 That's it. Okay. Thanks Mateo for your kind words I'm honored and and humbled But it's prestigious Invitation to come and give a series of lectures on Random Physics theory, which is the field. I entered about 12 years ago and which still surprises and amuses Me I have prepared a couple of handouts. There are some spare copies Here one is on Let's say the history of the of the subject. There are some original papers and some curiosities So and the other one is about numerical numerical simulations. So I hope that Together they will help us form No round view of of the of the field from from multiple and different perspectives I In preparing these lectures, I Assume that while most of you may have heard about random matrix theory Probably none of you had have none of you has received like a formal training on it And you haven't worked on on research problems in in there So if this assumption is not turns out not to be correct, then we can we can tweak the the pitch a bit So in in summary, although I my aim is to cover pretty advanced material in the end I will try to start as smoothly and and softly as as possible in In essence random matrix theory is the happy marriage between linear algebra and probability probability theory and the fundamental. So let me So random matrix theory commonly denoted RMT well, we can say is the sum of linear algebra and probability theory in in essence the the main problem of of random matrix Theory can be formulated in extremely simple terms if I give you Let's say a matrix x which is of size n by n this matrix is Described by a certain joint probability density for the entries So we have a matrix with with random entries described by a certain joint probability density and out of this input we would like to say something about the Eigen values Okay, so this is the the the general setting of random matrix theory I give you a matrix with random entries and I want to say something about the eigen values of this of this matrix good, so Of course, I could in principle spend a lot of time trying to convince you that this type of setting is very useful It it comes up naturally in many in many different fields But actually I'm Decided to follow another another route So experience has taught us that sooner or later everyone comes across a random random matrix Okay, so I will let time work for for me And I will just focus on giving you some some tools So no freeze account on how to do calculations on random matrices, okay I will skip all the Motivation part because it is boring and because you will learn yourself that random matrices are useful without me telling you this Good, so the first calculation that we can do The simplest and most basic calculation, which is at the same time very instructive has to do with two by two Sorry two by two Random matrices. Let's call it spacing Distribution, okay, so the the question is is there a simple we have a two by two matrix x1 x2 x3 x3 so this is a real symmetric matrix These two elements are identical and we take x1 and x2 as Gaussian random variables with mean 0 and variance 1 Okay, so they are taken from the probability density function 1 over root 2 pi exponential minus 1 of x square and We take the off diagonal. So these these elements are all independent these three elements and and the off diagonal element we take it as Gaussian with mean 0 and variance 1 half So it is taken The off diagonal element is taken from a PDF of this type Now you might already start asking, okay Why are we picking the the variance of the off diagonal element equal to one half the variance of the diagonal elements? There's a quite deep reason behind it and I will try to explain why we do this but other than that other than this simple Difference the setting is pretty pretty simple, okay Now, what is the what is the question? I want to compute the probability density function of the spacing The spacing s between the two eigenvalues So if this one is the largest eigenvalue and this one is the smallest eigenvalue I take the difference between the two and I want to compute the probability density function of this object clearly this This object s is a random variable because the eigenvalues are random variables since the entries of your random matrix of your matrix are random So you have the two eigenvalues. They are real because the matrix is Real symmetric and I want to compute the probability density function of this object Excellent, so well if we don't have many more sophisticated tools The only thing we we can do is to compute the characteristic equation Compute the eigenvalues of this object in terms of the entries Compute the characteristic function compute the eigenvalues in terms of the of the entries compute the the spacing and then try to work out the probability density function Can I raise here? Good, so the characteristic equation for a two by two matrix has a particularly simple form So it is a second-degree equation lambda square minus the trace of x times lambda plus the determinant of of x So in this in this case the trace of x is just x1 plus x2 and The determinant of x is x1 x2 minus x3 square from which you can compute the two eigenvalues Which are one half x1 plus x2 plus minus root of x1 plus x2 square minus 4 x1 x2 minus x3 square So from from this expression we can compute lambda 2 minus lambda 1 so the spacing between the two So the spacing between the two eigenvalues is Lambda 2 minus lambda 1 it will be just given by the square root Which we can simplify to be x1 minus x2 square plus 4 x3 square Okay, so from from this point onwards. We can forget That this object came from a random matrix problem Okay, so in essence what we have is a function of three random variables x1 x2 and x3 and we want to compute the probabilities Density function of this object given the probability density function of the original variables We can completely forget that this object came from a random matrix problem now how do we compute the PDF of This object well we we just apply basic properties of functions of random variables so what we have is That the PDF of s Will be given by what it will be given by the integral over the PDF of the three variables so this one as root 2 pi This one at a root 2 pi this one at a root pi Then we have exponential minus one-half x1 squared plus x and then we have an exponential or minus x3 square Without the one out because you remember there was a factor one out Enforces the fact that s the definition of that so the definition of s is root of x1 minus x2 square plus 4x3 square So this this expression gives the probability density function of a function of three random variables Given the joint distribution of these three variables So if we are able to perform this Integration we will have our answer Now this integration can be if you notice the structure we have a square a sum of two squares here so the the natural thing to do is to go to Polar coordinates in a certain combination of x1 x2 and x3 