 I'm going to continue with our treatment of the Ising model. We had some problems with the identification. Let's go back to the problem of identification. So I have Ci that they are annihilation operators. So Ci acting on vacuum gives zero. And now I want to make a connection with the polymatrices. So I set Ci equal to sigma i plus. So this vacuum state has to be identified with the up state. And sigma plus on up gives you zero. And sigma plus on down state gives you the up state. Therefore, I need the down state to be identified with one particle state. Hence, when Ci acts on the one particle state, it gives me the vacuum. It kills the only particle there and gives me vacuum. So I also have my one particle state identified with the down state. And Ci dagger will be sigma i minus. Ci dagger on vacuum gives me one particle state. And it is concurrent with sigma i minus on the down state to give me the up state. So this guy, 1 minus 2 Ci dagger Ci, this operator. If it acts on the vacuum, this term gives zero. So it's just the vacuum. If it acts on the one state, Ci brings that to zero. Ci dagger brings it back to one. So it is 1 minus 2 on one. Therefore, equal to minus 1. So sigma i z. Now, this has to be sigma i z. So sigma i z on the down state gives me minus the down state. And on the up state gives me plus up state. Finally, Ci dagger Ci anti-commutator is sigma i plus sigma i minus anti-commutator. And it's zero. Everything OK? Is that the line that sigma i is the opposite line? Thank you. So next calculation I want to do is this one. At Ci, I better clean this thing. So the problem is that sigma i sigma j, commutator is zero. i not equal to j. And this is not good, because I want CiCj, even if not on the same side, to anti-commute. So I use it in a different notation for Ci, which is equal to i of k is smaller than i sigma i z times sigma i plus. So for example, C1 will be sigma 1 plus. C2 will be sigma z1 sigma 2 plus. And so on. That's what this formula was. What's the problem? Nothing? I was asking why it is a problem that sigma i sigma j commutes with mu? I want to make a correspondence from the Ising model to the Fermion model. In the Ising model, sigma i sigma i plus, sigma i minus sigma i plus anti-commute on the same side. So do fermions. So this is OK. But this then is a problem, because here I have Ci dagger Cj. Or I should say, let's do this way, because this is what I calculated. Equal to zero i not equal to j. So I cannot just put Ci equal to sigma i plus everywhere. So this what I wrote there, which I did my calculations with, is incomplete, because this is now a problem. This is a commutator. I want anti-commutator to go from Ising model and Pauli matrices to fermions. And I solve this problem by putting this pre-factor here. This pre-factor, what it is doing is that on the lattice, you are at point i. I multiply by a string which starts at the origin and goes up to that point, but with sigma z direction. And I'm going to show you that this solves the problem, that they will now anti-commute everywhere. So I take this form. And now let's take Ci Cj, i a smaller than j, so that I can do the calculation. But obviously, it doesn't matter, because if it's the other way around, I put Cj first, since I have to rewrite this again for the replaced version. Could you choose to introduce Bosonic creator and integration operators? So that we have the second relation? You could. You could, but something very difficult then comes in. The thing is that in two dimensions, fermions and bosons are in some sense equivalent. So if you want to do that, then you have to, in the end, do that boson fermion transformation, which we can do. But I think this is simpler. However, there is another representation which can replace this in terms of just bosons. It's called the Coulomb-Gaz representation of Isimov. Another question? Yes. Sorry. You didn't get the original picture or this business of string? String. So here is the problem. I want this, but this is what I have. So the solution is that instead of saying ci is equal to sigma i plus, I say that ci is equal to a string times sigma i plus. A string comes from origin up to sigma i plus. The question is what should the string be? What is the string? The physics of a string or the mathematics of it. The mathematics of it I am about to explain. The physics of it is that in many physical problems, we find that having just a particle here doesn't work. You must have a string connected to it. It has started with the Dirac monopole and then parafermions and the twist field in the Isimov model. All work only with this string. So what I need to show is that now this combination, if I take this combination and just use the Pauli matrix rules, will give me an anti-commutating ci cj. This is what I want to show. So it's equal to pi k smaller than i sigma i z sigma i plus pi k smaller than j sigma k z sigma j. So here k is smaller than j, means from one up to j. So I can take the one up to i. Then there is a sigma i par z. And then another string which is k bigger than i smaller than j. So this string has three parts. This part commutes through that and becomes this squared. Then I have sigma i plus. Then I have this sigma i z. Then I have this string. This is one. So I have sigma i plus sigma i z the string kj to