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.

  • $\begingroup$ 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? $\endgroup$
    – John M
    Dec 8, 2014 at 5:00
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/68639/2451 and links therein. $\endgroup$
    – Qmechanic
    Dec 8, 2014 at 6:57

1 Answer 1


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 $\sigma(X)=\sigma_c(X)= \mathbb R$.

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

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