Consider the one dimensional Schroedinger equation: \begin{equation} (1)\left(-\frac{\hbar^2}{2 m} \frac{d^2}{d x^2} + \frac{1}{2} m \omega^2 |x|^\mu \right) \Psi(x) = E \Psi(x) \end{equation} By introducing $\lambda := \sqrt{m \omega/\hbar}$ and by writing $E = (n+1/2) \hbar \omega$ we transform it to the following form: \begin{equation} (2) \left(\frac{d^2}{d x^2} + \left(\lambda^2 (2 n+1) - \lambda^4 |x|^\mu\right)\right) \Psi(x) = 0 \end{equation} The question is now to find the quantum numbers $n = n(\mu)$ such that the solution is normalized, ie $||\Psi|| := \int\limits_{\mathcal{R}} \Psi(x) dx < \infty$. Of course in case of the harmonic potential $\mu = 2$ we have $n(\mu) = 0,1,2,\cdots$. The question is what is the result for $\mu \neq 2$?
I solved numericaly equation (2) on a grid $\left\{ -10 + j \cdot 0.1\right\}$ for $j=0,\cdots,200$ using initial conditions $\Psi(-10) = 10^{-11}$ and $\Psi^{'}(-10)=0$. For each value of $\mu$ I recorded such value of $n(\mu)$ that minimised the sum of moduli of the wavefunction $\Psi$. Below you can see the functions $n(\mu)$ for the ground state and for the first two excited states.
As you can see the ground state quantum number does not seem to depend on the exponent $\mu$. If $\mu < 2$ then the spacings between the consecutive excited states seem to diminish with $n$, whereas if $\mu > 2$ the spacings between the consecutive excited states seem to increase with $n$.
Now, my question is is it possible to find a closed form solution for the function $n(\mu)$ for all excited states? What tools do functional analysis offer to tackle such problems?