i sigma kz sigma j plus. So this is the only part of a string is left. The rest of it was identical to that. And its power vanished. Not vanished. It became one. Disappeared. And just left me with a sigma i z here, which does not come yet with that, so it's sitting. This calculation needs i smaller than j. Now keeping still i smaller than j, but writing it in opposite. Opposite way to that. It is equal to k smaller than j sigma k z sigma j plus. k is smaller than i sigma k z sigma i plus. Again, first part of the string goes away to here and cancels that I have i k smaller than j bigger than i sigma kz. Exactly the same trick as here. Then I have sigma. When I do that, here I have one extra term which I keep outside, sigma i z. Yes, this j is bigger than i, so it includes i. However, because i is smaller than j, it can go through all of this without problem. So it is equal to sigma i z sigma i plus j i, which is exactly this guy, but with the difference that this term is in opposite order to that. And because they anti-commute, if I put it on the other side, it will be minus. Hence if I add them, they vanish. And I have achieved that. Only using matrix-parallel algebra. So polymetrics is algebra. So c i can be represented as sigma i. If I take this complicated form, this complicated form is something like that. It's not bad to just check it for something easy, such as c 1 3 plus c 3 c 1. c 1 c 3 will be sigma 1 plus sigma 1 z sigma 2 z sigma 3 plus sigma 1 z sigma 2 z sigma 3 plus sigma 1 plus. So you see sigma 1 plus is next to sigma 1 z. It cannot commute through it. But here, sigma 1 plus can commute through this string. And it becomes. And now this term has opposite order to that. So the sum is 0. So this is the sort of formal way of doing this particular thing. Then I use these analyzing creation operators. I go through the mathematics I showed you on Friday. And eventually, we arrive at that action. It's only a question of really just do the algebra. And then do a Bogolibov transformation to bring me to the correct canonical form. Here I have to identify this action with a conformal field theory. And to do that, I have to calculate its central charge, which is a calculation I don't want to do. I don't really expect you to be able to do that sort of calculation. So there is a calculation that starting from that action will show what its central charge is. And just believe me, avoid the calculation that C is 1 half. So C equals to 1 half conformal field theory corresponds to 2Dizing. With this knowledge, I can look and see what sort of operators I have here. I have three primary operators for C equals to 1 half. C equals to 1 half is m equals to 3. So three primary operators. I can put them in a diagram. It should be the 2, 2 state on 1 state, 1, 2, 2, 3. Just three fields. And these operators are the unit operator, sigma operator, and epsilon operator. So weight 0, 0, 1 half, 0, and 1, 16. This guy is epsilon, and this guy is unity. We certainly have a unit operator for the fermion field. And I argued that the conformal weight of epsilon has to be 1 half. It's the engineering weight there. So epsilon z epsilon w equals to 1 over z minus that. Now, this is a little s. Yes. C is 1 half for which it was equal to 1 minus 6 divided by m times m minus 1. For this to become 1 half, m has to be 3. I gave you a formula for what fields I will have. I have fields phi pq, such that p goes from 1 to m minus 1. And q is smaller or equal to p. So I always will have a triangle for p. p will have, so I will have, if I draw p on this axis, it will go up to m minus 1, which is these boxes here. And then I build the pyramid up with p taking each. So this is the p equal to m minus 1 row. This will be p equal to m minus 2 row until p equals to 1 row. And then q for each value of p will be from 1 up to p. So it will be q1, p1 here, q1, p2, q2, p2 here, and so on. No, these will all have different conformal weights. There is, in some conformal filtering papers, you see they talk of degeneracy. That's because you can, in fact, imagine the whole table. And so these guys will be degenerate to these guys. But I am not drawing them. There is no degeneracy. They all have different conformal weights. And there is a formula for them as well, if I can find it here. I don't remember the formula by heart, and I cannot find it here. There is an HQP, which gives you the conformal weight of each box. But each of these have a conformal weight, because these are operators. Each box is a conformal operator, for which you have to calculate h and h bar. And for each box, I have a formula, but there exists a formula, but I don't have it here. But for the Ising model, I know that it's these guys. One half, and one over 60. You have those numbers in the last row. These numbers, yes. There are two numbers. These are? OK. OK, for a theory which is unitary and real, h must equal to h bar. So in fact, here I have a one half zero plus a zero one half, which I showed you that I have two fields, epsilon and sub-h bar, and they together form a doublet here. And I have a 16, one over 16 for this box, and I have a zero zero for that. The three? Yes, the three sets of different representations. No, they are the members. Yes, they are different representations, but they are members of the same Hilbert