 Zdaj. OK, zato nekaj... ...zapravljati, da je taj zelo. Tudi, da se zelo posledili, da vendem je tudi veliko več nekaj, da se te, da se vse zelo, načinjamo vstaj, je zelo sekma f. If you are able to compute this, then we saw how to do thermodynamics in the presence of an exponential ilargen number of states. Now I just gave you one expression for sigma, which was not derived explicitly. I told you that the computation can be done. So today what I would like to do is to show you a method to actually compute sigma via replica calculation, so we are going to use replicas today. And we will do the computation in a way which was suggested by Renimene Soni in 1995, kaj je pričo. Proste, je, suposte, da težete m kopiče s systemem, tako, s replikacim systemem, izgledajte, replikacim systemem, izgleda, na m kopiče, kaj je, izgledajte, replikacim systemem, izgledajte, izgledajte, m kopiče, s systemem, izgledajte, s systemem, izgledajte, kaj je, izgledajte, izgledajte izgledajte, izgledajte, izgledajte, izgledajte, izgledajte, s systemem, izgledajte, ima m kopiče s m kopičem, i, s vrkama izgledajte, izgledajte, s, i, s, i, vrkama. Zato vidimo, kaj je vzgledajte, s m kopiče, s vrkama. Zato vzgledajte, s kaj je vzgledajte, izgledajte, replikacim systemem, izgledajte, izgledajte, izgledajte, izgledajte, izgledajte, izgledajte, s vrkama v zeskade stato. Why do we need to do this? Because if you think about the main difference between this kind of model having infinitely many states in random directions in the Kunival model, which has only two states in well-known directions, in the Kunival model I can force the system to be in one state in vsega vsega zelo. Nekaj ne sem v tem vsega vsega vsega vsega, tako sem ne vsega vsega vsega vsega vsega. Vsega je zelo, na različenju, vsega je, da vsega vsega pide vsega cipa, zelo se začala, da se začala na vsega cipa, zelo v tudi svojo in. The second scenario is like self sustained construction, where using replicas, I favor the situation where replicas are in the same state. In this way, even if states are in random direction, this term should favor the situation where the replicas are in the same state. Všeč, kako bomo zelo vse pošličeno? Vse pošlične dobrali je malo, da se začeli pošličeno začeli pošličeno v semom veči del, bo sem bilo način, da se začeli za tvoje, da bo začeli za vse način, da se začeli za vse način, da se začeli za vse način, da se začeli za vse način, z štev njah se našli in ne zelo. Zato ne všli vč serijo odmarati. Zato njo vse je odmarati. Zato ne všli všli všli, da epsilon je ni vesitmann ali kama imače. Tako da v každeznih z Sono vzela tudi replikacja, načo je nekaj mora v povrstavne energije, v同 stati. Vsledajte, da je tako changer in energija, da v sej stati zveči, in tako da je vzelim, nekaj tudi replikacja izgleda, vzela je na razlih 0, if they are in the same state. So this quantity here is roughly zero if the two replicas are uncorrelated and is q if they are in the same state. OK. So this quantity is as two very different values depending on whether the replicas are in random states, randomly chosen states or in the same state. Vse složemo, da bin mojiči vz stealing z dalej, veselaj nam je bogli štešnji, nekaj ne izgleda. I ta slaj se je štešnj mjelo bave, da se jen tvoj replikas in zelo jeselje. In po glasbeni, kad imelančjati epsilon v druh homenkeje, je to pomembno je, da bo applianceshit, bo je tem način stavil, da so hovozjanje reacity in nič nič se sprača pre kidney. Dodaj način je začal modifirati tukaj što je z potenkem tukaj posled, da je to nekaj repotenek, kot z dela Vey. vs. n, sorry, delta e is of the order of epsilon times n, and so is much larger than one. So you see that, choosing the right coupling, you can actually do this trick of modifying the interactions such that two replicas are put in the two, or actually m replicas are put in the same state because it's highly favorable to be in the same state, without changing the free energy landscape. And so you are just counting free energy minima, which are the states, by this trick. And so if you use this construction, what you get is that now, so let me call this hm just to remember that this is the replicated partition function. Now, from the replicated partition, from, sorry, the replicated Hamiltonian, you can compute the replicated partition function, which is nothing but the partition function of this replicated system. And let me call this zm. And again, if, sorry, if these two conditions are true, when I do the computation of the replicated partition function, I can write this as the sum over the states with the exponential of minus beta n m times the free energy of state alpha. Why? Because for what I just said, the m replicas are forced, they strongly prefer to be in the same state, but the states are unaltered, so I have to count the same state as in the uncoupled system, but the m replicas will end up all in the same state. And so I just have a factor m here. Now, we do the same transformation we did yesterday by transforming the sum over the states as an integral over the free energy, which is a parameter labelling the states. So this can be written as, again, the integral between fmin and fth threshold, df. And here you have one term, which is sigma of f. But now this term is different, is beta n mf. There is this difference with respect to the completeness of yesterday. We had n equal to one. Now we have generic m. What is good of this is that once you introduce, and again you can solve this by a subtle point and say that this exponential of n times f star minus beta m f star, where f stars now satisfy the equation f star is such that solves the derivative of this with respect to f equal to zero. So beta m is equal to the derivative of sigma with respect to f. So if we define the replicated free energy, capital phi. I use the notation of the Zamponi lecture notes, m of t. This is the generalization of the free energy to the replicated case. So we just take the same definition of the standard free energy t over n, the log of Zm. And thanks to this expression is just minus t, the square bracket. So this is just minus t. Actually, since minus t cancels minus f, so let me write it minus this m f star minus t. Sigma of f star, where you have to remember that f star is a function of m and t. Because this depends on m and t, beta is one over t. Also sigma, you have to remember that it can be a function of beta, so beta enters is not just a function of the ratio m over t, because the temperature enters also in the expression for sigma. So what is good about this construction is that now if we are able to compute phi, I will show you that we can reconstruct sigma. Because what we are doing is essentially a Legendre transform between sigma of f and phi of m, where m and f are conjugated variables. So we are just doing a Legendre transform. So if you are able to compute phi of m, we can compute sigma of f. In particular, from the definition, we see that once you have phi of m t, you have to think that t is a fixed number. You are computing the function sigma of f at a given temperature. So t here, you have to think it as a parameter while we are working with the independent variable m and we want to reconstruct sigma of f. So from this, you easily realize that f star of m and t is just given by the derivative with respect to m of this expression, is the derivative of phi with respect to m. You may notice that actually m also enters in f star. So in principle, I should derive this expression. So let me do it completely. So I can do the explicit derivative of this term. So let's do it. So I will show you that we get f star. When you do the derivative of phi with respect to m, you get f star, but then in principle you also get the derivative of phi with respect to f star, which is m minus t, the derivative of sigma with respect to, so is m minus