 So, we argued in the previous class that the integration over the variables K 1 and K 2 can be done as an integration over a circle or within a circle and that because of the symmetry can be done more easily with polar coordinates. So, let us learn this trick it is very useful very often it is used in physics. So, we have now a circle we forget the square now we are working on a circle and this was our K 1 axis and this is our K 2 axis and we have to do d K 1 d K 2 is a small element of area and integrate the function over all the points on the circle that is what we are doing we are actually integrating a function 1 by 1 minus half of cos K 1 plus cos K 2 is being evaluated at every point right that is what integration is about. Now, instead of going in the cartesian coordinates we decide to go by polar coordinates that means I am going to take a element of some K value vector a radius vector whose components are K 1 and K 2 here these are the real values of the circle and I can as well perform this integration in polar coordinates and because of symmetry my shell area if I know I can evaluate it first in the shell and go on integrating for various shells. So, basically now K is a vector with the components K 1 K 2 that implies that the area element d K 1 d K 2 is going to go as K d K if you are keeping theta as a variable then it will be K d K d theta supposing the function does include the theta. So, it will be K d K d So, once we transfer this function and if you convert our 1 by cos K 1 or quantity now what happens in terms of K and theta variable cos K 1 plus cos K 2 that will go over now K 1 is going to be K cos theta if theta is this angle then K 1 is K cos theta and K 2 equal to K sin theta. So, it is going to be cos of K cos theta plus cos of K sin theta. So, if we now look at the value of the denominator which is 1 minus half of cos K 1 plus cos K 2 immediately see that this also has a singularity at the point K 1 equal to 0 K 2 equal to 0 that is origin because when K 1 equal to 0 cos K 1 is 1 cos K 2 is also 1 1 plus 1 2 by 2 is 1 so 1 minus 1 is. So, this tends to 0 at the origin which means this also is going to end up as a singular integral and to see the nature of that singularity we expand it for small values of K 1 and K 2 that is very close to this origin somewhere here then we note that this function will become 1 minus half of 1 minus half K K 1 square plus 1 minus half K 2 square higher orders we are neglecting. So, this becomes 2 minus so it will be 1 minus half into 2 minus half K 1 square plus K 2 square when we fully expand it is going to be 1 minus 1 is 0 minus minus plus. So, it will be 1 fourth K 1 square plus K 2 square 1 minus 1 plus 1 4. So, our integrand now will basically we are going to integrate within this approximation the function pi naught 1 2 D is going to be 1 by 2 pi square minus pi to pi minus pi to pi D K 1 D K 2 of 1 by 1 minus this expression that is 1 fourth of K 1 square plus K 2 square. So, this is going to be 1 fourth K 1 square plus K 2 square 1 fourth we take up let us say this is nearly equal to since we have made an approximation it is nearly equal to. So, that is going to be 1 by pi square minus pi to pi minus pi to pi D K 1 D K 2 by K 1 square plus K 2 square. Since in terms of the vector K D K 1 D K 2 will be supposing there is no theta dependence. So, we can simply say 2 pi K D K and K 1 square plus K 2 square is simply K square Pythagoras theorem. So, we have this whole integral therefore, becomes pi naught 1 2 D will become 1 by pi square K will vary from 0 to pi because it is a radius vector 2 pi K D K divided by K square which has 1 by pi square 2 pi by pi square integral 0 to pi D K by K. Now, we note that this integration domain includes 0 and near 0 there is a singularity again of the order 1 and in fact, this integral is log K and log K diverges at K equal to 0. Hence this integral is going to be also a divergent will be infinity. So, for 2 D we have the result pi naught 2 D 1 is infinity in 2 D also. Therefore, the return probability R equal to 1 minus 1 by infinity equal to 1 R 2 D implies hence even in 2 D the random walker definitely returns to the origin definitely returns to the origin. So, even though in one dimension we can understand his reason to return to the origin because the space is very constricted sometime he has to come back. In 2 dimensions we would have thought now there is enough space to escape. So, once having started out there is no need for the random walker come back to the origin, but some of the nature of random walk the dimensionality of the random walk and the dimensionality of the space in 2 dimensions almost become same. In fact, the space paths that the random walks describe ultimately make up an object with the dimensionality of 2 in any dimension. So, in 2 dimension it just happens to be the dimension of the system itself and however there is this kind of a unpredictable situation, but ultimately mathematics tells us that the random walker will return to the origin it will not escape. Although this return occurs far more slowly as a as compared to 1 dimensional random walk. So, now therefore, our curiosity level increases what will happen in 3 dimensions is a natural question and 3 dimension is a real dimension. In fact, that is the dimension when most events or most physical processes take place. So, it has also a physical significance. So, we now go to the problem of return probability R in 3D to ask a question. So, what is the return probability in 3D? Let us revert back to the same formalism. So, in 3D the generating function let us keep remembering for the occupancy probability which is a quantity of interest will be defined as 2 pi to the power d. So, it is 2 pi q