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\exnumber{8492}
\heading{Stochastic picture of sweep in 2-site system}
\auname{Shoham Eyal}
{\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.
%
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 $A_\epsilon$ and a power spectrum $\phi(\omega)$,
it follows that $n$ relaxes to an average value $\langle n \rangle_\epsilon$,
with fluctuations that are characterized by a power spectrum $C(\omega)$ and intensity ${\nu \equiv \tilde{C}(0)}$.
\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 $\langle n \rangle_\epsilon$ and Var$(n)$.
Additionally provide approximations for small $\varepsilon$.
\Dn
(3)
Determine what is $A_\epsilon$ such that $\langle n \rangle_\epsilon$
would be consistent with the canonical result.
Assuming small $\varepsilon$ deduce that $A_\epsilon \propto \epsilon$,
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)
Consider a quasi-static sweep process, namely, a process during which $\epsilon$ is varied slowly with constant rate $\dot{\epsilon}$. Use your result for $\chi(\omega)$ in order to express
$\langle n \rangle$ in terms of $\langle n \rangle_\epsilon$ and $\dot{\epsilon}$.
\Dn
(6)
Deduce from the fluctuation-dissipation relation
what is the correlation function $C(\tau)$ that describes the fluctuations.
Explain how your answer in item~(5) is related to the fluctuation intensity~$\nu$.
\Dn
{\bf Advice:}
Care about factors of "2" in your answers. Failure to
provide strictly correct pre-factors will be regarded
as an essential error.
Exploit item~(6) in order to double check your answer in~(5).
\Dn\Dn
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{\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_\epsilon=\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 Kubo formula for the dissipation coefficient (AC version)
\begin{equation}
\eta(\omega) \equiv \frac{Im[\chi(\omega)]}{\omega} = \frac{\gamma N\beta}{\gamma^2 + \omega^2}
\end{equation}
Hence, for the D.C limit we get
\begin{equation}
\eta_{D.C} = \frac{N\beta}{\gamma}
\end{equation}
So according the linear response theory
\begin{equation}
\langle n \rangle = \langle n \rangle_\epsilon - \frac{N\beta}{\gamma} \frac{\dot{\epsilon}}{2}
\end{equation}
\Dn
(6)
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}
C\Paren{\tau} = \mathrm{F.T.}^{-1}\Brack{C\Paren{\omega}}=Ne^{-\gamma|\tau|}
\end{equation}
And as expected we get
\begin{equation}
\nu \equiv C\Paren{\omega=0} = \frac{2N}{\gamma}
\end{equation}
\begin{equation}
\eta=\frac{\nu\beta}{2}
\end{equation}
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\end{document}