 in theory, we maintain unitarity. And either today or the next lecture, we'll also try to perform a more general analysis of unitarity and decoupling of negative norm states, so connected issues, a loop, and we get strict theory. And then we'll go on to the discussion of how we're going to sink in the beta function equations. Okay, so before we start the analysis of unitarity, there was this issue about the adjoining of operators that we have to clear out from this class. Okay, and so I've actually worked it out carefully, keeping all the mentions in it, but let me get this out of here. Okay, so suppose we're working on the cylinder and the coordinate of the cylinder is nothing. Now we define w to be equal to sigma one plus i sigma two in Euclidean space. And this is the same, and we'll define sigma two is equal to i sigma two less. Okay, so let's check this, this is the right thing. So remember that an action is minus i dt that's intended to look potential energy. And we want it to become minus potential energy in Euclidean space, therefore it must be that Euclidean, that Lorentzian time is minus i times Euclidean time, or Euclidean time is i times Lorentzian. Okay, great. So if we wrote w before I take a continuation, we would get w equal to sigma one minus, okay, the sigma two Lorentzian. Now, given an object O in Lorentz space, we expand it as O is equal to sum over m, O m is the power minus i m sigma two Lorentzian minus sigma one. Okay, we will take it to the analytic expansion that's equal to 90 analytic expansion. What? Right. Okay, so the positive guys has to have to be, remember, the way to remember this, the way I remember it, is that wave function corresponding to a quantum state, first one type wave function corresponding to a quantum state is obtained by putting the operator, so psi of x, first one type wave function, the one type wave function is zero x psi. So we want something which has positive energy, the state that has positive energy to correspond to a wave function that has e to the power minus i m t, okay? Now, something that has positive energy here has a dagger here, so it gets its time to represent those from the a x. Yeah, I suppose x was from how many possibilities. Okay, so the time to defend its attach to annihilation operators must be that of a positive energy wave function as you... Now, you remember that the positive energy was annihilated operators. So positive energy should have minus i, minus i. Useful thing to remember always is some of the implicit stuff in the notation used in quantum theory, yes? But it's useful to remember, though I generally have to work it out in my head, okay? That annihilation operators behave like positive energy of the functions, you know, the sign of the sign. Okay, so this fixes whether this should be plus or minus, and then the remaining is fixed by the fact that we had this combination, signal number, signal number. Okay, so this was the expansion of all in Lorentz's space, okay? Now, let's convert to Euclidean space so that it's simply sum over m, or m, e to the power minus i m, and then we put minus i, e to the power two, minus signal one, which is sum over m, o m, e to the power, and now what is this? Minus i m into minus i m, which is now plus i m into that. This is the analytic continuation, this is the analytic continuation of the Lorentz expansion of it, okay? Let's in parallel write down what the analytic expansion of the Lorentz expansion goes, okay? So, o star in Lorentz's space is simply m or m star, and then this becomes minus, so this side is negative, so we get o star is equal to sum over m, just in the w, then the object that in Lorentz's space was o has this analytic continuation, the object which in Lorentz's space was o star has this analytic continuation. Notice that then it's not the analytic continuation, the complex conjugate of the analytic continuation of o is not the analytic continuation of o star, and then we're going to change each term in addition to what we're making, that changing this i, sorry, in addition to this i we're going to change that, you bar, that's not what it is, and because these two operations are okay, because in between doing the complex conjugate we change something by nothing, so obviously it doesn't commute with logic. Okay, but this is the notion of m of o star that is useful for us, it's the notion of o star that is useful for us in doing calculations about Lorentz's space in analytic continuation, okay? So if we were interested just in the w plane, this would be the natural u, this would be our definition of complex conjugation in usually the interface, would be the definition that would reflect what complex conjugation was in Lorentz's space before we began, this is clear. Now that we understand this, all we have to do is to take this result and translate it to the second plane. Now that is this phase, because we know how to take an operator and translate it to the second plane. And del f by del w times del f by del w to the power h, and so we stimulate something to be h power in the track of that, is equal to o over time. Now what is n for time? n for time is because z, and we define z to be e to the power minus i w. I basically take, this is something we