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# 6.7: Appendix I- Potts Model in One Dimension


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## Definition

The Potts model is defined by the Hamiltonian

$H=-J\sum_{\langle ij\rangle}\delta\nd_{\sigma\ns_i,\sigma\ns_j}-h\sum_i\delta\nd_{\sigma\ns_i,1}\ .$

Here, the spin variables $$\sigma\ns_i$$ take values in the set $$\{1,2,\ldots,q\}$$ on each site. The equivalent of an external magnetic field in the Ising case is a field $$h$$ which prefers a particular value of $$\sigma$$ ($$\sigma=1$$ in the above Hamiltonian). Once again, it is not possible to compute the partition function on general lattices, however in one dimension we may once again find $$Z$$ using the transfer matrix method.

## Transfer matrix

On a ring of $$N$$ sites, we have

$\begin{split} Z&={ Tr}\,e^{-\beta H}\\ &=\sum_{\{\sigma\ns_n\}} e^{\beta h\delta_{\sigma\ns_1,1}}\,e^{\beta J\delta_{\sigma\ns_1,\sigma\ns_2}}\,\cdots\, e^{\beta h\delta_{\sigma\ns_N,1}}\,e^{\beta J\delta_{\sigma\ns_N,\sigma\ns_1}}\\ &={ Tr}\,\big(R^N\big)\ , \end{split}$

where the $$q\times q$$ transfer matrix $$R$$ is given by

$R\nd_{\sigma\sigma'}=e^{\beta J\delta_{\sigma\sigma'}}\,e^{1\over 2 \beta h\delta_{\sigma,1}}\,e^{ 1\over 2 \beta h\delta_{\sigma',1}}= \begin{cases} e^{\beta(J+h)} & \hbox{ if \sigma=\sigma'=1}\\ e^{\beta J} & \hbox{ if \sigma=\sigma'\ne 1}\\ e^{\beta h/2} & \hbox{ if \sigma=1 and \sigma'\ne 1}\\ e^{\beta h/2} & \hbox{ if \sigma\ne1 and \sigma'= 1}\\ 1& \hbox{ if \sigma\ne1 and \sigma'\ne 1 and \sigma\ne\sigma'}\ . \end{cases}$

In matrix form,

$R=\begin{pmatrix} e^{\beta(J+h)} & e^{\beta h/2} & e^{\beta h/2} & & \cdots & & e^{\beta h/2} \\ e^{\beta h/2} & e^{\beta J} & 1 & & \cdots & & 1 \\ e^{\beta h/2} & 1 & e^{\beta J} & & \cdots & & 1 \\ \vdots & \vdots & \vdots & & \ddots & & \vdots \\ e^{\beta h/2} & 1 & 1 & & \cdots & e^{\beta J} & 1\\ e^{\beta h/2} & 1 & 1 & & \cdots & 1 & e^{\beta J} \end{pmatrix}$

The matrix $$R$$ has $$q$$ eigenvalues $$\lambda\ns_j$$, with $$j=1,\ldots,q$$. The partition function for the Potts chain is then

$Z=\sum_{j=1}^q \lambda_j^N \ .$

We can actually find the eigenvalues of $$R$$ analytically. To this end, consider the vectors

$\phi=\begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0\end{pmatrix} \qquad,\qquad \psi=\big(q-1+e^{\beta h}\big)^{-1/2} \begin{pmatrix} e^{\beta h/2} \\ 1 \\ \vdots \\ 1 \end{pmatrix}\ .$

Then $$R$$ may be written as

$R=\big(e^{\beta J}-1\big)\,{\mathbb I} + \big(q-1+e^{\beta h}\big)\,\sket{\psi}\sbra{\psi} + \big(e^{\beta J}-1\big)\big(e^{\beta h}-1\big)\,\sket{\phi}\sbra{\phi}\ ,$

where $${\mathbb I}$$ is the $$q\times q$$ identity matrix. When $$h=0$$, we have a simpler form,

$R=\big(e^{\beta J}-1\big)\,{\mathbb I} + q\,\sket{\psi}\sbra{\psi}\ .$

From this we can read off the eigenvalues:

$\begin{split} \lambda\ns_1&=e^{\beta J} +q-1 \\ \lambda\ns_j&=e^{\beta J}-1 \quad,\quad j\in\{2,\ldots,q\}\ , \end{split}$

since $$\sket{\psi}$$ is an eigenvector with eigenvalue $$\lambda=e^{\beta J} +q-1$$, and any vector orthogonal to $$\sket{\psi}$$ has eigenvalue $$\lambda=e^{\beta J}-1$$. The partition function is then

$Z=\big(e^{\beta J} +q-1\big)^N + (q-1)\big(e^{\beta J}-1\big)^N\ .$

In the thermodynamic limit $$N\to\infty$$, only the $$\lambda\nd_1$$ eigenvalue contributes, and we have

