 So, we go on with our work and I'm going to tell you about the last components of conformal filtering, which we need to know. And then after that, we use this knowledge to analyze the various estatomic models, which we come across. So, conformal filtering, I had said this before, but I will recap. So, I construct the Hilbert space by taking some state of weight H, which has the property, it's the eigenvector of the L0 operator, and it is vanished by all N positive N operators. And then the Verma module is formed by estates like this, so that from this H you take, you go down by multiplying L minus N's, and you can multiply as many of them as you like. But each combination forms a level, the level is such that, so you'd have H and your level is such that you have an N here, N here could have been produced by L minus 1 to the power N, or by a simple L minus 1 or any combination in midfield. So that all these K's, which are used to produce a state of level N, they sum up to N. Therefore, the number of ways I can do it is the partition of N. Partition of N is defined as the total number of ways. You can add up digits to sum up to, so I have a Verma module starting with H at the top, the grandfather, and then his children. All his children come below, and at any level you have here PN estates. This I call a Verma module. It's a sub-space of the Hilbert space, which has been constructed like this. And I will therefore have one for H bar, treating H bar not as the complex conjugate of H, but as an independent number from H. So H and H bar are two numbers, both of them are in this case real. And then I take a direct product, so for H and H bar I have a direct quick and then add up all of them over all H and H bar, and this will give me my Hilbert space. So in this way, a Hilbert space is constructed, and I have the operators already acting on the Hilbert space, which I had already defined as the field operators. So that almost completes my work, yes. It's not irreducible. If they were irreducible, our work would have been finished, but they are not irreducible as far as I have explained. So if these were irreducible, I would have now had a fully functional Hilbert space. The problem is that if it is not irreducible, if it is reducible, then it's possible that you have sections of it which rotate among themselves. And if I have this situation, it really means is that I have another verma module which I have not considered. So I have to extract all these things out of them, and just I'll be then just left with irreducible modules, and then I sum over all the irreducible modules to get the Hamiltonian, to get the Hilbert space, yeah? I talked about this state operator formalism that presumably this H state is related to the vacuum by some field operator, phi of H, and then they will have a correspondence between H and phi of H. So if I construct a verma module for the H, correspondingly, there will be a set of operators associated with this structure. So I have all the fields, which are all the field operators which I need also, and basically solves my fundamental problem, which is a Hilbert space and the set of operators acting on it. Then I can go on to do my calculations. First calculation which we can do relatively easy is that we take the two point function of two operators like that, and I showed you that simply by symmetry, you can derive this. I didn't show it to you completely because the last step I left out, which was the inversion, but you can do that yourselves, and you get this four. The same trick can be used to get the three point function. The three point function is fixed up to this constant C, and this constant C will not be known unless you specifically tell me which conformal filtering I'm dealing with, and then I can extract the C's for that. The next step is the four point function. The same game is played with the four point function, and you get this structure which has the right scaling properties, but the problem with four point function is that you then have this cross ratio. This cross ratio you can show is invariant on the conformal transformations, and therefore it is possible to have a function of eta multiplying this set. I don't have any way of determining f of eta unless you give me something more about the theory, so some sort of differential equation that this must satisfy, and then I can derive f. Higher point functions will be much harder because when you have five points, you have more than one cross ratio, and then you have a function of two unknown variables in the front. Yes, two independent. I cannot. I cannot. It sort of needs a presence of mind, which I don't have. Fine. We come to the point of the problem that there may be irreducible spaces of this module, and the argument goes like this. You have here n vectors which arrive out of this structure, which I call a level. The n vectors sit here, and it's not, I didn't give a proof, and it's actually not true that all of them are independent of each other. So it's possible that some vectors within this set may be related to each other. To give an example, the best thing is to look at the level two, the simplest level is level two, and here for level two, I have just two vectors, minus two. Two of h and l minus one is squared of h, and if they are not independent of each other, there may exist a linear relationship among them. Suppose such a relationship can be found. Then what have I found? I have found an operator which annihilates h. So here it looks that I have an operator which, if it hits h, h is vanished, vanishes h. Therefore, h is a null state in the sense that there is an operator that kills it. So what does this mean? It