 Zna ki? To je da sem bojne, stejfano? Preprinujem, kaj je? Znam. Paredje vs. Enlega. Ok. Pozirajte konfiguracije na karanele, če je pozir enough. Oje obočno jaz li prachum si trik. V carane del je neko Stav, obkrat tega, če je vzgleda, obkrato so v spesih. Zato, da potožilo lokalne interakcije, lokalne in nezavrne interakcije. Zato, in odkrati, nekaj se dojel, odkrati nezavrne terapise. OK, so, there are three objects on this, on this model. It's plus pins, minus pins and defects, OK? OK, so these are walls from minus to plus, OK? So this is, for instance, a configuration with 12 plus pins, 8 minus pins, and u half is, so u is the number of antiferobons is 4. One, two, three, four, OK? So now you have to move around the walls, the spins, and so on, to count these states. OK, now I give you the result, and then I will do on the blackboard the calculation. The result is that the volume, a given number of plus pins, given number of minus pins, and given number of walls of defects is twice the combinatorial factor n plus over u half, times combinatorial factor of n minus u half, OK? This is true, as you can see, approximately for large n plus, large n minus, and large u, OK? And if you go to, but there is an exact expression that is not published in the paper, so it's worth that I give you two u. It's the first time. So moreover, in the paper, there is a wrong factor 2. So we don't put the 2 in the paper. That was a mistake. So I will do the calculation now for you for open boundary conditions. And you understand that this result depends on the boundary conditions, periodic, open, and so on. So I will do it for open boundary conditions. And for odd numbers, these are the conditions. So let me try if I'm able to reproduce it, because I did it last night. So I do it for u odd and open boundary conditions. So if boundary conditions are open, suppose I have a string of n spins, OK? I choose the first spin with a plus, OK? I choose the first plus. Then if I have a odd number of walls, for instance, 3, then if this is plus, this is minus, this is plus, this is minus. So it means that the n spin must be minus, if I have a odd number of walls. Now, OK, I number the walls, let's say k1, k2, and so on. And now there are, among the walls, there are u minus 1, because u is u minus 1, because the last wall is here, over two walls that have plus spins, there are to the right of plus spins, OK? Did you get this? So you can try a longer configuration and you will convince yourself that you have u minus 1, half walls that are to the right of plus spins. And there are u minus 1 over two walls that are to the right of minus spins. So u is equal to 3, right? Yes. And you have two walls, which are to the right of plus spins, no? This and this. Yes, there are two. Two. U minus 1 over 2 is 1, because I don't count the last one, which is to the right of... Oh, OK. OK. So in fact, I have also to subtract n plus minus 1. So I have to subtract one spin, one plus spins, OK? So the correct number is this, times n plus and minus minus 1, u minus 1 over 2, times 2, because I can flip all the spins and I get a new set of configurations. So the exact expression at finite n is 2 for open boundary conditions is 2 times n plus minus 1 over u minus 1 over 2, times n minus minus 1, u minus 1 over 2. Please check this, because it was... In the large n, the ones goes away and you get the expression that is on the blackboard. So this assumes that you can rearrange the n plus... So I can choose u minus... U minus 1 over 2. This is a way of choosing. U minus 1,5 volts. In certain... Among n plus minus 1, u minus 1 over 2 volts over n plus minus 1. So you are choosing the u minus 1 over 2 spins who have a wall on the right. On the right, yes. You have to do it yourself in order to convince that it is correct. In fact, the way we were reasoning for the fact in the paper was different. OK, then you can do it for close boundary condition. Try close boundary condition to see and fix boundary condition. You will get different numbers but in the large n, all of them will be reduced to this irrespective of the boundary conditions. OK, so do for exercise both the close boundary condition and the open boundary condition. I think you will have to convince yourself. So putting the pluses, maybe increasing the number of walls and play with configurations to convince you that if it is wrong, tell me. Because already we published the wrong result and I don't want to be wrong for the second time. So it was not so bad after all because the factory here was OK. Log 2 over n as Matteo was mentioning yesterday is in the large n is 0 so there is no problem for everything. So it was OK, finally. OK, so you know that to do long range interactions in this case you need to learn a little bit about combinatorics a little bit more than what you learn in the usual courses on combinatorics. It's interesting. OK, so what do you get as an entropy? So now you express n plus and n minus in terms of the total number of spins in terms of the magnetization and of the number of walls and you get this expression for the entropy where you see, here I forgot to say that this is also a function of u. It's an intensive quantity that is the number of walls divided by the number of spins. So I get this other entropy again in the same class of previous entropies and u