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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 $\pm \infty$?

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

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  • $\begingroup$ What is the tail of integral? Could you give a detailed proof? $\endgroup$ – Lucas Pereiro Jul 19 at 16:26
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    $\begingroup$ What is $u$? $\ $ $\endgroup$ – cmk Jul 19 at 16:57
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    $\begingroup$ @cmk: $u$ is usually a potential function of some kind, like potential energy. $\endgroup$ – Adrian Keister Jul 19 at 18:35
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    $\begingroup$ @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. $\endgroup$ – cmk Jul 19 at 18:38
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    $\begingroup$ @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. $\endgroup$ – cmk Jul 19 at 18:44
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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.

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