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An approximate method for the investigation of the Potts antiferromagnet in zero external field at zero temperature
Volume 123, number 4
PHYSICS LETTERS A
3 August 1987
AN APPROXIMATE METHOD FOR THE INVESTIGATION OF THE POUS ANTIFERROMAGNET IN ZERO EXTERNAL FIELD AT ZERO TEMPERATURE D. HAJDUKOVI~ Institute of Mathematicsand Physics, 81000 Titograd, Yugoslavia
and I.M. SAVI~ Department ofPhysics and Meteorology, Faculty of Naturaland Mathematical Sciences, P.O. Box 550, 11001 Belgrade, Yugoslavia Received 6 March 1987; revised manuscript received 26 May 1987; accepted for publication 27 May 1987 Communicated by A.A. Maradudin
We calculate the exact ground state degeneracy for the antiferromagnetic q-state Potts model in zero external field at zero temperature for two and three coupled chains. On the basis ofthese exact results the ground state degeneracy ofthe square lattice is expressed as an explicit function of q which is foundto be a good approximation.
The ground state of an antiferromagnetic q-state Potts model in zero external magnetic field (see for example ref. [11) is in general highly degenerate for q~3. However,~except for the one-dimensional lattice, the exact ground state degeneracy per site in the thermodynamic limit is known only in a restricted number of cases. These are the ground state degeneracies, in zero field, for the q = 3 square lattice [2], for the q= 3 Kagome lattice [3], and for general q in the case of a triangular lattice [4]. In the present paper, on the example of a square lattice, we shall demonstrate a simple method by which the ground state degeneracy may be approximated by an explicit function ofq. To this end, we are first going to obtain the exact ground state degeneracies for two and three
model we attach to every lattice site i a spin variable a, that takes values in the set { 1, 2..., q} and define the hamiltonian by
coupled chains as a function of q. After that, on the basis of this exact result we shall be able to obtain for the whole square lattice some approximate expressions for the ground state degeneracy as a function of q. To introduce the antiferromagnetic q-state Potts
of size 2 x I. Then, because of the fact that at zero temperature two neighbouring sites cannot be in the same state, for each of P1 configurations there are q 1 + (q ~2)2 possible states for two sites on the (1+ 1 )th vertical. Thus p1±1 [q 1 + (q ~2)2 jp,,
To whom all correspondence should be addressed.
0375-9601/87/$ 03.50 © Elsevier Science Publishers B.V. (North-Holland Physics Publishing Division)
~~‘=
~
a’), a <0.
(1)
The sum is taken over pairs ofnearest-neighbour sites and ôKr( a, fi) is the Kronecker delta. Let us restrict ourselves to the case ofa square lattice. An infinite square lattice with N=L x L (L—~oo) sites can be conceived as a set of L parallel chains. In the first step, instead of considering the whole lattice to be infinite in both dimensions let us consider two coupled chains with 2 x L sites (see fig. 1 a). Let us denote by P1 the ground state degeneracy for a strip
—
—
i.e. for a lattice with 2 x L sites 197
Volume 123, number 4
PHYSICS LETTERS A
__________________
~
I
I
By using eq. (3), eqs. (4) and (5) may be transformed into
_____
~
4
24q+5)p,+(q2)p~ba,
p~~a
___~
(6)
1=(q
(a)
_____
_________________
3 August 1987
p,b+ci=(qs.....6q2+l4q_.l3)pj+(3q)p~ba.
(7)
So,
=(q3—5q2+10q—8)P 1+P~. p1±1 p,~a1+~~1
(b) Fig. 1. Schematic representation of the two (a) and three (b) coupled chains.
2J~ P2XL
(2)
.
