 Okay, so because of any discussion of phytosophore theory, larger than phytosophore theory, let me remind you of the theory we were studying. It's usually an accident that's dealt by i, Nb by i, Nb by 2, that's M0 squared by i ratio, by i, N2. Okay, plus, and I think you've studied a different notation last time with the different names. Sorry, V4 by 4, some number that's kind of fixed in the last little bit. By i, I'm not sure if I got it by 2 by 4, I'm not sure if it was a gu, V4, something. Okay, let's use this notation. Okay, and then what we did was we found an exact expression, we found an exact expression for the 2-quad function, the propagator of the value. And let me remind you how we did that. We did it in two ways, one way was just diagrammatic, and the other way was completing a square. Let me do it by completing the square factor again. If we have some problems with factors of 2, let's try to get that straight. Okay, so completing the square factor takes the b4 by 8, blah, blah, and write it as N by 2b4 times sigma minus, sigma to become an additive shift 2. This gets this term right, but it adds to addition now. So we have to subtract those. The first thing we have to add to this quantity, we write this quantity as this, minus N by b4 sigma square by 4 plus sigma 5 by i. This is a simple algebra identity. This quantity is equal to this. Okay, for any field sigma. And now we take this object as what comes by integrating nodes. I want to write this as I want to write in a sigma by 2, 5, i, 5, i. Now plus some number, I'll do that straight. Wait a minute. Sigma's N sigma squared vector. Then if I now complete the square to this object. So let me call this velocity E. If I now complete the square to this object, plus 5, 5i, 5i. So this, sorry, sorry, sorry. So this is equal to this, minus 4 by e squared by 8, 4 by n. This is right, thank you. Yeah. So b4 by a is by a, n and a 4. So, but we wanted this to be b4 by 4n. So that works, but it is chosen to be minus 4. Right, so I'll put a plus here. Really? Suppose we take a minus of this. Yeah, I mean we want that to be like. We have to add this. We want that plus. That's quite a little negative. So I'll sit down and see what's going to happen. That'll be like this. Minus n by 4. Yes, so from the battery you have to write the sigma as 1. Yes, that's why I have to cancel it. And suppose we take this one and we integrate out sigma. The result will be this one. Because I read that this one is equal to this. The equation of motion for sigma sets this bracket to 0. So what remains is this. Is this clear? Okay, sorry. This should be faster. Okay, so the action to be written as del pi, del n pi i del mu pi i by 2 plus m0 squared pi i pi i by 2. Okay, plus sigma pi i pi i by 2 minus n sigma squared by 4. And then you remember the gymnastics we went through. We justified. The idea was to do the path detector over phi's. And then we would do the path detector over sigma. The path detector over sigma was one field. The action was going to be of order n. We could do that classically. If we did not do an integral over sigma. So if we expect the saddle point value, the mean field value of sigma would be translational invariant. Then we just had sigma translational invariant. And in doing that, then we got this quantity here. So we did the integral. That's an inverse determinant. Square root of an inverse determinant. N copies of that. So the action was equal to minus n times 2 log of n of p squared plus m0 squared sigma. Sigma was plus minus n of sigma squared by 2. The integral of the path detector over sigma. Well, it's the same as doing the path detector over sigma with just a classically minimizing concentration. The classically minimization total of n minus n times 2 integral of these would be divided by the cube. 1 over p squared plus m0 squared sigma. Sigma by 2 p4. It's suggested here that the sigma appears as m0 squared plus sigma. The sigma looks like the self-energy of the process. We verify this idea by computing the sequence of using that. So how do we recompute using diagrammatics? Well, it was basically this equation that sigma, the self-energy is equal to sigma is equal to minus 3 p by 2 pi u 3 p by 2 pi log of the cube times 1 over p squared plus m0 squared plus. Now we have to get the numbers and signs and so on the factors, right? So where did we get that from? We got this. We took a single 5 to the 4 vertex down. We took a single 5 to the 4 vertex down and then we closed off one of them. Taking the single 5 to the 4 vertex down gives us a factor of, if we just use this thing, p4 by 4 n. Then there's a choice of 2, but which kind close up? So that's the factor of 2. And about the sign. About the sign. Okay, let's also get the sign straight. It's equal to plus n by 2. Very good. Thank you. Thank you. It is Romain's point that we write the action as e to the power minus 6. We write the path integral as e to the power minus s. So that even though we got minus n by 2, I mean we got a determinant and the denominator, we got minus n by 2. Once we've explained that, we've tried that as e to the power minus 7. That's the plus, yes. This is the plus, yes. Why is this a plus? It's just because there are two minuses that can't see here. You see, when we pull this factor down, this factor down, we get it with the minus because we're doing e to the power minus s. But then we have to equate it to the mass, which also comes with the minus. Okay? So whatever we get is going to be written as sigma minus sigma by 2 phi squared. So whatever this diagram is, is equal to minus sigma by 2. Because sigma appears in the same way as a mass squared. That appears as mass squared by 2 with a minus sign because of e to the power minus s. The action. Okay? So this diagram came with a factor as we just said of minus e to the power by 4. Then there was a factor of 2 because the choice of which kind it goes up. That was the 1 over n. But we got rid of that because of the n different ways of contracting this. Okay? And so the equation of sigma is equal to p 4 by 2 d3 to e to the power minus s. Exactly the same equation. Oh. It's still the same factor of 2. It's still the same factor of 2. That's it. No. We actually know the same factor. You see? Because sigma by 2. That's what we're saying. This is 3 equal to the sigma by 2. You understand why it's... This is what we're learning. You understand why it's to be related. Equate to sigma by 2. Because when we re-expotential up in the action, we want to write us something by 2. Okay? So now there's no by 2 because this 2 cancels it. Okay? And there's no by 2 here. Okay? So it's exactly the same. d is 3 by 2 by w to the cube 1 over p squared plus m to the power minus s. Okay? This was our computation of the exact problem. We also, last time, did a calculation of the exact scattering. It was 2 to 2 particle scattering. I won't go through the calculation again. But you remember, the key point we took away from that was there were no divergences in that calculation. So