t, and the derivative of sigma with respect to its argument, computed in f star, times the derivative of f star with respect to m. But you see that as usual in the La Jean transform, this is zero because it's the definition of a star. So it's enough to take, when you have a La Jean transform, you can get the value of the conjugated variable, in this case f is conjugated to m by just doing the explicit derivative with respect to m. And then sigma, well, sigma is easy to get because you just have to bring this on the other side and you get that sigma of m and t is just given by, you have to bring that beta m f star, m and t as usual, minus beta phi of m and t. So with these two, you can plot, or you can compute, you can plot sigma of f once you know phi. You can do, for example, a parametric plot as a function of m. So you have phi, so you can compute this derivative, this expression here, these are two expressions, not m, t is fixed, and then you do a parametric plot, bar in m, and you can draw sigma of f. So this is just to convince you that once you compute the replicated free energy, your job is done. So you can reconstruct sigma of f, and so I will show you, you can understand what is the dominant value of free energy that dominates the thermal dynamics. Ah, one more important thing. Yesterday I used a very bad notation, so I will repeat something I did yesterday in a better notation. Ah, because, ok, so now the question is, I will show you what are the values of m which are relevant in order to reconstruct the full curve. In particular, yesterday we saw essentially that for m equal 1, we get the value of the free energy that dominates the thermal dynamics, and below a certain temperature, this value of f goes beyond, outside the valid range f mean f threshold, and so you have to, the thermal dynamics is dominated by f mean. Now, if you notice, this new equation allows you to compute the replicated partition function even below tc, because if you make m smaller than 1, you can still solve this equation for very low values of temperature. So, let me repeat something I already said yesterday, because I used very bad notation. Yes, I said that one interesting point of this function, sigma of f, is the way it goes to zero, so essentially you can show, well, in general, sigma of f goes to zero linearly here, and the slope is very relevant for the thermodynamic properties of the model. The problem is that I use a very bad notation for the slope, so I prefer to write the formula according to the notation of the nodes, otherwise we get confused. So, if you approximate linearly the function sigma of f close to the lower band edge, so close to the value where it goes to zero, you can write this in general as f minus f min. I just want to change the notation. Yes, I call it ms, but it was a very bad idea, because ms today in the nodes is reserved for another quantity. So essentially what I told you yesterday that this coefficient is identifying the temperature where you have the condensation, so essentially lowering the temperature, the point where the slope is equal to 1 over t moves down until you reach this point. At this point the slope is given by this quantity, and so at tc, when you have the condensation threshold, you must have that the complexity goes to zero, so you are at f min, and at f min the slope is just given by sigma zero of tc. And so at tc you have that sigma zero of tc must be equal to 1 over tc. So this in general the equation that yesterday wrote with ms, is another quantity, so it is better that we use this notation. So in this way if you know the linear behavior of the complexity close to the point where the complexity goes to zero, you can compute the condensation threshold, the critical, thermodynamically critical point. Now if you repeat the computation with the m parameter, you realize that in general depending on the temperature you have to vary the m parameter in order to visit all the sigma of f curve and so in order to compute it from a value that essentially brings f star to f min to a value of m that brings f star to f threshold. So you want to vary m. So in general we see that since this equation is telling us that the parameter m should be such that the derivative is better times m. So in general if you want to be here the slope should be better times m. So you see if you want to go to this point you have to make m larger. If you want to go to this point you have to make m smaller. In general you need to use a range of m parameter that belongs to a finite value where we call say m threshold the value that you need to be here and we call ms the value that you need to be here. And in general this will be larger than this one. So in general you will need to vary your m parameter in this construction in this range. We will compute explicitly this value. I will show you how this value changes with the temperature in general. I prefer to show how they change when we really solve one model. But just keep in mind m should be vary in a range and outside that range the construction is not very meaningful because either you are computing a complexity which is negative or you can show that if you insist in this construction for m which is smaller than m threshold and so you go beyond this point actually you come back in a branch which is unphysical. So you have to be careful that this construction is fine but as long as you have a sigma which is positive and a sigma which has the right curvature so in general you want the second derivative of sigma with respect to f twice to be negative. So in the right range otherwise if you use it in a broad range you may find negative sigma or sigma with the wrong complexity because this you can use m which is any real number. And notice that we started the construction using m being an integer number in the sense that we said we take m replicas and now if you see what are the values of m that you need in order to compute this curve these values may be even smaller than 1 actually they will be most of the time smaller than 1. So one more caveat we are deriving an expression assuming that m is an integer number because we started the construction with m real replicas we derive an expression an analytic continuation so we are using the same expression which is derived for integral values of m to any value of m this again is a delicate point but let me say when you work with replicas it's always like this in the sense that you compute something on the integers and then the powerful of this method is that the analytic continuation is still meaningful then you have to interpret but it's still meaningful so this just to convince you that once we compute this we are done now I will show you how to do the explicit computation in one case so you really see how we can do the computational of the replicated partition function I will do it in the case of the spherical again because this is the one that we already solved with TAPA question so what we want to compute is this phi of m and t which is minus t over n the log of the replicated partition function notice that this replicated partition function still depends on the disorder we have to average over the disorder at a certain point and what we discuss at the beginning of these lectures is that is worth taking the average of the log not the average of z so we are going to take the average of the log but again since the average of the log we don't know how to do it we use the replica trick so in practice what we are going to compute is minus t over n the limit of n going to 0 the derivative with respect to n of z so the replicated partition function with m copies to the power n average this in practice means that you have m times n replicas because what you are doing you are taking a system of initially m replicas coupled and then you are replicating again so you are doing m