where with a$_{0}$ and b$_{0}$ being the lattice constants without bond-bending. Another way of writing them should be − → a 0 = ( − → R$_{j}$$_{+1}$ − − → R$_{j}$ ) / 2 and − → b 0 = ( − → R$_{j}$$_{+1}$ − − → R$_{j}$ ) / 2 for site j + 1 situated at x- or y-direction away from site j . There are ten bands in our calculation. The main contribution comes from the topmost occupied band. We have also made the approximation E$_{n}$$_{+}$ ≈ ε$_{p}$ , taking the advantage of the fact that V ≪ | ε$_{p}$ − ε$_{d}$ | . If even number of bands are occupied, the polarization will be much smaller. The polarization produced by two bands tends to cancel each other. As a result, there is an extra factor of 4 V $^{2}$/ ( ε$_{p}$ − ε$_{d}$ ) 2 which comes from the denominator of eqs. (10): To see more clearly how polarization and − → q are related, let U$_{P}$ be the space inversion operator. We then have Under the space inversion, the displacements of oxygen atoms and hence, α , β and V change sign under inversion and the polarization is an odd function of − → q . The form of sine function in eqs. (14) and (16) seems to be a natural form. In the continuum limit ( a$_{0}$ ≈ b$_{0}$ −→ 0) − → P − q ≈ $_{32}$√$_{2 sin}$$_{βV}$ $^{3}$λ π ( ε$_{p}$ − ε$_{d}$ ) $^{3}$Δ E$_{cf}$ ( $^{eρ}$̂ e$_{z}$ )[cos φ sin( a$_{0}$q$_{x}$/ 2) + sin φ sin( b$_{0}$q$_{y}$/ 2)] (16) U$_{P}$ ψ − → k , − → q ( − → r ) = ψ$_{−}$ − → k ,$_{−}$ − → q ( − → r ) . (17) − → P − q = − e 〈 U$_{P}$ U$_{P}$ − → r U$_{P}$ U$_{P}$ 〉 = − − → P − − → q (18) 11 ρ ≈ − ∫ d 3 − → r ψ ∗ $_{zx}$( − → r ) zψ$_{px}$ ( − → r − $^{a$_{0}$}$̂ e$_{x}$/ 2) ≈ − ∫ d 3 − → r ψ ∗ $_{yz}$( − → r ) zψ$_{py}$ ( − → r − $^{b$_{0}$}$̂ e$_{y}$/ 2) (15)