 We now take up a detailed study of what is known as the random walk models. We introduced this model in the earlier lectures to illustrate specifically the central limit theorem. We now pursue it in great detail, obtain some exact solutions not asymptotic solutions and then of course, we pursue variants of this model explorations in many dimensions. So, it has lot of implications. So, why so much importance to random walk models? One is at it is a simplest level it represents a kind of a simplest picture of Brownian motion we all knew I mean we we said it before. So, we model like we have a lattice and lattice as many lattice points and an atom gets emitted either to the left or to the right with equal probability and at every time it lands on the site that site absorbs it then again emits it to the left or the right with certain probability and this we perform in steps and after a long time we examine the consequences compare it with the diffusion models and then we can extract diffusion coefficient. But that is not the only this is let us say a diffusion model of an atom on a discrete lattice. Well let us say lattice we can find many examples where similar things will happen. So, as for example, the price of a commodity it also will undergo random jumps increase and decrease. Sometimes the average may just remain steady, but it could have fluctuations up and down fluctuations. So, you can see it as in the price. So, this will be let us say cost space of price space you can say the price of a commodity jumps to the left decreases next time it again increases. So, you can similarly map it on a discrete lattice if let us say it is again a model. So, the change in price let us say is fixed and random walk models need not have fixed length of transition it can also have any transition possible and that also falls within the same framework as random walk that we are discussing. So, this could be a let us say price fluctuations and you can think of many fluctuations like whether fluctuations we can go more abstract. As I was saying supposing we are discussing a growth of a particle model the aggregation model nucleation then we know that the clusters in the cluster space now this will be the clusters say the cluster volume space. Then every time it absorbs an atom the cluster psi undergoes a transition to the right and every time it loses an atom by evaporation the cluster size undergoes a transition to the left and then this will be a random walk this will be a growth model growth modeling. So, this just these 3 examples tell us that studying a simple random walk has many implications and therefore, it is a very important problem it is not just about a person walking along a it is not just a hypothetical problem for a person walking along a along a linear stretch having lost his memory completely as we said in the last lecture. To be very specific let us consider the random the model let us describe let us describe it in the easy fashion of this lattice model. So, the random walk starts from some place origin which we call as origin and then takes a jump of fixed length L. So, we can call it as. So, we will denote all of them as 1 2 3, but so to either 1 or 2 minus 1 2 minus 2 and each of course, of length L. So, just to illustrate that we connect to physical space we keep a length L, but we can as well do it just in the integer space. So, it is an infinite lattice the process can go on and on as the step sizing increases. The first thing that we should now focus on is about the transition probability for this problem. Let us assume that the jump occurs only to the nearest neighbor at whichever site it is. As I mentioned it is not necessary you can jump from anywhere to anywhere, but right now let us focus on a nearest neighbor jump which means that the transition probability for landing on site m from another site m prime is non-zero only if m prime equal to m plus or minus 1. Alternately the transition probability can be considered as a sum of one forward jump from m minus 1 or one backward jump from m plus 1 that is this implies P m from m prime is half Kronecker delta m m prime plus 1 plus half Kronecker delta m m prime minus 1 Kronecker delta having the property that it is a 0 when the 2 indices are not equal it is 1 when the indices are equal. So, the probability for both the left and the right jumps are equal and so it is called symmetric random walk. This Kronecker delta guarantees that the transition occurs only to the nearest neighbor and the probability is 0 when m prime is away from m by a distance of more than 1 unit. So, with this we can go to the Markov chain equation which we remind ourselves can be written as W n plus 1 m. So, what is the meaning? W is the probability occupancy probability not the transition occupancy probability of a being at a site m at the n plus 1th step this should have happened by transitions at the nth step as we proceed along and more formally we write that transition if it is of any arbitrary length then it will be W n let us say he was at some site k and a transited p from k to m for all k's in this case theoretically minus infinity to infinity. So, if formally speaking if I had a transition probability p to m from k existing for all distances then this should be the equation to solve