 to your rotation that we have stuck on last time. The question went as follows. You see, I'm just reminding you where I was stuck and what I think the resolution is. You remember that we had, let's say, a Euclidean space correlation function, which was a T-E-Euclidean defined by G-Euclidean j and K-E-错. Let's say, I go down into 5. And you remember that what we wanted to do is to continue the Euclidean current joke, set it equal to, we want to maintain the substitution, T is equal to K-5, A-5, T is the limit of the sign. That's the continuation that would make e to the power minus h t, which is minus i h t, which is what we want. And we want to know what this continuation implies, what this continuation meant for the Fourier code. So, one way of thinking of this thing is as follows. You see, as we try to emphasize, as I sort of indicated in the last class, this whole idea of analytic continuation in time requires a certain ordering. And the evolution operator from t i to t e analytically continues to e to the power minus beta, that is t f minus t i under the operations that we were talking about. Only if t f is greater than t i. So, implicit in the sum procedure is this idea, that the final time is later than the initial time. Because otherwise, this would give you a plus, a positive sign. And then the path in technical over-intermediate states would diverge because you would have flowing up contributions from arbitrary energy states. So, that is why it is quite important to keep in mind. So, similarly, if we put some insertion here and another insertion here, the evolution from here to here is only well defined. If this time is later than that time. So, in this Euclidean code that we are talking about, we should think of this t e here as having a definite sign and we are choosing a convention so that sign is positive. Now, suppose we look at the definition of this Green's function here. That was key physical, that there is a positive t f t, everything else is not there. If we look at the definition of this object here, they say object is defined by a contour integral over a contour that runs over the radius. But if you remember what in our history, the pole structure of a Green's function was, and we had one pole here, so the poles were at omega square plus k square plus m square is equal to 0. So, omega equals plus minus i of square root of m square plus k square. So, the poles were here and here in the omega plane. Because t is positive, the contour integral that we wanted to do here can actually be of no cost complete with this. Because the integral over the imaginary part here just matches this. Because when omega is positive in the imaginary part, this guy here, negative is actually here. So, the contour integral that gives us the Fourier transform can be reverted because of the positive integral of this t as a contour integral. The continuation is like here, because now we have a contour integral, we can move the contour around. However, we want, provided we do not cross any of the singularities. One way that we will not cross any of the singularities is this. To any angle that we are interested in, as long as that angle does not exceed by 32, because once we exceed by 32, we will cross the singularity. Now, you see, now the point is that what we want to do is to set this equal to that. So, first even before setting this equal to that, the expression for this object as, you know, a function of t can be reverted as this integral over this contour. Now, we use this expression for the object. To do the, suppose we want to do the analytic continuation of t up to an angle nearby by 2, we will rotate this contour. Suppose we want to do the analytic continuation to each of the power i m, to each of the power i alpha times t. We rotate this contour pi f. That will basically, you know, cancel these two alpha factors here, because this was our positive rotation of t, but we have a negative rotation of t. So, we effectively rotating omega by a negative rotation to cancel the positive rotation. So, that this guy remains unchanged. That is, along the variable that goes around this contour, we will still have an e to the power i omega prime times t prime. Okay? Up to the overall factor that we picked up in the measure that we discussed was that where d omega going to is equal to minus e to the power, because we changed that one. Now, with this new contour and this new rotated t, you can easily check for yourself that this part of the integral continues to vanish. It is the same argument, sir. And therefore, we can just ignore this part. We can do this for any algebra in particular, but we are going to almost pattern. That the analytic continuation just implies an inverse analytic continuation of the frequency. Of course, you might find this very intuitive. You know, I think most textbooks would even bother to talk about it, right? If t goes to i t, clearly omega goes to minus i omega. Okay? Whatever that means, whatever that clearly means. I mean, this is one way about doing this, but even though these arguments usually work, once in a while they do. Okay? Okay? Okay? So, this was the clearing up from last time. Any questions about it? Now, since we talked about the analytic continuation of the path, I plan in this lecture to start our discussion of renormalizing. Okay? So, let me tell you how I plan to start this course. I'm trying to structure the course. Let's say we have about 25 lectures. I'm trying to structure this course as roughly half and half, in the sense that the first half, we've done maybe five or six. So, the first half of the course will be a relatively elementary material. It's like discussion of path attack was clearly very elementary. We're going to have two or three lectures discussing renormalization. Okay? Which is also elementary in the sense that it's just part of basic technology for that subject. And we will also then have some discussions of beta functions of particular theories and maybe of other technical issues like anomalies and instantials. Okay? That's what I hope will not take us past the first half of the class. Okay? In the remaining half of the class, what I hope to do is to turn to particular examples of quantum field theories and try to understand their dynamics. I hope to study the tough model for another in order to understand confinement of UCD in two places. That's what I mentioned. Some of the larger models to understand wills and fissure type of skating behavior and so on. Let's see how far we go. Okay? So, that's the plan. I hope we don't get, don't spend too much time with me. But let's go. Okay. So, that's it. So, in the last class we discussed this analytic continuation of the path attack. And the main idea of course was that the main idea was that we wanted to take this oscillatory path attack that was not terribly well defined and convert it into a damped path attack that was well defined. So, we decided to define e to the power minus ihd as the analytic continuation of e to the power minus hd. So, that was the basic idea. Okay? Now, I just talked a few words about how this analytic continuation works. Oh, even for that, there is actually a, there is actually a yellow with a self-unexpected payoff from this, from this exercise. It seems logically disconnected to what we're doing. It was very interesting for you. And that's the following. The quantity of power minus e to the power h, independent of any analytic continuation of quantum mechanics of the quantum mechanics illusion operator is of course quite a famous object in physics by itself. And the reason it's famous is that the trace of this object has a name that's called the partition function. Now, the partition function is the fundamental object of the equivalence to the statistical physics. If you compute the partition function, you can compute all thermodynamic quantities of an object. So, a basic object to study is statistical physics. In study, it will be properties of the phase structure, the theories, all kinds of things, as you know from your studies. Now, the payoff that we got from our little discussion is very interesting in the formula. You remember that we found a formula for what this was, between 92 states. Okay? So, suppose we have psi, that's chi e to the power minus beta h times psi. Then what we found was that this was equal to e to the power minus s infinity of the fields where they are. This