means that I don't really have two independent states. I just have one independent state at this level, and when I come down from h, I cannot create all that comes under h, but just only the ones which come out of one of the independent operator. Since existence of such a relationship gives me the ability to look for values of h which satisfies this state. So for all h, this is not satisfied, but there are special values of h for which this relationship is satisfied. This is the example for the Ising model. I will tell you in a minute that I can then, I can associate conformal filteries to statistical models, and each of these models is identified by a number of states. There is a state of weight one-half in the Ising model which satisfies this relationship. And therefore, finding values of h which satisfies the null state will also give me, will serve two purposes. One is that I will find my irreducible towers and complete my construction. And the other is that since it doesn't happen for all values of h, I will find a specific value of h for which this is satisfied, and hence identify my conformal filteries specifically. So once you do that, you find a whole series of h values which give you null states. And these are all the h values which give you null states for values of c between zero and one. And this proof was completed by Katz quite a while ago, and it's a rather involved proof. I don't really want you to know the proof. But we are looking at conformal filteries in an applicable way. So instead of going for the proofs, we want to know how we can use this knowledge. So there are states for each value of c, the central charge, then there are a number of weights which are given here which give rise to null states. And then these, in fact, identify each verma module which I have to construct. And when I construct all the verma modules for each c, I have the complete Hilbert space. And I have my conformal field theory. So instead of the proofs, let's just analyze the numbers which have come out here. I have a, I gave you a sort of half-baked argument that c has to be positive. Another argument is added to that that for this kind of construction, c has to be bigger, smaller than one. So I need to have c a smaller equal to one or bigger than equal to zero. This then tells me that c is equal to one minus six divided by m times m plus one, and m can be any number from two upwards. So at m equal to two, c is exactly zero. I gave an argument yesterday that this is an empty CFD or it's a trivial CFD. Trivial in the sense that it just has the operator one as the operator contact. Then I can take m equal to three and c comes out to be one-half. If, for that case, then I have one conformal rate of one-half and conformal rate of one-sixteen. If you work these out. Because when m is equal to three, p can take three values. And q takes values between, sorry, p can take two values and q takes values within one and p. So you have just three possible states for various values of p and q, and these three possible states are these guys. And hence, you just continue with other c values and other m's. Finally, we now try to identify what these states are. These states which we have here. So these numbers imply that for I have one state which is p equal to one, q equal to one. And that is the vacuum state. It has to be, because it has zero weight. Then the next state comes out of p equals to two, q to one. And then another state p equals to two, q equals to two. We don't know yet, but I will tell you that this is the case that these theory, this theory of c equals to one-half corresponds to the Ising model. So these two states have to correspond to the state to what you can construct in the Ising model. And it is easier to identify it with operators than states. So this, and this state are two states, one-half and one-sixteenth, which correspond to the operators of the Ising model. And we have no choice. We just have two operators in the Ising model which we can call upon. One is the spin and the other is energy density. And so on, we have to continue this game for every m. This is the simplest m. And we then find a table. The table continues for all m, I have only written the first four. So the following statistical models come out. The Ising model, the tri-critical Ising model, tri-state parts model, tri-critical tri-state parts model, and so on. What do you mean all include? All what? Not all statistical mechanical systems, no. Sorry, I can't hear you. I don't, sorry, I have to say that sentence something, somehow other, some other way. I cannot understand what your meaning is. I think the question is something like, would a three-dimensional Ising model also correspond to any question to conformal field theory? Okay, the three, that is the question. The three-dimensional Ising model, because it is three-dimensional, will not be in this table. So it is possible that there is some other conformal field theory, but a three-dimensional conformal field theory which will correspond to the three-dimensional Ising model, although it's not found yet. But another question could be that can you then, for every two-dimensional critical model, find the corresponding conformal field theory, and the answer is no. There are, and we have found for a large number of them, but not all. However, it's a matter of belief, I guess. Whether you believe that this procedure is robust enough to find everything. Central Church has a physical interpretation in that it's related to the energy density of the system, but it's a very loose interpretation. So here I can ask a very, very interesting question. Suppose I put as this top estate the vacuum. Then the