 So, armed with this Fourier transform technique, let us proceed to handle the problem of random walk in 2 dimensions or more specifically in a 2 dimensional lattice random walk in 2D. So, let us consider a lattice a rectangular lattice or a square lattice that is we have transitions in along the x directions and transitions along the y directions as well. So, the rule specifies that there is supposed to be equally spaced. So, let us consider the grid like this it goes from x equal to minus infinity to plus infinity. Let us consider the origin here and consider a general lattice point. So, since we are going to deal with integers and we are going to generalize to higher dimensions also, let us specify the lattice points through integers m 1 m 2 etcetera. So, when I say this point is having is described by 2 numbers m 1 and m 2 it is equivalent to x and y. So, if it is 3 dimensions we would say m 1 m 2 m 3 where m 1 comma m 2 they are all integers 0 plus minus 1 plus minus 2 etcetera. So, the random walker in this 2 dimensional problem through a certain process of deciding randomly either takes a left jump along the x axis or a right jump along the x axis or a upward jump along the y axis or a downward jump along the y axis. And the choice at any instant could be any one of these 4 options and we assume that each of them is equi probable that is a symmetric totally symmetric random walk. Hence, the transition probability is still a nearest neighbor transition probability that is here you can jump either from m 1 to m 1 plus 1 or you could come from m 1 minus 1 to m 1. Similarly, you could transit from m 1 m 2 to m 2 minus 1 or transit to m 2 to m 2 plus 1. So, these are the 4 directions each point has and corresponding transition probabilities can be assigned as if they it is an equi probable problem as one forth. We can then write the most general equation for this problem will in terms of the occupancy probabilities now which is a 2 dimensional quantity that is w at the n plus 1th step the probability of finding it at m 1 m 2 any of this it should have come from transitions with one fourth probability either from the left that is w n m 1 minus 1 m 2 or with one fourth probability from the right the backward transition w n m 1 plus 1 m 2 or with one fourth probability from below w n m 1, but m 2 minus 1 or with the one fourth probability from above w n m 1 m 2 plus n 1. So, it is now having 4 terms and not 2 terms and if you want to solve it we have to therefore, consider the fact that there are 2 variables. So, reverting back to our Fourier transform method we must have a 2 dimensional Fourier transform now that is we will define the most general characteristic function in terms of 2 variables now k 1 and k 2 as summed over m 1 minus infinity to infinity or m 2 minus infinity to infinity it will be e to the power i k 1 m 1 plus i k 2 m 2 w n m 1 m 2. So, this is definitional you can call it a definition. So, defining a 2 dimensional Fourier transform like this we can now apply to the equation above that is multiply each of these terms first term here for example, with this e to the power i k 1 m 1 plus i k 2 m 2. So, here we you will have it is a double sum everywhere sum over m 2 and sum over m 1. So, maybe we can do it as one exercise. So, the LHS by because it is going to be just the definition. So, the LHS is simply going to be w ok. So, upon applying the transform to the equation we will get w n carrot w n plus for carrot of the 2 variables here. So, if you consider now the first term here there is an m 1 minus 1 that is the only difference otherwise it would m 2 is just the same. So, when we multiply by e to the power i k 1 m 1 this term is going to be summed along with this term. So, if I make a transformation m 1 minus 1 as m 1 prime as a new integer. So, it will be then e to the power i k 1 m 1 prime plus 1 that is e to the power i k 1 plus e to the power i k 1 m 1 prime, but the argument here now will be m 1 prime alone and that m 1 prime will be summed from minus infinity to infinity thereby fully satisfying the definition. In other words this term is going to be there will be 1 4th e to the power i k 1 and the rest is going to be just w n carrot k 1 k 2 follow the same logic here for the second term 1 4th will remain. Now, when we are summing with this what matters here is again this term alone because m 2 there is no change, but m 1 plus 1 will be replaced with the m 1 prime. So, m 1 m 1 will be m 1 prime minus 1. So, e to the power minus i k 1 will hang out. So, that will be e to the power minus i k 1 will hang out and again it will be k 1 k 2. We can see that the terms along the y direction will now be shifted in the k 2 variable precisely the same way as we did for the x direction. Hence the other two terms will be e to the power i k 2 w n k 1 k 2 plus this will be 1 4th e to the power minus i k 2 w n tilde k 1 k 2. So, basically