 As we have been doing so far in these lectures, we focus on the closed string. In a few lectures, we'll start talking about things like open strings and unarmed strings and so on, but we'll first want to understand how we can go through a string with open strings. Okay, so the aim of today's lecture is to calculate three and four point functions of the dachion closed string theory. And in the next lecture, we will also do some calculations with the more interesting, but slightly more technically involved states, maybe the random ones. Okay, so, now, what do we need? So we have some of the general formulas that we have to do some string theory. The general formula is in terms of any integral over correlation functions in the open field. So the elements we need in order to calculate these correlation functions are correlation functions of the open field. Okay, so much of today's lecture will be devoted to computation of correlation functions of the kind of operators we're interested in in the open field theory. And the component integrates of interest are both free theories. The free scalar field, say, free BCCFD, so it's very simple, and anyway we have to do it. Okay, so let's first start with scalar field. You know, since this is the first time we're doing it, I'm going to give you two or three different methods of doing these calculations of the first level's critical, okay, all methods will actually be in the same answer, but this doesn't have to be doing it in many ways, and you can see all the time. Okay, so, so much of today. So our action here is equal to minus 1 by 4, sorry, so it's equal to... So we're calculating the X mu behind S, where S would be equal to 1 by 4, 5 on the sub-rank, D to sigma, where G is the last bar, X, L, X, G are from B. Okay, now, before we get into detailed, you know, really logical, completely satisfactory calculations, I'm going to first give you a rough and ready calculation, okay, more or less because when you do your calculations in real life, that's all you'll ever do. Once you've seen how all the subtleties cancel out with each other, it's not the way a textbook would present it, but it's not the way a good textbook would present it, but it's what you actually do. So let's see. So the first thing we need to do is to say that, well, see, we should be doing the calculation of the sphere. That's going to be interesting in three of them, please. But the sphere is conformally equivalent to the plane. It's not quite, you know, that's one very little thing we have to do. It's not funny at all. Let's see. Okay, so we do the calculation of the plane. So G is set to the flat dimension of the plane. Okay, so S is equal to D to sigma 1 by 4 by alpha prime D to sigma del alpha X, del alpha X, where you're contracting the del to the flat dimension. So what that means is that we want to compute. So let's start by saying what we want to do is to compute e to the power i k dot x k1. So let's say sigma 1 by k2 x dot x at sigma 2 by kn dot x. Okay, these are the tachyon vertex operators. Let's get into the identity, that's the moment. And we will soon turn to oscillators. And let's say first one. You can do everything you want. You can do anything you can do, so that we completely need to do this. Okay, so now how do we do the calculation? Well, here we use the fact that, you know, the fact that it's true of any Gaussian intake, that if you have, if you have, suppose you've got some Gaussian intake in the x-axis, then you want to do an activation like J i in this argument, exponential of J i x i. There's a general formula for this expectation value to the Gaussian measure. And the general formula is just e to the power J i J g by 2 x i x j. The x i x j is the two-point function of x i x j in this implicit table. Okay, you can think of this as, we can derive this in two minutes if anyone wanted, but I'm not planning to unless somebody's speaking on that. Okay, yeah, you could put an arbitrary m i in J, x i x j minus m i j x i x j by 2, Gaussian intake would be there, but we're going to complete in the square compared to two-point function. The simple one we're going to remember, the fact of it, is to expand this out to second order. See, if you expand this out to second order, it's 1 plus J i x i plus J i x i times J j x j. And this is the plus under y, J i x i times J J x j by 2. So the expectation value of the both sides, one is 1, this is the equal, it's Gaussian, okay? So I think the method of this form, the factor is easily derived by this function. Okay, good. This is what we want to do. You see, it's of this expected value of the form of interest. So the answer is simply e to the power, now this is minus because i squared is minus 1, then sum over i and J, k i dot k j into expectation value of, what should be more precise, this is k i alpha k j beta with the expectation value of x alpha sigma 1 of x beta sigma 2. But as we don't see, what's that expectation value? This is an expectation value we've already computed. We computed in an early class, just to remind you how we did that. We did that by saying, consider the integral dx mu, then put an explicit insertion of x mu. Okay? And e to the power minus s, and take this action functionally differentiated with respect to x. Because it's an integral of the derivative, we get 0. But that 0 has two terms. There's one term which is expectation value of delta of, so suppose this is x sigma 1, and we differentiate it with respect to sigma 2. We get delta of sigma 1 minus sigma 2. And the second term, we put it down functionally derivative with respect to the action, x of the action. That gives us a plus delta squared. Plus means we integrate by paths, which cancels the minus. So you get a plus, and there's a factor of 2, which is 1 by 2 pi alpha prime, delta squared, x of sigma 2. So this is equal to z. And in this term, sorry, this is also the x of sigma 2. Okay? So that's the super function of the differential equation. There's squared by 2 pi alpha prime x of sigma 1, x of sigma 2 is equal to minus delta of sigma 2. And we're measuring the law of the law for it. The derivative of log r, d by d r log r is equal to 1 by r. And then to the surface integral of this, we get 2 pi. 2 pi heart x. Because this is normal vector. We