 We may do it a bit fast, but let's try to finish that. Okay, great. So, you remember the last class? This is the goal. We were talking about larger planets, planets, diagrams and so on. Okay, so I'll give you back the rule. The rule was that the leading order in large end, the theory without fundamental matter, the only adjoints, receives contributions only from playground diagrams, genus G diagrams. More precisely, the rule was that graphs of genus G awaited by relative factor 1 over n to the power 2g. Okay, and the second rule was that an index, every hole, every index, fundamental index loop, was awaited by a factor of 1 over n, rather than 1 over n squared. Okay, so adding extra handles to the surface, or adding extra holes to the surface are both sub-dominant in the 1 over n expansion. Okay, now, of course, the whole 1 over n expansion is very interesting, but to be able to have any chance of doing that, you have to be able to solve a theory of leading order in the large end. We had many examples of theory of leading solving a theory of leading order in the large end, matrix integral, matrix quantum mechanics. But now we're in 1 plus 1 dimensions, and we want to know whether we can get a non-trivial theory of theory, and solve it in the large end limit. This problem was addressed by Doge. You know, Doge is a great theoretical system. Doge went through a fantastically creative period between 1970 and 1984, each opening 30 papers, but just individually outside of me. Each paper, you know, you wrote a problem to it. And, but in the 13th one. So, okay. So, you may have heard of the main loop of renormalization that they created. Many, many things that, and this little paper is a little gem. It's a kind of paper you could write in three days if you are there. As you can see, there's nothing to it today. So, it's so nice. Okay, so what I thought, was to study the problem. Okay? Study the model of cyber, slash, side, trace, okay? Okay? So, what's this model? The model is a model of fermions. Okay? Okay. Now, he allowed for these fermions to have a certain number of limits. Okay? So, first of these fermions were in the fundamental, in the fundamental, in the fundamental, in the fundamental representation. We generalized adding flavors a little later, but now, there's only one. Okay? So, the question about this model is, what is it? Physics. Now, suppose you were to treat this in free-field theory. In free-field theory, what would the answer be? Meaning, you know, keeping this action truncated to quadratic. Okay? So, at quadratic level, you would get you would get the quadratic part of the algorithm. But the quadratic part of the algorithm, of the algorithm itself, as we have seen before, has no degrees of right. Right? We argue this in one of the dimensions. The gauge field has no degrees of right. Okay? In general, we saw that in d-dimensions, d-2 is propagated. So, it's a freedom. So, at quadratic level, you would say that there's no propagated degrees of freedom. But, if you looked at this, you would just get free fermions. And free fermions have fermions in them. So, you would think that there would be one or an A-dependent if you got it. A-dependent, if you got it. And for me, I think it is a freedom. If you were going to do free fermions, you would get the spectrum of the theory. It would be a Hilbert space made up of a Fox space of any fermions. And then perhaps corrected to something. This is totally the wrong answer. Actually, that is the wrong answer we can see quite intuitively that there's still four dimensions but it's harder to see this intuitively, classically. But we'll come to that later. But now what we want to do is to just work in the large area. And try to see if we can just solve this theory without any intuition. It's just mathematics itself. And try to get in the spectrum. Okay? So, how are we doing this? So, first thing that both of us are analyzing this this model is to set a gauge. First, I'm going to define the location. So we're going to do two dimensions. We've got two coordinates. So let's say the dimension is a diamond space. Okay? And we will define X plus minus of X X plus minus of X squared. Yes. This whole thing is a problem. This D slash is D plus X squared. So this is what makes it an interactive thing. There are interactions between as you say it's a question. Okay? Now, we work in like-on coordinates. So X plus minus is X plus minus D. And with the lower X so suppose we take the momentum for instance, plus minus that is equal to right, is equal to let me switch to this dimension. So he calls this X. What I call the X is called X1. What I call D is called X0. Okay? X plus X1 plus nothing. But the P plus minus is equal to minus P0. Where P0, P lower 0 is minus of P upper 0. The metric this is plus minus so is this but this P0 is lower important. Which makes if you write in terms of P