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\begin{document}
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\heading{E5651: Ising in interaction with lattice gas}
\auname{Rotem Kupfer}
\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 $N'=\sum_{i=1}^{N}n_{i}$
atoms ($N'<\sigma_{2}|P_{2}|\sigma_{3}>...<\sigma_{n}|P_{n}|\sigma_{n+1}>\]
The transfer matrix for two spins without an atom between is:
\[
P_{i}=\left(\begin{array}{cc}
e^{\beta J} & e^{-\beta J}\\
e^{-\beta J} & e^{\beta J}\end{array}\right)\]
The transfer matrix for two spins with an atom between is:
\[
P_{i}^{'}=\left(\begin{array}{cc}
e^{\beta J(1-\lambda)} & e^{-\beta J(1-\lambda)}\\
e^{-\beta J(1-\lambda)} & e^{\beta J(1-\lambda)}\end{array}\right)\]
Both matrices commute:
\[
[P_{i},P_{i}^{'}]=0\]
So the multiplication order does not matter. After changing to the
diagonalized base we get for a single configuration:
\[
Z=Tr(P'^{N'}P^{N-N'})\]
\[
=Tr(\left(\begin{array}{cc}
2cosh(\beta J)^{N-N'} & 0\\
0 & 2sinh(\beta J)^{N-N'}\end{array}\right)\left(\begin{array}{cc}
2cosh(\beta J(1-\lambda))^{N'} & 0\\
0 & 2sinh(\beta J(1-\lambda))^{N'}\end{array}\right))\]
Assuming $N\rightarrow\infty$ we can neglect the smaller eigenvalue
so:
\[
Z=(\begin{array}{c}
N\\
N'\end{array})2cosh(\beta J)^{N-N'}2cosh(\beta J(1-\lambda))^{N'}\]
Notice that a combinatorial factor was introduced in order to count
the different configurations possible for a given N'.
The free energy is therefore:
\[
F=-\tau lnZ=-\tau ln((\begin{array}{c}
N\\
N'\end{array}))-\tau l(N-N')ln(2cosh(\beta J))-\tau N'ln(2cosh(\beta J(1-\lambda)))\]
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\Dn
(2) For any configuration with N' impurities the partition function
is:
\[
Z=2cosh(\beta J)^{N-N'}2cosh(\beta J(1-\lambda))^{N'}\]
The free energy for any configuration with N' impurities is (It doesn't
depends on the specific configuration):
\[
F=-\tau lnZ=-\tau l(N-N')ln(2cosh(\beta J))-\tau N'ln(2cosh(\beta J(1-\lambda)))\]
The average free energy for a given N' is:
\[
=\frac{\sum_{configurations}F_{conf}}{\sum_{configurations}}=\frac{(\begin{array}{c}
N\\
N'\end{array})\times-\tau l(N-N')ln(2cosh(\beta J))-\tau N'ln(2cosh(\beta J(1-\lambda)))}{(\begin{array}{c}
N\\
N'\end{array})}\]
\[
=-\tau l(N-N')ln(2cosh(\beta J))-\tau N'ln(2cosh(\beta J(1-\lambda)))\]
The entropy difference between the two calculations:\[
\triangle S=-\frac{\partial F}{\partial\tau}+\frac{\partial}{\partial\tau}=ln((\begin{array}{c}
N\\
N'\end{array}))\]
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\end{document}