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Orthogonality in discrete Hilbert spaces is straightforward - those encountered by typical examples of infinite wells of any type, spin systems etc. Continuous Hilbert spaces are fine too - we usually talk about free particles and such.

My confusion comes when I try to 'put the two together'. In systems that have both bound and free states (finite potentials), are bound and free states always orthogonal? Can this be shown easily?

My confusion stems from the fact that the continuous momentum eigenspace is the basis of plane waves, and yet bound wave functions exist in this space. How can this be reconciled? I feel as though there is something being swept under the rug.

anon01
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2 Answers2

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They are not orthogonal because planar waves do not belong to hilbert space. It is true that you can have hilbert spaces of any cardinality, but the one in quantum mechanics is $L_2$, the space of square integrable functions, which has a countable cardinality, $\aleph_0$. A consistent way to do the math would be to have atoms in a box of length $l$, where $l$ can be as large as you like, and in such a case the unbounded states will be countable, as close as you like to planar waves, and orthogonal to any bound state. Using the dirac delta (a generalized function, or distribution, that does not belong to $L_2$) is a slight of hand that works well in practice but is not mathematically rigorous.

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The unbound free states are not plane waves. And, in the extended (rigged) Hilbert space that allows for un-normalizable states, they are orthogonal to all the bound states.

mike stone
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