A Mathematical Journal
Vol. 7, No 3, (39 - 48). December 2005.

The exact solution of the Potts models

with external magnetic field on the Cayley
tree

Nasir Ganikhodjaev 1

Centre for Computational and Theoretical Sciences,Faculty of Science,
International Islamic University Malaysia,53100 Kuala Lumpur,Malaysia and

Department of Mechanics and Mathematics, National University of
Uzbekistan,Vuzgorodok

700095,Tashkent ,Uzbekistan
nasirgani@hotmail.com

Seyit Temir
Harran University, Department of Mathematics Sanliurfa-Turkey

temirseyit@harran.edu.tr

Hasan Akin
Harran University, Department of Mathematics Sanliurfa-Turkey

akinhasan@harran.edu.tr

ABSTRACT
The exact solution is found for the problem of phase transition in the Potts

model and the Potts model with competing ternary and binary interactions with
external magnetic field.

1This research was supported in part by the grant Uz.R.FTM F-2.1.56. The first named author
(N.G.) thanks NATO-TUBITAK for providing financial support and Harran University for kind
hospitality and providing all facilities.



40 Nasir Ganikhodjaev, Seyit Temir and Hasan Akin
7, 3(2005)

RESUMEN

Se encuentran soluciones exactas para los problemas de transisión de fases en
el modelo de Pott y también para el modelo de Pott con interacciones binarias y
ternarias en un campo magnético externo.

Key words and phrases: Cayley tree, Potts model, Competing interactions,
External magnetic field.

Math. Subj. Class.: 82B20 Secondary 82B26

1 Introduction

The Potts model was introduced as a generalization of the Ising model. The idea
came from the representation of the Ising model as interacting spins which can be
either parallel or antiparallel. An obvious generalization was to extend the number
of directions of the spins. Such a model was proposed by C.Domb as a PhD thesis for
his student R.Potts in 1952. At present the Potts model encompasses a number of
problems in statistical physics and lattice theory. It has been a subject of incresing
intense research interest in recent years. It includes the ice-rule vertex and bond
percolation models as special cases.

We consider a semi-infinite Cayley tree J k for order k ≥ 2, i.e., a graph having no
cycles, from each vertex of which, except on vertex x0, emanates exactly k + 1 edges
and from vertex x0, which is the root of the tree, emanates k edges.

The vertices x and y are called nearest neighbors, which is denoted by < x, y >,
if there exists an edge connecting them. The vertices x,y and z are called a triple
of neighbors ,which is denoted by < x, y, z >,if < x, y > and < y, z > are nearest
neighbors and x 6= z

Let V be the set of vertices in J k. We set

Wn = {x ∈ V |d(x, x0) = n},

Vn = ∪nm=0Wm = {x ∈ V |d(x, x
0) ≤ n}.

where the distance d(x, y), x, y, ∈ V is given by the formula,

d(x, y) = min{d|x = x0, x1, x2, ..., xd−1, xd = y ∈ V

such that the pairs < x0, x1 >, ..., < xd−1, xd > are nearest neighbors.
The set Wn is called n-th level of J k and the set Vn is called n-storeyed home with
root x0.

We consider models where the spin takes values in the set Φ = {0, 1, 2, ..., q}, q ≥ 2
and assigned to the vertices of the tree.A configuration σ on V is then defined as a



7, 3(2005)
The exact solution of the Potts models ... 41

function x ∈ V → σ(x) ∈ Φ;the set of all configurations coincides with Ω = ΦV . The
Potts model on the Cayley tree is defined by the Hamiltonian

H(σ) = −J
∑

<x,y>

δσ(x)σ(y) − h
∑
x∈V

δ0σ(x) (1)

where the first sum is taken over all nearest neighbors, δ in the first and second sums
is the Kroneker’s symbol,J, h ∈ R are coupling constants and σ ∈ Ω.

