 Okay, can you hear me? Okay, thanks so much for coming again. I'm happy to be back and to see all of you in good shape. I'm sure you couldn't wait. So the first part of the course, we went through or over a few techniques to analyze the spectra of random matrices. You remember one of the main point I insisted a lot was this Lehmann classification of random matrix models where we have this if restricted to matrices with real spectrum, the two main classes that we can consider are those with independent entries and those with rotationally invariance. And I told you that there is only one as a consequence of the Rosenweig and Porter theorem. There's only one class of ensembles in the intersection, which is the Gaussian ensemble in its three declinations, real symmetric complex admission and quaternion self-duo matrices which are basically indexed via the Dyson's index called beta equal to one, two or four. Now today and this week I wanted to go over a class of techniques which basically in essence we can group under the name Edwards-Jones formulas. So this class of techniques is in principle useful or valid for any matrix here. So it is incredibly general in principle. In practice it is not, but it's nevertheless mainly useful for matrices in this class for example, for independent entries matrices, which is the case where we have very few other weapons to tackle. So this offers us a good way to deal with problems on that side of the diagram. So in general the goal of this class of techniques is the following. We can write it as we want to connect the joint probability density of the entries of a random matrix model. Let's call it P of x11 xnn. So this is our input, the input of the formula, which can be anything, it can be very general, it can lie anywhere on this diagram with another object which we discussed extensively with the average spectral density, an object that we denoted I think row n of lambda in the first week. So this will be the output. Okay, so we are trying to connect two words. The input is the joint probability density of the entries and the output should be the average spectral density. So in principle one would like to establish the following chain if possible. So from the input, the joint probability density of the entries where the number of variables is of order n square, we would like to reduce this number of variables by considering the joint probability density of the eigenvalues which gives full information about the spectrum. Here we have, let's say a reduction in complexity. We go to order n variables and then from here we would be in principle able to compute our output because we know that the average spectral density is the marginal. We used this property a lot in the past week. So we could basically integrate over all eigenvalues but one and we would get from the joint probability density of the eigenvalues to the average spectral density. Unfortunately we know that the first step of this chain is not always possible. Why it is not always possible to carve out from the joint probability density of the entries the joint probability density of the eigenvalues. Well we know that the integration over eigenvectors cannot be always computed in closed form. So in some sense the Edwards-Jones formula and its relatives allows us to bypass this step entirely. So to go directly from the joint PDF of the entries to the average spectral density without knowing or without the need to compute the joint PDF of the eigenvalues. So in some sense it's a quite powerful tool because it doesn't require this step that we know can be very difficult or in some cases, in most cases even impossible. Okay, so let me give you without wasting further time. Can I erase here? So I will first give you the formula. And then we try first to prove it and then to see how it can be used. So the formula reads as follows. So the spectral density is equal to, let me get this straight, minus two over pi n, the limit epsilon to zero plus. The imaginary part of the derivative with respect to lambda of the average of the logarithm of something where this z of lambda can be written as a multiple integral over auxiliary variables y. It's an n-fold integral of the form exponential minus i over two y transpose lambda epsilon identity minus x into y. So lambda epsilon is lambda minus i epsilon and the vector y is a column vector of auxiliary variables. Well, I defined z of lambda as this, where lambda epsilon is in here. Here we have basically a full calculus course in one single formula. So we have a multiple integral. We have a logarithm. We are taking the average, where the average here is taken with respect to the joint PDF of the entries. So recall that this one, this object here is our only input. This is the only thing that we know. And then we have this multiple integral which contains the entries of our random matrix in here. In principle, if we compute this integral for lambda epsilon, so lambda minus i epsilon, then this would give the average spectral density at the point lambda. Now, the proof of this formula, well, can I erase here? So I can keep the formula there. Good, so first remark. So while the formula is in principle valid as written for finite n, it is in practice used or usable in the limit of large matrix size, several simplifications. Okay, so the proof starts from the definition of the average spectral density as the average over the ensemble of one over n, summation i one to n, delta of lambda minus lambda i. So this is the first ingredient, just the definition of the spectral density. And then the second ingredient is the famous Sokotsky and Plemerge formula that we used already in our lecture on the resolvents. So the Sokotsky-Plemerge formula reads, you remember, we have this in the notes. So why is this object, this identity interesting? Well, first of all, it justifies the presence of this limit here. It is interesting because it provides an interesting representation of the delta function. So we can now trade this