\documentclass[11pt,fleqn]{article}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% template that does not use Revtex4
%%% but allows special fonts
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%
%%% Please use this template.
%%% Edit it using e.g. Notepad
%%% Ignore the header (do not change it)
%%% Process the file in the Latex site
%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Page setup
\topmargin -1.5cm
\oddsidemargin -0.04cm
\evensidemargin -0.04cm
\textwidth 16.59cm
\textheight 24cm
\setlength{\parindent}{0cm}
\setlength{\parskip}{0cm}
% Fonts
\usepackage{latexsym}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{bm}
% Math symbols I
\newcommand{\sinc}{\mbox{sinc}}
\newcommand{\const}{\mbox{const}}
\newcommand{\trc}{\mbox{trace}}
\newcommand{\intt}{\int\!\!\!\!\int }
\newcommand{\ointt}{\int\!\!\!\!\int\!\!\!\!\!\circ\ }
\newcommand{\ar}{\mathsf r}
\newcommand{\im}{\mbox{Im}}
\newcommand{\re}{\mbox{Re}}
% Math symbols II
\newcommand{\eexp}{\mbox{e}^}
\newcommand{\bra}{\left\langle}
\newcommand{\ket}{\right\rangle}
% Mass symbol
\newcommand{\mass}{\mathsf{m}}
\newcommand{\Mass}{\mathsf{M}}
% More math commands
\newcommand{\tbox}[1]{\mbox{\tiny #1}}
\newcommand{\bmsf}[1]{\bm{\mathsf{#1}}}
\newcommand{\amatrix}[1]{\begin{matrix} #1 \end{matrix}}
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}
% Other commands
\newcommand{\hide}[1]{}
\newcommand{\drawline}{\begin{picture}(500,1)\line(1,0){500}\end{picture}}
\newcommand{\bitem}{$\bullet$ \ \ \ }
\newcommand{\Cn}[1]{\begin{center} #1 \end{center}}
\newcommand{\mpg}[2][1.0\hsize]{\begin{minipage}[b]{#1}{#2}\end{minipage}}
\newcommand{\Dn}{\vspace*{3mm}}
\def\Abs#1{\left| #1 \right|}
\def\Paren#1{\left( #1 \right)}
\def\Brack#1{\left[ #1 \right]}
\def\Seq#1{\left\langle #1 \right\rangle}
% Figures
\newcommand{\putgraph}[2][0.30\hsize]{\includegraphics[width=#1]{#2}}
% heading
\newcommand{\exnumber}[1]{\newcommand{\exnum}{#1}}
\newcommand{\heading}[1]{\begin{center} {\Large {\bf Ex\exnum:} #1} \end{center}}
\newcommand{\auname}[1]{\begin{center} {\bf Submitted by:} #1 \end{center}}
\begin{document}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\exnumber{8490}
\heading{Stochastic rate equation}
\auname{Kallos Itai}
{\bf The problem:}
\Dn
Consider $N$ classical particles in a two site system.
The two sites are subjected to a potential difference $\varepsilon$.
The temperature of the system is~$T$.
Define $n\in[-N,N]$ as the occupation difference.
%
In items (3-6) assume that the thermalization process
can be described by a stochastic rate equation
%
\[ \frac{dn}{dt} \ = \ -\gamma n + A(t) \]
%
where $A(t)$ is a noisy term that reflects
the fluctuations of the potential difference.
Assuming that it has an average value $A_0$ and a power spectrum $\phi(\omega)$,
it follows that $n$ relaxes to an average value $\langle n \rangle$,
with fluctuations that are characterized by a power spectrum $C(\omega)$.
\Dn
(1)
Write what is the interaction energy $H_{\text{int}}$ of $n$ with the field $\varepsilon$.
Later you will have to be careful with the identification of the conjugate variables.
\Dn
(2)
Using the canonical formalism, find what
are $\langle n \rangle$ and Var$(n)$.
Additionally provide approximations for small $\varepsilon$.
\Dn
(3)
Determine what is $A_0$ such that $\langle n \rangle$
would be consistent with the canonical result.
Assuming small $\varepsilon$ deduce that $A_0 \propto \epsilon$,
and find the pre-factor.
\Dn
(4)
What is the $\chi(\omega)$ that characterizes the
response of $n$ to the applied potential
in the linear-response regime?
Assume that the dynamics are described
by the stochastic rate equation;
care to identify correctly the conjugate variables;
and take into account your answer to item~(3).
\Dn
(5)
Deduce from the fluctuation-dissipation relation
what is the power spectrum $C(\omega)$.
Care to use the appropriate definition for $\chi(\omega)$,
else the result will come out wrong.
\Dn
(6)
Deduce what is the power spectrum $\phi(\omega)$
that is required in order to reproduce $C(\omega)$
from the stochastic rate equation.
\Dn
{\bf Advice:} In item (5) verify that your result is consistent
with the answer to item (2). Likewise you can debug the
numerical pre-factor in your answer to item (6).
Care about factors of "2" in your answers. Failure to
provide strictly correct pre-factors will be regarded
as an essential error.
\Dn\Dn
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}
\Dn
(1)
Let us consider a two site system, with $N$ particles. The potential difference $\varepsilon$ can be given by the following formulation: a particle in the first site has $-\varepsilon/2$ energy, while a particle in the second site has $\varepsilon/2$.
