\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}
\usepackage{graphicx}


% 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{Mean field approximation for a classical Heisenberg model } 
\auname{Shir Shenkar} 
\exnumber{5713}


{\bf The problem:}

\Dn
Apply the mean field approximation to the  classical spin vector model
\[{\cal H}=-\epsilon\sum_{\langle i,j\rangle}{\vec{\bf  s}}_{i}\cdot { \vec{\bf s}}_{j}-{\bf h}\cdot \sum_{i}{ \vec{\bf s}}_{i}\]
where ${\vec{\bf s}_{i}}$ is a unit vector and ${i,j}$ are neighboring
sites on a lattice with coordination number $c$. The lattice has
${N}$ sites and each site has ${c}$ neighbors.

\Dn

\begin{itemize}
\item [(a)]
Assume that $\; \vec{h}=(0,0,h)\;$, define a mean field ${{\bf h}_{eff}}$ ,and evaluate the partition
function ${Z}$ in terms of ${{\bf h}_{eff}}$.
\Dn

\item [(b)]
Define $\theta_i $ as the inclination angle of $\;  {\bf{s_i}} \;$ with respect to $ \vec{h}$.  
Assume that at equilibrium $\; <{\bf{s_i}}>=(0,0,M)\; $ where $\; M =<\cos{\theta}>$.
Find the equation for M and find the
transition temperature $\;{T_{c}}\;\;$.
\Dn
\item [(c)]
Write an expression for the mean field energy of the system assuming that $\;M(T)\;$ is known.
\Dn
\item[(d)] Identify exponents $\gamma$ and $\beta$ that describe the susceptibility
${\chi \sim (T-T_c)^{-\gamma}}$ above $T_c$, and the magnetization
$M\sim (T_c-T)^{\beta}$ below $T_c$.
\item[(e)] Find the jump in the heat capacity $C_V$ at
$T_c$.
\Dn

\end{itemize}

\Dn
\Dn

{\bf The solution:}
\Dn
\begin{itemize}
\item [(a)]
We make a rough Mean field approximation and get the Hamiltonian:
\begin{equation}
    {\cal H}=-\epsilon c {\langle \vec{\bf s} \rangle} \sum_{i} {\vec {\bf s}}_{i}-{\vec h}\cdot \sum _{i} {\vec {\bf s}}_{i}=-(\epsilon c {\langle \vec{\bf s} \rangle} +{\vec h} )\cdot \sum_{i} {\vec {\bf s}}_{i}=-\vec {h}_{eff}\cdot \sum_{i} {\vec {\bf s}}_{i}
\end{equation}
we define $\vec {h}_{eff}=\epsilon c M +{\vec h}$ where $M={\langle \vec{\bf s} \rangle}$.

we can see this is in fact a sum of single spin hamiltonians, each of the form:
\begin{equation}
 {\cal H}^{(i)}= -\vec {h}_{eff} \cdot { \vec{\bf s}}_{i} = -\vec {h}_{eff} \cdot cos \theta _{i} \;\;\;\;\;\;\;\;
\end{equation}


 \Dn
Notice that the Mean field approximation transforms the problem from multi particle problem to a single particle problem (without a direct interaction) and the spins are pinned to their positions so that ${ Z_{N}}={ Z_1}^{N}$.

\begin{equation} { Z_{1}}=\frac{1}{4\pi}\int^{1}_{-1}e^{\beta {\cal H}^{(i)} }d(\cos{\theta_{i}})\int^{2\pi}_{0}d\varphi=\frac{1}{4\pi}\int^{1}_{-1}e^{\beta{\bf h}_{eff}\cos{\theta}}d(\cos{\theta})\int^{2\pi}_{0}d\varphi=\frac{\sinh(\beta{\bf h}_{eff})}{\beta{\bf h}_{eff}}
\end{equation}

\begin{equation} { Z_{N}}=({ Z_{1}})^{N}={\left(\frac{\sinh(\beta{\bf h}_{eff})}{\beta{\bf h}_{eff}} \right)}^{N}
\end{equation}


\Dn

\item [(b)]

we define ${M(T)=\langle \cos \theta_{i}\rangle}$
For a single spin:
\begin{equation}
 M\equiv \langle \cos (\theta_{i})\rangle=\frac{1}{Z} \int ^{1}_{-1} cos (\theta) e^{\beta{\bf h}_{eff}\cos{\theta}} d(\cos \theta )\int^{2\pi}_{0}d\varphi=\frac{1}{Z} \frac{1}{{\bf h}_{eff}} \frac{\partial}{\partial \beta}Z 
 \end{equation} 
\begin{equation}
M= \frac{1}{{{\bf h}_{eff}} }\frac{\partial}{\partial \beta} \ln {Z } =\frac{1}{{\bf h}_{eff}} \frac{\partial}{\partial \beta} \left[    \ln\left(  \sinh(\beta{\bf h}_{eff})\right)  -\ln \left( \beta{\bf h}_{eff}  \right)          \right] 
\end{equation}
\begin{equation}
M =\coth(\beta{\bf h}_{eff})-\frac{1}{{ \beta { \bf h}_{eff}} }
\end{equation}