Along with this model, we will consider the Potts model with competing interac-
tions on the Cayley tree which is defined by the Hamiltonian below

H(σ) = −J1
∑

<x,y,z>

δσ(x)σ(y)σ(z) − J2
∑

<x,y>

δσ(x)σ(y) − h
∑
x∈V

δ0σ(x) (2)

where the first sum is taken over all neighbors tripples, and δ in this sum is the
generalized Kroneker’s symbol (see [1]-[4] for models with competing interactions).
Such model was investigated in [3], where for the neighbors tripple < x, y, z >the
generalized Kroneker’s symbol δ had a form

δσ(x)σ(y)σ(z) =
{

1 if σ(x) = σ(y) = σ(z),
0 else.

For the neighbors tripple < x, y, z >, we assume

δσ(x)σ(y)σ(z) =




1 if σ(x) = σ(y) = σ(z),
1
2

if σ(x) = σ(y) 6= σ(z) or σ(x) 6= σ(y) = σ(z);
0 else.

(3)

where x, z ∈ Wn for some n and y ∈ Wn−1. This definition is well coordinated with
the theory of quadratic stochastic operators, where the quadratic stochastic operator
corresponding to the generalized Kroneker’s symbol (4) is the identity transformation
[5].

Let

Sm−1 = {x = (x1, · · · , xm) ∈ Rm :
m∑

i=1

xi = 1} xi ≥ 0 ∀i = 1, · · · , m}

be the (m − 1)-dimensional simplex in Rm. The transformation V : Sm−1 → Sm−1 is
called quadratic stochastic operator , if

(V x)k =
m∑

i,j=1

pij,kxixj

where pij,k ≥ 0 , pij,k = pji,k and
∑m

k=1 pij,k = 1 for arbitrary i, j, k ∈ {1, · · · , m} .
Such operator have applications in mathematical biology, namely theory of heredity,



42 Nasir Ganikhodjaev, Seyit Temir and Hasan Akin
7, 3(2005)

where the coefficients pij,k are interpreted as coefficients of heredity. Assume pij,k =
δijk,where the generalized Kroneker’s symbol δ has a form

δijk =




1 if i = j = k,
1
2

if i = k 6= j or i 6= j = k;
0 else.

(4)

Then it is easy to show that the corresponding quadratic stochastic operator is the
identity transformation .

2 Recurrent Equations for partition function

There are several approaches to derive the equation or a system of equations describing
the limiting Gibbs measures for lattice models on a Cayley tree. One approach is
based on the properties of the Markov random fields on a Cayley tree [6, 7]. Another
approach is based on recurrent equations for partition functions(see for example [8]).
Naturally both approaches lead to the same equation(for example [9]). The second
approach, however, is more suitable for models with competing interactions.

Let Λ be a finite subset of V. Assume σ(Λ) and σ(V \ Λ) are the restriction of σ
to Λ and V \ Λ respectively. Let σ(V \ Λ) be a fixed boundary configuration. The
total energy of configuration σ(Λ) under condition σ(V \ Λ) is defined as

H(σ(Λ)|σ(V \ Λ)) = −J
∑

< x, y >
x, y ∈ Λ

δσ(x)σ(y) − J
∑

< x, y >
x ∈ Λ, y /∈ Λ

δσ(x)σ(y) − h
∑
x∈Λ

δ0σ(x).

in the first case and

H(σ(Λ)|σ(V \ Λ)) = −J1
∑

< x, y, z >
x, y, z ∈ Λ

δσ(x)σ(y)σ(z) − J2
∑

< x, y >
x, y ∈ Λ

δσ(x)σ(y) − h
∑
x∈Λ

δ0σ(x)

−J1
∑

< x, y, z >
x /∈ Λ, y ∈ Λ, z /∈ Λ

δσ(x)σ(y)σ(z) − J2
∑

< x, y >
x ∈ Λ, y /∈ Λ

δσ(x)σ(y) .

for the second Hamiltonian respectively.
The partition function ZΛ(σ(V \ Λ)) in volume Λ under boundary condition

σ(V \ Λ) is defined as

ZΛ =
∑

σ(Λ)∈Ω(Λ)

exp(−βH(σ(Λ))|σ(V \ Λ)), (5)

where Ω(Λ) is the set of all configuration on Λ, and β =
1
T

is the inverse temperature.

