 If I, gamma, v equal to 4 is squared k plus k minus the fundamental communes and the atmosphere is in two-dimensional theory. And you remember we had, we were looking at the, what we had to do was like on gauge, like on gauge we were working in was a lower minus is equal to 0. We were working in minus 0. And to find the great thing about that gauge we remember was that there were no self-interactions, but there were no self-interactions of the, of the gauge fields. And then we wrote this, there was that, we tried out an integral equation for the self-energy, which was schematically sigma was equal to 1 over k minus minus p minus squared, which is this gauge field propagator times the full propagator, which is k slash plus m plus sigma, with minuses and i's through this. Okay, so sigma p was this and that, then we did some matrix stuff. We realized, oh, then we did some matrix stuff. We realized that because of this matrix stuff all of sigma was purely in the gamma minus. That's like one comma plus direction. And so we called the coefficient of gamma plus wherever it was i's. Okay, and then we wrote down this integral equation for seeing this, this equation more precisely became that equation. You remember this? Any questions or comments about this? Okay, great. Fine, so now let's proceed with solving this integral equation. Okay, so now the first thing you notice down is that on the left-hand side we've got gamma of p. Now where does p appear on the right-hand side? Because i gives a gift from the problem, from the problem. Where does p appear on the right-hand side? Well, it appears only in the gauge field propagator. And it appears only as p minus. Okay. So at least, nightly, this expression this tells us that i with this gamma here is a function only of p minuses. It's not a function of p plus. Now let's see. You see we needed to a Lorentz invariant answer, because our age preserves Lorentz invariants, preserves roots. Now, you remember that we had something like this thing was proportional to gamma plus that sigma is equal to gamma plus times whatever. Because gamma plus into gamma minus is equal to 1 and we have 1 to gamma plus. Thank you, thank you. Well, gamma minus times whatever, thank you. So, this object should have the charge of a p plus or the charge of a 1 over p minus. See, you might expect that roughly our answer is going to be something like gamma minus by p minus, because what else could it be? What else could it be? It has to depend only on p minus, cannot depend on p plus. So, there cannot be an dependence on p square, which is a Lorentz scalar. It can depend on the mass. The mass can just come up here in the manner needed to fix up dimensions. So, roughly this might be our expectation and now we will see how that is going. Again, that is what we might roughly expect to start with. We will see how it actually works. Now, firstly the fact that this object here, the first thing that we do is this. We take this k plus k minus integral and separate out the integral. We first do the k plus integral. Now, k plus integral is very easy to do, because it appears only here. k plus appears only here. So, this part goes along to the right, this part goes along to the right and we are just doing an integral of the form 1 over a x plus b in the denominator. This is a trivial integral to do. So, you do that integral and let me tell you. Now, that attack is a bit tricky to do. So, let us focus on it for a little bit. So, the integral we wish to do is v k plus over m square minus 2 k minus k plus plus sigma of k. So, this looks like a completely straightforward thing to do. So, if you want once again, you see that this integral has one subtlety namely that at large k plus it behaves like d k plus by k plus. At large k plus it behaves like d k plus by k plus and therefore, it appears as if this integral that much. So, we are in company of first funny feature here. We are getting a divergent integral in this integral here. In our whole analysis of this model, we are just going to follow on this. You see it is true that there is divergence at large k plus, but there is also divergence at very small k plus namely k plus large, but negative. What is the divergence? You see if you plot the function 1 over k 1 over x, it behaves like this at positive x, it behaves like this and negative x. So, why you get the divergence here because you are summing things that are a rank of 2 slowly. So, you also get the divergence here because you summing things that are a rank of 2 slowly, but this divergence is a negative divergence, because this divergence is a positive divergence. So, these two things could well cancel. Now, we are going to do of course, properly deal with this problem we have to regulate. In our in this very simple problem we have to laugh because if you assume only one thing about the rank of 2, namely that is symmetric between positive k and negative k. That makes this integral finite. Are we saying gamma k is a function of k minus? Gamma k is a function of k minus. Gamma k is a function of k minus. Yes, we know this because gamma p was given by this and p there is no p plus. So, gamma p is given by the right hand side. The right hand side depends on p only so p minus. Sir, how do we know that right large k, large k, gamma would probably in the large k limit is just p k plus by k plus. This gamma is a function of omega k. So, that is the point that is why this k plus integral is easy to do. The k minus integral you cannot just do because it depends on going gamma k minus which you do not know. So, this that is why it is an integral equation, but it is an integral equation only in the variable k minus. In the variable k plus it is just something trivial. The integral on the right hand side is just something trivial. We can just do the integral. Again, all are known and assume following the output that the regulator we put is symmetric between large positive k and large negative k. Just to be a bit clear, you might think this is an extremely unnatural condition to put, but just to be a bit clear it is a bit dangerous. It is a bit dangerous because you see depending on what you do in this regulator because you have this infinity here and the infinity there and they more or less are cancelling each other, you are going to get something finite. That is clear, but depending on what you do in the regulator, you might get different finite terms. Actually, in this particular case it is a logarithmic divergence that is largely abbreviated. We will see that in a minute. The conceptual point I want to make was that when we say we are putting a regulator that is symmetric between large positive k and large negative k, the k here is the k that goes into the matter propagator, not the natural momentum of the problem. That is the momentum that goes into the gauge goes on propagator. The statement about putting a cutoff that is symmetric between positive and negative k for the gauge goes on propagator is different from the statement of putting a cutoff that is symmetric between positive and negative k for the matter propagator because the two k's are shifted by a b. Do you understand what I mean? No. So, you see I said that all we are going to do is to impose some sort of cutoff on this integral that was symmetric between positive and negative k. But for that statement to be meaningful, I have to give you what my definition of k is. If I take a propagator, I take some sort of cutoff which is symmetric around k is equal to 0, it will not be symmetric around k equals a. Now, you might think well, there is a natural k in the problem, there is a natural scale of momentum because there are two propagators. There is a gauge goes on propagator in the matter propagator. But remember that their momentum are relatively shifted by b. So, if you put a cutoff, a regulator that is symmetric for the matter propagator, for the momentum that runs in the matter propagator. It will not be symmetric for the momentum that runs in the gauge goes on. These two are in principle different things. So, the statement that I am putting a symmetric regulator, once you start examining it, is by itself a little ambiguous, symmetric for which momentum? Once you see that there is some choice, you might say why not some linear combination? It is a bit ambiguous. But in this case, we are just going to put the symmetric regulator because the divergence is so mild that we are not going to put the symmetric regulator for the matter propagator and the procedure. So, just to give you a caution, because if you try to do similar things in slightly more difficult problems, you will have to put the symmetric regulator. So, let us anyway proceed. So, now what do we have? So, what is the structure of this integral? The structure of this integral is this. So, the structure of this integral is this. It is of the form A x plus B minus I m 7 and we want to do this integral over here, where this x here is this k plus. So, let us just do the integral. Firstly, if you just do this integral indefinitely, what do we get? So, this gives us 1 over 8 log of A x plus B 7. Take this x here from large to a large positive, you know, a large positive lambda and a large negative lambda. So, this thing is going to give us 1 over A log of lambda A lambda plus B lambda x 1 minus A lambda plus B minus 1 over A log of minus 1. Stuff that is subleading in 1 by lambda. You see, that is the point here that if I shift this by something finite because this divergence was so mild, this logarithmic wouldn't have been a difference. Now, this problem that is basically no, don't beat it. And this log of minus 1 is either i pi or minus i pi. Now, you have to do this a bit carefully. You have to do this a bit carefully and when you do it carefully, the famous, what we get is i pi times mod of this i pi. Sir, I just understood that this is minus 1. So, we take the minus 1, it is A lambda minus B, where is it is A lambda plus. Ah, but lambda is very very much. Also, we do not. Everything else is 1 over lambda, right. You can expand them to lambda, lambda expansion, ok. Everything else is 1 over lambda. So, the clear because of the answer for this integral here is i pi divided by mod A. Now, how do you see that it is divided