 Okay. I found this report on the Internet. It's very, very good. In fact, I suggest you read it. Analyzing two-dimensionalizing model with CFD, Paolo Molinini or Molinini, et.h. Zurich. I prefer not to photocopy it. This is a long document with a lot of detail. Just find the electronic version, save some paper. It's very interesting to look at because it does all the detailed calculations and you also get to know conformal field theory. I will follow it in the class. So I will cover part of it on the blackboard to continue on the RG of Ising model. So we got this RG part out that the K coefficient renormalizes like this. What we now need to do is the edge, the edge component of Hamiltonian. Hamiltonian was like this. We have analyzed this part. I now want to look at this component as well. We need this for RG to be complete. It's an easier part. So in fact, if I look at minus beta H, this becomes K and let me just call this H, redefine H to beta H. So what happens is that I have taken spins in the triangles. Each triangular section has three spins. So this is now very easy to do because it divides naturally into sets of three. So it looks like this. By sigma 1i, I mean one of the spins which sits in the triangle. So it is this spin here. But we had worked out what the expectation of this is. It's equal to this factor times sigma i, sigma i being the spin which sits here. But there are three of them. So it is 3H, this guy sigma i, sum of this term becomes that. So I can easily find that H prime. H prime is H after renormalization. Is equal to 3H times E to the 3K plus E minus K A plus 3E. The difference with the K renormalization is that it mixes the two coefficients. In the K, you just had K to K so you could easily differentiate it which was why I did it first. But this one actually involves two. So I keep this here for a while so that you can take the note and continue down here. So I have a RG formula for K prime H prime which is equal to some sort of, which is equal to an expression of K and 3H. Something like this. So when I form my stability matrix, sorry? What is f of K here? It's that guy. I don't know who said it but I said the mark of a theoretician is laziness. So I'm just being lazy. So when I form this matrix, the K prime dK, the H prime dK, the H prime dH and the K prime dH. This is clearly 0 and these two are not 0. So this is my M which was the coefficients of the beta functions. However, so you see that it is triangular which means that the determinant of it is this term multiplied by that term. I know what this term is when diagonalized and the eigenvalue. So this is surely the other eigenvalue. I don't have to worry more. So dH prime dH at K star is equal to 3 times e to the 3K star e to the minus K star and that is a number which we have from previous calculation. So lambda to the power YH is equal to 3 over root 2. Hence YH is equal to log of 3 over root 2 divided by log of root 3 which is approximately 1.37. 1.857 is exact. So by application of Rg I can get both exponents, the other exponents we worked out yesterday, and having two exponents is enough to calculate all the exponents and that this is of course an approximation which has been which has been the first term of the expansion which I used. You can go to the second approach to the second term in the approximation which has been done and these numbers improve a lot. But for us it demonstrates the point that how does Rg look like for Ising model and what does this business of calculating the exponents involve? Which is a calculation like this. And the point of this part of the course was to see the ideas which were explained in abstract how they apply to the Ising model as an example. All right, next thing which we need to look at is how does conformal invariance apply to the Ising model? You are asking where does the exact number come from? Exact number comes from Onsager's exact solution. That's it. Yes? Is the approximation only when we define this big I? No, well yes in a sense. Hamiltonian was written like this. The difference between h0 and h interaction was not that there is a small interaction but that I separated the spins inside the triangle and spins outside the triangle. So spins inside the triangle went into h0. Spins outside the triangle which would be these two interactions. All these interactions exist between the lower triangle and upper triangle and so on with the sides as well. It's a lot of spins interacting but my point or in fact these two guys point of their observation is that these other spins they are thinner. They're not as many as the internal spins. So they probably play a less role. And this is what we did. We separated these guys from these guys. And the approximation is that now this h interaction will come in as corrections to the mean term. And this was the first correction. So I was calculating the average of h interaction and this went in as a correction to the main calculation. This guy which is giving me these numbers. The next term will be h interaction a squared minus h interaction a squared. This will be the next one. A little bit harder because when you square it not only you have to take account of these spins but also this guy interacting with the second triangle away. A little bit hard but can be done okay. So what I have done is easing. I have looked at it under spontaneous symmetry braking Rg and now I want to look at it under CFT. This is my claim that any critical point so it actually