The interaction Hamiltonian is described by
\begin{equation}
\mathcal{H}_{\text{int}}=-\frac{\varepsilon }{2}N_1+\frac{\varepsilon }{2}N_2=-\frac{\varepsilon}{2}n
\end{equation}
Where $N_1,\,N_2$ are the occupation of the sites, and $n=N_1-N_2$ is the occupation difference.
Let us identify our conjugated variables
\begin{equation}
-\frac{\partial\mathcal{H}_{\text{int}}}{\partial n} = \frac{\varepsilon}{2}
\end{equation}
\Dn
(2)
Let us calculate the partition function for a single particle
\begin{equation}
Z_{1}=e^{-\frac{\beta\varepsilon}{2}}+e^{\frac{\beta\varepsilon}{2}}=2\cosh{\frac{\beta\varepsilon}{2}}
\end{equation}
The particles don't interact with each other, therefore the sum over states can be factorized and we have
\begin{equation}
Z=\Paren{2\cosh{\frac{\beta\varepsilon}{2}}}^N
\end{equation}
This is the partition function for N particles. We calculate $\langle n \rangle$ in the following manner
\begin{equation}
\langle n \rangle=-\frac{2}{\varepsilon}\langle \mathcal{H}_{\text{int}} \rangle=-\frac{2}{\varepsilon} E =\frac {2}{\varepsilon} \frac{\partial }{\partial \beta}\ln Z=N\frac{\sinh{\frac{\beta\varepsilon}{2}}}{\cosh{\frac{\beta\varepsilon}{2}}}=N\tanh{\frac{\beta\varepsilon}{2}}
\end{equation}
We notice that in the limit $\varepsilon\to0$ we have $\langle n \rangle\to0$ and in the limit $\varepsilon\to\infty$ we have $\langle n \rangle\to N$, as expected. Using the same logic as before we substitute into the known variance equation
\begin{equation}
\mathrm{Var}\Paren{n}=\frac{4}{\varepsilon^2}\mathrm{Var}\Paren{E}=\frac{4}{\varepsilon^2}\frac{\partial^2 }{\partial \beta^2}\ln Z = \frac{4}{\varepsilon^2}\frac{\partial}{\partial \beta}\Paren{N\frac{\varepsilon}{2}\tanh{\frac{\beta\varepsilon}{2}}}=\frac{N}{\cosh^2{\frac{\beta\varepsilon}{2}}}
\end{equation}
For the approximation of small $\varepsilon$ we have
\begin{equation}
\langle n \rangle\approx \frac{N\beta}{2}\varepsilon
\end{equation}
And for the variance we obtain
\begin{equation}
\mathrm{Var}\Paren{n}\approx N
\end{equation}
\Dn
(3)
We take the average of the rate equation in steady state, and compare it with the canonical result to find the pre-factor
\begin{equation}
\Seq{\frac{\mathrm{d}n}{\mathrm{d}t}}=-\gamma\Seq{n}+\Seq{A\Paren{t}}=0\Rightarrow A_0=\gamma\Seq{n}
\end{equation}
Hence in the small $\varepsilon$ approximation using Eq.~\!(7) we have
\begin{equation}
A_0=\frac{N\beta\gamma }{2}\varepsilon
\end{equation}
\Dn
(4)
By Using the Fourier transform on our rate equation, and substituting the result from the previous paragraph we have
\begin{equation}
-i\omega n_\omega=-\gamma n_\omega + N\beta\gamma \frac{\varepsilon_\omega}{2}\Rightarrow n_\omega=N\beta \gamma \frac{1}{-i\omega+\gamma}\frac{\varepsilon_\omega}{2}
\end{equation}
From which we derive an expression for $\chi(\omega)$, one should be careful and remember the correct conjugate variables which we found in Eq.~\!(2)
\begin{equation}
\chi\Paren{\omega}=N\beta\gamma \frac{1 }{\gamma-i\omega}=N\beta\gamma \frac{i\omega+\gamma }{\gamma^2+\omega^2}
\end{equation}
\Dn
(5)
Using the fluctuation dissipation relation (FDR) AC version, under the canonical preparation assumption
\begin{equation}
\mathrm{Im}\Brack{\chi\Paren{\omega}}=\tanh\Paren{\frac{\omega}{2T}}C\Paren{\omega}
\end{equation}
In the classical limit, which is obtained by taking the small $\omega$ limit, we have $\tanh\Paren{{\omega}/{2T}}\approx\omega/2T$. Solving for the power spectrum we obtain
\begin{equation}
C\Paren{\omega}=\frac{2T}{\omega}N\beta\gamma\frac{\omega}{\gamma^2+\omega^2}=2N\gamma\frac{1}{\gamma^2+\omega^2}
\end{equation}
For consistency let us remember that the relation of the power spectrum to the variance is the inverse Fourier transform at $\tau=0$. From the well known Fourier transform of a Lorenzian we obtain
\begin{equation}
\mathrm{F.T.}^{-1}\Brack{C\Paren{\omega}}=Ne^{-\gamma|\tau|}
\end{equation}
This yields $C\Paren{\tau=0}=N$ which is consistent with our previous results.
\Dn
(6)
We solve this \textit{via} the rate equation in Fourier-space which leads to
\begin{equation}
\mathcal{C}\Paren{\omega}=\Paren{\omega^2+\gamma^2}^{-1}\mathcal{\phi}\Paren{\omega}
\end{equation}
Solving for $\phi$ we have
\begin{equation}
\phi\Paren{\omega} = \Paren{\omega^2+\gamma^2}C\Paren{\omega} = 2N\gamma\
\end{equation}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document}