Exact Partition Functions for the q-State Potts Model with a Generalized Magnetic Field on Lattice Strip Graphs

Abstract

We calculate the partition function of the q-state Potts model on arbitrary-length cyclic ladder graphs of the square and triangular lattices, with a generalized external magnetic field that favors or disfavors a subset of spin values $\{1,\ldots ,s\}$ with $s \le q$. For the case of antiferromagnet spin–spin coupling, these provide exactly solved models that exhibit an onset of frustration and competing interactions in the context of a novel type of tensor-product $S_s \otimes S_{q-s}$ global symmetry, where $S_s$ is the permutation group on s objects.

Appendices

Appendix 1: $Z(sq,2 \times m,cyc.,q,v)$

We review here the result for the partition function $Z(sq,2 \times m,cyc.,q,v)$ of the cyclic square-lattice ladder graph of length $L_x=m$ vertices presented in [1]. We include this in connection with our discussion in the text showing how our new result for $Z(sq,2 \times m,cyc.,q,s,v,w)$ reduces to $Z(sq,2 \times m,cyc.,q,v)$ in the zero-field case $w=1$. This partition function has the form of

$$\begin{aligned} Z(sq,L_y \times m,cyc.,q,v) = \sum _{d=0}^{L_y} c^{(d)} \sum _{j=1}^{n_Z(L_y,d)} (\lambda _{sq,L_y,d,j})^m , \end{aligned}$$

(8.1)

with $L_y=2$, where $n_Z(2,0)=2$, $n_Z(2,1)=3$, $n_Z(2,2)=1$, and the coefficients $c^{(d)}$ are given in Eq. (3.2), so

$$\begin{aligned} c^{(0)}=1, \quad c^{(1)}=q-1, \quad c^{(2)}=q^2-3q+1. \end{aligned}$$

(8.2)

As is evident in Eq. (8.1), to distinguish the $\lambda _{sq,2,d,j}$s in the zero-field partition function $Z(sq,2 \times m,q,v)$ from the $\lambda _{Z,sq,2,d,j}$ in the field-dependent partition function $Z(sq,2\times m,q,s,v,w)$, we suppress the subscript Z in the former. Explicitly,

$$\begin{aligned} Z(sq,2 \times m,cyc.,q,v)= & {} (\lambda _{sq,2,0,+})^m+(\lambda _{sq,2,0,-})^m \nonumber \\&+ \, c^{(1)}\Big [ (\lambda _{sq,2,1,1})^m + (\lambda _{sq,2,1,+})^m + (\lambda _{sq,2,1,-})^m \Big ]\nonumber \\&+\, c^{(2)} (\lambda _{sq,2,2})^m , \end{aligned}$$

(8.3)

where (in order of decreasing d) $\lambda _{sq,2,2}=v^2$,

$$\begin{aligned} \lambda _{sq,2,1,1}=v(q+v) , \end{aligned}$$

(8.4)

$$\begin{aligned} \lambda _{sq,2,1,(2,3)}\equiv & {} \lambda _{sq,2,1,\pm }\nonumber \\= & {} \frac{v}{2}\Big [ q+v(v+4) \pm (v^4+4v^3+12v^2-2qv^2+4qv+q^2)^{1/2} \Big ] ,\quad \quad \end{aligned}$$

(8.5)

and

$$\begin{aligned} \lambda _{sq,2,0,(1,2)} \equiv \lambda _{sq,2,0,\pm } = \frac{1}{2}(A_{sqd0} \pm \sqrt{R_{sqd0}} \ ) , \end{aligned}$$

(8.6)

where

$$\begin{aligned} A_{sqd0}=v^3+4v^2+3qv+q^2 \end{aligned}$$

(8.7)

and

$$\begin{aligned} R_{sqd0}=v^6+4v^5-2qv^4-2q^2v^3+12v^4+16qv^3+13q^2v^2+6q^3v+q^4. \end{aligned}$$

(8.8)

The reader is referred to [2] for our corresponding solution for the partition function $Z(tri,2 \times m,cyc.,q,v)$ of the cyclic triangular-lattice ladder graph of arbitrary length.

