 introducing this counting function n of x, we manage after a series of manipulation to write this object for large n in the form of a functional integral over counting functions that are positive or at least non-negative and normalized one of exponential of minus beta n square into a certain energy functional plus subleading orders that we are able to ignore. And due to the long range structure of the interactions between particles, this factor here is not n as in standard short range models which means that the free energy of this gas is not extensive as we are typically used to see but it is super extensive so it goes as n square not just n. In particular the entropic contribution to the free energy is subleading for large n. So the entropy is always extensive but the energetic component of the free energy is super extensive so we can neglect all the other terms and concentrate on this object here. So the functional here, I gave it to you yesterday, it is just given by nxx square minus one half double integral dx dx prime nx nx prime log of x minus x prime. So the form of this functional integral lends itself to a very nice subtle point approximation for large n so all we have to do now is to find the stationary profiles n star of x which minimize this functional. And in order to do this we need to find the solution of the formal solution of this stationary condition which is this equation here. Now performing the variation with respect to a certain profile n of x of this functional here we are basically led to the following equation. So let me explain it. So this term here is due to the functional differentiation of the first integral. This term here is due to the functional differentiation of this integral taking into account that when we differentiate over nx as a function we've got two functions here that are multiplied together. So we need to first differentiate one of this and then differentiate the other one which gives rise to a factor two overall that's why here in front you don't have a factor one half. And then there is a mysterious constant appearing here and the source of this mysterious constant is due to this constraint here. So remember that we need to look for a solution of the optimal density profile which is normalized to one. So we can enforce this constraint by simply multiplying this object here by a delta function that enforces the constraint. Then we can exponentiate this delta function and if we put this extra exponential in the action the old result after differentiation is to introduce another constant here. So this constant must be fixed in order to enforce the fact that n star must be normalized to one. Good. So now we can forget the problem this equation came from and we just need to solve this equation. So what kind of equation is this if you were to define it to a mathematician? Yeah. So this one is an integral equation because the function that we want to determine is inside the integrand. So this is the unknown that we want to determine in such a way that this equation is satisfied. So now the technical challenge is how to extract how to find out a function that when integrated over a logarithmic kernel will reproduce just a quadratic behavior plus a constant. Okay. Of course now I put like a suffix omega or w here because due to the fact that the range is up to omega this function will be a parametric independent on omega as well. So it will be a function of x and of the location of the barrier. So if the barrier is at omega we will have a function of x but also function of the barrier because if the barrier is here the profile will change. Is that clear? The way to solve so in the mathematics literature this type of equation of integral equation is known as Carlemann's equations and there is a standard technique to solve this type of equation. The standard techniques is well first of all we make an observation on the behavior of this object for large negative x. Okay. So the question the claim that I'm making now is that n star omega of x cannot have unbounded support. What do I mean by this? Remember we will have a barrier at omega and we argued that most probably something will happen here. We will have an accumulation of charges at the barrier but the question is what's happening for large negative x? Can this function have an unbounded support meaning can it extend all the way to minus infinity? Now the claim is that this is not possible. It is not possible because if you send x to minus infinity here in this equation then this object here will basically behave as log of x. Right? So if you take this out here log of x this object here will be equal to 1 the remaining object because the density is normalized. So you will see that there is no way you can ever balance an x square and the log x behavior. So if your density has an unbounded support then this integral equation will not be satisfied which means that we expect there should be a cutoff here. So your density should have a lower edge as much as we know the semicircle has a lower edge. So we expect that the density profile should have something like a behavior like this but it will not extend beyond a certain cutoff. Okay. So the idea to solve this equation is first to differentiate it with respect to x. So if you differentiate this equation with respect to x you need to be extremely careful from the mathematical point of view because here we have a non-analytic point, a non-analytic dependence on x. We have a logarithm of the absolute value of x. So there is a formal procedure to make sense of the derivative of this object with respect to x. It is in the notes and it is the notion of weak derivative. It is just a technical point if you are really picky on math. In the end the final result is basically what you expect. So you differentiate here and you obtain an x and then you formally differentiate this integral as you would do normally. So you would have an omega star x prime and then here you would have a 1 over x minus x prime. The constant here will give a derivative equal to 0. But then