 Okay So well, I'm going to do something that is in between the two previous speakers that is neither infinitely slow nor infinitely fast So this is Since this is an exactly solvable session, let me be bold and pause an exactly solvable question So consider the non-stationary Schrodinger equation Where h of t is some interesting interacting many body or a matrix Hamiltonian that explicitly depends on time and The question is under what conditions on h of t is a non-stationary Schrodinger equation exactly solvable So to Understand this question better. Let's start with Usual time independent Interability take for example the one-dimensional Hubbard model, which is a famous exactly solvable model So it is a tight binding model plus on-site interaction with strengths you or Equivalently you can think about any other Solvable model that has a parameter such as the xxz model or Libliniga model So there is an exact solution of the especially Schrodinger equation by that answer that gives you the eigenstates and eigen energies and There is an infinite sequence of parameter dependent integrals of motion HK that commute with the Hamiltonian and among themselves in the case of the Hubbard model. They are polynomial in you So now suppose we make you Time dependent a function of time In general this completely destroys the integrability. So for example The commuting partners are no longer integrals of motion So the while the commutator is not affected and vanishes This partial time derivative doesn't because HK depends on you and you depends on time Moreover the knowledge of the spectrum eigenstates and eigenvalues is basically useless unless you change you infinitely slowly or infinitely fast like in a quantum quench So while I can write my time dependent state in terms of this eigenstates which now have the meaning of instantaneous or adiabatic eigenstates the problem is that this coefficient Cn Depend on time. So this is just a rewrite of the Schrodinger equation in some basis and the physical reason behind this is Landau's inner tunneling between adiabatic eigenstates Which can which is generally highly non trivial So in this context the question is can we make Parameters of an integral model time dependent without breaking integrability in other words Can we have integrable Landau's inner dynamics and for those in the audience who are not expert in Integrability let me mention that this is something that is almost unheard of And not to keep you hanging. Let me give you the answer at this point already that Yes, we can at least for some integrable models and now I will show you two interesting examples that later I will show are Indeed exactly solvable so my first example is the famous BCS Hamiltonian So this is this are fermions in some arbitrary single particle potential with energy levels epsilon k plus pairing Interactions between fermions in this form the Hamiltonian was written by Anderson in 1958 and if we take zero single particle potential then we'll get The BCS Hamiltonian in terms of momentum states in its original form So like for the Harvard model, there is an exact solution for the spectrum and there are non trivial G-dependent commuting partners Now let's make G a function of time as we said in general this breaks integrability But we'll see that there are certain special choices of a time dependence when the problem remains integrable and one such choice is Inverse function of time proportional to 1 over t and for this case I'll show you an exact solution for psi of t and I will also show you an explicit answer for the fermion distribution function At t going to plus infinity when we start from the ground state at t Equals to a t at t zero so essentially initially we have an infinite superconducting coupling and we turn it off as 1 over t Now my second example is an important problem in the physics of BCS BC condensates of ultra cold fermions So in these systems the strength of interaction can be controlled by magnetic field So the plot here shows a scattering Length as we scattering lens as a function of magnetic field So on the right side of the Feshbach resonance you have fermions with attractive Interactions and on the left side they bind into bosonic molecules that repair each other So this physics is captured by this fermion boson Hamiltonian that has the fermions as a bosons in a mode that is condensing and the fermion boson interaction that converts a pair of fermions into a boson if this parameter gamma Dimensionless resonance width is much less than one than this model is known to be quantitatively accurate throughout the BCS BC crossover Otherwise, it's accurate for large gamma sufficiently far from the resonance and Similar to the BCS model. This is a better than that soluble model with G dependent commuting partners now Let's examine its ground state in two limits when omega naught goes to plus infinity having a single boson costs infinite energy So the in the ground state there are no bosons, and this is just this is just a Fermi gas With Fermi Dirac distribution in the opposite limit when omega naught goes to minus infinity There are no atoms in the ground state and everything condenses into a single mode So there are no atoms and the number of bosons is maximum However, what you can do in experiment you can change this omega naught Only with some finite rate new So let's say omega naught is minus new t's and at t minus infinity we begin in the ground state which is the Fermi gas and So now at t plus infinity because of the finite rate, we are not going to end up in the ground state We are going to end up with some number of fermions and the question is what is the distribution and the question is how many of them condensed into this bosonic mode So now notice two things first of all that you know as we as I mentioned this is an exactly soluble model better than that soluble model and We are making one of the parameters omega naught a function of time So this is an example like before and Secondly in this case the Hamiltonian is going to be linear in time So this is an example of what is known as a multi-level London's in a problem so in this problem we have