Appendix 2: $T_{Z,sq,2,0}$

Five of the $s^2+2s+2$ $\lambda _{Z,sq,2,0,j}$ terms, each with multiplicity 1, are determined as the roots of a quintic equation which is the characteristic polynomial of the transfer matrix $T_{Z,sq,L_y,d}$ with $L_y=2$, $d=0$, and the following entries:

$$\begin{aligned} (T_{Z,sq,2,0})_{1,1}= & {} {\tilde{q}}^2 + 3v({\tilde{q}} + v)\end{aligned}$$

(9.1)

$$\begin{aligned} (T_{Z,sq,2,0})_{1,2}= & {} 2s({\tilde{q}} + v)w\end{aligned}$$

(9.2)

$$\begin{aligned} (T_{Z,sq,2,0})_{1,3}= & {} s(v+1)w^2\end{aligned}$$

(9.3)

$$\begin{aligned} (T_{Z,sq,2,0})_{1,4}= & {} s(s-1)w^2\end{aligned}$$

(9.4)

$$\begin{aligned} (T_{Z,sq,2,0})_{1,5}= & {} ({\tilde{q}} + 2v)(v+1)\end{aligned}$$

(9.5)

$$\begin{aligned} (T_{Z,sq,2,0})_{2,1}= & {} {\tilde{q}}^{\ 2} + v(2{\tilde{q}} + v) \end{aligned}$$

(9.6)

$$\begin{aligned} (T_{Z,sq,2,0})_{2,2}= & {} w[2s{\tilde{q}} + v(q+v)] \end{aligned}$$

(9.7)

$$\begin{aligned} (T_{Z,sq,2,0})_{2,3}= & {} (v+1)(v+s)w^2\end{aligned}$$

(9.8)

$$\begin{aligned} (T_{Z,sq,2,0})_{2,4}= & {} (v+s)(s-1)w^2\end{aligned}$$

(9.9)

$$\begin{aligned} (T_{Z,sq,2,0})_{2,5}= & {} ({\tilde{q}} + v)(v+1)\end{aligned}$$

(9.10)

$$\begin{aligned} (T_{Z,sq,2,0})_{3,1}= & {} (T_{Z,sq,2,0})_{4,1} = {\tilde{q}} ( {\tilde{q}} + v)\end{aligned}$$

(9.11)

$$\begin{aligned} (T_{Z,sq,2,0})_{3,2}= & {} (T_{Z,sq,2,0})_{4,2} = 2{\tilde{q}} (v+s)w\end{aligned}$$

(9.12)

$$\begin{aligned} (T_{Z,sq,2,0})_{3,3}= & {} (v+1)(v^2+2v+s)w^2\end{aligned}$$

(9.13)

$$\begin{aligned} (T_{Z,sq,2,0})_{3,4}= & {} (2v+s)(s-1)w^2\end{aligned}$$

(9.14)

$$\begin{aligned} (T_{Z,sq,2,0})_{3,5}= & {} (T_{Z,sq,2,0})_{4,5} = {\tilde{q}} (v+1)\end{aligned}$$

(9.15)

$$\begin{aligned} (T_{Z,sq,2,0})_{4,3}= & {} (v+1)(2v+s)w^2\end{aligned}$$

(9.16)

$$\begin{aligned} (T_{Z,sq,2,0})_{4,4}= & {} [(s+v)^2-s-2v]w^2\end{aligned}$$

(9.17)

$$\begin{aligned} (T_{Z,sq,2,0})_{5,1}= & {} v^3\end{aligned}$$

(9.18)

$$\begin{aligned} (T_{Z,sq,2,0})_{5,j}= & {} 0 \quad \mathrm{for} \ 2 \le j \le 4 \end{aligned}$$

(9.19)

$$\begin{aligned} (T_{Z,sq,2,0})_{5,5}= & {} v^2(v+1) \end{aligned}$$

(9.20)

Appendix 3: Matrices $T_{Z,tri,2,d}$ for $d=1$, $d=0$

1.1 $d=1$

Five of the $s^2+2s+2$ $\lambda _{Z,tri,2,1,j}$ terms, each with multiplicity 1, are determined as the roots of a quintic equation which is the characteristic polynomial of the matrix $T_{Z,tri,L_y,d}$ with $L_y=2$, $d=1$, and the following entries:

$$\begin{aligned} (T_{Z,tri,2,1})_{1,1}= & {} v {\tilde{q}} + 2v^2\end{aligned}$$

(10.1)

$$\begin{aligned} (T_{Z,tri,2,1})_{1,2}= & {} swv\end{aligned}$$

(10.2)

$$\begin{aligned} (T_{Z,tri,2,1})_{1,3}= & {} (T_{Z,tri,2,1})_{2,3} = v^2\end{aligned}$$

(10.3)

$$\begin{aligned} (T_{Z,tri,2,1})_{1,4}= & {} (T_{Z,tri,2,1})_{2,4} = 0\end{aligned}$$

