 This is for the camera, which has come late for this class. We're just clearing up. We're just clearing up. Thanks, guys. It's no problem. We're just clearing up confusion from last class about the index structure of this integral equation. So we're defining M alpha beta to be psi alpha psi alpha beta in the presence of one insertion of psi alpha. G alpha beta, the exact property that we psi alpha psi alpha beta with the presence of zero insertion of psi alpha, just the exact property. And with these definitions, it's not difficult to determine for themselves. But the right structure integral equation is that M is equal to M0 G gamma minus M gamma minus G. That's the right exact way, where multiplication is just by matrix multiplication, where it attributes the lower index as the first index and the upper index as the second index. It's just straight forward matrix multiplication of this object, where an M0 is just G times G. As we look, when we get this index structure here, it's clear that the only part of M that can contribute is the part proportional to gamma plus. We call the coefficient of M that is proportional to gamma. M could be anything, but the only part of it that appears on the right-hand side is the part proportional to gamma plus. And we call the coefficient of that M minus. So the only thing that appears on the right-hand side is M minus. So if we want to solve the equation for M, let us first solve the equation for M minus. Once we have M minus to become all other components of M is trivial. Because we just have to work out the right-hand side. Once we know what M minus is, then M is determined from the same question. There is a complicated equation because the unknown appears both on the left and the right-hand side. But as we've seen, the only part of the unknown that appears is M minus. So once we know what M minus, our problem is easily solved. So it boils down to just determining M minus. So we are only interested in the minus component. So we are interested in the part of this equation that is proportional to gamma minus. That means it's the part proportional to gamma minus. Right. So we want the part proportional to gamma plus in this equation. You remember last time, the singularity we can figure out in that. We just need this homogeneous equation. So we want the part proportional to gamma plus. When you say this term, even if we take the part proportional to gamma plus here, for the whole thing to be proportional to gamma plus, we need the part from G that's proportional to gamma plus. And this G also proportional to gamma plus. So we need gamma pluses to gamma minus so that the net charge is plus. So the whole product will be gamma plus. And that gives us exactly the equation we wrote down last time. See, the part proportional to gamma plus here comes with K minus. Part proportional to gamma plus here comes with K minus. You look with your notes, that's the equation we wrote. We took the part of the propagator for the vermioles proportional to gamma plus. The part of the propagator vermioles proportional to gamma plus. And just multiply it through. And that gives us the integral equation that we wrote down last time. This integral equation that had a factor of P minus minus R minus and P minus in the numerator. Because the part P minus came from P minus gamma plus. P minus minus R minus came from P minus minus R minus. Is this clear? So this, OK. Once we solve vermioles, and what we find is that this is the only option we'll solve in the process. The only place where you have to invert something. And thereby the only place from which you can get a singular. Once we solve vermioles, all the other components of it will be easily determined. And the procedure in terms of it has no possibility to generate a new singular. So all singularities in the two point function can only come from singularities in the solution for a minus instance. And therefore, if we are interested in the spectrum of the theory, we need only to study the singularities of the operator of this M minus. Or in other words, the zeros of this homogeneous operator. Is this clear? Questions, comments? By the way, if I didn't want to throw away this part, it would be easier to keep. We just keep g, g, k minus, k minus. And just keep the part proportionately down to the gamma plus problem from the other side. We're not interested in that, because that would also be easy. So this just justifies the calculation we did in class last time. Everything else we said about the integral equation. There's something kind of confusing about the integral structure, which I hope we've cleared up now. Any questions or comments about this? Let's complete the analysis of this model. And then I'll come back to the second confusion we had last time about the derivation of the lemma representation. But just to not break the discussion, let's finish this discussion in a minute. Any questions or comments about, I'm sorry for the mess up last time, but it's not a good thing. Sorry about that. OK? Excellent. So now, at the end of the last class, we've achieved the following integrity. Remember that we called the coefficient of this gamma plus in m? We called that coefficient psi. And then the integral of the psi, an integral over p minus of p plus, or the integral over p plus. And we called that phi. Remember this? And we had an integral equation for this phi. And you remember that we processed that integral equation? And one very nice thing happened. The very nice thing happened was that there was a new singularity in that integral equation, a new IR singularity coming from an integral of phi with a 1 over p minus squared. That had a new singularity at the denominator of that kernel. And then we wrote that kernel as that integral, which was just something integral. It is a big set that we was actually looking for. People don't remember this? Yeah, yeah, yeah. You remember we had that number? Firstly, you remember it was phi by 4 impurations. OK, let's just remind you. So there's this exact m. It has to move into the constant, which we got now. And then the division moment. So if we call this momentum p, this was r. It was constant. And each step in the integral equation, r was just constant. r was not a variable of the integral equation. p was a variable of the integral equation. Because if you've got this guy here, then you look at this. If this is some p, this would be some other p. And p is being integrated. So it's like a matrix equation. So the matrix value is p. And r is an upper. What was the integral equation? Phi of p minus r minus k minus r is the interact top of. And just stopping it where the gauge was unpropagated, the momentum of the gauge was unpropagated, reached. This is the interact top of an exact combination of the propagator. What is capital M? m is where the expected renormalized mass is. The number will be solved for the gauge goes unpropagated. We've got one correction term that just renormalized the mass. And then the other correction term that was infrared divergent. So now everybody knows that this is renormalized mass. And then we have this additional really ugly infrared divergent piece. So n squared is written in terms of the bare mass squared where the sum takes. This was the integral equation we got. Where we get this theta p minus theta r minus 2? Somebody? Where we get this? r greater than r greater than r. Where the positive here? Yeah, got it. Yes, yes, got it. Since r is the total, and it means r minus is the total. And at each p minus you want r minus to be mean, it will be