 Thank you very much for the invitation and I would like to say something about the derivation of this environment, Oeds' measures for some non-linear Schödinger equation, in particular for Hartree type equation starting from many-body quantum mechanics. I'm going to say something to start about Hartree theory. Then I'm going to say something about many-body quantum mechanics and then I would like to try to explain the relationship tudi, ki je to teori, včasno, ki bom bilo prezent teori, ki je to da bomo počutiti. Dobro, teori hrti je basite na energijske, ki bomo vrte ovaj. Zelo, da smo teori, ki je energijske, za funkcion 5 v l2, v l2 rd, in bomo vse v dimension 1, 2 in 3, z valjom v c, v valjom v c, da je teori, ki je teori, ki je teori, ki je to potencič, ​​ this is the interacting part of energy ​​that we want to consider. So we will assume that the potential, the external potential V of x, is confining, Кстати, so it will Stra Foundation, so growing that infinity, and in a second I will be in little bit more precise about what kind of condition we need and I will also assume that the interaction potential W is of positive type, meaning that Fourier transform is positive, it is repulsive potential and that it is bounded, tako zelo več Sach-tvima. Tudja ste pošlični, tako pogledajo, da ne letila ki modijo. Sahel moh je energi. Ko če čavo počer na energij, vs. chi doma v bih trimveno materij, počel bo dousit nekaj zvršenih zemljenih, nekaj zvršenih z zemljeni hamiltoni. Če je nekaj zvršenih zemljenih, so je izbratitejo. Tudi časelo je dozvyk tudi in dočeje pakneh v tukamenu u vstavnih drževov, pravno objevki. Najs Jeżeli časelo ti, ki zdajte, da vsenje zelo odrčit, z njega, rečenje izgledajte, kak je tukaj, kako jel ga naredo, nega je vsezal, borrowed, podjel, potenci. In, da je, interakcij, potencij, ki je, da je, dejte, konvolusijo, taj. Ok, čas, pervajte, čas, pervajte vsezal, protvaruje, protvaruje, da je, protvaruje, energija. In, je, očist, konstrukta, obradi, tako, čas, So, you take the energy, which is, as I said, the preserve of evolution, you take the to norm, which is also preserve of evolution, you multiply with some kappa, with some number kappa and you take it into the exponential. And this D phi, here should be something like LeBerg measure of all, this is of course very, very formal. And yesterday we heard, in the talk of Andreia that it is, there has been a big effort in the PDE community to construct, ki prišli do boju drugi zelo, prišli do oblokov, z učinjenja in z pravom dovolj, ker je to prejelivo, da je ovo vsečne, da bojte vsečanje, da so neč vsečanje štah na vzelo, neč na vzelo, neč na vzelo, neč na vzelo, na počet, na rekačna, neč inična data, neč inična data. Mi ja poželim nekaj navodnih, kaj je vsečanj, u komuniti matematikalno, to je zelo potrebno, da je zelo v 70. V komunijne, v konstručnih delovih, in je zelo vseznačil, ali bila v Limenu, in je zelo vzelo, tako ne wažlja. Češa, ono vseznači, njeni, da vseznači, zelo je zelo veš zelo veš zelo vzelo, matematikalne počke, v seče, da je tudi nrčen izgledan je občan. So smo odrobili jeznik na taj nonlinaritivno, z kaj smo zrpustili in potenčili, ko je tega modnega vsema vsema, je tukaj zrpustila, ki sem dobroso vsema, če, da se brekaj ne potrešila, da se boš ne veči zrpustili, če ne se zrpustila. So to ne je točnog naprej. Ne bomo še, da smo vsema vsema, na veči zrpustili. Proto imam, da ste srečanje, v križu je, da entrepreneur. Vnešto ga se je, da bo bila boj, pa je to, da je smo sem zelo ujeliti na križa, naravno da se tu poznajo, da zelo povestimo 5 minut in prišlej mi, č Lady. Mi jeänniko se dovolj, da poplazim, da sem ujel TS in učil jelly, da jim videl, da spostel v tem, In je lač, da počekaj taj držav zmoni te vživaj do zelo. In je napriša, da dopliša naša zelo izgledaj jazom. Imamo na vse, da je ta visitsar, oz. S čutim, da bo včas XP, z gripovom, enoče prej.] Je to tudi blizno, da je jazom, ne bo, da čutimo, prijistim, počekaj. Ka bi ideje, da operator H jeistas, tako da poznamo, da potencija tkoeta je nakovna, tako tez tako da campsite izvršamo počk, tako da počkaj smo z艾ovna v ajdu, lambda n, h, v ajdu vektor v ajdu v h, imam jeznek, ki se o bar었nega nočnega, da ne tubej vstavljam, pro Hardr noz u n. To je zelo v hel. Daj da je postočno trijne všetje, kaj je tkoeta na svojo padneje, zelo nezaj nače, node je temjenostledil znom,所以 we will assume that the potential line is so confining that one of these two bounds holds. So the first one is what we assume for the one-dimension out-case, we assume that the trace of the inverse of h is bounded, which is the sum of the inverse of the again value S-Lamdahelcano and in dimension 2 and 3 we will assume the potential is so confining that the trace of 1 over h squared is bounded. Let me remark immediately, as we probably already know ta ste naredili vzve. In dimenzijon in vzvega na način podačanje vzvega, da način ne je vzvega, da najtej v po produit s hodim delom k bo çešel in domano. Zato da ne ljuda v metu p-bom. Da ne zelo, da vzvega h po metu poehlja k p squareda, jim je kompetu doljesta, še se do svoju b-z p squareda. Tudi vima je zelo integrable. Tk pa razvoj items, je piss square. One of p square is not integrable for large momenta, so it is not reasonable to assume that the trace of h to the minus one is bounded. Right? Instead if you take trace of h to the minus two, one over p square square is one over p to the four, which is integrable also in two and three dimensions, so this is reasonable assumption. Okay. It is however an assumption, right? Because not for all external potential v, this will be true. If you think, for example, about the harmonic oscillator, and you are in one dimension, for example, you know that the value of the harmonic oscillator are just the natural number, so it's n. And you know, of course, that the sum of one over n is infinity in one dimension, so the harmonic oscillator is not confining enough. You need to confine the particle with some little bit more than the x square will do it, right? As soon as you confine the external potential growth at infinity faster than x square, you will have enough decay in the x to make sure in one dimension that this is correct and in two dimension that this is true. In three dimension you need a little bit more than, you need a lot more than the x square, I didn't do the computation. Very well. So we have this spectral decomposition of the Newtonian, and we are going