by mod A? You see that basically, if you took A to minus A, if you also make the variable change, x goes to minus x, ok. The integral unchanged up to the big signs of these things, which never mattered, ok. So, the answer had to be invariant at the A goes to minus A, dimension. Yeah, it was 1, ok. You have to choose the sign of whether this was i pi or minus i pi coordinated in a way with A and so, it becomes a mod A. I will leave you to figure out exactly why this integral is this simple job, ok. Good. So, in the case of A, this integral here, this A was 2 k minus and so, we conclude the integral equation gives us an i pi is equal to 4 g squared by 2 pi divided by 2 k squared and 1 over k minus of minus i k minus divided by modulus of 2 k minus. Now, you see the very simplified feature of this. This integral equation here was a difficult equation, because when dama goes to the left and the right. So, it was some self consistent integral equation. You could just solve it by doing integral. But once we did this k plus integral, dama disappears from here. Is this clear? It disappears, because by the way, we put the limit on k plus and k plus plus dama didn't matter. It is not an integral equation. It is an expression for what dama means. It is still there, but it is just an integral that we have to do. It is not like we have to solve anything. So, that is the equation. Now, let us simplify this a bit. So, we get 2 g squared by 4 pi squared. So, let us call that square. This cancels and the i and pi cancels. One of these, that is pi, i minus i gives us minus. And then, yeah, that is this 2. And then we get 1 over k minus minus k minus. And here, this guy and this guy cancels. Just to give you a sign of this, because this was a mod. And this was just a k base. And this is a square. And we have, he has a dama, p is equal to, let it be, can I leave you guys trapped on the minus side? Okay. So, integral is r function. This integral is not, no, it is not an r function. This is k minus p. It would have been an r function that would have been k minus something. Yeah. So, p minus is equal to 0 to 0. Okay, fine. We will see about it. We will see how that works. Also, this integral together has a divergence. Except that the divergence at k minus is equal to p minus. Because you are doing an integral, which is 1 over x squared, dx. And that is divergent. This, unlike the earlier divergence, was an infrared, is an infrared divergence. It is a divergence when the momentum and sub-propagator, in this case, in the gauge goes on. It is going to 0. But it is a physical reason for this divergence. And I am not telling you what is the physical reason for this divergence. Very good. The gauge goes on, it is going on-shell, except that you remember there are two dimensions. There are no propagating gauges. So, there is nothing to go on-shell. But, as you would think, it is right that this divergence is occurring where the propagator of the gauge goes on is becoming infinity. Why is the gauge goes on propagator becoming infinity when the minus component of the momentum through it goes to 0? It is a very strange condition. It is not Lorentz invariant. It cannot be something physical. I understand what I am asking. You see, this was the gauge goes on propagator. It is blowing up. It is blowing up where p minus going through that gauge goes on is going to 0. And p minus is not a gauge invariant. Sorry, Lorentz invariant. There is now some reflection of physics, some reflection of formalism. What is it reflecting? Very good. Very good. You see, why was it so important in the perturbable calculations that we fixed a gauge? It was important because otherwise the propagator becomes non-important. Right? Because a propagator is non-invertable if there are quadratic fluctuations that leave the quadratic action invariant. So, if there are any gauge transformations left over that leave the quadratic action invariant and do not fix those, those will reflect non-invertibility of the propagator. Non-invertibility means the propagator will go on 1 over 0. Now, what gauge transformations had we not fixed? We had said a minus is equal to 0. We had said a minus is equal to 0. So, what are the gauge transformations that are left unfixed by this? That is gauge transformations are by functions that are functions only of x plus. The functions that are functions only of x plus in momentum space have k minus equal to, if something is not a function of x minus, its k minus is 0. They are also 0. The constant function is 0, 0 Fourier 1. These are residual gauge transformations. Residual gauge transformations are by functions only of x plus. But we expect where you are some non-invertibility of the propagator. So, we should have expected that there will be some interact problem at k minus equal to 0. Where the k minus and the gauge goes on goes to 0. And that will proceed. Is this clear? So, what is coming back to it? It has here is the fact that we are not completely fixed gauge. However, and you know the real cure of this problem is to complete that, to completely fix gauge. Now, if you try to do this carefully, it becomes a very complicated problem because actually when you do try to do carefully to ruins this beautiful fact that the problem, but the gauge, gauge interactions are, but there are no, no gauge interaction. So, both of himself as a master completely fixing gauge transformations is in a study of confine and diffusive difference. In this case, chose the pragmatic way. Let me, you see, I know that there is this, this singular idea and some sort of age artifact. Let me deal with this by mediating. I will put an infrared cutoff in this problem. I will put an infrared cutoff in this problem. And then in the end, I will take the infrared cutoff to zero. Any answer that depends on how I take the cutoff to zero and so on, I cannot cross. If for such things, for such questions, I will have to come back and carefully understand how to completely fix the rescue gauge. But if I get some answer in the end that does not depend on how I take my infrared cutoff to zero, the infrared cutoff cancels in some physical point, then I won't have to worry about it. This is my pragmatic way to work. So, firstly what we do is we write. So, what, what are we doing here? So, what are we doing? We say let's replace this propagator and over k minus minus p minus over k squared. Let's replace it by, let's say that the propagator v of p minus minus k minus. We replace this function by 1 over k minus minus p minus squared for mod k minus minus p minus greater than lambda. Lambda is some small number which we will actually take to zero. And it is equal to zero for mod k minus minus p minus less than lambda. A brutal regulation of the problem. No physics involved. The only justification will be that eventually we will find some physical quantity in which lambda will cancel. So, we will not have to worry about more careful, more careful treatment of this problem. And this we can proceed very simply. What do we have to do? We have to do this integral. So, let's see. So, let's, let's do this integral by changing variables. So, we will do this integral by, let's do this integral by changing variables. We come out v is equal to, and here I will switch to variables k minus minus p minus. So, g squared by minus g squared by 2 pi integral s g n of k minus minus p minus plus k minus p minus over k minus squared v k minus. When we must remember this is the right way to do it. Let's look at this, let's look at this integral. You see suppose k minus, suppose we look at the contribution of this integral from mod k minus greater than mod p minus. In that case, positive k minus is, contribute a positive number to this integral. But negative k minus is contributed to negative number. Those numbers exactly cancel each other. So, this is in line of your comment. That part of the integral from, from mod k minus greater than p minus just vanishes. So, this integral here is received, this integral here receives contributions only from mod k minus less than p minus. So, what we can do is come out v is equal to minus g squared by integral minus p minus p minus s g n of k minus plus p minus squared v k minus. But in this range, the sign function is just one, because minus is greater than mod k minus. So, I am going to move that you right. So, now we have got an even integral. For our regular integral, this is equal to minus g squared by pi, then integral lambda to p minus of 1 over k minus squared equal to minus g squared by pi times 1 over lambda minus 1 over p minus. So, this was true with the, shall we say it? Sir, lambda should be minus p minus. In fact, are you seeing this sign of here? No, no, no, in the limits of the integral. Yeah. The lower limits will be lambda minus p minus. Well, I was using that this was an even integral. So, I do it between 0 and p minus. I will get an extra factor of 2 and then 0 is not 0, because the propagator lambda. I am just wondering whether I have carefully taken into account the sign of p minus. So, actually I am not very careful about that, because you see I said that this was always positive, because that is not true. It is always the sign of p minus. You see this quantity is positive and p minus is positive. But negative would be minus. We shouldn't. So, what we should get here, yeah. So, in the end, if p minus is positive, we should get a positive answer, which is what we've got. If p minus was negative, we should have got a negative answer. Yeah. So, I think we just need to multiply the whole thing here by sign. What? Yeah, you are saying because everything, because there is a problem. No, we are not adding. This is the full answer. And so, one says, gamma of p is equal to minus g squared s g n of p minus by my lambda plus g squared by pi. It would more. So, I am getting a little bit better. Minus g squared by pi lambda sign p minus minus. So, you are looking at the region where more of k minus is less than more of p minus. Yeah. And the sign s a n would be less than p minus is positive or negative. Right. Yeah. Seems to be. Yeah. Now, okay, let me just do this carefully. Let me just do this carefully. Okay. Yeah, I'll just do it very carefully. So, suppose p minus is positive, what