Method+of+images

toc =Introduction= We are interested in the density matrix of a non-interacting particle subjected to various boundary conditions. =The free particle in a box= Consider a particle living in a box [0,//L//].
 * Class Session 06 : The Method of Images **

Case of the periodic box
math \displaystyle\sum_{n=-\infty}^\infty g(n) = \displaystyle\sum_{w=-\infty}^{\infty} \int_{-\infty}^{\infty} d\phi \:g(\phi) \: e^{-2i\pi w\phi} math > show that math \rho^{\text{per},L}(x,x',\beta) = \displaystyle\sum_{w=-\infty}^\infty \rho^{\text{free}}(x,x'+wL,\beta) math math \displaystyle\int_{0}^L dx\, \rho^{\text{per},L}(x,x',\beta) = 1. math > You can show this identity by direct computation and by using the evolution equation of the density matrix, together with the boundary conditions. Make the analogy with a particle diffusing on a ring.
 * We start from expression (3.11) of SMAC for the density matrix with periodic boundary conditions. Using the Poisson formula
 * Check that

Case of the box with hard walls (SMAC 3.1.3 pages 137-139)
math \displaystyle \rho^{\text{box},L}(x,x',\beta) = \sum_{w=-\infty}^\infty \Big\{\rho^{\text{free}}(x,x'+2wL,\beta)-\rho^{\text{free}}(x,-x'+2wL,\beta)\Big\} math > This writes: //ρ// box, //L// (//x//,//x//',//β//) = //ρ// per, 2//L// (//x//,//x//',//β//) − //ρ// per, 2//L// (//x//, −//x//',//β//).
 * What are the eigenvectors and eigenvalues of a free particle living inside a box [0,//L//]?
 * Compute the corresponding density matrix //ρ// box,//L// (//x//,//x//',//β//) at inverse temperature //β//.
 * Express //ρ// box,//L// (//x//,//x//',//β//) in terms of the free density matrix //ρ// free,//L// (//x//,//x//',//β//), using again Poisson formula, as

math \displaystyle\int_{0}^L dx\, \rho^{\text{box},L}(x,x',\beta) \neq 1 math
 * Explain why

Method of images for a single hard wall
Using a geometrical construction for the Feynman paths, it is possible to obtain the same result in few lines. We start by considering a simpler example: a particle living on the semi-infinite [0,+//∞//[ line with a unique hard wall at //x// = 0. [[image:image_construction_wall.png width="230" caption="Figure 1. The free density matrix as a sum over two class of path."]] math \rho^{\text{wall}}(x,x',\beta) = \rho^{\text{free}}(x,x',\beta) - \rho^{\text{free}}(x,-x',\beta) \qquad\qquad\qquad(1) math math \displaystyle \int_0^\infty dx\,\rho^{\text{wall}}(x,x',\beta) math
 * Analysing figure 1 and thinking a little bit we can write the density matrix of the particle at inverse temperature //β//:
 * Compute the following integral and comment.
 * By using the same trick and decomposing paths in different classes (see figure 2 and 3 below), it is possible to rederive the expression of the density matrix in a box (see SMAC part 3.3.3 pages 155-157).



Sampling in a box
We now sample the paths contributing to //ρ// box,//L// (//x//,//x//',//β//).
 * 1) First idea: start a Lévy construction and reject if the path is outside the box. Can you estimate the rejection rate?
 * 2) Smart Trick: it is better to sample the path on large scales first, and then to work one's way to small scale. Do you improve the acceptance rate?
 * 3) Imagine a rejection-free Monte-Carlo algorithm making use of this formula.

Idea number 1: code format="python" import random, pylab import math beta = 10 N = 500 Lbox = 10 dt = beta/float(N) B=[0 for n in range(N+1)] B[0] = 1. B[N] = 3. xlist = [n*dt for n in range(N+1)] itot = 1 ireject = 0 while itot < N:    DT = dt * (N-itot) mu = ( DT*B[itot-1] + dt*B[N] ) / ( dt+DT ) sigma = 1./math.sqrt( (dt+DT) / dt / DT) Bnew = random.gauss(mu, sigma) B[itot] = Bnew if Bnew < 0: itot,ireject = 0,ireject+1 if Bnew > Lbox: itot,ireject = 0,ireject+1 itot += 1 print beta, 'beta', ireject,'rejection number' pylab.title('Sampling in a box') pylab.plot(B,xlist, 'r-') pylab.plot([0 for n in range(N+1)],xlist, 'k-') pylab.plot([Lbox for n in range(N+1)],xlist, 'k-') pylab.axis([-5, Lbox+5, 0, beta]) pylab.show exit code

Analogy with standard diffusion of a one dimensional Brownian particle
There is an interesting analogy between our non-interacting particle in Quantum Mechanics and a Brownian motion. code format="python" from pylab import * x0 = 4. time = 500 steps = 200 xspace = linspace(0, 15, steps) dt = 0.7 rhob = [] for n in range(time): rhob.append([exp(-(xspace[i]-x0)**2/2./dt)/sqrt(2*pi*dt)-exp(-(xspace[i]+x0)**2/2./dt)/sqrt(2*pi*dt) for i in range(len(xspace))]) dt += 0.025 ion#interactive mode line, = plot(xspace, rhob[0]) plot(xspace, rhob[0]) plot([0,0],[1,0],'k-') axis([xspace[0]-5, xspace[-1], 0, .5]) for n in range(len(rhob)): line.set_ydata(rhob[n]) draw code
 * 1) Can you explicit this analogy? What is the signification of the spatial integrals of //ρ//(//x//,//x//',//β//) that you have computed?
 * 2) What are the boundary conditions conditions verified by the density matrix //ρ// wall (//x//,//x//',//β//) given in equation (1)?
 * 3) Find the equivalent of equation (1) for //reflecting// boundary conditions.

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