to use it. And to define the measure we're interested in, the free measure we're interested in, we decompose the phi in this basis. So we take, we call the coefficients with respect to the basis u n omega n divided by square root of lambda n. And we do it like that, because then it's easy to see that the free energy of phi is just the sum of the omega n squared. So instead of defining the free measure on the phi, we define it in the sequence of omega n. Because we see that, because of this form here of the energy, in the sequence of the omega n, in the omega n, the measure we want to define is just a product measure, a product of independent and identically distributed measures, if you want. So, to be more precise, we define mu zero on the set of sequences with complex entries, each omega n is a complex number, and we have one for every index n, equipped with a, with a typical, I mean the cylinder, with a sigma algebra, which is generated by all cylindrical sets, if you want. And we define the measure mu zero as the product of Gaussian measure, so we have a Gaussian measure for every omega n and this is the density of the Gaussian measure. I hope that I normalize it correctly. We want, of course, to get a probability measure, so we want that the integral of this guy is equal to one. Good, so this is a formal definition, this is a precise definition of a free measure mu zero associated with a free part of the Hartree equation. And once you have this measure, you can start to make, to make some computation, to ask questions about the properties of this measure. So, for example, you can compute what is the typical, what is the expectation for the two norm of phi squared, and when you do the computation, this phi is here, so the norm of phi squared is this sum here, the expectation of each one of this omega n is order one, maybe one, or maybe something else, it doesn't matter if it's a two here, but it's proportional to the trace of one over h, so the sum of the inverse eigenvalues of the Newtonian h is linear operator h. So, this is finite in one dimension because of the assumption that we made, but it is infinite in two and three dimension, because, again, you don't have enough decay momentum to make sure that this is finite in two and three dimension. So, the typical L2 norm in two and three dimension is infinite, so it means this measure lives on a space which is below L2. The measure mu zero of L2 is equal to zero in this case. You can do a little bit more general computation, and you can try to compute the hs norm, as a real number, and hs, of course, with respect to the Newtonian small h, it's not the typical hs norm, because we also have an external potential, and when you do the same computation and you find that this expectation here is the trace of minus one plus s. So, we assumed for d equal to two and three that the trace of h to the minus two is finite, so we see that the expectation of h minus one norm of phi is finite in two and three dimension. So, it means that the measure mu zero in two and three dimension lives somewhere between L2 and h to the minus one. So, this is a different way to say it, the measure of h to the minus one is the full measure, is one. The measure of L2 is zero in two and three dimension. In one dimension, the measure of L2 is one. Very well. So, you can be a little bit more precise if you make more simple assumption about this external potential. For example, the simplest case you can think about is instead of taking an external potential confining your particles, you can just put your particles in a box and you impose periodic boundary conditions. So, you have this small h, is now the Hamiltonian plus a constant on the torus, on the d-dimensional torus. And in this case, we know exactly what are the eigenvectors and what are the eigenfunctions of this operator. So, these are plane waves. So, if you take the wave e to the i px and you act with h on it, you get this number here. This is the eigenvalues. And p has this quantization condition coming from the periodicity condition at the boundary of this torus. So, in this case, you can compute exactly what is the expectation of the hs norm square with respect to the free measure mu zero. And you get this formula here and you see that this is finite in d-dimension, if and only if s satisfies this bound here. So, for example, you see that in two-dimension the measure mu zero lives just below l2. In three-dimension it lives just below h to the minus one-half. Just to be more precise statement with respect to the general statement that follow from our assumptions on the external potential. So, this is the free measure. Now, we want to construct the full, interacting hard three measures. So, how will we do it? The idea is, we would like to define h the interacting invariant measure as an absolutely continuous measure with respect to mu zero, with density given by e to the minus up to a constant, by e to the minus the interact. So, the idea is that e to the minus a hard three function, you write it as e to the minus epsilon zero times e to the minus w and the e to the minus epsilon zero you absorb it into the mu zero. So, what is left is this density e to the minus w phi. So, this is how you want to define the interacting hard three measures. So, how is it possible to do it like that? Well, let's think for a second. If you are in one dimension, typical phi have finite l2 norm. So, if you assume that the w is in an infinity, for example, but it is not needed, you see immediately that the double, the interaction w is almost surely finite. So, it is fine. You can really define the exponential, the measure on the full hard three measure by this formula here, by taking the mu zero and taking the absolute continuous measure with respect to it, with density given by this function here. So, this is well defined. It is no problem. If you are in two and three dimension, however, we just discussed that the two norm of phi is typical infinity. So, this interaction here where phi in the support, typical phi in the support of this measure mu zero is going to be infinity. And now you can tell me, well, okay, it's infinity. It's plus infinity. So, e to the minus plus infinity is still finite. But it's equal to zero. So, it's difficult to define a probability measure if the density is identical. It's almost surely equal to zero. So, this is not going to work. This definition here is not going to work if you