upper 0 to be P1, P upper 1 minus plus B. The H field in the class is equal to A1 plus minus A0. And now the point is this. What we didn't do is to try to solve the path integral of the theory. Now the theory, we're working in a particular gauge. And the gauge that we chose for this problem was the gauge A plus and A minus A minus equals A plus the path integral of the gauge theory. We made sense of the path integral. At least maybe we needed to insert an delta function. That fixed gauge. But associated with that delta function was a path integral. And the determinant when C bar times C times what A meant here was the transformation. The thing that you said to 0 is A minus. So we need to know how A minus transforms. This C So the general rule was that we got some condition. That condition is a matrix condition. Under a gauge transformation it transforms in some way involving some gauge transformation. The general rule was that the path integral of the determinant of this expression multiplied by C bar and replace lambda times the determinant of this. So how does A minus change under a gauge transformation? So the change in A minus is L minus lambda plus I minus the change in a gauge transformation. The covariate variable of the gauge transformation parameter. Okay. Do you remember this is happening when we have inserted a path integral of delta of A minus? So A minus in particular is 0. Okay. You do your path integral along with the delta function. It tells you that A minus is C. Do you remember the gauge? Yes. And therefore the full path integral of the determinant is just C bar delta minus C or the determinant of the operator D minus. Now the important point for this path integral of the determinant is that it does not involve angular side. It does not involve the gauge fields or the fermionic fields at all. Is this specific or is this gauge for this gauge? I will contrast this to what I have to learn. I will do that in a moment. But because it does not involve angular or the fermions at all, the path integral over C just gives you some overrun number. Does not affect any computation or correlation functions anything like that. And if we are interested in the path integral up to an undefined normalization, we can just forget it. So this is one of the great things about this gauge or any such gauge. Any gauge in which you set is particularly important to me. There are no ghosts in that gauge. Now what? You want to make life as simple as you can. So you are doing this in a way that okay, let me interpret you. It looks like a lot of notches. But let me read what your question is. You can ask the question, why is this a particularly useful thing? As you might know, this gauge is rarely used in analysis of four-dimensional equations. However, this gauge is extremely useful in analysis of four-dimensional equations. And you can ask why? And that is for the following reason. You see this f mu mu is a mu computed like a mu. Sorry, it is d mu a mu minus d mu a mu plus with some i, the mu computed like this. Now in two dimensions, there are only two possible mu's and mu's. If you set one of them to zero, the commutative term vanishes. And therefore, the f mu nu squared, the self-interaction of the gauge becomes effectively quadratic. This would not be true with any other gauge. Because if you have, let's say, three a's, you set one of them to zero, you can still get the commutative of the other two. So doing this is extremely advantageous in this dimension. Now you can ask, everything I said, both the lack of funny-poke as well as the kidding of self-interactions would have been true of, let's say, a zero is equal to zero for a one equal to zero. You can ask, why do I choose the set lack of future? This is because it retains boosting variance. That is, as you know, under a bool, under a bool, a minus, let's say, transforms to minus, that's right. Are you familiar with this? I'm saying something more, let's do it. Let's recall the Lorentz transformations. The Lorentz transformations are x zero prime, which is particularly significant. x zero prime is equal to cos theta x zero plus sin h beta and x one prime is equal to cos beta x one plus sin h beta x zero. You know this notation. Beta is 10 meters. You're familiar with this, right? Yeah. But it doesn't matter. Now, let's see what this means for x zero plus x one for x zero plus x one prime. How does that work? Well, we will get x zero is equal to x zero plus x one into x zero plus x x zero into cos theta plus sin h plus x one So then, 0 plus 1 into cos plus sin h. What is cos plus sin h? What is x? Simply, if you have x zero minus x one prime is equal to x zero minus x one So, the great thing about these light comb coordinates components is that they transform homogenous families. If you work in 0 and 1, you get a mixture of 0 and 1. If you work in the light comb coordinates, you