=q(q— 1 )[q— 1 + (q—2)
(8) From eqs. (6) and (8) after a few steps of a simple calculation we have 3—5q2+llq—10)P
P,~1=(q
1
—(q4—7q3+19q2—24q+ll)P 1_1
,
(9)
This result was obtained in a different way in a previous paper [5]. For three chains (fig. ib)state the degeneracy calculation P, is more complicated. The ground for a system with 3xL sites may be written as
or, in the—Sq2 limit+I—co, 2 —(q3 llq— l0)P p +(q47q3+19q2 —24q+l1)=0,
(10)
p 1 = p~bc+ p~ba
(3)
where
Here P~ is the number of configurations with three spins on the Ith vertical in three different states and p~bais the number ofconfigurations when two spins
~
on the Ith vertical are in the same state and the third one is in a different state. A straightforward but careful analysis shows that for every of p~ba configurations there q 1 + (q— 2)2with andthe (q (1+ 1 )(l)th q —2) + 2(q—are 3) configurations ver(q—2) tical in an “aba” and “abc” state, respectively. On the other hand for each of P~ configurations there are 2(q—2)+(q—3)2 and 2(q—2)2+(q—3)[q— 2+(q—3)2] configurations with the (I+1)th vertical in an “aba” and “abc” state, respectively. Thus
in the thermodynamic limit, for the lattice considered with 3 x I sites. For a lattice with 3 x L sites, up to a multiplicative factor which is not important, we have P3XL={P}’-’
—
p~a
—
21P~” 1=[q—1+(q—2)
+ [2 ( q— 2) + (q— 3)2]
(4)
p~bc
and P~
2(q—3)]P~ 1=[(q—l)(q—2)+(q—2)
+~2(q—2)2+(q—3)[q—2+(q—3)2]}P~.
198
ij~P,+1 IF,
(11)
is the exact ground state degeneracy per three sites
={~[q3—5q2+l1q—l0+(q6--l0q5+43q4
i o~c~3 + 1 45q2 1 24q+ 56)1/21 } L (12) Let us note that P~(q)~(P2~~)”21 and —
—
P~(q) (P3~L)1/3~.,which are determined by eqs. (2) and (12), are upper bounds for the ground state degeneracy per site of a square lattice. In the particular case q=3 these bounds are P~(3)=1.732...and PZ(3)l.658.... Comparison with the exact result p(3)= 1.5396... [2J for the square lattice shows that convergence is slow and that for a reliable estimate the ground state degeneracy for a larger number of chains should be found.
Volume 123, number 4
PHYSICS LETTERS A
Now, let us consider a square lattice. In principle we have to find P= ~ P,÷ 1IF1 where now P is ,
the ground state degeneracy per chain and P,=_ PLX1 F1~L~ However, such a calculation is extremely difficult. Instead of this limit, as an approximation we can calculate the ratio P1±1/P1 for small values of I. In fact, we have calculated F, for 1=2 (eq. (2)) and 1=3 (eq. (12)). On the other hand it is trivial to see that for 1=1 1
L FL~1~P1=q(q—l)
.
13Lxl
—
In the particular case q = 3 these approximations may be compared with the exact result of Baxter [3]. So, the exact ground state degeneracy per site forq= 3 is p = (4/3)3/2 = 1.53960..., whereas in our first and second approximation we have p’ =1.5 and p” = (5 +~.Ji7)/6 = 1.5205..., respectively. So, the second approximation is already in good agreement with the known exact result for the particular value q=3. In conclusion let us note that the same method without difficulties may be applied to other twodimensional lattices such as the triangular and the honeycomb ones. In addition if we want an accuracy betteraspects than 1%will we be need the third approximation. these treated in another paper. All .
So, in the first approximation the ground state degeneracy per chain (i.e. per L sites) is P =P2/P1 = PLX 21’ =(q— l){[q— 1 +(q—2)2]/(q— l)}~._I ,
3 August 1987
.
(13)
and in the second approximation (up to a nonimportant multiplicative factor)
References
F” = P3 IF2 —P = LX3 IF Lx2
[1] Mod. Phys. [2] F.Y. E.H. Wu, Lieb,Rev. Phys. Rev. Lett.5418(1982) (1967)235. 692; Phys. Rev. 162
={[q3
+ (q6
—Sq2 —
+
1 lq— 10
lOq5 +43q4
—
102q3 + l45q2
—l24q+56)~’2]I2[q—l+(q—2)2]}”
.