path integral is done over the fields from time 0 to time beta. And then at time 0, we convalent this expression with the wave function, the wave function psi. And at time time beta, we convalent this expression with the wave function chi, chi star. That was the time of these conclusions. In order to compute quantity of interest statistical mechanics, this is very close, except we want the trace. The trace can be written in a particular basis. The basis will be the field basis. Okay? So, we can take the trace and rewrite it as integral d phi. This d phi of x at one particular time, phi of x e to the power minus beta h, phi of x, a couple of this thing here was that if you had some phi of x, some particular, let's say phi 1 of x and phi 2 of x. And this path integral would be done subject to the boundary conditions that at time t to the 0, we will have 5, at what is 1, a phi 1 of x and t to the theta, we will have 5 phi 2 of x. Now you have to remember that phi of x has phi of x. So, that's the boundary conditions. At the time t to the 0, you have the same phi as at t to the theta. So, same phi, but which phi? You will see he doesn't get to be integrated over all This generates a path integral with such that time, the Euclidean time, has become a circle of circumference. The fields that require you have to be periodic on this time circle, beautiful formula, because the formula of the Z beta is equal to this, where phi is on this space, where time becomes a circle, space is whatever it was, and all of these are, there are no out-of-traditions that are satisfied for the path integral, apart from those of periodicity. Can you do it in the last part of the video? Period. Period. You see, in order to compute the trace, we want the initial phi, and the final phi to be the same, because that's how you compute the trace. They have to be the same, but what same value? That we don't care about, we integrate over all that. So you see, now look at what we're doing in the path integral. In computing this transition amplitude, we were doing a path integral in that. We integrate over all that into the phi, and this, that, this, that, this, that. But at this time, in this time, phi was fixed. To your initial value, you're fine. Now, however, the quantity that we want on compute has this initial value set equal but not equal to anything particular. Just absolutely arbitrary. So this just becomes another step of the integral in the path integral. So this phi in configuration is integrated over, just like this one was, and this one was, and this one was. Except that it's equal to this. So that's effectively doing the path integral, not on R3 times R1, but R3 times S1. Where the circumference of S1 is written, because that's the time difference. You see? So we, we arrived at a very simple, but very interesting, very useful formula. And the formula is that the partition function of our, the partition function of our quantum system of quantum field theory at inverse temperature between time is given by a very simple path integral formula. The path integral formula is take the, take the action, take, take a, perform the path integral of the system after Euclidean continuation. And on the manifold, whatever your spatial manifold is, R3, that's what you're interested in. Times S1, where the information of the temperature goes into the, into the circumference of S1. And the circumference is bigger. Since time is kind of closing in on itself. Yeah, well, this is euclidean time. Okay, it's analogous to space. Exactly. There's no issue of causality. You know, causality has to do with the light code. Light code is an inherently Minkowski instruction. All of the light code was your point in Euclidean space. Because those, those points that are, those distance from another point is zero. And Minkowski, Euclidean space, there's no R. Okay, so causality requires some discussion of the knowledge. Okay, yeah. Okay. Other questions about. Okay, excellent. Now, since we've got this far, I will go through one more little, little diversion. Which is to, which is to try to remind us about, try to reduce how the same object here, how the same this set, how this, this, this, this formula or an analogous formula of this sort, how this formula or an analogous formula of this sort works for, works for, for me. Now you might remind, remember, where I definitely, before the next class, had your problem set. So this was one of the problems. And we half worked out class, then I asked you to finish this derivation for how to do, for me on the part of the table. So you can find this IHT. So I'll ask you what the problem is like too. We'll do some new definitions for that. But you might remember that we could build it in class if we had a two-state system. You remember we have this operator's, this, this operator's psi and psi, which were raising and lowering operators respectively. And we had the special state psi, which was, now I have to remember which one it is, psi down and psi plus half volume. Yeah, we have the special state. So, this guy was the lowest operator. This guy was the reason. And our special state was psi, which was down plus psi times up. You remember that we also defined, this ket psi, which was not the ket corresponding, not the dual of the states. But it was defined so that psi times psi was equal to psi greater than psi. And the nice thing about this, this Kepler-Brahler notation was that there was a completeness relation. Because of this completeness relation, it follows that the trace of any operator, the trace of an operator A, the trace of any operator A is the same thing as trace of psi, psi integral psi. The trace we can take is that side. So this is just integral d psi psi d psi. I did this wrong because I took something through something else. In fact, that's the point. So, let me start this again. Okay, yeah. Let us, the point was to try to determine, the point was to try, this is true. So this is a pro-relation, let me insert it and say, the point of this exercise was to try to determine the relationship between the trace. Okay, what is, okay, the question, what is psi A psi? This is a very easy question to answer. Nice two states. So we know what psi is. And we know that this, we also had an expression for this side prime. But actually the only important thing about this side prime and the facts that we have checked was that psi prime on down was equal to one and psi prime on up was equal to one. So psi prime on down was equal to psi prime. And psi prime on down was equal to one. These two relations allowed us to, of course, dissolve what that psi prime was. I don't remember the solution. These are all that we made. Okay, so now what we want to do is to determine this object. So we just expand it out. See, this is d psi psi A acting on, A acting on the sides. This was down plus psi times up. And A acting on down. A acting on down could either give us down or up because A sum up if you operate. So A acting on down would give us up. Then we would have up times psi. But up times psi is one. So the term A acting on down gives us up. The term A acting on down gives us down. That is going to be because A, because psi on down is psi. According to this formula. And d psi times psi is one. So this whole expression here is equal to A down down. The matrix element of A acting on down gives us up. Is this clear? On the other hand, let's look at this object here. It's the reverse. A acting on up giving down will produce another psi when we take this in a product. And since psi is square is zero, that will not happen. Therefore, when this guy acts on this, the only thing that contributes is A acting on up to give up. But A acting on up to give up, produces, then gives us another minus sign. Because psi prime with up is minus A up. The interesting relationship is that A starting is equal to trace of minus one to the f times A. Where I define this operator f. Let the operator f be defined in such a way. Operator f is defined as f on a two-state system. Let's say we're acting on the basis of f on down and f on A. It's sigma 300 in this two-state system. Now why do I call it f? I call this operator f. Because f is supposed to remind me of something. Why do I call this fermion? This is because in our application of this path integral, we're going to apply it basically to the Dirac system. Now the Dirac system, let's say the free Dirac theories