you see that there is a problem here because when x prime becomes equal to x you might have a non-integral point here. You might have a logarithmic divergence at x. So how do you cure this logarithmic divergence? You take the Cauchy principle part. So there is a formal mathematical way to go from the derivative of this logarithm to the Cauchy principle part and it is in the notes. But now I don't think there's any reason to spend more time on this. This is finally the equation that you need to solve. This equation will be equivalent to this one. And now why is this trick interesting? Because this type of singular, this is a singular integral equation because it contains Cauchy singular value. The nice thing is that we have an explicit inversion formula for this type of integral equation. So there is a formula giving the value of this function as a function of the known potential here. You see we get the term x here because we started from a Gaussian model. Suppose that we started from a model characterized by a generic potential of V of x. So what we will have here? Something like the first derivative of the potential. Probably with a factor of one-half. So this object here will be the same for all models, this principal value integral, but this object will be model dependent. So if we start from another random matrix model which is invariant characterized by another potential, here we will have the first derivative of that potential. Good. So the inversion formula for this type of singular integral equation is due to an Italian mathematician whose theorem is if you have a singular integral equation of this form, principal part between a1 and a2, dx prime, divided by x minus x prime. Then rho of lambda is equal to 1 over pi root a2 minus lambda, lambda minus a1 into a certain constant minus principal value integral of dt1 divided by over pi root a2 minus dt minus a1 over lambda minus t times g of t, where c0 is a constant. So we have a singular integral equation. This is rho and this is g. And the final formula gives rho as a function of g. So all you have to do to solve this integral equation is to set here, instead of g of t, you just replace it with t. Excellent observation, except that it will not. But your observation is very good. So you see here we have that probably the support of our density, the upper edge of the support will be omega and we are sort of fine with this formula. But the lower edge of the support, we know that it will be finite but we don't know where it is. So what we really have to do is to solve this integral equation from a certain lower support up to omega, or if you want up to a certain value a2, which in the end will turn out to be omega. And then we will have to find a way to fix this a1. So you see in this type of integral equation problems the support of the density is another unknown on top of the density itself. So in some sense the edges of the support of the density arises as a phoenix from the ashes of this integral equation. So not only you need to find the density but you need also to fix the support of it. And how do you fix the support of it in the end? Well, remember that this equation came from a minimization problem of basically this function. So what you have to do is you solve this equation for generic a1 and a2 for example and then you plug the result back into the energy functional and then you minimize the energy functional with respect to a1 and a2. That fixes the... So in some sense the solution of this integral equation is not in one-to-one correspondence with the minimization problem we started from. The classes of solution, the class of solutions of this integral equation is much wider than the one we need. We only need one representative from that class, the one that minimizes the energy function. Okay, so all you have to do now is a lot or a bit or a lot of algebra just replacing t here performing this integral. But it is a long exercise but not particularly appealing. And I can just give you the result which is n star of lambda is 1 over 8 pi root of a2 minus lambda, lambda minus a1. So I consider that the upper edge is a2, just to leave it completely general. And then this is the result, 8 plus a2 minus a1 square plus 4 a2 plus a1 lambda minus 8 lambda square. So this object here is the general solution of the integral equation, of the singular integral equation we had before with edges a1 and a2. You can check, just take this function, plug it in here, compute the integral and you will find x. So what you have to do now, well what you have to do is to plug this object into the energy function. So the energy function was... So you plug this formula into this integral and into this integral, you just solve the two integrals. Of course this minus infinity here will be truncated at a1, right? And so this object, once you have computed the integral will be just a function of a1, a2 and omega. Because you are just computing the integral so the x dependence will be washed out. What remains is the dependence on a2 and a1, which are the edge points of your support. And then what you do is you minimize this function with respect to a1 and a2. So you find a way for your density to be supported exactly on the support that minimizes the energy function. So is the procedure clear? I've skipped several steps, but at least in essence the steps should be clear. Good. So all you have to do is a series of integrals and a minimization property. If you do that, well first of all you get the... Once you have minimized this function with respect to a1 and a2, well then you can go back here and replace the values of a1 and a2 that you found. So the optimal density has this shape. If omega is larger than a2 you obtain that a1 equal to minus a2 equal to minus root 2 and the density has the shape that we expect. So if the barrier is larger than the upper edge, a2, then the upper edge becomes equal to root 2 and the shape in between is the semicircle, which is the situation that we expect because the barrier is ineffective. So the gas stays in the same