 I wanted to have a brief discussion, a brief discussion of lattice methods, a brief discussion of lattice methods with these lattice models before we continue with our analysis of continuity. These lattice, you know, these studying these lattice models, in high energy this is very popular in the 1970s and 80s. The versions of it, of course, continue to be very popular in the 90s and 60s. It's a useful thing to do. Actually at some point, in ten years, I don't want to cause of this, these kind of things. So there's a lot of material, okay, and I will not try to, try to, pieces thematic about them. It's quite fun subject. But, yeah, so there's just, I will briefly indicate various things that happen. So let me first start with the appeal. Remember how it went, right? We know this lattice, okay? And we had a variable that looked at these links. Do you remember what we called these variables? Oh, I'm fine. So let's say that way, the links are, the link variables are able to, once you start out, in the direction, okay? And the final action, if you remember, is 1 by g squared, or back to 2, doesn't it? Okay? 1 minus cosine of pi of, let's say, let's say that we're working in the x, so 1, 2 direction. 1, 2 direction and we'll sum it up. Or the a, b direction, we'll sum it up. So pi of x, many, minus pi of plus b, minus pi of x plus n, a and b, minus pi of x plus b. Does this look familiar? It's slightly different. Sum over a, b. So we had this, plus this, plus this, plus this, and then plus complex conjugate. The plus complex conjugate is what makes it cosine. Okay? Then what's this, this, plus this? This was this, plus this, plus this, plus this. So I've written that as this minus this. So that's pi of x, so think of this as the a direction. And this is the b direction. But then this is pi of x, starts at x goes in the a direction. And then there was a guy that started at this kind of in this direction, but that's minus starting here with that. So that starts at x plus n, b goes in the a direction. Okay? And then minus, this is, this is pi of x plus n, a and going in the a and b direction. Minus pi of x, and going in the a and b direction. Is this clear? So this is negative. This, when discretized, is a negative derivative. Negative b derivative of a a. And this when discretized is the negative a derivative. It is the positive a derivative of a b. Okay? So this was our, this was our last stage today. Okay? Now in the last class we very briefly looked at the strong coupling expansion. Right? We very briefly looked at the strong coupling expansion. So I'm just going to think of this model as a model in its own right. Okay. And the connection to the continuum theory later. What are you talking about? It's a, it's a model itself. And we very briefly looked at the continuum, the strong coupling expansion. G was very large. The one by G was very strong. Okay? And you saw that for instance we got this Wilson area law for the Wilson loops. More generally if we wanted to, if we wanted to compute the free energy of this area. What we would do is bringing down all terms from the action. That don't give you zero on integration. But you can see that that is bringing down cliquettes and closed surfaces. So remember we completed the Wilson loops. We had an open surface ending on the Wilson loop. Because we had that insertion. But we have no insertion. If you bring down a cliquette you have to bring down the neighboring cliquette and so on. Until you make a closed surface. Okay? So you see that the strong coupling expansion here is organized in a sum over surfaces. Okay? For every surface that you can draw in lattice there is a term. And the surfaces that are smaller and smaller contribute. I mean the surfaces that are larger and larger contribute in increasing the subdominant way. In the strong coupling. So I'm not sure if anyone came in after I said this. We came to the decision that the evaluation in this course will be based just on assignments. I'm going to pad up the assignment set with 5 to 10 more problems. Okay? And 15th December is a hard deadline for submission. It's a hard deadline because I have to submit a grade soon after that. So I mean you're free to submit after that but won't come for it later. Okay. Okay. Yeah. So I'll give you a couple of problems from these lattice models seriously doing this strong coupling expansion. But now there's a qualitative feature here that I wanted to point out about the opposite expansion. Namely the weak coupling expansion. Suppose we take this model at the G to be very small. Then I briefly mentioned in the last class. This is the limit in which this cosine. Okay. Does not want that the argument of the cosine does not want to deviate too far from zero. Because anything where this is large is highly suppressed. That's high action. Okay. If the cosine is not going to deviate too far from zero. Then you can expand the cosine quadratic order. Because alpha is equal to one minus alpha squared two. Okay. The ones cancel. And the alpha squared by two. Then just give you the, you know, then you can, this is just the continuum, the discretization of the derivative. And so this just genuinely becomes the F squared action. Is this good? Okay. Except that I've done something wrong. One of my statements was wrong. Oh, was he that? Yeah. Can you identify which one? Okay. And then you would see. The statement was that when G is very small, you want to expand the cosine around zero. Now you know the cosine from zero. That was it. Why does that happen? We should expand it around zero. About? The T, G. No. The optimum. We're doing seven points basically. So you look at the, because the cosine is a derivative. It's a, zero is not different from two pi or minus two pi or two n pi. So really what you should do is to expand this around any of its two n pi's. Okay. So really, you know, apart from this one factor, which is the right approximation for this cosine. It's not this quantity squared. Just continue about this one. The alpha. But instead is alpha minus two n pi. The whole thing squared by two. Some over one. Okay. Now if your alpha happens to be in the neighborhood of one of the n's. The other terms are completely negligible. So they being there makes no difference to what you're doing. So we just pick out one down. We just have the whole series. We keep the whole series. Because any one of them may turn out to be important. We don't know which one. Okay. Those terms that are not, where alpha is not near one of them, will contribute extremely sub-dominantly. But we just keep it. There's no harm in it. Okay. So this whole one minus cosine is well approximated by this. Where n is the sum over all integers. Is this clear? That's at the side of one thing. In this problem, these phi is here. These phi's here are taken to be periodic. Phi runs from zero to two pi. Phi is a fence. It was an ideal body. Okay. Now. Yeah. Now, this fact, even in the coupling level, is very important. The fact that we are not allowed to take this object, and to expand it entirely around just alpha equals here. And I want to try to explain why this is so important. The point is like, how can we do a paradigm of all integers simultaneously? Yeah. The point is that if you just say this is the case. Okay. So you've got this cosine of, so what we have is an e2. The correct statement is this. e to the power minus 1 by g squared of 1 minus cos alpha. Okay. Is approximately e to the power minus 1 by 2 g squared sum over alpha minus n to the power minus x squared. Okay. Fine. Okay. You see, if alpha is in the neighborhood of one of the other, one of these points. The other ones are extremely subdominal. Yeah. Sir, I was just asking, like, if there is a mathematical value to be treated, that's not a vertical point of perspective. Well, this is the statement that was taken. Oh. Sir, as long as this is true, that's all I'm doing. I have this up in my exponential, and I'm replacing it with this. Yeah. Yeah. Now, there is something here that I will only qualitatively to get you, and I will actually set you a problem in the problem set to try to make this precise. You see, these n's here live on faces. This alpha is approximately FAB, you know, flux due to the A field. If you discretize your file, it's flux due to the A field. Okay. So what these n's are here, additional fluxes. Apart from the flux that you get from your A field, you can have additional integer values of fluxes running through each face. Okay. Now, you see, these additional fluxes, these additional fluxes have the following property. Now, suppose you take a little cube. Suppose I take a little hypercube, let's have three dimensions. Okay. So I take a little hypercube. Okay. The flux that you get from just alpha has the property that if you sum the flux over all the faces of the cube, you can see simply the discrete version of the statement that integral VA equals zero. Okay. The flux that comes from alpha is built out of an A. And it's built out of this. This is this VA. And if you can easily convince yourself, we can do it if you're interested. If you can easily convince yourself that if you just sum the flux over this, flux, when the flux is defined by this object, plus this face, plus this face, all the terms that you get here will just cancel. Okay. That's an easy exercise to do. And we could do it in two minutes, right? Okay. Yeah. That's an easy exercise to do. Where does it come from? It comes from the fact that every particular link is shared in between two faces. This link appears in the flux in this face. It also appears in the flux in this face. And so if you do that sum with the appropriate orientations, each link just cancels it. Is that clear? In this one it comes like this. And in this one it goes like that. Okay. So just one minute of thinking about it will make this totally clear. Then there's sum over fluxes. The orientations from this part is just same. Sir, all this holds only 12 periodic orientations. Periodic orientations to the lattice? Yeah. I mean, if it's not periodic, then the end is more necessary. Yeah. I assume you've got something funny happening. But even if it were not in the middle of the lattice, this is correct. Okay. However, there's no particular reason at all for the sum over fluxes for these ends to be equal. These ends are completely arbitrary integers. And the sum over ends over all faces of a cube can be anything but. Now, this whole stuff can be processed as follows. You can take these ends, these integers, and rewrite them. Rewrite them effectively as d of something. Okay. Plus a curve. Plus a part that contributes to basically this non-zero source frame. And the part where this n is d of something can just be absorbed into the others. And something can be just absorbed into the files. Because the alphas in themselves were d of the files. Okay. So, I'm not working this out in detail. I'm just indicating you about this. Because the sum over ends, sum over all, one integer associated in each face can be decomposed into integers that can be absorbed into this, into the 5pb. And integers that cannot. The only integers that cannot, as is perhaps intuitive, is the source of flux in each hyperbolic. Because