 Thus far, we explored the several aspects of one-dimensional random walk, free random walk, random walk with one barrier, random walk with two barriers. The latter we called as equivalent of a gambler's ruin problem. Let us now move over to the problem of random walk in higher dimensions. When we speak of random walk in higher dimensions, the way we define a random walk with the so called nearest neighbor or a constant length random walk itself has to be redefined. For example, one way of a random walk in two dimensions is so called Pearson's flight, Pearson's random flight in two dimension. We mentioned about it as one of the founding formulations of random walk, paper published in 1905 which is formulated as follows. Consider a random walker who takes always a constant step length of A. So, A is a length constant step length, this is does not change. Now at any point in a two dimensional plane, he starts let us say he is at the moment at some point x y x i y i. So, let us say x n y n, then he takes a step of length A. However, the direction he takes let us define directions with respect to x axis and this is y axis. So, the direction he takes is some angle theta let us say n in the nth step. Now even though he takes a constant step length, the randomness is in the directions he takes at each step. So, the for example, if theta n could be distributed uniformly over the entire angle 0 to 2 pi. So, p theta n the probability of obtaining an angle between theta n and theta n plus d theta n could be just 1 by 2 pi which means every time it takes a random step it includes a constant length A, but a random direction and hence his x plus 1th coordinate if you resolve the motion that is going to be the coordinate at the nth step plus A cos theta n. Similarly, the y value at the n plus 1th will be y n plus A sin theta n. Here we have also defined that theta n is measured from the x axis where theta n is a random variable belonging to 0 to 2 pi can take any value. So, when he proceeds like this the values of x n and y n have the same quality as the random walkers in our one-dimensional random walk. However, the values are now not a fixed numbers, but it can vary anything by a length A between 0 to A. So, this problem propagates into finally, into a distribution let us say what is the probability that the random walker at some position x n and y n at the nth step. So, 1 seeks an answer to this obtaining this distribution. In fact, it so happens that ultimately the radial symmetry has to be maintained and this problem can be mapped in an equivalent manner to seeking a question on what is the probability that the walker is at a radial distance r from the starting point. It has a closed solution somewhat complicated, but it has a closed solution and we will discuss at a later time. But basically the point here is that this is all this is a type of random walk in which the step length is radially fixed, but at any direction x or y direction it is a quantity which varies as the random walker proceeds. There is another way this is of course, method 1. There is another method by which I can execute a 2 dimensional random walk. That is it is almost an extension of the 1 d random walk, but constructing a 2 d lattice. For example, take a 2 d lattice we have let us say x and 1 method 2 of 2 d r w. Of course, we are discussing symmetric random walk. So, in that case let us say now the coordinates are fixed by coordinates are fixed by the Cartesian system. So, let us say he is at a position x n y n. Now both x and y are basically multiples of integers. So, they are just integer values now. Now we give him the equal probability of transition to the left or to the right to upwards or downwards. Upwards and downwards in the sense as we see the graph or as we see the map. So, and we the jump lengths are now constant along each direction. So, here rather than ask the question of radial constancy we maintain constancy with respect to each of the Cartesian directions. Thus if we say that the step lengths are unity then there is a probability that he will at any instant at any step the probability that he will jump to the right or to the left or upwards or downwards they are all equal in symmetric random walk. So, in symmetric random walk on a lattice the step length is constant along each axis is fixed along each axis and the probability of nearest neighbor jump nearest neighbor transitions transitions will be one fourth one fourth or one fourth on each direction towards each direction. There are four directions and hence there will be total it will be one. So, these problems we have to solve the the dimensionality as the dimensionality increases the number of terms in the random walk equation also increases and that calls for more and more sophisticated tools to solve them. If you now extrapolate to 3 dimensions we will see that in 3 dimensions there are 3 axis and on each axis there are 