 OK, welcome back everybody. So before the start of next lecture, let me make a very, very short announcement. As you have seen in the program, there are some computational hands-on plans, both in Python scientific computing and in Mathematica. Regarding Mathematica, in case you don't have Mathematica installed, you should have got a free license that allows you to use Mathematica for the weeks that cover essentially the school. In case you have not got an email with your permission to use Mathematica by tonight, please come to me tomorrow, and then we have to set it up. Instead, for the Python lecture, just bring your laptop, and we will most probably use application through server, so make sure that you have a terminal installed, but that should be more than enough. With that, I have the pleasure to introduce our next lecture, a deeper doc that will tell us about our sphere models. Thank you. So let me start by thanking the organizers for asking me to give you lectures here. So firstly, I'm not fully familiar with the background of the audience here, so I'm going to make some guess about what you know. And some of you will be more advanced than others, and so there will be always some mismatch of the level at which I'm going to present the lectures. But you can interrupt me and say, or privately later, you can tell me that this is going to slow or too fast, and I'm happy to adjust the pace to your comfort. OK, so then what I will discuss in these lectures is phase transitions in hardcore systems. So firstly, this is going to be mostly dealing with classical statistical mechanics. And of course, in the title of the school, it says quantum matter. So why am I discussing classical statistical mechanical models in a school on quantum matter? So I had discussed this with Professor Moshner earlier, and he said that one should not take the title too seriously. And whatever you think is interesting or useful for people to know would be useful to discuss. So what I would like to discuss are things which I feel are very important for you people to know, whether it is classical or quantum. That is the first thing. The second is that for macroscopic systems, which is what we discussed, the quantum matter and classical matter are not actually so different. When you discuss very small objects, then the difference between classical mechanics and quantum mechanics is very substantial. When you discuss bigger things, like cars of course, you know, it's quite obvious that classical and quantum descriptions should coincide. If you discuss something like superconductors, where things are, you know, there is this thing with superconductivity said to be quintessential quantum phenomena. But then you can hook up some classical version of it by just introducing a quantity called phase, which of course has to be explained quantum mechanically. But then you think of phase as a classical dynamical variable. And then the rest of the superconductivity can be very well explained by theory, which just involves one phase, the phase of the wave function, whatever that thing is. So classical mechanics can be thought of as an effective Hamiltonian for whatever system we are going to discuss. And it is a very good effective Hamiltonian. The second point is that for hardcore models, you know, the kinds of hardcore models we discuss are things like dimers on a lattice, fully covered dimers or other shapes of stuff. Hardspheres were mentioned several times in the lectures before. So all these are classical stuff. But they are quantum mechanical systems like RVB wave functions or something which are well described by these. Or the kinds of orderings one sees in systems of hardcore particles are the kinds of orderings you will find also in quantum matter. So that is the explanation for these. But there is, of course, another sort of justification, which is that suppose you want to discuss a d-dimensional quantum mechanical system. It can always be we discuss quantum mechanical systems usually at zero temperature, and that's the most commonly studied problem. Then the corresponding problem can be set up in some path integral formulation where it becomes a d plus 1 dimensional classical mechanical problem. So you sum over histories, which is like sum over configurations. And so whatever you learn in classical stat mech can then be applied to some other problem in quantum stat mech. So firstly, provides simple model for phase transitions, classical or quantum. Then I need to know some prerequisites, which I will assume. So I will assume only the minimum, which is that all of you are familiar, have had the first course in statistical mechanics. You are familiar with micro-canonical, canonical, grand canonical ensembles. And you know partition functions, ideal, Bose ideal classical gas, Bose inferring non-interacting gases, non-interacting. So then we ask that why are these hardcore systems interesting? So my sort of answer to this question, of course, is that they show very rich variety of phenomena. You find interesting non-trivial behavior in these systems. And since these are hardcore models, either configurations are allowed or not allowed. And so one can work with these models without defining ever something called energy or something called temperature. So these are purely geometrical systems. And they show interesting behavior. So they are models of what is called geometrical phase transitions. What are phase transitions here? You can have things called equilibrium phase transitions. So these are examples, hard spheres, which have been discussed a lot. It turns out that if you take a big box and you put in the only thing I say is that there are hard spheres in this box. I assume everybody knows what is a hard sphere. These are just, we imagine things like balls made of steel. And they cannot interpenetrate each other. And if they are far away, they have no