greater than p minus. You give a good physical interpretation. But we got from mathematics. How do we, what you say is actually completely right. But we made a big difference. We got a certain step. In the integral for psi, the integral equation for psi, we had no thetas. We got a theta function when we moved to the integral equation for phi. What did we do in which from the integral equation for psi to the integral equation of phi? We integrated over p plus. Now, in the integral equation of the right-hand side, it has these two fermion propagators. And we have p plus, it has only those two fermion propagators. Because the h goes on propagating, it has the entire function of p minus. So, we have to integrate the two fermion propagators over p plus. Remember, because we are working in light cone, in light cone variables, the denominators are linear. If we are working in ordinary time, the denominators would have been quadratic. d squared is p plus p minus. So, it is linear and p plus. Right? We have written that. So, what we have got is a product of two poles. It is the integral by contour. The equation of p plus was p minus. So, we got p plus in the p minus. And then we have r minus. That is it. p plus minus r plus in the p plus minus r minus. And this was the integral of the p plus. And we were doing integral in the p plus. Exactly. Now, the point is that p minus and p plus minus r minus are of the same size. Then both of the poles on the contour go either above or below the contour. And we can always evaluate the contour integral by closing the contour in the other way. So, we get zero. Of course, you could evaluate it by closing the contour in this way. So, what you would find is that the two pole contributions magically cancels each other. As you can prove by evaluating the contour in the other way. This thing is zero unless p minus and p minus minus r minus are of different sizes. We are interested in this integral equation for some fixed r minus. So, we could just concentrate on one particular we have to look at both possible signs for r minus. But of course, you know in the end we are doing for physical singularities. r minus in the center of mass infinity. r minus in the center of mass infinity. So, we want that to be positive in the right dimensions. So, we choose r minus to be positive. And then how can these two guys have different sides? The only way to put it at different sides is if p minus is greater than zero. But p minus is less than half. In which case this guy would be positive and this guy would be negative. And that gives us p minus p minus greater than zero. p minus less than half. That gives us those two integral equations. So, outside the equation that integral which is zero. Inside the region without something on zero. Clear? And then we evaluate the integral without this quantity. Then the next thing we do is take this guy to the left-hand side. So, this becomes times now is a basically integral equation for it. So, you get some homogeneous part of the notifying file is equal to theta theta times of integral. And then we notice that this quantity was a diagonal to the integral. Because what does it mean? K minus square result. It is not a integrable singularity. But you give meaning to that the quantity here K minus equals zero. And of course, since we give meaning to this quantity. We evaluate in the proper equation, give the same meaning here. We get the same meaning. We just stopped the integral over K minus at lambda. some result. But as we will see in two minutes, what is convenient for us is to give a different definition to this integral of k minus. So, what we need was to say well we can relate this definition what it physically is to what is mathematically convenient by some traction. What was mathematically convenient was to return this integral as the average over two contours. One which skirted the singularity from above and one which skirted the singularity from below. And we deduced last time that the difference between the physically irrelevant prescription and the mathematically convenient one was some quantity that went like 1 by 1. We deduced that by computing the contribution from this one to this one, adding them up. That we noticed that low and way old that difference which appears on this side exactly cancels the g square by pi r multiplied by 5. And so we finally got an integral equation that written in terms of this integral over average contours had no reference made no reference to this i r cut off. Yes i r cut off has disappeared in the final answer in the final expression for the integral function. Any questions or comments about this? As you guys see you spent five minutes before taking this class reading through your notes just to jog your memory. And the final integral equation we achieved through the sort of process integral equation we achieved through this process was minus r plus pi of b minus i r. Now we understand this equation. You find some new variables that are sort of convenient. Let's define the variable minus divided by r minus. Now remember that b minus range from 0 to r minus because of those two theta functions. I have not written the theta functions in this form yet. So you remember that I have not written the theta functions. This integral equation is only valid for that range of b minus the class written in this equation. Outside of this. b minus divided by r minus is this variable x. What is the x range over? x range is over 0 to 1. You see that every side of this equation scales like r has positive charge charge plus 1 plus plus 1 here and blah blah blah. Let's make the whole equation neutral just by multiplying through by r minus. r minus is 1 by x. It can also be written in terms of 1 minus 1 by x. 1 minus 1 by x. And here we make a change of variables to the x range. So we call r minus divided by b minus x and k minus divided by b minus we call that y. Now the k minus divided by r minus we call it y. When we make the change of variables since there was a d2k which is sorry a dk minus here divided by k minus square. This was also of 1 over r minus charge. So we can have 1 over r minus out when we change. So the whole equation becomes neutral and can be written as minus r plus r minus by equal to m square by 2x m square by 2 into 1 minus g square by 2 pi principal value of phi of. You see that every quantity here in two dimensions g squared has dimension of mass square. Can somebody explain this to me? Somebody explain why is g squared dimension of mass square? What is the dimension of the field strip? What is the dimension of A? Dimension of A is 1. A enters a covariate derivative. A enters like del plus a. So A has the same dimension as a derivative. This may be taking 1 over g squared to lie entirely outside the action. This is dimension 1. If m is dimension 2, f square is dimension 4, d2x is dimension 2, whole action has to be dimensionless. So it has to be multiplied by quantity of length squared to be multiplied by 1 over g squared. Therefore g squared is mass square. So g squared into dimension is dimension of mass square. Mass square is dimension of mass square and momentum square is dimension of mass square. So divide this whole equation through by g square of the excitation. A plus R minus is the mass of your state. It is the mass of the excitation in units of g squared. This quantity is the renormalized mass of the problem in units of g squared. It is whatever it is. It is an equation. It is like an eigenvalue equation. It will determine the allowed singularities, mass singularities of the