delta function using this relation. So we can rewrite rho n of lambda as what? As one over pi, so we need to kill this pi. Then we need to add this n, one over pi n. Then we will have limit epsilon to zero plus the imaginary part of what? So basically I'm taking, instead of using the delta function, which was here, I'm using this rational function here, but I will need to take the imaginary part of it in the limit epsilon to zero. So the net effect of this is that I don't have any more delta function in there and I replaced it with a rational function. I'm just trading delta for a rational function. So this explains a few of the terms that you see in the final formula, the imaginary part and the limit. Now we need to see how to get the remaining bits in the game. So first of all, I can change just a sign here. It's not really important, but just for convenience. So I'm just changing the sign here. Okay, now if you compare this bit with the final formula I gave you there, you see? So we got here, if you compare here, you see that basically the only step that is missing is to interpret this rational function one over something as the derivative of a logarithm, right? So this is what we are going to do. The only problem, I mean, the only thing that we need to be a bit more careful about is that this object is a complex number. So we will need to be a bit more careful on how to deal with logarithms of complex numbers. That's why I wrote a log with a capital L here. I'm forced to erase this. You have it in your notes, right? So a brief two minutes tutorial on complex logarithms just as a reminder. So the logarithm of a complex number is the number W such that the exponential of W is equal to Z. This just by definition. So the problem that we face is that the logarithm, sorry, the complex exponential is not injective on the whole complex plane because it does not map distinct values to distinct values. For example, if you take exponential of W plus two pi I, this is still equal to exponential of W. So in order to define the logarithm, which is the inverse function, we need to restrict the domain of our function, okay? Otherwise we would get a multi-valued function, which is what the logarithm is. So the solution to this problem is to restrict the domain of the exponential function to a region that does not contain any two numbers differing by an integer multiple of two pi I. If you do so, we get that for any complex number, we can define, so Z will be X plus I, Y. So we can define the so-called principal value that we call log Z with a capital log as the logarithm whose imaginary part lies in the interval minus pi pi. Now, the only problem that I wanted to flag and make you more aware of is that not all the familiar properties of the real logarithm, so not all the familiar properties of the real logarithm carry over to the complex log. For example, and these two will be handy, if you take the principal logarithm of exponential of Z, this may not be just equal to Z and also the principal value of the logarithm of Z1 times Z2 may or may not be equal to log of Z1 plus log of Z2. So we need to be a bit more careful on this. As an exercise, if you don't believe it, I ask you to compute log of minus 1 times I which will give minus pi I over 2 and if you compute log of minus 1 plus log of I, this will give 3 pi I over 2. So just keep these potential problems in mind. Anyway, we can still, can you read here in the back? So we can still write the object that we had was summation I1 to n of 1 over lambda I plus I epsilon minus lambda, which will be the derivative with respect to lambda of summation I1 to n logarithm of lambda I plus I epsilon minus lambda with probably a minus sign in front. So the property of the derivative will be the same. These properties might be different. Now we have explained the reason why we've got the derivative with respect of lambda into the formula that I gave you, but it is still not clear why we are doing all of this, right? The problem is suppose that the point, suppose for a moment that this log was just a normal standard real logarithm. Then we would have summation of a I of a logarithm of something. Summation of a logarithm, we can trade the summation for a product, so we could in principle write this as logarithm of a product. And now what would this product be? So it would be the product of lambda I plus I epsilon minus lambda. So it is the product morally of eigenvalues, right? So what is the product of eigenvalues? It's a determinant, right? So if this logarithm was real, we could trade summation of a logarithm for a logarithm of a product, and this product would be a determinant. Now logarithm of a determinant, and now you know that a determinant to some power, for example, minus one-half can be written in terms of a multiple integral. So this would be the logic that would lend onto this multiple integral that we call z of lambda. So this would be the idea. Summation of a log is equal to log of a product, is equal to log of a determinant. This determinant can be represented as a multiple integral, and then we are done. You see the logic. Unfortunately, we need to be a bit more careful due to the fact that this logarithm is not real. So the summation of principal logarithm is not necessarily equal to the product of principal logarithms. It's property here, which is clear, because if the argument of this object must be between minus pi and pi, and the argument of this object must be between minus pi and pi, this doesn't necessarily mean that the sum of two arguments between minus pi and pi will be between minus pi and pi. It is still true, though, that we can write a multiple integral representation. Can I remove here? You have it in your notes. So we can still use, well, we call the object z of lambda. So this would be the integral over rn of dy. So when I write this boldface, boldface y, I really mean the product of dy1 up to dyn of exponential of minus i over 2, y transpose lambda epsilon identity