see that log r has delta squared equals so log r, delta squared of log r is equal to 2 pi. Let's get that. Okay? So now what's the right answer? The right answer is x of sigma 1, x of sigma 2 is equal to minus alpha prime by 2 log r squared, which is more than z 1 minus z 2. Well, let's get a square of z. What do you mean by that? Oh, yes, I'll ask. x minus z. Well, we've got this two-point function in our hands now. It's the same thing we derived before. Just reminding you. Okay, let's do this one with the two-point function. Of course, at this point, you could ask, at this point, you could ask, how do we know this is the right function? Why don't we add any zero more to this square? To this function? Well, it sounds right. It's translationally invariant. Basically, it's a unique translationally invariant solution to that equation of the plane. What are the solutions below? We've come to such questions in more detail in 10 minutes. Okay? But this is a subject variation of the answer. So, so now there's more. Okay, so let's just use this answer and apply it. So, of course, if we put the mu nu in this easier, we get mu nu and we get delta nu nu as well. Okay? So, now we apply what we get. We get e to the power minus sum over i m h j k i dot k j. And then we have minus from here. So, we get alpha prime by 4 ln of z i minus z j delta nu. Let's do two kinds of terms. Terms of i is equal to j and the term of i is i is 9. Breaking this up into the two terms, we get e to the power plus alpha prime by 4 sum over i k i dot k i ln of z i minus z i the whole thing is squared. Okay? Now, the other stuff is just z i j which z i j z i minus z j with the power k i dot k j by 2 for all these well, that's 5 minutes. i less than j product over test. Is it even done except for this business? Except for this business? Because this thing gives us something to define. Now, what's the problem here? The problem is it's constantly looking at an opposite operator and we're going to renormalize the opposite operator for computing its correlation contract. We're going to come back to this in eight minutes. But at the first pass, we say, well, suppose we somehow renormalize it so that we don't allow self-contractions. Okay? Then we wouldn't get any of these terms. These terms all come from x is contracted to x is at the same point. We don't allow self-contractions to be normal or something like that. We'll get any of these terms for this business. Once we do the correct renormal, it's actually in this business. This is almost right. We lose property. But, okay, this is our first pass. Is this clear? Okay? So, and we basically have some alpha-primes here. Okay? This would already be already actually active on the calculation. Okay? Let's continue with our lessons already met. Okay? For computing correlation functions, on the other hand, there's a few things we need to do. For instance, suppose attached to some of these e to the power k-axis. Suppose attached to some of these e to the power k-axis. We also had, suppose we also had derivatives of x. Those are the other kinds of vertex operators in here. We'll be interested. Okay? So now I'll be in the back end. Back end generation, you use another nice property of Gaussian equation. Okay? And the other nice property of Gaussian equation is this. Suppose you've got something down in the denominator. You want to compute expectation values where it's just down in the denominator. And you've got these other x, you know, linear terms of the expression. Then, in order to evaluate the expectation values of such an integral, the first thing you do is to substitute whatever is in the denominator according to the solution to its equation of motion. We'll do this in one detail. So, it's like this right formula. Is this connected to this? Okay. On the compute, e to the power minus k i x i. k i plus k i x i, okay? Just in any Gaussian integral, times x product over m x n. But we can do this. That's what I mean down in the denominator. Just that, that, that. And now instead, it's not our case. Sorry. Okay? Now, if you want to compute this, you see, basically, imagine how you're going to do this calculation. The first thing you're going to do is to make this a complete story. And then you're going to compute expectation values in the compute. So you would write these x n's and tell them the new variables. So anyway, when you did that calculation, you would have the x n written as whatever you get from the solution to the equation of motion of the stuff that's upstairs, plus something else, okay? Something else can then can then be contracted, ignoring, interfering without using the power i k x i. Are you... Let me give you an example of this. Yeah, it should be an example. Just to make sure you understand what I'm saying. Suppose you wanted to compute e to the power minus a button of factor. No, no. This is totally exact. Oh, you mean like the determinant? Okay, a factor like j is saying j i, which is like j x i. Sorry, what's j? Like, okay. There exists to be shifted into some mindless something. Yes. Okay. Now, for some clear, what do you mean that there's something there? Imagine calculating the expectation value of one of those things. I mean that last example. Let me give you this example. Yeah. So suppose we have e to the power minus x square by 2, okay? Plus alpha x, just one minute. And I wanted to compute expectations value of x square. So let's first do it the way we were actually doing and then just interpret it. So what do we mean? We'd say, well, e to the power minus x minus a over x square by 2 plus x square by 2. So we're going to get whatever we would've got otherwise into whatever. So let's go through this example of that. Okay, wrong. So here, first, have I got the sands right? Probably have the sands wrong. The minus. Then we would've had, and we have, this thing has a with a minus. Yeah, sorry. Yeah. And then we have x square, which I write as x minus a plus a. So now the answer is e to the power plus a square by 2. Which is the answer we would've had without any of these instructions. The contraction of x of y square plus a square minus y plus a over x square into the power minus y square by 2. Okay? So now this new calculation involves, this course can be taken