did in the last class, but I'm not making sure that all the dimensions match to the position scheme. So if you look up the position scheme, you look up the notes from this, you get the same ones. So that is equal to e to the power minus i w. Let's check that that works. It works because, because w was sigma one plus i sigma two. So the modulus of this modulus there is equal to e to the power i, and e to the power sigma two, as we want, the length of the distance, and then the h. Okay, and the i is sigma one plus i sigma two. So this is a nice definition of z to the power. Now, we need del f by del w, so we need del by del w of e to the power minus i w, but that's simply equal to minus i w. We get that all of that, which is simply the insertion in the z coordinate system, okay? It's related to the insertion of the w coordinate system by a factor of minus i or z to the power h. Or z is equal to one over minus i z to the power h. And similarly, it goes south, right? But, and by our style, I mean the thing that we get from analytically continuing complex multiplication in the learning system, right? So, star of z is equal to one over minus i z to the power h. In the final step, we put in the expansions for o star of w, o star of, o star of w star of w. So, o star of w is this, and remembering that z was equal to e to the power minus i w, we get z to the power h, and then sum over n, o n z to the power h plus, because e to the power minus i n w is z to the power of z. Okay, whereas our star of z is equal to i to the power h, sum over n, o n star of z to the power of z to the power h minus i. Realizing the variance in space, then o n star would be equal to o of minus i. Okay? So, we'll be more general. I'll just say this in, is the coefficient of plus i in here. So, if you define o star, exactly, as the coefficient of e to the power minus i n w, so that's exactly as you said, it's o minus n star. Okay? So, o star n is o minus n star, so we put that in here, we have o minus n star, and then we can convert this sum over n into sum over minus n, to write this in a form that looks like this. Now, terribly sure I understand, Kulchinski had exactly, I mean Kulchinski had given him the motivation of this definition, but he had exactly this, with an additional factor of minus 1 to the power h, some funny convention he's made with, which I'm not terribly, terribly sure of. Okay, but in the, when I need complex conjugation, this is what I'm going to make. Okay? And it's probably somewhere, somebody on the other conformal field here, but I don't see the value of the value. Ah, no, wait, why does that change, you know? Let's say you wanted to, let's take this to the given value, and you wanted to bring the i to the same value. i inside minimum, you had h, it's going to be i to the right, meaning no. O star of k equals 0 to h plus n, that's what we're starting with. Well, let's write the form we had before this step. O minus m, yeah. O minus m star over 10 to the power h minus m, and another way of writing this was O, O star of minus m, which is O m star. Minus 1 to the power h, y, y, y. Yeah, we thought that would be minus 1. So, what did I do? The i to the power h just came from the factor that was the dimensionation. O equals star over h, is this better? Okay, if you had o of z is this. Yes. Okay, yeah, yeah, yeah. I actually don't see what, well, you know, there's a matter of definition here. You know, of course, just redefine what would be my complex multiplication. And as we can see, everything only goes to constants. You know, in any statement, we remain only out of constants. So, it won't matter anything physically, I think. You know, I think that basically he likes this because I think there's a formal thing that's nice with the other minus sign. That is that if you now take z to 1 over z, if you take complex multiplication plus the conformal transformation z to 1 over z, this rear field is self-adjoint, you know, goes into itself under that combined operation with that additional minus sign. I think that's the reason for putting it there. So, I think the bigger the work, the less the force to change the sum. The sum. Actually, in the z-coordination. Yes. You take the complex combination. The kind of O and O star as you did, that was not exactly the complex one. You were at each other. Right. And then all the things is transform them to z-coordination. Right. So, now you transform O, I mean, transform O to the z-coordination. Yes. And then you take a complex one. No, I don't take a complex one. I took this expression and transformed it. But why do you do that? Because that's not what we do. You see, we're not supposed to take complex multiplication after after you get the equation. Take the expression that you get, the two different expressions that were complex quantities of each other in Lorentz's case and then do all the transformation treating them as independent quantities. I never take the complex conjugation. But the star is really not the star. It's not the complex conjugation he uses. It's really the conjugation under which there is this, I mean, conjugation of the actual physical interface. Exactly. Exactly. It's the conjugation of the actual physical interface. That's exactly