a planar crystal on the xy-plane. For bonds along x(y)-direction, p$_{y}$$_{(}$$_{x}$$_{)}$-orbital of the oxygen atoms and the xy-orbital of the transition element ions form π - bond. The p$_{z}$-orbitals are ignored because they do not give rise to polarization in z-direction.orbitals are considered. Thus, we have the Hamiltonian: where V ′ = V ( pdπ ) and l = 1 , 2 for the oxygen atoms on x-axis and y-axis respectively. Making a transformation similar to that in eqs (5), we obtain the Hamiltonian in momentum space: Now δθ$_{l}$$_{=3}$$_{,}$$_{4}$ = q$_{x}$a$_{0}$/ 2 and δθ$_{l}$$_{=5}$$_{,}$$_{6}$ = q$_{y}$b$_{0}$ . We can solve for eigen values where ε$_{p}$ is two-fold degenerate. The eigen vectors are 18 ψ ′ 1+ , − → k , − → q ( − → r ) = ∑ j e i − → k · − → $^{R$_{j}$}$A ′ 1+ [ ψ$_{d}$ ( − → r − − → R$_{j}$ )+ ∑ l V $^{′}$( − 1) $^{l}$ei − → k · − → r l E ′ 1+ − ε$_{p}$ ψ$_{pl}$ ( − → r − − → R$_{j,l}$ ) ] ( cos θ$_{j}$ e $^{iφ}$sin θ$_{j}$ ) (A-4a) E ′ 2 ± = ε$_{d}$ + ε$_{p}$ 2 − √ ( ε$_{d}$ − ε$_{p}$ 2 ) 2 + 4 V ′ 2 ± V ′ 2 ∣ ∑ l cos 2 δθ$_{l}$e 2 i − → k · − → r l 1 ∣ (A-3b) E = ε$_{p}$ (A-3c) E ′ 1 ± = ε$_{d}$ + ε$_{p}$ 2 + √ ( ε$_{d}$ − ε$_{p}$ 2 ) 2 + 4 V ′ 2 ± V ′ 2 ∣ ∑ l cos 2 δθ$_{l}$e 2 i − → k · − → r l 1 ∣ (A-3a) H$_{0}$ = ∑ ε$_{p}$p † $_{lk,σ}$p$_{lk,σ}$ + ∑ ε$_{d}$d † $_{k}$d$_{k}$ − ∑ n.n. V $^{′}$( − 1) $^{l}$[cos δθ$_{l}$e ( − → R$_{i,l}$ − − → R$_{j}$ $^{)}$d † $_{k}$p$_{lk,}$$_{↑}$ − e $^{iφ}$sin δθ$_{l}$e i − → k · ( − → R$_{i,l}$ − − → R$_{j}$ $^{)}$d † $_{k}$p$_{lk,}$$_{↑}$ ] + H.c.. (A-2) H$_{0}$ = ∑ ε$_{p}$c † $_{pil,σ}$c$_{pil,σ}$ + ∑ ε$_{d}$c † $_{dj}$c$_{dj}$ − ∑ n.n. V $^{′}$( − 1) $^{l}$[cos θ$_{j}$c † $_{dj}$c$_{pil,}$$_{↑}$ + e $^{iφ}$sin θ$_{j}$c † $_{dj}$c$_{pil,}$$_{↓}$ ]+ H.c. (A-1)