correlation functions, but it can be taken correlation functions without this exponential information. Here, you can just replace things by weak contractions. So now the problem is what comes in the tables in the variable y. So what's the general rule? This is, I work out what variables totally general. What's the general rule? If you've got an exponent, a Gaussian detector with exponentials and start down the status. Then, firstly, you get the answer with just the exponentials. That's the model prime. Secondly, you replace the variables downstairs by what that variable is on the solution to the equations of motion. You see, what is the equation of motion here? The equation of motion was x is equal to 8. Okay? So you replace that by 8 plus something else. Now, something else is a fluctuation variable, this y, which then would be contracted according to usual weak contractions. The rule for having, including about the integrals with exponentials upstairs and stuff downstairs, but the stuff downstairs is replaced by what you get from the equation of motion plus a fluctuation variable which you contract by weak contractions. Okay? So, I'm going to implement that in the situation of interest to us. Okay? So suppose I interested in some number of, and this works for any number of variables, just for concrete result, but it doesn't matter. Okay? So, del x at sigma called sigma a, sorry, sigma tilde, one, del x at sigma, tilde, two, has a one, and then into the power i, k, x at sigma, one, into the power i, k, x, k, one, is k, m, x, sigma, one, is k, m, x, sigma, one, is k, m, x, sigma, one, and then, I'm supposed to look at what these del x's would become on solutions to the equations of motion. So what are the equations of motion? Equations of motion. The equations of motion were one by four pi, no, by two pi, it's two faces, two pi, alpha prime, del squared x is equal to plus i, k, one, k, m, dot, k, m, delta of sigma minus sigma, m, sigma, is equal to zero. Of these equations, the answer is simply, let me give you i to minus is right, it's i, alpha prime, minus log of sigma minus sigma prime, whatever the square, sigma minus sigma, m, and sum over n. And there's a k of course. k. k over n. And again, it's here. Minus alpha prime by two, sum over n, k, log of sigma minus sigma, sigma, m. This is the solution for it. So if we want to del x, now, just what this object is, all the solutions to the equations of motion, we depends on this. So what we get? We get delta del x mu. So we get del x mu, is equal to, I have sigma to the left. Is equal to minus i by alpha prime by two, k, mu, m, by one of these del x mu's. We replace the del x mu by del x of sigma one to the left. And the q's have the proper top, how many is that? We call it del little x, mu, of sigma one to the left. So we can replace this type, this thing minus i, so by this, stop, we can do it. So every time you have del x mu, so del x of sigma, j, little left, is replaced by minus i by alpha prime by two, k, mu, m, over 10 minus n, sum over n. Okay? And then plus del of sigma one. The rules, now to correct function, del little x, is just contract, then we're going to win contractions. But little x has the usual 2.5 function. The 2.5 function we derived at the beginning. Okay? So you can contract two of them, four of them, a non-possible win quiz, using the, using the 2.5 function we derived, minus alpha prime by two, log of 1, 6 squared, for the contractions. Of course, if you only have single del, this is effective contractions. We'll take the log and make it a 1 by 6 squared. If you're too bad at this. We won't give the rule in more detail than this. The, the description is to replace and insert the del x like this, and then do the, do the contractions with the del x, and again, you can try to write down a general formula for that, but it's not a good thing. In any case, you just do the contractions. Okay? So, there's just one thing I'd like to say about this, and that is, that, you see, and the other methods would make this, better methods would make this much clearer. But, that is, in doing this path integral over x, in particular, you do the path integral over the zero of all x. The path integral over, the path of x, that is constant integral. The path integral, where a part of everything we've talked about here, enforces a delta function for the, for the various moment time. How does zero over x? Yeah, so let's, let's take x and shift x to x plus constant. One time behind the constant, the sum over all k i's. The sum over all k i's appears later. And you know the integral over the constant, you have a delta function of sum over all k i's. Because, in any correct way, you would also find the addition to all of this, a delta function that imposes no value in constant. That has to be a delta function of this sort of square from any point of view. You see, x goes to x plus constant as a symmetric theorem. So if you take any of these correlation functions, and you send x to x plus constant, you say that you get that this correlation function is equal to the same correlation function times e to the power i, sum over k i, times that constant. The only way that correlated and then not be zero, is if that sum over k i is zero. It's sort of there being a sum that's proportionate. The delta function that imposes the momentum conservation. It does that. So we've seen all the main features of this theory. Basically, what we did is we got a momentum conservation delta function from zero mode. We get this power, this power set of z ij to power us from exponents. And we get this classical basis class with contractions from other insertions to the infinity. Now since this is the basis of everything we're going to be doing later on, and since this is a course, we want to do this in a few two, three different ways where we understand some of the rough things we've done in the future. So why don't we sweat down the property? The first thing that we sweat down the property is that you're actually looking at correlation functions in the sphere of momentum. Again, we're actually looking at correlation functions in the sphere of momentum. So does that correct things somehow? So now I'm going to follow what you should see and we need two different methods to