ят, in there is a generic. But I will give specific examples of what I have said. OK, suppose the model is more complicated than the Q-revize and the magnetization and energy imaš vse vsi nekaj vzve. Vsi nekaj zgodne pripravljenje, pripravljenje z našem solim vse. Zelo, ko je razpozoril, je, da sem spes, sem spes na magnetizacij in energiji. OK. Danes. OK. You know that not all values of manifestations are possible. OK. In a model. OK. So, for instance, if this is a ground state energy, suppose this is a ground state energy, for instance, you know that in the Ising model, I suppose you know the Ising model, there will be two magnetization possible, minus one and plus one. OK. OK. Because you are on the ground state, so all the spins are either in one direction or in the other direction. So, OK, if there are plus or minus, so the magnetization can either plus or minus if it is normalized. OK. And then you know, you would like to draw a curve that contains all the states of the Ising model. OK. You would like to draw a curve that contains all the states. OK. So, OK. I guess it is this one. I know, because it is. So, there are many, many states in the Ising model. And I would like to draw a curve that contains all the states of the Ising model. It will be more or less like this. OK. And, of course, the density in this below this dome will not be uniform. OK. For instance, if you are at very, very small energy near the ground state, you will have more states near the boundaries. You will have more states near the boundaries and less states in the middle. And as you grow, at some point there will be more states here near zero magnetization because you are close to the transition and then there will be no more states. OK. OK. The transition will be, it is not the usual transition because you are used to think of the transition in terms of temperature. OK. Here I am in a different setting. I am speaking about energy. So, transition will be here, for instance, at some given energy. It must be within the dome because otherwise it is not the state of the system. OK. First of all, does this dome exist? Are there states? You can imagine that there are not states of energy if energy is very, very large. There will be a boundary for this region. OK. But what is important for this type of description is that the shape of this region is convex. OK. So, there cannot be non-convex shapes that delimit the region of, there cannot be non-concave shape that delimit the region of accessible states in the thermodynamic limit. Why? OK. So, I will use that activity. OK. So, first of all, suppose it exists. So, suppose that I have a shape like this, which is non-convex. OK. So, then that's the type of reasoning. I have, oh, again I have troubles. OK. So, suppose I have a state here, m1 e1, and there is a state here, m2 e2. OK. And now I build a chunk of matter where I put together one and two with a surface effect, which is smaller and smaller as the size of the system grows. Then, I mean, you can do it even more rigorously, but you can understand very easily that if I construct the line of states that join these two points, OK, which is that line with parameter lambda, so this is the line that joins the two points, OK, which means I am distributing matter among the two parts in such a way to vary lambda. OK. Then, I can go for large n as close as I want to be the straight line. So, it means that if the system is additive, I fill the region that is inside. And, in fact, I can fill the full region here with states that are obtained with this trick. OK. It's not the proof. I'm not a mathematician, but you can understand more or less how it goes. OK. Then, if you are a mathematician, you can do it and prove it as a theorem. So, the space theorem, the given aditivity, hypothesis, the state of admissible energy and magnetization in the large n limit, in the large volume limit is bounded by a convex shape. So, I will show you an example, an explicitly computable example of a spin model, analogous to the Ising model, for which we get a non-convex shape of admissible states. So, we can compute explicitly this boundary curve, and we show that it is not convex. OK. So, now it's time to be more specific about the models. I will run a little bit faster because there is no concept here. So, both the models that I will study and the ones that will be studied by Nikola fall in this class. So, either we give a potential function, which is a function of the position of the particles, with eventually some external field that acts on each particle, which is, for instance, of a self-garbitating system, so of Coulomb point charges, which will be in this category, or, for instance, it could be an energy function, where instead of having the dynamical variables move in space, they are attached to a lattice. OK. This is, for instance, a lattice in one dimension. And then the couplings between the energy function for a pair of particles decaze with distance with power of. OK. This, for instance, is a setting which is very important for the quantum case. In fact, typically the variables that you attach to lattice are either spins or rotors, and this type of systems can be nowadays simulated in the lab. For instance, there are beautiful examples of Hamiltonians of this kind in cold ion experiments where the particles are aligned, are trapped, the atoms are trapped, the ions are trapped, and they are put on a line. And then because of the vibration of the lattice, they couple the spins and they, one can tune the interaction alpha from zero to higher values. And it's a very interesting lab for long range interactions. Unfortunately, when I, when this type of physics took off, I also became the director of CISA. So I had to cut down my research to essential