 The discussion of the problem in theories that we are doing on the reading last class. You remember we discovered that the operator product expansion was a platform that stands for hand and leg. OK. Now, I would like this class to complete our general discussion of the problem in theories with generalities. From next class onwards, we'll go on to study more detail specifically about it. OK. So the things I want to talk about in this class, center largely around the state operator map, which will be important for many purposes. But as a preliminary to do this, I want to tell you about, OK, so part of what we're going to be doing in this class is to understand the importance of this central charge testimony. So the first thing I want to understand is what it means for what this operator product expansion means, what it implies for how the stress states transforms under a conformal transformation. OK. So this is a very simple. We know that the generator of a conformal transformation parameterized by the function. So suppose we had z equals z prime plus epsilon z times z prime. OK. We have the conformal transformation of this state charge. We know that the charge that generates this conformal transformation, where we know that the charge that generates this conformal transformation is epsilon of z times z. OK. We also know that the change in the operator, OK, we also know that the change in the operator under such a conformal transformation is the residue of the form in epsilon of z, g of z, for the operator, I say no, in question. So residue. Now, let us see what this would imply. OK. The stress tensor with itself has an operator product expansion that has these things. Let me first start with the simpleture. Suppose we had an operator. Suppose we had an operator such that it's operator product expansion with the stress tensor. I lift the form, g of z with the form of 0 was equal to h of o divided by z squared, that's 0 divided by z. It's just the general form of any operator which we diagonalize this bit in the scale dimension charge of the operator. But what's special about this thing is that there are no hours in it. So it is to compute the residue of the pole in this expression. Well, what do we have? We have a pole here directly. And we have a double pole here, which we can get to make a pole if we get the first order term of the data expansion of this epsilon of z function. So we assume that epsilon of z doesn't have any pole to it. So with that assumption, we find that g of z of 0 is equal to epsilon prime of z of o, that's h, plus epsilon of z, that's h of o was the operator that we get in the path in table after making the change of variables. New operator, the operator o tilde. So the operator o tilde that we have to insert into. So if you had a path in table with an operator o inserted and the original z variables, then once you get the transformation to the z prime variables, you are supposed to insert the operator o tilde. My o tilde will be equal to 4 plus delta. My o tilde of z, but since this is z prime, that's called z prime, is o of z prime plus delta o of z prime. This is the insertion that we have to make in the z prime variables if you want to get the same result as the path in table with o inserted in the z variables. But let's expand this out. This is o tilde of z prime is equal to o of z prime plus h epsilon prime of z o plus epsilon of z delta. These two values can be combined into let o tilde of z prime is equal to o of z. Remember z was simply z prime plus epsilon of z prime. So this is just the first term in the table expansion of that. So this part of the transformation is very simple. It's just telling you that the insertion that was at the point z in one variable, if you make an insertion at a given point with a given value, let's say 1 in the first set of variables, the same insertion has moved to points with different values in the second set of variables. So the variables are different, not for the smallest set. It is very difficult. This part can also be promoted to a finite expression which has this term as its first term in the table expansion. It's simply del of z divided by del of z prime to the power h. z was z prime plus epsilon of z prime. So if you expand this out, this is just 1 plus epsilon of z prime, and then to the power h into the factor of h. There is a 1 plus epsilon prime of z. It's done a lot of stuff, so it's not h. I have to break the exercise. So this is a finite form of the transformation that will reduce to this infinitesimal transformation, 400 decimal epsilon. This is an exercise for you to convince yourself that you can get to this finite form by integrating infinitesimals and operating them such that the operator product expansion of the stress tensor with the operator had just these two severe instances. The 1 by z square piece, the 1 by z piece, then that edge of that operator transforms in a very simple way under the form of transformations. And we found that the form of transformation. We found that that simple transformation. Now, we could see that it's like this d z prime by d z to the power h over to the value of z prime. Simple transformation property under the stress tensor about primary operators. This operator product expansion would map to some nice property of representation theory of this. Well, this operator product expansion will imply that the state due into this operator, what that means, we can't do it in a minute. We can't do it any soon. But the state due to this operator is at the bottom of a tower of its representation under the video so go ahead and do that. But the real answer is we also can't do it in a minute. Sessions. I need to shift this. That is great. We see that this form of the operator product expansion implies that operators have a special role to play to enter that tension theory. That's basically the reason they call it primary. The primary or most major. OK, but at the moment, it's just me. We see that the stress tensor is not itself a primary operator. Because the operator product expansion for the stress tensor, while it has these two pieces, also in addition, but unless C is 0, the stress tensor also in addition has this effect, this is 0. Now, what does that imply for the transformation? What does that imply for the transformation of the stress tensor under the formula? So now, suppose you perform an infinitesimal formula transformation on the stress tensor in a second. Is the verification of the definitions? Yes. What is the difference between a primary and a quasi-primary? OK, a quasi-primary operator is an operator that has definite weights under state dimension and state. OK, I am state. And state. So any operator has a definite value of age in each part, OK? It's called quasi-primary. For an operator to be quasi-primary, all that is required is that its operator product with a stress tensor has this form, as well as 1 by z squared and 1 by z squared. We