problem. It is an equation which we want to solve for. We want to find solutions phi that await this equation. And as often as the case we will find. The question is, are there solutions for phi or phi? For every value of alpha is alpha. Only special values of alpha is alpha. So don't call this quantity as some name for this. You call this alpha because the mass squares because this guy, this guy you call alpha without the 2. So m square by g square is alpha. And the quantity that we actually interested in which is minus 2 R plus R minus square by g square. Alpha and mu are dimensionless. Obviously. And so if we take an integral equation, mu square phi is equal to alpha. It is 1 by x minus 1 by 1 minus x minus principal value of phi of y minus x. So, this is quite a nice need of factor form of this equation. We are looking for the, we want to know what are the solutions to these equations. Alpha solutions in every value of mu square. Only particular. Why are we calling that? You see, why are we calling that? You see because what we are interested in is the singularities of the 2 point. What we were interested in was the 2 point function. And we want to know what its singularities were as function of its center of mass. Center of mass for when it was r. From Lorentz invariance, the singularities of the 2 point function can only occur at special values of r square, r square is r plus r minus. And as we reviewed last time, singularities are of 2 points. They are poles or cuts. Poles occur at special values of r plus r minus. Cuts occur at ranges of r plus r. So, what we want to know is where, what the singularities structure these 2 point functions. You remember we argued that the singularities of the 2 point function only occur at the, when you have solutions to the homogenous part of the equation. But for solutions to the homogenous part, this is the homogenous part of the equation. So, solutions to this equation give us singularities of the 2 point function as function of r plus r minus. Is this clear? So, that is what we have to know. We want to know what we have to physically is what values of r square will the 2 point function have singularities at. And the answer to that question is exactly those values of r square such that this is a solution to the solution. So, that is why it is only a mass, but why, what excitation are we doing? What excitation? You see it is the excitation that goes, that runs in the middle here. If it is, if there are poles, these poles will reflect the fact that there exists propagating particles. If there are cuts, then this will represent will reflect the fact that there are propagating multipartial states. So, every singularity is associated with somewhat shell excitation. And you remember the whole, we stopped in this exercise because we wanted to check whether or not there are cuts. Can somebody tell me why lack of confinement would imply cuts in this process? What is the answer for this 2 point function and what is the answer to the singularity structure for this 2 point function in the field? Sir, you should have branch cuts. You should have only branch cuts. No poles only branch cuts because psi by psi produce 2 part, produces 2 part of the states. So, 9 perturbation theory would lead to the expectation that we have 2 particle states or cuts. So, 9 perturbation theory would lead to the expectation that this integral should have solutions and every value about a certain mass pressure. Now, should we check? Is this clear? Let me find the integral equation. Actually, this one question about the boundary condition, this x is only defined between 0 and 0. Some question are what boundary conditions to impose at 0 and 1. Now, this actually has been a subject of huge discussion in literature, but this discussion is entirely avoid by Tov to give some arguments, which we turned out to be right. And the simplest argument is that we have got the theta functions. So, 5 vanishes on the other side of 0 and 1. So, it would be very weird if it didn't vanish as a 10 to 0 1. So, if you just didn't think too much if there are natural boundary conditions for this 5, is that 5 at 0 is 0, 5 at 1 is 1. And in all this amount of discussion of this question, let them tell you the right. 5 at 0 and 5 at 1 are 0. There is an enormous amount of discussion in the literature of the wire. Tov even has a good note in his paper saying, this will not be the last word of this question and he was very right about that. We will at this level of this class will figure out that. Sounds reasonable. Sir, the first one on the right hand side also goes over at 0 and 1. So, the word that creates some problems is right. First on the right hand side also goes at 0 and 1. Yes, that's right. So does this? Yes. The sort of this because you see this thing here has 1 by x square seniority if you set x equals 0. Why equals 0? Why is an integration there? You set x equals 0 and not for a minute. Hang on for a minute. You see, you will see. It will become clear. Yes. For the reasons that, for the kind of reasons you mentioned there has been a lot of discussion of this. Hang on for a minute to see what we do with this. Now, the question is what are the right, how do you solve these integral equations? Nobody's matters to solve these integral equations like that. It's not difficult to solve the numeric equation. Tov himself did some numerical analysis on this paper. But the most elegant thing related to this paper is to just find out first that a w is very large. Suppose we go at the limit of where you can choose your own. Suppose we match. You know that there is a limit of very large marks. New square is very large. Now, if new square is very large, these terms here in case of any values of x, it means not. So alpha is something fixed. New square is very large. So this looks like a dominant term of comparison. So it's plausible that what you are doing is solving an equation where this guy is balanced. It's this guy at the limit of very large marks. Now, because at x equal to 1 or 0, no matter how large new square is, the first term is equally large. Let's look at every value of x. This equation has to be satisfied at every value of x, not just the 0 or 1. Look at values other than 0 or 1. Then it's fine. We have to worry about how the boundary conditions are. We have to worry about what happens at the end. But anywhere else, it's okay, right? And then you have to patch this up. So let's leave the considerations of boundary conditions aside. At the boundary, this is always the sum of values. The second thing that we have to understand, the other thing that's not pointed out was that look at this object here. Remember, we're defining it with this principal value. Now, one of the things I want to show you is that this principal value definition removes all the possible significance. So I told you that we got rid of intradial divergences. But we got rid of it by some sort of trick. The trick was rewriting what was an integral term for principal values. And this would not really be getting rid of intradial divergences. This is the principal value integral itself diverged. This is an example to see that this principal value integral does not diverge. As an example, let us evaluate the point. Let us evaluate the integral minus infinity. t by omega y. Try to understand how this principal value is. Let me just try and see what we get. Where did we get the point? It's some shift of where you were. Now, let's look at this. Whenever we got integrals of this sort, we got some nice contour prescription. We