we can see in this equation that w n tilde k 1 k 2 is a common fact. So, we can write more transparently the whole equation that is we are going to have w n plus 1 carrot in the new variable is going to be 1 4th of we will have e to the power i k 1 plus e to the power minus i k 1 plus e to the power i k 2 plus e to the power minus i k 2 of w n carrot k 1 k 2. We can simplify write in a simplified manner by identifying this to be 2 cos k 1 and similarly this will be 2 cos k 2. So, it is going to be 2 by 4 or half of cos k 1 plus cos k 2 w n carrot k 1 k 2. If we now iterate it by putting n equal to 1, 2, 3 etcetera we will it will just raise this factor cos k 1 plus cos k 2 divided by 2 to the power n and there will be a term and if the random walker starts at the origin when m 1 and m 2 are both 0 then this transform will also result in just one number which is unity. These methods are exactly same as we did for 1 dimension and hence in general hence after iterating we will iterating n equal to 0 may be 1, 2 etcetera and making use of the property or the initial condition that is bracket we will write this properties initial condition that is w 0 m 1 m 2 will be delta m 1 0 delta m 2 0. So, it is localized at the origin then we get we obtain w n carrot k 1 k 2 k 1 k 2 will be half of cos k 1 plus cos k 2 to the power n. It will just raise to the power n as it happened earlier. Well I will correct it it will be divided by 2 it will have this form. So, the form is cos k 1 plus cos k 2 divided by 2 to the power n and we must give some thought why this 2 came here actually because it was 1 4th here. 2 dimensions have 2 directions and 2 2 axis. So, total you have 4 options 1 4th 1 4th each and cos k 1 only demands one of the only half out of 1 4th. So, you get 1 by 2 remains here. So, that is why that 1 by 2 continues to remain. So, we have this it is interesting whether we can obtain actual solution for the random work problem even formally in 2 dimensions. It turns out that it can be done it may be more tedious, but I will I will just show you how that can be done. We go back and see that. So, in 2 dimensions 2D the solution is in the transform domain at least we have solved it fully w n tilde k 1 k 2 is cos k 1 plus cos k 2 to the power n divided by 2 to the power n. Now if you want it is inversion we have to use the 2 dimensional inversion integral that is in 2 dimensions w n let us say m comma l is going to be 1 by 2 pi minus pi to pi. In fact, it will be 1 by 2 pi square to be precise and again minus pi to pi e to the power minus i k 1 m minus i k 2 l of its transform w n karat k 1 k 2 is integrated over d k 1 d k 2. So, this is the way we define Fourier inversion. It is now a 2 dimensional integral each of them is 1 by 2 pi. So, it will be 1 by 2 pi square. Now if we substitute this expression we will have and 2 to the power n we can take out it will be 1 by 2 pi square 1 by 2 to the power n minus pi to pi again minus pi to pi e to the power minus i k 1 m minus i k 2 l of cos k 1 plus cos k 2 to the power n d k 1 d k 2. We further expand this cos k 1 and cos k 2 now in terms of binomial expansion. So, it will turn out to be 1 by 2 pi square 1 by 2 to the power n and if you do binomial expansion and take the both the integrals inside it will have the form integral say let us say some r equal to 0 to n it will be then we write the 2 sums 2 integrals minus pi to pi minus pi to pi keep this e to the power i k 1 m minus i k 2 l. Now if you see it will be n c r of course and cos k 1 to the power r and cos k 2 to the power n minus r d k n d k 1 d k 2. It is quite easy that all that we did here was to write this sum this power function cos k 1 plus cos k 2 to the power n as sum over r 0 to n n c r cos k 1 to the power r cos k 2 to the power n minus r. We have seen now in the previous derivation that we know the key result we have shown that that is 1 by 2 pi minus pi to pi e to the power minus i k 1 m when we dealt with the one dimensions cos k 1 to the power any say general power say any general power say n d k 1 was see here the there is an m index and an n index. So, it was 1 by 2 to the power n n c n plus m by 2 and let us always remember the restriction that n plus m here should be even because this whole thing should have to be an integer. So, that restriction we will always keep in mind and we will not state it every time, but for once we will write n plus m e 1 integer otherwise it will be 0. So, if we now go back we have a series each of them for example, in this cos k 1 to the power r this will multiply with e to the power minus i k 1 m and integrate and you are going to have r c 1 with the subscript r plus m by 2. So, likewise we can show that this whole integral will come to the solution will come w n m l will be 1 by 2 to the power n and it will sum over r equal to 0 to n and we had our n c r and it will it will then be 1 by 2 