saw last time that that was the same statement as having definite transformation properties under under state dimension. A quasi-primary operator could in addition have other additional singularities in its operator property, OK? A primary operator has no other, no additional states. So the T would be a quasi-primary. It would be a quasi-primary. OK, so now the question we're asking is what is the transformation problem? How does T set transform under control transformations? OK, so that's 1,000, 1,000 into the test method. What we want is the recipe of the poem in this expression. Well, of course, these two parts survive. So T till and out z prime is equal to, as these two pieces in it, so it's 2 epsilon prime z T plus epsilon z then T. But that's an additional piece. The initial piece comes from taking this epsilon and differentiating three times, OK? So that would give us an epsilon triple prime of 0 divided by 6 from the data expansion. And we're also going to do the z cube. The z cube, we can't set against the z to the 4 here. It'll give us a 4, will it? And then we will have the c by 4 and then c by 12. And then we'll have the c by 2. So you get c by 12 and epsilon triple prime, OK? Now, you say that this piece is conceptually different from this piece. I insert the second t at 0. If I insert it at w, it would be epsilon triple prime and w. You see, because what am I doing? Instead of epsilon, let's look at this, OK? I'm taking t of z over 0, so the analog of that is t of z, t of 0. I want now the pole in z of this expression. So what I want to do is expand epsilon and I'll see this expansion in z. So I take this way epsilon and I want the pole of epsilon by z times w to the power 4 prime. Exactly. So there's no w. We'll put the other epsilon here. OK, so we multiply it by epsilon and we want the pole of that expression when we divide by z to the 4. So no, if you want to cancel, so we've got a singularity at the end of z to the 4, if you want to cancel that, you have to do data series expansion in epsilon of z. It should be epsilon prime over z to the 4 prime. Yes, yes, I'm sorry. Do it back up by just putting this thing in w. So let me tell you, OK, suppose I put this in w, I put everything in w. Everything will be out of shape now, OK? Now let's put everything in w. Then this, everything here is w prime. Should I try to clear out everything here? Maybe I should. OK, so everything here would be w prime and w. So this is 4 of w prime. And it's just a switched z goes to the u. Again, the same, the translation would be equal to the same. It depends on what name you put on it. Sorry, sorry. OK, I'll just find the answer since w is w by u and it's the same. I won't change that. And this is all it's going to be. It's all in there. Maybe we should put w equal to 0 because we had a problem with the extra piece which we had. The last thing, yes? If it hits over the left square, even if there's a rotation, you're asking what they will be? t of z, t of w is always h. But o, let's say o prime, o prime, w prime is z minus w prime. Always true. It holds for every value of w, in particular w equals 0. So there's no problem in putting w equal to 0. It's just sometimes obstacles things. But it's not wrong. This relation has been terrible. What I should really do is to clean it out and get rid of it in unitation. But you understand what I've done, right? Logic is clear as you can fill out the notation. Now, everyone, I want you to look at this transformation piece and notice that it's logically different. I mean, it's a big difference between this piece and this piece. Because this piece, this piece is all homogeneous in stress terms. All of them, degree 1 in stress terms. Well, this piece is degree 0 in stress terms. But what's happening here is that this additional piece, because it's not proportional to the stress terms itself, it's just picking up and telling you that you pick up. When you go into the corner transformation, in addition to what you would have got, in addition to the transformation that you would have got, had t been a quasi-prime, had t been a completely primary operator, what you get is an additional constant piece. So t shifts by something that's proportional to t plus a constant. Oh, so this is the ratio of t under the constant. Under the constant. Exactly. I think your physical interpretation of this is one really good case in a moment. It's basically, the physical interpretation will be the vacuum energy. You see, since it corresponds to a constant, it's only the constant. It contributes only to the 0. So this basically is telling you that the zero order of t is the vacuum energy, not the vacuum, it's the vacuum energy. So this basically is telling you that a formal transformation can change the vacuum energy. But this is a very familiar event, because vacuum energy has a Casamine part. So in a problem field, in any one of these areas. I mean, we have not put anywhere that you know, like, we are both in refining. Well, it could be. We'll see in a moment. And an example of this, such as a formal transformation, where we want the texture from the cylinder to the plane. So it takes you from working in a finite space to an infinite space. So the infinite space you have no gas. The finite space you do have gas. So there should be a slip in vacuum energy. That will be the physical interpretation of this effect when we see it. But let's have more to see in the example. No, but it's true it sounds funny and puzzling. And the key point is that it has to do with the fact to get the energy, energies are generally infinite. To get them right, you have to perform a subtraction. That subtraction has to be local. You set it to zero in one space, it becomes something finite. That's essentially what's going on. That's what I was thinking about. Gasoline energy. That's right. But it's correct, and it's visited. Now, once again, I'm going to leave as an exercise for you people. I think we should make actually a list of exercises. So let me promise that by next class, I'll give you a set of seven-eight problems to solve. Once again, I'll leave as an exercise for you people. And it's not an entirely true exercise as far as I know. I'm not telling you shows. I don't think I've ever wanted to do that with you guys. To integrate this to a finite space. Yeah, you know how much it changes the system. There's plenty that you have to work it out a finite expression. If the final answer is given in Kuczynski's book, it's given in terms of a short shit there. The final answer is supposed to be the following. I can't remember it. I'm sure it's not difficult. But you will explain to us. No, but OK, it's the following. It says that remember this was, so once again, plus your del z, z prime squared, z prime, equals the t of z, then minus c by 12 times this funny object, z prime z, where z prime z, f of z, f times z, is equal to 2 del qf del f, then square f, then square