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As I'm studying quantum mechanics, it is said without any reasoning or proof, that a self-adjoint linear operator's eigenvectors make a basis for a Hilbert space $\mathcal{H}$. I am aware of the spectral theorems for bounded and unbounded self-adjoint linear operators and see how they can suggest using the continuous "eigenvectors" corresponding to the continuous spectrum. But also there is a theorem that all orthonormal bases in $\mathcal{H}$ are countable. So my question is: Is it true that only the discrete spectrum with it's eigenvectors create a total orthogonal system or we also need to incorporate the other types of spectra in some way? (I might add that this course I'm studying from isn't that math-oriented so they just say that eigenvalues are the solutions to $\hat{T}\vec{v}=\lambda\vec{v}$ and that the set of eigenvalues IS the spectrum which only adds to my confusion.)

Krum Kutsarov
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  • "But also there is a theorem that all orthonormal bases in $\mathcal H$ are countable." this is true if and only if $\mathcal H$ is separable – Lorago Oct 13 '24 at 16:37
  • @Lorago can you provide a proof because I thought what I said can be derived from the Bessel inequality which if I'm not mistaken works for all Hilbert spaces? – Krum Kutsarov Oct 13 '24 at 18:23
  • Ok maybe I'm a little mistaken. I equated the number of basis vectors and nonzero Fourier coefficients. – Krum Kutsarov Oct 13 '24 at 18:25
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    Here is a proof of my claim. You are, however, correct in that at most countably many terms in the basis expansion of any element are non-zero – Lorago Oct 13 '24 at 18:27
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    Your favorite self-adjoint operator, the position operator, has no eigenvalues ... not even on a bounded interval (where it is continuous) – jd27 Oct 13 '24 at 18:41

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