 Just like the casserole, the valentines, all of these. As you remember, we completed a discussion of scale-up game fairies, and we had begun to discuss gauge fairies. And we said we would start our discussion of gauge fairies in the simplest... So gauge fairies are more interesting and more complicated than scale-up game fairies. And we said that we would try to build up our discussion of gauge fairies by starting with the simpler scale-up game fairies. And so we started studying an example of a gauge theory in 0 plus 0 dimensions. Okay? So this is not even one of the mechanics. It's just a matrix integral. This matrix integral had a gauge invariance. So we had one matter field. This matter field was a commission matrix. This matter field transformed into the adjoint of the gauge group. So the gauge group was 0. But it is said that a gauge theory has aiming fields, but mu runs from 1 to d. Since in this case d is 0, there are no aiming fields. So all we had is a Lagrangian built out of the matter fields. The only reverent of gauge fairies is that the Lagrangian is invariant under UN gauge transformations. The Lagrangian invariant, S of n, is invariant under M, which is the general integral of this problem, though it's not too difficult. We'll bring in a specialised to the following. Notice that any function... So we said that we would specialise to evaluating the partition function, which was Z is equal to integral Vm exponential of minus n media trace. The problem was an arbitrary function. Arbitrary function was published from below. It does not remind me why we put the factor of aiming in here. We were interested in taking... In general, we won't be able to solve this integral. We'll be able to solve it in the larger element. So why would we... So we were interested in solving this larger element. Why would we put the factor of n behind the integral? Tell me your mind. So the naive reasoning that we used, which will be justified by our solution, is that the naive reasoning we used was that you get non-trivial integral with measure and the energy of the state. Now the measure was of order e to the power of n squared, because the n squared different matrix elements are going to be integrated over some defective volume. So we read to the power n squared, so we read to the power of n times n squared. So we wanted the action also to be of order n squared. A trace is the sum of n elements. So the trace already has n. That's why we put an extra element there. We'll see how that works out. You have class related element. So the value of the class is not coming. You have class related element. It's not going to be. Okay. Excellent. So now we want to evaluate the syntax. Now, we discussed last time that any matrix, any equation matrix can be written as you need the equation matrix and you need to have the diagonalist. Okay? So we wrote the summation matrix as you need the universe. And then we noticed that u disappeared from the action. Because for instance, if you take trace of u to the power n, it's just trace of the same function of the n. Okay? So we realized that basically this integral over n, which was an integral over u and an integral over d properly same. It could just be written as an integral over d with an appropriate measure of n. So you can do the integral over u once and for all. And that integral doesn't care about what's in the action because that's independent of u. Okay? So what we're going to do is effectively change the values from integrating over n to integrating over the eigenvalues of eigenvalues of this matrix. You remember we discussed the analogy with the integrals of spherically symmetric functions of three dimensions. If you have a spherically symmetric function what you're going to do is the integral dx dy dz of some function of only r. But then you just go to polar coordinates and that becomes dr r squared times f of r. So doing the integral over the eigenvalues just gives us an effective measure factor which is a function of r. The analog of that would be in the case of an effective measure factor which is a function of the eigenvalues. Okay? And so we needed to find out this effective measure factor. Now we said last time that we had a derivation of this effective measure factor at least a strong motivation of this factor, measure factor last time. And I'd asked you in an exercise to try to work it out using the funding point method. You see because this is just a gauge theory. It's the simplest gauge theory in the world. And by taking m to be diagonal what we're doing here is gauge fixing. We're fixing the gauge. The gauge is that m is taken to be diagonal. Actually it does this gauge by the way to fix. Completely fix all the gauge points. Once we take m to be diagonal we fix the gauge. Just completely fix all the gauge points. First, can this gauge always be achieved? Can we always, to give an arbitrary matrix to always make the matrix that? Of course, because every emission matrix is diagonal. So the gauge can always be achieved. Okay? Now, so when you've got a gauge condition there are two things you should think about. First thing, whether the gauge can be achieved. Secondly, whether the gauge can be achieved. Secondly, whether that condition completely fixes gauge points. Now there's this condition. How many gauge transformations that leave a matrix diagonal? That's the question, that's the question. How then do you actually make decisions such that there has to be some kind of a matrix that remains diagonal? You have our primary one somewhere. Maybe I can do a comedy then. Yeah, that's a good, complicated question. What's simple? We've got to come up with that. It's a simple thing to say. Which is just diagonal. You're a tricky person. You have two matrices. You just act on each element and then by the inverse. If you've got all the matrices diagonal this is just like u1. At the intersection of u1 is tricky. Okay? So this here, this gate fixing breaks. It doesn't completely fix it. It breaks down. It fixes un. Now u1 to the body. Okay? Times. Actually there are also some off-diagonal gauge transformations that are not fixed. This is what negators are going to do. It's the off-diagonal elements that permute the line. So this gauge fixing actually