We consider the configurations σ(Vn), the partition functions ZVn in the volume Vn
and for brevity we denote it as σn, Z(n) respectively.



7, 3(2005)
The exact solution of the Potts models ... 43

Let us first consider the model (1).We decompose the partition function Z(n) into
the following summands

Z(n) =
q∑

i=1

Z
(n)
i ,

where
Z

(n)
i =

∑
σn∈Ω(Vn):σn(x0)=i

exp(−βHn(σn)). (6)

Let θ = exp (βJ), θ3 = exp (βh) . From (5) and (6), the following system of
recurrent equations can be easily derived

Z
(n)
0 = θ3

[
θZ

(n−1)
0 + Z

(n−1)
1 + Z

(n−1)
2 + ... + Z

(n−1)
q

]k
Z

(n)
i =

[
Z

(n−1)
0 + ... + Z

(n−1)
i−1 + θZ

(n−1)
i + Z

(n−1)
i+1 ...Z

(n−1)
q

]k (7)
for i=1,2,...,q , where Z(n−1)i is a partition function in (n − 1)-storeyed home with
root located a vertex x ∈ W1 for which σ(x) = i.

After replacing u(n)i =
Z

(n)
i

Z
(n)
0

, we have the following system of recurrent equations

u
(n)
i =

1
θ3

(
1 + (θ − 1)u(n−1)i +

q∑
j=1

u
(n−1)
j

θ +
q∑

j=1

u
(n−1)
j

)k; (8)

for i=1,2,...,q and n=2,3,...
We describe the fixed points of this system recurrent equation (5). For this, it

suffices to solve the system of equations

ui =
1
θ3

(
1 + (θ − 1)ui +

q∑
j=1

uj

θ +
q∑

j=1

uj

)k; i = 1, 2, ..., q. (9)

Before we begin to solve this system of equations, we turn to the model (2). Here we
consider a slight modification of the Hamiltonian (2)

Definition 1 A triple of neighbours < x, y, z > is said to be two-level and is denoted
by ¯< x, y, z > if the vertices x and z belong to Wn for some n, i.e. they are located
on the same level and y ∈ Wn−1.

We consider the Hamiltonian

H(σ) = −J1
∑
¯<x,y,z>

δσ(x)σ(y)σ(z) − J2
∑

<x,y>

δσ(x)σ(y) − h
∑
x∈V

δ0σ(x) (10)



44 Nasir Ganikhodjaev, Seyit Temir and Hasan Akin
7, 3(2005)

where J1 6= 0 and in contrast to (2), the first sum inlcudes only the two-level triples
of neighbours. Such a model is called a two-level model (see [3], [4] and the references
there for the physical motivation underlying the study of these model).

It is not hard to derive, in this case, the system of recurrent equations is as the
following

Z
(n)
0 = θ3

[
θ1θ2Z

(n−1)
0 + Z

(n−1)
1 + Z

(n−1)
2 + ... + Z

(n−1)
q

]k
Z

(n)
i =

[
Z

(n−1)
0 + ... + Z

(n−1)
i−1 + θ1θ2Z

(n−1)
i + Z

(n−1)
i+1 ...Z

(n−1)
q

]k
for i = 1, 2, ..., q

where θ1 = exp(βJ1), θ2 = exp(βJ2) and θ3 = exp(βh).
Thus both models (1) and (10) are described by the same system of recurrent

equations.

