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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
\[ {\cal H}= -J\sum_{<ij>} \delta_{n_i,n_{j}}\]
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 $<ij>$ 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

${T_{ij}=<n_{i}|T|n_{j}>=e^{  {\beta}J{\delta}_{n_{i},n_{j}}  }}$

\Dn

Using this definition and the boundary condition ${n_{N+1}=n_{1}}$, we get

\Dn

\[ Z = \sum_{n_{1}=1}^p{\times}\sum_{n_{2}=1}^p{\times}...{\times}\sum_{n_{N}=1}^{p}<n_{1}|T|n_{2}><n_{2}|T|n_{3}>...<n_{N}|T|n_{N+1}> \]

\Dn

\[= \sum_{n_{1}=1}^{p}<n_{1}|T^{N}|n_{1}>=trace(T^{N})=\sum_{j=1}^{p}{\lambda}_{j}^N \]

\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 ${E=-NJ}$. 

\Dn

In this limit there is no external energy coming from thermal fluctuations, and thus the system will be in equilibrium when all atoms will be in the same state. Another interpertation - at low temperature the entropy has no effect over the free energy, and thus the minimum of the free energy $F$ will also be the minimum of the energy $E$.


\Dn
\Dn

In the limit of high 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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