doesn't completely fix it. It leaves a u1 to the body and then it's an s a, this permutation group of n n elements on things. That's just for information. When you do more certificate more you, more. Difficult for every such thing to do. Okay? All we did was to fix the edge and evaluate the final code of today. I left the evaluations in the exercise view. I motivated. I left the evaluation of the exercise view. And we said that suppose the eigenvalues of the matrix m by mi then the effective measure factor the effective final code of today was product over i m i let's say i less than j mi less than j. Okay? Square. That's a number. Actually using the value of a method it's not difficult to evaluate also. What matrix integral became that is equal to integral product over i m i minus m j square using the formula the trace of the function of mission matrix. It's the sum of the functions of the same thing. Is there any question about that? Any question about that? What is the formula? We've made a lot of progress. We have to n squared integral and we've reduced the problem to doing n integral. n in our mind is a large number. So that's progress. But it is also a large number. Actually doing these integral for now is basically impossible. Okay? So now what are we going to do is try to understand how to do this integral in the large angle. Now you see the large angle as you will see as we will now very explicitly see is very well suited for a saddle point. In the original form it was large. So what is saddle point? That's saddle point thing. Ideas basically whenever you're going to include an integral to this form that you've got exponential of something like this. This quantity is large. And so the integral is dominated with the integral in seconds minimum. However you have to be very careful. That's accurate if you see this quantity is large suppressed by we see here n squared it's the n times n times. But if the measure also has some n squared then the minimum action point need not dominate it. Okay? And that's how what was the situation here. But once we've got it in this form what are we going to do? Here we've got the number of integral that's now n. And we take this major factor which we also want to n squared. As you see exponential of some i less than j dot dot m i n squared. The major factor is now about x1 e to the power n. This quantity is not n squared. Both these are. This is because the function is sum over n squared. This is because the sum over n times n as we will see as we call it. The saddle point will be of the following form. In the larger limit what we'll have is a distribution of the item value there. Okay? So let me define a few things that will be useful for us. Suppose we define a row of n between one over n sum over l star out to m m i. It's an item value density function. Because it aint as finite this is very jangly kind of function. But in the larger limit you could imagine that there is a sense in which it becomes a nice smooth function. Very much like if we are we've got a collection of water molecules. And if the collection is a thousand and you've got a box you ask what is the density of water? The question is sort of ill-defined. The density of water is defined by a formula like this and the entire function is a positionally. And if you look at small enough regions you like to get zero or one, two. It's not a smooth function. But if you take the limit the number of water molecules is very large and you keep your boxes of size fixed. Then you get a smooth density function. So you do hide and relax. So there is sort of an idea as we will see that but in the larger limit there is density of water molecules which tends to a to a continuous function. In terms of the density of water molecules this guy can be really that is just for intuition. You can use that. Except writing i not equal to j i less than j i not equal to j divided by 2 and this square is basically identical to this missing out the terms i is equal to j. This one nth which is n terms out of n squared it's just small in the larger limit. You see because this term has this restriction i not equal to j. We have really wrote rho rho that will literally include the terms i equal to j. i not equal to j. i not equal to j. So we have to be careful to take them into account. So this is to be understood and that with some principle value in it. Principle value kind of prescription for the integral. Let's remove that. We will see how it works. The point of doing this rewriting we are not going to use it for the next 10 minutes. We are going to try to convince you that this quantity here was about n squared. Express in terms of the things that will stay finite in the larger limit. It's all n squared. This quantity now we could influence to really write this integral over i equal to j in that integral over the density of n squared. She made this a field theory with one field in density of n squared. Again, actually the measure factor for that rewriting is quite complicated and we will not worry over it too much except to note that that is all about n. I mean obviously this is product n. So to really order the larger limit you can just forget about it. So now this problem is well suited for the 7th one. All we have to do in order to evaluate this is to evaluate the stationary point of this action. So now we are going to try to evaluate the stationary point of this action. So let's evaluate the stationary point of this action. This was for motivation we will restart here. So what we want to do is to find the stationary point so what we have to do is take the subject and differentiate it with respect to each of the variables. They will be zero. So when we differentiate this with respect to each of the variables now we have to be let's just write this what we have to do is i. No When we differentiate let's say with respect to m1. This thing has m1 minus m2 the whole thing squared m1 minus m3 the whole thing squared and so on. And so when we differentiate if you forget 2 divided by sum over i or sum over j 1 over m i minus m3 I can say this with respect to m i that means there is something very simple that's minus m then we try it is the satellite point equation we wish to solve. We've got n equations for