 and they are characterized by robust power laws in correlation functions and we want to understand how these occur and the sort of the proposal made by Peierbach was that many natural systems, many natural driven dissipative systems. Threshold dynamics show steady states, having burst like relaxations, events of all sizes and power law correlations and this is the SOC hypothesis. So, let us just go through this because this is the main point of the whole talk. So, lectures, so these are natural many natural systems driven dissipative. So, these are not equilibrium systems, these stuff are driven from outside with some influx and they have under they have dissipation. So, laws of the equilibrium thermodynamics etcetera are not going to work explicitly and typically we work with threshold dynamics. Maybe I should put this like this, maybe you can imagine systems which do not have threshold dynamics which also show SOC, but we will typically work with systems which do have threshold dynamics and show SOC and then they show steady states and then they have these burst like relaxations means I drew a picture which was this was time and I forgot this axis. So, some activity of some sort just one second let me finish my sentence that I will use. So, this was some activity and many system show this kind of behavior and these are the ones we are going to describe yes sir. Threshold dynamics is something does not happen unless a threshold is crossed. If you apply a perturbation small perturbation does not cause any effect, but if you apply a big enough perturbation then something happens. So, that is the model which we discussed last time was this senpile model where unless the height becomes more than something nothing happens. So, in these systems you know there are two kinds of systems one is called linear response where you apply a little bit of perturbation and a little bit of response occurs. And these are non-linear response if you apply a little bit of perturbation nothing happens only when the perturbation exceeds some value will a response occur. So, threshold dynamics is very important in generating these kinds of events ok. So, then we discussed this B T W model of senpiles which we will recall model of senpiles. So, there was a square lattice and on each side you could have some grains which could be 0 1 2 3 4 and the rule was that this is that you should imagine that you the grains sit on top of each other, but if you try to put too many grains on top of each other the pile is unstable and it cannot stay and it topples. And so we said that in our case we said if more than 3 particles then it topples and 4 particles leave the system 1 to each neighbor. And then if something topples at the boundary then some particles leave the system. Now, you keep on adding particles one by one at a randomly chosen site then relax the system and then add again then relax the system each time you add something. So, sometimes nothing may happen sometimes a toppling event will occur something will leave and then you do it again and again. And there is a slow time steady state and in this steady state you see that kind of behavior for various observables. I can see number of things leaving the system or I can look at number of events generated in the toppling number of topplings when you add one particle and all of them will show some such behavior. And then we are calling that critical steady state and we want to understand the behavior. So, this is what we will study now, but references the primary reference is review article I wrote in 2006, Physica A it has been uploaded on the on the course college website. And you should you know that is a primary reference it all the references cited in the end there will be the ones which are the ones I will use mainly. We will go a little bit beyond this one, but at that time may be I will give you extra references. Otherwise the references cited in this paper are enough for our purpose there too much you know there are some 80 100 references I do not expect you to read all of them in the next two weeks, but if you happen to look at the problem further then you can perhaps have to look at them. So, this was the model which we defined. So, what one can do is that one can define such a model where we said the critical height is 3, but I can think of a different model where the critical height is not 3 it is 7. It turns out you had some question what do you mean by steady state very good. So, no I will try to explain what is meant by steady state. So, what happens is we have some system or the other I show it like this I have some input which I put in. And then something happens some relaxation happens inside and then I can watch the properties of the system later and see what happens I can monitor some behavior and I will find some output. So, for example, I can take a piece of metal and I put it in sunlight and I measure the temperature as a function of time. So, what will I see I will see temperature as a function of time. So, let me just to be sure that we are talking there is a thin sheet of metal or it can be glass I put it in sunlight and the thermometer is placed somewhere such that whether you put it here or here or here it does not make a difference you get the same measurement. And I measure the temperature as a function of time I might get a graph like this. So, there is a after some time the temperature does not change much and that is the steady state. Steady state is one in which when you observe on a macroscopic scale you do not see any noticeable variation in properties. And so in my sand pile for example, I can start with some part may be empty or some particles and I keep on adding particles and I may monitor the total mass in the pile as a function of time. So, what I will find is that initially the mass generally rises, but after some time it becomes kind of steady and does not increase much on the average it shows tiny fluctuations, but the mean does not increase. So, we will say this part is the steady state and this part is called the transient state. Transient is the one which dies away after some time then of course, there is a quibble about when is steady state reached and we will just say we wait long enough. So, that we do not see any difference. So, for us the notion of equilibrium is defined by a in a very narrow way. Equilibrium systems are