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348 H.Y. Shik et al. / Nuclear Physics B 666 [FS] (2003) 337–360 Fig. 3. The labelling scheme of the three-dimensional model. We draw the three-dimensional model considered in Fig. 3. Let the lattice separation be one and we include all couplings with separation no larger than 2 √ 2. There are total of twenty-two couplings. Since the couplings J 10 and J˜ 10 cannot be grouped in the bond representation, we have to discard them. The Hamiltonian looks rather complicated but it is basically built up by the bilayer net spin model. Couplings between the bilayers are specified with tilde (˜), while those lie on the xy-plane are labelled with parallel sign (‖). Specifically: Separation Coupling(s) included 1 J 1 , J˜ 1 ,J √ 2 2 J3 , J˜ 3 ,J ‖ 3 ,J 4, J˜ √ 4 ,J ‖ 4 3 J5 , ˜ 2 J 6 ,J ‖ 6 √ 5 J7 , J˜ 7 ,J ‖ 7 ,J 8 √ 6 J9 ,(J 10 , J˜ 10 ) 2 √ 2 J 11 ,J ‖ 11 J 5 The bond operator representations of the Hamiltonian Eq. (42) for the spin-1/2 case is too tedious to present here, readers who are interested could ask us for manuscript. Basically we combine terms according to combinations of creation and annihilation boson operators. Straightforward manipulations lead to the following conditions which make the completely dimerized state be an eigenstate of the Hamiltonian Eq. (42): 2J 2 = J 3 + J 4 , 2J 5 = J ‖ 3 + J ‖ 4 , J‖ 6 = J 7, 2J 6 = J˜ 1 , 2J 8 = ˜ J 3 = ˜ J 4 , 2J 9 = ˜ J 5 , 2J 11 = ˜ J 7 , J ‖ 7 ,J‖ 11 = 0. (43)
H.Y. Shik et al. / Nuclear Physics B 666 [FS] (2003) 337–360 349 As a demonstration, we present derivations for the spin-1 case here. Obviously, expression of the Hamiltonian in terms of bond operators is even more complicated. However, since the main concern is about how H 3D Eq. (42) acts on the completely dimerized state ψ D , only terms consisting of singlet operator s need to be considered. Hence, one can simplify the expression of the bond operator representation by keeping the terms with singlet operator s only, √ 2 S z =−S ′ z ( →− t † 3 (44) and 0 s + ) s† t 0 , √ 2 S + =−S ′+ ( → t † 1 (45) 3 s − ) s† t −1 , (46) S i =−S ′ i . S i · S j ψ D = S ′ i · S′ j ψ D =−S i · S ′ j ψ D =−S ′ i · S j ψ D . Thus, the bond operator representation for the four spin interactions, (S i · S j , S ′ i · S′ j , S i · S ′ j , S′ i · S j ) are now the same upto a sign. So we only need to consider one of them, say, S i · S j , (47) S i · S j = Si z Sz j + 2( 1 S + i Sj − + ) S− i S+ j , (48) when for i ≠ j S z i Sz j = 2 3 ( t † (49) 0i t† 0j s is j + t † 0i s† j s it 0j + s † i t† 0j t 0is j + s † i s† j t ) 0it 0j , S i + Sj − = 4 ( −t † (50) 3 1i t† −1j s is j + t † 1i s† j s it 1j + s † i t† −1j t −1is j − s † i s† j t ) −1it 1j . When i = j, some four-operator terms will reduce to two-operator terms, e.g., t † 1i t 0it † 1j t 0j → t † 1i t 1i(1 + t † 0i t 0i) → t † 1i t 1i. Thus, we have to keep terms with t operators but neglect terms with p operators, i.e., S z i Sz i → 1 2 t† 1i t 1i + t † 0i t 0i + 1 2 t† −1i t −1i + 2 3 s† i s i, S + i S − i → 2t † 1i t 1i + t † 0i t 0i + t † −1i t −1i + 4 3 s† i s i, S − i S + i → t † 1i t 1i + t † 0i t 0i + 2t † −1i t −1i + 4 3 s† i s i, (51) (52) (53) To simplify the expression, let us denote O i;j as O i;j ≡ 2 3 ( t † 0i t† 0j s is j + t † 0i s† j s it 0j + s † i t† 0j t 0is j + s † i s† j t 0it 0j − t † 1i t† −1j s is j + t † 1i s† j s it 1j + s † i t† −1j t −1is j − s † i s† j t −1it 1j − t † −1i t† 1j s is j + t † −1i s† j s it −1j + s † i t† 1j t 1is j − s † i s† j t 1it −1j ) . (54)
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