PHYSICAL REVIEW B VOLUME 56, NUMBER 17 1 NOVEMBER 1997-I
Singlet ground state of the bilinear-biquadratic exchange Hamiltonian
A. Tanaka and T. Idogaki
Department of Applied Science, Faculty of Engineering, Kyushu University, Fukuoka 812-81, Japan
Received 3 March 1997
We show that the ground state of the bilinear-biquadratic exchange Hamiltonian is always singlet on a
connected finite lattice with an even number of sites, while it is always triplet on one with an odd number of
sites, if coefficients of the bilinear term J and that of the biquadratic term J satisfy the condition
J J 0. We also find that the total-spin eigenvalue Stot of the lowest-energy state in each subspace classified
by M, an eigenvalue of z component of the total spin, depends on the parity of M, i.e., Stot M for even M
and Stot M 1 for odd M on an even number site lattice and vice versa on an odd number site lattice.
S0163-1829 97 04941-2
In some magnetic materials it has been recognized that state in a subspace classified by an eigenvalue of z compo-
the higher-order couplings play an important role as well as nent of the total spin and to the case of an odd number of
usual bilinear exchange interactions, and many theoretical sites.
and experimental investigations about the effects of these For a while, we restrict ourselves to a finite bipartite lat-
couplings have been performed. For the case of S 1, an tice A B, where there is no bond connecting sites i
isotropic model with these couplings is expressed by the fol- A and j A or i B and j B. This restriction will be
lowing bilinear-biquadratic exchange Hamiltonian: removed later. We denote the number of the site on X by X
and assume A B without loss of generality. We define
the total spin Stot j Sj and denote the eigenvalues of
(S z by S
H tot)2 and Stot tot(Stot 1) and M , respectively. We de-
J Si*Sj J Si*Sj 2, 1
i,j i,j fine the state m(M) , which belongs to the M subspace, in
the following manner:
where summation is taken over all bonds on a lattice .
Blume and Hsieh1 suggested that quadrupole ordering might
m M C Sj mj 0 , 2
occur on this system, and many people have discussed the j
possibility and the properties of this type of ordering.210 In x y
particular, Chen and Levy6 discussed the possible ordered where Sj Sj iSj is the usual spin raising operator at site
z
phases and phase transitions associated with the dipole and j, 0 is the eigenstate where all Sj's have eigenvalue 1,
quadrupole moments in detail by means of a mean-field ap- and C is a positive normalization constant. m represents spin
proximation and high-temperature series expansion. Their re- configurations m m1 ,m2 ,...,m with mj 0, 1, 2 pro-
sults strongly support the occurrence of quadrupole phase viding that j (mj 1) M.
transition for J J 0. Munro's argument is as follows. For J 0, J 0 the non-
As far as we know, however, the exact result about the zero off-diagonal matrix elements of the Hamiltonian H are
ground state is still rare, except for recent remarkable always negative if one uses the basis states U1 m(M) ,
progress on linear chain systems.1117 One of the exact re- where U z
1 exp( i j ASj). It is easily seen that within M
sults on general lattice systems is Munro's argument18 for subspace any two states can be connected by some applica-
J 0, J 0, which is the straightforward extension of the tions of H . Then, from the Perron-Frobenius theorem it is
Lieb-Mattis theorem.19 Furthermore, this result was extended shown that the lowest-energy state in this subspace is unique
to the case J J 0 by Parkinson.20 His argument is almost and positive definite. Following the argument of Lieb and
exact. However, to determine the total spin eigenvalue, he Mattis, it is also shown that this state has Stot M if M
needed one assumption that the ground state for J 0, A B and Stot A B if M A B , therefore the
J 0 is nondegenerate, which was confirmed only by the ground states of H have Stot A B with obvious degen-
numerical diagonalization restricted for the finite-size chain eracy 2Stot 1.