f divided by 2 del f. I did check, is that if you plug this in for n is equal to the z plus the decimal, and you work out the first order, the reason for it. The problem is integrated to a finite. If you take this expression, the claimers, that they're going to integrate this to a finite transformation this week. So this is how the stress tensor transforms under the form of transformations. It transforms like a primary operator plus a shift in its zero point in its constant bar. And the shift in its constant bar is given by this. That's all I wanted to say about transformation of properties of operators under the form of transformations. We come back to this in some middle part of this lecture. We come back to this in some middle part of this lecture. But before we continue now, and now I'm going to introduce to you a particularly looking form of transformation. That is a particular use of the study of the problem. So let us, this is, OK. When we were studying string theory, when we, the whole discussion of the problem in theory started because we were studying string theory. So when we were studying string theory, we were studying it on a cylinder. We had a variable sigma that went from 0 to 2 by. And we had a variable tau that defined a variable w. We're working in nuclear space now, unlike the discussion that we had in string theory, which was in Minkowski's space. Yeah, so metric of sigma and tau is the included metric. And let's define a variable w. So Kuczynski calls this guy sigma 2. Sigma 1 is sigma 2. Let's define a variable w, which is sigma 1 plus i times sigma. This is a complex variable that parameterizes where you are on the cylinder. This study of a particularly interesting conformal transformation of the cylinder. And the conformal transformation of interest to me is to a variable z, which is e to the power minus i times m, where which is e to the power minus i times m. Now, the first thing that we know is that the variable w are defined over the whole complex plane, because sigma 1 is equal to sigma 1 plus 2 by. So we get all the action that's happening in the w complex plane. It's happening between 0 and 2 by in this complex. w goes to w plus 2 by, z then comes back to itself. What a length. What sigma 1 is for the variable z is the fact that sigma 1 went from 0 to 2 by in the complex plane, where the z is not a restriction. That will restrict the region of the complex plane over which z. So we've already seen, we've already been able to understand what z is in terms of length. The phase of z is where you are on the circle. What about the radius restriction? The radius position on z is, well, let's look at it. So the minus i here, the minus i here and the plus i here. So the radius, we see that model z is equal to e to the power sigma 2. The radius goes to infinity, negative infinity. The radius goes to z. We've understood what's going on. We've got this w plane here, and we've got this z plane here. Line to constant w map the circles of constant modulus of z in the z plane. So higher up, these circles become much. When down to t equals minus infinity, the circle becomes extremely small. If you went up to t equals plus infinity, the circle becomes extremely small. The conformal mass cylinder is also an extremely useful thing. It's often an extremely useful thing to do. Firstly, you understand that there is a sense in which, if you know everything about the theory of the plane, you also know everything about the theory of the cylinder. Because these two theories can normally relate, because this theory is a conformal theory. And the map from the cylinder to the plane is a conformal. So you don't have to separately understand the theory of the cylinder and the plane. You can do whatever you want. Secondly, from the point of your complex analysis, the plane is a much simpler surface. Because you don't have to worry about any conditions. So it's often simpler if you're trying to use the techniques of complex analysis, which we will use a lot as we go on, to perform this map and work with it very soon, that we look up to very soon, that makes this map very good. We need that conformal transformation. This is a particular conformal transformation that will be of great interest to us. Immediately, we'll talk about the change of the stress tensor. What the change of the stress tensor is under this map. OK, first let's work out how primary operator of weight H would transform. OK, so primary operator of weight H, remember we had, which we didn't go up, we have that. Yeah, so there is z prime squared t of z, prime equals t of z, the one that's ready to see it. So let's choose z equals w, z prime equals z. And remember that z was equal to e to the power minus i. OK, so del z over del z prime, sorry, del, which is the subject here, is equal to minus i times z, which is the transient. This relationship becomes, remember z prime or z, minus i times z, all things squared, oh well, let's call this h, I think we're going to get it h. Minus i times z to the power h of o of z prime z is equal to o of w. That tells you what an insertion of an operator in the w brain would transform to as an insertion of an operator in the z plane. So suppose you had some function of, you were doing a path integral of the x, OK, and you inserted some operator with some function of x in the w plane. In order to get the same answer for the path integral, you have to insert a different operator and the same operator, the same function of the field times this one. So the operator would have to insert z, that way it would be of that. It would be the same operator as a function of the field. Divide that by minus i z to the power h. So look, the operator I said equal to 0 would, you'd only be able to take the limit because it would blow out, blow down. We will understand why that's the case. But the stress sensor, we have in addition a constant piece which you can work out by working out this expression on the exponential map. This is a question that if we have any operator, which is not the local operator, then we just replace the path integral in the cylinder with this particular touch. Is it always? This is true for any primary operator. You see, we derived this under the assumption that the transformation of that of a private operator. Any operator whose singularities with the stress sensor are simply those of the double pole in the cylinder. It will not be true if it's not a primary operator. But I think we'll see actually knowing everything for all primary operators in the theory is not to know everything about everything because representation theory allows you to go from a primary operator to anything. So primary operators are particularly important. We'll see this. But this statement's only true of a primary operator. Is this clear? Now, the next thing we want to do is to see what works out with the stress sensor. So the stress sensor, we're going to get the same reaction. So