Perimeter Institute Lecture Notes on Statistical Physics part 4: Diffusion and Hops Version 1.6 9/11/09 Leo Kadanoff 1

Transcription

1 Part 4: Diffusion and Hops From Discrete to Continuous Hopping on a Lattice Notation: Even and Odd From one step to many An example Continuous Langevin equation An integration Gaussian Properties Generating Function A probability Discrete A generating Function A probability calculation we get an answer! fourier transform formulation binomial theorem Higher Dimension continued probability density One dimension Current Diffusion equation Higher Dimension You cannot go back 1

2 Hopping: From Discrete to Continuous We are going to be spending some time talking about the physics of a particle moving in a solid. Often this motion occurs as a set of discrete hops. The particle gets stuck someplace, sits for a while, acquires some energy from around it, hops free, gets caught in some trap, and then sits for a while. I m going to describe two mathematical idealizations of this motion: discrete hopping on a lattice and continuous random motion. One point is to see the difference between the two different topologies represented by a continuous and a discrete system. One often approximates one by the other and lots of modern physics and math is devoted to figuring out what is gained and lost by going up and back. There is a fine tradition to this. Boltzmann, one of the inventors of statistical mechanics, liked to do discrete calculations. So he often represented things which are quite continuous, like the energy of a classical particle by discrete approximations, A little later, Planck and Einstein had to figure out the quantum theory of radiation, which had been thought to be continuous, in terms of discrete photons. So we shall compare continuous and discrete theories of hopping. 2

3 Hopping On a Lattice A lattice is a group of sites arranged in a regular pattern. One way of doing this can be labeled by giving the position r=(n1,n2,...)a where the n s are integers. If we include all possible values of these integers, the particular lattice generated is called is called the simple hypercubic lattice. We show a picture of this lattice in two dimensions. This section is devoted to developing the concept of a random walk. We could do this in any number of dimensions. However, we shall approach it is the simplest possible way by first working it all out in one dimension and then stating results for higher dimensions A random walk is a stepping through space in which the successive steps occur at times t=m τ. At any given time, the position is X(t), which lies on one of the a lattice sites, x=an, where n is an integer. In one step of motion one progress from X(t) to X(t+ τ) = X(t)+ aσ j, where σ j is picked at random from among the two possible nearest neighbor hops along the lattice, σ j =1 or σ j = -1. Thus, < σ j >=0, but of course the average its square is non-zero and is given by < σ j2 > =1. We assume that we start at zero, so that our times t =j τ. It is not accidental that we express the random walk in the same language as the Ising model. We do this to emphasize that geometric problems can often be expressed in algebraic form and vice versa. 3

4 Notation: Even and Odd We represent the walk by two integers, M, the number of steps taken and n, the displacement from the origin after M steps. We start from n=0 at M=0. The general formula is n = M k=1 σ k Notice that n is even if M is even and odd if M is odd. We shall have to keep track of this property in our later, detailed, calculation. We shall also use the dimensional variables for time t=m τ and for space X(t)=na. We use a capital X to remind ourselves that it is a random variable. When we need a nonrandom spacial variable, we shall use a lower case letter, usually x. 4

5 From one step to many steps We start from the two statements that <σj >=0 and that <σj σk >=δj,k On the average, on each step the walker goes left as much as right. and thus as a result the average displacement of the entire walk is zero M < X(t) >= a < σ j >= 0 j =1 However, of course the mean squared displacement is not zero, since iv.1 M < X(t) 2 >= a 2 < σ j σ k >= a 2 δ j,k j,k =1 M j,k =1 = a 2 M iv.2 Our statement is the same that in a zero field uncoupled Ising system, the maximum magnetization is proportional to the number of spins, but the typical magnetization is only proportional to the square root of that number. Typical fluctuations are much, much smaller than maximum deviations. We can see this fact by noting that the root mean square average of X 2 is a M, which is the typical end-to-end distance of this random walk. This distance is is much smaller than the maximum distance which would be covered were all the steps to go in the same direction. In that case we would have had a distance am. Thus, a random walk does not, in net, cover much ground. 5

6 An example: The difference between maximum length and RMS length (root mean square length) needs to be emphasized by an example. Consider a random walk that consists of M= one million steps steps. If each step is one centimeter long the maximum distance traveled is M centimeters, or 10 thousand meters or 10 kilometers. On the other hand the typical end to end distance of such a walk is centimeters or ten meters. That is quite a difference! M 1/2 6