has to be critical easing. You can analyze any critical model, critical point from these three points of view. Each of them has its own merits. So it's this guy which I want to do and this is an example of the easing model. It has to be possible for all other critical models in two dimensions. So let's do the CFT. How does this work? So this is a lattice calculation which you can find in many books. This does come out that the two point function of spins at i and j folds like the distance between them over a correlation length. But we know that the correlation length diverges with T as T tends to 0. Hence what happens is that this power law disappears. Again you can do this calculation for the 2D easing and you find that at the critical point we have this behavior. So two spins still have a reducing correlation but the reduction is much slower with power one-quarter of the distance. Connected green function, is that this question? Yes, yes, this is the connected green function. So I have not written the second part assuming that I'm above Tc and it vanishes. But what has to sit here is the connected green. I want to get this power over one-quarter out of conformal field theory. The whole point is that where does this one-quarter come from? This is one of the critical exponents. In fact, eta is equal to 1 over 4. So let's see how that can be done. The claim is that this Lagrangian for a Majorana field, epsi, corresponds to the easing model. What you should read that as is epsi and its complex conjugate and the derivative is now with respect to z bar. So it's actually saying that epsi is only a function of z bar. Sorry, only a function of z. This is the action. The equation of motion will actually be d bar epsi is 0. So epsi is just a function of z. The $60 million question is how do you know that this is the easing model? And this is what I'm going to do. Let's see. All on the blackboard. I am sorry. Too much mathematics and could not prepare a slide for it. It has to be explained. Why it should be a fermionic field? Why should it be a single component fermionic field? It's a single component fermionic field. It is not two. It's not a doublet. So for those who came late, this paper is in fact what I'm going to follow now, which you can find on the internet. Search for these keywords. It is not a paper. It's not an archive. It's just on the internet. Somebody wrote it and put it up. So what I want to do now is to go from lattice easing continuum quantum field theory. To do that, the best point of a start is quantumizing. Not classicalizing. So I start from quantumizing. And the other thing to observe is that if you start in quantumizing, you have to have d equal to 1. It is an observation which is old. And on my part, I don't understand it fully. But quantum critical phenomena in d equal to 1 correspond to critical phenomena, thermal critical phenomena in d equal to 2. And there is a calculation which shows, says that in fact, if you take quantum, there is a quantum mechanical theory, which is d equals to 0. And that will correspond to easing in d equal to 1. These calculations exist. I think maybe this one also is in that paper. So it's interesting that now there is a whole community of people challenging that statement who are studying quantum critical phenomena. They are saying that you cannot really get all of it out of thermal critical phenomena in a higher dimension. For example, Mr. Sachdev has a thick long book on quantum critical phenomena, which I would have thought it's unnecessary. You just have the two higher dimension and you come. So he has a reasoning and I cannot follow it. But here it applies. For the case of easing, it applies. So we can in fact do this. We take the one-dimensional quantum easing and from this we can go to thermal easing in d equal to 2, which is our aim. But first I will come down this way and show that quantum easing in d equals to 1 can correspond to this conformal field theory of Majorana of free Majorana phenomena. But how? How is the question? So the quantum Hamiltonian for easing in 1d is equal to 2. So what I now have is a chain of spins and now they are quantum spins. So I have the full set of Pauli matrices flipping them. So urban it is one-dimensional. So on each point of the lattice these polymetrics and the Hilbert space will be a Hilbert space of 2 to the n quantum spins which can be in the state up or down. The other things apply. Neighboring sites only interact. Let's write this carefully. i plus 1. Now this picture is okay for a statistical mechanics but because I want to go to field theory it's not okay. In field theory I need creation annihilation operators. So I need to connect these guys with creation annihilation operators and its question is how? So in fact in on the chain I have a spin up and a spin down to a states possible on every side. I want to change that picture to chain empty side full side. This is the quantum field theory image. If you are not familiar with quantum field theory the central operators in quantum field theory are creation and annihilation operators. So that Ci when it hits the vacuum kills it because the role of Ci is kill a particle at i. It's called an annihilation operator and a Ci dagger when it hits the vacuum creates a particle at site i. It's a creation operator. So I'm going to associate the concept of no particle with a spin up for example the concept of one particle with a spin down. And in this way the Hilbert space