get this answer, each of which has its own nice things in each of which will help you understand certain things about the calculation. The first method we use is to appear in the general properties of conformance. Again, we try to do the answer, we try to do the calculation in conformal field theory, using the techniques of conformal field theory now. But remember that we're learning of the sphere. Now how do we actually learn the sphere? We'll deal with it the way we did last time in yesterday's class. Remember what we said in yesterday's class. What we said was that we can cover the sphere by two patches, z plane and u, z patch and u patch. We're dealing with a theory which we can gauge by the geomorphisms and by the transformations. So, if you have some field in one patch and bear the overlaps with the other patch, it's equal to the field of the other patch after the appropriate b-phiomorphism is a plus-y function. So firstly, the two metrics in the two patches have to agree up with the geomorphism plus-y transformation. The way we did it was to make these two things conformally equivalent. You know, anything functions in each other. So that was guaranteed. The geomorphism transformation changed the metric by an overall factor which you can then absorb by a y transformation. And then we said if we do have every field in the question transform that or if you have a geomorphism plus-y transformation, then we know how to transform fields in one patch fields in the other. We know how to transform into both. Now, in every case, we looked at last time we were in the B and C system. Remember the B and C system was neutral under y transformation. So the transformation under the geomorphism plus-y transformation that takes you from one stage to the other is simply the transformation under geomorphism. As a principle, we use to determine if you remember how many zero-modes to see there were and how many zero-modes to be there were. We're just transforming them from the z-patch to the u-patch by geomorphism. There's also the y transformation that we threw that away because these things were neutral. Similarly, in the x field, more precisely, the things that we've actually used is that it transforms only under geomorphism. Again, it's an advantage of one operator and all those transformation properties are what you get by the transformation properties of death. So now, let's use this basic line here to try to see if we can pin down to try to see if we can pin down these correlation functions from another point of view. What I'm going to do is a little convoluted. It's a little tricky. This is the first principle of everything. It's a little tricky, but it's worth doing the forwarding here, guys. Finally, I'll give you another more serious part of the technique. To start with, to start with, we need to see if we can compute del x, del x of sigma 1, sigma 2 using the forwarding technique. So the singularities, we know all the singularities of del x del x because we know it's open. Okay? So we know that from the AOP, since we had xx with minus alpha prime by 2, so we get minus alpha prime by 2 1 by z1 minus z2, though, square a. We know that this is the only singularity. Next, we can add any function of z1 and z2. Next, we can add any function of z1 and z2 that is ended completely when we find the everywhere. Suppose we have del z of x. Let's see what it becomes in terms of del u. So we would play the same game we played last class. Okay? We have u is equal to 1 by z. So we get del u of z is equal to minus del z of u of x times d z by d u is equal to minus del del z of x times d z by d u is minus 1 by u squared. So it is equal to z squared minus z squared times z squared. So in order that del u of x be well defined at u equals 0, del z of x has to fall off at least as fast as 1 by z squared at least as fast as 1 1 by z squared and infinity. We see that the singular term does not exist. Also falls off as fast as 1 by z squared. Suppose you have analytic keys that also falls off as fast as 1 by z squared. No such term exists. There is no function that is analytic of the whole of the z plane and it goes at infinity falls off as fast as 1 by z squared. Is it clear? It is possible that analytic contribution must be equal to 0. So we find that this is equal to this in no further case. You see the vertex operators, remember what the vertex operators are. The vertex operators and derivatives of fields span to each part of things. That is what we want to calculate. Because it is at the technical level derivatives of fields and what happens? You have good conformity which involves properties. Correct. So for the physical reasons but it is nice. So far and this can easily be generalized to products of arbitrary numbers of their x's without any to the pi x's. At least that is an existence for you. Basically you will get all possible with contractions and anything else that you will try to add anything else you will try to add will have to be analytic and will be set to 0 by this in place of this department. Okay? You see this is easy by the way. It sounds tricky but it is not so tricky. So we are doing it correctly accounting for the fact that we are on a sphere. We are using the fact that the OP is correct in any local batch. And then we are allowing the fact that we are actually on the sphere and it is still getting interesting. And you can add exactly. You can add some additive pieces. Exactly. Basically you will have to apply boundary conditions and identity. The boundary conditions we have an identity is that we conformally at the sphere and that fuels our identity. That is basically basically the problem. Okay? Now let me use another method that is a little more clever to try the same kind of reasoning to get the correlation functions of insertions of del X along with the i t axis. Now this is a little tricky because X itself does not quite factorize into left and right and del X is purely there. But even for i k X does not quite factorize into left and right and left. Because of that we are going to use a slightly integrated method. Okay? To see what you need. Oh this is slightly trickery but okay let's go to see it once. Okay? The method goes as follows. Suppose if you were interested in del X times e to the