 Okay, so let us continue with the morning's lecture. So we have been discussing this, Lee Young theory of zeros of partition function. Don't chat now. If you want to organize something, please do it during the coffee break. Okay? Pay attention to the lecture. Okay, so we have been discussing the Lee Young theory of phase transition, so let me just recapitulate the main points. To bi smo se povrstili, da je zelo vsega pratiče, je omega z, ki je obrsta, da je zelo na volume sestim, ki je polinomir. Profesionalna še in na vsega, da bi smo se napravili na n- in tudi je prizaj, da ne, a zelo n- in po kdo se je zelo vzelo na polinomir, da bi je tukaj po nekaj obrstav, in tukaj n- je obrstav, na kako je zelo vzelo vzelo. se vzupil na dobro vzdne, vzupil na dobro vzdne. If this is my curve, I will be able to define rho of s, which is the fractional number of zeros in this interval. Ok. And then, and the omega of z, which is the Srečo should call something else. G of z is equal to log, omega of z is equal to log, omega n of z divided by n limit n goes koncernigsipa z pilovim voljem. Because we divided by volume. The volume I call the volume n. So let me say the number of sites or this will be the free energy per site. So let me just be more careful what we want to do is we said that omega z is the complex free energy per site. for the complex activity z. So how do we define that? It is defined like this, you take omega z defined on the real axis, positive real axis that is well defined and real positive number and then analytically continue it to other values of z and so we said g of z was equal to integral sigma s log z minus z s. So this is the log of z, now it is a complex log not the modulus of the log not the real part of the log. So this is the complex activity and it is defined on the real line, it is real number so imaginary part is 0 then when I get off the axis then the log z will pick up imaginary parts. But it is a multivalued function so I just continue it and I can keep on defining it, you come here there is some imaginary part I come here there is some imaginary part it is smooth here, but when it comes to a line of zeros then the imaginary part of log z will have a discontinuity and this discontinuity is equal to the change in the derivative of g z. So that is the precise definition of g of z because log z is a multivalued function it is defined by analytical continuation of the function over the real line and continue it everywhere. So this definition works well except if you have something like this then I do not know how to analytically continue from outside into this stuff. So for the moment let us not worry about it there will be some way of defining this, we will not worry about it just now. So that is the definition of g z uniquely requires analytic continuation the real axis positive real axis and then there are discontinuities along the discontinuities in imaginary part of omega of z along lines of zeros. So next comment is that there was one participant who explained the Lee and Young theorem saying that you said that it is hard to explain, but actually so long as the polynomial omega of z is equal to z to the power n omega of 1 by z it is a symmetric polynomial on two sides. If then all zeros on unit circle so it has some symmetry property which is that the partition function with n holes is equal to the partition function with b n minus small n holes with n particles is equal to partition function with n holes then there is this symmetry in the system and then you will have these zeros this was the claim, but this claim is incorrect and I will give you this example which I give actually in the morning omega n of z is equal to 1 plus 2 to the power n by 2 z to the power n by 2 plus z to the power n. This thing has this symmetry I can try to figure out what are the zeros of this polynomial. This is not so hard because I think of it as a quadratic in z to the power n by 2 then it has two values and then of course if z to the power n by 2 has two values then I can take the nth square root of it which will give me all the values. So the zeros of this are actually on two circles like this and they are uniformly dense along the circle. So let us without going into lot of formal theory it is good to get some practice with actual evaluation of zeros of polynomials. So if I have a million degree polynomial how to calculate is zeros that is the question and of course the answer is you cannot do it for a generic polynomial, but if the polynomial has some special structure then we can hope to calculate it. So let us do a practice problem that one is actually stupid problem but let us you know this is sort of to start with verma. So I have a lattice and there is no interaction. This is a lattice case at each side you can put either zero particle or one particle but there is no interaction omega n of z is equal to 1 plus z to the power n. So this problem is very easy there are n zeros all of them are z equal to minus 1. In the