we wanted the scattering amplitude to stay fixed in the limit lambda output stability. We had to hold g0 before fixed. Okay? So now we're ready set to do renormalization of the scattering. It was very simple yet, not completely trivial. Okay? So now let's try to see how we do that. What we're going to hold fixed is, let's say, the 4-point, the 4-prosan scattering to something. And so you'll effectively hold it before fixed. Okay? And we will also hold fixed the pole of the problem. So let's say that the pole of the propagator, whatever it is, propagator, whatever it is, it will be called m squared. But m squared clearly, as we've seen, is m0 squared plus c. So we can take this equation and write it as m squared minus m0 squared. Other than this equation, m0 is sometimes here, I don't know the function of lambda. And there's a g there. Before, but this is something simple. Okay? So before fixed, because it's required to be fixed in the scattering. Now, so this is an as-is unknown function of lambda 1. But now we can find what function of lambda 1 is. That's all of this equation. So let's try to understand what this function looks like. Firstly, it's a very large momentum. There's a dimension. Well, firstly, let's do dimension analysis. The dimensional analysis tells us that this quantity is a dimension loss. So if it was a convergent integral, it would be some number, I think. Right? That's what it's going to be. But it's not a convergent. It's d3 by p squared, actually. So the need for the behavior here will be d4 times some number which will be very easy to determine. Try d3 p4 of ordnance and divide by p squared. How does it do that? Times lambda. Okay. Maybe let's do a bit sophisticated. Let's do a bit sophisticated. Write this quantity here as d4 into d3 p by q by q in a 1 by p squared plus m squared minus 1 by p squared. What's the point? The point is that the divergent piece has isolated to make it m into m. So this quantity, whatever it is, the number times lambda as we discussed. This quantity on the other hand is not convergent because it goes like by combining fractions you'll easily see that the integral goes at largely the quantity to the fourth. Now it's a convergent integral and by dimension analysis it's some number times m. Okay? The two numbers are very easy to determine. But they're irrelevant to what I'm going to say. So this quantity has d4 times number 1 plus number 2. But it's number 1 and number 2 are very easy to determine. The numbers. How do you determine that by the way? How do you determine this guy? Well, we simply move to polar coordinates. This would be 4 pi at p squared divided by p squared. So 4 pi at p squared times dp. So in fact, the number is 4 pi by 2 pi though I think. Because it's that easy. This number has easy to determine how we do it. The limit is 2. Prescale I'm already thinking. Define p is extended. Okay? That rescale explicitly makes this quantity explicitly makes this object m times the dimension this pure number. That pure number is the integral 1 over x squared plus 1 minus 1 over x squared d3 x by 2 pi This is definitely integral over all x. Okay? Easy to evaluate. Okay? Whatever the number is before i pure number is pi. But b4 has dimension. B4 has dimension. You see? Pi has dimension r. So this has dimension 2. Okay? So the equation that we have is m squared minus m0 squared of lambda0 is equal to b4 into this number i times mass plus integral 3. What do we have to do to m0? We have to just have it solve this equation. We have to choose m0 to c to 2. We have to choose m0 squared which is equal to lambda0 times b4 and we take lambda0 to infinity. We just have to choose m0 and lambda0 to be this. Okay? And this will achieve we will achieve a theory in which we have headboard scattering as a benefit of mass. Okay? Now in terms of renormalization proof there was a free fixed point here. There was this interactive point that we are trying to reach. Okay? Now what we have done here is to find the theory with two parameters. The two parameters are b4 and n. What we have done is to find the quantity you can in the tap zeroes in a round fixed point that is two relevant equations. That is the free theorem. The flows from this free theory went like that. This is the mass squared direction. This is the b4 direction. And what we have done is to make sure that we are setting something up. By the way, if we wrote this m0 squared in dimensionless coordinates okay? So let's say that we have x2 is equal to m0 squared by lambda0 squared b4 by lambda0. Okay? The unit beta function equations the beta function equations that tell us how we have to scale the bare parameters in order to hold physical stuff fixed. What do we get? We get dx2 by v log of lambda0. We can write as x2 is equal to minus x4 by 2 pi squared y times x2 x4 we have to divide by lambda0 squared so that's x4 by lambda0 squared by lambda0. What do we get? Now if we differentiate this with respect to log lambda we get a term which was just a constant times 2 plus x4 times a constant times 1, okay? And we see what significance this one has what we have to do is to choose this x2 we have to choose x2 to scale it on this equation and it has this strange about this strange beta function So we've managed to define the theory of the free fixed point problem. Now, our main goal was to define the theory of this interactive instrument the fixed point that's going to be at the ISA. The magnet is an ISA. How do we do that? Well, you see we've got these two relative directions we've got the mass by direction it's a B4 direction okay? And what we want to do is to what we want to do is to define the theory when we continue to go to pole mass fixed because we want to keep the relative deformation of this new fixed point fixed itself We want to continue to keep the pole mass fixed but we want to flow as far as we can go in this direction oh, B4 is dimension that's a dimension problem this is a mass dimension problem B4 zero is not a renormal list it's just half fixed but that means that as we flow to the IR x4 becomes larger and larger because x4 is B4 divided by lambda and that will become larger and larger so how are we going to achieve flowing to this fixed point? You see, there are basically only two distinguished values of B4 there's zero and there's infinity since we found no other fixed point in the space of closer B4 it's not like there was an ultra-real beta-function equation for B4 there could have been some fixed point and some finite value but there wasn't so there are just two other stars from zero just infinity so this theory abounds me is defined by the limit is the only possibility is B4 to infinity now what we want to do is to define this kind of model in the limit before we get to that's the candidate for the second fixed point so let's first look at our equation let's first go back and look at our gap equation so first if what we understood it's not very interesting the question is how do we understand the second fixed point so let's