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\begin{document}
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\heading{E5971: Potts model in one dimension}
\auname{Yair Judkovsky}
{\bf The problem:}
\Dn
A set of N atoms, each with $p$ states, is arranged on a 1d chain with periodic boundary conditions. The atom at the $i$-th site is in a state $n_i$ that is chosen from the set $\{1,2,...,p\}$.
\Dn
Two neighboring atoms at sites $i$ and $j$ (where $j=i+1$), respectively, have an interaction energy $-J$ ($J>0$) if they are in the same state, i.e. $n_i=n_{j}$, and 0 interaction otherwise.
\Dn
The Hamiltonian is therefore
\[ {\cal H}= -J\sum_{} \delta_{n_i,n_{j}}\]
where $\delta_{n_i,n_{j}}$ is the Kronecker symbol, and the boundary conditions are $n_{N+1}=n_1$. The summation is being done only over nearest neighbours, i.e. over $$ satisfying $j=i+1$.
\Dn
(At all items consider the limit $N\rightarrow \infty$.)
\Dn
(1) Derive the free energy of the system for $p=2$.
\Dn
(2) Derive the free energy of the system for a general $p$.
\Dn
(3) Find the energy $E$ at the low and high temperature limits and interpret the results.
\Dn\Dn
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{\bf The solution:}
\Dn
(1) First, we shall calculate the partition function by summing over all states:
\Dn
\[ {Z}= \sum_{r} e^{-{\beta} E_{r}}=\sum_{n_{1}=1}^p{\times}\sum_{n_{2}=1}^p{\times}...{\times}\sum_{n_{N}=1}^{p}e^{{\beta}J\sum_{i=1}^{p}{\delta}_{n_{i},n_{i+1}}}= \]
\Dn
\[ = \sum_{n_{1}=1}^p{\times}\sum_{n_{2}=1}^p{\times}...{\times}\sum_{n_{N}=1}^{p}e^{{\beta}J{\delta}_{n_{1},n_{2}}}e^{{\beta}J{\delta}_{n_{2},n_{3}}}...e^{{\beta}J{\delta}_{n_{N},n_{N+1}}} \]
\Dn
We shall now define the transfer matrix $T$, and use the Dirac notation to denote its (i,j) element:
\Dn
${T_{ij}==e^{ {\beta}J{\delta}_{n_{i},n_{j}} }}$
\Dn
Using this definition and the boundary condition ${n_{N+1}=n_{1}}$, we get
\Dn
\[ Z = \sum_{n_{1}=1}^p{\times}\sum_{n_{2}=1}^p{\times}...{\times}\sum_{n_{N}=1}^{p}... \]
\Dn
\[= \sum_{n_{1}=1}^{p}=trace(T^{N})=\sum_{j=1}^{p}{\lambda}_{j}^N \]
\Dn
Where the ${\lambda}$'s are the eigenvalues of the transfer matrix $T$.
\Dn
In the case of p=2, we get
\Dn
\[
T
= \left( \amatrix { e^{{\beta}J} & 1 \cr 1 & e^{{\beta}J}}\right)
\]
\Dn
And by demanding
\[
Det\left( \amatrix { e^{{\beta}J}-{\lambda} & 1 \cr 1 & e^{{\beta}J}-{\lambda}}\right) =0
\]
\Dn
We get the eigenvalues:
\Dn
${{\lambda}_{+}=e^{{\beta}J}+1 }$
${{\lambda}_{-}=e^{{\beta}J}-1}$
\Dn
From which we can derive the free energy:
\Dn
${{\ln}(Z)={\ln}({\lambda}_{+}^{N}+{\lambda}_{-}^{N})={\ln} ({\lambda}_{+}^{N}(1+\frac{{\lambda}_{-}^{N}}{{\lambda}_{+}^{N}})) }$
\Dn
\Dn
In the thermodynamic limit $N\rightarrow \infty$ we get
\Dn
\Dn
${{\ln}(Z)={\ln}({\lambda}_{+}^{N})=N{\ln}{\lambda}_{+}=N{\ln}(e^{{\beta}J}+1)}$
\Dn
\Dn
Thus we get the free energy for the case p=2:
\Dn
\Dn
${F=-NT{\times}{\ln}(e^{{\beta}J}+1)}$
\Dn
\Dn
(2) For a general $p$, the problem is finding the eigenvalues of the ${T}$ matrix, which has now ${p*p}$ elements. Since we consider the thermodynamic limit only, it will be enough to find the largest eigenvalue of the ${T}$ matrix.
