 Now, yeah in the previous lectures I discussed some 1D models and the Lee Young theory and I showed how one can calculate the line of 0s in simple models or maybe not so simple. What I want to do now is to discuss again slightly different topics. So one will be to discuss some examples of face real phase transitions in hardcore models. The Lee Young H similarity which I discussed last time it is a phase transition of source it helps us understand the behavior of you know occurrence of phase transitions, but maybe one should justify the title of the lecture series better. So I will actually discuss some real systems and in the second part I will discuss some connection of this problem of hard particles to other problems like animal problem or the heap problem or the branched polymers. So let us see ok. So first these are examples real phase transitions. Of course we start with hard spheres. It turns out that there is an enormous amount of literature for everything that I discussed here. So I am not going to discuss any of them in great detail. I am not going to give you the references to all the stuff available because you type hard spheres on Google and look up all the references and you get you know 1000 papers and you can make your selection. But I will try to point out the key ideas which are important in the discussion and then you know you can pursue them further. So the hard spheres of course has been a very interesting problem for several centuries. So for 3D hard spheres it turns out that there is a phase transition 3D phase transition disordered gas phase to a periodic solid ok. So you just change the as you increase density that is the only thing we vary. So the this transition of course is very interesting was not people discuss these things called geometrical phase transitions and then they say that the first geometrical phase transition observed was predicted was by Onsager for long needles. But hard spheres have been studied much before Onsager's work in 1949 and so why was in this phase transition in hard spheres liquid studied. So people used to study this integral equations for hard spheres. Now the hard spheres were studied for a long time people wrote down approximate equations, hypernetic chain equation and all these kinds of stuff to study the equation of state of hard spheres. But they never predicted or they never realized that once all these equations break down once the thing becomes beyond the near ideal gas then you will actually get a periodic solid. So what they could say is that ok up till this density the my theory kind of works and the pressure as I calculate pressure as a function of density does something and now I do not know what happens beyond this. So the high density case was not well understood of course. So it turns out that it was not realized that the high density phase is actually a periodic solid and it was a computer simulation which suggested this first and I guess even now that is all we have. The evidence in favor of this phase transition which is fairly well established is only computer simulation will be nice to have some kind of a proof that this happens and it is not there it is a tough problem ok. So just how come the at high densities we get a periodic solid because what one says is that the configuration you the state you get is the one which has maximum entropy. So the how how how does it happen that the highest that ordered state has higher is the highest entropy state ok. So this is not really a very difficult question it is a question of what is meant by entropy and it is slightly abuse of notation of language to say that ordered state has lower entropy than the disordered state. Because what happens is that yeah you create a periodic density but you know I am drawing a 1D picture this is the density along some one direction in a solid. So each atom has some constraint that it is more likely to be here than there. But then everybody else gets more elbow room to move around. So you lose entropy in one mode but you gain lot of entropy in other modes. So in the end the ordered state has more entropy than the disordered state ok. This point should be well realized it is not a contradiction at all ok. So that is all I would say about the 3D heart sphere transition. So in 2D so in 2D you can make at high density the spheres kind of become this hexagonal packing like that. But because of the continuous symmetry of the system there is no preferred direction and these can these directions can be tilted no it can be that way or that way. And so there is a instability due to these phonon modes and you cannot get long range periodic order that can be shown rigorously. But let us just say that if I calculate the root mean squared deviation of a particle in a box of size L in some stuff like this. Then even in the harmonic theory one would get that rms sorry delta x squared average is turns out to be some integral over k squared. But there is a lower cutoff at something like 1 by L so it goes like log L. So if you look at a particular particle in a system of size L and make L bigger then the mean squared deviation keeps on becoming bigger because the low frequency there are more lower frequency modes and they have a higher amplitude ok. And so there is no long range order so you do not get a periodic solid. But you get a nearly periodic solid which in which if you look at these orientations even if let us take this one if you take this and the whole thing oscillates a lot the orientations are more parallel to each other than the actual positions. So if you look at the orientational order between different bonds then you will find that they are more often parallel. And so it turns out that the orientational order between different between bonds at different sides decays as 1 upon r to the power a where a is some power in the high density phase and it decays exponentially in the low density phase. So there is a phase transition from low density phase to high density phase in one phase the correlations decay exponentially in the other phase they decay like a power low. But that transition is not to a periodic solid it is what is called a coastal is thawless transition. So it is a this is a coastal is thawless transition ok. Then you can read more about this I will not go further ok. So now let us look at hard squares. So hard squares are defined like this you take a square lattice. So it is a square lattice with lattice case one square lattice with nearest neighbor exclusion we have already had this term before. So what happens is that you occupy some site then you cannot occupy all the four neighbors that is the only rule we use. Now you put in some density and see what is the configuration like ok. So if the density is low you have very few particles then one particle can be here the other can be there and you know there is not much order between them it is like a gas of some sort short not much correlation. However if you put them as many as possible the maximum possible density is maximum density is one half which will occur when all the what we call chess board order you occupy all the black sites and you know do not occupy all the white sites you know think of it as a chess board then that is a fully packed configuration. There are two such fully packed configurations one is when you occupy all the black sites or the other is when you occupy all the white sites. That much is very easy to see. Now suppose the density is a little bit less than the maximum possible then what we can do you can just remove a few particles out of the maximum packed state. That is a nice configuration you know there is so it has a lot of it has a lot of entropy because there are a lot of ways of removing a few sites out of it. So near maximum packing I can remove a few sites but I can actually sometimes I can put in some sites in the a sub lattice also. If I remove enough b sites then I create four vacancies then I can put some sites in the a lattice and so at not so low densities the configuration will be one in which there will be a lot of b sites occupied and a few a sites occupied. And so the configuration kind of looks like this I am taking time to but it is okay. This is fully