absence of interaction there is actually a delta function at z equal to minus 1 where all the zeros are there, but this was too trivial. So let us take less trivial problem. I take a line I want to take periodic boundary conditions and there are n sites and the interaction is next nearest neighbor exclusion, neighbor exclusion lattice case. So what that means is that on this line if you can occupy a site this is occupied, but the nearest neighbor of an occupied site cannot be occupied nearest neighbor exclusion. So if you have n sites you can occupy alternate ones of them that is allowed that is the maximum possible packing. You cannot put more than n by two particles on an n site periodic lattice. So you should imagine that each of these particles has a size like this then the next one cannot come here, but this one can come here and like so on. Of course you can imagine bigger radius, but let us work with n equal to 1 nearest neighbor exclusion. So I have this lattice case I want to determine omega n of z. What is it? Well I guess if you have two sites then I know what is omega n of z two sites it is 1 plus 2z because you can only occupy 1, 4, n equal to 2. So what is it? Suppose I put n equal to 3 then what is it? 1 plus 3z because you know you can put in one particle, but by definition you cannot put two particles adjacent to each other. So in there they are three ways of putting one particle. So for n equal to 4 this one requires more work, but not very hard work 4z plus with n you can put two particles on a 4 site ring in two ways and so it is plus 2z square and I can determine the zeros of these. If you put 5, n equal to 5, of course still you can only put 4 particles and I am doing ok. If you put 6 then you get a cubic and if you put 10 you get a quintik and so on and then it becomes a little bit theory. Ok, so but can I determine omega n of z for arbitrary n, n equal to 500? The answer is yes, I have done it in the past or you have done it in your class using transfer matrices. So I define, I usually prefer to, there is a transfer matrix t. So that sort of the partition function up to n and then I add one more. So this is a 2 by 2 matrix. You look at the last site, it can be occupied or empty, this is the vector, but if it is occupied, occupied you cannot put it there. If it is occupied and then it has to be empty, sorry. So if it is empty then you can occupy it, but you put in a bit z. If it is, in either case you can put one and so this t is a matrix like that. Now I erase this and z n is equal to trees t to the power n. This formula is familiar to everybody. 2 by 2 matrix and no, ising model in a field or some such thing. So now this looks much more tractable because this is a 2 by 2 matrix. I can diagonalize it and the answer is lambda plus to the power n plus lambda minus to the power n and I can write down this matrix. This is a small exercise, I can do it here. Zero z, one, one minus lambda minus lambda determinant equal to lambda square minus lambda minus z equal to zero. So this answer is actually just very simply 1 by 2 plus square root 1 plus 4 z, 1 by 2 plus z to the power n plus 1 by 2 minus square root 4 plus z to the power n. You can check of course that for any n when you square n, add them up, all the square roots will cancel out and you will get a nice polynomial. Which will of course match with these numbers when 1, 2, 3, 4, but it is a general simple formula. So this is my lambda, this is z n of sorry, we had called it omega z, omega of n of z. Now, what are the zeros of this function? In this case it is actually easy to see. So I want to solve this equation, omega n of z equal to zero. So this should be equal to this with a negative sign. So I get 1 by 2 plus square root 1 by 4 plus z over 1 by 2 minus square root 4 plus z to the power n equal to minus 1. So that is easy to solve because you know this nth root of unity I can write down and then this should be equal to nth root of unity. Sorry, can you? T matrix, very good. I should explain. So usually when I want to write the T matrix, I work with let z omega r plus is equal to partition function of r side line, open line with actually let us call it 1, left most site occupy. So it is a restricted partition function. So I take a r site ring, I put occupy this site and allow all possible positions, configurations of the rest, put in a Boltzmann weight z to the power n, if there are n sites in the cluster add them up that is the partition function. Then I define similarly omega r zero is equal to everything is the same, but unoccupied. The left most site is unoccupied. Now it turns out or it is easy to realize that omega r of 1, omega r of 0, r plus 1. Suppose I have determined all the functions up to r and I want to determine them to order r plus 1. So omega r plus 1, 0 is you want this one to be occupied, but then can be do anything or the rest. So I can choose two choices for this, it can be 0 or 1, correspondingly there will be a partition function