is not... I don't see u... OK, now it's better. And u satisfies the equation for the energy so the energy is minus j half m squared. This is the q revised there plus k, u, the number of walls. OK? And so you can express u as a function of epsilon in m and then you substitute and then you maximize with respect to m and you get the entropy as a function of energy and you do the same, you do the same and the interesting part of the phase diagram is for k negative, for k positive, this was known also to Kardar and Nagel, for k positive you have a second order phase transition line and the second order phase transition line in the canonical ensemble ends at the canonical tri critical point and then you have a first order phase transition line which ends here. Similarly, as yesterday you can compute this value exactly at zero temperature comparing the paramagnetic to the ferromagnetic energy. OK? You can get exactly 0.5, minus 0.5. Antiferro? Antiferro, yes. Sorry, sorry. You compare the ferro with the antiferro because here at zero temperature you have the antiferro state. As soon as you increase the temperature the antiferro state disappears. There is no antiferromagnetic order at finite temperature in one dimension. So as soon as you... This is different two dimensions. For instance, Kardar and Nagel also studied the two dimensional problem. By the way, it was the PhD of Kardar with Berker. He did this PhD with Berker. OK. In fact, I knew this model because at that time I was interested in renormalization group and the effects of long range interactions in the renormalization group and I found this work by Kardar when he was a student in MIT and I wrote a letter to him by saying I did an approximate renormalization group and he replied to the letter and he said I did the exact solution. And I understood that Kardar was someone very strong. So I had an approximate method for some model for which he had an exact solution. So I remembered these after many years and so I decided to study this model. OK, we found the micro canonical solution. So you see that the shape of the phase diagram is very similar to the other one. It's amazingly similar, no? You have the micro canonical tri critical point which breaks into two lines of two temperatures, the upper temperature and the lower temperature. Here there is the temperature jump and the two lines join at zero temperature. So it's very similar. In fact, one could think, OK, you will find always the same, no. There are other models and the other good thing here is 0.6, 0.5, 0.4 is no more restricted to that small area of numbers. So it's easier to study this model because the parameters move order one when you vary the couplings. So you can get better. OK, so and we could have done it for the other model also but we did it for this. We realize that you see that in this entropy there are a lot of logarithms and these logarithms should have a positive argument and these put bounds on m. OK, so clear, no? For given energy, given m, not all values of m are accessible, because the argument of the logarithms should be positive. Now you can see that, just an example in this model is very simple. So suppose you take a positive magnetization state, so it means that the number of pluses should be larger than the number of minuses in order for magnetization to be positive. So you are on this side. Then what is the maximum of the energy? The maximum of the energy, sorry, what is the maximum of the number of walls? There is a maximum for the number of walls and is when we put walls side of one of the two species. So at maximum you can have twice the number of minuses as number of walls. So if you rewrite twice the number of minuses as the total number of spins minus the magnetization, so you see that this is bounded by one minus the magnetization. So this bound reflects into a bound in u. See, minus magnetization, it cannot be negative. And it gives you a bound on m, fixed in epsilon. And if you compute all these bounds, the bounds in m, fixing energy, you get this plot that is very artistic. So this is the line m below minus the square root of epsilon, so the maximum energy is zero, and the upper line is m less than minus square root of two epsilon that you get from this bound when u equals zero. And then these other bounds are bounds in m plus and m minus that result from this quadratic equation when you bring m on the left side. So you see that in this model the accessible states are not in a convex-shaped domain. These are the bounds for the states, and I can get them exactly. These are the bounds for the states in this model. And as I have said in the first talk, and this is the proof, the shape of the area of accessible states in systems that are non-altitude can be in non-convex. So in the Ising model, all these states are filled. There is a probability, a non-zero probability that these states are filled, while in this model you will not found states here. You will not found states here. So in order to check this, I will go slightly faster. So we did a study of dynamics in this model using the Kreuz algorithm, and maybe if you want, I can say more about the Kreuz algorithm. The idea of the algorithm is the following. It's a micro canonical algorithm, but there is a demon in the game. And the demon works this way. You try to swap a spin, and if