Chapter 12: Problem 10
Insights into the rotational phase transition in the perovskite structure can be gained by treating the octahedra as nearly rigid objects. If each octahedron is assigned a single variable to describe its orientation, \(\theta\), two neighbouring octahedra have lowest energy if they rotate by exactly the same amount in opposite directions. Show that the harmonic energy between two neighbouring octahedra can be written as $$ E(i, j)=\frac{1}{2} K\left(\theta_{i}+\theta_{k}\right)^{2} $$ For a phase transition to occur, it can be presumed that each octahedron also experiences a local double-well potential of the form $$ E(\theta)=-\frac{\kappa_{2}}{2} \theta^{2}+\frac{\kappa_{4}}{4} \theta^{4} $$ Show that when the harmonic term is combined with the interactions between neighbours, a soft mode is obtained at \(\mathbf{k}=\left(\frac{1}{2}, \frac{1}{2}, 0\right)\).
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