(a) The measurements of A, B, C and D yield a_{1}=-1,a_{2}=3, a_{3}=5, b_{1}=-3,b_{2}=1,b_{3}=3, c_{1}=-1/\sqrt{2},c_{2}=0 ,c_{3} =1/\sqrt{2}, d_{1}=-1,d_{2}=d_{3}=1; the respective eigenvectors
of A, B, C and D are
|a_{1}〉=\frac{1}{\sqrt{2}}\left(\begin{matrix} 0 \\ -1 \\ 1 \end{matrix} \right), |a_{2}〉=\frac{1}{\sqrt{2}}\left(\begin{matrix} 0 \\ 1 \\ 1 \end{matrix} \right), |a_{3}〉=\left(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right), (3.212)
|b_{1}〉=\frac{1}{\sqrt{2}}\left(\begin{matrix} 0 \\ -1 \\ 1 \end{matrix} \right), |b_{2}〉=\left(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right), |b_{3}〉=\frac{1}{\sqrt{2}}\left(\begin{matrix} 0 \\ 1 \\ 1 \end{matrix} \right), (3.213)
|c_{1}〉=\frac{1}{\sqrt{26}}\left(\begin{matrix} 3 \\ -\sqrt{13} \\ 2 \end{matrix} \right), |c_{2}〉=\frac{1}{\sqrt{13}}\left(\begin {matrix} 2 \\ 0 \\ -3 \end{matrix} \right), |c_{3}〉=\frac{1}{\sqrt {26}}\left(\begin{matrix} 3 \\ \sqrt{13} \\ 2 \end{matrix} \right), (3.214)
|d_{1}〉=\frac{1}{\sqrt{2}}\left(\begin{matrix} 0 \\ i \\ 1 \end {matrix} \right), |d_{2}〉=\left(\begin{matrix} 1 \\ 0 \\ 0 \end {matrix} \right), |d_{3}〉=\frac{1}{\sqrt{2}}\left(\begin{matrix} 0 \\ 1 \\ i \end{matrix} \right). (3.215)
(b)We can verify that, among the observables A, B, C, and D, only A and B are compatible, since the matrices A and B commute; the rest do not commute with one another (neither A nor B commutes with C or D; C and D do not commute).
From (3.212) and (3.213) we see that the three states |a_{1}, b_{1}〉,|a_{2},b_{3}〉,|a_{3},b_{2}〉,
|a_{1},b_{1}〉=\frac{1}{\sqrt{2}}\left(\begin{matrix} 0 \\ -1 \\ 1 \end{matrix} \right), |a_{2},b_{3}〉=\frac{1}{\sqrt{2}}\left(\begin {matrix} 0 \\ 1 \\ 1 \end{matrix} \right), |a_{3},b_{2}〉=\left(\begin {matrix} 1 \\ 0 \\ 0 \end{matrix} \right), (3.216)
form a common, complete basis for A and B, since \hat{A}|a_{n} ,b_{m}〉=a_{n}|a_{n},b_{m}〉 and \hat{B}|a_{n},b_{m}〉=b_{m}|a_{n},b_{m}〉.
(c) First, since the eigenvalues of the operators \left\{\hat{A} \right\} , \left\{\hat{B}\right\}, and \left\{\hat {C} \right\} are all nondegenerate, each one of \left\{\hat{A} \right\} , \left\{\hat{B}\right\}, and \left\{\hat {C} \right\} forms separately a CSCO. Additionally, since two eigenvalues of \left\{\hat {D} \right\} are degenerate (d_{2}=d_{3}=1), the operator \left\{\hat {D} \right\} does not form a CSCO.
Now, among the various combinations \left\{\hat{A} ,\hat{B} \right\} , \left\{\hat{A}\hat{C}\right\} , \left\{\hat{B}\hat{C}\right\} , \left\{\hat{A}\hat{D}\right\}, and \left\{\hat{A}\hat {B} \hat{C}\right\}, only \left\{\hat{A} ,\hat{B} \right\} forms a CSCO, because \left\{\hat{A}\right\} and \left\{\hat{B}\right\} are the only operators that commute; the set of their joint eigenvectors are given by |a_{1}, b_{1}〉,|a_{2},b_{3}〉,|a_{3},b_{2}〉.