we did was completely right. Okay. This was just okay. Now, let's take the second game. Suppose p minus is negative. The suppose p minus is negative, then what we get is that we always get minus. Okay. So, the whole integral is minus. And it's been integrated from minus mod p minus to plus mod p minus. Okay. So, if p minus was negative, we would get an overall minus sign. But then the integral will go from mod p minus to mod. So, the right answer is, if this times sg n of p minus, these are placed by mod p minus. Therefore, if this times mod p minus, sign of p minus divided by mod p minus is p minus. Yeah. So, I don't want this p minus. Okay. Is this okay? Yeah. Sir, I was thinking like, when we did the, when we did the vk plus integral, we chose to symmetrize it around on a matter. Okay. If we had instead done it around the gate, we wouldn't have got that divergence which we have eliminated. No. Actually, in this case, there may be no difference because any finance shift will change the law. You see, what do we get? We have log of lambda plus something over minus lambda plus something. If we symmetrize around the matter at the propagator, sorry, around the gate, goes on propagator, what about shifted is this constant. But that didn't enter the final answer. In this case, it was so minor. Namely, logarithmic. It didn't matter precisely around where we shifted. As long as we shifted, because as long as we symmetrized around any finite point, you would get the same answer. So, in this case, we got, we didn't have to worry too much. Sir, if you had symmetrized it around p minus, wouldn't that, like, if you shifted this p, c by p minus. Yeah. So, you see, because we would have got lambda plus p minus of the lambda minus p minus. Yeah. But that didn't matter. We didn't lambda infinity. Okay, good. No, first you understand. Log of lambda plus a over minus lambda plus p for any finite a and b the level of lambda goes to infinity is log of minus p. You understand? Yeah, sir. Okay, so suppose we symmetrized around some other point. Lambda would have been placed by lambda plus that point. Yeah, but that point wouldn't have mattered. It doesn't matter. It was just insert shifting in. Yeah. So, in this particular case with the divergence, it's so mild that it didn't matter. I just want to caution you about the general principle. The general principle is a bit ambiguous. In this case, it's anything reasonable you do. It's the same answer. You don't have to worry about it. Final answer. This is the answer for the self-energy of the, of the Fermi. The answer for the self-energy of the Fermi. So, the Fermi or the Propagator. The Fermi or the Propagator as an additional piece of it. As an additional piece of it, just proportional to gamma minus that is actually to the usual p slash. p plus, gamma minus, plus, p minus, gamma plus, this nk. In addition to all of that, okay, it has this, this additional piece. Now, what does this mean? What does this mean for this proper theta problem? So, something non-gauge invariant. The Propagator is something non-gauge invariant in a, in a gauge state is non-physical quantity. Because it's, it's the inverse of the two-point function of something inserted here, something inserted here, there. And that's not a gauge invariant. It's gauge transformations here that transform it by, by the gauge parameter at this point. There's gauge transformations here, transform it by the gauge parameter at this point. So, I'm canceling each other. I'm canceling each other. So, the propogator by itself is an auxiliary quantity. Okay, that goes into computing something like this. So, the fact that we've got a propogator that manually violates Lorentz and Lorentz. For instance, we've got gamma minus term, but no gamma plus term in the propogator. It's not by itself a matter of concern. However, there is a part of the propogator is poles in the property. So poles in the propagator represent physical propagating particles. So it's important to know where the poles in this propagator are located. So let's see how the dispersion relation, the dispersion relation is modified by this factor, by this change in the, by the change in the property. So what is the usual dispersion relation is? 2 k plus k minus plus n squared equal to 0. That follows by solving the finding solutions of the Dirac equation with the unmodified propagator. So that's this piece plus this piece. This piece gives you an effective shift to k plus as we say before. So what this whole thing becomes is 2 k plus k minus, plus k minus into this square. Minus g squared is g n of k minus by pi lambda, plus g squared by pi k minus, plus m squared is equal to 0. So the poles in the propagator appear with that equation as a sin square. Let me see if I got the sin of my m squared right. My k plus and k minus have been defined following those so that the space part was part of it possible. So tan part is negative so this is right. Now let's look at these two terms. This term is very civilized because k minus cancels k minus and all it does is renormalize the mass. So you can write this as 2 k plus k minus, plus k minus into minus g squared s g n k minus, m squared plus g squared by pi lambda is equal to 0. So if you had only this term, this would just be a finite renormalization of the mass of the gate of the, of the propagator, right. But we also have this term. This term is multiplied by something that is going to be diverged. Because this term is multiplied by something that is going to be diverged, lambda is going to be equal to 0. What we see here is that there are no poles in the propagator except that k minus is equal to 0 and k minus is equal to 0 poles as we have seen before are associated with lack of, some lack of complete fixing of gauges there. There is no, if this, this object here and another way of saying is this that suppose we think of this as ignoring the fact that it was, it is not the lens in general. Suppose we think of this as, yeah, maybe the next way to say it is suppose we think of this as something that allows us to solve for k plus. So the pole in the propagator does not, what value k plus? Because there are two k plus is equal to minus g squared S g n of k minus over pi lambda plus n square plus g square lambda, lambda over k minus is equal to 0. In the limit the lambda goes to to, the lambda, the lambda goes to 0. The pole only occurs at k plus equal to infinity, okay. So this pole here is non Lorentz invariant and secondly it only occurs at some value of k plus. Do you want to think of this as some sort of light code energy going to infinity? Although it is a bit hazy at this point, you may wonder what this is saying. What this is really saying and we will see this very clearly in 10 minutes. What this is really saying is that there is no physical pole in the propagator. The poles have been pushed to infinity. Quite a dramatic thing. We will see it much more cleanly. It is very unsatisfying because it comes with this lambda and this one. We will see it much more cleanly in a completely physical observe. Completely physical observe. We will see this as you will see. Sir, I did not take this. What constraint are you putting on k minus? No, sir. Okay, yes. k minus is any arm train. Okay. k plus goes to infinity. k plus goes to infinity just because lambda goes to infinity. So, the poles in this propagator occur only at infinity k plus. Okay. So, it is like very heavy poles. So, there are no finite poles in the propagator. What does that mean? But that tells you that there is no physically propagating Fermi on parties. We have to solve this condition. It does not matter. We have to solve this condition. This is something. But this is larger than it because lambda's width is here. Sir, here you said that there is lambda goes to 0, k minus goes to also go to 0. Yeah. You put k to poles if you cancel this with this. Okay. A finite pole if you cancel this with this. Okay. But suppose we do the following. Take k minus to be any number that is fixed. Non-zero and fixed. Any quantity. And then we take lambda goes to 0. Okay. In that case, this guy just goes up to you. We could do the same thing for k plus and k minus. No, not so easily. I mean, k plus is not special because between this and this, the k minus cancels. Should we get this? You want to solve for k minus this function of k plus, you say? Yeah. Can we do that? I suppose you could. Yeah, I suppose you could. You get two solutions. One would be k minus equals 0 or one would be something else. I suppose you could. But I is clear as to say in this way that the only poles that occur unless perhaps k minus is 0 is at k plus equals infinity. Okay. So firstly, this pole condition is not Lorentz-Inviting. Secondly, it only occurs at infinity. There is only one reason for the interpretation of this fact. The only reason for the interpretation of this fact is that there are no poles. There are no physical fermions. Now, what is this statement? The statement that there are no, that there are no propagating fundamental fermions. That statement has a name. Can somebody tell me what the name is? If we prove the same thing for 410, what would we claim we have proved? Confinement. Confinement. The fact that there are no asymptotic states that are color charged. Okay? So if this conclusion is borne out by future, by further analysis, then it would be the statement of confinement in this problem. Okay? Fine. Any questions or comments or others? Because it is slightly more, more sophisticated and more satisfying analysis. But because this, you know, doesn't feel very satisfying. It depends on some, it depends on how I take my lambda to 0. I mean, however you take it to 0, in the end it goes to 0. But it's not very satisfying because it's an answer that's not very clear. It would be nice to have a finite answer for something that's short. And that's what we can do. Yeah, please. No, no. It means masses infinity. Means that there are no states that if you ant with the fermion problem, fermion operator on the vacuum, that creates no finite answers. Okay? Then you only get finite answers. See, let me remind you, all of you know this, let me remind you of the situation in QCG. In QCG we've got quarks and we've got