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\exnumber{5645}
\heading{Potts Model in One Dimension} 
\auname{Alon Ishakbayev}


{\bf The problem:}

\Dn
A set of $N$ atoms is arranged on a one-dimensional chain.
Each atom has $p$~possible 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:}

\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 Kronecker's delta:
	\begin{equation}
	\delta_{\sigma_i,\sigma_j}=\begin{cases}
	1, & \sigma_i=\sigma_j\\
	0, & otherwise
	\end{cases}
\end{equation}

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}
Instead of summing over spin orientations, we can sum over bonds. The partition function is given by:
\begin{align}
Z=\sum_{\sigma_1}...\sum_{\sigma_N}{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}}}=\sum_{\sigma_1=1}^p\prod_{i=1}^{N-1}\sum_{s_i\Paren{\cdot}}e^{\beta\varepsilon{\delta_{s_i,0}}}
\end{align}
 The exponent in the product is 1 if $s_i\neq0$ and $e^{\beta\varepsilon}$ if $s_i=0$.
The final expression sums the energy contributions from the bonds relative to the first spin which can be found in $p$ orientations.
For each $s_i$ there is one state which contributes to the energy and $p-1$ states that do not contribute to the energy.
\\* 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}
Therefore: $A=p$ and $q=p-1+e^{\beta\varepsilon}$.

\item
For periodic bounday conditions $\sigma_{N+1}=\sigma_{1}$.
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:
\begin{align}
Z=\sum_{s_i\Paren{\cdot}}e^{\beta\varepsilon\sum_{i=1}^{4}{\delta_{s_i,0}}}=\text{Tr}\Paren{{T^N}}\,.
\end{align}
Diagonal matrix elements - spins in the same orientation (contribution of $e^{\beta\epsilon}$)
\\* Off-Diagonal matrix elements - not in the same orientation (contribution of $e^{\beta0}=1$)
\\ Therefore $T$ 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}
=\text{Ones}_{4\times 4}\,+\Paren{e^{\beta\varepsilon}-1}I_{4\times4}$
\\

\item
In general (p orientations):
\begin{align}
T=\Paren{e^{\beta\varepsilon}-1}I_{p\times p}+\text{Ones}_{p\times p}
\end{align}
The first term is diagonalized \\*
The second term is ones matrix with $Dim(Ones)=p$, so the eigenvalues are:\\*1. $p$ (Degeneration 1) \\*2. 0 (Degeneration $p-1$)

 Diagonalizing $T$ we get the following eigenvalue matrix:
$
\\
\\
 \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}
\, $
\\
\\
The partition function:
\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}

\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}
Where we defined: $\mu=\ln{\Paren{p-1}}/\beta$ and $f\Paren{\epsilon-\mu}=-\Paren{1+e^{-\beta\Paren{\varepsilon-\mu}}}^{-1}$.
\end{enumerate}


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