first. It's just a bunch of these fermion and harmonic oscillators. One or a few depending on how many damage are there. Is it minus one to the s2f or just f? I think the eigenvalue is minus one to the s2f. Minus one to the s2f and being on down is down. Yeah, you're probably right. Yeah, I think you're right. So I suppose that means... What does that mean? So you write f on time. This would be the definition that f on down is equal to 0. And f on down. The eigenvalues of f... Yes, exactly. The eigenvalues of f are 0 on down and 1 on down. This is what we want. This is what we want and this is what f is. Exactly, thank you. Now as we said, we're interested in working this out with the Dirac system. And the Dirac system is just a bunch of fermion and harmonic oscillators. One for a few for every moment. One for every moment. As you know from your study of the system, at least for the positive energy states. Yeah, for the positive energy states. So when you've got no fermion there, this is a down state. The up state is when you have a fermion. You say, why was two state systems... Why were two state systems of interest a quantum field theory? Two state systems were of interest a quantum field theory. Because all fermionic field theories are a bunch of two state systems. So for every momentum board, you either have a fermion or you don't. Is this clear? So from this way of looking at it, f is the usual fermion. It's a kind of... just counting how many fermions you have. Okay? And so the trace that we want is trace of minus one to the fermion number. Because you can easily check for yourself that now we're not doing a two state system. But two two state systems. But here the hypothesis thing occurs because of the absorption of fermion. No, no, no, this is not real spin. This up and down has nothing to do with physical spin, if that was the answer. Up and down should just be the way the gardener is. Occupy or unoccupy? You see, when we quantize the fermion theory, we've got this creation of that. It's fermion and creation of that. We've got c. That way a, a, a, k prime is equal to that of k prime. Okay? And the vacuum is the state in which for every positive energy, and all these are positive energy, they're all annihilated by a k. Then we can make states by acting with a k dagger. So every k, we've got one of these fermions in two state systems. Okay? For each momentum state, the unoccupied state is what we have here, we've got it done. And the occupied state is what we've got done. Nothing to do with, with spin. Is this clear? At point of view, this is of course the normal definition of fermion. Just how many fermions you have. Why fermion number in general in the quantum field here is not a consultant model. These are processes that take two fermions to form. Minus one to the f is consultant. This is because fermions are always destroyed or produced in place. Just because all fermions have half intact integers. Well, the things that they can decay into the bosons, always have integers in them. It's never possible to have one, just angular momentum to be conserved. If one fermion decays into bosons. It says it's only a conserved quantum number at three quantum fields. Minus one to the f is a conserved quantum number in all quantum fields. Basic formula here is very important. This basic formula here is very important. Okay? This basic formula here is very important. It tells you that when we try to compute this quantity. Psi A Psi by identifying the initial and final. So what we conclude is that the analogous thing. The analogous thing. That is the Euclidean qualsentation for fermions. With periodic boundary conditions on a circle of circumference beta with periodic boundary conditions. Does not give you trace into the form i as beta h. It in fact gives you trace minus one to the f times e to the power n. I understand. If the acting on the states tells you whether there is an occupancy or there is no occupancy. Right. Raising into minus one to that power. Minus one to that power is just because you know what we wanted for some operator. That gives sin plus one to and sin minus one to. So let's see. Minus one to the power zero is one. But minus one to the power one is minus one. Right. Okay? So but there is of course a much simpler way of thinking on this. This operator minus one. It's just an operator which is one if the total number of fermions in the system is. And minus one if the total number of fermions in the system is one. Now this two state system that's equal to what I said. If this is the total number of fermions can only be either zero or one. But now we can do two state systems. For each of them this derivation will go through. And we would get minus one to the power f one plus x two. But f one plus x two is two total fermions. Okay? So we will always look out think of it as a total fermion in the system. Is this clear? Yes. So minus one to the power f is just something that grades inverse piece. Into even and odd parts. And multiplies the odd parts by minus one. Is this clear? So here it's clear that now minus one to the power f just. But why would it not appear in trace? Why would it appear? This is a matter of you know between you and God. All we can see is what does happen. Why is it a tough question? Okay. How is the physics of different if you have an odd number of fermions or an even number of fermions? See, see, see. That's it. So what have we seen? We've seen that there is a formula relating this integral to trace of minus one to the power f times a. Is the algebra that gives you that clear? Yes. Oh, the right answer. The right answer? Yes. Fine. So all that we have is the uses of all that. If you do a path in time. You see it's just a technical observation. Okay. A technical observation that can actually be done. That's a technical observation. Okay. A technical observation is that while path in time flows with periodic boundary conditions. Yes. Are you completely uncertain? Compute the thermal partition function for both ends. Path in time flows with periodic boundary conditions. You know how to compute the thermal partition function for both ends? They compute something a little different. This is not to say that, you know, there's some inherent difference. I mean, there are differences whether they're from your numbers or a net spin can be zero. Can't be zero, for instance. But that's not for important, of important. It's just a simple technical observation. So can I, I mean, this is also just an observation. So can I say that when the number of fermions in the system is even? Yes. When the partition function is similar to that of bosonic system? You can't say that because the partition function sums over all fermions, you see, even and not. Okay? So we, we're not, if the total f was fixed, then it's a normal resistance. Okay? And this f was fixed, number of electrons was fixed. Then this would be trivial factor. It would be an overall factor. But the point is it's much more serious. Because you see, when we look at the partition function we've summed up those days with one fermion or two fermions. Three fermions or four fermions. And we add all those contributions. And so it's relative signs between these contributions. That is much more serious than the overalls. Is this clear? Okay? Well that's, that's supposed to be considered n for where we pair as like, like, like what we did in statistical mechanics was like, and for, for an n particle, it's still, it was z1 raised to the power n. Right? I mean, the single fraction, partition function was raised to the power and so does the same thing follow over here, isn't it? Well firstly, the single particle function part of the power, part of the partition function raised to the power n is overly correct for non-identical particles. Okay? If you had identical bosons, it's more complicated. As opposed to distribution. Okay? Similarly this will lead to conformity. Okay, excellent. But now, of course, the thing that is often of interest to us the thing that is often of interest to us is computing not minus 1 to the power f times e to the power minus v dih, trace of that. Although if you look to people who do supersymmetric quantum field queues, there is including myself, you might think they've forgotten. They often are computing this. This is easy. But the thing of physical interest is this guy. And in the thing that somebody is to this