 Thank you very much for inviting me. I'm very happy to participate in the celebration of the anniversary of Professor Shadash Srilay. Many of his achievements were mentioned earlier in the conference. Today I want to emphasize the role of Samson in Integrability and specifically in Birthday Gauge Correspondence. I think this photo was not shown before. This is a new one. I'm just taking pictures. Well, but new for the conference. Your sister Nana is like Georgia Got Talent. So let's start with non-linear Schrodinger equation. First of all, the model has many, many different names. It's called Leap-Leninger, non-linear Schrodinger, Bosie-Gas, with delta interaction, and some people call it Tung-Schirard-Do-Gas. So there are also several different ways to write down Hamiltonian. This is the representation in terms of many quantum mechanics. Kinetic energy and delta potential, coupling constant will be positive for this lecture. Later I will write it in terms of the quantum field, Bosie field, like by Bosie field Kappa, standard Bosie commutation relation, delta is the positive number, which I will interpret later the step of the latest. So at first the model was solved by coordinate batanzas and stayed that way for a long time. Many interesting physical properties were extracted from coordinate batanzas, but later it was reformulated in terms of quantum inverse scattering method. That's our matrix, which is i squared of minus one, permutation, and this is lambda and mu, we call spectral parameters. So this is a l-operator, and somehow in the original paper it was written approximately in the sense of the step of the latest. Step of the latest was supposed to be small, and the l-operator was written approximately with this precision delta squared. But later we were able to restore all the next orders in the step of the latest, and this is exact l-operator on the latest, I mean no corrections. So the step of the latest, and under the square root, rho is the square root of this one, coupling constant, side by side, number of particles in the latest side number j, all the way index. And that's mainly modification of off diagonal elements by the square root, but also some correction on the diagonal, this number of particles. So this is exact l-operator, this is latest non-linear Schrodinger, l-operator for latest non-linear Schrodinger. It has involution, it has involution, so sigma x is off diagonal matrix with identities on the off diagonal, and then this is dagger is a conjugation of quantum operators. And lambda, later I will argue that lambda can be kept real for the purposes of this lecture. Excuse me, what is lambda? Oh, the coupling constant. No, no, no, no, no, no, no, no. This is coupling constant which enter Hamiltonian. Lambda is a spectral parameter which is a new parameter which was not in the Hamiltonian. This is like artificial new parameter designed in order to solve the model. model. There's something new which was not in the Hamiltonian, but appear, you know, the trick of solution. Also, the dimension 2 by 2 of the operator was not mentioned in the Hamiltonian. This is yet another auxiliary object, you know, designed to solve the model. Okay, so this is entries of this operator. This is two of diagonal and diagonal, and I can calculate the commutation relation. This is SU2 algebra, representation of SU2 algebra. But spin is negative. Spin is minus 2, kappa divided by delta, and I'm very happy to present this in the presence of Professor Karchemski, who's somewhere here because he likes negative spin, minus 1. So I mentioned minus 1 specifically. Complex even better. Well, for the purpose of this lecture, it will be negative spin. So once again, kappa is positive. Okay, SU2 representation was negative spin. Then quantum inverse scattering method works very similar to the continuous case. So I can multiply elliperes in the auxiliary space. I mean like matrices 2 by 2, row by column, and then the entries, the quantum operators in different letters such commute with one another. We call it ultra-locality. So this product I can write as explicit matrix 2 by 2 in the auxiliary space. But a, d and b and c are complicated combination of quantum operators, all of these s plus minus in all of these letters sides from 1 to l. I'm sorry, for the degeneration of notation. I mean, length of the letters is also l. Trace of one of the matrix, a plus d, is important, and the Hamiltonian will be extracted later from this trace of monodimetric by means of tracer densities. This involution still works for the monodimetric. So dagger means Hermitian conjugation of quantum operators. Zero means this is an explicit matrix in this auxiliary space of 2 by 2 and sigma s is of diagonal matrix 2 by 2 and multiplication that auxiliary space. So it's easy to invert this matrix. I kind of can use Kramer's formula, but with some shift of spectral parameter. So here, I mean, I can divide t inverse and, I mean, can multiply by t inverse and t inverse will be given by Kramer's formula almost with some shift of spectral parameter. And the determinant also given like ad minus bc with some shift of spectral parameter. So it's like we call it quantum determinant. It's the center of the algebra. It can be