we've seen that the source of flux for anything that can be absorbed into some shifted alpha. The shifted phi is 0. However, the sum over ends does not in general have that integral over flux equal to 0. So that part certainly cannot be absorbed. And one can show that's the only part that cannot be absorbed. Okay. Now, this absorbing some integer shift into phi A. And summing over those integers turns the integral over periodic phi into an integral over each phi from minus infinity to infinity. Okay. So the path that can be absorbed in phi A actually simplifies it. Because it's easier to do integrals from minus infinity to infinity. Then from 0 to 2 pi. How did this happen? Basically what's happening is that when the sum of these fluxes here can be written as a sum over shifts of phi by integer. Something else which is basically the plus, plus the source for this sum of integral of flux for each hyperbolic. Now the shifts of phi by integer. That integer, when you get integral of phi from 0 to 2 pi. And then we also win the sum of the shift by an arbitrary integer. So that sum plus integral just becomes integral over minus infinity to infinity. Really simple. So first we have integral dx. Okay. From 0 to 2 pi. Okay. So we actually have this integral dx plus m and sum over m. This is the same as integral minus infinity to infinity of some dx. Is this clear? Okay. These m's do not affect what goes. Yeah. Why can't we do this with the deviations? Like if you could observe all this stuff into the phi integral. Wait, but we can't absorb everything. This is precisely the part. There's a part of it that you can absorb it into the phi integral. Okay. But there's part that you can't. It's exactly the analog of what happens. So what this is then? So there once again you can absorb a part into the phase. But you cannot absorb the part that gives you what this is. Okay. You can't absorb other variables. Yeah. So let's say, you know, so there once again. Maybe we can talk about this later. It's very similar. Okay. It's actually very similar. Okay. In fact, it's a bit easier. I don't know. But this analysis was first done by B of BKD in this PhD thesis. Yeah. Exactly the way. That I remember. Because nobody remembers what it is. Of course nobody remembers. But very similar, yes. Okay. So now let's go. You see, so what the end result is, is that if G is very small, what you get is that this part, you know, these, these phi is just, we didn't even think that for minus infinity did you know. Plus you have to sum over sources for integrals of F over each possible hyperbolic. Now the whole F, the whole, you see, and then what, as long as G is small enough, this part, you know, the alphas can, we've already made this square. So this just becomes like an F and F give you minus this extra source kind of F, the whole thing squared. Now what is the source kind of F, the whole thing squared? The source kind of F, the whole thing squared, is precisely the field strength of a magnetic monopole of every possible integer source located at the center of one of these hyperboles. It's a magnetic monopole because the integral of F around the cube is non-zero. That's what defines a magnetic monopole, integral F, non-zero. And in fact the strength of that integral of F is the net integer, the sum over N's. No, this hypercube is the, the hypercube is in lattice, in the lattice. So just imagine that I'm in three dimensions. I don't put four dimensions. That's even very easy. So imagine that I'm in three dimensions. So we've got a lattice, that is a union of hypercubes. I'm in three dimensions, it's a cube. It's a union of cubes. Something, one of those cubes. It's a cube. It's not a blanket. A blanket is a two-dimension. This cube is what I draw. Do you understand? It's just a cubic lattice. It's a bunch of cubes. It's called something like unit cells. Okay, so it's a bunch of cubes. It makes up a whole lattice. So the net result, and this you will do carefully in a moment. So the net result is that what we have to do, what we have is a discretized version. Overdoing the integral over all A mu's. You know, just A mu field continues. But also summing over the field strength of magnetic monopoles. Of arbitrary in a teacher's plan. Okay, placed at the center of every high quality. Because we take g very small and now we move to continuum notation. Okay, so this integral over this alpha part can just really genuinely be approximated by f. And then you get f squared plus the magnetic monopole thing squared. Plus f cross term of the magnetic monopole. Sir, is this the same? Or is it the same? It's like Gauss's law, but for magnet. Okay, you see Gauss's law of magnetism in the genuine continuum framework. Just gives you that integral f over any closed surface is zero. Because if f is da, it's a well-defined field. But there's nothing left to say. Integral f over, it's just zero. But you see the point here was that f being equal to da was something that emerged out of the weak coupling limit. Precisely equal to da. You see it was equal to da plus the centrifugal shift. And that fact allows us to violate the nine Gauss's law of magnetism. And in fact have sources of magnetic field strength. Is this clear? So now where do you process this whole problem? When you process this whole problem. Okay? Sir, is there any physical way to think how this weak coupling describes to magnetic molecules? Is there a physical description for this? Mathematically I get it that this shift gives us the magnetic molecule. But if you take it, was it happening physically? Yeah, well somehow very important is the period. I'm not sure if this is what you want, but let me say one more thing. You see, what was very important was that this u1 was period. Okay, that was crucial for everything. Now, you can ask me. And actually everything I'm saying will work whenever the u1 is periodic. Where that comes from is a matter of fact. Okay? So you can ask, is there some other situation that you are aware of? In which you get a periodic u1. Clearly in the continuum. Can somebody tell me about the situation? Clearly in the continuum. Where you get naturally a periodic u1. It appears in the standard. Well, you get the u1 or the spontaneous breaking of the non-abidance structure. You see, non-abidance and actually it's a very nature periodic. Okay? So if you take SU2, there's no option of making any of the u1s inside the non-abidance. It just is periodic. Okay? So if you take it, if you have an SU2 model, and you spontaneously break SU2 down to u1. Okay? Then you will automatically have a periodic u1. But you will also automatically have a monopole. Why? Because you got the top model. The top model. So this is like breaking of symmetries. No. I'm saying that every time you have a periodic u1, you also have a monopole type configuration. Okay? In that situation, in that situation, I'm not sure how much you guys know about monopoles and so on. But anyway, in that situation, what are these magnetic monopoles? They are genuine, smooth solutions of equations of motion. But how does Gauss's law get violated? Gauss's law gets violated by the fact that in the score of the monopole, you don't have a u1 theory. You have an SU2 theory. Okay? So in many cases, what happens is that you've got an effective u1 theory at long distances. But at very short distances, there's some new physics. That violates your analysis of u1. Here, what is the new physics? It's a philacus. There, what was the new physics? It was the acetylcate gauge theory. Okay? But any time you've got this continue u1, you have in some model that can make sense all the way to the other way. You get new phenomena. Okay? Then allow this model. Okay. Fine. But let me continue. So how does this proceed? This proceeds at the bottom. So what you had was the field strength of the A field. You had the monophilic field strength, and then you have the whole field square. So you've got the right field strength, the cross term, and the monophilic strength. Okay? Now, but the A means because that's just a Gaussian problem. What does this do? What this generates is some effective interaction between the models. What is the effective interaction between the two? How do you solve the distance? So it's like something like a J mu in two kinds of places. It's sort of like a J mu in here, right? It's not quite J mu because magnetic. Otherwise, it's J dual. Okay? So that's good. So what interaction does it generate? One over r, right? It's the duality between electrical and magnetic field. So just like if you had sources of electric charge interacting with maximum field and you integrate it with maximum field, that generates one over r in tension between the sources of electric field. There is a middle way of the structure of maximum field. If you have sources of magnetic field, you generate r at the end. You generate one over r at the end. Okay? So what this problem becomes, because now you can just do the integral, the part of the integral over the continuum gauge fields. That's a trivial thing. Okay? You can just do that integral. And what this becomes is an effective problem for interactions of these models. So what you have to do is to sum over all possible model model strengths. So you've got a gas of model models. It's like a gas of more disease for the beginning transition. Okay? Where there would be some sort of long-out of the beginning. That's replaced here by one over r in tension. Okay? So you get a gas of model models. The model model numbers are not conserved. It could be anything. Because at each point, you could have model model strength minus one, you know, minus infinity to infinity. Okay? And these guys, these guys interact with each other. See, the following is... Okay? It turns out that, firstly, this gas of model models costs energy. Because you've got one over r potential, but you've also got the core energy of each model. Which is basically just this n squared energy. Okay? So you've got core energy for each model. So you've got these model models being produced. G is very small. So they're produced very diluted. So they're very diluted gas of model models. Okay? But there's this really interesting fact. The interesting fact is that even if this model of gas is produced very diluted, if you've got an infinite space, there is some density of this model of gas. That density will go like into the power of the n squared. But there is some density. So if you're computing correlation functions in this problem, at very large distances, at extremely large distances, G is very small for n squared. And you've got very large distances compared to e to the power of n squared. These models will affect your correlators. And it turns out that what it does is affect your correlators in a way to give the photon a mass. A very small mass. A mass of order e to the power of n squared. Okay? So that this theory does not actually have the masses part of this. In a similar way, you can actually show. As I said, I will formulate a problem but I'll help you set my step. So I'll show these things. This will be one of the problems. Okay? In a similar way, you can show that the Wilson loop, even in this weak couple of elements, has an area. With