omega r of 0 or 1. So there is a linear set of recursion relations between omega r plus 1 of 1 and omega r plus 1 of 0 in terms of omega r of 1 and omega r of 0. And this is a 2 by 2 matrix, the set of coefficients and that matrix is this one. So very good. So how does this work out? Well, if z is positive, then this thing is bigger than this or this stuff will never be equal to 1. There is no hope of this equation being satisfied if z is positive. If suppose z is negative, but less, bigger than minus 1 by 4, then this quantity is positive and this will be smaller than this, it would not work. However, suppose z is minus 1 by 4, then of course the equation is satisfied, because the modulus of this is equal to the modulus of this. I think I should write. Suppose there is a little bit less than minus 1 by 4, then this quantity is negative and square root of a negative quantity is imaginary. So this stuff looks like, I can erase this part. This stuff looks like 1 by 2 plus i epsilon and the denominator looks like 1 by 2 minus i epsilon. So both of these are the same magnitude, they are like exponential i phi and I want exponential i phi to the power n equal to 1 and so phi can take some real value. If n is large, then phi can take any real value, otherwise it takes some discrete set of values. Just one more step, which I have worked out in my notes. So I write this as 1 by 2 plus square root 1 by 4 plus z equal to 1 by 2 plus i epsilon right. So z is equal, actually I write 1 square root epsilon z is equal to minus 1 by 4 minus epsilon and then I get e to the power 1 by 2 plus i epsilon equal to 10 exponential i phi phi is equal to 10 inverse root epsilon. And then I guess, all I need to do is I put exponential i n phi is equal to minus 1, which has n roots. So it has n values of phi and I will get the roots. More importantly, for arbitrary n the condition for the roots is that the mod value of lambda plus of z divided mod value of lambda minus of z equal to 1 or n goes to infinity. So this is the condition for the zeros of the polynomial. Now this condition will not be satisfied everywhere. In this case it is satisfied along the negative real line going from minus 1 by 4 to minus infinity. So as phi varies from 0 to pi by 2, this function will vary from 0 to half plus i infinity, because there is a tan. When epsilon goes to infinity tan inverse will go to pi by 2. So the roots of this polynomial lie on this negative real line from minus 1 by 4 to infinity, sorry all the way up to here. That is where the zeros of this polynomial lie. I can also determine the density of zeros of this. How? Because this phi is actually, the solution involves different values of phi, but all values of phi are equally likely. There is a root for each value of phi and phi are uniformly distributed on the circle. So in the larger limit you should think of phi as uniformly distributed on the circle as a continuous variable, not a discrete variable. And then this transformation tells me that rho of phi will look like this. There is a pile up of zeros near the end, because even when phi is very small, epsilon is very small, phi is much bigger, because it is a square root of epsilon. Phi is proportional to square root of epsilon. So the integrated density of zeros from here to some distance epsilon will be root epsilon, which says that the density of zeros will diverge as 1 upon root epsilon. So in this case rho of z is equal to non-zero for z on the negative real line, real line less than minus 1 by 4. And then the density of zeros behaves like this. This is z, this is minus 1 by 4, the density of zeros behaves like this. There is a pile up, blows up, blows up near z equal to minus 1 by 4 as 1 upon square root z minus z. 1 by 4 plus z is negative, so I should write the other way. Or maybe I write 1 by 4 plus z mod, not so good. I think the whole idea was to rely on the fact that omega n is actually looking like this. It is sum of nth powers of simple functions. And then evaluating all the n roots became very easy. And then we found that the result is that the zeros lie on the negative real line and they blow up near the endpoint of zeros as a square root singularity. So now I can try to see if I can extend this result to some other problem. So suppose I had a same model, but it was not the nearest neighbor exclusion. It just had a repulsive interaction with next neighbors. I can still write a transfer matrix. Just it does not have a zero here. It has an exponential minus beta u or some such thing. And I can do the whole thing. And everything works out. Except that the position will of course depend on beta u. So which comes to the point that as you vary beta, the zeros will change. The position of the zeros will change. But the general character of the solution remains the same. At the edge, the density of zeros diverges as the same square root power. Yes, there was a question. So I put repulsive interaction. So get out of the lean, but I can also put attractive