 Je vais vous présenter un joint-work avec Mathieu Levine et Femme Thamnam, qui est aussi, pour dire, ce joint-work avec Youf Rolish, Anthony Norris, Benjamin Schlein et Vedran Soinger, par qui je veux dire qu'ils ont obtenu exactement, sur le site web, le même résultat que nous, à exactement le même temps, avec une méthode complètement orthogonal. Si vous aimez le résultat, je vais vous présenter, mais ce n'est pas la méthode de prouve. La meilleure partie d'advice que j'ai, c'est que vous allez vérifier leurs papiers. En fait, ils ont annoncé des nouveaux résultats très récentes, donc il y a un talk par Anthony Norris sur l'IMP Worldwide Seminar. Il y a aussi un joint-work par Andy Deutschert, Robert Seringer et Jacob Inverson, qui, ce que je vais dire, peut être comparé, mais la comparaison n'est pas exactement straightforward, donc je ne vais pas faire ça dans le talk. Donc, depuis que je n'aurai pas oublié de les mentionner, je les mentionne directement. Ok, donc je vais vous parler de... Je fais tout en grand canoniquement. C'est l'ensemble grand canonique pour les bosons dans les unités flat tours en 3D. Donc, ce que nous faisons, c'est beaucoup plus général, mais pour avoir une notation raisonnable, sous contrôle, je vais juste focuser sur ce cas-ci. Nous pouvons faire des gaz homogénaux et nous pouvons aussi faire des 2D, mais nous allons le faire pour un exemple particulier. Donc, je vais me rappeler un peu ce que Mathieu a dit dans ses lectures. Donc, le space d'Hilbert, sur lequel tout le monde sera vivant, est le space bosonique. Donc, le summe direct de tous les espaces pour beaucoup de bosons. Donc, le petit n dénote le nombre de particules et je ne peux pas considérer qu'une position supérieure des states avec différents nombre de particules. Et depuis que les bosons, les states basiques, ou les fonctions, les valeurs complexes et les exchanges symmétriques de leurs arguments. Comme vous l'avez dit dans plusieurs lectures. Donc, là, il y a un mithonien qui s'appelle Big H. Par contre, je mets le lambda pour le constance de coupling et ça commence comme d'gamma de minus laplacien. Donc, c'est utile pour cette notation parce que je ne peux pas l'aider, mais je vais l'utiliser. Donc, je vais le définir. Mais c'est ce que vous avez vu dans les lectures. Et puis, le constance de coupling, et l'interaction W. d'gamma est une sorte de notation standard pour dire que je prends le minus laplacien qui est une condition périodique. Et je le mets dans cet espace naturel. Donc, il s'agit diagonallement sur les secteurs avec différents nombre de particules et sur un secteur avec un n particulier il y a un j de 1 à n minus laplacien dans le variable xj. Et l'interaction est aussi l'interaction naturelle pour faire de paire sur cet espace naturel. Donc, le W est le grand somme pour tous les secteurs et un somme de paires d'un potentiel paire qui acte comme multiplication sur les fonctions de cet espace naturel. Bien sûr, je suis en train d'y aller sur le torse. Donc, ce x, y minus xj est de la distance sur le torse. Mais c'est un n qui est resté, un n, c'est un n, un n, un n, un n, un n. Et donc, il y a un n qui est resté, il y a un n, un n, un n, un n. Comme l'on l'a déjà vu, je l'ai notationné par Gamma lambda. Donc c'est 1 over z lambda, e to the minus 1 over t d'opérateur minus nu, qui est un autre numéro, un numéro particulier. Right, so t is temperature, nu is a chemical potential, so nu must be here negative. So this, you see, e to the minus 1 operator, so this is a trace class operator, positive trace class operator over this space, and the z lambda is the partition function, which makes the trace equal to 1, so the usual convention. Nicolas? Yes? Is nu really negative in your case? Because we have the interaction. Sorry? You have an interaction, which match it? Yeah, that's right. So maybe I would say this. I will put this later. I think you are right that it's better too. But my notes says it's actually negative, but anyway. Perhaps I can squeeze in here some assumption that I will have all the way through the talk, but that I will always forget to recall, so I will assume positive Fourier transform for this operator, for this guy, which is convenient for us. And I will assume some regularity. So let me record the exact notation we need. If you take the Fourier coefficients of this guy, this sum, weighted sum, must be finite. But for the purpose of the talk, just think that w is smooth. I don't really care too much about the optimal assumption. Ok? A bit of notation that I will use all the time or so is that I will denote expectations in a state. So in this state, expectation of dot, which is so, if I take some operator, bounded operator, for example on the fox space, this will always mean trace of this operator against gamma lambda. So the standard quantum mechanical expectation value. Good. And for the purpose, or maybe not the statements, but the proofs, it's important to recall that this gamma lambda, this is to the minus something, it doesn't just come out of the blue. The reason why we're interested in this form is that this minimizes the free energy, which I think physically should not really be called free energy, but I will use free energy anyway. So maybe it's a functional. Over states of gamma is a positive self-adjoint operator with trace one on the fox space. So