Square-integrable eigenfunctions of the Schrödinger operator decay $\pm \infty$

Question

Why is it necessary for an eigenfunction of $H=\frac{d^2}{dx^2}+u(x)$ that is square-integrable that it tends to zero at $\pm \infty$?

Answer 1

Mainly because of the probability interpretation. If $\psi$ is an eigenfunction of the Schrodinger equation, then the statistical interpretation says that $\int_a^b|\psi(x)|^2\,dx$ gives the probability of finding the particle between $a$ and $b$. By the rules of probability, we must have $\int_{-\infty}^{\infty}|\psi(x)|^2\,dx=1<\infty.$ This certainly cannot happen unless the function itself decays to zero.