 Whatever you have on the right hand side is now going to be an intersection of this number. So, the operator O is expanded in the series of normal order expressions, we call it the sigma. Is this clear? For instance, let's see it at work at the quadratic order. Of course, we have x of sigma 1 times x of sigma 2. This equation will tell us that it is equal to normal order of x of sigma 1 plus x of sigma 2 plus, now, these variables get to add. Since there are only two operators on the right hand side, we can expand this explanation linear order. We get plus alpha prime times 2 log of z1 minus z2 square times nothing sigma 1 times z1. And this is the same thing as what we defined before, because this is going to be minus of the sigma function. Is this clear? Expression like this and equation like this. Once you have such an equation, it's very easy to invert. Once you have such an equation, it's very easy to invert. Suppose you want to know what is normal order of all. It's simply exponential with minus alpha prime times 2. Okay, look, this gives us an inversion, telling us how this is binding definition now or not. Now, we're good. I got this sign here. If I wanted the operator itself, I wanted to, sorry, I got this sign here. You see that, right? The operator is normal order plus 2 point function. So the operator is normal order plus 2 point function. Normal order is operator minus 2. And the 2 point function has minus alpha prime. Is this clear? I have a very confused view of the signs. Okay, sorry, sorry. Okay, we said this is yes, plus 49.0. What it tells you about is a formula that allows you to expand the product to normal order of operators in terms of normal order. Okay? Let's say it's intuitive. You see, inside normal order, you subtracted away all the contractions between operators that you have. But if you were normal order of the whole family, you would have to subtract away also cross contractions. So intuitively what's going on? Suppose you've got the product, you've got the product to do normal order expressions. Then if you put the whole thing normal order, you know, if you put the whole thing normal order plus whatever you would have got in the normal order procedure that involves the cross contractions. Let's put that intuitive statement into an equation. Okay, so suppose that the normal order is normal order if, then what I need to do is to, and I want to write the sum of the chain. That's normal order. It has additional singularities coming from contractions. So we get a minus alpha prime. Actually I have no two now. You see, minus alpha prime now is z1 minus z2. Then by del x is z1. Then by del x. Subscribe, z1 acts only on O and z2 acts only on O. Either we don't have a good time, first act and second act and second act. Yeah. Or we have one half of first acting on second. What the problem will be then? Right, right, right, right. And then if you have one way to, then you just do two kinds of terms. First acting on the second acting on O. Because all the product that you was in the, in the two part function. Yes. Okay, come on. Okay, so that's fine. Then there, now there will be two kinds of terms in which the first one acts on O, second one acts on N. That is one kind of a term. Yes. The second is the first one acting on N. Yeah, can I say that the, that this guy acts, you know, z1 acts only on O, z2 acts only on N. Because we are, then we are done with that. Oh, let me just check this out a little bit in a case. Suppose we want x of sigma1 then x of sigma1. Okay, I want, I, this is normal order method, this is normal order. I want to expand it. The right answer is normal order method x of sigma1, x of sigma2 minus alpha prime by 2 log of z minus square sigma1 and sigma2 squared. That's the correct answer. Okay, now square one. Sorry, now let's, let's, let's use that. Suppose we use x of minus alpha prime by 2 log of sigma1 and sigma2 on square times del by del x of sigma1, but this one acts only on this guy. Times del by del x of sigma2, but this guy acts only on this guy. Okay, acting on x of sigma1, x of sigma2. Okay, so we get the right minus alpha prime by 2 log. There would be ways of writing even different factors depending on what you, if you say that this is an algebra on both, then we would require an additional half. Then we would require an algebra, therefore. Right, so fixing one dimension. Other dimension, because the first element acts only on the second operator. Second element acts only on the second operator. Then we just have this. Okay, great. This, this is the, is the boundary line that allows you to take a product of a normal operator and expand it into cities of normal. Okay. Okay, so the last 10, 15 minutes of this class, where I'm now, okay, so we finished this four, yeah, so we finished our whole discussion of normal. Okay, whenever I write down, whenever I write down a product of operators from now on in this area, I would always mean normal order. Okay, if it's confusing, please ask me. The gene type that I write down is me normal order product and often now we take use the normal order example just because it's okay. We go back to the question of what is the stress sensor? One that consistency will show is correct. It is the stress sensor. This is a sensible expression. This is a sensible expression. And, you know, okay, this is, this is our, this is our, a possible stress sensor. If we're off by a constant, that is, sometimes I have the menu, we'd see it when you do for six dimensions. If you divide things a little differently, you might be off with it. Okay, this is our definition of the stress sensor. But, stress sensor, let's try to compute the TTO. Okay, so what we want is T of z tends T