always evaluate the integrals by count. We got an integral like this with a singularity here. And y is equal to x. It doesn't matter. It's positive and negative. It will not matter. Positive and negative will not matter. There is some x here. And we got one contour that goes like this. And another contour that goes like this. And we evaluate. Now, let's look at the various cases for the sign of omega. What is the sign of omega as positive? Then we are allowed to close the contour like this. So, this part of the contour does not matter. But this part does. So, suppose sign of omega is positive. Then we get e to the power i omega y integral minus infinity to infinity and we get only this contour, y minus x, so x square. Now, we are going to use Cauchy's theorem. Cauchy's theorem picks up only the pole. This is without the pole. So, to get the pole, we need to differentiate this into theta expand this function here, y equal to x. So, when we theta expand this, this is half of e to the power i omega x plus i omega e to the power y minus x e to the power i omega x over y minus x to the power e to the square. Again, this part is configured. This part does because it gives you a pole. And the contribution of this is simply 2 pi i times e to the power i omega. No, I'm doing the integral over here. So, you pick out the contribution at the pole. So, here it is. So, this is equal to minus omega and there is a half minus omega pi minus pi omega x. Okay, good. This was when omega was positive. Here, when omega is negative. Exactly. When omega is negative, the only thing that changes is we have to choose the other contour. The other contour, well, this contour is clockwise and not the clockwise. The other contour is clockwise. So, it picks up a additional minus sign. So, we get exactly the same thing as this except of the additional minus sign. So, the pole answer is just this to the mod. This quantity further thing to notice is that it's finite. How can it fail to be? You see, it's got a polar light singularity somewhere in a contour. No matter how close that polar approaches the contour, there's no possibility of a singularity. It just has a contribution of 4. To say it again, in complex analysis, if you've got a contour like this that hugs this with some singularity that is power. It's polar. Polity is a power. What's this theorem? You can always... So, it's clear that the contour approaching singularity is not a singularity. Because it's meaningless. That is the reason why we change this integral from the integral over... this complex integral over contours to ensure that the integral will not be singular no matter how... no matter if we look at the steps, not the same. You understand, this is a very important point. It's a trick to use all over many areas of physics. Very important. And of course mathematics. But it's very important to think that contour integrals are generically not singular. By the way, just as an exercise, when can contour integrals get singular? Because contour integrals are always not singular. No. Then it's singular. You know what? When two powers approach the contour from different sides of the contour. Because now we can no longer move the contour away. We want singularities. That's called pinch singularity. Contour integrals get singular basically in two situations. A, if they get pinched by singularities from opposite sides. And B, if their contour is not over it's an infinite contour over the boundary. And the singularity approaches the boundary. Why don't we just take it and move it here? When this pinch is showing out, can't go through a singularity. No, no, just go over there. Go over there. And this is my contour. Now how should I move it? I can go like this. If I go about through here, I can go through a singularity. That is your original contour. I'm just not making it here. Yeah. This is the contour. We can now use Cauchy's theorem to turn it into some other mathematically equaling idea. So it's the way we move for this thing? Yeah. No, no. You can only go above it. You can only deform the contour over a region that the function is analytic. So you're allowed to deform the contour through legitimate analysis. You're not allowed to deform the contour. Through any singularity. So you can take it about when we left you the contribution of the singularity in this pole. And this pole will be to a singular contour. Yes. Because the pressure will be singular. Because you get the contribution from this nearby pole. Is this clear? Yeah. So pinch singularity. I can't even say singularity is a contour of thickness. And boundary singularity is why you've got a fixed boundary. But you're not allowed to change. We're intact from here to there. A singularity approaches that boundary. That can often be a genuine singularity. But singularity is approaching from one side. Approaching the contour from either above or below. That's just fakes. That's what though has used. Okay. So first thing we say is that it's not a singularity. But the second thing that we see is that this function here e to the power i omega y. Okay. If we were talking about the integral not from zero to one. But from minus infinity to infinity. It's actually an eigen function of this boundary line. That is you take the function e to the power i omega y. You act with this integral expression. And you get back the function e to the power i omega x. If this integral goes from minus infinity to infinity. Because these contour methods mean that. It's not true for the integral from zero to one. But let's look at this integral. In the limit omega is very large. In the limit that omega is very large. This integral here is going to get contributions you can evaluate by saddle points. In saddle points it will only get contributions where the derivative of this function. Which is omega. It balances against the derivative of this. Exponential. This is e to the power minus log. Balance against the derivative of that log. Since omega is very large. We get contributions where y is very near x. So in the limit that omega goes to very large. This function. Yes it's dominant contribution from the region y near x. If that is happening. Provide that x is not near zero or one. Once again we're staying away from the boundary lines. Provide that x is not near zero or one. It means there is a difference whether you're integrating y. From zero to one. Or minus infinity. Because let's say that x is somewhere here. This is zero. This is one. All of the dominant contributions come from here. So if you integrate from here to here. Here to here. Here to here. Or minus infinity. In the large omega limit. In the large omega limit. Okay. This integral equation which now I will write. The integral equation which now I will write. mu squared pi of. x. is equal to alpha. 