it is a very complex equation because it depends on all states infinity of them. So, we have simplified this our p m k into a Kronecker delta. So, it this becomes instead of the infinite number of terms it results into just two terms. So, in our case this becomes the probability half the probability that he has come to mth state from a by jumping from a m minus 1th state or with the probability half he has jumped from m plus 1th state to mth state. So, we must note that there are two indices the state index m and the step index n as far as possible we try to maintain these notations and again the step the site index or the state index I am keeping it in the argument right now although it is not necessary since it is discrete that also could be subscript. Now, let us understand this equation. So, what are the range of m values possible? So, m is there are n in let us say in n steps the maximum one side range that system can have is m maximum can only be plus n. So, in other words total length possible is either plus n or minus n. So, in other words it cannot move more than that. Hence, my transition at nth step is expected to be bounded from minus n to n this is of course, a physical information and the and this random work equation allows for the possibility of a collocation of events in which all the steps have been taken to the right and what would be probability of such an event. Since each step the probabilities half the if it has conspired to take all the n steps to one side then the probability will be 1 by 2 to the power n. It increases very rapid it decreases very rapidly as n increases, but in principle it can happen. So, it would almost look like a deterministic motion very very rare probable events and that is the beauty of a random work equation because it is gives us hint into very rare possibilities of very rare events and the probabilities for them. So, with this let us now explore how to solve this equation. So, what was the equation we have W n plus 1 m equal to half of in the nth step he was at m minus 1 and or he came from m plus 1. A smooth way of solving it is the generating function method generating function. So, we define that function g we simply call it as g as g equal to sigma W n m z to the power n this is definition over all m. So, it will have a subscript n and it will be a function of z. So, we call this as site summed generating function. This is the there are many ways one can define a generating function you can have step summed you can have site summed somewhat analogous to the distinction we make with respect to Fourier or Laplace transform. So, similarly we can make a distinction between step summed or site summed. So, right now we are talking of site summed generating function sorry this is yes site summed not step summed. So, z is of course, just a number can be a complex number its domain will depend upon the convergence and all that. So, we will leave it at the moment undetermined only thing we should remember that it should exist this operation generating function should be defined at least over a narrow strip of z some space that is sufficient. So, once we have this definition we can multiply the random walk equation here by z to the power m. We can multiply this equation throughout by z to the power m and then do the summing process. So, then we arrive at. So, we can write. So, from this equation it is going to be sigma z to the power m W n plus 1 m m equal to minus infinity to infinity and here this will be I am multiplying by z to the power n and summing over m. So, z to the power m which I can write actually as z to the power m minus 1 plus 1 and W n of m minus 1 m of course, is minus infinity to m runs from minus infinity to infinity. Similarly, the second term summed over m over its set of values again here also I have multiplied basically by z to the power m, but I will write it as z to the power m plus 1 minus 1 and this becomes m plus 1. So, I have done no changes just multiplied by z to the power m the left hand side I keep it as z to the power m and the right hand side in the first term I write it as m minus 1 plus 1 equal to m here I write m plus 1 minus 1 is equal to m. You will see the advantage because the left hand side now by definition above is going to be just g n plus 1 g n plus 1 of z. In the right hand side we can make it small change replacement define a new variable another dummy variable because m is a dummy variable it is going to be summed anyway. So, I can as well redefine m prime now wherever m minus 1 I will replace it with m prime. So, this will be m prime plus 1 now the domain of n m prime will remain the same since m varies from minus infinity to plus infinity. So, m prime also will remain minus infinity to plus infinity and this will be m prime plus 1. So, z to the power plus 1 I take out. So, this will be half z into sigma z to the power m prime w n m prime similar thing if we do here here I am going to put just same way I am going to define m prime equal to m plus 1 here which means again the domain of m prime will remain the same, but the term m plus 1 will be replaced with m prime and instead of z it will be z to the power minus 1 here. So, I will get 1 by z and the sum m prime varying from minus infinity to infinity z to the power m prime w n n