 Okay? Can you hear me? Yes. Okay. I think we can start. And now I will start with a short recap of what we did the last lecture. So you can, since we are following many different lectures, I just want to remind you where we are. Okay. So last lecture we did a high-temperature expansion for the free energy of a generic disorderly system, which up to the second order reads, we call this A beta of M, is the zero-order term, which is essentially an entropy term, because for beta equals zero you have uncorrelated variables, or roughly uncorrelated variables, as we will see in the example we will discuss today. And then the second order term is the energy part computed on the measure with beta equals zero, so the uncorrelated variables. So this is the mean field term. And then you have the on-sugar reaction term that looks like this operator U that we introduced completely with beta equals zero, and again average, this is unobservable, average with beta equals to zero. And this last term, just to remind you that you can write it also as this one, okay? So the average of U squared is equal to the average U times H. Then we realize that this expansion up to the second order is fine for fully connected models. In particular, we realize that the first order, the zero-order term and the first order term for quibis provides the naïve mean field equation, while considering the three terms, so up to the second order, for the Sheraton-Kirpatric model, which is the fully connected version of the spin glasses, provides you with the TAP-Taoles-Anderson-Palmer equation, okay? And then we have to solve this equation. Actually, we discussed a little bit the problems of the solving this equation for the SK model. Essentially, the main problem I point out to you, and this essentially is an open problem in the field, is that when you try to solve the TAP equation for the SK model, you have one solution which is the paramagnetic solution, so all the magnetization equal to zero, which is unphysical for two reasons. Essentially, you can easily prove that the entropy becomes negative for T slightly below the critical temperature, so you have to abandon this fixed point. However, this fixed point is very attractive. If you compute the second derivative of this free energy, you realize that the n equals zero solution is always a minimum, far from exactly at the critical temperature, and since very close, so when you are below the critical temperature, but not so much, the solution you are looking at have very small magnetization because the magnetization becomes different from zero at the critical temperature, and then if you look at the Q, which is a quantity measuring how large are the magnetization in the solution, you expect this to become different from zero at the critical temperature and then grow up to zero temperature. If you are looking solution here, you expect solution to have very small magnetization, and if this solution to the TAP equation is very attractive, it's very, very difficult to find solution to the TAP equation in this regime. Indeed, up to now, essentially, we are able to find solution only at very low temperature, but we are still unable to find TAP solution for the SK model at a higher temperature, and so obviously we cannot count them, and in particular, when you do a replica calculation, you can understand how many solutions you expect very close to the critical temperature and the quantity that counts those solutions that I will introduce today, the complexity grows very, very slowly, like a fifth power of Tc minus T, so it's really a tiny number and so it's very difficult to find solution for this model. This is the reason why today I will move to a different model, which is a PSP model, because I would like to show you the first computation of the complexity, so counting the number of TAP solutions in a model where TAP states are much more stable, are much better defined, rather than in this SK model where the TAP solutions are very hard to find. And so I will choose a model that has essentially the random version of the first-order transition that you have already seen in the Curivice model just to finish this reminder. In the Curivice model, if you do a plot in there, so let me remind you the Hamiltonian of the Curivice model, so if you do a plot in the HT plane, you have learned in the first lecture that there is a second-order phase transition at this point, which is Tc equal 1 and h equal to 0, but along this line you have a first-order transition. What I mean is that if you cross along the orange line the h equal to 0 axis, what you have is, as a function of h, the order parameter, which is the global magnetization, has a jump. Along the orange axis you have something like this, a jump changing h. Enough light? Okay, this is good. No, that was good. But if you cross this second-order phase transition along this blue line, what you have is that the order parameter, the magnetization, now becomes different from zero in a continuous way, okay? This is the critical point. So this is what we call a first-order phase transition and this is what we call a second-order phase transition. Here the order parameter, which is given by the first derivative of the free energy is discontinuous. So you have a discontinuity on the first derivative of the free energy. Here the order parameter, which is the first derivative is continuous, you have to look to the second derivative of the free energy, which is the first derivative of the order parameter, which is discontinuous. In this case, it has a divergence because if you do the plot properly, this has a square root of t minus tc, tc minus t, okay? But what is more interesting for our purposes of counting the minimum of the free energy is that in this case, the states, the minimum of the free energy are very well defined even at the critical point because in this situation, if you look at the free energy as a function