1 by x. 1 over 1 minus x. y of x. Mypress principal value of integral. pi of y. Y minus x double grey. Square. Has very simple. Has very simple. x Satur structure. Because if we deliver a plan to omega, we can take any function which obeys the boundary conditions that it manages at 0 and 1, and expand it in a Fourier c axis. Okay? So, any function 5x can be written as sum over n, sine pi n x as equal to pi i n pi x minus pi minus i n. What this integral operation did? In each of these functions, it just gave us the factor. What was it that somebody reminded? So, it slips into like a loop. Hang on. Let's just evaluate this. We plug this in and evaluate it. We just plug in that 5x is equal to this. Sorry, what could we do then? Because we're looking at the basis function. We have functions that are 0 at 0 and 0 at 1. There is a function here. Right? Yes. Okay? Now, if we plug this in, can somebody remind you what pi is? Yeah, it's minus omega, the model omega pi. Well, was there some i? No, no, no, no. Minus pi mod omega, attach itself. This function, what we get? We get minus mod omega, mod omega is n pi. For both of these, because mod doesn't change. Okay? In this basis, what do we have? We have that this equation becomes, let's take the coefficient of sine n pi x. We get mu squared is equal to, we're ignoring this. Because that's smaller than 10 to infinity. Yes, we know that. It's equal to n pi times pi times minus, but minus goes away because that's minus. Plus pi squared, mod n pi. So, if we just plug this in, what have we got? We've got alpha n times mu squared minus n pi squared sum over n is equal to 0. That's our equation. This equation has obvious solutions. That's a diagonal equation now. So, the solution is mu squared is equal to mod n pi squared. These are the eigenvalues and the eigenfunctions. That's why the sine n pi is... Sir, he's interested in what you do in quantum mechanics. But it's a bit unusual. Except that it's a bit unusual. Perhaps it's not the perfect equation. Yeah, but it's very special. It's very special. It's not even that unusual. Yeah, it's just... Yeah, this is basically a Schrodinger equation. This is basically a Schrodinger equation with some effect. So, can we take the first term as a perturbation on the... Right. So, as we take omega away, the only thing you have to be careful about is dealing with, as you have pointed out correctly, dealing with a lot of things. You have to be a bit careful about it. Because all things systematically expand away from this, and, in fact, I think we can have that. So, I won't try to work this out. The thought is, try to find the first directions. So, I made a large... So, you get n squared, n pi squared, plus 2 alpha times log n, plus... Okay, so it's not just some extreme large. You can systematically... Okay, now there are many things to notice about this answer. The first thing to notice about this answer... The first thing to notice about this answer is that a spectrum is continuous and discrete. Discrete? Mass squared is proportional to n. This is what's called a regeal spectrum. It's... Mass squared is an integer. This is a regeal spectrum. This happens all the time instantly. Second thing to notice about... So, what is the spectrum of psi bar psi acting on the right? This theory. A bunch of single particles. There are no multi-particle states in psi bar psi acting. Contrast this with what happens when you act... psi bar psi on the right in those two states. You get only multiple particles. Secondly, how many single particle states do we have? Infinity. In fact, first in logic. This spectrum of multi-particle states will truncate at some number that has to do with it. Capital N. But we were interested largely in that. We don't say this. For us, it is called a... There's mod n which is inside the summation over n. How do you know if it is u squared or n? Let me just present this whole thing a little bit more systematic. There's something I'd like to say. On the right-hand side, you have a summation n mod n. That's here. Yeah, yeah. This is the final equation. Sorry. Right? So, I think that's the condition. Either or minus 0. Yeah. Oh, mu squared minus mod n pi squared is equal to... This is the structure of eigenvalues here. The conclusion is that this... This integral equation has singularities only and mu squared is equal to pi squared times n. Pair this to what would happen in the free theory. In the free theory, it's very technical. And it won't be an integral equation. Because the two-point function would have singularities. And every value of mu squared. Larger than 4 m squared. That's the spectrum of two particle states. And we do not have a continuum that we have solutions to this equation. Therefore, singularities are the two-point function. Only and discrete values of mu squared tells you that there are no... That cyber-sides is not going to create multi-particle states. That tells you... But that is consistent with the fact that we also saw when looking at the particle states. That there are no single particle states created by cyber. Because in the world, ensuring cyber-sides would create multi-particles. So it's just a manifestation of... It's just a manifestation of... Okay? So what's going on? What's going on is that the free theory that this particle... But in the interacting theory, there's a force between particles. And the force between the particles is like that of a rubber band. It's a force that grows larger and larger without limit. As you take the distance between the particles to infinity. And therefore, you consume the squeeze factor. All this is familiar from non-realistic quantum fields. Suppose you've got some sort of... In non-realistic quantum mechanics, you've got some sort of... Potential. What is like this? What is your spectrum? We have a few ball states. And then you've got static states. So you've got continuum of some ball states. Okay? On the other hand, if you have a tension that's like this, what do you have? Only ball states. So that's what's going on here. What's happening is that there is some effective force between these quads. Being developed keeps going off to infinity. Give me your only ball states. This sounds very dramatic, okay? And it's very... It's very beautiful. It's a beautiful model. But I just want to remind you. I want to emphasize something in this. It's clear from the beginning. But it's less dramatic than it may have seemed because in the particular case of 1 plus 1 dimensions, the Coulomb force. So in... In 1 plus 1 dimensions, the... Okay. So let's remind ourselves. The force law in four dimensions, the force law between two, the Coulomb force law in four dimensions, is a force law that goes like 1 over r squared. In three dimensions, the Coulomb force law is the force that goes like 1 over r. In two dimensions, the Coulomb force law is the force that goes like... That is like 1 over r. In terms of potential, in four dimensions, The Coulomb force is like 1 over r. In three dimensions, the Coulomb potential is like logr. In two dimensions, the Coulomb potential grows linearly like modern. The Coulomb potential that is there between quarks and empty quarks is the rowing potential. The phenomenology of a hydrogen like atom in one plus one dimensions would be very different from our dimensions because even classically there is this force that grows without limit. So it's not too surprising that the same feature persists quantum mechanically. Now we see that with much more precision and much more sophisticated way than the rough density analysis would allow. But this is not too terribly surprising. The point however is that it is very strongly expected that the same thing happens in four dimensions in the quantum problem even though classically the four-dimensional force falls out like 1 over r. The four-dimensional potential. So this is a template for how the four-dimensional quantum theory might work. But it's not that surprising by itself because you know if you were going to be able to cynical about this, you would say, well, okay, classically I see this potential rising. Should I be surprised? You shouldn't be surprised. But it's nice that you can work it out in such great detail and such. The other thing to notice about this is that the spectrum of masses, the reading order is independent of the mass