thing, somebody who is turning on a temperature in a lamp, we want you to compute. Is this without that? So, what do we do? Suppose we're going to pass in that room that computes the guy without this minus 1 to the power f. And I'm going to be... Okay, what do we have to do? Well, we make progress because we know that we know that this object is completely near. So, why don't we do the following? Why don't we try to compute... Suppose instead of trying to compute trace a, we try to compute trace a times minus 1 to the power f. Suppose we compute the psi psi d psi from the minus 1 to the power f times minus 1 to the power f is minus 1 to the power 2f. And 2f is always an even integer. That's 1. Minus 1 to the power 2f is 1. So, instead of computing the trace of the equilibrium evolution operator, we want to compute the trace of the equilibrium evolution operator times minus 1 to the power f. We would have computed what we're physically interested in. Okay? Who is to compute this object? We want to compute V times minus 1 to the power f because it's really equal to the partition power f. Psi wasn't equal to... We need to remember what's on the side. So, down to this. Minus 1 to the power f on psi. So, minus 1 to the power f on psi doesn't change down but changes up to minus 1. So, the important minus beta action tells us that if we want to compute the partition function at finite temperature of fermions, all we have to do is to compute the Euclidean partition function. All we have to do is to compute the Euclidean partition function. Okay? On a circle of size, but to make sure that the fermions that we're almost on have periodic boundary conditions on this one. All fermions have that make partition functions in terms of quantity-theority-type boundary techniques. You compute a partition function on a circle but flipping boundary conditions of the bosons and the fermions. The fact that you have to flip boundary conditions for those interested in super symmetry means, of course, that even if we underline theory of super symmetry, the thermal partition function will not be computed by supersymmetric boundary techniques because the fields that go to each other goes out to the fermions and different boundary conditions to create the symmetry. This, of course, is necessary for getting supersymmetry to protect things if I like them to protect things. Sir, this is the aspect of the anthropometry as described by Probably, it's empty. That's probably true. You know, the the many minus signs, some some but this is the algebraic way I have I have to be right with it. I'm sure you can find some. Use that minus one to be correct so then I will have the periodic Yes, if you want to I can compute thermionic potentials with periodic boundary conditions as well as with different quantities, exactly. You know, any quantity that is sensible, you can compute any part. The question is, what is it? What are you computing? So, you know, from the point of view of parking technicals, or the question usually is what is the Hilbert space interpretation? So, if you compute with periodic boundary conditions if you compute with periodic boundary conditions then it's a fair computation and it's actually often very interesting this one is called the Witton index which is very interesting but it what it is is not it's minus quantity if you are interested in that quantity that the path of technicals that computes is the one with periodic boundary and there are often reasons to be interested in that quantity okay, there's nothing wrong with that it's just not the one that computes the thermal partition function Is this correct? The thermal partition function is computed by this one Is this correct? That's a good question It's a continuation of fermionic multi-technology Before we start our discussion of perturbation theory in the green organization Oh Sorry Cleaning up from last class I may be late let me complete this and I'll just go Let me discuss fermionic multi-technology in class We came to the following conclusion We came to the conclusion that that our evaluation operator T is equal to minus quantity the matrix elements of this operator were given by this path This is what we have to do state system but of course it generalizes once you've got the basics n copies of this n copies of this This was our basic fundamental formula for the connection between evolution operators the connection between evolution operators and the ferromagnetic If we wanted to compute If we wanted to compute e to the power minus h minus d i just like repeating the derivation repeating the derivation will very simply give us this part won't change because it comes from inner products between states that we inserted all this integral but this part basically minus i will be minus minus h and go and show you understand all this very well I thought there are two things okay path integral expression for the matrix elements of this operator I thought once I would say this just take it explicitly through how this works how this applies to a particular Dirac system just so that you it's clear right okay let's see how this works for the simplest Dirac system namely the Dirac equation the Dirac system for massless real in two dimensions okay I'm looking at one plus one equals two dimensions just so that I can show you everything without writing any matrix any gamma matrix because it's so simple to write and then the physics becomes totally clear okay so let's remind ourselves for the Dirac system the Dirac equation and so on but you know very well that I'm teaching you this more I'm reminding myself okay okay yeah so we've got this there is the equation del slash minus m psi zero now there is either this equation or some i's from the place and that depends on what your size signature space times let's check the way to check is one way suppose I've got this this would be i t slash i z psi zero and then you multiply this equation by i p slash minus m plus since this is equal to zero it's also equal to i p slash plus m on size now to expand this out the cross terms the non cross terms are i p slash that's i p slash so you can get minus space and then p slash takes p slash but then using the fact that dama mu anti complicated dama mu is equal to two g mu nu you can of course this is minus p squared and this is now minus is this clear what part is this this slash is p mu p mu so it's p squared it's p squared pi the minus side it's just minus from the i p mu gamma p mu gamma mu gamma nu then you do because this is something symmetric you write this as gamma mu gamma nu plus gamma nu gamma nu by two so that's g mu nu nu by two two g mu nu by two this equation here in our side convention is the correct because it's minus p zero squared plus p zero squared minus p i squared minus n squared in size zero is the right muscle equation so we've got the right equation we need to get the right Lagrangian now we need to get the right Lagrangian so the Lagrangian is either the x slash minus n psi or this with an i basically the equation of motion has to be the right one so when you knock off psi value you should get the right equation of motion this comes with i to 1 over i 1 plus i to psi after some psi in the line okay now let's try to figure that out this of course might depend the dimensions one let's try to figure this out in a particular case dimensions for the particular real two-dimensional programming system we'll understand it two dimensions one plus one dimensions and we have to remember we have to work with gamma matrices such that gamma i dagger is equal to gamma i and gamma zero dagger is equal to minus gamma so we want Pauli matrices or Pauli matrices times i okay let's obey this relation so now so now there are this is really one plus one dimension there's only one choice I mean one of the natural thing to do gamma one is one of the sigma matrices gamma zero is i times z on the sigma matrices but you see one of the sigma matrices is special namely sigma two because it's purely imaginary so we can make Dirac's equation pure completely real of gamma matrices let's suppose we choose gamma zero is i times z and the gamma one is this nice choice then then Dirac's equation is completely real which basically is the observation that is consistent to choose your spin over here because your equation wasn't real you demanded your spin over is real you find that there are no solutions to the