7 Continuous Random Walks: Langevin Equation The physics of this random motion is well represented by the two equations <X(t)-X(s)> = 0 iv.3 <[X(t)-X(s)] 2 > = a 2 t-s /τ iv.4 We would like to have a simple differential equation which appropriately reflects this physics. The simplest form of differential equation would be a form described as a Langevin equation, dx/dt = η(t) where η(t) is a random function of time that we shall pick to be uncorrelated at different times and have the properties <η(t)> =0 iv.5 and <η(t) η(s)> =c δ(t-s)δ j,k This looks like what we did in the random hopping on the lattice, except that we have to fix the value of the constant, c, to agree with what we did before. Equation iv.5 has the solution X(t) = X(s) + t s du η(u) 7

8 X(t) = X(s) + t s du η(u) Using this result, we can calculate, that as before <R(t)-R(s)> =0. A new result is that for t>s, we find: (what happens for s>t?) More specifically, < [X (t) X (s)] 2 > = du dv < η (u) η (v) > t s t t = du dv c δ(u v) s = du c = (t s)c s s t s < [X(t)] 2 >=c t t iv.6 so that a comparison with our previous result indicates that we should choose c= a 2 /τ so that ` iv.7 < [X (t) X (s)] 2 > = a 2 t s /τ iv.8 8

9 Gaussian Properties of Continuous Random Walk In the continuous case, X(t) is composed of a sum, or rather integral, of many pieces which are uncorrelated with one another. According to the central limit theorem, such a sum or integral is a Gaussian random variable. Hence, we know everything there is to know about it. Its average is zero and its variance is a 2 t/τ. Consequently, it has a probability distribution 1/2 τ < ρ(x(t) = x) >= e x2 τ /(2a 2 t) iv.9 2πa 2 t Now we have said everything there is to say about the continuous random walk. As an extra we can exhibit the generating function for this walk: < exp[iqx(t)] >= e q2 a 2 t /(2τ ) 9

10 A generating function: discrete case In our calculation of the statistical mechanics of the Ising model, or of any other problem involving the statistics of large numbers of particles, the traditional approach to the problem is to calculate the partition fiunction, Z, a generating function from which we can determine all the statistical properties. The same approach works for the random walk. It is best to start on this calculation by a somewhat indirect route: The spirit of the calculation is that it is simple to find the average of exp(iqx(t)) where q is a parameter which we can vary and X(t) is our random walk-result. If one knew the value of < exp(iqx(t))> it would be easy to calculate all averages by differentions with respect to q and even the probability distribution of X(t) by, as we shall see, a Fourier transform technique. We start from the simplest such problem. Take t=τ so that there is but one step in the motion and < exp(iqx(τ))>= < exp(iqa σ 1 (t))>= [exp(iqa) + exp(-iqa)]/2 = cos qa Since X(t) is a sum of M identical terms, the exponential is a product of M terms of the form < exp(iqa σ j (t))> and each term in the product has exact the same value, the cosine given here. Thus the final result is <exp(iqx(t))>= [cos(qa)] M. iv.9 As we know from our earlier work such a generating function enables us to calculate, as for example <exp(iqx(t))>= [cos(qa)] M at q=0 <1>=1 d/dq { <exp(iqx(t))>= [cos(qa)] M } at q=0 <ix(t)>=ma sin qa =0 (d/dq) 2 { <exp(iqx(t))>= [cos(qa)] M } at q=0 -<[X(t)] 2 >=-Ma 2 d/dq sin qa = -Ma 2 10

11 A Probability We wish to describe our result in terms of a probability, ρ n which is the probability X will take on the value na. We can use this probability to calculate the value of any function of X in the form (see equation i.3) < f (X ) >= f (na) n ρ n This is an important definition. It says that if X takes on the values na, and the probability of it having the value na is ρ n, then the average is the sum over n times the value of the function at that n times the probability of having that value. The formula that we shall use is only slightly more complex. We define X as being time dependent: the probability that X(t) has the value na at time mτ is then defined to be ρ n,m. The formula for the average is then < f (X(mτ)) >= f (na) n ρ n,m iv.10 so that the average is once more the sum of the function times the probability. So tells us everything there is to know about the problem. ρ 11