of the magnetization, what you have is that you go from a situation where the free energy, let me draw it here, the free energy is something like it has two minima, but the dominating one has positive magnetization, you go from to a situation where you still have two minima, but the dominating one has negative magnetization. But even at the critical point, the two minima are very well defined. So in a first order transition, the minima are always very well defined and what changes is the relative weight among the two minima. If you want to count minima, here you have no problems. So it's there. Even in a situation where this minima is metastable, it's there, so you can count it. In a second order phase transition, which is a continuous phase transition, like the one which happened in the SK model, the minima arrives at the critical point. You go from a situation where the free energy has a unique minimum in N equals 0 to a situation where the free energy minima arrives at the critical point. So if you want to count minima, obviously it's much more complicated because at the critical point, the minima are just arising. So let's move to a model where the phase transition reminds because we are in the realm of disordered system, reminds the first order phase transition. For this purpose, we move to a p-spin model where p is larger than 2, a 3-spin model because here from the solution of the fully connected ferromagnetic model, we know that if spins interact in 3, the phase transition is first order. Yes. This is the magnetization is H because I'm going along the orange line. So I'm in a temperature which is smaller than Tc and moving along this line while moving along this line the magnetization, which is the order parameter, changes from positive to negative because if you remember the free energy, the thermodynamic canonical free energy along this line it looks like F of H it looks like this magnetization is minus the derivative of F with respect to H you have a positive magnetization up to here and then when you move to the negative H part the magnetization becomes negative and so you have this jump in the first order derivative of F maybe I don't get your question you haven't never seen this picture in temperature I have to change the model because in the Curivice model you have this first order transition only when you cross this line you can cross this line only in a field in temperature I can only cross this point and so you have a second order phase transition if you want to see a first order phase transition changing the temperature because maybe you like more changing the temple than changing the external field you can move to another model which is this ferromagnetic interaction but ok you can also it's much better to have minus and 3-spin interaction now this model in temperature has a first order phase transition so if you look at the free energy of this model as a function of the magnetization because again this can be written apart from sub leading terms as minus nm to the cube where m is 1 over n the sum of the spins so again this Hamiltonian being fully connected you can simply write it in terms of one global parameter which is the magnetization again if you do the standard computation that you have done in the Curivice model the energy of this model always have a minimum in m equals 0 and then let me do it always has a minimum in m equals 0 but eventually if the temperature is low enough it develops a second minimum at m different from 0 if you keep the temple going down this becomes dominant so eventually you will have at a certain point this situation so this is the critical point so the minimum first arises at a higher temperature you decrease the temperature and you eventually make the paramagnetic solution and the ferromagnetic solution of the same weight so the partition function is not dominated only by m equals 0 but also by m larger than 0 states and this states becomes more important and so this is happening in temper so this T larger than Tc T smaller than Tc and T equal than Tc so here you have again the first of the transition in temperature also the position of the minimum moves right towards decreasing the temperature so the plot is not perfect here in this medium should be slightly on the on the right but still this is the same first of the transition in temperature and you see again that the minima are well defined even before and after the transition even after the transition there is nothing becoming flat and making the definition of the minimum more difficult to relate to the state so again this is a situation where if you are here and you use any local update dynamics jumping here takes a lot of time so on any reasonable time you have electricity breaking and you visit only this part of the phase of the configurational space and so any dynamics on a reasonable time remains trapped in this state very large times you jump out and you go to the if you are in this situation you jump out if you are in the other situation you jump here ok you go from the metastable to the stable state so let's take ok so let's now take the random version of this model that we call p-spin model I want to keep this because I will use this expression I am still following the Francesco Zamponi notes ok so I will consider this spherical p-spin model which is defined by the following Hamiltonian minus 1 over p factorial the sum over all I1 Ip so you have p indices running from 1 to n j I1 Ip s I1 s Ip actually let me rewrite this in a in a formula which is more convenient because every people I want to appear just once so I don't want to consider a coupling 1, 2, 3 and a coupling 3, 2, 1 so the the triplet S1, S2, S3 must appear just once so it's better to rewrite this expression as 1 minus I take out the p factorial I order the indices so I1 is strictly smaller than I2 strictly smaller than Ip again running from 1 to n of the same expression ok so in