and then the minus pi integrals there will be 3 such integrals yeah we will write all of them minus pi 2 pi d k 1 d k 2 d k 3 and the denominator is going to be 1 by cos k 1 plus cos k 2 plus cos k 3 by 3. Now, it is interesting to know whether this will have a singularity or not. If you look at from the Cartesian perspective when k 1 equal to 0, k 2 equal to 0, k 3 equal to 0 that is at the origin 1 plus 1 plus 1 by 3 is going to be 1. So, 1 minus 1 is going to be once again 0 and hence there is a singularity in the denominator and from the experience we gained so far that singularity will have a form of 1 minus. In fact, it will have a form of k 1 square plus k 2 square plus k 3 square type. Now, we are not sure in the Cartesian system of integration what does it mean whether the overall integral exists or not. To answer that we have to go to once again the equivalent spherical polar system under the assumption that this function is a positive function we are integrating over a cube we are actually like earlier we are integrate first let us draw a circle that is the difficult thing to do take a circle of diameter 2 pi then there is a hyper cube let us say touching this and let us make it touch so that there is no misinterpretation and this is 2 pi is a hyper cube in 3 dimensions I am showing it as 2 d there is one more dimension to it. So, this is a sphere. So, these are sides of a cube 2 pi here that is minus pi to pi etcetera minus pi to pi minus pi to pi etcetera. So, we say that since the function is positive the value of the true integral in this cube will be more than in any case the integral within the circle. So, if we get some value in the circle and if it is singular already in the circle then it will remain singular definitely in the cube that is the logic it is already infinity for a circle it is going to be infinity. So, converting the square to a circle is just to make sure that if we get infinity we have a result, but if we do not get infinity finite then it is a matter of numerical precision only qualitatively we have understood the problem. So, first let us do it over the circle because it takes care of the origin origin is the most tricky point. So, in this case we now consider k as the radius vector this is a boundary of course, but you can see k radius vector having components k 1 k 2 k 3 and hence k square the norm square of the norm of the vector you should write it as actually square of norm it is going to be k 1 square plus k 2 square plus k 3 square and the volume element d k 1 d k 2 d k 3 since actually the denominator in the first order does not depend on theta phi coordinates other coordinate systems this is going to be the surface area volume element 4 pi k square d k that is the volume element of a sphere. So, then the integral within this framework will turn out to be pi naught 3 d 1 will be now it will be 1 by 2 pi cube and the entire 3 dimensional integral will be 1 dimensional integral of the radius vector 0 to pi d k 1 d k 2 d k 3 is 4 pi k square d k and the and all the other things will be 1 minus 3 into if you see this will be 1 minus k 1 square by 2 plus 1 minus k 2 square by 2 plus 1 minus k 3 square by 2 etcetera and this whole thing 1 plus 1 plus 1 3 cancels. So, it is going to be the first term cancels all the others will be positive. So, 1 sixth of k 1 square plus k 2 square plus k 3 square equal to 1 sixth of k square I am referring to the denominator only. So, which means this integral is going to be 0 to pi 4 pi k square d k divided by k square by 6 most important thing is we notice that the k square cancel with each other. So, it is going to be 4 pi into 6 into 2 pi into 6 into 2 pi into pi cube 8 pi cube we can say in any case integral 0 to pi d k which is actually a finite value. We are not going to be very sticklers about the exact numerical value, but this demonstrate that it is finite. So, once the integration within the circle is finite the singularity is taken care of we know that the integration around the edges of the square are also going to be finite because there is nothing wrong that can happen the denominator does not have a 0 there. Hence we are now prompted to say that E in 3 dimensions there is hence pi 0 3 d 1 is finite prompting us conclude. That the return probability in 3 d that is r will be finite will be finite, but I would not say for it will be non-zero finite that is more important non-zero or less than unity or non-zero or less than unity or and now exact calculation is a matter of finite numerical detail one can actually integrate these days this whole thing numerically and we can actually evaluate the integral numerically we know that it has no singularity now. So, we can do it. So, the result turns out to be 1 by 2 pi cube minus pi to pi minus pi to pi minus pi to pi d k 1 d k 2 d k 3 divided by one third of cos k 1 plus cos k 2 plus cos k 3 if you do this integral it actually turns out to be 1.51639 goes on. So, we can have r. So, we can obtain hence the return probability in 3 d r 3 d will be 1 by 1 1 minus 1 by 1 point let us say 5 2 just I am routing off. So, this definitely less than 1 and it turns out to be 0.3405. So, there is a 34 percent probability that a random walker starting from the origin will return back to the origin somewhat 66 percent probability that he will escape we will never return to the origin this is a very very important that is a random walker in 3 d will return to the origin with 30 percent probability 34 percent approximately or he will escape forever he will escape with 60 approximately 66 percent probability. And since origin is only a special point we considered this is true of lattice points in general because wherever he lands from there onwards that can be