- #1
jostpuur
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It seems to be true, that if some eigenvalue of a Hamilton's operator is an isolated eigenvalue (part of discrete spectrum, not of continuous spectrum), then the corresponding eigenstate is normalizable, and on the other hand, if some eigenvalue of a Hamilton's operator is not isolated, then the corresponding eigenstate is not normalizable (not a vector of a Hilbert space) (like plane waves).
This starts to become intuitively clear once one has seen sufficiently examples obeying this pattern, but I have never encountered any general proof for this. Anyone knowing something about the proof?
This starts to become intuitively clear once one has seen sufficiently examples obeying this pattern, but I have never encountered any general proof for this. Anyone knowing something about the proof?