power i k 1 X 1 e to the power i k n X by the exponential operator insertions we can always understand conformal normal law or anything and whatever whatever so that everything is totally indescribable. Okay? Suppose I was interested in this object. So let's say that this is the point sigma. This is at sigma 1 this is at sigma i. Okay? Now the first thing I can do is easily calculate all the terms in this expansion okay? So I can do the OP of this operator So let's do that as an exercise. Let's look at the del X mu of sigma where the e to the power i k let's say 1 X k 1 mu X mu of sigma 1 and let's compute all the singular pieces of e. Okay? So one way to do it is to expand the exponential in X mu sigma sum over m i in mu X mu sigma to the power m by Gaussian that is the OP written by subtractions So we get minus i in log that goes away z minus z i we get a factor of m because we have n possible ways in which the contract cancels the effect area and this just becomes e to the power So I'm clearing that let me just find the result and say the del X mu of sigma times e to the power i k mu of sigma the only singular piece in this OP is equal to i k mu alpha prime by 2 z minus z 1 the sigma 1 e to the power i k mu X mu of sigma and you see how it works, right? we look at the exponential expanded it looked at the singularities term by term and then re-sampled it should be Is it clear that when we contracted this we're done with the m Yeah but yeah there are n such terms we get a factor of m which cancels the n factor will make it an m so what we have left is X mu to the power n minus 1 by n minus 1 factorial so that re-explanates yes but there's an extra factor of i k mu which you have pulled out that's this and here we've got alpha prime by 2 and then of long as you say okay fine so okay let's start here this now if you look at the singularities with the with the other you have the same kind of thing so the singular part of the OP you know the thing that the singular in the OP is so this thing includes a piece that is sum over m i k mu over 2 z minus z m k mu m over 2 z minus z m alpha prime times e to the power times z times expectation value of e to the power i k m z that's not the angle taken set anything else we have we just seen that we just did it that the function it is going to be well defined in the u plane del z has to behave like 1 by z squared and you can okay now you might be confused because of the old 50 singularities beginning do not behave like 1 by z squared and you should correct the coefficient 1 over z but you see the coefficient 1 over z is simply sum over the k mu so the first thing to conclude is that we have momentum that the only way this amplitude will be non-zero that is this thing here will be non-zero is it some of the k times okay once we ensure that this thing does indeed go like 1 over z squared is this correct you can write down a Taylor series that was z squared and so on so that term is fine but this down here has to be and I think it has to go like 1 over z squared and there's no such term and so we conclude that this correlation function we have to be able to calculate it but what we have to be able to do is to relate it to the correlation function without the insertion of a case so we've got a nice equation that relates the correlation function with the LX insertion and a correlation function without the X what we want to do is to turn to turn this equation into a differential equation for this code and do that is to notice that if I take I take this object okay if I take this object and multiply it by K mu okay so I take this object and multiply it by K mu then it's a piece okay so let me say suppose I was interested in del X mu K mu K1 mu K1 mu K1 of X so this is a sigma and this is a sigma we just compute the most similar part of the OP but what remains once you remove the OP once you remove the OP it's simply so we've got the sigma piece we just compute it plus normal order of K mu into pi K1 X mu and now if we want to compute all the ratio of the OP all you have to do is to K doesn't need to expand so to write sigma is sigma1 plus sigma1 minus sigma plus sigma minus sigma1 and K has to expand this in sigma1 the order of zero term the term that is minus sigma1 over zero S X goes to X1 it's simply the ratio times del X mu at sigma1 that is pi K1 X mu at sigma1 everything is written as normal order of sigma1 so multiply it means the same point is over so if you are writing it as sigma1 plus and the moment I just started looking at this OP the first idea is this operator that is this operator in the first few lectures we understood how to compute such a piece okay all we have to do is to use weights here into normal order okay so here is sigma1 plus whatever remains is the product of these two things within the normal order that's so that's whatever done sigma piece plus whatever remains within the normal order this is still a point sigma this is still a point sigma now if you want to generate the full OP not just singular every third OP now it's so literally all you have to do is del X times no singularities anymore there's just nine del X times just correct so all you have to do is del X times sigma sigma1 plus sigma sigma1 and X times sigma sigma okay but the lowest level of del X times is just sigma sigma okay so the lowest level is sigma so when we get something singular plus this plus order sigma by sigma so this piece here into the power i k1 X mu sigma1 has an i so let me go over there and you have to get del X this object i k1 and del 2 X mu suppose suppose okay so what what do we do what our aim is to compute this formula is the following relationship we say that well what we have is that the derivative what we have is is the following claim we have as the sorry what we have is the following claim we've got that the derivative with respect to sigma of the thing we want to calculate part of the OP E the term that is order 1 when you take sigma to sigma in the sigma minus sigma 1 expansion the derivative of the thing we are interested in with respect to sigma 1 is simply that part of this expectation value that when you take expressing as an analytic function in sigma minus sigma 1 is the constant piece of it compute it the sum second it has a 4 in