\documentclass[11pt,fleqn]{article}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% template that does not use Revtex4
%%% but allows special fonts
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%
%%% Please use this template.
%%% Edit it using e.g. Notepad
%%% Ignore the header (do not change it)
%%% Process the file in the Latex site 
%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


% Page setup
\topmargin -1.5cm      
\oddsidemargin -0.04cm   
\evensidemargin -0.04cm  
\textwidth 16.59cm
\textheight 24cm 
\setlength{\parindent}{0cm} 
\setlength{\parskip}{0cm} 


% Fonts
\usepackage{latexsym}
\usepackage{amsmath} 
\usepackage{amssymb} 
\usepackage{bm}


% Math symbols I
\newcommand{\sinc}{\mbox{sinc}}
\newcommand{\const}{\mbox{const}}
\newcommand{\trc}{\mbox{trace}}
\newcommand{\intt}{\int\!\!\!\!\int }
\newcommand{\ointt}{\int\!\!\!\!\int\!\!\!\!\!\circ\ }
\newcommand{\ar}{\mathsf r}
\newcommand{\im}{\mbox{Im}}
\newcommand{\re}{\mbox{Re}}

% Math symbols II
\newcommand{\eexp}{\mbox{e}^}
\newcommand{\bra}{\left\langle}
\newcommand{\ket}{\right\rangle}

% Mass symbol
\newcommand{\mass}{\mathsf{m}} 
\newcommand{\Mass}{\mathsf{M}} 

% More math commands
\newcommand{\tbox}[1]{\mbox{\tiny #1}}
\newcommand{\bmsf}[1]{\bm{\mathsf{#1}}} 
\newcommand{\amatrix}[1]{\begin{matrix} #1 \end{matrix}} 
\newcommand{\pd}[2]{\frac{\partial #1}{\partial #2}}

% Other commands
\newcommand{\hide}[1]{}
\newcommand{\drawline}{\begin{picture}(500,1)\line(1,0){500}\end{picture}}
\newcommand{\bitem}{$\bullet$ \ \ \ }
\newcommand{\Cn}[1]{\begin{center} #1 \end{center}}
\newcommand{\mpg}[2][1.0\hsize]{\begin{minipage}[b]{#1}{#2}\end{minipage}}
\newcommand{\Dn}{\vspace*{3mm}}

% Figures
\newcommand{\putgraph}[2][0.30\hsize]{\includegraphics[width=#1]{#2}}

% heading
\newcommand{\heading}[1]{\begin{center} \Large {#1} \end{center}}
\newcommand{\auname}[1]{\begin{center} \bf Submitted by: #1 \end{center}}


\begin{document} 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\heading{E8481: Mass on a spring } 
\auname{Michaele Epshtein and Amir Bernat}



{\bf The problem:}

\Dn

A balance for measuring weight consists of a sensitive spring which
hangs from a fixed point. The spring constant is ${K}$. The balance
is at temperature ${T}$ and gravity acceleration is ${g}$ in the ${x}$
direction. A small mass ${\mass}$ hangs at the end of the spring. There
is an option to apply an external force $F(t)$, to which $x$ is
conjugate or apply an external vector potential $A(t)$.

\begin{itemize}

\item [{{(a)}}] Find the partition function $Z$.
 
\item [{{(b)}}] Find $\langle x \rangle$ and $\langle x^{2}\rangle$ and $\mbox{Var}(x)$.

\item [{{(c)}}] What is the minimal mass that can be meaningfully measured? 

\item [{{(d)}}] Write a Langevin equation for $x(t)$, with friction $\gamma$, and a random force $f(t)$. 

\item [{{(e)}}] Assuming $\langle f(t)f(0)\rangle=C\delta(t)$, find $\mbox{Var}(x)$, 
and deduce what is $C$ by comparing with the canonical result.  

\item [{{(f)}}] Now, assuming a different force autocorrelation function $\langle g(t)g(0)\rangle= D\frac{sin(at)}{at}$, find the position power spectrum, 
under what condition will the result found here and (e) agree?

\item [{{(g)}}] Describe the external force $F(t)$ by a scalar potential and demonstrate FDT. 

\item [{{(h)}}] Describe the external force $F(t)$ by a vector potential and demonstrate FDT.

\end{itemize}

\Dn

Note: $\int \frac{d\omega}{(\omega^2-\omega_0^2)^2+\gamma^2\omega^2}=\frac{\pi}{\gamma \omega_0^2}$.

