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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 ($N'<N$), i.e. on average $\langle n_{i}\rangle=N'/N$.

\Dn

(1) Evaluate the partition sum by allowing all configurations of spins
and of atoms. Calculate the free energy.

(2) If the atoms are stationary impurities one needs to evaluate the
free energy F for some random configuration of the atoms and then
average F over all configurations. Evaluate the average F. Find the
entropy difference between the result calculated from the averaged
F and from the F calculated in (1).

\Dn\Dn

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\textbf{The solution:}

\Dn

(1) The partition function is:

\[
Z=\sum_{\sigma_{1}=\pm1}\sum_{\sigma_{2}=\pm1}...\sum_{\sigma_{n}=\pm1}e^{-\beta(-J\sum_{i=1}^{N}\sigma_{i}\sigma_{i+1}(1-\lambda n_{i}))}\]


Using the Transfer matrices formalism we get:

\[
Z=\sum_{\sigma_{1}=\pm1}\sum_{\sigma_{2}=\pm1}...\sum_{\sigma_{n}=\pm1}<\sigma_{1}|P_{1}|\sigma_{2}><\sigma_{2}|P_{2}|\sigma_{3}>...<\sigma_{n}|P_{n}|\sigma_{n+1}>\]


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:

\[
F=-\tau lnZ=-\tau ln((\begin{array}{c}
N\\
N'\end{array}))-\tau l(N-N')ln(2cosh(\beta J))-\tau N'ln(2cosh(\beta J(1-\lambda)))\]


%%%%%%%%%%%%%%%
\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:

\[
<F>=\frac{\sum_{configurations}F_{conf}}{\sum_{configurations}}=\frac{(\begin{array}{c}
N\\
N'\end{array})\times-\tau l(N-N')ln(2cosh(\beta J))-\tau N'ln(2cosh(\beta J(1-\lambda)))}{(\begin{array}{c}
N\\
N'\end{array})}\]


\[
<F>=-\tau l(N-N')ln(2cosh(\beta J))-\tau N'ln(2cosh(\beta J(1-\lambda)))\]


The entropy difference between the two calculations:\[
\triangle S=-\frac{\partial F}{\partial\tau}+\frac{\partial<F>}{\partial\tau}=ln((\begin{array}{c}
N\\
N'\end{array}))\]


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\end{document}
