 We're heading up really to the study of the four dimensions of crucial importance to the real world, because we've been the strong interactions in the real world in a long time. Now, one of the things I want to do is to compute the beta function, the one loop of the beta function. And while this calculation is in principle about as simple as that for the 45 to the 48, remember we've got this factor of 2 by 16, 16 by square. I think there's nothing really more to it than that little much of a calculation. But it's mild and extra technical because of these issues. So in preparation for the study we're going to perform, I'm going to have a little technical lecture on this thing called the quantum effect of action. This lecture is really taken almost entirely out of chapter 16 of Weinberg's, yeah, chapter 16 of Weinberg's of Weinberg's quantum effect. So which is very nice to the chat. Sir, why do you think the question is not about the quantum effect? The exact normalization equation is not valid to gauge theories because the way that we set up the calculation, we had a cut-off momentum space. Now, that's a very dangerous thing to do for the theory, for all the theories. You see, the gauge transformation is delta a mu is equal to del mu suban plus a mu. Suppose we put a cut-off momentum space. Suppose you try to put a cut-off momentum space in saying that all our gauge fields have momentum less than a certain amount. That's not a gauge in that instance. Because if in one gauge, the gauge field has momentum restricted assertion, it's not the gauge depending on what you're doing. It can have momentum outside assertion. So all those up-to-flow and everything, do you have to think about this? No, no, up-to-flow is correct. It's just a technical issue. Nobody has managed to formulate as beautiful a set of exact normalization group equations. Of course, you think they exist. This is for the lattice by some block spin and something like that. It exists. Well, it would be very nice to have something as lovely as the Poichet's key equations for gauge theories. I don't think that's been achieved, but you should think of it as conceptually existing. So the problem is really putting a cut-off. The problem is in regulating gauge theories in a manner that preserves gauge theories. So we have this beautiful finite regulator for the scalar field just by taking the propagator and modifying it. And that's another way of saying it. If you're just trying to modify the propagator of a gauge, instead of without, you can't, the propagator is part of f-milk. f-milk, you can't fool around with the propagator by itself. That will violate gauge theories. So the good gauge theorem regulator, the really useful gauge theorem regulator we have is this thing called dimension theorem, which is very nice but doesn't have this finite version. We'll discuss this as we study gauge theories. But that's the main issue. The main issue is that it harder to regulate gauge theories in a manner that preserves gauge theories. That's one of the most main motivations by introducing a large gauge theory. It's an explicit gauge theory of the gauge theory that we discussed. Okay, fine. Sir, when you're relying on this exact interpretation, you ensure the partition function was in the end of the state. So when you're not doing it by hand, when you're cranking it, it's just the one move type. Sir, how are we sure that partition function was in the end of the state? Because we don't want anything by hand. We don't demand any condition over there to make partition function in the end. Well, you have to understand what you're doing. It's a question of what you mean by the randomization. But if you are referring to the kind of thing people normally do in textbooks for randomization, what we do is to tune the bear action. So you put some sort of scale in the bear action. Either that's a cut-off or it's a very regular scale. And you have some bear parameters. And then you try to tune the bear parameters as a function of the scale so that something finite is unaffected. So it is the same thing because, you see, what was not important was not that the partition function was unaffected, but that the generator correlation function was unaffected. The correlation functions are the finite things you mentioned. So what you normally do is the same thing except that what is normally done is only at infinity. The great thing about the Ponchinsky denomination loop equations is that it allows you to go all the way down to finite states. The high-end tree textbooks normally do is just that infinity. The thing that they do is what we proved using the Ponchinsky equations. If you remember we had these finite equations and then we demonstrated that it was possible to choose bear parameters as a function of lambda part so as to keep renormalized parameters at some limit in those states. With that second part, that is normally called this. So it's slightly different, but related. The precise relation is what we tried yesterday. Okay, good. Anything else? Okay, so as I discussed today, we were in preparation for a discussion of the computation of the beta function and some physics of the Canvas series. We're going to have a technical side of this lecture to discuss what's called the quantum interaction. So what is the quantum interaction? It's a general notion of quantum field theory, like to gauge theory, scale of theory, and the notation I will use in describing it is that of the scalar theory. It's very general. So consider we have, for instance, some theory which is described as focused. Now, as you all know, this is the object of the quantum theory organization. It's often convenient to have a source to this action. It's often convenient to have a source to this partition function. If you can compute this part of the table in the presence of this source, you can then compute all kinds of correlation functions of the field that is differentiated. In particular, this is clear, right? So you take the log of that differentiated with respect to many j's and you get done with the correlations of those parts. So this is, of course, a useful option. In particular, okay, now over the world result. Obviously, it is useful to write this z of j, whatever it is, as e to the power w. Why is this useful? It's useful on diagrammatic. It's useful mainly for many reasons. But the first motivation is diagrammatic. The point is this. This is a partition function in the presence of some source. Now, a partition function gets contributions from all kinds of diagrams. However, the contributions from diagrams that are disconnected to each other can be dealt