minus x. So now this is an exercise for you. I'll just give you the exact result of this integral. You can check it with Mathematica if you want. So this is a sort of correct version of the identity for the determinant in terms of a multidimensional integral. So normally we would have here no complex variable. We would have just a Gaussian integral, exponential of minus one alpha, let's say, of y transpose a certain matrix A times y. And here on the right-hand side, apart from constants in front, we would have one over the square root of the determinant of A. Now here, the problem is that this object is not really a Gaussian integral because we've got complex variables, complex values all over the place. So to be more precise, this would be what is called a multidimensional Fresnel integral. And this matrix here is a weird object because it is a complex symmetric matrix. So if h is real symmetric, this object basically alters the diagonal with a complex number. So it is still a real symmetric matrix but with the diagonal form of a complex number. So this is not an Hermitian matrix. It's a complex symmetric. And here on the right-hand side, we have some sort of one over square root of the determinant because we have the exponential of a logarithm with a minus one alpha in front. So this would be morally, if you took this logarithm as a real logarithm, this morally would be one over the square root of the determinant. Except that written in this form, there are no ambiguities because this is the principal logarithm of the complex exponential. So that's an exercise for you on how to prove this identity. Okay, now we are almost there. Why? Because this object here is the chunk that we need for our formula. So if we manage to extract this chunk from this formula and replace it in here, our proof would be complete. Do you agree? You want a coffee? Either you agree or you don't agree. So just... You agree. So how would you imagine that all these logarithms were real logarithms? How would you extract this chunk from this formula? We have some integral equal to exponential of something that we want. Yeah, you would take the logarithm, right? Except that I told you that the logarithm of exponential z may not be equal to z. Well, it depends on the argument of z. So it may be equal, depending on the argument of z, or it may not be equal, so it may differ from this object by a certain multiple, for example, 2 pi i or something. So we cannot be sure that this equality will be true. But it is still true that if we apply the logarithm on both sides, we can still write summation i1 to n of the logarithm, principal logarithm of lambda i plus i epsilon minus lambda will be equal to what? Suppose that we take this logarithm, will be equal to what? There is a factor minus one-half that will go on this side as a minus two. So this would be equal to minus two, the principal logarithm of z of lambda plus a certain number of terms, and here what saves us is this. Remember that in the formula we will have to take the derivative with respect to lambda. So whatever other terms we have there, these terms will be killed. So in some sense the difference between the real logarithm will become ineffective or less effective due to the fact that we have to take this derivative with respect to lambda which will kill extra phases that might arise from taking this logarithm. So now this minus two explains the minus two in the final formula. That's the origin of it. So you can just then write if you collect all the terms rho n of lambda will be minus two over pi n which is exactly what we had in the formula. Limit epsilon to zero plus which comes from the Sokosti-Plemming formula, the imaginary part again. Then we have the derivative with respect to of lambda and then we have the average over the ensemble of the logarithm of this multiple integral which is complete. Now what we need to understand is whether this formula can be useful or not. So clearly the crucial object is this object here because we need to take the average over the distribution of the entries of our matrix of the principle logarithm of a multiple integral. Now if we write it explicitly what we are really after is the following. So let's write it explicitly. This would be the integral over the entries let's say the entries in the upper triangle of the probability of x11, xnn which is our input, remember this is the only thing that we know about our ensemble. There's no other information and then we have inside the integral the logarithm of what? Of this multiple integral. Do we agree? So that's the object given our input which is only the joint distribution of the entries that's the object that we should in practice compute. So here we have this average and here inside here we have the entries of our matrix. Now this is what it is but of course if this was our final final situation our final point this would be lethal, right? So it would be the end of it. Why? Well, because of course there is this logarithm in the way which is inside the integrand so we have a multiple integral the logarithm of another multiple integral. So how do you proceed with this? The only thing you could do would be to compute this inner integral inside the argument of the logarithm then try to massage it a bit and then perform the second integral. But if you do this then basically what you would do is you would run the Edward Jones formula backwards, right? If you did this you would get the identity rho n of lambda equal to rho n of lambda which is hardly useful, right? So this is like an obstacle that we need to solve. So the only way forward would be if we were able to carry out the integration over the x's before the integration over the y's so if we were able to exchange the order of the two integrations otherwise we would be running the formula backwards. Now how do you exchange these two integrals having a logarithm in the way, in between? That's the main technical challenge. This brings us to the two strategies, so just