So the the the right change of variable is probably x1 minus x2 will be our cos theta We will have 2x3 should be our sin theta and then we can Use x1 plus x2 equal to psi this will Help us in the in the calculation. Okay, so we we make this this change of variable and this integral so this change of variable Gives rise to this inverse change of variable so now We all know that we need to compute the Jacobian of this change of variable and Express the integrand in terms of the new of the new variables if we do that the Jacobian is Given by the partial derivative of x1 with respect to our partial derivative of x1 with respect to theta partial derivative of x1 with respect to psi and the same thing for the other variables so You can you can work this out. I just give give you the answer this cos theta over 2 minus r of Sin theta minus cos theta over 2 r theta over 2 one half Sin theta over 2 R cos theta Over 2 and 0 you can check it easily by taking the partial derivatives of the of this scheme here And then all you all you have to do is to compute the determinant of this of this object So if you compute the determinant of the Jacobian you get That the determinant of J is minus r over 4 So all the angular dependence drops out and We will have to take of course the absolute value of this of this object Okay, so if we do that. This is just simple algebra You need to take derivatives of this object and compute a three by three determinant, so we'll not Spend more time with okay, so What we have here is That p of s is equal to what is equal to 1 over 2 pi Root to pi times root to pi and here we have a root pi okay Then we have a factor one fourth Coming from the from the Jacobian So we put in here a factor one over four and then we have the integral over r Which is the radial variable The integral over theta Which goes from 0 to 2 pi and the integral over psi Which remember was the sum of x1 and x2 and so the range of this object is minus infinity to plus infinity Then what we have here is the absolute value of the Jacobian, which is R because one fourth has been already taken care of So we stick in here an extra r Then we have a delta function here of s minus This combination here is root of r square. So it is just r And that's where the simplification comes comes from and then we we will need to rewrite this object in terms of the new integration variables So if we do that you get exponential of minus one-half so we have x1 square which is r cos theta plus psi over 2 square plus Psi minus r cos theta over 2 square and then we have we've taken out a factor one-half. So we add 2 times r square over 4 sin square of So this one is x1 square x2 square and this one is x3 square But there was a factor one-half missing. So I'm correcting for it. So now this is the integral we we have to solve We are of course helped a lot by this Delta function. So using this that the function we can kill one of the integrals the r integral and exchange every occurrence of r with an occurrence of s So if we do this If we perform this calculation Yes No, because you need to take the absolute value of the of the jaco of the determinant of the Jacobian matrix Right because they the change of map the change of measure between two integration region cannot cannot have a negative signature Right because otherwise you are mapping positive regions into negative regions So the determinant is minus r over over 4, but you need to take the absolute value. Okay, so From here we have equal 1 over 2 pi root pi times 1 over 4 and Then we use this delta to kill the r integral. So we get a factor of s here But we need to be careful because the integration over r is between 0 and infinity Okay, so when we use the delta function to kill this integral We also need to impose that our s should be between 0 and infinity. So s should be positive This is clear because s Physically is the difference between the largest and the smallest eigenvalue. So it has to be positive So we can put a theta function here, which means that s must be positive So net then we kill the theta. Sorry. We killed the r integral And we are left with two integrations of this type exponential minus one-half and then just write s square cos square theta plus psi square plus 2 s psi cos theta over 4 This is the first the first bit here where I replace s Here and expand the square plus psi square plus s square cos square theta minus 2 s psi cos theta over 4 plus s square over 2 sine square theta Okay, so I'm just expanding the squares there and replacing every occurrence of r with an occurrence of s and And and then it's just a matter of algebra to just sum these bits up For example, you see that there is a cancellation like this guy here and this guy here Cancel cancel out and and then we have yes, so sorry Any question? Can I raise here? So After a bit of algebra and just we get 1 over 2 pi root pi 1 over 4 Then we have s theta of s Then we have an exponential of minus s square over 4 Here and this this comes from from these bits So we have s square cos square theta plus s square cos square theta So this is a 2 s square cos square theta divided by 4 so it is one-half which Combines exactly with this object so cos square theta plus sine square theta is equal to 1 and the theta dependence drops out Note that this can only happen because we took the variance of the off diagonal elements to be one-half of the variance of the diagonal elements So had we taken the variance equal between the two then the theta dependence wouldn't have dropped out and We would we would get still an integral over theta to perform which would give rise to a best of function Okay, so only because of this factor of of 2 we can Simplify the theta dependence, okay, but this is not the end of the story. So this factor of 2 has much deeper Consequences, okay, we will we will see later Okay, so then we have the theta dependence becomes trivial Okay, so this is just a factor of 2 pi and then the psi dependence is Also easy so we have exponential of minus sine square over 4 Which is just a Gaussian integral giving rise to root 4 pi so the integral is Done and if you absorb all the constant we get that p of s is for basically s positives s over 2 Exponential or minus s square over 4 So this is the final the final result This is the probability density function for the spacing of a 2 by 2 Gaussian random random matrix Okay, so let's let's plot it. So you see that for s going to zero this object goes to zero linearly So it goes like this Goes up and then the exponential decay wins and so it goes to zero So this is the shape of the probability distribution probability density function for the eigenvalues of a 2 by 2 Yeah, yeah so if you if you pick if you pick the variance of all the elements like Different from from from each other then you wouldn't get this this the simple the simple form Okay, but there would be a way to rescale the final result in such a way that that all the results would fall on