 There's some exercises for you people to do. Has anyone tried to do exercises? So let me ask the first one. Has anyone tried using the conservation of stress in this conservation of the stress sense? So using the fact that we have del z bar tz, del w bar tz, del w bar, is equal to del z del w of tz. Is there anyone tried to use this relationship to pull out how t alpha into 2, 0 into 2, and we find this. Sorry, right. Has anyone done this and got the constant? How did you get the right answer? Anyone else? OK. I won't ask you to repeat once, but OK. Let's check in with the minus sign, I think that is. We'll go through it in a moment, but OK. I won't ask you to come to the board. I'll just take too much time. OK. OK. Just in my experience. What about second exercise, just computing the matrix that I've found? That comes out to be something like C by 2, 0, 0, 0. 0, 0, 0, 0. If you can't consider it, we can go on the right. On the right, please. Yeah. It's some copy of computer, at least. And it's where? Exactly, exactly. So already, that is sufficient to tell you. Because there will be a vacuum always, and this has to be bigger than 0, that the C is too big. Very good. So dirty boys give the answer. When I did the calculation for the most general thing, by the way, I got the following. The most general matrix I've been following, I got the 4h plus C by 2, 4h into h, with 2h plus 1, as the matrix of all the popular products. OK. As dirty boys says, all you need to know to get the result, C must be greater than 0 for unitarity. All you need to do is apply this special case to the identity of the material vacuum, in which case, positive 3 of C, is simply the positive 3 of this matrix. OK. But this general result from a general operator will also be useful later on, in determining, for instance, the identity of normal states. It's a sort of interesting exercise to compute the determinant of this matrix. I'll leave that as an exercise for you to do. I'll just give you the answer. The answer turns out to be, suppose you compute the determinants. I got the other thing. This was 6h, this was 6h. And this was 4h. If you compute this matrix, you get the following answer. You can get the x minus equal to h minus h1, h minus h100, h minus h2 into h minus h3. So h1 is equal to 0. h2 is equal to h2 and h3 are equal to 16 in a 5 minus c plus minus plus 1 by 16 to the square root of 1 minus c into 25 minutes. It's just solving for some quadratic equation. It becomes a quadratic equation because you get a factor of h immediately outside, so then it's quadratic. So you get a factorized as a fraction of the determinant. The point to find is that you see that certain values are fine. You see, for instance, that the determinant is now always positive for these two different values. And if h is in between h2 and h3, h is greater than 0. But in between h2 and h3, then the determinant is a product of three factors, two of which are positive and one which is negative. The determinant is negative. It's impossible for all eigenvalues of the matrix to be positive. So this already gives you some restriction based on unitarity on the operator content of a theory with central charge c. OK? Do you see the restriction saying that h of the theory cannot lie in a certain range, depending on what c is? The constraint is the only question. The constraint is the only question which is not imaginary now. Exactly. So that also tends to, for instance, Yeah. So if c is less than 1, or if c is greater than d is, but no, just c is less than 1. Then that condition that Logan Adler pointed out is met. This is some positive greater than 1. So when the c is less than 1, you find four models to which c is less than 1. You find this constraint. Given a value of c, already the central calculation is told to us that not all the values of scaling dimensions of primary operators are allowed. There's some restriction. Now, this is a restriction that we got from level 2. You could try to do the same calculation in level 3. And so on. And try to satisfy all the constraints that come from unitarity. And if you go ahead and do that, you find, though I don't think in this class we'll actually go through this calculation, but if you go ahead and do that, it turns out that you find that when c is less than 1, unitarity severely constrains the possibilities. It turns out that you can have unitary theories if and only if c takes one of a set of discrete values. An infinite set of discrete values. And then given that particular value of c, the possible allowed values of h in that theory are some finite double values. There's a set of two integers in terms of which all the values of c that are less than 1, that are consistent with unitarity, and given that particular set of integers, the range of possible h's in some other value, which runs over finite set of values, is also specified. So just unitarity in a conformant view of theory with central charge less than 1 tells you a great deal about the structure of the theory. It tells you that the central charge can't be anything, it has to be one of the discrete set of values. Once you know the central charge, the operator scaling dimensions can't be anything. They have to take a finite number of values. So unitarity gives you important constraints on the physics of conformant view of theory with c less than 1. Also, for every value of central charge allowed by this unitarity analysis, there exists at least one conformant view of theory. This has been constructed by direct construction. With that value of the central charge, there are some of these operators. These theories are called the minimal models. And it turns out that these values of, that all, you see, not every operator is allowed in this sense. Only operators are then distinguished, but then distinguish the scaling dimensions. Now, the reasoning of every scaling dimension in the game is that if you move the scaling dimension a little bit, the normal sub-state will go negative. If that's going to happen, it must have been that the normal sub-state was zero at the distinguished values of h. That is in neither case. So these minimal models, the models in which you have the representation has some null states. Because of which, for the reasons we talked about in the last class, you can write down differential equations that help constrain the behavior of correlation functions and essentially solve for these problems. So although this is not, none of what I said is going to be terribly important for the direct line of the gap of our class, it's an interesting point that