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pdf, 9 MiB - Infoscience - EPFL

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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)

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