(14)
(1967) 162. [3] R.J. Baxter, J. Math. Phys. 11(1970) 784. [4] R.J. Baxter, J. Phys. A 19(1986)2821. [5] I.M. Savié and U. Maurer, Phys. Stat. Sol. (b) 129 (1985) K39.
199
PHYSICS LETTERS A
3 August 1987
AN APPROXIMATE METHOD FOR THE INVESTIGATION OF THE POUS ANTIFERROMAGNET IN ZERO EXTERNAL FIELD AT ZERO TEMPERATURE D. HAJDUKOVI~ Institute of Mathematicsand Physics, 81000 Titograd, Yugoslavia
and I.M. SAVI~ Department ofPhysics and Meteorology, Faculty of Naturaland Mathematical Sciences, P.O. Box 550, 11001 Belgrade, Yugoslavia Received 6 March 1987; revised manuscript received 26 May 1987; accepted for publication 27 May 1987 Communicated by A.A. Maradudin
We calculate the exact ground state degeneracy for the antiferromagnetic q-state Potts model in zero external field at zero temperature for two and three coupled chains. On the basis ofthese exact results the ground state degeneracy ofthe square lattice is expressed as an explicit function of q which is foundto be a good approximation.
The ground state of an antiferromagnetic q-state Potts model in zero external magnetic field (see for example ref. [11) is in general highly degenerate for q~3. However,~except for the one-dimensional lattice, the exact ground state degeneracy per site in the thermodynamic limit is known only in a restricted number of cases. These are the ground state degeneracies, in zero field, for the q = 3 square lattice [2], for the q= 3 Kagome lattice [3], and for general q in the case of a triangular lattice [4]. In the present paper, on the example of a square lattice, we shall demonstrate a simple method by which the ground state degeneracy may be approximated by an explicit function ofq. To this end, we are first going to obtain the exact ground state degeneracies for two and three
model we attach to every lattice site i a spin variable a, that takes values in the set { 1, 2..., q} and define the hamiltonian by
coupled chains as a function of q. After that, on the basis of this exact result we shall be able to obtain for the whole square lattice some approximate expressions for the ground state degeneracy as a function of q. To introduce the antiferromagnetic q-state Potts
of size 2 x I. Then, because of the fact that at zero temperature two neighbouring sites cannot be in the same state, for each of P1 configurations there are q 1 + (q ~2)2 possible states for two sites on the (1+ 1 )th vertical. Thus p1±1 [q 1 + (q ~2)2 jp,,
To whom all correspondence should be addressed.
0375-9601/87/$ 03.50 © Elsevier Science Publishers B.V. (North-Holland Physics Publishing Division)
~~‘=
~
a’), a <0.
(1)
The sum is taken over pairs ofnearest-neighbour sites
—
—
i.e. for a lattice with 2 x L sites 197
Volume 123, number 4
PHYSICS LETTERS A
__________________
~
I
I
By using eq. (3), eqs. (4) and (5) may be transformed into
_____
~
4
24q+5)p,+(q2)p~ba,
p~~a
___~
(6)
1=(q
(a)
_____
_________________
3 August 1987
p,b+ci=(qs.....6q2+l4q_.l3)pj+(3q)p~ba.
(7)
So,
=(q3—5q2+10q—8)P 1+P~. p1±1 p,~a1+~~1
(b) Fig. 1. Schematic representation of the two (a) and three (b) coupled chains.
2J~ P2XL
(2)
.