first look at our gap equation okay even before that let me do one minute you see when we did this completing of the square we saw the thing action came with a B4 by two sigma squared four n sigma squared do you remember when we wrote this thing in terms of sigma you see so if we took the little B4 to infinity in this language it looks pretty singular four infinity in this language it doesn't look singular at all it just looks like the sigma squared term is tangent so now let's do this properly including the effect in the class let's do this calculation properly what we have done here was to rewrite this term as B you have to help people with the numbers sigma squared n times four B4 plus sigma phi i phi i we replaced this term by these two terms I'm saying here it looks very easy to take the little B4 because we know at infinity it's just dropping the stuff whereas B4 was infinity in this language looks like about a strange and singular thing to do once you do this how much the language should you see that before going to infinity is rather simple effectively it's effectively dropping this the square term now I want to do this a bit carefully I want to do this a bit carefully including this term so I want to do a completing of the square in a way that will also eliminate this term yeah what if we try to observe the infinity of the B4 in the end in what in n no but n appears to mark a place that is the number of i in this so yeah so this is not the only place I have let's proceed let's do this rewriting in terms of sigma in a way that eliminates this object as well as this object in favor of a linear determinism so what we're going to try to do is the following we're going to try to arrange an action so that it has so that it has what we have here which was n by B4 by 4B4 4sigma squared can you somebody tell me what's the sign right for this minus minus plus some number that I don't care for times sigma and plus sigma phi 9 phi 9 okay I'm going to try to to replace both those terms in the action then I don't have to put it to try to eliminate this okay now in order to see how that goes let's do this as usual I just complete this way okay so I write this as minus n by 4B4 sigma and this term will come with the class that's right so sigma minus minus 2B4 a by n B4 by n phi i by i 4B2 plus stuff so what is my plus stuff my plus stuff is whatever was not here so whatever being involved was sigma okay so it's this stuff so plus n by 4 B4 into so 4 cancels B4 squared by n squared term into this is precisely this that's what originally an engine now what I will do is to choose this A so that this term into this is precisely this okay that will work if B4 B4A ah thank you very much where where is n yeah this is correct and this was so the n's now I just take the n's out ah so I just get now 1 by n so well so this original term is B4A by n phi i by i we equated to m0 squared by 2 so we choose A is equal to n squared by 2 this will also include this constant piece but the constant piece is a vacuum in addition it doesn't involve anything we don't care because it's a case of like that way from here if you want it minus this quantity I'm not even going to bother you okay so what we have done here is to determine what this A is so this A I'll just write it down ah A is n m0 squared by 2 okay so these two terms in the action so now if we finish the algebra I'll just write that okay and by realizing the simple it's simply that this action so these two terms can be replaced by this so the action can be equal to the d mu phi i d mu phi i 2 ah plus sigma phi i by 2 plus n m0 squared by 2 a to 4 sigma minus n by 4 okay we define this critical theory we want to take before to infinity we have to be a bit careful because now if we take before to infinity this term will also go and that then we'll define a master's theory or maybe define a theory with no parameter with no parameters rather than one parameter we want to define a theory with one parameter we want to keep this term fixed okay and the way we do that is to take the following limit we take the limit before to infinity m0 squared by 2 for the ratio m0 squared by 2 d4 which we'll call some a name we'll call it ah A we'll see it's interpretation of it now this fixed you know what's going on here in terms of re-homologization of the flow of the coordinate what we're doing is taking a theory that was suppose we took one of these theories that was here and flowed it for a very long time we wouldn't reach here in terms of before we would also have zoomed off in terms of this so we're doing this thing that we've talked about a lot that if we want to define a theory about this fixed point we have to scale our mass so that it approaches this critical surface while we're waiting to flow for a long time and that's basically that's achieved by this funny scaling of it okay so this limit here before goes to infinity m0 squared goes to infinity m0 squared by 2 d4 fixed defines the sets of the particle now in this limit we can just simply drop this term and replace this term here by e and this is now a new path integral that is a candidate definition for the new fixed point okay this term here is sometimes called the nonlinear this theory is sometimes called the nonlinear sigma why is it called nonlinear sigma it's called nonlinear sigma because we now look at the equation of motion for sigma written like this it's a very simple equation it tells you that phi by phi when 2 is equal to e it's like we're doing a path integral over any different vectors subject to the condition that the vectors we're authenticrating over have a fixed modulus so the first lesson is that they will simply show fixed points is defined as deformations around the nonlinear sigma let's work out the gap equation here now we don't need to work out scattered it's only one path so in order to understand how we have to so there's only one object in the game that's the same okay and in order to try to do renormalization of the theory we just have to work out how this a scales with lambda not as we did with lambda not so now we work with the simpler the gradient so we do all the same things that we did before okay the same thing we did before well we integrate out this the phi field so s is equal to n by 2 log of v squared plus m is p squared plus sigma only sigma no n completely wrong p squared plus sigma plus the gap equation I did this and the difference here is to this okay so I get n scale n is all over all of them the gap equation is 1 over 2 times by the whole k cube 1 over p squared plus sigma this is if I really go to as such there would be a bias in this view I have to a is the bare parameter a is the parameter limit that I have to arrange for this equation to be satisfied okay and now once again I do the same analysis that I did before so this left hand side here this is equal to half into what was it 1 by 4 pi squared 1 by 4 pi squared 1 by 4 pi squared number or not this is a wonderful minus ns I just changed my definition I read if I