are in dimension two and three. And the solution of this problem is pretty clear in the 70 by words of Glim and Jaffens and so on. And it is to replace the interaction by weak-ordered interaction. So, let me try to explain what this means. So, we fixed a cut-off k, positive cut-off. It's like, you can think of it as being a cut-off in the frequency, but it's actually, it's in the energy spectrum of this operator h because the energy is not just p squared in this case. It means that we take the field phi to the k by summing not over all n, but only over n smaller or equal to k. Okay? Good. So, we can define, you see that as k goes to infinity, this is going, the L2 norm of this guy here is going to diverge. Okay? So, you want to compensate for this divergency. So, you define rho k as the expectation of the square of this field here. Okay? So, it's given by this formula here. Okay? So, the L2, this is a deterministic quantity and it's L2 norm is going to diverge as n. It's integral, it's not L2 norm, sorry. It's integral is going to diverge as k goes to infinity because it's just the sum of lambda n to the minus 1. Okay? So, what you want to use is the fact that you have a divergence in the phi, you have a divergence on the rho and then it should cancel. So, you define a cut-off and weak-ordered interaction by taking, we call it Wk, but you see that instead of taking phi phi phi square phi square we take phi square minus rho k at the point x and we take the same thing at the point y minus the corresponding rho at the point y. And if you do this definition here now, let me repeat it once again. If you don't have the rho k, this is going to diverge as k goes to infinity and this is going to diverge as k goes to infinity, but if you subtract the rho k, you subtract exactly the right quantity in order for this guy to remain bounded as k goes to infinity. And when you can prove as this Wk defines as q sequence, you should think of a Wk as a function of a phi or if you prefer as a function of the omega, because in principle we defined our measure on the set of all possible sequences omega. So, the Wk as a function of the omega is a Cauchy sequence in Lp, for all p and then it has a limit and the limit is independent of p and we denote the limit by W, which is maybe not the best notation but I just wanted to remind that this is the weak-ordered interaction. And using this new Wk, we can define the new weak-ordered measure in hearty measure and denoted the measure with mu Hw and the corresponding expectation with this symbol here. And then you can prove that the measure is invariant with respect to the time evolution. OK, so this is the construction at the heart 11. Now let me switch to the many body quantum mechanics. Let me say something about this different theory here and at the end I will try to establish a relationship between the two theories. Good. We are interested in system of n particles and at the quantum level at the many body quantum level a system of n particles can be described by a wave function which is an element of this Hilbert space here. The S in the bottom means that we only look at wave function which are symmetric with respect to permutation. So in the physics language we are looking at system of bosons so we assume that the wave function we are interested in are symmetric with respect to all possible permutation. And another important assumption we only are interested in wave function which are normalized because in quantum mechanics wave function is a probabilistic interpretation if you take the absolute value squared of psi n is the probability density for finding the particles in the different regions so we always assume that wave function is normalized. Good. We are interested in this system in this quantum system in the so-called mean field regime and we characterize the mean field regime by taking an Hamiltonian here. The Hamiltonian is a self-adjoint operator on the silver space here and the important remark or the important thing property is here this coupling constant proportional to 1 over n. This is what characterize the mean field regime. Mean field regime means that you have many particles n particles each particles interact with almost all other particles we have n order n collisions if you want but each one of this interaction is very weak it is of the order 1 over n so that the total effect of this many weak interaction can be approximated by an average mean field potential. This is why we take this 1 over n here. If you prefer to think in different term you want to put 1 over n here to make sure that the three part of the Hamiltonian this operator here which you see is a sum of n particle so it's an object of order n this guy is a sum of all pairs so it's an object of order n squared but if you divide by n they are of the same order so they can give rise to some non-trivial limit as n goes to infinity. Okay, so this is the Hamiltonian we want to look at so what are the properties of this system and let's start to see some relation with the heart that you have in the previous part of the talk. So the first remark that you can make to see a relation with heart relationship with heart is by looking at the ground state vector. The ground state vector is the eigenvector of Hn with the smallest possible eigenvalue and the ground state energy is the corresponding eigenvalue. So for this type of potential so this type of repulsive potential it's quite easy to see to check that the ground state exhibits condensation meaning that in the ground states you can approximate the ground state by a state where all particles are in the same one particle with the ground state vector by a product of n copies of the same one particle where function phi and when they say one particle it means that phi is in L2 of Rd not dn and you take the product of n of m and then you get something which is in the right space. Now if you believe for a second that you have condensation that this is correct in the ground state then you can compute the ground state energy just by taking the expectation of Hamiltonian so you have to take this operator here you have to take the expectation with respect to this state here and it's a very simple computation you get exactly the Hartree function you get the Hartree functional for the wave function square root of n phi or if you want you can take out the n and well one of these two formulas in this formula here is not completely right