space. So the Hilbert space of the Ising model is the v zero zero plus v one half zero plus v 16, 16. And what are these v's? These are the v's which you obtain by ladder operation, which I told you each one of them forms a triangle, and it's the sum of them. Yes? One half zero and zero one half. Because I have two fields there, one fc and one fc bar. fc is one half zero, and fc bar is zero one half. Yes, we have the same one. Yes, we have one field which acts as both of them. The three sets of boxes are the representations, which we have in the same. They are. They are the basis there. They are. So here I have written only the pq. I have written their names here, which I call operator one, operator epsilon, operator sigma. Here I have written their conformal weights. So these three sets, these three boxes are equivalent. The problem which I have now is the final identification. Final identification, I have, this is the unit operator, so it must exist, and I have no problem with that. Now it means that I have two non-trivial fields in the sigma model. But in the sigma model, sorry, in the Ising model, I have two non-trivial fields. Two non-trivial fields is a problem, because I see here only one non-trivial field. So this box is fc, which you see on the screen. What is this guy? And this fc does not make me happy because epsilon z epsilon w is equal to 1 over z minus w. And this is just the wrong color rate correlator for Ising. So there has to be another field, which corresponds to this box, the sigma field. And where does this field come from? So there has to be another operator in the conformal filter of the Ising, such that when it acts on the vacuum, it creates the 1 over 16 state. So I assume the existence of this field sigma, which is equal, which gives me 1 over z minus w to the power 1 8. And this means that eta is 1 over code. Sorry, box is epsilon. What is epsilon? Do you know what this is? This should be the Ising model. This should be the spin of the Ising model. So I must have another operator in the Ising model, which I can correspond to this. So that's I have. But you have another operator. The triangles are two different maps of the same concept. So this shape here. It's doing the same thing in two different ways, right? Doing the same idea in different ways. So here I have the PQ or QP. Here I have the conformal weights, which I have some complex formula for HPQ. Here, as he pointed out, I have to identify them against the Ising model. This is the unit vector, the unit operator. Obviously, this is the spin. So what is this? This is, in fact, equal to sigma i sigma i plus 1, the energy density. So, in fact, the Hamiltonian is the sum of energy densities. This operator does exist in the Ising model, but you really have to do this calculation to then see it. And then it does have that 1 over r correlator. Yes? So this is the same thing. But epsilon is in the field theory. Epsilon is in the lattice model. And the final result is important. That I have used CFD to obtain the critical exponent. And, in fact, using this diagram, I can calculate all critical exponents of the Ising model. So this c equals to 1 half theory is, in fact, exactly the same as the 2D critical Ising model. Because we wanted to identify the 1 over 16 and 1 over 16. They filled it with weights 1 over 16, 1 over 16. It's near that it was the spin operator, right? I know that this operator must exist. So I claim that there is some operator which, when acting on the 0, gives you the 16 operator. To be more precise, I have to explain a lot more about Ising conformal. I don't really want to ask you to know all the details of conformal field theory. I want you to just have a little knowledge about it. Yes? So what I did in these last two lectures, I showed that the Ising model has Rg as conformal field theory, which was the start of my talk was that these lattice models, these two dimensional models which show critical behavior, thermal critical behavior, they can be looked at from another point of view. One is the Rg point of view, and the other is CFD. And from both of them independently, you can derive the critical exponents, which is the whole point. M is equal to 3, yes? So our C is 0. M should be 4 here, right? We need to add 1, 1 up. Ah, plus 1. Thank you. This also means, and it's a little deeper, that if these two are the same descriptions of the same model, they must be related. And there is a very deep relationship between Rg and conformal field theory. I don't want to go into it now, but they are connected. Now I want to give you a second example, percolation that shows you that this holds. So we can look, we want to now look at the percolation model, the percolation problem, to see how Rg is defined for it, and also to see how conformal field theory is defined for it. It's a second example where these two concepts come in. So what is percolation? So the problem is that, is the image you see here, the physical problem is the image you see here. You have a porous media, and water can percolate through it to the other side, very slowly, but surely. It does percolate. And it percolates if the holes and spaces in the percolating medium are connected from one edge to the other. They could be very small, so a small amount of water will flow, but over a long time, you can