used with all other operators actually equal to some function of spectral parameter with quadratic function. So algebraic betanzot works in the form which was designed by Professor Tartajan with quarters. And then it leads to betanzot equation which we shall analyze in a moment. The equation, so lambda is those spectral parameters. Later, they will play the role of momentum. Momentum is a simple function of this spectral parameter. And then they're even more convenient than momentum because if I write two-body scattering matrix as a function of momentum, it will depend on both momentum of two colliding particles. But if I write this in the terms of spectral parameters, then scattering matrix will depend on the difference of spectral parameters. It's like more convenient parameterization from the point of view of study of two-body scattering. So, and then next we have to design Hamiltonian starting from a leperator. There are several different ways to construct Hamiltonian on the letters. The most popular was constructed by Tarasov, Tartajan and Fajiev. Later, I will mention other Hamiltonians. So, this one, TTF, people call it F, well, TTF. The Hamiltonian is written in terms of the logarithmic derivative of gamma function. And this J, depending on the local spins, here is, here is this, I mean, this is spinning one letter side, spinning the neighboring letter side. And then in this equation, everything commutes in this equation. So, I can, this quadratic equation, which I can solve by, you know, standard formula. And then, substitute this into here. So, this is famous, Tarasov, Tartajan. Hamiltonian, which we shall study for a while, and then later we shall switch to another Hamiltonian. So, for spin minus one, specification for Professor Karczemski, and then spin minus one, this is better than this equation and this is expression for the energy. Let's study better than this equations for a while. We can prove three different theorems for this better equation. So, let's prove the theorems. Do I have time for proof of the theorems? Okay, 15 minutes. Theorem number one, this system of equations, first of all, I have n variables and lambdas and also an equation. So, for K here, supposed to run through one to n, so it's not one equation, system of equations. So, theorem number one, if solutions of the system exist, all of them have to be real numbers. So, how can possibly prove this statement? Well, I have to study these factors in the right-hand side. So, in the right-hand side, I can simple elementary calculation show that the models of each and every factor in the right-hand side is larger than one only if imaginary part of this lambda is greater than zero. If it's smaller than zero, then it's smaller than one. In the left-hand side, it's the other way. So, if imaginary part of this lambda, of this lambda is greater than zero, then smaller than one and the other way around. So, how I can prove it? First of all, let's, I mean, consider this set of lambdas. Maybe they're imaginary. Let's pick the lambda with the maximal value of imaginary part. If there are several of them, I pick one of them. And then, I plug this in the right-hand side. So, in the right-hand side, I will have this difference. This lambda has a largest imaginary part. So, the difference actually will have a positive imaginary part, this one. So, the models of the right-hand side is larger than one, as well as the left-hand side. So, left-hand side of imaginary part, I mean, is larger than one. Then, this means that imaginary part is smaller than zero. So, all of this show that maximum lambda with the maximal imaginary part, it's imaginary part is smaller than zero. So, all others, lambdas also have imaginary part smaller than zero. So, all lambdas has imaginary part smaller than zero. Then, I can go through similar consideration, can pick up lambda with the minimal imaginary part. And then, I can prove that all of the imaginary part is greater than zero. So, this means that they're equal to zero. So, that's like real, right? Okay. So, I proved is that... George Poirier was using same arguments in 1918 in his break on Riemann-Zieterfams. Poirier, 1918. I will have reference for 1856 later. But for the different subject. Not about this, some other stuff. Okay, so, we proved that it's the solution. If they exist, they're real. Do they exist? So, take logarithm. Take a logarithm of this equation. So, L is the length of the letters. Theta is this. The logarithm of what was left-hand side. And theta is of the difference is this logarithm of the former right-hand side. So, built equations, famous. So, how I can possibly prove this solution? So, I want to prove next theorem that solution exists and unique. So, after I choose integer numbers. So, first I pick set of integer numbers, which is the subject of the third theorem. Well, the second, let's pick set of integer numbers, this N. And then, after we do that, solution exists and unique. So, how can I possibly prove this? Actually, this equation has young action, has action. So, that's action was suggested by young and young. So, theta 1 is integral of the theta of the previous transference integral. And then, this is integer numbers. And this is theta integral of those