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\exnumber{5645}
\heading{Potts model in one dimension}
\auname{Eli Gudinetsky}
{\bf The problem:}
\Dn
A set of $N$ atoms is arranged on a one-dimensional chain.
Each atom has $p$~possible {\em orientations}, labelled by ${\sigma=1,2,...,p}$.
Two neighboring atoms $\sigma_i$ and $\sigma_j$ have a negative interaction
energy $-\varepsilon$ if they are in the same orientation, and zero otherwise.
It is useful to define bond variables ${s_i=\sigma_{i{+}1}-\sigma_{i} \mod(p)}$.
\Dn
(1) The partition function $Z_{\text{chain}}(\beta)$ of an open chain
can be written as ${Z=Aq^{N-1}}$. Write what are $A$ and $q$.
Tip: the partition sum factorizes in the "bond" representation.
\Dn
(2) The partition function $Z_{\text{ring}}(\beta)$ of a closed chain,
with periodic boundary conditions, can be written as ${Z=\trc(T^N)}$.
Write what is the matrix $T$ for ${p=4}$.
\Dn
(3) Find what are the eigenvalues of the transfer matrix~$T$ for general $p$,
and deduce an explicit expression for $Z_{\text{ring}}(\beta)$.
Tip: The $T$ matrix is diagonal in the "momentum" representation.
\Dn
(4) Find the energy per atom at the $N\rightarrow\infty$ limit.
Write the result as ${E(T)/N = \epsilon f(\epsilon-\mu)}$.
Provide expressions for $\mu$ and for $f()$ using $p$ and the temperature $T$.
\Dn\Dn
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{\bf The solution:}
\Dn
{\Large Alert: over complicated perpective on trivail diagonalization in item 3}
\begin{enumerate}
\item
The Hamiltonian for this model is:
\begin{align}
\mathcal{H}=-\varepsilon\sum_{\Seq{i,j}}{\delta_{\sigma_i,\sigma_j}}\, ,
\end{align}
where $\delta_{\sigma_i,\sigma_j}$ is the usual Kronecker delta and it is 1 if $\sigma_i=\sigma_j$, and 0 otherwise.
Introducing the bond variables $s_i=\sigma_{i+1}-\sigma_i$, we can rewrite the Hamiltonian in the following form:
\begin{align}
\mathcal{H}=-\varepsilon\sum_{i=1}^{N-1}{\delta_{s_i,0}}\, .
\end{align}
The partition function is therefore,
\begin{align}
Z=\sum_{\sigma\Paren{\cdot}}{e^{\beta\varepsilon\sum_{\Seq{i,j}}{\delta_{\sigma_i,\sigma_j}}}}=\sum_{\sigma_1=1}^p\sum_{s_i\Paren{\cdot}}
e^{\beta\varepsilon\sum_{i=1}^{N-1}{\delta_{s_i,0}}}=
\sum_{\sigma_1=1}^p\sum_{s_i\Paren{\cdot}}\prod_{i=1}^{N-1}
e^{\beta\varepsilon{\delta_{s_i,0}}}\, .
\end{align}
Notice that the exponent in the product is 1 if $s_i\neq0$ and $e^{\beta\varepsilon}$ if $s_i=0$, so in a sense the product `counts' the number of bonds which contribute to the energy. The sum over $s_i$ factorized as $s_i$ and $s_j$ are independent. There are $p-1$ exponents which do not contribute to the energy and one which does (per $s_i$). So we get
\begin{align}
Z=\sum_{\sigma_1=1}^p\Paren{p-1+e^{\beta\varepsilon}}^{N-1}=
p\Paren{p-1+e^{\beta\varepsilon}}^{N-1}\, .
\end{align}
We then see that $A=p$ and $q=p-1+e^{\beta\varepsilon}$.
\item
As already stated, the Hamiltonian is:
\begin{align}
\mathcal{H}=-\varepsilon\sum_{i=1}^{4}{\delta_{s_i,0}}\, ,
\end{align}
which means that the partition function factorized, again, and it is:
\begin{align}
Z=\sum_{s_i\Paren{\cdot}}e^{\beta\varepsilon\sum_{i=1}^{4}{\delta_{s_i,0}}}=\sum_{s_1=-3}^{3}\Paren{{T}^N}_{s1s1}
=\text{Tr}\Paren{{T^N}}\,.
