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\begin{document}

\begin{center}
$\ $
\vspace{-4\baselineskip}
\\
\textbf{\Large Advanced statistical mechanics II}\\
{\large (SS 16, FU Berlin)}\\ [0.5em]
\textbf{Problem sheet 9}\\
Due date: June 28th, 2016 
\end{center}

\subsection*{Problems}
\begin{enumerate}[leftmargin=-0ex]
\setcounter{enumi}{20}

%%%=====================================================================
%%%================================ 2 ===================================
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\item{\bfseries Correlations in the 1D Ising model (2+2+2+1+1+1*=8+1*)}\\
In this exercise we consider the classical Ising model in 1D, as introduced in the lecture. 
For the purpose of the exercise we first make the generalization that the interaction between neighboring spins depends on the location. The model is then defined on the state space $\{-1,1\}^N$ and its Hamiltonian can be written
\begin{align*}
H_N(\boldsymbol{\sigma}) &= - \sum_{k=1}^N J_k \sigma_k \sigma_{k+1} - B\sum_{k=1}^N\sigma_k ,
\end{align*} 
with $\boldsymbol{\sigma}\in \{-1,1\}^N$. We also assume open boundary conditions, i.e., $\sigma_{N+1}=0$. Note that this defines a sequence of Hamiltonians $\{H_1,H_2,\ldots\}$ where we assume $H_1=0$.
The goal of the exercise is to find out how the correlations between two spins depends on the temperature and on the  distance $r$ between them. Naturally, we work in the canonical ensemble.
We first study the case $B=0$.
\begin{enumerate}
\item Show that the partition function $Z_N(\beta,J_1,\ldots J_{N-1})$ fulfills the recurrence relation
\begin{align*}
Z_N(\beta,J_1,\ldots J_{N-1}) = 2\cosh\left(\beta J_{N-1}\right) Z_{N-1}(\beta,J_1,\ldots J_{N-2})
\end{align*}
and deduce that the partition function is given by
\begin{align*}
Z_N(\beta,J_1,\ldots J_{N-1}) = 2^N \prod_k \cosh\left(\beta J_k\right).
\end{align*}
\item Show that the expectation $\langle \sigma_k \sigma_{k+1}\rangle$ can be calculated as 
\begin{align*}
\langle \sigma_k \sigma_{k+1}\rangle = \frac{1}{\beta} \frac{\partial}{\partial J_k} \ln Z_N = \tanh\left(\beta J_k\right).
\end{align*}
Show all of the equalities!
\item Use $\sigma_j^2=1$ to show that $\langle \sigma_k \sigma_{k+r}\rangle = \prod_{l=k}^{k+r-1}\tanh\left(\beta J_l\right)$. 

\hint{$\sigma_j^2=1$ implies $\sigma_1\sigma_2\sigma_2\sigma_3=\sigma_1\sigma_3$.}
\item Now assume that $J_k =J$ for all $k$.  Use the result from above to express the correlation function as
\begin{align*}
\langle \sigma_k \sigma_{k+r}\rangle = \exp(-r/\xi),
\end{align*}
with the \emph{correlation length} $\xi$. How does $\xi$ behave as $T\rightarrow 0$ while $J$ is kept constant?
\end{enumerate}
It turns out that $\langle \sigma_k\rangle =0$, so we could have written the last result as 
\begin{align*}
g(r) := \langle \sigma_k \sigma_{k+r}\rangle - \langle \sigma_k \rangle \langle \sigma_{k+r}\rangle = \exp(-r/\xi).
\end{align*}
Now consider the other extremal case $J=0$ and $B>0$.  
\begin{enumerate}[resume] 
\item Show $g(r)=0$. What does this imply for the correlation length?
\item \bonus{Show that $\langle \sigma_k\rangle = \tanh\left(\beta B\right)$.}
\end{enumerate}

