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\exnumber{5440}
\heading{One dimensional XY model} 
\auname{Yuval Friedman}


{\bf The problem:}

\Dn
Polymer in two dimensions:  Configurations of a polymer are described
by a set of vectors ${\bf t}_i$ of length $a$ in two dimensions (for i = 1,...,N), or alternatively
by the angles $\phi_i$ between successive vectors, as indicated in the figure below. The energy of a configuration $\{\phi_i\}$ is
\[{\cal H}=-\kappa \sum_{i=1}^{N-1}{\bf t}_i\cdot{\bf t_{i+1}}=-\kappa a^2 \sum_{i=1}^{N-1}\cos\phi_i   \]

\begin{itemize}
\item[(a)] Show that the correlations $\langle {\bf t}_n\cdot{\bf t}_m\rangle$ decay exponentially with distance and obtain an expression for the "persistence length" $\xi$; you can leave the answer in terms of simple integrals.

Hint: Show  ${\bf t}_n\cdot{\bf t}_m=a^2\re\,\{\eexp{i\sum_{j=n}^{m-1}\phi_j}\}$.
\item[(b)] The end-to-end distance  ${\bf R}$ is defined as illustrated in the figure. Calculate $\langle R^2\rangle$ in the limit $N\gg 1$.  
\end{itemize}

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\Dn\Dn


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{\bf The solution:}

\Dn

\begin{itemize}

\item[(a)] We begin by evaluating $\langle\cos\phi_i\rangle$ using the partitition function:

\begin{equation}
\bf{Z} = \int_{-\pi}^{\pi}\eexp{\beta\kappa a^2\sum_{i=1}^{N-1}\cos\phi_i}\text{d}\phi_1\,\dotsi\phi_{N-1} = \bigg( \int_{-\pi}^{\pi}\eexp{\beta\kappa a^2\cos\phi}\text{d}\phi\bigg)^{N-1}
\end{equation}
\begin{align}
\langle\cos\phi_i\rangle = &\frac{1}{{\bf Z}}\int_{-\pi}^{\pi}\eexp{\beta\kappa a^2\sum_{j=1}^{N-1}\cos\phi_j}\cos\phi_i\text{d}\phi_1\,\dotsi\phi_{N-1}
=\frac{\int_{-\pi}^{\pi}\eexp{\beta\kappa a^2\cos\phi}\cos\phi\text{d}\phi}{\int_{-\pi}^{\pi}\eexp{\beta\kappa a^2\cos\phi}\text{d}\phi}\label{eq:1}
\end{align}

We continue by caculating the angle between the vectors: ${\bf t}_n$ and ${\bf t}_m$. To do so, we should use the fact that we are given only the angles between two succesive vectors, so the angle $\phi_{nm}$ between vectors ${\bf t}_n$ and ${\bf t}_m$ will be a sum of all angles between those indexes:
\begin{equation}
\phi_{nm} = \phi_n +\phi_{n+1}+...+\phi_{m-1} = \sum_{i=n}^{m-1}\phi_i
\end{equation}

So the correlation between those vectors will be:
\begin{equation}
\langle {\bf t}_n\cdot{\bf t}_m\rangle = a^2\langle\cos\phi_{nm}\rangle = a^2\langle\cos(\sum_{i=n}^{m-1}\phi_i)\rangle = a^2\text{Re}\{\langle\eexp{i\sum_{i=n}^{m-1}\phi_i}\rangle\}
\end{equation}

Knowing that different $\cos\phi_i$ have the same distribution function gives:
\begin{equation}
\langle \eexp{i\sum_{i=n}^{m-1}\phi_i}\rangle = \langle \eexp{i\phi_n}\rangle\cdot \langle \eexp{i\phi_
{n+1}}\rangle\cdot ... \langle\eexp{i\phi_{m-1}}\rangle = \prod_{i=n}^{m-1}\langle\eexp{i\phi_i}\rangle = \langle\eexp{i\phi_i}\rangle^{|m-n|}
\end{equation}

Because $\langle\sin\phi_i\rangle=0$, $\langle\eexp{i\phi_i}\rangle$ becomes $\langle\cos\phi_i\rangle$, so:

\begin{equation}
\langle {\bf t}_n\cdot{\bf t}_m\rangle=a^2\text{Re}\{ \langle e^{i\phi_i}\rangle^{|m-n|}\} = a^2\langle\cos\phi_i\rangle^{|n-m|} = a^2e^{-|n-m||ln\langle\cos\phi_i\rangle|} = a^2e^{-|n-m|a/\xi}
\end{equation}

Where the persistence length is: 
\begin{equation}
\xi = \frac{a}{|\ln{\langle\cos\phi_i\rangle}|}\label{eq:2}
\end{equation}

Persistance length is a parameter quantifying the stiffness of the polymer: the polymer is less flexible when the persistance length grows.\\

Substituting $\eqref{eq:1}$ into \eqref{eq:2} we get:
\begin{equation}
\xi = a\Big|\ln{\big(\int_{-\pi}^{\pi}\eexp{\beta\kappa a^2\cos\phi}\cos\phi\text{d}\phi}\big)-\ln{\big(\int_{-\pi}^{\pi}\eexp{\beta\kappa a^2\cos\phi}\text{d}\phi\big)}\Big|^{-1}
\end{equation}

\item[(b)] 
Substituting $r=n-m$ and taking the sum to infinity for $N\gg1$ we get:
\begin{align}
\langle R^2\rangle = \sum_{n=1}^{N} \sum_{m=1}^{N}\langle {\bf t}_n\cdot{\bf t}_m\rangle = a^2\sum_{n=1}^{N} \sum_{m=1}^{N}\langle \cos\phi_i\rangle^{|n-m|}= Na^2\sum_{r=-\infty}^{\infty}\langle\cos\phi_i\rangle^{|r|} = \nonumber\\
Na^2(1+2\sum_{r=1}^{r=\infty}\langle\cos\phi_i\rangle^{r}) = Na^2(1+\frac{2}{1-\langle\cos\phi_i\rangle}) = Na^2\frac{1+\langle\cos\phi_i\rangle}{1-\langle\cos\phi_i\rangle}
\end{align}

And in terms of the persistence length:
\begin{equation}
\xi = \frac{a}{|\ln{\langle\cos\phi_i\rangle}|}\implies \langle\cos\phi_i\rangle = \eexp{-\frac{a}{\xi}}
\end{equation}

So:
\begin{equation}
\langle R^2\rangle = Na^2\frac{1+\eexp{-\frac{a}{\xi}}}{1-\eexp{-\frac{a}{\xi}}} = Na^2\frac{\eexp{\frac{a}{2\xi}}+\eexp{-\frac{a}{2\xi}}}{\eexp{\frac{a}{2\xi}}-\eexp{-\frac{a}{2\xi}}} = Na^2\coth{(\frac{a}{2\xi})}
\end{equation}

\end{itemize}


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