# 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? – John M Dec 8 '14 at 5:00
• Possible duplicates: physics.stackexchange.com/q/68639/2451 and links therein. – Qmechanic Dec 8 '14 at 6:57

## 1 Answer

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. – Ajay Shanmuga Sakthivasan Mar 3 '20 at 21:09