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\heading{E8490}
\auname{Eran Avraham}


Consider molecules A and B in a box. 
The molecules are subjected to weak electric field $\mathcal{E}$ 
that modifies the binding energy of the B molecules 
such that ${\varepsilon_B = \varepsilon_B^{(0)} + \alpha \mathcal{E}}$, 
where $\alpha$ is a constant.
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In the lack of  electric field the molecules reach 
chemical equilibrium  A$\leftrightarrow$B, 
such that their fractions are ${f_A=\langle N_A\rangle/N}$
and ${f_B=\langle N_A\rangle/N}$.
%
In item (c) it is assumed that the system can be described 
by the rate equation
%
\[\frac{dN_A}{dt} = k_B N_B - k_A N_A + A(t)\]
%
Where $A(t)$ is a stochastic term with zero average 
and correlation function ${\langle A(t)A(t')\rangle=C\delta(t-t')}$.
In item (e) assume that only~$k_B$ is affected by the weak electric field.

\begin{itemize}

\item[(a)] Express $\langle N_A\rangle,\langle N_B\rangle$ by $f_A$ and $f_B$ and $\alpha\mathcal{E}$.

\item[(b)] Express $\mbox{Var}(N_A)$ by $\langle N_A\rangle$ and $\langle N_B\rangle$.

\item[(c)] Determine the constants $k_A$ and $k_B$ and $D$ in the 
stochastic rate equation such that $\langle N_A\rangle$ and $\mbox{Var}(N_A)$ 
will agree with the canonical expectation of items (a) and (b).

\item[(d)] Determine how $k_B$ is modified by the weak electric field.
In other words find the coefficient in ${\delta k_B \propto \mathcal{E}}$.

\item[(e)] Find the generalized susceptibility $\chi(\omega)$ 
that relates the variation $\delta N_A$ to $\mathcal{E}$.

\item[(f)] Find the power spectrum of $\delta N_A$ in steady state.


\end{itemize}



\textbf{The solution:}

(a) In chemical equilibrium (without electric field) we have:
\[\frac{f_A}{f_B}=\frac{Z^1_A}{Z^1_B}\]
Where $Z^1_{A/B}$ is the single partition function.

 In the present of an electric field the Hamiltonian for a single type B molecule  becomes:
\[H_B=\frac{p^2}{2m}+\epsilon\]
Therefore the single partition function change:
\[Z^1_{B}\rightarrow e^{-\beta\epsilon}Z^1_B
\implies \frac{\tilde{f_A}}{\tilde{f_B}}=\frac{Z^1_A}{Z^1_B}e^{\beta\epsilon}=\frac{f_A}{f_B}e^{\beta\epsilon}
\]
Using $\tilde{f_A}+\tilde{f_B}=1$ and after a some algebra we get: 




\[\langle N_A\rangle=\frac{f_A}{f_A+e^{-\beta\epsilon}f_B}N,\langle N_B\rangle=\frac{f_B}{f_B+e^{\beta\epsilon}f_A}N\]

(b) The full partition function Z is:
\[Z=\sum_{N_A,
N_B=N-N_A}\frac{(Z^1_A)^{N_A}(Z^1_B)^{N_B}}{N_A!N_B!}=\frac{(Z^1_A+Z^1_B)^N}{N!}\]

\[\langle N^2_A\rangle=\frac{1}{Z}\sum_{N_A,
N_B=N-N_A}\frac{(Z^1_A)^{N_A}(Z^1_B)^{N_B}}{N_A!N_B!}N^2_A\]
\[=\frac{1}{Z}Z^1_A\frac{\partial}{\partial Z^1_A}(Z^1_A\frac{\partial Z}{\partial Z^1_A})=
\frac{1}{Z}Z^1_A\frac{\partial}{\partial Z^1_A}(\frac{Z^1_AN(Z^1_A+Z^1_B)^N}{N!})\]
\[=\frac{1}{Z}[Z^1_A(Z^1_A+Z^1_B)^{N-1}/N!+(Z^1_A)^2(Z^1_A+Z^1_B)^{N-2}N(N-1)/N!]\]
\[=\frac{Z^1_A}{Z^1_A+f_B}N+\frac{(Z^1_A)^2N(N-1)}{(Z^1_A+Z^1_B)^2}\]
 
\[\implies Var(N_A)=\frac{Z^1_A}{Z^1_A+Z^1_B}N+\frac{(Z^1_A)^2N}{(Z^1_A+Z^1_B)^2}=\frac{Z^1_A Z^1_B}{(Z^1_A+Z^1_B)^2}
=\frac{\langle N_A\rangle \langle N_B\rangle}{N}\]