\begin{eqnarray}
\coth{x}\approx \frac{1}{x}+\frac{x}{3}-\frac{x^3}{45} \;\;\;\;\;\;\;\;\;\;
\tanh{x}\approx {x}-\frac{x^3}{3}\;\;\; \rightarrow \;\;\coth{x}-\frac{1}{x}\approx\tanh{x}
\end{eqnarray}
\begin{equation}
M \approx \tanh{ (\beta { \bf h}_{eff})}\;\;\Rightarrow  \:\:  \tanh^{-1}(M)=\beta(\epsilon c M+h)
\end{equation}
 To first order $\; \tanh^{-1}(M)\cong M\;$ and to find $T_c\;$ we take $\; h\rightarrow 0$.
\begin{equation} 
M=\frac{ \epsilon c M}{T_c}\;\;\;\;\;\rightarrow T_c=\epsilon c
 \end{equation}
 
 
 \Dn
\item [(c)]
 
 
In the mean field approximation the spins are independent of each other, and therefore \\
$ \left\langle {\bf  s}_{i}{\bf  s}_{j}\right\rangle=\left\langle {\bf  s}_{i}\right\rangle\left\langle {\bf  s}_{j}\right\rangle   $. When calculating the energy we take $ h\rightarrow 0 \;\;$. 
\Dn

It follow that the energy is:
\begin{equation}
E=\left\langle {\cal H}\right\rangle= -\epsilon\sum_{\langle i,j\rangle}\langle{\bf \vec{s}}_{i}\rangle \cdot \langle{\bf \vec{s}}_{j}\rangle
\end{equation}
\begin{equation}
E=-\frac{1}{ 2} N \epsilon c M^2
\end{equation}

 \Dn
 

\item [(d)]
From section b we get:
\begin{equation}
{M =\coth(\beta{\bf h}_{eff})-\frac{1}{{ \beta { \bf h}_{eff}} } \approx \tanh{\beta { \bf h}_{eff}}}
\end{equation}
so we get the known result for M:
\begin{equation}
{M =\tanh(\beta{\bf h}_{eff})  \Rightarrow \:\: \tanh^{-1}(M)=\beta{\bf h}_{eff}}\nonumber\\
\tanh^{-1}(M)\approx M+\frac{M^3}{3}
\end{equation}
\begin{equation}
\frac{1}{T} (\epsilon cM+h)=M+\frac{M^3}{3} \:\:\Rightarrow\:\: h=(T-T_c)M + T_c \frac{M^3}{3}
\end{equation}
For the last term in the final expression we replaced $T$ with $T_c$, since we are looking only at temperatures close to $T_c$. 

First we look at $T>T_c$ :
there is only the trivial solution $m\rightarrow 0$
\begin{equation}
h=(T-T_c)M
\end{equation}
\begin{equation}
M=(T-T_c)^{-1} h = \chi h \;\;\Rightarrow \;\;\;\; \gamma = 1
\end{equation}

Now for $T<T_c$ we take $\;\;h\rightarrow 0^+ \;\; $
\begin{equation}
(T-T_c)M=-T_c\frac{M^3}{3} \nonumber \\ M^2=3\left(\frac{T_c-T}{T_c}\right)
\end{equation}
\begin{equation}
M=\sqrt{3 \left(\frac{T_c-T}{T_c}\right)} \;\;\Rightarrow \;\;\;\;\;  \beta=\frac{1}{2}
\end{equation}

 \Dn
\item [(e)]
We consider the mean field energy $\;E=-\frac{1}{ 2}N \epsilon c M^2 \;\;$:
\begin{equation}  
C_{v}= \frac{\partial E } {\partial T} =\frac{\partial } {\partial T}(-\frac{1}{2} N \epsilon c M^2)
\end{equation}
The only dependence of T in the energy is $M(T)$. The dependence is different below $T_c$ and above it. 

for $T>T_c$:
\begin{equation}
M=(T-T_c)^{-1} h \;,\;\;\; h\rightarrow 0 \;\;\Rightarrow \;\;\; M(T)=0 \;\; \Rightarrow \;\; C_v(T)=0
\end{equation}
and for $T<T_c$:
\begin{equation}  
C_{v}= -\frac{1}{2} NT_c  \frac{\partial M^2 } {\partial T} =-\frac{1}{2}NT_c\cdot \frac{\partial  } {\partial T}\left[3\left(\frac{T_c-T}{T_c}\right)\right]\cong \frac{3}{2}NT_c\cdot \frac{1}{T_c}=\frac{3}{2}N
\end{equation}
Since we are used $T\approx T_c$, we got a constant.
So the jump in $C_v$ is:
\begin{equation}
C_v(T_c^+)-C_v(T_c^-)=0-\frac{3}{2}N=-\frac{3}{2}N .
\end{equation}


\end{itemize}


\end{document} 