it. The commission conjugation. That's exactly it. That's exactly it. Okay. In fact, that's the next point. It's the complex conjugation in actual physical space. The thing that we would have, if you use this star pattern, then you will compute the correlation functions of states that you get by taking the dagger of a state in Lorentz's case. But that's what we mean. We trace it through what we would have done by taking the star of Lorentz's case and then doing the analytic integration. That's the key point. Okay. You can try to convince yourself from this, which is a formal argument that you can go through. But it's more or less clear from the logic. Okay. That this process of complex conjugation gives you a relationship, gives you a relationship between the pairing of states that we defined in the previous class, the zomological pairing of states. You remember we had this i, I think we defined the three in the other order like that, i and j, which meant that the i operator inserted at infinity in the u equals one by z, chant. Okay. The two-point function is that with the j operator inserted at zero. That gave us, that was well defined, you remember, you know, the powers of z cancels between the translation of the operator from the z-chart to the u-chart and the explicit powers of z that appear in the two-point function. Those gave us a pairing of states, at least a pairing of operators. Now that was a natural question to ask. Okay. It's a natural question to ask. That's not an oi energy. It's a natural question to ask. How does this thing relate to the pairing of states that you get from the inner product? The pairing of operators that you get from the inner product is the state operator map. And again, the answer is simply that this is oi star. Where on the right-hand side, there's an illusion in the program. Or maybe another way of saying it. That's a very good way of saying it. Oi or j. Chinsky chapter 6, it gives us formers, sort of quote-unquote, demonstrate, not a very convincing demonstration of this statement. And you need to find a more convincing to just think through the logic of what this oi star works. Okay. You'll see that putting this thing and using the state operator map defines the insertion of the state in the same sense on the other side. I think it's more useful to just sit and think through it yourself. If you have questions about it, you can ask them. But certainly let's leave this to the next slide. What is the discussion that I plan to have about general properties of operator insertions? General properties of operator insertions on the sphere interpreted in Hilbert's case. You see, if you think about it now, you put together what we discussed last class and the details class. We're a complete Hilbert space understanding of any correlation function that you like to write down in the sphere. You can translate that in the space language. So just to review for you, the correlation function of two operators, one at zero, one at infinity, we've related to the two-point function of two states. The correlation function of two operators, one at zero and one at some point, at two different finite points, is also related to the inner product between two states. Because we saw that we did that we saw that oi of zero, oi of say one, oj of say two, to be written as gij over say one, two to the power. And gij was the two-point function of oi with oj star. Again, great. Then we also have a Hilbert space interpretation of three or four points. So let's look for this. It's a three-point function. Suppose you had A i prime. Remember, prime says it refers the operator to the one-by-u chart. But that's the same thing because these are usually answered operators. That's the same thing as A i, but with that additional power of z, you can make all this very fine. This is what you have A i prime of infinity, A i prime of one. What is this in operator language? You write this in operator language? No, sir. The answer is certainly this is this. Now this seems in terms of all the inner products you know as this. It doesn't have to be this state or this state or this all of it. z, z, yeah. You have this z, which you could pull out by how z bends in operator language. Basically, I've z to the power L naught on the left and z to the power minus L naught on the right. Okay, so you just get the L naught dependencies. So by the way, so I'm saying that A k of any operator of z to z bar is equal to z to the power L naught, z to the power L naught bar, A k of one to one. So I need to use a statement of time and evolution and then I'm going to quickly continue adding space here in Lorentz's face and then I'm going to quickly continue adding space. Again, you just get this. But this operator, you know, we evaluate this state by the state operator map. Remember that we evaluate codes. According to our conventions, we evaluate a state-by-state operator map by inserting and operating as the origin of doing the author table of the universe. That's why I was completely interested in evaluating an operator itself in the universe because it acts on that state, which is the usual universe-based operation of an operator on the state. And some other time it would be acting on that state. But it easily