12 To calculate probability Start from < f (X ) >= f (na) n ρ n choose the function to be an exponential, since we have had, by this time, considerable experience dealing with exponentials < exp(iqx(t) / a) >= exp(iqn) ρ n,t /τ n We have already evaluated the left hand side of this expression, so we find < [cosq] M >= exp(iqn) ρ n,m iv.11 n I m going to spend some time on this result. First, let s say what it is. It gives the probability that after a random walk of M steps a walker will find itself at a site n=m-2j steps of its starting point. The pattern is slightly strange. For M being even the possible sites are M, M-2,..,0,... just before -M+2, -M. For M being odd the possible sites are M, M-2,,...,-1,1,...,-M+2, -M Thus there are a total of M possible sites which could be occupied. Is it true then that an average probability of occupation among these sites is 1/M? or maybe 1/ (2M) What s a typical occupation? What is that going to be? Recall that the RMS distance covered is... 12

13 to expand the left hand side as ( eiq + e iq ) M = 2 j to get... We get an answer! < [cosq] M >= exp(iqn) ρ n,m iv.11 n Here M is the number of steps. We do not even have to do a Fourier transform to evaluate the probability. We simply use the binomial theorem n (a + b) M M! = j!(m j)! am b M j ρ M 2 j,m = j=0 M! j!(m j)! M! j!(m j)! 2 M e iqm 2 M e i2q j iv.12 iv.13a ρ n,m = M! ( M+n 2 )!( M n 2 )! 2 M iv.13b We are a bunch of theorists. We have an answer. A formula. Now what can we do with that answer? What can we figure out from here? 13

14 p= Fourier transform ρ p,m e iqp = [cos q] M iv.11 To invert the Fourier transform use To obtain π π dq 2π ei(p n)q = δ n,p π π dq 2π e inq p= ρ p,m e iqp = ρ n,m = π π dq 2π e inq [cos q] M Because n=m (mod 2), the integrand at q is just the same as the integrand at q+π. Therefore one can also write ρ n,m = 2 π/2 π/2 dq 2π e inq [cos q] M We have come across this kind of integral before. Now what? 14

15 Result From Binomial Theorem ρ M 2 j,m = M! j!(m j)! 2 M ρ n,m = M! ( M+n 2 )!( M n 2 )! 2 M We are a bunch of theorists. We have an answer. A formula. Now what can we do with that answer? What can we figure out from here? sterling approximation, for large M, M! M M e -M (2 π M ) 1 /2 what happens for n=m, n=-m, n=0, n close to M, n close to -M, n close to zero? 15

16 Higher Dimensions Reason for putting many different calculations on a lattice is that the lattice provides a simplicity and control not available in a continuum system. There is no ambiguity about how things very close to one another behave, because things cannot get very close. They are either at the same point or different points So now I would like to talk about a random walk in a d-dimensional system by considering a system on a simple lattice constructed as in the picture. The lattice sites are given by x = a(n1,n2, n3,...)/2. The n s are integers. There are two possible kinds of assignments for the n s: Either they are all even e.g. x = a(0,2, -4,...)/2 or they are all odd, for example x = a(1,3, -1,...)/2. If all hopping occurs from one site to a nearest neighbor site, the hops are through one of 2 d vectors of the form ξ α = a(m1,m2, m3,...)/2, where each of the m s have magnitude one, but different signs for example a(1,-1,-1,...). We use a lattice constant, a, which is twice as big as the one shown in the picture. We then choose to describe the system by saying that in each step, the coordinate hops through one of the nearest neighbor vectors, ξ α, which one being choosen at random. This particular choice makes the entire coordinate, x, have components which behave entirely independently of one another, and exactly the same as the one-dimensional coordinate we have treated up to now. 16

17 RANDOM WALKS dspace.mit.edu/.../coursehome/index.htm 17

18 Higher Dimensions...continued We denote the probability density of this d-dimensional case by a superscript d and the one for the previous one-dimensional case of by a superscript 1. As an additional difference, the d-dimensional object will be a function of space and time rather than n and M. After a while, we shall focus entirely upon the higher dimensional case and therefore drop the superscripts. We have d ρ d an,mτ = α=1 ρ 1 n α,m However, we can jump directly to the answer for the continuum case, If the probability distribution for the discrete case is simply the product of the onedimensional distributions so must be the continuum distribution. The one-dimensional equation answer in eq iv.9 was < ρ(x(t) = x) >= τ 2πa 2 t 1/2 e x2 τ /(2a 2 t) so that the answer in d dimensions must be iv.9 < ρ(r(t) = r) >= τ 2πa 2 t d /2 e r2 τ /(2a 2 t) Here the bold faced quantities are vectors, viz r 2 = x 2 +y