need to minus n that's it it's a parameter just to have a last second so lambda not by 4 pi squared plus this I1 by 2 sigma is equal to a of that which side sigma equals 0 no problem and so a of lambda not but simply lambda not by 4 pi squared that would define the criticality okay the deviation of a of lambda not away from the critical so this we call a lambda not critical a at the critical theory fit me 0 a at the critical no the thing that has to be 0 at the critical thing is the poles you see what these a's are the parameters are huge you have no idea what they are supposed to be okay what you have to do is due to the scale in there okay m squared times the effective action has to be 0 which is the pole okay so a of lambda not minus a lambda not critical I sigma by 2 yeah where is I squared where is I squared previously thank you thank you sigma is over the mass squared okay excellent very good so now what we see is that we have determined how the bare parameter has to be scaled to the critical value in order to get a finite mass this helps something interesting you see the scale goes like square root of the mass parameter there in the last class we had found a relationship between the scaling dimension of the new operator about the scaling the dimension of the operator about this IR fix point and how you have to scale the parameters of your bare the ground and how you have to scale the parameters of your bare the ground as in order in in in order to get with lambda not in order to keep something fixed so this scaling behavior allows us to read off the scaling dimension of the operator 5 squared the relevant operator of the scaling and if you go and look at the formula we derived last time you will find that the operator dimension this operator dimension we had a formula for how the correlation length scales in terms of the scaling dimension of the operator sigma is just the inverse of the operation so we can just read off that formula from last time what we got here is that sigma is scaling like l basically sigma is scaling like 1 over l times l times 2 and this is l times 2 the correlation length is the inverse of the mass which is just 1 over the square root of sigma so sigma is scaling like 1 over l times 2 and we have a formula for how sigma scales in terms of the dimension of this operator is an easily accessible subject you have to derive the game sigma is like sigma is 1 over 1 over l times 2 to the path nu 3 minus nu what we have sigma going like 1 over l times 2 that means nu is equal to 2 now notice that nu in the free theory was 1 it was the dimension of the operator 5 squared 5 is the dimension of half of the operator 5 squared dimension of half of the operator nu is equal to 2 is far from free game and this is basically the square root the square root that you will be interested in it's connected to the fact that this a guy girl had a b4 in it and expressed in terms of a 1 squared and that b4 as far as this theory is concerned is a uv scale which you can trace this fact back to that this comes up to be interesting already our solution of this theory has already told us that the operator 5 is better than the relevant operator among this free things point as nu is equal to 3 there are many other things one could do with this theory okay so given that we have nu is equal to 2 we can now compute for instance we want to know how the specific heat of this theory scales as a function of temperature difference as you scale to the critical phase point we have the formula for that which is nothing new normalization root flow did not require us to do any wave function in normalization it wasn't like we needed to scale the l5 squared like some function with some function of lambda not in order to keep everything finite okay this fact as we have discussed the wave function in normalization is linked to the scaling dimension of the operator 5 the fact that there is no wave function in normalization tells us that in this theory the operator 5i by itself continues to have its free scaling dimension okay so the operator 5i dimension of 5i we need a half the dimension of 5i 5i is equal to 2 some of these strange things that can happen the dimension of 5i 5i is not the same as dimension of 5i dimension of the relevant operator even to say that this is 5i by n is an inactivity the dimension of the one relevant operator around the interacting phase point these two numbers enable us via the discussion we had in the last class to compute all critical components okay so if we were in this discussion that we had was to be applied to our discussion of statistical physics for phase transitions we're more or less doing of course viewed in terms of high energy physics there's a lot more one can do in this theory it's interesting for instance the details of scattering in this theory in the last class we found an explicit formula for the scattering amplitude let me let me just show you what happens in the formula before we found the s matrix minus d4 by n symmetry factor remember d3 p by 3 5i by 3 q times 1 over d3 r r squared plus x squared in a b plus r squared where the m is square root of s m square root of s the formula for scattering in this theory at any value now to get scattering in the critical theory scattering in this non-linear sigma model about the non-trivial phase point what are we supposed to do we're supposed to do all mass things take before infinity and at first you might think wow that thing scattering blows up and certainly in motivation theory we should do before infinity scattering amplitude because each term in motivation theory is put on a b4, b4 squared, b4 q so but if you're really finding a new good intermediary scattering should not blow up how does that work beautifully see because the dosage we call the numerator that's one of the denominator so the before goes to infinity and then it's well defined and in fact we have some interest oh and you remember the scalar p was by the way p p1 p2 p3 p4 and p was this is i so in this limit s 1 over a with a minus sign now it's integral d3 p by 2 pi q 1 over r squared plus n squared into p plus r it's what ah the whole okay and you know this is one last thing because we're discussing this in a couple of lectures but how do you have the constraints and we're just going to do this one last thing the last thing I'm going to talk about is how the how the s matrix behaves at large moment at large moment we can basically forget about this okay so this quantity here is d3r by r to the 4 dimensionally okay so it's 1 over s goes like 1 over n times p now p you can view the scattering process in this channel it's simply the center of mass energy so p is squared and this goes like n squared to s okay and as we will see as we will see as we commit to a series of special scattering the scattering energy this is good high energy behavior for the scattering energy this special organic to make corrections is the is the fastest growth of scattering energy is that what you're going to do okay fine