interaction. So this kind of a system, you just put n balls in a volume. Or let us say right now they are all equal size. And all configurations which are allowed have equal statistical weight. That notion we still will take. And then I just ask what is the typical state of this system? If I look at these, these may be molecules in a box. These kinds of models were studied by Boltzmann. So it's very clear that at low densities of these objects, particles can move all over the place. And you have something like a gas, but it's not an ideal gas. It's a nearly ideal gas. If the density is very high, this goes into a state in which particles become localized to some small regions. And each particle can move locally. But the whole system shows a crystalline order. This we will not prove, but that is experimentally observed. So just the fact that you have hardcore interactions can show a phase transition from a gas phase to a periodic solid phase. I think that's very remarkable. And we would like to understand this. And we would like to understand similar properties of other systems. What can we understand about hard sphere system? You know, if one studies phase transitions in the standard way, then there is this PT phase diagram. And there are two variables, two independent control parameters. And it is hard to discuss. Here you have only one control parameter, which is the fractional volume occupied by the spheres. And so the problems are expected to be easier to understand. And to that degree, they are more useful as a starting point. So this is a hard sphere and undergo a gas to solid transition, periodic solid. This is an equilibrium transition. But you can also have non-equilibrium transitions, which is very simply described like this. Suppose I take up some shape. This time I take a triangle and put two spheres inside it. Now these spheres can move around. So they can go all over the place, wherever they can reach. But now I squeeze this triangle a little bit slowly. Eventually, you will go to a state in which the triangle is smaller. Spheres are of the same size. But now one of the spheres stays in one part in the other space, in the other part. And they cannot cross each other. This sphere cannot come here and this cannot come here because there is not enough space. This is intuitively obvious. Maybe I can take change the shape of the box, but you know. So initially, the phase space available is fully connected. You can go from any configuration to any other allowed configuration by continuously moving the spheres. If you change the density, then you get to a stage where the phase space breaks up into two parts. One sphere is on this side, two is on this side, or the other way around. And these two cannot be exchanged. So you have a ergodicity breaking phase transition. It's a non-equilibrium transition. When you study time-dependent property, you will find that, oh, this particle always stays on this side, this particle always stays on this side, like so on. So you can have non-equilibrium phase transitions. This one is called localization. OK, so then that was sort of the general preliminaries. This I can show in the lecture today. I have a limited plan. And what I would like to discuss are three things. One is what is called a tongs gas. The other thing is, and the third thing is, OK, so that's all that we want to do in the remaining one and a half hours. And so I will go slowly. I mean, I'm not going to cover too much material in this time. OK, so what's a tongs gas? Tongs is, of course, a name of a scientist. The gas was already introduced earlier. You have a one-dimensional line. And you have hard spheres. And they move on the line. But they cannot cross each other. And we want to calculate the equilibrium statistical mechanics of this system. OK, so this is sometimes represented like this. But I prefer a picture like this one. So because it's a one-dimensional system, the shape doesn't matter. All it says is that there is an excluded interval in which the other particle cannot come. And then they can slide around. Is this picture clear? They are very simple notions. But I just want to ensure that everything is clear to everybody. I expected an answer, yes. And all of you are very quiet. So I think. So I see that you are reluctant to speak out. But as I said, since I don't know your background, it will help me if you give a feedback about what I'm asking. I just want to ensure that these things are obvious and trivial. And I don't have to explain further. OK, very good. So tongs gas, this is the definition of the model. So this is the simplest interacting particle system that we can discuss. And in this case, we are able to get the exact answer analytically simply. And I will do that calculation in front of you. So it's easy to follow. So this is, let me write this usual Hamiltonian, which is summation pi squared plus 2m plus v ij xi minus xj summation over ij, which are pairs. So this is the kinetic energy. That is the potential energy. This is very standard, but let me do it once. v want to calculate z is equal to integral dnp dnq. These are the momenta coordinates. These are the position coordinates. And e to the power minus beta h. And so here the momenta integrals separate from the configuration integrals. And I can just do all the momenta integrals together because they are separate. Right now I have only one dimension. So there is a p1, p2, p3. But even if I had a vector p1, xp1, yp1, z, they will all be separate integrals. And it will be some number e to the power minus beta p squared by 2m dp integral minus infinity to plus infinity to the power n. Bigger than this. I thought I was writing very big. So because the momenta integrals separate out, you can just put them away. And this is called configurational integral, which is what we discussed. Hence on the momenta will not be mentioned because they are all integrated