scale. It depends. New squared which is the mass square of particle of the excitation divided by g squared was 1 n times 5 squared was independent of the alpha. As you see the first correction starts depending on the alpha. So the thing this emphasizes is that the generation of this dynamical mass scale, the mass scale that sets the scale of masses of at least very heavy particles, is the gauge company. It has very little to do with the fact that it quarks for original mass. It's irrelevant. It's a detail, irrelevant detail. The genesis of the mass is in gauge interactions, not in the masses. That's another fact, a feature that's very expected to be true of ordinary QCD. If ordinary QCD you have massless quarks, we would have some massless speaking of the violence. But all the other mind, me is also thinking a little less. Or maybe a more dramatic way of saying it is that people sometimes go to this hard to understand statement that up quarks have a certain mass and up quarks have a certain mass. But then you add up those masses and you try to get an estimate of the mass of the proton from there. It doesn't work right. It's because the mass of this moment state has its origin in the QCD company. Not in the sprintative masses of QCD. That's the other thing we see from this model. I mean, it's a very beautiful model. Every time you look at all of the quantum field theory, it's always very nice. We saw this quantum field theory. What should it want to be true? It's so nice. So I propose to end our discussion now for two-dimensional QCD. With this we discuss zero plus zero-dimensional QCD. Matrix integrals with because zero plus one-dimensional QCD. It's a general matter field with two-dimensional quantum mechanics of free fermions. And we discuss one plus one-dimensional QCD. Now this is another solvable problem in one plus one-dimensional QCD. We discuss a little bit English with infinity. There's another problem that's pretty simple. It's just n equals 1. Not trivial, okay? It's not a non-tracked theory. This is solved by Schringer. It's called the Schringer model. Now if we're going to have these presentations, I would suggest the Schringer model as one of the big, one of the big plus one-dimensional presentations. So we help out. I would love to talk to the Schringer model I like, but we need to perform it. So I think that this ends our discussion of one plus one-dimensional QCD. We have to graduate to two plus one and three plus one-dimensional QCD. Any questions or comments? The student can be expressed for as per, which is like the denominator squared by g squared. So n squared is how we name it. It's how we say it. It can't really depend on how we name it. It depends on n squared as well. But at least if we go to energy scales large compared to this mass squared, that's a sum of things. The first term is independent. So please stress on the first term. At large, at large angle, n is much larger than log. So what we are seeing basically is that if we're in two energy scales in the mass, the scale sets by the gauge coupling and the scale sets by the mass. The scale set by the mass is irrelevant and energy is much larger than the mass. But the scale set by the gauge coupling is relevant to all energies. The analogy of string theory, more than analogy, the similarity of this solution to what you get by quantizing string is quite dramatic, quite spiky. And of course it's part of the motivation for it. A thought to propose the large-engaged theories, learn the string theories, proposal that was made of the same, typically prescient way in the 1970s and then was made concrete, I must say. Any other, any questions? Let's move on then. We had one more clearing of operations from last time. This is a Kalan Lemma in Representation Theory. So to draw my memory of the derivation, I first have to try to look up textbooks but then I realize that we're in the modern age. Modern age we need only to look up Wikipedia. So Kalan Lemma in Representation Theory from Wikipedia and that's a beautiful one page derivation. It's remarkable. I'll start by amplifying the likelihood of that derivation. I think it's a bit different from the derivation I studied as a student. It's nice. So this gives me the opportunity to tell you something. A fact that is very nice and actually used a lot in the analysis of unit energy in this category which I would love to cover. Maybe if we really continue this course next semester maybe. So many interesting things about field theory that would be nice to discuss. But this is just for the structure of free propagators. So let me, a propagator which is p squared minus f squared is mostly positive. This is a propagator for free. The propagator in position space is e to the power i v to the power x. The structure of the propagator. Something we discussed a lot when we were discussing analytic constituentation. Let me remind you of the pore structure. This is what the propagator is. The pore, if this gas positive minus suppose that p0 have a negative imaginary part when p0 is positive and a positive imaginary part when p0 is negative. We're computing this Fourier transform. Now suppose I'm talking about p0. So everything I've drawn is p0. So that e to the power i p to the power x is e to the power i p is 0 x 0 plus. Suppose we are looking at this Fourier transform. Let's look at it for many cases. So suppose we look at it for many cases. For first case. No. This is all. So there should be a negative minus sign. Where I put p0 and go. The minus sign is in the relationship with p0 to 1 and p0. Just to avoid having to draw with it. Yeah. So suppose we had x0 greater than 0. This integral here can be completed same over and same over here. We close it in control. The positive imaginary axis. If we close the integral the positive imaginary axis. We pick up only this contour. Now picking up this contour gives us what? Pick up this pole gives us what? It gives us the following. All we have to do is to all we do is to say p0 equals minus square root of p lower 0 would be negative. For x0 greater than 0 we'll close it in the lower half way. Suppose x0 is greater than 0. Yes. We're doing the integral over p lower 0. This is integral d4 0. d4 p. p0 is just an integration. And this contour is singular. We get p0. So what I meant was when we put in a negative sign even for x0 greater than 0 there's a negative sign sitting in front of p0 x0. When we are doing the integral for p0. I understand. So again, where is the negative sign sitting? In the exponent? Yes. p dot x is equal to plus p dot x. p dot x is equal to p mu x mu. It's equal to p0 x0 plus p1 x1. This is just true. Now we write this in terms of p upper 0. This is equal to minus p0 x0 plus p1 x1. Because p upper 0 and p lower 0 are related by a minus sign. Actually a trivial thing won't matter. It's just a change of variables. I've got an integration variable. For p upper 0 you can call it p lower 0. It will not matter for anything. I'm just happening to use it. For convenience of using p lower 0 as the integration variable. It doesn't change. And of course, the two poles that get it suppose we were working the p upper 0 plane. This pole would be this one. This one would be that one. But they were just totally in it. So if x0 is greater than 0, if x0 is greater than 0, then p0 is this kind. So what we get is an x of u. So we get e to the power i. Let's call this omega p. It's minus i omega p. This quantity I've got omega p. Minus i omega p x0. What do we get here? Here what we get is we plug. We're supposed to... So how are we supposed