so I think I'm going to stop doing one thing okay we've solved this theory in the large envelope we've solved it well enough to understand it's critical for us we've solved it well enough to understand scattering nutrients okay and for many little exercises what to do with this theory it's a beautiful example of a solvable what to do with it where you can illustrate general ideas in a very simple way of course because it's solvable it's simpler than what's going to be so not every idea gets illustrated it's a nice story model to play around with if you ever get confused fine I think that's all I wanted to say about this five to four theory and I think that's all I'm going to say about beyond this theory in this course we're now going to tell our attention to much more interesting theories about engagements any questions about comments was it with the the four evidence before because in the UV before we know how do we know that if we hold the protrusions when one starts scattering we have to do the same things before mass scattering it's a very good question because I have a value right before it's going to be a theoretical equation I should have called it before or not okay so in the first theory we were trying to define the non-critical theory we had M0 and B4 and the gap equation told us in order to know how M0 and B4 are not scaling as a function of lambda we need two equations okay so we need to calculate two things exactly two physical things exactly and hold them fixed we computed the two things the first thing was the pole and the problem the second was the four-parted section now holding the four-parted was actually having to do fixed required us to hold it before fixed okay so we just held that dimension so if you're expecting in terms of x4 it scales in the usual way but this was the dimension fully where the scale goes to x4 and this is the dimension scaling no no no x4 does scale you see suppose there's no running in the UV then every parameter scales according to its naive mass equation and that's what's happening in this case because there's no running in the UV for B4 x4 means B4 by lambda just scales like 1 by lambda okay so x4 is running but in the trivial way that is just by its dimension okay just because we're choosing to look at x4 is this clear okay excellent, other questions about this yes can you please put it under our symmetry in this here symmetry is a four-hand symmetry what are the symmetries it's symmetries under phi i it goes to lambda ij, phi ij where lambda ij is any matrix that will not change theogram but any matrix that is just will not change anything to the contraction of ice which will not change theogram and those are O n okay so this is called the O n long denials okay and n can be anything down to like of course in some sense down to yeah but if you take it down to 2 then you've got you know just a vector in the complex plane those modulus is fixed if you take it down to 1 now you're restricted to 2 points this thing becomes base this is the isic see the isic model is a model where your spins have fixed modulus but they live in one dimension so there is some sense in wengie it's a bit similar the isic model is something like the isic model O n long denials is completely straightforward okay by the way I looked at the isic model but there are many actual systems in n chart which have an O2 that is the same as u1 solution so even magnetic systems even a classical magnetism in which the fields are vectors that live in a plane it's sometimes called a planar isic model okay and the phase transition from that system are governed by this auto-dermatism right you'll write out as a classical magnetic system where the fields in this way natural are a vector that live in three dimensions space that model is governed by the O3 not the isic so this problem is of use and different values of n in its own right so it's a good thing to generalize I was studying 5 to the fourth theory into this O n theory just in its own right but it's also of use as some way of you know adding the parameter to the isic model hard to find examples in nature of generalizing any who say okay so where we can solve it it's hard to find instantiation in nature but that's okay you know theorists will never get beyond this that's the structure of mathematics some mathematical problems are so difficult that human beings likely will never find exact solutions theorists have never been able to find exact solutions or approximations around those you know that's the limitation of this kind of theory the kind of theory that doesn't say you don't have any good it's a limitation that you're not going to do everything so the job of the theorist is to find the solve of the points and to make a qualitative picture based on what he can he or she can so we do it very well do we study n times zero? n times zero or what? n times zero maybe you can have some f**king way but sir I know for a fact that the n vector model does n times zero limit physically I don't know anything about this for me n is an integer you could formally study it then you say that this also has a physical instantiation I don't know okay that's interesting this thing has more supplier at some point than the theory of time or the sigma model yeah how does it relate to time? let me okay so this is a bit of a diversion most of you know what is being taught in this course it's okay the theory that we are referring to as it's originating interesting fact about the QCD vacuum and that interesting fact about the QCD vacuum that there is a condensate of a binary okay so there is some sort of cyber and in the QCD vacuum now what do you see? it's like a both sides this expectation value is now important in the QCD vacuum that's the thing okay the fact that this is not zero at least one thing is breaking of some of the cement trees of the QCD vacuum what I'm saying is that only approximate quarks have mass so then we work in the person's idea element by quarks are massless even a little bit by quarks are massless as in suppose you have three quark flavors the up to the down and strange which are approximately massless on the scale of the QCD vacuum or we could go to two very very very accurately massless scale of the QCD vacuum okay if you have a theory of two or three years some number of massless quarks then as you know as you know this theory has an SU but as classically let's fix that number let's call it two let's call it three it has a U3 times U3 cement tree which you can independently rotate the left and right quarks because the only time the Lagrangian mixes them the only time the Lagrangian mixes them is the master the master zero okay this is the symmetry of the theory and that's a long story about most of which we will go through most of which we will go through but the art of the story is that you will take this U3 and U3 can be thought of as acting on left and right quarks but you can also think of it as acting on in a diagonal and off diagonal do you understand what I mean equal