scattering phase. So, if I consider extremum of this action, that derivative of S with respect to lambda, I will come back to this equations, to this equation. So, this equation has action, which was used in many papers by Professor Shadashvili. So, let's prove that this is, it's convex. So, let's prove that it's convex, then it has unique minimum, then solution exists and unique. So, let's calculate the second derivative. Second derivative of this is diagonal element K. It's positive, right? Because K is small, coupling content is positive. And this is of diagonal element, which is also with a minus sign. So, let's calculate quadratic form. So, V is some vector, a real vector with real components. So, after some massaging, it's equal to L, this K multiplied by the square. And this of diagonal will be combined to this square of the difference, square of the difference. So, anyway, it's positive for, if V is real. So, the young action is convex. So, unique minimum. Nice. So, theorem number two, solution exist and unique, given by the set of integer numbers. By the way, the determinant of this play will pop up again later, when we shall construct eigenvectors of this Hamiltonian, which was mentioned in the beginning, then the square of the norm of the better way function is actually given by determinant of this matrix, determinant. It was conjectured here first by Michel Godin from Sacle and then I proved it ten years later. So, now the third theorem, which I promised, Pauli principle, Pauli principle. Let me come back to this equation. So, Pauli principle, the third theorem about this equation. So, third theorem is the following. I can prove that if all ends are different, then all lambdas are different, which solutions exist. But if two ends coincide, then corresponding lambdas, corresponding means with the same index, also coincide. This is not good. If two lambdas coincide, that construction of algebraic betanzots will have this B square, and then all these calculations of algebraic betanzots will lead to one additional extra equation. So, the system of betanzots equation will be overfilled, will be too much equations. And this is the third equation, looks like positive number equal to zero. So, there is no solution. So, if two ends coincide, there is no corresponding eigenvector. It does not exist. So, it means that all ends has to be different. So, that's the end of the theorems, and the implication for physics means that the ground state of the hemispheres will look like a thermosphere. It will look like interacting electrons. So, all ends will, at zero temperature, will feel some interval, and this will minimize the energy. So, I started with bisonic theory, but in the momentum space, it looks like fermions because of the Pauli principle. So, let's move further. Maybe something interesting. So, this is thermodynamic limit. I have this length of the box, total length of the length, which I assume go to infinity right now, and also number of particles also go to infinity. So, number of particles go to infinity, length of the box goes to infinity, ratio is fixed, density, and then it can be this lambda's solution to bet equations. They distributed along some curve. Curve described by Roa, and Roa is a solution of integral equation. Q is a value of spectral parameter on the thermosphere, which people sometimes call it thermosphere. And as long as Q is finite, then one can prove that this linear integral operator, I mean I can move this in the left-hand side, is not degenerated. So, there's a unique solution of this equation. It's somewhat similar to a Leap-Leninger equation. For a Leap-Leninger equation, I have a similar equation, but instead of this one, I have one divided by two pi. So, this is kind of Q deformation of Leap-Leninger, because I moved to the lattice. So, in the continuous, I had this equation, but this K was one divided by two pi, and now in the lattice, I have this K. So, these equations, I mean we cannot solve it analytically. We don't have an analytical expression. There is very simple decomposition. When coupling constant kappa go to infinity, the integral becomes small. But at small kappa, this is very, very singular. And I mean Leap-Leninger are studied right now for small values of coupling constant and the coefficient given in terms of our own Riemann z-ade odd arguments, which further Smirnov likes so much. This is even more singular at small kappa, which we didn't study it yet. Okay. So, in thermodynamic limits, there's no bound states, right? Because all lambda's are real, no bound states. And any energy level can be interpreted at a scattering state of several elementary excitations. So, elementary excitation can be described, well, it has some energy and momentum. So, the energy of elementary excitation is, this is original energy. This is chemical potential. And this is integral equation, which we already saw. Later, I mean, as I said that lambda is not really momentum. This is formal spectral parameter. So, momentum as a function of lambda, I will write later, a little bit later. So, this is the picture of this elementary excitation. This is my firm sphere. So, meaning that all the states here are filled in. So, I cannot put any more particles