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\begin{document}
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\heading{Ex 5958: Mean field approximation for the Model of ferroelectricity } 
\auname{yair massury} 
\Dn

{\bf The problem:}

\Dn

Model of ferroelectricity: Consider electric dipoles ${\bf p}$
on sites of
a simple cubic lattice which point along one of the crystal axes,
${\pm \langle 100\rangle}$. The interaction between dipoles is
\[\ U =\frac{ {\bf p}_{1}\cdot{\bf p}_{2}-   3({\bf p}_{1}\cdot {\bf r})
({\bf p}_{2}\cdot {\bf r})/r^{2}} {4 \pi \epsilon_{ 0}r^{3}}\]
where ${\bf r}$ is the distance between the dipoles, $r=|{\bf r}|$
and $ \epsilon_{ 0}$ is the dielectric constant.
\begin{itemize}
\item [(a)]
Assume nearest neighbour interactions and find the ground state
configuration. Consider either ferroelectric (parallel dipoles) or
anti-ferroelectric alignment (anti-parallel) between neighbours in
various directions.

\item [(b)]
Develop a mean field theory for the ordering in (a) for the
average polarization $P$ at a given site at temperature T: Write a
mean field equation for $P(T)$ and find the critical temperature
$T_c$.
\item[(c)] Find the susceptibility
$\chi =\left(\frac{\partial P}{\partial E}\right)_{E=0}$  at
$T>T_c$ for an electric field $E||\langle 100\rangle$, using the
mean field
theory. \\
\end{itemize}
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{\bf The solution:}

\Dn
\begin{itemize}
\item [(a)]
\begin{eqnarray}
 U =\frac{ { p}_{i}\cdot{ p}_{j}-   3({ p}_{i}\cdot { r}_{ij})
({ p}_{j}\cdot { r}_{ij})/r^{2}_{ij}} {4 \pi \epsilon_{ 0}r^{3}}
\end{eqnarray}


For $  r_{ij} || p_{i},p_{j}   $ which mean's that $    r_{ij}||\hat{x}  $ we get:

\begin{eqnarray} U_{ij} =\frac{ -2{ p}_{i}\cdot{ p}_{j}} {4 \pi \epsilon_{ 0}r^{3}}
\end{eqnarray}

The lowest enrgy is when $ \: { p}_{i}={ p}_{j} \: $ so that $ \: U_{ij} =-2U_{0} \:$ where $\: U_{0}=\frac{ p^2} {4 \pi \epsilon_{ 0}r^{3}}$

For $ \ \:\: r_{ij} \bot p_{i},p_{j} \ $ which mean's that $ \ \:\: r_{ij}\bot \hat{x} \ $ we get:

\begin{eqnarray} U_{ij} =\frac{ +{ p}_{i}\cdot{ p}_{j}} {4 \pi \epsilon_{ 0}r^{3}}
\end{eqnarray}

So $\:{ p}_{i}=-{ p}_{j} \:$ (in opposite direction),which mean's that $ \: U_{ij} =-U_{0}$
\\
\newline
\newline
\\
The ground state:\\
\includegraphics[scale=0.5]{pic5958.jpg}\\
\small FIG.1
\textbf { We have a chain in $\; \hat{x} \;$direction as each chain four \\
different neighboring chains with opposite polarity.}
 \normalsize
\Dn
\item [(b)]
Lets define two sublattice: 
\newline
A=positive moment +, which means that y+z=even.
\newline
B=negative moment -, which means that y+z=odd.
\newline
we get the Hamiltonian:


\begin{eqnarray}  H=-2 U_{o}{\sum^{{r_{ij} || \hat{x} }}_{i\in A , j\in A} }\frac{{ p}_{i}\cdot{ p}_{j}}{p^2}+ U_{o}{\sum^{{r_{ij} || \hat{y} }}_{i\in A , j\in B} }\frac{{ p}_{i}\cdot{ p}_{j}}{p^2}+ U_{o}{\sum^{{r_{ij} || \hat{z} }}_{i\in A , j\in B} }\frac{{ p}_{i}\cdot{ p}_{j}}{p^2}-E_{elctric}\sum_{i\in A , i\in B}{ p}_{i}
\nonumber\\= -2 U_{o}{\sum^{{r_{ij} || \hat{x} }}_{i\in B , j\in B} }\frac{{ p}_{i}\cdot{ p}_{j}}{p^2}+2 U_{o}{\sum^{{r_{ij} || \hat{y},\hat{z} }}_{i\in B , j\in A} }\frac{{ p}_{i}\cdot{ p}_{j}}{p^2}-E_{el} \sum_{i\in A}{ p}_{i}-E_{el} \sum_{i\in B}{ p}_{i}
\end{eqnarray}
\Dn
we make a mean field approximation (not curfully) to get:
\begin{eqnarray} H_{A}=-2 U_{o} P_{A}\sum_{i\in A  }\frac{{ p}_{i}}{p^2}+2 U_{o} P_{B}\sum_{i\in A  }\frac{{ p}_{i}}{p^2}-E_{el} \sum_{i\in A}{ p}_{i}
\end{eqnarray}
\begin{eqnarray} H_{B}=-2 U_{o}P_{B}\sum_{i\in B  }\frac{{ p}_{i}}{p^2}+2 U_{o} P_{A}\sum_{i\in B  }\frac{{ p}_{i}}{p^2}-E_{el} \sum_{i\in B}{ p}_{i}
\end{eqnarray}

