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Let $A$ be a self-adjoint operator. Then, we can define

  1. the pure point spectrum of $A$

$$\sigma_{\mathrm{pp}}(A) := \{z\in \Bbb C \ | \ \mathrm{ran(A-z)} = \overline{\mathrm{ran (A-z)}} \neq \mathcal{H}\}$$

  1. the point embedded in continuum spectrum of $A$

$$\sigma_{\mathrm{pec}}(A) := \{z\in \Bbb C \ | \ \mathrm{ran(A-z)} \neq \overline{\mathrm{ran (A-z)}} \neq \mathcal{H}\}$$

  1. the purely continuous spectrum of $A$

$$\sigma_{\mathrm{pc}}(A) := \{z\in \Bbb C \ | \ \mathrm{ran(A-z)} \neq \overline{\mathrm{ran (A-z)}} = \mathcal{H}\}$$

These form a partition of the spectrum $\sigma(A)$, i.e. they're pairwise disjoint and their union is $\sigma(A)$.

While defining the spectrum of self-adjoint operators along the above lines, my professor made a remark that there's a theorem that the only intrinsic information in a self-adjoint (or Hermitian) operator is its spectrum. (The spectrum of a self-adjoint operator doesn't necessarily contain just the eigenvalues, for infinite-dimensional operators. And I'm primarily interested in the infinite-dimensional case.) I tried hunting for the theorem in textbooks but couldn't spot anything so far, though it certainly seems quite intuitive. (I am aware of the fact that to every observable in quantum mechanics we may associate a linear, Hermitian operator $\mathcal O$, whose spectrum of eigenvalues correspond to the spectrum of possible observations of said observable.) Could someone provide me an information-theoretic or mathematical argument as to why the only essential information in a self-adjoint (or Hermitian) operator is its spectrum? I'm looking for something rigorous.

P.S: I'm asking this question on Physics Stack Exchange rather than Mathematics Stack Exchange as I feel physicists would be more acquainted with Hermitian operators than mathematicians, but in case this is the wrong venue, feel free to let me know in the comments.

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    $\begingroup$ The spectrum doesn't tell you whether eigenvalues are degenerate. $\endgroup$ Commented Dec 26, 2019 at 16:20
  • $\begingroup$ Isospectral $\endgroup$ Commented Dec 26, 2019 at 17:02

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All information is contained in the eigenvalues, which appear along the diagonal, with all off-diagonal elements being zero. Diagonalisation is achieved by two unitary, information-preserving transformations. Hence all information in the original matrix must still be present in the diagonal representation, and hence in the eigenvalue spectrum.

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  • $\begingroup$ The spectrum of a self-adjoint operator doesn't necessarily contain just the eigenvalues (for infinite-dimensional operators). And I'm primarily interested in the infinite-dimensional case. $\endgroup$
    – user199113
    Commented Dec 26, 2019 at 9:40

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