a bit of jargon. So this average here over the joint distribution of the entries is called in the literature over disorder. So the disorder means the randomness which is in the matrix entries. So when you average over the disorder you are just meaning this. We are taking this integral over the joint distribution of the entries of whatever it takes. Good, so there are two ways to overcome this obstacle. I just tell you this and then we erase it. Was it fine? Well, it needs to be. So there are two possible strategies which corresponds to, well, let's call it like the annealed so an averaging over annealed disorder or an average over quenched disorder. I'll try to explain you the way I understand this. So first a bit from the dictionary so it's quenched or quenched. So this means from the dictionary like made less severe or intense subdued overcome and then there are several other acronyms. For example you can quench a fire you can quench your thirst you know it gives the idea of something that is sudden. You've got something erupting and you just quench it. To anneal is a metallurgical term which means to heat or glass and allow it cool slowly in order to remove internal stress and toughen it. So now one hour lecture on metallurgy. First a break. Can we start again? Good. So we have these two terms which apparently mean nothing until I will explain it. Good. The problem that we are facing is the following. We have one multiple integral. We have one multiple integral of the logarithm of another multiple integral over let's say dynamical variables y running over rn and this object here we call it z of lambda. So whenever I use the letter z well this is intentional, right? So I could have called it in many different ways but calling it z reminds you of what? Yeah, so we add always for partition functions, right? So if you forget for a moment the fact that we have complex numbers all over the place here this is a multiple integral of an exponential of something, okay? So we would like to interpret this object as a need so we are integrating over all degrees of freedom of an exponential of something. So it is tempting to interpret this object module of the fact that we have complex numbers in there as a sort of canonical partition function where we are summing over all degrees of freedom which corresponds to a sort of Gibbs Boltzmann distribution of the type p of y1 yn so this is the distribution of positions or degrees of freedom of an associated dynamo dynamical model which is in equilibrium at inverse temperature one, let's say and with a Gibbs Boltzmann distribution of this type so here is the total energy and this depends on y the degrees of freedom on x which is our random matrix and on lambda so if you forget the complex nature of this object which stays here it would be tempting to interpret this object as the integral over let's say all the positions or all the degrees of freedom of an object which is distributed according to exponential of minus something and now if we take the logarithm of this object which is a partition function then what do we get? So here basically what this formula suggests is that we are basically averaging the free energy of an associated thermo dynamical system over the disorder so you have a free energy corresponding to an Hamiltonian which includes a term that is random so inside the Hamiltonian we have the disorder coming from the randomness of the matrix entries okay so we have if we interpret this object this way then we are basically taking two different averages so we are taking an average here over the Boltzmann distribution so average over the Boltzmann distribution and here we are taking the average over the disorder so we have two conceptually different things that we are doing an average over the Boltzmann distribution and then we take the free energy and then we average the free energy over the disorder so this means that if we perform this operation the way it should be done so by taking this object taking the logarithm and averaging over the disorder well this is equivalent to considering that the two levels let's say levels of randomness so act on let's say different time scales what does it mean? so first the let's say dynamical variables y thermalize for a fixed value random matrix x so for a fixed let's say instance of the disorder so we take one instance of the disorder we let your system thermalize and then you change the instance of your disorder you let the system thermalize and then you take the average of all so it means that the dynamical variables y are thermalizing because we are averaging the free energy so this is an equilibrium problem once they have thermalized for a fixed value of the instance of your random matrix you change the instance you draw another instance you let the system thermalize and then you take the average of the free energy over the different instances of the disorder that's the picture that you should have in mind this corresponds which is the right way to proceed because this is an exact formula that we obtained so this corresponds to taking the so called quenched average so the disorder is fixed there is one instance that is fixed you let the system of the y thermalize at this fixed value of the disorder and then you average over the disorder so this is what is called a quenched average what is the correct way to proceed unfortunately it is quite complicated because we still have the problem of having to exchange the two orders of integrations to get something meaningful otherwise we are stuck we would just be running the formula backwards so this is correct but difficult now there is another strategy which is the so called annealed strategy and this interpretation that I gave you makes it maybe more clear what we are trying to do so the annealed strategy so if you treat the disorder as annealed well basically what you are assuming that the associated statmec model is described in terms of the joint of variables x and y so you are not picking one