# Discrete vs Continuous spectra of operators [duplicate]

Why is it that if an operator $Q$ has a discrete spectra, that the eigenfunctions are all in Hilbert space? Why is it that if the spectrum is continuous we automatically know that the eigenfunctions are not normalizable?

I was reading this in Griffith's but I didn't quite understand it. Is it because in the discrete case, the expectation value of the operator for a system in that eigenstate returns the scalar q? I can see how that would mean the inner product converges and so doesn't present a problem, but I'm having trouble seeing what goes wrong when the spectrum is continuous.

• Sometimes, if the spectrum is continuous, you can represent the solution as plane waves, which happen to be cosine and sines at some frequency. Can you integrate these over the entire domain? Dec 8, 2014 at 5:00
• Possible duplicates: physics.stackexchange.com/q/68639/2451 and links therein. Dec 8, 2014 at 6:57

Consider an operator $$A$$ on a Hilbert space $$\cal H$$, say $$L^2(\mathbb R)$$ in order to deal with QM of a particle on a real axis without spin. Let $$D(A) \subset \cal H$$ be the domain of $$A$$.

The spectrum $$\sigma(A)\subset \mathbb C$$ of $$A$$ is defined as the union of the following three pairwise disjoint subsets $$\sigma_p(A)$$, $$\sigma_c(A)$$, $$\sigma_r(A)$$. The first one is that you call discrete spectrum, but its name is point spectrum (the discrete spectrum is a special case of it).

Point spectrum, $$\sigma_p(A)$$. It is made of the complex numbers $$\lambda$$ such that $$A-\lambda I :D(A) \to {\cal H}$$ is not injective.

Consequently, by definition if $$\lambda \in \sigma_p(A)$$ there must be $$\psi \in \cal H$$ such that $$A\psi = \lambda \psi$$.

Continuous spectrum, $$\sigma_c(A)$$. It is made of the complex numbers $$\lambda$$ such that $$A-\lambda I :D(A) \to {\cal H}$$ is injective, $$(A-\lambda I)^{-1} : Ran(A) \to D(A)$$ is not bounded, and $$Ran(A-\lambda I)$$ is dense in $$\cal H$$.

Consequently, by definition, if $$\lambda \in \sigma_c(A)$$, there are no $$\psi \in \cal H$$ with $$A\psi = \lambda \psi$$.

Residual spectrum, $$\sigma_r(A)$$. It is made of the complex numbers $$\lambda$$ such that $$A-\lambda I :D(A) \to {\cal H}$$ is injective, $$(A-\lambda I)^{-1} : Ran(A) \to D(A)$$ is bounded, and $$Ran(A-\lambda I)$$ is not dense in $$\cal H$$.

This last component of the spectrum is always empty for normal (generally unbounded) operators. In particular self-adjoint and unitary operators have no residual spectrum. This is the reason why the residual spectrum does not appear in QM barring exceptional cases.

If $$\lambda \in \sigma_c(A)$$, for every $$\epsilon >0$$, there is $$\psi_\epsilon \cal H$$ with $$||\psi_\epsilon||=1$$ and $$||A\psi_\epsilon - \lambda \psi_\epsilon|| < \epsilon$$ In other words there is a class of approximated eigenvectors, though no proper eigenvector exists for $$\lambda \in \sigma_c(A)$$.

Under suitable further hypotheses on the Hilbert space (rigged Hilbert space) the set of $$\psi_\epsilon$$ admits a limit outside the Hilbert space. The distributions $$\delta(x-x_0)$$ are typical examples of this situation when referring to the position operator $$X$$ in $$L^2(\mathbb R)$$, since $$\sigma(X)=\sigma_c(X)= \mathbb R$$.

• suggested edit Paragraph 2 should contain σ-p, σ-c and σ-r. Mar 3, 2020 at 21:09