have this situation where the water from the surrounding field has collected in a sinkhole. Or you have seen it in buildings where you have a wall, water this side, and water slowly seeps through, and you get dampness in the room, yes? No, it does not determine them, because I am not clever enough. But if I can do it completely and exactly, it is exact. The calculation I did for you was because I'm not clever enough, I do first-order calculation. Somebody is smarter, he will go to the second order. But if you are very smart, you can calculate all the terms, then it becomes exact. It determines exactly. It determines exactly, yes. Now let's look at the second example, percolation, which I told you what the physical problem of percolation is, so let's go forward. So I make an abstract mathematical model for percolation. That is, this is what I do. I take a lattice, and I randomly place some bonds or some nodes as on. So I've placed this bond as on, this bond as on, but they are completely uncorrelated. I just choose a bond randomly and put it on. If you have a finite lattice, like what I have here, in some time, the whole lattice will be covered. So it is not much fun. It has to be an infinitely large lattice so that these guys, some of them will be connected, some of them will be alone. This is called bond percolation. This is called site percolation. And the question is, what is the probability p at which if I do this, I will have a global cluster? Global cluster is tantamount to water seeping through. If there is no global cluster, water will only go halfway. So pictures of bond percolation, you see, as the probability of placing bonds increases, you get a denser and denser image. Hence, it is very probable that you have a global cluster here. It is almost equal to 1. The probability of finding a global cluster is less than 1. You do not find a global cluster here. When the lattice is large enough, you can feel the problem that it is actually a difficult problem to solve. And the main question is, what is the critical probability with which if you place bonds, a global cluster will fall? Another way to look at it is, I set a boundary condition, all of them on here, all of them off here. So percolation cluster has to start right here. And then it moves on the edges only. See, this red line is moving such that whites are on its right hand side and blacks are on its left hand side. And it will just give me a path through the percolation cluster. If it was percolating, this path would have to go to the other side. But as it is, it is not going to the other side. Let's start with the easy problem, a study percolation in one dimension. So when you want to look at percolation in one dimension, the problem is very easy. So I'm very slow in explaining percolation because it's a less known model than I think. So maybe there are people in the audience who don't know what percolation is, those who know can rest it. So I have a one-dimensional chain. And with some probability, I fill the bonds. Just take bonds randomly and fill them. Question is, when will I have a path which goes from here to there? And the answer is obvious. You only will have a total path if probability is one. Because it's one-dimensional, it has no way of going around. It has to go through every link, so the probability of the percolating probability is one. So this is a easy problem. We can leave it and not very interesting. Now, if I go into two dimensions, then I have a lot of variety coming in. Because in two dimensions, I can change the shape of the lattice as well. So I can have 2D honeycomb, or 2D square, or 2D triangular. And the problem changes because each of these lattices has different coordination number. On each lattice, I can ask, what is the difference between site percolation and bond percolation? They are not the same problem. And finally, in this last line, you can see that in three dimensions, the problem is very different. So fortunately, we are only dealing with two dimensions. And in many instances, the problem of percolation gives critical percolation is 1 half. The reason is that I am looking at percolation from bottom to top with black lines, as I explained. However, you can always reverse the problem and say, I'm looking at percolation of left to right with white bonds. The two must give the same answer. So if bottom to top is percolating, definitely left to right is not. So it's only at the critical point where only critically you have one global cluster. In one, you may have a global cluster in the other. So in many situations, 2D percolation has the answer of 1 half. But however, you also have a difference between bond site and bond percolation. They are not the same because this dual point does not exist on the site percolation. Because, for example, on the 2D square lattice, I can use this dual picture of black and white for this bond percolation. But I cannot for the site percolation. Now, the next question is, what is the order parameter? Just like icing, I need that. Yes? Yes? It should not matter whether it's a triangular or a square lattice because the criticality of the long, very less boards, less are right. So how the microscopic scale is so important that you get different levels for different lattice. That's a good question. But this probability would