of l. So that's the same thing as minus 1 by alpha prime T of, so that's x in the minuses. T here is del x of z, del x of w, where del x is z times 1 by alpha prime, del x of w, del x of w. Remember, these guys are normal order to do themselves. Okay, let me write that. These guys are normal order to do themselves. Now, what we're going to try to do is to re-expand this product of normal order to operate us in a series of operators, each of which is completely normal order. Why are we going to try to do that? You see, we have already argued that inside normal ordering, there can be no singularities. Okay, so anything that appears inside normal order, you can honestly take an example. Other singularities will appear in the process of normal order. How do we complete the singularities in the OV? It's still normal order, so it's fresh. So how do we do that? Well, the rule is choose all possible pairs and contract. So first, let's look at a contraction that we get in a vector of the contractions. This with this, this with this, or the other way around. There are two such terms. Each of us will ask for it. See, x with x was minus alpha prime by 2 log. Now we take one derivative with respect to w, one derivative with respect to z. The first derivative, if it was taking with respect to z, gives us no minus, right? The second derivative taking with respect to w also gives us no minus. Because it's 1 over x differentiated with this minus sign but we differentiate one over minus sign. With respect to minus sign. Therefore, we find that del x, so let's note down this thing as unusual. Remember, del x with z, del x with w is equal to minus alpha prime by 2, 1 over z minus w. This one. The terms in which we've got all contractions. What do we get? We get these factors. We get alpha prime by 2 with a minus sign of the whole thing squared. We have 1 over alpha prime squared. And we have 2 of them, so we've got a factor of 2. And we get 1 over z minus w. This simplifies to half of z minus w. So the term in which everything is contracted away is half of z minus w. It's a step. It's just 2 of the x's of alpha prime. We've got a 4. Because we have 2 ways of choosing which contract here and 2 ways of choosing which of alpha prime here. So first, we have a factor of 4. We have 1 over alpha prime squared from the overall factor. We get a minus alpha prime by 2. And then del x of z, del x of w, and then minus w. Let's simplify this. This is equal to minus 2 by alpha. Del x of z, del x of w, del z minus w. We return to this term, but before doing that, we've got terms in which we have everything contracted away. That means that nothing is contracted away. It happens. Because this correlation, this is the normal ordered operator product, is genuinely analytic in insertion points of the operators. There is a thing that's completely opposite. So we've already identified all terms that give us singularities. What have we got? So let's rewrite the singularities. We've got half of z times w, the whole thing. 4 plus 2 into minus 1 by alpha prime del x of w, del x of z minus z minus w. Let's take it normal. I think that this is it. And this thing looks awfully like twice the stress answer. He might be wondering, why was the other term that has 1 over z minus w? And the point is that why are these expressions completely analytic? Of course, they do have potatoes in the expansion. So remember the one of these on the w, the other one was z. So if we expand the a to write it in the form where everything is a w, we need to take this del x of z and rewrite it as del squared x of z, del x of w plus z minus w del squared x of z. Because I'm going to examine it honestly before I take it in expansion in side note. The singular part of the e is half z minus w or plus 2 times d of w z minus w, where and then plus, you see, this, the fact, we've got a second derivative acting on one of the x's in this term. But we can select write. So we write this as plus and remove the fact of 2. Del of del x of w del x of w over z minus w minus 1. Which is half of z minus w or plus 2t of w z minus w squared plus del t w. But you're going to have a product expansion with a stress tensor. Except this standard is specifically valued at the center chart. C is equal to 1. That's for a few more manipulations of this with the free scalar at the bottom field. It's so much fun. Basically, it was so easy to create here, you can make any kind of it anymore. The next thing I'm going to try to do is try to discover some primary operators. It is completely the operator product expansion of the stress tensor of the operator. Let's do that. We have minus 1 by alpha prime del x del x del z times del x a double. Let's do the calculation. In fact, once, in fact, once we have a factor of 2 from which one of these two. So we get minus 1 by alpha prime by 2 and minus w the whole thing squared times del x and z over z minus w the whole thing squared like del x times this. In the standard form we should have operators that are only in w. So this is going to be written as del x and w minus w with h equals 1. I'm missing so many classes that I want to try. By the way, I'm going to get out. It's almost like I did one class weeks. I need to have this class. So now let's try to demonstrate that e to the power kx is a primary. e to the power kx. Now e to the power kx well, you can use a formula that actually simplifies matters in this case. Let's just do it by hand. Let's sum and let's try to state this in expansion as i k for m by m factorial times x to the power k. So this is at z. I need to go and I've got action for you. At this guy, I need to go out again. First guy has a choice of m that goes to which you want to write. And