 So, far we ah considered various cases of random walk on an infinite lattice. The lattice points were extending from all the way minus infinity to plus infinity. So, in principle the random walker could access any lattice point. This problem of course, we focused on one dimension to illustrate ah the basic features. Although the problem acquires much greater complexity as we increase the number of dimensions associated. So, we will go to that a little later. Right now we will continue our excursion in one dimensional random walk, but however progressively ah more ah more and more complex or more and more varied problems. To bring home a fact that random walk many a time is not just about walk it had yielded many other perspectives in fundamental science of physics itself. For example, the idea that the probabilities the occupancy probability that sites are a result of various paths contributions from various paths and this perspective or looking at that way had a lot of impact in fundamental ways of interpreting quantum systems as well as as a tool for solving differential equations. The the concept that the net effect is a superposition of effects from various paths suitably weighted paths are weights and that is exactly what we have been doing. The occupancy probabilities are a result of various paths. So, in an n step random walk symmetric random walk we had 2 to the power n paths and the net contribution was what we were calling as occupancy probability. Now, we take up ah different problem on in one dimension, but related to semi infinite lattice. The particle is not free to move to let us say to other negative axis. This is very ah special or relevant problem called the restricted random walk problems restricted domain. So, random walk in let us say restricted or semi infinite in this case that is I have my domain is let us say r plus real line, but lattice points we again discuss lattice points to etcetera etcetera and the walker now is not allowed to move back. So, his entire motion is limited on this itself. That means, if a walker lands at the point 0 which is allowed we have to specify something what will happen to that. So, we can disallow transition to the left, but then we must say something about transition either to the right or onto itself. So, these are discussed under 2 different cases. One is case of reflecting barrier this is let us say case 1 which is a semi infinite domain problem. So, it is also a restricted domain problem. So, what is meant by a reflecting barrier? So, reflecting barrier is we as earlier let us talk of only ah symmetric random walk which means p equal to q. However, this is true everywhere equal to half everywhere for m greater than or equal to 1. However, p becomes 1 and therefore, q will be 0 for m equal to 0. So, what it means is whenever a particle lands at side 0 at let us say nth step it lands at side 0 at n plus 1th step it lands back at site 1 from which it came with the probability forward probability 1. This will now affect significantly the occupancy probability at some site starting from some other site. We now consider a specific case a specific case in which we have particle starting from some site m equal to m naught. And we are interested in finding the occupancy probability at some site m and 0 is where we have a barrier this line is the artificial there is actually in a one dimension there is no barrier it is just a site where reflection takes place. You can imagine a random walker meeting a wall on the path from where he turns back the left side is not at all accessible now for the walker. So, the probable the question we ask is so, what is the probability that the walker will be at site m after n steps subject to the constraint that he will be reflected at site 0. Here we note the fact that the contribution to site m for example, if we consider there is a barrier at site let us say this is the barrier problem I now show it as this. So, this is the first site and he has started from some m naught and we are interested in some site m where we want the occupancy probability whether there was a barrier or not. So, there will be many paths he would have executed to reach m which did not come through the barrier. So, the. So, these paths will be similar to the probabilities without the barrier. However, the total contribution now would also involve some paths which would have crossed the barrier at some step we can consider now this as a step actually. Whereas, if it was a free case by free we let us let us specify our nomenclature the bound problem and a free problem or a barrier problem and a free. Free is the case in which was allowed to move over the entire real entire lattice without the barrier. So, whenever we is free we refer to that case. So, if it was a free motion some trajectories which would have crossed the barrier would also have come back and made a contribution here along this say dotted line and that is how the total probability arises many paths like this. However, because of the barrier the path that touches the barrier in the next step it is reflected back here. In other words at this point 50 percent would have otherwise penetrated to the non accessible region in the free case, but in the bound or barrier case that path is reflected back. As the result of which you would expect now that