we're going to get T of, well, that's right. So T of w is equal to, in the case of stress sensor H2, so that's minus z squared T of z. Plus some constant, which will work out by working this thing out on the exponential map. And I'm not going to do it, I'm going to quote the result, which is k is C by n. It's the error operator will be the mode of this T. So this is relatively minus sine wave. So this is not going to work out okay, actually. But it will imply a shift with a minus sign when we do a more expansion. So the first thing we'll notice is a minus sign between these two, okay? Obsolve that minus sine wave, what we will get. And we will see in more and more detail as we go on, is simply the gas energy. I mean, the fact that the stress energy on the plane has no gas mean energy. And the stress energy then cell on the cylinder, that's what makes it work. Okay, we'll see this in increasing detail. Now, let me go. So this can form a map in the cylinder of the plane. We've understood how the stress transfer transforms and this can form. The second thing I want to understand is, suppose I was interested physically in some path integral on the cylinder, right? So I was interested in calculating some path integral which is from some state, sorry one here, some initial state, sorry one, to some final state, sorry two. Okay, with various operating answers. We've got initial state, final state, and operating answers. And I wanted to do the path integral. I want to understand what path integral this is and what it transforms to, what it transforms to on the safety. We have a question of what path integral calculates a transition amplitude with operating answers. What if you know very well from the general formulae. So suppose the fields that we were doing the path integral over are denoted by x, we have dx, and then we have exponential I times the action, suppose we were in the Minkowski states just to be homophobic. I times the action, then you put the operator as four, one, two, three, four, one. Four, one, two, three, four, one. That represents this part of the evolution. But now we're going to put in the fact, we want to put in the fact that we're making a transition amplitude from this state to that state. How do we do that? Even state is represented by a wave function. So let's say that this state is represented by the wave function psi of x. x, as in this case it is, that is the function of sigma, this would be psi of x of z. A Schrodinger wave function whose variables are x of z. And the final state, I got the phi one, so let's go to the psi one, the final state is sum psi two of x of z. And then this transition attitude. We want the time t two of the conditions that act time t one value of the function of x of z. So let me call, do you know that by time t one x of sigma is frozen to somebody here. We freeze it to some other value, let's call it y of z, not completely modified. With the boundary conditions not completely modified. So this is some function of, which depends on x of sigma, it arrives. And then in order to get the path into, to get the attraction amplitude in question, what do we do? We take this function of, and we integrate it against psi one of x of sigma one, of sigma, and psi two star of y. And then do the integral over all values of state to expression, of the transition attitude between state psi one, state psi two, where the operator is excited. Is this correct? I mean, if quantum mechanics has a familiar analog, the path integral where you demand that the initial particle is at x one, and finally the particle is at x two, is the transition amplitude for a particle from point x one to point x two. But of course, if you've got some wave function, you have to convolve that amplitude with the wave function. Yes, you take that, take what's fixed entities, and convolve it against the wave function, the two wave functions, and that's how we do it. If you want the transition amplitude, you know what I mean, right? I mean, the path integral gave you x one, e to the power i h minus i h d x two. But if you are interested in psi one and psi two, what, let's call this one and two. Okay, what you will do, is to say that if I was interested in psi one, e to the power minus i h d psi two, and then instead of complete set of states, and get psi two x two, x two e to the power minus i h d x one, x one psi one d x one d x two. This quantity is the complex conjugate of the wave function, the final wave function, this complex quantity is the wave function, and these two entangles are the entangles of it, d x sigma d y. Right? So, if you were interested in computing some state-to-state transition amplitude with some operating sessions, this is the part of the amplitude, if you do. That's a scare, right, everyone? If you were interested in calculating a state-to-state transition amplitude, I chose my states to be eigenstates of, because we want everything to be nice and complex, let me use it to be eigenstates of both the Hamiltonian and the angular momentum. Initial state, for instance, I'm going to do the same with the final state, but first, for concrete conceptions, my initial state, okay, to be an eigenstate of the Hamiltonian and the angular momentum operator, and therefore, L0 and L0 bar, the way we do it. Okay, so this linear combination of left and right, 0.30. The next question I want to ask is, suppose I do this path entanglement, and I compare it with this path entanglement. Wherever the state of the final state, same operating insertions, in a state at an earlier time, I wanted to get the same answer for the path entanglement. What should the state at the earlier time be? The question is clear, because the path entanglement just generates that revolution. So if you start with that state, which if evolved, would have reached the state, and then do the path entanglement over this original region, you get the same answer for the path entanglement. We just have to say that what we will have to do is that we just have to say that if we started with a state psi, it was e to the power minus i h d of my heart. This state at time t is the same. So within the path entanglement, when you have done with convolution with that state at time t, and by that n time t, the final answer would be the same. That was true in the Minkowski space. When we are calculating the Euclidean space. Okay? You know that when you break the transition from Euclidean to Minkowski to Euclidean space, the time that we should operate have become e to the power minus h d. Okay? So in Euclidean space, if you wanted to get the same answer for the path entanglement, if you had a state that was my entanglement, and you wanted to get the same answer for the path entanglement, you made the insertion at different times, you should start with a state, Euclidean, that varies depending on your time of insertion, like e to the power minus h d, that's not the same. So the state was also