packed and then I removed some from here and from here and from here that was easy but I can actually do the following I remove these and then I can put one here which was so these were a sites and this is a I actually call this one a and call these b sites okay. So then if you increase the number of holes or you decrease the density further then the size of these a sites will grow you could find another a particle adjacent to the b y and so at lower densities you go up to a state which on a large scale looks like this. There are small islands of a clusters buried in a sea of b clusters okay and that is the picture and then as you decrease density further the size of these a clusters will grow and at some state they become similar in size to the b and then there is disorder. So the point is at this stage I realize that this is starting at large scale this system looks like the ising model in which you have two competing states and so the phase transition by the standard expectation would be in the ising universality class okay. So two competing states competing ground states so you get a ising like phase transition from this phase to another case where a and b will be randomly scattered all over or the fractional area of a and b will be equal in the end okay very good so now we can these kinds of arguments you can extend to other problems and you know I am just giving an example of the kinds of arguments. So let us take something else this is called hard hexagons hard hexagons are defined like this you have a triangular lattice yes what is the analog of magnetic field it is the sub lattice magnet magnetization so it you know it is the field suppose I were to give an activity to particles which sit on a sub lattice is distinct from the particle which sit on the b sub lattice. So I will have to define z a and z b and then the log of z a by z b will be the corresponding field okay. So in hard hexagons one works on a triangular lattice so it is a triangular lattice and you still have this nearest neighbor exclusion gas so I did not finish drawing my triangular lattice like that if I occupy this site I cannot occupy any of the six neighbors then I can put one side here I guess and then I can put one side here and so on okay. So this one looks a little bit difficult to see so it is visualized as follows you think of this as hexagons so instead of occupying the vertices of the lattice I occupy the areas and I occupy them with hexagonal tiles like this and these tiles cannot sit on top of each other it is much better to see it like that and then everything else goes through at maximum packing maximum packing here will be one-third of the sites will be occupied because there are six of that oh I should yeah so if the number of sites is n then you can break them into a b c sub lattice a b c sub lattice I should be careful so on a triangular lattice the three sub lattice decomposition is well known to everybody or I should just write it down the way it is done is as follows write a b c a b c a b c like that along a line in the next line I write b c a b c a b c a like that in the next one I write a b c a b c like that all the time making sure that the neighbors of a are only b and c you do not put down an a below an a ok so then all the neighbors of a are b and c all the neighbors of b are c and a and so this is a three lattice decomposition and of course I can occupy all the a side so I can occupy one third of the sites possible in the maximum density is one third ok so now I try to use the same argument here and so I will say that if I disordered this lattice a little bit what will happen is that you will have in the ground state everything will be let us say b but then if I decrease the stuff and put in other stuff maybe you can get an a but you can also get a c you can also get an a adjacent to a c like that these kinds of disordered structures will show up as you decrease the density. So, these are not ising like configurations there are islands within islands or whatever you want to call them but they are not ising like because you have to realize that you also get structures like this were three different stuff but there are only three different stuff so this will not be in the universality class of the ising model it will be in the universality class of what is called the three state parts model there are three competing ground states and you know you can have a state in which every spin is type a and then b and c are distinct and so on and so this thing will be in the universality class of three state parts model so I am doing this in 2D but you can also do this in this argument you can extend to 3D even particular shape of particles I tried to put them in maximum density hopefully that is a crystalline state then I see how many distinct crystalline states I can get by translating this in different ways and that is the number of ground states and there will be the corresponding in the continuum description I will have to say the system can exist in state a or b or c and there are that many states and that is the number of parts states and all these states are equivalent so the interaction between them will be like b likes to be near e b but if you put b and a together then there is some mismatch and there is some empty region and you have to pay some cost for it so that will be the to find out what is the order of the transition you look at the ground state to look at the degeneracy of ground states and then that is the corresponding parts model and then somebody else has studied that what is the phase transition in parts model like so this one will be three state parts model in 2D so people suppose I was studying some such model in 3D and I have a 3D parts model so it is known what happens to that 3D parts model what kinds of phase transitions it undergoes and so on what happens to the 3D ising model what is the phase transition I know that okay so like that okay so next so what I would like to do for the hard hexagon is that these are sort of just finding the order of the transition by looking at the qualitative feature of the problem but I want to do a little bit more and I want to calculate the free energy per site in one of these models let us take this hard hexagon model at nonzero densities what is free energy free energy was E minus Ts but is always 0 so it is just entropy I want to calculate the entropy of the system at densities which are not one third but near one third this thing is called high density series expansions high density perturbation expansion so I start at the ground state very good so there are many ways of doing these high density expansions but the most convenient one for us will be the following that we actually assign three different activities zA zB zC to the three lattices sub lattices okay so the full partition function omega is a function of zA zB zC is summation over all configurations zA to the power NA zB power nB zC to the power nC where nA nB nC are the occupancies of the sub lattices okay so this is a polynomial in all these variables and then I have to take the log divide that by the number of sides that gives me the partition function per site and if I take derivatives with z then I get the densities okay very good so what I realize is that suppose I put zA finite but zB equal to 0 zC equal to 0 then B and C sides cannot be occupied and the A sides can be occupied independent of each other so this partition function is easy to write this is 1 plus zA to the power n by 3 okay so now I expand this stuff in powers so I write omega zA zB zC equal to omega zA 0 0 1 plus terms of order zB zC plus order zB squared zC like that okay so now what are these terms of order zB is a term which is linear in zB in the full power series expansion there will be the one there is only 1B particle so if I put 1B particle so this this is my A particle which are occupied but I put 1B particle here then it says that the corresponding 3A sides which were near it will have to be empty only then that only those configurations where 3 adjacent A particles are unoccupied are allowed so the answer is that this thing looks like omega zA 0 0 1 plus zB times the fraction of configurations where 3 neighbors