with in a rather trivial way in the sense that there's a theorem that says that the set of all diagrams is simply the exponential of the sum of all connected. Now, all of you know this. But I just illustrate this in an example just so that you see how it works. Suppose we had pi to the power 4 theorem and we had a diagram like this and another diagram like this, for instance. So what I'm going to do is to first compute. It's just an issue given to symmetry factors. So what I'm going to do is to first compute. I'm going to be able to do this first compute. Maybe I should take a moment. Let me start. Let me start with three again. So suppose we wanted to look at this diagram. Suppose our action had lambda 4, 5. So what is this graph? So firstly, what is the value of this graph? So what do we have to do to get this graph? Just one of one of these. We bring down a factor of minus lambda 4, 5. That comes with the factor of 1. And then we have to contract two 5s to two other 5s. So there are four 5s in the game. We choose two 5s to contract with each other. We choose any two of these 5s to contract with each other. These can be chosen four into three. Then we contract the remaining two 5s. There's no choice. So if I have this as lambda 5, 4 by 4 factorial, this graph would appear with factor 4 into 3 by 4 factorial. So it would appear with a factor of minus lambda 4. And then with the final integral of this graph, which of course you know very well. 1 over x squared plus p squared integrated. So this would be the factor for this graph. Now suppose I wanted this graph. Suppose I wanted this graph. So then I would have three factors of this graph. What does that come with? Because when expanding an exponential, it comes with 1 over 3 factorial. So it comes with minus lambda 4 u by 3 factorial. And then each lambda 4 comes with this 1 over 4 factorial. Now for each of these, I have to do the same thing as I did before. So for each of these, I will get the factor of 4 into 3. So this will make this lambda 4 by 2 cube by 3 factorial. All think perpetrators. All think perpetrators. See, there is this way of thinking of these things using symmetric factors using these permutations. I am not doing this, I am just counting them. So out of this for each set of vertices that we would for one graph, but because of the 1 over 3 factorial that from expanding the exponential, we got this 1 over 3 factorial and that tells you that the whole thing was expanded. Different graphs, actually it is the same thing for the sum of all that. I think a particularly simple example because I had three I think you could rework this thing if you had. So suppose we do an example where we have one graph like this, one graph like this and then we looked at all the cases this plus minus double. If we looked at all of that, you can convince yourself and leave this in the exercise for you. But you would just get this plus this double because you think about it is not a different and the standard lies the same. It is a simple combinatorial problem if a connected graph, even if an initiated produce with the right combinatorial factors the sum of all graphs connected is not a different. This w j, this w j is an object, this w j is wj is an object that can be computed by summing all connected graphs, by summing all connected graphs that contribute to this function. And that takes all the information of z of j. For many purposes, it's a more useful object, you know, z is, it's the logarithm of z of j. As you remember, whenever we compute correlation function, we differentiate the log of z of j. So, we differentiate w of j. Now, w is a more, more useful object than z. And so, should it be s to minus wj to compute the function that is the nature of minus h? Yeah. Now, this is a method because of convention. Certainly, when we talk about the effective action, we'll call it minus gamma. I was trying to follow a white work. Unfortunately, white work works in Minkowski space. So, I'm not even sure I'm using the same as its connections. I mean, I'm using some convention. This is up to you. Maybe it should have been minus. Okay, excellent. Now, sir, we get the similar graphs in the series. So, from there, we can see that w j is some of the connected graphs. How do we use that? Okay, in this particular example, you saw that if you took all graphs, which are this thing tied to the power curve, in computes of this, that will contribute simply what one of these graphs contributes to the power end development of the curve. But that's exactly what you would get by taking this particular connected graph and exponentiate it. So, this particular example you see how it works. So, now you want to see how it generalizes. Well, I did not celebrate. And then, like, what I can understand is that you're taking these similar copies. So, as a whole, this is the disconnected directed graph. Yes. So, what I'm saying is that this w j is computed by summing all disconnected graphs. I mean, sorry, summing all connected graphs. So, sum of these, which is one connected graph, plus this, which is another vector. So, you draw all connected graphs that you can. Okay, excellent. So, let's go on. Cross disconnected, no? Product of the two disconnected graphs. Just of all the connected graphs, you see. Yes. And then you do this thing where you add one connected graph plus another connected graph. You say that this will be just like the sum. So, you will have cross graphs. You will have cross graphs. So, suppose the answer for this graph was eight. The answer for this graph was, okay. Now, you definitely have graphs that perform square of two copies of this times one copy of this. This is there. It's not zero. You want to know what is the other? How does it come to this? It doesn't matter. It doesn't matter. Because the full answer is a plus b. So, the expansion of this gives you all the cross graphs. For instance, if you wanted the graph with two a's and one b, you expand this. So, you will get a squared by two b. Yes, yes. So, that's the point of this whole thing. Summing all connected graphs, you see, to compute a partition function, you need all graphs connected as well as this graph. But if you sum all connected graphs and exponentiate the answer, that generates all disconnected graphs. Okay. All graphs with connected, disconnected. Is this clear? So, wj has all graphs. Wj has all graphs. It's a sum of all connected graphs. Z of j has all graphs. Z of j is the exponential of wj. Excellent. There's an interesting object. In another object that is built out of wj, that is, perhaps even equally interesting, maybe more interesting. Also very interesting. And