be? Yes. It depends on the shape of the underlying lattice. And it's a mathematical fact. I don't have an answer for you. Why? You would expect that if you look at it from a large scale, it should not matter. But it does. When I come to the renormalization group theory of it, you'll see how exactly it operates. The shape of the lattice directly affects the RGS structure. How that course will depend upon whether the triangular lattice or the square lattice, it will not matter that much Of course, then you will get the same rhythm. Yes. Especially that I will write down a conformal field theory for it, which does not see the shape of the lattice. So that question is difficult to answer. So order, the order parameter. So I need a parameter which changes at the point of percolation, probably at the point of critical percolation from 0 to 1. So on this side, I have a global cluster. Below it, I don't have a global cluster. And you can do a computer simulation. And on a very small lattice, it's 20 by 20. You will see this change over. Sorry. However, this is not quite satisfactory, because I want to claim that I have a continuous phase transition. So I want an order parameter. This cannot be the order parameter. I want an order parameter which slowly changes. So I would like to have an order parameter, PC, which doesn't jump, but changes like that. So the sentence on the top says, take the mass of the largest character. Don't take the probability of connection. The mass of the largest cluster is that you have a percolation problem, and you take the biggest cluster. You calculate its mass being the number of on points. Divided by the total area, this density will be a better order parameter. It will have this behavior, rather than what I've drawn. This is a much better order parameter. Now you can define exponents for the percolation problem and you see that they change near the critical point. They change with a set of exponents. We are exactly dealing with a continuous critical phenomenon. So percolation, this is a standard percolation, because now there are many models of percolation where they adjust the way that you choose the next on point. But this is the very first and simplest model, and these are the exponents you get. Now I want to get these exponents using RG. So comparing to what I said in the general description of critical phenomena, I don't see any symmetry breaking. To see symmetry breaking, you have to do a trick and approach percolation from a more complex theory, the Q-State POTS model. And of course, there is no ergodic theorem breaking because these two go together. We have a scaling variance. The scaling variance is shown in the fact that my physical quantities near the critical point are power loss. And I have a conformal field theory describing percolation, but it's a problematic conformal field theory because it has central charge equal to 0. Central charge theory is either null, meaning that it has only one operator, or it is non-unitary, and therefore it has to be non-unitary to be a little more interesting. And if it is non-unitary, it's a logarithmic conformal field theory. A little explanation here is good. So in conformal field theory, I claimed that OPEs are of this shape. Sorry, the correlators are of this shape. But another shape is possible where you have a logarithmic term multiplying it. So under a scaling, you see that something strange will happen. If you scale them, you don't get just the same theory, but this pre-factor will repeat itself. So if I put lambda here, this will now then end up looking like plus a term which is log of lambda times. It turns out that this is allowed. You can have this kind of conformal field theories. And in fact, this percolation CFD is of this form. It has this strange way of transforming. But then it has to be non-unitary. Everything is gone. OK. Yes, just a second. Yeah. If C is equal to 0, we should have n equal to 3. Yes. So shouldn't we have only one field, so we are identical? Yes. So obviously, it doesn't fit into that picture. And that is why when CFD was invented, all this was left aside. So here is the question. When you have this formula, y start m equal to 3. What happens to m equals to 2 and m equals to 1? This gives you c equal to 0. This gives you c equal to minus 2. So when the first CFD was developed in 1982, these theories were left aside. They said m doesn't take value 1 and 2, because these theories are non-unitary. But now we know that they exist. And these are logarithmic CFDs. But they are non-unitary. They were left out because unitarity implies that this m has to start at 3. When CFD was first developed, it was developed by people for particle physics. So unitarity was very important because it had probability of observation, et cetera. But we want to use it for statistical mechanics. And we don't need to interpret the norm of states as probability of observation. The only thing we need is a good correlation function. And it produces the right correlation function. Yes? I told you that phi phi is like that. But in LCFT, in logarithmic conformal field theory, this is 0. So in fact, you need another field to make it non-zero. I don't want you to learn logarithmic conformal. I just want you to know that it exists. If we start on logarithmic conformal filtering, we will