minimal models that control the field theories with C less than 1 are completely classified and essentially all solved. Any conformity theory in two-dimensions with central charges of 1 can complete classification and any modelist can do it. This illustrates the power of these techniques. You can't generally solve a quantum field theory in these cases you can. But it's all that we can give you. I mean, find the full operator spectrum, find out what it is. That's all I wanted to make about this normal calculation. If you would read a little more about this kind of stuff, these lecture notes by Ginzpa called of mind of homogeneity is, I think, an illusion lecture notes from 1980s. Then there are good sources of stuff that we can give you. So that's all I wanted to say about this first thing. Now, that's completely both of the second calculation. You know, there's a way to understand that there had to be an h in this one. Can anyone tell me about that has to be an h in this formula? What is 0? It follows from calculations. Is there a more structured way of understanding this? You see, it's simple way to understand this fact is that at level 1, we did the calculation of norm for states. And we found that the norm was equal to h and a state with 0 norm at level 1. That implies the existence of a state with 0 norm at level 2. Because it takes a level 1 state and act with it, act on it with some some basic operator. And you do the norm calculation for that other state that would relate the norm of this new state to the norm of the level 1 state with 0 norm. So you get some number times 0 and therefore 0. So this is a general factor. Any state that is a descendant of a state with 0 norm at h equal to 0, a state with 0 norm at level 1, there must be at least one state with 0 norm at level 2, the descendant of that 0 norm state. This will be a general factor about this. This is a general factor. By the way, it has a name. The German name was in the products of states at level n. It's called the Cauchy determinant. And Cauchy is given an explicit formula for this determinant at every level, which is what is used to give the analysis of allowable values of central charge and allowable values of scaling dimensions of operators that are in dimension. Okay. So this was a bit of a diagram, but about something very interesting about the properties there. It will not form the heart of what we talk about in this industry. So now let's get back to the second exercise. So let's do the same exercise. So the second exercise was this thing. Let's take it back. So we have this nice equation which says that the del z of tz z bar times del w of p w w r is equal to del z bar del w bar tz z p w w. Okay. Now, suppose I define tz z bar p w w bar and define that as x. Then I conclude this as del z del w of s every function of del z of tz. Then I conclude that del w of del z del w of s is equal to del z bar del w bar of c by 2 of z minus w. Now, let me rewrite this quantity in terms of 1 z and 1 w derivative of something. Okay. So I can rewrite that as as del w del z del z bar del w bar of 1 over z minus w is the whole thing squared except that I need an additional factor of 2 times 3 and a minus sign because one when I differentiate I get 2 minus sign but the third one because one of these is the w of the other one just like I said. So the final answer is minus c by 12. You don't worry too much about what I get as constants of integration. So this derivative n is minus c by 12 times del z bar del w bar of 1 over z minus w of the whole thing squared. Now the z minus w the whole thing squared as equal to del w del z of long above z minus w into z bar minus w. The z bar minus w bar is going all the way to the right. I just want to put it there to get a real expression. Because if I say it's 0 I'm going to have to put it there so I get a nice real expression. Okay. The minus signs work out okay. Because we've got one minus sign from differentiating 1 by 8, 1 by x and one minus sign from one of the derivatives is the w. So finally we've got here that is equal to minus c by 12 del z del z bar del w del w bar of of law of r squared where r is more than the more of z squared. Is this correct? So what if we concluded that d z z bar d w w bar is equal to minus c by 12 del z del z bar del w del del w bar of law of r. Now it was convenient to do the calculation of this complex relation but it's convenient to interpret the answer okay. So if these things are each proportional to t alpha contracted whatever proportionality constant is needed to deliver these two those things, same as proportionality constant we need to deliver this to del squared. So if you rewrite this as t alpha of z and c beta beta of w is equal to minus c by 12 del squared in the variable z, del squared in the variable w of law of z, mod z minus w squared. Is this correct? Alright if we say about del squared of law of r squared okay, what can we say about del squared of law of r squared? Now if you do that mentioned I will pretend that del squared of law of r is equal to 2 pi times the del squared. How do I know this? Well I know this using Stokes' theorem. You see because del of law of r is equal to r hat divided by r therefore integral del law of r over a circle is equal to 2 pi. You see there is a volume integral of del squared of law of r over the circle. That means that del squared of law of r must be 2 pi times the del squared. Of course the z-injected del squared of law of r is 0 everywhere except perhaps a volume of 0 but there is a singularity there from this application of the source. Alright so this formula here adds t z z and t alpha of z t beta of w del squared minus c by 6 del squared of the delta function I got an answer to that. It turns 6 because this was r squared and this relation has been to a lot of relation actually. You might value a minus sign from the fact that some of the derivatives are z derivatives and some of them are w derivatives but we always have 2 derivatives so this is no issue. Now what does that tell us about the stress in an arbitrary method? You remember as I remind you last time the change the change in t alpha of alpha was equal to 0.45 like the neighborhood of 2 omega let's say for x and t alpha of alpha x d beta y let's call this an x of y d 2y but this was the change in the value of t alpha of alpha d beta was equal to 2 omega and let's call it as follows from the definition of our stress tensor and what the change in the metric was this small change fine so now let's use the two function of the stress tensor okay so we get delta of