3 The proof of existence of phase transitions for
zero external field

In this section, we let J > 0 for model (1), that is we consider model (1) as a
ferromagnetic Potts model and J1 + J2 > 0 for model (10). Then θ > 1 in the first
case and θ1θ2 > 1 for second case. We consider the system of equations (9).Assume
ui = exp hi, i = 1, 2, ..., q Then

h′i = k ln
1
θ3

(
1 + (θ − 1)hj +

q∑
j=1

exp(hj )

θ +
q∑

j=1

exp hj
); i = 1, 2, ..., q (11)

is the transformation Rq into Rq. Evidently the line l0 : h1 = h2 = ... = hq in Rq is
invarinat with respect to transformation (11) and the restriction of (11) on the line
l0 has the following form

h′ = k ln
1
θ3

(
(θ + q − 1) exp(h) + 1

q exp(h) + θ
)

where h ∈ R. Again, after renaming u = exp(h), we have

u =
1
θ3

(
(θ + q − 1)u + 1

qu + θ
)k.

The following Lemma is a generalization of the Proposition 10.7 from [8].

Lemma 1 The equation

θ3u = (
(θ + q − 1)u + 1

qu + θ
)k (12)



7, 3(2005)
The exact solution of the Potts models ... 45

(with u > 0, k ≥ 2, q ≥ 2) has a single solution if

1 < θ < θcr =
−(k − 1)(q − 1) +

√
(k − 1)2(q + 1)2 + 8q(k − 1)
2(k + 1)

If θ > θcr then there are numbers η1(θ, q, k), η2(θ, q, k) with 0 < η1(θ, q, k) <
η2(θ, q, k) such that equation (12) has three roots, when 0 < η1(θ, q, k) < θ3 <
η2(θ, q, k) and it has two roots if either θ3 = η1(θ, q, k) or θ3 = η2(θ, q, k) or θ = θcr.
The numbers ηi, i = 1, 2 are defined from the formula

ηi(θ, q, k) =
1
ui

(
(θ + q − 1)ui + 1

qui + θ
)k (13)

where u1 and u2 are the solution of the equation

(θ + q − 1)qu2 − [k(θ − 1)(θ + q) − θ(θ + q − 1) − q]u + θ = 0 (14)

Proof. Assume f (u) = ( (θ+q−1)u+1
qu+θ

)k. It is easy to check that equation (12) has more
than one root if and only if the equation uf′ = f (u) has more than one solution. The
equation uf′ = f (u) is no other than just equation (14).

Although there are three solutions for the system of equations (9) for θ > θcr,
one cannot claim that there is a phase transition. Among these solutions, only one of
them is a stable solution. It is necessary to find other stable solutions. This problem
is rather complete for arbitrary k and q when θ 6= 1. The case with θ 6= 1 will be
considerd separately when k = 2 and q = 2. We shall now solve this problem for
θ3 = 1, that is, h = 0. Then, the system of equation (9) has the following form

ui =
( 1 + (θ − 1)ui +

q∑
j=1

uj

θ +
q∑

j=1

uj

)k
i = 1, 2, ..., q (15)

and the transformation Rq into Rq (11) has the following form

h′i = k ln(
1 + (θ − 1) exp hj +

q∑
j=1

exp(hj )

θ +
q∑

j=1

exp hj
); i = 1, 2, ..., q. (16)

Then, apart from the invariant line l0 we can find other q invariant lines, namely the
line lj : h1 = ... = hj−1 = hj+1 = ... = hq = 0, j = 1, 2, ..., q. The transformation (11)
reduces to the following transformation of R:

h′ = k ln(
θ exp h + q

exp h + θ + q − 1
)

on each invariant line lj , j = 1, 2, ..., q.



46 Nasir Ganikhodjaev, Seyit Temir and Hasan Akin
7, 3(2005)

Now we will solve this simpler equation

u = (
θu + q

u + θ + q − 1
)k (17)

Let us consider the function φ(u) = (
θu + q

u + θ + q − 1
)k. With the help of the

Lemma, it is not hard to show that the equation (17) has three solutions when

θ > θ∗cr =
k + 2q − 1

k − 1
. In this case only one of these roots is stable, namely, largest of

them. For equation (12), when θ3 = 1, we showed above that it has three solutions
when θ > θ∗cr (see Lemma) and only one of them is stable. It is easy to check that
θ∗cr > θcr .

