PHYSICS 653 HOMEWORK 1 P.1. Problems 1: Random walks (additional problems)

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1 PHYSICS 653 HOMEWORK 1 P.1 Problems 1: Random walks (additional problems) 1-2. Generating function (Lec. 1.1/1.2, random walks) Not assigned 27. Note: any corrections noted in 23 have not been made. This exercise is to show (or remind) you what a generating function is. (It is related to the partition functions in thermodynamic ensembles.) I didn t have time to find an easy application of a generating function. Consider a Markov process (e.g. random walk) happening in discrete time t =, 1,.... Let X represent the initial state which may recur during the process (e.g., the walk revisits the original site). Let p t Prob({X happens at step t}). (1a) f t Prob({X first recurs at step t}). By the definition, p = 1; what is f? It is often comparatively easy to obtain p t : for example, the probability for a onedimensional walk to be on a certain site is given by a binomial distribution, or approximated by a Gaussian. But it can be hard to obtain the distribution for first-passage to that site, f t. So, the object of this exercise is to find a general relation for p t in terms of f t. (Notice we really want f t in terms of p t ; hopefully, the functional relation we get is invertable.) a. On the way, it helps to define (1b) f k t Prob({X happens for the n-th time on step t), (1c) thus f t ft 1 Clearly, p t = k=1 f t k. So, since small numbers are easier, let s first figure out how to get ft 2 from ft 1; maybe that will reveal how to go to higher f t k, which in turn gives p t. Show that p 2 t = t t = f tf t t. (What is p 2 1, incidentally?) b. A very useful technique is the generating function that function whose Taylor series coefficients are the sequence of interest. So, let P (z) = p t z t ; F (z) = t= f t z t ; F k (z) = t= ft k z t. (2) t= What s the meaning (in terms of the walk) of P (1) and F (1)? (Since F (1) is a probability, we know F (1) 1. Thus F (z) is absolutely convergent for z 1, at least.) What s F (z)? (trivial). c. Show that F k (z) = F (z) k. (Hint: induction/recursion on k.) What, finally, is the relation between F (z) and P (z)? This should be a special case of the relation, derived in class, between the mean total number of returns, and the probability of never returning at all.

2 PHYSICS 653 HOMEWORK 1 P Random walk with random traps On a one-dimensional line, immobile traps are placed at random. Thus, the intervals free of traps have varying lengths L; the probability distribution of L is proportional to exp( L/L ), where L 1 is the typical spacing between traps. Say also that the walker walks with diffusion constant D.) Now, a random walker is started, at time t =, at a random place on the line; we are interested in the distribution P sur (t), of the survival time, i.e. the time until the walker s first encounter with a trap. Such distributions are measured e.g. in NMR (nuclear magnetic resonance) experiments: here the traps represent sites of magnetic ions bearing an electonic moment. (That scrambles the direction of the precessing nuclear spin in a liquid molecule, if its random motion happens to brings it near the magnetic ion). It turns out that P sur (t) has a slowly decaying tail out to long times: the question here is what is the asymptotic time dependence? (Keep in mind, it s more important to get the right functional dependence which power law, or exponential or Gaussian than the coefficients.) Preliminary Discussion. (1) This is an example of quenched disorder, an enormously important topic. (However, I ve left to the last week this year; since might not get to it, I ll mention disordered systems casually in a couple other places.) It means the Hamiltonian (or whatever else defines the rules of the game, e.g. the transition rates of a random walk) is not translationally invariant, depends on some prior random ensemble, and stays fixed (while some other ensemble depends on it) In real life, disordered systems come about when different degrees of freedom change with very different time scales: for example, you might make a metal crystal which is a random alloy of Cu and Zn atoms, and these move around much slower than the electrons moving through the lattice. Anyhow, the model in this exercise is another realization of the general notion of having two random ensembles, or of building one ensemble on another. (2) Before jumping in, think: where does the long-time tail of P sur (t) come from? Corresponding to the three random ensembles in this problem, there are three factors that could help a walker can survive an exceptionally long time * (i) it might be lucky by starting in a place where there are fewer traps than on average in a really wide interval; (ii) it might be lucky to get dealt a string of random steps such that it walks back and forth visiting comparatively few sites, so it has fewer chances to hit a trap. (iii) there s a third random ensemble, which is the initial position: the random walker would be lucky in having its initial position near the center. My intuition tells me either (i) or (ii) is the important factor, since the best cases are exponentially rare; as for (iii), any of the sites near the middle of an interval is nearly as good, so the tail of the distribution isn t exponential. (3) What do we know? Firstly, since this is 1D, the traps completely disconnect the system into a collection of the intervals between them: a walk can never pass from one interval to the next. Secondly (see class notes), we already know the probability distribution for a walker in one such interval. * Incidentally, I think these correspond to the two ways you may survive an exceptionally long time!

