%% LyX 1.5.7 created this file. For more info, see http://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\documentclass[11pt,english,fleqn]{article}
\usepackage[T1]{fontenc}
\usepackage[latin9]{inputenc}
\setlength{\parskip}{0cm}
\setlength{\parindent}{0pt}
\makeatletter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands.
%% LyX 1.5.7 created this file. For more info, see http://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\makeatletter
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%% 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
% Fonts
\usepackage{latexsym}
\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}}
\makeatother
\usepackage{babel}
\makeatother
\usepackage{babel}
\begin{document}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\heading{Ising in interaction with lattice gas}
\auname{Ori Hendler}
\exnumber{5651}
\textbf{The problem:}
\Dn
Consider a one dimensional Ising model of spins $\sigma_{i}=\pm1,i=1,2,3,...,N$
and circular boundary condition ($\sigma_{N+1}=\sigma_{1}$). Between
each two spins there is a site for an additional atom, which if present
decreases the coupling $J$ to $J(1-\lambda)$. The Hamiltonian is
then \[
H=-J\sum_{i=1}^{N}\sigma_{i}\sigma_{i+1}(1-\lambda n_{i})\]
where $n_{i}=0$ or $1$ and there are $M=\sum_{i=1}^{N}n_{i}$
atoms ($M<\sigma_{2}|T_{\sigma_{2}\sigma_{3}}|\sigma_{3}>...<\sigma_{n}|T_{\sigma_{n}\sigma_{n+1}}|\sigma_{n+1}>\]
The transfer matrix for two spins without an atom between is:
\[
T_{i}=\left(\begin{array}{cc}
e^{\beta J} & e^{-\beta J}\\
e^{-\beta J} & e^{\beta J}\end{array}\right)\]
The transfer matrix for two spins with an atom between is:
\[
T_{i}^{'}=\left(\begin{array}{cc}
e^{\beta J(1-\lambda)} & e^{-\beta J(1-\lambda)}\\
e^{-\beta J(1-\lambda)} & e^{\beta J(1-\lambda)}\end{array}\right)\]
Both matrices commute So the multiplication order does not matter:
\[
[T_{i},T_{i}^{'}]=0\]
\\* Reminder: The trace of matrix doesn't change upon diagonalization.
\\*After changing to the diagonalized base, we get for a single configuration:
\[
Z=Tr(T'^{M}T^{N-M})\]
\[
=Tr(\left(\begin{array}{cc}
2cosh(\beta J)^{N-M} & 0\\
0 & 2sinh(\beta J)^{N-M}\end{array}\right)\left(\begin{array}{cc}
2cosh(\beta J(1-\lambda))^{M} & 0\\
0 & 2sinh(\beta J(1-\lambda))^{M}\end{array}\right))\]
Notice that a combinatorial factor is introduced into the partition function, in order to count the different configurations possible, for a given M.
Assuming $N\rightarrow\infty$ we can neglect the smaller eigenvalue
so:
\[
Z=(\begin{array}{c}
N\\
M\end{array})2cosh(\beta J)^{N-M}2cosh(\beta J(1-\lambda))^{M}\]
The free energy is therefore:
\[
F=-\tau lnZ=-\tau ln((\begin{array}{c}
N\\
M\end{array}))-\tau (N-M)ln(2cosh(\beta J))-\tau Mln(2cosh(\beta J(1-\lambda))) =\]
\[-\tau(ln(n!)-ln(m!)-ln((n-m)!))-\tau (N-M)ln(2cosh(\beta J))-\tau Mln(2cosh(\beta J(1-\lambda)))\]
%%%%%%%%%%%%%%%
\Dn
(2) For any configuration with exactly M impurities, the partition function
is:
\[
Z=2cosh(\beta J)^{N-M}2cosh(\beta J(1-\lambda))^{M}\]
As the number of impurities is fixed, the combinatorial factor is not needed.
\\The free energy for any configuration with M impurities is:
\[
F=-\tau lnZ=-\tau l(N-M)ln(2cosh(\beta J))-\tau Mln(2cosh(\beta J(1-\lambda)))\]
The average free energy for a given M is:
\[
=\frac{\sum_{configurations}F_{conf}}{\sum_{configurations}}=\frac{(\begin{array}{c}
N\\
M\end{array})\times-\tau l(N-M)ln(2cosh(\beta J))-\tau Mln(2cosh(\beta J(1-\lambda)))}{(\begin{array}{c}
N\\
M\end{array})}\]
\[
=-\tau l(N-M)ln(2cosh(\beta J))-\tau Mln(2cosh(\beta J(1-\lambda)))\]
The entropy difference between the two calculations:\[
\triangle S=-\frac{\partial F}{\partial\tau}+\frac{\partial}{\partial\tau}=ln((\begin{array}{c}
N\\
M\end{array}))\]
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\end{document}