a delta function integrated by parts of the other side on the omegas so we finally get the change of t alpha of alpha of x is equal to the change of t alpha of alpha of x is equal to so we get c by 6 let's see let's check let's see let's see where there's a minus sign I'll give you the final answer I think that's mr minus sign okay so now you see but we had general considerations we had general considerations that d alpha of alpha was equal to some number now we know how the alpha of alpha changes under a change in a conformal transformation so all you have to do is to see how are the right hand side changes under a change in the conformal transformation and it's easy to check that this thing changes by a that is then square of 2 this is a standard formula that we pull out of we ran in the investments so you were right so comparing these two that a has to be equal to c by 12 unfortunately we should get a minus c by 12 so if mr minus sign is somewhere okay let's not try to track down the minus sign here just wait a second we'll get it through the next class okay it's not worth the time so I'll track down the minus sign carefully next time sorry for that so up to the minus sign um okay so we've concluded although we've considered the minus sign wrong that t alpha g2 minus c by 12 completely universal result it's universal in the sense that the only thing about the conformal field theory that it depends on is the central charge by the way all we use here is the next movement the quantity that appears is the operator product expansion of t with t left moving but we could have done the same calculation with the right moving central charge but we were going to get the same answer for this so if the left moving central charge by theory was not equal to the right moving central charge by theory then something is wrong we do a calculation and we get one answer from the left moving central charge do another calculation and we get another answer from the right moving central charge this conclusion is correct actually the thing that we've used importantly all over the place the important thing is to maintain difumofism in variance what we see is that while you know there's nothing wrong with the conformal field theory just in flat space with left moving right moving central charge is different however if you want the theory to be difumofically there that is you want to be able to define it in a difumofically varying way on an arbitrary line the only way that can be done is that the left moving and right moving central charge theories with different left moving right moving central charges have what is called difumofism or sometimes the gravitational anomaly and that anomaly serves to the statement that it's not consistently possible to lift and something great started to try to lift the theory to an arbitrary space that's not understood okay now I want to point out one consequence of this fact a consequence of the fact that the trace distress instance don't just have set the direction that's good this statement is more powerful than you might first think it's more powerful than you might first think for the following suppose you had interested in concomitant theories where the method is the alfamida because you put it out of the alfamida and use it to find it a small change in finding you can compute how the partition function of this theory changes because under the small change in find the change in the partition function is going to be proportional to the trace of the stress tensor in that theory you have to look carefully if you've got the result of this it's a set up by the definition of distress instance but we already know what the trace is in terms of other functions form with some constant delta z is proportional to integral delta 5 times minus c times constant delta r where r itself can easily be written down in terms of phi we've got functional differential equation which is actually possible quite easily it's possible quite easily some of us I want to give you an exercise show that z is equal to some number which is determined times 5 del squared 7 z is equal to e to the power some number that you will determine times 5 del squared 5 solves the function differential equation that you get from probabilities times z is equal to 5 times z is equal to 0 which is 3 plus 1 that is exactly that the partition function is a conformity of theory that is related by wire in flat space both we consider the arbitrary metric and if you are always an invariant up to possible discrete global problems we've already used the statement that we can always put use the few moments of invariance put the metric into such a form then if you are always an invariant as we've always been doing then this information is enough to determine the partition function for the theory and the arbitrary method and the formula that is the formula related to flat space is every formula related by the formula plus a few moments of transformation in flat space up to 7 global then we use the search for covalent expression that reduces to this expression for metrics of that form so this is part 1 the possibility of the exercise is to show that the covalent expression z is equal to 1 meter across another number in fact r is 1 by 10 squared and then the right squared will be the basic point is that r is dead squared 5 once you think about it but essentially r is dead squared 5 so that that's where 5 is that's where 5 is that's where 5 is this number, of course, is determined by the central charge with central charge but either the other way around the right would be central charge all of these variations only make sense in the theory in which we both are and this is what we've concluded we've got what we've concluded is that we have incredibly incredible power this is two dimensions of the bomb of new theories if you know the central charge of the theory and you know the partition function of the theory in flat space you know the partition function of the theory of any arbitrary function the difference between these two for a partition function is just in fact that it becomes only on the central charge isn't this quite incredible and it's such a simple variation it gives you such powerful results this is part of the answer to the question why is the central charge important there's one other calculation that I wanted to outline for you one other calculation I wanted to outline for you that says the same thing but in different words and words that you may find physically more more real okay I want to again to really derive the statement that the trace of