2 directions and you will have equi probability on each of these directions and hence the probability will be one sixth for each direction. So, likewise now we can extrapolate to higher than 3 dimensions for certain theoretical purposes. Before we actually go to 2 dimensions we can revisit 1 dimension because we are going to apply a Fourier series technique to solve these problems rather than the generating function techniques which we adopted generously in the 1 dimensional cases. So, accordingly we revisit the 1D problem gain confidence that it can be also solved with the Fourier series technique and then move ahead to higher than 1 dimensions. So, revisiting 1D use of Fourier series or transform technique. Let us go back to our lattice 1D lattice problem. So, the walker has started from 0 and it is a desire to seek the occupancy probability at some site m and the transitions occur between the nearest neighbors and we have by now familiarized ourselves with the rules of transitions. So, accordingly we wrote for symmetric random walk unless specifically mentioned we will discuss only symmetric random walk 1D. Our equation is in the n plus 1th step the probability of being at m comes from transitions with the probabilities half each from m minus 1 and m plus 1 at the nth step. So, we had written it down. Now let us carefully develop a Fourier series method of solving it. Of course, we had an initial condition at let us say 0 m probability of finding the random walker was delta m 0 it was at the origin. So, Kronecker delta is the value. So, let us define the characteristic function or Fourier transform or Fourier coefficient. Let us call it Fourier transform, but via a sum formulation. Let k be the conjugate variable then by definition w let us say n in 1 dimensions let us say k be the conjugate variable. Then the definition would be a series since m is integer actually Fourier's integral becomes a series that is the basic logic e to the power i k x becomes actually i k m and w n. So, this is going to be the definition of the Fourier coefficient or Fourier transformed quantity. So, this carrot denotes that it is in the k space. We multiply both sides by e to the power i k m just like generating function. Then m minus 1 will lead to 1 term e to the power i k when we put m minus 1 as m prime here those techniques we have used. So, it easily will turn out to be easily we can see that this will lead to a solution w n carrot k will be say w n plus 1 from the left hand side. And the right hand side we will be left with half is the probability which anyway pops out and this will be e to the power i k will remain in the first and e to the power minus i k will remain in the second term w n carrot k for all n. So, if we now iterate then we will get w n k is going to be cos k to the power n let us say w naught k w naught k. It just raises itself to the power n where we have identified cos k by definition is e to the power i k plus e to the power minus i k by 2. So, now w 0 k by definition we have to strictly go by the rules then it is by definition e to the power i k m e to the power i k m and w 0 at the value m and it is summed m is summed this is a definition. And w n 0 m is a chronicle delta. So, this whole thing is nothing, but delta m 0 which means the sum exists only for m equal to 0 at m equal to 0 e to the power 0 it is 1. So, the answer is 1. Hence the solution is w n carrot k cos k to the power n. Now, we have to invert it. So, for inversion we need actually in definition of inversion for the Fourier series. First we write the inversion formula and then we can see it how it is arrived at. So, in general any function w at an integer value let us say j that is nothing, but the Fourier inversion in a finite domain from minus pi to pi of its Fourier series well if you can keep the subscript n w n at k, but e to the power minus i k j d k. The exactly like the Fourier inversion, but it is now not minus infinity to infinity, but minus pi to pi because my j values occur on regular lattices. So, they are integer values. So, how to prove this? Let us consider the right hand side. The right hand side we will now consider the series it is 1 by 2 pi minus pi to pi e to the power minus i k j d k and what is w n k by definition it was e to the power i sum m k w n m. This is the way we define m varying from minus infinity to infinity. Since the integration is over k value we can take it inside the sum. So, it is going to be sum m equal to minus infinity to infinity and w n m does not carry any k parameter. So, it can be taken out w n m and this integral is now 1 by 2 pi e to the minus pi to pi. Now, we combine e to the power terms together it is i k m minus j difference between 2 integers d k. We can easily note that if m equal to j the integral 1 by 2 pi minus pi to pi e to the power i k into 0 d k which is e to the power i k 0 is 1. So, it is simply d k and 2 pi which is 2 