(c) For the average  $\langle N_A(t)\rangle$ eq(*) becomes:
\[\frac{<dN_A(t)>}{dt}=k_BN-(k_A+k_B)<N_A(t)>\]
 
With the general solution: 
\[<N_A>=De^{-(k_A+k_B)t}+\frac{k_AN}{k_A+k_B}\]

And for long times $t\rightarrow \infty$
\[<N_{A}>=\frac{k_A}{k_A+k_B}N\]
Comparing with the canonical expectation $<N_A>=\frac{f_A}{f_A+f_B}N$ , $<N_B>=\frac{f_B}{f_A+f_B}N$ we get the relation 
\[ k_A/k_B=f_A/f_B\]

In order to determine $C$  we first need to solve Eq(*)
\[N_{A,h}=D(t)e^{-(k_A+k_B)t}\]
\[\dot{D}=(k_1N+A(t))e^{(k_A+k_B)t}\]
\[D=\frac{k_AN}{k_A+k_B}e^{(k_A+k_B)t}+\int_0^tA(t')e^{(k_A+k_B)t'}dt'\]
\[N_A=-\frac{k_AN}{k_A+k_B}e^{-(k_A+k_B)t}+\frac{k_AN}{k_A+k_B}+e^{-(k_A+k_B)t}\int_0^tA(t')e^{(k_A+k_B)t'}dt'\]
\[\implies N_A(t)-<N_A>=\delta N_A(t)=-\frac{k_1N}{k_A+k_B}e^{-(k_A+k_B)t}+e^{-(k_A+k_B)t}\int_0^tA(t')e^{(k_A+k_B)t'}dt'\]

\[ [\delta N_A(t)]^2=
(\frac{k_AN}{k_A+k_B}e^{-(k_A+k_B)t})^2-2\frac{k_AN}{k_A+k_B}e^{-2(k_A+k_B)t}\int_0^tA(t')e^{(k_A+k_B)t'}dt'+\]
\[e^{-2(k_A+k_B)t}\int_0^tdt''\int_0^tdt'A(t')A(t'')e^{(k_A+k_B)(t'+t'')}\]
Taking $t\rightarrow \infty$ and using $\langle A(t)A(t')\rangle
=C\delta(t-t')$ we get:
\[\langle [\delta N_A(t)]^2\rangle=e^{-2(k_A+k_B)t}\int_0^tdt''\int_0^tdt'e^{(k_A+k_B)(t'+t'')}C\delta(t'-t'')=\frac{C}{2(k_A+k_B)}\]
From  (b) we know that :
\[\langle [\delta N_A(t)]^2\rangle=\frac{<N_A><N_B>}{N} \implies C=\frac{2k_Ak_B}{(k_A+k_B)^2}N\]



(d) If the electric field is modified  $\mathcal{E}\rightarrow\mathcal{E}+\delta \mathcal{E}$ the single partition function becomes:
\[Z^1_Be^{-\beta\epsilon}\rightarrow Z^1_Be^{-\beta\epsilon-\beta\delta\epsilon}\approx Z^1_Be^{-\beta\epsilon}(1-\delta\beta\epsilon)\implies f_B\rightarrow f_B(1-\delta\beta\epsilon)\]
Keeping in mind that $k_A$ stays unaffected  by the electric field we get:
\[\implies k_B\rightarrow k_B-\beta\delta\epsilon k_B\]

(e) The rate eq(*):
\[\dot{N_A}=k_BN-(k_A+k_B)N_A\]
Setting $\tilde{N_A}=N_A-\langle N_A\rangle$ and adding the variation  $\mathcal{E}\rightarrow\mathcal{E}+\delta \mathcal{E}$ we get:
\[\dot{\tilde{N_A}}=-(k_A+k_B)\tilde{N_A}+\beta\delta k_B\tilde{N_A}\mathcal{E}\]
Taking into account that the field is weak we get:
\[\dot{\tilde{N_A}}\approx-(k_A+k_B)\tilde{N_A}+\beta\delta k_B N^{(0)}_A\mathcal{E}\]
Fourier transform the eq gives:
\[{\tilde{N_A}}(\omega)=\frac{\beta\delta k_B N^{(0)}_A}{(k_A+k_B)-i\omega}\mathcal{E}(\omega)\]
\[\implies \chi(\omega)=\frac{\beta\delta k_B N^{(0)}_A}{(k_A+k_B)-i\omega}\]

(f) We calculate the power spectrum $\tilde{C}(\omega)$ using the FD relations:
\[\tilde{C}(\omega)=coth(\frac{\omega}{2T})Im[\chi(\omega)]=coth(\frac{\omega}{2T})\frac{\beta\delta k_B N^{(0)}_A\omega}{(k_A+k_B)^2+\omega^2}\]

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