for six years. And in these six years, there have been a tremendous development of this area of research. And I'm not aware of all what has been done in the last six years. So I asked the Nikolao, who is an expert in this field. He has been my PhD student and is now an ATH duration. He wrote also a review paper on this type of models. So I thought it was good to leave to the experts the discussion of these kind of systems. Ok, so, so let us, so to finish the lecture, I have five minutes. Dyson, Dyson died, Mateo, this year, no, last year. Last year, I think, then last year. So Dyson wrote this model in 69. These are ising spins in one dimension. And they are coupled with long range distance. And you see, I've introduced this sigma here. This is alpha minus one, which I am in one dimension. Ok, so this model is cannot be solved. There is no solution of this model. So if you find one, you will get a prize. After so many years, there is no solution. So we don't know the free energy. We don't know the... Ok, so I decided to study this model, but to work in a comfort zone where I can do the solution. I'm not as clever as Dyson. So Dyson avoided to enter the region of sigma negative. Which is what can be called the strong long range region. Sigma was always positive and the energy was always growing within. We know, because if alpha is larger than D, then the energy is extensive. On the contrary, if sigma is less than zero and larger than minus one, the energy grows super linearly with the number of particles and with the volume, with power one minus sigma. And now I ask my friend Mark Katz to do the scaling and they get the free energy. Ok, so this is a very simple model for which I can do stat mech. Ok, in this region sigma between zero and minus one. Which was not considered by... That's a very interesting model. I will give you an exercise on this model. One can prove that for sigma less than one, there is a phase transition. And there is even a more interesting case which is sigma equal two, which corresponds to the condo problem studied by Phil Anderson and by Dave Thawless, where there is a Kosterlitz-Thawless phase. So it's a very, very rich model. But I decided to go in a region where all these mamma santissimas were not working, which is the sigma negative. And to do this, already you need some work. So there is a paper that I published some time ago in, I think, JSTAT fees. It was 15 years ago, where we could prove that I can write a functional for the local magnetization of the model. So I divide the lattice in k boxes, each with n over k sites, and I introduce the box average magnetization. So I take the average over the magnetization in a box, and then I take the limit n goes to infinity, with k goes to infinity, and the number of boxes goes to infinity, in such a way that k over n goes to zero. So this is called the mean field limit. Then I can prove that the n-body Hamiltonian of Dyson, this one, can be written at n, the number of particles, of spins, times this functional, where you see I have to do this double integral in x, between zero and one, I normalize the lattice to zero, one, plus correction, which is o, small o in n, so it goes to zero, goes to infinity, weaker than n. So it can be neglected in the large n limit. In the large n limit, this little o n can be neglected, and you get this function. What do I do? I can do many things with this functional. I can study this functional, but I want to do stat mech. So what I did, I did write an entropy for this model, and the entropy requires a little bit of more work. One has to use large deviations. I will speak about large deviations in my lecture course. As a result, I have an entropy functional. I have an entropy functional. OK? So I have to be fast because, ah, no, 10-45, OK? 10-40, I have 15 minutes. OK. So how do I construct this entropy functional? So first of all, I have my k-boxes, and I can argue that there is a large deviation principle for the magnetization in each box. And this is the entropy of the ising system, 1 plus m divided by 2, 1 minus m divided by 2. This is the entropy for the ising, you can see. And then there is no interaction. You know how statistical mechanics is built, I think Matteo can tell more important things on that, but it's based on the principle of maximum ignorance. So the thing can fluctuate as it likes. And then when you fix the constraint, you optimize the entropy given the constraint. So, given the entropy of the ising model, which is nothing but a cointoss entropy, so you cointoss the spins, and you compute the binomial distribution, you use sterling, and you get the entropy of the ising model. You will see, I will do an exercise on that. And you will get this entropy, OK? And in all boxes they are independent, and there is a large deviation principle in all of the boxes. And when I do the limit, I get the exponential in n of the total entropy. And the entropy is a functional of this, OK. So what I could prove is that I have written here in words so that you understand that, because I give reference to the papers. Maximizing entropy at fixed energy, one can obtain the magnetization profile in X, and for periodic boundary conditions it can be proven that the variational problem is solved by constant magnetization profile, m of X equal m, and m is obtained by solving the Q-revised consistency relation, tanh of beta j m equal m. OK, you know everybody knows this Q-revised thing, so it's a tanh, and then there is a line, magnetization is given by the intersection of the tanh with the line. So if the interaction is sufficiently long range, so if sigma equals zero, so alpha is between zero and one, I am in the so-called strong range, strong long range region, the mean field is exact for this model. The