gauges. Okay? And so if you look at the theory classically or chronologically, you would say the particle spectrum of the theory is that of quarks and gauges. Therefore, you might think that, you know, when you're doing milk and oil drop experiments, sometimes you will get charged one third because the oil drop would have a quark. Never have that. Because in actual QCG there are no asymptotic states that carry any that are non-gauged by it. Both the quarks states as well as the beyond states are non-gauged by it. They transform some representation of each one. But in actual QCG, all asymptotic states are hallucinates. What are the actual spectrum of QCG? Isons, baryons, gluons, a neutral thing, QQ bar, or SCU-3 neutral things, because of confinement. Okay? So the Lagrangian looked at with your bare eyes is lying to you. It looks like the spectrum has all these charged objects, but they're not that. It's trying to see something similar in this very simple toy-to-dimensional model. Nobody knows how to prove these statements I made about QCG in foreign countries. In this toy-to-dimensional model, if you allow me to do a little hand-waving, which then 500 papers later on have justified it. Okay? You can see it very simply in a couple of images. That is the attempt. What is the phenomenon? Well, if you looked at the Lagrangian with your bare eyes, you would think that the only path-propagating particles were the fermions. Because in two-dimensions, there are no propogating nature presence. What do you see? The first indication of what they see is that these fermions are not. They're not propogating. They are waiting for the masses of particles. What does it mean? Yes, it's a very fuzzy concept. Because of current algebra of masses, whatever it is. You see, in a particular regulation speed, once you take the QCG attack and you regulate it in a well-specified manner, the amount you add to the action as M psi bar psi is a well-defined physical quantity. And that's what you mean by the core mass. But you must distinguish this from the pole in the power propagator. Which normally is what we normally talk about when we talk about the electron mass. We're not talking about some term we have in some imagination. We're talking about the physical electron mass, the pole in the electron propagator. The products have no mass in that sense. They do have a mass in the first sense. The term is the Lagrangian. Is this clear? Yes? I'm sorry, I've seen a very clear notion between poles of propagator and propogating modes. Why do they have propogating modes? Why do they have propogating modes? If you have poles of the propogator. Let me take a ten minute digression. The cleanest way to say this is from the Lehmann representation theorem. Which I would have imagined QFT1 would have dealt with. Sorry, so I should have told you about this. Let me take five minutes. Sir, don't the poles like this represent the equations of motion. So that is just saying that they are propogating modes. Very good. So that is the intuitive thing to say, as I'm sure understands. That if you've got a quadratic term in an N. So let me first say this intuitively in the way I'm doing it. Let's say you've got some quantum effective action. And if there is a quadratic term in the Lagrangian, such that that quadratic term in the Lagrangian can be written as field equation of motion field. And there is a solution to the equation of motion. Then when you invert that, it will become a pole of the propogator. Solution to the equation of motion of propogator. Every time you've got a solution to the equation of motion, you've got a pole. But there is a more rigorous... This is one of the few... Let me take ten minutes to say this a little bit. This is a very one. Let me... You'll have to help me get all the signs. Let's consider a scalar field. You can do the same for fermions, just to add more indices. Let's consider a scalar field, such that the phi of x, phi of y, time-ordered, time-ordered, phi of x, phi of y. Consider this propogator in an arbitrary interactive order. Phi is any scalar field. Not necessarily one of the Bayer fields. Any scalar field. Now, let's write this out in more detail. So this is theta of x0 minus y0 of phi of x, phi of y. Plus, vacuum theta of y is 0 minus x0, phi of y. Writing out what time-ordering means. If you've got earlier times up to the right. So this is later. This is kept... What we do is to take this object in Fourier transform. So this is equal to g of x. g of x minus y and Fourier transform. So suppose g of phi is equal to phi of phi dot x, phi dot x minus y. So everything becomes only n of g of x. Each of the terms in those... in that these objects are processed in the following way. Let's realize that g of x0 minus y0 times this matrix element. And now this is the key step. In the matrix element, I insert a complete set of states. In the Hilbert space of the field theory. In the Hilbert space of the field theory, I'll insert a complete set of states in the Hilbert space of the field theory in between these two. So we've got 0, phi of y, and n, n. This is the full set of states of the field theory. n is phi of x. We use something. We use the fact that in any other quantum field theory, phi of x is equal to g to the power minus i of beta x, phi of 0, the momentum of it. Translation of it. I hope I've got the signs of it. In fact, I've got the signs. Okay, but you've forgiven me on this. So we have both for this and for this. Then we use the formula that p, the translation generator in recent time, annihilates the vacuum. So this quantity becomes theta x 0 minus y 0, times 0, phi of 0, times e to the power minus r p dot x acting on n. Now let's choose our basis to be a basis to diagonalize this p. We can always do that with complete set of computing variables. So this becomes n, n, phi of 0, times e to the power i p n times y minus x. Sir, that's on the relative space, not on the universal phase theory. But what's that? No, p is full translation generator in the field theory. Sir, what do you say? Quantity. This quantity is some matrix element. We'll keep it. So this is what we get from the first term. From the second term what we have here is something very similar. We get plus theta of y 0 minus x 0. And then the only thing that's different is that we get an additional minus sign. So i to the power i p n dot x minus y times the same matrix element. Is that clear? Sir, do you mean that the n states are the n states of the p of p? Yes. I'm choosing a basis in which p mu's are tagged. And we can always do that because they commute to that side. Right. I'm assuming that. Exactly. We are transformers. If I didn't have these theta functions here, the Fourier transformation to base on the job, we would just get some delta functions p would be replaced by p n. But I do have the theta functions. So in order to deal with that, what am I supposed to do? Let me put an integral representation for this theta function. Sir, let's see. What is the error representation? I want the representation of theta of a is equal to e to the power i a x. So contour integral. Right. Yeah, i here. That's right. You see. Look. Suppose a was a positive number. Then this contour integral. This is an integral from minus infinity to infinity. But if a is positive, I can always close it at the positive infinity. Because that's exponentially set up. Okay. Now, so if a is positive, I can close it at the positive side. Then I use Cauchy's theorem. It gives me the function of a pole. But there is a pole at positive. There's a pole at x equals i actually, which lies within our top two. So you get 2 pi i. But I'm going to divide it by 1 over 2 pi i, so I get 1. On the other hand, if a is negative, we can't close it up there. We can't close it down. But there's no pole, so you just can see. Okay. This is a purely integral representation for the theta function. So now let's put that together first. So we get integral e to the power i x naught minus y naught times something. Let's call that thing zeta over zeta minus i epsilon. 1 by 2 pi i times... Now let's call this object 0, 5, 0, n, n, 5, 0, n. Let's call that a pole. This whole thing. This is just something. So I don't have to divide it into 0. Okay. Times e to the power i pn into y minus x. 1 by 2 pi i e to the power i y 0 minus x, 0, zeta over zeta minus i epsilon. 5 pn into 5 minus x. And we have this minus x minus y, plus i pn into... What was the point of using this integral representation here? The point was that now... You see, I want to do a Fourier transform. Now since all x and y dependence is up in the exponent. Since all x and y dependence is up in the exponent, we can now do the Fourier transform really truthfully. So here this shifts pn 0 by zeta. And here it also shifts pn 0 by zeta. Both of these have minus x, right? So what do we get? We get 1 by 2 pi i times rho n squared. Of course, sum over i. Sum over n. Rho n squared times 1 by zeta minus i epsilon minus pn plus zeta. Plus zeta, they'll... Denominator. It's a denominator, I'll give you a minute. I really expect that. Are you saying I've got something wrong here? Yeah, I mean... Yes or no? Is it certain that the point is true? I'm telling you. Why do we have a delta function? There's a function because of the formula e to the power i a dot x. But there's an integral over zeta, which is sitting in the denominator. There's an integral over zeta, that's right. I just have this integral over zeta. So... Thank you. That's what I'm saying. Yeah. But this is correct, all right? Yeah. Right. Good. Right. Yeah, yeah, yeah. So that's right. Now let me do the integral over zeta. Right. The spatial integrals give us these delta functions. So you're shifted by zeta. I shifted by zeta. Exactly. Well, sorry. Shift is only the calculation. Yes. Yeah. Yeah, it's... What I mean is saying, I mean, where is the p square? I mean, where is the p square? I mean, where is the p square? Where is that? Just a minute. And there it is. I'm waiting for you to put this down. Just a minute. Why do you want dcx or d4x? What? No, this thing is dcx. So we actually... If you want to put d4x, you have to put a... this thing. The delta function... Why was the Fourier function dcx? Because... No, no, no. I