T is minus C by 12 times but from another point that's the same point but slightly under the conformal transformation we know how the stress states are transformed as we are saying the transformation for analytic excellence is given by that that meaning is equal to the pole in epsilon on the side thanks to that it's analytic which seems that there are three terms in the recipe of the pole there's there's epsilon prime of Z there's epsilon prime of Z times del T there's plus 2 epsilon epsilon of Z times del T plus 2 epsilon prime of Z times T and then there's negative just do this okay okay let's look at the problem z times 1 C by 2 Z minus W 4 plus 2T of W Z minus W square plus del T of W okay we go to the recipe we get del T of W plus the first term the del T of W here so that's plus 2T of W times epsilon prime of 0 and W let's say W is 0 and then plus we get the term that has 3 poles of Z minus W so we get epsilon triple prime of W by 6 so that's C by 12 C by 12 times 10 by 8 these two terms we understand what they mean basically the transformations of T under the coordinate change that was part of the conformance because it's only here in the other stressors however the conformance transformation is not just the coordinate change it's a coordinates change plus a winding scale this is a winding scale by what see if we rate the transformation Z is equal to Z prime plus epsilon of Z then this is actually the first order epsilon prime of Z prime plus epsilon Z bar times dz prime dz bar prime so the coordinate change changes the metric length is in order to get the forward transformation that is to get the metric back to the flat space in the new coordinates we have to perform a wire transformation because wires are this fact so conformance transformation is the sum of coordinate change plus a wire transformation the wire transformation need in this case is that the need to the power minus epsilon prime of Z plus epsilon bar prime of Z bar so this third term here represents the change in the stress tensor under the wire transformation let's try to rewrite what that change in the stress tensor under the wire transformation is in terms of the wire factor the more you get it quick so the wire factor was down here twice the wire factor was equal to epsilon prime so this change which is super derivative of this thing here there is just like change in Tzz minus C by 12 times del z del z del z cubed ah no, let's call it del z squared of del z of epsilon which is the same thing as del z squared of 2 that's what's given by the situation so the stress tensor transforms under only very specific kind of wire transformation it's easy to guess the gender of formula which would be valid by any so formula if this was write down formula then you plug in this thing for the special case of and then it has to be something that depends only locally on the wire factor so I don't have to try to do it can't you give me the problem here I'll give you that okay, that's great but now we use we use we use we use a coordinate we use a few options with values so that the equation that Tzz del z bar is equal to del z of Tzz bar with the minus sign with the minus sign with the minus sign and so let's let's put a del z bar here so I put a del z bar here so that tells me what the change in del z del this thing is I cancel one del z by by so the final confusion is that del of Tzz bar is equal to anyway I'll clear up both of them C by 12 times it must be wrong whether we want to minus this or plus this okay so this is del z del z bar of del z del z bar of 2w this is T alpha alpha is equal to C by e to the power of times del squared of 2w so this is how the closure is so that tells me that T alpha if T alpha has to form a r a is C by 12 I'll clear up these minus signs there's some minus sign problem I have on my hand so I'll clear up the next ones I'm working consistently and getting positive there must be something consistently wrong sorry so once again we've cleared the confusion that is equal to C by 12 the nice thing about this derivation is that it makes itself really clear what this anomalous term what this anomalous term the stress tensor transformations it's not always straight simply the transformation of T under a while transformation and this leads to the fact that the trace of stress tensor okay good the last thing I want to say about this this reasoning about this stuff is something I should start with a little bit but didn't because I think we were at the end of the class because this equation you see we keep saying that T alpha is equal to A times r and then we try to determine that you're A but how do we know that this is the right form for T alpha no I couldn't be something else so can anyone help me with the argument can anyone come up with a general argument that says that if you like this then the first thing we know is that T alpha alpha is equal to 0 in fact then next what is it could T alpha alpha depend on the state of the thing actually we've been doing it by expectation value but this is the strongest statement it's actually an operating statement T alpha alpha is this because you see this the theory is T alpha alpha classically is 0 so quantum mechanically the way you get some non-zero value for T alpha is from regulating the theory some measure which comes from short distance in humanities that is always incensed in any one way it is incensed into the state of the thing something that happens at very short distances okay is determined just by the fact that the theory is by the energy state it's like an actual extreme short distance okay so some effect that comes from some chemical regulation happens at very short distances cannot depend on the state of the area okay so so on the other hand it could depend on some parameter of the theory but we are not informed of the theory so there are no if there are any parameters of the theory it doesn't depend on some parameter of the theory that parameter of the theory it just cannot have any dimensions this is the theory we can form okay so we have to make up our dimensions from there to the metric now what can we get this thing here is dimension 2 so we want some scalar on the right hand side that has dimension 2 what could it be it's the metric the colors basically there is no other candidate you might say well suppose it didn't have dimension 2 you might say well okay you are telling me that everything here has come from short distance effect so suppose what we call was not something to dimension 2 but dimension let's say 4 and it's made up by some power of the short