I have this very simple question.

Premise: Let $A$ be a linear densely defined symmetric/self-adjoint operator in a complex separable Hilbert space $\mathcal H$ (typical example in Quantum Mechanics).

Definition: The set $\sigma_{\text{pp}}:=\{z| z\in\mathbb C,~ \nexists \left(A-z I\right)^{-1}:\mathcal H\to\mathcal H\} $ is called the pure point spectrum of A. (Definition from Stone, M.H. "Linear Transformations in Hilbert Space and their Applications to Analysis", AMS, 1932, page 129.)

Then is the following result true?

Statement: $\sigma_{\text{pp}}$ is countable (or an empty set).

Kindly provide me with a counterexample if not true, or with a rigorous proof if true.

Thank you,

  • $\begingroup$ What you have written sounds like an imprecise definition of the entire spectrum, not of the pure point spectrum. $\endgroup$ Oct 15 '19 at 22:11
  • $\begingroup$ I copied that from my QM notes. Here is the definition from Blanchard & Bruening: $\sigma_{\text{pp}} (A) = \{\lambda\in\sigma (A)|\text{Ker}\left(A-\lambda I\right)\neq 0_{\mathcal H}\}$. So it is really the same. $\endgroup$
    – DanielC
    Oct 15 '19 at 23:13
  • $\begingroup$ How is that the same? $\endgroup$ Oct 16 '19 at 0:15
  • $\begingroup$ @DanielC: no, it's definitely not the same. $\endgroup$ Oct 16 '19 at 3:31
  • $\begingroup$ My mind is blown, since I have three definitions: the one in Blanchard & Brüning, then the one in V. Moretti: the point spectrum of A, $\sigma_p (A)$, made by complex numbers $\lambda$ for which $A−\lambda I$ is not injective. And the definition in Prugovecki which is my definition. $\endgroup$
    – DanielC
    Oct 20 '19 at 14:08

If $Av=\lambda v$ and $Aw=\mu w$ with $v,w$ unit vectors and $\lambda\ne\mu$, then (using that $A$ is selfadjoint) $$ \lambda\langle v,w\rangle=\langle Av,w\rangle=\langle v,Aw\rangle=\mu\langle v,w\rangle $$ (note that $\lambda,\mu\in\mathbb R$). So $\langle v,w\rangle=0$. Thus the eigenspaces corresponding to distinct eigenvalues are orthogonal. If $H$ is separable, it can only have countably many pairwise orthogonal subspaces, so the set of eigenvalues of $A$ is at most countable.

  • $\begingroup$ So it's essential in the proof that the operator be self-adjoint. OK, I understand the argument. Is the continuous spectrum of a self-adjoint operator always uncountable? $\endgroup$
    – DanielC
    Oct 20 '19 at 14:30
  • $\begingroup$ @DanielC you should probably ask that as a separate question. $\endgroup$
    – cmk
    Oct 20 '19 at 14:44
  • 1
    $\begingroup$ @DanielC: if you ask another quesiton, I'll answer. Meanwhile, no, the continuous spectrum can even consists of a single point. $\endgroup$ Oct 20 '19 at 14:50
  • $\begingroup$ Just done it: math.stackexchange.com/questions/3401536/… $\endgroup$
    – DanielC
    Oct 20 '19 at 15:01

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