was just going to mention space is the problem. No, no, no. Just... Just, yes, man. I could just go and run... call it dcx. But just one... I mean, it's a standard relation, but I don't know where I can see it. What do you have written? It's just fine. No, that's fine. It's fine. That's fine. The propagator has to have a... Give me that. Please proceed. Like this. Okay. Let me just proceed. So... Okay. So... If you want four dimensions to get this zeta by... by a... Sorry, maybe I should prepare this rather than trying it. But... Still, give it... More dense. More added? Give me one little more, otherwise it gets... I think you are saying... Yeah. Let me select which equation I move from now on. Is there a plane? There's no plane. There's a plane. There's a plane. There's a plane. So there's a plane. There's a plane. There's a plane. There's a plane. There's a plane. There's a plane. There's going to be a plane. There's a plane. There's a plane. There is a plane. There is a plane. There is a plane. I'll tell you what I'm after with the right answer. This is that... Can you remember any spectral derivation? I'm sorry. That's what I was trying to do. Give me one more guess. Yeah. Sorry. I complete this. I've answered. First is, I've answered before. I've... I've... I've... I've... I've... I've... I complete this for you next class. I'll tell you what the final answer is. The final answer when you do it, in some way or two, many of you just hear the events and variants of this formula correctly. I'm not sure. But the final answer is rho n squared over p squared minus n squared by dm squared. Where the rho n squared is what? It's this rho n on squared times delta of n squared minus n squared. This is the final answer. I'll show it to you next time. I'll show you that. What do you have this? We have this? We have a straight delta of p minus p0 plus p minus pn plus pn plus pn plus pn upon the corresponding p minus p minus p plus pn. I'm just going to say that, I'm just going to confuse, right? Because we get delta of n squared. And yeah, right. You're going to comment on the p squared by the x squared. Yeah, yeah, yeah, yeah, you're absolutely right. Let's, let's, let's, let's, let's, let's complete. So yeah, you're absolutely right. Wait, wait, wait. But this one's just p0. My confusion is why do we get pn over the spatial part? Where is that coming? Yeah, okay. That's somehow coming from Lorentz-Indian. In the sense that the sum over all states with pn0 and a particular value of p has a p in it because of Lorentz-Indian. That's where it's coming from, okay? So that when you, it's, it's, it's somehow coming from there. You have to process that. There's a standard trick to process this phenomenon, okay? But, right. So, so when you, so what, what, what, what? The claim is that you get this, rho m squared, so that's how I write it. Then you get this p squared minus m squared minus i m, so on. Where this rho m is what I wrote it down. Okay, now, so the first, I'm going to prove this to you with in the next class, okay? I'm sorry if that's not possible. But the, the first thing about this is the following, okay? You see this thing here is this full object. Rho m plus this whole rho n mod squared, okay? This whole rho n mod squared times 10 times of pn squared minus m squared sum over. This n goes over the full spectrum of the theorem. This n goes over single particle states, multi particle states, and so on. Okay, the next thing you can show is the following that if you look at the contribution to this quantity from only single particle states, okay? Then this gives you a problem. This gives you a delta function. And therefore, plotting here gives you a problem. This is totally clear. Because what you get here is just the spectrum of states formed by single particles. They are only at a particular value of the mass, okay? So the whole thing here will be your delta function at the value of mass of the particle. But you use this. Is this clear sir? This n becomes an integral over the d3p of the particle, okay? But it only takes one value of the mass. So it's clear that there's some delta function, the delta function setting there, right? So if you take the contribution from single particles, the contribution to this row of n includes a delta function at m, right now. I'm just going to say this way. However, if you now take the contribution of two particle states to this quantity, because if you've got a two particle state, you can get a center of mass energy, a p squared, that's absolutely anything within a certain range. You don't get a contribution. This thing is not just localized in one particular value of n squared. It's a broad contribution here. It gives rise to a cut in the two points. It's a cut because there's some discontinuity between the i itself. But the discontinuity is not localized in a point, it's everything. Is this clear? Just to say this again, suppose I had a formula of this sort. And what I tried to do was to compute. So suppose that I've got this taken up on the complex plane. And I try to compute, I regard this as a variable. This p squared, let me call this s. And on the s complex plane, I compute a contour integral that encloses some region. I compute a contour integral that encloses some region of the real axis. Then because I've got a pole in here, so I do the n integral last. But any given value of n, I've got a pole in here. So I get some contribution. And then I get a contribution to every value of n here. So you see that this object here, which this contour integral over the real axis, will be effectively integral to one aspect and some numbers. But if you just regard this whole thing after you've done the n integral as an analytic function in the variable p squared, then this integral just gives you an idea of the discontinuity across. So it's clear that there's a cut. And then the strength of the discontinuity across the cut is proportional to rho. In the particular case that rho m squared was a delta function, the cut becomes a pole. So the important point here is simply that every time you've got a pole in the propagator, there are single particle states. Cuts in the propagator on the other hand represent multi-particle states. And this is one of these. This is one of the few results that I've been trying to derive. Any more questions about our analysis of pole in the propagator? The answer that we found out is that there are single particles probably in the state. It still does not tell us that there are bound states. That there are no bound states. That there are bound states. So that's what we're going to analyze next. So the comment is perfectly banged on. All we know is that there are no single fermions states in the theory. But we don't know anything else about the state. So now the next thing that Dov does is to try to compute something more physical. And the more physical thing that he wants to compute is the two-point function of the operator side-by-side with side-by-side. Now, why do we want to compute the two-point function? A, firstly, it's KG pair. So this is an answer that should have a finite answer. This should have some finite answer. It should have a finite answer. We won't be satisfied with getting some infinity. But how does it have bearing on the question of confine? It has bearing on the question of confine, but precisely in the way that we were trying to understand. You see, because for any operator, this Neymar representation theorem, once you get it right, tells you that there's the spectral decomposition of the propagator. So the supply is also your side-by-side. Now, if there are... So the question is what is the analytic structure of this two-point function? If there are poles in this propagator, that will tell us that there are single particles. Single particles created by the action of cyber-sci on the back. That was vacuum operator state non-squared. Single particles created by the action of cybers. Now, suppose there were single fermions states. What should this robot, this cyber-sci, create? What would it create in free with further? Suppose we did the calculation in the free theory. What states would this operator act in the vacuum create? Well, it goes on. But how many particles states? In the free theory, what's the spectrum of the particle? The theory free particles. Cyber-sci acting on the vacuum creates what? And the zero particle is unique. That's a vacuum. But when it creates anything, it creates two particles. In the free theory, when we compute this thing and we analyze its analytic structure, what will we get? We get cuts. And only cuts. We have no bound states. So we have no poles. And we get only cuts. Right? All in the other hand, what we are going to show in this theory is that we get the reverse. We get only poles and no cuts. In fact, there are no cuts. See, cuts are multi-particles of single-quark particles. So if there are no cuts, it means that there are no single-quark particles. Because if there are single-quark states, then inevitably created by sin, then inevitably cyber-sci would create a two-particle state. So the fact that there are no cuts and not two-particle states of cyber and sin, that is the statement of confining. Moreover, you also find that there are poles, which are like the piezoids. That's what we are going to try to do. And this is something like the clinching. The clinching thing. You know, as you understand the spectrum of this theory, also you clearly understand that the clinch knows you can find it. So that is the goal. Is this clear? We are going to proceed diagrammatically. We are going to proceed diagrammatically. And how do we proceed? So if you want to complete this two-point equation, what would we do? Well, we are only exposed to complete planar graphs. So how do we do these two insertions, these lines in which we have to connect up these lines and then address this gauge problem? Gauge problem. What diagram do we have? Well, you can put whatever you want here, in each of these lines. And we are placing each of these by the exact property. If we allow, and if we already determined the exact property, we have actually already found the exact property. By this line, by this line allow ourselves to meet the exact property. We