a very small coefficient. The coefficient will be e to the power of n squared. Okay? So we conclude the following. And by the way, this is in three dimensions. Everything I said is true in three dimensions. So we conclude the following. From this analysis, we conclude that the theory of strong coupling and area of the Wilson loop and also as a mass gap. Okay? The mass gap is just the fact that when you compute free energy, it's dominated by vacuum plus some small stuff, some small stuff. The vacuum dominates. That was not a surprise. But there was a lot of room in the problem for there to have been a phase transition as you went from strong coupling to weak coupling. Strong coupling is some lattice artifact. It has nothing to do with the real problem we're trying to study. The real problem we're trying to study is trace s squared. And that only emerges in weak coupling because it forces this cosine to be the other side. But the point we've seen is the following. It's very beautiful. The point we've seen is the following. That, the point we've seen here is that even at weak coupling, this particular lattice model displays a mass gap and displays an area of force. This happens even at weak coupling. It happens very weakly. The area of mass gap, both scale like e to the power minus 1, which is a very small mass. What do you mean by the area of mass? Area of mass. This is the thing that we talked about last time that the expectation value of Wilson do. Wilson do is very large. Scale is like the area of mass. This, by the way, is an order parameter for confinement. If you put a classical charge and an anti-charge, the cost for energy for separating that over distance L, scale is like L. Isn't it? So, there is this little lattice model in three dimensions. This little lattice model in three dimensions. This little lattice model in three dimensions. Both the weak coupling and strong coupling displays a mass gap and every local Wilson. So, there is no reason to believe that the model undergoes a phase transition at any value of the coupling. There appears all the ways to display confining times. This will include that our attempt to... If we use this as the attempt to make a continued field theory. It is a field theory that has a mass gap and a mass gap and area law, area law, area law for the Wilson loop even as we approach the continuum. Okay? And this in some way is the work of Bola also. And it is very nicely reviewed in his little book, Gage Field and Strails, which you may want to consult for help in solving the problem. I will give you. Okay? Bola, as I explained to you if you take the U1 theory, if you take the U1 theory with the spontaneously broken gauge symmetry instead. Purely in the continuum. Okay? Then various number of analysis also are done. Because you are supposed to do the path detector summing over all instance. And the solitonic configurations of all instances are instant ones for the three dimension. Okay? This is just for the people who know what this means. In the lectures that follow this course and the next one here. In the next semester, I will explain all these things in more detail. Okay? But the solitonic configurations, which are instant ones with three dimensions, they have to be summed over. And doing the summation of the instant ones produces a mass gap and produces the area. Okay? Even in this continuum. Okay? So in three dimensions, therefore, we find that periodic U1 theories. U1 theories where the gauge group is periodic. Typically appear to confine. Typically appear to confine. Even in the continuum. And the thing that is responsible for the confinement is instant ones. Or these, in the opposite way, these magnetic monopole type configurations. Proliferation of magnetic monopole type configurations. Okay? Deep to mass gap. The analogous statements since you brought it up for the Berenstein, Berenchinsky, Kostlowski's transition is that when you were a periodic U1, if you allow, you know, configurations with vortices, proliferation. Actually, there it's really more subtle. Because there's a coupling constant in theory. Okay? And after a critical value of the coupling constant, these vortices proliferate. And once the vortices proliferate, you know, before the critical value is reached, in that situation, you don't have a finite density of vortices. Can somebody remind you why? Do you know the model I'm talking about? Shall we have a 5-10-minute diversion in studying the model? We could do that. Let's try it. Okay. Last time we tried to spot the easiest diversion in the world we might have a little problem. Probably the same. Let's try it. Okay? The BJP model is simply a model of spins. Okay? So in that case, let's say we're going to do that. And in that case, what we have is we want to model this S-I-A state, right? So the model is 1 by g squared in 1 minus cosine of 5i minus 5j. 5i minus 5i plus ni i plus na sum over all a sum over all i. You've got to do the next lattice. Our phi is given to the sides. Okay? And what we have is that there's an interaction associated with it. Various different sides. But you've got this phi minus this phi cosine of that 1 minus that the whole thing squared. And you've got the sum of this this way as well as that. Okay? One? I mean all the units here. What? In our case, phi is different mix. So this is a completely separate problem. Let's call these S's. This is a spin. Okay? It's got matter fields. Not gate fields. These matter fields live on sides. Okay? This is a model that if you take the naive continuum limit like we did, we replace this by s i minus s i plus that whole thing squared which is simply the derivative. Simply the derivative of the spin. So the naive continuum description of this model. Okay? The naive continuum description of this model is delta s times squared. That of a free scale of it. Okay? So the width of the limit of this model looks like it's just going to become the free scale of it. Okay? So what's very important is that we then treat this model as s is periodic. So s is the same as s plus u. So what it really becomes is like a continuum description of free scalar field with periodic s. Free scalar field with periodic s admits configurations that the free scalar field with unperiodic s does not work. These are winding configurations. Configurations so that you start somewhere and you go around the circle and you come back and s has not come back to itself but has come back after a winding of 2 pi or 4 pi or something like that. Such configurations are called vortices. And this continuum model has these vortices. This discrete model has these vortices. Very similarly described. This cosine should not be approximated by the difference squared. It should be approximated by difference squared minus integer. We've got an integer associated with each lake. So the integer is a vector field. Then we use the analog, the discrete analog of the theorem that every vector field can be written as a gradient plus a gradient. Again in two dimensions, any vector field can be written as Li of pi plus epsilon i nj nj of pi. As a continued statement this is true and obvious. And there is a discrete analog of the statement that you can easily manufacture. Now this set of integers that is in this Li of is just the difference between an integer here and an integer there and that is just absorbed into shifting s. So the integral over periodic s plus the sum of these integers just changes the integral over s equal to minus infinity. However there may be these integers. What is the property of this thing? The property of this thing is that it will allow for winding it will allow for winding of this effective file. Okay? So these ties here are effective morticity markers. Then much like the integers that we couldn't shift of the files in our gauge model it was effective monoprotos. Now this model does become del phi plus the effective phi for a vortex. So phi vortex plus del phi vortex so the thing is square. And once again exactly like in our problem this part is just integrating it is gauging in techno can be integrated out. Generates an interaction between two vertices. Okay? What is the interaction strength for a vortex? It is just one over del square in two dimensions. Just like one over r goes one over del square in three dimensions. Okay? And simply gives you a lot of two dimensions between two vertices. So you get a coulomb gas an effective coulomb gas but in two dimensions. Now this is the point. The point was that let's ask firstly let's ask can you generate a single vortex? And this we can estimate by energy versus energy. A single vortex has infinite energy not because of the coulomb because that's regulated by the matrix. But because of what happens at infinity? Because of the log. The energy of a single vortex is of order one over g squared okay? Times of log log of r is the size of the system. But how many places are there to put a single vortex? How many how many places can a single vortex be in? Number of number of plugins which is r squared. Okay? So what is the entropy for a single vortex? It's log of r squared. Okay? So the energy associated with having such a vortex is this if we have a finite temperature the free energy associated having such a vortex this is the energy and then there's a minus p that's log of because x is u minus ts this is u this is ts you understand this log r comes from the log of r squared and ignore the factors of it. Now you see that there is critical value of g of order one at which this becomes negative. Still that happens you do not produce single vortices. You can produce vortex anti-vortex pairs but vortex anti-vortex pairs do not produce the mass cap. Why do you produce a mass cap? You produce a mass cap because right it's been uncountable to say all these things anyway we'll let you just say. Yeah you see what does a vortex or a monopole field it takes the vacuum and randomizes it a bit that is you know when you do not have a mass cap you don't have a mass cap in a theory when correlation functions fall off at infinity very slowly like a power sorry when you do have a mass cap yeah when you're not having one when your mass is part of this when correlation functions fall off at infinity very slowly like a power okay in this situation what happens when you've got a whole lot of vortices randomly sprinkle all over the place because the randomness is in this what's happening here becomes effectively uncorrelated with what's happening there okay that causes exponentially but if you know vortex, anti vortex pair as far as far away is concerned it's like having nothing so vortex, anti vortex pairs that are bound to each other don't cause this randomization of the phase don't produce more the kind of thing that will produce the mass cap is a gas of single vortices and we've seen that in this problem that happens there are only few people in finite energy and only an higher half temperature when the entropy beats the energy so an anti vortex is something that is the current in the other way okay on the other hand in our problem it was very important that the potential didn't go like long but went like one over hour one over hour is highlighted in infinity okay so the energy of a