that other object goes as follows. That other object goes by saying, let us consider wj minus j times phi cn. To be a bit careful with notation. We had the integration variables phi in the original action. Those were some integration variables. We integrated them out and w is a function only of the sources j. If phi cn, logically, has nothing to do with the integration there. I'm using the same symbol for that cn. Okay. It's just some number, some function. This object here, as it stands, is a function of j and phi. It's a function of 2x. Interestingly, a trick in mathematical physics, that all of you know, or anyone studying thermodynamics, knows about it, let's call it the Legendre Transforming, which is another way of saying, take this function and regard it as an action for the variable j. So, extremize it with respect to j. So gamma of phi cn, this quantity, w minus j phi cn, okay, w minus j phi cn. It's a function only of phi cn, because you think of this as an action for the variable j. Extremize j and evaluate w on the solution to the equation of momentum. Let me show you this is what we normally, maybe these are not the words that are normally used to describe the Legendre Transform. Let me show you how that works. So what do I mean in particular? . We may or may not have wanted to do it. For the purpose of regular correlation functions, it's true, you have to reflect it. But on the other hand, w of j itself is an interesting function whose another way of saying this is that Taylor's expansion coefficients of that are the coordinators. But I am now treating w of j as an object in its all right. It's functionality. . Why? You see, that's the point. No, this stops you from doing what you want. The question is, is it used? It's so easy. So what do I mean by treating j as a variable? What I mean is after, differentiate and set the derivative to 0. So the equation I get from this is that there at w by l of j is equal to 5 here. So this business about treating this object as an action project and then it I mean on its minimum, on its selection of the equations of motion is equivalent to saying that we have to, we want this object once we substitute for j by solving the equation. The usual statement of the genre. The next question that I want to ask is, hey, this is nice. We've got a way of getting this 5 c l from j with a drama transform by doing this. This is the relation. I want to ask is that an inverse of this relation? Okay. j w of j once and no 5 c l. Once and no drama of 5 c l. As you see, because you see, suppose I take the object, drama of 5 c l, j tilde, this once again is a function of j tilde and 5 c l, function of 2. But I am going to define a new function of w tilde to be this object as a function of j tilde evaluated on the solution to the classical equations of motion. Of that object regarding 5 c l as a relation. Okay. That is this way. Now I'm going to show you that this process gives w tilde as equal to w. And so it was obvious. Why? Because what was the definition of this? It was this. So this is w of j minus j 5 c l. So minus j minus j tilde of 5 c l. Minus j minus j tilde of 5 c l. Where, what are we supposed to do? Okay. Where we are supposed to take this object and vary it with respect to 5 c l and with respect to j. Once I vary this object with respect to 5 c l, it gives me the equation j is equal to j tilde. Once I put j is equal to j tilde of 5 c l. Moreover, since I put j is equal to j tilde, I can substitute this with respect. As j tilde, the varying with respect to j does nothing. And I've got the answer. Okay. So what I'm trying to do is that it's obvious that this object here, as a function of j tilde l, is the same thing as w of the function of j. Okay. So, Lejeune the transform should be called as following objects, add a source, type something, and then make equation of motion with respect to that source. Okay. And when you think about it this way, many of these, you know, variational kind of properties become totally obvious. See, variational. Okay. This is our definition of gamma of 5 c l. Any questions or comments? I can't do that, but the point is that it's the only thing. Now this is a function just with this j. Yeah, right. The new definition is that it's a Lejeune transform of 5 c l. So now let's move on. So this gamma of 5 whatever it is, just a Lejeune transform of 5 c l. But there's something very interesting about this Lejeune transform. And that goes for this. You see, what was w of j? W of j was what you got, I mean, it was the log of what you got when you did the full partition function here. Okay. Now I'm going to ask a scenically strange question that may be even interesting. Now the scenically strange question is, suppose I computed some other z-tilde of j. Okay, suppose I computed this z-tilde of j. Okay. Why? And I defined this object as exponential of minus gamma of 5 plus j is 5. The whole divided by some j. A partition function. A partition function. So I did this very strange thing. That is, I considered a path integral generated by the effective action gamma 5. I consider the path integral generated by this effective action gamma 5. That will place in the action. And then I also put a one of a g here, you know, like this is like h part. Okay. What I want to know is, what if any is the connection? What if any is the connection between this partition and the original partition function? That's the question that I ask. Is the question clear? The original partition function there is equal to each of the power w of j to any limit in which we will get, right, a path integral. That's a function of j and g. Is there anything we can do with j or g in order to get something like w of j? G goes to zero. You see, because when g goes to zero, several points work. Then you have to differentiate this thing with the strength of 5. But that is the definition of w of j. I think I want, you know, I want gamma to be like an action. I guess this is going to be my action. I want gamma to reduce to s. Okay. Now you see, therefore, okay, in the g goes to zero limit, this thing would simply be, it's simply exponential of w of j by g plus higher value than g. Graphs contribute to the limit. G goes to zero. Suppose we're doing a path integral like this. We're computing the process. What graphs contribute to the limit of g goes to zero? Only three graphs. You see, now, right? G is a loop count. If we sum all three graphs for this action and compute the generating function, we get the same answer. We get the exact answer. This 1 by g, of course, is a formula for our meter. It would appear in all the three graphs. We don't