interaction. And then it is very interesting. All that happens is that as soon as you, you know, in order to go from attractive to repulsive, you have to go through zero. So as you decrease the interactions between the zeros pile up, but this point moves out. And at u equal to zero, there is a single point at which all the zeros occur minus one. That is what we discussed first. And then when you decrease u further to make it attractive, then the zeros go like this. You can check. That comes very, it is the same equation. And now the zeros are on the unit circle. Again as you vary beta, the zeros will move. Or in our way of saying, we will say that the zeros move and the density of zeros will change. But it always has an edge point and its square root singularity at the edge point. In this one dimensional problem, what happens if you decrease beta more and more, it keeps on decreasing. But it never actually closes up. For any finite beta, it remains at the finite distance. Only at beta equal to infinite, it actually pinches on the real axis, which is consistent with the fact that in 1D there is no phase transition. So all our general phenomenology was working, except that the zeros never pinched in on the real axis. No sweat. So then, can I do this problem? I take a ladder rough like that. And again, there is a nearest neighbor gas. So if I sit at this side, this side, this side, this side are excluded. But I can put one here, I can put one here, I can put one here and here and so on. Now, can I find the partition function for zeros on this ladder? Well, we can still try to use the transfer matrix. Now, the transfer matrix is bigger because it has to start with, you have to allow the possibility that both of the, you know, I have a ladder built up to here and I add two more sides and then try to set up recursion relations for the partition functions at level r with r plus 1. What is the dimension of the transfer matrix? Yeah, in principle 4, but actually only 3, because you can have at most one side on one column. So, you write down a 3 by 3 transfer matrix. T is some 3 by 3 matrix. I will not write it down. And then z n or omega n will be equal to lambda 1 of z to the power n plus lambda 2 of z to the power n plus lambda 3 of z to the power n, where lambda 1, lambda 2, lambda 3 of z are the three complex roots of the cubic characteristic equation. And now, I want to solve this polynomial equal to 0. So, the point is this, suppose lambda 1 modulus is bigger than lambda 2 modulus is if lambda 1 modulus is bigger than lambda 2 modulus is bigger than lambda 3 modulus. So, suppose this is, we can always assume, we can assume no problem. This is just ordering by hand. So, if lambda 1 modulus is strictly bigger than lambda 2 modulus, then I look at this equation. This term is much bigger than this term or this term. No hope that nth power of that term will be able to cancel the other ones. So, it will not work. You will not get any 0. So, for a 0 to occur, for a 0 of omega n z for large n, for large n, we must have modulus of lambda 1 of z over lambda 2 of z equal to 1. So, that is a condition on z. And that condition determines all the places where the zeros will occur. They may not occur, you know, they, but they cannot occur outside this. So, this is the curve on which the zeros can lie. In this case, I guess, in this case, we are in a lucky case. Actually, it turns out, it is not so bad because this is a 3 by 3 matrix, but it can be converted into a 2 by 2 matrix using the symmetry under reflection. And up-down goes to down-up. But ignore all that stuff. This one, you can write the expression fairly easily. And again, the qualitative features that, well, there is a, the zeros lie on some line. Actually, again, they lie on this line. And there is an endpoint. And there is a pile up near the endpoint. All of these are the same. So, that is some interesting, nice feature that in spite of the, so, I can do 3 by, you know, I can do this problem. Now, the matrix will be even bigger dimensional, but it does not matter. I will just take lambda 1 and lambda 2 and ratio. That equation is true for whatever is the size of the transfer matrix. And so, that is the condition for the partition function to have zeros. And that is the equation of the line of zeros. It turns out that for the case with Li and Yang discussed, this equation is just mod z equal to 1. But in general, it can be much more complicated. For all these models, which have this one-dimensional transfer matrix, it turns out that the solution is always of this form. And the zeros are like that. You know, whatever this equation. Then people can study numerically this stuff even in 2 dimensions. And so, numerically, people can generate partition functions with 1 million degree now and get 1 million zeros and then plot them and see, you know, what the density of zeros looks like. And so, let me do that. Just one minute. So, I am actually giving some simple examples, where you can see the