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\begin{document}

\heading{E7020: combined with E7488, Oscillators in FDT} 

\auname{Oren Rosenblatt, Michael Chapsky, Nimrod Cohen}

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\Dn

Question

Consider a mass $ M $ on a harmonic oscillator of frequency $ \Omega $ ,The oscillator experiences dumping with coefficient $ \eta $, and external force $ f(t) $. The system is at temperature $ T $.
\begin{enumerate}
\item Write the generalized susceptibility that describes the response of $ x $ to the driving by the external force $ f(t) $.
\item Using FD relations find power spectrum for $ x $ and for $ v $.
\item Find the general expression for the autocorrelation function $ \left\langle {v(t + \tau )v(t)} \right\rangle $ (in the integral form), find the exact expression for high temperatures for both the classical case, and quantum case. How is it consistent with the canonical result ?
\item Evaluate the driving force power spectrum in the high temperature limit.\\
Hint: $v(t)=v(0)e^{-t/\tau}+e^{-t/\tau}\int^t_0 e^{u/\tau}f(u)du$
\item Find the fluctuations in the limit $ \gamma \to 0 $ the fluctuations  $ \left\langle v^2 \right\rangle $ and  $ \left\langle x^2 \right\rangle $ (quantum and classical).
Show consistency with the canonical results.

\end{enumerate}
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Solution\\

1.
 The equation of motion, defining $ \gamma = \frac{\eta}{M}$, (from now on we will use this notation for convenience):
\[\ddot x + \gamma \dot x + {\Omega ^2}x = \frac{1}{M}f(t)\] 
Therefore the response of $ x $ to the external force is given by the equation:
\[x_ \omega = \frac{1}{M} \cdot \frac{1}{{{\Omega ^2} - {\omega ^2} - i\gamma \omega }} \cdot f_ \omega \]
and the susceptability itself is:
\[\chi = \frac{1}{M} \cdot \frac{1}{{{\Omega ^2} - {\omega ^2} - i\gamma \omega }} \]

2. 
FD Relations states that ( $ \tilde C( \omega )$ relates to power spectrum of $x$ unless otherwise stated by an index):
\[\frac{{{\mathop{\rm Im}\nolimits} \left( {\chi (\omega )} \right)}}{\omega } = \frac{1}{{\hbar \omega }}\tanh \left( {\frac{{\hbar \omega }}{{2{K_b}T}}} \right) \cdot { \tilde C}(\omega )\]
it is then  straightforward  to get:
\[{\tilde C^{xx}}(\omega ) ={ \hbar \omega } \cdot \coth \left( {\frac{{\hbar \omega }}{{2{K_b}T}}} \right) \cdot \frac{{{\mathop{\rm Im}\nolimits} \left[ {\chi (\omega )} \right]}}{\omega } = \frac {1}{M} \cdot \frac{{ \hbar \omega  \gamma }}{{{{\left( {{\omega ^2} - {\Omega ^2}} \right)}^2} + {\gamma ^2}{\omega ^2}}}\coth \left( {\frac{{\hbar \omega }}{{2{K_b}T}}} \right)\]
since $ v_ \omega =- i \omega x_ \omega $, in the same way we get (can also be obtained by Winner-Kintchin Theorem):
\[{\tilde C^{vv}}(\omega ) =\frac {1}{M} \cdot \frac{{\hbar {\omega ^3 \gamma}}}{{{{\left( {{\omega ^2} - {\Omega ^2}} \right)}^2} + {\gamma ^2}{\omega ^2}}}\coth \left( {\frac{{\hbar \omega }}{{2{K_b}T}}} \right)\]