want to take the resistor loop count in the end of z. 1. It reads the conclusion that if you use gamma of five in place of the action, but some only three graphs from gamma of five, then you produce the exact generator of correlation. Let's take this again. Gamma of five is the object with which if you do classical physics, you get the exact word. So gamma of five is called the quantum effect of action, and we've already seen one rule used on it. It is the object such that if you do classical physics, that is, you sum all three graphs from it. You will produce exact generator. Let's imagine we're computing some w of j of five in the book. So suppose we've got j, j. So that this is five. No, this is five. Five legs. It doesn't matter. So I've got some graphs like this in fact. You don't have to contribute. This is a nice connected term. All these external lines are these j's. We conclude it that we would get the same answer from w of j from gamma. If we consider only three graphs, do you say that this implies that gamma is generated by the sum of all one particle in reduced programs? So this example, suppose I shrunk this whole graph here to some vertex. I shrunk this whole thing to some vertex. Then three diagrams in this vertex would be, okay, so this can be thought of as a three graph where this guy as a vertex and this guy as a number vertex. Any graph can be regarded as a tree whose once you shrink all one particle irreducible graphs, graphs that cannot be cut into a sum of two disconnected graphs by cutting one way into a reactive vertex. So since we have concluded that we get the full exact answer for w of j by doing trees with vertices coming from gamma j. But we also conclude that we get the full exact answer from w of j from doing trees whose effective, with a process of effective vertices of one particle irreducible graph. It follows that gamma of j, that gamma of phi can be computed by summing all one particle irreducible graphs and whose external legs are now interpreted as just five bits. Prescription for how to compute gamma of phi more effectively than actually computing the Legendre transform. We draw all final graphs. Keep only those that are one particle irreducible. Allow external lines. But the external lines are just are associated with the phi classic. And the sum of all such graphs is arbitrary numbers of insertions of phi classical. But phi classical is an arbitrary function. Again, gives you gamma of j. In my final diagram, we often think of just this phi classical momentum space. That's if you grab the momentum, you grab the function. Is this clear? In particular, you know, in the process of computing phi of gamma, you do not insert propagators on extra. You see, because the propagators have been taken into account. But now these vertices are no more psycho vertices. They are something else. There's something totally different. There will be in general all orders, whatever. But it's the object, okay? The vertices of gamma of phi is that object that you get by looking at all one-partically irreducible graphs, allowing external lines. And for those external lines, you do not put propagators. Those are just external lines. You put a propagator at the level of gamma. What? You put a propagator at the level of gamma. And you connect two lines to external lines. You connect this. You say that's it. From gamma. The vertex of gamma times the propagator. Times the vertex problem. Because the propagator, it would be overcount and you would have propagator ink. Okay? Also, when you connect, when you compute w of j from gamma, you will have the j times phi. And that process will put a propagator here. So gamma itself does not have that propagator. Okay? So we can... But those two vertices are not same. What? Maybe different. All things are different vertices. Okay? So what you have to do is to sum over all one-partically irreducible graphs with external insertions of phi's. That gives you gamma of phi. Even more elegant way of saying this. And even more elegant way of saying this. Suppose in your original action here, in your original action here, you put phi plus phi c. Okay? So suppose we do... Exponential minus phi plus phi c. Now, we genuinely did such an unse... A path integral. How would the answer depend on phi c? At least if it had no... Find the infinity. The answer would be... This... This thing that I've written out here, the answer would be independent of phi c. Because you've got to make the change of variables. Phi tilde is equal to phi plus phi c. And absorbs that into the path integral. Anything else? However, suppose we compute this thing subject to the restriction. That we keep with x. We regard this as the exponential of only 1pi at the distance. Connected 1pi at the distance. So we do this... We don't really do this path integral. We do this at the graphical level by exponentiating all the connected 1pi reducible... The 1pi reducible is totally crucial. Statement. 1pi reducible graphs. Okay? We'll find out regardless of the external variance. Those 1pi reducible graphs are exactly what we said we should be computing. Because we remember what we said we should compute. All the graphs with arbitrary numbers of external insertions. And those external insertions should be replaced by phi c l. Okay? But when you expand this, that's what you get. For instance, in phi c, the fourth year, this vertex is a vertex where these two are phi's. But these two are phi c l. These were all phi's. These were phi's, but this was a phi c l. These were... These three were phi's, this was a phi c l. These two were phi's, this was a phi c l. These two were phi c l. Okay? In any place that you... The substitution phi plus phi c l automatically puts all possible external insertions of phi c l. Exactly. And therefore, the conclusion is the following. Compute the path integral with phi replaced by phi plus phi c. You compute this path integral, you don't really compute a path integral. It is a strange thing. Of computing the exponential of connected one-partically irreducible graphs. Because you're not really doing an integral, you can't shift away phi c l. Okay? And that process, graphically, obviously generates all the graphs that compute phi c l, gamma of phi c l. And therefore, these objects compute gamma of phi c l. So gamma of phi c l, the most convenient way to compute it, to find the diagrams, is to pretend we're just integrating phi out of this thing. But restrict attention only to one-partically irreducible. This is the calculation of the mathematical shape. Mathematically sharp. Meaning it's not... You want to ask, does it go beyond multiplication