top of the same universal curve, okay so that's Yeah, that's that's a bit a bit trickier You're talking about the mean right the mean value. Yeah, so the mean value is not that important the variance is Excrucian Yes, yes, you can you can you can do it Normally, you don't you don't do it just because it makes the calculation more difficult without adding anything But the variance is excrucian so the choice of the variances is more crucial Now this this curve that that we have just computed is so important in random matrix theory that it has a name So it is named after one person who contributed a lot on On the field of random matrices. So this is called a Wigner's surmise Wigner's surmise. Okay, and now here comes the first Well, you can find you can find something on the Wigner's surmise in the in the handbook in the first pages of the of the handbook of the So here's a paper by Wigner on page two of the handbook where he basically Did exactly the same calculation that I that I was doing here. So it's page two. Yeah Well, I mean this one was just we're just toy model to It's hard. It's hard to tell what kind of practical problems You would like to solve with a two by two Gaussian Gaussian random matrix I chose this example just because it is very instructive and it tells you a lot about how to perform Calculations and what type of information you get about eigenvalues of larger matrices. So typically when you when you cannot do a calculation for an n by n Matrix the best thing you can do probably is to go to a two by two case try to work it out And then try to infer what would happen for for larger for larger matrices So I wouldn't I wouldn't take this example as a true representative of the type of problems that we encounter every day But this this Wigner's surmise tells us a lot about what happens in general. That's why I chose this this example Okay, I wouldn't read too much into it Okay, so The the reason why it is called Wigner surmise. So this is one of the first Examples where the the name given to things is not really Appropriate and and in random matrix theory. This is this happens all the time. So surmise in in English surmise means to think or infer without certain or Strong evidence this is just taken from the diction Without certain or strong evidence So it is it is hard to speculate why it is called as a surmise given that it comes from an exact calculation Okay, now it is called the Wigner surmise because the the story was the Wigner was attending a certain conference on neutron scattering and And people ask the question about what would be the typical spacing of resonances in in a scattering in a neutron scattering Experiment and then he just walked up to the blackboard and he wrote Basically this this formula and was like I think it should be like this And then from from that moment onwards it is called the Wigner surmise even though there is no surmise I mean it comes from from an exact calculation Okay, but you will you will read it all the time So I thought I should you know I should point this point is out Now what is this? Picture telling us this is a very instructive point You see the really important region is is this one? Why because for s going to zero The probability density function goes to zero So it it means that the probability of finding two eigenvalues that are very close to each other goes to zero It means that the two eigenvalues don't like to stay too close to to each other So one eigenvalue feels the presence of the other This is quite striking because we started off from a matrix with independent entries So the entries are independent They don't talk to each other but the eigenvalues do because the second eigenvalue doesn't like to stay too close to the first one Okay, so this is a very Very important point that survives also for larger n and it is called level repulsion So the eigenvalues of random matrices generally not just for two by two and not just for Gaussian Measure don't like to stay too close to each other They also don't like to stay too far away from from each other But this is less, you know less less striking the most important thing is that they talk they talk to each other Even though the entries are are independent if you want an analogy you can think of the eigenvalues as for example birds that are perching on an electric wire or Parked cars on on a lane. You don't like to stay too close to the car ahead of you You don't like to stay too too far away or when when you when you park Of course, you might think okay, this is totally crazy This analogy doesn't doesn't work But if you go on page three actually people went people measured the beginner surmise in the position of birds that are perching on on an electric wire and In the in the distribution of spacing between parked cars and they found a very good agreement Okay, so I don't read too much into into it I think I've heard a Story that the guy who did this experiment with park cars was actually questioned by the police Because he tried to measure the spacing of park cars and we're like what are you doing? Oh, I'm trying to demonstrate the big new surmise and they were like, yeah, yeah come with us good So this is a very this is a very important point a very important feature of random matrices is that eigenvalues do repel Okay, they talk to each other even if the entries are are independent Excellent now just to Okay, well Just to complete the picture. Can I raise here? Of course, I forgot to tell you in the clearly in the in the other handouts The one on numerical simulations you can of course test this prediction very very easily So I did it like in Matlab. You can easily generate a two by two matrix diagonalize it compute the eigenvalues and put and and computer histogram of the of the spacing and and what you've got is is exactly the The the the beginner surmise shaped is on page three of the of the other handout So so you see that the number of events Where two eigenvalues are very close to each other is is negligible. It's vanishing a small now to Of course, we we should compare actually this situation with what happens in the case of IID random variables, so if you take independent may Possibly identically distributed random variables, which is not the case of the eigenvalues Okay, so the I the independence here is on the entries not on the eigenvalues But if you take truly independent random variables, what would be the distribution of the spacing? Let's let's compare the two the two cases. Okay, and and we will appreciate the difference. So in this situation The calculation so we can perform a Calculation that is quite general so it doesn't depend on the distribution of the individual random variables and So I must say just to motivate you a bit that although I search for a long time For this type of derivation the one that I'm going to do now I failed to find a single reference where this calculation was was performed in full details from top to bottom so So what