factor on the time dependence, using time evolution, and that happens. So this quantity is that and last class we also saw that we evaluated that we evaluated to C i j k or i k j must be right again. Times those powers of state. In particular, this object, the 3-point function at 1, 1 is simply the matrix element of the operator a k 1, 1 between these two states. And that quantity is C i j k. With everything in Lorentz, as I mentioned, if we're lowering the disease using these, this is our metric function. Two-point functions in this one. There's double-point functions. And so on. So let's see how our similarity works for four-point functions. That would be the clock of our discussion. Any questions or comments? So by following this, we'll have to take expressions that have been written in positive language and give these expressions an inverse-based interpretation. Now we're going to use that in the inverse-based interpretation to understand the structure of singularities in three-dimensional diagrams. A diagram from the sphere of the string. But any questions or comments about this? But do you remember that the amplitude of... Okay. Supposedly, we're interested in four-point scattering of four different states. Let me make sure I stick to this presentation. So I'm interested in the four-point scattering of the states 1, 2, 4, 3, 4, 5. So let's close that issue. Now, I'm interested in the scattering amplitude that is evaluated by 2z of 1 and infinity of 3. And I'm going to put a hat in the sphere of the hat. That means we do a hat and 1, 1, 3 hat at 0. And we have a hat at 0. And we have a hat at 0. Okay. You remember that our general evaluation of the four-point function of particles 1, 2, 3, and 4 evaluated to this expression where hat implies also put the scene sessions. Include also the scene sessions. And no hat implies no hat. Now, these scene sessions are 1, 1 operators. The operators with the scene sessions are 0, 0 operators. The many nice things about 0, 0 operators. Okay. For instance, you know, one nice thing about the 0, 0 operators is that the insertion of an operator in any frame is the same thing. The z-frame of the w-frame is the same thing because you have powers of 1 over z squared to the power 8. If the h is 0, that's a man. Okay. So for the reason that we look at the field, I'm going to transform this expression into an expression involving only 0, 0 operators. Similarly, we draw an operator with a hat. How do I do this? Well, in order to do this, I use the relationship. I use the relationship that v4 of z, z, power is equal to 1 over z, z, power and the combinator of v0 and the combinator of v0 to the power v4 of z, completely accurate. Now, this is literally a statement for the purpose of insertion into this relation. Not as a general statement about the operator, just for the purpose of insertion into this relation. Now, you see, for the purpose of inserting into this relation, this v4 is dressed with Cs. But we know that the only thing that we have to worry about is the part of C that involves the 0 for all such insertions. So it's very easy to just plug into what C is here. So C is equal to Cm z to the power m plus 1 1 over z to the power m minus 1 which is weight minus 1. Operator? Okay. So, what are you going to do? Even without the other statement. You just plug this in. Actually, you see, this is an identity by nature. Identity by nature where? Yeah. This definition in this expansion is equal to here if I expected an identity. Oh, yes. How would you rate it? I mean, should we rate it? Okay, good point. Let's see. Yeah, that is the part I just observed. And I'm trying to figure out the minus sign. Maybe. Maybe that's it. You're right that we're going to have expected an identity. Which would have squared much more than minus sign. Maybe the minus sign. Let's see the calculations. Okay. Let's make it come out of this without that. Let's see the calculations. So if we had, this is equal to V of z to the power times C of z times C of z to the power z. Okay. Let me put the expansion for C. Okay. So this is equal to V of z to the power sum over m Cm. I will put a arbitrary constant. Cm over z to the power m minus 1. Cm star or C bar. m over z bar to the power minus 1. Okay. Now we take this as a commutator. This commutator pulls out the C0 term. The other active commutator. But okay, so let's do the sign here. So maybe it shows sign. Listen. I think we do get a minus sign. Okay. Let's see. Let's take this commutator. So this is supposed to be written in this way. Yeah. You see. Let's work about this commutator. This commutator is B0 C of z. Let's say B0 to the power C of z. And only C0 matters. C by z bar C0 bar minus C of z. C of z C0 bar B0. Okay. Now we have to take this B0 to C. So that gives us minus. So it's minus of the commutator of B0 bar and C0 bar. That anticommutator is plus 1. So this term is equal to minus C of z. Okay. So that's the minus C of z. And then B0 with C of z. Just because you see. Just because you want. Clicking with only the C of z. So we've got minus sign. So we get that this thing here is equal to minus of B of z. Z bar with R. Okay. Now the term that you wanted was an I to the power H. H was 1. There are two such terms because