a Hamiltonian that is linear in time a plus b t where a and b are some Time independent and by an Hermitian matrices So Let's say we are given the initial state at minus infinity Then the state at plus infinity because equations of motion are linear is going to be some matrix Times the initial state so this matrix is known as a scattering matrix And the question is the problem is to find it and in particular transition probabilities from state I at minus infinity to state k at plus infinity our modular square of elements of this scattering matrix So it is customary to work in a basis where B is diagonal and this is known as diabetic basis so for any cool to This problem was solved independently by london Zener and my Rana my Rana and Stuckelberg in 1932 and published in four different journals You can show that without loss of generality. You can bring any two by two problem into this form Then there is a solution for of the non-stationing Schrodinger In terms of parabolic cylinder function while the transition probabilities are given by some elementary functions of g and lambda And here in particular is the probability to remain in the ground state Survivor probability is given by the famous lambda or in a formula So as a function of this interaction g and the rate lambda and you see that it goes to one as lambda goes to zero indicating adiabaticity Now already for three states This problem has no general solution. So for n greater than two Solution exact solution is known only in certain special cases So so my question in this context is under what conditions on A and B is the multi-level Landau's in a problem exactly solvable and what is the solution so solvable in this context I will define as if you can find the transition probabilities as elementary functions or matrix elements of A and B Okay, so I'm going to take a bottom-up approach meaning that I will list all solvable examples that I know And then I'm going to ask what do they have in common and try to generalize from there So first of all as there are So-called reducible multi-level landaus in the problems and one example is a driven quantum ising model Where we take the transverse field to be a linear function of time So after Jordan Wigner followed by Fourier the Hamiltonian breaks up into a sum of independent Hamiltonians for each K and each of them involves only two states. So this is reduces to the 2 by 2 Landau is in a problem. So I mean not to say that it's not interesting You can calculate, you know some interesting stuff with it But from the point of view of the multi-level landau in a problem. It's it's trivial and there are many more Reducible multi-level landaus in the problems like that but since you know the original work of Landau's in a moirana and Stuckelberg in 1932 only three Irreducible multi-level landaus in the problems have been identified. So now I'm going to list them So the first one is Demkov-Osharov model So what you see here is a plot of diabatic states. This is a diagonal plot of diagonal part of the Hamiltonian So what you see is that there are many levels that have the same slope You know set to zero here and one level one time-dependent level that crosses them all So this level interacts with each of the other levels with Interaction what is called interaction Gk while the other levels do not directly interact with each other So the second case is a so-called bow tie model because a diabetic energy diagram resembles a bow tie So here you have all levels have different slopes. They all cross in one point and again Only this privileged level with zero slope here interacts with all of them. Well, they do not directly interact with each other and The third example, which I will show that is integrable in this talk is This model that we had and I think I forgot to mention on that fly on that slide on BCSBC Condensate that in quantum optics this model You know has appeared and is known as inhomogeneous decay model Okay, so the third case is inhomogeneous decay model So here is for example the solution of the demko-bosher of model the simplest case So you see the scattering matrix here written in terms of p and q which are elementary functions of this interactions gk and lambdas and one feature of the scattering matrix is that it factorizes Into product of scattering matrices each of them corresponding to a two by two scattering Of level one with two one with three, etc That's a general feature of most solvable multi-level and I was in a problems Okay, so what is special about these models? What sets them apart from any other Hamiltonian linear in time? So, you know the first thing that my student Aniket Patra found was that the demko-bosher of and bow tie model have Non-trivial commuting partners of this form linear plus inversely proportional to time You know that commute with demko-bosher and bow tie Hamiltonians at all t now the Decay inhomogeneous decay model is a known integrable model. So we know its commuting partners already and Let me know that if you take a generic linear matrix like if you randomly generate a and b It's not going to have any commuting partners Of this form except a combination linear combination of itself and identity So for example here are the commuting partners for the inhomogeneous decay model So it is convenient to rewrite This model in terms of Anderson pseudo spins spin one-half which is an identical rewrite So in this case pin lowering and raising operators are pair Creation relation operators and the Z component is related to the occupation number Then the decay model takes this form of a spin boson Hamiltonian So it's a bunch of spins in an inhomogeneous field along the axis The boson and the spin boson interaction and here I already took the model with time dependence Omega naught is minus nu t so the commuting partners are kind of central spin Hamiltonians where you have this privileged spin SK which interacts with all other spins and with boson so these guys commute Among themselves and with the decay Hamiltonian and you know writing that omega naught is minus nu t Of course doesn't spoil this commutation relation So it also has this non-trivial