to the power r r c of r plus m by 2. This formula that we are applying n has to be replaced with r this formula n has to be replaced with r for the first term and the second term will also turn out to be 1 by 2 to the power n minus r and n minus r c n minus r plus l by 2. This is the almost it follows axiomatically from the way we have derived the various relationships, but we should remember that our m and l should be such that or our r value should be such that r plus m is also a even integer and n plus l minus r is also an even integer. So, it is it is kind of even on both axis. So, this actually now can be simplified and it can be written as we can give it an explicit form for example, as n factorial divided by 4 to the power n r equal to 0 to n of 1 by r plus m by 2 factorial r minus m by 2 factorial 1 by n plus l minus r by 2 factorial and n minus l minus r by 2 factorial. We have actually formally solved this problem. We have to basically sum a series now subject to the constraint that each of the factorial terms has to be an integer. What we should note here is in one dimension we are just one term which was equivalent of an ncr. Now, we have a series a kind of a progression that is not a series, but a n terms as n increases the number of terms increases, but one has to sum over this all the terms in order to obtain the solution. But the point that we should understand here is the method applies and a close form solution for 2 dimensional random walk also can be obtained to find the occupancy probabilities. So, armed with this we reworked back to the fact that the problem was very neatly solved in the conjugate space at least. So, in the conjugate space the solution that we obtained was cos k 1 plus cos k 2 by 2 to the power n. So, we can now we can now move forward to higher dimensions this is in 2D. So, higher dimensions when we go to higher dimensions let us say we have 3 dimensions. If you have 3 dimensions we would have basically 6 terms each with the probability of 1 by 6. So, we would have had for example, we would then have 3 variables W n plus 1 3 integers m 1 m 2 m 3 which would carry 1 6th of probability W n say m 1 minus 1 m 2 m 3 plus 1 6th W at the nth step of course, m 1 plus 1 m 2 m 3. So, this kind of shift processes like this there will be 3 terms plus plus and the second one will be a shift in the m 2 and the third one will be. So, here it will be m 2 minus 1 just as an indicator plus 1 6th this will be m 2 plus 1 m 3 likewise. So, if we now do the same method take a 3 dimensional Fourier transform by the construction that is W n carrot now we will have 3 variables k 1 k 2 k 3 it will be triple series e to the power i k 1 m 1 plus i k 2 m 2 plus i k 3 m 3 of W n m 1 m 2 m 3. So, likewise it will take the solution compact solution for the characteristic function we obtain using a very similar method or by following the method similar to the 2D problem we will get W n carrot k 1 k 2 k 3 will be here now we will have cos k 1 plus cos k 2 plus cos k 3 divided by 3 whole to the power n. The 3 will remain because of the 1 6th half will be absorbed in cos k and one third will remain. So, in general we can now visualize that in general in D dimensions it is a hyper lattice we will have W n considering all the D dimensions it will have D variables k 1 k 2 k 3 they are all conjugate to m 1 m 2 m D and the solution therefore, can be written as sigma say i equal to 1 to D cos k i divided by D if it is D dimension whole to the power n. So, formally we have solved the problem of at least obtaining a conjugate function or the characteristic function of a random walk on a hyper cubic or hyper lattice in D dimensions and D can be 1 2 3 etcetera and we can easily verify that it agrees with the derivations we have done for 1 2 3 1 and 2 dimensions earlier. At this point we take a pos the pos is ok why are we driving ourselves to higher order dimensional problems. Let me introduce that the concept of what is called as a return to the origin problem return to the origin basically is a kind of variant of the absorber problem that we did for 1 dimension. So, if a particle what is the probability that will be absorbed eventually similarly if a particle starts from the origin what is the probability that he will revisit the same place given infinite time is there a chance that a random worker will not visit the same point in some dimension at all only 1 visit occurs I mean only he starts from the point and 0 further visit occurs. It implies that there is a tendency to escape the point from where a worker started it is a gain deep implications in random worker diffusion based reactions etcetera. So, to understand how that probability depends on the dimensionality of the system even 1 2 and 3 dimensions are sufficiently diverse, but a hyper lattice gives you more general understanding we proceed to that problem in the next lecture. Thank you.