prime. Each of this we can see is basically definition of g n this also will be g n. So, what is it that we are going to have can write it whole thing neatly in the next step comes to saying that g of n plus 1 the step is going to be half of z plus 1 by z g n. This is now amenable to iteration first let us say put n equal to 0. If you put n equal to 0 we get g 1 equal to half z plus 1 by z g 0 g 2 if you put n equal to 2 it will be half z plus 1 by z g 1, but that is already this. So, you will basically have 1 by 2 square z plus 1 by z whole square g 0. So, very quickly we can see that it all leads to like that if you go very general solution g n z should be half to the power n of z plus 1 by z to the power n g 0 of course function of z. So, we have solved the generating function problem only unknown now is g naught z. So, going back to the definition we will soon see that since g n z was using the definition g naught z will be w 0 m because it is always a step sum this is a side sum. So, it is a minus infinity to infinity z to the power m where w 0 is in the zeroth step before we began where the where the particle was. So, before we began let us say that the particle was at the origin. So, we just a starting. So, initial condition initially walker is at m equal to 0 side. So, which means that is w 0 m with your Kronecker delta m 0 hence g naught z will be when you sum only one term will remain corresponding to 0 and that its p factor is 1. So, g naught z is going to be 1. So, that finally, yields as a complete solution g n z equal to z plus 1 by z to the power n into half to the power n. Now, to extract back say our objective is to get how to get w n m our aim is to obtain the occupancy probabilities of state at any step. So, for this we know that w n m is a coefficient of z to the power m in the g expansion right. So, w n m is the coefficient of z to the power m in the defining equation for g. So, we must then compare the expansion for g n z with the definition. If you look at this result that we obtained we can write it as n z equal to 1 by 2 to the power n z plus 1 by z to the power n and we consider this is a binomial expansion basically a two terms a plus b to the power n type and we know the binomial expansion this is 2 to the power n its going to be some let us say arbitrary term r varying from 0 to n maximum n terms will be there it will be n c r combinatorial term and z to the power r a to the power r and b to the power n minus r. So, 1 by z to the power n minus r since both a and b are related to z we can combine them here and rewrite it as 1 by 2 to the power n 0 to n n c r and z to the power r and this is z to the power minus n plus r. So, it is going to be z to the power 2 r minus n. So, let us now just replace 2 r minus m n with another number m both all are integers. That means, this implies that my r which always should be an integer can be written as m plus n by 2. Now, r of course, has to be maintained as an integer, but there is a factor 2 dividing. So, obviously, it means that there must be some symmetry relationship between n and m minimum. That is that if n is even that is if the steps are even then the locations where the particle will be found also will be should be at the even locations and if the steps are odd then the locations should be odd. So, basically it is intuitively consistent that when you take the first step the particle will be at the first site it and the second step it will be at the second site and only in it will be in the even sites only when it takes even number of steps. So, that fact constraint is maintained in this. So, here we should remember that my r has to be integers. So, that is the way m should operate. Besides we can see that since my r domain when r equal to 0 my lower bound of m will be minus n. When r equal to plus n the upper bound of that binomial term then m will be plus n. So, my m index therefore, will run from plus n to minus 1. So, which basically means my m is a set which runs from n like this to So, going back to the previous equation here the 2 to the power n are going from 0 to n n c r z to the power of 2 r minus n we replace r with the m now and write it as z to the power m. So, that then becomes we can write g n z as g n z will be 1 by 2 to the power n will remain as such my sum now will be over m variable varying from minus n to n and it will be n c n plus m by 2 z to the power m. This lends itself for comparison with the definition of the g by which we identify this whole thing of course, 2 to the power n together being the coefficient of the occupancy probability W n m will therefore, be 1 by 2 to the power n n c n plus m by 2 which you can write in more explicit form as 2 to the power n n factorial into r fact this is r factorial which means n plus m by 2 factorial multiplied by n minus r if you do it will just come out to be n minus m by 2 factorial and we have already noted that n and m have a symmetry relationship if n is even m is even and if n is odd m is odd. So, if n is odd then m is odd and y c versa. So, this come brings us to the end of derivation of the expression for the occupancy probability for the problem of symmetric random walk. We have to examine the properties of this relationship and that we will do in the next lecture. Thank you.