3 PHYSICS 653 HOMEWORK 1 P.3 (a). We can break the problem down into (i) what is the probability of being in an interval of length L? (ii) given the interval has length L, what is the survival probabilty S L (t)? We focus on (ii) first. Let S L (x; t) be the probability, given we started at x: we computed that in lecture. We decomposed S L (x; t) into normal sine modes, just like a vibrating string clamped at both ends, and the longest-lived mode was with a exponential decay rate p 1 (x) sin(πx/l) (1) τ 1 (L) 1 = D(π/L) 2 (2) Make the approximation of ignoring the other modes. That approximation may be bad at short times, in that the second-slowest mode in a long interval is surely slower than the slowest mode in a short interval. But we are going for the dominant longest time contribution. The distribution of starting points in the interval [what I called random ensemble (iii) in this problem] is uniform, P L (x; ) = 1/L. To write this as P L (x; ) = c 1 p 1 (x) +... (3) use orthogonality c 1 = dxp (x, )p 1(x) dxp 1(x) 2 (4) where the denominator is normalization. (So what s c 1? ) (b). Now P L (x; t) = c 1 p 1 (x)e t/τ 1(L) (5) since (by def n) a normal mode decays as a pure exponential. The survival probability in this interval is S L (t) = dxp L (x; t) (6) Your result should be dominated by the exponential factor e t/τ(l). This is the first desired result: mean lifetime, given that we landed in an interval of length L. (c). Meanwhile, what is the probability of being in an interval of length L? Imagine first calculating how many intervals there are of each length. March from the left, where some small fraction p is a trap. The probability of going for L steps w/o a trap is (1 p) L which we ll write as exp( L/L ), with e 1/L 1 p. If we toss a point randomly into one of these intervals, an interval of length L has a bigger cross section it has L times as many chances for points to land in it, as in an interval of length 1. Hence, the final distribution of interval lengths is dp (L) = (L/L ) 2 exp( L/L )dl (7)

4 PHYSICS 653 HOMEWORK 1 P.4 This is the second desired result. No work required in (c). (d). Putting it all together, we have S(t) = dp (L)S L (t) (8) in which the contributions from both kinds of randomness [(i) and (ii)] give an exponential factor. Finally, we deploy a simple method from asymptotic analysis. Write (8) as S(t) = dl exp[ F (L; t)]; (9) What is the function g(l; t)? The result of (9) is dominated by the value L (t) where F (L; t) is minimum. (You needn t check the validity for this exercise, but you need to in real life. It depends on the fact that, the larger t gets, the steeper that minimum is.) Plug L (t) into (9) to obtain a final formula for S(t). (This should be a stretched exponential, exp( constt x ) where x < 1.)

5 1-4. Mean time of a 1D random walk PHYSICS 653 HOMEWORK 1 P.5 Not assigned 27. Note: new problem watch for errors. This follows up on Prob 1-1. Let T (x) be the expected number of steps of a walk starting from x, to first reach the origin (or the location of a trap.) How can we compute this? (a). Take the model of Prob [Interval (, R).] I think you can see T (x) is related to Φ +. Use the same reasoning as in Prob 1-1: that is, if you re on site x at time t, then at time t + 1 you re on site x 1 or x + 1 with equal probabilities. This allows you to express T (x) in terms of T (x + 1) and T (x 1). Show this equation is 2 T (x) = 1 (1) where 2 f(x) f(x + 1) + f(x 1) 2f(x), as defined in class. (b). Eq. (1) is an inhomogeneous second-order linear difference equation. If it was a differential equation, the solution would be trivial for you. Show that this solution, in fact, works here. (c). Given that you know T () and T (R) [that was the first thing you asked right??], what are the coefficients of the homogeneous term in the solution to (1)? (d). The mean duration till the next return to the origin is T ret = 1 + T (1). (Why: any return to the origin starts by going one step away). The situation for a line without a trap at R must correspond physically to the limit as R. What is lim R T (1)? (e). Now let s consider a different approach. In lecture 1.1 (which section?) we used the method of images and the continuum diffusion equation to find the total survival probability S(t), given that we started at x. [Actually I should write it S(t; x) but I will suppress the x argument for a few steps.] We found 1 S(t) = (2) 4πDt for t > x 2 /D. (For smaller t you can use the approximation S(t).) Now, the function 1 S(t) is the probability that the walk died before time t. Therefore, its derivative is the probability density that the lifetime was exactly t: ds(t)/dt = p X (t) (3) The mean lifetime is Does this converge? T = p X (t)tdt. (4)