there? Yes. Yes. So since I'm giving you this restriction to one-partically irreducible graphs, it's a problem. Otherwise you would have phi c l squared terms everywhere. What? Otherwise you would have phi c l squared terms, right? We have phi c l. We have all powers of phi c l. Yeah, phi c l squared are the same vertices. You would have... No, no, we do. This goes to phi c l. Yeah. At the graphical level. At any order of the loop expansion, this is totally precise. Okay? If you want me to give you a definition of a graphical analysis, it may be possible to do. But this basically is a spectacular state. Yeah. How does it change the nature of the vertices? Like in this one, I always have a 5-4 thing there. Even if I shrink the level diagram, it's a 3-level diagram. No, but this... We have an extended line also in the lower graph, right? Because now we have that. Okay? But this particular vertex was a 5-2 vertex. This was a 5-4 vertex. No. 1, 2, 3, 5-6. And this particular vertex was also 5. Let me draw you a 5. Let me draw for you a 5 to the 10 vertex. That's 5 to the 10 vertex. Did you see? But these are 5s. And these are 5 scales. This is an effective 5 to the 10 vertex. So if you could start from 5-4, let's say, and then water was... water vertices were being generated in the RG. Those vertices belong to the RG. They belong to the RG. For example, 5-2 vertex is one vision there. For example, 5-2 vertex is one vision there. Am I not... Okay, maybe... Yeah, that is actually 5-4, right? The external length? Oh yeah, you're right. I forgot about this. You're right. Actually, 5 goes to minus 5, so we do from generating 5-3. You generate 5 to the 6, 5-2, 5 to the 10. Yeah, I said wrong. Yeah. Sir, another thing is that I could understand that I understood that you could stream that loop level to a 3-level and that thing, but how making the action going... making 5 to 5 plus 5c l? How this gives me the same thing? Okay. So let's take an example. Okay? Let's take the following effective 5-4 connection. In 5 to the 4. Okay? What was our prescription? Our prescription was that we should compute this graph where these external lines were 5c ls. Okay? But now suppose I had 5 plus 5c l, some of the big score. Am I correct like this? And another 5 plus 5c l, some of this way. There is a term in this, which is 5 square 5c l. I could contract these two 5s together. You see that generally it's exactly this part. Are you understand? Yes. Sir, you have also the option of 5 plus 5c l. No, 5c l is regarded as a background field. Okay? The positive end result is being done only over 5. So you take 5c ls, set it to some functional value. 5c l is some function. Substitute s is equal to 5 plus 5c l. And in the graphs, you do the integration only over 5s. The answers are functional. Okay? Providing you some only over one particle, irreducible, connected graphs, the answer to that problem is gamma of 5c l. Is this clear? Okay? This, by the way, since Mungage brought it up, is very similar to its own universe. How is it similar? Remember, in the renomalization group, when they bring the integral over a shell, we broke up 5s into 5s that have momentum inside the shell and 5s that have momentum below. So the 5s that have momentum inside the shell are like these 5s. And the 5s that have momentum below are like 5c ls. By integrating out the momentum inside the shell, we got a functional only of low momentum 5s, like that we get a functional only of low momentum 5c ls. In fact, it's a precise relation. Provided you're in a theory with no intraride divergences. Okay? Intraride divergences get spoiled in the same way. But if you're in a theory with no intraride divergences, then gamma of 5c l, okay, is the same as the Wilsonian effective action, run down to arbitrary new energies. Okay? Why do you know that's true? You see, a Wilsonian effective action run down to very low energies. Once you've got a Wilsonian effective action at some scale, in order to compute correlation functions, you need to do the integral over all momentum up to that scale. If you run down to very low energies, you have to do basically no momentum integrals ever. And therefore, you get more or less by classical answer. But we see that gamma of 5c l is the object whose classical physics dimension is exact coordinates. Okay? So there is a tight relationship between the... between... Okay? It's imprecise in a couple of directions. Okay? But conceptually, there is a tight relationship between the Wilson, between gamma of 5, which is called the quantum effective action, and what's sometimes called the Wilsonian effective action, integrating it out, everything out, quite easily. There are a couple of inaccuracies. That's not... In fact, there's a famous study of the relationship between these two. You'll be able to run to it. Okay? Great. Any other questions or comments about this? Sir, can you give us, in fact, these points? Yes. You have to calculate some integral. Yes. What is the way you are doing it? Minus is complete, right? You see, here we're not doing anything, sir. Okay? If you put a cut-off in your theory, a large unit cut-off, you do it with that cut-off. We're just doing the path integral. Define the path integral. Like we do a quantity. Okay? And then you do it. I mean, we're not touching on that. So this is a much less subtle discussion than that of renormalization. It's trivial manipulations for combinatorial interpretation. Okay? So let me summarize what we've learned. Okay. In the second year of the year, how do you generate five-year propagators? How do you generate five-year propagators? Well, for instance, this term will contribute to... That's fine. Diagrams where you have five-year as the internal diagram. But a two-parted diagram would have five-year internal diagrams. No, we don't. There's no such thing as a five-year algorithm. That's fine, sir. It's complete. You see, how do you generate these effective vertices? And then we will later in order to generate, let's say, some correlation function, we will later get all three graphs using these vertices. So this propagator, I think, will come in that process. Three drafts for CMF, for contracting one, five, one, five... So if we had such a... You would have some effective renormalized... Yeah, you would put the... Then you would put the propagator. So you're putting it on a supplied step. Exactly. We can't... Should we go on with this? Sir, for example, in this diagram, when we shrink the loop, then the number of vertices is reducing by one plate. I mean, for example, that entire loop thing won't continue. All four, five. Yeah, there's a whole thing, six to