but look at the last formula here it's n times the Hartree well to get the ground state energy which I call en the smallest taken values of Hamiltonian and you divide by n you just have to take the phi which minimizes the Hartree function so the ground state energy is given in the limit by the minimum of the Hartree function and the ground state vector the corresponding psi n exhibits condensation in the minimizer of this Hartree energy so this in some sense a first relationship it's the first way in which Hartree teoric is an approximation to many body quantum mechanic in this main field regime another example is if you look at the dynamics so look at the dynamics in many body quantum mechanics the time evolution is governed by the Schrödinger equation which I wrote down here this is a many particle many body Schrödinger equation so on the left you have the time derivative of wave function at time t times i and on the right you have the action of Hamiltonian on the same derivative function psi n of t ok, so this is a linear equation but if n is very large you can imagine it's not so easy to solve it I mean you always have a solution because you just you can just solve it by applying the unitary group e to the minus i ht on the initial data and you always get a solution but to say something more about the solution is very difficult and Hartree helps you in this respect in the sense that you can prove the following convergence to the Hartree equation and this is the following statement so if you assume that the initial data you start with n particle wave function which exhibits approximate condensation and now we're not assuming here that phi is the minimizer of art can be any phi because if it is if psi n0 is the ground state then it doesn't move it's not so interesting but the point is that here phi can be any one particle wave function you just have to assume that at time 0 the many body states approximate by product of this phi you let it evolve with respect to this full many body dynamics when at time t you still have approximate condensation you still have you can still write approximate the psi solution with the product of n copies of a one particle wave function phi t of course the one particle wave function is not the same as at time t equal to 0 you have to evolve it and the evolution is described by the time dependent Hartree equation ok so again we have a relationship between many body quantum mechanics and the Hartree and the Hartree here and also here I I I wrote some of the names of the people who have been involved in this war so the first work are back in the 70 by Hep and by Gene Bramvelo we are here and then there was a worker by Spohn and then many other people and I also have some work in this direction but maybe maybe let's move on and I would like to switch now to the to the main question of this talk and now we have we have seen that there is a relation between many body at the level of ground state energy and at the level of dynamics so the natural question that you can ask is is there a way to understand this invariant measure that you have in Hartree theory this invariant measure that we constructed in the first part of the talk as seeing this measure as emerging from many body quantum mechanics and what are the right object what does correspond to this invariant measure in quantum mechanics and for the case of one-dimensional system a quite precise answer to this question has been given by by Mathieu Levine Nam and Nicolas Rougeri who proved I think it was about two years ago now that many body give state in this mean field regime can be approximated by the Hartree invariant measure that I constructed in the first part of the talk so there is a relationship with this invariant measure if you look at this many body give states ok so let me try to explain what are this many body give state so mean body give state describe thermal equilibrium at positive temperature and in order to get this correspondence you have to take high enough high temperature as I will see in a moment now in order to explain what are this thermal equilibrium state I have to tell you first what are mixed state in quantum quantum mechanics because this is something that I didn't mention yet maybe not everybody is familiar with it so when I told you that you can describe by a quantum system by a wave function it was correct but it is not the most general description if you are at positive temperature it is also important to know what you can describe your state by mixed state mixed state means you don't know in which state you are you only know that you have a certain probability to be in several possible states ok and then the states you are in is described by a density matrix density matrix is a trace class non-negative operator on the Hilbert space where we worked with trace equal to 1 ok so because of this assumption it means that the density matrix can always be written as a linear combination of orthogonal projection with some weights pj that you can interpret as probability because all pj are between 0 and 1 and the sum is equal to 1 ok so if your rho is given as a formula here the interpretation is that your state your system is in the state psi j with probability pj ok it's very different in quantum mechanics of course quantum mechanics are linear theory so if you have several psi j you can also take a new wave function but take a linear combination of psi j but this is not the same thing as taking linear combination of a projection ok because if you take linear combination of a projection you don't have interferences between the difference wave function that's the that's called the incoherent superposition of states instead of a coherent superposition of states anyway if you have your rho then if you have this interpretation it's also clear how you can compute the expectation of an observable in such a state because if you have an observable a and you have your state rho well what is the expectation well we are in the state psi j with probability pj so we better take the expectation of a in the state psi j and then we take the weighted average according to this weights pj and if you think for a second you will find that this sum here is just the trace of the product between a and rho so this formula here tells you how you take expectations with respect to mixed state ok so why did introduce mixed state because equilibrium states at positive temperature are mixed