19 Higher dimensional probability density: again one dimension d dimensions is a product of ones < ρ(x(t) = x) >= τ 2πa 2 t 1/2 e x2 τ /(2a 2 t) < ρ(r(t) = r) >= τ 2πa 2 t d /2 e r2 τ /(2a 2 t) Notice that the result is a rotationally invariant quantity, coming from adding the x 2 and y 2 and... in separate exponents to get r 2 = x 2 +y This is an elegant result coming from the fact that the hopping has produced a result which is independent of the lattice sitting under it. We woulod get a very similar result independent of the type of lattice underneath. In fact, the result come from what is called a diffusion process and is typical of long-wavelenth phenomena in a wide variety of systems. 19

20 Diffusion Process Diffusion here is a result of a conservation law: a global statement that the total amount of something is unchanged by the time development of the system. dq/dt=0. In our case the Q in question is the total probability of finding the diffusing particle someplace. Diffusion has a second element: locality. The local amount of Q, called ρ, changes because things flow into and out of a region of space. The flow is called a current, j, and the conservation law is written as ρ(r,t) conservation law + j(r,t) = 0 t The time derivative of the density is produced by a divergence of the curent flowing into a point. On a one dimension lattice, the rule takes the simpler form: ρ n,m +1 ρ n,m = I n 1/2,M I n+1/2,m Here, I m going to visualize a situation once more in which we have a discrete time coordinate M and a discrete space coordinate, n. I shall assume that the initial probabilities is sufficiently smooth so that even and odd n-values have rather similar occupation probabilities so that we can get away with statement like ρ n,m +1 ρ n,m ρ n,m / M conservation law This equation says that the change of probability over one time step is produced by the flow of probability in from the left minus the flow out to the right. The conservation law then takes the form ρ n,m / M + I n,m / n = 0 20

21 The Current We do not have a full statement of the of what is happening until we can specify the current. An approximate definition of a current in a conservation law is called a constitutive equation. We now write this down. Our hopping model says that a In+1/2,M is given a contribution +1 when the site at n is occupied at time M, and σ n is equal to +1. On the other hand it is given a contribution -1 when the site at n +1 is occupied at time M, and σ n+1 is equal to -1. Since there is a probability 1/2 for each of the σ-events the value of I is I n+1/2,m = (ρ n,m ρ n+1,m ) / 2 constitutive equation Once again, we write the difference in terms of a derivative, getting I n,m = ( ρ n,m / n) / 2 constitutive equation This can then be combined with the conservation law to give the diffusion equation ρ n,m / M = 2 ρ n,m / n 2 / 2 which can be written in dimensional form as ρ t = λ 2 ρ diffusion equation x 2 with the diffusion coefficient being given by λ=a 2 /(2 τ). 21

22 Higher Dimension In higher dimensions, the current is proportional to the gradient of the current j(r,t) = λ ρ(r,t) so that the diffusion equation becomes t ρ(r,t) = λ 2 ρ(r,t) 22

23 Diffusion Equation This equation is one of several equations describing the slow transport of physical quantities from one part of the system to another. When there is slow variation is space, the conservation law guarantees that the rate of change in time will also be slow. In fact this is part of a general principle which permits only slow changes as a result of a conservation law. This general principle is much used in the context of quantum field theory and condensed matter physics. The idea is connected with the construction of the kind of particle known as a Nambu-Goldstone boson, named for two contemporary physicists, my Chicago colleague Yoichiro Nambu and the MIT theorist Jeffrey Goldstone. 23

24 Solution to Diffusion Equation Cannot be Carried Backward in Time The wave equation is ( t 2 - X 2 )F = 0 Its general solution is F(x,t)=G(x-ct)+H(x+ct) This is a global solution. It enables you to look forward or back infinitely far in the future or the past without losing accuracy. Find solution from F(x,0) = X F(x,0) =1 for 0<x<1 and F(x,0) = X F(x,0) =0 otherwise. Use c=2 A global solution to the diffusion equation is ρ(x,t)= dk exp[ikx-λk 2 t] g(k) with g(k)= dx exp[-ikx] ρ(x,0)/(2π) as initial data. Small rapidly varying errors at t=0 will produce small errors for positive t and huge errors for negative t. You cannot extrapolate backward in time. Information gets lost as time goes forward. Solve for ρ(x,0)=1 for 0<x<1 and ρ(x,0) =0 otherwise. Use λ=2. Plot solution for t=0,2,4. What happens for t= -2? Boltzmann noted that equations of classical mechanics make sense if t is replaced by -t. But diffusion equation has a solution which does not make sense. Where did we go from sense to nonsense? 24