I haven't seen this before. Okay. I'm going to say I give up here. Okay. We have to take this back. Maybe there's a sign. Sorry. I'm not sure. Okay. If you want to do it for real, you have to be very careful. Okay. Okay. Sorry. Okay. So here can be written as the coordinate function of this. Of this active object along with these other intersections of B. I take this object and realize it in inverse direction. Okay. What about this object? So suppose we had, let's first deal with V of z and then we substitute it once we did that. Suppose, what about this object? Suppose we have V1 hat of infinity, infinity, V2 hat of 1, 1. And then I'm going to be familiar with signs that is, you know, this is V4. I'm going to be keeping track of signs that I really want to add. If these were for the other guy, I should be keeping track of signs. The Busani's should be there. Same thing. V3 hat of 0, 0, 0, 0. Okay. And then I'm going to do it again. And that's basically the interpretation of this. It's very similar to what we did with V1 functions. So that's a 4-y function. Because it's a 4-y function. It's a coordinate function of some operators. But remember the operator's time order. A time order of that, if you continue. It should amount to a radius. Remember quantum mechanics is always great statements, you know, like Hebrew, Arabic. So I think this first appears to the right. Okay. So when z is greater than 1, mod z is greater than 1, this guy is first. So we have theta of mod z minus 1 times. Then we get since this V4 z, z above V2 hat 1, 1. And then between wages plus theta of 1 minus 1. V1 hat V2 hat 1, 1. V4 z, z above. And V3. Now the next step what I do is to substitute these guys. So let me deal with one of these two expressions. You can really be representing one in exactly the same way. Let me deal with the first one. So the first is two expressions where we get a theta of mod z minus 1 times V1 hat. Now we substitute this in. So we get divided by z, z above V0, V0 hat 2 hat of 1, 1. And then V3. That's complicated. But now the next step we need to remember that all of these states are physical states. These states are more than 18 than V0. So between more than V0, at least one of the V0, V0 power on the left is just 0. Because that highlights the sweet one here. So the only part, the four terms that contribute when we write this thing out, is the terms which both V0 and V0 power are to the right. So now keeping track of the signs of where we saw it. This is the same thing as theta on z minus 1 over z, z bar times V1 hat, V4 hat of z, z bar V0, V0 bar times V2, 1, 1. And then we... So I evaluate the time dependence. I evaluate the time dependence of the subject. I evaluate the time dependence and let me get which one is to the left and which one is to the right. So, that V4 hat of z bar is equal to z bar divided by the amount of bar V4 hat of 1, 1, times z bar minus n naught, z bar to the power of naught. N naught equals n naught by equal to 0. Remember physical set of string theory. So the matter part of it is n naught equals n naught by equal to 1. But that's also the cost part of it, which is minus. Full n can't power all physical states. Remember every operator knows the hat of an inch. It has the C stuff inside it. That's the great thing. Now it makes you 0. So n naught equals 0. So because of that, the part in which this guy acts to the left just gives you 1. But the part in which it acts to the right survives. So this term is eta of 1 z minus 1. And now we write this as V1 hat of V4, 1, 1. Then z to the power n naught minus 1. Where the minus 1 is the z of the denominator. Z bar to the power of naught bar minus 1. And V1, 1. This thing evaluates to in the special which is pretty complicated. Because of this z to the power n naught minus 1 and z. Oh, I'm delaying the sign off versus the mic. It's pretty complicated. In order to understand what we did, we had to do an integral over z. So we want to understand the z dependence of this object. But this seems pretty complicated because this z multiplies an operator. So all of the simplifying what we do is to insert a complete set of states in the machine. So let's take this object here. And insert the complete set of states. So we put z to the power n minus 1 times V1 hat V4 hat 1. And I'm sorry, I'm not with the B zero. There's a B zero, B zero. And I can choose either of them and insert the complete set of states after this. So we've got B zero, B zero bar. And then we get, let's call it psi, psi, psi, psi. And we get z to the power minus n of psi. So let's give the exact set. Sum over all the states psi. z to the power n of psi minus 1. z to the power n of psi minus 1 times V1 hat V4 hat 1. V, B zero, B zero, B zero bar, psi. And then psi times V2, 1, 1. The full state of psi is, the full, the set of all the sides are given by all vertex operators, sincerely. Dressed with an even pi kx. I mean, that side will be associated with the particle of mass n. So let's suppose that this psi is some vertex operator and in pi kx. And suppose the weight of psi was such that it would have been associated with the particle of weight m. Then we know the dimension of this operator is equal to, yeah. It's k squared plus n squared by 4. The