systems up to eight sites. Since this region is assumed to be Parkinson extended Munro's result to J J 0 by
a quadrupole ordering phase at a sufficiently low tempera- using other basis states U2 m(M) with U2
ture, it is desirable to discuss the ground state in detail to exp i( /2) z
j (S j)2 . These basis states make all the
clarify the nature of this ordering. off-diagonal nonzero matrix elements of H negative, pro-
In this paper, we show the lattice where Parkinson's as- vided J J 0. Since H within M subspace is still irre-
sumption is not satisfied and prove that the ground state of ducible, the lowest-energy state is again unique and positive
the Hamiltonian H is always singlet in the region J J 0 definite from the Perron-Frobenius theorem. However, one
on general finite lattice systems with an even number of cannot determine Stot of this state immediately because it is
sites. We also extend the discussion to the lowest-energy positive definite on the basis states U2 m(M) which are
0163-1829/97/56 17 /10774 4 /$10.00 56 10 774 © 1997 The American Physical Society
56 BRIEF REPORTS 10 775
where 0 and 2 has Stot 0 and 2, respectively. This
result means that Parkinson's assumption is not satisfied on
the lattice 1 .
It is interesting to notice that Eqs. 5 and 6 are rewritten
in a positive form
1 2
FIG. 1. Four site bipartite lattice 1 with A 3 and B 1. 0 2
0 a U a U
) m 2 m 0 m 2 m 0 ,
m 6 m
different from those for J 0, J 0. To determine S 7
tot of the
ground state, Parkinson made one assumption that the
ground state is nondegenerate for J 0, J 0. As stated in 2 1
a 0 U a 2 U
Ref. 20, there appear obvious eigenstates, and also defined 2 6 m 1 m 0 m 1 m 0 .
m ) m
by z z
j A(S j)2 j B(S j)2 becomes the conserved quan- 8
tity in this case, which means that H within M subspace is We know that the ground state of H with J J 0 can be
reducible and one cannot prove the uniqueness of the lowest- 1
energy state within this subspace. If this assumption is satis- expressed as a positive definite state on the basis states
fied, however, the ground state will automatically be the U2 m(0) .20 This implies that the ground state for
eigenstate of (S J J 0 is not orthogonal to
tot)2. In addition, since this ground state will 0 and has Stot 0. On the
be a positive semidefinite state not only on the basis states simple lattice 1 , we can obtain the eigenstates of (Stot)2 for
for J 0, J 0 but on those for J J 0, it will not be J 0, J 0 and find that these states are expressed as a
orthogonal to both ground states for each region. Therefore positive form. On general lattice systems, however, it is im-
one may be able to conclude that S possible to know what spin states are realized for J 0,
tot for the ground state for
J J 0 is the same value as that for J 0, J 0, i.e., J 0 because of their possible degeneracy. So we have to
S go another way to determine S
tot A B . There really exists, however, a lattice on which tot for J J 0.
the assumption is not satisfied, as discussed below. Hereafter, we consider H on a connected finite lattice
Consider the Hamiltonian H with an even number of sites, which is not required to be
with J 0, J 0 on the
1
lattice bipartite. Since U2 does not depend on the bipartite structure,
1 composed of A 2,3,4 and B 1 , whose
structure is described in Fig. 1. In this case, M 0 subspace the lowest-energy state within M subspace is still unique and
is found to be decomposed into two disconnected subspaces, positive for J J 0. Here we consider the lowest-energy
one of which corresponds to 0, the other to 2, where state G(0) only within M 0 subspace. Since every en-
is the eigenvalue of . By simple calculation one can find ergy eigenstate with a given Stot always has a representative
the lowest-energy state in M Stot subspace, the global ground state has the same
Stot as the lowest-energy state in M 0 subspace. Specifi-
cally, if this lowest-energy state has Stot 0, then this is the
0
0 am m 0 , 3 unique ground state. If within M 0 subspace there exists a
m state
(0) which is an eigenstate of (Stot)2 with eigenvalue
S tot(S tot 1) and is a positive semidefinite state on the basis
2
2 am m 0 , 4 states U2 m(0) , then we are able to conclude Stot S tot
m by using the nonorthogonality between G(0) and (0) .
in the 0 and 2 subspace, respectively. The coeffi- In fact, we can find a positive semidefinite state
(0) with
cients a(0)
m are 3/ 15 for the spin configuration m S tot 0, as discussed below. Therefore it can be proved that
m1 ,m2 ,m3 ,m4 1,1,1,1 and 1/ 15 for m the ground state for this region is always singlet on general
2,0,1,1 , 2,1,0,1 , 2,1,1,0 , 0,2,1,1 , 0,1,2,1 , and lattice systems with an even number of sites.