state in quantum mechanics and in particular if you are at a temperature beta the the equilibrium state is given by a normalizing constant times the exponential of minus beta times hn ok so if you have an Hamiltonian hn with eigenvector psi j and eigenvalue vj it means that you have this formula here it means at equilibrium you are in the state psi j with probability e to the minus beta vj ok this is what you need to know when you are at positive and you are not knowing in exactly which state you are you only know that you are in many possible state which some probabilities this is only the only information that you that you have well if you look at this formula in particular you see that if beta tends to infinity so the temperature tends to zero then the probability are going to collapse to the one with the smallest energy right so at zero temperature the equilibrium state is just the ground state and in the other limit when beta goes to zero so when the temperature goes to infinity well this goes to one so it means that all states have the same probability ok good so these are the types of states which will lead in the limit as n goes to infinity to the non-linear Gibbs state or to the Gibbs state associated with Hartree theory that we defined at the beginning of the talk ok and this is what we want to understand but there are two if you think about it you will see immediately that there are two very simple objection why this cannot be true for this rho beta here so the first remark is that if you fix beta positive so it means if you are at a fixed positive temperature then it is easy to see that the state rho beta still exhibits the same type of condensation as we had for the ground state ok remember in the ground state all particles are described by the same one particle difficult to understand that if you fix the beta also at positive temperature this is the same because you see you have a system where the ground state energy is of order n ok now if you take excitation of this order n energy by something of order 1 right which is what you see if you take a fixed temperature if you fix the temperature which means you are not only in the ground state you have a combination of states with energy which is a little bit above the ground state it is still going to be very small and you don't see it ok but if you excite the ground state by order 1 it means that you can only move a finite number of particles out of the ground state ok and if you only excite a finite number of particles most of the particles we have n particles in the system the bulk of the particle is still in the same one particle state ok so because of this argument it is clear that if you take beta fixed function exactly on the minimizer of the Hartree function ok the bulk of the particle are all in the same one particle state which is a trivial measure it is certainly very different from this invariant measure that we wanted to understand ok so we have to do something else and the thing that you have to do of course you have to take the temperature to grow together with n as n goes to infinity you don't want only the number of particles to go to infinity but also the temperature and the right choice turns out is given by beta equal to 1 over n which means temperature is equal to n ok so this is the first remark now there is a second remark and if you think for a second back to the what other thing about the ground state we saw that there is a relation between the number of particles in many body quantum mechanics with n and the l2 norm of the corresponding Hartree state of the phi ok so it means if we consider a system with a fixed number of particle with n particle then in the limit we will have a measure at the Hartree level where the l2 norm is fixed ok we fixed l2 norm and this is not what we have what we had in the previous part of the talk so we have to switch to a different representation of this quantum system where the number of particle is allowed to fluctuate you don't want to fix the number of particles you want to take combination of states with possibly a different number of particles so this means so your j there goes up to n or is your sum of all energies this is all energies so you have so as beta varies the measures don't become mutually singular as you change beta as you no, no, no it's always the largest weight is always on the ground state that's where you have the most energy but then you have larger tails so it will move up in energy in a sense if you increase the temperature ok, so in the physical language it means that you want to switch to a canonical ensemble canonical ensemble means the number of particles fixed to a grand canonical ensemble where you allow the number of particles to fluctuate so you will take we will take averages not only overstate with different energy but also overstate with different number of particles ok, and to switch to the grand canonical setting the right way to do it in quantum mechanics is to a fox space representation of the system so let me try to go through this this formalism second quantization formalism ok, so the fox space the bosonic fox space it means that instead of looking at the fixed L2 space we take the direct sum of all possible L2 spaces of all possible number of particles you see we are summing overall m and we take the symmetric product of m copies of L2 and if you fix the m this space here or which is the same as this space here describe states of the system with exactly m particle and now we are summing overall possible m because we want to allow the number of particles to fluctuate we don't want to fix it ok, so you have the fox space so vectors in the fox space are sequences of wave function psi j and for different values of j you have different number of particles in your system ok, so if you are a fox space this is very useful to introduce creations and analysis operator so this is the formula it's not so important how they are defined but the interpretation is somewhat important so the A star of f you see f is an L2 function on rd so it's a one particle it's a one particle wave function and acting with A star of f means that you are creating a new particle with that wave function you are increasing the number of particle by one and if you act which is the adjoint of A star you are decreasing the number of particles by one ok good and when a simple algebra tells you that these creations and analysis operator satisfy