systems where equilibrium stat mech works by which is meant what is equilibrium stat mech there is a Hamiltonian which is defined for the system. If you look at an isolated system with a given energy all configurations with same energy occur equally likely that is the micro canonical ensemble. And using this ensemble you can predict all kinds of properties of the system what is the color of the system what is the mean and mean temperature or I do not know all other properties what is the susceptibility what is the magnetization. So, whenever these prescriptions work that is called equilibrium stat mech that is called in equilibrium. However, if you have systems in which there are external fluxes like as I said there is I shine light and I guess there is a steady state. So, some energy comes in some energy must be going out there must be some scattered light may be not at the same frequency may be at a different frequency or whatever. I do not know all these details, but there is a notion of a steady state. This system there is not an equilibrium state of matter because the laws of equilibrium thermodynamics do not apply to metal sheets put in sunlight. So, that is the sense we are using a steady state as a more general notion than an equilibrium state. And typically in this system there is no Hamiltonian you can write down which will describe the steady state behavior. So, we have I was saying that you can take a different model in which. So, we had a in the BTW model Z C which was the critical value of Z was 3 and instability occurs if Z is bigger than Z C, but suppose I work with Z C equal to 5. Then what happens the answer is nothing changes it is all the same what happens is after sometime the pile will get height any site you observe will have some height. Once it becomes 5 then it decreases 2 4 particles are lost sorry Z C is 5. So, 5 is stable 6 is unstable. So, when you start with 6 then it goes to 2 then you add 2 3 4 5 6 is unstable 2 3 4 5 6 is unstable. So, at the site the allowed values of height are only 2 3 4 5 in the original model the height set the site was 0 1 2 3 4 was unstable. So, it does not matter all you have the allowed heights are maximum and 3 less and no other. So, you can if I work out the problem with Z C equal to 3 I can also work out with Z C equal to 6 yes sir yes of course, no it is the detail balance condition is not satisfied. So, I can have a state which is called 0 0 and you can go to a state which is called 0 1 and I am not sure how I go from 1 0 to 0 0 because we are allowed to add particles, but we are not allowed to remove particles. So, detail balance does not work in our problem. So, so this point is clear then we can make some other changes and so one can actually work with the more general model. Yes, so the rules are the same every time a toppling occurs 4 particles leave 1 to each neighbor that is the rule. So, even if Z C is 6 4 particles leave 1 to each neighbor that is that is how we are defining it. So, I can define a more general sand pile. So, you can take any graph you like there are nodes and we say that z i is equal to height at node i which is the number of sand piles or the height of the local column of sand at the site and this is the number integer which is bigger than or equal to 0 and then there is a instability condition instability condition is that z is bigger than z C and then toppling occurs and in a toppling site j gets some number of particles which we will tell gets minus delta i j. So, I define a matrix delta matrix delta is n by n matrix n is the number of sites i is equal to 1 to n n by n matrix and so site i z i goes to z i minus delta i j for sorry z j goes to z j minus delta i j for toppling at i. So, if I topple at site i then all the heights at all the sites are updated and the height at z j becomes z j minus delta i j and delta i j is a specified matrix given that is the rule. Once I give you the matrix delta then the and I give you this value z C then everything is good, but we can choose without loss of generality delta i i equal to z C plus 1. So, in my problem my graph was just square lattice and this delta i i is 4 the site where the toppling occurs 4 particles are lost and each neighbor gets 1. So, delta j i j will be 1 for sites which are sorry delta i j will be minus 1 for the neighbor sites and 0 for all other sites ok. So, this is this is all, but this matrix delta has to have some good behavior properties. So, requirements for delta where we should have delta i i greater than 0 the point is this there is an unstable site. Now, after toppling if the height is going to become delta i i minus j minus delta i i. So, unless delta i i is positive you will still be unstable keep on being more unstable or equally unstable and that is no good. So, delta i i should be bigger than 0. So, that the height decreases it becomes less unstable ok. So, that is the first condition the second condition is that delta i j is less than 0 for j not equal to i by which is meant that sites other than the site where you are toppling the height there can only increase because some particles go there it cannot decrease. So, less than or equal to 0 and the third condition is that summation over delta i j over j is greater than equal to 0. Now, delta i i is the number of particles which leave and delta i j minus delta i j is number of particle which reach j. So, the number of particle which leave should be more than equal to the number of particle which reaches other sites because some particles may leave the system that we allow and. So, this should be is true there is no particle creation in the toppling process that is the meaning of this equation no particle creation during toppling sorry above. So, no particle creation in toppling. So, this is summation over delta i j over j is greater than equal to 0 for all i this is the third condition and we actually need one more condition which is that summation over i n j of delta i j is strictly greater than 0. So, on each site they may not be any loss of particles, but eventually there should be a loss of particles somewhere. Otherwise I keep on adding particles and there is no way they can leave then the number of particles will keep on increasing and they would not be any steady state. So, this is an obvious condition for the requirement of a steady state do I need anything else? Yes