n variables solving these equations plugging back the solutions into the integral sum of all the variables that is 2 2 that's 2 it's just log but you see the term comes in both there are 2 how many factors of law m1 minus m2 are there m1 minus m2 and m2 minus m1 that's very easy to miss clear? that doesn't matter there is a log of minus or something okay it's i pi take log of that and this is not going to it's not going to it should be some overall constant see the difference between log of m1 and log of m1 and m2 there is some overall constant term we actually don't care about it these are actually complicated equations actually the many many tricks these equations were first solved by media by the famous mathematical physicist William and I don't think many many now there are zillions of ways to solve these equations once somebody solves the equation once once problem solved I'm going to use one that's sort of slick and elegant maybe not sometimes it should be suspicious of slick and elegant but I'm going to use that the slick and elegant method it doesn't matter actually beautifully but how do you think of it it's it goes like this let me define this object p of x z will do sum over 1 by n sum over i that's an arbitrary difference I find a function of x given the eigenvalues of the matrix and use that to find some function of x it's a nice function is the part actually you can make it a part maybe make it a part is to take this p and p of z square so let's say you go to 1 over square analytic structure both of the function p of z and the function p square where is the p of z obviously it is a meromorphic function with inputs there may be special points where it has to be generated it's true let's assume for these manipulations that the are equal it will not matter for the following if the eigenvalues are regenerated then the contribution will be taken in this the measure factor kills so the saddle point will not have eigenvalues equal that's all so now suppose we take these mi's and we get the distinct this object has a sum of n different poles again now let's look at the object p square p square has double poles and single poles let's isolate the double poles so p square that's just when i is equal to g and in the single poles let's just evaluate the residue the terms are i not equal to g those are the terms that are going to be the single poles and i have already evaluated the coefficient of the z minus mi all we have to do is to multiply that expression by z minus mi and in that expression sets z equals mi since we are multiplying by z we only get contributions from one of the terms which has a factor z minus mi there are two of those because mi m1 will be here this is going to give us two sum over 1 over mi minus mj we will evaluate the residue of 1 over z minus mi is 2 1 over j not equal to pi 1 over mi minus mj because this whole function dies off at infinity like 1 over 10 squared there is no there can be no constant this is so what i am saying is that this is an algebraic identity this is an exactly right simple algebraic identity that you can check every one the best way to think of it though is to decompose it in the singularity structure of z it says here this nice expression for p squared over z oh and i forgot my problem i am going in here and i am here i have to sum you got a pole for each now on where to make the only approximation of where to make the only place where i am going to use the library and really is the fact that this guy because this sum over n divided by 1 over n and sum over i divided by 1 because this guy is 1 over h because it is for 1 over h which is a single sum all i am saying is that the product to do p z p z double pole is right because z minus mi z minus mj what are the coincidences for that in my integral mj so i am going to just throw this guy away then we throw it away and see what happens is equal to equation that was true for every matrix for that distribution of mi's then obeys the starting point equation that's my equation this object here 2 times j not equal to i 1 over mi mj is simply n times 3 prime over mi p squared over z this factor of 2 goes into that sum over n sum over i sum over this is very close now i am going to do something here then write this as sum over i sum over mi minus mi prime over z nobody can stop me at this point the point of the starting subtract subtract thing is that now this normal integral can be taken as x and so this so we have d squared of z is that you know this and this are i's equations quadratic kind of equation so there is an unknown which means that the unknown function is that the unknown function this v prime of mi minus v prime of z over z minus mi that has a simple analytic structure if the original function v was a v of z was a polynomial degree polynomial degree this v prime of mi minus v prime of z over z minus mi it's polynomial for v minus 2 because v prime itself is of degree m minus 1 and then it has a 0 this kind has a 0 so there are no pauses that's the point and if i use a polynomial of degree m minus 2 it works only for v prime being a polynomial more complicated functions it doesn't easily handle there is a more powerful Russian school method that solves these equations without these assumptions if this was of course a matrix model i would teach you all methods but we'll just hurry up and we'll answer this in this way in this way i just want to warn you though something very different so if you have a problem that you want to solve in a research project and it doesn't fall in the class of the elegant method then look for the Russian okay okay this happened to me by the way and he solved the matrix model in a paper and i tried to try to analyze this it didn't work and i hope this Russian takes a look and i can't see it okay great this you know what this thing here is in terms of a few unknown in terms of few unknown numbers okay so the most general polynomial of degree m minus 2 it's going to be 0 plus a1 then plus a m minus 2 z prime minus 2 so in terms of n minus 2 unknown numbers we completely don't know let us try to see how how we can determine those unknown numbers let's first take the simplest case m is equal simple as n is equal to case v was equal to a m squared v was