to do this? What we're supposed to do is to write this guy as p0 squared minus omega p into p0 p0 plus omega p with an overall minus. Because this whole quantity does minus p0 squared plus omega p squared. Up to the i epsilon. Then to the i epsilon by taking the right one. Taking the right one. So we have to find the residue of this pole. So we need to take p0 equal to minus omega p and plug it in here. So we just get minus 1 by 2 omega p into the power minus i omega p x0. Then we will get exactly the same answer if instead of this quantity here we use the quantity delta of p squared minus m squared times theta of minus p. Of course we do this. This is technical. But it's not critical. We do it by i theta x. We've got 2. The delta function kicks in 2 places. Only one of them is not vanishing given the theta of p0. So you get only the negative room. Then you get a Jacobian factor which is one over derivative of p0 squared to 2 p0 modulus. So you get this quantity. And this minus sign is also 1. You plug minus in here. Okay. And maybe we need a 2 pi. Maybe we need 2 pi. And for x0 greater than 0 this Fourier space propagator is just the same as the statement that you stay on chair but you take only negative lower p0 or positive upper p0. Similarly, now let's look at x0 less than 0. Maybe for x0 less than 0 we have to close the contour again. Take this guy up. Okay. So we get a minus sign and we only get the contribution from the propagator from p0 positive. Okay. Now minus sign here which will kill that minus sign because now we just get maybe put omega p here 2 omega p. So once again what we will get is 2 pi i times delta of p squared minus n squared of the strength eta of p0. This fancy propagator that we are so used to writing this fancy propagator that we are so used to writing now out here actually has an alternate representation that takes its physics model aspect. It says that if you are interested in the Fourier transform but it's not greater than 0 all you have to lose will be on-shell delta function but take the negative p lower 0 series. If you are interested in the Fourier transform where x1 is less than 0 what you need to do is take on-shell momentum but you flip in it is that energy always flows in one direction always flows from depending on which energy you are talking about earlier through later times. This is actually very useful it's clearly this is true in every dimension it's clearly a very useful way of thinking physical way of thinking of the problem and as we will see it will have as we will see in our analysis of unitary scattering has very useful formal implications. That's a useful factor. Is this clear? Now what we are going to do once we got this in our hands the Lehmann representation theorem is very easy to forget just by working with two special cases that's worked separately for x0 greater than 0 and x0 absolve ourselves of dealing with those horrible theta functions which is the cause of all this. Sir, how is physically more transparent? It's physically more transparent because propagation is entirely on-shell you see this see what was the ordinary problem with one of these square presents coming in it has a peak when you are on-shell when it has this tail everywhere and that's correct but once you look at it in terms of x0 greater than 0 and x0 less than 0 that tail is entirely a consequence of the fact that you are switching which one solution you choose see this function here is 0 away from the solution of the classic equation it has no tail away from the option this thing here changes discontinuously depending on what x0 is see if you want to write a uniform function that is the Fourier transform and all values of x0 you cannot do it entirely we see that these are the loops here because they are not on-shell this way of writing that's the point this is only valid for x0 greater than 0 this is only valid for x0 less than 0 now when you are going to loop you only do your momentum entirely so there is one function whose Fourier transform gives both these results for both x0 positive and x0 negative and that is the thing that but it's useful to know that there is a way of thinking of it that makes it very close to what's going on on-shell now we have this very much representation here we want to look at phi x by y we want to look at the time-order thing but let us first suppose that x0 is greater than y time-order is just the same thing as this object now we repeat the analysis we did in the last class we put a complete set of states here we get this in terms of 0, phi is 0 and I'm assuming phi is the machine otherwise we take phi x by y so that is what's going on mod square times t to the power of i I can't remember the sign this quantity here suppose I define the following I define p minus p i and we make function of p square that changes of p square now what I do is I define this rho of p square times theta of p0 let's see what's going on here because pn's are energies of physical states physical states are always defined as energies greater than 0 this quantity only has support when p0 is greater than 3 that's why there's this p0 theta of p0 and this quantity whatever it is some function of p square the next thing I do is the problem so this quantity here again can be written as d d d p times rho of p square theta of p0 d d p rho of p square times theta of p0 times phi p dot x minus y just by plugging in the definition the next thing we do is to define rho of p square and the next thing we do is to write this as d d p d mu square theta of p square minus mu square I just introduce the variable and put a delta function so it does nothing because it's the upper zero this tells us that the Fourier transform for x0 greater than 0 is what? it's integral rho of mu square d mu square times theta of minus p0 delta of mu square minus but this is exactly the propagator of the free boson propagator as we just did is this clear? we can do the same thing with y and x suppose x0 was less than y we would just get everything reversed a change is that this theta will become theta last piece here since we've done it for both cases and we conclude with something time-powered propagator phi of x, phi of y is equal to rho of mu square d mu square of the propagator for the free boson and therefore mu square minus m squared plus m squared this is the answer that will have the correct answer for both x0 positive and x0 p0 okay okay how you can get conflicts with unitarity if you try to understand this okay it's an interesting ongoing problem, not completely understood but very likely the end story is that you can't do anything about it okay basically there is some problem it's okay the problem is what kind of you could say that another way of saying it is is yeah another way of saying it is that we want to associate a Hilbert space with the path integral if you go back to the first lectures of this course we found that anything that was x dot squared plus something we are very easy to have associating a Hilbert space you try to generalize that to x double dot squared but I don't know path integral you have to associate a Hilbert space structure a cut in that like we try to do very carefully in our first lectures and even 9 is very difficult to do anything like that isn't that like saying not able to quantize anything not being able to build a Hilbert space picture for that path integral is not being able to quantize well the path integral could be a conjecture over the quantization yes right let's say we have got an action and we define a quantum theory by the