of course, I need something else which is connected graph by which is meant the following you cannot have a situation where there is a graph like this and there is a graph like this. These two parts are not talking to each other, but suppose there is a site where dissipation can occur is here, but these sites cannot go there then even though in the full system my condition number 4 is satisfied it is not satisfied in this part and so they would not be a steady state. So, all we are trying to ensure is that whatever rules we put in should allow for a steady state to be there and you can check that these conditions are enough to ensure the existence of a steady state. We will prove, but you should be able to convince yourself that this is true even without a proof given on the board ok. So, now the key result which makes all these further analysis of this problem possible is called the Abelian property and the Abelian property is rather easy to derive it says in an unstable configuration toplings commute by which is meant that you have an unstable configuration there are two or three sites which have height more than three ok. So, now I can topple this one or this one or that one in some order, but take any two particular sites this one and this one both of them have height more than three. So, I can first topple this one and then this one or first topple this one and that one and the final result will be the same independent of which way you do it ok. Yes sir. Yes. So, in some cases they may be particle loss. So, you know there is no particle creation, but they may be particle loss. Yeah, yeah because if at all sites the sum is 0 then you can lose anywhere and then they cannot be a steady state. There have to be some boundary sites where the loss of particles can occur ok. For a general graph I do not need to define a boundary site I just say at some site or the other the inequality should hold ok. So, two topplings commute and then if I have some site. So, this proof is all clear if. So, here is my configuration here is a site 4 here is a site with height 5 I topple here I will decrease this by 4 and send one particle to each neighbor I will decrease this by 4 send one particle to each neighbor clearly it does not matter in which order I do this ok. If these sites are adjacent to each other that is slightly more tricky, but it is easy to convince yourself that even then it does not matter in which you order you topple them and then it is true ok. So, now we take this condition and we have a stronger result. So, we say that if you take a system like this which has several unstable sites this one is 5 this is 4 this is 4 this is 4. Now, you topple anything in any order and you will get the same result in the end. The final stable configuration will be the same whichever order you toppling the proof is a repeated use of the previous lemma two toppling commute. So, if I produce toppling in this one by some way T 1, T 7, T 8, T 19 I topple one and say topple 7 side 8 side 19 side and so on. And in the second one I topple in some other order, but I have said that sometime or the other T 1 must occur here in the second sequence because the site is unstable and other particles can only add to it. So, this T just let me finish the argument T 1 will occur somewhere, but these topplings commute. So, I can pretend that this T 1 occur you know I commute it across this sequence and it will occur in the beginning and then I look at the rest of the sequence. And at induction says that it will happen for everything ok. Now, there was a question yes yes. There is no threshold. No, no, no all the unstable sites all the unstable sites you can topple in any order and you will get the same result. No, so in our case let us stick to this convention the stable heights are 0 1 2 3 4 is unstable. So, 4 is unstable that was my rule, but whatever you said continues to hold even in this convention ok. So, now of course, no so I. So, there have been lot of cent pile models proposed in literature. For example, there is a model in which which is called the Zhang model where you add something and all the height is distributed equally amongst all the neighbors and the height becomes 0. It is easy to check that even with pile with 2 sites in this model is not a billion. There is another model in which the toppling condition depends on slopes and not on heights ok. So, what can happen is that if you topple this site you cannot topple this site because the condition for stability is now modified. Suppose, the condition of stability is a slope condition it says the height difference should be bigger than something and you topple here then it adds something here and this height difference is no longer bigger then you cannot topple at the second site. And so, the abelian property is not a generic property which is obviously true for all models. In fact, it is not true for most cent pile models and most other models of self organized criticality as well yes. Because sometimes what can happen is that this site 4 is unstable, but its neighbor was also unstable and I topple this one first and then this becomes 5. It is not stable, but I have in the middle of the toppling process they may be sites which get more height than 4 because I did not manage to topple them ok. So, very good. So, now, one more step is that now I can do the following here is my pile and I add a particle here and let us say it becomes unstable then I can topple it and relax, but maybe I then I add another particle here ok, but then I can do it the other way I first add this one and then add this one and relax and do not look at this one for a while and then do it again. So, the addition of particles is also commuting with the toppling process because addition only adds to height and it does not change the criticality condition it does not make stable heights sorry it does not make unstable sites stable ok. So, then it follows that you can take any configuration C you add a particle at I add a particle at J and relax or you do it in the reverse order and relax and you get the same configuration in the end ok. So, that is the final result which says that starting configuration is C. So, C you can add at I and relax it goes to C 1 and then you add at J and it goes to C 2, but you can do in the other order you can add at J it will give to some other configuration, but