times n copies what you have to think is that you have this group of m replicas which are coupled among them ok so these are coupled among them with epsilon and since you are not able to take the average of the log what you are doing you are again replicating it but the couplings are always between the groups of replicas different groups are not coupled they will become coupled when we take the average over the disorder ok so we have n groups of m replicas coupled replicas weekly coupled and now essentially we want to compute this average of the of the m times n replicated in order to do this we use an index a which runs between 1 and m times n but we have to remind that replicas are not all equivalent these are n groups of m replicas and within a group there is a very small coupling while between groups there is no coupling ok because we are replicating the original system which was made of m weekly coupled replicas so if you take this index you have the z m to the n average is just the integral over the sigma 1 the sigma m n exponential of minus beta sum over a 1 to m times n the Hamiltonian of sigma a where in the integration I also put this spherical constraint so I am integrating over the sigma variables according to their measure so if they have a spherical constraint each system has a spherical constraint everything is here and all this needs to be average this average we are able to take it because if you remember the Hamiltonian is linear in the J so we can do this integral and so if we do this integral this is proportional I am interested in the exponential term in capital N of this expression so often I will put proportionality because I am forgetting some pre-factor I am interested in the leading exponential term so this expression is proportional to the integral over the all the spins of all the replicas A is an index running over all the replicas and I is usually an index running over the spins of the each system and now in order to take the average over the disorder what you have to do is the product over all independent couplings so you have to integrate over the J i1 ip each coupling is independent and distributed according to a Gaussian measure so here you have the Gaussian term minus is J square let me avoid bringing all the index here J is this J and the variance is such that here you have mp minus 1 over p factorial you remember J square it was 2 p factorial over mp minus 1 and the Hamiltonian is the same coupling and the sum over all the replicas because the system is replicated okay sigma i1a ipa so this is the Hamiltonian when you take the product over all the J this is exactly the replicated Hamiltonian for the moment I am doing the computation set the epsilon to zero so the computation is a little bit shorter at the end I will tell you what is the additional term that you will get keeping also epsilon different from zero and you realize that this integrates a Gaussian integral okay so you integrate over J and you get apart from the prefactor something the product over all the pupils of the exponential beta square p factorial for mp minus 1 the sum and here so this when you do this Gaussian integral you get okay this square divided by this and also this square so if you square this you get a double summation over a and over b so a and b from 1 to n times m sigma i1a sigma i1b sigma ipa sigma ipb okay so what happen essentially in taking the average here you have a replicated system where each system was independent you take the average over the disorder over the coupling and what you get here is a system a measure where different replicas so this is what usually happen when you do replica calculation by taking the average over the disorder you are essentially coupling different replicas even replicas that at the beginning they were not coupled because in this setting we have a subset of replicas that have a real coupling epsilon and other subset of replicas that were uncoupled at the beginning now taking the average over this disorder you couple all the replicas among them okay if you take this product you bring it in the exponent so it becomes a sum and remember that p factorial times the sum over ordered indices is equivalent to the sum over unordered indices you bring this there you take the p factorial you convert in a sum over unordered indices you can rewrite this term as the integral over all the variables the exponential of beta square over 4 and p minus 1 and now the sum over this p indices you take the sum over i1 and you sum this term all together and you write this as the sum over i from 1 to n sigma i a, sigma i b to the p so you see you repeat p times the same sum so the sum over i1 is the sum over i1 of sigma i1a, sigma i1b and you say you do the same for i2 up to ip so you are doing this sum p times ok and finally you just put a factor 1 over n here and this cancels the mp minus 1 and you get a factor n there ok, so we got as an effective model of interacting replicas and what we notice is that I am at the action so we have to integrate this over the variables but the action mainly depends on this quantity here and we give to this quantity the name q of sigma a sigma b and this is the overlap that we already we already met the self overlap but now these are generic overlap what does it mean essentially is the amount of similarity between two replicas essentially is a scalar product between the configuration if they are very similar actually what you can say is that the spherical constraint implies that the self overlap is equal to 1 because if you put the same index here essentially this is the spherical constraint the sum of the squares of sigma is equal to n so you divide it by n and you get 1 so what we get is an action over which we have to integrate that depends only on the relative overlaps between replicas so it depends only on how the n time n replicas are similar ok among all these overlaps only one is fixed by one this is for any a n times m overlaps are fixed by by the spherical constraint ok so now we have to compute this integral this is not easy we take some some algebra erase here ok so now we want to encode so you see that these are an action which depends this quantity of order 1 because it is a scalar product between two vectors of size n divided by n so this order 1 everything is extensive in n so we hope to be able to do this integral again by a saddle point integral but we have to integrate over this variable and once we arrive that the action depends only on the overlap we would like to encode these numbers in a matrix which depends on two indices so it is very comfortable to put all these numbers in a matrix and then make a change of variables from the sigma to an integration of the elements of the matrix so we want to change variables from sigma to what we call QAB which are essentially these numbers here ok so in order to make the change of variable I can introduce in the integral a delta, a delta function so again I can write that integral this you have to as the integral of all the sigma variables now we can introduce in order to change variable the following delta for each a different from b from 1 to mn we introduce this quantity QAB with a delta forcing QAB to be equal to 1 over n the sum over i of sigma i a sigma i b ok I want to make a change of variable from the sigma to this matrix QAB the matrix QAB as the following properties that diagonal elements are equal to 1 and is symmetric because if you exchange a and b this is symmetric and then the action now can be more easily written as n times beta square quarter the element QAB to the p so you just have to take the sum over a and b of QAB to the p so you see that it is irrelevant but if you want to follow the nodes let me use this notation it is practically the same so now since changing this factor n it just changes a pre factor and so you see that