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\begin{document}
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\heading{E5966: Long Range Interaction Ising Model 1D, Domain Walls}
\auname{Yaakov Kleeorin}
\textbf{The problem:}
\Dn
Consider the one dimensional Ising model with the hamiltonian
\[
H=-\sum_{n,n'}J(n-n')\sigma(n)\sigma(n')
\]
with $\sigma(n)=\pm1$ at each site $n$ and long range interaction
$J(n)=b/n^{\gamma}$ with $b>0$. Find the energy of a domain wall
at $n=0$, i.e. all the $n<0$ spins are down and the others are up.
Show that the standard argument for the absence of spontaneous magnetization
at finite temperature fails if $\gamma<2$.
\Dn\Dn
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\textbf{The solution:}
\Dn
The standard argument for the absence of spontaneous magnetization
is that in $1D$, it is always favourable (for any finite Temperature)
to create a domain wall at some location (lets call it $n=0)$ thus
gaining entropy at the expense of a small energy increase.
~
$F=E-TS$ is the free energy we need to minimize. Since the energy
gain (in comparison to the ground state) in creating a domain wall
is $2J$ (in the standard argument) and the entropy becomes $\ln(N)$
where $N$ is the number of spins and also the number of positions
the wall can assume. the change in free energy is $\Delta F=2J-T\ln(N)$
and since $N$ is huge, this is negative for any finite temperature,
making this transition favorable.
~
%%%%%%%%%%%%%%%
\Dn
In our long range interaction model, the ground state energy is the
summing of all interactions in any pair of spins, all being the same
sign. We will count over all $N$ spins, making sure we dont double
count by demanding $n'>n$ in our sum:
\begin{equation}
E_{0}=-\sum_{n=-\infty}^{\infty}\sum_{n'=n+1}^{\infty}\frac{b}{(n'-n)^{\gamma}}
\end{equation}
We will not solve this sum.
%%%%%%%%%%%%%%
\Dn
When creating a domain wall at $n=0$ we flip the sign of the interaction
between every pair on opposite sides of the wall, thus paying a double
price (-1 goes to 1, the difference is 2). We sum over all $N$ spins
(we assume $N$ spins on every side, but obviously it doesnt have
to be and it doesnt matter):
\begin{equation}
E_{1}=E_{0}+2\sum_{n=0}^{N}\sum_{n'=1}^{N}\frac{b}{(n'+n)^{\gamma}}
\end{equation}
%%%%%%%%%%%%%
\Dn
In order to go to an integral form, we make $n$ continuous by going
to new variables representing spin location. We introduce spin separation
$a$ and define $x=na$ and $y=n'a$. $dx=dy=a$.
\begin{equation}
E_{1}=E_{0}+2ba^{\gamma-2}\int_{0}^{Na}dx\int_{a}^{Na}dy\frac{1}{(x+y)^{\gamma}}
\end{equation}
This integral is solved differently for different $\gamma$.
%%%%%%%%%%%%%
\Dn
For $\gamma>2$:
\begin{equation}
\Delta E=2\frac{b}{(\gamma-1)(\gamma-2)}
\end{equation}
%%%%%%%%%%%%%
\Dn
For $\gamma=2$:
\begin{equation}
\Delta E=2b\ln(N)
\end{equation}
%%%%%%%%%%%%%
\Dn
Under the value of $\gamma=2$ the divergence in $N$ is even stronger.
For example the $1<\gamma<2$ case gives:
\begin{equation}
\Delta E=2N^{2-\gamma}\frac{b}{(\gamma-1)(\gamma-2)}
\end{equation}
%%%%%%%%%%%%%
\Dn
for the cases of $\gamma\leq2$ we see that it is not favorable, in
terms of free energy, to create a domain wall:
\begin{equation}
\Delta F=-T\ln(N)+2b\ln(N)
\end{equation}
\begin{equation}
\Delta F=-T\ln(N)+2bN^{2-\gamma}/(\gamma-1)(\gamma-2)
\end{equation}
in the first case it depends on the value of $T$. In the second case
it is never (as $N$ is huge) favorable to create domain walls, thus
breaking the argument.
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\end{document}