then the second guy has a choice of m minus 1 that goes up right. Then m into m minus 1 times let's put some here. Then we get i k squared times m, well, let's try that same 10 minus 1 by m factorial. And then two of these guys are moved to the next by minus 2. And then we have z minus 1 to the power k squared with a no, plus r by 2 to the power k squared. Next to the power minus 2 and that's m into m minus 1 minus m minus 2 by m. i k squared but out here I can reach some of the six. So let's just get the factors, right? So we have alpha prime minus alpha prime by 4 times i k squared so that makes it plus k squared. Then we get e to the power i k as one contraction that will give us the derivative of exponential. So this is indeed the operator product expansion stress tensor in the primary field and we identify that h to the weight is equal to alpha prime by 4 k squared. What I want to do in this theory is to talk a little bit about the interspace interpolation and the state of it case of pleasure the interspace interpolation of the state. Go back to your notes and review how we introduce oscillators in the quantization of these x's. You'll find that we introduce oscillators which had funny combination relations alpha n times alpha minus n was equal to n rather than 1. You take that the expansion they have performed in w, the cylinder plane and converted an expansion in the z plane. In the cylinder plane we had something rather simple just a standard Fourier series expansion with square root alpha prime by 2 on the side. In performing the map from w to the z plane we pick up a factor of minus i and a factor of 1 over z. So the expansion that we done in class previously went back to the z plane simply becomes a volume expansion and then we get factors correct becomes the expansion del x from there will be equal to minus i square root of alpha prime by 2 sum m is equal to minus infinity to infinity alpha m by z. Then the plus one here is simply the fact that primary operators transform with the additional factor of weight. The minus i was the same as when del z had was minus i by z. This anyway is our definition of the mode of operators alpha m. So del x is an analytic operator and we get the most general authentic expansion. This completely general thing to do but we chose a factor carefully so that it agrees with these alphabets all the creation and annihilation operators that we don't know for. It just reminds you again alpha m which means positive are annihilation operators alpha m are creation operators and the combination annihilation is alpha m with alpha n is plus m times less than less than m. That's all we do. We work to have less than m. Let's actually check directly that these combination annihilation. We use the kind of imaginations we use to derive the result of that. Remember what we want. So suppose you want alpha m is equal to 1 by 2 power i times alpha prime times i less than m get x. That's everything. That's all I've got. I'm going to get alpha n. First expression I have m of first so inside so this is z and w and the next expression I have m of first and so inside so this is w of z. 2 to the power w this is z integral. Sorry. So in the first expression I had yeah. The first expression is alpha m alpha n. So alpha n is the operator that acts first. And so now we do this. And the next expression is z integral. Okay. So we do this we choose some when we're done we do the z integral. So what we get is so we get the i let's take each of these i's and square them so we get minus 1 times 2 by alpha prime now all that remains is the 2 pi i's. So integral d w by 2 pi i times the residue in z of the pole z to the power m del x w to the power m del x. Okay. So this will be so what we get is z to the power m w to the power n and then 1 over now we have del x del x which is minus alpha prime by 2 so that can serve as alpha prime by 2 so we get yeah. So we have d w by 2 pi i and then z to the m w to the n over z minus w in the whole thing square that's what we had but we want the residue of the pole in it since we've got the square down here we want to expand this once. So whatever we have d w by 2 pi i and then m times w to the power n plus m to the power minus 1 which is equal to n times del down to the power n plus k alpha m alpha n del down m plus n to the power n which was a combination of relations if you remember we used we're doing a context session it's always by the way a safe thing once you write down your relations just in z-coordinates you can directly check the combination of issues you might have made a fact or mistake but easily tell you it's much easier than tracing through all the conventions that go from translating from one picture to the other great. So we've we've got this nice expression we've got this nice formula now we've got a more expansion of x in terms of these oscillators that are based on standard computation relations okay now the next thing I want to try to understand is the state operator map the next thing I want to try to understand is the state operator map and our industry you remember we have this nice general theory which tells us for every operator insertion that's a corresponding state in my subversa what I want to do is to discover what I want to discover and lose to discover what the operator knew to any state is or what the state knew to any operator is completely the spectrum of states was what was the spectrum of states was firstly you had some very functions of 0 you had a very function for 0 in addition to that in addition to that you had in addition to that you have a little bit exhaustive the most general state was alpha minus 1 to the power m1 alpha minus 2 to the power m2 alpha minus n