Given the second order linear homogeneous differential equation $$ -\dfrac{d^2}{dx^2}\psi_m(x) + V(x)\psi_m(x)=E_m\psi_m(x) $$ with eigen-functions $\psi_m(x)$ and eigenvalues $E_m$, what information about the eigenvalue spectrum can be extracted if we know the eigenfunction of the zero eigenvalue?

A common occurrence with many second order linear homogeneous differential equations that cannot be solved analytically for all eigenvalues, is one can find the eigenfunction of the zero eigenvalue. $$ -\dfrac{d^2}{dx^2}\psi_0(x) + V(x)\psi_0(x)=0 $$ My gut tells me their should be away to extract more information, but I do not know of such a method.

The eigenvalues depend on the boundary conditions imposed on the eigen-functions. For example: $\psi_m(0)=\psi_m(1)=0$ or $\psi_m(1)=\psi_m(\infty)=0$.

One possible method is to solve for the ladder operators for $V(x)$ and its pair $\tilde{V}$. For example $$ A_{\pm} = \pm \dfrac{d}{dx}+W(x) $$ and \begin{align} A_{+}A_{-} = -\dfrac{d^2}{dx^2} + V(x) && A_{-}A_{+} = -\dfrac{d^2}{dx^2} + \tilde V(x) \end{align} But this depends on knowing the lowest eigen-function, not the zero eigen-function.

The potential has the following limits: $V(x=\pm \infty)=0$.

in principlepossible to find the eigenvalues, given $\psi_0$, the answer is of course yes because you can recover $V$ as $V=\psi''/\psi$. $\endgroup$anddown to get the full spectrum. $\endgroup$youhave in mind. $\endgroup$