general I will switch between the sum over all indices with the p factorial term in front and the sum over ordered pupils of indices these two are equivalents so every time I use this I'm assuming that the j's corresponding to permutation of indices are the same so j1, 2, 3 is equal to j2, 1, 3 j3, 2, 1 is the same ok and this is the p-spin model the objective spherical means that these variables are not plus or minus 1 but in order to make the model more easy to solve analytically I will take the spin that belongs to reals and they satisfy the spherical constraint the spherical constraint imposes that the sum of all the spins of the square of the spins is exactly equal to the number of the spins for this reason I want to work on a model where the minima are much better defined than models undergoing a second order phase transition in a second order phase transition it was drawn here essentially the free energy go from a single minimum to two minima and at the critical point you just have a very flat minimum so this makes the definition of states more difficult because states are rising they are growing while decreasing the temperature and so this makes states much less stable they are much what we call marginal so they you can leave one state much easier than in a situation like this where states form again with a mechanism one of the mechanism of bifurcation analysis so at a very high temperature you have something like this and then eventually you get something flat and eventually you get a minima but this happens well above the critical temperature and you want to count states when they start being thermodynamically dominant so when you are interested in counting states which is the orange situation the states is very well formed the barrier is very high and so we hope that it will be much easier to count there are many more indeed I will show you that now we will have in this model exponentially many states and so counting something which is exponentially in n is much easier than counting something which is very tiny now we will see when we solve the dynamics that this model is much better for describing structural glasses and ergodicity breaking in a way different from models having two spin interacting which hand goes a critical slowing down at the thermodynamic transition we will see that this model we have a dynamic ergodicity breaking transition before the thermodynamic transition so this model described a different physics but it's the good one for learning how to count states and so how to understand the effect of this Hamiltonian which at the end are the dynamics I want to understand what a system does when I let it relax ok so this is the definition of the model and the first thing that we want to do is to derive a free energy an approximated free energy again and sorry no I have to tell you one more important ingredient these are random numbers and we assume them to be p so j i 1 i p are i i d random variables of 0 mean and variance p factorial over 2 n to the p minus 1 ok this is the right normalization in order to have the Hamiltonian actually the average value of the square of the Hamiltonian which is extensive ok so this is the definition of the model and now we want to derive a free energy an approximated free energy and we are going to exploit this expression that we derive in the first lectures so we have to compute the entropic part the energetic part and the on saga reaction term because we are in the in the disordered case this term now as you notice even at let me write it so the entropic term what we call a not given the magnetization I want to fix the magnetization as usual this the integral overall the overall the spin variables with the spherical constraint that must be satisfied at any time on any configuration ah let me tell you something about this spherical constraint you may notice that the spherical constraint is satisfied if you take an easy configuration so if you take spins that takes value plus or minus one it is satisfied so what we are doing essentially we are making the space of feasible configuration compact convex and continuous ok because otherwise the space of physical configuration for easy variables is an hyper cube are the vertices of an hyper cube which always is a discrete set so it is more difficult to sample it is more difficult to when you solve the model you always have this summation of a discrete set while in order to make the analytic computation easier we like to integrate over continuous variables but you want to bound these continuous variables in a compact set that at least contains the easy configuration so what you take is the hyper spheres that passes through all the vertices of the hyper cube and you hope that ah the physical behavior of spherical variables is not too different from the physical behavior of easy variables a priori is not completely imposed by this constraint because this constraint is also satisfied but very nasty configuration like a configuration where you have one spin having square root of n length and the other zeros so this is a configuration that satisfies the spherical constraint but it is very far from an easy configuration which is a configuration of plus or minus one ok so there is that as long as the all the variables play similar rules so you don't have a condensation on a single variable then the physical behavior of spherical variables which are much easier to deal with analytically is not too far from the easy one indeed you have to be careful because for example if you define these spherical models on a diluted graph you realize that condensation on a single component phenomenon happen while in fully connected models you can do the computation you can see that the energy the weight concentrated in a single component has an energy which is much higher than the configuration where the weight on the components is equally partition on the n components so you can check