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How integrable are solutions of the Schroedinger equation?

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What can one say about the integrability of solutions of the time-independent Schroedinger equation $H\psi=E\psi$, where $H=p^2/2m +V(x)$, with $V(x)=W(x)^2$?

a) For bound states, $\psi$ is square integrable. Is the same true for $p\psi$? For $W(x)\psi$?

b) For scattering states? Here $\psi$ is not square integrable. What are the requirements on $\phi$ such that $\langle\phi|\psi\rangle$ exists? Is the square integrability of $H\psi$ enough? Or that of $p\psi$ and $W(x)\psi$?

Where can I find such questions discussed?

This post has been migrated from (A51.SE)
asked Apr 20, 2012 in Theoretical Physics by Arnold Neumaier (11,395 points) [ no revision ]
Good question on the references problem. I was taught by blackboard and never really used a textbook for this... quant-ph/9907069 is a short paper, and references textbooks. As far as the "best" space goes, I think that depends on application. My position is that all this is usually devoid of physical content, in that the physics will be independent of the mathematical machinery employed. In cases of ambiguity, one usually can find a completely finite truncation and take the limit carefully.

This post has been migrated from (A51.SE)
@genneth: I know that paper; Gieres doesn't discuss the eigenvalue problem at all. But he shows the importance of getting the spaces right. That I am looking for the best spaces is of course just curiosity.

This post has been migrated from (A51.SE)
@genneth: why? = references to where I can read more. Of course, indicating the (mathematical) arguments is also welcome. But I am also interested in getting to know the best possible spaces.

This post has been migrated from (A51.SE)
Then it seems I have missed the heart of the question. What kind of answer to "why" are you looking for? The proof of these things are textbook material, but you are looking for a more physical argument?

This post has been migrated from (A51.SE)
Minor correction to above: the parenthetical remark "since they are really eigenvectors of the adjoint $H^\dagger$" is incorrect).

This post has been migrated from (A51.SE)
@genneth: It is not clear to me why the eigenvectors for the discrete spectrum have to be in the domain of $H$, or why those of the continuous spectrum are in the domain's dual. And in the latter case, I hope for somethng better. E.g., if $H=p^2/2m$ then the domain is the Sobolev space $H^2$ but the eigenvectors are already in the dual space of the space of continuous functions.

This post has been migrated from (A51.SE)
Isn't this standard functional analysis material? Sorry if I'm not understanding the full thrust of the problem --- I suspect you already know this: Usually we pick a domain for $H$ which is not the whole Hilbert space but only a dense subset, and the discrete eigenvectors will be in this subset. The continuous eigenvectors will be in the dual of this subset (since they are really eigenvectors of the adjoint $H^\dagger$). So the answers to your questions are "it depends" --- specifically on the form of $V$.

This post has been migrated from (A51.SE)

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