commuting partners Okay, fine. I mean there are all these commuting partners, but what's the point? I mean what is their role? How do they help us to solve for the dynamic of the system as we discuss them not even conserved, right? So in this case DHK DT for example is this expression that is non-zero So what's the point and here comes? the main idea So it turns out that this HK Determined the evolution of the state of the system with respect to Parameters xk which are parameters of the problem other than time I mean time you can also regard as one of the parameters But you have other parameters as well like matrix elements of a and b and it turns out that while the Hamiltonian determines evolution in time These guys determine the evolution with respect to other system parameters in other words the non-stationary Schrodinger equation can be Consistently embedded into a system of multi-time Schrodinger equations like this Okay, so at this point let me you know make notation more uniform So nu t I'll call x naught and h I will call h naught and the derivative with respect to xk as DK And then the system you know compactly Rewrites like this and Now if I take the derivative of this with respect to xj and then take the derivative of J's equation with respect to xk and Equate the mixed derivatives. I get a consistency condition like this and in most cases that We are interested in Hamiltonian is real meaning in appropriate basis matrix Hamiltonian matrix elements are real Then this part is real. Well, this part is purely imaginary. So the condition separates into two conditions one is This condition which I call cross derivative condition and the other is a commutation condition that You know signals integrability of the underlying model So for example, let's check this for the DK Hamiltonian So we have these integrals we know or is that they commute now let's check the cross derivative condition Let's take hk and differentiate with respect to epsilon p and only this term contributes because epsilon p is not equal to epsilon k And so I have I get this expression and it's Symmetric in k and p so it's equal also to DHp to the epsilon k So you see that this condition is satisfied and also for x naught Which is minus omega naught? I have to check it and indeed the derivative of the keg Hamiltonian respect to epsilon k is SKZ and the derivative of this guy with respect to minus omega naught is a SKZ now notice that if I take any other so here any other time dependents This will not work. So here Omega, you know omega naught is basically rescaled time if I take t squared then in this derivative DHk dt I'll get a T-dependent coefficient in front of SKZ and it's not going to be equal to that. So only linear Time dependence of omega naught will work Okay, so we have this multi-time Schrodinger system How does it help us to solve the problem? So you can write a formal solution You know in this multi-dimension real parameter space as an ordered exponential of this form and the consistency condition Means that a non-Abelian gauge field with components minus i hk has zero curvature Which means that this? Exponent is pass-independent. I mean it better be So that this system have a unique solution for any initial condition Which means that what we can do is we can deform the path To our convenience. So here is for example the solution of demko-bosher of model with this method. So In this problem we need to determine The evolution from minus infinity to plus infinity along the real line, okay? And it's relatively easy when the time is large But when we approach the origin at finite time, we have multiple non-adiabatic transitions So it's Difficult to handle it But now what we can do, you know having this enlarged space and the freedom to deform the path We can force first evolve along this ace and you know Ace is a diagonal part of this matrix a and in this case they're equal to xk So and we and if we do that evolve along some ray like at minus infinity In ace we can achieve a situation when all ace are well separated it So which means that all levels are Infinitely separated from each other which means that the evolution is going to be adiabatic Unless there is a degeneracy and the degeneracy only occurs when this time-dependent level crosses level to level 3 etc Which also implies that the scattering matrix is going to factorize like this Okay, so now I'm going to switch gears a little bit and talk about you know About this from a different angle So there is a famous multi-time Shooting gear system of Schrodinger equations called can you think them a logic of equations? So they have this form Where hj are? central spin Hamiltonians like that known as go down magnets and So the question is if there is any relationship relationship between the multi-time shooting gear equations we derived and Can you think them a logical equations? You know it's natural to ask this question because for example the decay model in how much is decay model is known to be related to go down magnets So for our purposes we want to generalize this can you think them a logical equation by adding a magnetic field on the central spin like this now the advantage is that in this Formulation is related to the BCS model indeed if I take linear combination of this go down magnets up to a constant I'm getting a spin Hamiltonian like this with infinite range interaction and in images magnetic field along the z-axis And this is nothing but the BCS model that I showed you earlier in understand to the spin representation where G is identified with 1 over 2b and this kind of generalization of new geomological equation has been you know considered by You know various people including our chairman so However One thing that apparently was not Noticed was that we introduce this a new parameter B. So the state of the system now depends on an additional parameter B so We can ask how does it it depend on B? And it turns out that the evolution of the state of the system with B is governed by nothing else But the BCS Hamiltonian and you can indeed check that this equation is consistent with the rest of the equations by Checking the cross derivative