answer immediately. You can always construct much more complicated examples, where something else will happen. So, I am giving the typical behavior. Or I am sort of saying that this is what happens here. And I give you that, if you change the problem a little bit, the answer does not change so much qualitatively. But if you change it a lot, if you put some funny interaction, something, something, then the answer will change. Yeah, sure. I am not, right now, deliberately, I am not proving any theorem. I am only giving you examples of systems, where you can see the zeros come and how they behave. So, now, let me just say that, ok, suppose you do it in two dimensions, with only attractive interactions, then, of course, Lee and Young have already shown that the zeros occur like this only on a unit circle. But now, like in the 2D Ising model is an example, 2D Ising model in an external field has not been solved. So, omega of z as a function, you know, in a non-zero field is not known analytically, ok. However, numerically, you can determine the partition function for 20 by 20 lattice or some such thing. And so, you find that at t equal to infinity, where there is no coupling, of course, all the zeros are sitting here. Then, if you decrease the temperature, the zeros move like this. And typically, the density of zeros has this structure. This is my variable s, and I do not know, from s min to s max seems to go like this, seems to pile up at the end, some little bit. Then, as you increase the beta or decrease temperature, the zeros spread out more, but the behavior is the same, now they look like this, ok. So, there are regions, there are lines of zeros, and there are endpoints of zeros. The lines of zeros have a finite density, endpoints have sometimes a pile up, ok. And the pile up has some exponent typical divergence, which does not depend on the temperature at which you are working. So, now these zeros, you decrease the temperature, they spread out like this, they come like this, but at a finite temperature, they meet at the real axis, and then below that temperature, there is a finite density of zeros throughout this circle. So, if you go to t less than t c, and you calculate the discontinuity in the derivative of the potential, which is the discontinuity in the electric field, which is the discontinuity in the charge density, which is equal to the charge density at that point, it is finite. For the Ising model, it says that for t less than t c, as you change field, the magnetization jumps by a finite amount, ok, which is our definition of phase transition. So, everything is working fine, but I wanted to know what happens at t c. So, what happens at t c? Here, the condition is that lambda plus equal to lambda minus, for the modulus of 2 is the same, and so that says that the gap vanishes. If the gap in the spectrum vanishes, that means that the correlation length becomes infinite, so that is where the criticality occurs. The correlation length has to diverge at this point. So, suppose we are at t less than t c, then I have this circle, and suppose I am here, or even suppose I am here. Again lambda plus will be equal to lambda minus, the modulus will be same, the gap will vanish. So, all along this line, correlation length suitably defined will have to be infinite. So, the picture is very nice, you have a complex z plane, there are some lines, where the free energy function is singular, everywhere else it is well behaved, and there is this electrostatic analogy, you can imagine there is a charge density of zeros, and the density determine the potential and the singularity and everything, so far so good. So, then, so what was realized is that, so you can change the potential now. Suppose I started with some model with Leonard Jones potential, this one, I can make a lattice model with this potential, and then there will be some set of zeros, it does not have to be critical point, it is at some value. Now, I change the potential here a little bit, continuously deform the potential, what will happen to the zeros, they will also deform, they will move continuously, all the eigen values of the transfer matrix will change in some continuous way. So, this line may become like that line, it will change a little bit, maybe this point will also move like that. The topological features of the transfer matrix will not change much, oh sorry, topological features of this picture will not change much. The positions of the lines will shift as you vary temperature, as you vary interaction, and so on. But keep points which are for us important is that, there is a density here, and there is an endpoint, there is some power low behavior of the density at the endpoints. So, it turns out that this power low behavior is independent of the potential, you can change the potential and the power low does not change, it is a universal behavior for all models in which you get this kind of structure, repulsive hard core singularity. If you have a model with repulsive