just rescale the light comb. X plus stretches out and X minus stretches in. So, the light transformation as after any object in the same way with similar structure. So, this is true for X 0 and X plus and X minus such a true for X plus and X minus. And therefore, this gauge A minus equal to 0 is boosted by A. So, A minus equal to 0 in 1 light strain. That is equivalent to setting A minus equal to 0 in any other light strain. So, you retain more of the symmetries of the problem by doing this, than you would by setting A 0, A 1 to 0, which is always in the day, about the distance. One good reason actually that I have stressed something please in front of the couple. Doesn't matter. Now, if you caution you about not being careful. And the caution is that look, we have not been very careful about equipment because setting A minus equal to 0 does not completely fix gauge. It does not fix those gauge transformations that are functions only of X plus. Because under the gauge transformation, A minus changes at A minus or something. At A minus, any function of X plus is same function of only X plus. Therefore, any gauge transformations function only of X plus. It is not fixed by this gauge. This is the residue of gauge transformations. And if you try to fix that residue of gauge transformations, light will get quite complicated. Now, how? You see, doing complicated problems in physics is basically not. So, the thought has a good taste to ignore the problem. Just proceed and see what you are capable of. After 500, this way goes inside it may be a problem. People have already filled in the details. Don't work just to buy everything is done. We are going to follow it. Forget this. Forget this unfinished gauge intelligence. Small thing. Usually, if ignoring something matters, you will see it in your condition. You will see an inheritance. And you will see we run into some assumptions. So, what we are going to do is we follow just those solution paper, not try to worry too much outside of this. Follow our nose and see where we are going. Now, we ignore the fact that some unfinished gauge intelligence is just a procedure. If the unfinished gauge intelligence is really bad by the way, it would show itself in the propagator not being important. It is that with main problem of the gauge intelligence. Main problem of the gauge intelligence. So, unless we run into problems with the inertiality, non-invertiality of the propagator. Okay? We gauge of not worry about things too much. It is a hit and miss approach that can sometimes hit you badly. If it works, it works very well. Okay? Fine. By the way, we do the same thing in the range gauge. We ignore unfinished gauge and variance, at least in the gradient space. Usually, quantum field queries are so complicated that they have to think through a very last subtlety and try to solve a quantum field. It is a realistic and unfeasible task. What is feasible may be put on the machine. And, you know, that is part one of the truth. This is not an addition. You proceed. It makes sense. Okay. Okay. So, you know, in that spirit, we fix this gauge intelligence. Now, once we fix this gauge, as we said, there are no self-interactions in the range gauge. Okay? Um, there are... Has it ever become purely non-interactive? No. It is what you said. There is a side bar, side e. Interactions. So, there are interactions. But, the theory is much simplified. Diagrammatic. So, we make all the diagrams this multitude. Okay? So, let's try to imagine a particular result. The example that we look at was not quite observable. The thing that we will try to compute in this case, to start with, is the propagator of the effort being fixed. Okay? Now, see, in an arbitrary gauge, what kind of a diagram should come up? All right. We will continue. We should be worried only about plane and radius. Okay? We don't want the gauge to be still. Don't talk to me like that. Now, an arbitrary gauge is what we could have planned for all categories. We could, for instance, have worked with this. Where there is a new fermion group. We could have worked with graph numbers. We are very interested in leading order and target. We are not going to allow new index groups. The feminine main index groups. Because there is a press button. Okay? So, we are not going to allow any new fermion to pop up. Anywhere on the side of the graph. Such graph, such graph. On an arbitrary gauge, this would still be a hell of a tormenting script. Just some of these main index groups. You know, some crazy mess. Compute. Now, you start with the new way. Okay? But what special about our gauge? Special about our gauge is that no gauge goes on set. So, in our gauge, what graphs can be? This, a graph like this, such as the green circle by something like this. Well, right by diagrams. But these are quite similar to that. Okay? Now, we encountered this class of diagrams