condition. Yeah, DH, BCS, D epsilon K is going to be 2s kz and which is the same as DHk Db right, which is also to you know SKZ Okay, which means that this system of equations is consistent So we have this enlarged now system of equations which I call kz BCS equations and this immediately gives me a Solution of non-stationary Schrodinger equation for the BCS problem because now I can make this be a function of time So in particular if I take B equal to new T I'll get this BCS Hamiltonian with coupling that goes as one of a T which I already Advertised and if I take it some periodic function of time I'll get some integrable Floquette BCS Hamiltonian and the solution You know by construction the solution of the Schrodinger equation is just going to be the kz function where I You know make B a function of time so Now what about You know this exactly soluble multi-level lambda was in a problem. I mean what what about them? Well, it turns out that there is a mapping from Godin magnets to all of them so in the case of the decay model this Mapping was noted already by Godin So if you take one of these pins and send it You know write it in terms of Holstein-Primakov bosons and send the magnitude of the spin to infinity You'll get the decay model from one of this HG Let's say H1 while the rest are going to be its commuting partners and similarly you can show that you can That the dim core washer and the bow tie model also map to Godin magnets So for example here is the mapping for the dim core of Osher of model So all this central spin Hamiltonians Conserves the total z component of the spin which means that in the basis where as total z is diagonal They break up into You know into they have block diagonal form So the block that corresponds to all spins Let's say up and by the way the magnitude of the spin here can be arbitrary not necessarily one half So all spins polarized up That's just one state. That's one one one by one block and the next block One spin flip block is n by n if there are n spins and this is and this block exactly corresponds to dim core Bosher of model So here is the mapping and the mapping is rather you know involved it involves in particular the magnitude of spin Which is just some parameter in this block And so then the n by n block of h1 maps to the dim core washer of Hamiltonian while n by n blocks of the rest of hj map to commuting partners of the dim core washer of and similarly for the bow tie now the only thing that this mapping ensures is That you get commuting partners the second condition is Unrelated to the mapping and it is some mystery and you know fortunate Choice of parameters and luck that the second condition also holds and only the second condition You know together with the first guarantees the Integrability of the resulting model and there is no general way or at least I don't know of any general way of You know coming up with a mapping that is going to also preserve this cross-derivative condition so Okay, yeah, I'm almost done. So here is you know like some off-shell bet on that solution for the finishing the logic of equations and then you know as I said if you make this be here a Function of time you'll get the solution of the for the BCS a time-dependent BCS Hamiltonians And You can I mean use a similar technique off-shell bet on those to solve all other problems So here is a solution for the dim core washer, which is particular particularly simple But let me not you know dwell on that further and let me show you some physical Result instead so So again, so so so let's take this BCS Hamiltonian and Let's start in the ground state at t You know equal to zero plus so this is an infinite For a magnetic interaction between these spins which corresponds to an infinite attraction between fermions So the ground state is just the you know a lowest eigenvalue eigenstate of this Of this term, which is basically like s squared So, you know in the BCS problem all spins are one half and it's natural to work in the sector where the total z projection is zero That's what you have when you have particle whole symmetry for example, so So the you know for s squared as a Maximum value of s is achieved in a symmetric combination of all these spins. So this is then the ground state Okay, and now we You know Turn the interaction off the coupling goes to zero as one of a new T and The question is what is a spin distribution at? T going to plus infinity and by the way the spin distribution, you know translates into the fermion distribution easily because s Jz is equal to nj minus one over two Okay So now this distribution Was found not by me but but my but by my collaborators on an early award And so here is the answer, you know, it's a rather simple answer and You know one thing to note is that If all levels are equally spaced Epsilon j is equal to j delta then this is a thermal distribution with a temperature That is the rate times a level spacing divided by 2 pi but for any other Distribution of epsilon j which is arbitrary in principle. This is not a thermal distribution so, you know an interesting question here is How to explain the failure to thermalize So, so I mean normally it is explained by some conservation laws and here we have no conservation laws because of explicit time dependence and In particular you can also turn off the interaction using some non integrable time dependence and an interesting question Is whether it will thermalize in that case? Okay, so let me conclude here Okay, so Yeah, I guess I don't even you know want to read this summary So, let me acknowledge my collaborators my student Aniket Patra Nikolai Sinitsen from Los Alamos Vladimir Cherniak and also Chentsan who is a student of Pokrovsky, you know who did some work on one of the Papers and so this are the papers on which the talk was based And let me also, you know acknowledge one other important collaborator Which is Pierce you see the thing is that my office and Pierce's office are adjacent and the walls are very thin and Pierce if you know him speaks very loudly, right? So effectively we collaborate on all projects Right, so I always have to acknowledge Okay, thank you very much