since it's a complexe Hilbert space, if it's positive, it's automatically self-adjoint. So nice lemma. This is free energy, so it's energy, then minus the action of the chemical potential. So this is, if you want expected value of the energy corrected by chemical potential. And then minus temperature times entropy. The entropy, I mean, it's the usual formula, minus integral rule of law. So I warn PDE people amongst you that I'm using the physics convention that the entropy is really what gets produced and dissipated. And this is just the quantum mechanical analog, so trace gamma log gamma. And that will be used a lot in our proof. OK. So next, this workshop is about scaling limits, right? So now I will scale my parameters. And I think in the abstract, I announced that it's a toy model for looking at the BC phase transition. Rather than toy model, I would like to say toy scaling, because you see the model is the Bonafide exact model. But what we will play around with are these three parameters that I've put. So there is lambda, there is nu, and there is temperature. So what I will do is that I will just write down the scaling and then spend a few minutes explaining what it does. Because it's not supposed to be immediately obvious that what I'm going to write down is the relevant thing to do. And it's not the unique relevant thing to do anyway. So I'm going to call this a mean field limit because that's what it is, which in this situation will mean to me T temperature to infinity. But the reason I do this is that the particle number is not fixed. So taking temperature to infinity, in fact, is the way we impose actually you should really think of it as particle number going to infinity. So I will explain this in a moment. And then as you might expect, there is a scaling of the coupling constant so that things balance in a nice way. So it's lambda will be like 1 over T. So this goes to 0. And then the funny part is the chemical potential. Well, as Matthew explained at length, when you do grand canonical ensemble, the choice of the chemical potential, I mean, you have to be a little bit careful about what you take to achieve what you want. So it will be a funny number. Well, not funny for those who have already heard about VEC. So zeta 3 1⁄2, don't worry. So it's the Riemann zeta function, but it's just a number. Over 8, if I'm not making any mistake, pi to the 3 1⁄2, then I need to put here Fourier transform the interaction at 0, lambda minus 1⁄2 minus some new 0. Fixed. So this is fixed. And I claim, and I will explain that this means that in some sense, I'm zooming before the VEC phase transition. I accept, of course, I mean finite volume so it's not a proper phase transition. You can call it a quasi phase transition like they do in condensed matter physics. May I ask, so there is a minus in front of the chemical potential? Yes. So this tends to minus infinity? This tends to minus infinity, yes. Ah, okay. Now I have a doubt. Yeah, it's probably a plus here. Or probably it's a plus over there. Okay, so. It's the way that makes sense. So what I want to explain is oh yeah, definitely you should subtract something from the Hamiltonian, not add something. So the sign that makes sense is a plus here. So it's the idea that the mu really goes to infinity or minus infinity and not to 0? Oh, sorry, no, 0. What did I do? Oh, it's infinity. Yes, okay, it's infinity. Okay, so to motivate that I'm zooming before the phase transition I will consider the free-gip state and I start to motivate that this is really like taking n to infinity. So look at the free-gip state. Namely, so gamma 0 will be the guy with W equal to 0, not just lambda equal to 0 because if I put lambda equal to 0 here I'm in trouble. So the notation I hope is not too confusing when I say gamma lambda it means it has a W and I will say gamma 0 it means it doesn't have the W. So let's look at this one and so it does neither what this guy has it has this mu 0. Okay, I think I'd better write it down. So gamma 0 is 1 over a partition function exponential minus 1 over t d gamma of the Laplacian in the notation I had above plus mu 0 particle number. Okay, then what do we want to do? You want to look at the expected number of particles in this guy so I would note expectations in this guy with 0 so it doesn't have a fixed particle number. So you would like to know what is the expected particle number and that's where this funny number comes from is that it's this if my notes are correct at least t to the 3 half okay let me write it down once zeta 3 half 3 half it's really a dreadful constant 2 to the 3 half whatever that's leading order so that goes to infinity with the scaling it should have and then there is a correction which is t times some function of mu 0 that I will not define precisely because it's not really relevant what it is but it's like that and then you have still a lower order correction which is if I made no mistake is t to the 1 half so this if you remember Matthew's lecture I don't think he had the expression t to the 3 half but this is the expression when we do the calculation from the formula this is the expression of the critical density for BEC so let me call it C of t so which really means that BEC maybe I recall it means that note that I am on the torus with a fixed size so particle number and density is the