this part now is easy because we can rewrite this as an integral over Q by integral over Q means the integral over all the elements of the matrix so this this dQ you have to intend as the product over a smaller than b the QAB since the matrix is symmetric on the diagonal you have once the only independent parameters are those with a strictly smaller than b the action which is exponential of beta square quarter the sum over a and b QAB to the p and now all what remains I will call it the Jacobian which depends on Q if you look at this expression is just the integral over all the sigma variables of this delta function enforcing the equality between QAB and the scalar product between the sigma so this the product over a smaller than b delta of n times QAB minus and let me introduce this notation for the scalar product so I don't have to rewrite the sum every time so you see that this part of the action is pretty simple and it corresponds to the energetic part in some sense the fact that variables interacting peoples produces in the effective replicated action these terms Q to the p but now you have to compute this Jacobian which essentially will give you the entropic part so in some sense this Jacobian is telling you what is the volume in the space of configuration corresponding to a given matrix Q you are imposing a set of constraints in this huge space so remember that these are m times n vectors of length n so this you are integrating in a huge dimensional space because you have a is an index running up to n times n index running up to capital n so you want to do an integration in this space according to this constraint so you fix the matrix and so this is giving you the entropy in how many so actually not how many ways because that is fine for discrete value what is the volume corresponding ok, so let me I don't think I will be able to do all the computation in detail anyhow, I am following the also in case you find the computation so so the simplest way to do this computation is the following is to notice that the matrix QAB is real and so it can be diagonalized and it can be diagonalized and so you can write the matrix Q as lambda transpose a diagonal matrix times lambda where lambda is an orthogonal matrix so you satisfy that the inverse is the transpose and it is actually the matrix of all the eigenvectors of the matrix Q while D is a diagonal matrix where you have all the eigenvalues on the diagonal ok, once you realize this so remember that Q is a matrix so remember what are the sizes so remember that Q is a matrix NM times NM ok and the same holds for for the others ok, so if you realize this then you can rewrite essentially you want to solve you want to impose a constraint that looks like lambda transpose d lambda is equal to so you see that for each element of the matrix you want to impose a particular scalar product between the sigmas so if you write a new matrix sigma which is made of all the vector over which you are trying to integrate so each one is a vector of size N you take all the vectors that you put in a matrix you can rewrite this equation here as sigma transpose sigma so essentially each element of this matrix is this scalar product you are taking the external product of you are taking the scalar product of each element so in this matrix form if you multiply by lambda on the left and suppose on the right this cancel because it's an orthogonal matrix and if you call you call this suppose and you call this new then you realize that imposing these constraints is equivalent to so you want to solve this integral what you can do you can change variable from sigma to new which is defined this way and since lambda is an orthogonal matrix the determinant of lambda is one so changing the variable from sigma to mu as a Jacobian which is one because it's the determinant of the matrix lambda so you can go to variables mu it's like doing a rotation in this space in this space of in order to have these constraints that now becomes much easier because now the constraint becomes a diagonal matrix is equal to the product mu transpose mu and so the constraints is much simpler because you have almost all the constraints which are zero so they are imposing the product of two mu equal to zero and so for jq after the change of variable from sigma to mu can be written in this way you have to integrate over all vectors mu which are related by this expression to the vector sigma and now you have to impose the diagonal constraint so this diagonal constraint now reads now you have d and let me call lambda a the the eigenvalues so it's a diagonal matrix with elements lambda 1 to lambda nm k minus the scalar product between mu a and mu a and then you have all the others all the other constraints which is the product over a smaller than b of delta of mu a scalar mu b so you see that by doing this rotation now the computation is it looks and it is much simpler indices a and b runs always over all replicas and we have to do one further simplification which is splitting this integral which is still an integral over a huge space by passing to polar coordinates so you can write this vector mu a as a absolute value times avertso the norm of mu a is 1 so essentially this is a direction and this is the absolute value of the vector if you do this change of variable let me jump let me skip essentially you do this change of variables and you rewrite is always proportional product over a you have the integral over the absolute value mu a as usual when you do when you go to polar coordinates you have a factor m minus 1 because these are vectors in n dimensional space as the sigma r and let me forget about the angular part because you can prove that the integral over the angular part does not depends on the eigenvalue of q but I've write it but we'll forget it in a while so now this term here when you do the change of variables is a delta function between lambda a and mu a square and if you want to convert it in a delta function of a mu a you just have to to use the properties of the you have to remember just this delta of f of x if f of x has a root in x not say the unique root you have to remember this can be written as the absolute value f prime in x not you remember this is a rule for so since here you have a quadratic function of mu a you can rewrite it as 1 over square root of lambda a of mu a minus square root of lambda a here you get 1 over 2 mu a and then you substitute mu a with the value and the factor 2 you don't care everything goes here in the exponential term and then these other terms again they have some mu a times mu b times the scalar product and again you take this mu a mu b you bring it outside the delta function say they go down as here and so you get something which is square root of lambda a lambda b and here you have a delta function between the the angular parts that we don't care so the integration of the angular parts you can prove that it does not depend on the eigenvalues of q which is what you are interested in one expression which depends essentially on the spectrum of q on the eigenvalues of q so forget about the angular integration and you end up with this expression that now is very simple because you see you have to integrate over every mu a using this delta function so it's trivial you just have to substitute to mu a square root of lambda a so what you get at the end is j q proportional to the product of a from 1 to m times m of where it is here lambda a to the power n minus n m minus 1 divided by 2 and just put in together all terms lambda a so this one is the main term so this is much larger than small n and small m so this essentially is proportional to the product of the eigenvalues is just the determinant so this is the determinant of q to the power n half because in the larger limit these are finite numbers while these diverges so this means that