question is, what are the estates which I can construct as the descendants of the vacuum estate? So I come down, the first thing that can act on the vacuum is L minus one. So level one, you only have L minus one of zero. And this has to be zero, because L minus one is just D by DZ, and what this says is that the vacuum is translationally invariant. But also it tells me that this is my knowledge state in this value, in this Verma module, because I now have a relationship like that of the knowledge state, but it's a very simple one. An operator acting on H is giving zero. Next question is, suppose the next estate level will be created by L minus two on vacuum, and note that L minus one squared gives you zero. So at this stage now I have just one estate. And the reason that this is happening is because on top of it I have a knowledge state. So the Verma module is somehow reduced. Now this estate L minus two is an interesting estate because yesterday I did a simple calculation which I showed that the central charge is related to the norm of this estate. So in fact this central charge is given by this, the norm of this estate. And worse than that, if this vanishes, see vanishes and then the entire conformal filter will be empty which will be this possibility when see vanishes. So in some sense see is telling you that there is some content in your theory. Now it's a little overstating to say it's the energy density with some content in your theory. Sometimes see is negative. Then we have a non-unitary theory. If you have a negative see, this means that the norm of this guy is negative and having a Hilbert space with negative norm of the states means that you have a non-unitary theory. But doesn't mean that it's not physical. So in this box values of see lower than one half are not written. But there are two other values which are famous and of course there are other as possible estates. This gives me the percolation model and this gives me the a billion cent part. They are both respectable models in statistical mechanics and very well studied. They come out, one comes out of a negative norm CFD and the other one comes out of a theory which should have been trivial but is not. So this is not the only possibility for see equals to zero. I can construct a see equal to zero theory but non-unitary. This emptiness is imposed if you impose unitarity as well. And there are other see equal to minus one see negative theories and these are non-unitary. In general it's a Pandora's box. Once you open the door to the possibility of having non-unitary estates, then you have removed one of your conditions views which you used for proof of things and then it is very difficult to work out the operator content and so on of these theories and therefore these theories are still, there is still a space for work to work out the operator content of these theories. I will not go into that, it's a very difficult problem. But the nice and clean result is this, that I have a number of models which connect up to CFD. Okay, the good news is that I'm not going to tell you anything more about the CFD, but I said in this lecture and a half is all we are going to need for the purpose of this lecture. So next step is to analyze a few systems which we are familiar with. So two very famous models. I'm going to analyze with all these machinery which I have introduced. And it has two purposes, this part. This is to doing these two examples. What I will get is a working knowledge of the concepts which have been given so far and also see how they are implemented in case of these two models. The good thing about these two models is that we already know a lot about them. So let's try and do this. The first point is an Ising model. The 2D Ising model is a network of spins. Spins are set on the nodes and they can be parallel or anti-parallel, parallel has a lower energy. So the tendency is to go to parallel state. I can write a Hamiltonian for it in this shape. Sigma i are spins sitting on nodes. Each i is a compound index. It refers to a specific point on the plane. And the way it is written is that two neighboring states, two neighboring spins, when they multiply each other and if they are in the same state, it will be positive. So sigma i can be plus or minus one. So if they are sitting next to each other in the same direction, two plus ones, this will be a one and lower energy. Sometimes I need to add an external magnetic field which is given in this shape. Something which, a model which I'm sure you know, so we don't need to spend a lot of time on it. At h equals to zero, the Hamiltonian is invariant under the action of the group z2s when you change the sign of all spins. The order parameter is the mean magnetization given in this form. And it's clear that this guy vanishes due to the z2 symmetry because you take z sigma to minus one, sigma m goes to minus m and the only solution for m therefore is zero. But we know that for h equals to zero at temperatures below Tc magnetization is non-zero. And this is the approximate, this is the exact form and it's approximately 2.26. So what happens? M magnetization again of the Ising model is zero above Tc and non-zero below Tc and it deviates from zero near Tc by this exponent beta. Okay, here is an example of ergodicity breaking and the thing is that at below Tc the entire phase space is not available to your calculations and you have to take him in over only half of the phase space. So if you are, for instance, in the domain that spins are directed up you just use the states which can be obtained from this state by a finite number of flips. Or in other words, I shouldn't say a finite number by a finite energy because each