action on the left and right quarks and opposite action on the left and right quarks that's another basis for this U3 tension now as we will discuss in this course but as you already know now the every symmetry of a classical theory is the symmetry of the motor okay it turns out that the diagonal actions it turns out that the diagonal actions of these of this U3 and U3 continue to be symmetries of the water but the inverse diagonal which it acts opposite to the left and right quarks okay of that U3 okay SU3 continues to be a symmetry of the quantum theory but the U1 does not it's anomalous okay so as we have done in the quantum theory we just forget about the U1 for this discussion okay now so there is a global symmetry of the theory which is the times SU3 the U3 is sometimes called the vector and the SU3 is sometimes called the axial axiality now it turns out that this condensate okay it turns out that this condensate this condensate so now wherever you got something condensate and you got a global symmetry it's very important to know how the condensate transforms under the symmetry in particular it's important to know which symmetries of the theory are annihilated by the condensate you know which symmetries are symmetries even in the presence of the condensate don't change the condensate okay and the U3 the vector symmetries continue to be symmetries even in the presence of the condensate condensate has that character okay how about the axial symmetries now wherever you got a symmetry of an energy that's not a symmetry of your vacuum you have goldstone muscles you have muscle spheres but what what is the good way of parameterizing the muscle spheres the good way of parameterizing these muscle spheres is simply by the action of the symmetry group on the vacuum the space of vacuum of the theory rotated into each other by the action of this this broken symmetry is some group coset the theory of the low energy theory is the goldstone bosons that describe fluctuations on the symmetry that theory is a sigma model on some appropriate space whose physics is very similar to what we discussed except that is a 4 much of this we will try to discuss in detail when we discuss this dcd but this was just an appetizer any other questions no because it's 4 dimensions nothing at all sigma models and 4 dimensions are more than anything sigma models and 3 dimensions any other questions okay so let's move on we are also going to move on to now studying Youngman's theories okay but since we discussed these large energy bits um I was thinking of introducing the study of Youngman's theory um through the study of the solvable model of Youngman's theory very much like we discussed here this is a beautiful solvable model solved by Toft it's called the Toft model it's a model of Youngman's theory in 2 dimensions okay just like this theory of scalar fields were solvable in in the large end of it and the Toft model was also solvable in the large end of it and this preparation for that I want to live more discussion okay so let me clarify I said that we're going to turn to now to the study of Youngman's theories what do I mean in the beginning of the course of course we did some kinematics we talked about the path and technology of Youngman's theory we talked about it's in that space interpretation we talked about the path we pop off and how it takes fixing and so on but there's very big difference between the kinematics of the theory and the dynamics of the theory like we've seen in the study of these Youngman's theories you know there's very interesting stuff in these scalar theories new fixed points, critical exponents and so on it's not obvious when you just look at it it's not obvious from kind of kind of facile analysis we did in that space that's like okay that's just yeah that's just kinematics how the theory actually behaves is an interesting question now let me start by introducing the questions okay the theory we can study is plus some matters if we want to study the other theories that we've seen in the other world we'd have this is Ranjit here he describes Youngman's theory interacting with Farnian's and there's a very good description of the physics of the strong interaction quarks interacting with SU3 gauge fields very good description I say exactly because I'm going to be talking about weak interactions and so on but for many processes you can ignore them okay there's physics there's simple one line two term Lagrangian that's just a huge amount of physics okay and in the beginning of our course we discussed we discussed some the structure of not only the gauge advances we discussed the Hilbert space of the theory and so on now we're interested in more interesting things that can form in the classes we're interested in how this theory means one of the things we're going to do in the next few lectures is to compute the beta function of this gauge function okay and as you know we're going to find that the beta function is such that G grows to a theta therefore we will conclude that this theory is defined as a flow away this theory is defined as a flow away from a friend it's a theory let's say if I let me first in our mind throw this away let's say that we have no books this theory now is a theory with exactly one marginally relative value this is the situation the flow of that theory is very simple this is zero towards the higher as far as we know the flow goes up okay question question what is the behavior of this theory how does it behave now this question is a question about very elementary things by looking at this theory you might think well maybe what you mean by that question is can you compute the scattering amplitudes of the view ones because when you do free field analysis this theory is a theory possible scenario view ones maybe that you would think is the correct question to ask or one of the correct questions but it turns out that even this is even this is assumed to you see because you are assuming that the particles that this theory has a particle spectrum okay when the particles are view ones the theory was free that would be the case but because so if we were there is no question if we were any way away from the fix we would ask you questions about arbitrarily long time arbitrarily long distance behavior as necessary when you do an S matrix action it prepares something in infinity we have very long time calculation clear you know the answer it is not clear that even the question you are asking is not clear that the particle spectrum is the particle spectrum of the theory includes view ones I mean of course you have not got used to the state but as far as we can tell it is true the angle theory appears to have this this dynamical property or sometimes referred to as confinement and that property goes that probably basically is the statement that there are no asymptotic states in the theory that transform non-trivial under the SUN gauge group on the global part of the SUN gauge group in particular view ones which transform non-trivial