go to another year. And I'll be talking here until next meeting next year. So some things I just need to talk about but not continue. Yes? Sorry, I didn't get the state of your system. You know, phi of z does something at point z this much I know. But what it does at the moment, I don't want to specify. But what I will, as I go ahead, what I will say is that what it does at this point is that it changes the boundary condition. So it becomes important as operators acting on the boundary of the problem. Whereas in the Ising model, which is a proper conformal field theory, sigma of z created a particle at z. Or I should say fc of z. Here I don't want to say that. This will do something but let's wait for it. But it's not an obvious interpretation. OK, so this paper was not the first paper but was an important paper in getting the results for the RG of percolation. And I'm essentially going to follow that paper. It's also an example. This is Eugenia Stanley, which is a very famous name, as you know, in aesthetic. It's an example I always tell my students that publishing in PRL is not the aim. You can publish it in less known journals with good results. The result that matters, not the journal. Since they get very depressed, they submit to PRL and the editors, which is not the good quality for our journal and rejects it. Nature is even worse. So it's a very nice, very nice paper. It says it asks, how can we implement RG for a percolation problem? So for RG, as you know, we have to block some the lattice. So suppose I'm doing RG on this lattice and then according to Kadanov block summation, I want to take a combination of sites as my block. And now the new block is this. How am I supposed to redefine my percolation problem so that it maintains the meaning in the resummation? In the Ising model, I had a Hamiltonian. So it was very easier in some sense, very non-trivial, as to how I do it, because I sum and then I have to work out the interactions. Here I just have a geometrical picture. So what I do, what these guys suggest you do is that you make this site, when you have this big four site, so you have four sites involved, you make the big one on if the sub-sites are in some sense connecting below to above. So for instance, if you only have these two, you cannot call this on because there is no connection from here to there. But if it is enough to have this guy on, then I can call the big block on because there is a root from bottom which goes to the top. And in this way, in the new lattice, I will have a new probability of sites being on, which is a function of old probability of sites being on. And the shape of this function will depend on the microstructure of lattice intimately. Because if it was triangular, a different criterion would give it on as it is rectangular. Another criterion is giving it on. So I have to work out R given a particular microstructure of the lattice, yes? Either. In fact, this paper says, transition group for site and bond percolation. Until this paper was written, people were thinking that it cannot. So in this paper, they solve this problem. Site and bond will give you the same, can be treated with the same mechanism. All you need to do is to define what you mean by connectedness and stick with it. So now I'm going to say something which is a little difficult to justify. If you give me r of p, and p is smaller than pc, the summation process will take you to p prime equal to 0. You agree with that? Because the lattice is filled in a dilute way, and codon of resumption will make it dilute tab. At each step, it becomes diluter and diluter until p becomes 0. So if I, in fact, this would be p prime of r of p, this is 0. Yes? This is the lambda. Lambda is the re-escaling. So in this case, lambda is 2. Lambda is the re-escaling. I allow A, the lattice spacing, to go to lambda. OK, so if you agree with this, then the next statement you will also agree with. If p is bigger than pc, and you do that, then lambda tending to infinity, p prime will tend to 1. So in fact, if you're above percolation threshold, you have global clusters. And repeating this resumption process will make the lattice darker and darker until it is totally dark. So under this p, in percolation is in fact playing the role of temperature. And I have a critical percolation. I have a critical value of p for which percolation happens. Above it, I go to infinity. Below it, I go to 0. And this is a repulsive point. These are attractive points. Let's. It means we iterate the normalization and infinite number of times. Exactly. So because it is 2 in one iteration, I multiply it again. It comes 4, 8, and so on. So I hope you have taken this down. I'm going to clean it. So I expand R of p on the critical point. I did exactly that for the Rg equations in general. But here, I have a specific case. This is equal to pc by definition because this is the fixed point. And I have p prime minus pc is equal to the Rdp at pc times p minus pc. So this is delta p prime is equal to lambda to y times delta p. And this y is equal to the log of drdp at pc divided by log of lambda. And this is my critical exponent. And I expect, of course, for this to be positive. This is positive. So this has to be bigger than e. OK. I can call it. OK. You are right. This is a scaling exponent. But from this, I