10
this term becomes
− λ2
6Jcluster
· (3/4)[3/16 + ( τx/2 − 1/4)2]
= − (λ2)/(32Jcluster) · (2 − τx
k ).
Another second order perturbation term r2λ2 Pjk Sk2 ·
(Sj3 × Sj4)(1 − P jk )[0 − Hcluster j − Hcluster k]−1(1 −
Pjk )Sk2 · (Sj3 × Sj4)Pjk can be computed in the similar
way and gives the result −(r2 λ2)/(32Jcluster) · (2 − τx
j ).
For one of the cross term
r λ2 Pjk Sj2 · (Sk3 × Sk4)(1 − Pjk )
× [0 − Hcluster j − Hcluster k]−1
× (1 − Pjk )Sk2 · (Sj3 × Sj4)Pjk
We can use the previous argument for both cluster j and
k, so (1 −PAB)[0−Hcluster j −Hcluster k]−1(1−Pjk ) can be
replace by c-number ( −2Jcluster)−1. This term becomes
− r λ2
2Jcluster
Pjk [Sj2 · (Sk3 × Sk4)][Sk2 · (Sj3 × Sj3)]Pjk .
Spin rotation symmetry again helps to separate the terms
for clusterj and k, and we get −(r λ2)/(32Jcluster)·τz
j τz
k .
The other cross term r λ2 Pjk Sk2 · (Sj3 × Sj4)(1 −
Pjk )[0 − Hcluster j − Hcluster k]−1(1 − P jk )Sj2 · (Sk3 ×
Sk4)Pjk gives the same result.
In summary the second order perturbation from λ[Sj2 ·
(Sj3 × Sj4) + r Sk2 · (Sj3 × Sj4)] is
− r λ2
16Jcluster
· τz
j τz
k + λ2
32Jcluster
(τx
k + r2 τx
j − 2r2 − 2).
Using this result we can choose the following pertur-
bation on z-links,
λz Hperturbation, z
=λz [Sj2 · (Sk3 × Sk4) + sgn( Jz ) · Sk2 · (Sj3 × Sj4)]
− |Jz|(Sj3 · Sj4 + Sk3 · Sk4)
with λz = 4
√
|Jz|Jcluster, r = sgn( Jz ) is the sign of Jz.
The last term on the right-hand-side is to cancel the non-
trivial terms (r2 τx
j + τx
k )λ2
z/(32Jcluster) from the second
order perturbation of the ﬁrst term. Up to second order
perturbation this will produce−Jzτz
j τz
k interactions.
Finally we have been able to reduce the high order
interactions to at most three spin terms, the Hamiltonian
Hmagnetic is
Hmagnetic =
∑
j
Hcluster j +
∑
x−links <jk>
λxHperturbation x
+
∑
y−links <jk>
λyHperturbation y
+
∑
z−links <jk>
λz Hperturbation z
where Hcluster j are given by (2), λx,y,z Hperturbation x,y,z
are given above. Plug in relevant equations we get (16)
in Subsection IV B.
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