\Dn
\[
T
= \left( \amatrix { e^{{\beta}J} & 1 & 1 & ... \cr 1 & e^{{\beta}J} & 1 & ...\cr 1 & 1 & e^{{\beta}J} & ... \cr ... & ... & ... & ...} \right)
\]
\Dn
In order to find the eigenvalues of the $T$ matrix, we write down the vectorial equation
\Dn
${Ta={\lambda}a}$
\Dn
where $a$ is a column vector consisting of p components ${a_{1}, a_{2}, ... ,a_{p}}$.
\Dn
\Dn
Multiplying the column vector ${a}$ by $T$ from the left, we obtain $p$ linear equations:
\Dn
for each $j$, we get \[{(e^{{\beta}J}-1)a_{j}+\sum_{i=1}^{p}a_{i}={\lambda}a_{j}}\]
\Dn
Assuming that the vector $a$ is not the trivial zero solution, there is at least one component $a_{j}$ NOT equal to zero. Thus, we can conclude that there are two ways of solving the eigenvalue equation:
\Dn
(a) A solution where \[\sum_{i=1}^{p}a_{i}=0\] and \[{\lambda}=e^{{\beta}J}-1\]
\Dn
(b) A solution where \[\sum_{i=1}^{p}a_{i}{\not}{=}0\] which implies that $a_{j}{\not}{=}0$ for ANY $j$. Thus, we get
\Dn
\[{\lambda} = e^{ {\beta} J} -1+ \frac {1} {a_{j}} \sum_{i=1}^{p}a_{i}\]
\Dn
Since ${\lambda}$ cannot depend on our selection of $j$ here, we deduce that for this solution all $a_{j}$ are equal, and hence we get
\[{\lambda}=e^{{\beta}J}-1+p\]
Which is the LARGEST eigenvalue.
\Dn
\Dn
Solution (a) (${{\lambda}=e^{{\beta}J}-1}$) has a $p-1$ degeneration, and solution (b) (${{\lambda}=e^{{\beta}J}-1+p}$) has no degeneration - only a vector with $p$ equal entries has this eigenvalue.
\Dn
\Dn
After finding the largest eigenvalue, we can find ${{\ln}(Z)}$ in the thermodynamic limit $N\rightarrow \infty$:
\Dn
${\ln(Z)=\ln((largest-eigenvalue)^{N})=N{\times}\ln(e^{{\beta}J}-1+p)}$
\Dn
\Dn
And thus we obtain the free energy for a general $p$ in the thermodynamic limit:
\Dn
\Dn
${F=-NT{\times}\ln(e^{{\beta}J}-1+p)}$
\Dn
\Dn
(3) The energy $E$ will be found by differentiating $\ln(Z)$ with respect to ${\beta}$:
\Dn
\Dn
${E=-\frac{\partial \ln(Z)}{\partial \beta}=-NJ\frac{e^{{\beta}J}}{e^{{\beta}J}-1+p}}$
\Dn
\Dn
In the limit of low temperature ($T<>J$) the system approaches the energy $-\frac{N}{p}J$ .
\Dn
The thermal fluctuatuions will be so strong that the system's behavior will be totally random - there will be no significance to the interaction energy $J$, each atom could be at any one of the states with equal probability $\frac{1}{p}$, and the number of $J$-interactions will correspondingly be $\frac{N}{p}$. Thus, the energy of the system in this limit will be $-\frac{N}{p}J$.
\Dn
\Dn
\Dn
As a general comment, it should be noted that the properties of the largest eigenvalue and of the corresponding eigenvectors when T is a non-negative real square matrix is being described more generally in the Perron-Frobenius theorem.
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\end{document}