flip needs an energy to go to change and then a finite energy means that you can only change so many of these spins to the down and when you just add up on that magnetization will be non-zero. So think of it in this term, that we take the phi plus state as our ground state and then we assume some sort of dynamics which takes you away from this phi plus and the phase space is created by the long-time action of this dynamics. It's a sort of practical phase space which is connected to one of the vacuum. And when you just sum over this phase space magnetization will be non-zero. And this is a case of ergodicity breaking in the 2D ising model. When I talked about the 2 ergodicity breaking, here we see that below Tc the ising model phase space is broken into two parts, those which are connected to the positive ground state and a completely different section connected to the negative ground state. And hence if you add up magnetization only on this section it is non-zero and it loses its symmetry. You can no longer change all the sigmoids to negative because those fall into this part of the phase space. So this is a clear example of what I mean by ergodicity breaking. I can also calculate the critical exponents for the ising model and they are given in this table. There is an exact calculation in two dimensions. The 3D ising doesn't have an exact solution yet. So numerical simulations have produced exponents for the 3D ising model. And in four dimensions we believe that the ising model has a good approximation by the mean field theory so these are the exponents of the mean field theory approach to the ising. The other concept which I introduced was renormalization group techniques. So let's see what we can do in this case. For Rg what we can do in a spin model is follow Kadanov's block diagonalization. That is you take a number of spins, in this case four of them, you add them up and set them in the middle of the square. So in this way the number of spins which you are dealing with is reduced because four has gone to one and the size of the unit cell has also increased because in now these two spins are two A apart whereas these were only one A apart. This gives me a procedure by which I can implement the renormalization group because I'm changing the scale from one A to two A and I can see what happens to my theory as a result of that. What I have to do is then look at the interactions in the Hamiltonian and see how the interactions in the Hamiltonian change as a result of this block resummation. The problem is that it's actually very hard to do. Any of this in practice. And so there is this famous sentence by Michael Fischer which actually doesn't say anything, it just says it's difficult. It's Michael Fischer was one of the instigators of RG and what is actually saying that the actual process of explicitly doing this is non-trivial. So I'm sorry? It is non-trivial. Okay. So what I'm going to try and do is to actually do it now for a simpler model or not simpler for the Ising model sitting on the triangular lattice. And it's a process which is very interesting that one can actually see explicitly how this resummation process happens. And it's a very good, it gives you a very good understanding of RG. So let's take Ising model on a triangular lattice. So I have a triangular lattice spins are sitting here. What I then do is that I take a resummation process which means that I take these three spins, add them up and then take a sign. What is usually a difficult thing to do is that in your resummation you want to at least ensure that your dynamic variable takes on the same range of the original dynamical variable. So if I call the new spin this thing, it will have the same range. So I define a new spin which is the sign of this sum. And all the various combinations which you can take here from three ones down to three minus ones will definitely have a specific sign. Note that if I had an even number of them, a zero would have become possible which would have thrown me out of the Ising model. That would give you three values. This is why this trick is so neat that when you take just three spins, you always have either plus or minus one. And therefore what I do is that I take this rule and then set a spin in the middle. Which I will color separately and this is sigma i. So in this graph, I hope it is, or in this shape it's, I hope it is obvious that this has happened that three spins are added up and one spin sits in the middle which is on this new point and also one at this new point. So what happens is that you have a distance a prime between the new spins, which is in fact this line here as opposed to a distance between all the spins which was a and a prime is related to a. I think that's an error must be multiplied by root of three. And the other thing which happens is that the number of plaquettes that is faces which have the complete faces changes by n divided by three when you do this. It's obvious that the number of plaquettes decreases because these guys occupy bigger plaquettes. And then what happens is that as a result of this process, the Hamiltonian changes. And so I should be able to write it as interacting terms for h i's which forms the new Hamiltonian and then terms which forms the if you like the magnetic field action on the Hamiltonian which is this guy here. And then I can write the old Hamiltonian as the sum of two different types of spins. S i and s i, small s i, I have to clarify something here. You see Hamiltonian is written as the sum over, my notation from slides to here has changed minus beta h of sigma i to start with. But then I sum some of these spins and I should be able to then write it as a