are 0 3 neighbors are empty plus higher order terms so but this fraction is easy to write because at each side these are independently occupied so it is empty with probability z over 1 plus z yes yes I should write that bigger or I should write less yeah so let us write plus alpha cubed plus other terms where alpha is equal to 1 over 1 plus zA okay because that is the probability similarly of course there will be a term with zC which will be the same alpha cubed n by 3 and there will be other terms but this is a very nice and systematic way to evaluate the perturbation series in this system and then I can take the log of this and get the cumulant so zB times alpha cubed times how many places can I put zB in n by 3 anywhere so the number here will be n by 3 sorry this is zB this is alpha cubed is the probability that the 3 neighbors will be empty in the unperturbed state where zB is 0 sorry zC is 0 zC is 0 okay and then you can the higher order terms are a little bit more complicated but they are very systematically evaluated there is no problem and this actually is a good way to think about this system with the little bit of work without too much you should be able to write first three or four terms in this perturbation expansion and that gives you an idea that oh yeah the entropy varies like this in this system okay so in fact these kinds of series converge very well up to say 10 percent away from the critical point if you are not at the critical point and you stay away then these things are converge very well they give you very good feeling of what is the actual entropy like suppose I want to know what is the entropy for this hard external gas at density 0.9 of the maximum packing density that is an question an experimentalist friend has asked me they are not interested in precisely the critical exponents of the problem always okay and this method will give you the full answer to you know there is some error because you have forgotten some terms but they are usually within a few percent and so it gives it is very good for understanding the qualitative behavior of the systems involving hard particles okay so we have done the high density expansion so let us take some other shapes number 5 I guess so I want to be very convoluted so I took my particles to be looking like this I take these kinds of particles on the square lattice and put them in and I want to calculate what is the entropy what is the what can I say about this system okay so I realize firstly I have to I have to see how I can put this together to fill the entire lattice and that is easily done maybe it is better to do it like this so you get the picture there is a there is a full styling of the lattice using these styles so what is the maximum packing density what fraction of sites will be covered by the sorry yeah so I think of course all the sites are covered by tiles but what we do is that we identify one particular site on this tile I can identify the middle site as the typical marked site then I just say that this tile is sitting at this site it specifies the position completely instead of giving five coordinates I give one coordinate okay and so I will say what fraction of sites are occupied by the centers of tiles so I am sure everybody of you knows the answer but let me ask one of you just you sir louder one fifth very good so maximum density here is 1 by 5 so there are actually five distinct states because the system can broken up into five sub lattices here I think I call them what are the five sub lattices let me call a b c d e and now these tiles are covering the whole thing I will call everything a b c d e and then the square lattice has been broken into five sub lattices okay and so there are five distinct ground states and correspondingly when this system undergoes a phase transition it will correspond to a five state parts model critical behavior will be that of the five state parts model somebody else has told me that five state parts model undergoes a first order phase transition so I realize that oh yeah in this model there will be a first order phase transition okay then what so then let us consider two by two hard squares so the squares look like this these are my tiles they are different from the hard squares I discussed in example 3 because there what happened was that these were called hard squares but the squares are kind of looking like that each occupied site is like occupying a diamond and then half the sites can be occupied and then you know then the maximum density is one half here these are two by two hard squares of course they can feel the whole space that is obvious but now the maximum possible density is one by four so this is a different problem so maximum density is one by four so some question okay then what so does this system undergo a phase transition if I increase the density of squares so I would say let us guess that it does so we will go into this high density phase which looks like this filled with squares so the actual lattice size is smaller you know the lattice looks like that and the tiles are two by two step so now I want to see if this ground state is stable or if I can do so I do the usual which we discussed before high density expansion we can pick I have to identify one side of this tile as the marked side so we can take the upper left corner and I you know identify those squares as marked squares so then I will define this now I have four sub lattices z a and z prime equal to z b equal to z c equal to z d and then I write omega of z a z prime equal to 0 equal to 1 plus z a to the power n by 4 this much is good you know I took four sub lattices d prime c prime have no activity you cannot put anything there then you get independent contribution like this but now I want to evaluate for not equal to 0 so omega z a z prime not equal to 0 is equal to 1 let us write it like this omega z a 0 1 plus I write z prime times something z prime times term plus z prime squared plus z prime cubed and then you know this is the series expansion then I take the log of this and I will be able to see what is the order by order in power x series expansion in z prime so I look at the first order term in z prime so what can I do just for a minute z prime is the activity of a tile which is not kept in the standard you cannot read z prime well ah oh here yeah yeah yeah okay so okay so what I do is that I start with the fully packed state so omega z equal to z a first term is z a to the power n by 4 this is the four large z a z b I assume both of them are large maybe one of them is much larger than the other but both of them are very large then the leading term is z a to the power n by 4 that is all the a sites are occupied now suppose I remove an a site z a to the power n by 4 minus 1 so I have all these tiles occupied and it is a different picture so I am trying to so I remove these are occupied tiles and I remove this one everything else is occupied there is an empty hole but now there is a something very interesting happens you have a sliding instability this square can be slided to here okay and then this one can be slided next one to here and next one you know you can create an arbitrary big disturbance by just removing one hole okay so the effect of this in this case is there plus with some coefficient here and so this looks like z a to the power n by 4 1 plus I am looking at contributions to order 1 by z a and how many terms are there how many configurations are there if I remove one one particle so the answer is this single tile breaks into two half tiles and each of these half tiles is called a half defect instead of a full defect and these half defects unbind they just go away from each other okay and then so one will be here and the other will be there the total number of possible positions when I remove the square in this row will be l into l minus 1 by 2 or some such thing okay so and then there are choices of these different stuff here so you get order l cubed plus 1 plus 1 by z a squared actually it is l cubed by 2 right so let us write it like that l cubed by 2 plus order now if I take the log of this log omega z a z b equal to n by 4 log z a that was fine no problem but then I have to take the log