single monopole was finite and therefore monopole gas is produced and once a monopole gas is produced the randomization of stuff happens and the mass cap is all but infinity so I'm sorry this has been very qualitative it's been very qualitative but it was meant mainly to be an introduction to give you the ideas and you're going to work this out in detail in your homework okay any questions or comments about this at in D equals 3 before we turn to D now the integral from 0 to 2 by 1 sub n over n you see what we have let me stay with this half extra it's a bit easier to say okay we've got these integers associated with links okay so we've got n i j for each of them now we use the fact that any n i j can be written as n i minus n i j plus something else that can't trace the water water city okay let me note that something else because that's not important okay so if your n i j was just n i minus n j what happened to this this n we use the periodic approximation for the cosine so this cosine is replaced by sum over s i minus s i plus n a okay plus m i minus m i plus n a to the extent that n i j is n minus n j this is true and what we're doing is the integral is 1 minus 1 by g squared d s d s i but we're also summing over all n i j now you see that the summation over it now define a new variable just s i plus s the integral over s from 0 to 2 pi plus the summation over n is effectively an integral from minus infinity to infinity of this new variable s squared is this clear that's what happens also in the case any other questions or comments what we saw in the 3D model is that what the thing that we saw in the 3D model is that there is no de-confined phase at all that there is no de-confined phase at all that's right it's all very clear to me why exactly is that I mean the model that we started showed this particular problem is it mimicking nature correctly I look well, like I tried to say if there is some other way of producing a compact u1 theorem one easy way of producing one easy, purely, continuum way of producing a compact u1 theorem what was the name? what was compact u1 theorem if you just take the continuum u1 theorem by itself it's a free model and certainly does not combine that's just del nu any of the whole thing square that does not interfere but we want to put in the compact this somehow because the u1 was compact so suppose we now look at the model which is s u2 spontaneously broken to u1 and we can take the scale of the spontaneous breaking to be very high how does the path integral over u1 is how does the path integral over this the u1 h field differ from integral twice s squared it differs from it in the fact that we have to in the full model include instantaneous that is there is some region when you go to very small distances you see that the theory is no longer u1 this happens at very small distances doesn't really matter very much like highly suppressed processes but the important point is that that produces these monopole configurations exist that monopole configurations exist and with some finite action no matter how once that's the case okay once that's the case this deconfinement is basically inevitable the confinement is essentially because the path integral includes the sum over all these monopoles that randomizes the cases that is the basic physics so if you are dealing with any u1 theory in which monopole configurations exist okay then the three dimensions confinement is basically inevitable monopole configurations exist and contribute with finite action no matter how large okay then confinement is essentially inevitable that was the essence of our lesson of our analysis is this clear suppose you read this whole analysis in four dimensions so you read this whole analysis in four dimensions okay then what would you do you have this in any three-dimensional cube you get this monopole like plaquettes but perhaps you would not be too surprised to hear that these monopoles turn into particles that you've got this magnetic monopole because they're actually four dimensions so there's a world line for these monopoles now what you get is that when you read this whole analysis you get these monopole world lines you get to the integral over all these monopole world lines okay you get to the integral over all these monopole world lines interacting with the potential that you get by integrating out by basically a one over del square potential in four dimensions monopole world lines are transmitted inside the cube in which they are arranged no so now we've got you see in every three-dimensional cube we get a monopole but that monopole extends that entire so the monopole world line lives in a sequence of hyper cubes so it's like a tube kind of thing it's like a wire yeah right these monopole world lines that you sum over are closed surfaces okay and so their effect on their effect on long-distance cooperators just like the vortex-antiportex pair is not large unless monopole world lines like this infinite long world lines unless infinite long world lines condensate the vacuum and when you analyze this problem it turns out to be similar to the BKT transition in the sense that this question whether this happens or not depends on the coupling okay a very strong coupling these monopoles do condense that gives rise to affinement an area lower for the as we've already seen from the strong coupling expansion which was damaged anyway but there is a finite coupling at which this condensation of monopoles stops after that coupling the coupling becomes smaller these long world lines they do not exist in the theory you just have these smaller little loops and it's like the other phase of the BKT transition and you recover a massless skater so in four dimensions this lattice model has very different phenomenology in four dimensions this lattice model has a very different phenomenology okay it describes a problem and now you actually you can see this very clearly you see there is a problem of particles interacting by some sort of potential imagine that writing down a field theory for these particles so it's a field theory of these monopoles interacting with some sort of potential and here clearly you've got a choice if you've got a field theory of some monopoles either the field condenses or it doesn't the condensation of that field leads to the confine phase that's the phase in which these long world lines of these monopoles proliferate so what do you think the condensation has a monopole in it? no so the condensation in terms of a field is exactly what the theory is just phi has non-zero expectation so there is a macroscopic number of monopoles in the vector I mean exactly that the classically field has an expectation so in that lattice model what happens in that lattice model what happens is that as you tune the coupling as you tune the coupling from strong to weak there is a phase transition at some point for weaker couplings you get a theory which has no proliferation of these monopole body lines and you get in the extreme weak coupling limit you recover pure free x squared here that can consist of no particles that can consist of these monopoles I mean the scalar field for the monopoles has a non-zero expectation value okay? now that statement is a bit what is condensate? what is condensate? it's like this it's phi okay it's a value okay now it's a bit too you have to be a bit careful about what this means because for instance it is not true that there is charge in the boundary suppose you have scalar field that condenses the charge operator the charge operator is phi star phi dot minus phi dot phi star dot phi since this condensate has no dot both terms in this charge operator as in okay? so this condensation is not like it prefers monopoles over anti-monopoles it's not like there is a next charge but in some classical configuration which as many monopoles as anti-monopoles but it is just proliferate all over the vacuum it's the same charge condensation as used in the hexagon okay? the same word is used there it's the same thing okay? so in the a billion problem okay? in three dimensions and four dimensions for the exact problem we know what happens the summary was that the a billion problem in three dimensions okay? that the a billion problem in three dimensions that the a billion problem in three dimensions has confinement and a mass gap for every finite value of the coupling whether weak or strong the a billion problem in four dimensions has confinement and a mass gap and sufficiently strong coupling but no confinement, no gap, mass gap for weak coupling and a phase transition in the middle and that phase transition is associated with monopole condensation like inks binominal for the monopole field okay? for the non-a billion theory I'm not even sure if my knowledge about this is up to date but in the papers I had looked up when I studied the subject which was in twenty-five years ago maybe somebody knows more about this now I haven't carefully searched but in the non-a billion theory it was strongly expected that in three dimensions everything that was said for the a billion theory would be true also of the non-a billion theory at some stage the situation was there was no proof of the statement but it seemed very, very likely to be true you have to know more this whole story about that okay? the expectation, the strongly expectation perhaps by now there's a proof because it was such an intuitive expectation you know roughly the idea is that if you've got configurations a subclass of configurations that disordered the vacuum you see if you've got an S2 theory then you want configurations that are subclass of such configurations so you definitely have the you want instantons you are making more books so that's sort of disordered the vacuum now you may have more but it sounds like it should disordered the vacuum that was roughly the idea okay? so in three dimensions the non-a billion theory was strongly expected to behave and by now I'm pretty sure somebody has to prove this it was strongly expected to behave like the a billion theory now this is the non-a billion theory strongly expected and perhaps proved to display confinement in any law and every value of the country not just in the stock market so the next lesson is that in three dimensions the strong expectation what you see in the strong coupling expansion is a good qualitative guide to physics and all these finite dimensions in the strong public in four dimensions we have seen that the a billion theory as strong coupling behaves exactly like the non-a billion theory you know, it displays the area of our displacement confinement as we saw in the last class but we've also seen that the a billion theory has a phase transition at finite values of the of the model and so by the time you actually get to the field theory it displays completely different behavior it displays free behavior no mask at the moment now the expectation is that it's different for the non-a billion theory that the a billion and the non-a billion theory will be named differently that is that the non-a billion theories that the strong coupling expansion is a good guide of good qualitative guide for the behavior of non-a billion theories at every value of the