configuration where it would say as if the wall was not there at all. But when the wall is inside the C, then the profile gets modified and we get a non-trivial result. The non-trivial result, well I just give it to you. L of omega is a non-trivial function. So you see that indeed as we expect we have a square root divergence of the spectral density at the position of the wall. So our density really looks like this. It has a square root divergence, so an integrable divergence here at the position of the wall because all the charges are accumulating there. They would like to hop over if they could to reconstruct the semicircle, but they can't because they are forced to stay to the left of this impenetrable wall. If omega is equal to root 2, then we expect expression becomes equal. And so we are exactly at the marginal situation where the semicircle is reconstructed from both sides. So the problem is in essence solved. All you have to do now is to plug this expression back into the energy functional. And this energy function will be basically the free energy of your problem, right? Because we know that the original integral goes as exponential minus beta n square into this object evaluated at the southern point. So rare deviations of the largest eigenvalue to the left are described by a large deviation function which is nothing but this functional evaluated on this object. It is just a matter of computing two integrals. Good. So all this derivation in full with all details is in the nodes that I will scan. And I also ask Erika to upload a couple of papers, like some review papers, where they basically redo this derivation in full from top to bottom. So you should be able in principle to reconstruct all the missing steps. If there are things that are not clear, just drop me an e-mail and we will try to search it out. Okay? If the number of e-mails gets extensive, then I will give you my Skype, then my phone number, then my home number. Okay? Then you can organize individual meeting, tutorials, stuff like this. Okay? All free of charge. Just for you. Okay. So this was, in essence, one application of the Coulomb-Gas technique for a specific problem. And I wanted to change, as I promised, to change the setting completely and discuss a new branch of mathematics, which I find particularly appealing, and its connection to random matrices. So as you see, I'm trying to give you some tools and techniques, some general tools and techniques to tackle problems that you might and you will get in your professional life. Okay? So the tool that I'm describing now is, well, it's something that I would have liked, I would have loved to invent myself. It's another one of the various things that people have done before me. And it's called, it has a fancy name, so free probability. Okay? So I will describe the problem in, again, in very simple terms and see what this construction, why this construction of free probability allows us to solve it, at least in some special cases. So the problem is as follows. You have two matrices, A and B. Let's say they are N by N and, well, they are Hermitian, but the problem can be formulated in more general terms. This is just to set the stage. Okay? So the problem that we have is, we want to find the eigenvalues. So A and B are not random. So they are, for the moment, fixed matrices. So the problem is to find the eigenvalues of the sum of A plus B. Now, this is a linear algebra problem and we know that already at this stage the problem is extremely complicated, right? Why? Because to know the eigenvalues of A plus B, it is not sufficient to know the eigenvalues of A and the eigenvalues of B, right? So the ingredients that you need are the eigenvalues of A, the eigenvalues of B, and the relative position of the eigenspaces. Because if the matrices A and B don't commute, they cannot be simultaneously diagonalized. So you need to know how the eigenspaces of the second matrix are oriented with respect to the eigenspaces of the first one. So in order to solve this problem, we need to have much more information than we would like to. We would like to solve this problem by saying, well, we know the eigenvalues of A and the eigenvalues of B, then we are done. This is not the case. Okay? Now, what if we want to circumvent this problem and we want to define an operation? Let's call it a non-standard addition of matrices that depend only on the eigenvalues of individual matrices and not on the relative position of the eigenvectors. Now, this seems like a crazy idea, but what we can do and this is where random matrices come into play is we can define a sum. Let's define it like this with a plus inside a box A plus B. And we define it in this way, U dagger A U plus V dagger B V. U and V are random unitary matrices, let's say, drawn uniformly from the unitary group. So let's think about it for a second. What we are doing here? We are taking the first matrix and we are randomizing its eigenvectors. We are taking a second matrix and we are randomizing the eigenvectors and then we are summing the two. So we are obtaining now a random matrix because one specific instance of this ensemble depends on which U and V you picked. So this object is not a deterministic object. This A plus B is a random matrix. But the good thing is that we have washed out completely the information about the eigenvectors because we have randomly rotated, we have randomly shuffled them. So the information that prevented us from computing the sum of the eigenvalues before, which was the relative orientation of the eigenspaces, now we have reshuffled the system completely and at the price of introducing randomness into the game, we now have a hope of performing this summation and that this summation only depends on the eigenvalues and no longer on the eigenvectors. You see? Is the general philosophy sort of clear? Now, of course, originally the matrices A and B were fixed, were deterministic but we can redo the same procedure in principle with A and B random themselves. So they might be random themselves and we randomize