of this plus you get l cubed by 2 1 by z a so the leading correction to the partition function log of the partition function is coming out to be proportional to l cubed and not l squared which is what it should have been so what is happening is that this expansion is diverging the first term in the series expansion diverges because of all these half defects unbinding so this is to be plus so the first correction diverges first correction term to the ground state diverges so what what is going on I hope this one was right nothing wrong with this one if you are very careful you will say no no no there were four ground states so I can put enough four here the partition function when z is infinite is exactly 4 z to the power n by 4 nothing can be done about that but the next term is actually l cubed by 2 and now what do I do about that if I take the log is the free energy of the full thing going to be proportional to l cubed no way we know that it cannot do that right so the series expansion is breaking down and how do we fix it that is the question and the way to fix it is to realize that the half defects are now picking up their own life because they have become free they have become unbound and you have to take them into account in a proper description of the system if you try to perturb about this unperturbed state it will not work but let me consider a slightly different problem which is the following I take a 2 by l row and I put in squares in this one so these are the squares and very long and there are this is a half defect this is another half defect they might even be two half defects adjacent to each other I will try to write the partition what is the weight of the half defect well if the weight of this stuff is z I can renormalize I can say that weight of the ground state is 1 because that is what that is what we did here we pulled out this factor then the weight of a full hole is 1 upon z and weight of a half hole is 1 by root z so then this becomes a guess of half holes because half holes can sit anywhere in this state I guess they cannot sit adjacent to each other no they can sit adjacent to each other but you cannot have a configuration of half defect and then empty space and a half defect the space between half defects must be an integer number by definition but there is not a difficult problem I can solve this problem this is a one-dimensional problem I can write some transfer matrix or the other and I can get the partition function of this row what does that look like the partition function of this row let us write it here partition function of a row equal to some transfer matrix I will not work out the details here it is very similar to what we have done yesterday that it is the same as what we did yesterday so this is equal to some transfer matrix calculation so but what is the answer log of omega for one row will be equal to n times log z this was the leading term plus n times 1 upon z to the power half suppose z is very big then even the half defects have a very low activity so then we have dilute gas of half defects and the partition function will be n times the dilute activity which is 1 by z to the power half so the true log z in a row actually has a term like n times 1 upon z to the power half and so this is what is happening and the reason I got a problem here is I assume that the thing will have a Taylor expansion in powers of z or powers of 1 by z in for log z but it is actually the power series in powers of 1 by z to the power half so I become smarter I so I assume so I assume the I do the following I take the system break it up into rows so I say within a row you can put anything anywhere but don't put anything between rows then that problem can be done that problem is equal to the partition function of one column sorry I should have said column 1 column to the power number of columns ok now I so I will assign a weight z to this row vertical row vertical column and z b or z prime two things where the center lies on this column in the middle and then I expand in powers of z prime and now things work out a bit better at least you don't get a divergence like 1 by z cubed sorry divergence like L cubed this one it is a well behaved expansion but this so you can develop the series further actually it encounters additional problems which I do not want to get into just now my time is you know I have to do other things so I don't want the stuff can be seen in some published paper or the other actually it was done by a student called Kabir Ramula that was part of his thesis to develop a series expansion for hard squares where you this is a singular expansion and it requires additional fixing later this this this is required but this is not enough but anyway if you work enough then you get a decent series expansion so what do we learn from all this we have learned that the state which is the sub lattice ordered state is not stable here because of the sliding instability so but what happens at a finite density known one so you won't get a sub lattice order will you get any order whatsoever the answer is yes you get what is called a columnar order I will just define the columnar order and then we will stop for five ten five minutes I am eating into my ten minute break so columnar order so what happens is that if you look at finite density you will find that okay there are a few holes here but most of the time in most of the space the system breaks up into this non-interacting rows where in a row you can slide at random and different rows don't talk much to each other once in a while it may happen that there is a particle here then in this row you know there has to be some vacant space but because I have lot of holes there it's allowed I can accommodate a few particles there by making space for them so but what happens is that most of the time you will find that one of the vertical columns will be occupied and then adjacent one will be nearly empty and the next one will be occupied next one will be empty so it has broken translational invariance in one direction in the horizontal direction but in the vertical direction everything looks similar of course you could have done just the opposite all the order could be like this the things could slide like that okay so you have the low density state is what is called as it is not a sub lattice ordered state it is a columnar ordered state where you the system breaks up into columns and these columns can be either horizontal or vertical I say oh but you know if I go to higher disorder lower density what will happen you will have a stuff in which there is this stuff is vertical ordered and then there is this law which is horizontal ordered and so on and then what will be the universality class of this phase transition eventually it will again turn out to be icing like but it is because of these two competing ordered states which are with two different columnar order and the ground state has columnar order sorry not the ground state the low high density state away from perfect filling has columnar order okay so let us stop here and we will restart in five minutes seven minutes is the argument of how you first of all what is L here L is the size of the lattice L by L lattice L by L lattice so could you please repeat how did you get this term yeah sure so if you have an L by L lattice and I only put one defect that can be coming as two holes each hole can be anywhere on this size L so the number of ways is L into L minus 1 by 2 and then there are L possible layers so even you take the log of this so in the usually what one does is take this function and you have to take the log of this and then the leading term is 1 by z a and the coefficient is proportional to n the volume of the system and then you know next term will also be proportional to the volume that is called the cumulant expansion and here it is coming out to be a higher power of L than volume yes so volume is a power yeah so the correction is to be all right but but here it is okay because correction is okay here the correction is now becoming okay precise and how to see is it okay how to see that this diverges and this is not I mean you calculate the correction term here and you will find that