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I am having trouble identifying CSCOs, and I would like to make sure what I have understood is correct. I do understand the first requirement

All observables must commute in pairs,

but what I do not understand clearly is the following statement:

If we specify the eigenvalues of all the operators in the CSCO, we identify a unique eigenvector (up to a phase) in the Hilbert space of the system.

I would give some examples to try to explain what I understand. (Suppose that all observables commute in pairs).

  1. If an observable, $\hat{A}$, has no degeneracies, it forms a CSCO itself and with any other observables that satisfy the first requirement, independent of their degeneracies. (In other words, one observable with no degeneracies and the first requirement gives a CSCO automatically.)

  2. If all observables in the set have degenerated eigenvalues, we must write these observables in their common eigenbases and identify combinations of the eigenvalues. For example, if $$ \hat{A} = \begin{pmatrix} 1 & 0 & 0 \\ 0 &-1 & 0 \\ 0 & 0 & -1 \end{pmatrix} $$ and $$ \hat{B} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} $$ $\{ \hat{A}, \hat{B} \}$ form a CSCO since the eigenvectors are specified by the pairs of eigenvalues $[1,1],[1,-1],[-1,-1]$ which are unique, but $\{ \mathbb{I}, \hat{B} \}$ does not form a CSCO since the pairs of eigenvalues are $[1,1],[1,1],[1,-1]$ and the first two repeat.

Are these two statements correct?

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The whole point here is to uniquely label the states, using eigenvalues, pairs of eigenvalues, triples of eigenvalues etc.

For your operator $\hat A$, its eigenvalues are not enough to uniquely label the eigenstates $\vert 2\rangle$ and $\vert 3\rangle$ with the eigenvalue $-1$ since both of these are eigenkets with the same eigenvalue. Hence you need another operator - like $\hat B$ - where the eigenvalues of $\hat B$ on the eigenkets $\vert 2\rangle$ and $\vert 3\rangle$ are different. Note that you could have $B_{11}=-1$ if you wanted: as long as $\hat B$ has distinct eigenvalue on the $\vert 2\rangle$ and $\vert 3\rangle$, that wouldn't change the fact that the common eigenkets of $\hat A$ and $\hat B$ would be uniquely labelled by pairs of eigenvalues $\{\lambda_A,\lambda_B\}$.

Note that $\hat{\mathbb{I}}$ doesn't help has it has the same eigenvalue in the subspace spanned by $\vert 2\rangle$ and $\vert 3\rangle$.

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