lattice boundary now of course there is a huge amount of practical practical evidence for this thing which is a success of the lattice gauge theory program there's a huge number of lattice simulations performed in lattice gauge theory all over the world using this action when they take it to weak coupling with C effectively they can find it is of course very strong evidence for the correctness of this connection but there is nothing like analytic understanding of this see the kind of thing that people like think that would be satisfactory is an identification of the mechanism for producing the mask in three dimensions we have this mechanism it's these monopoles this gas of monopoles the proliferation of the monopoles produces the mask and this happens with a certain inevidentity as long as these monopole configurations exist randomizes these things these monopole configurations have IR finite energy itself produces the mask this is certain inevitability about the whole thing what is the mechanism for that the billion theory suggests what happens is that there is condensation of these monopoles and the billion theory of this condensation stops at a finite value of the coupling ok so the proposal first made by Toft is that what happens in the non-linear in theory in some way of looking at the problem pardon difficulty of making this precise finding exactly what that way of looking at the problem is Toft worked in this max maximum of a billion gauge I can tell you about that then some way of looking at the problem there is also condensation of monopoles in the non-linear theory but that this condensation persists all the way down to very big content van is a conjecture and proving it is one of these clay practices of you know proving it will be a million dollars it's one of these clay foundation practices for a while it used to be as doubted as an outstanding problem of theoretical physics proof confinement of course you know it's often the case with such things that it's not so clear to me at least that it is one of these great problems of theoretical physics there may just be an issue of detail that is if it's true and there's a lot of evidence that's happening by monopole condensation it could be that if you change the parameters of the problem in some way maybe we should discuss this it doesn't matter that the proof does not teach you very much at the end once you have it the proof that teaches you stuff is something with elegance or something with a new physical mechanics if we already know the physical mechanics and somehow just can't argue that the parameters are such that it complies I'm not sure we will learn that much okay so this was meant to be a very brief bird's eye summary the behavior of these lattice models that define lattice gauge theory as a function of coupling the summary was that strong coupling all of them did make confinement weak coupling is what relates to actually what the field is the important question is is that a phase transition as you go from strong coupling to weak coupling in three dimensions strongly believe that both in the non-abelian as well as the abelian theory there's no such and for the abelian theory we will have problems in the problem sense they will do this okay in the in four dimensions we more or less know for sure because we can work out the strong coupling and weak coupling conventions that there is a phase transition strong coupling and weak coupling expansion is very different in non-abelian theory we will never be able to work out the weak coupling so we do not know for sure whether or not the non-abelian theory confines but there's strong difference that it does there is no phase transition in these lattice models from strong to weak coupling in four dimensions okay I'm sorry for the very qualitative later of this lecture there have been almost no equations but you know to do this I'm just assigning this one part that will become qualitative I'll put it into problems part of the reasons that this is qualitative is because we don't understand the time for the two three the two three lectures that would take the other part that is qualitative is that it's just not very well understood like at least when I last looked at it when I last seriously looked at this the non-abelian theory in three dimensions was not that well understood and certainly the non-abelian theory in four dimensions is not very well understood okay so partly it's my heliporty partly it's the heliporty in the field in these lectures are qualitative but just to give you a bird's eye overview of this any questions or comments before we how do we say that well let's look at the gauge group SU2 okay the gauge group of SU2 is a three square okay oh maybe even more SU2 can be thought of as unitary transformation unitary 2 x 2 unitary matrices every 2 x 2 unitary matrix can be written as alpha i alpha i sigma i where sigma i is okay let's take the u1 subgroup that comes with that sigma i just being sigma 3 okay so it's e to the power i alpha i sigma 3 now let us remember that sigma 3 may be one sigma 3 by 2 now let us remember that the eigenvalues of sigma 3 are plus or minus 2 therefore if alpha i equal to alpha i by 4 pi this matrix goes back to c therefore the parameter of this the sketch transformation is automatically written that if you take a u1 subgroup within the space of unitary matrices it's always written that the whole distance is compact so how are you going to produce a u1 that because we have e to the power i 3 that's the rule now when you have you see that the point is that the u1 composition rule is x and y that's that x when you compose x just goes to x plus y that composition rule is consistent whether x and y live on an influence phase ok so group theory doesn't force you to be KPI some physics might but here group theory forces you to make this it's just a fact a mathematical fact of the word logic that unitary groups all the u1 subgroups of unitary groups are compact ok i will have my last 10-15 minutes before getting back to the continue last 10-15 minutes there's one really nice piece of work that the stage 10 is by Poliakov and Sassi on the deconfinement transition for these field theories as a function of temperature in the strong object expansion which i want to show you just because it's so nice and it introduces a new piece of physics in a controllable setting that's far from the beginning this way so the 4 dimensional version of these Yang-Witz the super symmetric version of the 4 dimensional the Yang-Witz series is well understood in the strong coupling the 4 dimensional version of the 4 dimensional the Yang-Witz series in the strong coupling the operation of the 4 dimensional the weak coupling is not does it have such does it have such an analogous understanding? No. If one could develop an understanding from the gravity, is it at all possible to develop an understanding from the gravity itself? But please, what do you want? Where are you going? No, I'm just trying to ask a very simple question. Why? Whether at all, there wouldn't be one who would be able to understand what we're talking about from the gravity itself. And now, to get a sense before I answer, is this question concerned with confinement or not? No. So the very important fact that the super-symmetric version of the problem that I've well understood is the conformity of the theory for everybody of the country, that you can prove without a gravity. And conformity of theories cannot have a mass cap, because the mass cap, the size of the mass cap would be a scale, it's not a scale. So there is never was a confinement issue. OK. Now, it's true that the maximally super-symmetric n equals y-angles theory is understood completely through gravity. Could we imagine a gravitational description of we take up n equals y-angles? Answer is by gravity being really Einstein gravity, on student theory, on student theory, OK? You can see this in a Zillier-Grey's movie. He is the easiest way to see this. You see, a theory of gravity has the feature that it has particles of spin 2. If you take a theory of gravity, there's something distinguished about particles of spin 2. On the other hand, let me answer this question for confining. That's a QCT, for instance. Why QCD could not be possible? On the other hand, QCD has global states. Large and QCD's global states are wall states. And spin 3 blue walls are not parametrically higher in energy than the spin 2. Already, that makes it clear that it's impossible that gravity describes, gives you a new execution of QCD. Because at least in the templates that we have, the global of QCD has simply graviton wave function, graviton excitation. And just the fact that gravity stops the spin 2 makes it clear that a qualitative feature of the actual spec in QCD is not being utilized. However, stream theory has particles of spin 2. And almost certainly, it does give you a new rhythm of QCD. It's when the set goes forward. It's not going to be possible to complete. But the issue is not whether QCD's description is useful. For that, you need to know the description. You see, NDSCF is great not just because of the conceptual thing, but because one side of the correspondence, you can actually do calculations. Unless that happens, it's not useful. OK. Our question's gone. So I'm going to now introduce for you the Hamiltonian version of lattice-cage theory. What do I mean by that? So far, what we've been doing is taking the path integral of lattice-cage theory. And making all bonds discretize in both ways at a time. Now what I'm going to do is to follow. I'm going to do something A-selection. Break the symmetry between space and time. So I will keep time as a continuum direction. But discretize only space. So I'm going to keep time as a continuum direction. But discretize only space. OK. Now what do I do? So I need a spatial direction. So let's do the UR problem first. So we wanted to generate F mu nu F mu nu. Now there are two kinds of F mu nu's. There's Fij and there's F0i. So in order to generate Fij, I do exactly what I need to do. I get 1 over G squared times 1 minus cosine of... Just after we have a different time. This minus, this, this, this. This is the whole thing square. This is as far as the logical structure of the problem is concerned. The potential of the problem. The other part. The other part is where to be continued. So we also had in the lattice description, we also had A0. Some sort of phi field. We also have some sort of phi field here. Living on the time limits. Now there are no time limits. There will be a scalar field at each lattice set. Time is continuous. Now let's just write down Fij... Sorry, F0i is the whole thing squared. Now what is F0i? F0i is del i A0. Minus del 0 ai. Let's call it del 0 ai minus del i A0. F0i. Okay, del 0 ai. We'll just create a script by... Each of these, there was a link variable here. We'll just get the time there if you go back. So suppose I have a link variable. I'm going from the i to the j. Phi, i, how everything is. So there will be a dot here. But remember there is also now a new scalar field on each side. Which was A0. The new scalar field, let's call it di i. So do you see that F0i, the whole thing squared, is simply