in normalizing our expression for the replicated partition function now we have it explicitly is given by so z m to the n average is proportional to the integral over the q exponential of n let me call x of q x of q as energetic term which is beta square quarter sum a b q a b to the p in the case of the spherical p-spin plus you have to raise d to the exponent and so you just have one half the log of the determinant so you derive the explicit expression for the action now we have to extremize it we have to maximize it or minimize it over the elements q a b so we still have to do the integral but at least we know what is the action now before going to the explicit computation that we do after the break let me just comment that we are computing a function an action over a matrix which is of size n times m n times m and we are interested in the limit n going to zero so this again is one of those fantastic things that happen when you do replica calculation so you compute an action over a matrix of this size you do the computation and at the end you have to send the size of the matrix to zero in some cases you can prove rigorously that the analytic continuation is the right result but in general you cannot prove it so you have to be careful and actually the best way to realize that so first of all even clear that you are computing the you are computing them just want to check the sign is not even clear that you are computing the maximum because you are computing an integral over well m n times n m minus 1 divided by two variables and when you send n to zero this becomes when this is smaller than one it becomes a negative number so it is really not clear what you are doing so you have to so you have to assume that in some sense what you compute for any integral value of n and m can be safely continued analytically to any value of n and m and in general this is a delicate point if you want to understand whether the point that you are choosing saying that it dominates the integral is the right one in principle you have to take you have to compute the action so you have to take the second derivative of x with respect to qabqcd you have to check whether that tensor because now is a tensor 4 indices is positively defined so you have to compute all the eigenvalues again you have to do this computation for finite values of n and m then you send n to zero and you check what happened to the eigenvalues and in general you can recognize the region where these eigenvalues are all of the right sign and when one of the eigenvalues becomes zero so this is the right way of doing the computation in particular for the spherical peaceful model we are studying one of the eigenvalues becomes zero exactly at the threshold states so the part of the curve sigma of f that we exclude from the computation because of the wrong concavity it actually can be excluded also on the basis of the analysis of the derivative of this action replicated action with respect to q and a last comment is on the fact that there is one more delicate point that actually the function phi of m and t that we are computing in principle is defined as the limit n going to infinity 1 over n n going to zero the derivative with respect to n of z m to the n average unfortunately when you plug this expression here the only way you know to do the integral is to assume that capital n is very large so actually what we are computing is a result that you obtain by inverting these two limits because what we know is how to take the large and limit of this expression so do the saddle point so what we actually compute is that we first do the saddle point in the integral and later we do the derivative with respect to n and the limit n going to zero so this again is a thing that from the mathematical point of view is not obvious that it is correct because we are interchanging two limits that may not be interchangeable but it is the only way we know how to do the computation otherwise we are not able to do the the computation so the regular calculation has been proved rigorously in many settings especially fully connected models all the Parisi formula has been proved so that solution which has been obtained by an analytic continuation from integers to real value interchanging the two limits are rigorously proved but when you use this approach in other context you have to keep in mind that you are doing something that from a mathematical point of view is a little bit nasty so let's do five minutes break and then we come back so we got this expression now we want to extremize it over this space of matrices obviously if you want to do it something in general is very hard so let me let me choose well first of all let me say that if you add the epsilon coupling here you have to put a term which is this the sum of the groups over the blocks sum over a and b belonging to the same block of q, a, b ok so if you redo the computation will also the coupling well you see essentially you are saying that within each group each block you are putting a rewarding term because this is the action a rewarding term proportional to beta epsilon according to the overlap q, a, b between replicas in the same group which is reasonable and now we want to extremize this expression but we don't want to take matrix which will be very, very difficult so let me make a choice which is physically suggested by the physical construction that we made and so the choice for the matrix q will be the following the matrix you remember is n times m times n times n so we can take for sure one on the diagonal because we have the spherical constraint you have to remember that the physical meaning of q, a, b is 1 over n sigma a scalar sigma b and even at the maximizer this will hold so for a equal b must be 1 then if you divide this in groups of m replicas and let me suppose that we have 2 groups of m replicas each so this case is for n equal 2 we replicate twice so if you want to break the symmetry among all the replicas the most simple thing that you can do is to assume that you have no overlap at all between replicas in different groups and a non-zero overlap in q between replicas in the same group why is this? because in a group we are putting a coupling so we expect this replica to go in the same state and in the state I hope I expect to have a non-zero overlap because even from the TAP solution we know that the self-overlap system with itself or two copies of the same system in the same state is non-zero while on the contrary if you have two replicas belonging to different groups they will end up in states which are uncorrelated and so I expect q and b to be equal to zero so this is an answer that as you see reduces enormously the space where you have to maximize that function with m squared n squared divided by two parameters to a space with just one parameter k, which is q so this simplifies simplifies the computation so if you want to write explicitly this action as a function of q we just need to compute this term which is trivial and the determinant of this type of matrices which is not difficult so in partime I let you as an exercise compute the determinant of a matrix of this type here so one on the diagonal then q of diagonal and easy because this can be written as the matrix of all q's so let me write this for the matrix of all q's plus one minus q the identity matrix this is a very simple exercise this matrix has one eigenvector which is m times q sorry, the size of this is m so this has one eigenvector which is m times q and all the other which are zero because all the rules are equal and the eigenvectors of these are all q and so the eigenvectors of this matrix is one minus q plus mq and then one minus q you take the product and you get the determinant so the determinant of this matrix redo it more carefully as an exercise but is one minus q m minus one times one plus m minus