one, five, four. But each vertex is associated with some lambda. So shouldn't the two diagrams have same parts of lambda? But from here, they are not... So in gamma, we will generate vertices with all orders in a couple of minutes. So that means something... Yeah, so this object, for instance, is a five to the four vertex in gamma. That is order lambda squared. This object was also five to the four vertex in gamma. But at order lambda, this object is the same thing in order lambda to the four. So we will generate the contemplative action. We will generate vertices, a five to the four, five to the six, and all orders in it. They will be same. What should... What is... The powers of lambda... No. Why? Why is this? It is the same thing in the same diagram. It makes... Just look at this. This is a five to the four process. It contributes to four, to both of them, to both of them, scattering. But this occurs at lambda to the four. So gamma, for instance, can be used to compute scattering processes by using... a classic. Okay, all of this, something else. Of course, in the end, we are not teaching much less in the evening, too. But if we really have this extension in the next semester, we will go through all of it. Like the LXZ procedure, you understand. Okay? We will do this kind of work. This is all classic, for instance. Okay? If you know coronators, you can get SMA process. Since you can compute all coronators from gamma, you can compute all SMA process. Okay? So gamma is another thing that generates all SMA processes, just by doing classic processes. Does anyone know how to get SMA process from classic processes? What is the classical? It's a classical analog of an SMA process. Suppose I want to... let me start to make a question. First, I have five to the four general. And I want... if you do compute SMA processes at three levels, you're never doing more than classic processes. So the three level five to the four SMA techniques or the three level five to the ten SMA processes is the answer to some classical question in classical things. Then suddenly, can you work that classical question that you have to answer to compute? Third, we need compute SMA processes. Like, at the field level, we essentially contract the files with the momentum, right? So if you think of it classical, it will, like, put in some initial conditions and some final conditions and place something like that last equation... Yes, very good. What I say is more or less right, but what do you compute? The SMA process is a complex number. What do you have to compute? Well, when you compute a transition amplitude, quantum transition amplitude to go from here to here in quantum mechanics, what is the classical? If the SMA goes to limit, it reduces the classical problem. What is the classical quantity you have to compute? Action. Action is a function of our mechanics. Okay? So, as you said, this classical computation that gives you the SMA techniques is the action, field theory action is a function of boundary values. But you have to put some boundary conditions. Okay? And the boundary conditions you put are final boundary conditions. That all of you have read about a quantum field theory and have never understood these negative energies here or positive energies there. You know what I'm talking about? No. We are in different generations. When we first studied quantum field theory, we had read about this. It was some crazy intuitive stuff. What? Where did it come from? It's in the modern field theory. But, with some appropriate boundary conditions that I won't get into it. It's computing in action as a function of stuff here and stuff there and some appropriate boundary conditions. So, if you did that action calculation but with gamma that would be exactly exactly the same. Okay, we'll discuss some of this stuff. We'll have a proper discussion of scattering. We really continue. We'll have a proper discussion of scattering. Unitarically scattering. Interesting. Okay. But let's go on. Okay. So there is one more formal property of quantum effective action that I want to tell you about, which in some ways is one of its most important properties. And there is an energy interpretation of the quantum effective action, selecting it. Let me first take the statement, and then we write it. The statement is as follows. So suppose I have gamma of phi, phi is some classical function. And I evaluate gamma of phi, OK? Let me restrict my attention to time independent, phi is. So I've got some spatially varying field, right? And I evaluate gamma of phi. That is a function, so you get a number, OK? You know, we've thought about how gamma as a function of phi, it's a agenda transform, blah, blah, blah, blah, how its vertices are important in computing, various processes. But the question that you might ask is, well, that Rona alluded to at some point. You know, what is the meaning of gamma of phi itself? And this is very interesting interpretation, OK? Oh, for phi. You see, suppose we ask the question, for a time independent field configuration, the classical, what is s of phi? The answer to that question is minus the potential time to time, OK? So if you divide by the volume of space time, because you've been translation, you know, right? If you divide by time, you get the energy of the configuration that has that value of gamma, that value of phi, you know, right? I'm trying, because that's why I try to restrict myself to time independent things, we're going to make a difference. OK, let me remind you the quantum mechanics. Suppose I have actually, I can say that either you did it, as in, it starts where I drew my video, OK? I compute what is the interpretation of this action as a function of x of x is time independent. It's a video of x up to some sign, which is the energy of the particle. That's the thing that I'm trying to say, OK? So now you might wonder whether a gamma of phi c, at least for time independent, also has some interpretation in terms of energy. But of course, now we're going to do quantum physics, classical physics. So energy of what? Energy of which, with state? States are not labeled by phi c l by some function. That's a basic difference between classical and quantum physics. That sounds like, first, it cannot have an energy interpretation of a particular state. But the answer is that it does. And the answer is that this is the interpretation. Let me tell you the answer, and then we try to prove it. The answer is that, suppose it takes all states in the Hilbert space of the particle, we restrict the attention of all the states, OK? And we look at those states such that the expectation value of phi in