now you add at I you will get back to the same configuration which order you add particles does not matter you will get the same result if you relax everything and this is true for all starting configurations. Yes firstly the proof is rather you know there is a proof I gave it of course, but you should imagine that you are a high school student and you have to construct the proof yourself and you can do it nothing more than addition of integers is involved in the proof and you should be able to get the result once the statement is given ok. So, the proof said that I do not want to repeat the proof I am sure you will be able to construct the proof yourself it is not a difficult proof once the basic point is known ok yes yeah you can do that clearly ok. So, now we just want to formalize this result a little bit. So, define an addition operator A sub I is equal to add at site I and relax ok. So, if you have a configuration C and you add at site I and relax this configuration is called C prime that is the definition of operator A ok yes yes yes no all of them have to reach a stable configuration relax as long as you needed until you reach a stable configuration and then look C prime is the final stable configuration once you add at site I to C and relax ok. So, what are the allowed configurations C which are stable allowed configurations at each site let us take with the square lattice L by L. So, their number of configurations allowed is 4 to the power L square number of stable configurations ok. So, I can think of this is n 4 to the power L square 4 to the power L square cross 4 to the power L square matrix it looks like this. So, C is a configuration C 1 C 2 C 3 you make a list of this 4 to the power L square configurations and this matrix will be some matrix like 0 0 1 0 0 0 0 0 it takes this configuration takes it to that one and this one goes to that one ok yes. C is a stable configuration C is a stable configuration A is a matrix which applied to a stable configuration gives you another stable configuration it will have exactly 1 1 somewhere and 0 everywhere else yes. C's do not form any group C's are the configurations they are the vectors the a's we are getting there ok. So, what we have proved we have shown or argued A i C A j is equal to A j A i A j C for all i j C because each side can take 4 possible heights 0 1 2 3 in a stable configuration and there are L square possible choices because I remember working on L by L square yeah yeah because that is right no no no so that is that is all the vectors are 0 0 0 1 in 1 place and 0 0 0 and this matrix applied on them gives you another such vector ok. And now the claim is that this now equation can be written as A j A i commutator equal to 0 because these are just matrices no I do not worry about C because it is true C ok. So, this is called the abelian property the key point here is that actually in the proof somewhere we involve the looking inside the relaxation process what happens when you topple but now the a's do not look at the toppling they just look at stable configuration goes to another stable configuration we do not see what happens inside ok alright. So, this is the basic abelian property but what can I learn from this one well just once again ah very good. So, now starting with this result one can get some non trivial results rather easily ok. So, firstly yeah so suppose I start with some configuration C and I apply the addition operator at psi type. So, I get to another configuration by applying with A i I get C 1 then I can apply A 1 again on this one I get C 2 then apply A 1 again on this one I get C 3 and keep on doing this. So, what will happen well the space of allowed configurations is only finite. So, it eventually this thing will come back on itself and it might do something like this. So, what happens is that if you keep on applying it will go and then after some period T it will come back to C 3 and then you know it is just a periodic repetition of the same thing ok, but it is possible that in this process I have not exhausted all possible configurations. So, I can take another configuration which is not in this set and that will go like this and it will also form its own loop and another one which may merge with this one and so on ok. So, finally, the set of configurations will have the cyclic states and non-cyclic states states which are in the cycle and states which are outside the cycle and the obvious result is that at long times all the non-cyclic states occur with 0 probability because you know sometime or the if you are here then sometime the other you will apply A 1, but you can never get back to C 2 once you have gone to C 3 by applying A 1 ok maybe by applying some other operator you could get there ok. So, let us see what happens if I take this state and I apply A 2 some other operator which produces a state. So, A 2 can produce a state which is here in the same then it does not produce a new state then I do not care I want to produce a I want to go outside this cycle. So, I go to a state A 2 which produces a state here and now I want to see what is the cycle produced by A 2 ok. So, cycle produced by A 2 will give you now apply A 1 here you get here and if you apply A 1 here you get here and this point is that if you take this state and apply A 2 you get to the same state of course, by a billion property. So, this one will go to A 3 and this will go to A 3 and there is a shift. So, the whole procedure of applying A 2 is sort of like taking this graph and shifting it together and. So, in particular there is a result which is that all these different cycles you will produce by A 1 will have the same period because if I apply A 2 I eventually go to the same cycle. Let us produce a formal proof if there is a configuration C such that A 1 to the power t C is equal to C. So, C is a cyclic state with period t C is a cyclic state with period t then C prime is equal to A j C then A 1 to the power t C prime is equal to C prime. So, all the periods will be the same that is a very strong result you know we started with this just this phase space and saying that look eventually it will have to come back to itself nothing else can happen. And now we are saying that you can produce all these kinds of cycles, but all of them have to have the same length. So, now let us produce this picture you have this cycle and you apply