\end{align}
Here $T$ is the following matrix:\\
$
\\
\begin{pmatrix}
e^{\beta\varepsilon} & 1 & 1 & 1 \\
1 & e^{\beta\varepsilon} & 1 & 1 \\
1 & 1 & e^{\beta\varepsilon} & 1 \\
1 & 1 & 1 & e^{\beta\varepsilon}
\end{pmatrix}
=\Paren{e^{\beta\varepsilon}-1}I_{4\times4}+\text{One}_4\, ,$
\\
\\
where $\text{One}_4$ is a $4\times4$ matrix full of 1's, and $I$ is the identity matrix.
\item
In the general case:
\begin{align}
T=\Paren{e^{\beta\varepsilon}-1}I_{p\times p}+\text{One}_p=\Paren{e^{\beta\varepsilon}-1}I_{p\times p}+|\psi\rangle\langle\psi|\, ,
\end{align}
where $|\psi\rangle$ is the vector $\Paren{1,1,\dots,1}$. Notice that $|\psi\rangle$ is an eigenvector of $T$ with an eigenvalue $\Paren{e^{\beta\varepsilon}-1}+\sum_{i=1}^p1=e^{\beta\varepsilon}+p-1$. Notice that this matrix is symmetric, which means that every other eignenvector which has different eigenvalue is orthogonal to $|\psi\rangle$\footnote{Quick proof: $\lambda_1\langle \psi,\phi\rangle=\langle T\psi,\phi\rangle=\langle \psi,T^{\text{T}}\phi\rangle=\lambda_2\langle \psi,\phi\rangle\Rightarrow\Paren{\lambda_1-\lambda_2}\langle \psi,\phi\rangle=0\Rightarrow\langle \psi,\phi\rangle=0$}
so if $|\phi\rangle$ is another eigenvector we have $T|\phi\rangle=\Paren{e^{\beta\varepsilon}-1}|\phi\rangle$. This means that every other eigenvalue is $e^{\beta\varepsilon}-1$. In fact, diagonalizing $T$ we get:
$
\\
\\
\begin{pmatrix}
e^{\beta\varepsilon}+p-1 & 0 & \dots & 0 \\
0 & e^{\beta\varepsilon}-1 & 0 & 0 \\
\vdots & & \ddots & \vdots \\
0 & 0 & \dots & e^{\beta\varepsilon}-1
\end{pmatrix}
\, ,$
\\
\\
which means that:
\begin{align}
Z=\text{Tr}\Paren{{T^N}}=\Brack{e^{\beta\varepsilon}+p-1}^N+\Paren{p-1}\Brack{e^{\beta\varepsilon}-1}^N\,.
\end{align}
In the case of $p=4$ we get:
\begin{align}
Z=\Brack{e^{\beta\varepsilon}+3}^N+3\Brack{e^{\beta\varepsilon}-1}^N\,.
\end{align}
\item
In the limit of $N\to \infty$ we can neglect the smaller eigenvalues and have:
\begin{align}
\ln{Z}\approx N\ln{\Paren{e^{\beta\varepsilon}+p-1}}\,.
\end{align}
The energy per atom is:
\begin{align}
\frac{\langle E \rangle}{N}=-\frac{1}{N}\frac{\partial\ln{Z}}{\partial\beta}\approx -\frac{1}{{ e^{\beta\varepsilon}+p-1}}\varepsilon e^{\beta\varepsilon}=-\varepsilon\frac{1}{1+e^{-\beta\Paren{\varepsilon-\mu}}}=\varepsilon f\Paren{\epsilon-\mu}\,.
\end{align}
Here $\mu=\ln{\Paren{p-1}}/\beta$ and $f\Paren{\epsilon-\mu}=-\Paren{1+e^{-\beta\Paren{\varepsilon-\mu}}}^{-1}$.
\end{enumerate}
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\end{document}