Constructing soluble quantum spin models - Department of Physics ...

H.Y. Shik et al. / Nuclear Physics B 666 [FS] (2003) 337–360 345
K−1,M
∑ 1( h 2a = s
†
k,m
4
t αk,ms † k+1,m t αk+1,m + s † k,m t αk,mt † αk+1,m s k+1,m
k,m=1
+ t † αk,m s k,ms † k+1,m t αk+1,m + t † αk,m s k,mt † αk+1,m s k+1,m
K,M−1
∑ 1 (
h 2b = s
†
k,m
4
t αk,ms † k,m+1 t αk,m+1 + s † k,m t αk,mt † αk,m+1 s k,m+1
k,m=1
(35)
)
,
(36)
+ t † αk,m s k,ms † k,m+1 t αk,m+1 + t † αk,m s k,mt † αk,m+1 s )
k,m+1 ,
(37)
K−1,M
∑ i
h 3a =
4 ɛ (
αβγ s
†
k,m t αk,mt † βk+1,m t γk+1,m + t † αk,m s k,mt † βk+1,m t γk+1,m
k,m=1
− t † βk,m t γk,ms † k+1,m t αk+1,m − t † βk,m t γk,mt † αk+1,m s )
k+1,m ,
(38)
K,M−1
∑ i
h 3b =
4 ɛ (
αβγ s
†
k,m t αk,mt † βk,m+1 t γk,m+1 + t † αk,m s k,mt † βk,m+1 t γk,m+1
k,m=1
h 4a =
h 4b =
K−1,M
∑
k,m=1,α≠β
K,M−1
∑
k,m=1,α≠β
− t † βk,m t γk,ms † k,m+1 t αk,m+1 − t † βk,m t γk,mt † αk,m+1 s )
k,m+1 ,
) , (39)
(40)
1
4 t† αk,m t (
βk,m tαk+1,m t † βk+1,m − t† αk+1,m t βk+1,m
1
4 t† αk,m t (
βk,m tαk,m+1 t † βk,m+1 − t† αk,m+1 t )
βk,m+1 .
Then, based on the arguments we have in previous sections, the conditions for the
completely dimerized state to be an eigenstate are
2J 2 = J 3 + J 4 , J 6 + J 7 = J 5 + J 8 .
(41)
Again, as in the previous section, same conditions hold for cases with spin S other
than 1/2. Conditions Eq. (41) could have various forms. For example, we can let J 6 = J 7
due to symmetry and let J 8 = 0togetJ 5 = 2J 6 ,orletJ 5 = J 6 = J 7 = J 8 and study their
energy specturms. In fact, then the model could be maped into coupled spin chains with
spatial anisotropy. Quantum phase transitions are expected. We also remark that one can
obtain soluble bilayer net spin model with inter-bilayer couplings under similar conditions.
5. Three-dimensional exactly soluble model
In this section, the bond operator method is used to determine the condition(s) required
for the completely dimerized state to be an eigenstate of a three-dimensional model H 3D ,
defined by the Hamiltonian,
K,L,M
∑
K,L,M−1
H 3D = 2J 1 S k,l,m · S ′ k,l,m + 2 ∑
J˜
1 S ′ k,l,m · S k,l,m+1
k,l,m=1
k,l,m=1