is not enlarged. I had a Hilbert space of 2 to the power n and again still I'm keeping it as 2 to the power n. If you have a background in solid state you may be able to understand this much easier than me because in solid state people are used to this kind of Hamiltonians. This kind of Hamiltonian kills a particle at i and creates a particle at i. It's the aim is to derive such a shape before we can turn it into a field theory. Once we have that shape then there is a standard method to turn it into field theory. So what I have to do is to go from this spin up and down to existence of particles or non-existence of particles. So in the spin formalism sigma i z gives me the sign of the spin sigma i plus flips the spin of three operators in the spin formalism. These plus and minus are x plus y sigma x plus sigma y for example. So what are the corresponding operators in this picture? So corresponding to these operators I have the ci operators. So ci operator annihilates. So this picture is opposite to this picture. Up is existence, down is not not existent. And ci sigma i minus one brings it back. And then I have this operation sigma z just gives me the sign and you can see from here that if I have a spin up ci takes the spin up to spin down ci plus picks it back there and one minus two will give me one. Alternatively if it is down this will annihilate it and I get a plus one. So these are the right operators. Now yes. Yes. The general answer to that general question is nothing. There is no algorithm for every Hamiltonian that works. It's like the other question that came up the other day. A lot of theoretical physics is creativity. You create a solution for a particular problem. To be able to create a general framework is pretty hard. So in general to go from a statistical mechanical model to the corresponding field theory is no easy task. I don't know how to do it. For instance your question can be applied to Q state POTS model which has a bigger symmetry and the corresponding conformal field theory. Although I know what it is I don't know how to arrive. At least I don't know. Maybe somebody does to arrive from it to it from a given discrete model. So even for icing as you will see it's pretty involved. And this is the simplest model. I set you a research problem if you like. You know the Abelian sand pile model that Professor Dar is talking about. I know the field theory that it corresponds to. I have it. I can write it on the paper. But I cannot go by similar calculation from the discrete statistical model to the quantum field theory. And that's an excellent paper if you can do it. Huh? Okay. So the next question is that we need to have commutation relations. We need to have commutation relations which are C i dagger C i equals to one. And you see that it holds here sigma i plus sigma i equals to one. So far so good. The same answer I gave him. It's magic. There is no how to it. Here. So I have a model. I have a model which a spin up is existence of particle. A spin down is non-existence of particle. The sigma fields what it does is that it flips the spin. So if I have this as existence of particle it flips it to a spin down. So it must correspond to an annihilation operator. It's a particle and I kill it. Sigma i plus on the minus state should kill it. I see what you are you saying. It doesn't do anything because sigma i plus squared if I'm right is equal to sigma i plus. So if you apply another one nothing will happen here. And it just leaves it the same. Yes. Yes. Yes. Exactly. And the same is true. And these operators exactly do the same thing. They annihilate the particle and then if applied to a site which has no particle nothing happens. Yes. So if I apply c i into minus which is the state zero I should obtain zero the number. If you apply c i to the state zero it does it is it just gives you zero again. It doesn't kill it. Yes. There is a problem here. I'm confused. Let me think about it a little tell you. Yes. You mean it's you mean you see I see I dig it. No. I see that. If I accept that it's not true but I mean apart from his question why this works. If I accept that one minus two c i dagger c i on zero will equal zero zero is a spin down zero is a spin down. However if I apply this on a spin up so which is one particle here this will equal to one minus two I see I dig a zero and then this will put a one in it again one minus two one which will equal to minus one. Okay. I agree with you. Okay. I agree. Thank you. Because if you if you apply sigma i plus two c plus you obtain minus but if you apply it again you obtain zero right because if you go down two times and you go to the this is what I'm this is what I cannot remember exactly sigma i plus is this so sigma i plus squared seems to be zero. So if sigma i plus squared is zero then then this works out right because sigma i plus and minus will give us zero. So I have to and then one of the two of the two times two ones at the same stage should be zero. Yes. That's right. So sigma i minus and and the plus must give you zero based on this definition. So I have to I have to interpret the no state with positives. Yes. This one. Yeah. That's okay now. Thank you. Excuse me. These plus minuses are confusing. After the lecture I walk it out and tell you otherwise we will spend half an hour doing this. Let's come back here. So I want this anti commutation relation for my operators and it's okay because these guys also have these guys also have this anti