instance of your randomness equilibrate your y system and then pick another instance of your x variables equilibrate your system and then taking the average instead what you are doing is assuming that x and y fluctuate together so on the same time scale so you have basically what you have is a certain partition of lambda which will be basically integral over x your random variable and y together so this is not the object that we should compute the object that you should compute is this one but if we make this let's say approximation that x and y fluctuate together can you tell me what is the effect on that integral there so how are we approximating that integral there what is the operation that we are doing there we probably take the logarithm out so doing this interpretation or this approximation is tantamount to removing the logarithm from inside the integral and taking the logarithm of the partition function where you are integrating simultaneously over x and y okay I find it very ironical because this is something that when you were maybe 19 or 20 pulling a logarithm outside the integral would have costed you a fortune with your teachers now we are promoting this incredible mistake to a fantastic trick so basically we are revolutionizing the math let's spread to a way to describe this operation so sometimes in textbooks you find this expression so instead of taking the average of log of z you take the log of the average of z in essence this is not really true in this object here what you are taking is the logarithm of the annealed partition function so there is no there is no average of the disorder left because you have just promoted the disorder as the dynamical variable so now x and y are on the same footing so this is an approximate way to proceed which stands from the mathematical mistake if you want the problem is that it often works fabulously well as I am going to describe to you so although it is not really great from the mathematical point of view it can still give you some information on your random matrix model furthermore the computation is much easier because you don't have this problem of having a logarithm in the way anymore you are just kicking him out so I wanted to show actually a top to bottom calculation well that's hard to tell at this stage I would rather skip this question and duck it and just consider it as a mathematically improbable way to get the true results if you allow me to if you allow me this okay the question was in terms of our problem what is the physical interpretation of the annealed approximation okay good so now I will apply this formalism to a case where we don't actually need it so we can do quick and dirty calculation as the annealed calculation for GOE so we can carve out the semicircle in a few steps obviously in general we don't need the Edward Jones formula in the Gaussian ensembles why so the Gaussian ensembles are here so we've got our full arsenal available we've got orthogonal polynomials we've got the resolvent we've got many different tools at our disposal but just as an exercise we can try to recover the semicircle from the Edward Jones formalism in the annealed approximation changing the order of integrations between y and h carry out all the integration take the log at the very end and then we will see that this gives the correct solution and then my plan was the next lecture to try to do the full quenched calculation using replicas again for the GOE case again a case where we don't need any training ground to see how this type of calculations are carried out top to bottom the idea is that we start from the joint distribution of the entries of our GOE now by now you should be able to write it down to some diagonal elements and I have rescaled I've just rescaled the variance so I have rescaled the variance by 1 over n to ensure a good limit for large n otherwise we would have to rescale the eigenvalues by 1 over the root of n ok good so what is the for the annealed calculation so we need to compute the annealed partition function which is what well it is the integral over degrees of freedom y so remember that the logarithm has been kicked out of the way now we are treating y and the disorder on the same footing normalizing together then we take the logarithm so we have the free energy of the joint system and then we apply the Edward Jones formula so we have to and then we have an integral over the variables in the upper triangle times the joint pdf of the entries which is this object here minus i over 2 y transpose lambda epsilon the identity minus x times y so this is the object we should compute before we had this integral of what? of the logarithm of this other integral but we just kick the logarithm out of the way so we can ignoring some prefactors here just in the larger limit we can pull this first term outside because it does not depend on the xij so this will be the integral over dy over r to the n of what? of exponential of minus i over 2 lambda epsilon i over 2 lambda epsilon and then we have summation i1 to n yi square so I am taking this diagonal term this is a column vector so this is minus i over 2 lambda epsilon summation over i yi square and then we have the product of the averages over the diagonal entries and the off diagonal entries so I can write this as the average of the exponential there is a minus and minus plus so i over 2 summation i1 to n xii yi square and the average is taken over the diagonal entries they are all independent so we can factorize the integral and then we have the average over the off diagonal entries which is e to the i summation i smaller than j xij yi yj so this is the average is taken over diagonal entries and this is taken over off diagonal so now we can perform these averages keeping in mind that we want the large and limit so we can be a bit quicker a bit sloppier if you want so that's the average that we need to do for the diagonal entries and for the off diagonal entries you have them in your notes so I can erase so we