into here because of Pauli principle, but I can make a hole. So, here I have excitation, which is a hole, and this is like energy. And out of this interval, there is no particle, so I can put particle into here. And then, this is my dispersion curve, dependence of energy on spectral parameter of this elementary excitation. My momentum will come momentarily. So, it's gapless, the energy of vanish. And then, later, we shall see that the velocity of sound, sound velocity, is given by the slope of this curve. Well, I have to differentiate with respect to momentum, but we'll do this later. Okay, this is formula for momentum, like physical momentum. This P0 is a log, this is log of the left-hand side of mybert equations. Theta is a logarithm of the right-hand side. And the rho is something which we saw before. So, this is momentum of the particle. This is momentum of the hole. But all of this together describe elementary excitation, give the dependence of energy on the momentum of this elementary excitation. This is scattering symmetric, these two elementary excitation. This scatter, and the scattering is elastic. The resultant transition and phase shift is given by the similar integral equation. In the right-hand side, I put this theta. Theta actually has a physical meaning of the phase shift in the bare vacuum, zero density. And then, when I have non-zero density, I have to dress up this phase by this integral equation. So, phase shift. V is a derivative of energy with respect to momentum of that elementary excitation on the Fermi H. And it describes, well, by now, it's called quench velocity. Quench velocity means many different things. Among other things, if I make some local quantum mechanical measurement in one latest site, then it will cause the entropy wave. And entropy wave will spread with this velocity, quench velocity. In a moment, I will go to, we'll describe thermodynamics, young and young thermodynamics. Thermodynamics has an entropy, thermal entropy. But before that, at zero temperature, I have another entropy, which is entanglement entropy, which should not be confused. So, first I talk about entanglement entropy, and then we close up zero temperature and move to the positive temperature. Entanglement entropy behaves standardly. So, ground state is unique. So, entropy of the whole ground state is zero. But there is some entropy in the block of spins. This is quantum mechanical phenomena, right? Because in classical, if I can see the classical random variable, if the total entropy of the classical random variable is zero, then there is no entropy in any subsystem. One can prove the theorem. Quantum k is not so. Total entropy can be equal to zero, but there is entropy in subsystem. Quantum fluctuations, entanglement entropy. And then, of course, entropy is a complicated function of the size of this block of spins. But for large size of the spin, it's logarithmic dependence with a coefficient one-third, and it agrees with safety. It agrees with quantum field, conformal field theory, because central chair is equal to one. Rainy entropy. So, rho is a density symmetric of block of that spin. So, alpha is some fractional number from zero to one. And then, I take this density matrix, raise in the power, alpha, take trace, log. This is a rainy entropy. Rainy entropy also depends logarithmically on the size of the block, but coefficient in front of the log depends on that alpha. So, this is entanglement entropy. I mentioned this. Well, Professor Verlinde mentioned this in his lecture. So, it might be appropriate. So, now, let's consider thermodynamics. In principle, for continuous nonlinear Schrodinger, thermodynamics was constructed by Young, seeing Young as Cp Young. In here, in the latest version, it's similar. Construction is similar. Also, equation is not different from the continuous case. Difference in here, in the homogeneous. In the continuous case, it was square of lambda square minus chemical potential. In here, I have this my bare energy. That's the one which I saw when I wrote that equation. This integral looks really, really similar to a Young-Young equation. So, this is famous Young equations. There's some notation. So, epsilon is the ratio of the density of the holes to the particles. I mean, at zero temperature, I have a concentration of particles for a small moment of minus q. There was no holes on the particle. But, at positive temperature, everything mixed up, particles and holes. So, the ratio is epsilon. Epsilon also has other meaning because this model is integrable. There's infinitely many conservation laws. This leads to consequences that even at positive temperature, there's stable excitations which does not decay. And then the energy, this epsilon is actually energy of this stable excitation which exists for positive temperature and does not decay. So, thermodynamics, free energy. Free energy can be described by, well, that sounds in terms of this epsilon. I wrote equation for the epsilon. This equation, the previous Young-Young equation was analyzed. I mean, the only one mathematical theorem which was proved that if I started erasing this, like in the first approximation, epsilon is given by this expression. And then I put this into here and keep it