commutation relations. So everything so far is okay. Now there is a there is however a problem. It's just on the face of it is okay but there is a problem. The problem is that these guys at i not equal to j must anti commute but these guys at i not equal to j commute is that true? So here we have a problem. We need to we need to solve this problem and this is solved by doing what is called a Jordan Wigner transform method string various names. What Jordan Wigner suggest suggested actually this was used to solve the Ising model in 2D but it's also used here is that you assign a string to solve your problem. That is you take sigma i plus is equal to pi if it is about that to know. Okay. These are fermionic operators so they anti commute no matter where they are. This is always true when i is not equal to j. These are operate these are matrices which act on the Hilbert space and if they are at two different sites they won't see each other so they commute. This commutation is inconsistent with this anti commutation. I need to have that. So this simple definition a simple definition that this is equal to that will not be acceptable. I have to find something more complex and what I find which is more complex is this guy. I take this pre-factor multiplier c i width and now despite the fact that c i is anti commuting this guy will become commuting. This is called the Jordan Wigner transform or construct. So at the site i you are in fact taking a string coming in from infinity ending up at the site i. To see how it works it may be easier to invert it and then look at the relations. So when even inversion happens so if I invert those relations I'll get c i is equal to pi sigma i z sigma j z. See when I if I use the correspondence this is sigma z my this power two is equal to one so I can multiply it by that side and get an inversion. Now this guy c i c j equals to pi of k smaller than i sigma j sigma k z pi k smaller than j sigma k z sigma j plus. We want to see what this looks like. In fact if I take i bigger than j I can start calculating. If i is bigger than j this has no problem going through that so it goes and sits here. Now if I do the reverse this will go and get a stock this will this time it will not go through because somewhere here sorry this is better. Now this cannot go through that string because somewhere here j equals to k and sigma plus with sigma z does not commute. So I have a string and here I have to use the commutation relation between sigma plus and sigma z which is this. This from here to go to there. See this sigma j goes through this string until it arrives at i. It commutes with everybody until it arrives at i. Yes so at j equal to i I pull it out and then the the rest of the string goes here. You are right I have missed something. So when it hits this string there is a term for which the index of this is equal to one of these sigmas. I take that out which is this guy and then the others are either smaller than j or bigger than j which which go in here. So whereas this here could commute through the whole spring when you reverse it and you write it like that it cannot commute through the whole string. There is a sigma in here that it has to it has to get past what? Yes because this thing a squared gives gives you one and these are sigma z's. There's multiply by this factor from the left hand side but use sigma instead. If you get this you get that expression. Yes. You should control that when we act to plus and minus we get the right. Sorry I didn't understand. We should control that when you act to the Jordan beginner operators. Yes. You act with the right you get the right state when you act on a plus and minus. Yes but you do because all these states are plus or minus one all these operators and only this one does an annihilation of creation now with that problem sorted out. So this doesn't create this doesn't give me anything wrong because these are just plus minus ones and see either annihilates or creates and the same is true here. What I need what I really needed to check is that these guys then now anti-commute at unequal sites which is this relationship. This is what I'm trying to test which is more important than the sign problem there. So this is somehow stuck here. I did this morning and it worked. I don't know why it's not working now. Okay I don't seem to be able to do this but anyhow. So this calculation also I will do later and give it to you. So with this formulation now we have C i C i equals to zero C i C j dagger equal to one. However this was done for simplicity. What we now need is to rotate the spins a little to get there to the Ising model. So I actually need to instead of sigma z use sigma x and instead of sigma x use minus sigma z that is to get back to my Ising model. So what I now have to do is take h which is equal to minus j minus g sigma i x plus sigma i i plus one of z put in this transformation and now this is really long calculation to see what this goes to and also a very careful calculation. It's not solved. And Hamiltonian comes out to be okay now this calculation is going to take a while. So let's just trust this paper and I know from other sources that this is right that you get a Hamiltonian in terms of C i's which is now looking very much like condensed matter Hamiltonians with C i killing at i psi psi i and C i dagger creating. So this is actually saying that it jumps to the next. So a particle if exists at psi i jumps to the psi i plus one and here