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Suppose that A is a self-adjoint (possible unbounded) operator from a separable Hilbert space H to itself. I would like to know if the following statement is true:

A has pure point spectrum (i.e., the continuous spectrum of A is empty) if and only if the set of eigenvectors of A is complete (in other words, there is an orthonormal basis of H whose elements are eigenvectors of A).

I know that this result is true if A is self-adjoint and compact, but I would like to know if it holds for the class of second-order differential operators that typically appear in quantum mechanics.

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  • $\begingroup$ The operator $A:\ell_2\to\ell_2$ defined by $A(x_n)=(x_n/n)$ is compact, self-adjoint, has an orthonormal basis of eigenvectors (the unit vector basis $(e_i)$), but $0$ is in the continuous spectrum of $A$. $\endgroup$
    – M.González
    Commented Dec 20, 2023 at 19:26
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    $\begingroup$ Yes, this is true. I think your question would have been better suited for math.stackexchange.com , though. $\endgroup$
    – Christian Remling
    Commented Dec 20, 2023 at 20:15
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    $\begingroup$ @M.González: I don't think this is what the OP means by "continuous spectrum" (a vector is in the subspace corresponding to continuous spectrum if its spectral measure is a continuous measure on $\mathbb R$ (= has no atoms)). What you call continuous spectrum is more commonly referred to as "essential spectrum". $\endgroup$
    – Christian Remling
    Commented Dec 20, 2023 at 20:17
  • $\begingroup$ @Christian Remling: For me, $z$ in the continuous spectrum of $A$ means $A-zI$ injective with proper dense range. $\endgroup$
    – M.González
    Commented Dec 21, 2023 at 10:28
  • $\begingroup$ @M.Gonzalez: Since $R(A-t)^{\perp}=N(A-t)$ for $t\in\mathbb R$, this last condition repeats the first one, so this set is $\sigma(A)\setminus\sigma_p(A)$ (with $\sigma_p$ denoting the eigenvalues), and it's not even a spectrum (closed) in general. $\endgroup$
    – Christian Remling
    Commented Dec 21, 2023 at 15:41

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