distance regulator so that indeed of you in which you regulate it it will be true it gets our contribution with the trace of the stress let's say r to the 4 but that is not mass dimension 4 so it comes with an explicit distance of square then you get to continue with nothing matters so basically you might ask why was something the dimension 1 or 0 well there is no covariant scalar with dimension 1 or 0 that there is just no covariant scalar with dimension 1 okay actually the last argument was not quite needed but anyway it's true so the only thing that's going to have been as an operator state in our system is 8 and 3 and now you've got lots of facts to erase in the general case of the minus signs that I need if you have a theory that you can have you would have to do it in a way that that was 0 in flat space okay so you see the curvature scale already is the lowest dimension scalar okay that has dimension 2 if you add additional things you only get higher you could try to do something with fields without the state curvature scale but then you get a non-zero outside flat space there are no there are essentially no other okay what he was getting at was is there any argument why they should be you know like even if you're doing this here I can understand that it should be different to the state right but why is this true for you know all theories what do you see is it once we argue that it's of this form then we have a derivation that this parameter is c so the derivation is wisely of this form and then we just see what else will it be I'm asking you see there's probably something better to say than what you're asking see there's an indication that this is not the best answer to your question but I don't want to just say something inaccurate if you have a problem let me stop there's something more to say let's discuss this okay look other questions for this now there's one last thing that I wanted to talk about there's one last thing that I wanted to talk about that was general to conformity in general before going out to study the specific conformity theory of the free skin and this last thing is this carniform that many of you have heard of and it was very simple it was very interesting of course suppose we consider conformity the theory suppose we consider conformity on the torus the partition function of conformity on the torus the space direction of magnitude pi and the time direction is named beta we did this partition function in utility so though I point one of the directions basically at one time these two things are completely different but specifically now I focus on conformity theories with no fermions in them so we don't have to make boundary equations there's an extension of what we say fermions will get that let's be simple now suppose the first thing I want to remind you of is that this partition function has a simple interpretation in Hilbert's case this is simply equal to the trace of e to the power minus beta h trace of e to the power minus beta h of the theory h Hamiltonian of the theory h is Hamiltonian of the theory answer to this partition function calculation is now, you know, given this interpretation one can easily compute what this partition function is in the middle of any argument why is that? because this partition function is dominated by the lowest energy state in the theorem but we've already seen that any utility conformity theory the lowest energy state is the state due to the vacuum and it has energy 0 in the plane but that translated to L0 and L0 bar both equal to minus c by 24 and therefore energy is L0 plus L0 bar plus minus c by 12 on the cylinder now there is a vacuum scaling dimension 0 but there was a shift between energies of the cylinder and scaling dimension c on the plane the shift was by a factor of minus c by 24 for L0 similarly there is a shift of minus c bar by 24 for L0 bar c bar is the center right, no big center yes, as you can see this is one of these different models so you get a total shift of minus c by 12 minus c by 12 in L0 plus L0 ok, so the lowest energy state of the theory of vacuum is unique and its energy is minus c by 12 ok, so when V star is large we have this partition function even if the dropping terms are exponentially suppressed in V to the power minus V star so what happens to this because the unit comes and we generate this all that we said is that at very low temperatures the partition function is a little bit of a matter who didn't know that however now let's look at this let's look at this let's take this this partition function now let's take this partition function and view it slightly differently you see at first let me just take this torus and flip it around in my mind ok, so it becomes a torus one of those sides is V star here and the other side here is V I'll make a part of that into the torus which I call the x side and which I call the y side so this is also the answer to that following the next crucial step I realize that by a scale transformation I can shift this to a torus that's 4 pi square divided by beta and to find merely by performing a wire rescaling which is everywhere uniform performing a uniform wire rescaling the partition function of my theory does not change we just derived that all changes appear with derivatives of that then square is the wire factor when you perform a wire ok, so this partition function of this theory is the same partition function of this theory and therefore is the same so we have that z of 4 pi 4 2 pi squared by beta equal to e to the power c when beta is very large but all these manipulations are exact but we only know the answer to this but this side is the is true of beta you have already said that there is no signature idea exactly that is just a property of the path that's what I'm talking about ok, so that z so now that comes first of this we already made a statement z of beta is equal to z of 2 pi squared by beta that's true but we only know the left hand side of beta let's rewrite this in terms of the argument of this quantity so let's define beta prime equals 2 pi equal to squared by beta yes we will find beta prime this is 2 pi equal to squared by beta and then substitute this in to this expression so what we get we get z of beta prime is equal to e to the power c by 12 beta prime and then we divide it by beta prime we've got 2 pi squared by beta by 2 pi or have I got 2 pi squared no sorry I've got that beta prime divided by 2 pi squared is 1 and there's no more the law of the function is equal to c pi squared by 3 beta prime or if we interpret beta prime as the most temperature c