some other operator you get another cycle just once again. Yes. That is reachable from C. Yeah. So, I could have a disconnect the region of states that are. We will get there no you will see the good objection. So, you can have something which is connected and this goes to this I apply A 2 to this one you produce another some similar stuff and A 2 to this one. So, now I can think of this cycle as a simple object and when I apply A 2 to it it produces a next cycle and when I apply A 2 to it produces a next cycle and but now my argument can be repeated you start with this cycle it goes to this cycle goes to this cycle, but eventually the cycle have to produce come back to themselves. So, you will have when I take two operators A 1 and A 2 then I will produce something which is like a torus two dimensional torus at best it may be a one dimensional torus if A 2 was the same as you know if the cycle produced by A 2 was the same as the cycle produced by A 1 or it will be a torus of size two then I take one more variable A 3 and apply the same procedure again then I get another torus of a bigger size and so on. And I keep on doing this until I exhaust all the possible states all the states which are reachable from my initial state and all of them will either be inside the cycle or will be outside if they are not within the this cycle all these states these are also transient because under A 1 they come back to themselves, but when I apply A 2 they do not come back to themselves. So, the recurrent states are only the states which occur in the final torus so that is the statement there is no randomness yeah yeah yeah yeah of course, there is no stochasticity in the relaxation process the A's are fixed operators given that my graph and given my rules of relaxation. So, the recurrent state has the topology of a torus what is meant by topology of a torus you have these states I construct the states these are the points and join them by arrows if you can go from one to the other and the state you will in the end will be an n dimensional torus that is what we said and then on a torus then what happens is that from this site there are n arrows which go in various directions and from this site there are n arrows go in various directions, but on a torus or sites are equivalent. So, in the steady state where you can choose to go this way or that way or that way it will be the steady state will be the state which has uniform weight on all the recurrent states in steady state configurations have equal weight. So, there are some configurations which have 0 weight in the steady state and there are other configurations which all have equal weight in the steady state. So, we have defined I have to tell given a configuration is it recurrent or is it transient, but at least we know that all the recurrent configurations will have equal weight, but in addition now I have the extra property that if you take us any site on the torus it has only one arrow going out and it has one arrow going in because we constructed it by hand you know for each operator A i there was a cycle. So, there was one arrow coming in one arrow going out. So, then you can define an A inverse operator A i inverse is well defined on the set of recurrent states. So, now I have a much richer structure I have a group because these A i operators you can you know you can write you can multiply them because A 1 A 2 A 7 A 1 A 3 A 8 is a product of operator it says it is my C occurs here. So, I apply A 8 then I apply A 3 then I apply A 1 then I apply A 7 then A 2 and A 1 and so on and I want to know what is the final result. So, the answer is that in this you can flip the order you can you know commute these operators any way you like and you get all the powers of A 1 I will put first and powers of A 2 I will put next and powers of A 3 I put next and that will give me the final state and I can put right now I do not need A 1 inverse. So, very good, but actually turns out that that is not enough because there is some extra properties which these operators A i have which makes them satisfy an algebra. So, the extra property A i have is that suppose I take a site on the lattice this site is called 0 this site is called 1 2 3 4 then the rule is or the observation is that if you add 4 sites at the origin 4 particles at the origin then whatever be the initial height it will become unstable and it will topple at least once. So, I just topple it once what happens then 1 particle goes here 1 goes here 1 goes here 1 goes here ok. So, I get A i to the power 4 A 0 to the power 4 is equal to A 1 A 2 A 3 A 4 adding 4 particles at 1 site is the same as adding 1 particle at each neighbor on any configuration. So, this is my equation a similar equation holds for each site yes I get the proof you start with some configuration and look at my site it has some height I add 4 more particles one after another I can first add them and then topple them you know that we have agreed. So, I add them now the height will be more than or equal to 4. So, I can topple them I topple it once only what do I get is the same as you added one particle here 1 here 1 here 1 here acting on C. So, A 4 C is equal to this time C 1 any C in which case I can get rid of the C is this clear. So, this is my equation and similar equation for other sites now I have a group algebra because now I multiply all these operators and I get very high powers of A suppose I get A to the power 15, but then I can use these equations to reduce the 15 power to lower values then I generate something which will then make maybe A 2 becomes B then I reduce the power of that one and so on. So, let us do this once on a small size. So, this is 1 2 3 4 5 6 and I write A 1 to the power 12 A 2 to the power 7 A 3 to the power 8, but A 1 to the power 4 is equal to I take 4 this. So, you know add here is here you will add something here something here something here there is this quality 7 and 8 this is A 7 A 4 A 2 A 8 right. So, this power 12 will be broken into A 3 to the power 7 A 4 to the power sorry A 2 to the power 3 A 8 to the power 3 A 2 to the power 7 A 3 to the power 8 then I collect powers of 2 together A 2 to the power 10 and use the same thing and you can see that this is actually the original toppling process I have not done anything think of these as the height configuration the powers as the height configuration. Once