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: $V(x=\pm \infty)=0$.

  • $\begingroup$ The question is rather vague in this form (also, what about boundary conditions?). If you are asking if it is in principle possible to find the eigenvalues, given $\psi_0$, the answer is of course yes because you can recover $V$ as $V=\psi''/\psi$. $\endgroup$ Jun 24, 2015 at 15:43
  • $\begingroup$ If you can indeed factorize your ODE as described, then knowledge of any eigenfunction will be enough to construct the spectrum. You have ladder operators, so you'll need to step up and down to get the full spectrum. $\endgroup$
    – josh
    Jun 24, 2015 at 16:49
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    $\begingroup$ It's probably also worth mentioning that, a priori, the zero eigenvalue $E=0$ is not distinguished in anyway by the form of the equation, without further conditions on $V(x)$. That's because it is always possible to shift all eigenvalues by $\delta E$ by shifting the potential $V(x) \to V(x) + \delta E$. However, for instance, if you require $V(x) \to 0$ at infinity, the such shifts are no longer allowed and the notion the $E=0$ eigenfunction may have some special properties with respect to the equation. Of course, such a condition could be very different from what you have in mind. $\endgroup$ Jun 24, 2015 at 16:51
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    $\begingroup$ well, you know that the other eigenfunctions are orthogonal to the one you've found; hoping for more seems like asking what you can tell about a unitary matrix if all you know is one column. $\endgroup$ Jun 24, 2015 at 21:47


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