this object here. Is this clear? Sir, why did you get... So chi i is del i A0. So chi i is A0 at side time. What? So x0i is del i A0. And that's why I got minus and plus. This difference gives me a del i A0. And this thing was just del 0 of ai. So now we're going to write down the final model. That the Lagrangian is equal to what? So we'll write down the final model and then get factors of the lattice coupling correct by taking the time. So first I'll write it down without factors of the lattice coupling. So the Lagrangian is phi i and... So it's difficult to write this notation. So i plus nj i plus nj dot plus... Okay, so this is del 0 ai minus del i A0. So minus phi chi i plus i i plus n. Minus... Minus this thing here gives us... This thing here gives us the correct infinity description. Now remember that phi was to be associated with k thanks to this little thing. If we take a middle of a lecture now, could I go back in maybe half an hour? Okay. Yeah. Fine. So if we plug this in here, what does this guy become? This guy becomes approximately... This guy becomes A... This guy becomes A to the 4 because the derivative requires a 1 by A. So there was an A with the gauge field. And then we wrote the difference but to turn it into derivative we need a 1 by A. Okay. Okay. Whereas this guy here becomes... This guy here becomes ai ai dot squared E squared. So this chi should be identified... This chi should be identified with... But this is correct. This chi is just identified with A0 because then this is minus li A0 squared with an overall A squared because the difference converting it to the derivative picks up a factor of A squared. Okay. Since we want a uniform factor of A squared outside the whole thing, what we should do is to either multiply this guy by A squared. Okay. And so that this whole thing now becomes A to the 4 by g squared summation. And then if we want to convert the summation into an integral, summation times A to the power D is integral. That's A to the 4 minus B integral is squared by g squared. Okay. So if we want the actual Yang-Mills company, the actual... We want to parametrize it in terms of the actual Yang-Mills query. You know, this object here, g squared by g squared is g squared by e to the power 4. Okay. But we'll keep this g squared because g squared is the dimensionless object. Okay. This action as I've written it now is manifestly dimensionless because manifestly dimensionless if you assign this guy dimension 1. This guy has a face that has no dimension but has a time derivative. So that's dimension 1. A squared is that up. And then you assign this mass dimension 1 because it's A0. A0 at mass dimension 1. This is the manifestly dimensionless object. So this is the... Actually, we're going to do that. Okay. So this is the asymmetric version of lattice-cage theory that brings time to fluid space. Okay. Any questions or comments about this? Why? Basically, we have seen that we need to put an expect value of time value. This is my... Yeah. It's just to make it dimensionless. Yeah. It's by 5 out squared. That would be what makes it. You see, this guy's clearly a dimensionless object. Okay. Every time the Lagrangian has to have the same dimension, so if you put 5 out squared, you need something like that. So this is the Lagrangian we're going to be dealing with. Okay. But this continuum lattice-cage theory is useful, not in the Lagrangian, but in the Hamiltonian. Okay. So in order to understand this theory, what we're going to do is to move to... We're going to move to the Hamiltonian. Other than that, we're actually going to closely parallel something we did in the early part of our course. When we wrote down the path in temporal for reactants here, we moved to the Hamiltonian framework to understand what universe space this was described about. Okay. This, in fact, is a pretty precise version of what we did there. Because everything's really concrete. There's no continuumism. And yeah, it's really... If you put it on a torus, it would actually be completely correct. It doesn't move back and forth. Okay. So it's going to be very similar to what we did there. So let's do it very briefly. Suppose we've got some phi ij. Okay. We want to find the canonical dimension of the free. So this phi ij is simply the derivative of the Lagrangian with respect to this. We've got a factor of 2 here. So it's the derivative of the Lagrangian with respect to this. And so that phi ij is equal to a squared by g squared. So phi ij is equal to a squared by g squared. Phi ij dot minus chi i into 6. What is this? Canonical moment. Oh. Let's complete the Hamiltonian. So the Hamiltonian is h is equal to p ij phi ij dot. And by the cost of canonical momentum for chi was 0. So we're going to give it a Hilbert space of interpretation without trying to, you know, quantize chi. Chi would be Lagrangian multiply, which exactly like we did in the Hilbert course. So we've got p ij phi ij dot minus Lagrangian. Before we write this as p ij p ij. So we just invert this. So we'll write phi ij dot is equal to g squared by a squared p ij plus chi i into 6. Okay. So we write this as g squared by a squared p ij into chi ij a squared by g squared by a squared p ij plus chi i into this part here. This part combined with this term as we did before is just simply trips the sign of this space. I mean it just gives you a factor of half of this space. This minus this. So it just is p ij p ij g squared by a squared. What? Yeah. Plus two factors you see over here. Okay. Plus whatever we got from this potential term. This is one minus cosine of. Okay. Which comes with the same one by two g squared Now this e squared is a plane in the naik. And we get that rid of it by a rescaling of time. And a rescaling of the sky. Okay. So imagine we do that again. We rescale time. So time is now dimensionless. T is t by a. And we rescale chi. So chi is chi bends. A. So we got rid of this a field. From now on we are going to have main contact with physics later on which we instated. But because of that I am just going to give you an example of this a squared looping discussion. You are going to easily instate it. Underneath where the image is. Is this clear? One thing that is really important here now is that in this Lagrangian, in this Hamiltonian. And then of course I have got this part. There was class pij chi chi minus chi. This is the Hamiltonian. Firstly we see this fact that I need to pick the law from the beginning of the course. That is what this is very straight forward simple Hamiltonian system. Plus this Lagrangian. Doing the integral over chi's. Generates a constraint. Okay. And we examine the nature of that constraint. What is this Lagrangian? What was this Lagrangian? This Lagrangian is just some pretty simple. You know it is like a Lagrangian of some spins. There is some kinetic term here. And there is a potential term. The one thing that we see immediately is that the strong coupling limit for this model is extremely simple. Because the strong coupling limit zero is to infinity. So the Lagrangian formulation, the strong coupling limit. The leading term is one. Now it is no longer the case. The leading term is this object. This does not go away but this is not. So what is the strong coupling theory? In the strong coupling limit, in the limit chi goes to infinity. The strong coupling limit of this Hamiltonian lattice gets there. It makes it a free theory. Remove the potential. This was electric field. This was magnetic field. You throw away the contribution of the magnetic field. And it becomes just a free theory. What do you know about the constraint? You are saying that the free theory, because magnetic field used to have the additional limit of strong coupling just from this interaction or... It just looks like a non-linear place. There is potential. See, in any quantum mechanics, but kinetic terms with potential terms. But when you expand with some alphabetic terms. What? When you expand cos, you will again get some alphabetic terms. Yeah, alphabetic terms. So in the limit chi goes to infinity, all those terms can be ignored. So the strong coupling limit of this model is an extremely simple model. Okay? Let's say, suppose you don't have a Schrodinger equation now for this problem. Okay? What is the Schrodinger equation? The Schrodinger equation has... The Schrodinger equation has is a wave function. It's an equation for a wave function whose values are phi x, these linked variables. Okay? Pij, by the ordinary rules of quantum mechanics, is i del by del phi x. This just gives you del squared phi x. We know the solutions, right? We know the solutions for del squared. The plane waves. Phi, however, are angular variables. So they're plane waves with quantized... with quantized piece. So if we ignore this constraint for a moment. The Hilbert space of our problem is extremely simple. Okay? The Hilbert space of our problem is given by e to the power i to pi n i j, pi i j. Okay? Product of such things. Well, you choose a separate n i j for each link. Sir, but Pij also consists of this guy's link. So it's not a pi... No, no, no. It's the canonical momentum conjugate to phi. According to the rules of quantum mechanics, the canonical momentum conjugate to a variable is implemented by the operator del by del x, that operator. Doesn't matter whether it's phi or the operator. Well, it's... Did I have to correspond with this one? The the object whose Poisson bracket with your variable is 1 becomes the object whose operator is i d by d, that one. No matter what it was. You see, now we've gone to the Hamiltonian framework. We don't even know what p was in terms of i. That's not our business. This product overall, for some choice of n i j's, gives you the basis set of wave functions for the space. Okay? So we work in the spaces. We work in the basis of n i j's. The momentum basis. Okay? So in the basis of n i j's, what's the Hamiltonian? It's just n i j squared. Because when we took the cutting model, we got a very simple example. So, you know, we got that by taking the cosine and approximating it by its quadratic. Here we just throw them in the cosine. Yeah, it's not like that. I mean, it's not like that. So, when we take the model, we get this plane game solutions as the basis. In the very, in an appropriate limit? Presumably, yes. Yeah, I think the answer is yes. I'm strong on it. Okay? Now, the only question here is, what does this constraint do? What the constraint does, well, what we have to do is just do the integral over all chi i's that gives you some delta h. Can you see that that tells you, remember chi is little points. What this tells you is that the sum over, what are these n i j's by the way? These are electric fluxes that liverly is. Because n i j's are the ideal values of f 0 i. Moment of conjugate to phi i j was just f 0 i. Okay? So, these n i j's are the quantized values of electric fluxes that liverly is. Okay? And these five, okay? So, what we get here is that the sum of all n i j's at a particular point must match. Just to keep track of the signs. The signs are such that if you draw all the fluxes inwards to the same point rather than summing them somehow. The sum of all electric fluxes ending in the same