I'm telling you is so I'm giving you a Gift that you will not find this derivation anywhere Or probably you will and that was just stupid and I didn't couldn't find okay Okay, so Just for comparison. What is the the law for the spacing of IID? Random variables So the the setting Everybody knows what IID means no D means independent and identically Distributed So the setting is you have a set of random variables x1 xn Which are described by a joint density Function which is of the form p of x1 p let's say px of x1 px over xn So the the joint distribution Of the n random variables is equal to the products of individual distributions for each random variables, which means that the random variables don't talk Don't talk to each other and They individually follow the same distribution So we also define So this is the probability density function for one single random variable and associated to a probability density function we can introduce the cumulative Distribution function Let's call it f of x which is basically the integral up to x of the pdf So the cumulative distribution function Is basically equal to the probability That one any of this random variable is takes values smaller or equal To x okay And you obtain it by integrating the probability density function For all the values up to to x okay, so We want to can I erase here? So this is the setting and So in order to perform the calculation of the spacings So what what I what I'm after is the following thing You have a certain domain on the on the real axis for example the full the full real axis and you start throwing darts independently Okay, so you get like one here one here one here with a certain probability distribution which is the same for every toss And then after a certain number of of tosses you you ask what is the distribution of the nearest neighbor? Spacings between the tosses okay Clearly if you think about it suppose that the pdf has this type of shape Okay Then this means that with high probability you will always toss your dart around this region Right the region where the pdf has its its highest value So it is highly likely that you will get a situation like this A situation of clustering where all the darts just end up on the same region Which means that most likely the probability of adding Small spacing will be high right because Because the the the the tosses will cluster Around the position where the pdf is is higher So this this situation is markedly different from what what we've seen in the case of random Random matrices where two the two tosses would end up spaced apart Is that clear? This is just an idea. We'll try to formalize it Good So in order to formalize it We can define an object Let's call it pns Given that xj is equal to x So this is a conditional pdf so given that One of the random variables Let's say xj takes The value x So this is a conditional probability Given that one of the random variables xj takes the values x that there is another random variable xk So with k different from j At the position x plus s and no Other variables In between so what i'm saying is the following we We condition the probability to the fact that in the position x you have one So one one dart has landed at the position x already. So it is there Now we ask given this fact What is the probability that there is another dart? Which has landed at the position x plus s And all the others have landed either to the left of x or to the right of x plus s So there is an empty space in between So the the claim is that so i'll give you the formula for this object and then we'll try to To understand why it is like this So the formula is like this The probability density function of one single variable computed at x plus s Times this object One plus f of x minus f of x plus s Race to the power n minus two so f of x is the cumulative distribution function for each individual Random variables. So why is this formula true? We we know that there is a random variable sitting at x. Okay, so we can forget about about this one. It's it is there Now we want a spacing of size s between two Random variables. So we want that another random variable has landed in the position x plus s And this happens with probability p of x of x plus s Just by definition. This is the probability density function of any random variable to land at x plus s But this and then we want this space to be empty So we want the n minus two random variables that are remaining To be located either to the left of s of x or to the right of x plus s So the probability So f of x is the probability that any of of these variables Has landed to the left of x Because it is the cumulative distribution function And one minus f of x plus s Is one minus the cumulative distribution function. So it is the the probability That this random variable has landed to the right of x plus s And all this is true for all the remaining random variables, which are n minus two So we need to multiply the probability together So this the So this is the final The final expression. This is an exact expression which is valid for Any pdf and any number of Random variables. So it is an and it is a completely exact Result, okay Note that this is not the object that I promised that we should we should compute Okay, because this is a conditional pdf. We assume that one random variable is at at x We we we want now to lift this this condition. We don't want any specific random variable to sit anywhere specifically Okay But but this expression this way of proceeding is is quite clear at least to my To my eyes Good. So now how do we move forward from from here? Well, then we can We can say, okay from here we can compute the probability The conditional probability of having a spacing s given that any Of the x is at the position x Not just the the jth Random variable And this one we can compute it using the law of total probability So this is the sum over all random variable of the probability. We've just computed Times the probability That x j is equal to x Okay So if if we want any of the random variables to sit at x not not just the x jth Then we need to sum over all the possible possible occurrences of this event for all The random variables x one x two x x n But this calculation is is simple because this probability is just the pdf So it is px of x and this summation just becomes, you know, we are summing Identical terms. So it just becomes a factor of n So what we have here is n times p n s x j equal to x Which is the object we just computed times the pdf of a single random variable So now we are getting closer to the object. We want we want to to compute except that this We are still not done because this is still a conditional probability. We have one guy sitting at x Which we we don't want we want to compute the histogram No matter where the Initial particle is the initial seed is Well, so What shall we do? Any idea just to to remove the x to remove the dependence on on x Sorry Yeah, okay. So how