As uj =
P (x0=j)
P (x0=0)

for some limiting Gibbs measure P with θ > θ∗cr =
k+2q−1

k−1 , we
have q + 1 differences translated invariant limiting Gibbs measures. The same way as
in [9], it is possible to prove that all of them are extremal.

Theorem 1 For Potts model (1) with null external field, a phase transition occurs
when,

θ >
k + 2q − 1

k − 1
.

Similar assertation is also valid for the two-level Potts model with competing ternary
and binary potentials with null external field.

Theorem 2 For the two-level Potts model (10) with competing ternaty and binary
potentials with null external magnetic field, a phase transition occurs when θ1θ2 >
k+2q−1

k−1 .

4 The case of non-zero external magnetic field when
k = q = 2

Here we consider Potts models both (1) and (10) with external magnetic field h 6= 0,
when k = q = 2 and θ > 1 for model (1) and θ1θ2 > 1 for model (10) respectively
(The case h = 0 was considered in [9]for model (1) and for model (10) in [10] ). Then
the system of equations (9) reduces to the following

x =
1
θ3

(
θx + y + 1
x + y + θ

)2

y =
1
θ3

(
x + θy + 1
x + y + θ

)2
(18)

where x = u1, y = u2 for brevity. As

x − y =
1
θ3

(θ1)(x − y)[(θ + 1)(x + y) + 1]
(x + y + θ2)

,



7, 3(2005)
The exact solution of the Potts models ... 47

then some solutions of (18) can be found from equation

u =
1
θ3

(
(θ + 1)u + 1

2u + θ
)2 (19)

where x = y = u and other solutions can be found from equation

θ3z
2 − (θ2 − 2θ3θ − 1)z + θ3θ2 − 2θ + 2 = 0 (20)

where z = x + y. First of all, let us consider equation (19).Then the equation (13)(see
Lemma)has the following form

2(θ + 1)u2 − (θ2 + θ − 6)u + θ = 0. (21)

This equation has two roots u1, u2 if θ >
√

73−1
2

. Then by Lemma, equation (19)have
three roots if θ >

√
73−1
2

and η1(θ) < θ3 < η2(θ), where ηi = 1ui (
(θ+1)ui+2

3ui+θ
)2, i = 1, 2.

Now we consider the equation (20). Again with the help of elementary analysis
it is not hard to show that the equation has two solutions for θ3 > 12 with θ >
2θ3 − 1 + 2

√
θ3(θ3 + 1) and for 0 < θ3 < 12 , with θ >

1+
√

1−2θ3
θ3

. By virtue of
symmetry of equations (18) we have two stable solutions.

Assume A = {(θ3, θ) : η1(θ) < θ3 < η2(θ) ; θ >
√

73 − 1
2

} where η1(θ) and η2(θ)

as above and B = {(θ3, θ): 0 < θ3 < 12 ; θ >
1+

√
1−2θ3
θ3

} ∪ {(θ3, θ): θ3 > 12 ;
θ > 2θ3 − 1 + 2

√
θ3(θ3 + 1) }. Then for arbitrary (θ3, θ) ∈ A ∩ B there are three

stable solutions of the equations (18). We have thus proved the following theorems.

Theorem 3 For Potts model (1) with q = k = 2 and non-zero external magnetic
field, a phase transition occurs when (θ3, θ) ∈ A ∩ B

A similar result is valid for model (10).

Theorem 4 For the two-level Potts model (10) with competing ternary and binary
potentials q = k = 2 and non-zero external magnetic field a phase transition occurs
when (θ3, θ1θ2) ∈ A ∩ B.

Received: July 2005. Revised: August 2005.

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