the eigenvectors. So if your A and B are fixed, then you draw the first U. Yeah, you can... Well, but how does this make a difference if you are averaging over the unitary group? Well, I mean, you and V are fixed in the sense of, you know, you pick an instance of U and an instance of V and you get an instance of A plus B. So this will be one random instance of the sum. So you pick a sample, a first sample, so a first unitary matrix and you do this plus a first sample of V, right? So this would be one sample of a unitary matrix drawn from the unitary group. And so you get a matrix C, first sample, then you redo this operation, sorry, this is B, with the second draw from the unitary group and you obtain a second sample. So it's an instance of random matrix. Yes, yes, yes. Yeah, so the whole point is that now you have a probability distribution on these matrices C which is induced from the probability distribution that you have on U and V, right? So all these elements, all this matrix here will be different from this matrix here, but they will have some statistical property. Yeah, exactly. No, no, you will have made the two, I mean, the two eigen space, there will be an angle between the two eigen space, but in the end it will not matter which angle this is. While if you are summing two deterministic matrices without doing this reshuffling, there will be an angle between the two, but it matters which angle this is. No, no, you are not integrating. What you are in the end, the goal of this process is to show that in the limit n to infinity, the density of eigenvalues of this probabilistic of this random object can be determined only knowing the density of eigenvalues of A and the density of eigenvalues of B. So you don't need to know any information about their relative eigenspaces. So the price to pay is that you are introducing randomness in the system, so you are talking about density of eigenvalues, not a precise correspondence between the first eigenvalue of the sum and the first eigenvalue of each of the two individual summands. But the advantage is that you have washed out completely any dependence on the relative positions of the eigenspaces. And this brings us to the concept of free-ness. So free-ness, when I first learned about it, I thought that it was an excellent and brilliant idea, and I'm still convinced about it. So free-ness. Free-ness is the, we can define it as the generalization of the concept of independence between random variables for non-commutative objects. So you see, when we think about random variables, we never think typically about whether they commute or not, right? So if we take two random variables, two standard random variables x and y, then we assume that x times y and y times x give the same result, right? This is of course not true for matrices. So the notion of free-ness generalizes the concept of independence between random variables for non-commutative objects. So we are injecting this extra ingredient into the game, into the probability theory game of whether they commute or not. So free-ness is defined for random matrices. Of course it is a larger concept, so it applies more generally than to random matrices. But specialized to random matrices, it means that free-ness is the combination of three ingredients. The three ingredients are n to infinity. So it is very, very hard to define free-ness for finite n, the size of the matrices. Free-ness is an intrinsically infinite dimensional concept. Then we have that a and b are statistically independent, and this is expected. So the matrices, the way you sample the entries of your matrices a and b should be in a statistically independent way. So the entries of one should not know anything about the entries of the other, and this is expected in some sense. So you don't want to have built-in correlations from the start. But the third ingredient is the fact that eigenvectors are in generic position. So the precise way the eigenvectors of a and the eigenvectors of b are located is not important, should not play a role. Another way to achieve this is by this randomizing operation. Now if this was just a definition or a fancy way of describing a new type of probability theory, then we wouldn't care. But the problem, the point is that you mean the mathematical definition? Well, it's a bit more complicated. But it's essentially a condition on the independence of certain combination of trace moments of the sum. But in essence, it boils down to these three properties. Now if two matrices are asymptotically free, then we can apply a theory that gives us the spectral density of this object here in terms of the spectral density of A and the spectral density of B without any reference to eigenvectors. Now what is this theory? Let's draw a parallel to the standard probability. So in standard probability you have an object, so in classical or commutative probability you have that if you have random variables that are independent and are drawn from certain PDFs p of xi of x, and we construct the sum of these random variables. This means that you have washed out by uniformly randomizing the eigenvectors any information about what the real eigenvectors of your matrix are. So your matrix will have some eigenvectors, but by combining it with random matrices drawn from the unitary group, you are performing random rotations. And then you are summing the randomly rotated versions of your matrices and you obtain a matrix that is not deterministic it is a random matrix because for each choice of your random rotation matrices you will get a different sum. But the problem is that this one will create an ensemble of random matrices and we are interested in the eigenvalues of it. Good. So if you have the sum of independent random variables there is an object that we can use to compute the statistical properties of the sum which is, there is Matteo there. Well I just recognize this object. So this is the characteristic function. So if we are summing standard random variables we know that the characteristic function of the sum will be the