that on fully connected models like this one the typical configuration of satisfying the spherical constraint and having low energies are those where all the components are on an equal footing if you move to diluted models you have to be careful because p-spin diluted models have this phenomenon of condensating on a given component so then the physics is very biased by this it's not meaningful so let's do the computation of the entropic term we want to integrate this with the term sum over i lambda i not s i minus m i so you see in the definition of a beta there was a terms exponential of minus beta h now we are setting beta equal to 0 so the interacting term disappears still contrary to easy variables here we still have this term that makes variable to interact slightly so in order to solve this integral we just use the just use the integral representation of the delta function and we call mu this is the integral over all the spin variables so this is an n dimensional integral this now we write in an exponential form the delta function so we have minus mu sum over i of s i square plus mu n and then again that term which is sum over i lambda i not s i minus m i and now you realize that thanks to having opened this delta function each variable s i is just a Gaussian integral for each variable s i square plus lambda i not s i and so you can integrate each of these n variables by Gaussian integral and so what you get is an integral minus i infinity plus i infinity then mu over 2 pi and here you write the term exponential of mu n the term sum over i lambda i not m i and now you have the outcome of the n Gaussian integrals that gives you a factor n half log of pi over mu which is the normalization and the quadratic term which is plus 1 over 4 mu sum over i lambda i not square okay so each of these Gaussian integral gives you a lambda i square term and now you want to maximize this integral over mu because you want to solve this by the seven point method but also on lambda i if you remember the value of lambda i that you want to impose in order to have this relation is equivalent to write the derivative of a i with respect to lambda i equal to zero so essentially you want to maximize this expression also with respect to the lambda i and so if you maximize this expression with respect to lambda i you realize that taking the derivative with respect to lambda i you get something which is lambda i not is equal to you derive here and you derive here and you get 2 mu mi so you impose this value for lambda i in order to have the mean value of sigma equal to mi and if you do this you finally get that let me write the result here so you substitute this value of lambda in this expression and you have a subtle point method by extremizing the exponential the exponent and so you get that a not of m is the extremum of the exponent so you get a not of m which is divided by n is equal to the extremum of this expression where I just substitute the lambda i not with that and this mu 1 minus q 1 half log where q is defined here is just 1 over n sum over i mi square and again if you do this extremum with respect to mu and let's write here the result you finally get a not of m which is n times let me write this way n times s infinity plus n half log 1 minus q where s infinity is the what we call the entropy a beta equal 0 so a temperature equal to infinity and is equal to 1 plus log 2 pi divided by 2 ok so in this way we got the yes here I'm just using the representation of the delta this the integral if you want let me write this way from minus infinity to plus infinity in the mu over 2 pi exponential of i mu x minus x not this the integral representation of the delta function just not to have the imaginary part in the expression I integrate over the imaginary axis I just do a rotation on 90 degrees and so the mu here so the mu here I'm using is just i mu hat using here ok and at the end when you do the saddle point method on the complex plane you have to choose the point along the real axis that is where this expression is is extreme also the derivative with respect to the real part is 0 ok this is why it's called saddle point because ok I think that if you want to make a 5 minutes break now is the time to do it so so this expression essentially is giving you the entropy of n variables which are independent apart from having a spherical constraint so in particular you see that if you set q equal to 0 so the parametric solution you get this number which is what we call the entropy at infinity which is not log 2 as it is in spin is a different number and then obviously if q becomes greater than 0 then you get an entropy which is reduced by the fact that you have non-zero magnetization and for these variables what we said is that with beta equal to 0 variables were really independent and with variables which are really independent not only you can easily compute the entropy of n variables you can also compute any correlation and so for these variables we were really independent and we said ok it's clear that if you compute any correlation this is the product of the mean values and other variables are not completely uncorrelated they are not completely independent but you can prove that this is correct up to correction 1 over n ok so variables that have only the spherical constraint are almost independent are very really correlated and so this is still true for any correlation at the leading order is no longer true only at the 1 over n correction this allows us to compute the 1 over 10 the energetic part very easily because now thanks to this for any higher correlation also we have that the average value of h at 0, beta equal to 0 is nothing but the Hamiltonian where we write m instead of s so you can simply write the sum of i1 smaller than ip j i1 ip m i1 m ip ok and finally we have the on saver correction term I will not do all the detail computation otherwise I will not have time to