the state is phi c. Now, vary over all such states and find the minimum energy, the minimum energy in the infinite number of states that whose expectation value of the operator phi c is phi c. The answer, the minimum of that energy is gamma phi c. Sir, so the fact that we wrote the gamma phi c as a long term, it's essentially like our interest theory, thinking in terms of expectation. Sir, we can write the Hamiltonian as a Legendre transform. So, the Legendre transform is a classical analog of the converse of what we did today. We did the Legendre transform because the Legendre transform is just a classical physics, path integral is doing quantum physics. So, the point is that gamma is like the object whose classical physics gives you the quantum answer variable. Now, if that is the same as what you are saying about Aaron first theorem, that is right. Do you understand my claim? Before we go on to try the claim. The claim clear to everyone? It's a very powerful claim. The claim is that take all states in the quantum field theory whose such that the expectation value of the operator phi in that state is phi c. There are infinite number of states. For instance, the quantum mechanics. This would be the state. Take all states such that the expectation value of x is a particular number. The infinite number of such states are constant. All wave functions such that the average value of x is not very much. It's x naught concrete is clearly an engine number. It's much greater than t here because it's very good space. Now, so different states in this class have different energies. Now we minimize the energy among all such states. Look at the lowest energy state among all among this class of states. And we ask what is its energy? And the answer is gamma of x. What? This is a constraint. We're looking at both sides. Now obey this principle. So we're doing a constrained maximization. Is the claim clear? Now the proof is an almost trivial one. It's kind of proof where you wonder whether you're cheating yourself. That is very beautiful. See, suppose we wanted to solve this maximization problem. So what do we want to do? We want to extremize subject to the constraint that this is true and that's how it's normalized. Now, all of you know of a way of doing constrained extremization. That's the method of Lagrange multiplies. This is another way of saying you impose the constraint as a data function. That's another way of saying it. Let's say it. So what is the method of Lagrange multiplies? The method of Lagrange multiplies says that you add a Lagrange multiplier times the constraint. So suppose we add a Lagrange multiplier, alpha times the constraint, side by side, because I'm putting the addition side by side. And another Lagrange multiplier, that's a beta x times the constraint, side by side. Now, the method of Lagrange multiplies, we can do it either way. We can use the method of Lagrange multiplies or put the data. So we've done this in minus one and this minus five C L. But the way Lagrange set the method of Lagrange multiplies, just take this object, differentiate it with respect only to side and adjust alpha and beta so that the constraints are the same thing. So you take this object, differentiate it with respect to side, that gives you an equation. And then you have to solve that equation for those values of alpha and beta so that the constraints that you're interested in are met. It's clearly the same thing, right? Because you see, if we put this minus one here, or the minus five C L here, the derivative of this action with respect to five doesn't work. Okay? And then the remaining equations that you get when differentiating with respect to alpha and respect to beta are just simply the conditions of the constraints are met. So you just adjust alpha and beta so that the constraints are met. Okay? So what is the equation we get? We get eight sides plus alpha beta of x. Okay? It's times side. They put a zero where alpha and beta have to be chosen so that these two constraints are met. So we have to solve this equation plus the equation plus an equation like this. We have to solve this equation from Dover's alpha and beta that in short of this and this. Observation is that we already know of a solution amount to this problem. Let us try to understand how. Of course, we consider this exponential, the z of things, equal to exponential of minus s of pi, the j five. If our j is taking to be time independent. Okay? If our j is taking to be time independent, then we can think of this whole thing here as some, the whole thing as a quantum mechanics of some time independent Hamilton. Okay? Where j is somehow, is a new coupling inside the curve too. Okay? This Hamilton, whatever it is, has a ground state. Clearly, w of j is equal to time or is equal to time, minus time times E of j, where E of j is the ground state energy of that ground state. Because the z of j is, you know, it is just, it's e to the power minus beta h over an infinitely long time. So only the ground state energy contributes. Think of this for instance as the limit of time being on a circle and the size of that circle going to infinity. So the time being on a circle as we have explained in the class is stress over e to the power minus beta h. Beta is the size of the circle when beta goes to infinity. Only the lowest energy state contributes. Okay? So you get, so you get just the energy of that thing times time. So if it's equal to the power minus beta h, it should be one over the other because beta h. So t was like time, not temperature. t is the size of the time circle. It's called that beta h. And the size. Yeah, okay, imagine we're doing this on a very dark circle whose the circumference is beta. Okay? So what do we know? We know that the Newtonian of this problem, okay? So that there is some state which is the ground state of this. This quantity is there with this j. Okay? Let's call that state psi k, psi 0 j. Such that the Hamiltonian of that on psi 0 k is w of 2. Okay? So the Hamiltonian of that problem is the Hamiltonian of this problem plus minus, what is this? Minus, right? Minus j times y. It's just an additional term for the potential. So I'm saying that the Hamiltonian that we're really interested minus j times y. On, there exists some psi 0 j such that this object is equal to w of j. What else do we know? Of course we can choose the psi 0 j to be normalized. Let's do that. And what else do we know about psi 0 j? What else do we know about psi 0 j? Another thing that we know about psi 0 j is that the expectation value of phi of x in psi 0 j which is the same thing as d phi exponential of divided by d phi