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\exnumber{8490}
\heading{Stochastic rate equation} 
\auname{Eli Gudinetsky}

{\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)
Determined 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 is 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

{\Large Wrong factor of 2 throughout, and last item cumbersome}


\begin{enumerate}
\item
Let us consider a two site system, with $N$ particles hopping between them. Assume that the potential difference $\varepsilon$ is given by the fact that a particle in the first site receives $-\varepsilon/2$ to its energy, while a particle on the second site receives $\varepsilon/2$.
The interaction is described by the following interaction Hamiltonian:
\begin{align}
\mathcal{H}_{\text{int}}=-\frac{\varepsilon }{2}N_1+\frac{\varepsilon }{2}N_2=-\frac{\varepsilon}{2}n\, .
\end{align}
Here $N_1,\,N_2$ are the occupation of the first and second sites and $n=N_1-N_2=2N_1-N$ is the occupation difference.
\item
There are a few different ways to find $\langle n \rangle$. Let us calculate the partition function for 1 particle:
\begin{align}
Z_{1}=2\cosh{\frac{\beta\varepsilon}{2}}\, .
\end{align}
The indistinguishable particles don't interact with each other, so we have
\begin{align}
Z=\frac{1}{N!}\Paren{2\cosh{\frac{\beta\varepsilon}{2}}}^N\, ,
\end{align}
which is the partition function for N particles. The simplest way of calculating $\langle n \rangle$ is by noticing that $\langle n \rangle=2\frac{\langle E\rangle}{-\varepsilon}$, so
\begin{align}
\langle n \rangle=\frac {2}{\varepsilon} \frac{\partial \ln Z}{\partial \beta}=N\frac{\sinh{\frac{\beta\varepsilon}{2}}}{\cosh{\frac{\beta\varepsilon}{2}}}=N\tanh{\frac{\beta\varepsilon}{2}}\, .
\end{align}
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. \\
In the same way we say that $\mathrm{Var}\Paren{n}=4\mathrm{Var}\Paren{E}/\varepsilon^2$, so
\begin{align}
\mathrm{Var}\Paren{n}=\frac{4}{\varepsilon^2}\mathrm{Var}\Paren{E}=-\frac{4}{\varepsilon^2}\frac{\partial E}{\partial \beta}=-\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{align}

In the limit of small $\varepsilon$ we have:
\begin{align}
\langle n \rangle\approx \frac{N\beta}{2}\varepsilon\, ,
\end{align}
and (first non-zero term in the expansion)
\begin{align}
\mathrm{Var}\Paren{n}\approx N\, .
\end{align}
\item
Taking the average of the rate equation in accordace with the canonical result (steady state), we have
\begin{align}
\Seq{\frac{\mathrm{d}n}{\mathrm{d}t}}=-\gamma\Seq{n}+\Seq{A\Paren{t}}\Rightarrow A_0=\gamma\Seq{n}\, ,
\end{align}
so in the small $\varepsilon$ approximation we have:
\begin{align}
A_0=\frac{N\gamma\beta}{2}\varepsilon\, .
\end{align}
\item
By inspecting the rate equation and taking the Fourier transform we get:
\begin{align}
-i\omega n_\omega=-\gamma n_\omega +\frac{N\beta\gamma}{2}\varepsilon_\omega\, ,
\end{align}
which immediately implies that $\chi\Paren{\omega}$ is:
\begin{align}
\chi\Paren{\omega}=\frac{N\gamma\beta}{2}\frac{1}{\gamma-i\omega}=\frac{N\gamma\beta}{2}\,\frac{\gamma+i\omega}{\gamma^2+\omega^2}\, .
\end{align}
\item
The fluctuation dissipation relation reads:
\begin{align}
\mathrm{Im}\Brack{\chi\Paren{\omega}}=\tanh\Paren{\frac{\omega}{2T}}C\Paren{\omega}\, ,
\end{align}
in the DC limit we have $\tanh\Paren{{\omega}/{2T}}\approx\omega/2T$ and the power spectrum is:
\begin{align}
C\Paren{\omega}=\frac{2T}{\omega}\frac{N\gamma\beta}{2}\frac{\omega}{\gamma^2+\omega^2}=\frac{N}{\gamma}\frac{1}{1+\Paren{\omega/\gamma}^2}\, .
\end{align}
A nice way to check this result is to see whether $C\Paren{\tau=0}$ coincides with $\mathrm{Var}\Paren{n}\approx N$, in order to do so we calculate $\mathrm{F.T.}^{-1}\Brack{C\Paren{\omega}}$:
\begin{align}
\mathrm{F.T.}^{-1}\Brack{C\Paren{\omega}}=Ne^{-\tau\gamma}\, ,
\end{align}
which gives $C\Paren{\tau=0}=N$ as it should!
\item
Let us multiply the rate equation for time $t'$ by the same equation for time $t''$. The result is as follows:
\begin{align}
&\frac{\mathrm{d}}{\mathrm{d}t'}\frac{\mathrm{d}}{\mathrm{d}t''}n\Paren{t'}n\Paren{t''}+\gamma n\Paren{t'}\frac{\mathrm{d}}{\mathrm{d}t''}n\Paren{t''}+\gamma n\Paren{t''}\frac{\mathrm{d}}{\mathrm{d}t'}n\Paren{t'}+\gamma^2n\Paren{t'}n\Paren{t''} =\\
&=A\Paren{t'}A\Paren{t''}\, .
\end{align}
Taking the average of the above equation, and going to a new variable $\tau=t'-t''$ we have:
\begin{align}
-\frac{\mathrm{d^2}}{\mathrm{d}\tau^2}C\Paren{\tau}+\gamma^2C\Paren{\tau}=\phi\Paren{\tau}\, .
\end{align}
Taking the Fourier transform we get:
\begin{align}
\Paren{\omega^2+\gamma^2}C\Paren{\omega}=\phi\Paren{\omega}\, .
\end{align}
This gives $\phi\Paren{\omega}=N\gamma$, meaning that $\phi\Paren{\tau}=N\gamma\delta\Paren{\tau}$, both of them have the appropriate dimensions.
\end{enumerate}



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