# Is the point spectrum always countable?

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,

• What you have written sounds like an imprecise definition of the entire spectrum, not of the pure point spectrum. Oct 15 '19 at 22:11
• 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. Oct 15 '19 at 23:13
• How is that the same? Oct 16 '19 at 0:15
• @DanielC: no, it's definitely not the same. Oct 16 '19 at 3:31
• 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. 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.