\input twocolumn
%\magnification\magstep1 \nopagenumbers%\parskip=12 pt plus 4pt minus 4pt  
%\advance \hsize by .4in\advance \hoffset by -.2in
%\advance \vsize by .4in \advance \voffset by -.2in
\nopagenumbers
\def \chapno{9} %roman for chapters; arabic for problem sets; 
\newcount\notenumber \notenumber=1
\def\problem#1{\par\medskip\noindent \chapno .\the\notenumber \global\advance
\notenumber by 1 ~{\sl #1}\quad}
\def\?#1{ } 
\line{Physics 352 \hfill Spring, 1995 \hfill Problem Set \chapno \hfill Due
date: 
Th. June 1}\smallskip\hrule\bigskip

\head{Monte-carlo linear response} In class we showed that the response to a
weak perturbation could be deduced from the fluctuations of the unperturbed
system.  If the perturbation changed the configuration energy $E(c)$ by an
amount $\Delta E(c) \definedas f(t) A(c)$, then the average of another system
variable $B(c)$ could be expressed as 
$$
\expectation{B}_f(t) - \expectation{B}_0 = \integral_{-\infinity}^t dt' ~ 
\chi_{AB}(t-t') f(t')
,$$
where
$$
\chi_{AB}(t) = -\beta {d\over dt}\expectation{
(A(0)-\expectation{A}) (B(t) -\expectation{B})
}_0 .$$
We showed this for systems where the time development between a perturbation at
time $t'$ and its response at time $t$ was {\it deterministic}: Given a
particular initial configuration $c'$ at time $t'$, the later configuration $c$
appearing at time $t$ was uniquely determined.  You might well be skeptical
about this requirement, since we seem to be using it over times where we would
not expect to know the final configuration uniquely.

Happily, this is not a worry.  The fluctuation-response theorem above is true
even when the dynamics is not deterministic.  In particular, the theorem is
true if the dynamics is {\it Monte-Carlo} dynamics.  In non-deterministic
dynamics, if we are given an initial configuration $c'$ we don't know the
later configuration $c$.  But we do know the {\it conditional probability}
$q(c', c, t)$ (My convention, following Handout D, is to list the earlier
configuration first.)  Knowing this $q$, we are able to find the {\it average}
of an observable $B$ if it is measured a time $t$ after the system was known
to be in configuration $c'$.  We call this conditional average $B(t, c')$. 
Using this definition of $B(t, c')$, the rest of the derivation of the
fluctuation-response theorem as given in class follows.  In particular, if a
small, constant $f$ is applied the probability that the system is in
configuration $c'$ is given by its equilibrium value $p_f \definedas e^{-\beta
E(c') - \beta fA(c')}/\sum_{c''} e^{-\beta E(c'') -\beta f A(c'')}$.  If the
field
$f$ is removed at time 0, the average
$\bar B(t)$ at a time $t$ is given by 
$$\eqalign{
\bar B(t) - \expectation{B}_0 =& \sum_{c'} p_f(c') \quad \quad \sum_{c} q(c', c,
t) (B(c) - \expectation{B}_0)\cr
 =& f \sum_{c'} \left .{\partial p_f\over \partial f}\right
|_{f=0} (B(t, c')- \expectation{B}_0)\cr
=& \beta f \sum_{c'} p_0(c') (A(c') -
\expectation{A}_0) (B(t, c')- \expectation{B}_0) \cr
&\definedas  \beta f \expectation{(A -
\expectation{A}_0) (B(t) - \expectation{B}_0)}. }
$$
 
\problem{Monte-carlo random walk} Consider the monte-carlo simulation where a
particle on a one-dimensional ring lattice of length $L$ is moved randomly
one step to the left or to the right in a time step.  On this ring, the site with
$i=L$ is the same as the the site with $i=0$. The probability that the particle
is at site $i$ at time step $t$ is $p(i, t)$.
\item{a)}
Show that at any time 
$$
\partial p(i) /\partial t \definedas p(i, t+1) - p(i, t)
= {\rm const} [p(i-1, t) + p(i+1, t) - 2 p(i, t)]
.$$
What is the prefactor?
\item{b)}
If $p(i)$ is initially a sine wave, $p(i, 0) = \sin(2\pi/L ~ ki)$, show that 
$p(i, t) = p(i, 0) \exp(-t/\tau(k))$.  (In addition to this sine wave, there
should be an added constant, so that the total $p$ is positive.  Since the
equation for $p$ is linear, we are free to add a constant in this way.)  Find the
decay time $\tau(k)$ as a function of the mode number $k$.

\problem{Monte-carlo correlations} We define the density $n_k$ as $\sum_i n(i)
\sin(2\pi/L ~ ki)$.  Since there is only one particle in our simulation, $n(i)$
is simply 1 for the occupied site and 0 for the rest.
\item{a)} Using a lattice of length $L = 16$, find $n_k(t)$ for a series of 10000
configurations $t$ spaced by 20 time steps each.  Find the average value of 
$n_k(t') n_k(t' + t)$ over all $t'$'s from 0 to $10000 - t$ and for $0 < t <
500$.  Do this for two mode numbers: $k=1$ and $k=2$.  
\item{b)}How does the average
$\expectation{n_k(t') n_k(t' + t)}_{t'}$ decay with time?  Is this consistent
with the diffusion law discussed in Chandler, 8.4 and Problem 9.1? 
\item{c)} How does the decay time depend on $k$?   Does it double, quadruple,
cut in half or what when $k$ goes from one two?  Is this behavior consistent
with the behavior of Prob. 9.1?

\problem{Monte-carlo response function} Add a potential $\mu(i) = \mu_0
\sin(2\pi/L ~ ki)$ to the simulation in the previous problem, using the
Metropolis dynamics of Problem 4.2.  Take $L=16$, $k=1$ and $\beta \mu_0 = 0.1$.
\item{a)} Find $\expectation{n_k}$ for this static $\mu_0$.  Does it agree with 
$-\beta [\partial^2 A(\mu_0)/\partial \mu_0^2] ~ \mu_0$ as linear-response theory
predicts?
\item{b)} Now take $\beta \mu_0 = .1 \sin(t/\tau)$, where $\tau$ is 100 time
steps. Find the average power dissipated in 10000 time steps: $P\definedas \frac
1 {10000}\sum_t\sum_i
\mu(i, t) [n(i, t+1) - n(i, t)]$
Does it agree with the fluctuation-dissipation theorem as stated in Chandler,
page 260, top?  You don't have to perform the fourier transform numerically. 
Just assume that the correlation function has the expected exponential time
dependence, with the relaxation time and amplitude you measured in the
previous problem.  Note that perfect agreement is not expected, since the
perturbation amplitude
$\mu_0$ is not infinitely small.
\bye