 $H=H_{A}+H_{B} \;\;$to find the enrgy for one site .we took $\;\;{ p}_{A}=-{ p}_{B} \;\;$ (in opossite direction)
\begin{eqnarray}
E=\frac{1}{2}\left(-2U_0\frac{P^2}{P^2}+2U_0\frac{P_B (P_A)}{P^2}-E_{el}P_A-2U_0\frac{P^2}{P^2}+2U_0\frac{P_A (P_B)}{P^2}-E_{el}P_B \right) \nonumber\\ \Rightarrow \;\; E=-4U_o
\end{eqnarray}
In zero filed we look at average filed $\;P_{av}\;$
\begin{eqnarray}
H^{MF}_{E=0}=-4U_0 P_{av}\frac{p_i}{p^2}
\end{eqnarray}

\begin{eqnarray}
P_{av}=\frac{+p e^{\beta4 U_0\frac{P_{av}}{p}} -p e^{-\beta4 U_0\frac{P_{av}}{p}}  }{ e^{\beta4 U_0\frac{P_{av}}{p}} +e^{-\beta4 U_0\frac{P_{av}}{p}}  }=p \tanh\left({\beta4 U_0\frac{P_{{av}}}{p}}\right)
\end{eqnarray}
To find $T_c\;\;$ we take the first order in $\;\;\tanh \left({\beta4 U_0\frac{P_{{av}}}{p}} \right)$
\begin{eqnarray}
\frac{P_{av}}{p}=Tanh({\beta4 U_0\frac{P_{{av}}}{p}})\;\;\;\rightarrow \;\;\frac{P_{av}}{p}={\beta4 U_0\frac{P_{{av}}}{p}} \;\;\;\; \Rightarrow T_c=4 U_0
\end{eqnarray}
To see the behaviour near the critical temperature:
\begin{eqnarray}
\beta4 U_0\frac{P_{{av}}}{p}=Tanh^{-1}(\frac{P_{av}}{p})\cong \frac{P_{av}}{p}+ \frac{({\frac{P_{av}}{p}})^3}{3} \rightarrow \;\;3(\frac{T_c}{T}-1)={({\frac{P_{av}}{p}})^2} 
\end{eqnarray}
\item [(c)]
To Find the susceptibility
$\chi =\left(\frac{\partial P}{\partial E}\right)_{E=0}$  at
$T>T_c\;\;$.
we take first order in Taylor Series for  $\; tanh^{-1}(x)  \;$
\begin{eqnarray}
P_A=p\tanh \left( \beta2 U_0 \frac{P_A}{p}-\beta2 U_0 \frac{P_B}{p}+\beta Ep   \right) \nonumber\\
P_B=p\tanh \left( \beta2 U_0 \frac{P_B}{p}-\beta2 U_0 \frac{P_A}{p}+\beta Ep   \right)
\end{eqnarray}
\begin{eqnarray}
\tanh^{-1} \left(\frac{P_A}{p}\right)=\beta2 U_0 \frac{P_A}{p}-\beta2 U_0 \frac{P_B}{p}+\beta Ep \nonumber\\
\tanh^{-1} \left(\frac{P_B}{p}\right)=\beta2 U_0 \frac{P_B}{p}-\beta2 U_0 \frac{P_A}{p}+\beta Ep \nonumber\\
\rightarrow \frac{P_A}{p}=\beta2 U_0 \frac{P_A}{p}-\beta2 U_0 \frac{P_B}{p}+\beta Ep \nonumber\\
\rightarrow \frac{P_B}{p}=\beta2 U_0 \frac{P_B}{p}-\beta2 U_0 \frac{P_A}{p}+\beta Ep
\end{eqnarray}

\begin{eqnarray}
P=\frac{N}{2}(P_A+P_B)\;\; \Rightarrow \;\;\;\; P=N\beta Ep^2
\end{eqnarray}

\begin{eqnarray}
\chi=\frac{\partial{P}}{\partial E} |_{E=0}=N\beta p^2
\end{eqnarray}
\end{itemize}
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\end{document}