are only able to use to sum this class. Can you see that there are no other leaning on a larger planar diagram to contribute to this problem? Is this clear? So the thing that we actually wanted to compute, this computing this fermion propagator, we think of as a technical step in the procedure of computing this physical property, in the process of computing this physical property. That technical step we have achieved, and now we are going to use it to compute this physical property. Is this clear? What happens when we are going to compute this? Yes, we are in a light-home gauge. In two dimensions, there are blue and blue-on vertices only because of a mu commutator, a mu dot experiment. Okay. But in two dimensions, one of these is zero. They are gone. Yeah, okay. That was a great thing about life. Okay, that's why we are working in this game. This is the thing we wanted to do. So we want to sum these graphs. Then infinite set of graphs, it's a pretty simple set of graphs. Pretty infinite set of graphs. Simple infinite set of graphs. Okay. Now, you understood the tricks for summing these graphs. You write down some equation that the sum of all the graphs should be. And then... So, you see, instead of thinking of this process as a whole, it's better to try to consider these graphs with these two lines open and then swap them. Okay. So let me look at this set of graphs with the lines open. Okay. Now, this set of graphs obeys the following integral equation. See, suppose I... Okay. Suppose, let me first write down all the graphs. Firstly, there's this. And I'm including in this the propagator as the curve. Whereas, how would you see that this whole thing has to follow in form? That if you put this whole sum, let me call this whole sum, let me call it sum of all these graphs. Okay. Right. Then, blob is equal to... Well, the meaning of that. Blob... I should... this blob, whatever it is, there's two lines coming out. There's two edges. These are not propagators. These are just things emerging out of it. Okay. That's this... Sonar pair... cross. What? It should be a cross. No, this is a cross. This is like a bear diagram. You see, every time in this thing, you can take all of these, except this. Because each graph here, that every edge goes online. Okay. It's contained in there. So, leaving out the bear diagram the rest of them are in here. What? Leaving out the bear diagram, the rest of them are in the edge of edge equation. Yes, exactly. That's that. We can solve this integral equation. We'll write it down in a moment. This will become an integral equation. Okay. But if we can solve this integral equation, then what we will get is something that depends on the momentum that goes into it. Let's call that... what do we call that? r. And then the momentum that goes up out here is some r minus p and p. And then what we will do, what we will do is show up these two n's and integrate over p. That could be also cool too. So, what we will do is solve this in the end what we are most interested in. In the end what we are most interested in is not the answer to the two-point function, but just its integral. Okay. So, this integral equation together. Okay. This integral equation together. Okay. It's going to give us some... some... it's got a homogeneous part and a non-homogeneous part. Now, the structure is going to be where we are interested in the analytic structure in r. This variable r. Okay. So, we've got some homogeneous term and a sauced term. Okay. So, when will we get singularities of the full equation? Only when there are solutions of the homogeneous equation without the sauced term. Of course, we've got a general linear equation. It doesn't matter what the details. We've got a linear equation which is a, which is some function of r. Okay. On... on the variable we are interested in is equal to some sauced term. If we're looking for singularities of... of this equation in r, there are two possible sources. A, that the sauced has a singularity. In our case, that's not the case. The sauced is just a free line. There's no singularity. So, you only have a possibility is that A has a zero. Then at any non-secular value of the sauced when we invert this, appropriately said, we'll get a singularity inside. The solution of such equations, the solution of equations, homogeneous equation of the sauced, are always associated with the zeroes of the homogeneous one. Is this clear? Because the zeroes will drop because the singularity inverts. Because... we'll put that down. Okay. So, if we're interested, not in the full problem, not in the full 2-point function, which we could do, but only in its singularities, then we don't even have to bother about solving this equation with the sauced. We can just take the homogeneous equation and try to solve for the homogeneous equation. The solution, the non-trivial solutions of the homogeneous equation will give us all the singularities of the 2-point function. That we're after. Is this clear? This is sometimes called the equation here without the sauced term. And with this open and not closed, it's sometimes called Bethe-Salt-Peter equation. Okay? Bethe and Salt-Peter basically use the kind of logic that I'm saying. I'm not sure what logic they use. They argue that this solution, the solution of this equation gives you the spectrum. You see, we argue the same thing because we argue that singularity are obtained by this. The singularities as we've from the Lehmann representation theorem are associated with the spectrum. So, that's such an infinity thing. What? That's such an infinity thing. It cuts to sooner or later. No, no, no. Even if we had to cast this, it would work. Yeah. An analytic function. If this was an analytic function, then this would be completely analytic. So, the only possibility is that you've got some zero of this thing if you've got a non-analytic thing. Now, the point will be that these... You see, the weight will happen is that you have many zeroes integrated and that will ameliorate the extent of the non-analytic. So, basically, of a cut as being an integral over points. And that's how it's going to work. Okay? So, either you will get isolated zeroes which may give rise to poles or you will get a continuum of zeroes which may give rise to cuts. And that's just... You should just give it this now like in the sense of a Schrodinger equation. It's a little like this guy, the homogenous equation, is the bound state Schrodinger equation for a problem. Or you will find solutions at isolated points in energy. In this case, s squared. Or you will find continuous equations. And that will basically determine when you get poles or cuts. So, just focus on this homogenous equation and try to take this homogenous equation and for the solutions. Okay? So, let's think about this. Yes? Suppose we had this object, this block which I mean at the side of P and R. This is R. And this is P. R is the center of mass and the moment that goes through the process P is the division of the moment between up and down. Okay? I here, really speaking, really speaking, we've got some index here and some index there. Right? Some Dirac index up there and some Dirac index down. Okay? But, we've got a gamma minus because this interaction okay, comes with a gamma minus. Okay? Only one part of that that can matter. So, let's say this is a little better. So, this psi really was some let's put some alpha index here and some beta index here. So, let's write down now, let's put an alpha and a beta here and let's write down the integral equation and let's put this here. Okay? So, what if we get a psi, alpha, beta? Then let's first put this gate which was our propagator first. This gate was our propagator can contain some momentum. Let's call it L. And that will change the effective P and R. Okay? So, what we will get is, so let's draw them diagrammatically. So, so on the left-hand side because that's psi P R, P R, here what we have is psi P minus L plus R and let's say L plus R. Sorry, P minus L. Let's say both of them are going out. So, this is R minus P plus L and this is P. Then we had the momentum L going up here. So, this becomes momentum R minus P and this was P minus P minus L and this becomes P. Okay, so psi alpha beta psi alpha beta of P plus L is alpha index and the beta index multiplied by gamma minus. You see, because we got this gamma minus factor. Okay? So, we will get one of those gamma minus transpose. So, what we will get is a gamma minus here and a gamma minus here and then the propagator. So, let me write it down and fix the transpose. So, we get psi alpha beta gamma minus beta beta gamma minus alpha beta times one over L minus O d squared this is of P plus L integrated d and then of course we have to attach we have to attach this to these propagators. So, these propagators are one over phi slash plus gamma plus m theta a and one over R minus P slash plus gamma minus P plus m probably this is just a bit of a problem. So, I need this. No, alpha is contracted here beta is contracted here. Yes. Theta is contracted with this and that gives us an A. That is all the P is actually here. And phi contracted here that gives us a B. The first term is one over L minus what? The first term is one over L minus that is the propagator one over L minus X minus. Exactly. We have to deal with the matrix structure which is basically a triviality we have to deal with it and then we have to write down the equation for the scale just the scale equation. So, now look at this matrix structure the matrix structure takes the psi alpha beta and sandwiches it between two gamma minuses. So, it takes the psi alpha beta sandwiches it between two gamma minuses. Now, we wrote this in Eto. Let me write this as a matrix product. So, let me write this as a matrix product. So, this beta the only thing is that this guy means that means that one is supposed and that means so, I think that when you write it in L minus the whole thing squared. So, suppose I write this psi as a matrix it will become psi gamma minus one over beta, beta, theta and theta A one over E slash plus gamma plus M and then this is a this is the consequence that gamma minus transpose and this is one over R minus beta over A slash plus gamma plus M also transpose. So, psi as a