exponential of x. Okay? It's simply del w of j by del j. Del w of j by del j. That's right, yes. And therefore, we also know that phi 0 j, phi of x, psi 0 j is equal to phi c. Now, can we have equation one? And we've chosen phi c. This was the trigger. Phi c l, phi c l, phi c l to be 1. Okay? So, this is equation two and this is equation three. Therefore, this phi 0 j is an object that satisfies our constraints, this constraint as well as this constraint. And moreover, quite intuitively, satisfies this equation. Why didn't we identify, you know, we have to satisfy this equation for both alpha and beta which satisfies us, right? So, we found a solution provided we identify alpha with w of alpha equal to minus w of j and beta is equal to minus j. It should be integral j phi. When? In the each minus j phi equation. It will be integral j phi. Yeah, right, right, right. This is integral j. No, well, let's see, let's see, let's see, let's see, let's see. Wait, wait, wait. Let me think about that. It was integral here as well. It's because we've got the because we've got the sum over all those numbers. So, there's an integral here and there's an integral here. Yeah, phi is the star operator. I think so. So, we found one solution, right? I think that this is the only solution or the right solution, the right solution. Very plausible because this was the state of minimum energy, right? Assuming that it's the unique solution, the right solution. Yes, sir. Sir, what is psi j 0 again? Psi j 0 is the lowest energy state for this Hamiltonian. The Hamiltonian that's generated by this, by the path of the Hamiltonian with j treated as part of your, okay? Psi j 0 is the lowest energy state for the Hamiltonian h minus j. We conclude psi j 0 is what we want and now what we're interested in is not psi j 0 but the expectation value. So, now we use this equation and we take the expectation value with respect to psi j 0. i j 0, h i j 0, okay? But we have minus w j there. Minus w j times 1 because this is normalized. Now beta of phi goes what? Minus j of x phi c of x because the expectation value of phi and phi j 0 is phi c f. Therefore, this expectation value, phi 0 j, h j 0 is equal to w of j plus integral j of x, yeah, of x. So that, you get my minus. Why is this incredibly in your prediction? That's quite incredible, you know, seems so far away from what we're doing. Extremize over all states, it's all right. I mean, you almost never think like that. This little formal interpretation allows you to see that gamma measures the lowest energies, the lowest energy of all states such that expectation of value of phi in that state is passive, okay? So, you see, there is value to computing not just gamma as a power series expansion of phi, something that helps you to generate time and data. There's value to actually know the number. One way, one thing this is useful for is to understand when it's symmetric. Okay, because if you find, suppose you find gamma as a function of constant values of phi, that is going to use the potential of your problem. If that potential has a dip somewhere away from phi equal to 0, you know what the reason is? Because that means that there is a state which has lower energy when the expectation value of phi is a value than when the expectation value of phi is 0. So, the question about whether you're working around the right vacuum of your theory is effectively addressed by computing the shape of gamma of phi even for constant values of phi, assuming that you're a vacuum strategy. Then many kinds of theories and truths like you just got translation in there. See, you characterize whether symmetric making is happening or not, whether where the true vacuum of the theory is by computing gamma of phi for constant phi. It's actually a very useful thing because while gamma has a formal object, it's very complicated object. It's as an object as a function of constant phi is often not very complicated. It's just a number. It's a function. It's not a functional. Okay. Now, at this point, what I wanted to do was to how hungry are you people? Let me give you an example. So, what I wanted to do is to give you an example of computation of gamma of phi and phi to the fourth theory at one loop to obtain what is called the Coleman-Weinberg prediction. But they messed up. If you don't mind it. So, let me actually complete the the equation. Yes, we did that. You'll see. The final question. Maybe you should start the equation. Yes, let's just put it down. Okay, so this is unrelated to our discussion of gate series, but it is of interest in its own right. And the last question is the use of the content. So, what is the use of the content? So, what is the use of the content? And the last question is the use of the content. It's a simple question. It's coming. I was just thinking maybe you should postpone it. Why do people feel that? Do you have the energy to start a new calculation? I'm scared that in 15 minutes we want to finish. And then we have a week's break and you won't remember. Let's postpone it. Next week there's no class. Next week I'll put it down. The Monday we got the next. Actually, couple of weeks decided to stop. Sir, we can actually, there is time. There's time to remain on the date. Two minutes. Okay, that's more than one. Okay, thanks. Okay, so the reason maybe you could just go into my question. This is usually an illustration of how the compute and gives you an illustration of the questions you're asking. That's all. This is an illustration of how the compute. This is one loop with active action. And one loop in a very nice physical context. Actually, it's also very important. It's the kind of calculation that is used to, as a lot of you know, the Higgs field has this double bed potential. Higgs condenses, but it's spontaneous symmetry. But, you know, I'm not sure if you know this, but if you take the standard model and heat it up, the actual active action of this Higgs field gets modified as a function of temperature. And there's a temperature that is large enough so that the Higgs stabilizes around this, around 5 to 0. For days of natural gradient, it does not happen. Okay, and the kind of calculation that shows you is just a finite temperature value of the calculation you're doing. So it's a very simple, very simple kind of thing. There's a lot of real physics. Okay, so we'll discuss this. And then, in the next class, we'll discuss this in either the subsequent class or the one after that. We will then do the same computation of the one loop quantum active action for an automobile engaged. And that will help us understand the beta function precisely. Okay, thanks.