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Aug 1, 2020 at 13:21 vote accept Mr. Palomar
Aug 1, 2020 at 13:12 comment added Mr. Palomar Thanks, that makes sense.
Aug 1, 2020 at 12:32 comment added Vercassivelaunos You can always find a basis of the Hilbert space made up of eigenstates of the Hamiltonian, and you can then just invent a fitting set of operators by specifying how they act on these eigenstates, so yes. Wether they are physically relevant is a different question, though.
Aug 1, 2020 at 11:21 comment added Mr. Palomar Thanks, I think I'm starting to understand better. Is this situation unique for the hydrogen atom? E.g. if we take an atom with more electrons, is there still a finite set of operators whose eigenvalues uniquely specify a basis of $\mathcal{H}$? (I am aware that the Schrödinger equation for atoms with more electrons cannot be solved in terms of exact formulas, but that doesn't exclude such a basis from existing, I presume.)
Aug 1, 2020 at 11:05 history edited Vercassivelaunos CC BY-SA 4.0
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Aug 1, 2020 at 10:35 history answered Vercassivelaunos CC BY-SA 4.0