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Sakurai states that if we have a complete, maximal set of compatible observables, say A,B,C... Then, an eigenvector represented by |a,b,c....> , where a,b,c... are respective eigenvalues, is unique. Why is it so? Why can't there be two eigenvectors with same eigenvalues for each observable? Does maximality of the set has some role to play in it?

I asked this question on Physics SE and was not satisfied with answers. Hope that I get help here.

Yes, since it is the maximal set of compatible observables, it includes all observables for which $|a\rangle$, $|b\rangle$, $|c\rangle$, etc. are the eigenvectors (I'll use the notation $|\psi_1\rangle$, $|\psi_2\rangle$, $|\psi_3\rangle$ etc instead). Hence this includes the observable $D = \sum_k k |\psi_k\rangle \langle \psi_k|$ . However $D$ has a unique set of eigenvectors, and hence so does the any compatible set of observables which contains $D$.

You can and you do sometimes have degenerate eigenvalues.

This might just be a question of definitions: "a complete set of commuting observables (CSCO) is a set of commuting operators whose eigenvalues completely specify the state of a system (Gasiorowicz 1974, p. 119)." [1]

That said, if you give me some cooked up Hamiltonian with at least one degenerate eigenvalue, perhaps one might be able to prove no observable commutes with it.

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