canonical commutation relation which I wrote down here so if you have a commutator between A and A star it's just the scalar product in L2 between f and g and all other commutator are going to vanish it is also useful to introduce operator value distribution which I call A of x and A star of x which create or annihilate a particle at the point x ok, these are only distribution because you have to you have to smir it out for integrating against a function f otherwise we don't we don't quite make sense and the definition is so that this relation holds true and in terms of these creations and analysis distribution you have you can rewrite the canonical commutation relation in this term here but more importantly you can define other operators and the two operators that I'm going to need are the number of particles operator so now remember that in our system the number of particles is not fixed it's a variable so you can measure how many particles a state has and the right way to measure is to apply this operator n if you think for a second about it A star of x A of x is the density of particle at point x so it's clear that the total number of particles is given by integrating this thing here ok so this is the number of particles and the other one is the Hamiltonian of the system and to define the Hamiltonian where we have this part here we integrate over all x this is the second quantization of this operator here now this is just a notation if you prefer you can check that the Hamilton operator is defined in such a way that it commutes the number of particles it does not change the number of particles it's equal to zero in this term the number of creation is the same as the number of annihilation operator ok so the number of particles is not changing and if you restrict it to a fixed number of particles you get exactly the same operator as we had before right this thing here which may look scary is just the sum of the operator minus laplacian plus external potential acting on all possible particles from one to n and in the same way this is the interaction is equivalent to this thing here ok so now we have an annihilation operator so we can construct the grand canonical equilibrium state so this is a mixed state so it's a density matrix it's a positive it's a non negative trace class operator with trace equal to one and the trace equal to one is guaranteed by this normalization constant here and then you see that it is proportional to the exponential of minus one over n and then you have again a combination of energy and number of particles right these are the two things that preserve the dynamics the energy and the number of particles so you put it in the exponent here and kappa the constant kappa is called the chemical potential right when you switch from canonical to grand canonical the number of particles is not fixed so you have a new variable in a sense it's very joint transform of this number of particles and the idea is that you have to fix the kappa so that the expectation of the number of particles is the right one right you don't fix the number of particles but you can fix the expectation of the number of particles so this is the state that we are interested in and to write it you see that we divide by one over n because we want the temperature to be equal to n the beta is one over n and that's why you have this extra one over n otherwise it's just e to the minus beta times h plus number of particles so to rewrite it since we divide by one over n it is useful to rescale the creations and the relations operator by dividing it by one over square root of n because you remember that we have the hn maybe I go back one slide it's a sum of a quadratic and a quartic part but in front of a quartic part there is already one over n and then we multiply everything with one over n in front so the first guy is carrying a one over n factor the second guy is one over n square factor so if you put one over square root of n in each a and a star all this n get absorbed into the definition okay so we define these new operators with these extra variable here and then the same stator of n the constant is the same as before and then it's just a way to rewrite it now I absorbed all the n inside this a star and a factor so this is a measure that we want to look at positive tempers this is a state which describes thermal equilibrium at temperature n good now what can we say about this state well let's start by looking at the free part so we turn off the interaction double equal to zero we try to do the same steps as we were doing for the art the first look at the free measure so first look at the free state by removing this interacting part here so if you do so when you have this quadratic expression in a and a star that you can diagonalize by switching to the remember this is exactly the operator h that we had in the first part and you remember that I called the eigenvalues lambda j right maybe in this way it's easy to understand because a star uj a uj is measuring the number of particles in the state uj and lambda j is their energy the free energy that they have so if you sum over all j of number of particles in the state uj times their energy you get the total energy of this many particle system okay so we use this notation here and this is our free state with the free partition function the free normalization constant here so what can we say about this state here let's try to measure the number of particles in this state let's try to measure for example the number of particles in the state ui ui is a one particle state you can't you have n particle you want to know how many of these n particles in the state ui good so this is the expectation of this guy here expectation means that you have to trace of this operator against this density matrix here so I put the density matrix here and in the bottom and you see that the density matrix actually it was a sum over several j right but all the terms with j different than i they simplify the cancer from the denominator and the denominator because they are independent the particle because there is no interaction in this setting here so you cancel everything and you are you end up just with this expression here and then and then well you have to know that this operator here measured in the number of particles it's again values are just n and then the expectation that you get is exactly this thing here okay so now let's try to compare with what we