0,1,1,2 . The coefficients a(2) From here we prove the existence of a positive semidefi-
m are 1/ 30 for m 1,1,2,0 ,
1,1,0,2 , 1,2,1,0 , 1,0,1,2 , 1,2,0,1 , and 1,0,2,1 and nite state
(0) on the basis states U2 m(0) with S tot
2/ 30 for m 2,2,0,0, , 2,0,2,0 , 2,0,0,2 , 0,0,2,2 , 0. We denote by a set whose elements are pairs of sites:
0,2,0,2 , and 0,2,2,0 . These two states have the same en-
ergy 8J and are the ground state. Since and (S j
tot)2 do 1 , j2 , j3 , j4 ,..., j 1 , j ,
not commute each other, these states are not the eigenstates where j
of (S k and jk jl for k l. We denote by ( jk , jl) 0
tot)2. The linear combinations of these states become the the singlet state on the site j
eigenstates of (S k , jl :
tot)2, and of course we can obtain these states
without effort in this simple case, i.e., we have jk , jl 0 1 j 1 0 0 1 1 ,
k jl jk jl jk jl
9
1 2
z
0 where S . Now, using these singlet states, we
) 0 6 2 , 5 jk jk jk
define
(0) as follows:
2 1
0 1 /2 j
2 k , jl 0 . 10
6 0 ) 2 , 6 jk ,jl
10 776 BRIEF REPORTS 56
Since ( jk ,jl) 0 is the singlet state, the relation Sztot M M M . 24
S
j S j
l jk k , jl 0 0, 11 This implies S tot M. Furthermore, by substituting Eqs. 15
is always satisfied and therefore and 20 into Eq. 21 we can express
(M) as a similar
form to Eq. 16 . Therefore, the lowest-energy state in even
S M subspace has Stot S tot M.
tot
0 Sj Sj 0 0. 12
j l k In the case of odd M, we decompose into two dis-
k , jl joint subsets
By noting (S z z 1 ( j1 , j2),...,( j M , j M 1) and 2
tot)2 (Stot)2 Stot StotStot , we obtain ( jM 2 ,jM 3),...,( j 1 ,j ) and define the state
S
(M) as
tot 2
0 0. 13
On the other hand, if we use U
2( jk , jl) defined by M i 1 /2
U z 2 Sz 2
2 jk , jl e i /2 S jk jl , 14 jk ,jl 2
j
Eq. 9 is rewritten as k , jl 1 jk,jl 1
jk ,jl 1
jk ,jl jk ,jl
jk , jl 0 U2 jk , jl 1 j 1 0 0
k jl jk jl jk ,jl 0 . 25
1 j 1 ). 15 j
k jl k , jl 2
By substituting Eq. 15 into Eq. 10 we can express
(0) In this case, by making use of the relations
as
S
j S j
l jk k , jl 1 2& jk , jl 2 , 26
0 U2 bm m 0 , bm 0, 16
m S
j S j jk , jl 2 & jk , jl 1 , 27
with the definition of l k
m( M ) given by Eq. 2 . From Eqs.
13 and 16 , we found that
(0) has S tot 0 and is posi- we obtain
tive semidefinite on the basis states U2 m(0) . Now, the
proof of the singlet ground state on a lattice with an even S
totStot
M 2 M 1 M . 28
number of sites is complete.
We extend the discussion above to the lowest energy state Combining the same relation as Eq. 24 , we find
G(M) within M 0 subspace. We define the states
( jk , jl) 1 and ( jk , jl) 2 as Stot 2 M M 1 M 2 M . 29
jk , jl 1 1 j 0 0 1 , 17
k jl jk jl This signifies S tot M 1. By substituting Eqs. 15 , 19 , and
20 into Eq. 25 ,
(M) can be rewritten as the positive
jk , jl 2 1 j 1 . 18
k jl semidefinite state again. Therefore, the lowest-energy state in
It is noted that these states can be rewritten as odd M subspace has Stot S tot M 1.