equations because to do the equations real you choose your spin over here consistently the question yeah size of two size of two you need at least two because you need two matrices that don't become real to each other and certainly two dimensions to be given so you need only two matrices so with this choice let's say that size is equal to size one and size two let's work out the Dirac's equation okay because our our tangent was real our spin aside was real psi bar is just psi 10 so all we have to do is to work out the matrix so we've got psi transpose gamma zero gamma zero plus gamma one this is the thing we want to work out okay but gamma zero is i times sigma two gamma zero is square it's minus one, because sigma two is square it's one so this is psi transpose that's minus one that's gamma zero and gamma zero times gamma one so that's i times sigma two times sigma one now sigma two times sigma one is minus i times sigma three so this is minus i times sigma three so that's sigma three so that's plus sigma three times that one that's i now this one to be written as a two plus two guy is minus there is zero plus there is one zero zero minus there is zero minus since it's diagonal this can be more simply written psi one minus there is zero plus there is one psi plus psi two minus there is zero minus there is a master let me get back to this good so we've got the master as well because how do we have to okay so the master is now the master is just minus i sigma okay so it's plus i it's i times minus i times sigma two that's sigma two okay so the master gives us minus i if you have a master if you have a master you have psi one right so psi one minus two i m psi one okay this comes to the minus sign this comes to the plus sign that comes to the opposite order so you re-order them let's find the fact that the fact that without the master this splits into two different advantages the advantage of psi one and the advantage of psi two is the fact that you know very well that a master's programs can be kind on the other hand the master invites you to couple one chirality that this is the chirality by the way is very clear why is it the chirality that somebody explained why this psi one and psi two are the two chiral chiral anti chiral systems how do you judge that what does it mean for something to be chiral anti chiral the moment we're not even talking about moment maybe just but you know there's this notion of chirality even before we go on it's ideal value under gamma five gamma five is the product of gamma zero, gamma one, gamma two, gamma three now in two dimensions the handle of that is the product of gamma zero, gamma one now gamma zero was proportional to the sigma two gamma one proportional to sigma one so gamma zero, gamma one is proportional to sigma three up to some plus minus sign sigma three of course is die so the two the two chiral anti chiral the upper component of the bottom so this is a very simple decomposition in two dimensions where you've got the upper component of the psi one where some convention is the chiral one the bottom component psi two by some convention is the anti chiral one it would be consistent to have the grunge into the only the chiral and not the anti chiral one if the chiral is massless then a mass of is the two now we work out this this grunge in here without this i here wait you're saying this formula is wrong no i think i should not come this formula should come without an i let me check or there should be overall i could tell you let me check oh you're saying i put an overall i here i did right what i was computing was this and this her point is minus i sigma two minus i sigma two it is minus i into minus i so that's minus one here so that's minus one one minus one, thank you it's about this that i wanted you to wanted you to see we would argue this but anyway now the first thing our convention should there be an overall factor in fact they should should be hermit so if you consider the mass stuff it's cybar mcybar which is used in the definition of cybar is anti hermit so there should be an exactly exactly so in fact they should but that's going to be a failure another way of saying this is that we have to make contact with this two-state system the two-state system path in length because a path in length is minus kind of but no i minus, this has to do with the path in length we're going to get out of the Lagrangian is e to the power i times s time derivative term then should have an i in the Lagrangian so that the i goes away between e to the power i times s okay so so actually the Lagrangian should have an i here the actual Lagrangian to get perfect matching with this i had a minus sign so minus sign okay so that the thing that appears inside the path integral is e to the power this because what appears in the path integral is e to the power i s okay this minus sign cancels the i s so we just got what we had so we got integral d side of this object now is exactly of the form that we had here we can make this more advertised by now with the Fourier transformant space so suppose we suppose we do a Fourier transformant space so that we said psi 1 e to the power e to the power i k psi 1 of x psi 1 of k suppose we said that and similarly to psi 2 i get to the formula of psi 2 so psi 5 this and formula that comes d psi 1 psi 1 of k d psi integral d k by 2 pi psi 1 of k minus psi 1 of minus k psi 1 of k dot okay and then this is i k psi 1 of k this k is a special one you understand that k by 2 pi similarly with psi 2 of k the sine root psi 2 minus k psi 2 of dot of k plus psi 2 of minus k suppose this which is plus the d k psi 1 of minus k psi 2 of k so this is the Lagrangian that generates that generates e to the power minus i for you to see that it's exactly of the form that it's exactly of the form that may have written out unfortunately I forgot to show I'll write that again the time was e to the power minus i h was equal to e to the power minus i minus i what do we have to do because what was what was minus i minus i minus i psi dot minus i h minus i exactly of this form if we take these k's and divide them up into positive and negative psi is with positive k psi 1 with positive k are playing the role of these signs psi 1 with negative k are playing the role of the ants is this clear you know that the commutation relation is psi k psi minus k is equal to something so one of them is like the creation the other is the destruction right and we got two state systems for every k one associated with psi one associated with psi 2 and then we also got some interactions between these two state systems quadratic interactions we could really diagonalize them diagonalize them by working on a better basis but now they're the positive right so you see that the dirac path integral gives you exactly the kind of path integral that we were that we were that we were interested in computing is this clear okay now there was a question of the Euclidean continuation so if we were interested in the Euclidean continuation you remember what we had to do is what was the rule what we had to do is take this e to the power minus i h I mean replace this stuff here by by minus h replacing this stuff by minus h as as you said as the shop was mentioning is automatically achieved by the replacement i times the h e to the power minus h integral te will become minus i h integral tm okay but this guy is not supposed to be replaced okay so let me say it again let me say it again what is the Euclidean path integral I'm going to give you a set of rules and then you see it again the Euclidean path integral e to the power minus h minus minus h times t is generated by e to the power minus s Euclidean where we have to get some rules how do we get s Euclidean out of s minus s Euclidean so the rule is t goes to t goes to i t minkowski go from s Euclidean to s minkowski what we have to do is t Euclidean is equal to i times t minkowski e to the power minus x minkowski now what will that mean that will automatically account for giving you this part correctly but for this part okay that i start out there in the rates that's where integral i start out in t so your t gets transformed and side out gets transformed in the other way because it's divided t by 4 dimensions the first part turns into integral exactly but you see there is this I want to know how you go from Euclidean to minkowski in action now it's through that easy