minus b over h s i when I do that. So let's rub this off. If I now do this summation and whenever I have three spins then something is summed up here but not everything. And you have some left over variables still here because you have summed them up with some imposed condition. So you now have a new Hamiltonian in terms of new spins. And if I now sum all of these with respect to s i, I should get the same partition function. So what now I have to do is to see what the Hamiltonian will look like after this re-summation. And this is really the non-trivial part of the statement by Michael Fisher, that to actually get out the new Hamiltonian is not an easy thing. But in this case, there is a trick which makes it easy but in some sort of an approximation. Exactly, exactly, exactly. So what you say is true. I have now two types of interaction. Interactions which are among the spins within the same triangle. Interactions of spins within the same triangle. So these spins have interactions with each other. And this is easier to deal with because I am summing up over these three spins and put a spin in the middle. Then there are interactions between neighboring triangles from here to there. So for the moment, I'm going to ignore the interactions which are inter-triangular. I will just look at the inter-triangular interactions and that's easy. So I mark these guys by i, if you like. And for each of these set, I have interactions between these three spins. Let's work that out. Now to rest your mind, this means that the Hamiltonian is divided into two parts. One Hamiltonian which comes from the intracell interactions plus a Hamiltonian which comes from inter-cell interactions. And this is the trick which makes this problem solvable. That is, you make a division which we call the main Hamiltonian and the interaction around it. The interaction around it is not necessarily a smaller interaction, but it is somehow a simplifier of the problem. So I can in fact now forget about the interaction and just do this guy and this is easy to do because we can just count the number of states and work out the Hamiltonian for it. No, at the moment before the renormalization. This Hamiltonian is the usual Ising Hamiltonian except that I have broken the interactions into inter-cell and intracell interactions. So that's the rearrangement. But now I'm going to forget about that, keep this one only and that is not the usual Ising interval. So now for one cell, I have all possible spins. So it's sigma one, sigma two, sigma three, all belonging to the cell i. You can write them down and they will have possible states. Corresponding to this, I have a sigma i which is using this rule. So these three sigmas can take various values and then sigma i can take a value which is in this case plus, plus, plus, minus, and so on. And now I have e to the power minus beta h zero because I'm keeping only this term and this gives me some values three to the power k, e to the power minus k, minus k, minus k. No, sorry, this last one is three k. k would be beta j. H zero is minus j sigma i sigma j. So h zero is the sum of spins, neighboring spins, but I am only keeping spins within the same cell. So it's not all of spins. Last line is what? You are right, this is minus k and so on. So this z zero guy now is equal to e to the minus beta h zero, somewhere all spins, but now because the spins have decoupled from each other, it actually is just somewhere all i, e to the minus beta h, but this is over only one cell. And this happens because the cells are no longer coupled to each other as in the normalizing. They are now decoupled. So I can actually calculate this simply by looking at this last column which is e to the power three k plus three e to the minus k, which is just the sum of these terms here. I should really organize this a little bit better. So if you complete the table, you will see that this is repeated. This will then be repeated for the other four. Hence, this partition function is really this guy and two sum power. So this is the value of z zero. Now what is left is what happens to the h interaction. H interaction will have terms which are a multiplication of two different cells. So this guy h interaction is equal to k times spins from different cells. You'll be exactly the same as the upper half. So we should have it distributed twice. We should have what, Salih? We should have a coefficient equal to two in the partition function. You mean we should have a two to the power n over three here? Okay, that's a possibility, but it doesn't change much because these coefficients don't matter in the calculation. All the lockspin. Yes. The weight is only two parts, k plus three to the minus k. Yes. Then there will be no factor of two. Then there will be no factor of two. So now what I want is this e to the minus beta h of sigma i, which is equal to e to the power minus beta h zero minus beta h interaction. And I can write this down in the following way. I now actually interpret this as interaction and expand it. So it gives me an e to the minus beta h zero for the first time plus or minus beta h interaction, e to the minus beta h zero, et cetera. Let us also divide by z zero. This is the z zero. So it comes out to be one plus the expectation value of h interaction. And then of course I have second order terms, et cetera. As you can guess, I'm not going to calculate these. Since this is a perturbation, I may be able to get a good enough answer by calculating just the first interaction. Of course, the inventors of this method, which are these guys, I think I should write their name, however, that and van Luywen. These two guys invented this method to actually see that is