We briefly comment on the case of a connected finite
jk , jl 1 iU2 jk , jl 1 j 0 0 1 ), lattice with an odd number of sites. In this case, we select
k jl jk jl 19 a site, say j, and make a set , except site j. In M 0
subspace, we set the state
(0) as
jk , jl 2 U2 jk , jl 1 j 1 . 20
k jl
0 1 1 /2 0 j
In the case of even M, we decompose into two j k , jl 0 . 30
jk ,jl
disjoint subsets 1 (j1 ,j2),...,( jM 1 ,jM) and 2
( jM 1 ,jM 2),...,( j 1 ,j ) and define the state It can be easily seen that this state has S tot 1 and is positive
(M) as semidefinite. Therefore, we can prove that the ground state
on a lattice with an odd number of sites is triplet.
M 1 /2 jk ,jl 2 jk ,jl 0. In the case of odd M 0 subspace, we decom-
jk ,jl 1 j pose into
k , jl 2 1 ( j1 , j2),...,( j M 2 , j M 1) and 2
21 ( jM ,jM 1),...,( j 2 ,j 1) and set the state (M)
We have as
Stot 2 M M M 1 M , 22
M i 1 1 /2 1 j jk ,jl 2
j
by noting that k , jl 1
jk ,jl 0 . 31
S
j S j
l jk k , jl 2 0, 23 jk ,jl 2
56 BRIEF REPORTS 10 777
In the case of even M 0 subspace, we decom-
pose into 1 (j1 ,j2),...,( jM 1 ,jM) and 2
( jM 1 ,jM 2),...,( j 2 ,j 1) and set the state
(M) as
M 1 1 /2 0 j jk ,jl 2
jk ,jl 1
jk ,jl 0 1 1 /2 1 j
jk ,jl 2
FIG. 2. The ground-state phase diagram of the bilinear-
jk ,jl 2 biquadratic exchange Hamiltonian. In the region J 0, J 0, the
jk ,jl 1 jk,jl 1
j ground state has S
k , jl 1 tot A B the proof is restricted to a bipartite
j lattice . In the region J J 0, the ground state is always singlet
k , jl jk , jl triplet on a lattice with an even odd number of sites. In the
jk ,jl 0 . 32 region J 0, J J, the ground state is ferromagnetic.
jk ,jl 2 first exited state for J J 0 is not triplet assumed to be
Following the same way in the case of an even number of quintet on an even number site lattice.
sites, we can find that these states have S tot M or S tot M We have not discussed the ground state for J 0, J 0
1 in odd or even M subspace, respectively, and are rewrit- on general lattice systems in detail. We assume that in the
ten in positive semidefinite form. Therefore, we reach the
conclusion that on a lattice with an odd number of sites the case of a bipartite lattice with A B O( ) there is
lowest-energy state in odd or even M subspace has S multiple degeneracy on the ground state for this region be-
tot M
or S cause (Stot)2 may connect the two states, one of which has ,
tot M 1, respectively.
In Fig. 2 we summarize the known results. The ferromag- the other of which has 2 or 2.
netic ground state for J 0, J J was proved by Aksamit21 Recently, it was shown by Tian22 that in the case A
by means of the variational method, though we paid little B O( ) there are both ferromagnetic and antiferro-
attention to this region in the present paper. We have proved magnetic long-range orders once the ground states are ex-
that in the region J J 0 the ground state of H pressed as the positive definite states on the basis states
is singlet
triplet on any connected finite lattice with an even odd U1 m(M) . Therefore, on the ground states of the
number of sites. We also obtained the total-spin eigenvalue bilinear-biquadratic exchange Hamiltonian with J 0, J
of the lowest-energy state in M 0 subspace. The lowest- 0 there coexists the ferromagnetic and antiferromagnetic
energy state in M 0 subspace can be treated in the same long range orders if A B O( ). In the present paper
way. These results are summarized as follows. In the case of we found that the ground state is always singlet on an even
a lattice with an even number of sites, the lowest-energy number site lattice in the case J J 0, although the bilin-
state has Stot M or Stot M 1 in even or odd M subspace, ear exchange interaction term is a ferromagnetic one. In this
respectively, and in the case of a lattice with an odd number region it is a very interesting problem determining what type
of sites, that has Stot M or Stot M 1 in odd or even M of long-range orders occur associated with dipole or quadru-
subspace, respectively. From our results, it turns out that the pole moments.
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