equal to alpha m squared so v of m we kept on from this point is that little m degree was equal to 2 whatever it is we don't even know what it is this is the one here independent of that in this case we have that for the pi of 2 so v prime is equal to alpha the difference is our discriminant equation v squared is z alpha z is c this case this case we don't even know this case we actually don't know we don't know this case because of the constant it doesn't depend on m i we just know that okay so this is a minus 2 alpha all n should have gone let's see you see there was a sum of n such as n that's the point of these equations there should be no ns left now we can solve maybe I'll call this guy v squared I'll call it v squared I'll just put it in the way that it's supposed to be I'll do a very nice equation it's an equation any of us can solve sorry I'm sorry to keep it just it will be a good thing the equation is the nicest most convenient for 1 by a squared by 2 a squared minus square root of v squared 2 is gone so 1 by a squared z squared by a squared z squared by a to the 4 for a c when 4 goes away it doesn't do anything e of z is equal to 1 by a squared you know 1 plus minus z squared let's check one thing you see we want to know whether the answers are also minus remember the basic definition of v of z p of z was sum over i 1 over z minus z 9 how does this behave at long z 1 over z let it go like 1 over z and sum over i each of these quantities n such quantities one side so this v of z whatever it should go at long z to 1 over z if we chose x of z plus sign of course it goes at long z to um assuming we're choosing you know the usual large number square root if we choose the plus sign this goes at long z to 2 z 2 z by a squared that's not good we choose the minus let's check if we choose the minus n numbers so at long z this is 1 over z squared I have missed a 2 somewhere ok ok 1 over z this is z minus z into 1 minus half a squared minus z squared into 2 cancels a squared a squared cancels and this is equal to 1 cancels what if we look at this in your friends ok so I got this p of z now some of you are thinking ok so you computed p of z but who cares about p of z I want to know what that integral was and so but you see we actually solved the problem how we solved the problem we solved the problem because let's remember again what p of z has p of z we should put 1 over z minus m i 1 by n what are the locations of all m i's of the z ok but this is a very easy way of getting it which is this that suppose I take this p of z I have partnered with the complex plane it's got a bunch of poles not any poles that are enclosed in a certain region where do I do it I just take that region and impose it in a country not sure if they so this pi r i integral over the contour z is equal to integral rho of alpha p alpha over the corresponding region integral of is this clear integral rho of alpha p alpha ok if I take some this then m of z m of z this is when you look at this thing but actually this p of z was a nice analytic function how can an analytic function give you some non-zero integral if you integrate over some region of the real axis only it has a cut to the radius when this quantity this quantity integral will be twice the discontinuity will be equal to the discontinuity percent ok so we come to the conclusion that rho of alpha is equal to discontinuity of p of z because this is the case because this is the case we can easily evaluate what our rho is now let's look at this analytic function how it is ok is greater than z squared is greater than 2 a squared when z squared is greater than 2 a squared then it's non-discriminate there's a discontinuity only but the argument of the square root will take it ok so let's take a little minute to study the analytic structure branch kind of such that when we go this way we get square root of modulus and plus i5 when we go this way we get square root of modulus minus i the other function is smooth continuous yeah smooth continuous yeah that's the function because p have discontinuity i have to subtract square root of modulus square root of modulus only between z is equal to minus square root of 2 a plus square root of modulus in the statement of the problem where we would have the discontinuity of square root of modulus i'm sorry stays the problem that has a good answer ok and then what is the value of the discontinuity what's the value of the discontinuity well i 2 pi times a so what is it suppose you can square root of minus 1 it's r or square root of minus a squared it's i times a above and minus i times a below ok so the difference is i times a that's the problem so here this thing is i times square root of modulus here it's minus i times square root of modulus ok so the discontinuity is times modulus of the squared ok so we get that this rho of alpha is equal to so what is the square root which is square root of 2 a squared minus i squared over a squared square root of 2 a squared minus z squared over how much we were in the problem so we have when v when squared by 2 a squared so when the 2 a squared would become v squared this will become 2 a squared by 2 so the rho of alpha for this problem is equal to v squared minus z squared so suppose we were equal to z squared by v squared this is the rho of alpha I need you to check that this integral rho of alpha integrates to this is the same thing the integral of rho is 2 by pi integral 0 to minus 1 to substitute x is equal to z so no I am just getting pi I messed up factor 2 I am getting 2 you see the integral would just be pi the sine cos I messed up a bit just be careful to go through this find the right factor somewhere along the way I messed up it has to be by the way going from this to this because the fact that p of z at the right was the same statement that rho integrates to so I don't even have a formula to do a z on the board because this right what was the problem of p of z something like this no I am trying to find the analytic for it this was v squared x squared because 2 a squared is equal to this does have the right behaviour so now I am making the discontinuity there is a factor of the discontinuity x is equal to cos that is x theta this becomes cos theta sorry this is cos squared