would you compute that? Okay, we're we are very close Thanks to your inputs So we we are here now. This is still a conditional probability and we have computed one of the crucial ingredients Which is which is here now all all we have to do is To compute the spacing pdf we can just integrate This conditional probability over x Because the initial point can be anywhere. Okay anywhere on the support of the original So sigma is the support of your initial p px of x Okay, so in summary So this is equal to n integral the x p n As given that x j is equal to x times the probability So this is the probability basically that the first the first there dart has landed at the position x So this is the completely general formula for the probability density function of the spacings between two consecutive random random variables irrespective of where the first one was um the first one landed Now it is uh, it is interesting. I haven't put it in the in the handout you can write like a short. Yeah Um Well, um, this is um, this is uh, no wait No, I don't think so right because um here here you're saying that x x j So one one of the x's is staying at x, but you don't you don't know which one which one this is You you're you're fixing one, but you don't know which one it is So it can be number one it can be number two or it can be number n and you've got n of them n of them not not n minus one this one no why Well, you you're you're you're putting one at at x plus s No matter which one which one this is and then and then the n minus two I mean you can do it with the with the permutation with the then you would have a sum over with with a binomial coefficient But you can you can show that this sum Turns out you can re-sum this expression to get this this one This is just a quicker a quicker way of course You can you can compute the permutation because you can have the second one To the left and the fifth one to the right or or you can change the two So you would sum over the binomial, you know coefficient times blah blah But then you can re-sum this expression to get exactly this this binomial It's uh, if you if you expand this this binomial you will you will see Okay, so um I haven't put it in the uh in the handout, but actually this this final expression you can Now put uh on on a computer for for example in matlab you can generate Random variables taken from from from a Gaussian distribution or from an exponential distribution You can compute the histogram of all the nearest neighbor Spacings and then compare this object with this expression For example for an exponential distribution you can really compute these objects and you can compute the integral analytically if you want Okay, because uh, you just need to put This object in in in here and perform just an extra Next integration and then you can you can really you can really see how the histogram and the theory Would uh would match this is an exercise that that I suggest you to do and if you are if you are interested We can we can sit down and do it together at some at some point in in matlab It's really a one line for for exponential distribution you sample random variables You take the diff and then you histogram the the diff and then This integral you can perform analytically so you will you will find the perfect agreement between the theory So this is the the object we we were after in the end except that the expression is not really Informative it is not as evocative as as the big nurse bigger surmise So maybe we can do something better here so we can take the large n limit to find a more Striking and universal result which does not depend on the pdf of each single toss In order to to oh, okay. So first of all, uh, this this is a pdf. So it should be normalized So I ask you as an exercise to check that if you compute this integral It is correctly normalized to to one This is a pdf. So it must be it must be normalized so sigma Sigma is the is the support of the individual pdf So for example, if the the pdf of each single random variable is exponential Then this integral would run from zero to infinity if it is a gaussian It would run from minus infinity to plus infinity if it is uniform zero one then it would run from zero to one So the beauty of this derivation is that it does not depend in So there is no It's completely universal in the sense that it is valid for any pdf Any original pdf of your of your random variables. You are not assuming that it is a gaussian or an exponential in fact if if it has if if it has A fold it is that it is too general problem So that's why we are we are trying to compute the large n limit to extract some universal universal feature So this is correct, but it is correct for Everything so we cannot really read off what what it looks like if we were to plot it Okay But we are almost We're almost there and then we can make a break This is just to give you another Like a reference point Yeah, sure Okay, so just just do this exercise It is very instructive to show that that your pdf is correctly normalized the pdf of the of the spacing This is an exact result for any n and any pdf So it is a very strong very strong result very important so important that you don't find it anywhere excellent Good. So in order to extract some some information here, we want to make what is called a local Local change of variables, so I just write it and then I explain what what I mean So I want to go from our random variable or better Our variable s here To another random variable that I write s hat over n px of x So I want to measure the spacing between two consecutive Random variables in unit that are measured according to this You know unit length, which is capital n times the local pdf. So the pdf at the point x So why are we why are we doing this? because suppose that we have A lot of random variables, so we just toss a lot of darts So clearly if we toss a lot of darts For example in this domain, so if we toss two darts The situation will be like this we toss three four five and then You know we toss millions of darts Then as you may imagine the typical spacings between two darts Will go down right So the typical spacing between two darts two tosses Goes down if n increases But it also goes down if Around a certain position x the probability Of you know throwing a dart there is is high Because in this in this type of situation Then you will get many more tosses around here than you get here So it is it is natural to scale Your spacing between two consecutive tosses To make it clear that these two Objects the number of tosses and the local density of tosses are somehow washed out So this is the idea of measuring the the spacing in in in some units taking out the Purious effect of the fact that if you if you toss too many darts or if the local profile of the of the pdf is too high then you would have Smaller spacings not not because of of the of the fact that the spacing is intrinsically smaller But just because you have too many