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90 CHAPTER 4. HONEYCOMB LATTICE<br />
make some predictions about the spinon excitations. In fact this simplified RVB<br />
theory predicts d–wave superconductivity on the square lattice [20] and was also<br />
applied to the Shastry–Sutherland [110] and triangular lattice [77]. It is based on<br />
a decoupling of the superexchange term in singlet operators. Given the identity<br />
S i · S i+α − n in i+α<br />
4<br />
= − 1 2 B† i,α B i,α (4.10)<br />
a decoupling in ∆ ij ≡〈B i,α 〉 is a natural choice. To handle the bosons one assumes<br />
that all the Lagrange multipliers λ i are equal to a single value λ which plays the<br />
role of the boson chemical potential. 〈 We 〉 replace the boson operators by a static<br />
value which is chosen such that b † i b j = δ for all i, j [109, 20, 110, 77]. Thus in<br />
this approximation the effect of the slave bosons (or equivalently of the Gutzwiller<br />
projection) is included in a very simplified way which effectively multiplies the<br />
kinetic energy of the spinons by the hole concentration δ and redefines their<br />
chemical potential µ → µ 0 + λ:<br />
H MF<br />
tJ<br />
= −tδ ∑<br />
〈i,j〉σ<br />
(<br />
)<br />
f † iσ f jσ +h.c. − µ ∑ n i<br />
i<br />
− J ∑ (〈 〉<br />
B † i,α B i,α + B † i,α<br />
2<br />
〈B i,α〉<br />
〈i,j〉<br />
−<br />
〈 〉 )<br />
B † i,α 〈B i,α 〉<br />
(4.11)<br />
In addition we use the simple ansatz of equation (4.2) for the RVB order parameter<br />
∆ ij . The mean field solution is obtained by minimizing the corresponding<br />
free energy density<br />
φ = −µ + 3 4 J∆2 − 1 ∑<br />
(<br />
ε j (k)+ 2 )<br />
L<br />
β ln [1 + exp(−βε j(k))]<br />
kj<br />
with respect to ∆, φ ij and under the conservation of particle number constraint<br />
∂φ<br />
∂µ = −N L<br />
⇒<br />
δ = − 1 L<br />
∑<br />
kj<br />
( )<br />
∂ε j (k)<br />
∂µ tanh βεj (k)<br />
.<br />
2<br />
ε j=± (k) are the dispersions of the single particle spinon excitations above the<br />
ground-state. They are given by<br />
(<br />
ε ± (k) = µ 2 + J 2 |B(k)| 2 + δ 2 |ξ(k)| 2<br />
√<br />
) 1/2<br />
± 2δ J 2 |B(k)| 2 (Im ξ(k)) 2 + µ 2 |ξ(k)| 2 (4.12)