the powers become too big then I reduce them using the reduction rule which is now this operator algebra rule ok very good. So, that is all there is two operator algebra ok. So, in my case for my square lattice. So, for my square lattice the matrix delta looks like this it is 4 4 4 along the diagonal and it is minus 1 minus 1 minus 1 at the four neighbors ok. So, I this is sort of schematic here there will be some minus ones in each row there are four minus ones at most at the corners there are fewer. So, this matrix delta is actually kind of a familiar matrix it is called the Laplacian matrix it is 4 in the diagonal and minus 1 at the neighbors. How many of you have seen the Laplacian matrix before discrete Laplacian matrix large number of seen it actually it is about 30 percent, but I think this is a good place to be familiar with it because it turns out to play an important rule in what we are going to show ok. So, so that is my Laplacian matrix. So, now I wrote these rules as a i to the power delta i i is equal to product over j not equal to i a j to the power minus delta i j because minus delta i j were positive numbers for i not equal to j. So, this was well defined, but now since the inverses are defined I can actually simplify this equation and write this as a i to the power delta i j sorry a j to the power delta i j product over j is equal to identity for all i ok. I am allowed to simplify by a to the power delta i j on both sides because that is now well defined and so then this equation follows ok. Just so that you are not scared of this if I take a 3 by 3 board like this 1 2 3 4 5 6 7 8 9 there are 9 operators a 1 a 2 a 3 and I get a 1 to the power 4 equal to only 2 neighbors will come here and 2 particles will leave to a 2 a 4 a 2 to the power 4 is equal to this a bit complicated can somebody of one of you will help me with this one I pick one at random know that one sir would you be able to tell me what is a 2 to the power 4 on this graph Laura a 1 a 3 a 5 a 3 to the power 4 equal to a 2 a 6 I will not write all of them know a 9 to the power 4 is equal to a 6 a 8 yeah. So, these equations together define the group algebra for the a billion group generated by these operators. So, that is very nice what is the user for this well the use is the following. So, we have learnt in some math fees course that if matrices commute then they can be simultaneously diagonalized ok. So, all these A I matrices they are all commuting with each other. So, I can simultaneously diagonalize all of them ok. So, then there exist there. So, if phi is a simultaneous eigenvector eigenvector A I A I with a bracket around it is a set of A I which are A 1 to A all of them right. So, I write A I phi is equal to A I phi sorry for the bad notation this is an operator and this is an eigenvalue this is a number and I am deliberately using some confusing notation a little bit ok. Then these equations now become apply A 1 phi they become equations for the eigenvalues themselves ok. So, the and then the eigenvalues A I satisfy the equations A I to the power delta I j A j to the power delta I j product over j equal to I ok, but these equations are any sort of messy looking equations can you given suppose now I know that these A I's are complex numbers they are not matrices can you solve this set of equations I can see that if you put everything 0 that is a solution, but suppose I ask you for all possible solutions to these equations how hard is that well the answer is it is not. So, hard because we realize that on the set of recurrent states A I's will be unitary operators. So, they will keep the norm fixed and so they A I small A I which are the eigenvalues are e to the power I phi I phi j phi j real and then this set of equations becomes delta I j phi j product over sum over j is equal to twice pi m I for all I yeah. State by a number no no no A I's are some matrices I no we are not forgetting, but we will come back to it eventually in between I am trying to forget I think A I's are just some matrices I want to diagonalize ok yeah no no no so using the operators you can get some information about the original problem we will get back to the original problem, but right now if I am not I might be able to say that oh, but all the A I's are operators they have these values the eigenvalues are these. These are the eigenvalues of operators A I. So, what does that mean I guess if all so one of the one of the solutions of these equations is all A I equal to 1 vector phi is equal to 0 0 0 0 is the steady state vector A I acting on phi gives you phi because e to the power 0 if phi if I am sorry suppose there is a solution like this there is a vector phi such that all A I acting on phi give you back the vector then that will be the steady state vector. So, I want to find the eigenvectors of A I or the simultaneous eigenvectors of all A I equal to 1 or some such thing because that gives me the steady state vector in the original problem ok. So, so getting back what was my original problem. So, let us define a probability vector at time t is a vector that is equal to summation over c probability of configuration c at time t ok. We already introduced c you know there was a 2 to the power n dimensional vector like this which was 1 if that state was there and 0 otherwise now we will give a vector p 1 p 2 p 3 this is the probability that there is a vector configuration 1 this is the probability there is a configuration 2 there is a probability configuration 3 at time t there is a distribution of probabilities that gives me the probability vector at time t this probability vector evolves in time how does it evolve probe t plus 1 is equal to some transition matrix times probe ok that is the general formalism for Markov processes we started with the discrete space there was a choice you know when I add a particle something happens it goes to a new state and. So, there is a transition probability to go from state c to state c prime all that information is coded in w w is the transition probability to go from c to c prime and then this probability of t plus 1 is w times probability of t from the general theory of Markov processes there is a transition matrix right everybody is familiar with this idea ok very good. So, now I got to write what is w yes no no no ok very good. So, what we are saying is that we add a particle and something