matrix P R is equal to this P to L squared and psi at R and because we got this this object here we can deduce what the what the index structure of this sign was. Why? Because as before we can take this propagator and make it a simple denominator and some matrix in the numerator. If we do that if we do that you see there are two possibilities here. You could get the contribution you could get the contribution that was proportional to identity which will make this gamma minus which will leave this gamma minus or you could get the contribution that was proportional to to gamma plus which would make it identity. So, this thing is either gamma minus or so if you So, what we got here is psi and then we get either the identity which is gamma minus gamma plus or gamma minus and on the right hand side you get something very similar with the transposers. So, either you get gamma minus transpose or you get identity. We are interested in in the end this thing coming about and contracting with itself. So, we are interested in the final contribution of this which is identity. So, we want something here that in the end we give us identity. Now, how will that work? How will that work? Isn't it? Sir, why do you want identity? Why sorry? Why do you want identity? Because we are interested in this two point function. So, we are interested in tracing the indices. Interested in taking this to the psi by psi to trace and if it was proportional to any gamma structure we would get same. Sir, you want to take the same subject like we should have given that can be Yeah, the index structure should be proportional to identity. Just one more question. Now, gamma minus. No, gamma minus transpose. Now, gamma minus transpose is gamma plus just one more question. The rule is taken. I am sorry. I am sorry. Sorry. You see, gamma minus transpose is just gamma plus. You see that. Gamma minus is in two dimensions. It is just sigma minus. This is 0, 1, 0, 0. The other one is sigma plus which is 0, 0, 0, 1, 0. So, these are transpose. So, you see what we got here is psi sandwiched between gamma minus and gamma plus. And we also have the gamma minus we also have the gamma minus squared is 0 and gamma plus squared is 0. So, the only thing that contributes on the right hand side here is the part of psi that is proportional to identity. Because if you have part of psi that goes away because gamma minus squared is 0. If you have part of psi that is proportional to gamma plus that goes away because gamma plus squared is 0. The only thing that contributes is the part that is proportional to identity. Whatever appears on the left hand side on the right hand side we only we only get contributions from the part that is proportional to identity. Now, the left hand side may have other terms as well. But we are only interested in the part of psi that is the identity because we want to complete this 2 point. So, we just simply set psi to be proportional to identity. That would have been inconsistent if we had the left hand side contribution mixing from the parts of psi that were not proportional. But because that does not happen, because we have this gamma minus transverse psi, we just simply set gamma minus to be proportional to identity. But just take the tracing equation. So, let us do right. Therefore, what we have is psi of p r is equal to 1 by L minus the square trace of this part has been this psi minus psi plus e to the other of sigma minus sigma plus just 1. Okay. And let us write that. 1 by L minus the square 1 over this r minus p slash plus gamma plus m okay. Then transpose then pick up this gamma minus transpose gamma minus and 1 over p slash gamma plus m. So, this is the only part that got confused. The part here that was proportional. Now, next step we simplify these denominators sorry, I will get that in a minute. Trace of 1 by L minus square. So, then there is this identity part but gamma plus is going to be the identities. What should now happen is the only part in these denominators that contribute is part proportional to gamma plus. So, I say gamma minus right. Yeah. Gamma plus here and gamma minus transpose there. Yeah. I mean just if you remove the transpose it is only the part that is the k plus part k minus. Gamma minus is k plus. No, the one that is proportional to gamma plus is k minus. The one that will contribute is the one that will contribute. Yeah. On the right hand side the one that will contribute is gamma plus. On the left hand side will be gamma minus but because of the transpose that gets that is the contribution from k minus. See it is just a transpose of this. Okay. So, the only terms that should contribute when we do this are this. Now what we have to show here. It is clear that the terms proportional to gamma minus here is just because gamma minus square is here and similarly the terms proportional to gamma minus transpose here can contribute. So, the only terms that could contribute are the terms proportional to gamma plus or the terms proportional to identity. And we have to show that the identity contribution that the identity contribution goes away. Now let's see. Let me just check the values this way. Yeah. He does have this contribution proportional to both numerators. Here. So, what? Okay. Let's work it out. So, here on the right hand side what we get because the psi of B plus L okay. We'll get various of why don't the identity contribution because we did both identities. Why don't they contribute? You know, one identity and one gamma plus it cannot contribute. Just from coverage up. What I am not saying is that both identity cannot contribute. Okay. So, the only terms that if you take gamma plus gamma plus gamma plus transpose and identity identity. But for some reason the identity identity part does not help. Let me tell you about this next time. It's true for some reason the identity identity term does not help contribute. I cannot say that at the moment. I'll make it. Continue to be inconsistent because then there will be a part which is proportional to 1 0 0 0 0 which is not proportional to identity and the left hand side is proportional to identity. No, but we think that now we just isolate it. We know that the equation is valid. Right, the equation is valid. Yes, yes, yes. Go on. And then if it contributed there would be an inconsistency. And tell me why? Because gamma plus gamma minus part is 1 0 0 0 0 not 1 0 0 1. So that is a bit different. I didn't understand. Because of identity identity part proportional to 1 0 0 0 What is 1 0 0 0? 1 on one of the titles and the element 0. Identity identity part. Oh yes, gamma this this term. We'll make it. So this is like 0 0 1 I see. And times 0 1 0 0 something like this. And you are saying that this has an expression only in the first row and last column which is proportional to 0 1 0 0 this one you are saying? It is proportional to 0 0 0 1 1 on only one of the diagonal elements. Oh, on the diagonal elements. Thank you. It is 2nd one like this. Here. Yes, yes, yes. And why would this be an inconsistent scale again? Because the left-hand side is proportional to identity. No, but we are only looking at the part of the left-hand side. You see if the left-hand side had a at other terms we don't care about that. Okay. Let me just drop this. But if the part of the Yeah. Okay, let me give you a clean answer next time since I want to finish this. I will just do things together. If we somehow assume that the identity part is not going to be, I can't see that. But if we do that, then we get the equation side of p d r is equal to integral 1 by l minus the whole p squared in the same minus square twice. Okay. And then it just becomes r minus p the whole thing squared. Okay. Let's be clear this is the full-propagate effect. So r minus r minus p minus r minus p plus then this r minus p plus was shifted also by gamma. So plus r minus p minus gamma plus m squared plus i epsilon then we get with p. So 2 p minus p plus plus p p minus gamma plus m squared plus i epsilon. And then because the numerators which were p minus and r minus p. And then of course we've got psi of p plus l. Okay. I'll explain by the idea. Okay. And then there's some factors of g squared because of the 2 vertex factors of g squared which is 4g squared minus 1 by l. Okay. This is the equation. Okay. This thing is integrated d2l. So we were interested in the integral of this the right hand side d2l. Now notice that here we have l minus is in the answer. But we have no l pluses in this curve. We just take the left hand side and integrate it over l plus. Then both the left hand side and the right hand side will depend only on not on psi which is a function of both p. By the way, r here is a fixed way. r is not changed. So as far as the structure of this equation is concerned r is a parameter. The variable in the equation is this p. Do you understand what I mean? I mean the mixing is happening because psi of p obeys an equation which depends on psi on all p primes. But r on the left hand side and r on the right hand side are the same. So this is like a matrix equation. You should think of this like a matrix equation. Where the matrix index is what value of it. And it's like an eigenvalue matrix equation. Psi is equal to m times psi. But the matrix indices are the p indices but now the r index. The r index is just enough. Is this clear? No. It's not like psi r depends on psi and different values of r. But psi p depends on psi and different values of p. So now this p index will have p plus and p minus. But if we just take this equation and integrate it over dp plus. The thing that appears here also is the integral of psi over dp plus. So now we can reduce it and next with only one matrix index. Namely psi p minus. Is this clear? So let's do that. So let's take this equation. So let's define phi of p minus and r is equal to psi of let me use the same. So we do the integral here. So then the equation we get is phi of p minus and r is equal to integral of the whole of the left-hand side. Now the integral over L plus simply turns this psi into phi. I'll change the variables. The key is that there is no L plus that appears anywhere in this group. There isn't L minus. We can't do this in terms of L minus. So we are going to add this dp plus times