=q(q— 1 )[q— 1 + (q—2)
(8) From eqs. (6) and (8) after a few steps of a simple calculation we have 3—5q2+llq—10)P
P,~1=(q
1
—(q4—7q3+19q2—24q+ll)P 1_1
,
(9)
This result was obtained in a different way in a previous paper [5]. For three chains (fig. ib)state the degeneracy calculation P, is more complicated. The ground for a system with 3xL sites may be written as
or, in the—Sq2 limit+I—co, 2 —(q3 llq— l0)P p +(q47q3+19q2 —24q+l1)=0,
(10)
p 1 = p~bc+ p~ba
(3)
where
Here P~ is the number of configurations with three spins on the Ith vertical in three different states and p~bais the number ofconfigurations when two spins
~
on the Ith vertical are in the same state and the third one is in a different state. A straightforward but careful analysis shows that for every of p~ba configurations there q 1 + (q— 2)2with andthe (q (1+ 1 )(l)th q —2) + 2(q—are 3) configurations ver(q—2) tical in an “aba” and “abc” state, respectively. On the other hand for each of P~ configurations there are 2(q—2)+(q—3)2 and 2(q—2)2+(q—3)[q— 2+(q—3)2] configurations with the (I+1)th vertical in an “aba” and “abc” state, respectively. Thus
in the thermodynamic limit, for the lattice considered with 3 x I sites. For a lattice with 3 x L sites, up to a multiplicative factor which is not important, we have P3XL={P}’-’
—
p~a
—
21P~” 1=[q—1+(q—2)
+ [2 ( q— 2) + (q— 3)2]
(4)
p~bc
and P~
2(q—3)]P~ 1=[(q—l)(q—2)+(q—2)
+~2(q—2)2+(q—3)[q—2+(q—3)2]}P~.
198
ij~P,+1 IF,
(11)
is the exact ground state degeneracy per three sites
={~[q3—5q2+l1q—l0+(q6--l0q5+43q4
i o~c~3 + 1 45q2 1 24q+ 56)1/21 } L (12) Let us note that P~(q)~(P2~~)”21 and —
—
P~(q) (P3~L)1/3~.,which are determined by eqs. (2) and (12), are upper bounds for the ground state degeneracy per site of a square lattice. In the particular case q=3 these bounds are P~(3)=1.732...and PZ(3)l.658.... Comparison with the exact result p(3)= 1.5396... [2J for the square lattice shows that convergence is slow and that for a reliable estimate the ground state degeneracy for a larger number of chains should be found.
Volume 123, number 4
PHYSICS LETTERS A
Now, let us consider a square lattice. In principle we have to find P= ~ P,÷ 1IF1 where now P is ,
the ground state degeneracy per chain and P,=_ PLX1 F1~L~ However, such a calculation is extremely difficult. Instead of this limit, as an approximation we can calculate the ratio P1±1/P1 for small values of I. In fact, we have calculated F, for 1=2 (eq. (2)) and 1=3 (eq. (12)). On the other hand it is trivial to see that for 1=1 1
L FL~1~P1=q(q—l)
.
13Lxl
—
In the particular case q = 3 these approximations may be compared with the exact result of Baxter [3]. So, the exact ground state degeneracy per site forq= 3 is p = (4/3)3/2 = 1.53960..., whereas in our first and second approximation we have p’ =1.5 and p” = (5 +~.Ji7)/6 = 1.5205..., respectively. So, the second approximation is already in good agreement with the known exact result for the particular value q=3. In conclusion let us note that the same method without difficulties may be applied to other twodimensional lattices such as the triangular and the honeycomb ones. In addition if we want an accuracy betteraspects than 1%will we be need the third approximation. these treated in another paper. All .
So, in the first approximation the ground state degeneracy per chain (i.e. per L sites) is P =P2/P1 = PLX 21’ =(q— l){[q— 1 +(q—2)2]/(q— l)}~._I ,
3 August 1987
.
(13)
and in the second approximation (up to a nonimportant multiplicative factor)
References
F” = P3 IF2 —P = LX3 IF Lx2
[1] Mod. Phys. [2] F.Y. E.H. Wu, Lieb,Rev. Phys. Rev. Lett.5418(1982) (1967)235. 692; Phys. Rev. 162
={[q3
+ (q6
—Sq2 —
+
1 lq— 10
lOq5 +43q4
—
102q3 + l45q2
—l24q+56)~’2]I2[q—l+(q—2)2]}”
.
(14)
(1967) 162. [3] R.J. Baxter, J. Math. Phys. 11(1970) 784. [4] R.J. Baxter, J. Phys. A 19(1986)2821. [5] I.M. Savié and U. Maurer, Phys. Stat. Sol. (b) 129 (1985) K39.
199