We know that in one dimension if ##E>V(\infty)## or ##E>V(-\infty)## then the resulting wave function will not be normalizable. The basic argument is that if ##E>V(\infty)##, then a stationary solution to the Schrodinger equation will necessarily have a concavity with the same sign as the solution itself for all ##x## greater than some value ##a##. So if the resulting wavefunction was positive after ##a##, then the wavefunction would also be concave up and curve away from the x-axis. Then the integral of the squared wavefunction would tend towards infinity since the wavefunction would never again go to ##0##. A similar argument can be made if the wavefunction was negative after ##a## and again for ##E>V(-\infty)##. We don't have this problem with the infinite square well or simple harmonic oscillator because the potentials go to infinity at ##\pm \infty##.(adsbygoogle = window.adsbygoogle || []).push({});

In short, my question is this: What are necessary and sufficient conditions for a potential to generate normalizable solutions?

I would prefer to keep things simple with a one-dimensional treatment that doesn't need to be as rigorous as say a functional analysis proof (so maybe constrain ourselves to potentials that are at least C2). But after that's accomplished, I definitely wouldn't mind a more advanced discussion.

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# I Constraints on potential for normalizable wavefunction

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