okay if you put in a hole put in one particle here this can be done in n order n ways which is the number of sites but once you put it there then there has to be some vacancy here right so then this is followed so this is z b divided by probability that you have enough vacancy here to put this particle in so there will be z a squared sorry I beg your pardon this is good but I should write plus n over z to the power half this was the single row case even when z b is 0 you get this term but now if I put z b then I get a correction which is only z a squared because these things have been taken care of in this term thank you so much you are welcome sir this fact that we're getting a icing tension again is it related to the fact that the hard squares like this two plus two squares are like the brown one portion of the one plus one square like the same shape after all no it is it is exactly only this argument that there are two competing states I mean this is the reason the main reason no the other reason is not good because if you take three by three square no because if you take three by three squares they don't give you icing okay no no no so the answer is they don't give you icing okay very good no no so that's what I that's I didn't draw too many pictures because it becomes hard but I will draw a schematic picture like this there is an a so here you put sites on the a sub lattice and in this here there is an island in which you put sites in the b sub lattice the center but then there is some mismatch here where you have to leave some empty space because you cannot put overlaps okay so that is taken into you know we assume this is the same this is the same I have drawn it draw something we draw further let us see can you extend this further this is the same this is I start with this doesn't give you the possibility to add another cross here because yeah so this one will have to be looking like this there are five distinct states but besides that there are extra two degeneracy maybe there are ten distinct states what do you think this state is clearly distinct between this one it's a kind of another degree of I mean you can you can but this is not translation ah yeah yeah there is another there is a different choice of five sub lattices you are right right so there are yeah so the huh so it will become a ten state parts model but actually maybe it's a kind of icing transition and below its five states no so the ten state parts model undergoes a first order phase transition and five state parts model also undergoes the first order phase transition so here there is a center yeah yeah yeah this is a two one yeah yeah so one two is also loud so if you take uh reflection of this it's a different tiling yes yes so don't you think I mean I mean just get an orientational one I mean no all we are saying is that there are ten possible nine other choices of if I fix the tiling here but you see that if you decorate domains of these guys and these guys then it will be very I mean very high energetic domain ball it would be much high energetic yeah yeah yeah yeah so the ten states are not all equal yes yes so this means I mean huh there is a kind of possibility for I mean uh confrontational phase transition icing like and then low at lower temperature what's like five okay so yeah that is right now that is certainly possible so sometimes you do get symmetry breaking in sequence you know partial symmetry breaking then more and more and so on and so that may be done here I don't know the answer I just you know I was making up the lecture so I picked up this one I didn't really I didn't appreciate that yeah yeah yeah there should be actually ten states which is correct yeah and so this may undergo two phase transitions as a function of density send me an email and I'll send you a good reference right now I cannot give a very good reference which covers this material quite well so you kind of made this lecture by from your head or you used some particular references or you right now I made it from no I mean of course this some of this stuff we have worked on earlier but to make this particular lecture I didn't use any other text yeah I mean you know everything by heart not everything but you know like this basic stuff when it's not so hard as you can see I had not actually done something very complicated some similarities you see that I mean yeah yeah yeah no but there are some basic ideas which you can use for a large number of different problems and that is what I'm trying to convey it's not focused on any one particular hardcore model same question very good you have done this before you are studying yeah my study is related I will be presenting this the poster on the third week I don't know whether you will be staying through the third week or week no no maybe we'll can chat yeah I would love to okay so you have actually studied this model I mean a kind of I study maybe okay I would love to explain you okay first one 13th model related to some particular material chart density waves but the statistical model it looks like the same okay very good the power of l to the three yeah sure yeah so suppose you look at one row in which you remove one particle yeah then you create two half holes right yeah and they can be anywhere on this line right yeah so the number of ways of putting these is l squared yeah and but the number of choosing rows is l you can put it in this way okay we were thinking out of the fourth because I double counted the horizontal direction it makes the yeah okay no this one has to be in the same row as this yes so it is only a cube you to describe the hardcore molecules right so if you get a two by two hard squares you could just say okay instead of taking this lattice I will take a double spaced lattice and then it will be exactly the same as one by one square no because the main distinction is that if you take one by one square that is like tipping only these sides but there is a technical possibility that the square can be put in between the I have a continuum space I make a lattice okay now a square if I have a lattice spacing like this then I can put a square here but let me imagine that the square is of this size then I can put it here or here or here or here or here that is one case but this can be expanded to make a bigger square that doesn't matter it says this space is allowed but in my case it's also possible to have a square shifted halfway the big square can be shifted like this yeah the entire lattice was just auxiliary right it's not really in the system no no no it is at least in the model we described we start with a lattice and tiles sit on the lattice so it's there it's not starting with the continuum there is a tile the background space is there okay it's like saying we don't have any I please do that yeah on a lattice extruded volume problem on a lattice actually one of the previous models was like this one I had continuum models of exclusion and also lattice models of exclusion so like the problem we did on the young leaf zeros was on the lattice gas lattice gas of excluded models so you start with the lattice some size can be occupied some cannot be occupied and there is some extra constraint right so these are lattice models okay there is an extra problem if you work with continuum models so I was it's a half shifted square this t here this symbol this was just I was just saying a term I because in there was no space I was just saying there is a something z prime if you say order z prime that is one thing but I didn't define it further okay we can start I usually want to ring your bell once more you know because I thought the audience has disappeared I should start okay okay yeah so there is a correction you know I prepared this lecture last night and it was pointed out to me that actually there are no five competing ground states here but 10 competing ground states because you can take these five states but you can also XY reflect them and you get different five different ground states okay so then you have a 10 state post model to worry about turns out that qualitatively quite often the five state and 10 state model behaves similarly but in this case you might even get a two stage