path integral associated with the action that is something that will allow you to compute all computations question does that definition give you answers within a Hilbert space does that definition give you does that path integral define a transition amplitude in a Hilbert space with the structure of a Hilbert space this is now a mathematical question the answer is either yes or no and there is no reason to suspect the answer is yes in general maybe special case is the answer in good and n theory case only for like because we could do that the reverse like tangent to Hamiltonian because we knew that we were like quadratic in p that was the main I mean the reasoning behind being able to write the like tangent of the Hamiltonian it's much more than that it's that the whole structure of Hamiltonian mechanics is built out of position and momentum there are two pieces of the issue data because the differential equations are setting up if you work with the Rangian let's say we've got a second quantum mechanics we've got position the differential equations of the Rangian quantum mechanics the x double dot is equal to something so the initial data we need to specify evolution is x and x dot at some time because the equation gives us x double dot so there are two pieces of data we trade those two pieces in building canonical framework of quantum mechanics for x and p so we go over face to face parametrization of initial conditions so x and p has this beautiful structure that easily lifts to the operator structure of quantum mechanics it was very important that they would do suppose we had x triple dot is equal as your equation the whole structure canonical structure of classical mechanics collapses okay the more cross on structure collapses there seems no natural operator structure to build now none of this is a theorem that we cannot build so if we had three initial conditions it takes triple dot yes we would need three initial conditions yeah maybe the structure would be different but you said we needed enough in order to make the loop did not correspond to the classical and quantum mechanics we need the structure in face to face we need a bracket that is very strongly tied to the equation of motion mechanics okay so none of these words are a statement that it cannot be done it is just a statement that is not clear that can be done the sharp question is you have got some path integral you want to interpret that path integral you want to find a Hilbert space within which the path integral gives you an overlap of states by an inventory evolution operator in the case that it is second derivative we can often do it in the case that it is not I am not saying this is a theorem we cannot do it but I think it is very likely that in general this this thing here though is theorem it is telling you that the unitary quantum field theory the propagator has to follow us like one more piece what that means in the end is up to you to understand but certainly it seems somehow inconsistent with having higher derivative terms some of these are tied things you should be wary of them the point is in the same direction this by the way was tried a lot in that this is part of the reason why it is tough to make sense of quantum theory of gravity by just putting out the terms so one of the attempts that some people did was to take to try to take one another thing with r plus r squared now r squared has four derivative terms in the end editor okay and so tames the divergence of how far your propagator is going to be like one over here okay because the propagator falls out faster your divergence is better and you can actually define the renormalizable theory of gravity with r and r plus r squared but the point is that it tames it in precisely the wrong way it tames it by making the propagator one over p to the 4 and therefore the model is guaranteed to be inconsistent with the editor as detailed analysis shows that it is okay if you now try to remove this fact somehow it's very likely that the divergence is going to come back you see so taming divergence is by propagators falling off faster it's a very dangerous thing to do an interesting general subject which I don't think the last word has been written about at all but let's let's move on are the questions or comments okay so this was the oh sorry I want to continue to talk about this rehan representation this is the structure of the rehan representation now the last thing that we want to understand is what is rho of u squared what is rho of u squared that what rho of u squared you get from a spectrum of single particles we have a spectrum of single particles okay suppose we have a spectrum of single particles here then what is the end level the end level is just the momentum just be the momentum and there will be a state here for every value of the momentum okay so this thing here would just be 0 phi of 0 times momentum and the sum over n would just be d3t and then we would have to put delta of okay so let's call this some q q and delta of b minus minus q where q is the form when the energy part of q is on check for this particular single particle this quantity we understand this quantity we understand either by Lorentz invariance or just by doing the calculation in a trivial theory this quantity is just a portion of 1 over all in my term in a trivial theory for instance okay so we can just compute this plug this into the definition of of rho of p you see that actually what we are going to get is just rho of p is equal to delta because all this thing is doing here is putting the energy part on shell with this 1 over omega q factor which is precisely what you get if you had a delta function a delta function saying that p squared must be in the mass square condition of your free particle okay this would be a Jacobian and this would be this is what you get okay so I went through that very fast you can easily do it more carefully and just check that if you insert the spectrum of a free particle in here what you will get as a contribution for rho of mu squared is a delta function at the value of the mass of the free particle rho of mu squared would be mu squared minus m squared delta of mu squared minus m squared okay so we have seen that this general structure arises and we have also seen that the contribution to rho of mu squared from n being a single particle spectrum is false and even to verify that if you take a 2 particle spectrum you get a cut you can just try to work out what rho of mu squared is for instance for example of theta okay so this is all I wanted to say I am sorry any questions or comments about this I am sorry this last part is a bit fast but it is good okay any questions or comments about this okay fine so in the next class I will turn to a study of n-theories and border equations I don't know much I don't know much to say about n-theories and theta equations I don't understand the theory there are three dimensions this work of the guy called Nair who has tried to guess the ground state of the theory using a variational approach and it is quite interesting it is totally uncontrolled and it is a guess it is not like there is a parameter that justifies it it is only worth presenting in a class it is sort of interesting okay fine so maybe we don't have much to say anyway but do you understand this I love to you know I love to you know I want to discuss this instant terms maybe just maybe it is something I am working on so I would love to talk about it but I think you should there is some basic foundation of the material that one should get through