Harmonic Approximation

The rotation term $\frac{K(K+1)}{2\mu r^2}$ is assumed as small, and the potential $U_\alpha(r)$ is expanded around a minimum $r_\alpha$,

$$U_\alpha(r) = U_\alpha(r_\alpha) + \frac{1}{2}\frac{d^2}{dr^2} U_\alpha(r=r_\alpha)(r-r_\alpha)^2 + ...$$ (1.29)

Here, $r_\alpha$ can be considered as the equilibrium distance of the two nuclei which clearly still depends on the electronic quantum number $\alpha$. If the higher order terms in the Taylor expansion are neglected, and $\frac{K(K+1)}{2\mu r^2}$ replaced by $\frac{K(K+1)}{2\mu r_\alpha^2}$, the approximate SE becomes

$$\left[ -\frac{1}{2\mu}\frac{d^2}{dr^2} + \frac{K(K+1)}{2\mu r_\al... ...rac{1}{2}\mu \omega_\alpha^2 (r-r_\alpha)^2 \right] P^{\rm harm}_{\alpha;Kv}(r)$$

$$= \varepsilon_{\alpha;Kv}P^{\rm harm}_{\alpha;Kv}(r),\quad \omega_\alpha^2= \frac{1}{\mu} \frac{d^2}{dr^2} U_\alpha(r=r_\alpha).$$ (1.30)

This is the equation of a linear harmonic oscillator apart from the fact that $r\ge 0$. However, $\vert r-r_\alpha\vert$ has been assumed to be small anyway and within this approximation, the energy levels are therefore those of a linear harmonic oscillator shifted by $\frac{K(K+1)}{2\mu r_\alpha^2} + U_\alpha(r_\alpha)$,

$$\varepsilon^{\rm harm}_{\alpha;Kv} = \frac{K(K+1)}{2\mu r_\alpha^2} + U_\alpha(r_\alpha) + \omega_\alpha\left(v+\frac{1}{2}\right).$$ (1.31)