discuss the interesting results so I will leave you as an exercise but you can first write u0 which is the observable u with beta equal to 0 and this tells how to be equal to minus the sum over the ordered pupils of j i1 ip then you have s i1 s ip minus m i1 m ip you recognize this h of s minus the mean value of h and then you have the other term that you can write as s i1 minus m if I write it ordered I have to I have to be explicit m i1 times m i2 m ip all the other terms where the terms s minus m is put on the 7 index up to the p index so you have the other terms and the last one will be m i1 m ip minus 1 s ip minus m ip you have also what this is irrelevant about once you have this I will leave you as an exercise to compute u0 square which is equal to u0 times h use this one which is much much easier you realize that this is equal to n half 1 plus p minus 1 q to the p which comes from essentially this term and here you have m appearing p times one here one in each term and so in total you have p minus 1 terms of type m to the p so because when you take the square only the terms having the same j counts all the other you can prove they are sub leading and then you have one more term which is p to the p minus 1 that essentially comes from all the terms where you have 1s and p minus 1m and when you take the product with the Hamiltonian which are all s's s square is 1 and then s times m is like m square and so you get this term here a little bit of an exercise so finally we have our TAP free energy let me so you just have to take this term and this term and multiply by beta and beta square half to get the the TAP free energy for the spherical p spin model let me rewrite it here since we are going to use it to discuss the physics so minus beta f TAP of m is equal to 2 ns infinity plus n half log of 1 minus q plus is irrelevant but we define the Hamiltonian yes minus beta ok so here is a plus sum yes sum over i1 smaller than ip j i1 ip m i1 m ip and then you have this term with a beta square ah sorry there is a beta factor here and finally you have this term with a beta square half factor in front so beta square fourth 1 plus p minus 1 q to the p minus p q to the p minus 1 ok this is what we need in the rest of the lecture so now we assume that this free energy is the right one actually one should check that higher terms are irrelevant and this is easy because the couplings j goes like 1 over n to high power like p minus 1 half so the j cube term will be really relevant and you should check the PLEFCA condition so that this series is converging and so you can keep the first three terms and you get something meaningful and I will tell you later what is the PLEFCA condition in this case and I will tell you what are the the solution that we are counting but do actually corresponds to physical states so let me check yes, right, thanks obviously the notation is such that in Francesco denotes there are minus is f and so there is a minus t in front of everything and then if you take the intensive f you have to remove this this put factor 1 over n in front of this and so as usual there are different notation and one gets so let me first rewrite the free energy for the parametric state so if you take the parametric solution that always exists because you can either prove by taking the second derivative that n equals 0 is always a minimum of this free energy or if you want you take the first derivative so you write the TAP equation and you check that n equals 0 is always a solution to the TAP equation so the parametric solution is this one for any e and you can write the free energy and the free energy now you see I have removed the minus beta divided by n so I can write minus beta over because plus t as infinity because I essentially this term goes to 0 this term goes to 0 all this term goes to 0 and the only one which survives beta square over 4 then you multiply by minus t and you divide by n so you cancel this minus t produces minus beta 4 and from here minus t as infinity so this is the parametric free energy first observation is that if you draw it you have that the parametric free energy does something like this because here goes linearly with minus t times as infinity while here it goes like minus 1 over 4t as diverge this may seem unusual because we know that the entropy is the derivative of the free energy with respect to t so in general you have that the entropy is minus the f in the t the f I will try to use lower case letters for intensive quantities and uppercase letter for extensive quantities let's see if I manage and so you notice that in this part of the curve actually there is a negative entropy you don't have to worry about this because we are not using variables that take values in a discrete state space but we are using continuous variables and entropies for continuous value distribution are not positive defined so you can perfectly have a negative entropy and it's fine ok so we cannot say the parametric state will become unphysical because the entropy is negative no, not in this case we are going to I will try to convince you that the parametric solution is no longer the physical one because there is an ergo-easy breaking transition somewhere but you cannot say it from the entropy of the parametric solution ok so a nice property of this pure p-spin spherical model and this is the reason why I am using this model to show you the physics is that you can so you like to solve the equation which are the derivative of the TAP free energy equal to zero but this is not easy to count this solution so let's try to have a physical picture of what's happening by noticing that this term here is essentially very similar to the original Hamiltonian but in variables M that don't satisfy the spherical constraint but actually they satisfy a different constraint no constraint the sum of the squares of the variables M is equal to q which is not one in general so what