out. We will only deal with this stuff, which is qn, let us write down. qn is actually, I think, Werner wrote it. Integral dnq e to the power minus beta v, which is the potential energy function. And this was equal to integral dn. This is just change of notation, 1. This function is 1 if allowed equal to 0 if disallowed. This is a function of xi. No problem so far. Now we want to calculate qn for arbitrary n. I know that large number of you can do this, but let me do this still. Sometimes it is helpful. So I am not sure how to do this integral. It's kind of hard. Let's see if I can do q1. That's very easy. There is a line here. There is a line of length there. There is a single particle, which we do like this. And it moves around. And I have so the position is x1. And I have to integrate over the allowed values of x1. And the integral is 1 over the allowed values and 0 outside. And it is very clear. So I should 1 minus, sorry, for change in notation, sigma will be the diameter. So vij is equal to 0 if xi minus xj is greater than sigma by 2 is equal to infinite if xi minus xj is less than sigma by 2. So in this thing, the lowest value of x1 allowed is sigma by 2. The left end is sigma by 2. Right end is l minus sigma by 2. So the allowed range is l minus sigma. And that is my partition function. q1 is l minus sigma. That was easy, q2. It's a little bit more work. Now I have two particles. So I have the same range l. And there is one particle, which comes from here to here. The position is at x1, there is a second particle at x2. And I got to do the integral over the allowed range of integration. So I will redraw this picture in the qx1, x2 play. So whenever x1 is less than sigma by 2, that is not allowed. So this region is forbidden, sigma by 2. Whenever x2, actually, yeah, if this particle was here on this side, even then it shouldn't be beyond l minus sigma by 2. So this region is also forbidden. This is l minus sigma by 2. Similarly for the x2, this value is forbidden. This region is forbidden. This is sigma by 2. This is l. This is l minus sigma by 2. This is sigma by 2, 0. OK? Also, whenever x1 minus x2 is very small, that region is forbidden. So x1 equal to x2 is this line. This is not allowed. In fact, they are not allowed to come close to each other like this. This region is also forbidden, and the rest is allowed. So the problem is very easy. You got to calculate this area and this area together, and that's the allowed region, and that is the integral. So I got to determine the length of this side, length of this side, and the area of the triangle. That is easily done, because this length, if you fix x1 to be at one place, the value of x2 minimum is 3 sigma by 2. The best I can do is to put this particle here, and then the next one here, and this is at sigma by 2, and this is at 3 sigma by 2. So this coordinate is 3 sigma by 2, then you can figure out the area of the triangle. It is not very hard. So the answer is L minus 2 sigma whole squared. The way to do this is to realize that I can actually work a little bit better if I define a variable called delta 1 and delta 2, which are the spacings between the, this is spacing from the left end to the first particle, and this is the spacing between the two particles. There are two variables, so I can write delta 1, delta 2. So there is a change of variables. x1 equal to delta 1 plus sigma by 2. x2 is equal to delta 1 plus delta 2 plus 3 sigma by 2. Oh, yeah, that is correct. But I would like to write it as x2 minus x1 is equal to delta 2 plus sigma. So obviously, this is a linear change of variables. dx1 dx2 is equal to d delta 1, d delta 2. Jacobian of this transformation is 1. And then I will draw the same picture in the delta 1, delta 2 plane, sorry, delta 1 here, delta 2 here, and it becomes simpler. Because the condition is delta 1 is bigger than equal to 0, delta 2 is bigger than equal to 0. And delta 1 plus delta 2 is less than equal to L minus 2 sigma. So that didn't work out very well. L minus 2 sigma, this is L minus 2 sigma. This is the allowed region. This is the forbidden region. That was pretty good, except that I missed out a factor of 2. This is the allowed region, no? Delta 1 plus delta 2 less than L plus 2 sigma. But that gives a triangle, which will have some, you know, it looks like this. So what is going on? Some mistake was made, obviously, perhaps deliberately. So what I did here, which I should have said or didn't say, before making this transformation here, let us impose x1 less than x2 less than L. I impose an ordering on the variables, x1 less than x2. This is not necessary. And the original problem in the phase space, x1 doesn't have to be less than x2. But in one dimension, I can do this. In two dimensions, it will not be very obvious how this can be done. So if I make x1 less than x2, then I define delta 1. Delta 1 is the distance of first particle from the left end, minus sigma by 2. And then the partition function evaluated here will be, I am just trying to make sure of my notation. This one is good right here. So this partition function, partition function with fixed ordering, this is too low for people. It is ok. I should work above. partition function x1 less than x2 less than m equal to L minus 2 sigma squared by 2 factorial. That is not inconsistent with this other result, which I wrote here. The difference is just a factor of 2, which comes because of the orderings. If you take another with x2 less than x1, that will be the same result. And when you add those two together, you get this full partition function q2. Is that clear to everybody? So without further ado, now I can do for n bigger, arbitrary n, n bigger than 2. You define delta 1, delta 2, delta 3, so on delta i is equal to x i plus 1 minus x i minus sigma again the Jacobian d x i by d delta i is equal to 1. So then the conditions on delta, hardcore conditions on delta i