can obtain the critical exponents. They must come at exactly function of this. Now, when I do this operation, the correlation length changes. I do this operation. I go from p to p prime. And this is given by this function R of p. And as a result, I have a new correlation length over the lattice, which is differently filled, depending on whether p is below or above pc. So the correlation function changes. Hence, the correlation length changes. And the correlation length changes by a simple factor. Or I can interpret this as the length of the size of the biggest cluster instead of correlation length. Maybe easier to understand it that way. But xz of p is equal to p minus pc to power minus nu. Or xzp prime is delta p prime to power minus nu, which is equal to this guy. So this has to be, this is R. This has to be just that factor. So this y of t here has to be equal to 1 over nu. Of course, we knew that. Because when we did RG calculation, we knew that y of t, or the exponent of scaling, exponent as correctly pointed out, in the temperature direction is just 1 over nu, the exponent of the critical length. And the critical length here can be interpreted as the correlation length, or easier, perhaps, to interpret it as the size of the biggest cluster. So next step. Let's do it in one dimension. Always the simple problem first. So I have in one dimension, I take three sites and make it into one site. What is the probability that, so, sorry, I would call this site full if I can already go from here to there. So I must have three sites on before I can go from here to there. Hence, p prime is equal to p cubed. In other words, the form of function R is fixed. Now clearly, p is smaller than 1. Critical p here is 1. p is smaller than 1. This will always go to 0. And there is no other section except p equal to 1, which will definitely go on to 1. Fixed point is p equals to p cubed. So p equal to 0, or p squared equal to 1, which gives you p equal to 1, p equal to minus 1 is not admissible. So two fixed points. One dimension of percolation is finished. Next problem, let's look at site percolation and triangular lattice in 2D, because it's easy. So site percolation, triangular lattice. So I have this formation. And this can be on or off. So this is off. This is off. Now, the criterion was that I set a whole triangle to on or off, depending on whether I can go from one side to the other. Here I have three sides, so I have to reinterpret this demand. So in fact, I will accept it from connecting any side to any side. So it's enough to have just two to make this on. So in fact, two positions, two. I have three of them on. Definitely the triangle is on. And if I have just two, also on. So p prime equals to p cubed plus 3 p squared, 1 minus p. So 2 on, 1 off. This is r of p. So dr dp, sorry, let's say r of p is equals to p is the fixed point. So I get p critical equal to 0, p critical equal to infinity, sorry, 1, p critical equal to 0.5. Solution of that equation. Three fixed points. These are trivial fixed points. And this is the interesting fixed point. And that's exactly what we know from other calculations. Now dr dp, I need dr dp at PC. Yes. But the thing is that all the fixed points of the RG equation, this is the RG equation for percolation. All the fixed points of the RG equation somehow are critical. They are special cases. They are that scale dependence disappears. And from a physical intuition, it is obvious that these are fixed. These are points which are, in some sense, critical. The partition function has problems at this point. Although you don't even have a partition function for percolation. So this is equal to 3 p squared plus 6 p. I want to calculate that at PC equal to 0.5. Three halves. And as you know, in the triangular RG, since each triangle is reduced to one point, lambda is square root of 3. Hence, nu is log of square root of 3 divided by log of 3 halves, which is equal to some strange number. And we know this to be true from simulations, exact calculations, et cetera. So this is a very, yes? In many cases, maybe this class is too easy. Too easy. I don't know. We can determine the solution of percolation for n. Yes, you can. I think what you are saying is exactly what this guy was saying. Yes. The point is that this method is OK. Whether I can do it or not is another question. Since I am very foolish, there are many things I cannot do. But maybe there is a smart guy in Princeton who can do it. So I don't throw away the method, because I cannot do it. We keep the method. Maybe somebody can do it. The RG equation will give us a speech point, 0, 1, which are these three points, and then we will give a critical. An unstable point will give you the critical. In general, without the structure of the lattice, in general, without paying attention to the structure of lattice, I argued, and it's an acceptable argument, that the critical point is unstable, the two points are stable. This we know. Now for each particular lattice, I get a different R. And the shape of that R will give me the exact value of PC. Now, as Nahid says correctly, you may not get the shape of R right, but if you get it right, it will give you the piece. So percolation has very nice, easy organization group action. What about this conformal field theory tomorrow? There are two sizes.