before. In somebody's mind, is where we can come. Well, in English theory, except, the large end limit of the scalar theory in three dimensions. Remember, we solved that. And we found that, not exactly this, it was like, like this. But very similar classes of diagrams were invented. Okay? And we used the same method to solve that problem that we're going to use here. Actually, that we used two or three different methods. You can use the same term. Yeah, I'm just going to look at the diagram. You see, all of these graphs, whatever they do, are a sum to give you a self-energy connection for the fermi. Now, any graph here that is separated by a fermion line x, or something else, that just is taken care of with the geometric product. So, the self-energy is just all graphs that cannot be cut by cutting a single fermion line. So, what are we interested in? Something else? All such, we'll get out of this. Okay? So, sigma is the self-energy. I'm going to write down the answer first, without any eyes that might be the sense. Okay? Then I'll just look at the order here. Okay? Okay, you can look at it. Okay? So, sigma is what? You see, all such graphs have the feature that there is an overline end-of-the-pink thing. And inside is whatever you want. If you look at the set of all such graphs, you see, all the sum of the graphs that contribute to the self-energy, every one of them has an overline guide here. Because if it didn't, then there would be a way of cutting through this. And inside is whatever you want. But inside being whatever you want means that this is the exact propagator. The propagator along with the self-energy. You've got the sky for all outside. And here we could have all kinds of things there. Each of these by itself is arbitrary. So, it's the self-energy collection. So, it's free propagator, self-energy, free propagator, self-energy, free propagator, self-energy. What is that? That's the exact propagator. Let me say this again. What is the self-energy collection? It's the term that you add to the effective action. The quadratic term that you add to the effective action. The exact propagator is what you get when you invert the quadratic term including the bare part plus the self-energy. Exactly. So, if now I'm going to write down an equation and you tell me this takes action. Let's know the momentum of this R to follow that is coming from the equation. And let's call this the momentum that goes in here and then the momentum that goes in here here is sigma of P. So, sigma of P of course will be true because I mean a matrix in the speed of this is going to be what's added to the quadratic term for the following argument. So, what is sigma of P? Well, here that is an interaction matrix but the only non-zero a was a lower class or a a upper minus so that comes with a gamma minus is this propagator. This propagator is what? Let's look at f mu into square because a minus is the only so a plus is the only non-zero guy l minus a plus so the only propagator in gauge field is a plus 6 is only 1 gauge field left in a gauge. What is this propagator? 1 over 3 minus. 1 over 3 minus square after some i's and minus s we will try to explain. So, this guy is one of what b minus k minus square. What about this? This is the fermion propagator. The fermion propagator but in tuning signal and therefore this is i p i k slash plus n. What does k have to be k to be anything? Plus minus sign some i's some factors which have to be a little careful about getting explained but that at the end is horse cook. This is the equation. Is this totally? Sir, just one thing I have to understand what is this gamma minus and I will learn another one. There is a gamma minus here there is an interaction here there is an interaction. Gamma minus is the vertex. You see let's look at this. See what do the interaction will take some time. The interaction will take some time for gamma minus. Another way of saying this is the verbeno is contracting this at this but with all interactions that gives us the same that property and we are contracting this in this. Also with all interactions that however gives us just a big bear a propagator. Why? Is the a propagator as an exact signal that connected? No, not just exact signal because you can have this because that is done by one operator. So the larger limit as in this game bear the exact a propagator. It is the same as a bear propagator. The exact psi propagator is not the same as the exact psi propagator. And that is our best a propagator. But it is what it is. These are the two that have to overcome one thing. Is this clear? Yes, I am making. Actually the thing that is written around this equation is through what is called the equation. I will give this to you as an exercise. You take the path you are going to take and then sort of in the bank we need a very similar to that that we used to play the equation. You write