3.
 Autocorrelation function is $ C( \tau)= \left\langle {v(t + \tau )v(t)} \right\rangle $ and since $C(\tau)$ is the inverse Fourier Transform of $ \tilde C(\omega)$ we get:
\[{C^{vv}}( \tau ) = \int\limits_{ - \infty }^{ + \infty } {\frac{{\hbar \omega ^3 }}{M}} \frac{\gamma }{{{{\left( {{\omega ^2} - {\Omega ^2}} \right)}^2} + {\omega ^2}{\gamma ^2}}} \exp \left( i \omega \tau \right )\coth \left( {\frac{{\hbar \omega }}{{2{K_b}T}}}\right ) \frac{{d\omega }}{{2\pi }}\]
in the limit of high temperatures ( $ \Omega \to 0$) we get $ \coth x \to \frac {1}{x}$ and the correlation integral becomes:

\[{C^{vv}}(\tau) = \int\limits_{ - \infty }^{ + \infty } {\frac{{2{K_b}T}}{M}} \frac{\gamma \exp \left( i \omega \tau \right ) }{{{\omega ^2} + {\gamma ^2}}}\frac{{d\omega }}{{2\pi }}\]
and then using contour integration and residue theorem we fonally get:
\[{C^{vv}}(\tau ) = \frac{{{K_b}T}}{M}\exp \left( { - \gamma \tau } \right)\]
so we got the brownian particle as a result.\\
the result can also be compared to the canonical result by using $\left\langle {{v^2}} \right\rangle  = C(0) = \frac{{{K_b}T}}{M}$, and
therefore we get the consistency with equipartition:
\[\left\langle {\frac{1}{2}M{v^2}} \right\rangle  = \frac{1}{2}{K_b}T\]

4.
The formal solution of Langevin equation in the high temperature limit ($\Omega \rightarrow 0$) is
\[ v(t)=v(0)e^{-t/\tau}+e^{-t/\tau}\int^t_0 e^{u/\tau}f(u)du \]
Multiplying both sides of the equation by the random force yields
\[v(t)f(t)=v(0)f(t)e^{-t/\tau}+e^{-t/\tau}f(t)\int^t_0 e^{u/\tau}f(u)du\]
This is nothing but the momentary power spectrum. The exception value of the power is
\[\left<v(t)f(t)\right>=v(0)\left<f(t)\right>e^{-t/\tau}+e^{-t/\tau}\int^t_0 e^{u/\tau}\left<f(t)f(u)\right>du\]
The first term in the right wing tends to zero since the average of a random force is zero. In order to deal with the second
term we make the following substitution 
\[ x=t-u \]
so the equation takes the form
\[\left<v(t)f(t)\right>= -\int_t^0 e^{-x/\tau}k(x)dx\]
where \[k(x)\equiv \left< f(t-x)f(t) \right>\]
When $t<\tau$ a rough approximation can be made - $e^{-x/\tau}\approx 1$ and for the other case, $t>\tau$, $k=0$ so we can
change the boundaries of the integral to the whole space and divide by two (because of the parity of integrad). 
\[\left<v(t)f(t)\right>= -\int_t^0 e^{-x/\tau}k(x)dx=\frac{1}{2}\int_{-t}^t e^{-|x|/\tau}k(x)dx\approx\frac{1}{2}\int_{-\infty}^\infty k(x)dx=\frac{C}{2}\]
The next step is to evaluate C from thermodynamic considerations. Squaring the formal solution of $v(t)$ and taking its exception value gives the following equation
\[<v^2(t)>=v^2(0)e^{-2t/\tau}+C\frac{\tau}{2}\left(1-e^{-2t/\tau}\right)\]
in the limit of $t\gg\tau$ last equation becomes 
\[<v^2>=C\frac{\tau}{2}\]
On the other side, from the equipartition law we get 
\[<v^2>=\frac{kT}{m}\]
and equating the last two equation we find
\[C=\frac{2k_BT}{m\tau}\]
substituting it back to the driving force power spectrum expression gives
\[\left<v(t)f(t)\right>=\frac{k_BT}{m\tau}=\frac{k_BT\gamma}{m}\]

5.

For the limit $ \gamma \to 0$ the expression for $ \tilde C^{vv} ( \omega)$ is (using the lorentzian definition of delta):
\[ \frac {{\tilde C^{vv}}(\omega )}{  \coth \left( {\frac{{\hbar \omega }}{{2{K_b}T}}} \right)     }= \frac{{\hbar \omega }}{M}\frac{\gamma }{{{\gamma ^2} + {{\left( {\omega  - \frac{{{\Omega ^2}}}{\omega }} \right)}^2}}}  \to \frac{{\hbar \omega }}{M}\pi \delta (\omega  - \frac{{{\Omega ^2}}}{\omega })\]