first use the expansion of the exponential 1 plus z plus z square over 2 plus so these are the ingredients to make a quicker calculation is everything all right and then we use the fact that the average of xij is equal to zero so these are Gaussian random variable with mean zero and the average of xij square so the second moment is 1 over n times 2 minus delta ij so depending on whether i is equal to ij we have a variance 1 over n or 1 over 2n so now we can perform the average over the diagonal entries for example exponential i over 2 summation i1 to n xii yi square so this is the this is the average here so we have the exponential of a sum we can rewrite it as product i1 to n of the exponential of i over 2 xii yi square and then all of these variables are independent so we can put the product outside the average so we can write product i1 to n and then expand the exponential like this so you will get 1 plus i over 2 xii yi square minus 1 8 xii square yi to the fourth plus and so on and so forth which I would really like to keep this one but I can erase this one so then we can take the average term by term so the average of 1 is 1 the average of xii is 0 and the average of xii square is given here so in summary what we get is that this object is roughly speaking the product i1 to n of exponential minus 1 over 8 n yi to the fourth because I just rewrote this was a product over i of 1 minus 1 over 8 to the n yi to the fourth and I just re-exponentiate this object as long as we are interested in the large and limit we are making a negligible error on this approximation and similarly so this was for the diagonal part so similarly for the off diagonal part I'll leave it to you as an exercise we get product i1 to n exponential of minus 1 over 4n yi sorry this was an exponential correct this one so we have our two results and we are going to go into this expression right so we plug this object in here we plug this object in here and then we have to perform the integration over y now have you copied everything can I erase again so if I put these two results together here just after a few couple of steps minus 1 over 8n summation i and j so without i different from j yi square yj square or in a form that is a bit better summation over i yi square all to the square so I'm just putting this object multiplying it by this object and then symmetrizing this product is based over i smaller than j and then I'm symmetrizing so you get the factor of 2 that you need here to rewrite this object in this form so in the end you have your beautiful integral your beautiful y integral which gives you the annealed partition function ignoring constants of course keeping only the leading exponential term dy exponential of minus i over 2 lambda epsilon summation over i yi square which came from here and then we have this exponential here so we have times exponential of minus 1 over 8n summation over i yi square okay now we would like to perform this y integral and as you can see here we have an annoying term because we have a sum of the integration variable all squared so we would like to get rid of this extra square to perform the integration in an easy way because this term is coupling my integration variables so do you know how to get rid of it in an exponential and reduce it to a first order term sorry okay which in essence is what so in essence what we can do is introduce a Gaussian identity so if you have the exponential of something square of this gamma square then rewrite it as an exponential of gamma so you lower the degree from 2 to 1 the price to pay is that you need to integrate over an extra integration variable q it's nice we have a square an exponential of a square in here we can reduce it as to an exponential of a degree one term which will allow us to carry out the y integration the price to pay is that there will be one extra integration over the variable q but we are going from an n fold integration to a single integration that's better right maybe not you don't appreciate okay but I will use it anyway good so we use this identity with gamma equal to summation over i y i square and alpha equal to 2n because if alpha is equal to 2n this object becomes 8n which is exactly this term here we use this identity so the annealed partition function becomes the integral over dq which is just a single integration variable of what exponential minus alpha q square but alpha is 2n so minus 2n q square right and then we have i gamma q but gamma is integrated because it's a function of the y's so what we have to put in here is the integral over dy of what of exponential minus i over 2 epsilon summation over i y i square plus i q summation over i y i square so now the square here has disappeared the square has disappeared the price to pay is that we have an extra integration variable over q but now you should be happy right because we know how to do this integral this is just n copies of a single integral it is the exponential of a sum of integration variables that appear in a factorized way so what is this object here just the integral over the n-fold integral here is just n copies of a single integral which is the integral between minus infinity and plus infinity of dy just a single integral exponential of minus i over 2 lambda epsilon y square plus i q y square good so now you solve this you solve this integral as an exercise it is of the form exponential of alpha x square between minus infinity and infinity should be doable we rewrite this object which is the integral to the power n as exponential of n log of this integral so if you do that what you get for your annealed partitioned function you get a single integral over q of something that we can write in this form exponential of minus n then we have 2 q square which comes from here minus one half the logarithm of 2 pi divided by epsilon plus i lambda minus 2 q so all you have to do is perform this simple integral and then rewrite the result to the power n as exponential to the power n times the