relaxation occurs and then I wait long enough and then I add another particle. So, for the given size of the system size l all the relaxation occurs in 1 milliseconds I am only adding a particle every second. So, I wait long enough such that all the relaxation processes have died down by now and so, I get into a new stable state and then I add a particle I get to a new stable state and I do not see what happens inside I just evolve using finite discrete time additions yes sir. Stable state means that is over state. Yeah all the avalanches have died down. How it always. No no no you do not come back to the same configuration ok. But the steady state is a linear it is a stochastic state. So, if you look at long time you will find that some configurations occur with some probability and after some time you will again see the same thing. So, this probability vector will tend to a steady vector for large times. Like a period. No it gets to a unique vector. So, it says that. So, this is from the general theory of Markov chains is that this w is a probability matrix and for large times probe t which is a vector tends to a unique vector which is called probe infinity. So, that is all we are saying this is a steady state vector then applying w on it further gives you back the same vector and we want to know what is this probe infinity vector that is the steady state vector that is my first concern in the study of this system because we were looking at steady states of this pile yes please. Oh this one it says phi 1 equal to 0 equal to phi 2 equal to yeah yeah yeah. So, that was given here it a j equal to e to the power i phi j ok other question ok. So, very good. So, this is w what is w in terms of my operators a w is equal to summation p i a i sorry. So, new notation let us consider a sand pile in which you choose to add at site i with probability p sub i ok. So, that is the p sub i here. So, let me write for a driving adding at i with probability p sub i w is just summation p i a i because with some probability you will add a 1 and adding a 1 on this vector will give you the configuration with c goes to c prime ok. So, this is my time evolution operator for the stochastic yes please no it no it has only n components because phi 1 phi 2 phi n there is a steady state vector can which is labeled by phi 1 phi 2 phi n or eigenvalues of operators a i or phi i e to the power i phi i ok. So, I am just taking one such vector each choice of phi i which is allowed will be one vector ok what choices of phi i are allowed they have to satisfy this equation where m i's are integers because these are these equations yes. So, phi no no no it is very good I think the operators maybe I should use some notation like this a hat is the operator in this matrix 4 to the power l square times 4 to the power l square. Then there is a it has a eigenvector which is e to the power i phi i times phi this vector will be labeled by phi 1 phi 2 phi n. So, phi i are just labels of eigenvectors each phi phi 1 phi 2 phi n and that will label an eigenvector in this of the big matrix no it has a 4 to the power n eigenvectors no that is the next task we have some operators we can diagonalize them. Now, what do you do with all these diagonalized operators they have so many eigenvectors 4 to the power l square ok. So, that is what we want to do we want to get something related to the Markov matrix and the Markov matrix is specified by this evolution operator W. W is a linear combination of the a i's. So, if I have diagonalized the a i's I have also diagonalized W which is very good for me ok all right. So, W equal to p i a i. So, here how do I find the solutions of phi i for a given choice m i you choose these integers m i by yourself whatever you like and then phi i summation over g. So, now, I have I can generate as many vectors as you like because I just choose different values of m i for each choice of m i this is a linear couple set of equations for phi j just involves the matrix inversion. So, given any choice of m i we just apply delta inverse on the vector m and you get a set of phi i which gives you the eigenvector of the operator c and different choices of m give you different eigenvectors. So, I can get an infinity of such eigenvectors, but I only wanted 4 to the power l square you know there were the only that many. So, how where what is going on how do I get an infinite number of eigenvectors. The answer is that sometimes you can take two different m's and they will correspond to the same vector because phi j is the same as phi j plus 2 pi there is no difference because e to the power i phi is e to the power i phi plus 2 pi. So, if you change phi j by 2 pi you do not get a new vector you get the same eigenvalue and hence the same vector. So, while it may appear that with all choices of m I get an infinite number of distinct values I actually do not get so many different ones. Very good let us see what is the time like I have 20 no I have less than 20 minutes. So, I think let us not go too fast we will stop here, but I asked yesterday for I asked a home assignment problem which was to find you have a square of size l by l you add a particle at random it does a random work and it till it leaves and. So, in the process of leaving it has some steps it takes and what is the mean number of steps taken by the walker that was the home assignment how many of you managed to do it one one that does not sound very promising ok no no no very good. So, I expected that all of you will not be able to solve it, but I thought that at least half or 30 percent will get there. So, I have to give some hint ok. So, the hint is like this suppose you do the restricted problem I add the particle at some site r and what is the expected number of steps you need to take to exit the lattice ok. So, let t of r. So, what we this is sort of a standard random work problem you imagine that the walker takes one step, but you need time and t of r is the time required for it to exit that is the way it is usually posed. So, I will use the same notation t of r is the expected time when it starts at r ok. So, then it turns out or it is easy to see that suppose the walker is here. So, what it can you do in the next time step it can just jump to one of the four neighbors then it is here then what is the time required to exit it is t of r plus 1 ok. So, with some probability