(10.4)

$$\begin{aligned} (T_{Z,tri,2,1})_{1,5}= & {} (T_{Z,tri,2,1})_{2,5} = v(1+v)\end{aligned}$$

(10.5)

$$\begin{aligned} (T_{Z,tri,2,1})_{2,1}= & {} v {\tilde{q}} + v^2\end{aligned}$$

(10.6)

$$\begin{aligned} (T_{Z,tri,2,1})_{2,2}= & {} wv(v+s)\end{aligned}$$

(10.7)

$$\begin{aligned} (T_{Z,tri,2,1})_{3,1}= & {} (T_{Z,tri,2,1})_{4,1} = v {\tilde{q}} + 3v^2 + v^3\end{aligned}$$

(10.8)

$$\begin{aligned} (T_{Z,tri,2,1})_{3,2}= & {} (T_{Z,tri,2,1})_{4,2} = swv\end{aligned}$$

(10.9)

$$\begin{aligned} (T_{Z,tri,2,1})_{3,3}= & {} v {\tilde{q}} + 4v^2 + v^3\end{aligned}$$

(10.10)

$$\begin{aligned} (T_{Z,tri,2,1})_{3,4}= & {} swv\end{aligned}$$

(10.11)

$$\begin{aligned} (T_{Z,tri,2,1})_{3,5}= & {} (T_{Z,tri,2,1})_{4,5} = v(1+v)(2+v)\end{aligned}$$

(10.12)

$$\begin{aligned} (T_{Z,tri,2,1})_{4,3}= & {} v {\tilde{q}} + 3v^2 + v^3\end{aligned}$$

(10.13)

$$\begin{aligned} (T_{Z,tri,2,1})_{4,4}= & {} wv(v+s)\end{aligned}$$

(10.14)

$$\begin{aligned} (T_{Z,tri,2,1})_{5,1}= & {} v^2 {\tilde{q}} + 3v^3 + v^4\end{aligned}$$

(10.15)

$$\begin{aligned} (T_{Z,tri,2,1})_{5,2}= & {} swv^2\end{aligned}$$

(10.16)

$$\begin{aligned} (T_{Z,tri,2,1})_{5,3}= & {} 3v^3 + v^4\end{aligned}$$

(10.17)

$$\begin{aligned} (T_{Z,tri,2,1})_{5,4}= & {} 0\end{aligned}$$

(10.18)

$$\begin{aligned} (T_{Z,tri,2,1})_{5,5}= & {} v^2(1+v)(2+v) \end{aligned}$$

(10.19)

1.2 $d=0$

Five of the $\lambda _{Z,tri,L_y,d,j}$ with $L_y=2$ and $d=0$ are the roots, each with multiplicity $s-1$, of a quintic equation which is the characteristic polynomial of the transfer matrix $T_{Z,tri,2,0a}$. This matrix may be obtained from $T_{Z,tri,2,1}$ by the replacements $s \rightarrow q-s$ and $w \rightarrow w^{-1}$ and then multiplication by $w^2$.

Six of the $\lambda _{Z,tri,L_y,d,j}$ with $L_y=2$ and $d=0$ are the roots, each with multiplicity 1, of a degree-6 equation which is the characteristic polynomial of the matrix $T_{Z,tri,2,0b}$ with entries

$$\begin{aligned} (T_{Z,tri,2,0b})_{1,1}= & {} {\tilde{q}}^2 + 4v{\tilde{q}} + 5v^2 + v^3\end{aligned}$$

(10.20)

$$\begin{aligned} (T_{Z,tri,2,0b})_{1,2}= & {} sw({\tilde{q}} + v)\end{aligned}$$

(10.21)

$$\begin{aligned} (T_{Z,tri,2,0b})_{1,3}= & {} sw({\tilde{q}} + 2v)\end{aligned}$$

(10.22)

$$\begin{aligned} (T_{Z,tri,2,0b})_{1,4}= & {} sw^2(1+v)\end{aligned}$$

(10.23)

$$\begin{aligned} (T_{Z,tri,2,0b})_{1,5}= & {} s(s-1)w^2\end{aligned}$$

(10.24)

$$\begin{aligned} (T_{Z,tri,2,0b})_{1,6}= & {} (1+v)({\tilde{q}} + 3v + v^2)\end{aligned}$$

(10.25)

$$\begin{aligned} (T_{Z,tri,2,0b})_{2,1}= & {} {\tilde{q}}^2 + 3v{\tilde{q}} + 3v^2 + v^3\end{aligned}$$