mean field is exact, and if the boundary conditions are periodic, the mean field profile is the solution in each point. And if you try to perturb this profile, so for instance you add a Fourier mode, you add a wave, the wave will dump. This is part of the proof of stability. So the profile is constant, and if I perturb the profile, the perturbation is damped, and the optimal value is... There is an open problem, still open problem, in fact you see here it's a numerical solution. What happens if the boundary conditions are not periodic? In essence, suppose the boundary conditions are free, and so free means that the right spin will see nothing, and the left spin will see nothing. This is free boundary conditions, and I am between zero and one, and this is a solution that we obtain in computer simulations. The boundary conditions are not clear, but there are different profiles for energy 0.1 and alpha is zero, two, zero, five, and zero, eight. The dotted line is the closest one to alpha, I call one, is one, and you see that the magnetization profile is not constant. And it's hard to see a constant part even in the center of the magnetization profile. So it means that the solution depends on the boundary conditions. And this is no surprise at all. So I got this by optimizing the Dyson functional, so the entropy functional, at fixed energy numerically. And I see that the optimal profile is not constant. So if you try with fixed boundary conditions, also you will get that the optimal profile is not constant. So no one knows the value of the magnetization here, here, here, here. I don't have results on that. So even in this regime, and this allows me to go back to the first comment I did today. When solving the energy in a sphere, when giving a formula for the energy in a sphere, it was clear that for alpha less than D, there was an effect of the boundary because I should have said what happens at distance r, what are the particles doing when they reach the boundary. And it will be very important what happens in the boundary for long-range interactions because what happens in the boundary propagates in the bulk. There is no disconnection of what happens in the boundary to what happens in the bulk. So this is another difficulty of long-range interactions. OK, let me skip maybe this, because it is... But then I will have to enter in a more difficult. OK, so it's time for the exercises. So five minutes. So I prepared a few exercises, and then if you do them, you will be able to do the exam next week. It's a very simple one. It's just a stat mech. So first exercise is compute the entropy and the free energy of the classical perfect gas. No interaction. The simplest case. And show that the Le Jean transforms. In the next talk, I will add a name. This guy, Fenchel, was sort of forgotten. So I will tell something about the Le Jean Fenchel. But for the first thing, it's enough Le Jean. And show that the Le Jean transform of the entropy does not coincide with the free energy. It does not coincide with the free energy at finite energy. However, better than E. So the usual Le Jean transform for... You can open a book in stat mech. OK, not of the density. Of the energy. Le Jean transform in energy inverse temperature. So I would like... If you want, I can... OK. I can write it. So you have the entropy as a function of epsilon. And you have the massier potential, which is better, the free energy. OK. And this is the Le Jean. OK. This is Le Jean. Ah, yes, it's on the slide. It's on the slide. Yes, beta F is beta epsilon minus S. OK. I have derived these by the Laplace method. The usual Laplace method. OK. So, however, the functional dependence, dependence on beta in V, V is the volume, is the same. And then, second part of the exercise, using the steering formula, show that they coincided Le Jean. This is true for the perfect gas. It is true for all short range interactions. It is not true for long range interactions. Unfortunately, this is the first exercise, exercise number one. Second exercise, exercise number two. Argue. I like exercise like this. Argue. Why? There cannot be phase transitions in the Ising model in 1D with the nearest neighbor, nearest neighbor interaction. And then, second part, guess, this is very hard. I will do it, the solution in the lecture. What happens? Sorry? Ah, yes, sorry. I don't want to break everything. So, argue why there cannot be phase transitions in the Ising model in 1D with nearest neighbor interaction. Guess what happens? What happens for the Dyson model in 1D, where the interaction decreases like 1 over i minus j to the 1 plus sigma for sigma larger than zero. This is the exercise. I will say this is very hard. It is a taules paper. So, you can try to think what happens. And this is very simple. You find in books, so if you want, you can. But you have to understand. I checked in the books, there are two ways of reasoning. One way of reasoning is dynamic. So, defect is diffusing and does not attract. Another way of reasoning is static. So, you look at how many defects are allowed and so on. And what is the entropy and so on. So, this is, instead of here, you have just to write the perfect gas in a fixed volume and check. I think it's interesting because you understand something that not often you pay attention to. So, the fact that you have this thermodynamic scheme of the Legend transform, but at the same time, you do not have this type of equivalence at finite n. So, there are finite n corrections, which are important. If you try to use it at finite energy, at finite volume, the relations are not yet in the thermodynamic limit. So, you don't get what you expect. So, I finish in perfect time with two exercises. OK. So, you will work out the solution tonight. To no stafno naj tempi.