that exit time is t of r plus e y otherwise it is this one. So, t of r is equal to 1 plus 1 by 4 summation over neighbors t of neighbor is this argument clear. So, now this is also the same del squared kind of matrix I think. So, you can solve this equation by this is a linear set of coupled equations. So, you solve it and you get t of r then you have to just solve it for each value o and the there is an initial condition t is t of r is equal to 0 for r on the boundary because then you do not have to do anything you know you are already there boundary is this outside region and. So, if you solve for all r. So, if you start with initial condition r equal to 0 here in at originally for the particle is here then you can solve this equation and you get an equation which is del squared t is equal to 1 there is no information about r 0 in this the starting point. So, this equation is good, but it depends on the starting point in some way I am only supposed to give you a hint no I am not supposed to solve the problem now. So, I am giving you a hint there is some equation like this of course, it has to depend on r 0 in some way where you start. So, you put that in you solve this equation and then you get the answer then you average that answer for all r 0 to get the final answer it is a two step process is the hint clear yes sir yes please. The second step is that I will find t of r 0 for all r 0 and then I have to sum over r 0 1 by number of sites is equal to average time when you add at random that was the original question that you know I added at a random site. So, r 0 is also a random variable. So, I have to average over r 0 ok. So, I think we should stop now no I have 15 minutes left ok very good. So, in the next 15 minutes I can assign you two more homework problem no no no they are problems are chosen to be instructive and not infinitely hard and if you work them out you will understand the next lecture a bit better that is where in ok. So, problem 2 ok. So, it says there are just these two sites and I have a sand pile the toppling matrix is 4 4 minus 1 minus 1. So, if the height is more than 3 a toppling occurs here and three particles leave and one particle goes to the other site and that is all there is. So, now all the matrices there are only two of them a 1 and a 2 ok. So, you write down the matrices a 1 a 2 write down the algebra for I think it goes like this a 1 to the power 4 is equal to a 2 a 2 to the power 4 equal to a 1 as simple as this ok. And then what is the corresponding group that is the question you figure out you know it is not a big deal, but you work out the group see what you can say about the group generated by these two generators with this algebra of course, if you are very ambitious you can change the take three sites and do the same thing. This question number 3 is a little harder it is not harder mathematically, but conceptually there is a little bit of a subtlety there which I will of course, want you to figure out. So, the answer is not being given now. So, I take a problem like this 1 2 3 4 there are four sites, but again the rules are the same you topple drop 4 particles 1 2 each neighbor at each site to leave. So, now the operator algebra is like a 1 to the power 4 is equal to a 2 a 4 a 2 to the power 4 is equal to a 1 a 3 3 to the power 4 is equal to a 2 a 4. And a 4 to the power 4 is equal to a 1 a 3. So, I can write these matrices these are I guess not so easy to write these are supposed to be 4 to the power 4 dimensional matrices right that is not so easy I do not want you to actually write. Imagine that you have written this 256 by 256 matrix then you can still construct the group corresponding to this and what can you see about the group. So, the question is that subgroup generated by operators a 1 a 3 suppose I only have these two generators a 1 a 3 and I keep on using them then what is the group I will get will it be the full group or it will be less the answer is clear you know you have this thing you can only adhere or adhere. So, if you start with the symmetric structure you will remain on a symmetric structure. So, that is some kind of a subgroup of the group you will have to encounter and the idea is to figure out what exactly is the subgroup here and you know what can one say about I think we will stop here now even though I have 10 more minutes yes sir sorry can you speak louder please can you be quiet yeah yes yes yes. So, actually the firstly I had tried to explain in the first lecture the motivation for doing all these problems we want to address the experimental systems the systems we observe in nature you know like coastlines whatever then we make some models about these the models are necessarily a little bit simplified because if you take a too complicated a model you cannot do anything with it the advantage of having a simple model is that you can analyze it you can understand something about it some of the properties of the model will be shared by the actual experimental system some other properties will be specific to the fact that you have chosen a simple model I think that is not a bad thing you know there is a statement by Einstein which said that you should make things as simple as possible, but not simpler and when Einstein says something most people not their head of course you know that he cannot be wrong, but in this case I think he is not fully right in some systems it is necessary to make further approximation you cannot describe it in the most complete way some simplification is necessary in order to understand complex systems. So, we simplify further we say oh let us consider the kinds of sand piles, but with this extra property which is called the abelian property then it helps us analyze the problem you take some other sand pile I will not be able to find the steady state vector. So, using the fact that you have some special property you get some results which help us understand the original problem in some way. However, it turns out that you can actually preserve the abelian property even for what is called stochastic sand piles. So, we can get there and you know it is not way out of place maybe this model is too simple, but slightly more complicated models can still be handled analytically or they can be understood in some way and so that is what we would like to do ok. Other questions complaints ok then let us stop here.