melting where you start with the ground state then first the XY symmetry which was broken here may get restored and then the other one or vice versa or maybe they work together but anyway it's a 10 state model that the number of possible tiling so the full prepacked case is 10 and not 5 yes yeah the argument is like this that if you look at just the this B sub lattice it has this kind of structure okay but then you can you can reflect this and you get a different tiling if you reflect about the y-axis then there will be different way the lines will go so it's a different tiling and so the total number of tiling sees these times two and so you know so the rest of the argument goes through but that's actually rest of the argument doesn't go through because the now it becomes a 10 state post model and then you know you have to ask what happens there and it is a unstudied problem I think in detail okay so now I wanted to take just a little bit of time to describe one more um so this one number and this is long needles the point being that all these show different behavior depending on the shape of the molecule shape of the excluded particle okay so I am not going into the details of any of these models they are all studied in great detail but I am just pointing out the richness of the kinds of phase transitions you can see in hardcore models okay so the long needles is defined like this you have a right now we are working with the square lattice and a long needle is a tile which looks like this it's k adjacent squares this is of size k that k can be 2, 3, 5, 6, 7 and then using these tiles you just put them in they cannot sit on top of each other and what is the statistic what is the system at various densities that the question we ask and well the k equal to 2k is called the dimer model that has been well studied a lot and 3 and 4 are not so well studied k equal to 2 it is known that if you put any density at k equal to 2 full tiling you look at the orientation orientation correlations between different dimers they undergo a power load decay but if you put any density of holes then that decay becomes exponential and you don't get a power load decay okay so k equal to 2 at non one densities non full packing is not doesn't show long range order k equal to 3 also it turns out that for non non one densities it doesn't show order but for k big enough in our case it turns out to be bigger than equal to 6 you actually this system actually shows ordering and the high density phase is different from the low density phase okay so this is what I want to just explain in five minutes so what is so this transition was actually the first transition which was studied uh of geometrical phase transitions by own Sagar own Sagar I think it's 1947 49 so he pointed out that if you have long needles and you put them together in three space or in two space usually he was working in three space then they all become parallel or you can pack them much more easily if they are nearly parallel than if they are not so he argued that if you have long molecules like these rigid molecules like this in solution then they will become parallel and this is what is observed with some virus particles at low density the virus particles are randomly oriented at high densities they become oriented together it's called nematic order and then you can see this using optical activity or some you know the liquid becomes an isotropic okay so so this model is a model which shows the transition from isotropic liquid to disordered sorry to nematic order which is an isotropic you know so the molecules are preferentially oriented in one direction rather than uniformly distributed all over okay so this is a lattice model of the same thing so in a lattice model the rods can be either fully horizontal or fully vertical there is no intermediate orientational out and so but you know own Sagar's argument was very qualitative but very good and so he calculated some virial coefficient and showed that if at high densities it's much better to align sometimes it is cited as a rigorous result he did some very interesting calculation he took cylinders and he said that this is the sigma is the radius and l is the length and l squared sigma is the volume sigma squared l is the volume and he studied this problem in the limit when sigma squared l is finite so a finite fraction each rod has a finite volume let us say and it feels a finite volume of the there is a finite fraction of such things which fill the liquid but then it took the aspect ratio limit goes to infinity so l by sigma tends to infinity but if the volume remains finite l by sigma goes to zero it becomes like a very long line and that is sort of a very unphysical limit so I do not know if you derive some result in this very unphysical limit it should be considered as a proof that some physical system will undergo the same phase transition the results as stated by own Sagar are of course correct but I think there are some strange limit which is not fully comfortable all right but anyway the reason for mentioning this was not so in the lattice model also it turns out that if the needle is big enough if you work at low densities you get a phase which looks like this okay there are needles placed at some distances and they are in random positions and they have no orientational order half of them are horizontal on the average half of them are vertical if you increase the density of these needles they kind of go into a state which looks like this where most of the needles are parallel to each other they might be a few needles in the perpendicular orientation but they are very few okay but then we were asking okay so what happens in this model in the high density limit so in the high density limit full packing limit you can have a system like this you know I can just fill the space with these needles like that which all with parallel orientation and these things can be slided against each other a little bit but it doesn't contribute significantly to the bulk entropy of the system but now it tends it's obvious that this state which is fully ordered doesn't have any bulk entropy the number of possible ways you can do this is k to the power l where l is the size of this system because each vertical column can be ordered in l possible ways sorry each vertical column can be ordered in k possible ways and there are l columns and so it's k to the power l however I can construct a different state in which I break this system into k by k squares and each row I can each square I can independently occupy either horizontal or vertical that number of ways you can do this is true to the power l squared by k squared so that is much bigger number it's extensive so there is a very large number of these states and there is only very few of these states so these states will and these states don't have any orientational order so at high density you don't get orientational order then for a long time it was believed that at low density you don't get orientational order high density you don't get orientational order maybe there is no orientational order in 2D in this lattice model maybe in the continuum model which Onsager discussed there is orientational order but in the lattice model because of the peculiarity of the model there is no transition but it turns out that actually what happens is the following as a function of density this is nematic order which is the fractional orientational you know fractional number of vertical minus horizontal at low density it is disordered then you increase the density becomes ordered like this when you increase the density even further it becomes disordered again it becomes like this okay so you get a two phase transition disorder and this in between is nematic order or orientational order the nature of this disorder phase is not fully understood because it is not as simple as this this is just a you know one of the some of the states which are like this but there are many other states which are more complicated and it doesn't have orientational order but is it fully disordered that kind of question is not yet answered okay if you study problems like more complicated in 3D so I just once mentioned 3D 2 by 2 by 