Normalizability of continuous and discrete spectrum

[Sunny Singh]

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?

 

[PeroK]

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?

It's not obvious. In linear algebra you have the finite dimensional spectral theorem, which guarantees an orthonormal basis of eigenvectors for any normal matrix (which includes Hermitian ones).

For infnite dimensional spaces, you have the same property for compact, self-adjoint operators.

In QM it's assumed that the Hermitian operators associated with observables have this property.

A continuous spectrum of eigenvalues takes you out of the realm of traditional linear algebra and into the shadowy realm of "rigged Hilbert spaces". In which case the eigenfunctions won't be the usual normalisable functions, but something more exotic.

Again, I would simply take it as a postulate of QM that the eigenfunctions of an operator representing a continuous observable have this property.

 

[A. Neumaier]

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

By definition, a number $E$ belongs to the discrete spectrum of an operator $H$ if there is a normalizable eigenvector $\psi$ with $H\psi=E\psi$. The full spectrum is the set of $E$ such that $E-H$ is not invertible. Except in the accidental case where a discrete eigenvalue is embedded in the continuum (which is unstable under perturbations and leads to resonances, in physics responsible for particle decay) the continuous spectrum is disjoint from the discrete spectrum, hence there are no associated normalizable eigenstates.

but still both span the hilbert space

This is misleading. Given a fixed Hermitian operator, the discrete eigenstates span the full Hilbert space only if the spectrum is discrete only, and the continuum eigenstates span the full Hilbert space only if the spectrum is continuous only. Both these cases occur in the same Hilbert space but for different operators.

 

[dextercioby]

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?

There is no "B" level proof that the eigenvectors of an operator with continuous spectrum are not in the Hilbert space. One needs (big chunks of) the full machinery of functional analysis to write a comprehensive proof.