Quantum Mechanical Observables
Complete Sets of Commuting Observables
\(A|\varphi_n\rangle = a_n|\varphi_n\rangle\)
For non-degenerate eigenstates, \(a_n\) relates to one \(|\varphi_n\rangle\). Then \(A\) is a Complete Set of Commuting Observables (CSCO).
For degenerate eigenstates, some \(a_n\) relates to some set of states, \(\{|\varphi_n^i\rangle\} \in\mathcal{E}_n\)
\(L_z^2 \doteq \begin{pmatrix}1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}\), \(|L_z^2=1,1\rangle,|L_z^2=1,2\rangle\). Then \(\{L_z^2\}\) is not a CSCO.
If \(A\) is not a CSCO, look for \(B\) such that \([A,B]=0\).
Example
Given \(A\Rightarrow a_1=0,a_{2,3}=1\). \(|a_1\rangle = (1 0 0)^T,|a_2\rangle = (0 1 0)^T,|a_3\rangle=(0 0 1)^T\).
Let \([A,B]=0\) with \(b_1=0,b_2=1,b_3=-1\), \(|b_1\rangle = (1 0 0)^T,|b_2\rangle=(0 1 0)^T,|b_3\rangle=(0 0 1)^T\).
Then, \(BA|\psi\rangle = AB|\psi\rangle = \sum_i c_ia_ib_i|b_i\rangle\). We can also then write, \(|a_2\rangle = |a=1,b=1\rangle\) and \(|a_3\rangle=|a=1,b=-1\rangle\).
Alternatively, if we had \(b_1=1,b_2=1,b_3=0\), \(|b_1\rangle = (1 0 0)^T,|b_2\rangle=(0 1 0)^T,|b_3\rangle=(0 0 1)^T\). Then, \(BA|\psi\rangle = AB|\psi\rangle = \sum_i c_ia_ib_i|b_i\rangle\). We can also then write, \(|a_1\rangle = |a=0,b=1\rangle,|a_2\rangle = |a=1,b=1\rangle\) and \(|a_3\rangle=|a=1,b=0\rangle\).