to answer Michael Fisher's dilemma that the process is non-trivial. It surely is non-trivial, but they managed to calculate h squared as well. I will only present h because it's much easier and takes less time. However, it can be taken to the next level. Right, and there is the one after the normalization. Yes? And why can you say that it is small that you can do perturbation? Okay, that's a good question. Why can I say that it is small? It's small because there are really, look, the number of interactions which are formed between cells is not many. And I am hoping that all these interactions which are formed between these two cells, somehow in this black resumption will have a smaller effect because it becomes thinned out. Not all of them will be relevant. Some of them will be simply summed up. But it's a hope. It really comes out of the smartness of these guys that they managed to show that this actually makes a small contribution. And you can only see that at the end. Here, I'm only making a hand, at this stage, I can only make a hand-waving argument that each of these terms which come out of that expansion are smaller in respect to the previous one. You kind of can see it because all the, a lot of interactions are taken into account here and then there are inter-spin interactions between here and there, here and there, which determine, so there are two interactions here between these two, which determines this cell with that cell. The interaction term before doing renormalization, because you dropped on there. No, renormalization has not happened yet. All I have done is that I have defined a block resummation. We don't know anything about renormalization yet. So the general plan of this calculation is like this. In any resummation, what you do is that you define a resummation rule, which I have defined here. In the previous slide, a different resummation rule was defined and if you have ever tried to do that, you see that this is a very difficult rule to implement. People have done it, but on a blackboard and in front of a class, this one is easier to do. So the question is, can I have a simpler resummation rule such that I can actually do the calculation? I define a resummation rule, which is this guy here and then what I have to do is to then show that the new Hamiltonian is of this form. It hasn't lost its ising personality or identity. It's almost the same shape as the previous ising. It will not be exactly the same. The interaction terms will become more complex and there will be more coefficients of coupling constants, but that is good for me because that will give me, then that will give me the renormalization group. The renormalization group equations are equations which connect the new coefficients to the old coefficients. You choose the cells because you can rotate the cells and those interactions. Yes, you are right. Most of the interactions actually are in the part you are neglecting. No, that's not true. Because when you are neglecting four out of six, isn't it? No, these spins are interacting with each other very strongly and then there is this interaction of this spin with its neighbors which will determine the interaction of this triangle and that triangle. Just counting them for one triangle appears to be less than this. So I'm taking most of the interactions in. However, you are right in the sense that this is completely arbitrary. Completely arbitrary and there is no reason to assume that this will always work. It just happens to work so that we can see the effect of it. It's an example for in practice where you can see how this resummation works and you can derive the renormalization group equations finally as you connect the coefficients of this guy to that guy. You can make the calculation more precise. That is, you include more and more effects by taking the higher-order perturbations, which people have done. And you then hope for the higher-order perturbations to be smaller than the first-order perturbation. And that hope is only justified after we do all the calculations, not before, because before your argument is correct and you can say, okay, this is not true in all statistical mechanical models which I can separate in this sense and you'll be right. Just a smart example for seeing an effect where you can actually perform all the calculations. Yeah? Go on. Now the interaction term is like that and now I want to take the mean of this thing where i and j each one belong to neighboring cells. But this mean is taken over z0, which is this guy. So I should put a z0 here as well. Now, here is the beauty of the trick and that is this guy will now, the mean, because it doesn't connect, because h0 doesn't connect the different cells, it will just be like that. The Hamiltonian doesn't connect the different cells. So the mean of operators which are in different cells will be independent of each other. So what I therefore need to calculate is sigma i averaged over 0 and this is what it comes out to be. So if sigma i is equal to 1, I go and look at the table which I rubbed off and see where does sigma i equal to 1 for which combinations and then calculate that combination. So it's e to the 3k and this time plus e to the minus k. Better write this side. In fact, in all those combinations I just had e to the 3k and e to the minus k except that now because of sigma one of them picks up a negative sign. Previously, because there was no sigma none of them would become negative. One of them becomes negative. So I will get that and of course this has to be divided by the value of z0 which is e to the 3k plus 3e to the minus k. And when sigma i is minus 1 it gives me a minus e to the power 3k plus e to