An atom is situated in a simple cubic crystal lattice. If the potential energy of interact...

An atom is situated in a simple cubic crystal lattice. If the potential energy of interaction between any two atoms is of the form cr−α, where c and α are constants and r is the distance between the two atoms, show that the total energy of interaction of a given atom with its six nearest neighbors is approximately that of the three-dimensional harmonic oscillator potential

where A and B are constants.

[Note: Assume that the six neighboring atoms are fixed and are located at the points (±d, 0, 0), (0, ±d, 0), (0,0, ±d), and that the displacement (x, y, z) of the given atom from the equilibrium position (0,0,0) is small compared to d. Then  where

with similar expressions for r2, r3,…, r6. See the approximation formulas in Appendix D.]