$F(T,N,h=0)=-N\kT\ln\big(e^{J/\kT}+q-1\big)\qquad\hbox{ for N\to\infty}\ .$

When $$h$$ is nonzero, the calculation becomes somewhat more tedious, but still relatively easy. The problem is that $$\sket{\psi}$$ and $$\sket{\phi}$$ are not orthogonal, so we define

$\sket{\xhi}={\sket{\phi}-\sket{\psi}\sbraket{\psi}{\phi}\over \sqrt{1-\sbraket{\phi}{\psi}^2}}\ ,$

where

$x\equiv\sbraket{\phi}{\psi}=\bigg( {e^{\beta h}\over q-1+e^{\beta h}} \bigg)^{\!1/2}\ .$

Now we have $$\sbraket{\xhi}{\psi}=0$$, with $$\sbraket{\xhi}{\xhi}=1$$ and $$\sbraket{\psi}{\psi}=1$$, with

$\sket{\phi}=\sqrt{1-x^2\>}\,\sket{\chi} + x\,\sket{\psi}\ .$

and the transfer matrix is then

\begin{aligned} R&=\big(e^{\beta J}-1\big)\,{\mathbb I} +\big(q-1+e^{\beta h}\big)\,\sket{\psi}\sbra{\psi}\nonumber\\ &\qquad +\big(e^{\beta J}-1\big)\big(e^{\beta h}-1\big)\,\bigg[(1-x^2)\, \sket{\xhi}\sbra{\xhi} + x^2\>\sket{\psi}\sbra{\psi}+x\,\sqrt{1-x^2}\>\Big(\sket{\xhi}\sbra{\psi} +\sket{\psi}\sbra{\xhi}\Big)\bigg]\nonumber\\ &=\big(e^{\beta J}-1\big)\,{\mathbb I} +\Bigg[ \big(q-1+e^{\beta h}\big) + \big(e^{\beta J}-1\big)\big(e^{\beta h}-1\big)\bigg({e^{\beta h}\over q-1+e^{\beta h}}\bigg)\Bigg] \sket{\psi}\sbra{\psi}\label{RPotts}\\ &\qquad\qquad+\big(e^{\beta J}-1\big)\big(e^{\beta h}-1\big)\, \bigg({q-1\over q-1+e^{\beta h}}\bigg)\, \sket{\xhi}\sbra{\xhi} \nonumber\\ &\qquad\qquad\qquad\qquad+\big(e^{\beta J}-1\big)\big(e^{\beta h}-1\big)\, \bigg({(q-1)\,e^{\beta h}\over q-1+e^{\beta h}}\bigg)^{\!1/2}\,\Big(\sket{\xhi}\sbra{\psi} +\sket{\psi}\sbra{\xhi}\Big)\ ,\nonumber\end{aligned}

which in the two-dimensional subspace spanned by $$\sket{\xhi}$$ and $$\sket{\psi}$$ is of the form

$R=\begin{pmatrix} a & c \\ c & b\end{pmatrix}\ .$

Recall that for any $$2\times 2$$ Hermitian matrix,

$\begin{split} M&=a\nd_0\,{\mathbb I} +\Ba\cdot\Btau\\ &=\begin{pmatrix} a\ns_0 + a\ns_3 & a\ns_1 -i a\ns_2 \\ a\ns_1 + i a\ns_2 & a\ns_0-a\ns_3\end{pmatrix}\ , \end{split}$

the characteristic polynomial is

$P(\lambda)={ det}\,\big(\lambda\,{\mathbb I}-M\big)=(\lambda-a\ns_0)^2 - a_1^2-a_2^2-a_3^2\ ,$

and hence the eigenvalues are

$\lambda\nd_\pm=a\ns_0\pm\sqrt{a_1^2+a_2^2+a_3^2\,}\ .$

For the transfer matrix of Equation \ref{RPotts}, we obtain, after a little work,

\begin{aligned} \lambda\nd_{1,2}&=e^{\beta J}-1+\half\Big[q-1+e^{\beta h} + \big(e^{\beta J}-1\big)\big(e^{\beta h}-1\big)\Big]\\ &\qquad\pm\half\sqrt{\Big[q-1+e^{\beta h} + \big(e^{\beta J}-1\big)\big(e^{\beta h}-1\big)\Big]^{2} -4(q-1) \big(e^{\beta J}-1\big)\big(e^{\beta h}-1\big)\>}\ .\nonumber\end{aligned}

There are $$q-2$$ other eigenvalues, however, associated with the $$(q\!-\!2)$$-dimensional subspace orthogonal to $$\sket{\xhi}$$ and $$\sket{\psi}$$. Clearly all these eigenvalues are given by

$\lambda\ns_j=e^{\beta J}-1\qquad,\quad j\in\{3\,,\,\ldots\,,\,q\}\ .$

The partition function is then

$Z=\lambda_1^N + \lambda_2^N + (q-2)\,\lambda_3^N\ ,$

and in the thermodynamic limit $$N\to\infty$$ the maximum eigenvalue $$\lambda\ns_1$$ dominates. Note that we recover the correct limit as $$h\to 0$$.

6.7: Appendix I- Potts Model in One Dimension is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Daniel Arovas.