are simpler. They are each delta i is greater than 0 and summation delta i is less than L minus n sigma ok. And then this part is easy now, delta i are bigger than 0 and some over delta i is less than n sigma and the corresponding figure in three dimensions is some stuff like this. It is a n dimensional pyramid in n dimensions. And the partition function, the volume of this pyramid, if you have a pyramid of size L, L size, each side L, what is the volume of this, L cube by 6. I hope you remember. Now, there is one of those things one has to remember you know not ok, so it is it is here. So, q n is equal to L minus n sigma to the power n divided by n factorial with fixed ordering. So, that is the main result it was rather straight forward that was sort of end of our discussion, but one should get a feel for what this q n does. So, I will recall this is from the stuff you have read. If I know the partition function and its dependence on volume and I change the volume that is equivalent to pressure. So, given the formula for the partition function I should be able to calculate the pressure of this gas as a function of the density ok. So, I write here pressure is equal to d log z divided by dv ok. Because in our case it becomes n times d log L minus n sigma divided by d L ok, so that is equal to n over L minus n sigma. So, this can be written as rho over 1 minus rho by rho star where rho star is equal to n sigma by L ok. So, rho star is a scaled density or rather rho is the is equal rho star is equal to sigma by L and rho by rho star will be the, so n sorry just one second n sigma by L is equal to rho by rho star that is the scaled density ok, it is proportional to n n sigma by L that was the problem ok. So, the pressure as a so P is equal to k B T rho divided by rho star over 1 minus rho by rho star. It is the Wanderwald's form you know the pressure as a function of density initially grows linearly and then it diverges near rho equal to 1 and rho can never become bigger than 1 all very reasonable ok, but we could get this result directly from very elementary considerations. So, that is the Tongue's gas yes sir V is equal to which V or this is the volume of the gas this is the sort of formula I expect you to remember whatever notation there is is the one which is given in your textbook when I interpret it in my problem it becomes this one ok. In our case the volume is just length ok any other question ok very good. So, that was easy then I can do the second one I hope I will finish that one if not we can still continue in the next one, but I hope to finish in the next 15 minutes because it is an easy once we have set up this much machinery then the rest becomes easier. So, somebody will come and say that oh, but this was rather trivial you know you had these hard spheres and there were no you know other interaction everything became known interacting in the new variables delta I and then suppose you had extra interaction then what so Takahashi considered a slightly more complicated V ij was of this form it had a hard core part, but then let us say there is some attractive interaction and then it is 0 beyond some range. So, he said that V of r is equal to infinity for r less than sigma is equal to minus f of r for r between sigma and 2 sigma and is equal to 0 for r bigger than 2 sigma. So, this function here can be any shape you can take this one or this one is just a function f of r and for arbitrary f of r we will try to get the exact solution to this problem ok. So, what is the point of this suppose you have hard particles this was our system you have hard particles. So, the closest the third particle can come to the first one is distance to sigma it cannot come closer ok. And then by this condition the interaction will be 0. So, in this case by construction the gas becomes a nearest neighbor interacting gas we did not assume this we just chose a form of the potential, but for this form of the potential the interaction only occurs between nearest neighbors ok. So, then I would like to again calculate q equal to integral d n x e to the power minus beta summation d x i plus 1 minus x i ok. And we have already learnt. So, we will use the same change variables delta i which are the spacings ok. And then q is equal to integral d n delta e to the power minus beta summation of delta plus sigma ok sorry this is from 2. So, what is going on well I had this gas. So, there was this first particle, but the first particle I had this spacing delta 1 from the left wall, but I do not know what is the interaction with the left wall maybe I should specify that it is a boundary condition the interaction between the left wall and the first particle has to be specified in addition you can put some other function there you should put some other function there. We are going to take the easy way out and we will just put it to 0 it turns out that it does not you can check or you can verify convince yourself it does not matter the details of what is the interaction with the left wall does not matter if the system is big enough the effect of walls will only matter to order 1 by volume. So, we will we will not worry about that thing and we will just set it to 0 we will not we will say there is no interaction with the wall ok. Similarly, on the right side the n minus one-eth particle will have an interaction with the nth particle no, but nth particle does not have any interaction with the n plus one-eth particle ok. So, this was delta n and there is no delta n plus 1, but so there is no term here sorry this is n, but there is no term corresponding to the spacing between the last particle and the right wall that is by our constructs that is our definition of the model. If we change the definition you can put that in you can worry about it I will not ok. So, the statement is that the energy in our case v simplifies to summation delta i i is equal to 2 to n f