a good eigenstate of angular momentum. It was also a good eigenstate of angular momentum at a particular eigenvalue and the eigen, and the angular momentum. We could say not just how the state varies with time, but also how the state varies with time. Because any eigenstate of angular momentum of eigenvalue l, which is a function of where you are on the circle, into the path by l sigma. Okay? So suppose we put in that condition as well, you know, we've got a star, which is e to the power minus h d, plus i is sigma. Sigma, sigma. And let's call this sigma two, taking your own notation. Yeah, this is just, okay. Well, this is on some fixed state. What I mean is that suppose I wanted to look at what the state became, if I changed my zero mode of what sigma was. Yeah, okay, this could be still, let's forget about it. It's not, it's not important. Okay, let's just deal with the time. Okay, now, now, is to understand what all the states that we've been making translate to the state. Translate to something that is simple. If I wanted to do a path integral with various operator insertions between one time and another, what happens to the z-plane integral with various operator insertions between one time and another? What operator do you get taking the operator that you start with and then performing the component? States, it's states, the functions of, you know, i is the function of where you are. It's just that the states become boundary condition of the path integral, not in time, but in radius. A lot of the second questions we have. That is, what path integral can I do where I change this state to be defined on at only a time to the cylinder? Cylinder picture, which means a smaller radius in the z-plane. What state do I have to define in order to achieve this? Well, the answer is this e to the power minus h sigma two on that state, which just translates to one over, over more z to the power, right? Because we had more z to the power z. More z became very large at sigma two large. More z was smaller. What's sigma two? Okay. We soon get an answer in your question, but what's the definition? The singularity of z, we'll get into that in a second. Okay, but you see that as long as we don't know all the way down to z, makes that any non-zero value ever it is. We have a well-defined path integral that's a path integral that was of interest to us. In the standard also we're considering states at different values of power, right? The states at different values of power. We are considering wave functions. We are considering wave functions? But how are we going to scale the wave function as we change this radius? Yeah. In order to get the same answer, it's totally well defined. If you take the path integral with one wave function, yeah, with more z to the power of h, that is the same wave function. Remember h is just a number, because we need a variety of states to have it. We didn't need to, but let's say this is a mistake. Okay, so it's just some, if the energy was something, we're more z to the power of that thing. But we hate to do that. Hate to do that. So if you repeat this, then I'm going to get off of scale introduction. It would be wherever this h was, under transportation. It would be the same way. It would not be the same attitude in this case. Yeah, okay. So it would pick up that extra fact. Yeah, yeah. The great thing about z plane is that there's a path integral. I could make z to be as small as I can. It's got a possibility that t equals minus infinity to z to the power of h. In that case, what does it make up? This becomes an arbitrarily small insertion. A wave function insertion arbitrarily smaller than an arbitrarily small insertion. That is the same thing as a local operator insertion in a path integral. A local operator insertion is simply a weight associated with a path integral that depends only on the value of the fields at that point and its name. Which is pretty different. Apparently, certainly the cylinder's very different because it depends on the value of x finite nature. But if you take this t to infinity, you say what we're doing is weighting this path integral by what x is doing by x of sigma, by what x is doing only in the name of a particular point in the meteorite, okay? And therefore in the limit that z goes to zero, this object, whatever it is, is some local operator. Yes, so this circle had a magnet radius. Then basically we are talking about the same path integral that you have given, but evaluated on that thing. Yes. With the initial and final shites. Yes, now. I'm currently thinking that as you are taking that, one of the weight functions out there that just becomes one of the operators in operation. Exactly, exactly, that's going to be it. You say the weight function, what is the weight function? It tells you to do the path integral in particular boundary conditions on the circle, give it different, and give the path integral different weights depending on what these boundary conditions are. Now, the place where the wave function is shrinking is zero. So what you're doing is doing the path integral with weights depending on what your field is like in the neighborhood of zero, then operate and search. Any function that weights the path integral, any weight, the path integral that depends only on the phase of the data, whatever, given points, whatever. If it is little local operation, insertion and weight, you can say that. We'll talk about that. We'll give you a say. We'll talk about that first. Sir, a particular point is mapped to a particular point there. I don't understand that. Just say that. Second? Okay. Okay. I can understand that the first one is equal to minus. Right. This is mapped to a point. Okay. But if it happened at that point. Wait, which point? This point? Yeah. Okay. Let's say I insert an operator at some point. Yes. Which is not zero, you're asking. Which is not zero, let's say. Yes. Then it doesn't, I mean, I mean, it doesn't become a way for me because it remains operator in session. That's correct. But what we're doing is establishing a map between operators and their functions. Okay. What I'm going to do is tell you what the processor given to you does is given an operator, given a way function, it defines a local operator. Once you've got the definition of local operator, you can insert it anyway. Okay, because what is a local operator? It's some function that feels locally at that point. Once you know what an operator is, you can insert it wherever you want. Wait. But the same function, not if it's at that point, but at some other point. Yeah, but what in mind about this? At 10 equal to zero, this mapping is not 1-1. There's the whole surface map to a point. That's why this good stuff is happening. See, you know, suppose, perhaps your body, it's that what's been defined is a singular. You see, because we're doing a bottleneck all over the place. So the generic way function is to put it at equal to minus 