one times q then this matrix block matrix the determinant of the rule matrix is just n times the determinant of a sum matrix ok, so this is easy and so once you plug now this term also is very easy I will write it here so the sum of a and b of q, a, b to the p is very easy because this expression is the same for any rule and as you see when you go from one row to another one you are just permuting the elements so this quantity is row independent so is just n times m which are the number of rows times what you get in a single row what you get in a single row is one, one m minus one q to the p and then the rest are zeros so once you put all together this you take the determinant of the matrix and you put everything here you finally get the following expression for x you get x of q is equal to I put minus beta n m in evidence and I will call this phi one rsb which depends on m, q and t I will write explicit expression for that and then you have this coupling term that essentially we will disregard in the following m, n m sorry n, m, m minus one because you see the number of indices is number of groups n number of pairs of indices m, m minus one times q and the explicit expression for the one rsb is the following so phi one rsb depends on m, q and t and is given by minus t half times is written in the notes so you can even avoid writing it one plus m minus one t to the p which you see is this term here plus m minus one over m log of one minus q that you see is essentially this term here when you take the log you have minus m minus one divided times log of one minus q and this m is the one we put in front here and then one over m we put in front here log one plus m minus one times q which again is the log of this term here so it's very simple to get it and the function the function so and if if you realize that the the relation between phi capital phi the replicated free energy and this one rsb replicated free energy just a factor m you can immediately recognize that this phi of m and t is just you take the delivery notice that everything is linearly in n this particular this model is particularly simple because the n replicated system are not interacting we are using an answer they don't interact so everything is linear in n so under these answers you can even do the computation for just one subsystem it would be equivalent in general this is not true in more complicated models you have a matrix qb which is full of non-zero elements so all the n replicas interact and you really have to take the derivative with respect to n and send n to zero here is very simple because everything is linear in n so you just have to cancel n and get the result which is m times phi 1 rsb minus epsilon m m minus 1 q sorry I forgot the argument of this q star because this was the expression for x now we have to take the saddle point and so even here we have to write this phi must be computing m q star and t because we have to take the saddle point and the saddle point fixes the value of q and so the replicated free energy is this 1 rsb free energy which plays the role of the of the Landau free energy that you have in the Curivize model so in the Curivize model you have a Landau free energy with a double well and then you fix the value of the magnetization by saddle point and here we have a dependence on q you fix this by q star of m and t is such that the derivative of all this expression and remember that epsilon is eventually sent to zero so essentially is the derivative of this expression with respect to q equal to zero so this is such that the derivative with respect to q of phi 1 rsb of m, q and t is equal to zero now we discuss the epsilon term so I want to forget about this epsilon coupling but before doing it I want to understand what would be its role so we want to make epsilon very small so what does it mean as long as there is a unique solution for this equation you can send epsilon to zero and you get the right solution but suppose that this solution has multiple solution so you have maybe three solution as usual when you have bifurcation so it has three solution among the three solution which one should I select and usually two solution have a better phi value than the third one so one I can disregard it because it has a lower value for the action but usually the other two have the same value which one should I choose in that case well in that case if in the two solution the phi takes the same value I can choose among the two thanks to this term so this term is saying among the many solution to this saddle point equation take always the one with larger overlap so in case this equation has multiple solution the rule induced by the fact that you have a very small but non-zero coupling is to choose the one with larger value of q so you want to bring replicas closer essentially so and now from now on we remove this term but we keep in mind this rule ok and in terms of this new one well in terms of this phi one RSB if you rewrite the no let me comment one more thing before so there are some limits of this phi one RSB that are very simple to understand in particular it is easy to check that phi one RSB with m equal one is equal to the paramagnetic free energy easy to check if you set m equal to one this term disappear this term disappears and this term disappears so you essentially get minus t half times beta square half ah, here again the infinity entropy term is missing I'm sorry beta over 4 and then if I didn't forgot the infinity term you should have also s infinity times t ok, but I forgot it so it is not in this expression and another trivial limit is where you set q equal to zero again is easy to check that phi one RSB any m q equal to zero t is again f para again easy to check if q equal to zero this is zero if q equal to zero this is zero ok so this also tell you that if the action is maximized with the value m equal one you will get the paramagnetic free energy not the paramagnetic solution because q may be different from zero but what this expression is telling you is that once you set m equal to one independently on the value q star that maximized the action you get the paramagnetic free energy not the paramagnetic solution because q may be different from zero ok, this is why between td and tc we have exactly the paramagnetic free energy but the solution to this equation may be non-trivial actually in general when you solve this equation this is a suggestion this equation can be always written as one minus m times something which depends so let me say a g which depends on m, q and t ok so this general you take the derivative you realize this so if you want to look for the q star maximize the action at m equal one you have to remove this term otherwise the equation is trivial is always satisfied what is saying to you is that when you are very close to m equal one this action essentially is almost constant in q and if you want to identify which is dominating the action say for m equal one minus delta you have to remove this term and solve this equation so this general when you want to solve the one or sv equation you take the real respect to q you divide by one over m otherwise you don't get any sensible equation and then you can set m equal to one and so solving the equation m equal to one q and t equal to zero can be done if you set m equal to one here you always get zero equal to zero which is not very useful ok so let's look how the solution looks like I don't want to take derivatives so believe me I will show you the solution and that you get by taking derivatives of this equation before doing this I will write the expression for f star in sigma in terms of this phi one or sb because there are some general properties that is worth understanding immediately so if you take the definition of s star and sigma in terms of capital phi and you rewrite it in terms of this m times phi