pi by 2 pi equal to 1. So, if m equal to j the integral is unity. However, if m not equal to j that particular integral 1 by 2 pi minus pi to pi e to the power i k say m minus j d k it can be written as 1 by 2 pi into m minus j. Since since it is not 0 we can write denominator it is just an exponential integral. So, it will be i into 1 m minus j and it will have e to the power i k m minus j and value of k has to be put first 2 pi first pi and second term is minus pi m minus j. And basically e to the power i pi function each of them they are exactly identical whether it is plus i pi into some integer or minus i pi into some integer. So, this function is 0 if m not equal to j. So, to be to summarize we have that important result that 1 by 2 pi minus pi to pi e to the power i k m minus n integer it took j here m minus j d k we can write it as delta m j it is 0 if m not equal to j, but unity if m equal to j. So, with this property we can now go back we can see now this equation say this this equation 1 by 2 pi minus pi to pi this integral will be 0 for all values accepting when m equal to j and m is running through all the integers. So, somewhere it will meet j quantity that is required. So, the RHS will be simply W n j. Hence RHS will be we will go back and the RHS will be W n j which is equal to LHS that is how we started if you see we started with the left hand side as W n j we wanted to prove that it is true. So, that proof is now attained because it comes out to be this same as LHS. So, this proves that our method of inversion by using a finite domain integral is quite valid. So, if we now go back to our identity W n k cos k to the power n W n k cos k to the power n we have to we can use therefore, the formula. Now returning back to the result W n tilde k equal to cos k to the power n the occupancy probabilities can be obtained the inversion W n tilde k that is we have W n at some lattice point let us say L all the time let us not think of m, but L is also as good. So, that is going to be 1 by 2 pi by definition minus pi to pi e to the power minus i k L it will be and then cos k to the power n because it is the nth step and lth location. So, that is the way it is where L 0 plus minus 1 plus minus 2 any integer values and of course, n is also steps. So, but it is only one sided. So, we can expand cos k to the power n we revert back to the definition and write it in the following fashion that is W n. So, that is W n L will now be 1 by 2 pi minus pi to pi e to the power minus i k L again we write cos k as e to the power i k plus e to the power minus i k divided by 2 whole to the power n. Insert the definition back we can take out 2. So, then it becomes 2 pi minus pi to pi e to the power minus i k L and 2 to the power n. So, we take out that 1 by 2 to the power n it can be outside the integral or it can be even inside, but it is outside the bracket. So, here now it takes the form e to the power i k plus e to the power minus i k to the power n d k everywhere of course, we should we should not forget to put the variable over which we are integrating. Now we use the binomial expansion that is e to the power i k plus e to the power minus i k to the power n is going to be sigma some r equal to 0 to n binomial. So, n c r first term to the power r e to the power i k r and the second term to the power n minus r. So, it is minus i k n plus i k r which can be written as r equal to 0 to n n c r e to the power i k 2 r minus n. So, upon inserting this into the integral here that is we now insert this sum to the integral here. So, then we obtain W n L and go back. So, 1 by 2 pi and 2 to the power n we can take out 2 to the power n and 1 by 2 pi then we have e to the power minus i k L integral minus pi to pi we combine all the e to the power terms. So, e to the power i k 2 r minus n minus l. So, we are then left with e to the power i k 2 r minus n minus l. Now we use the orthogonal theorem and of course, we should remember that this is summed over all r. So, we should insert the sum here. So, we will r equal to 0 to n. So, when you sum this will exist only when r equal to n plus l by 2 or it will be 0. So, it has the property that this is delta of 2 r n plus l or r when becomes n plus l by 2. So, if we use that chronicle delta property this just all the terms disappear accepting the term corresponding to n plus l by 2 and of course, we have also missed here n c r term. So, we had n c r. So, the answer is going to be n c n plus l by 2 subject to the condition that n plus l is always even because it has to be an integer r has to be an integer. So, n plus l has to be always even. So, this was this is completely consistent with the result obtained with the generating functions obtained by the generating function method. So, the purpose of these exercise is to demonstrate the usefulness of the Fourier series method which we will apply as we move over to higher dimensions. Thank you.