path integrals map to each other and the path will take us before by an i in the definition okay so we've got e to the power i s minkowski e to the power i s minkowski has to be e to the power minus s minkowski i times s minkowski give us this so if i want to know what is the rule for computing s euclidean it's going to be there will be the initial formula gamma 0 goes to plus minus and up to itself gamma 0 you see why is that what we want to do is to make the rule that gives us the euclidean the minkowski action from the euclidean action the this part of the rule is working for the non time derivative terms it's not working for the time derivative terms the time derivative terms need to pick up an extra plus or minus i'll leave you to figure out where the plus or minus is okay so what we have to do is basically take this gamma 0 plus 12 times gamma 0 now this is a very satisfying thing it's a very satisfying thing because if you want to work out the path integral in euclidean space you would imagine that you would need to use gamma 0s gamma matrices that obey the euclidean clip on that gamma mu, gamma mu is equal to euclidean cubic but of course multiple times gamma 0 by 9 precisely makes about that okay so the analytic continuation of the fermionic path integral and though we've worked it out in this dynamical case you can see immediately that the lesson is completely tender involves also a replacement of gamma 0 by a euclidean gamma 0 is this clear? okay the last thing I should say about these analytic continuations the last thing I should say about these analytic continuations is that operators that were complex conjugates of each other do not continue to complex conjugates of each other under analytic continuation why is that? you see suppose you have an operator A under time evolution this Heisenberg sense how does it behave? A of t is equal to e to the power minus i h 0 had a dagger 0 as its dagger under time evolution A of t dagger b equal to e to the power minus i h during the rule that when you dagger you change, reverse the order A 0 dagger e to the power i h which is exactly equal to a dagger at t time evolution of the dagger Kowski's maze of course hermiticity is preserved by Heisenberg evolution in order to to employ this trick this trick of analytic continuation to compute things we replace this e to the power i h we looked at this Euclidean operator A Euclidean which was e to the power minus h t e to the power A 0 e to the power i h we take the dagger A Euclidean this is equal to e to the power i h t A over 0 e to the power minus h t which is not equal to a dagger plus minus is equal to what? A dagger e to the power h t is not exactly right exactly right but what this means is that two operators that were related to each other by complex conjugation in the Minkowski path but two insertions that suggests that complex conjugates of each other in the Minkowski path will not be complex conjugates of each other in the Euclidean path this is something you must remember because it causes no interference the only the only relevant notion of complex conjugation is the notion in real physical Minkowski space the notion of complex conjugation in Euclidean space is the correct notion should be that inherited from Minkowski space so that objects that are complex conjugates of each other should not just be complex conjugates of each other in Euclidean space rather should be a relation that you will deduce from from what we just seen just a small aside for those those of you who know what I am talking about when we start in two dimensions we can form a real theory of quantization when we take an operator a primary operator of dimension delta we expand it as 4n by z to find first delta and then the correct notion of complex conjugation is not that this is related to the anti-analytic part when it is that O-n a complex conjugates of each other this comes from tracing through the right notion of complex conjugation in Minkowski space and then seeing what that means for these coefficients rather than blindly working incorrectly in Euclidean space this is one example of how you can easily go wrong there is just a portion for those of you who don't know what I am talking about we might actually talk about this at the end of course but for now it is a portion that we must be careful of not being too naive about what complex conjugation means in Euclidean space it should just work the way naively should of course in Euclidean Minkowski part of the technique everything should just work the last last bit of cleaning up from the last class we have this problem I wonder we will talk about these conserved currents I want to I wanted to I wanted to show you two different equivalent notions of conserved currents for translations the first is to do what we did in the last class the second is to couple the theory with respect to a metric and then vary a metric and these two turn out to be the same but we have this in the problems so I think this is all I wanted to say about in my first first brush with particle angles and quantum field theory I thought at some point that we would discuss realistic quantization but reflecting about it it's a better place to discuss to talk about to discuss it when we start discussing beta functions vocationaries so let's postpone that discussion any questions, comments or any questions, comments or anything we would like to discuss the problems we would like now this basically at this point at least we can play around with how the particle angle is complete and we move on unless there are questions we now weigh the depth slightly less you know let's a dagger of t which by definition is e to the power blah blah blah acting on a dagger of 0 will be related to e to the power minus 2 ht a t dagger e to the power plus 2 ht may be about the minus just take a dagger and e to the power that's exactly what exactly so this a dagger is e to the power minus we can do it separately but there is no you can use this to find a relation without going back by going the notion should be so that after you analytically continue back to Minkowski space the relevant matrix that's basically the principle okay and the point is that there is an i going from Minkowski space so that's not the same as being complex conjugate that's basically it other questions or comments let me just stop this discussion okay so first let me summarize what have we learned so far we have learned that given a quantum system there is a way to represent its evolution in terms of a path integral the Hilbert space is obtained by slicing the path integral along lines of constant time we've learned that the that the that insertions of fields of that path integral inside this integral compute time-order operator insertions we've learned that in edge theory this slicing operation produces a Hilbert space the slicing operation produces a Hilbert space with which is to be thought of as when one way of thinking of it is made up of fewer degrees of oscillate wave functionals of fewer fields that you have in your path integral class at the straight we've learned that formulaic systems can also be represented in path integrals but as integrals of a grassland variance we've learned that that there is a euclidean continuation of all these path integrals that often is mathematically very defined than the Mikowski path integral often the thing we were actually compute but there are clear rules for going from the euclidean and we've also learned that the path integral for gauge theories we've learned that the path integral for gauge theories at least perturbatively seemed hard to make sense of in the continuum because of these gauge zero modes but there was a way of curing that with this path integral point of click finally we've seen examples of interesting relations that we introduced using path integral tricks these interesting relations were stringerizing equations or identities for conserved amounts and we see more of these as we continue there's a summary of what we've discussed but now one of the problems with all of this discussion is that why this sounds quite interesting I mean who would deny that it sounds interesting transition amplitudes represented as an integral over all possible paths an amplitude for each path it sounds interesting suddenly it was very interesting to find this one of the mechanics and actually this formulation finds its real