the minus k minus 2e to the minus k divided by e to the 3k 3e to the minus k. So I can combine all of them and write them like this because sigma i just multiplies a minus 1 in the bottom. Probabilities or I don't understand what those factors or fractions are? It's a fraction because I have a z0 in the denominator. It's a probability but only in the sense of h0. So I'm calculating the average of an operator in the h0, in the h0 model. So one has to be negative. Yes. So let's go back here. I broke my Hamiltonian into a free Hamiltonian plus interaction. So using this Hamiltonian I will now do my calculations as though it is a statistical mechanical system. So I have a z for this and then I have averages of operators for this Hamiltonian. So if I have minus beta h interaction it means that this is equal to minus beta h interaction e to the minus beta h0 divided by z0. So in this sense you can say that each configuration is happening with a probability and I'm summing them over. When I look at b minus beta h interaction it's of this shape. This is h interaction. When I want to calculate the average value of this I notice that i and j belong to different cells and h0 does not connect the different cells. So each of them will have a separate independent or their averages can be calculated separately which is what I have written here. Now when I look at one of them which is just the average value of sigma i the problem is that what is the value of beta e to the minus beta h0 for each combination which I had a table for. I write those values but now with the difference that because sigma sometimes becomes negative and positive some of them will pick up negative and positive factors. It simplifies into that. So this term will therefore be k this ratio sigma i sigma j. However for each neighborhood of a cell there are actually two spins connecting this cell to the other cell and each connection is equally valid. So I have a 2 to add here and this is the this is now the average of this interaction. There is a square power because each sigma i has one of these factors. I still don't understand the meaning but you say that the average of sigma i is equal to something that contains sigma i I might be confused on that. No, the average of sigma i equals to this if sigma i is equal to 1. It is 1, I know that it is 1 so its average is the 1. It's the term. No, it's not the this term is not the average of sigma i itself. It has minus beta h0 in there. This is what we are calculating. In a single cell in the block we calculate every sigma of that block given that the block on the hole is plus 1. Yes, so what is happening is that I take one of these blocks and there are three spins here which can have therefore eight possibilities. Out of these eight possibilities one of them gives you sigma i plus 1 and the other one gives you sigma i minus 1. So four possibilities go into the calculation of sigma i plus 1. And therefore this factor has to be calculated for each of the four possibilities. And there are four possibilities calculated to 1, 2, 3, 4. And then when sigma i is minus 1 again these four possibilities have to be calculated. And as professor Dar pointed out four possibilities as you sum over the little zikmas. So this i is not summed over. We have just for this sigma i connected by sine of sigma 1, sigma 2, sigma 3 but all these zikmas 1, 2, 3 belonging to block i. There are four possibilities. Four possibilities has gone into this. In the denominator that factor of 2 that this guy pointed out. Yes, that factor of 2 is taking care of us. There is a factor of 2 if you want to keep everything right. Now what we do is that we get go of that factor of 2. We just work with this value of log of z0. Eventually all I want what I want to do is to take log of this and it will not be necessary. However this other guy here is important in that the interaction, each interaction actually has two terms in it because you get two spins connecting one cell to the other. But these two spins are just exactly the same and therefore the calculation doesn't affect it and it doesn't affect the calculation. I just get a factor of 2 here and so if this average is like that I can place it in this edge interaction. I just place it in there. There is a k outside. This is the k outside. This is the term which comes out of 1 sigma so a squared for the whole term. So now this is interesting because it is telling me that hi is of the same form as h0. So if I can rub this off I tell you what has happened as a whole. I have a partition function which I wrote as a sum over i and then a sum over i, a small i with the condition of sin of 3 spins bringing me sigma i. Sigma bigger. This was broken into two parts so I formed it into this shape and I have now and I then went on to calculate the value of sigma i by performing some of this calculation and expanding in the powers of h interaction and this is what I got. So I see that beta h sigma big i looks just like beta h sigma a small i which is the point of realisation group. Now I can hope to extract my RG equations because this has coupling constants this has coupling constants and I can relate them to each other. So by writing e to the power k sigma i sigma j. The other thing which we need to notice here which is implicit is that if in the big i, big j picture if they are not neighbouring there is no interaction. So this cell and that cell don't have an interaction. Only neighbouring cells have and it's a property of taking the first power. If you took the second power of the h interaction then you might have interactions between further away cells. So this is now in this shape. Hence I can say that the new coupling