product of the characteristic functions of the summands. And if we take the log of the characteristic function, so we define g of xi of t as the log of phi of xi of t, then this relation becomes additive. So g of s t is g of x1 of t plus g of x l of t. So this object here is the generating function. The characteristic function is the generating function of moments. The log of it is the generating function of cumulants. So this is just a crash course on classical probability if you have the sum of classical random variables. Now what happens in the context of free random matrices? Can we draw a dictionary between the two fields? Can we find some objects that behave in the same way for random matrices? Then the answer is yes. So the main object in the game is what we have already seen at the beginning of the course. Remember this object which is called either resolvent or for some funny reason it is nicer to talk about the Green's function in this context and I will explain you why in a second. So let's call it the Green's function. Why is this object important? Well we can rewrite it as lambda rho a of lambda divided by z minus lambda. So now there is a very interesting paper that I also reproduced in the handbook by Tony Z Low of addition in random matrix theory and he is a very funny guy so what he wrote in the paper is for the sake of convenience due respect to Green somewhat fancifully define Blue's function. So we have the Green's function this object here and now we define the blue the blue's function. The situation now is taking a turn that is almost surreal because now we have papers with titles like 50 shades of blue. So the situation has gone slightly out of hand. So what is the blue's function? The blue's function is defined very simply it is the functional inverse of the Green's function. So you have g a computed in b a of z is equal to b a computed in g a of z in this equal to z. So the definition of the blue's function is just the functional inverse of the Green's function. And remarkably the blue's function satisfies the free addition so the blue's function of the sum of L free matrices is equal to bh1z plus bh2z plus bhlz plus 1 minus L over z. Now this one is a standard summation. So this is a standard summation between complex functions. So this h1 hl are random matrices which are summed according to the sum operation. Or alternatively they are random matrices that are rotational invariant. The two things basically coincide just because matrices that are rotational invariant have their eigenvectors already washed out. So this operation here can be also, okay this term here might be a bit disturbing because there is this. So what we can do is we can define the r function as the blue's function minus 1 over z. So this is called the r function and in terms of the r function r function is strictly additive. So the r function is for, yeah, so there is a theorem. So if, let me write it here. There is a theorem which is basically a sufficient condition for freeness is let n by n random matrices an and bn such that an and bn have the asymptotic eigenvalue density for n to infinity. This is just technical, we don't care. An and bn are independent meaning that the joint distribution of the entries of a and the entries of b factorizes into the joint distribution of the entries of a times the joint distribution of the entries of b. And bn is a unitary invariant so a rotational invariant ensemble. So at least one of the two is a unitary invariant is drawn from a unitary invariant ensemble. Then an and bn are asymptotically free. So we can use an plus bn in the same way as we would use an plus bn. These two things are interchangeable. So they produce an ensemble, the ensemble sum has the same statistical properties in the two cases. So you see that the r transform here has the same property for random matrices that the cumulant generating function has for classical random variables. So it is additive. So what is the usefulness of the r transform for example? And then I will give you an example. So the r transform is the generating function as the cumulant generating function is the generating function of cumulants for standard random variables. The r transform is the generating function of objects that are called free cumulants. So it is a complex function that you can expand in a series. So r for example property 1 r of z can be written as summation n1 to infinity k nz to the n minus 1 where k n are called free cumulants. So n to infinity n to infinity is in here. The fact that the eigenvectors don't matter is basically inside here because one of the ensembles is unitary invariant. So the eigenvalues don't matter there. Because the statistical weight is unchanged if you perform any rotation. And the third element was independence. The fact that the first matrix and the second matrix are not trivially correlated through their entries. This is obvious, right? If you sample the entries of the first matrix in a way that is correlated with the entries of the second one then you are breaking the free-ness just from the very beginning. These are sufficient conditions. It's not a necessary condition. But if this happens then automatically the standard sum reproduces an ensemble that is exactly equivalent to the ensemble that you would get taking a free sum. Now the concept of free cumulants is a bit complicated. But we can use the same intuition we have in classical probability to try to find a way to assign a meaning to these objects. So for example, in classical probability if you take a Gaussian pdf like a standard Gaussian, what are the cumulants of a standard Gaussian? The classical cumulants. The cumulants, please. The cumulants of a Gaussian. Sorry? So the first cumulants are identical to the moments. The first moment, which is 0. The second moment is the second cumulants would be equal to the second moment if the first moment is 0 and it is equal to 1. And all the other cumulants 3, 4, 5 are 0. So all cumulants are 0 but the second classical cumulant which is not the same object as this