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

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 $$\pm \infty$$?

closed as off-topic by cmk, RRL, mrtaurho, Xander Henderson, The CountJul 29 at 1:28

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• What is the tail of integral? Could you give a detailed proof? – Lucas Pereiro Jul 19 at 16:26
• What is $u$? $\$ – cmk Jul 19 at 16:57
• @cmk: $u$ is usually a potential function of some kind, like potential energy. – Adrian Keister Jul 19 at 18:35
• @AdrianKeister You're right, of course, but I was encouraging them to add more context to their problem, like their assumptions on $u$. Otherwise, they're risking having their question closed. – cmk Jul 19 at 18:38
• @AdrianKeister I don't disagree with you, but I'd say that if someone is looking at the problem statement and sees "show $L^2$ eigenfunctions of $\frac{d^2}{dx^2}+u$ vanish at infinity," they might like to know what assumptions are present on $u$ before providing a rigorous argument. FYI, I agree with your physical interpretation of the problem and +1'd. – cmk Jul 19 at 18:44

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.