point must match. This is a continuum version of, a discrete version of what famous continuum equation. Because del dot e is equal to 0. The handball space is this, subject to the constraint that sum over n i's in the sum over the n's ending on any given point is here. Okay? So, this is very simple. This is very simple. What a beautiful, simple, beautiful theory. Now, we buy a, of course, stroke up lift. Okay? So, it's very far from the active continuum. I guess. Far from turning this into a i dot square. It's really around this flat rate over everything. It's very far from the active continuum model. But anyway, let's take this simple continuum model and ask, so it's simple, the last model, and ask, is there an interesting calculation? There is an interesting calculation. And the interesting calculation is to compute its thermal free energy as a function of time. Okay? So, what we have to do, what we have to do, what we are trying to do is to compute this of, you can define this, we can h over this there. For this h, we understand very well. You got it on the board. We understand the handball space very well. Okay? Had it not been for the constraint, this would simply have been sum over n i j product over j e to the power minus beta n i j squared g squared by 2. That's relatively, okay, very important. But there is a constraint. And the constraint is, okay, and the constraint is that this implements values. So that we've got e i, okay, e to the power i p i j i i squared k. That implements the energy. That implements the constraint. That implements the constraint that the sum of fluxes coming to any side is here. The sum over n i j. No, it's just this, I'm just rewriting to the grand mind client. Yes. Okay, rewriting. The physical constraint was that the sum over n i j is going to a point this here as we saw. So the sum over n i j is going to a point this here as we saw. Doing the integral over i i is, forces that. This is clear? This is clear, right? Now, this is the mathematical expression for our, sorry. This is the mathematical expression for our partition function, okay? There is a mathematical expression for our partition function, but this mathematical expression involves the sum over all integers. Now, whenever you've got a problem in physics, where you've got some particular function, summed over all integer values of that function, it is often useful to use Poisson's equation. To reconvert, to look at that sum of integers, in terms of a sum over another second. So let me just remind you of the statement of Poisson's. Sir, this Hamiltonian actually is a free Hamiltonian with a certain. With a constraint. But maybe it's just to differentiate that in a large, rich, complex element, can we have sum over the second term because there's a whole difference in the first term? No, but you see, this constraint sums the allowed values of energies. Constraints the allowed values of energies. Do you see? Yeah. You see, we could do it, just to remove that constraint, if you only summed over those energies. Which allow. Which allow. But you see, that affects it on a leading order, right? Because configurations which the energy don't satisfy that. When you contribute on the same order, those that do. So this is a leading order thing. Remind you of the statement of Poisson's. The statement is that, suppose you got f of x plus, so you got some function f of x. The summation over f of x is equal to the summation over f tilde of x. Where f tilde is a Fourier transform of f. Once you define the Fourier transform of f of x. Okay, it's got some name. Personal. Some kind of purpose. It's a simple term. This result to this point. So this is of this form, when we identify. So let's, for any particular energy. This object is of this form, when we identify this f as e to the power minus beta g, g squared by 2 x squared. Plus i times x, chi i minus chi g. That's clear? So, if I want to use this theorem, which I do, all I have to do is to compute the Fourier transform of this object. Fourier transform of this object is added each of the part a dot x, do the Gaussian integral. So, what does that do? That has chi i minus chi g plus k. Then the whole thing squared. So, we complete the square and we also get a Gaussian. So, you will, do you see that after some factors, this will be 1 over beta g squared. It's important that we get this weight. When we complete the square, we get 1 over beta g squared. Times, get minus i, chi i minus chi g plus k. This person, the person who has got the name, right? Applying this person, this person converts it. So, this is the integral factors of 2 and so on, ok? That you can look up in the papers also, in this section. This is d chi i. This is d chi i. Exponential of minus 1 by beta g squared chi i minus chi j minus 2 pi m i squared sum over m i. Because you have to take the Fourier transform, replace the argument of the Fourier transform by integers. There's a 2 pi over the place. I'm not in charge of this 2 pi. It's dramatic, ok? And, yes. Is it clear that the partition can be rewritten in this form? Remember, g is very large. Earlier in the lecture, we encountered this object. Sum over this object. Where this g is very large? Where we encountered this? Large. What? Stop. Large in. Yes, yes, yes. We'll be taking some time. I just wanted to ask, would you see this expression before? This is an approximation for something simple. When g is very large. Approximity, no one. Come on. We were looking at si, si, si, si, si. That kind of thing, more universally. It's an approximation of 1-cosine. This is an approximation. In the larger element, this is a very good approximation. b, chi, i, as we said. This is a very powerful lecture. An explanation of 1-cosine of chi, i, minus chi, j. The whole thing, I'm going to put it to be of one select. Do you see the beautiful thing? The thing is, the difficult thing in solving this problem was the sum of integers. Sum of integers are a big thing. Integrals, we can do sum of integers in every class. That's why number theory is a hard subject. Maybe father was in this one. Okay. So, okay. But, now in the approximations, we got, this is going to be an amazingly simple model. And the amazingly simple model is in fact one that we wrote down earlier in this class. It's called what? We wrote it down in two dimensions. We'll be getting there. When you say that this has turned into the partition function of a spin system. We don't have any of the sums in A and I. Yeah. This is, no. Everything is not. This is now just, it's exactly this. Okay. This makes perfect sense. Spin is laid at sites rather than at links. We've integrated all the limits of freedom by doing the sum of integers. And we've got an effective action for the side-degree of freedom, which was the current model. By integrating all the physical stuff, we've endowed dynamics to the current model. And so the grand multiplier has turned into a spin. Okay. Now, now what do we see? This is exactly the partition function of a spin system. It's said that the temperature has come downstays rather than upstays. That's why we could just use what we know, right? Since we're all experts in physics, we all know that at high temperatures a spin model has no... Let's have a look. Okay, let's have a look at doing this in four space-time dimensions. Then how many dimensions was the spin system living? Three. The matches was three dimensions. So this is just a discretized spin model. This is just a discretized spin model in three dimensions. And we all know that at high temperatures if there's one phase, at low temperatures, that's the other phase. Tell me what the two phases are called. Order and temperature. Order and dissolve. Which is the order phase? The low temperature phase. Low temperature phase is the order, where there is... Spotting is made for symmetry and some alignment of fives in some direction. It's called planarizable. The fives are all oriented. At high temperatures, there's randomness order. What? In the high temperature, there will be order. Exactly. I'm just talking about the ordinary spin system. Oh, right. Okay, so now in this model, and... Yeah, now let's write beta as one of the temperatures will be greater temperature. Okay? So we can talk about temperature and beta. Okay? So there's a system at high temperatures. The system is order. At high temperatures, phi gets this expectation. The chi's get the expectation. The chi's get the expectation. At low temperatures... It'll be zero. Yeah. So because high temperatures, the exponential factor is very large, this thing doesn't want to stray very far away from... So all chi's are in the same direction. What? No, no, I'm just telling you about ferromagneticism. I'm just asking you. We could analyze this model because I'm appealing to you to use your knowledge of ferromagneticism. You know that for a ferromagnet. At high temperature, you've got disorder. At low temperature, you've got order. This model is not an easy model to study. It's ferromagneticism. You can understand the phase transition in general. But this we know, right? Let's not get into analyzing the model. Let's just use what we know. This all of us know this, right? It's just that here, the temperature, the role of temperature has been reversed. So what we get is a phase transition in the system at some value of T of order G square. So we get a phase transition in the system at some value of the temperature of order G square. And this phase transition basically is a transition between the phase in which we've got chi ordered and called chi disorder. Now let's look at this phase in this transition a little. What was chi actually? Chi was actually a Lagrange multiplier. Now when is a Lagrange multiplier effect? You see, when is a Lagrange multiplier effect? It's effective in enforcing this delta function. Only when it runs over all the values it's supposed to have. If for some reason chi is effectively frozen over some range of values, chi i minus chi t is effectively frozen, then this Lagrange multiplier is not doing its job. Now this may seem like a facetious argument, but that's actually correct. And so you see, where high temperatures this chi becomes ordered, the Lagrange multiplier, effectively because of dynamics, it's been frozen to near a particular value. And what that effectively means is that the model, approximately, this term is not playing, that constraint is not very important. I don't know, it sounds like inverting logic in various ways. To see this clearly, what you should do is a calculation. To see this clearly is what you should do is a calculation. Let me just say it again, but it's some effective potential for this chi. This effective potential is some steep potential. Sharply localizing chi to some value, then we shouldn't make too much of an error by taking the original model and replacing chi while sharply localizing value. Because in the end the dynamics is not getting substantial contribution away from those locations. We were well into the magnetized phase. So very high temperatures. So the chi, chi minus chi j is very sharply localized to zero. Or to some integer multiplied by 2 pi. If this figure just replaced by zero or some integer multiplied