erasing. So, this iteration converts, so one of the solutions exists. But, it's probably unique, but it's not proven. There is no theorem that it's unique. But, one solution exists. So, this is free energy. This is pressure. And this is entropy. So, but this is thermo entropy, right? Thermo entropy of the whole bulk. So, it's not entanglement entropy. I mean, it's, thermal entropy also exists for classical thermodynamics. It doesn't describe quantum mechanics. So, while entanglement entropy is a difference between classical and quantum. So, it should not be confused. So, all of this can be done for other values of negative spin. So, here I decided to deviate for Professor Karchemsky and consider not necessarily minus one, but maybe some other number. Everything goes through. This is the basic equations. This is thermodynamics. Entropy. Everything works. Okay. So, this was so far analysis of the Hamiltonian of Tachtajan, Fadeev and Taras. So, there is other Hamiltonians of the latest non-linear Schrodinger. So, I keep our matrix. My matrix is the same like it was in continuous case. And for the latest, I never change it. Like it was discovered by Young actually. So, I keep Young for me. Eloperator is the one which was in the beginning. But by now, I want to change it a little bit. I want to make it different in the order and even latest sites. Well, the purpose of doing this to get some other Hamiltonian. So, this is the shift of the spectral parameter, which is a little bit different in the order and even latest sites. So, and I think it was J before. So, I'm sorry for changing notation. So, it was J. Now, it's N. And it's evidently different. All diagonal elements looks the same. But this in homogeneous, it also appears under the square root. And kappa now becomes size. So, I just copied from the other paper. So, and then later we shall see. Delta is still like step of the letters. And this is number of the total number of particles in the whole latest set. So, with this modification of Eloperator, we can design new Hamiltonian. This is expression in terms of trace identities. So, tau is a trace like a plus z of the product of all of this Eloperator. And this is a special value of spectral parameter where Eloperator become one dimensional projector. The quantum determinant vanish at this point. So, that's the reason I can write these Hamiltonians looks even more complicated than Fadi if Theras if it's an octogen. So, this T we express in terms of alpha. And alpha is relatively simple function of the local bosid field. So, this Eloperator actually describe interaction of eight latest sites. So, people might argue that Fadi if Theras of octogen is better. Okay. So, now I'm actually moving to form factors. Before that, well as I mentioned the square of the norm is given by this golden formula but the determinant. Form factors in the behaves similar in the continuous in the latest case. So, I will kind of explaining first I will remind what happens with form factors in the continuous case. I mean they don't exist. The answer is like negative. But, I mean the behavior is similar on the latest in the continuous case. So, this is some one of form factors. So, this J is the side I guess I for the continuous case. This is local density of particles in the X and T space standpoint. This is standard canonical bosid operator. And this Q is number of particles on the interval of zero X. So, I can take this Q and taken matrix element between two birth states. This is like one and this is another and normalize it. It's actually was long story. First, some determinants representation was written for this. And determinants representation I mentioned determinant of the matrix of a large size. The size of the matrix equal to n capital number for particles which goes to infinity. And later I will mention the similar determinant representation also exists for the latest non-linear Schrodinger. Not identical but similar. And then this determinant representation was studied and the answer was negative. Because when the lengths of the letters go to infinity this form factor decays goes to zero as some fractional power. So, f is some solution of integral equation. Actually, this is a phase scattering phase of those two elementary excitation divided by two pi. This is definition in terms of the lambdas. We call this shift function, but it also coincides with the phase shift of two scattering matrix. So, the form factor is vanish in the continuous. This is just other form factors also was analyzed. They also don't exist. This is the determinant representation, but maybe it's published somewhere. Maybe it's not interesting to the people right now. Okay. So, this last couple of transparencies. I was explaining problems with the form factor the continuous non-linear Schrodinger. Similar problems occur for latest and Takeshi Otto from Japan. He wrote the term and representation for some operator or for latest non-linear Schrodinger. The operator somehow has correct continuous limit and maybe I just come back. So, and then Karl Kazlovsky. Karl Kazlovsky here analyzed the thermodynamic limit of this determinant of Takeshi Otto and then it also goes to zero as a fractional power of the length of the latest actually