pi squared by 3 into the temperature which is minus t times the law of beta function is c pi squared by 3 t squared minus you can compute both the entropy density and the energy density I'll leave this as an exercise for you but I want you to show that up to some numbers in this formula there is that you've got c times some number times the entropy density of your theory scales like the square root of the energy and the square root of c times some number which follows in each of these formula what we've done is charge to compute the rate of growth of states in the theory at high energies. You see, this is another interpretation of the central charge. It's a definition that makes clear that the central charge tells you how much stuff you have in your theory. More precisely, it tells you how fast the density of states in your theory goes with the energy. Theories with larger central charges have more units of free energy. As you can see, most of the energy from here, the free energy function of temperature is proportional to the central charge. So here, any non-interacting fields of free energy is any type of that sort of function. See, it's like the number of degrees of free energy in a few states. Again, the knowledge generally concluded that the input has used this unit energy. It used the unit energy in computing. We know the energy in the lowest energy state. It's used to monitor the energy. It's just a fact that the torus oxygen function is very small. Provided that you have done the form of the theory at high and very authentically on a torus, you know, it's just a way of using basically very few requirements, we've got a very powerful conclusion on it immediately. Okay, so a lot of people in many different ways are thinking of the central charge. The central charge is that object that governs the density of states of the theory at high energies. It's that object that tells you how the theory, the function theory changes under a wide observation. It's that object that tells you what the trace of the stress tension is in a theory and an object. It's that object that tells you how stress tension changes under a wide view. So many different rollers at the central charge phase, they're all linked to each other in this beautiful way. You see that the central charge is an extremely important number. That characterizes the problem. Okay, well, unless there are questions or comments that's it for my general discussion of the problem theory, I want to now start the discussion of the free-spare theory in detail to flesh out some of these ideas, also because we've got a new theory heard. We're going to question the problems. Not a study of a particular component in theory, which is found in the very true universe. Yet one that is extremely useful in the study of the strict theory, namely the theory of free-spare theories. So this is the theory that appeared in the World Sheep of the Strict as we saw in our first few lectures. And the fact that, as you remember, was S equal to 1 over 4 by 1, right? The table square root of g, there are alpha x mu, nu, g alpha beta. Now, the study of strict theory, we will mostly focus on this conformity theory on the background metric, G. I wanted to understand, so what's the part, the action of our theory? Our driven boundary of our theory is exponential of minus the actions, minus 1 by 4 by, right? 10 alpha times mu. And let me, since we've got a bunch of decoupled conformity theories, one for each mu, to start with a study of the theory of just 1 x. It is the idea of copies of the theory. It's to derive the Green's function for all the three scalar fields of this theory. You know of many ways to do this. Anyone who's familiar with study quantum theory knows of many ways to do this. I'm going to use, you know, so as to get all the factors right, I'm going to give you an, and a first principle, this derivation, using as we would always try to do in this course, a path in there, you know, a path in there. So, what is the Green's function? The Green's function is the value you get when you act the equation of motion, del, del squared from master's point of view, on the two-point function of a field. Okay, now, you get something known as zero, only because, as usually comes time, because the field is a way of the equation of motion. So, you might think, why am I, when I act del squared on a two-point function, why don't I just get zero? It's the same as the discussion we have for, for the, for the, for the carbon, exactly, carbon conservation. You don't get the zero, you just get them, there's no functions of time, okay, this way. Okay? So, what we want to do is to try to understand how the equations of motion act a correlation function, not del. What is the equation of motion? Equation of motion is the derivative of the action with respect to some field. Okay? So, let's first try to understand why the equation of motion is true, at least as an appropriate x-axis. Okay? Now, that's true, it's obvious. It's obvious from the fact that the integral of a total derivative is zero. So, consider the following. This is the integral of del by del of x at some sigma of e to the power minus 1 by 4 by alpha, right? Uh, del alpha by x, del alpha by x. This should be zero as long as the path integral is very defined. This should be zero because the integral of the total derivative, we shouldn't have a boundary temperature. The formula is infinity. Okay? So, this zero is equal to this. Now, let's take the derivative. So, we get the exponential of the gates, and take a look into the power minus 1 by 4 by alpha prime, del x squared, times 1 by 2 pi alpha prime, two minus signs. The minus sign here and the minus sign that I get when, uh, the integration by pulse, del squared, okay? So, our first confusion is that it's going to know what the operators inserted, and since no other operators inserted at the point sigma, then, okay. So, we could have any another operator inserted here, o1, on, as long as the insertion points to these operators are not sigma. Okay? And, uh, then this manipulation will go through the o1, on, because the derivative doesn't act at o1, on, because local is the local function of the p is x sigma, uh, and the insertion point is not sigma. Okay? So, that the, uh, that the correlation function of the equation of the function with any other operator is zero, provided these operators are not at the same point. That's the statement that as an operator equation, the equation of the function is true. But we haven't yet got what we get out of this delta function the time-ordering delta