\Dn\Dn


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
{\bf The solution:}

\Dn


(a) The Hamiltonian of the system is: $H=\frac{p^{2}}{2m}+\frac{1}{2}Kx^{2}-mgx$, so the partition function Z is

\[
Z=\frac{1}{2\pi\hbar}\underset{-\infty}{\overset{+\infty}{\int}}e^{-\frac{\beta p^{2}}{2m}}\, dp\underset{-\infty}{\overset{+\infty}{\int}}e^{-(\frac{\beta Kx^{2}}{2}-mg\beta x)}\, dx
\]

This integral can be simplified  by changing the integration variable $x$ to $y=\sqrt{\frac{\beta K}{2}}x$

\[
Z=\frac{1}{2\pi\hbar}\sqrt{\frac{2m\pi}{\beta}}\underset{-\infty}{\overset{+\infty}{\int}}e^{-(y^{2}-\sqrt{\frac{2\beta}{K}}mgy)}\, dy=\frac{1}{\hbar\beta}\sqrt{\frac{m}{K\pi}}\underset{-\infty}{\overset{+\infty}{\int}}e^{-(y-\sqrt{\frac{\beta}{2K}}mg)^{2}}\, e{}^{\frac{\beta m^{2}g^{2}}{2K}}\, dy
\]

Finally:
\[
Z=\frac{1}{\hbar\beta}\sqrt{\frac{m}{K}}\, e^{\frac{\beta m^{2}g^{2}}{2K}}\]

\Dn\Dn
(b) $\langle x\rangle$ is defined as:

\[
\langle x(t)\rangle=\frac{1}{Z}\frac{1}{2\pi\hbar}\underset{-\infty}{\overset{+\infty}{\int}} \underset{-\infty}{\overset{+\infty}{\int}}e^{-\frac{\beta p^{2}}{2m}}\, x\,e^{-(\frac{\beta Kx^{2}}{2}-mg\beta x)}\, dxdp
\]

This is solved using integration by parts

\[
\langle x(t)\rangle=\sqrt{\frac{\beta K}{2\pi}}\, e^{-\frac{\beta m^{2}g^{2}}{2K}}\left[-\frac{1}{\beta K}e^{-(\frac{\beta Kx^{2}}{2}-mg\beta x)}\underset{-\infty}{\overset{+\infty}{|}}+\frac{mg}{K}\underset{-\infty}{\overset{+\infty}{\int}}e^{-(\frac{\beta Kx^{2}}{2}-mg\beta x)}dx\right]
\]


The first term of the R.H.S. goes to zero and we get

\[
\langle x\rangle=\cfrac{mg}{k}
\]


Applying the equipartition theorem $\langle x\frac{\partial H}{\partial x}\rangle=k_{B}T$ we can find that:

\[
\langle x(Kx-mg)\rangle=K\langle x^{2}\rangle-mg\langle x\rangle=K\langle x^{2}\rangle-mg(\frac{mg}{K})=k_{B}T
\]


\[
\langle x^{2}\rangle=\frac{k_{B}T}{K}+\left(\frac{mg}{K}\right)^{2}
\]

The variance is:
\[
\mbox{Var}(x) = \frac{k_{B}T}{K}
\]
\Dn\Dn

(c) For a meaningful result we want the fluctuation of the spring should be much smaller than the spring length
itself, Therefor $\langle x\rangle^{2}\gg\langle\delta x^{2}\rangle$. We define the fluctuations $\langle\delta x^{2}\rangle = \mbox{Var}(x) = \frac{k_{B}T}{K}$. Which results in the following condition for a meaningful measurement:

\[
\left(\cfrac{mg}{K}\right)^{2} \gg \cfrac{k_{B}T}{K} \;\;\rightarrow\;\;\ m\gg\sqrt{\frac{Kk_{B}T}{g^{2}}}
\]



\Dn\Dn

(d) We now add friction, denoted as $\gamma$ that interact with velocity
and a random force $f(t)$ such that $\langle f \rangle_t =0$. The Langevin equation for such a system
is:

\[
m\ddot{x}+m\gamma\dot{x}+Kx-mg=f(t)
\]

\Dn\Dn
(e) We redefine $x$ to be about the mean value so that $X(t)=x(t)-\dfrac{mg}{K}$. Writing Langevin's
equation for $X(t)$ and using Fourier transform functional relationships for time derivation
%$\frac{\partial^{n}}{\partial t^{n}}x(t)=(i\omega)^{n}X(\omega)$
we get $\left(-\omega^{2}-i\omega \gamma+\frac{K}{m}\right)\tilde{X}(\omega)=\frac{1}{m}f(\omega)$. 