form in our quantum field theory because every other way of looking at quantum field theory mutilates the underlying Lorentz invariance of the structure but the path integral preserves so beautifully because it's given in terms of the action of the Hamiltonian the action is a beautifully Lorentz invariant quantity so the final path integrals which are quite an exotic theme in the study of quantum mechanics it's perfectly possible to go through a very good quantum mechanics course without really hearing about final path integrals because no where do you use that become more than an exotic curiosity when you study quantum mechanics they become for many purposes the best way of thinking there is a problem and this problem is not in the form of path integral formulations it's in a parallel problem of the quantum field theory and the problem as you know is the problem of technologies the problem is when you try to compute any of the natural observables when you try to compute any of the natural observables that we've discussed in insertions of fields inside path integrals we often find the answer to be infinite in a very puzzling and very strange way okay so it's that makes the same that makes the same first sight while what we started out trying to do was a good attempt and had some form of beauty that actually made no sense you know a lot of people concluded that for a very long time this famous quote from Landau physicist who arrived old in the highest regard but of course he made very great mistakes listening one of them and you know I actually announced at some point that quantum field theory is dead I think in the 50s and 60s and should be buried but with appropriate honors Landau in fact makes significant contributions to study of quantum field theory it's not like he said it up in the air okay and certainly this problem of the infinities was at the was at the back of many people's minds and things because Landau said this after the normalization of community but there was a problem with that if we discuss this one more so so before we do that we have to see what the problem is so this class I just tell you about the problem we need the solution okay so let's see the problem in a simple context that is 8 of 8 say 5 to the 4th line you've seen how this before but 5 minutes this will also serve to illustrate how to generate the perturbation ceilings from the path of data the question suppose we want to compute the problem e5 the insertions of 4 operators I quickly slipped into Euclidean calculation so much somehow easier to keep track of all sciences and I need you to give me the answers okay m squared by 2 plus Landau by less than 4 factorial suppose you have a theory with a single scalar field it's actually given by this and you want to compute this object this object you can usually say it where it means this object divided by the expectation value of 4 1 that's the object that is as the interpretation of being the partition function of the insertions of evolution operator of the insertions of experience divided by the evolution operator of the insertions suppose we were interested in computing this object in fact before theory suppose we will know how to do it in general all references compute an integral but integrals are not so easy to compute integral this is an integral of an integral number of variables so we need to develop some techniques to compute such things now the simplest technique is perturbation perturbation theory involves if the action in the theory is quadratic with a small correction that is the case here if we assume that lambda is a small number ok aha that is the case here if we assume that lambda is a small number so let's try to understand ok so suppose lambda is a small number we take this path and take it just let's take the example in lambda ok so what we have got is d phi phi of x1 x2 phi of t minus minus d phi we will take like that so we have downstairs integral of i4 integral of i4 the point we want to make will already be visible in order to go by computing these things ok in order to go by computing these things we use a marvelous theorem of Gaussian and that marvelous theorem of Gaussian integration sometimes goes weeks and it goes this way suppose you get Gaussian integral it goes into the path xi ui a number of insertions of x's downstairs these one and two need to be the theorem says the following the theorem says the following it says that if you want to compute this Gaussian integral ok there is a very simple way of computing ok that this Gaussian integral given is the sum over all contractions of these things so let me take it off suppose I want to compute x1 and x2 ok so this Gaussian integral here is the sum over all contractions of the subject just so it's equal to x1 x2 x3 I'm assuming I can suppose I can get a little bit more insertions ok when x1 x2 is simply e to the power minus x1 x2 to derive this formula and we leave the derivation for homework exercise ok but I just want to emphasize this is a very simple straightforward statement about integration we have a couple of derivations this is in our homework exercise ok I'm not going to spend time ok ok the simple thing ok is the statement is the statement x7 so this one is now very simple to compute which theorem was derived using more greater stuff look back following the easiest way to ok this is now very easy to compute using all of this ok in this particular case what will we do we would go to the Fourier basis so as usual we would have 5 over x 5 plus 5 into here so the quantity that we wish to compute is we would have e to the power Ip1 x1 is by now p insertions 1, 5, p to 5, p to 5, p to 4 also in terms of momentum the part of the action here is just e to the power minus 5 minus p in a p square plus m square of course you have to use the power minus a x square by the expectation value of x squares ok running over the expectation value of x squares that is simply d by dA d by dA of e to the power minus a x square by 2 suppose we are interested in this this quantity divided by this then this becomes so the expectation value this becomes this that log of integral e to the power minus x square by 2 now log of e to the power minus a x square by 2 we can get by dimension of this because clearly e to the power a x square by 2 dx goes like x x goes like 1 over square root of d so this is log this is log of then differentiating that gives us that gives us d by dA of log A minus 2 x square e to the power minus a x square by 2 divided by integral by a x square by 2 is just one way this tells us here that this object that this contraction is equal to 1 over p square plus m square times 2 pi e to the power d integral which actually in fact is 2 pi e to the power and so we apply which theorem after writing everything moment in space with this rule for log fractions let's quickly do that let's talk to this class let's quickly do that so what do we have here we have 5v1 5v2 5v3 5v4 here we are supposed to take this pi of x1 x2 x3 x4 right at the moment of space that just becomes pi of let's write it out once dA is d let's say 2pi 2pi d dk 2pi 2pi d that's delta function k1 that's where inserting definitions is this clear any questions by the way please talk to me I'm going fast here assuming you've this is more or less ordinary this is too fast then you are not understanding please talk this is okay great what we of course have is where we have contractions of these two guys with each other you get such contributions in order lambda to the 0 so in this expression we will get terms like delta of p1 plus p2 1 by p1 squared n squared 1 by p sorry delta of p3 plus p4 and three such expressions because we can contract p1 with p2 p1 with p3 and p1 with p4 these are the expressions that occur at order lambda to the power 0 and what it's telling us is that there is that at leading order 4 point functions that is products of 2 point functions then we have a new contribution add order lambda to the power the new contribution at order lambda to the power 1 is when we take one of these guys and contract these with the 4 but nothing is allowed to stay on contract you also have contributions actually when you contract when you contract two of these with one another and these with two of those so 1 2 point function must correct 2 point functions many such contributions set these focus on the one that's going to be most important that