it goes in the opposite direction if it exists sits on site i plus one goes back to site i. This is also okay. These two terms are not what we want. The reason is that this term for instance just destroys a fermion. There is no conservation of fermion number here and there is no conservation of fermion number here. This kills a fermion completely and this one creates two out of nothing and this is the problem I have and how can I solve that? If you have an open system where fermions disappear on the edge it's okay. That's acceptable. These are fermions actually dying on site but it's a local conservation. So what is the solution? Yes I need the G. You can throw away the G but the thing is that this has this sort of gives you a number. That's actually yeah that's right. If you remember that was my sigma z. It is I need the G to keep this zero. That's all. So the solution to this problem is that what we do which is called the Bogolibov transformation. Bogolibov transformation is that you go to another set of spin operators which are related to these guys by a rotation. I'm always allowed to make such transformations which mixes up my operators on the Hilbert space. It's a linear combination of the old operators giving me a new operator and clearly gamma K dagger has a has a corresponding expression. The reason I'm you see the reason I'm allowed to do that is that maybe that operator which is written there is not in the correct diagonalized form. So all I'm doing is that rotating it a little bit and in fact that's all Mr Bogolibov has said that you are allowed to make such transformations provided it's just a rotation. So UK has to be taken some sine and cosine. UK is taken as cosine of theta K divided by 2. VK is the sine of theta K. And this is now just an angle not an operator. And if you choose now theta K carefully you can end up with the correct shape for the gamma operator for the gamma Hamiltonian. See this is you can see that this is a quadratic form and therefore the fact that I can find the transformation to transform it into this shape is not is not a strange. Say from beginning again yeah actually express your concern let's say I don't understand what is your concern. You see when it comes out like this I have started from some definition of annihilation creation operators and I've come up with this. What it does is that it does not conserve fermion number. Now this could be a physical problem as you say that is my theory is non conserving. It could be just a mathematical problem that this is not actually fermion number it's some other number and it doesn't have to be conserved. The question is can I rewrite it in terms of operators which gives conservation. However those operators will not be these operators. So you can you can imagine it like this for example if you have two chiralities comes you may have a situation where you don't have conservation of chirality but you have conservation of fermion number. The way you have actually is that a fermion operator is a linear sum of the two chirality operators or vice versa. The thing is that what you call here fermion number may not actually be fermion number. This is fermion number and it is conserved. Yeah so if we now claim I now claim that these gommas are the creation and annihilation operators for the fermion field which is that and if you know field theory as soon as you see this you know that I can do it because this is a standard form in field theory for the energy of a free particle or in other words I have interpreted my fermion field as a large number of harmonic oscillators and this is a harmonic oscillator energy but in all excitations k. There is some calculation which I've missed up in the middle of all this calculation I've left something undone and that is to derive these coefficients epsilon k which clearly would have to come out in terms of these tetras and the way these tetras are related to g and h which I have not written. So epsilon k is equal to twice j squared plus h squared minus 2hj cosine of k the square root of 4. So you see that for different values this will never become negative and may have a may have a zero of the energy or it will have a minimum of energy and I can subtract that minimum of energy. So there is a state zero which is the minimum state energy and it therefore has to have the property that gamma k on this vanishes. Now to go to the continuum limit what I need is a Fourier transform or some thing here is necessary. So I need a lattice spacing here to get the continuum limit relatively easy. So what I have to do now is to put this thing back into my Hamiltonian. Again a long page of calculation is necessary and I end up with this Hamiltonian. So this is a continuum field theory now by doing a Fourier transform and arriving from this Hamiltonian. Here is an observation if j equals to h delta will vanish. If delta vanishes then epsilon will be a scale invariant. So on the epsilon to lambda x you see that this part of the Hamiltonian is a scaling variant. This part is not because of the delta. So in fact delta is playing the role of the mass or I can say delta is proportional to the reduced temperature. So if you have a mass or if you have the t in here it's obvious that you don't have a scale invariance because that gives you a scale. Once this vanishes you have a scale invariant Hamiltonian which is almost what I have there. It's almost what I have there because it doesn't have a time component but I can