tosses or too high a local density profile Okay, so you want to take out this this ingredient from from the From the problem And you can do it Simply because this is just a change of variable okay So a change of variable in probability Is performed like this for example We want to Going back To the conditional distribution You know the first object we we computed the conditional distribution that one of the uh random variable xj was at position x But this time we computed it we computed In terms of this new random variable of this new variable. Sorry, it's not a random variable So we just take the formula we had before and we compute it On this Uh object So the the formula we had before was p of x of x plus s And this becomes x plus s hat over n px of x times one plus f of x minus f of x plus s hat over n px of x To the power n minus two So I just I just took the original formula that we derived and I'm computing it in this new In this new variable not s, but s hat over n times the probability density The local one which depends still on x Okay, and then what I do is I try to compute the What happens for for large n to this to this formula? So if if I throw a lot of dots Then uh, well what I can do here is I can expand this object for large for large n using a Taylor expansion This will be f of x Plus s hat over n px of x f prime of x plus And then you see that that we are we are uh on Something because f of x cancels with with this guy here So this f of x and this f of x go away Because there's a minus sign here And then we have one minus this object so F prime of x is it is the derivative of the cumulative distribution function So it is the probability density function So this object here cancels with this one And what we are left with is one minus s hat over n to the power n minus two Which goes to One minus s hat over n to the power n minus two for large n Exponential of what? positive or negative Exponential of s hat positive So one minus s hat Because if if I add a plus here Yeah, so I'm almost done. So Is everybody happy with this limit? One minus something divided by n to the power n coffee So I assume that we are all happy with this And here what's what's happening here if we retain just the the first the leading order in in n This is just px of x So for large n this object is just px of x times exponential or minus s hat Okay, this is valid for large n. Let's say Okay, now we are basically done because we need to yeah We need to inject this object Into this formula here To find what what's happening in the case of Large n for the pdf of the spacing Not conditioned on the position of any of any particle Okay, so if we do that we have that let's say pn hat Of s hat. So this is the probability density function of the new variable s hat Which is equal to the pn Of s equal to s hat over n px of x Multiplied by ds over ds hat This is the low of change of variables for probabilities And if you do that by inserting this formula here So we have n Integral over dx px of x which is this this object here Then we have 1 over px of x which comes from here There is also factor 1 Over n. So this factor 1 over n and this object here comes from this derivative And then we have an extra px of x exponential of minus s hat Which comes from here So this guy goes away with this guy this guy goes away with this guy dx in the interval of dx px of x is equal to 1 because it is a normalized pdf So the final result is that in the scaling limit n to infinity and much larger than 1 let's say The space the pdf of the spacings in unit Of n px of x Is just an exponential So which means that if we compare it So So this one is the vigner vigner surmise this one is the scaled low for the pdf of the spacings of iad random variables. So you see that there is a massive difference In the region of small Spacings the vigner surmise tells us that the eigenvalues don't like to stay too close to each other For the case of iad random variables no matter what the pdf of each individual random variable is So this is a completely universal result in the scaling in the scaling limit the probability of finding two Particles very close to each other is the largest And and the origin of this is the fact that if you have a certain pdf that is peaked around a certain point Most of your darts will fall In there so they will cluster they will not repair So this this type of scheme is what is normally uh called like the difference between vigner dyson statistics and poisson Statistics so vigner dyson statistics is A fuzzy name for the statistics of of stuff that repair And poisson statistics is the statistics of stuff that that attract Each other, you know Of course, that's another example of a badly probably badly designed name because poisson statistics Has nothing to do with a poisson distribution Okay, it is the distribution of a poisson process, which is an entirely different An entirely different thing But as as we can see This derivation Has nothing to do with stochastic processes. It is a completely, you know universal first principle Derivation which does not Require any assumption on the specific pdf of the individual random variables. Yeah Excellent question The vigner surmise in the form I I gave you so s exponential minus s square over four or Or whatever is only valid for a two by two Gaussian Random random magics. So it is strictly valid In the setting I gave you The essential features are valid for general n. So the fact that the eigenvalues repel So if you are asking, okay, suppose that I take an n by n Gaussian matrix, not a two by two So exactly as I gave you but three by three or four by four Can we compute the distribution? The answer is yes The the formula becomes well, it is exact but not very very useful It is given in terms of an infinite product Of eigenvalues of a certain freedom determinant operator So the formula is so complicated that actually you can only evaluate it numerical But we can in principle we have we have a formula for for it But the essential features are the same and actually what what people claim is that The vigner surmise for a two by two is an excellent approximation for the spacing of of larger larger and Matrices so there is a deviation of a few percent Close to the close to the top But essentially it is a very good. It is a very good approximation But strictly speaking it is not true that the vigner surmise holds for for larger matrices well, I mean the The the fact is that what you're what you're doing is you're saying I want to separate two effects the fact that the two random variables want want to stay close Closer to each other and the fact that they are forced to stay close to each other Just because there are too many of them or they need to all to land There just because the probability distribution is is larger So in some sense you want to unfold Your set of random variable and make sure that that they are stretched In such a way that their mean level spacing is equal to