had in the heart in the heart in case you remember that we had 1 over lambda i well it's not so different because if you expand the exponential it's 1 plus lambda i over n plus something else so at least for lambda i which is much smaller than n this looks exactly like 1 over lambda i so you start to see some relationship some similarity with the heart okay so if you want to know the total number of particles then you have to some of the all possible modes right and so far I just measure the number of particles in the mode ui now in summing of the all i and I dividing by n because I want to look at the risk actually I should not divide by n because we are already looking at this operator a n which carry the 1 over square root of n so forget about this one over n anyway we can do the computation and you find this sum here in 2 and 3 dimension we assume that the sum of 1 over lambda i is finite so expanding this guy here you easily you easily see that you get something of order 1 in 2 and 3 dimension on the other hand well the sum is still finite because even if lambda is very large you get the cut off at n as soon as lambda is much bigger than n the denominator is going to be very large is going to make everything converge but it does not converge which is finite for every n but it diverges as n goes to infinity ok so again we get a similarity with the Hartree case the only difference with respect to Hartree in Hartree the number of particles was just infinity the 2 norm was just infinity in here you have a natural cut off given this n here so you don't see the divergence for finite n you only see the divergence as n goes to infinity ok so these are the properties of the free measure now if you want to pass to the free measure when you can do the same type of thing as we were doing in the Hartree case look at the interaction this is the interaction right now the interaction is going to the expectation of the interaction or of powers of the interaction is going to be finite as long as n is finite because we have this natural cut off but if you look at what happens as n goes to infinity again you get a divergence ok you get something which is finite in one dimension but this is going to be infinite for dimension 2 and 3 so also here we called it it means that we replace the interaction now we don't need to introduce this cut off because it is this natural cut off given by hand so we just take the interaction and you subtract from these two guys and these two guys this quantity here which is just the free expectation of the product of n star so which is something whose integrats or one or more this guy here diverges as n goes to infinity in 2 and 3 dimension ok now if we are using this weak order interaction we can define a weak ordered canonical a grand canonical state which is what to would expect and this is again a state on this on this fox space on this many body fox space ok so now now we want to compare this state here with a heart measure that we constructed at the beginning and the theorem should be actually measure ok now maybe this is just a comment maybe we can skip the comment let's go so how do you compare this state this many body point on state with the measure you look at the correlation function and you look at the moments of the heart and you want to prove that they are the same so what are the correlation function you take expectation with respect to this weak ordered many particle system can be written like that and of course here we only look at the case where the number of creations is the same as the number of annihalations operator just because all other expectations are equal to zero and then we want to compare this gamma and k this correlation function with the one constructed with the help of the invariant measure as n goes to infinity for every final k this limit here for example measure this norm here for example measure in Hilbert Schmidt topology should be equal to zero so it should be the same in the limit you can approximate the many body quantum give state with the invariant heart restate here at the level of the expectation of the view of this correlation function and this is exactly what was proven by Levin, Rugeri for the one dimensional for two and three dimension but for a different type of potential so I'm not mentioning it here so what we want to do what we wanted to do is to prove the same thing in dimensions two and three where you need the weak order in one dimension the simplification is of course you don't need this weak order technology unfortunately we cannot prove this conjecture here yet but we have to modify the little bit this many body quantum state and instead of looking at the state e to the minus full Hamiltonian we have to look at this modified state with this parameter e that which is anything just bigger than zero where we take we take out two eta terms from here and we put it back there is a mistake here the one minus two eta should only multiply this h zero not the not the interaction interaction carries a one in front so in particular you see if this operator commuted with a full operator you can put everything together and you get exactly the same measure exactly the same state as we had in the previous slide the fact that this state here is not the same as the as the Gibbs state that we had before is the consequence of the fact that you cannot put the exponential of a times the exponential of b is not the exponential of a plus b for non commuting operators ok, so this is the state that we can look at for any fixed eta and for this state we can prove what we expect so that the k particle correlation function converge to the non-correspondent to hertr theory right so you see as n goes to infinity these things should converge to the same you see maybe this is the remark that I didn't make before but so why do we expect this convergence to hard the point is that in terms of this a and a star you see that the commutation relation are not zero they're still non commuting operators but they have this one of a n factor so they are almost commuting fields and as n goes to infinity they become commuting fields the same the phi that you have at the r3 level ok but this remark about commutativity and in particular as n goes to infinity it doesn't matter what you write whether you write like that or you put everything together because in the limit everything commutes and so you have only one limit so there is no modification