function differentiation. And I'm from this point of view. So, now let's look at, okay. Let's look at, uh, let's look at a particular case. Let's look at the case where we have, where we have, okay, delta by delta x of sigma, right? We have x of sigma prime, even far minus, minus s. We have x of sigma prime, and then o1, on, such that all these other operators, uh, are not coincidental in sigma, but x of sigma prime could be at the same point. So, now we get delta of sigma plus sigma, minus sigma, right? Let's look at far minus s and o1, on. The second term we've already evaluated. So, let's write it as delta of sigma plus sigma prime, minus sigma prime, plus, uh, 1 by 2 pi alpha prime, delta square of x of sigma, x of sigma prime. All of this, in fact, you can. This is equal to z. This was here. Look at that as, you know, as an operator state, delta square of 2 pi alpha prime times x of sigma, x of sigma prime, and any other other operator, such as, how do I say, not coincidental? Right? Is it equal to minus delta of sigma prime, right? That would be all right. We can care about coordinate function. One, this is just an operator state. Because if any d, the reason that we aren't going to have other operators to be banged on top is, why would other data functions? This is just an operator state. This is true. Okay. Now, if we want to do, we can use this to compute expectation value of x of sigma, delta square by 2 pi alpha prime of this, is equal to minus delta of sigma prime. We also know that this state is translational invariant, and so we've got only a difference of sigma and sigma prime interpretation invariant, and so it must be some function that depends on modulus of sigma and sigma prime. This is the usual green function problem, and we've already in this class solved the problem. We know that the solution to this is simply log alpha prime minus alpha prime. There's delta square and log of 2 pi, cancels this, minus delta of sigma, we need the minus. We write this in terms of z and z power. So instead of r, we write z times z power, so that's r squared, so we need the pi 2 function of our theory of the expectation value of 2-point functions of operators, is given by minus alpha prime by 2, log of z times z power. This is z minus 1, and z minus 1. Is this clear? This is the theory. It's a very important calculation, it's one that we will use on that. That's just a very important problem in our head. Okay, so we've all got conclusions. Right at the beginning, when we do extreme theory, we have the stress tensor simply concentrating on this. And you remember that it's del x, that tz z is del x, del x, tz z bar is del bar. It doesn't make an instance into the dimensions we've got to get the constants, right? Okay? And I'm not going to go through a check, just del u, and write values t is minus 1 by alpha prime, del x, del x, and the similar problem in our first problem. Okay? We'll trace through the definition of what the stress tensor was, and apply to this action. It might have a more minor thing, because, you know, why did the quantum field del u, which is leaving out the fields, that makes a lot of sense, by the years, don't make a lot of sense. Because, if you're sure, this can save you a lot of time. Okay? So before proceeding with our discussion, before proceeding with our discussion of, so, as is generally true in the study of quantum field, whenever you give a complex of operators, namely, operators that are made out of products and basic fields, you need to be fine working. Okay? You need to be fine working, because, if, by a composite operator, you just meant, phi at x, plus phi at x plus epsilon, limit epsilon goes to 0, inserted into the path integral, the answer to that question is infinite, almost always. Because of the short distance similarity, in phi at x, meeting phi at x minus epsilon. Okay? So, insertion of that quantity into a path integral makes no sense. By opening up, we need an insertion into a path integral. And, if I really meant, really say, you know, literally meant to insert this at the same point, or more precisely inserted, you know, in the limit of the particle, 0, the difference in the limit of the particle is 0, I would just get a lot of sense. You're getting too many. That insertion is not going to be fine. So, if we take x and x prime, okay? And we look at the vacuum expectation value of that object. It has a singularity, when the two points are together. Okay? And then, if I took two derivatives here, this is log singularity, it becomes a 1 over z squared singularity. Yeah, two transpose, that's singularity. Okay? So, by this object, I certainly don't mean, I certainly don't mean, just inserting this expression into the path integral, right? Well, there's always an inside of quantity very, very good question. What you have to do, basically, is to define, how well you define insertion into the path integral. Often, one uses dimensional regularization, sophisticated techniques to define this. How great is it? However, in the case of hand, the thing is essentially a normal object. It is basically a normal object, but also carefully leading to the zeroes. See, a normal object is an unimpeded prescription for oscillators, but it won't know how to deal with zeroes of the problem. Yeah, it doesn't immediately know how to deal with zeroes. So what I'm going to define for you is how to make sense of operator insertions. In a method, there is more or less a normal object for those of you who don't know how to deal with that. If you don't have an issue, never learn a normal object. I'm just going to define it for you, and we use that as an efficient system. Okay? Now, I'm going to define an issue. Look here, X of C, that's an X of C, half-right. And then you find this simple question, which you see, because it's a formula, right? Actually, it's a slightly souped-up version of normal object that deals also with zeroes. Okay? And that is equal to the sigma X of C, half-right. Minus, I'd like to make. Minus is nothing. This insertion into the path detector, the thing I actually put in the path detector is this. What you'll notice is that the equation of motion is a true equation in cyclical formula. Why is that? When I del square and X of this, I get something normal zero here. And something normal zero here, the actual operator product. You know, if you try to use the equation of motion naively on path detectors or operator insertions, you get, that's wrong, because you're not taking account of the data functions. But when I do it with normal or operator insertions, where