\[
\tilde{X}(\omega) = \cfrac{f(\omega)}{m(-\omega^{2}-i\omega \gamma+\frac{K}{m})} 
\]

The expression for $\left| f	(\omega) \right| ^2 = \langle f(\omega) f(-\omega) \rangle = F.T. \left[ \tilde{C(\tau)}\right]$ which is the Fourier transform of the given correlation function. 

\[
F.T. \left[ C \delta(t) \right] = C
\]

The power spectrum of the $X$ coordinate is:

\[
\left|\tilde{X}(\omega) \right|^2 = \cfrac{C}{\left(K-m\omega^{2}\right)^2+\left(\omega m\gamma\right)^2} 
\]

The relation of the power spectrum of $X$ to $\mbox{Var}(X)$ is the inverse Fourier transform at $\tau = 0$ (remember this calculation is done at the mean value of $x$).

\[
\mbox{Var}(X) = \langle X^2(\tau) \rangle = F.T.^{-1} \left[\cfrac{C}{\left(K-m\omega^{2}\right)^2+\left(\omega m\gamma\right)^2} \right] = \dfrac{m \pi}{2 \gamma K} C
\]

This result can now be compared to that found earlier in section $(b)$ that $\langle X^2 \rangle = \dfrac{T}{m}2\gamma $ to find that $C = \dfrac{T}{m\pi}2\gamma$ 
\Dn\Dn


(f)  We keep $X(t)=x(t)-\dfrac{mg}{K}$, the Langevin
equation for $X(t)$ and use the same methodology as in (e), however the Fourier transform of $sinc$ is a window function. 
\[
F.T. \left[ \frac{sin(at)}{at} \right] = 
\left\{
	\begin{array}{ll}
		\frac{2\pi}{a}  & \mbox{for } \omega < \frac{a}{2\pi} \\
		0 & \mbox{for } x > \frac{a}{2\pi}
	\end{array}
\right.
\]


The power spectrum of the $X$ coordinate is:

\[
\left|\tilde{X}(\omega) \right|^2 =
\left\{
	\begin{array}{ll}
		\frac{2\pi}{a} \cfrac{1}{\left(K-m\omega^{2}\right)^2+\left(\omega m\gamma\right)^2}  & \mbox{for } \omega < \frac{a}{2\pi} \\
		0 & \mbox{for } x > \frac{a}{2\pi}
	\end{array}
\right.
\]

One way for realizing a delta function is by the $sinc$ function. For the limit $a\rightarrow \infty$ the function's appearance in the time domain becomes very narrow (resembling a delta function) and in the frequency domain becomes very broad (resembling white noise). If in addition to the condition for $a$ we have $C=\frac{\pi}{a}D$  then the position power spectrum found from $f	(\omega)$ and $g	(\omega)$ will be the same.
\Dn\Dn


(g)  An external scalar potential creates the force $F(t)$, with conjugate coordinate $x(t)$ and the interaction term $-F (t)x$ as in the previous sections we will use $X(t)$. The DC response will change the mean value of $x$, the solution will be the same as in the (b). The AC linear response relation is $\langle X(\omega) \rangle = \chi_X(\omega) F(\omega)$ where we have:

\[
\chi_X(\omega)=\cfrac{1}{m(-\omega^{2}-i\omega \gamma+\frac{K}{m})} 
\]
The imaginary part of $\chi_X(\omega)$ is:

\[
\Im \left[\chi_X(\omega)\right]=\cfrac{\omega \gamma }{m^2(\omega^{2}-\frac{K}{m})^{2}+\omega^{2}\gamma^{2}}
\]

In general, the autocorrelation relation here can now be written as:
\[
\tilde{C}_{XX} (\omega)= \langle \tilde{X}(\omega) \tilde{X}(-\omega) =   \left| \chi_X(\omega) \right|^2 \langle \tilde{F}(\omega) \tilde{F}(-\omega) \rangle
\]

We now calculate the linear response for the coordinate $V = \dot{X}$ for which we have $X_\omega = \frac{i}{\omega}V_\omega$. We will find $\chi_V (\omega)$ from the modified force equation:
$\langle \frac{i}{\omega} V(\omega) \rangle = \chi_X(\omega) F(\omega)$, such that the velocity autocorrelation function is:

\[
\tilde{C}_{VV} (\omega)= \langle \frac{i}{\omega} \tilde{V}(\omega) \frac{-i}{\omega}\tilde{V}(-\omega) \rangle=  \frac{1}{\omega^2} \left| \chi_X(\omega) \right|^2 \langle \tilde{F}(\omega) \tilde{F}(-\omega) \rangle
\]
	