70. Ah, okay. I mean, when I map it, I will get something which depends on which... Which direction? Okay, but you see, we're also waiting by something singular. You know that the net effect is not singular. Because the answer to this process is the same thing as doing a pure, which is clearly well defined. You see, so if your body is that by this process where defining an operator is singular, that can't be. Because its correlation is a well defined. Its correlation function is a well defined. Are we using any property of asymptotic states that we make in states that are equal to minus 70? No, we're not using that. So, you know, it's conformal to the case. It's not going to happen. Ah, you know, in a subsequent lecture for the case of free bosons, work is a way of switching. Okay? Some of your body, I think, will act. Because you'll see how it works. Okay? Maybe I should do one. No, let's not. Let's leave that for when we get to the free bosons. Okay? For instance, we're sure that waiting by the operator del X to the power n is the same thing that lets you, is that if you act with the first oscillator excited eight times, you take that state and you do the procedure at the top. We'll show it explicitly that what this is is the same thing as waiting by the del X, del y. It's totally not singular. I think some of your body will act. All your questions would arrive in this context, you'll see how it works. But, even from the point of view of general logic, I think the algorithm is totally subtle. It's not subtleties. But things like normalizability of the states. You must start with well defined states if you want to. You want to go through this argument. When you're telling about the reverse, think in a moment. It's obvious. But you always have to go, other than to perform an argument that if you insert an operator, you can always produce a state. What we've done is understand how you can get a state from an operator. You always have to worry about whether you can do something formal. If you've got some state that doesn't like you, it's not normal. Those things are always possible and you have to be careful when you respond. But what we've done is what we've done is a map from states to operators. States to local operators. Yes, reverse. Because I think the path in general is actually like this. I insert the operator at the origin and I do the path in general subject to boundary conditions that x of sigma is something. The answer to the path in general will depend on what the x of sigma is. They're going to be functional of x. That functional is a way of functional. And it's clear in the root. It is the way of functional of the state operator map. It does function psi. And then you act it by the way of function that is 1 over the mod z to power h. You brought it down here, you got some operator. What was the property of that operator? By construction, it's supposed to be no operator, no other operator. And I chose this to be mod z into 1. Especially in this case. By definition, it is that operator is that operator which when inserted and with no other operators, it's the state. Is this correct? This operator which actually corresponds to the state at minus, at t equals to minus infinity. Or it depends on the state psi at the you know, I mean at this because you're trying to go to that state. Let's look at the logic. What I wanted to show you that if you start with the state, you can get an operator. Now I was supposed to forget about this. Give you an operator which you know nothing else about. You don't know how it's done from the state. But to ask, what state had I got this operation to move? Would I even need this operator? The question is clear. All you have to do is to do the positive integral with the insertion of that operator at the order. And you can give you all your functions. How does it give you all your functions? But the answer to the positive integral depends on what the values of your fields are on the subject. But it is a set of operators of the theory. Now you know, for instance in the theory of free state of field these operators would be operators like del x del square del x all the local operators in theory they're not quite creation operators because they look not the creation operators of particles which are not but they won't be local in the sigma 1 and sigma 2. But they're local always you see it up. It's the local weighting of the positive integral. So no matter which values you're looking at local is local. So, you see what we've done is define a map from a state to an operator well defined operator which is very insertive so we can just do the your functions of the transmission that on the operator and we can use that operator Exactly, exactly. That's good. You've got the operator, how it is some function of x is you know, reverse from that I think you wait, if you wanted to work with it that's it. That's perfectly right. If your question is how are you going to use this let's wait for me to use it for every statement. And the statement is there is a one to one map between the local operators and the state. What are your problems with normalization? You know, the second map is not so clear because it could give you a normalization of the state. It sometimes happens not on the things we're going to deal with. This formula we just have to be looking at. What are your possibilities of basically the stabilization? There is a one to one map between operators and theory, local operators in the theory and states. Now, I first want you to see how surprising this is because in quantum mechanics a state is a problem and operators are matrix. There are many more matrices in the world. Number of matrices in the square is the number of words. How can there be a map between number of operators and between states and operators? See it's clear that you can get starting with any operator you can get a state. Act on that state. If you stick that's a nice denominator map and that's actually what we're talking about. It's because you see there are a lot of annoying people but enough to do it. It's not, you know for instance then that would establish the person one to one. Because if you add it to the operator any operator that I added to the map should give you the sense of it. So your point is basically chopping across. It's not a local it's not a map between all operators of the theory and states. It's only a map between local operators and states. A map between operators and states would just be a manifest operation. Can't be true. It's clearly a map between local operators and states. It's not true in every theory. It is true in conformity. In fact it's true in conformity. It's in every dimension. The plane goes through more or less uniform. We're interested in that at the moment. Two dimensions. You can ask him. Okay. Very good. Your question was your question was why is this thing essentially acting with the operator on the back view? But you see