the energy goes down, you give the energy to the demon. You give the excess energy to the demon. You start the demon with zero energy, so the demon gets energy from the system. And if you need the energy to go up, you ask energy to the demon. If the demon has energy, you pick the energy from the demon. So this algorithm visits the energy shell for all energies that are below the energy that you had initially given to the system. But as the algorithm gets on, goes on, the number of energies of the system that are sampled are very close to the, if the system is large, the energies that are sampled are very close to the limiting energies. So if you check during the time process, which are the energies that are sampled in the system, they are very close to the micro canonical shell at the top energy. But it's also good if you sample energies below, because you know that the micro canonical shell only imposes that the energies below a certain limit. And also how can you extract the micro canonical temperature? You go and look at the, you cannot see it here, I don't know why. Maybe I eliminate. You go and look at the statistics. If I go out, I cannot eliminate these black things. I cannot eliminate these black things here. I don't know why. So if you check the statistics of the energy of the demon, the statistics of the energy of the demon is Boltzmanian, but the constant is the micro canonical entropy. So it's exponentially distributed and the constant is the micro canonical energy. So you can get the micro canonical energy by taking the inverse of the average of the energy of the demon. OK? Because it's exponentially distributed. So it's a very good algorithm. It gives the micro canonical ensemble and it gives access to the micro canonical temperature. So what we did with this algorithm, we ran the Kardar-Negel model. In a situation, for instance, where the entropy has this shape, there are four, three maxima, one in zero and two are the maxima at positive and negative magnetization. And if you start here, which is this state here, you see that from time to time it jumps to the other states. OK? But suppose now you go to a region where there are nonaccessible values of the magnetization. So do you see that the entropy here disappears in nothing because there are no states here, which are these cuts that appear in the at given energies. And this is a similar cut. And you see if I start here, I remain here. And if I start here, I remain here. There is no jump. Because there are no intermediate states that I can jump to. Of course, I could do global moves. I could swap all the spins, OK? But with local moves it will be impossible to move from one state to the other. So if you do also launch and dynamics, it will be the same. So this is a breaking over Godise. The phase space is broken. And this breaking over Godise is not the usual breaking over Godise which appears in the canonical ensemble or in other systems. It is due simply to the fact that there are some states that are not accessible to the system. So you cannot go to those states because they don't appear. And of course, if you are in the canonical ensemble since you have intermediate states you can jump even with very small probability. But in this case the probability of jump is zero. OK. So what's next? I think you can easily program that this is something that you could do. Try to write a code for the Croids algorithm. It's not very hard. OK, what happens when it is possible to jump from one state to the other so there is a barrier of height delta then what happens is that if you are in the canonical ensemble and the interaction is in field the time to jump is exponential n times the energy barrier and in the micro canonical it's exponential n times the entropy barrier and this is to be compared with the jump in d dimensions which is d minus 1 over d this is the scaling of the so we tried this for the model and you see that the lifetime this is the exponential increase of the lifetime with n and this instead is the logarithmic increase of the lifetime when there is no barrier so when the system goes from one state to the other so if there is no barrier there is a logarithmic increase if there is a barrier there is an exponential increase maybe it's too much OK, I just want to tell you that we have a theory that since it's unusual we have a theory that explains this log n which is based on Fokker-Plank equation for the probability of the magnetization and the theory you can solve this equation and the theory is based on the physical ingredient is the fact that the diffusion constant in long range goes down like 1 over n so it's the diffusion in long range system is smaller and smaller as the size of the system increases because the mean field sort of freezes the system and if you assume this it's a very natural assumption you can solve this equation and you get this solution at time p and if you fix the time to reach something of order 1 you get the logarithm of n so which is a direct this log n is here because there is a diffusion that goes like 1 over n so this is another feature of long range system that the motion is diffusive of the single particle but the diffusion constant is smaller and smaller as you increase the system size and if you assume this this is again for the evolution in the case in which there is no barrier