we can say is that all these terms here if we do the following change of variables and we write M i equal to square root of q times sigma i now the sigma i is not the original variable S i is a new variable but it actually plays practically the same role then with variable sigma i satisfy the spherical constraint so I am asking the variable sigma i to satisfy the spherical constraint I can rewrite this term here like the original Hamiltonian in terms of these new sigma variables times a common factor which is actually square root of q to the p so you can rewrite the TAP energy thanks to these change of variables as a function of these new sigma variables and the overlap I will call this the overlap q and this can be written as q to the p half the original Hamiltonian computed on the sigma variables divided by N because ok I am using the intensive free energy plus a reminder that depends on q and beta but don't depend on this configuration of the sigma variables so this r is just this term, this term and this term ok it's a function of q and beta so what does it mean? it means that so I start with this saying that I am looking for the magnetization that identify the states and now I am realizing that I can express this magnetization as a direction because this is a vector of fixed norm not unit norm but it is fixed on a sphere on a hypersphere so this is the direction and this is the length of this vector what I am saying I am splitting the magnetization in a direction intensive in a length now what I want to convince you is that this TAP equation are equation for the direction that don't depend on the temperature let's see, so when I write now the TAP equation becomes the derivative of this with respect to q and with respect to the sigmas and when I take the derivative with respect to the sigmas with the spherical constraint what I get is that the derivative of F TAP with respect to sigma i equal to 0 is like the derivative of H which is the only thing that depends on sigma with respect to sigma i so these are equivalent equal to 0 but these are nothing that the equation define the minima of the Hamiltonian have an Hamiltonian in n variable, continuous variable if I am asking this to hold for each sigmas for each i such that the spherical constraint is satisfied this define the minima of the Hamiltonian so what I am finding is that the solution to the TAP equation at any temperature they always point along the direction that they identify by the minima of the Hamiltonian and this equation don't depend on temperature because the temperature is entering here so essentially what's happening in this model which is particularly simple is that you have an Hamiltonian defining in terms of continuous variables the minima of this Hamiltonian define some directions and the states that identify the TAP equation always point along this direction now, how large are the minimization so it depends on the value of q that depends on the temperature so we have another equation for fixing the value of q that will depend on temperature so the states will have higher minimization or lower temperature and lower minimization at higher temperature but the direction is fixed from the Hamiltonian this is very peculiar that doesn't happen in all the other systems but this allows us to give you a simpler picture of the physical system we are studying so I don't write now the equation for q I will do in the following so for the moment I am just interested in understanding this so all the states determine uniquely by the Hamiltonian so the following step is I am interested in counting these many states so instead of minimizing the free energy at any temperature I can just try to minimize the Hamiltonian and try to count the minima of the Hamiltonian how to do this is very well explained in the lecture notes by Cavagna and Castellani because Andrea Cavagna worked on this problem of counting the minima of the Hamiltonian for some time so I will just want to sketch you how the computation is done without doing it because it's a very long computation but I just want to point out what are the problems people face and how did they solve it so I want to count these I will keep it because I will use it maybe yes so I want to count minima of this Hamiltonian over the hypersphere the spherical constraints is really not very relevant at this point I just want to count the minima of that Hamiltonian and and in order to do this I will introduce a quantity which is very is really central in in statistical mechanics of disorder systems which is called the complexity so I will call n the number of minima and I will say that the number of minima is proportional to exponential n the size of the system times sigma sigma is called the complexity in a spin glass community or the configurational entropy in structural glasses community so the aim is to compute this function sigma so since I expect the number of minima to be exponentially large in n this quantity of sigma has a good thermodynamic limit look at this, this diverges and is meaningless to count it ok so in principle what you would like to do is you would like to do an integral over the now I will be very sketchy but just to give you so this quantity n in some sense is the integral over the space of configuration where I put in here also the spherical constraints the sum of the whole solution of the TAP equation so where sigma solution of TAP equations ok so suppose that you are able to pinpoint all solution of TAP equation you would like to do something like this and then exploiting the fact that if you have and let me call the since let me call by this symbol the derivative of h with respect to any variable ok so if you want to impose so if you want to impose a delta function of all the points where the derivative of the Hamiltonian is 0 so the h is streamed this corresponds to summing over all solutions of TAP equation this you have to remember that