We can now see that the relation between the position ($X$) and velocity ($V$) autocorrelation functions is $ \omega^2 \tilde{C}_{XX} (\omega) = \tilde{C}_{VV} (\omega)$. We can calculate the fluctuation of $X$ and $V$ using the FD relation:

\[
\tilde{C}_{XX}(\omega) = \hbar cotgh \left(\frac{\hbar \omega}{2T} \right) \im \left[\chi (\omega) \right] = \hbar cotgh \left(\frac{\hbar \omega}{2T} \right) \cfrac{\omega \gamma }{m^2(\omega^{2}-\frac{K}{m})^{2}+\omega^{2}\gamma^{2}}
\]

We can now transform back to the time domain to have:
\[
C_{XX}(\tau) = \hbar \underset{-\infty}{\overset{+\infty}{\int}} d\omega cotgh \left(\frac{\hbar \omega}{2T} \right) \cfrac{\omega \gamma e^{-i\omega \tau} }{m^2(\omega^{2}-\frac{K}{m})^{2}+\omega^{2}\gamma^{2}}
\]
And the fluctuation of $V$ are simply:
\[
\tilde{C}_{VV}(\tau) = \hbar \underset{-\infty}{\overset{+\infty}{\int}} d\omega  \hbar cotgh \left(\frac{\hbar \omega}{2T} \right) \cfrac{\omega^3 \gamma e^{-i\omega \tau} }{m^2(\omega^{2}-\frac{K}{m})^{2}+\omega^{2}\gamma^{2}}
\]

\Dn\Dn

(h) An external vector potential creates the force $\epsilon(t) = -\frac{d}{dt} \vec{F}$, with conjugate coordinate $V(t)$ and the interaction term $-\vec{F} (t)V$ as in the previous sections we will use $X(t)$ which is $X(t) = x(t) - \langle x(t) \rangle$. The DC linear response $(\omega = 0)$ relation is 
$\langle V \rangle = \mu \epsilon$ which describes the drift motion. In general we have the linear response relation $\langle X(\omega) \rangle = \chi_X(\omega) \epsilon(\omega)$ where we have:

\[
-i\omega V(\omega) + \gamma V(\omega) + \frac{K}{m} \frac{i}{\omega} V(\omega)= \frac{-i\omega}{m} \epsilon (\omega)
\]

The susceptibility is:
\[
\chi_V(\omega) = \frac{i \omega}{\frac{m}{\omega}(\gamma\omega - i\left(\omega^2 - \frac{K}{m} \right))}
\]

And the autocorrelation function is:
\[
\tilde{C}_{VV}(\omega) = \langle \tilde{V}(\omega) \tilde{V}(-\omega) \rangle=  \left| \chi_V(\omega) \right|^2 \langle \tilde{\epsilon}(\omega) \tilde{\epsilon}(-\omega) \rangle
\]
We now use again the relation $V = \dot{X} \rightarrow V_\omega = -i\omega X_\omega$, the autocorrelation relation here is thus:
\[
\tilde{C}_{XX} (\omega)= \langle (-i\omega) \tilde{V}(\omega) (i\omega)\tilde{V}(-\omega) = \omega^2 \left| \chi_V(\omega) \right|^2 \langle \tilde{F}(\omega) \tilde{F}(-\omega) \rangle
\]

%\[
%\Im \left[\chi_V(\omega)\right]=\frac{\gamma}{\gamma^2 + \left(\omega - \frac{K}{m\omega} \right)^2}
%\]

Once again we relate the autocorrelation functions of the position and the velocity by the $\omega^2$ factor.

In the case where the external force creates a DC component we can relate the mobility of a particle to the spectral density of the force. this is gives by the relation : $\langle \tilde{F}(0) \tilde{F}(0) \rangle = 2D = 2\mu T$ where $D$ is the diffusion coefficient and $\mu$ is the mobility. 

%The variance in position and energy can now be calculated easily:
%
%\[
%C_{XX}(\tau=0) = \underset{-\infty}{\overset{+\infty}{\int}}\tilde{C}_{XX} e^{-i\omega \tau} (\omega) \frac{d\omega}{2\pi}= \frac{D}{\pi %m^2}\underset{-\infty}{\overset{+\infty}{\int}} \cfrac{d\omega}{(\left(\frac{K}{m}-\omega^{2}\right)^2+\omega^2 \gamma^2)} \langle \tilde{F}(\omega) %\tilde{F}(-\omega) \rangle
%\]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document} 