If the spectrum of a hermitian operator is continuous, the eigenfunctions are not normalizable. I have been told that these eigenfunctions do not represent possible physical states, but what exactly does that mean? Is there a good interpretation of the physicality of these eigenstates?(adsbygoogle = window.adsbygoogle || []).push({});

If a measurement is done by an operator with either continuous or discrete spectrums, doesn't the wave function in either case still collapse to one of the eigenfunctions?

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# I Implication of eigenstates that are not normalizable

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