l naught dimension is the standard dimension of each pi kx, the same with computer. And we know that m enters, I mean, quarter less by definition. m enters into this relation by, okay, this is case by case. So we can label the operators by the dimension. But instead I choose to label them by the mass of the particle they would have corresponded to per a physical state. Just by using the relationship between l naught and mass. Right, the relationship being, the relationship we see many times is right there. There's a thing that n squared is equal to m by alpha prime, l naught into the l naught of the operator, not including the c hat. That's at minus 1, not minus 1, but I'm writing it like there's a full l naught including the c hat. Okay, so this n squared if you like is just some label for the dimension of the operator. I'm just labeling the l naught of that operator and some sum. And why is it this way? Because we want to interpret in space sets. Is this correct? But this sum over all the states includes an integral over all the means. So what was the Hilbert space? It was specified by occupation numbers of the oscillators. You want to do the full, complete this over Hilbert space sets. Sum over all occupation numbers and occupation number basis of the oscillators. And also sum over the zero. So sum of the zero over the function is some integral over k. So we get some integral over k, some d where gk, d minus 1k by 2 pi 2 pi k. I'm actually not going to give you kind of numbers too carefully here. But you should see that in this checklist get very careful to drive along with these things. Just to see that it works out, you know, really perfectly. Let you give your sense for the structure. Leave you to work out the coefficients. So we've got d, dk and then we also have a sum over all states, over all particles. That's this sum here. Then the z dependence is z to the power alpha prime over k squared plus m particles. Square over 4. So there it is, alpha. And then we have these three different, these two different objects. We have one hat, we have four hat, minus 4 d squared. This one. What is this one? Ah, and this is minus 1. I had minus alpha here. Minus alpha, minus 1. This should be minus inside. Sorry, sorry, sorry. Can you confuse? That's the important point here, isn't it? Minus 1. The final answer is no. Yeah, there is a much one. Yes, you're completely right. Thank you. So there's this, minus 1. Now, happy? So this minus 1, so we're going to use it. Okay. And then there's this nonsense. Minus alpha, minus all this. And then there's a p, p. Actually, we should say p, p, p prime and then some g, p, p prime. But okay, not if we chose it that way. Okay? So the sum over all of p, integral over all of k, and this. So this was the matrix. We have to integrate it over all this thing. So what we were supposed to do is, now it's very easy to take this off, get an integration like this. Because it's written out as sum over terms. Each of those z and z part of the matrix is totally explicit. So any of these terms behaves like z to the power of a, z to the power of a. Okay, so z, so mod z squared to the power of a minus 1. This is the integral we need to do in order to evaluate area. Where a is simply case, alpha prime a squared plus m squared minus 1. a is minus alpha, k squared plus m squared. Yeah, yeah. Yeah, how do you call this minus alpha prime? Let a equal minus alpha prime k squared plus m squared. So this is there. First let's do it in a completely naive way. And then return and think about what? So this quantity was, remember this integral was supposed to be n over all of z. But, which tells us that we're only interested in doing the integral. We're only interested in doing the integral in the outside given disc. Let's see, so now we can do this integral. Basically, if it's convergent, we can do this integral by, what do we have? Let's write it down. So this is a proportion, that's some 2 pi and so on. But it's proportional to dr, r to the power 2 a minus 1. What was proportional to? 1 over 2. Into 2 pi. That's actually, it produces dimensions 4 pi. Because his dz, dz bar is twice the usual measure. Yes. Let's try to do it if you would like. Let's see if we can get back to that. So 4 pi divided by 2 a. Because the integral of this is half the part 2 divided by 2 a. And then you evaluate it at infinity n over 1. And you're assuming the value of a is such that in infinity we have convergence. Oh, you're evaluating it. I'm evaluating it. Okay, from 1 to infinity. So the answer is 1 over 2 a. We were playing the problem, exactly. You know, when this a is not in the range where it's convergent, we're going to define it by idealistic continuation. So if you remember, in the form of the expression that we had, even for 4, that came out scattered, the integral was not in general convergent. But what we did is we moved to some range of case space where it was manifestly convergent. You know, to go over to some very large space like momentum for this k. So for any given mass, you can always find this space. The momentum that's large enough so that this thing is convergent. It evaluated there and said that the actual expression is what is defined by analytic continuation of that