2 cubes so I have space and I have cubic lattice and my tiles are these cubes which are of size 2 by 2 by 2 8 cubes 8 tiny unit cells in 1 cube and I put these cubes at random and then you know then we are asking what is the phase what is the high density phase like what is the phase transition like in this system and it turns out well of course you can fully tile the space with this so that is not a problem it will have sliding instability I guess what else can I say not much off-hand but some of my friends have done some Monte Carlo simulations of this stuff and it actually shows four distinct phases so it starts with a disordered state at low density disordered and then it goes to something which is called layered state layered state is one in which the cubes kind of occupy layers so there is one layer and then there is a second layer which is separated by distance 2 and is like that so if you look at the center of masses of these cubes they will have a vertical periodicity of order 2 of 2 not ordered 2 just 2 they will be occupied layer and unoccupied layer and occupied layer and occupied layer when I just mark the so the way we think of this is that we always take the cube like that and take one one small cube and mark it all the other sevens are not marked we do not worry about them they just tag along so this mark cube will always lie let us say on odd values of z and the even values of z will not be occupied or maybe 1 percent of them will be occupied but mostly not occupied so there is a broken symmetry in the z direction that is layering okay of course it could have been in the x or y but then if you increase the density further it goes into this sub lattice order so it actually becomes 2 by 2 by 2 periodic lattice of width holes there are plenty of holes but still the density has a periodicity in all three directions okay and then you increase the density further then this doesn't survive and it goes into columnar order even a very elementary system like this basic cubes no 2 by 2 by 2 cubes can show a wide variety of interesting phases if you construct more complicated shapes you get many more kinds of possible orders and phase transitions between them and perhaps trying to understand them is interesting for some people or it is mental gymnastics for others no because this is an important question to ask you know why are we studying all this is why what is the interest in this you you just like some stupid problem that is why you are studying them right that is one possible attitude however there are actually experimental systems with liquid crystals where people have some complicated shapes of molecules which they dissolve in some solvent and they undergo phase transitions from one phase to another to another and people have seen up to nine distinct phase transitions in some particular liquid crystal just as you vary the density so these kinds of models are representative models to describe these kind of physical systems you may not like lattice models you can work with continuum models so suppose I had these cubes in a continuum I have finite size cubes I put them in continuum what will happen it is quite possible that the phase structure looks the same not so different that is the idea behind studying lattice models okay so this is all I wanted to say about phase transitions in hardcore models now what I want to do in the next few half an hour is to discuss the connection of the Lee Young age singularity to other problems connection of Lee Young singularity other models and we will write down the names of the models first so these are branched polymers and directed branched polymers also called directed animals I will define these terms so it turns out that if you know this if you have managed to solve the Lee Young age singularity problem then you have also understood these other models which are somewhat different in detail so you know of course there are a lot of equivalences between models you know like the Ising model is like the lattice case and so on these are just called languages you know you can describe the same language in one language or the other Bengali or Spanish or Greek right what's the big deal why is this more interesting than some other connections so this is an example of what we what is known as dimensional reduction and what it says is that the d-dimensional core lattice hard core yes suppose you understand this problem in d-dimension just hardcore spheres in d-dimension that's equivalent to d plus one-dimensional directed animals so some branched polymer problem in d plus one-dimension one-dimension higher and it's also equivalent to d plus two-dimensional undirected so this is actually a perhaps the best known case of dimensional reduction and it is very powerful and actually I don't know of many other cases which are equally tricky okay so I want to describe this equivalence of course the given the short time I have it will not be possible to give the full proof of all the statements so I give you a reference note it down the reference reaches an embryo got lost somewhere pj's and embryo annals of mathematics 2002 page number I have forgotten okay so you can find the details here it's a very nicely written paper but let me define let me explain what the various things mean here so the d-dimensional hardcore spheres is very well understood you have a space in d-dimensions d can be 1 2 3 4 5 6 whatever you like and you have hard spheres in d-dimensions and there is a gas of hard spheres and what we like to calculate is the pressure as a function of activity z and it is well known that this has a mere expansion which is z to the power n bn bn are called the virial coefficients or the mere coefficients and this goes from 1 to infinity and you know they depend on dimension and you have to understand what happens this kind of stuff so if you look at this function pressure of z pressure is just the log of omega so the lee young theory is working here this series will have radius of convergence which is governed by the closest singularity the closest singularity is on the negative real line and that is the lee young edge singularity so here this goes so singular as z tends to z lee young and so if you have solved this problem you know the lee young edge exponent sigma which is function of dimension okay so now we define this problem directed animals so I will define it in two dimensions you have a lattice and you construct clusters on this lattice actually maybe I should define the undirected animals first you take one site and occupy it and then you occupy another site which is connected to it or neighbor of it so you form a cluster like this this is a cluster of four or five occupied sites I can make other shape of clusters I can make a cluster in which this is not occupied may have had it here or here so how many distinct shapes of different clusters I can make that the question people can ask sorry can you here so there are sites which are occupied and they are all nearest they can be reached from each other by nearest neighbor bonds if I do not allow a site here and a site there with nothing connecting them then it is called one cluster otherwise it is called two clusters so I am like taking only one cluster the cluster should have n sites then if I take a cluster of one site then it looks like this wherever it is it can be here here but it looks like this a two site cluster will either look like this or it will look like this nothing else I cannot do anything else what about three site clusters the three site clusters can look like this or they can look in okay they can look like this or they can look like this and some rotated versions of this but nothing else right so there are two of these and four of these and so the number of possible clusters of n sites is just a n it is defined it is called a notation a n and a n is equal to 1 2 6 I actually worked out last night what was a 4 16 maybe may be let me not write it down it may be mistake there but it is a countable number it is a finite number you cannot have too many such configurations and for a given lattice this number is fixed I want to do this problem in 3D then again there is a corresponding series and of course