logarithm of the result and the exponential of n times the logarithm of the result is this object here is the logic clear? good now you call this object as pi lambda of q and now for large n you tell me how to evaluate this object here so you have exponential minus n this integral will come to the neighborhood of points q where this object attains its minimum value so if we do that then we have the annealed approximation for this partition function will go as exponential minus n pi lambda of q star where q star satisfies the stationarity condition of this action so what we have to do is to compute the first derivative evaluated in q star of pi and set it equal to zero so if we do that we get well I did it for you you can check so that's the that's the saddle point condition so for q star and then you take the derivative of this with respect to q so this is the equation that you get for q star plus one over two q star minus lambda epsilon and then you solve this is just a quadratic equation for q star which you solve and what you obtain is q star is equal to one over four lambda epsilon plus minus root of lambda epsilon square minus two here what you are what you are witnessing here is what I call the birth the birth of a semicircle so there's still some work to to do we need to apply the adverse Jones formula but in essence the semicircle is hidden here already so it's it's screaming hard to crop up good so now the only thing that we have to do is to apply the adverse Jones formula in this annealed approximation so where we have taken the logarithm out of the integral so I will remove everything okay so we apply now the adverse Jones formula in the annealed in the annealed approximation so we have the end in the limit n to infinity so we have that rho n to infinity minus two over pi n the limit epsilon to zero plus the imaginary part of the derivative with respect to lambda of now what we had the average over log of z and now we just taking the log of z annealed right so we take log of z annealed of lambda now we know that the logarithm of lambda will go for large n as exponential minus n into a certain function bingo so here we need to put exponential of minus n into phi lambda of q star right then you have the logarithm of the exponential we can kill we can kill them both and then we will have a factor minus n and the factor minus n will cancel out so we will get a nice large n limit so if we do that we will have a factor of two which is still there a factor of pi and then we have the limit epsilon to zero plus of the imaginary part of the derivative with respect to lambda of phi lambda of q star phi lambda is this one and we need to evaluate it in q star now the derivative with respect to lambda of q sorry of phi lambda of q star will be in general using the chain rule will be the first derivative of q star with respect to lambda times the derivative with respect to q of phi lambda of q evaluated in q star plus the derivative with respect to lambda of phi lambda of q the explicit derivative evaluated at q equal to q star and now this object is equal to zero because we are at the stationarity condition so all we have to do is to take the derivative with respect to lambda of phi lambda of q explicitly so not through any dependence of q star on lambda so what we have to compute is the derivative with respect to lambda of the only part of phi which depends on lambda explicitly which is this one so let's say one-half the logarithm of epsilon plus i lambda minus two q where I took the minus away so I put this one in the numerator all this evaluated for q equal to q star so if you take this derivative you get one-half i over epsilon plus i lambda minus two q evaluated for q equal to q star and now we have q star here so we can just plug it in here and what we get is one over two lambda epsilon minus four q star yeah this was an extra step that I was taking well this is one over lambda epsilon minus plus root of lambda epsilon square minus two so I'm taking the derivative with respect to lambda of the only bit of phi which depends explicitly on lambda so not through q star because we are at a stationary point so the first bit is equal to zero now we are taking this derivative and we evaluate the derivative at q equal to q star and if we do that we obtain this object here and now it's just a matter of computing of computing this object here so rho n to infinity of lambda is two over phi limit epsilon to zero plus of the imaginary part so we have taken the derivative and we obtained this object here so the imaginary part we only need to extract the imaginary part of this object taking into account that lambda epsilon is lambda minus i epsilon so we can rationalize this denominator and we get lambda epsilon plus minus root of lambda epsilon the imaginary part so this goes upstairs with the sign change and there is a factor of two that cancels this object here so this one over phi limit epsilon to zero plus now I let you carve out the imaginary part of this object using the fact that root of a plus i b is p plus i q where p is we use this identity several times root of a squared plus b squared plus a and q is sine of b over root two root of root a2 plus b squared minus a so if you carry out this imaginary part of this object here what you get is precisely one over pi root of two minus lambda squared plus of lambda is smaller than root two and zero otherwise so even using this screwed approximation of kicking the logarithm outside at least for a fully connected model like the GOE we manage using the Edward Jones with this approximation to carve out the exact result that we know is a semi circle of course this might be an accident it might be non convincing so in the next lecture I wanted to redo this calculation but using the quenched so using the exact formula in connection with the replica replica trick and again we will after a much longer calculation we will land on the semi circle so I think I am going to have a shower I am done for today