ok, so this is was an additional comment I don't know how many slides I have but ok so this is I put these slides just to help you in solving these models so I did it for the Q revise so in order to solve this model in the canonical I told you that you have to do the Q revise the uberstartenovic transformation so I put it here for those who don't know how to do it ok, this is the square ok, this is the Q revise square I use this integral this is the Gaussian integral to rewrite it as an integral over x ok and then you get the free energy and then you have to minimize over x and you get the free energy as a function of b if there are other terms ok, it's easy to find but in this slide you find all the, what you need to do the transformation that brings you to the canonical solutions of the two models that I introduce both the Blume-Capelle and x is this what is called auxiliary field and you see that you have to solve it for you have to take the minimum on the auxiliary field if you want the free energy yes, it would take some time to derive the Fokert-Plank equation yes, yes, yes suming global dynamics and you can get the Fokert-Plank equation for all macroscopic quantities also energy for instance you start from for instance you start from the n body probability distribution function and then you have a mean field so you go to you can go to the single body and then you have an equation consistency equation for the field there is a way of getting it it's not simple, but okay yes, I have already marked to put the paper on classification because someone asked about these invisible things also I will put the paper on azeotropy if someone is curious of this strange type of phase transition and I can put two papers one on the Ensembling Equivalence which is a 2001 paper with David Mouhamel and the other one again with David Mouhamel 2005 on the Cardard-Negel model so I don't want to invade the slack with with papers just references, okay just references and then you have access to the library okay, so I would like in the last few minutes and then I stop just because you are I think you are okay, just introduce the the method it's an interesting problem in again in non-concave functions so the question is the point is the following, suppose you have you have a partition sum can be expressed as a function of X you don't X could be the Hubble-Stratonovich auxiliary field could be other things so suppose you know that the partition sum is this, okay so usually you know it because you have the Hubble-Stratonovich suppose you have other methods that give you the partition function in this form and U is a differentiable function of beta and X and X is some dummy variable, okay then of course the massive potential which is beta times the free energy is the inf over X of U, okay now let us introduce the Legend-Fension transform of of U, okay which is the inf over beta of beta epsilon U beta X okay so it is a function of epsilon and X okay then the statement of the theorem I have a proof of this theorem but it is not rigorous and I think it is when some mathematician should prove it okay so suppose I take the soup of S then then I get the soup of X so so the inf gives me a function of X and then I take the soup of X the soup of X is the is the micro canonical energy entropy while if I invert the two so I take the soup of X and then I take the inf of beta I get the canonical so I get the Legend-Fension transform of the free energy so this shows that if you can express if you can give more freedom to the function S in order to explore different values of this dummy variable you can by inverting the soup with the inf you can get the micro canonical entropy so I will show tomorrow some examples although it is limited but it goes in the right direction which is the one forbidden by Kersen Wang of solving models in the micro canonical ensemble in the right class of partition sums for which you could get the micro canonical entropy by inverting the maximization problem and as you know inverting a maximization does not give the same result and in particular you know that if there is a convexity in the entropy the canonical entropy is always larger than the micro canonical entropy so indeed there is a relation that if you invert the soup if there is an order soup inf is less than inf soup I will prove it tomorrow also this inequality it's a two lines proof proving that this is the micro canonical entropy is harder you have to work with at least the proof that I know you have to work with the plus representation of the delta and then to have to do to argue I will try to give a simple idea of why we believe that this is the micro canonical entropy but I think for instance one should prove it with large deviations I think there should be some proof using Kramer or some other theorems from large deviations it is of course restricted to a specific class of cases but I will show that there are plenty of them and it's interesting you get other forms of entropies and you don't have to count that's very important because for instance with this trick you can solve models with continuous variables and you can get the micro canonical entropy for such models with continuous variables so tomorrow I will give a proof of that and I will show applications to class of models that have this feature ok thank you very much