probability of the finding at m would be increased because of reflection some additional contributions would come because of the barrier effect because of the reflection effect. You can imagine that easily because earlier 50 percent would have gone now with probability 1 he has returned it closer to the point itself. In order to account for those extra 50 percent term we can now imagine a virtual point at minus m naught. So, if m naught is let us say 10 units let us imagine a walker starting from minus 10 on the other side of the barrier obviously, because my barrier is at 0 this is the barrier this is the starting point at n equal to 0 this is the point of interest or site of interest where we wish to calculate the contributions probability occupancy probability. So, we imagine now a virtual walker we call him it as a virtual because you not a real walker or real walker has started from m naught, but a walker who starts from minus m naught is all the trajectories to the left of the mirror left of the reflector are completely identical to the trajectories of m naught to the right of the mirror. So, in so far those trajectories are concerned there is a complete symmetry. However, the trajectories free trajectories of the virtual walker we call him as a virtual the free trajectories of the virtual walker reaching the point m will not be of course, symmetrical with respect to the free trajectories of the real walker reaching the same point m. However, at the point where the free trajectories of walker from m naught has touched the line the barrier point and if up to that point there is a complete symmetry between all the trajectories of the left and right. So, at this point the free walker has a 50 percent probability of tracing this path which the walker from m naught had got it by reflection. So, the 50 percent extra that the m naught has gained is nothing, but the 50 percent coming from the virtual walker undergoing free random walk in the unbounded domain. So, since once reaching this point subsequently the random walk is completely dictated by that point. So, all the paths that the minus m naught walker would have created by crossing the barrier point are the same as the paths would have come from m naught. In other words the probability of finding the walker with the barrier is the sum of probabilities of walkers the both from m naught and minus m naught contributing to the point m in the absence of barrier. So, the free random walks of the real walker and the virtual walkers contributing to m is the same as the contribution of m naught under the reflecting barrier. So, we write W m under reflection of a walker starting m from m naught under reflection at m equal to 0 is the same as W n without reflection I call it as free of a walker starting being at m starting from m naught plus the probability of a gain a free walker being at m, but starting from minus m naught. So, this in a sense mathematically shows all the points that I was making about the symmetry as well as supplementing for the extra path that is introduced because of reflection from the virtual walker. At this point we note that W n free m m naught should be equal to W n free m minus m naught. So, earlier we had start our derivation was for a walker starting from the origin. So, because of translational symmetry of the free lattice random walk the solution for a walker from any point is nothing, but that obtained by shifting by a magnitude of m minus m naught that is all. So, it is very simple wherever m is there you must replace it with the m minus m naught. So, with that we have the expression we can write down to be specific W n free of a random walker being at m starting with the m naught that will be 1 by we have seen it is n to the n factorial divided by 2 to the power n divided by 1 by earlier it was n plus m now it is m minus m naught just a translational invariance has to be observed. So, divided by 2 whole of it factorial and here it is because n minus m. So, it will be n of m minus m naught. So, n minus m plus m naught it will be divided by 2 similarly W n free of walker starting from minus m naught is going to be just the same with m naught replaced by minus m naught. So, it will be n plus m plus m naught divided by 2 factorial multiplied by n minus m minus m naught divided by 2 factorial these expressions simply follows by replacing m naught with minus m minus m naught. So, we actually have now the final expression by just adding these two. So, we for completeness sake we write it W n with the reflection m starting with let us denote m starting from m naught is going to be n factorial by 2 to the power n into 1 by n plus m minus m naught divided by 2 factorial into n minus m plus m naught by 2 factorial plus 1 divided by n plus m plus m naught by 2 factorial n minus m minus m naught by 2 factorial. So, an expression which can be written down from symmetrical symmetric considerations. The effect of reflection sort of can be demonstrated like this. So, if I had our barrier if we had our barrier here now and if y axis is now this is the point m equal to 0 barrier and let us say our random walker started at some m naught and after some steps if you look at how he is distributed. So, if this is probability W n the entire left side is not accessible. So, the walker is now entire probability