sorry this becomes integral cos squared theta from minus pi to pi so I will just do half so this is my right but so we are happy we are happy I will plot the answer this formula by the way has a name it is called weakness semicircle it is called weakness semicircle rho alpha the answer is a semicircle as you can see from the formula rho alpha is equal to alpha squared from all things squared is equal to pi squared v squared after rescaling you take this right side after rescaling of so suppose I plot this by alpha by the way so it is a alpha by p into x this becomes the equation rho squared plus x squared is equal to this rho of alpha function of the my is going to be goes like this maximum at 0 minimum at the edges and 0 away from p let's see what is going on what is going on here is the formula look you have got this potential function and have the minimum at c all the eigenvalues would like to sit at 0 because that minimizes the energy but there was this entropy the entropy factor suppressed the integral it suppressed the integral when eigenvalues came together so entropy is pushing the eigenvalues apart so there is a competition and this other point value so that extremizes that gives you the compromise between these two pushes gives you the semi-center of width B as B goes to 0 the potential becomes very steep when the potential beats the entropy and this things crunches down to where they think semi-center almost all the eigenvalues as B goes to infinity the potential is very broad when the entropy is big the eigenvalues are very uniform this is this there is a phenomenon this phenomenon that the integral is dominated by a particular configuration a particular configuration of the dynamical fields we saw before while studying the Euler there is a particular value signal which dominates in the integral but it is much more non-trivial and it is a generic feature of large n it is a generic feature of large n theories and the configuration that dominates the dominates is it the dominates the integral is called the saddle point so this theory will be termed the saddle point of the of the theorem this competition would have been seen from the variability scenes from the path you did when we started the competition of quality we had measured the action which was both for n squared but the action was n squared by v squared so we went to zero level we wanted to do more comments you see the fact that a particular configuration in the large n universe dominates the integral is very reminiscent of another limit of quantum mechanics of quantum particles namely the classical classical limit of quantum particles is dominated by the classical trajectory fluctuations away from this classical limit are small so interestingly enough the large n limit is a new kind of classical limit it is a new kind of classical limit because the quantum table is dominated by a particular configuration fluctuations away from this configuration ok now how am I going to vary non-trivial kind of classical limit so non-trivial kind of classical limit because you cannot obtain the classical equations of motion just by studying the action of the theory in the case I had the classical equations of motion just by studying the action of the theory would just be prime is equal to zero that would have put all i values to zero because just the right classical equations have to account both for the action as well as for major factors and the major factors will contribute significantly change the effective classical equation what is the parameter that controls the classicality of the problem one over n not h bar it was not classical because it was like b square because the path integral comes like one over h bar the action was not classical the theory was not classical because b square was in a capacity for all i values against one over n that is controlling it ok so one over n gives you a new kind of classical limit for large elements are a new kind of classical limit for large elements it has been known for a long time in the 1970s but it has really been really spectacular prominence by the study of the large elements in the AESA in which we have discovered that the effective classical description of some large and aged theories is in fact a theory of gravity it is totally unexpected one of the most remarkable discoveries in the last 30 or 40 years we will try to talk a little bit about that ok fine one last thing to say about this now as we have repeated the emphasize we are already interested in the value of the integral we are interested in values of expectation values ok so you might be wondering look you have done a lot of work to compute this integral the integral is a number but the integral is a number you don't care about the number so what have you really learnt from this exercise of course in this theory we want to compute not just z integral dm trace m to the power k e to the power s divided by integral dm divided by z now make a claim I want to see if you can understand ok what I am making is that in order to compute this all you have to do is to compute trace m to the power k which is equal to rho which is equal to m i to the power k on the saddle point distribution so the answer to this is simply given by n times rho on power why is this true ok so you understand the claim the claim is that the saddle point is that we computed this not so much just to have to be able to really understand but to know the saddle point configuration that saddle point configuration now is sufficient to determine expectations of anything but it is evaluating on the saddle there is a claim very clear ok I want to first of course remind you the analogy of classical physics in classical physics once you know the trajectory if you want to know what was x at some t you just evaluate that x of t of that to take you you don't even think about it it is obvious ok the same thing is true here but I want to see this one suppose we have taken our original action here e to the power minus n we trace again and it plus a source of zeta some zeta n trace m to the power if we take this integral differentiate it with respect to zeta the log of this integral differentiate it with respect to zeta and then set zeta equal to 0 that of course gives