constant is related to the old coupling constant by this form. This is my RG equation. What does it say? It says by the bellocary summation of Kadanov and the triangular lattice the shape of the i's in Hamiltonian doesn't change but the coefficient the coupling constant changes and the coupling constant changes by this law. So this calculation although a little involved gives me an answer to Michael Fisher that it's not really all that non-trivial. Although I have used a lot of tricks. He meant without using tricks. So I'm going to rub everything off so that we now have an RG equation. And this RG equation has happened in making A go to root 3 times A. So a scale has been enlarged as a result coupling constant has changed. If you like you can say that these are coupling constants at different scales and this is the result of changing the scale coupling constant changes as a result of the change of a scale. It is not a differential equation as I showed you before because I have changed the scale by a finite amount not by an infinitesimal amount. So the first thing I have to find the fixed points. Why do I have to find the fixed points? Because the fixed points correspond to the points of the flow where the Hamiltonian doesn't change. And if the Hamiltonian doesn't change then I have a critical point. So what is the obvious fixed point? There is an obvious face fixed point at K star equal to 0. K star equal to 0 0 equals to 0 is a fixed point. And then there is another fixed point at infinity of course. For at finite interaction strength these correspond to 0 and infinity temperatures. So this is T equal to 0 and this is T equal to infinity. So these are not interesting. So we now look for a fixed point in a finite fixed point somewhere in the middle of these two extremes. So there must be a point where this equation is satisfied at a value between 0 and infinity. The little algebra shows you that you can find this fixed point. Let's not do the algebra. K star I have it here if somebody wants to see the algebra. I'll give you later. You take logarithm and then you juggle it. Now, you can immediately ask is it correct or not? Because we know what K has to be from on Zager's solution. Of course on Zager's rectangular but similar solution for triangular we know the exact K star. So you see it's not that bad. We have about 0.06 of difference between the obtained result and the exact result. It is to actually validate this method you have to do the next kind of perturbation which these guys did the guys who I wrote their name down. And if you go to the second order of approximation then this K star approximation comes out. So if I call this K star 1 that will be K star 2 will come out to be 0.3 something. And the approximation I think 0.301 something. So the approximation improves which is the answer to your question. Whether this is a perturbation or not only now I can say that it is because you can see that the approximation is apparently converging to the exact solution. Okay, what I can do next is to take a derivative of this term let's do this company precisely. So K prime is a function of K. What I write it as K prime as a function of K star at an expand around here. Our K prime at K star is K star itself because that's the fixed point and this equation can be reorganized in this shape. So what does it say? It says that the distance that the coupling constant has to the fixed point the distance of the coupling constant as a result of each renormalization step changes by this factor it changes by this factor so if I calculate this factor it will give me the factor change of the coupling constant near the fixed point. So dK prime dK at K star this has to be calculated what is negative? Maybe you can tell me afterwards I don't see any problem. Okay, this is a calculation which eventually gives you 0.883 So how do I interpret this 0.883 how? I'm asking you it's a repulsive point but what is the exponent of repulsion? What I need to interpret this is as if you if this is exactly like what I obtained in the abstract in the abstract rG I said that this m, a stability is e to the power y1 e to the power y2 00 so this is y1 sorry not e thank you very much lambda to the power y thank you so this is the delta T is the repulsion in the direction of delta T so that it's actually T prime is equal sorry, beta prime is equal to e lambda to the power minus 1, beta this but this is interpreted as that now I have exactly the same shape but K prime delta K prime is related to delta K with a coefficient so this has to be equal to lambda to the power yt this is the amount by which this coupling constant is re-escaped as a result of Arch exactly I just want them to do a little thinking they are not supposed to just listen it's not a theater what is lambda lambda is given by this relationship that a prime is root 3 times a so this is my lambda so yt is equal to log of 0.883 divided by log of root 3 which is approximately what sorry a lot of errors here a lot of errors here 1.624 1.624 0. 0.883 and yt exact is equal to 1 so this is not this is off by 12% which is of course reflected in this as well now you can calculate one of the exponents nu is 1 over yt and that is equal to 1.13 and it should have been 1 and alpha is equal to 2 minus 2 over yt which comes out to be minus 0.26 and it should be 0 so you can if you are if you are if you are able to do the next order of approximation these things all improve the next order of approximation is to keep the second power of interaction Hamiltonian interaction in there now to complete the calculation I have to also do the h hsi which goes to an h prime si prime we will do it tomorrow ok thank you