is only on the right side whereas, if it was a free in the former case you would have leaked like this whereas, presently it would be a distribution like this. So, this dashed line is without for free this is free walker distribution and this is for reflecting case. So, the his probabilities are enhanced at the reflection site. So, that you will see the effect near the site maximum as opposed to far from the reflector site. The 2 will merge systematically 2 will merge as m goes to infinity. The difference here this area should be exactly the leak area that has occurred in the free case in order to conserve the total probability. Total probability will be always conserved in the domain 0 to infinity. One can go a step further carrying forward some of the results that we had derived earlier. We had shown that in the limit of n tending to infinity and also m tending to infinity such that m square by n is fixed the free random walk distribution approaches a Gaussian. Upon converting probability masses to probability densities by treating the x axis as a continuous variable that is by assuming a jump length l and writing x equal to n l and tau equal to I mean n t equal to n tau and defining a diffusion coefficient d equal to l square by 2 tau. Under these assumptions these were done in the lectures dealing with the unbiased random walk. We have shown that free distributions will become a Gaussian and will have the functional form given by W then W becomes now probability density function instead of m it is going to be x instead of n now let us say it is a time variable and the particle is starting from x naught instead of m naught it will be x naught this will have the form 1 by square root of 4 pi d t e to the power minus x translation in invariant allows us to replace x with x minus x naught divided by 4 d t. So, this is free upon using this approximations to the expression we derived here where all these factorials now will disappear and we will have a Gaussian expression. The probability density takes the form W x t x naught with reflecting case here 1 by 4 pi d t will be a factor and this will be e to the power minus x minus x naught whole square by 4 d t plus it will be plus here minus x minus x naught whole square again by x plus x naught by 4 d t always let us remember it was virtual walker from minus m naught who made the contribution. So, minus x minus of minus m naught is x plus m naught and in the continuous case it becomes x plus x naught. An interesting corollary is when we differentiate these expression that is if we differentiate the expression differentiate W reflection x t x naught with respect to x it is actually a hint site that the derivative of the probability density is going to be equivalent of a probability flux or a probability current. So, to get an idea of that we anticipate this result we just differentiate now we do a mathematical operation. So, when you differentiate it is going to be d W by d x and this term will remain 1 by 4 pi d t will remain and here it is going to be minus of 2 x minus x naught the first term minus of 2 x minus x naught divided by 4 d t e to the power minus x minus x naught whole square by 4 d t. The second term is going to be plus, but I am going to have another minus. So, I will write it as minus of it will be x plus x naught twice divided by 4 d t and again e to the power x plus x naught whole square by 4 d t. Now, let us put x equal to 0 if you put x equal to 0 the exponential terms now will be both of them will be the same term e to the power minus x naught square by 4 d t. So, that factor can be taken out. So, I will take out e to the power minus x naught square by 4 d t. So, let us say that is set let x equal to be 0 we will rewrite this let x equal to 0 then we get d W by d x at x equal to 0 that is the gradient of the probability density near the reflector. So, r is going to be e to the power minus x naught square by 4 d t divided by 4 pi d t and these are 2 by 4 d t also is common. So, I will take out 2 by 4 d t and what remains here when x equal to 0 is plus x naught here since x equal to 0 and what remains here is minus x naught which is 0. So, we have a very interesting corollary that comes from the reflector case when we extended the random walk problem to a random continuous random walk problem by continuous variable that for reflection x equal to 0 the boundary condition d W by d x x equal to 0 become 0. This is a deep and very useful corollary which one utilizes as a mathematical boundary condition for solving transport equations for involving random random motion in the continuum variables. It cannot be derived otherwise these boundary conditions have to be assigned a priori based on logic. The so called Newman boundary condition follows or at least can be argued out from this process that we have carried out through first formulating a discrete random walk process arguing it out why there must be a contribution addition contribution from a virtual walker and then moving over to continuous variables and then differentiating and arriving at this. We continue with this restricted random walk problem to a complementary case where the boundary is not reflecting, but it is going to be an absorber and even more interesting results emerge in that case and we discuss it in our coming lectures. Thank you.