to the power mn acting on vacuum all the way acting on vacuum times that way function for times e to the power i kx0 and this is the basis then of course you can any of your combinations x0 to the 0 quantization in the system that we have arbitrary very functions of 0 so when you have these states you need a new operator let's try to discover the new operator this discovery in several ways today we only have time for one this first way is by experimentation so the first question we are going to ask is what is the dual of the identity in other words should be the vacuum state in other words should be the vacuum state but let's try to understand this in a way that is but it's so nice okay and then the generalize to do all this okay if the duality of the identity of the operator was the vacuum state then what was in particular visual it should be annihilated by every annihilation by every annihilation so of course we take identity the question we are going to ask is what is alpha m for all n positive whether I identify as a vacuum state apart from the process we need to find x how do I what is alpha m for all n identity but apart from irrelevant constants that is simply z to the power m del del x maybe that's nothing okay z to the power del x but we know that by the equation of motion del x is an identity conflict z to the power n is analytic as long as n is critical say no therefore this quantity is completely analytic this quantity is singularity in this contour but since it's added to every identity you can see the identity operator is the is the vector we will come back to the zero mode portion up to zero mode so please check that the identity operator is the vector what about a nice primary operator that we have nx in fact what about the operator that is k of x k there k there k there whatever that let's try this in trick let's try to check what we get when we take z to the power m del x so this operator whatever it is let's calculate alpha m acting on del k so all you have to do is to take this operator surrounded and then put del k of x at zero and see where we get yeah and we do constants because all of that I want to see is when it's zero or you know what we can work it out with constants okay but what we get here z to the power over z z to the power k plus one k plus two k plus two so k plus now k plus one okay now this thing is non-zero only if m minus k is zero because now we need to come to the table and we need to come let's use the pole this is the pole of n minus k is zero let's some constant times delta of n minus mk because x with x is a log okay so what we find that every annihilation operator except the k to the power annihilates the state what the k is state gives you a state proportional identity therefore this state must be proportional to alpha minus k see all these states that has this property annihilated by order annihilation operators except the k to the state k to the power and the action of the k annihilation operator gives you identity for that okay proceeding in this manner okay we really must stop now about one hour over time so for proceeding in this manner I'm going to ask you to check before me I'm going to ask you to check that delta k x to the power mk product of the k is due to the operator alpha minus k by k product of the k that makes an s field see if you can argue that so we found the due the operator the state cost fund for all the states except for the zero part what are the zero part if you want to add the zero part what you should be doing is put the operating part of k that is zero remember each of the i k's was nice primary operation because this is the full state operator in this area okay I'd like you guys to check this in any way that has much detail as you can before we meet next time okay that it completes at least one way of thinking of the state operator in this area now the second one of these things one can do with the three books you know something that people do a lot that you never might be able to use for is to write down the birasoro generators in terms of the three books on oscillators and then do a lot of manipulations using oscillator algebra somehow I don't find that very useful but it's a game that you play perhaps it's sometimes useful okay the first thing that's more useful than my opinion for you to try is to try some direct technical computations in this area okay so I'm going to suggest one more exercise for you again the way this starts to be viewed the next class just directing my awesome technical analysis compute the awesome technical with del k of x inserted at the origin okay and show that it is a way far for k of oscillators once you've done all these exercises you'll know enough about the three books I'm hearing for my purposes this concludes our discussion this essentially concludes apart from discussing the exercises we talked about this essentially concludes our discussion of the three books on performance theory in the next class I will describe some slightly more complicated performance theories okay we'll talk about some slightly more complicated performance theories and in the class the next class will go to Wednesday in the class after that which I hope will be a next Friday we'll then go back to taking all that we've learned about performance theory and applying it to our study of performance theory so our long interview into the study of performance theory will be finished we're also finished by one at the end of the next class we will cover every bit of every bit of the material chapter two of the specialty so if you will be like reading for you chapter two now everything except the part next class I also suggest set of exercise problems in the back which is