one across the spectrum Okay, so you want to to make particles that are too close to each other not because they wanted to be There but just because there are too many of them Or because of the dense of the local density there is larger than the local dense So you want to make the local density somehow uniform across the spectrum Yeah, exactly Yeah, exactly So you calculate you calculate the spacing such that after you have made the the density of your particles uniform So you are basically washing out the the fact that two particles are Need to be too close to each other not because they want they want to but just because there are too many of them Is that sort of clear? Okay, so, um, well that's time for um For break Yeah, we start again good so We've seen something about eigenvalues of random matrices then we We went one step Back to the situation of iad random variables now we come back to the to the topic of Random matrices. So what I wanted to do now is to offer You what I call a layman classification random matrix models Again, this is something Which I believe is is very important. It is so important that you don't find it on on textbooks. That's That's the usual the usual thing. So this is my second gift of the day So what why I call it a layman classification because if you google classification of random matrix models then you get into a domain of mathematical physics and It becomes Increasingly and immediately very complicated. I just wanted to give you like a map Conceptual map to find your way On this field without getting too much Tangled in in technicalities so we have Matrices let's say n by n Which are characterized by a certain joint probability density Of the entries so in in this This series of lectures I will impose a requirement And the the requirement is that x has real spectrum so the eigenvalues lambda one lambda n are real if this is not true then we have a second branch of random matrix theory which would Take me probably the same number of hours to to discuss So I don't I just don't have don't have time So we will discuss only matrices with real spectrum. So eigenvalues that are real The entries can be complex not only that they can be also something else, but the eigenvalues must be real numbers And well actually to be even more Restricted x we will take it as belonging to one of of these three classes So x can be real symmetric So the entries will be real number and the matrix will be symmetric This ensures that the spectrum is is real by the fundamental theorem of algebra And to real symmetric matrices, we will associate a number beta equal to one Okay, so when I will talk about beta one ensembles we make I mean I will mean matrices that are real and symmetric Okay, beta is is called dyson Index of the ensemble. Okay, just the definition Or they can be complex Hermitian beta equals to two Complex Hermitian so matrix with complex entries, which is Hermitian Now give me an example of a two by two complex Hermitian matrix Sorry Yes, another one What do I put here? Sorry zero Okay What do I put here? My side What do I put here? Okay So or you can have like So an Hermitian matrix is a matrix with complex Entries, which has real entries on the diagonal and complex conjugate entries on the off diagonal Okay I'm sure everybody knew this but you know, maybe not good Or it can be x can be Can have quaternion elements It can be a quaternion self-dual matrix Which I personally hate so I will not discuss I will not discuss that but you can have matrices with quaternion elements, which are characterized by A Dyson index beta equal to four Okay, for the beta equals to three k's Then you will need to look up Frobenius theorem On division Algebras Which will tell you why we have real numbers complex numbers Quaternions, but we cannot have turnions Okay So we cannot we cannot define a division algebra with three imaginary units Okay So we will only need to Deal with with this Three cases, which are beta equals one two and four Okay now If if this is the universe we are We are working on which is not Which does not exhaust all the possibilities, but it is just to be Now this is the universe That we are working on The universe of random matrices that belong to these categories Real spectrum Now within this universe we have two classes of random matrices that are of paramount Importance so we can We can draw a diagram like like this So the first The first class of random matrices Lies here. It is the matrix matrices characterized by independent Entries So here we have Random matrices Whose joint distribution of the entries factorizes into the product of individual Distributions one for each entry so the p Of let's say x 11 x n n Is equal to a certain p 11 x 11 times p n n x n n Okay Modulo, of course the symmetry the symmetry requirement So if we have like a real symmetric matrix matrix Then we only need To consider the upper triangle because all the remaining random variables are are just defined automatically by the symmetry requirement Okay, so independent entries. This is a a nice class of of random matrices characterized by this this property The second the second class of important random matrices to consider Well, the defining property is a bit less Obvious to to define. I will try to do my best So the second category is called a rotational They are characterized by rotational invariance So rotational invariant models They are characterized by this By this property that p of x The x 11 the x n n So the joint distribution of the entries Of of these matrix models Is equal to p of x prime V x 11 prime V x n n prime If x prime is equal to u x u minus 1 So this is the definition now. We're trying to to explain what what this means So we have one matrix x In the ensemble We perform some sort of similarity transformation or our rotation and we obtain another random matrix And this random matrix has the same statistical weight As as the previous one. So the two matrices occur in the ensemble with the same probability Okay, so this is this is a very Involved properties because it involves like a non-local transformation of all the entries Okay If you if you imagine to write the entry ij of x prime in terms of the entry ij of x Well, it will involve all the entries of Of x So it is a very complicated transformation property of the joint distribution such that It turns out After all this operation has been has been made that the probability distribution of one one random matrix and The rotated version of it are the same no matter what the rotation matrix is So it is a it is a pretty strong constraint okay I will I will show you like with with few examples How how this Is is put in into practice Now you've seen uh here that I put like an intersection between these two Classes so you might wonder okay, what? Um, what lies in the intersection and here the the story becomes quite Um quite interesting. So first of all Remember that x can be could be a real symmetric complex Hermitian