transform the agile representation are not particles in fact it is strongly believed to not prove in any in the way that will menu a million dollars because it is a clay foundation that it offers a million dollars to somebody who can prove for a great thing prove something it is not been proved but it is strongly suspended I think we should say no but this theory in fact has a vast gap what does that mean the first excited state about the vacuum is not an arbitrary go-end as it would have been if glue onto a box but in this end is that a fixed energy above is that some fixed energy above the vacuum second claim the spectrum of particles in this theory is entirely made up of sequence in fact think of this of this being a two-glow bound states that are arranged so that this bound state is colored in we are doing SU2 you can easily imagine that SU2 the adjoint is just a vector of SU3 and a bound state of the two vectors where the indices contract is a scalar so it arranged bound states of two adjoints to give you a sequence this is a general theory product of two adjoints the spectrum of this theory is roughly we think of as bound states of two real ones maybe in other ways the spectrum of these particles are referred to as glubons so pure annual theory we strongly believe in the theory of glubons now when we add quarks to the theory the statement is that gluons are not good asymptotic states in the theory carries on to the statement that quarks also are not good asymptotic states in the theory because quarks are also common adjoints in fact the spectrum of particles in the theory includes glubons and by the years of the quarks which are supplied through the mesons and then more complicated objects we will form symbols in SU3 by taking three fundamentals using the absolute elements and these objects are called balance so the claim is that the spectrum of this annual theory with the including quarks is made up of glubons mesons and marigolds it's not the same but this is the kind of thing we are going to try to understand this is the kind of thing we want to understand so you see the extent to which one can be a theoretical surprise not only is it a hard task to compute the estimate of the objects of the free field theory once the theory starts to fill up sometimes that's even the wrong question the particle spectrum the basic organization of the Hilbert space can then be very different from what perturbation from what the free theory or perturbation theory around it can test this of course can happen if the coupling constant would be quite large for the relevant question that you are asking this is the kind of thing we want to try to understand to the extent that we can in the annual theory for the annual theory everything that I have told you about has some qualitative understanding some pictures and a lot of numerical information but in two dimensions in the large element this you can prove to be true okay and so for this reason the large elements of the theory involving n cross n vectors okay and it turns out that the study of large elements of n cross n matrices is much more interesting than that of the n cross n matrices this is what we are going to do and we did an approach to the study the first part of the study was simple it is about models question yeah for instance all of these fields here the annual theories are adjoining values so at UN you should think of n n matrices okay that is why I am studying larger elements involving n cross n matrices now this will be relevant to the study of n n okay so in order to keep this discussion as concrete we will start with a simple the simplest example simplest solver example and yet the study of of an integral involving n cross n matrices and the secret in the study is about considering the integral over n cross n Hermitian matrices times exponential of minus exponential of minus trace we have where which is an arbitrary real potential of potential function what is the theory? it is quantum field theory at one space time point it is just a plain example now even quantum mechanics already this this problem has some interest we will start with quantum mechanics in the first lesson in the life okay now how I should scale can you guess whether this guy here would have a good larger element I could you see the rules are always you should scale entropy and energy in this equation like we did now if you look at entropy can you see like n square so we should want this energy to scale against square it is a sum of n you might expect that this trace scales like n for this reason let us study this we need to determine the same trace try to understand how we compute such integrals of energy let us call this to this we are going to see if we can find a way to compute in generality such integrals this is actually an extremely easy task if you think about it a commission matrix any commission matrix can be thought of as can be utilitarian in life can be thought of as u v and he is a guy commission matrix here can be thought of as u v universe where he is a guy and there is u here sorry and this expression here does not care about u it is invariant under similarity transform because it is built in some trace of n square percent u percent so the integral over u we can do it once and for all where this is doing the integral over the eigenvalue such an integral is the same as an integral over so let us suppose that the eigenvalues of this matrix are alpha 1 such an integral is the same as d alpha 1 d alpha some major factor that we are yet to do okay times exponential of minus n sum over i v over is to determine the major factor is this clear this is very similar to the following example suppose I was doing an integral over a two-dimensional integral and I was going to do an integral over some function that integral would be alpha you should just do the integral over the integral over the eigenvalue so this would be 2 pi that is r v r that is alpha 1 and this major factor is the analog of this 2 pi r okay the other analogy with that closer is to doing this in three dimensions suppose we were doing this in three dimensions this would be a function of this and this would be 4 pi r squared v r f of r okay the analogy here is that we are doing an integral over a lot of variables but the integral k is only some of them k is only some of them you just do the integral of the the variables it does not care about first and use an effective major factor the variables that does care about so now let's go let's go let's let's go let's go on steps okay so suppose this m bigger was and suppose the m bigger was a traceness 2 cross 2 okay then n could be written as alpha i sigma sigma is other three public matrices so one way to parametrize the space of these n's is by these three n's what do similarity transforms by s you do to these three sigma vectors of s or 3 so in fact this particular example the space of 2 cross 2 traceless emission matrices is parametrized by three n's with s you do invariance which is rotations in