(10.26)

$$\begin{aligned} (T_{Z,tri,2,0b})_{2,2}= & {} w(v+s)({\tilde{q}} + v)\end{aligned}$$

(10.27)

$$\begin{aligned} (T_{Z,tri,2,0b})_{2,3}= & {} sw({\tilde{q}} + v)\end{aligned}$$

(10.28)

$$\begin{aligned} (T_{Z,tri,2,0b})_{2,4}= & {} w^2(1+v)(v+s)\end{aligned}$$

(10.29)

$$\begin{aligned} (T_{Z,tri,2,0b})_{2,5}= & {} w^2(s-1)(v+s)\end{aligned}$$

(10.30)

$$\begin{aligned} (T_{Z,tri,2,0b})_{2,6}= & {} (1+v)({\tilde{q}} + 2v + v^2)\end{aligned}$$

(10.31)

$$\begin{aligned} (T_{Z,tri,2,0b})_{3,1}= & {} {\tilde{q}}^2 + 2v{\tilde{q}} + v^2 \end{aligned}$$

(10.32)

$$\begin{aligned} (T_{Z,tri,2,0b})_{3,2}= & {} w{\tilde{q}}(v + s)\end{aligned}$$

(10.33)

$$\begin{aligned} (T_{Z,tri,2,0b})_{3,3}= & {} w(v+s)({\tilde{q}} + v)\end{aligned}$$

(10.34)

$$\begin{aligned} (T_{Z,tri,2,0b})_{3,4}= & {} w^2(1+v)(s+2v+v^2)\end{aligned}$$

(10.35)

$$\begin{aligned} (T_{Z,tri,2,0b})_{3,5}= & {} w^2(2vs-2v-s+s^2)\end{aligned}$$

(10.36)

$$\begin{aligned} (T_{Z,tri,2,0b})_{3,6}= & {} (1+v)({\tilde{q}} + v)\end{aligned}$$

(10.37)

$$\begin{aligned} (T_{Z,tri,2,0b})_{4,1}= & {} (T_{Z,tri,2,0b})_{5,1} = {\tilde{q}} ({\tilde{q}} + v)\end{aligned}$$

(10.38)

$$\begin{aligned} (T_{Z,tri,2,0b})_{4,2}= & {} w{\tilde{q}} (s + 2v + v^2)\end{aligned}$$

(10.39)

$$\begin{aligned} (T_{Z,tri,2,0b})_{4,3}= & {} (T_{Z,tri,2,0b})_{5,3} = w{\tilde{q}} (s+v)\end{aligned}$$

(10.40)

$$\begin{aligned} (T_{Z,tri,2,0b})_{4,4}= & {} w^2(1+v)(s + 3v + 3v^2 + v^3)\end{aligned}$$

(10.41)

$$\begin{aligned} (T_{Z,tri,2,0b})_{4,5}= & {} w^2(s-1)(s+3v+v^2)\end{aligned}$$

(10.42)

$$\begin{aligned} (T_{Z,tri,2,0b})_{4,6}= & {} (T_{Z,tri,2,0b})_{5,6} = {\tilde{q}} (1+v)\end{aligned}$$

(10.43)

$$\begin{aligned} (T_{Z,tri,2,0b})_{5,2}= & {} w{\tilde{q}} (s+2v)\end{aligned}$$

(10.44)

$$\begin{aligned} (T_{Z,tri,2,0b})_{5,4}= & {} w^2(1+v)(s+3v+v^2)\end{aligned}$$

(10.45)

$$\begin{aligned} (T_{Z,tri,2,0b})_{5,5}= & {} w^2(s^2-s+3vs-3v+v^2)\end{aligned}$$

(10.46)

$$\begin{aligned} (T_{Z,tri,2,0b})_{6,1}= & {} v^2 ({\tilde{q}} + 3v + v^2)\end{aligned}$$

(10.47)

$$\begin{aligned} (T_{Z,tri,2,0b})_{6,2}= & {} (T_{Z,tri,2,0b})_{6,4} = (T_{Z,tri,2,0b})_{6,5} = 0\end{aligned}$$

(10.48)

$$\begin{aligned} (T_{Z,tri,2,0b})_{6,3}= & {} swv^2\end{aligned}$$

(10.49)

$$\begin{aligned} (T_{Z,tri,2,0b})_{6,6}= & {} v^2(1+v)(2+v) \end{aligned}$$

(10.50)