for the limit ok, so maybe one minus today is a couple of words about the proof proof is based on perturbation expansion so you have this big quantum state where there is a free part and an interaction and then also you have the r3 measure which is the free part and the interaction and when you expand both of them with respect to the interaction which means you can do it by ok, let's keep this part here so that's important maybe you do the duamel expansion so you expand everything in terms of the free Hamiltonian and then the interactions are here in the bottom right, this is the role the interaction that you have so then you construct this strange state where you have modified state where you subtract this guy and you add it back on the other side and you get something like that you see that the effect of this modification is that the integrals in the duamel series they do not start from zero and go to one but they start from eta and go to one minus theta ok, so this is the expansion we look at and then we have we look at all these terms here and we prove that each one of these terms here converges to the corresponding term to the corresponding term in the expansion for the Hartree mesh ok so well, how do you do it well, forget about this part here you see when you are here each one of this double each one of this interaction is a quartic operator is A star, A star, A and A so you have a lot of A star and A just in the denominator and you have to take very expectation to compute this correlation function and to do so well, to do so you use this vik tier in which tells you that with respect to the free measure the expectation of any product of A and A star is the sum of the all possible pairings of products of couples of A and A star it's the same it's the same kind of algebra that you have for Gaussian measures and it is exactly the reason why this guy at the end converges to the series for the Gaussian measure for the V-Gaussian measure that you get in Hartree theory ok so you know what are this expectation for the quadratic part you see you have this thing here that we already had before but in the limit converges to the to the classical to the Hartree to the Hartree measure if you have it in the opposite order remember they don't commute because we are in quantum mechanics so A star is the different than A star A but the difference is not so much it's only with one over n because of this almost commuting property so as n goes to infinity you can get rid of all this guy here you can compare each one of this guy with the one over h that you have Hartree expansion and you can prove that the two expansions are the same and to do so is this technique with diagrams so my collaborators are very skillful with drawing this graph so each one each one of this Vigoli line is a potential w of x minus y right and then you can the lines they describe the pairing right you have to some overall possible pairings so you can pair this guy here with this guy here this one this one and so on the only thing that you cannot do because of the V-coordering is to pair one of this guy with this neighbor with the same x right and this is what you don't want to do because if you pair this guy with the same x with A with the same x if you go back for a second you see it means that you have to take the expectation of this but at the point xx and then you integrate over x but if you take xx and you integrate which you know is infinite in two dimension in three dimension ok so best what you want to avoid you don't want to take the only one operator the trace of only one operator you have to take at least two but this is what you have because of this structure of the pairing here ok so using this diagram you can first prove bounds you can prove that the expectation of all this pairing here is bounded and then you can you can compare you can prove that the expectation of each one of this pairing here converges and goes to infinity to the corresponding term in the expansion for the high tree measure there is one last thing that I wanted to say one last final obstacle is that if you think for a second what is the number of pairing you have you have your 2n operators and you have to pair them and the number of possible pairing is of the order of 2n factorial ok so to be more precise you have 4n because for each term in the expansion you have one more interaction and interaction is squatting so you have 4n but it's only 2n creations and 2n annihilations operator so it's going to be proportional to 2n factorial now you have something which helps you which is this integration integration over this simplex which gives a 1 over n factorial that's fine but still 2n factorial divided by n factorial is behaving like n factorial so the series is not converging ok so we can compare each term in the series to converge to the corresponding term in the classical in the series for for the high tree measure but we don't have a convergence so it's not so trivial to get the results that we have and what saves is this this idea of Borrel summation which which use some additional information you have to prove some additional analyticity of this function you have to take the coefficient in front of the coupling constant of interaction to be complex and you have to prove analyticity in some in some appropriate region of this function and then this idea of Borrel summation allows you to eat up this additional n factorial that you had in this expansion and to get the convergence that you want ok so thank you for your attention in the convergence of reduced density matrices do you have an estimate in terms of n of the difference? no because of this Borrel technique this Borrel summability technique it gives you convergence but with no rate at all to get that you would need the series to be convergent and when you could you could estimate the error but here you cannot just get convergence pointwise convergence is not can we expect the next order correction like a Bogolubov theory for Gibbs state because it makes sense? I don't know I would have to think about it you see it at the level of the ground state when you have a finite excitation I think that you are so up because to take this temperature to be very large so you are very much up in the energy so I'm not sure that over there you can still have something like Bogolubov theory quite but I think not but maybe I'm wrong take a 5 minute break but let's thank you thank you