or paternal self dual So if x is real symmetric Then u is an orthogonal Matrix If x is complex Hermitian, then u will be a unitary matrix And if x is quaternion self dual, then u will be a symplectic matrix Which by the way, I also hate due to the fact that I hate the original Okay Questions, sorry, uh, you can't read problem So if uh here if x is real symmetric Then u will be an orthogonal matrix If x is complex Hermitian u will be a unitary matrix And if x is quaternion self dual, then u will be a symplectic Excellent You don't need to know I will absolutely don't waste Not even three minutes of my life describing what a symplectic matrix is Because it's a total mess and it's completely boring But I'm happy to discuss it later Okay Sorry It's just that I really it gives it gives me physical pain I'm just joking But I have more more interesting stuff to to tell you so that's that's why Okay Good Um But there is a an interesting point here Uh, you see this in variance property. So x This you have this matrix x you rotate it And and the rotate version of of this matrix has the same statistical weight as as the first one So what what does this mean in terms of the eigen vectors? Of the two of the two matrices you see if if by rotating it Like once or twice or three times you don't change the the statistical weight This basically means that the eigen vectors of of this type of ensembles are not that important, right? They are they are not important because a rotation by a matrix that contains the eigen vector of it Would leave the statistical weight unchanged no matter what type of rotation. So what type of eigen vectors you have So this this class of random matrix models is interesting because the eigen vectors are not that important The eigen vectors don't don't play a very very interesting role because you can always rotate it So forget about the the original eigen vectors and go to another basis and you leave the statistical weight unchanged Okay, so there is nothing specific about that specific set of eigen vectors of the matrix x Because by rotating it you are washing out all the information about the eigen vectors And still the statistical weight remains the same Okay So in for this class class of models the eigen vectors are not important For this class of models the eigen vectors are very important Why because the joint probability density is factorized This property depends on which basis you are in If you if you make a change of if you make a rotation in the space of matrices here It will no longer in general be true that the joint distribution of the rotated matrix Will have this factorization property. There's no reason to believe that this should be the case right so so the the Here you see the interplay between probability theory and linear algebra Here we have a property that is a probability with proper probabilistic property That the joint density factorizes But this property is intertwined To the algebraic property that you have this specific Set this specific eigen space If you change if you make a rotation, then it will no longer be true in general That distribution of the entries will factorize This is this the beauty of random matrix theory that you have this interplay between algebraic properties The eigen space and probabilistic properties That you have this factorization only in this eigen space This this p let's say this this p of p of x and p of x prime So the joint distribution so the joint distribution of the entries is unchanged So if you look at the two you cannot spot the difference between the previous one and the new one Okay, now, uh, we wanted to to understand what is in uh in the intersection here I'll give you a hint based on our first And then I'll conclude We can resume Later and the hint is the very first calculation that we did Remember we did a calculation for a real symmetric matrix two by two and the probability density The joint probability density of this object Was the product of three gaussians remember Remember the the very first example we had We had independent gaussian variables But the off diagonal element had a variance that was one half the variance of the diagonal Elements, so the joint probability density is the product of three gaussians now We can rewrite this object as one over two pi root pi Exponential minus one half x one square plus x two square plus two Again this factor of two x three square Now this object x one square plus x two square plus two x three square Can we rewrite it in a more clever way? Yeah, so this object here Is the trace of x square Just do it x 1 x 2 x 3 x 3 you multiply it by x 1 x 2 x 3 x 3 so you'll find x 1 square plus x 3 square x 1 x 3 plus x 3 x 2 x 1x 3 Plus x 3 x 2 x 1 square plus x 3 square And if you take the trace, this guy plus this guy, reconstruct the factor of two that we have here. So this object here, so the joint distribution of the entries of our matrix, we can write it in terms of the trace of x squared. And now bingo. So we started off from a matrix in this region because it has independent entries, but we have rewritten the joint distribution of the entries in terms of the trace of x squared. So if we now perform a rotation in the space of x, the p of x prime becomes proportional of exponential one-half trace of ux u transpose squared. And this guy here is exactly equal to exponential of minus one-half trace of x squared. So by performing a rotation in the space of matrices, we have left the original weight unchanged. And this only because of this factor of two, again. If this factor of two was not there, then we couldn't have written the joint distribution of the entries in terms of traces. So this hint tells you that this type of matrices lies in the intersection. So it has independent entries, but it has also the property of rotational invariance. So if you have a p of x that is proportional to exponential of minus one-half trace of x squared, this lies in the intersection. It has independent entries, but also the property of rotational invariance. So now the question is, are there other models in the intersection? And unfortunately, the answer is negative. So there is a theorem, which is a theorem by Rosenzweig and Porter, a very old theorem. You can find it reproduced on page six and seven of the handout. So it is published in a basically unknown Finnish journal. So this theorem is basically proving that the only ensemble that has independent entries and rotational invariance is the Gaussian ensemble. So an ensemble of this four for any n. This is bad news, right? It means that you have to make a choice. If you want to have independent entries, you will lose typically rotational invariance. If you want to have rotational invariance, then the resulting ensemble will typically not have independent entries. There is only one ensemble in the intersection, page six and seven of the. I think it's time to wrap things up, and we continue today at 2.30, right? Okay, thank you.