on a schindler writing equation for the property I will give you the give this to you as an exercise. Okay. There is a claim there which is this thing that does not stop. But even diagrammatically you see it. Is this clear? Yes. Is this equation clear? So now we are going to take this equation and process it. Okay. So we are going to take this equation and process it. How are we going to do that? Well what we are going to do is the following. Firstly we are going to take this object here and multiply it by you know this identity that we got p slash plus m into p slash minus m that gives you p squared. Is to multiply this guy here the sigma here or whatever it is is sum 2 cross 2 matrix. We don't yet know what it is matrix structure is but it can be written in terms of linear combinations. Unfortunately that's not for us. Linear combinations of comma matrices as well as identity. Okay. So whatever it is we improve the identity of the part at the end. We improve the gamma matrix part of the k and we use multiply and divide by such an object to make the denominator simple to get all the matrix structure in the numerator. But now whatever matrix structure we have is sandwiched into the gamma matrices. Now gamma minus square is what can be written. Okay. Exactly. Exactly. Gamma mu and gamma mu are more. It will be 2 give me a name. Now let's you can do it by what you know the simplest way is to just use what J mu nu is in plus minus matrices. In plus minus matrices the metric G plus plus is equal to 0 G minus 0 G plus minus. Okay. So anyway so gamma minus square is 0. If there was a term here of the identity that's 0 of course. If there was a term here gamma minus we know the 0 would be a term here of the gamma plus. Because then gamma plus and gamma minus is now 0. And then that would give us an answer proportional to the left over gamma minus. So do you see that we have argued that whatever sigma is it is all sigma is equal to something that's got little sigma and gamma minus. Because whatever this matrix is the right hand side is proportional to gamma minus and there is no left hand side. Okay. So we started by expressing ignorance about the matrix structure of sigma. But we end up I know the matrix structure of sigma. It's proportional just to gamma. Is this clear? Is this clear? Yes. This matrix equation can now be turned into a scalar equation an equation that's to the coefficient. This little sigma. Okay. So what is the equation turned into? Now I'm writing down all factors of i and so on today and we'll see that i is sigma. Okay. Let me use this. So i gamma is equal to what I think put there one by k squared over k minus integral sigma. Just this and k minus m squared minus k minus and here is purely proportional to what term here is purely proportional to gamma. Okay. The term here purely proportional to gamma minus k plus. That's the sky problem. So we had you see the free term is k plus gamma minus plus k minus gamma plus because the matrix is g plus minus. Are you understanding? Because you should contract its k mu gamma mu with a g mu mu. It's not proportional gamma minus k plus. We've got another gamma minus. So whatever k plus here gets shifted by this sigma. Okay. In fact, I think it's an idea. So it gets shifted by I take sigma. Okay. And therefore wherever you had k plus, let's write this and let's write this as minus k plus plus. It says that k plus gets shifted by a sigma. And you can ask where does it get shifted by 2 sigma? Does it get shifted by 2 sigma? So what is this? Basically, what is this? Let me check because it's 2 heat glass shifted by sigma. That's somehow to some reason it's very self-active. Okay. So up to such things you see that this is just the shift. Okay. Now when you may when you do the plus minus on top you do the strict denominator numerator this just becomes m squared minus k squared ordinarily. But k squared is a factor of 2 is needed because our square root 2 is okay. But k plus was shifted by sigma. So you get this time. The i epsilon will be discussed extensively. Is that then you have the numerator. But in the numerator only the term proportional to gamma plus contribute. Okay. That did not include our new sigma because that was k minus. Sorry. It did not include k plus because that's proportional to gamma minus. So the only thing that survived was the k minus. And therefore you see there are some factors of two you see that we that this that this 2 pi squared 2 pi squared So this is the integral equation. Now we the integral equation just for the a single single component just check Okay. Okay. So that's I'm sorry actually. Hopefully this discussion on Friday we try to go a little extra maybe. We complete this discussion. Oh sorry sorry about that. I've booked the room by the way booked for I'm sorry to say that actually it's going to be in the US. Yes it is. Let's go. Okay.