one site cluster will be still one but two site clusters will be three of them and these clusters are said to be distinct if they are not if by translating you cannot get from one to the other that is my definition I will consider these two as distinct so you can get from here to here by rotation so I do not include rotation some other people may want to consider accounting scheme in which rotations are considered equivalent but for the moment it is not considered equivalent and so this number is called a n oh I was getting worried because I know where is columnar order coming here so this is the definition of a n so a n for any lattice with translational invariance increases exponentially with n but it also has a pre-factor which is a power law which is called theta and this theta is an exponent which is function of that dimension this animal problem is a sub case of the more general problem which is called the percolation problem which is very well studied if you have percolation clusters then the probability that you have a cluster of size n with p goes to 0 limit goes as p to the power n n to the power minus theta so this theta is the same exponent sorry not p to the power n well actually it is correct so they I wanted to say more so I do not want to say p going to 0 p less than pc goes as exponential minus alpha p s alpha p n n to the power minus theta so if you are for p less than pc all clusters are finite and this is large n clusters probability of large clusters is exponentially small but there is a pre-factor which is a power law and you know you can detect this experimentally or you can so people would like to know what is this exponent theta it is also function of dimension but it turns out that if you know this pressure then you can actually determine if you know this exponent sigma then you know this exponent theta if you know the full pressure as a function of z then you can get this done for a particular see this is a particular lattice or continuum this is a continuum but a n depend on the lattice so they cannot be an exact equivalence between this pressure and arbitrary lattices in that dimension okay but it turns out that as we have written here that you can take a d by plus two-dimensional animal problem like this you determine theta there and it will be related to the corresponding exponent in d minus two dimensions which is much easier to do like if I start with a two-dimensional problem seems quite tough but the corresponding zero-dimensional problem for hard course is easy to do the just few lattice sites and so I know what is the answer okay so that is the definition of this part the directed animal problem is the same problem except that we say that if you have a site connections are only allowed in positive x or positive y direction they are no no no we do not allow connections in the negative direction it was originally as a simplification of the other problem the other problem was too difficult there were too many configurations too many class you know if you want to count them on a computer it takes too much time and so we said okay let us consider the simpler problem in which you only allow only allow two connections like this and then this can go here and go here and it can branch like that these are directed animals again there is a number of distinct clusters of size n and there is a so this way n also goes as lambda directed to the power n and n to the power minus theta directed and this exponent is different from this one because this directionality is an important constraint okay but it turns out that this very remarkable result holds in you can go across dimensions and solve the problem hardcore problem in the dimensions let you solve some much more complicated problems like this in higher dimensions okay so now given that one I have only five minutes I will not actually explain this better you should just read that if you are that is kind of pedagogical article by Digi's and Embry okay let me emphasize that I don't know of many other cases where you can get these kinds of results easily and so I don't know when they are useful next they are they haven't been found very useful so far in other contexts like I cannot apply these results to some other problem today but that is for you to do now you should find use for this and solve the Li Young gauge theory in four dimensions using some dimensional reduction techniques okay so let me just take the last five minutes to define a different problem which also I found strikingly non-trivial and interesting use of hardcore models so the problem is defined like this it's called heaps and the context is like this let me imagine I work in two dimensional space and I have these hard objects which are of different shapes maybe there are several of them some of them are like that now so there are tiles called t1 this is t1 this is t2 this is t3 this is t3 this is t4 each tile has a shape and for different tiles there are different activities so there is a partition function omega of z1 z2 z3 z4 which will give you the partition function for these using these tiles and then of course I take the large volume limit and I calculate the partition function per side I can vary z to vary the relative fractions of these tiles looks like a very tough problem given the general setting that you know I didn't tell what is the shape and so on and so forth so solving this problem in general is going to be very hard obviously however one can make some very remarkable connection to other problems for general shapes so this is what I want to describe so the connection to other problem is called heaps this word is due to VNO the reference is given in the references cited in the abstract of my lectures so the heaps are defined like this you have all these tiles you imagine there are planar tiles of equal width and you drop one tile so this is my floor and I drop one tile on it and then I drop another tile so if this tile falls somewhere disconnected then I don't I erase it but I drop it then it may sit on top of the previous tile and then I can drop another one which may sit here and the third one may sit here and then I drop another one which may sit on top of this and another one which may sit on top of this and so on so each heap is defined to be a set of tiles which are supported by each other from you know there is a bottom tile and we support the things above we support the things above and so on that's how it is defined they are all connected by this relation they have something has to overlap something below okay and then I will define some animal these are called heaps so heap generating function using the same heaps same tiles so heaps occur in three dimensions because I have two dimensional tiles the heaps will be three dimensional structures and I'm studying the generating function of these three dimensional structures and it turns out that h of z1 z2 z3 equal to 1 upon 1 minus omega of minus z1 minus z2 minus z3 okay there is a small mistake here I will not fix it the reason is that I don't want to get confused with the legalistic details the heap generating function is related to the generating function of tiles but only on a 2d plane nothing to do with 3d it's much smaller problem and but the activities you have to put a negative sign on so actually I if I want to make it correct I will write 1 minus 1 then it's very remarkable result that you know some rather complicated function here becomes related to this is a 2d problem and this is a 3d problem but this correspondence actually works in any dimension d goes to d plus 1 to do 4 dimensional space then you can have 5 dimensional heaps okay so I think I will stop there I'm sorry that the discussion has not been sufficiently detailed but you should look up these articles in any case my experience is that you know you can listen to lectures and then you learn something but you don't learn everything in order to learn the rest you actually have to read the original papers so the task of the lecturer is to only to show you that something interesting exists and some general idea of what what it is about and then you can go and read the actual paper and be enlightened okay so I will stop there