one, which is equal to 1. Now we could, we can ask the question, what is the, if we ask that the second free cumulant is equal to 1 and all the other are 0, what is the corresponding corresponding r of z and what ensemble does it correspond to? Yes. But now we need, you know the r function is related to the blues function which is related to the which is the inverse of the greens function. So from this chain of connections we may go back and find out what the greens function corresponding to this r function is and so we will have some information about the ensemble that is analog to the Gaussian in classical probability, right? You see the logical series of steps. Ok, so we want the second cumulant to be equal to 1 so r of z is equal to z. Now we need to know what relation exists between r function and the greens function. Well we know that the blues function is the functional inverse of the greens function. So it means that if we compute the r function with an argument, using as an argument the greens function we have, so this equation and the definition of the r function implies this relation here which implies that g a of z is equal to 1 over z minus r a g a of z. This is the fundamental relation between the greens function of a given ensemble and its r transform. Now people who are expert of quantum field theory might recognize something in this are there experts in quantum field theory? Just keep this no, I will tell you. So this looks very much like a first Dyson-Schwinger equation in quantum field theory where here you would typically have what is called the self-energy contribution. If you haven't heard of any of these just forget it. I'm just flashing it for the expert. There is actually a deep connection between the two things, but if you haven't heard of it, don't worry. Ok so all we have to do this is a general relation and now we apply this general relation to the r function which is exactly equal to z. Second free-cumulant is equal to 1 and all the others equal to 0. So the equation that we have if r of z is equal to z this implies 1 over z minus g a of z. Do you remember this equation? Well it's funny because it's the only equation involving the greens function that we ever saw. So this gives g a of z equal to 1 half z minus root z square minus 4 and if we apply the Sochotsky Flemish formula we obtain that rho of lambda is the semicircle. Ok so we have obtained by this formalism that the semicircle low and the neutral density, the semicircle low plays the same role for the addition of random matrices that the standard Gaussian plays for the addition of classical variables. In particular the semicircle low is stable upon free addition as much as the Gaussian PDF is stable upon classical addition. So if you sum classical variables that are all Gaussian distributed you get a variable that is Gaussian distributed and if you sum random matrices that have the semicircle as the spectral density you obtain random matrix that has the semicircle as its spectral density. So there is some sort of interesting parallel. Of course this does not happen in general. So in general you can sum different random matrices and you will obtain a profile for the density which is neither of the summands. So what is the algorithm using the R function? What is the algorithm that you can use to compute the spectral density of a sum? Yeah, that's true although there are the classes of free stable, like of free stable of stable free lows is larger. So this argument would not work for other free stable distribution. It is consistent, yes. Actually I think well as a curiosity I think the only low which is stable both in classical probability and in free probability is the Cauchy low. It is true that if you sum classical Cauchy distributed random variables you get a Cauchy distributed random variable and also if you sum matrices whose spectral density is Cauchy you will get a matrix whose spectral density is Cauchy. But that's the only one where the two things, it's the only low which is stable in the two cases which is stable and free. So suppose that you have like a matrix A1 drawn from a certain rotational invariant ensemble and the matrix A2 drawn from another rotational invariant ensemble just to consider the simplest case plus maybe A taken from a rotationally invariant ensemble and you sum these matrices producing you sum every instance of this matrices producing an instance of the matrix sum and this matrix will be a random matrix. So if the matrices are in the limit n to infinity you can compute the spectral density rho s of lambda from the spectral density of rho of A1, rho of A2 of Al. How you do that? Well you know the spectral density of these objects so from each individual spectral density you can compute the Green's function. So you can compute the Green's function of the first matrix the Green's function of the second matrix and the Green's function of the Lth matrix. This is a somehow easy operation because you only need to perform one integration from the average spectral density in the limit n to infinity. Then what you do from the individual Green's function you can compute the individual Blue's function which are the functional inverse of the individual function. So you compute the Blue's function of the first one the Blue's function of the second one and the Blue's function of the third one and of the Lth one. Now from each individual Blue's function you compute the r transform by simply subtracting 1 minus z from each of them. So you compute the r transform of the first one and the r transform of the last one. Then what you do you sum them all up and you obtain using the result that I gave you without proof you obtain the r transform of the sum which is r of a1 of z plus r of aL of z. Now you have already hopped over on the other side, on the side of the sum. Now you can proceed backwards from the r transform of the sum you can compute easily the Blue's function of the sum. From the Blue's function of the sum you functionally invert it so you get the Green's function of the sum and