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Suppose you have a simple 1D problem with a potential which is such to allow for bound, quasibound, and free states. Are the quasibound states orthogonal to the bound states, or is there some slight overlap?

Qmechanic
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Brian Enoch
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1 Answers1

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Mathematically, only the bound states, which have square-integrable wave functions, are "states". The bound states span a subspace of the Hilbert space so, by definition, everything else is orthogonal to them.

"Quasibound states" are technically part of the continuum spectrum. (There are some potentials, which go to $0$ slowly at $\infty$, that have more complicated spectra.)

Keith McClary
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