one or sb in q star you realize the following that f star of m and t can be written as phi one or sb of m q star remember q star is always a function of m and t t plus m the derivative with respect to m of phi one or sb m q star and t is the explicit derivative with respect to m because if you take the derivative with respect to q by this other point equation is zero so it is useless and then sigma of m and t can be written as beta m f star of m and t minus phi one or sb m and t sorry m star so once you derive this now what you can do you can actually solve the this other point equation and draw this function f and phi as a function of m because now we want to understand what is the range of m value which is meaningful in order to derive the right shape sigma versus f so you know the drawing so as a function of m you draw this phi one or sb of m q star and t obviously the equation for q star will be a nonlinear equation so if you are smart enough you plot this not solving the equation but doing parametric plot in you don't really want to solve the nonlinear equation now if you plot this and you exploit these two equations you already can understand what will be the behavior of f star look the different f star is phi plus the derivative of phi with respect to m so if you choose the point m where phi is maximum here the derivative with respect to m is zero so f will pass through this point first information second information sigma is the difference between f star and actually sorry second information then you have to add to phi something proportional to the derivative so in this part the derivative of phi is positive and so f will be above here the derivative of phi is negative so f will be below this is generic I just drawing you understand that f star of m and t t is always fixed as a parameter ok will pass through the maximum of phi and will reach another maximum finally sigma the complexity is proportional to the difference between f star and phi so at any point if this difference here is essentially t sigma over m I just put beta m on the other side so we already realize that this region has negative sigma so this region we want to exclude it because has a negative complexity all this region here has a positive complexity ok but the behavior now we want to restrict even more the region so for the moment we are saying that beyond this point we are not interested the complexity here is negative on the other side we want to understand up to where these curves provide a physical interesting result so remember the following when you draw sigma of f at a given point the slope is beta m so in general when I increase m at a fixed temperature what happen I increase m f and sigma decrease and vice versa if I decrease m f and sigma increase so in this plot I want let me do also the reverse because we are going down so decreasing m starting from this point I want f increasing so here is fine once I reach the maximum is no longer true ok if I go beyond this point by decreasing m I will get f star decreasing which is not what I want so this part I exclude it because the derivative of f star with respect to m would be positive which is unphysical physically I want m and sigma to decrease when m increases ok in principle there should be another thing to check that also sigma increases in this regime but this is not hard to prove because if I leave you as an exercise to show that take this expression derivative of sigma with respect to m bla bla bla and very easily you find that the derivative with respect to m is equal to beta m the derivative of s star with respect to m so what does it mean that when this is 0 this is 0 so when f star has a maximum sigma has a maximum so at this point here both f star and sigma have a maximum so as long as f increases sigma increases when f stop increasing also sigma stop increasing so in this region of value of m everything is working nicely because we have a positive complexity and we have this relation which are well satisfied in this values of m I will call ms of t this value in order to be consistent with the notes and m threshold of t this other value and you will see that these are is the range that we I were claiming before that is worth looking at in order to have the entire sigma of f curve so these are related to the slopes in these points they are related we have beta m equal to the slope at the end point and beta m equal to the slope at the end point on the other side fine so in general this is general I am not using the explicit expression for phi it is more of this shape but it is what happen in models that have a non-zero complexity so in general you can look for ms and m threshold has the value where sigma or f star is maximum and the value where sigma goes to zero these are the two definition of m threshold and ms ok now let's go and look to the solution so once we understand that these two parameters are important and remember that as long as n is equal to one the free energy is equal to the magnetic free energy now we want to understand well what is how m threshold and ms change with temperature and so what are the solution that dominates the thermodynamics so in general we have that m threshold must be strictly smaller than ms you see it by definition and if you for example if you take the spherical model you can draw the m values as a function of temperature so you compute this phi 1 rsb you compute m threshold ms for any temperature and then you draw this this function and they look like this so I'm not interested in what happens above m equal to one so here I have m equal to one I'm not very much interested in what happens above ok so they have this shape what does it mean? it means that there is a range of temperatures so let me sorry I didn't write what is this it's obvious from the inequalities m threshold and this ms and they usually go linearly to zero ok and they grow in temperature notice that if sigma not of t were constant so if the slope at the end of the complexity curve were constant this would imply this ms equal to t times sigma not so you see that these values of m that maximize the free energy or find the minimum value of the free energy are roughly linear and then they become a little bit nonlinear but this means that the sigma not of t is not strongly t dependent ok in this plot we also got two more information the information is that this value of f is f min and this value of f is f threshold ok so we not only have the values of m but also the value of the range of free energies and this is the only range we are interested in so essentially what is telling us is that if you want to have a non-zero complexity so here inside there are states while if you do the computation with an m value which is outside this range you don't get anything meaningful ok so now let's come to and one more important observation is that when you search for the solution of the point equation so when you search for q star is not obvious that well there is always one solution which is q equal 0 if you plug q equal 0 here the equation is always satisfied so what you are interested in is in understanding when a new non-trivial solution arise obviously this will depend on the value of m so for each value of m you can draw a line here which is such that if the temperature is too high q star equal to 0 is the only solution below this line you have a non-trivial solution this is true in general but you see it depends on m so in some sense when you use m close to 1 the non-trivial solution when you use a small value of m in non-trivial solution to this equation arise at a lower temperature ok so this is the scenario and now what we want to now we want to identify this two temperature with td npc why? at the first observation as I already told you the fact that there is an ergolyzity breaking at td we will see tomorrow by solving the