G_0^{1/2}$, to obtain $$\begin{aligned}
\bigg\{
\frac{\partial^2}{\partial \beta_2^2} 
\!&\!+\!&\!\frac{\partial^2}{\partial \beta_3^2}
+ \epsilon - V(\beta_2,\beta_3)-\frac{J(J+1)}{3(\beta_2^2+2\beta_3^2)}
\nonumber \\
&\!+\!&\! {V_\text{g}(\beta_2,\beta_3)} \bigg\}\ \Phi (\beta_2,\beta_3) =0
\label{E:5.3} \end{aligned}$$ with $$\begin{aligned}
V_\text{g}\!=\!\frac{1}{4g^2}  \bigg[ \left( \frac{\partial g
}{\partial \beta_2} \right) ^2\!+
\!\left( \frac{\partial g}{\partial \beta_3} \right) ^2 \bigg]
\!- \frac{1}{2g} \bigg[ \frac{\partial^2 g}{\partial \beta_2^2}
\!+\!\frac{\partial^2 g}{\partial \beta_3^2} \bigg]
\label{E:5.4} \end{aligned}$$

With the choice of $f(\beta_2, \beta_3)$ given in the Eq. \[E:2.4\], from Eq. \[E:2.4b\] one obtains $$\begin{aligned}
g\propto \frac{(\beta_2^2\!+\beta_3^2)(\beta_2^2\!+\!2
\beta_3^2)^2}{(\beta_2^2+5\beta_3^2)} .
\label{E:5.5}\end{aligned}$$ and, for $|\beta_3| \ll \beta_2$, $ g \propto \beta_2^4 \big[
1\!+\!4(\beta_3/\beta_2)^4\!+...\big] $. Therefore, the first and second derivative of $g$ with respect to $\beta_3$ tend to zero when $|\beta_3| \ll \beta_2$ and, at the limit $\beta_3\rightarrow 0$, ${V_\text{g} = -2}$ as in the original Bohr model.

![\[F:3.1\] (Color online) Calculated energies of excited levels of the ground and first excited band, in units of $E(2^+_1)$, as a function of the ratio $b=\beta_3^w / \beta_2^w$. ](biz_f5.ps){height="7.5cm"}

With the substitution $\Psi_0 =g^{-1/2}\Phi$, and assuming $V(\beta_2,\beta_3)=0$ inside the potential well and $=+\infty$ outside, the differential equation to be solved takes the form $$\begin{aligned}
\left[ \frac{\partial^2}{\partial
\beta_2^2}+\frac{\partial^2}{\partial \beta_3^2}+\epsilon 
+V_\text{g}(\beta_2,\beta_3) \right] \Phi (\beta_2,\beta_3) = 0
\label{E:5:6}\end{aligned}$$ with $V_\text{g}$ given in the Eq. \[E:5.4\] and $\Phi=0$ on the contour of the 