in doing this change of variables from h to sigma you have to put also the Jacobian that here the Jacobian of this quantity is the second derivative so this somehow is the Hessian ok so it's the derivative of h twice with respect to sigma i and sigma j so this just to say that when you want to do this notation so you want to integrate the whole space of configuration counting solution in practice you end up doing an integral over the space of configurations but you have this term here so you want to select all points where the derivative of the Hamiltonian is 0 but you also have this term so you have to wait each point by the Hessian actually this is the determinant of the Hessian computed on the solution so this just to say that it's not enough to count all points where the derivative of the Hamiltonian is 0 the gradient of the Hamiltonian is 0 because you have to count each point with a weight which is the determinant of the Hessian because otherwise minima and also maxima having different curvature have different weights in doing this integral and you want them to have weight 1 counting all the solution so this is a problem because writing the absolute value of the determinant is not easy so what people is able to compute actually is this integral without the absolute value because without the absolute value you can rewrite the determinant in terms of it's complicated but you can do it the problem is that if you do the computation without the absolute value what you get is something like this so you are counting points where the Hamiltonian is extreme weighting with the determinant of the Hessian actually you realize that this is not very meaningful to compute because this is a topological invariant it's called the Morse constant why? I just make you here an example for you suppose that you do the minus 1 every count a maximum plus 1 minus 1 plus 1 minus 1 plus 1 and the integral is 1 you modify the function you put another okay you get another minus 1 another plus 1 you modify the function you always get a constant so this integral is a topological constant it's invariant it's called the Morse constant so it seems useless to compute it what is the way out? the way out is to realize that the Hamiltonian has indeed many minima many maxima many saddle points is really a messy function in an n dimensional space with n going to infinity but luckily enough at low energy you have mostly minima at high energy you have mostly maxima so now if you constrain the energy you say I want to count this object that adding one more constraint I want all the solution of the TEP equation all the stationary point of the Hamiltonian having a given energy you add one more constraint so now you can call this the number of fixed point at a given energy fixing the energy so you fix the energy here for example you get a number which is mainly dominated you get an integral which is mainly dominated by minima and so this determinant of the action is mostly is almost always in particular in the large n limit positive defined and if you do the same computation at a too high energy essentially you get a determinant of the action which is almost always negative defined in that case it is not it is not clear because when you have samples you have many eigenvalues which are positive many which are negative and so is enough one eigenvalue change sign and the entire determinant change sign so let me say that the real computation is very well defined as long as you have strictly minima no eigenvalue of the action is zero or negative so as long as the spectrum of the action is positive defined this computation is fine and indeed people did the computation and get a reasonable result ok so this just to say you what is the way you can do the computation I will tell you what is the outcome so you can just keep in mind the outcome is what you know what are the difficulties in doing this computation so the outcome is essentially the following if you define sigma of E like the determinant limit of 1 over n the log of the number of solutions of energy E times n in other case E is the intensive energy what you get is that sigma of E is this function I have read it for you but just to show you that is not so complicated is one half this has been first computed by Grisanti somers in 95 you find all the references in the the two lecture notes this minus log of p z square half plus p minus 1 over p z square minus 2 over p square z square plus 2 minus p over p where z I define it here is defined as minus E minus square root of E square minus E threshold square divided by p minus 1 and E threshold is defined square is defined as 2 p minus 1 over p and so if you plug it here you realize that also z threshold is equal to square root of 2 over p p minus 1 ok so apart from you can use your prefer plotting program and plot this function and what you realize is that if you plot this function what you get is indeed something which has this shape if you plot this sigma of E you get something which has this shape so the energy range now in this expression is between E minimum and E threshold first of all we realize that if E is larger than E threshold the argument of the square root is negative so remember that all energies are negative so it's worth writing this axis here so you don't get confused the energies are negative so E threshold is the smallest in absolute value among all the allow energies so this expression here is positive as long as the absolute value of E is larger than the absolute value of E threshold so it must be below E threshold because all energies are negative and then if you plot this expression you realize that at another value of the energy 0 so it becomes negative and having a negative entropy doesn't make any sense because you are counting minima so if the entropy becomes negative it means that you have an exponential is more than minimum at that energy level so in the larger end limit you