delta i yeah f of delta i plus sigma ok and now it looks independent again. So, it was not such a big deal no if in hindsight what Takahashi did was actually rather obvious now, but only in hindsight at that time when he did this problem it was a highly non trivial problem one of the first interacting one dimensional system all that kind of stuff and people did not realize that such in this you can construct a potential v i j such that the problem can be actually solved exactly simply ok. So, very good. So, this is now this looks more tractable this is now these integrals are more or less. So, what are the conditions on these delta i is bigger than equal to 0 summation delta i is less than equal to l minus n sigma. So, all these integrals I could do if they were independent, but right now they are not independent because there is this funny condition that some of them has to be less than something. So, they are all coupled the delta i variables are not independent they are coupled variables ok it is obvious ok. So, how do we make them how do we solve this problem even when the variables are coupled and so the answer is that well these variables are coupled, but only by this condition. So, if I get rid of this condition then the variables will be not coupled. So, how do I get rid of this condition? So, I do not know I guess Tauka has he was very wise to realize that that thing to do. So, this is called q l is the partition function for size l. So, here do not worry about q l you take the Laplace transform of q l define q tilde p define is equal to integral q n of l e to the power minus p usually there is a beta written. So, let me write it l 0 to infinity it is a Laplace transform of q n of l you would like to determine this function for all l and right now I am not able to do it, but if I could if I knew q tilde of p then I can take inverse Laplace transform of this function and determine q of l. So, what we will do is we will first determine q tilde of p and then determine q of l. So, q tilde of p my time is nearly up is equal to integral d delta e to the power minus beta summation over i minus p beta delta i summation over i. So, the p term here we wrote a Laplace transform, but what is this p? This p sort of says that I imagine an ensemble in which the l can be varied, but if l varies it occurs with the weight e to the power minus beta p l this is called a constant pressure ensemble. The idea is not to define a new notation, but to make clear what is the assumption? If you work in a system in which the pressure is constant instead of the volume then l will be variable, but different values of l will occur with different weights you know like if you go up then you do work and this is the cost of that stuff and this is the partition function for this system. So, now if I do this then l is no longer fixed each delta i can vary independently of the others and so in the constant pressure ensemble all the delta i become independent variables and so this answer becomes equal to product over i integral d delta i e to the power minus beta f delta i plus sigma minus beta p delta i to the power n minus 1. I guess I should worry about what happened to delta 1 delta 1 will also have an integral which I did not write down, but it is e to the power minus beta p delta 1. So, I know how to do that one perhaps there is also an integral related to the last spacing because you know now the piston wall is variable to move then I have to integrate over all possible positions of the right wall, but again that integral is easy to do. So, my time is up so a few minutes. So, I cannot do all those integrals in gory detail here, but it is easy to see that they factorize and except for the boundary terms which we discussed orally the answer is just n fold multiplication of a simple one dimensional integral. So, you give me this function f this is just one integral I can do this integral it becomes a function of p. So, let us call this g tilde of p to the power n and then there are some boundary condition dependent pre factors which I will not write down in full detail here. And so then this partition function is done then if you want to determine q of l you have to take an inverse Laplace transform, but you know there are all these Maxwell relations and stuff if you know the answer in constant pressure ensemble then you can determine them in the other ensemble d by dp. So, that is my last comment d by dp of log q tilde of p is equal to beta. So, that is the condition if you give me l average and you can determine the corresponding value of p and vice versa these are conjugate variables if you think of p as a Lagrange parameter then initially its value is not known, but using this condition you know what is the best value of p and that is the corresponding result. If you do not like this semi hand waving treatment then you do the full Laplace transform we did not actually do the Laplace transform we said instead of doing the Laplace transform you just take the most likely value where the function is maximum and that gives us the most relevant answers. And so that is where I will leave this discussion about the Takahashi case are there any questions yes sir. So, this function f of course, the function should be well behaved for the limit for the well behavior to exist I did not specify in great detail the well behavior conditions on f, but the simplest condition is that f should be bounded from below that is enough it does not have to be bounded from above if it is infinite in some range it will just give 0 integral not a problem. If the interaction energy becomes minus infinity for some spacing that will be a problem that will mean that energy per particle will be minus infinity that is not what we want to discuss. So, that is not a good model to study choosing an f in which it goes to minus infinity is not a good model to study. So, we break for lunch.