expression. So that's what we do. So what we handle here is just the answer for the expression. We just replace this by the problem. So what we do now is that this space, this business here, so this first question is, v1 at v4 at vb0, v0, v0 above, vb, v3, it was a, it's precarious to keep track of numbers with the level of detail in between. We can write in the same situation here. A, dc, we had 4 pi here, because there's one over 4 here. But there's a half here. So 8 pi over alpha prime in the k squared plus m squared. And you know, when we do the analytic continuation right, we should get the right minus i episode. If we want it to come to this, we have something. We're analytically continuing approaching the axis where there's a curve from the right side. Please. We now have to learn to sub and draw on the bottom. What you see, the aim of this is to try to show that the stream 4-point amplitude is consistent with unit. Which we can show by using the usual propagator arguments. Now, the usual propagator arguments. Now, is it basically what we've done is to show that the amplitude is of the usual field theory form. Some 4-point insertion and the integral of a body like this. And what does that have to do with exchange of particles? What we've done is taken this expression here and cast it into a form such that it's manifestly the product of 3-point interactions with the propagator of that intermediate particle going into treatment. So what are the things you would have expected that you know, like, that might be 4-point vertices, but those don't need singularities. You see, the only, what we're going to do as you see, is to isolate all imaginary pieces. In the, you see, for instance, from this point of view, we've only done a part of the integral. There's the other term where we have to do the integral from mod z from 1 to 0. There's more stuff going on. But what we're interested in is not full answer, but the singularities of the answer. The imaginary part of the answer. There's only a clear but particles run when there's an intermediate particle going on chain. There was a direct forward interaction which is generally there, wouldn't you? Let's say that it doesn't contribute to the imaginary particle. The only contribution of 3-level, the imaginary part of amplitude in a 4-point function, is 2-3-point functions with a particle in it. That's what we're trying to isolate. What we have done here, let's look at the other stuff. In the other stuff, what would the difference have been? The difference would be on the other side. Exactly. So now this k here, is the k of v1 plus v4. So this is giving you only the terms which have singularities in k1 plus k4, that is t of v1. This part isn't exactly this sum. The other stuff, it's totally non-singular. We have singularities in some other... some s. The treatments are different. You can just give all the criteria. Yes, so if you were to isolate the correct fields here, the amplitudes, this would, at 3-level, just be given by the Feynman graphs of this. You probably already know all the correct fields here, the amplitudes. This is the way in the term. This is the same way that you just... Yeah, no, no. You know, it's just reconstructing these curves. But from the other stuff, you get the moment. You know, suppose you wanted to... in the gravity vector action, you wanted to compute some 6-grapes. You would take the full 6-grapes, the amplitudes, throw all the parts that came from lower amplitudes propagating in trees. It's the 6-grapes. That's how you would actually do it. You would use this expansion for it. You'd use the pull-down function on itself, subtract out all the stuff running it, and then evaluate the remaining thing. That would be finite, and that would be the 6-grapes. That certainly exists. It's there. It's just that here what we're trying to do is isolate the singularities in T. Is this clear? Fine. Now, Kulchinski goes through a song and dance after this to show that this answer is... Okay, so now, we're going to evaluate all singularities in this diagram as a function of T, if you want to do it. Simply by replacing this 1 over K squared by I, I times the death function. Okay? And then you could go ahead and show that these ways could cause K's rules. Why is it happening by the series I've given but since this is already now the final diagram for it and since final diagram has this property that they obey in these scattering amplitudes, there's really nothing more to this thing into the form of a field theory with certain three-point functions and a field theory propagator. So the unitary relations will be manifest, provided you appropriately choose the three-point function, provided the three-point function that you chose is appropriate, but these are the three-point functions of that function. Yes? Good. Let's start with that. If you know what I mean by that, for a moment, okay, a... Yeah, yeah, yeah, yeah. But now the view is basically no, no, we just use the physical state for the diagram. This one looks like that's the wrong idea. Zero should be doing is it projecting onto the physical space? Is it the physical space? Something like this. Okay. Forward or counter-accident? Counter-accident, yes, yes, yes. In