suppose you do an arthenthector. Suppose you do an arthenthector and it's a little more it's actually acting the lowest energy state that survives when you act. The operator. Let me say this thing. Suppose you do an arthenthector and let the cylinder think already clear. In any space depending on what boundary conditions you put here it could have any state in one. But in any space it could have any state in one property. But if you're doing this in euclidean space then no matter what boundary conditions you put down below as long as the boundary condition isn't it's not so funny that it projects out the back view. If you put the boundary condition far enough below the only state that survives at late an arthenthector is the back view and the boundary condition of it is e to the power minus hd. You can do an euclidean arthenthector and allow the arthenthector to go down to times t equals minus infinity. What you're doing is stopping with the back view in order to get our state is acting with so we've got the disc which is the angle on the cylinder that went all the way down to minus infinity. It's back view state because we can transform it translate it in some sense but it's hard to do because that's the singular point. Okay? The operation and definition we will always use is the pass and take. So the the corresponding operative of the back view is the irq in some theories in all unitary theories is the irq. Because it may be something else but just like the logic that you hear right now the operative acting on the back view I mean the back view is also an operative right? Right. If you want to do that operative at this point Right. So there has to be the back view at that point also. Which we are acting Right. Unless something funny to be used to spoil that logic what you're saying will be true and it will basically be generically true like what you're saying is totally right. There will be funny theories like what you just said which we will discuss when we get to this point. But basically if you are interested in unitary the operative everything is nice what you said is exactly right. We come to caveats when we get to that level. Basically right. The thing I said was not totally right. Let's get back. Essentially right the not always come back with me. But it's a good comment. It's good intuition also. Good. Any other questions or minds about this? I know this is a very subtle idea and it took me as a student a very long time to understand. I think the technology explaining it has improved basically because Pozinski wrote this book to me. Right. But you know the explanations pre Pozinski's book made absolutely no sense at all. But this is the language in which I think it is the best, easiest to understand. And to my taste. Many of the ways of saying it many people have said many strengths and weaknesses. Okay. Fine. Now. All that's very nice. Now we want to use this. Use this fact nor do you do two things. Okay we've got much slower in this class than I hope to. But it's okay. You can't get the idea straight. I want to use this fact in two ways. Let me first use it for calculation. And I'll next use it in a concept. Maybe I'll look at the concept. Since you'll get less time, you won't run. And then the calculation. First the conceptual thing. I don't understand the operator product of this action. Important thing. Say you operate a map. But you're now making a clean statement of what the operator product is. And I have two operators. I say 4-1 and 3-2. 4 operators at the point 4. Even if the operator product is action cells. In this case. Make a clear statement. Draw a circle around a 1 whose radius is which passes through it. All operators inserted within the circle. There can be any one inserted inside the circle. All operators inserted within the circle. Okay. Then this product of operators can be replaced by by OI or JJ. Some C, I, J, K, O, K inserted at let's call it Z1, let's call it Z2. I'll say it's 2. And this this replacement gives you the same answer for the path in the table. No matter what other insertion you perform. The claim of the operator product expansion is the statement clear. That insertion of this operator and this operator expires by an infinite sum of insertions on the edge, the point Z2 of different local operators. The correlations of this expansion are independent of what insertion happens. Of course, for any human insertion you know it's not so surprising that you can do some sort of data expansion of the answer. Because the answer depends on Z1 depends on Z2. So you could take the answer for the difference in Z2 and Z1 The key point is that it's independent of what these other insertions are. Is this clear? Just in order to see how he understood the statement, there's a general quantum field theory reasoning. You can read the one verse explanation in his book on quantum field theory. I never quite understood it but maybe it's a complicated explanation. No, that's correct. But I'll just see how simple it is. What's the act? What's the act? Well, you see doing the insertion with this operator insertion and this operator insertion we can do the path integral to some radius just outside O2. The integral is a well-defined path integral which depends on the boundary condition of X of sigma on this radius. And therefore define some state. Why just make an operator map? I should have drawn better pictures. Let me draw the pictures then. I'll start it. So let me just do simplicity. It's not essentially the argument at all. It's a translation. But for simplicity, I'll put this first guy at the origin. When you call the origin, it also totally repeats up. So I'll write an idea. This is O1. Let me draw myself. And I can always circle the operator O1 with the circle that lies outside both O1 and O2 and it plays no off. So whatever I put this correlation function is the one path integral. The path integral is two paths. The path integral over fields here and the path integral over fields here. There are also path integrals over fields here. So three paths are just integrating over fields in the wave function. So doing the path integral in the other disk computes over the function. And then doing the path integral on the other side is the other path integral weighted by that wave. It's some wave function with no reach in completion of the path integral that I have of interest. But that wave function is by the state operator map. The same thing as some operator. I don't know which but some operator when we call that operator is going dark with nothing here on it. In the disk. And these operators play instructions. What this operator is depends only on the map between between that has replaced the insertion of these two operators the insertion of the single operator and which operator it is depends only on the O1 and O2 one and the one on the other are the insertions in the which I choose to expand my operators in the basis of operators