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\exnumber{8032}
\heading{Sub diffusion of Brownian particle} 
\auname{Ori Hendler}


{\bf The problem:}

\Dn


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:
\[f(t) =  - \frac{{\partial A}}{{\partial t}}\]
\[\eta \dot x   = \frac{{\partial {H_{bath}}}}{{\partial x}}\] 

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:
\[\frac{{{\mathop{\rm Im}\nolimits} \left[ \chi  \right]}}{\omega } = \frac{1}{{\hbar \omega }}\tanh \left( {\frac{{\hbar \omega }}{{2T}}} \right) \cdot \tilde C(\omega )\]
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:
\[\tilde C(\omega ) = \frac{1}{\eta }\frac{{\hbar \omega }}{{1 + {{\left( {\omega {\tau _\eta }} \right)}^2}}}\coth \left( {\frac{{\hbar \omega }}{{2T}}} \right)\]
 
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 ($  T  \gg \frac{1}{2} \hbar \omega $):
\[\tilde C(\omega )= \frac{1}{\eta }\frac{{2T}}{{1 + {{\left( {\omega {\tau _\eta }} \right)}^2}}}\]
and for the low temperatures ($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:
\[C(\tau ) = \int\limits_{ - \infty }^{+\infty}  {\frac{2 T}{\eta }\frac{{d\omega }}{{2\pi }}\frac{{\exp \left[ {i\omega \tau } \right]}}{{1 + {{\left( {\omega {\tau _\eta }} \right)}^2}}}} \]
and setteing $ \omega \tau _ \eta =x$ we get the integral:
\[C(\tau ) = \frac{{T}}{{\pi m}}\int\limits_{ - \infty }^\infty  {\frac{{\exp \left[ {ix\tau /{\tau _\eta }} \right]dx}}{{1 + {x^2}}}} \]
using the residue theorm the integral is easily solvable:
\[C(\tau ) = \frac{{T}}{m}\exp \left[ { - \frac{\tau }{{{\tau _\eta }}}} \right]\]
b) For low temperature the FT gives:\\
\[FT(\tilde C(\omega )) = \int \frac{{\exp \left[ {i\omega \tau } \right]}}{ 2\pi \eta }\frac{{\hbar \left| \omega  \right|d\omega}}{{1 + {{\left( {\omega {\tau _\eta }} \right)}^2}}}\]
We recognize the expression above is a multiplication of $|w|$ and the Lorentzian.\\
The Fourier transfrom of $|w|$ is:\[FT(|\omega|)= -\frac{1}{{{\pi \tau ^2}}} \]
And, the Fourier transfron of the Lorentzian is: \[FT(\frac{{1}}{{1 + {{\left( {\omega {\tau _\eta }} \right)}^2}}})= e^{-\gamma \tau} \]

  Now we would like to make use of the convolution theorem, so:
 \[FT(\frac{{|w|}}{{1 + {{\left( {\omega {\tau _\eta }} \right)}^2}}})= -\frac{1}{{{\pi \tau ^2}}}*e^{-\gamma \tau}\]
  Taking into account the contribution of the tales of $e^{-\gamma \tau}$ are non significant compared to $-\frac{1}{{{\pi \tau ^2}}}$ one concludes: 
  \[C(\tau ) = -\frac{\hbar}{\pi \eta \tau ^2} \]
  \\
5. since $ T=0 $ we can take the low Temp. approximation in (4) and we have:
\[{C_0} = \frac{\hbar }{{\pi \eta }}\]
 
6.
The average displacement $ x(t)-x(0)$ is zero. However the second moment is:
\[{S(t) = \int\limits_0^{{t}} {\int\limits_0^{{t}} {\left\langle {v(t'')v(t')} \right\rangle dt'dt''} }  = \frac{1}{\eta^2}\int\limits_0^{{t}} {\int\limits_0^{{t}} {\left\langle {f(t'')f(t')} \right\rangle dt'dt''} }=
\frac{1}{\eta^2}\int\limits_0^{{t}} {\int\limits_0^{{t}} { {{C(t'' - t')} } dt'dt''} } \] \\

The last equality is valid due to (as seen in class section 15.3)  $f(t)$ being a stochastic variable and satisfying ${\left\langle {f(t)} \right\rangle=0} $.\\
\\ After a change of variables we receive:
\[ {S(t)= \frac{1}{\eta^2}\int\limits_0^{{t}} {\int\limits_{-t'}^{{t'}} { {{C(\tau)}  dt'd\tau } }} = \int\limits_0^{{t}}dt' {\int\limits_{-t'}^{{t'}} { - \frac{\hbar }{{\pi \eta  }} \cdot \frac{d\tau}{{{\tau ^2}}}}} \]
So the derivative is:
\[\frac{d}{{dt}}S(t) = \int\limits_0^t {C(\tau )d\tau  = \frac{\hbar }{{\pi \eta }}\frac{1}{t }} \]

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


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