do have minima of the energy function only in this range and the function sigma is monetarily increasing actually reaches a maximum here and so what you see is that you have many more metastable states than ground state because now if you see the physical picture what comes out is that your Hamiltonian is a disordered Hamiltonian and what you are getting a ground state at value a minimum but there are exponentially many more metastable states this will have some effect on the dynamics on the thermodynamics because now you have to live with exponentially many minima so we started from the Q device model having two minima and now we are ending up with a model which has exponentially many minima of the free energy and we have to understand now what happened to the physics of a disordered model that has exponentially many minima of the free energy one more information that comes out from this computation done only on the on the energy function which is very interesting is that since you are counting stationary point and you are computing the action on any stationary point and you can also ask yourself how many negative eigenvalues the spectrum of the action has because we like to be sure that we are really counting minima between here and here and we are not mixing minima with with saddle, with maxima indeed another really very nice result that you can find in a paper by Cavagna Giardini in Paris in 2001 is that if you look at the at the number of the fraction so the fraction of negative eigenvalues of the action as a function of the energy E and here we have E threshold that the number the fraction of negative eigenvalues goes to zero exactly at the threshold energy this means that this complexity is counting minima because below E threshold the mean number of negative eigenvalues is zero so it means the action is positive defined you are really counting minima but as soon as you go up you have what we call topological phase transition topological phase transition means that with high probability if you take a stationary point below E threshold it is a minimum and if with high probability in the thermo-limit if you take a stationary point above E threshold is not a minimum has some negative eigenvalue so it's a saddle has some positive many positive this very complex energy landscape which is made let me draw a drawing here which is something that you can try to keep in mind because actually this the typical picture that actually happen also in other more complicated disordered models the picture you need to have in mind is this one you have this let me say that this is the the space of configuration whatever it means is a n-dimensional space I'm drawing a two-dimensional blackboard and this is the energy here are the threshold energy value so what the computation is telling me is that below this threshold energy value I have with high probability if I pick up a stationary point but the number of minimum the number of minimum is very very different so I will have many minima here much less here and even much less and even we will see when we study the overlap which says how large is the minimum also much tighter narrow minima decreasing the energy to the minimum value above what we get is that with high probability if you pick up a stationary point there is no longer a minimum it's a saddle and so here you have to imagine to have saddles like this and the saddles if you go higher in energy they have more and more negative eigenvalues so here the saddle will have more negative eigenvalues so now suppose that you run the dynamics even at zero temporal dynamics so you take your system this is an energy function in continuous variables and you compute the gradient and you go down in energy what do you expect well here even if you come close to a saddle which is a point where the gradient is zero so in principle you stop but it's very unlikely that you really go on the saddle you go very close to the saddle but you can leave the saddle quite easily the number of negative directions are many so you come close to here and you leave the saddle very easily then you go here at this level eh you already spent a lot of time because the number of negative directions becomes very few and so you take a lot of time to leave the saddle ok once you are approaching the threshold states it takes more and more time to leave a saddle the number of negative directions becomes very very scarce ok very very slow and eventually what do you expect you expect to stop the threshold energy because if you want to go below now you enter into a minimum and leaving a minimum takes an exponential time so if you are interested in looking to what happened to a system only on times that grows say linearly with N or polynomially with N but not exponentially in N because times exponentially in N cannot be explored in a laboratory because systems have a huge number of degrees of freedom so times which are exponentially in N are unreachable on human time scales and so you expect a very simple dynamics slower and slower in this part because the number of decreasing the direction going down becomes scarcer and scarcer and eventually the dynamics get stuck at the threshold energy we see that this picture that is only suggested by the topological transition that you can find by counting minimum of the Hamiltonian is actually very robust and not now because but next lesson we will put the temperature we switch the temperature on so we go from the energy to the free energy and I try to convince you that all this picture is preserved at finite temperature and finally we will see that the dynamics is actually going to the threshold states in a system where you can actually solve the dynamics so before leaving since we still have few minutes let me just start switching on the temperature so now maybe it's time to stop let's do it next lecture ok so let's stop here and we switch the temperature next lecture ok yes