- #1

Sunny Singh

- 19

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- TL;DR Summary
- I need a proof which says that continuous eigenvalue spectrum of hermitian operators are non-normalizable

I was reading introduction to quantum mechanics by DJ Griffiths and while discussing the formalism of quantum mechanics, he says that if for a hermitian operator, the eigenvalues are continuous, the eigenfunctions are non-normalizable whereas if the eigenvalues are discrete, then they can be properly normalized but still both span the hilbert space. He doesn't really explain why this is so or maybe i missed something. Can anyone please help me see why this is so obvious?