from the Green's function of the sum you work back and you obtain the spectral density of the sum using the Sochotsky plumage formula. So the problem is solved. Just to show you how it may work in practice in the handout there is a small mistake, there is a part that is missing so I will send an updated version just copied, it just got cut out. So in the numerical handout on page 26 I will directly do this thing numerically for the addition of two random matrices one GOE and one random matrix which belongs to the so-called Wishart ensemble you don't need to know the details. So what happens is that the GOE has the semicircle as its spectral density which has a spectral density which is positive so the eigenvalues are all positive and it looks a bit like this. So I wanted to know what is the spectral density of the sum of a GOE and the Wishart matrix and they applied this algorithm. I take the semicircle and this function here I compute the Green's function, I functionally invert it I transform sum them up and then go backwards and I obtain numerically, analytically the spectral density for the sum which I then compare with numerical simulations these are given in the final page of the handout so you see the solid curve is the spectral density that you obtain from this algorithm results are results from numerical simulations, so diagonalization of large of the sum of large matrices. So you see this basically is a hybrid, the spectral density comes out as a hybrid shape of the two so it is a slightly elongated semicircle so yeah but the point is that you can compute the spectrum correctly, it comes out from a calculation so why is this method interesting? Well because sometimes in real life you might have to compute the spectrum of the sum of random matrices or of matrices that can be approximate by random matrices or you might have one matrix you need the spectrum of which you cannot compute but you can decompose this matrix as the sum of two simpler matrices and then maybe you have a chance to compute the spectral density of the original matrix by composing a free addition of the two building blocks that are simpler. So this root was actually taken in a few papers that are reproduced in the handouts. So for example on page 40 there is the Edelman's group paper who published quite interesting PRL where for example they consider the Anderson Hamiltonian which is, do you have the handouts with you? The equation 8 is the Anderson Hamiltonian so you have basically a three diagonal matrix with random elements on the diagonal and hopping terms of diagonal and so what they try to do is to decompose the Anderson Hamiltonian as the sum of a diagonal part and a hopping term part with zeros on the diagonal. So this is the composition of line A in their paper. So they were like let's suppose that these two matrices are asymptotically free so we have a diagonal matrix whose spectrum is simple plus three diagonal but with zero on the diagonal whose spectrum is also simple because all the elements are equivalent so it can be diagonalized using Fourier modes. And then they applied this free addition algorithm which in principle should not work because there's no reason to believe that the two matrices should be first of all they are not rotational invariant because they are sparse but okay let's, we don't know what to do just try it. And actually they found that there is a very good agreement between the procedure I described here for the spectral density of the Anderson Hamiltonian and the one that is obtained using a free convolution of the two spectral density of the diagonal plus hopping term. This is the figure one in this paper. So although this is not widely known and maybe it will not work in general but it gives an idea of one possible way to use this result when you want to compute the spectrum of a matrix that you don't know what to do with you might try to decompose it as a sum of two matrices that are easier whose spectrum is easier to compute and then try to you know to apply this algorithm to put the two together. Yep, so this object here is the average spectral density so this is not a statement on the probability distribution of the eigenvalues but this is the statement over the spectral density which we have defined in that way averaged over the sum ensemble in the limit n to infinity and provided that this is the definition then it is an exact statement. So we can reconstruct the average spectral density in the limit n to infinity of the sum from the average spectral density in the limit n to infinity of the sum months with no reference whatsoever to the eigenvectors. Yeah, yes. Yes, but unfortunately it has been done. Okay, so there are two parts of your questions. So the first one is yeah, so the first one is suppose that you have a correlation, let's say an empirical correlation matrix which can be decomposed as the true correlation matrix of your data plus a noise and if you assume that your noise is rotational invariant so if you assume that your noise can be modeled in this way then in principle you can use the tools of free probability to compute the spectral density of the sum in terms of the spectral density of the unknown signal and the spectral density of the noise about which you assume to know something. Yes, yes, because you can factor out, since this measure is uniform on the orthogonal group or unitary group and the measure as the composition property so the product of 2R measure is a R measure matrix you only need one of these. So the second part of your question is what information can extract from the spectral density well actually this is not probably the best thing to go and look at, you can do much better than this because you can go and look at the relative position of the eigenvectors between the unknown signal and the empirical correlation so for example can you reconstruct the leading like the principal component from the position of the principal component of the empirical matrix can you reconstruct the true one from the empirical one now this type of problems