three dimensional space so it's really the example we talked about with three dimensional space the the the ideal values are the radial distance can you see this it's going to be one alpha let's say how we see mathril but the ideal values would be plus or minus alpha 3 since it's rotation in there it can only be square root of alpha 1 square plus alpha 2 square plus alpha 3 square root of plus or minus that right so the ideal values care only about the radial location so this integral here is only about the radial location whereas the angular locations you can just do the integral in that case the analysis that we did previously tells us what the right measure is up to some number the measure factor simply is okay up to some number the measure factor simply is alpha square was alpha this alpha which says the ideal values so we have two we have matrix once we diagonalize the ideal values alpha and minus alpha okay so we can write this as that's called alpha 1 and alpha 2 alpha 1 minus alpha 2 in fact you can easily convince yourself that if we generalize to you to make to not trace but obituary to obituary obituary matrices now the two ideal values are not necessarily equal in all these because there is a u1 basis so please provide identity okay but we can easily convince yourself that in that case it's just a factorization of this of the three dimensional space that's a one dimensional point and so the integration measure becomes d alpha 1 d alpha 2 alpha 1 minus alpha 2 that's the number which is easy to keep track of to the number of the deductions now we want to understand the n dimensional generalization of this is this clear is this clear if we claim between integration over a 2 cross 2 emission matrices same as doing integral over the ideal values with the measure factor alpha 1 minus alpha 2 to the whole things alpha 1 and alpha 2 now we want to generalize this before doing that before doing that let's intuitively understand where the square what was the intuition for the square the point was the following you see the point was the following the point was that 2 cross 2 matrix okay and this ideal value happened to be equal then the space of rotations by the similarity transforms change by matrix n because u times identity times u inverse is just identity on the other hand the ideal values are different then the similarity transforms move you around okay the measure factor is the volume of that space in which you moved around think again of the 3 dimensional 3 dimensional space and rotations there is a fixed point on the rotations okay and as you go away from there so at the origin rotations don't change the origin which is even fixed but if you go away from there rotations cause spheres and the size of the sphere which is our square is the measure factor is a generic feature of these measure factors and if this measure factor comes because we got some symmetry action and there is a fixed point of that symmetry action then you have to know the co-dimension of the fixed point how many how many parameters we have to tune in order to reach that fixed point in this case we saw the number of parameters we have to tune was 3 because of 3 there was alpha 1 alpha 2 alpha 3 to reach the fixed point all ideal values had to be equal not only was it alpha 1 is equal to alpha 2 is equal to alpha 3 is equal to 0 so it's a co-dimension 3 fixed point then the measure factor is r square just like at least in the neighborhood of that one it's r square just like in flat space so this the fact that this there is a vanishing measure point and alpha 1 equals alpha 2 is because the unitary transformations have a fixed point at alpha 1 the fact that there is that square was because the co-dimension of that fixed point was equal to 3 parameters that gives you the parameter the quantity square now suppose we can increase the dimensions if any two ideal values become equal the symmetry group and as you do some group of the full SUL symmetry group of rotations becomes fixed like that any two ideal values become equal that requires a co-dimension 3 so this tells you that the measure factor whatever it is there is a measure factor must have a 0 of order 2 supported but any two ideal values become equal that this measure factor is the alpha 1 the alpha 2 the alpha n that is product over i i and j alpha i is equal to alpha j that is i and j square times something else times something else which is not 0 or singular as i goes to j that is anything so it is some function of i alpha i and alpha j not as alpha i and alpha j something else to be this measure factor has a certain homogeneity in mind how do we judge the homogeneity the original integral that we wanted to do was an integral over n squared n squared matrix elements ok so it was of dimensionality matrix element to the power n squared the final integral that we are going to do is over n ideal elements so the measure factor has to have dimensionality matrix element or ideal value to the power n squared minus n this object here you can easily check dimensionality i can manage the flight script why how many numbers are there for i less than j n minus 1 by 2 the dimensionality of this object is 2 so this exhausts the this exhausts the homogeneity so anything else that you get here would have to be some function with zero dimensionality ok such things would are basically possible to arrange without having singularity sum for instance suppose you have alpha 1 by alpha when alpha 2 went to zero it would be singular but that's crazy ok so arguments of this nature can convince you why certainly with a great deal of possibility that this can sound number is a correct measure but that's going to be never completely ok so suppose you're actually going to compute it there's a very simple way and that very simple way i'm going to leave this as an exercise for you is to apply the fadim pop-off trick ok this is a very simple example i'm going to give you the gates the gates are actually just unitary so i'm going to leave this as an exercise for you apply the fadim pop-off trick you remember the point of the fadim pop-off trick was to change the integral over all gauge configurations to integral only over gauge orbits with the right gauge of it that's exactly what we're doing in this case the gauge orbits are labeled by the eigenvalues the original eigenvalues are over all gauge configurations we're going to integrate only over gauge orbits it's a measure factor fadim pop-off it was a beautiful trick to do that i'd like you to blindly stick to the right if you know that including the ok i'm going to leave this as an exercise for you people ok i think we've exhausted that right by the clock so i'll finish the analysis of this model and then we'll move on to really study the gates ok well yeah that is an excellent class of questions so when do we next not have all the bits if Wednesday is fine right let's let's study the class of the things after you have a class i have to just check with various things it wasn't as easy alright