the same power. So, this is more or less the end of what I was going to say about non-linear Schrodinger on the latest. Maybe the last transparency that similar calculation can be done for Sainte-Gordon. This is continuous Sainte-Gordon and then continuous Sainte-Gordon was sold by Aldi Bragg with Anzac in Quantum University, recommended by Professor Tahtajan and Fadeev. But we have our own latest version which is interesting by itself. But maybe, maybe it's a good time for me to stop my lecture and wish happy birthday to Professor Shatashvili again and here he was my best student. So, that's the end of my lecture. Any questions or comments? Maybe, maybe just some references. The main reference. Here, the main reference is this and this was our inspiration. This was Professor Shatashvili related to dimension topological gauge theories to non-linear Schrodinger to continuous non-linear Schrodinger. This was inspiration for our work. So, if you look at Sainte-Gordon do you, on the left, do you see something of the Kostel et Thales transition? Is your Aldi Bragg method? I didn't see that phase transition. I mean, our contribution was kind of a boring mathematics. I mean, that answers for Sainte-Gordon, I mean, this is a relativistic model. There is one Feynman diagram in Divergent. For that answers, we have to learn how to make ultraviolet renormalization. And here, for Sainte-Gordon, for specific, like for rational values of coupling constants, there is an anisotropy parameter which is coupling constants. For the rational values, we can compactify the local Hilbert space. We don't have to have this infinite dimensional Hilbert space which we had for latest non-linear Schrodinger, but we have a finite dimension. I mean, dimension is equal to the denominator of this fraction, so we just use it more like for mathematical justification, everything is rigorous, but Schrodinger's is no to your question. There is another, I learned sometimes ago, Latis version of the delta function. Schrodinger operator? Ablovitz logic? There's two people in the United States, they live in the Midwest. One is Ablovitz, that's the name of a person. And Latik is another person. The author of that paper I was told is some Dutch person living in Brazil. That's the James. I don't know, well I don't know that paper, but Ablovitz-Ladik, you know Ablovitz-Ladik, so Ablovitz-Ladik has another discretization of non-linear Schrodinger. I mean, we kind of insist on ours because in their case, our matrix is different, so our matrix depends on the Latis step. And in our case, like going from continuous to Latis is like changing the representation of the same algebra, right? The algebra is given by our matrix, so we just keep our matrix, change the representation, so we think like this is an intelligent way to discretize. But I mean, Ablovitz-Ladik, they change our matrix. Sorry, and your question was? That one is actually this second order differential operator, delta function potentially replaced by some differential equation, which in the limit becomes that. Well, I mean, the way function there is what's called Hall-Litzwood polynomial. Yeah, I don't know, if I take non-linear Schrodinger, it replaces the second derivative by the difference, it won't be integrable, so one should be careful with this. I mean, it's just straightforward discretization would just keep, and then there will be problem because the phase transition will appear when I send the Latis step to zero, then it's not integrable, then it's not integrable, and this is kind of problematic. So it's not regularization, it's like falsification. Right, I think that at least this person. I don't know that paper, so maybe you show me the later. Well, maybe I should also mention that on the Latis non-linear Schrodinger, the model is equivalent to XRK with negative spin, which has multiple application to Latis gauge, you know, Professor Karchemski, and Lipatov, and Fadeev, but also Kazakov, in his fishnet, has this XRK with negative spin. So the Latis version has direct application to four-dimensional young males. We'll talk to him. Also, I have a question just about, you were saying that when, let's say, going to infinity, this land has become uniformly distributed with a certain density. Did young and young, or can it be proved rigorously? Oh, actually, yes, yes, yes, yes, yes. Sure, the answer is yes, because remember, well, maybe I can answer your question if you permit. Okay, so remember, when I wrote Bertrand's equations, I had three settings. The third, I kind of crumbled, you know, I didn't say clearly, but the third one was like this. I have logarithmic form of Bertrand's equation. I have this n, so I can subtract this equation, and I can estimate the difference between two neighboring lambdas in terms of the difference of n's from above and from below. So when l, total length of the way to go to infinity, I can prove that the difference between lambda go like one divided by l with some finite coefficient. So we can prove that the difference one divided by l, and then this is the basis of the proof that rho exists. So the answer is yes, it's not from young, young is like in my book. He didn't prove this. The idea, it's in the textbook. Thank you.