fact... Sorry, sorry, sorry. It's going the other way. So let me tell you what... Of course, many people are... The physical states are annihilated by one... Sorry, sorry, what's that? Let's say this again. You see, this is why it was confusing. Remember, there's this, there's this, okay, and there should... Okay, let me do it more. There's p, p prime and gp. When you insert a complete set of states, or it's actually a problem here, when you insert a complete set of states, we're doing it in this... In order to get not, you know, this Hilbert space of products, but insertions of operators because we want scattering hand-to-deals. Of course, these things are simply related to each other, right? But it's important that we come to this fact with gp, p prime, okay? Okay, now let's see why that makes a difference. You see, suppose p is a physical state, okay? p is a physical state, so it's annihilated by v0 and v0. To find the... If you want to find a state with which this connects using this inverse propagator, there is a state that is not annihilated. You see, there are the up states and the down states. We have up, up equals 0, right? Okay, okay. Since we've got two minutes, let me make what I think is the statement and I'll check it and tell you about the next time. But what I think is the statement is going to be this. You see, this p here will be a physical state that's annihilated by propagator operators, okay? Now this guy, that will force this guy to be it's a partner state, so suppose those are the down states, okay? That will force this guy to be the up states. Uh... The only non-zero gp, p prime elements will be between up and down. Okay? So that this pp0 on the up will give you another okay? I understand the statement and I can't see why this is thrown at the moment. But this, oh... This is because g of that is the both. That's right. Because the vc which will be the vc where that's exactly right. Because it's the vc okay. That's basically the point. Yeah, so exactly. Exactly, exactly. Actually you just said it. The reason that we were looking at the down-down states is that we were addressing every operator with a c. But if you want that non-zero two-point function then you want to have this guy dressed with a b. You know, the other way around. So this character which annihilates you know, one set will... Uh... I won't ever have to work this out to be given but it's exactly what you said. You see, suppose we're looking at operators that are annihilated by b. The corresponding guys which they have non-zero two-point function will be annihilated instead by c. Okay? By c0. So bc though, often our operator will not be annihilated but instead will give you a physical operator. That is in the down-down states. Give you a formula for these things. But I think this is exactly what you want. And this, because of the presence of the bb0's operators it's also annihilated by this. It's clear that you're annihilating by b0 because there's a factor here. But this is non-zero. It's a scare. But I think the idea is completed. Because of this factor we're... This would be an ordinary inner product of something with the same thing you know, mod squared of a three-point function. But now there's another important point which I'm not sure we did the other important point is that using what we used here was the summation over all states in the denominator in strength here from the point of view of space-tap you should only have a bearing inside this propagating set of states. Those states are states of the inner space of strength here. Only those states have lying to you for more than a few seconds. Yeah, probably a little. Let me at least quickly tell you the idea. So this is almost what you want, but not quite yet. Because we have to show that we don't get spurious falls from unphysical states propagating in time. This is the key point of your impact. Exactly. Only physical states contribute in this in this round, in the middle. Now that is true. That is true. I probably can't give you the proof before I start the seminar. But let me tell you the main idea. The states that you could want to have. The states that are Q-exact all the states that are not are just not too close. Now the states that are Q-exact clearly know how to do it. Because suppose a state is suppose VP Q wants VP right. Let me write this and Q and I write this and I write that. You take the Q2 and you get 0. So the Q-exact states are clearly not a problem. About the states that are not Q-closed. But the point is that since Q-squares to 0 Q has a very neutral hazard. The representation of Q on Hilbert's face is very the representation clearly of Q on Hilbert's face is very simple. Those states that are annihilated by Q and the states that are not annihilated by Q appear in pairs. The pairs being the states that are annihilated by Q and the next one that's exact. So any you know the representation here of Q is made up of those guys that are annihilated by Q. Those are the physical states. As I'll show you in popular's next class basically what you can do is to write this complete set of states. Yeah, we're doing it. You write this complete set of states in terms of a projector on which one side of the project is always Q-exact states.