hard core interaction with anything else, then typically it has this structure, it has endpoints of lines of zeros, at the endpoints of zeros there is some kind of a pile up, or maybe actually there are models known where the density of zeros goes to zero at the endpoint as a power. So, density at epsilon distance will go as epsilon to the power a, a may be a positive number, not a negative number. So, this power remains robust with respect to various changes, so what is it? So, it turns out this power is called the Li Young edge singularity exponent, there is density of zeros near the endpoint of zeros of line of, endpoint of line of zeros goes as epsilon to the power sigma. And then it turns out that if you put any model which is finite dimensional, so it is called d equal to zero models, then you know, then sigma is equal to minus one, there is a, just a single point, so that is, we said that the partition function went like 1 upon 1 plus z, sorry, the partition function went like log 1 plus z, so this is a limiting case, actually all the zeros are at one point. So, you can say that the zeros are like this, but this power epsilon, sorry, this power sigma, if it tends to 1, the pile up becomes bigger and bigger, the stronger divergence in the limit, everything is at the end, so that is a point distribution. So, in this sense sigma is minus 1 for all d equal to 1 models, sigma is minus 1 by 2 and this thing has been worked out exactly for d equal to 2 models, and then the sigma is minus 1 by 6 and it is not known, the exact value is not known in 3, 4, but in d equal to 6, which is sort of the upper critical dimension, so I said d bigger than equal to 6, sigma is equal to plus 1 by 2. This we have not shown, but this is possible to show, because you can show what happens in the spherical model, you can calculate something like this and it gives you sigma equal to half. So, that is sort of the interest in this model, usually there are things called critical exponents, and they depend on two parameters, one is called n and one is called d, n is sort of some vector model index, you know n vector model and d, but this repulsive hardcore singularity is very nice, because there is no n, it only depends on d. You can construct all kinds of models with different behavior, but they will have the same young singularity exponent. So, my time is kind of up, so there is a article I can give you the reference to shoes and fissure, did I write the shoes and fissure reference? No, I wrote the lie and fissure reference, journal of chemical physics, 25, 1, 0, 3, 8, 1, 4, 4. This will give you lot of other references on this topic, you know this is just the starting reference. And I think that is the end of what I want to say, there is one thing which I forgot to say earlier, which I should have said, so this particular theorem of Lee and Young, which says all the zeros lie on the unit circle, it is actually a very cute theorem and maybe the reason I did not give you the proof is that I hope you will look it up. It is a nice short proof and Young in his 80th birthday was asked to comment on his work and of course, he is a very modest man, so he said oh my one, you know he was to select some of his most interesting work and he said this theorem is a minor gem, but of course, he that is an understatement, you know it is a very profound result, which comes from very elementary analysis and it has a wide applicability and so all the students should be aware of it, all the students of statistical physics should be aware of that particular theorem. So, I will stop there, if you have some questions you can ask, otherwise then we meet again tomorrow. Yes, I do not know topological phase transition is too advanced a topic, can it be dealt with in this, no no in this setup, where I am dealing with some particles interacting with some potential, you should identify what are the degrees of freedom in your model, you know I can allow all kinds of interaction in mind, maybe not one species of particles, maybe five species of particles, then I will put z1, z2, z5, then I will perhaps have to study this Lee Young theory in the z1, z2, z5 complex multidimensional planes, already a little bit difficult, because the theory of several functions of several complex variables is not quite as advanced as the theory we are discussing. Things like topological phase transitions start with some extra variables, which are not easily taken into account in this kind of structure. So, I am not able to answer it immediately for sure, I do not know if I think about it for one day, can I still answer it, maybe not, but you know one can try, I should mention one more thing. Sometimes it happens that there is a line and then it at one point it develops side arms, so it can undergo topological changes, which occurs when two of the zeros suddenly become complex, instead of two real zeros, you have two complex zeros. So, at some point the lines may split and they may develop side lines.