always create a time component by going into the path integral representation go from this to the Lagrangian and that's a relatively artificial thing to do and I can put that in. The constant v is appearing here is not, does not cause any trouble. It's just sets the scale or the coupling constant or something. Yes, yes. So this is, I'm sorry this comes from my background. I sometimes talk about mass sometimes or temperature but they are really interchangeable. So if I go to the path integral representation what I need in here is f psi dagger d f psi dt to get that exact action. So once I get that exact action I have a two-dimensional action. It is a scale invariant and also conformally invariant. So f psi d bar f psi and psi bar d psi bar and a variation with respect to f psi gives me the equation of motion which implies that f psi is only a function of c. Let's do something. Let's say f psi lambda z is equal to some power of lambda times f psi of z to test a scale invariance. What I get then for this step is lambda to a minus one. From here I get lambda squared and this one's, I want to be one. I want it to be a scale invariant therefore a comes out to be minus one. Is that right? Yes. X and t, yes. Yes, so you have to make a definition for example of t plus ix. Yes, but this is something which I jumped over. The thing is that quantizing one dimension in presence of a magnetic field is equivalent to thermalizing in two dimensions without a magnetic field. Okay, this coming out to be one-half. Now I can ask what is the green function? Did I write bar or f psi with itself? Now this has to be something which respects this scaling so it has no choice other than being z1 minus z2. Yes. So they rescale the field. We define that with h, lambda to the h, right? Yes. So we can think about it of the exponent that keeps our action invariant. Yes. So if we get the Lagrangian of our theory, we can say directly what's the exponent do that we need to... If it is free, if it is free. Okay. If the action, if you don't have a free field theory, see these are called, here this is called the engineering dimension. This is the engineering dimension because you can just read it off the action to keep, which keeps this action invariant. However, if there you have interaction, the engineering dimension will pick up an anomalous part which comes from quantum interactions. So if it is free, you can depend on this calculation. If not, you have to do a renormalization group. A very good point. Okay. C or if you like h. So I have a field epsi which has h epsi, h epsi bar is equal to one half zero. And for epsi bar, h epsi, h epsi bar is equal to zero one half. To make correspondence with my notation of how you build fields for CFD. So this is a conformal field theory with this h h bar structure. Yes. No, it can have except that I have only written the real part. There is a complex part as well, which multiplies this. I have to, if I want to be completely careful, I have to. No, those things it cannot have because it has to be, it has to be conformally invariant in the sense that this tells me that epsi is only function of z. And if you have such combinations, it breaks the, it is just a function of z. So it has to be, it has to be able to transform in a meaningful way under transformation of z to say w of z. So, so if you have in this form, in this form, if you make such a transformation, each of these will transform independently. But if you mix them up, then you cannot. Now this has to have this transformation for z. So if you, if you, if you say I want a green function that's real, I have to have multiplied two things together. So this is now, yes. Yeah, from the action, the action. So I assume that epsi lambda z is equal to lambda h epsi z. So I put it in here. It gives me lambda 2h from here lambda minus one. And here lambda 2d2x. Sum them all up is lambda 2h minus one times that. Thus h is one half plus one is minus one. But my definition of conformal weights had lambda to minus h. So I should be careful really say lambda minus h minus 2h minus 2h plus one half. So this is true now. Can you conclude what? Sorry? I don't get the question. Lambda minus h psi of z. Yes. Yes. Okay. No. Why not? No, you can, you cannot interpret that because, because these, I am, when I think of these things, these are not a smooth functions. They are densities or if you like distributions over some space. So it doesn't, it's like the discussion we had about the, if the, the wire stress function. It has to be something mathematically really horrible. It is not a smooth function that, but you, what you can conclude is that it's expectation. So if you have this guy, this expectation can exactly equals lambda to minus h psi of z because these things are a smooth functions. Now, of course, two options are possible either that this is z to minus h and the other is that it is zero. And of course it is zero because you have a single field, two vacuums on either side. But for, in here it is much easier to interpret. So these guys now are a smooth functions. And if I multiply them by lambda, then it is true that this is equal to lambda minus two h, f psi, f psi. And since these entities are a smooth functions, then it follows that it has to be of this one. The psi itself is an operator. It's a very nasty operator. It's a density itself. Sure, of course, you know that better than. In the, it is, it should be valid in the sense of distribution. Okay. Let's stop here.