phi at p plus L minus and r and times this whole curve. The curve that comes from this propagator, this propagator, this propagator and these two. So this trick is very similar to what we did with gamma, right? I mean we used the same. Right. Once again we did anything. Exactly. So what we are using here is that the propagators gauge propagator depends only on p minus and not on p plus. That's what we are using. Okay, that's basically the same. That's why we are going to do the integral over L plus. Okay. Now what remains is to do the integral over p plus. p plus also appears rather simply because it appears here and here. I want to understand how to do the p plus again. Right. p plus integral is more needed. The coefficients of these p plus is here. So we got two guys. One is an r minus p minus times r minus p plus by epsilon. And then we add 1 over minus p plus now you see what we are going to do is do it by contour method. We are going to do this p plus integral by contour method. And if the sine of r minus p minus let me get the sine. 2 p minus minus r minus p minus Okay, it doesn't matter. So we are trying to have p minus minus r minus. I don't understand. Okay, good. So you see, suppose the sine of p minus r minus was the same as the sine of p minus. Then both poles in the complex plane would either be above or would be below the rails depending on what the common sign is. Now the integrand dies off as large p plus like 1 over p squared. So you have justified and closing the contour at infinity because anything faster than 1 by p follows just by closing the contour. So if both poles are above or both poles are below we just complete the contour in the opposite direction. And we can see so the only situation in which we can get something non-zero in this integrand is if one pole is above and one pole is below. Which means the only situation in which we can get something non-zero in this integrand is the sine of p minus is different from the sine of p minus. The end of p minus has been different from the sine of p minus r minus because this part of something is fixed here. So what that tells us is that this only works if p minus is greater than r minus is greater than this. So let's for definite this assume that r minus was positive again. Suppose the sine like this was never negative. R minus was greater than 0. Then in order to get a singularity what would be, in order to get a non-zero contribution this would only work if p minus was positive, p minus was less than r minus. If r minus was negative there would be a different time difference. Let's make one choice. So the answer whatever it is would be proportional to theta of p minus and theta of minus minus p minus. Once we've got this condition then we can choose to the contour in either direction. Either direction will give us one pole general principles of complex variable theory which direction we choose to close and it won't matter. We do that, it's a simple matter to do the technique. You just pick up the recipe of that pole like I did. Closing the contour we get the answer. You just go with the answer. So the answer is what? It's this times 1 over m e squared by 2 p minus plus m e squared by 2 r minus minus p minus plus g squared by 2 times integral and then what remains is the integral over this l minus. So we had we did the l plus integral with l minus integral left and we had p plus integral. P plus integral we've done. What remains is the l minus integral. This guy is called k minus. I'm just using a notation. So i of p plus k minus what we call k minus squared p squared. So the full integral equation now full to g squared by 2. Where do these denominator factors come from? They come from the poles. Because the poles let's say we choose to do this integral by giving this pole. This would have p minus r plus at something divided by p minus r minus minus squared by 2. Notice here we see this we see this we see this factor of lambda. Notice also here that this whole answer here was independent of this l minus. This l minus appear only. So we can reword this equation as follows. We can reword this equation as m squared by 2 p minus just take this up to the left answer. Plus m squared by 2 r minus minus p minus plus g squared by 5 lambda plus r plus on 5 p minus r theta of p minus theta of r minus minus p minus and integral 5 of p plus k minus p a minus over. Okay is this okay? This way. And now a key point is coming up. The key point is that you see so far we've been carrying along this infrared divergence. So far we've been carrying this infrared divergence along with us. Okay. But now you see that this integral here, this quantity here okay also has an infrared difference. So remember that our regulator was that when we took this gauge it was our propagator which just stopped it at some point. If you do the following. So what we're doing with such integrals, integral, this integral here is what we're doing is taking this integral and doing it up to from minus lambda you know we're just excluding this part from here. Okay so we're doing our contours is this, then this. Now it proves very useful to have complete contours in solving the problem. Not to have this broken up. So suppose I have this object here and I wanted to write it in terms of this symmetric order. In terms of half of this of this. With the contouring in that little region I do a semi-subtle integral okay. In this little region I do another semi-subtle integral here I do another semi-subtle integral here I want to see what is the relationship between this object and this object okay. So clearly the relationship between this is equal to this apart from the contribution of this plus this. Now let's evaluate the contribution of this plus this right. This is just one over z squared integrated over this region plus one over z squared you see if it was both in the same direction because then it would be the pole but it's in opposite directions okay. So it's one over z squared integrated over this contour plus one over z squared integrated z over this corner okay. So we can write we can write z is equal to pi alpha okay. So dz is equal to i d alpha pi alpha okay. And then this integral here this integral here is the integral from minus pi to zero and this integral here is the integral from let's say pi again we get i times d alpha by e to the power i alpha and each of these so the first one is the integral from minus pi to zero plus i times integral well instead of pi to pi you can take like zero to pi maybe this one is let me just be safe so this is from pi d alpha pi alpha okay. So what we get we get so this guy is i so this is e to the power minus i alpha divided by minus i because i is minus one integral minus pi to zero minus e to pi alpha minus i alpha integral from each equal okay. And they each equal to up to some minus a okay. So this quantity here and what we also ignored was that there was an overall there was an overall one by mod lambdas I was doing only the angular part of the integral there was a modulus, there was one overall one by mod lambdas where in the denominator and then there was a d lambda there was a lambda in the measure of the numerator so the whole thing was going to one over lambda and something finite something finite was 2 plus 2 divided by 2 because we supposed to take half of this and half of this okay so it's a factor of 2 we have to keep track of the minus signs but you do that carefully this quantity here precisely cancels this quantity so that ambiguity with lambdas is gone this is a key point in our program okay that if you take this quantity which is also infrared divergent okay re-write it in terms of this symmetric contour kind of integral then the quantity with which we with which we regulated the problem is this quantity plus a divergent piece and that divergent piece precisely cancels this quantity okay so now if by this integral we mean this regulated contour integral half down half up then we just remove this part then we just remove this part and now we are perfectly well defined perfectly well defined perfectly well defined perfectly well defined divergence free perfectly well defined divergence free integral equation for the spectrum what we do in the rest of this analysis is to analyze this integral equation and to determine the spectrum of this integral equation since it's 10 to 1 I really wanted to complete this I'm sorry for both the flasco with the lemma representation theory I should not apply to the regular at this point anyway it's how it took half an hour to 40 minutes no more class what we do is stop this class okay let me just tell you when you analyze this integral equation carefully for this this is like an eigenvalue now all the eigenvalues will be discrete but we will determine that in a little more detail next class maybe 10 minutes of next class before turning to this study there is in 2 plus 1 2 plus 1 okay so sorry so what I owe you next class is a problem of derivation of the lemma representation theorem and the statement about why this identity does not contribute which way we will continue our analysis of this integral equation okay the punch line is going to be here though when you look at this integral equation and look for the solutions you will find solutions only at values of r plus r minus that are discrete and in particular asymptotically at large at large r plus r minus the discrete values turn out to be fine tangent so we have got an infinite number of power states in this problem but no cuts that will be there that will be here that will be the punch line we will completely we complete this analysis completely sir roughly when we are trying to cancel this divergence we will get this theta function or is it because of very small lambda it's like we don't detail in the right hand side has 2 theta functions right has 2 theta functions right so you see this 5 whatever it is is only non-zero in this equation that's what it's telling you then when it's non-zero then it can exist yeah so right so we will put it in that we draw from this equation already that 5 whatever it is only is non-zero in this equation where we will then find the the equation for 5 in that it's just like a particle in a box and in that region okay sorry okay