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\begin{document}
\heading{E8032: Sub diffusion of Brownian particle} \auname{Michael Chapsky}

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\Dn

Question\\

The motion of a brownian particle in 1D is given by the Hamiltonian:
\[{H_{Total}}(x,p;A(t)) = \frac{1}{{2m}}{\left( {p - A(t)} \right)^2} + {H_{Bath}(x)}\]
and the equation of motion is:
\[m\frac{{\partial  <v> }}{{\partial t}} =  - \eta  <v>  + f(t)\]

\begin{enumerate}
\item Define $ f(t) $ from $ A(t) $.
\item What is the generalised susceptability $ \chi $. (of $ v $ driven by $ A $)
\item Find the power spectrum  $ \tilde C(\omega ) $ of the velocity $ v $. 
\item Find the explicit expression for the correlation function $ C(\tau ) $ in the limit of high and low temperatures.\\

For the following questions assume $ T=0 $.
\item Find the coefficient $ C_0 $ in $ C(\tau ) \sim  - C_0 / \tau ^2 $.
\item Given the definition $ S(t) = \left\langle {{{\left( {x(t) - x(0)} \right)}^2}} \right\rangle $, Find an expression for $ \frac{{dS(t)}}{{dt}} $, using the correlation function  $ C(\tau ) $.
\item Given $ S(t_0) =S_0 $, find $ S(t) $ for $ t> t_0 $.


\end{enumerate}
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\Dn

Solution\\

1.
Hamilton equations give:
\[\dot x = \frac{{\partial H}}{{\partial p}} = \frac{1}{m}\left( {p - A} \right)\]
\[\dot p =  - \frac{{\partial H}}{{\partial x}} =  - \frac{{\partial {H_{bath}}}}{{\partial x}}\]
They give a straightforward solution (equation of motion):
\[m\frac{{\partial \dot x}}{{\partial t}} =  - \frac{{\partial {H_{bath}}}}{{\partial x}} - \frac{{\partial A}}{{\partial t}}\]
The two terms on the right can then be easily identified (taking an ensamble average):
\[f(t) =  - \frac{{\partial A}}{{\partial t}}\]
\[\eta  = \frac{{\partial {H_{bath}}}}{{\partial x}}\] 

2. Fourier transforming the equation of motion results in:
\[\left( {\eta  - i\omega m} \right){v_\omega } = {i \omega A_\omega }\]
and if we define $ {\tau _ \eta} = m/\eta $ we get the we get:
\[\chi (\omega ) = \frac{1}{\eta }\frac{{i\omega }}{{1 - i\omega {\tau _\eta }}}\]
 
3. The FD relations are:
\[\frac{{{\mathop{\rm Im}\nolimits} \left[ \chi  \right]}}{\omega } = \frac{1}{{\hbar \omega }}\tanh \left( {\frac{{\hbar \omega }}{{2{K_B}T}}} \right) \cdot \tilde C(\omega )\]
and from (2) it is known that $\frac{{{\mathop{\rm Im}\nolimits} \left[ \chi  \right]}}{\omega } = \frac{ 1}{\eta }\frac{ 1}{{1 + {{\left( {\omega {\tau _\eta }} \right)}^2}}}$ we therefore get the power spectrum of $ v $ to be:
\[\tilde C(\omega ) = \frac{1}{\eta }\frac{{\hbar \omega }}{{1 + {{\left( {\omega {\tau _\eta }} \right)}^2}}}\coth \left( {\frac{{\hbar \omega }}{{2{K_B}T}}} \right)\]
 
4. In order to obtain the high and low temperature limits we note the behaviour of  $ \coth \left( x \right)\ $ at the requested limits:
\[\coth \left( x \right)\mathop  \approx \limits_{x \to 0} \frac{1}{x}\]
\[\coth \left( x \right)\mathop  \approx \limits_{\left| x \right| \to \infty } sign\left( x \right)\]
we then can get for high temperatures ($ K_B T  \gg \frac{1}{2} \hbar \omega $):
\[\tilde C(\omega )= \frac{1}{\eta }\frac{{2{K_B}T}}{{1 + {{\left( {\omega {\tau _\eta }} \right)}^2}}}\]
and for the low temperatures (${K_B}T \ll \frac{1}{2}\hbar \omega $):
\[\tilde C(\omega ) = \frac{1}{\eta }\frac{{\hbar \left| \omega  \right|}}{{1 + {{\left( {\omega {\tau _\eta }} \right)}^2}}}\]

The correlation function is the inverse Fourier Transform of $ \tilde C(\omega) $.\\
a) For high temperatures we get:
\[C(\tau ) = \int\limits_{ - \infty }^{+\infty}  {\frac{2K_B T}{\eta }\frac{{d\omega }}{{2\pi }}\frac{{\exp \left[ {i\omega \tau } \right]}}{{1 + {{\left( {\omega {\tau _\eta }} \right)}^2}}}} \]
and setteing $ \omega \tau _ \eta =x$ we get the integral:
\[C(\tau ) = \frac{{{K_B}T}}{{\pi m}}\int\limits_{ - \infty }^\infty  {\frac{{\exp \left[ {ix\tau /{\tau _\eta }} \right]dx}}{{1 + {x^2}}}} \]
using the residue theorm the integral is easily solvable:
\[C(\tau ) = \frac{{{K_B}T}}{m}\exp \left[ { - \frac{\tau }{{{\tau _\eta }}}} \right]\]
b) For low temperature the FT gives:
\[C(\tau ) =  - \frac{\hbar }{{\pi \eta }} \cdot \frac{1}{{{\tau ^2}}}\]
 
5. since $ T=0 $ we can take the low Temp. approximation in (4) and we have:
\[{C_0} = \frac{\hbar }{{\pi \eta }}\]
 
6.
Since the relations between the autocorrelation of $v$ and $x$ is (for $t=t_2-t_1$):
\[{C^{xx}}(t) = \int\limits_0^{{t_2}} {\int\limits_0^{{t_1}} {\left\langle {v(x)v(y)} \right\rangle dxdy} }  = \int\limits_0^{{t_2}} {\int\limits_0^{{t_1}} {\left\langle {{C^{vv}}(x - y)} \right\rangle dxdy} } \]
after some calculations we get:
\[\frac{d}{{dt}}S(t) = \int\limits_0^t {C(\tau )d\tau  = \frac{\hbar }{{\pi \eta }}\frac{1}{t }} \]

7.
Using the result od (6) we get:
\[\int\limits_{{t_0}}^t {dS}  = \int\limits_{{t_0}}^t {\frac{{dt}}{t}\frac{\hbar }{{\pi \eta }}} \]
and finally:
\[S(t) = S({t_0}) + \frac{\hbar }{{\pi \eta }}\ln \left[ {\frac{t}{{{t_0}}}} \right]\]

\end{document}