by 2 pi, why in the beginning we shouldn't be making too much of an error? Integer multiplied by 2 pi or zero is the same thing, multiplied by integer. That's like not having this term at all. It's clear. There is a constraint on the energy because the issue is, is that constraint affecting the answer or not? The constraint is always there. What we are seeing is that, in a particular regime, there will be very high temperatures, that constraint is a small change on the actual answer. So the final answer, you get more or less the right value by ignoring the constraint. The constraint is a small change on the actual answer. So why is the right value by ignoring the constraint? The constraint is a small loop because the energy constraint happens in a lot of configurations. Whether the higher load energies are the same. So like, some set of configurations are disallowed. Some set of configurations? No, no, the set is not changing. It's just the following question. It's that, though we are disallowed, suppose you do two problems. The first problem is some of them are all configurations. Second problem is some of them are problems giving you more than the same answer. So the configurations that are being disallowed, are they contributing significantly to the answer or not? That's the question. What we are seeing is that in the high load pressure phase the constraint is not making a problem. Now I should emphasize that we have to go to pretty high temperatures for this to be really true because just having a potential you still have some fluctuations. On the other hand, in the low temperature phase chi i minus chi j is completely unconstrained. So there ignoring there's no way, there's no sense in which we can set chi to even approximately some particular value. So in the low temperature phase physics is dominated by the constraint. In the temperature phase the constraint is not making much of a difference. This is a famous this is the lattice strong coupling lattice gauge version of a famous phase transition in continuum Youngville series. Can somebody tell me what the name of the transitions is? That's a punctual question. This is a class of strengths. This is a famous transition. So you are talking about transition gravity. Transition gravity. It's called the deconfinement transition. The deconfinement transition. The transition in QCD above which quarks are liberated. Actual quark states can propagate around without having invented it. And you understand why this is the case. You see if this what is the reason that a quark and anti-quark pair have an energy proportional to the length the reason is in the Gaussian because the quark produces a flux and because of the Gauss law flux cannot has nowhere to go. So it must continue, it must continue until it needs its activity. Since it's going to have therefore flux tube of order length that gives you the linear confining potential between quark and anti-quark. If the Gauss law is not important these are even no longer works which is really to say that the stuff can end somewhere in the thermosy or something like that. So in the phase where the Gauss law is not dominant in phase numbers you might suspect and be correct in suspecting. That this linearity of potential is no longer correct. There are many other ways of saying this let's leave it at that. On the other hand in the lot of countries this Gauss law dominates the physics and the theory is confined. So this is a transition confined with the linearity. And the really convincing way is to compute a quark quark is to add a quark to this quark add a charge to it and then compute correlation functions of the charge and the charge object in this model. This can be done. I'm not going to go through the details you really see in this transition between quark quark and linearity. Sir, is it somehow possible to explain the Ising model by taking this model instead of like kai i minus kai j if you can get kai i dot kai j then is it possible to study the Ising model? This is like the O2 version of the Ising model. If you just restrict your kai i is to be either Paso minus 1. This is a continuous version of the Ising model. But I just want to say that we have not in this class done any independent analysis of these of these global models. I'm just appealing to your your deep understanding of these models. To quickly reach the answer. I'm not claiming to have enriched your understanding of these models. I'm using the understanding of these models to understand these things. I said that we generalized for SU2. That's going to be one of your own problems. We're going to read out this calculation for SU2. You'll see that it's not different. It's a non-Amedian top. Non-Amedian spin top. It's quite fun. So understand one of the utility... I think I will not touch your lattice cage theory again in this class. In the next lecture. I'll return to the continuum and tomorrow one day in BRSC quantization. It's all boring stuff. But I had this two lecture in terms of lattice cage theory for two reasons. First reason is because it's very important. Lattice cage theory is the principle tool to obtain numerically accurate results in attenuation. You know let me say this again. What is the job of a theorist if for this purpose saying excluding numerical work from the ambient theory? By theorist I mean pencil paper. What is the job of that theorist? A, to produce the right equations for calculus. Perhaps some idea of how these equations behave. However, in the real world it's often the case that actual physical systems and equations whose behavior is so complex that pen and paper work we never learn. Not till the end of the universe. Produce exact results for most questions we're interested in. You have a purest notion of the idea of a theorist. You should get the exact answer with pen and paper. For most problems we're interested in that match. The job of a theorist is first to find the right equations. But secondly, this is another very important thing is to populate the space of mathematical models with solvable points. So that by looking at every solvable point in the space of theories, you can get qualitative expectations for some non-solvable point. And then, if you're really interested in the numbers, do you manage to get it? Okay? So this lattice gauge theory is played both ways. Okay? Firstly, of course, its primary use is that it is the principle technique of actually numerically stimulating a very hard problem. But we can never do otherwise. We've got the spectrum of examples in the real world. The spectrum of protons. And I ask the proton, how do you ever compute? It's really important, really. That's one thing. But there is a second use of lattice gauge theory that I wanted to emphasize in my two lectures. As giving a system which you can deform away from the theory that you're interested in, then with the actual continuum, to solvable points like the strongly coupled limit lattice gauge theory. Then when you go to these sorts of solvable points, you can ask in those definitions the analogs of questions that have great interest but of great difficulty in the actual theory. Okay? This problem, the problem of deconfinement is a problem of extremely interest. It's a vague that actual M's theory is a function of technology that goes phase transition. From confine to deconfinement. It's very important to beautiful phenomenon was key to, say, early universe physics. Okay? It's a very, very important thing. Great importance to understand the real world. Have you seen by taking the lattice gauge theory? Going away from the end of interest. The end of interest is a weakly coupled, because that's what approximates theory. But on the lattice, there is a matter of the limit. The other end is stronger. There, you see, in fine lines we could solve the problem of our elements. Find the deconfinement transition and put equations to this phenomenon in this problem and so understand it more deeply. Okay? This is a great, this is the kind of thing a theorist should be doing. Okay? To understand deconfinement, how else will you proceed? This is the kind of thing you should be doing. So, that was the other use of lattice gauge theory. There is somehow other emphasizing in education. Producing classes of models which can be deformed away from the problem of interest and then solved to give you great intuition. For what might actually be happening where the problem is of interest. Okay? This is the angle that I tried to emphasize in my tour. I didn't try to abuse you with my numerical abilities after these observations. This was something that, an activity that happened with great figure in the 1970s and the 80s, especially in 1980s actually continues in the tradition of condensed matter model in physics. Now, it was very more fun. But anyway, from the point of view of biology, this is largely an exercise for fun now. It's hardly ever used. I'm going to move away from this discrete model of studying back to the continuum next time. Any questions or comments? No. We only spoke about the strong coupling element because that is the element that we could easily understand. Now, the weak coupling element then the first term would be the weak coupling element would be a limit the weak coupling element would be a limit where the kinetic term is a perturbation on the potential. And that, of course, becomes quite difficult. Yeah. Why can't we just throw it away? We use the same logic to throw away the potential term in the strong coupling element. So, in the weak coupling element, why can't we just not consider the kinetic term because no, because that's similar. It's like this. And we had potential. Suppose we had quantum mechanics and we had some potential that was very steep compared to the kinetic term. Then we could get some wave function that is very highly localized. Okay? And in some extreme event, this thing will go to some effect of delta function. We'll go to that effective delta function. P squared evaluated really on a delta function is in there. So, it will never really be a delta function. Let's there's some there will be some smearing. And in that smearing P squared and the potential will always contribute comparatively. Let's look at harmonic oscillators very largely. Let's look at ground state wave function. It's always true as an exact statement that half of the energy comes from the kinetic term. And half of the energy comes from the potential term. So, you can never really know the kinetic term because what this does is it starts squeezing the wave function. And as you squeeze it more, the momentum starts contributing more and more. And the size of the wave function sucks that there is a balance between them. Actually, the weakly couple problems will be complicated problems. Because all of these problems in the 80s have been analyzed some attempted analysis. But I'm not aware of it. I could really argue that. The extraordinary problem on the other hand was very simple because it made a free counter to this. That's why they analyzed this. Very simple. Other questions? Very good. So, in the next class we will try to turn to the renormalization of the optimization of the beat up. Somebody knows what to do with this thing.