sigma minus sigma 1 that comes from z minus z1 but it also has all the other pieces which was z minus z where n was not even a 1 those are the order 1 part of the correlator as far as as far as the expat so what we are supposed to do we are supposed to set z is equal to sigma 1 so we are supposed to set z is equal to sigma 1 plus epsilon and there will be a piece that's 1 over epsilon there's a piece that's constant and there are positive powers of epsilon we are supposed to isolate the coefficient of piece that's constant the term that was just z minus z1 was simply 1 over epsilon the term that was z minus zj j not equal to 1 was z1 minus zj plus epsilon expansion of the epsilon gives you the higher order terms we don't care about this so we concluded that there is sigma 1 is what we want alpha prime by 2 k m not equal to 1 over z1 minus z m this is what we call z1 we have a similar differential equation with respect to every other thing because we chose 1 at random because of chosen variable k the solution to this equation is simply this completely determines z1 to n and it's simply that we have is equal to z 1 but z ij to the power i alpha prime by 2 i oh and I'm sorry I should have also you see we needed to multiply by sorry let's get this straight we needed to multiply by i k1 in order for this to be true so we should multiply by i k1 so we may have the minus sign off and I'm getting k1 dot k8n alpha prime by 2 and the multiplication and we have the minus sign I've done something wrong but you see this was exactly if you forgive me the minus sign this was exactly the answer to the answer we've done it with respect to the analytic part but this is similar to the the analytic part so the ij mod squared to the power alpha prime by 2 k i dot kj this was the correct answer somewhere I've lost the minus sign I'm going to ask you to forgive me in this differential equation we have this equation without plus okay I'm going to ask you to track down the minus do it carefully sorry next up the minus sign we did it carefully okay let's see that's a review let's see happy happy OP right if you never convince 29 with another i k but that gives a minus given by yeah there's a 1 by z but that restores it to positive sign yeah yeah yeah but actually put your keys OP just had a minus sign OP OP he had sorry his relationship here this sign that is minus that's minus sorry I messed up in calculating that I don't know why I messed up messed up somehow the minus sign is just a minus from the minus of the of the 2.5 the 2.5 comes as minus alpha prime let me ignore that the calculation the calculation is OP again you see what are we going to get we're going to get a factor of i times k mu that's left behind basically it takes the equation that we have in the index that's right that's right that's right that's right then the derivative of minus of alpha prime alpha prime by 2 that's right okay so that's the minus sorry okay we got a minus here okay a minus here and now everyone's happy it's the answer the slick method this was though it would be hard to think of yourself perhaps but you see using only conformal key theory methods never calculating a positive integral in your life using only conformal key theory methods and similarities from the OP we completely managed to reconstruct all the correlation functions that will be of interest to us it's pretty cool and this method you think about it all my previous methods totally satisfied it's completely correct now there's one more way I want to this is really going on I know but still that's one more way I want to I want to I want to introduce you to calculating this that's to do the part and take you know once we start doing the part and take a look then we do it in a slick short way that's sort of you want to do it in the right way okay so that's how do you do the part and take the correct I'll let me give you all the logic one one one one one one one one one one one one one one one one one one one one one one one one one one one one one one one one one one one one one one of our mxm, the zero is the constant root, and then square on xm is equal to omega i. Square is to take the action that is of interest to us, we can take the action that is of interest to us, maybe square root g. Will it be? Will it be something? Will it be something? We're not going to say much about it. No, some, some, some. Square root g, del of far x mu, del mu, here we have a contract number that we don't make sure. And by integrating by part, you can realize this action here. You also choose this of xm to be normalized. Choose this xm to be normalized in the sense that square root gx m mod squared by xm xn. We should operate as we know this limit. We should operate as we know this limit. The norm of the eigenfunction should be normalized. So now once you do this, you take this, integrate by part, because it m squared. And you see, of course, what this becomes is minus, so this thing is equal to minus sum over m, omega m squared. Now it becomes actually plus because it's omega m squared. It's minus plus omega m squared times alpha. This action here that you're writing down, this action here that you're writing down isn't purely in terms of the non-zero ones, of course. Okay, then the next thing you do is to ask, what is the gap? Suppose we were trying to compute e to the power j of sigma x of sigma expectation value in this manner. Yeah, this is just that. But it's not. In that sense, this is a derivative. We're just normalizing it as the usual norm. You know, this method is very abstract and it works in arbitrary manner for it. Okay, let me quickly run through this. It's interesting just to show that you can do it properly. You see, now the next thing you do is that you get supposed to work with the expectation value of something like this. You realize you're substituting x in terms of x n's. Okay, and then define j m to be the integral of j of sigma with x n's. What this becomes is alpha m to j m. Okay, plus alpha 0 x. This runs over all non-zero. This is the signal. Okay, now we do the integral. You see, the key part is that the method was written with a unscreened derivative. The maximum spectrum of n squared is discrete. So all of these things are discrete modes. Yeah, so there's no