since the delta function is of composite function we use the identity
\[\delta (f(x))  = \frac{1}{{\left| {{f`}({x_i})} \right|}} \sum {\delta (x - {x_i})} \]
where $X_i$ are the solutions of $f(x)=0$\\
we therefore get:
\[\delta (\omega  - \frac{{{\Omega ^2}}}{\omega }) = \frac{1}{2}\left( {\delta (\omega  + \Omega ) + \delta (\omega  - \Omega )} \right)\]
and the fluctuations becomes
\[{C^{vv}}(\tau  = 0) = \int\limits_{ - \infty }^{ + \infty } {{{\tilde C}^{vv}}(\omega )} \frac{{d\omega }}{{2\pi }} = \int\limits_{ - \infty }^{ + \infty } {\frac{{\pi \hbar \omega }}{{4\pi M}} \coth \left( {\frac{{\hbar \omega }}{{2{K_b}T}}} \right)\left( {\delta (\omega  + \Omega ) + \delta (\omega  - \Omega )} \right)} d\omega \]
calculating the integral gives the result (quantum case):
\[\left\langle {{v^2}} \right\rangle  = \frac{{\hbar \Omega }}{{2M}}\coth \left( {\frac{{\hbar \Omega }}{{2{K_b}T}}} \right)\]

to check our result we look at the Hamiltonian of harmonic oscillator $\hat H = \left( {{a^\dag }a + \frac{1}{2}} \right)\hbar \Omega $ and since the bosonic occupation for temperature $ T $ is given by :
\[N = \left\langle {{a^\dag }a} \right\rangle  = \frac{1}{{\exp \left( {\frac{{\hbar \Omega }}{{{K_b}T}}} \right) - 1}}\]
we get for the harmonic oscillator (for writing convinience $ \beta = 1/K_b T$):
\[E = \left( {\frac{1}{{\exp \left( {\beta \hbar \Omega  } \right) - 1}} + \frac{1}{2}} \right)\hbar \Omega  = \frac{{\hbar \Omega }}{2}\frac{{\exp \left( {\beta \hbar \Omega /2 } \right) + 1}}{{\exp \left( {\beta \hbar \Omega /2} \right) - 1}} = \frac{1}{2}\hbar \Omega \coth \left( {\frac{{\hbar \Omega  }}{{{2K_b}T}}} \right)\]
now if we take into account that $\frac{1}{2}M\left\langle {{v^2}} \right\rangle  = \frac{1}{2}E$,  we see that the result is consistent with the quantum case solution and we have the equality:
\[\frac{1}{2}M\left\langle {{v^2}} \right\rangle  = \frac{1}{4}\hbar \Omega \coth \left( {\frac{{\hbar \Omega }}{{2{K_b}T}}} \right)\]
For the classical case, we take $ \hbar \to 0$ and we get (same as in the previous section) :
\[\frac{1}{2}M\left\langle {{v^2}} \right\rangle  = \frac{{2{K_b}T}}{4}\frac{{\hbar \Omega }}{{\hbar \Omega }} = \frac{{{K_b}T}}{2}\]

since the relations between power spectra of $ v $ and $ x $ is $ {\tilde C^{vv}}(\omega ) = {\omega ^2}{\tilde C^{xx}}(\omega ) $ we then have the fluctuations of $x$:
\[{C^{xx}}(0) = \int {\frac{{d\omega }}{{4\pi }}} \frac{\hbar \pi }{{\omega M}}\coth \left( {\frac{{\hbar \omega }}{{2{K_b}T}}} \right) \cdot \left( {\delta (\omega  - \Omega ) + \delta (\omega  + \Omega )} \right)\]
calculating the integral we get:
\[\left\langle {{x^2}} \right\rangle  = \frac{\hbar }{{2M\Omega }}\coth \left( {\frac{{\hbar \Omega }}{{2{K_b}T}}} \right)\mathop  \to \limits_{classical} \frac{{{K_b}T}}{{M{\Omega ^2}}}\]
which is consistent with the equipartition :
\[\frac{1}{2}M{\Omega ^2}\left\langle {{x^2}} \right\rangle  = \frac{{{K_b}T}}{2}\]

\end{document}