dynamics for the moment the only thing that we can say is that all this computation with generic value of m if we set m equal to 1 we get the computation with it yesterday so the states with m equal to 1 if they exist they dominate the thermodynamics so the question is when do they exist the states with m equal to 1 in this region beyond tc there is nothing there is sigma negative so you don't have any states if you count with m equal to 1 so what happened and also here there are no states at td at m equal to 1 you don't have any states and so essentially the parametric solution dominates and notice moreover that the dash line ends exactly at td so on the right of td essentially you only have the parametric solution so for t larger than td you have a parametric solution then between td and tc you have a m equal to 1 sorry I haven't told you what 1RSB means ok, it is just a name but it has a meaning because in the matrix that we draw we put a structure with just two values of the overlap one value is 0 and the other value is q so we broke the symmetry between the replicas just once in this case it was an explicit breaking because we build the replicated system with a coupling within a group and so this is not a spontaneous replica symmetry breaking we impose a breaking between the replicas which are not all equivalent but for, let me say, coherence with a case where radius symmetry breaking is broken spontaneously and this solution is called 1RSB ok, so between td and tc the parametric has the same free energy as the m equal to 1 1RSB solution so the solution which has a q star larger than 0 so here we have two solutions which in principle may compete and we can understand from the dynamics that actually the one with q star larger than 0 which means you have a state if you start within the state you don't go away is the one actually useful for the dynamics so in this range the two solution paramet and this have the same free energy in practice we have to choose this one which is the one more relevant so in that below td the ergodesis is broken so the parametric solution is no longer relevant what we have to do below well, we have to choose not m equal to 1 state by doing the computation so we have to maximize the free energy it's very simple to convince you that the free energy is maximized along the line this one so when the the parameter m is the largest as possible so you have a condensation so the thermodynamics is dominated by the few lowest free energy states so this is the thermodynamic picture that we get from this computation and before leaving just in the last five minutes yes I just want to show you that everything is very qualitative although the process now is I just want to show you that from this one of these free energy we can actually compute the value of td and the overlap at td which is what I gave you for free yesterday so now you get it very easily in doing this you will also understand why the transition is first order on this line when a new solution arrives arrives with a Q which is strictly positive it's not continuous it's really discontinuous in order to understand this it's quite easy because I take this derivative and I remove the 1 minus m and I write the expression for g actually for g m equal 1 so I'm here I'm looking for this point so I want a point so I'm interested when the first solution at m equal 1 arrives so I write this one and believe me so you have to take one derivative in the equation g m equal 1 q t equal to 0 is equivalent to this beta square 1 minus q p half q to the p minus 1 equal to q you have to solve this equation how can I solve this equation well let me do it graphically so you understand why the transition is first order I plot as a function of q this is q and the left hand side actually the left hand side is multiplied by beta square so there is a pre-factor that grows when you decrease the temperature and so let me just write this expression here and this expression here if you this is not a straight line going to 0 but okay this this expression here like this okay this 1 minus q I use I use p equal 3 so 3 halfs q square is something like this now you take this you multiply by beta square so you raise it you see that here you always have a solution q equal 0 what happen when you raise it this and this is the first non-trivial solution these are first order transition because as soon as it appears the rule says choose the solution with the larger value of q so you have to pick up this this q star so we see that the transition is first order why essentially because for low values of q you have q to the p minus 1 which is q square and so you start like a parabola so when you start like a parabola when p is larger than 2 the first solution cannot be here you have a first order transition okay now okay so this is the first solution that should correspond to this point and if I want to understand what is q and beta at this point I notice one more thing when the first solution arise the derivative of this function is exactly the derivative of q which is 1 so you have to take the derivative of the left hand side and the right hand side and put them equal and this corresponds to right you take the derivative of this you simplify a little bit and you get q to 1 minus q square equal to 1 because the derivative of the right hand side is 1 if you satisfy both conditions you find the first solution to this nonlinear equation and even if the equation is nonlinear you can do it very easily why? because it is true that it is nonlinear but you realize that you can remove the nonlinearity by essentially taking for example this divided by this and the nonlinearity cancels because qp minus 2 divided qp minus 1 is 1 over q and so you remove all the nonlinearity and you end up do this divided by this you do the ratio and you get p minus 1 you see p minus 1 doesn't cancel p half cancel beta square cancel 1 minus q square divided by 1 minus q is 1 minus q and then you get qp minus 2 divided qp minus 1 is 1 over q but you also on the right hand side you get 1 over q so you have 1 over q and 1 over q you cancel this and you solve it very easily and you get that q let me call qd because now we are computing what happened at td is just equal to p minus 2 over p minus 1 this is your linear equation in one line ok and if you want the value of td you just have to take this value and plug it for example here or here is the same do it for exercise and you get that td square actually you have beta square so it is much easier to write td square turns out to be this I leave you as an exercise but it is much easier than the exercise I gave you yesterday is p minus 2 to p minus 2 divided p minus 1 p minus 1 so you see that with this replica computation something that yesterday I have to give it for free because otherwise it would be very hard to compute from the TAP free energy now we have it from one line computation so it is definitely much much easier and so we get exactly the same thermodynamics that we got with TAP equation so parametric phase 1RSB solution with n equal 1 means the same free energy of the parametric state and then I knew 1RSB solution with a free energy which is different and you follow these states that actually is easy to so at MS you always have f min while at m threshold you always have f threshold so along the blue line there are states of minimal free energy because they correspond to the minimal of the complexity so at the end this picture is perfectly consistent with the TAP solution but you derive it well once you do all the computation for deriving this x of q then is very simple the analytic computation because you just have to plug the answers into x of q obviously the right answers in more complicated situation is more complicated but still allows you to do computation in a very compact way so for today we see you tomorrow