that were dual or two I mean states of the Hamilton at any good state can be expanded in a convergent fashion in I mean states of the Hamiltonian you can even argue stronger properties of this operator you can argue to convert this problem that's not just actually convergent. This is true only of these properties of this operator product expansion basically by replacing it about statements about expansions of states in a basis in which it is and so using the state operator map that's what the operator product expansion map. So this is the first thing it's very simple argument not even very subtle all the subtleties land the operator that the statement of the operator product expansion is really clear. So if you understood this then you know what I mean I could make these statements like t or z or 0 or 0 or 0 or 2 blah blah blah blah I could make these statements not quite simple I mean there's a no no of course when I make such a statement I need make it in the sense of the operator product expansion and the operator product expansion but in a way it denies to manage the field in the stronger sense no I have to come out the same it's conformal in that basically what's important is the state operator that exists for that that exists for that see basically what was important was that we take finance as a thing and shrink it to a point once you can do that you have your statement okay right okay after you use last class that we want to testify last class we talked about how it was always possible to choose a basis in the state of operators that were eigenstates that are definite eigenvalue unestated dimension and rotations using the state operator map for this state you remember last class we borrowed with these particular operators for quasi-primary quasi-primary operators are definite scaling to the scale dimension and definite rotational quantities but under the state operator map the scale dimension goes to energy end I should have set this more clearly you see what is the generator of time translations on the cylinder what does it map to on the z-prime see time is a radius precisely radius is e to the power time which is the same thing as x mu d by dx it's the generator of scaling transformations on the z-prime of the cylinder maps to the generator of scaling transformations of the z-prime so with given values of scale dimension and rotation about a statement of states it maps to the statement that there are any states in an interface in a possibly expanded in a basis of state of item scales of the item scale which is a manifestly true statement about any mechanical system in which the handle got inactive because we are choosing a basis in the space of operators just have good scaling dimension and good rotational quantities this thing is not a heuristic thing it's just true and we see the power of the state operator maps operators are such a large unnamed class of states are much simpler local operators are the same states so any statement of operators becomes simpler, easier to understand what does the statement of operators mean? to t ok we see it's you see it essentially it would depend on the theory in theory the free pose of the the bilinear in in creation operators because t is a very special operator yeah it's you're right you're right in representation theory it's very special you see the best way to say it is this in words that we understand that as we go on but all states of the theory can be arranged in representations of the the formal group representation of the the formal group are labeled by primary operators there is one very special operator namely the identity which is a primary the operator corresponding to the state corresponding to the operator t is a particular descendant of the identity so it lies in the module generated in a very particular way so from the point of view of representation theory it is of course very special state what it looks like in terms of some other discussion yes it will depend on the theory especially theory but from the point of view of representation theory it's very special all that much of this is coming from other questions but the state operator why we always do state things in some basis now let me quickly before we end this class let me quickly use we're going to study the representation theory in a little bit let's handle it the basic point is that you generate states by the action of t on the primary operator but how do you get the operator t you get by acting on t on the primary operator which is the identity because that's t that's the basic point we'll get rid of it in more detail now however what I want to do is to use these ideas to understand the Villas alright part of this lecture as compared to the last lecture was to try to understand things now until the states and Hilbert spaces and so on you know you should want to build theories of operators but they're not only they also have this nice Hilbert space structure isn't it and it's the interplanetary between these two structures that often gives you the most interesting things in the theory okay you remember that we had this nice Villas sort of argument I mean we had this nice pk operator problem t of z, t of w t by 2 over z minus w 4 plus t of w square root of z 2 by 2 which says that this is a function only of z and w because we derive it inside the particle table the variation of all correlation functions as we move the right hand side in these arguments let's say we make the following condition okay I'll define t of z is equal to sum over n equals minus infinity to infinity l n divided by z what this what this statement means is always understood inside the particle we can compute correlation functions of t at every value of z we know how to solve the topic so that correlation function is we define a new operator l defined by this by this equation computer correlation function if you want computer correlation function of t of l what you do is compute the correlation function of t and isolate the dependence on z that was like 1 over z by n plus 2 what is it from 2 over infinity to infinity well it is it's true it's true outside it's statement about a Heisenberg operator evolving in time it's completely true fine we will only use it if you know new correlation functions you can compute it's only true within correlation okay so what do I mean more precisely suppose I want to compute something involving the correlation function of t how do I get a correlation function of l well that's very simple suppose I want a correlation function involving an l acting at some particular time the time might be important because this analyticity is only true if there are no operator insertions you know this analytic behavior of t that t is an analytic function it's true provided you're not sitting on some other operator because suppose you've got some operator here and some operator here then the dependence of t in this region is given by the expression