us the expectation value of trace at t ok now just like for this whole problem this new problem here is dominated by some saddle point this is changing the volume of the we could influence to find that we could influence to find that so when we find what we want to because we can differentiate it with respect to zeta and then setting zeta equal to 0 we only care about suppose we can evaluate this integral z of zeta what we need is d by d zeta zeta divided zeta divided by zeta 0 this is at zeta since we already need the derivative of zeta at zeta equal to 0 it's sufficient to work out what is it what the integral is here at first in zeta now two things happen in the saddle point equations for this problem two things happen firstly the saddle point shifts to order zeta and secondly the oldest saddle point this quantity gets competition because this quantity is evaluated at the old saddle point in order to evaluate this quantity of the old saddle point we don't need to worry about the shift in the saddle point this already has zeta that would be zeta square all that remains is evaluating the classical action the change in the classical action because the saddle point is shifted now what is the change in the classical action because the saddle point is shifted and what is the derivative that's how we found the saddle point the particular saddle point was an extremity so the saddle value shifts a bit the change in the action so we get second order zeta so the change in this whole action the change in the integral the part of the integral that is what is zeta is simply this quantity evaluated on the old saddle and therefore there's a rule for us that evaluating an engagement expectation as value of an engagement you just have to evaluate on the saddle point in the output is this clear it's a very beautiful thing it's a very beautiful thing of course life at the heart of classical physics is a quantum physics the last thing I want to say about this integral before we go on to gauge theories 0 plus 1 dimensions we graduated away from the integrals that's what the plan for our discussion is going to be build up in that dimension before we get to the equals 4 which is complicated the equals 4 you understand how we turn to the discussion of gauge quantum mechanics again before we turn to the discussion of gauge quantum mechanics there are 3-4 things I want to say firstly in this particular case of of of quadratic potential we very easily were able to find the solution but in this case of quadratic potential we have just done the integral exactly okay because this is just the real power of this that is like its potential is not quadratic suppose we had for instance a Q okay there in that case in that case things would not be as simple as they were because we have this one thing we prime of n we prime of z divided by m i now the quadratic terms here would just be nothing like what we pointed out would just be the coefficient of the quadratic so there would be a term that was just a constant but the coefficient of z would involve sum over m i's sum over z minus m i's okay and beyond if I only know that before we solve the problem okay so what how would we proceed how we would proceed there would be the same as suppose v prime v was equal to 2 2 by b squared m squared plus m cube by c we would have said that this one today we know sorry by 2 is 1 by b squared that comes to this contribution because sum quantity we don't know it's called theta times z then we could have solved the whole problem solved the quadratic equation of v in terms of theta because we've got a higher order function here we could get square roots of a higher order polynomial and we actually know the solving and typically we have a solution with branch cuts over two separate elements now I want to explain why that's the case you see once we add this this v this v function here our potential looks depending on the sign on the c potential looks something like this now what are the extremum points what are the extremum points of this potential this is and it will turn out the problem that the saddle point of this problem has an eigenvalue distribution around here and around here that is upside down doesn't matter it's unsable I think okay and there's one parameter what set of saddle points how many of the eigenvalues are there how many of the eigenvalues are there if you want the true middle of the problem of course you want all the eigenvalues to be on this theta okay and that by the way we have two cuts we have cut here and cut here if you want the true middle of the problem you have to tune this theta to ensure that there's no cut here that all of the cut is here it's always possible to do that that's how you solve this here okay similarly we have the end of the four piece we will have typically something like this three saddle, three extreme most general solutions are three cuts if you want to find the real middle of the problem you have to find the real the one that's of all these four solutions about the next key the minimizing key the fact that you have in this method these traditional unknown parameters is not a bulk of the method it's a feature it's just telling you that the problem actually has many different saddle points labeled by this labeled by these additional values and if you want the true example of the one that really minimizes the action you have to find it is this clear? the comment before we go on to the mechanics is that the last comment is that I have solved for you a matrix a gauge matrix in particular in a very simple case the case being that there is one matter you can imagine a gauge matrix in particular which is too matterful one is special because gauge transformations can be used to diagonalize any one matrix but once you diagonalize one matrix you're stuck with the same one you can't diagonalize the same so the problem of evaluating matrix integrals matrix integrals subject to unitary gauge of matrices with two matrices even this problem the potential for these two matrices very difficult because you cannot reduce the problem of ideal values larger limits are easy when you reduce matrices to ideal values but you cannot you're off the model so though I presented this very nice mean solution to you