same form so it means that if I take the same value of the number of operators in the 0 Fourier mode say in the free state this would this is supposed to be larger than a fixed constant that's the total expected number of particles if so particle if rho is less than rho sift is larger than rho sift so that's the number you get from the calculations in the thermodynamic limit this N0 is the guy which acts as on each sector it acts as if projects on each particle on the constant function so this is an orthogonal projector on the constant function ok so the leading order of the particle number is really this critical density and then I have a correction here which is non positive so basically all you need to know about this function phi so phi I don't really want to define because it's really just a remade it's really the error you make when you approximate the remand sum matured yesterday by the corresponding integral so it's really relevant what is important that this is non positive that phi is I guess it's positive and it's decreasing ok so that I hope explains why I take t to infinity because it's convenient to set so t to infinity with this new zero fixed so to say sets the system just before the phase transition so now I want to explain interaction in fact if I switch on the interaction and do nothing about it if I don't take a new of this form I will see nothing interesting the reason why this is that so why the choice of new is that if you just consider the bare interaction by which I mean this lambda w when you take the expectation say in the free state just to get an idea well of course this is a two particle operator so this will look like the expectation of the particle number squared which will look like well if you believe this it will be like lambda t to the 3 so it's not so much that it's too big but this is independent of new zero so I've set a parameter to help me look at things finely but if I look just at the leading order of the interaction then you don't see the parameter so it's probably you could expand to next to the leading order but we find it much more convenient to just remove something which would make the interaction talk not to this guy but to this guy which is the relevant one to go to the phase transition so what do you do when you see something like that what you want to do is renormalise so to say I will be more precise later renormalise interaction which will be like so the bare one and the chemical potential is used to so to say dump this guy in such a manner that if I take the expectation yes this will look like some energy reference that I don't care and that I will obtain just by completing a square so like namely something like I want this to look like so at this level of precision I want it to look like this I want it to look like a variance so quotation marks this is just heuristic but that's what I want so this energy reference is a constant I can just drop it each time I see it it doesn't set the state it just sets the energy reference but what is important is that I want this is that I want to have the variance and if you compute the variance of the particle number you will see that it will really talk to this guy and then this is going to be like lambda t squared and now it depends on nu 0 so that's good so then I can tune nu 0 and see something can I still ask a question about the scaling so you have the 1 over t in front of the Laplace Y and you take it to 0 so if you had the if you put in all physics constants you would have 1 over the usual temperature and then you would have an h bar squared instead of Laplace Y so is this in some sense semi-classical what you're doing? I think you can think of it like that although I will not obtain a really bona fide classical model so I can play around with parameters a little bit you're just trying to to make a connection and it seems to me is it very important for you that the coefficient of the Laplace Y term is really vanishing in your limit I'm not sure I can put it like that but it's what it says but you see that I will very soon sort of get rid of the Laplace Y so I'm not sure I really want to enter this discussion but if you can really write this as you said you can also take h bar to 0 you can do different things it certainly looks like a limit h bar yeah I think you can formulate it like that but it can be confusing because it's not going to be now a limit theory on phase phase so it's not going to be an extra limiting theory so that's why I prefer not to take h bar to 0 because to me h bar to 0 really means that you get to the phase phase and no longer go on to mechanics ok so the last choice I have to motivate is the choice of lambda and this one is ready for convenience so usually when we choose an interaction constant it's just to set things right on the same scale I just want to explain a little bit what it is that we set right let me put it like that if you take the difference between free energies on the interacting states and the free energy in non-interacting states so I hope the notation is clear this means the free energy without interaction and it's really a difference that you should think of it so both terms diverge very fast in my limits so it's really a difference that makes sense and this you can write exactly as t d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d d