continuum like it seems when you bring the problem on the plane. Once you do it really in the proper way. So if you want to compute this correlation function, all you have to do is compute the squares. Get whatever you get from the non-zero modes. And you get clearly now a delta function for j 0 from the zero. The only place where alpha 0 appears is in this instruction. Okay, suppose we just try to compute this kind of stuff. Okay, now the next thing you can do is to write down exactly what you get from these j m's. So what do you get? You can come to a green function. Now it's almost, okay by the way. Okay, but not just with that. But let me just finish. Okay, so what do you get? So what you get is j m. That's j m. Now you're completing the square. There was an omega m squared. Yeah, so you get 5 omega m squared. Roughly, maybe some 2. Okay, now you can rewrite. You remember what j m was? j m was integral of x m with j. So you can rewrite this as integral j of sigma, j of sigma prime, x of sigma, x of sigma prime. The problem is the square of the g is thrown in the place. Okay, m x m divided by omega m squared. Is what you get by completing the square. e to the power of the history. e to the power plus minus, minus sign of time. So you get e to the power of the history. Now you see that this thing you write as integral e to the power of sigma, j of sigma prime. Times the Green's function, sigma minus sigma prime. But now you have a precise definition of this. The precise definition of the Green's function is x m of sigma, x m of sigma prime, over omega m squared. Now that's what you get when we take del squared, when we take del squared of this. If we take del squared of this, that answers the omega m squared and almost gives us a delta function by completing this. But not quite, because that term doesn't have the x zero square. The Green's function will really open this equation. Del squared of g is equal to delta function. Exactly. Minus x zero squared. This is just a constant right there. Remember x zero is just a constant. Okay, and if you keep a different kind of a square root of g, you get something. Square root of g is you get a one over square root of g. If you do it carefully, you keep a different kind of a square root of g. Basically, you get del squared as a one over square root of g. Okay. Sorry, sorry, you get one over square root of g. We can do it more carefully. Okay. So this is just a precise definition of the Green's function. This is the Green's function. The Green's function that obeys the equation. Okay. Now, there's basically no predation to the x zero square root of g. Can somebody tell me, can somebody tell me what it is? I'm clarifying that if you were really working on the sphere of any compact manifold and every day with this, we had this equation without x zero square, there would be no solution to the equation. What? Yeah, it's like a background charge. Yes. Now, can you tell me mathematically firstly why every day with x zero square, this equation is impossible to solve. Suppose we had an equation with this on a compact manifold. I'm clarifying there's no solution to that equation. Why? It's like a point charge in a compact manifold. Completely. Completely. It's like having, it's like potential of a point charge, you know, of a, of a single charge of a compact manifold. It's mathematically related to the equation that you get. Because the total derivative of the left hand side gives you zero. Yeah. Yeah, it gives, it doesn't give you zero. And in times x zero square is carefully chosen. It's a constant term such that this integral is cancelled. Okay. And then square root g can take zero square into x to one. That gives the left hand function. Okay. So what? The electro-static problem that we're really solving is the potential of a point charge somewhere and uniformly distributed negative charge around the rest of the space. So as to allow, so as to allow the solution consistent with health. So now if you really find the green's function on the grid, okay, where does that go? So we know what equation we really have to solve. Okay. So you can honestly now find the green's function on a basis. It has a piece we worked with as a log, but it has an extra piece now we call it xz. However, now if you go through the whole equation that we did before, including this extra piece, you would find because of momentum, this is essentially a constant term. Okay. And you'll find that this constant term couples only to the, you know, couples to the kis. And because if the n for all k i must sum to zero, this constant term drops. We can give you that as an exercise. So how do we do the calculation we did in the beginning of the class? And maybe I should just write down the, maybe I should just write down the green's function. That's very trivial. Remember that this is equal to 1 by square root g and then del squared. Suppose the matrix. Suppose your matrix is g alpha beta e to the power 2 phi times delta alpha beta. Then the square root g times the g alpha here which goes against it. So you get del squared and the flat plane. Then it's 1 by square root g which is e to the power minus phi times del squared. Times del squared on the, on the plane. So the equation we want to solve is this. It's equal to e to the power minus phi delta function. Okay. Minus x zero squared which remembers constant. So we want to solve the equation del squared is equal to delta function plus e to the power phi x zero squared. Now, at the next up, there's a 2 phi. Okay. Now, we know how to solve the first, how to solve the first guy. That's the law we've been dealing with. So the solution is minus alpha prime by 2 log of z minus z 1 by square root. But then you also have to solve the second guy. That's really easy because it's a constant. We know this great function. So all you have to do is to integrate e to the power 2 phi times x naught squared times minus alpha prime by phi of sigma prime alpha prime by 2 log of z minus z prime root. Now we take del squared which we get a delta function and we get this delta function integrated over this which is this delta function. Yeah, so that's a real great function. But now if you go, really look at the calculation that we did. Really look at the calculation that