you should be wary of the fact that I presented this one class of problems to you of the infinitely many that I could have written down precisely because of itself even the simple problem of matrix integrals in the larger limit with two matrices in general not sorry so we should appreciate both two things first how much progress we've made in our work the genuine view of the human mind is making progress in pressing analytic questions how much progress we've made you know there's so many problems we cannot solve it and yet by studying problems that we can so we're trying to understand what physics are like it's a fantastic ambition okay okay that's each for my discussion of gauge theories in zero plus dimensions any question or is this the side of the confusion what's the care please come to me if there's something now we're going to study gauge matrix what do I mean by that so I have a gauge theory gauge quantity once again I'm going to do the solvable case is built on the following s is equal to n trace that same once again m is a commission matrix it's a commission matrix that's a function of time because an emission matrix that's a super problem of quantity can be actually we'll get the gauge version gauge version this is why it's not where we have to replace it's by b0m because m is a matrix it's a gauge matrix model some of you are thinking what about the action for the gauge fields the action for the gauge fields is f mu and mu squared but in order for f mu and mu squared not to be real we need a mu and mu but here we've got just an f a0 so that's nothing I can write down to the action that's okay because you remember our quantization of gauge quantization of gauge fields so that's what we had was a square integral of functions functions of the matter fields times the spatial gauge fields subject to the Gauss-Fielder subject to the strength of these wave functions and to be invariant under under gauge transformation so a0 is not power by heliopolis there are no a's there are only a's so what does that mean? what functions are m i? m i means the matrix elements of m subject to psi of m is equal to psi of u m u m so quantization of the problem that you're not facing the problem in this case is what it would have been if this was just m dot squared wave functions of the m of the matrix elements of m except that these wave functions have an issue of strength and that's the end of the new experiment this is clear so gauge point of mechanics is all very common mechanics subject to a work on strength is to find the Schrodinger equation for this problem what we're going to do is to determine the Schrodinger equation for this wave function restricting attention to the Schrodinger equation that's one of the wave functions that are neutrally event so before working this out let me work out an analogy the analogy is suppose we had actually it's not even an analogy the case n equals 2 the s u 2 case myself suppose we have a Schrodinger equation of a particle in three dimensions d2 by dx squared that's d2 by d2 by d2 by squared that's d2 by dz squared minus h bar squared by 2 v of and let's say it's a radius of potential on the side in three dimensions subject to a radius this is the kind of thing we do when we solve a problem okay now in general of course there are all kinds of wave functions that's all of this p-wave wave function all can vary in a moment suppose we were for some reason interested only in the s-wave that reason will be then of course we can replace this by r squared d by dr r squared dr psi bar is this correct? I mean all agree on this right? for a quarter c to the angle of dependence so we get an extra centrifugal value again in the term notice that this quantity here the volume of the symmetry the unit sphere at radius r is the statement here now we have got three dimensions at some radius there's we've got a problem with the rotational symmetry here you can ask what is the volume of space swept out by the orbit of the rotational group at the at radius r we know that the volume is 4 pi by 4 pi by the radius r so this is quite interesting of course that's not a coincidence it follows from the formula it follows from the formula for r squared on any space that has a problem with functions the space of the metric r squared plus r squared times d omega squared some other internal action then the formula of r squared in general rather than everything d by square root g d by d where only interested in the radial values d by d r square root g g r r d by d r on the side in this case we've got d r squared is one d r r is of the unit metric in this symmetry what is the energy here in what we want let me see what the volume factor what the volume factor for unit to the transformations of a matrix with eigenvalues m 1 to m n that's this this f of the term we don't know at last time simply this product of m product i not equal to j m i minus m d by d r squared and so this matrix the Schrodinger equation of this matrix problem simply reduces to the analog of this which is m h plus square plus m m i minus m i same what am I claiming I am claiming that the operator del squared on the space of unit on the space of Pavichy matrices is this written in terms of eigenvalues is this written in terms of eigenvalues if you assume assuming that the wave function is a function only of the eigenvalues is not a function of this I am claiming a that this is true and b that it's not difficult to see actually a clean matrices write down the matrix on the space of Hermitian matrices in terms of eigenvalues okay I leave it as an exercise to be careful the Schrodinger equation is a very complicated Schrodinger oh and I forgot the plus b plus sum over d of m i dimension of the field theory and half of the dimension of the population we did quantum mechanics so we are working in 0 plus 1 this was an analogy the analog of this x, y and z are the matrix are the 3 matrix elements of an SU 2 matrix okay quantum mechanics is effectively n particles because n eigenvalues with this complicated Schrodinger is this clear clear enough I have to believe today somehow carry on spaces of the book up into understanding okay now this is very complicated but there is a trick to simplify it but the trick goes as follows