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\begin{document}

\heading{E5966:Ising model on bipartite lattice, domain walls} 
\auname{Hamody Agbaria}
{\bf The problem:}\\

\begin{itemize}
\item[(a)] Consider the Ising spin model on a bipartite lattice, i.e.
it has two sublattices A,B
such that each site on lattice A has its nearest neighbors on
sublattice B, and vice versa. The Hamiltonian is
\begin{equation}\label{H}
{\cal H}=-J\sum_{{\bf n}, \bm{\delta}}\sigma({\bf n})\sigma({\bf
n}+\bm{\delta}) -h\sum_{{\bf n}}\sigma({\bf n})
\end{equation}
where  ${\bf n}$ are the lattice sites, $\bm{\delta}$ labels the
nearest neighbors, $h$ is proportional to a magnetic field and
$\sigma({\bf n})=\pm 1$. For $h=0$ show that the free energy
satisfies $F(J,T)=F(-J,T)$, hence the critical temperatures
satisfy $T_c^F=T_c^{AF}$ for the ferromagnetic ($J>0$) and
anti-ferromagnetic ($J<0$) transitions. Define the order
parameters at $T<T_c$ and the magnetic susceptibilities to $h$ at
$T>T_c$ and find their relationship, if any.
\item[(b)]
Consider the one dimensional Ising model with the Hamiltonian
${\cal H}=-\sum_{n,n'}J(n-n')\sigma(n)\sigma(n')$ with
$\sigma(n)=\pm 1$ at each site n and $J(n)=b/n^{\gamma}$ is a long
range interaction and $b>0$. Find the energy of a domain wall
(i.e. $n<0$ spins are $-$ and $n\geq 0$ are $+$) and show that the
argument for the absence of spontaneous magnetization at finite
temperatures fails when  $\gamma <2$.
\end{itemize}

{\bf The solution:}
\begin {itemize}
\item[(a)]
Ferromag \rightarrow AF:

\[Z=\sum_{\sigma_{n}=\pm 1} e^{\beta J \sum_{n} \sigma_{n} \sigma_{n+1}} \underbrace{\longrightarrow}_{\sigma_{n} \rightarrow - \sigma_{n}} \sum_{\sigma_{n}=\pm 1} e^{-\beta J \sum_{n} \sigma_{n}\sigma_{n+1}} \tag{2} \]. 

so that bi-partite lattice give:

 \[F(J)=F(-J)\tag{3}\]
\item[(b)]
\[{\cal H}=-\sum_{n,n'}\frac{b}{|n-n'|^{\gama}}\sigmma(n)\sigma(n')\tag{4}\] where $\sigma(n)=\pm 1$

\[E_{wall}=\underbrace{\downarrow \downarrow \downarrow ...}_{\frac{N}{2}} | \underbrace{\uparrow \uparrow \uparrow ...}_{\frac{N}{2}}\] where $\frac{N}{2}$ spins is up
and $\frac{N}{2}$ is down

\[E_{wall}=2b\sum_{n=-\infty}^{-1} \sum_{n'=0}^{\infty} \frac{1}{|n-n'|^{\gamma}} \backsimeq 2b\int_{-\infty}^{0}dn \int_{0}^{\infty}dn' \frac{1}{|n-n'|^{\gamma}}\tag{5}\]
 
The convergence at $n\rightarrow \infty $ needed, hence, $\sum_{n} \rightarrow \int dn$ is valid

so we got:
\[E_{wall}=2b \int_{-\infty}^{0}dn\frac{-1}{\gamma -1} \frac{1}{(n'-n)^{\gamma -1}}\bigg|_{n'=0}^{\infty}\tag{6}\]

\begin {itemize}
\item[(i)]
If \gamma >1 :

\[E_{wall}=\frac{2b}{\gamma -1} \int_{-\infty}^{0}dn\frac{1}{-n}^{\gamma -1}=\frac{2b}{\gamma -1}\int_{0}^{\infty}dn \frac{1}{n^{\gamma -1}}=\frac{2b}{(\gamma -1)(\gamma -2) \frac{-1}{n^{\gamma -2}}} \bigg|_{n'=0}^{\infty}\tag{7}\]

\item[(ii)]
If \gamma <2:

Where wall entropy $\backsim ln(N)$
now : \[F_{wall}\backsim N^{2-\gamma}-K_{B} ln(N)>0 \tag{8}\]
no walls are created when there are long-range intraction!  
\end {itemize}
                                    
\end {itemize}
\end{document}
