content_image/308990.jpg content_image/308986.jpg content_image/308987.jpg content_image/308988.jpg
where 

<img src='content_image/308990.jpg'>

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): 

<img src='content_image/308986.jpg'>

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) 

<img src='content_image/308987.jpg'>

<img src='content_image/308988.jpg'>