firstly we notice the following l i squared d i squared of random is a harmonic function on Cartesian space this is an easy to check identity let me check it for the case okay so suppose writing the square this will be important this one check suppose now I defends it twice to the second angle all I can do is hit this guy and this guy so the first term is minus of the second term gives me it's this guy and this guy gives me minus of the third term it's this guy and this guy gives me minus of okay it's easy to check instead of this object is here and of course of course del m i of so let's call this object delta so delta squared is there for the same thing as del m i squared of delta x both derivatives are delta but that would be too easy it has a term derivative of delta and one derivative acts on psi but the two ways which derivative acts on delta which derivative acts on psi two ways in factor of 2 and this is an example where both derivatives act on psi so the right side is 2 del m i delta del m i psi by delta plus del m i squared psi but left hand side is obviously the same this is also equal to 2 del m i delta del m i psi del m i squared delta is equal to 0 this complicated equation if you read it as 1 over delta I'll do a factor of course I'll just do the factor you know by 1 by r squared d r d r squared d r squared I'll just start with the equation now this is that algebra factor in special case of SU we depend chi is equal to delta x just multiplying this equation multiplying the equation through we just get the equation that i chi dot squared I forgot my minus if I add it to it the usual Schrodinger equation a bunch of free particles n free particles moving in the potential m i can't come together that's what they say there's something interesting which may be related to us the interesting thing is this let me ask first did the original matrix model okay whether there was any statistics for the different MIs remember that our original problem was psi of M they made psi okay a function of the eigenvalues which is what we used no because eigenvalues can always be polluted among each other by unitary transformations is this fact do you all know this for instance m 0 1 1 0 m 0 1 1 0 which is the first and second eigenvalues since you can do any two you can do anything okay so that constraint that this was true a ensured that we had a problem only by the wave function and b ensured that the wave function psi was symmetric under combination of eigenvalues so so we were dealing originally with a Schrodinger equation for a wave function of n identical bosons because the wave functions were invariant under permutations that's what most of this used to do but now we look back to the actual wave function that was the Schrodinger the wave psi was a wave function of n identical bosons with a very complicated Schrodinger it's a wave function with very simple Schrodinger equation but it has this factor of delta in addition to the psi now delta was what product of let's say i less than j mi minus mj notice that delta is anti-sylactic under the flip of any two eigenvalues for instance delta has this if we interchange m1 and m2 if we interchange m1 let's work it out let's interchange m1 and delta if i interchange m1 and delta so this is equal to minus m1 minus m2 minus m3 minus m1 minus 2 minus m3 so minus let me work it out with 2 there is m1 minus m2 obviously 2 is 2 and 3 isometric property under the interchange of m's times anti-sylactic under the interchange is a wave function anti-sylactic that takes up a minus sign if you interchange any two particles in a free Schrodinger equation so what do we do what do we produce we produce a connection of n non-interactive fermions moving in the potential moving in the potential quite an incredible thing would you say that you take a problem of matrices in crossing matrices again quantum mechanics gauge it or the matrices which is ordinary there is nothing anti-permerizing there is nothing that gauging the problem turns into a problem of n identical free fermions moving in the potential beam so the solution now that we have said this there is nothing else we need to do we want to know the ground state we will work out the eigenvalues of the potential beam to last the further we see so we know how to deal with the infertility on any one of the other courses undergraduate course on statistical mechanics we want to begin inside of the somewhere we all know how to deal with free fermions we don't it's not that you have to say we used it to solve the problem this is the this is the simple solution this is the simple solution here we didn't need to use large anti-permerized this is the exact state for the quantum mechanics of n cross n matrices of n cross n matrices gauge quantum mechanics okay once again this simple of course is very simple very general once again the calculations I solve the problem of one matrix one adjoint try solving the problem of two adjoint fields the general interaction will help okay there is particular problems that are easily solved okay there is a particular problem which is also one it's by the way interesting also to work out what the large n in the middle of this problem it's inside of me distributions in phase space labeled by the position of the Fermi surface phase space there is a very interesting thing there is an interesting story in there but we will not have the line we will not have the line to this extent so we leave our discussion where we've solved at least one problem in phase and we move on to one of those one dimensions okay but that's probably the next one okay next time so after that next time people solve a particular gauge there is one of those one dimensions the solution was first done by Doft and it's a remarkably simple beautiful little problem which can be solved in a remarkably simple it reveals a lot of physics it reveals a lot of physics after we discuss the one-plus-one dimension theory you must know if there is anything I can use to make a statement and think about if there is anything I can use to make a statement maybe we can find maybe we can discuss poliakoff's proof of confine in two-plus-one dimensions and then we discuss which one next question are people around yeah next question