considered as origin. So, this could be true for every lattice point in the space in space. So, this result of capability to escape return to the origin is not limited to the rectangular or kind of lattices that we considered. This is true in general for random walks of other types also such as the Pearson walks and others numbers may be different, but the qualitative important results that in the first to first and the second dimension the random walker necessarily returns to the origin, but in dimensions 3 d and 3 d he escapes. We can easily now visualize that higher the dimension probabilities of return are going to become less and less and probability of escape is going to increase. This result specifically is what Polyas theorem is all about that the first two dimensions are the ones where there is a definite certain return to the origin and in higher dimensions there are only probabilistic return to the origin. To complete this discourse we can borrow some results will not be able to prove it, but it follows the same logic that in higher dimensions for example, in 4 d you can do that integral they are called Watson's integrals is the name for those multiple integrals the lattice integrals. So, basically one would obtain for example, in 4 d the pi 1 1 gets something like 1.239 or. So, this resulting in the return probability in 4 d becoming 0.1932 in 5 d the return probability the pi function it is finite it has some value some number, but it implies the return probability in 5 d turns out to be 13.5 percent 0.135 and so on. So, one can now numerically evaluate in any dimension and then obtain the return probabilities. So, this completes our derivation of symmetric random walk in generalized dimensions. The formalisms that we developed is valid even for asymmetric random walk or random walk with bias. So, all that we need to therefore, know is the generating function for the occupancy probability at the origin which we know how to do it even if p and q are not exactly same and from there on construct the generating function for the W and then evaluate it at z equal to 1 obtain the return probability. Just to complete the topic we evaluate the question of return probability in 1 d return probability with bias we do it for 1 d in 1 d it is a we have shown that the occupancy probability with bias W n m 1 d plus bias this was 1 by 2 to the power n n factorial n plus m by 2 factorial into 1 d. So, this is 1 by 2 into n minus m by 2 factorial 1 plus gamma to the power n plus m by 2 1 minus gamma to the power n minus m by 2 where gamma is the asymmetry factor or drift factor you can call it as a drift it is an asymmetry between p and q. So, from this we can calculate that at the origin with m equal to 0 and further noting that it can return to origin only in even steps. We can say that W of 2 k instead of n we write an even number 1 d plus bias probability of return to the origin will just be 2 to the power minus 2 k we just put m equal to 0 and m equal to n equal to 2 k that is all it is going to be 2 k factorial divided by k factorial whole square. So, once m equal to 0 it is n by 2 and n by 2. So, it is 1 minus gamma square to the power n by 2 which is k for all k equal to with this we can evaluate the pi function pi naught 1 d plus bias 0. So, z by definition it is going to be the sum and we can show that it just sums into a we can use generalized binomial theorem and we can show that it will come to 1 minus 1 minus gamma square z square square root of the whole thing use generalized binomial theorem to fractional powers binomial expansion we should say binomial expansion to fractional. So, with this therefore, the generating function for the 1 d occupancy probability at the origin with bias is going to be pi naught 1 d with bias at z equal to 1 will be 1 by square root of 1 minus 1 minus gamma square which is equal to 1 by modulus of gamma modulus because it is a square root of gamma square we take the positive sign. That means, hence the return probability r for 1 d with bias will be 1 minus of 1 divided by pi naught at z equal to 1 that is 1 divided by 1 by mod gamma which is 1 minus mod gamma. This quantity is for all gamma not equal to 0 it is going to be less than unity. Hence, we have the return probability to origin with bias is less than unity regardless of the sign of gamma. So, what it means is if you have some bias the random walker has a certain probability of escape the return is less than 1 which means there is a probability of non zero probability of escape. If gamma is 0 the return probability is unity escape probability is 0 consistent with our previous discussions of symmetric random walk as gamma increases the escape probability increases and it is independent of the sign of gamma because here we are not talking of an absorber. So, whichever way there is a bias in that particular direction there will be an escape it can be plus infinity direction or minus infinity direction. Hence escape probability exists for non zero gamma one can actually carry out such exercises for other than a constant bias site dependent biases or others that is an advanced kind of exercises. With this we have demonstrated that in dynamic dimensions less than 3 D that is in 1 and 2 dimensions the return probability is unity for a symmetric random walker and for a symmetric random walker the return probability is less than unity for 3 dimensions and above. And in the case of one dimension specifically considering bias there is a finite escape probability. With this we conclude this discussion of return probability or Paulia's theorem in the next lecture we examine how the first time visit probability that is the F parameter actually depend on the step size the step length. Thank you.