has been studied using a combination of free random variables and replica trick by Bouchot and Potter's. There is also a review on this recent review on the archive and they call this like oracle estimation because it seems that in the limit n to infinity you can any dependence on the unknown things which is the true correlation matrix basically drops out so you can although in principle this ingredient must be there to say anything actually there is some miracle that happens in the limit n to infinity that any dependence on the true signal is basically drops out so you can get basically a universal I mean well there are some caveats but the combination of free probability and replica theory has been used already to address the problem of cleaning correlation matrices in presence of rotational invariant noise actually the there is something that could be interesting so how do you prove I gave this result without proof so how do you prove the law of addition for R transforms so there are essentially three proofs that three ways to prove this one so one is combinatorial which is which is messy the second one is using diagrammatics and it is even even messier the third one is using replicas and this is really beautiful it's a really elegant proof of the law of addition for random matrices using replicas you can find it in the appendix of this paper and actually it was so beautiful that I really wanted to do it for you so I started reproducing all the proof and so there is only a slight problem that there is one step in between that is not completely correct and I can say it loudly because it has been confirmed to me by one of the authors so I didn't want to to give a lecture containing one step that is not correct so the final result is correct but if you want to fix so it is not just a typo it is a bit more complicated so if you want to fix it the problem becomes slightly more delicate so I just didn't want to it would occupy too much time but the general philosophy of the proof is really beautiful so you can go and have a look at it but just bear in mind that one formula in the derivation is not correct as stated and if you look at it for five minutes you should be able to understand why pretty clearly so there is something on the left hand side that cannot be equal to something on the right hand side it is pretty obvious and I can say it although I have these three people in the highest esteem but one of them confirmed that there is an obvious glitch there so I hope they will fix it in the final version but the derivation becomes a bit more not so straightforward in fact when I first saw it I was like wow it is like one page calculation and then you get to this beautiful result but if you try to redo it properly it is about eight pages so it is yeah so you have to study spectral density in the product of random matrices so it is possible to make a link with the joint probability distribution in classical probability and say that the probability if we look from a quantum point of view the probability that the probability that we have A is in the probability where B is so it is possible to say that from I am not sure if I got exactly the whole point of the question the general answer is there is a theory for free products of random matrices as well as there is for free sums the free products thing is more complicated for one simple reason because even if you take two matrices A and B which individually have a real spectrum in general it is not true that the product A times B will have a real spectrum so the theory is more complicated because it needs to allow for the possibility of having eigenvalues that are complex so you want to have a theory instead in all this setting that I described all the eigenvalues are real and sums of real eigenvalues or matrices with real eigenvalues will have real eigenvalues so there is no problem but the theory is developed so you have a nest transform instead of an art transform and you have a different low to compose to recover so you will not have a simple addition low like this the low is slightly more complicated but if you are interested in the theory I mean there are like in four pages or five pages you have everything there is actually a review online on free products which is with examples so you will immediately understand how the system works it is by Gislof Burda it is called something like free products of random matrices a short review or something if you are interested I can ask Erika to give you the exact so the theory is developed also I just used the addition the sum case because it was it is simpler but the theory is there it is developed also for some and products of non-ermition matrices it becomes more and more complicated but you can deal with this type of problems very efficiently now so it is by now it is well established but it is something that happened in the last I don't know five to six years so it is really I mean maybe I made a mistake in the choice but I just wanted to show you something that might be relevant for your own research interest or might bring a bell in ten years time if you will ever face a problem like this so then you can look it up just to know that it exists you know it is not something that that is normally taught in any undergraduate or even graduate courses because it is something really that happened in the last five six six six years so it is good for you to know that there is something out there ready to use if you'll ever need it so I'm not sure if maybe I just botched it badly but I just tried to give you this this type of flavor of what people are doing today okay so with this I think I'm going to conclude so it was my privilege to teach you this course I hope you enjoyed it the second part will happen in ten days from now I'm jetting off just immediately and we will deal with sparse random matrices and replicas next in two weeks so to give you a couple of example calculations top to bottom with all the factors of pi and square root two and yeah that's it oh yeah that's also the example yeah you can go