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Let $T$, $S$ be two self-adjoint linear operators on a Hilbert space $\mathcal{H}$ with pure point spectrum.
Then $T$ and $S$ commute if and only if they have a complete set of common eigenvectors.

This is the full statement I want to prove. In the case of bounded operators this would be fine. My problem now is the unbounded case. I may tell you my thoughts for the first implication:

Let be $\phi\in D(T)$ an eigenvector from $T$ with eigenvalue $\lambda$, then we know, since $T$ and $S$ commute: $$ TS\phi=ST\phi=S\lambda \phi=\lambda S\phi $$ Now, here is my problem: Can I assume (and why) that $\phi\in D(S)$ just given that $T,S$ commute and $\phi$ is an eigenvector of $T$? Maybe this is quite simple but I don't get it at the moment...

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  • $\begingroup$ What is $D(S)$? $\endgroup$
    – Yossi Lonke
    Commented Aug 22, 2018 at 9:29
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    $\begingroup$ What is your definition of when unbounded (self-adjoint) operators commute? $\endgroup$
    – Matthew Daws
    Commented Aug 22, 2018 at 11:55
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    $\begingroup$ The spectral Measure of these Operators commute, or $[e^{itT},e^{isS}]=0$ for all $t,s\in \mathbb{R}$, or $TS=ST$ including the equality of the domains. As far as I know, they are equivalent. $\endgroup$
    – Tim S
    Commented Aug 22, 2018 at 12:11
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    $\begingroup$ Okay. Similarly $e^{itT}\phi = \sum_i e^{it\lambda_i} (\phi|\phi_i) \phi_i$. If this commutes with $e^{itS}$, which has a similar form, for all $t$, then you can perform calculations with bounded diagonal operators which will lead to the result you want. (Also $\lim_{n\rightarrow\infty} |\lambda_n| = \infty$). $\endgroup$
    – Matthew Daws
    Commented Aug 22, 2018 at 15:48
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    $\begingroup$ @MatthewDaws: You certainly can't conclude that $|\lambda_n|\to\infty$. For example, the set of eigenvalues could be $\mathbb Q$. $\endgroup$
    – Christian Remling
    Commented Aug 23, 2018 at 20:07

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This is an extended comment. This has been proven as proposition 6.6.2 on p. 233 of the book Hilbert Space Operators in Quantum Physics, 2nd ed. by Blank, Exner and Havlíček.

A set $\mathcal{S} = \{A_j\}_{j=1}^N$ of commuting self-adjoint operators on a separable [Hilbert space] $\mathcal{H}$ with pure point spectra is a CSCO [complete set of commuting operators] iff $\operatorname{dim}P^{(n)} = 1$ holds for all $\{k\}_n \in K_N.$

CSCOs, $P^{(n)}$, ${k}_n$ and $K_N$ have been defined in the text.

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