we did. Using this great function, I wanted to go and check that none of this lost conservation. Every term that we get from here in the end drops out of the formula. So that everything that we did at the beginning of the class was actually correct. But we won't be there yet. Okay. That's one last thing I want to say. This is my whole lecture on correlation function for free scale. It's a little much. But okay, that's all. Now, there's one last thing I wanted to say about doing this correctly. And that's this. You see, if we were to use a normal field theory in the formal, normal, normal way, then there's another set in the thing that we did at the beginning of the class, that's question. And that step is ignoring this inspiration. You see, what we should do is to honestly regulate and then to normalize the theory. So suppose we were, in some sense, to put the theory on the lattice. What does it mean to regulate the theory? What it means to regulate the theory effectively would be to take the two-point function of the theory and to replace it by something that looks like the same two-point function long distances, but is different at short distances. So that doesn't work. Our theory was log of z1 minus z minus z1 1 log of x squared minus z1 minus z2. This is log of short distances. So you might be very tempted to say that this is equal to, so g is equal to this, but z minus z1 is greater than x1. That is equal to minus of log of z minus z1 less than x1. So you could do something more smooth. But morally speaking, you might be tempted to say this. As we are giving a regulator, we still have to re-log it. First, it's regulating your theory. You might be tempted to say that if you did exactly this, you would be doing the right thing. You would be violating something. Can you tell me? Yes, in a way. Yes, you're right. But in some sense, from the point of view of the underline theory, it's something even worse. Remember what the metric was in our system? The metric was this. It was p1, p2, p5, etc. And we're putting a couple of, if you want, I could be there, and if you want, okay, very good. So this, okay, and then what would we have to say? We would have to say that if it was equal to, then it becomes a value that z minus z1 becomes at this point, which is when e to the power of 5 z minus z1 is less than 0, we get z minus z1 times, so epsilon times e to the power minus 2 5. So now this is going to be very fine. You don't want to jump. Once we've done this, we can then renormalize our operators. You know, we can then renormalize our operators by removing the appropriate power of x. Later, every cell contraction will give us e to the power minus alpha prime by 2 kk1 squared. That's really the enormous dimension of the operator, this alpha prime k1 squared. So we define the true operator with the appropriate power of alpha prime. We define the true operator with the appropriate power of epsilon that removes this epsilon. So what we're left with is this final difference. So the left result is that every insertion of an operator gives rise to findings. Every insertion of an operator gives rise to findings. And let's calculate the final difference. And this is what? The final difference was we had, remember, e to the power minus k i squared 5 2. And then we had k i squared by 2 times this to whatever. So we have minus alpha prime by 2. And then we get a minus 2 5. Yeah, minus 2 5. So what's the square root? Do you see where the sequence mod squared is? What do you mean? You're right. And this was squared. No, no, no. Now this one is squared. That's got this epsilon, epsilon squared. We've got this dependence each time. Now let's see. We have e to the power minus k i squared 5 by 2. With an alpha prime. So for every e to the power i k x, we get some final difference. If we were to think correctly, a vertex operator also had this insertion of squared g. So we were to add a squared g and take the pi k out of it. Integrated over the whole bunch. Integrated over the whole bunch. This object transforms that into the positive 5. So if we were to look at your vertex operator to get rid of the 5 elements, we need this object to transform that into the power minus 2. So if k i squared alpha prime k i squared by 4 was equal to 1. So it's precisely seen in the large-shake condition of the data. The main conclusion, is that if k is the positive variable, just in an honest way, regulating it in a diffeuropically invariant way, defining your operator in a diffeuropically invariant way, by not being allowed to remove powers of epsilon, defined value, invariant length, not value. Then you find that you get wild dependence from the normalization of the operator. Which precisely answers the wild dependence of the normalization that comes from explicit factors of the metric. If you are exactly of what machine. You remember the lecture I did when I promised you, that I would give you a better amount of physical way of understanding why the vertex operators have to do one more. This is essentially my point. You see, if the factor is vertex operator, really that scale dimension 1, that was the heart of this this pi dependence appearing when you normalize it. Of course, now it doesn't have to appear to be normalization. It could happen just automatically. You see, when we have the vertex operator del x and del x and even pi k 5, then k square, the master addition k square is equal to 0. So no function should be normalization. But in that case the vertex operator is instead classically wild operator. Because there is square root g and there are two dels. So in that case, the vertex operator must just honestly be one more. So the sum of all e to the power 5 factors that come from classically contributions, plus from quantum contributions, have to be equal, have to add up. So as we cancel the vertex e to the power 5 and this square root g. And you see how that works if you do it honestly. This is all I wanted to say about coordination function of the free state of being. So today we will have a discussion of all of this at the beginning of chapter 1. Can I take 5, 10 more minutes to talk about the correlation function of the DC system? Then in the next class we will really be ready to get into it. OK. Maybe 15 minutes. OK. OK.