it's correct? Is this clear? Obviously, what does it need you to do? Want to do it because the equation of motion applies to it. That applies to the del square of same one, that is, X of sigma prime equals zero, which tells you that this normal operator product is a genuine analytic, well, genuinely the sum of an analytic analytic function in the argument mix. Now, why is zero not allowed to see new lattice? Okay? Okay? This would be clear if I, well, del X, del of X prime, right now, actually, del R on this, actually, has zero. So, del X, the functions of del X, inside normal order of expressions, are genuinely analytic expressions in the point of insertion of del X. The effect that is an expression, which is normal, or they expand, you know, some correlation function that's normal order. And we try to expand the correlation function in one of the arguments of the operators. We only get analytic terms and no sequence. That equation of motion is a genuine equation inside normal order. So, del bar of del X is just zero inside normal order. Which means that the correlation function has a function of the insertion point of del X. It's an analytic function of that insertion point. It's a genuine analytic function. Where it's not. Inside normal order, the correlation function is inside normal order. A genuinely analytic function of del X inside normal order. But, that's correct. But, the statement I'm making is for normal, okay? And there is one analytic function of the insertion point of del, of the operator. When you open it, and cannot have a genuine answer. You see, we design normal ordering as a product of the product of two operators. Okay? By the way, now, any number of derivatives of X, it's just if I am saying, okay? Then you, you form the same derivative operation on the, on the subtraction, on the subtraction. Okay? In a way, you don't really need to define no derivative of X's, because derivatives are just differences between, between these two insertions. So once you have the derivative insertions of X's, not the derivative insertions of X's. Again, the fact. Now, products of three X's. X, X, X of sigma 1, X of sigma 2, X of sigma 3. I need to define normal ordering by the following recursive expansion. Okay, to define normal ordering by saying that this is equal to normal order X of sigma 1, X of sigma 2, X of sigma 3, X of sigma 2, sorry, plus X of sigma 1 times contraction. This is bit scary, but it would be bit scary if we had an independent definition. At the moment, it's like, it's a definition that agrees with X's. Right? Plus X of sigma 2 times. Okay, we are not defining normal ordering for any operator like above. I'm just defining for product of two operators. Right? We are defining the indents of the oscillator. This is just a definition. Have you defined the indents of the oscillator and then like this, it would have been the... Exactly. You see, this was a definition. So what it means to insert normal order of product of two X's? Why should we be careful about this? This is the definition we would take the three operators as that minus dimension. What? Three operators as well? This is the definition. Sorry, you were suggesting an alternative definition? Then, why can't we just define the normal ordering of three operators? Right? Also, that minus... That minus... No, then you would get the... Oh, come on, that minus expectation, that would be possible. You want to get the whole singularity. You want to get the singularity even when two operators come near each other, one is far away. That would be good. You want an expression which is totally normal. And that would get sensitive. Okay? So what we have done here does that. You see, what we have defined is that you take it, right? And then, what it does is separate out the singularity from two operators from here by times the soft break which is where it is. La, la, la, la. Okay. So this is our definition of normal. This recursive procedure defines what we need by normal order expression of three operators. Same thing is going to go through for... So by normal order, if you got product of N operators, you expand that in some overall contractions times whatever is left behind. Now some of our contractions base, you understand what I mean? So let me say, hey, suppose you have the product of N operators. Just the ordinary product. Now in my definition, this product is equal to normal order product of N operators plus normal order product of N minus two operators times contractions of the remaining two operators with a sum taken over our possible... Normal order of N minus four operators times contractions of two operators in pairs with some taken over our possible operations and so on. That is my recursive definition that defines what normal order of N operators means. N minus two, I already knew what N minus, what it was by N minus two. So this equation has only one undetermined element. Maybe the guy that's normal order. Okay, is this clear? I'm still going to write a formula for what this normal order operator is, but just, this is generally the most convenient way to calculate rather than use a formula. So you should understand what the definition is. You know, intuitively about that. It's performed. I'm going to leave the following exercise to you. Exercise. Check that with this definition of normal ordering. Okay? The equation of motion holds inside normal ordering for an arbitrary product of operators, not just products, two operators. The equation of motion holds as a genuine equation. And therefore, for instance, inside normal ordering, the correlation functions of del X are genuinely analytic. Are genuinely analytic inside an arbitrary normal order. These words may sound a little confusing. It's always nice to have an equation to replace words in a definition. The words that I just said can be written down in the following equation. Suppose I've got some operator O, some product of operators. This is not mostly local. It's product of operators at X, Y, X, X, X, X, X. Okay? I'm going to expand that in the normal order of C, I'm going to expand that in the normal order of C, I'm going to expand that in the normal order of C, I'm going to expand that in the normal order of C, this. Okay? So what do I do? What I do is what I have to do is what I have to do is take this 2-point function and subtract it for any contraction. Now, the 2-point function was minus alpha by 2 log of mod Z squared. So subtraction of the 2-point function is plus alpha by 2 log of Z1 minus Z2 log of Z2 squared d2 Z log of d2 Z2. Okay? And let's go delta by delta.