is contract this guy, this guy, this guy let's compute that object now to compute that object we're supposed to sum over all contractions what how many ways are there of making the contractions this fellow 5p1 so now we're in momentum space let's p1 p2 p1 p1 has a choice of 4 values of which to contract p2 now has a choice of 3 guys p3 p4 p3 p4 so we get 4 factorial ways of choosing the number of ways of doing this all these contractions give us identical expressions because though they contract with different k's k is a dummy variable so we just get an overall factor of 4 factorial there was one over 4 factorial in our action I rubbed that out you know this the 4 factorial can sense the one over 4 factorial this is the reason I wrote the one what retro was a sign of minus lambda and then what we get is 4p1 squared plus x squared how do we represent it? it was on a final level that's an insertion of 5p1 it goes into a vertex operator the insertions external insertions meet at an interaction because each of these lines represents a contraction which comes with a one over p squared plus m squared the factor of one over p squared plus m squared has names for a property so these lines can represent these properties and each of these factors are associated with something that knows about the nature of the interaction in the case of our very similar interaction was just this factor of lambda with the appropriate case sometimes the factor of numbers is called a vertex factor okay let me look 4 minutes I said let's represent the 0 for a value in a final 0 for a value was this p1 it only works since the contraction is only between equal momentum it only works if p1 is equal to p2 p3 is equal to p4 that was this guy and then there were three other final values p3 okay this was the insertion p1 and p2 but you could have connected them the diagrams at first one that I did not talk about were these like well let's say this was p1 to p2 and then p3 to p4 the reason I'm not talking about this will become clear a little later let me let me move on to one higher the contribution that you get when you take two of these vertex factors then you connect things in the following equation you've got one vertex from which emerges four lengths because it's an interaction vertex and another vertex from which emerges four lengths it's an interaction vertex and in a sense I will make precise as we go now we're in non-trivial graph of this order it's the one we're going to connect one of these lengths to this one one of these lengths and these ones to the external let's put this in p1 this is what's called one loop final graph let's study this one loop final graph let's compute what it is what's going on we have these four 5's from each vertex and we're supposed to contract them in the way that I described okay firstly we're going to have a vector minus lambda by four factorial squared by two by expanding the exponential and we're going to get lots of factors which count how many identically lost terms we have and then we're going to get the factor for each of these terms okay the symmetry factors let's just do the counting okay how can we count this now this p1 guy okay we had each of these two vertex we had each of these vertices p1 guy the p1 guy has two choices will it connect to something in the first vertex or the second vertex it's a factor of two then once it shows which vertex it wants to connect to it has four choices of which kind to connect to now in the graph that I've drawn p2 has no choice about which vertex to go to because it must not track the same one but it still has a choice of three of which left over five is to contract with so we've got two and a fourth p3 has no choice about which vertex to contract must contract with the other one but in the other one it could contract with four and p4 that could contract with three now what's left is the self-contractions take two ways of self-contractions of the two left over guys here this guy could contract with this and that factor of two now you see that this is four factorial into four factorial divided by four factorial is squared for the left over half so the factor for this graph the factor for this graph is lambda squared by two so lambda squared by two we also have the expression for each of these so we have each of these propagator factors left so in all such graphs of course we have the overall factors of one over p1 squared plus n squared one over p2 squared plus n squared one over p3 squared plus n squared one over p4 squared plus n squared that's this this propagator but in addition we've got this guy traveling with some momentum that's np remember there's a delta function of each of these propagators the momentum here is not independent the momentum here, if it's going in is equal to minus of p1 plus p2 plus p3 plus p2 similarly there would be a formula to relate the momentum here to the sum of these three and the fact that these are consistent is only true because there's an overall momentum conservation okay, so we get an overall momentum conservation of two bytes of p that's delta of p1 plus p2 plus p3 plus p4 let's say that p1 plus p2 is equal to minus p3 plus p4 is equal to q that's just the same name so then times we get pd prime let's call it pd r by 2 pi d by 5 r squared by m squared into 1 over r plus q down so this whole graph this whole term with the expansion of this propagator is equal to this times this times this times lambda squared by 2 where? okay, so this times this times this and that's the expression usually it's quite simple for these simple theories alright why do people invent fancy rules to try to understand what this factor would allow to be in there these are called symmetry factors I've never understood their rules nor have I I find by far the easiest way in any situation is to do the I follow the security okay so right okay so this is very simple looks very simple, very beautiful and actually if you compare to all the perturbation by which I mean trying to compute the same kind of thing using second order perturbation theory the way we do quantum mechanics okay using creation and annihilation operators exchanging it's actually much, much better textbooks nowadays don't even have that so you guys don't see how much better actually it's you know it's like a mutilated me okay and this looks like a very beautiful answer to me but this this is okay say whether this is an integral but this is an integral and you know okay, so we can just put this integral up but there's a problem suppose d actually there's no problem in d in this case okay but in d is 4 the problem is this integral doesn't converge it doesn't converge and all equals infinity as you can see a very large r this integral goes like d dr divided by r to the fourth okay and in d equals 4 that's not divergent the greater than 4 that's worse than much we have an answer that just is infinity and you might wonder have I computed everything in the same order that these infinities it's infinity and it's infinity, you know it's nothing to do with it you might wonder okay so the question is what's going on why are we getting infinity isn't because one of fields is dead it means to be better perfect on us oh is it because of something else and the answer is that if you actually But this is not the right way to define it. That's the problem. The point is, one of the fields here is defined as a limit. It's defined when you take a Lagrangian like this with what's called a at-off, which we'll describe in the next class. We take the parameters of the Lagrangian, this lambda, as a function of that cutoff. The cutoff is a cutoff in momentum space. So the problem is coming when we go to a very hard way. Take the parameters in Lagrangian as a function of the cutoff, and take the limit cutoff going to infinity not by holding the parameter fixed, but by taking the parameter scale as some appropriate function of that cutoff. That defines a good theory. It's this fact and the understanding of this, that is the question of renormalization, which we will turn to discussing in the next class. The good reference for the next class is a paper written by Pulchinsky called Effective Lagrangian, renormalization and defective Lagrangian. Renormalization and defective Lagrangian is 1983, nuclear physics. It's a good reference. You might want to bring me to the next class. Okay, thank you. Next class on 5 years.