\item{\bfseries Classical Ising model in 1D and the renormalization group (1+2+2+1+1+1+1+2+1=12)}\\
In this exercise we want to study the large-scale physics of the classical Ising model in one spatial dimension with \emph{periodic boundary conditions}. We will use a method called ``renormalization group''. The idea is, that by blocking together  a number of contiguous physical sites of the lattice and integrating out degrees of freedom, we can ``zoom out'' in our description of the system and obtain an effective description of a ``coarse grained'' system.
In our case we assume that the number of spins in the system is given by $N=2^M$ and that we integrate out half of the spins with each step. For the purpose of this exercise we will add a constant to the Hamiltonian and assume that all the interaction strengths are the same. Then the partition function is given by
\begin{align*}
Z_N(K_0,K_1,K_2) = \sum_{\boldsymbol\sigma} \prod_{j=1}^N\exp\left(K_0 + K_1 \sigma_j \sigma_{j+1} + \frac{1}{2}K_2\left(\sigma_j + \sigma_{j+1}\right)\right).
\end{align*} 
with  $K_0,K_1=\beta J,K_2=\beta B$ constants. We now start with the renormalization procedure. 
\begin{enumerate}
\item By executing the sum over all spins with even index, show that the partition function can be written as
\begin{align}
\label{part_integrated}
Z_N(K_0,K_1,K_2) = \sum_{\boldsymbol{\sigma'}} \prod_{k=1}^{N/2} \exp(2K_0)2\cosh\left(K_1\left(\sigma'_k + \sigma'_{k+1}\right) + K_2 \right)\exp\left(\frac{1}{2}K_2\left(\sigma'_k + \sigma'_{k+1}\right)\right),
\end{align}
where the variables $\sigma'_k$ denote the remaining half of the spins, i.e., $\boldsymbol{\sigma'} \in \{-1,+1\}^{N/2}$.
\end{enumerate}
We now wish to express the renormalized partition function in the same \emph{functional form} as in the original coordinates, that is,
\begin{align}
\label{part_renormalized}
Z_N(K_0,K_1,K_2) = \sum_{\sigma'} \prod_{j=1}^{N^{(1)}=N/2}\exp\left(K_0^{(1)} + K_1^{(1)} \sigma'_j \sigma'_{j+1} + \frac{1}{2}K_2^{(1)}\left(\sigma'_j + \sigma'_{j+1}\right)\right).
\end{align}
\begin{enumerate}[resume]
\item Using that the summand has to be the same in eq.\ref{part_integrated} and eq.\ref{part_renormalized}, express \\$\exp(K_0^{(1)}+K_1^{(1)}\pm K_2^{(1)}),\exp(K_0^{(1)}- K_1^{(1)})$ as functions of $K_0,K_1,K_2$.
\item Show from your result that this implies
\begin{align*}
\exp(K_0^{(1)}) &= 2 \exp(2K_0)\left(\cosh\left( 2K_1 + K_2\right)\cosh\left( 2K_1 - K_2\right)\cosh^2( K_2) \right)^{\frac{1}{4}}\\
\exp(K_1^{(1)}) &= \left(\cosh\left( 2K_1 + K_2\right)\cosh\left( 2K_1 - K_2\right)/\cosh^2( K_2) \right)^{\frac{1}{4}}\\
\exp(K_2^{(1)}) &= \exp(K_2)\left(\cosh\left( 2K_1 + K_2\right)/\cosh\left( 2K_1 - K_2\right)\right)^{\frac{1}{2}}.
\end{align*}
\item Even for $K_0=0$ we have $K_0^{(1)}\neq 0$. Interpret! 
\end{enumerate}
We can now view the parameters as a vector $\vec{K}=(K_0,K_1,K_2)$. The coarse graining then provides  a map $R:\R^3\rightarrow \R^3$ so that $\vec{K}^{(1)} = R(\vec{K})$. If we coarse grain again, we get new parameters $\vec{K}^{(2)} = (R\circ R)(\vec{K})$. Renormalizing twice gives another renormalization. That is why this procedure is called ``renormalization group'' (although it is not a group). The evolution of the parameters is called ``renormalization group flow''.
Note that  $K^{(1)}_{1,2}=R(\vec{K})_{1,2}$ do not depend on $\vec{K}_0$. We will now ignore $K_0$, so that our parameters are vectors in $\R^2$. 
\begin{enumerate}[resume]
\item Show that vectors of the form $\vec{K}=(0,K_2)$ and $\vec{\tilde{K}}=(\infty,0)$ are fix-points of the renormalization group flow, i.e., $R(\vec{K})=\vec{K}$.
\item Use your results from the previous exercise to determine the correlation lengths at the fix-points.  Interpret your results, remembering that the parameters depend on the temperature.
\item Can there be a finite correlation length at a renormalization fixed point?
\item Setting $K_2=0$ and changing the remaining variable to $t = \exp(-4 K_1)$ show that the renormalization map reduces to
\begin{align*}
t \mapsto \tilde{R}(t):=\frac{4t}{(1+t)^2}. 
\end{align*}
\item Sketch $\tilde{R}(t)$ and decide graphically whether the fixed points are stable or unstable.
\end{enumerate} 

\end{enumerate}
\end{document}
