young lad asked about the missing episode. David Silverman, the man speaking at the time closed down the presentation hours before they intended the presentation to end. Another lad, who was researching on the episode at the time, stalked Matt down at the end and asked about the episode. In his dumbfounded state, Matt wrote down deadbart.complete and exploited .net-slash-p-w-true-slash-7g-0-6 on a piece of paper, and handed it to the inquirer. He then begged him to never speak of the episode again. That man was the original writer of the deadbart creepypasta. Although only one VCR of the episode was ever made by the studio, some interns had stolen early footage in the form of Betamax tapes they brought in from home. You would think a major studio would keep their resources safe. Another man has documented his experiences with a bootleg tape of deadbart. Legend has it the start of the tape contained an aired episode of The Simpsons, and he had watched this from start to finish, but hadn't realized that the tape also contained deadbart. While making coffee for himself, he heard a loud shouting from the room he had been in, something along the lines of Homer shouting Bart's name out. When he arrived back in the room he was able to witness Bart being plummeted from a plane, 50 feet up in the air, to be followed by the crunching sound of broken bones. That's when the VCR player caught fire, and the tape was initially ruined, except for a scene right at the end, showing Bart's dead body. With both known tapes of deadbart ruined, the latter of people have assumed that deadbart was just a story, nothing but fancy words, ridden to a news and scare. But it wasn't. The truth is that the episode was produced. I saw it when I was eight. I haven't had a good night's sleep since. I now spend my life researching the meaning and the cause of deadbart. Whenever I hear a name being shouted or screaming from afar, I panic, my heart starts racing.  Feeling mutual.