(a) A measurement of A yields any of the eigenvalues of A which are given by a_{1}=0 (not degenerate) and a_{2}= a_{3}=2 (doubly degenerate); the respective (normalized) eigenstates are
|a_{1}〉=\frac{1}{\sqrt{2}}\left(\begin{matrix} 0 \\ -i \\ 1 \end {matrix} \right), |a_{2}〉=\frac{1}{\sqrt{2}}\left (\begin {matrix} 0 \\ i \\ 1 \end{matrix} \right), |a_{3}〉=\left (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right). (3.201)
The probability that a measurement of A yields a_{1}=0 is given by
P(a_{1})=\frac{\left|〈a_{1}|\psi(t)〉\right| ^{2}}{〈\psi(t)| \psi(t)〉} =\frac{36}{17}\left|\frac{1}{\sqrt{2}}\frac{1}{6}\left(\begin {matrix} 0 & i & 1\end{matrix} \right) \left(\begin{matrix} 1 \\ 0 \\ 4 \end{matrix} \right) \right| ^{2}=\frac{8}{17}, (3.202)
where we have used the fact that 〈\psi(t)|\psi(t)〉=\frac{1}{36} \left (\begin{matrix} 1 & 0 & 4\end{matrix} \right) \left (\begin {matrix} 1 \\ 0 \\ 4 \end{matrix} \right)=\frac{17}{36}.
Since the system was initially in the state |\psi(t)〉, after a measurement of A yields a_{1}=0, the system is left, as mentioned in Postulate 3, in the following state:
|\phi〉= |a_{1}〉〈a_{1}|\psi(t)〉=\frac{1}{2}\frac{1}{6}\left (\begin{matrix} 0 \\ -i \\ 1 \end{matrix} \right)\left(\begin{matrix} 0 & i & 1\end{matrix} \right) \left(\begin{matrix} 1 \\ 0 \\ 4 \end {matrix} \right)=\frac{1}{3}\left(\begin{matrix} 0 \\ -i \\ 1 \end {matrix} \right). (3.203)
As for the measurement of B, we obtain any of the eigenvalues b_{1}=-1,b_{2}=b_{3}=1; their corresponding eigenvectors are
|b_{1}〉=\frac{1}{\sqrt{2} } \left(\begin{matrix} 0 \\ i \\ 1 \end{matrix} \right), |b_{2}〉=\frac{1}{\sqrt{2} }\left(\begin{matrix} 0 \\ -i \\ 1 \end{matrix} \right), |b_{3}〉=\left(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right). (3.204)
Since the system is now in the state |\phi〉, the probability of obtaining the (doubly degenerate) value b_{2}=b_{3} =1 for B is
P(b_{2})=\frac{\left|〈b_{2}|\phi〉\right| ^{2}}{〈\phi |\phi〉} +\frac{\left|〈b_{3}|\phi〉\right| ^{2}}{〈\phi|\phi〉}
=\frac{1}{2}\left|\frac{1}{\sqrt{2}}\left(\begin{matrix} 0 & i & 1\end{matrix} \right)\left(\begin{matrix} 0 \\ -i \\ 1 \end{matrix} \right) \right| ^{2}+\frac{1}{2}\left|\left(\begin{matrix} 1 & 0 & 0\end{matrix} \right)\left(\begin{matrix} 0 \\ -i \\ 1 \end{matrix} \right) \right| ^{2}
=1. (3.205)
The reason P(b_{2})=1 is because the new state |\phi〉 is an eigenstate of B; in fact |\phi〉=\sqrt{2}/3 |b_{2}〉.
In sum, when measuring A then B, the probability of finding a value of 0 for A and 1 for B is given by the product of the probabilities (3.202) and (3.205):
P(a_{1},b_{2})=P(a_{1})P(b_{2})=\frac{8}{17}. (3.206)
(b) Next we measure B first then A. Since the system is in the state |\psi(t)〉 and since the value b_{2}=b_{3} =1 is doubly degenerate, the probability of measuring 1 for B is given by
P^{′}(b_{2})=\frac{\left|〈b_{2}|\psi(t)〉\right| ^{2}}{〈\psi(t)| \psi(t)〉}+\frac{\left|〈b_{3}|\psi(t)〉\right| ^{2}}{〈\psi(t)|\psi(t)〉}
=\frac{36}{17}\frac{1}{36} \left[\left|\frac{1}{\sqrt{2}} \left (\begin{matrix} 0 & i & 1\end{matrix} \right)\left(\begin{matrix} 1 \\ 0 \\ 4 \end{matrix} \right) \right| ^{2}+\left|\left(\begin{matrix} 1 & 0 & 0\end{matrix} \right)\left(\begin{matrix} 1 \\ 0 \\ 4 \end{matrix} \right) \right| ^{2}\right]
=\frac{9}{17}. (3.207)
We now proceed to the measurement of A. The state of the system immediately after measuring B (with a value b_{2}=b_{3} =1 ) is given by a projection of |\psi(t)〉 onto |b_{2}〉 and |b_{3}〉
|\chi〉=|b_{2}〉〈b_{2}|\psi(t)〉+|b_{3}〉〈b_{3}|\psi(t)〉
=\frac{1}{12}\left(\begin{matrix} 0 \\ -i \\ 1 \end{matrix} \right)\left(\begin{matrix} 0 & i & 1\end{matrix} \right) \left(\begin {matrix} 1 \\ 0 \\ 4 \end{matrix} \right)+\frac{1}{6} \left(\begin {matrix} 1 \\ 0 \\ 0 \end{matrix} \right)\left(\begin{matrix} 1 & 0 & 0\end{matrix} \right) \left(\begin{matrix} 1 \\ 0 \\ 4 \end{matrix} \right)
=\frac{1}{6}\left(\begin{matrix} 1 \\ -2i \\ 2i \end{matrix} \right) (3.208)
So the probability of finding a value of a_{1}=0 when measuring A is given by
P^{′}(a_{1})=\frac{\left|〈a_{1}|\chi 〉\right| ^{2}}{〈\chi |\chi 〉}=\frac{36}{9}\left|\frac{1}{6\sqrt{2}}\left(\begin{matrix} 0 & i & 1\end{matrix} \right)\left(\begin{matrix} 1 \\ -2i \\ 2i \end{matrix} \right) \right| ^{2}=\frac{8}{9}, (3.209)
since 〈\chi |\chi 〉=\frac{9}{36}.
Therefore, when measuring B then A, the probability of finding a value of 1 for B and 0 for A is given by the product of the probabilities (3.207) and (3.209):
P(b_{2},a_{3})=P^{′}(b_{2})P^{′}(a_{1})=\frac{9}{17}\frac{8}{9}=\frac{8}{17}. (3.210)
(c) The probabilities P(a_{1},b_{2}) and P(b_{2}, a_{1}), as shown in (3.206) and (3.210), are equal. This is expected since A and B do commute. The result of the successive measurements of A and B does not depend on the order in which they are carried out. (d) Neither \left\{\hat{A}\right\} nor \left\{\hat{B}\right\} forms a CSCO since their eigenvalues are degenerate. The set \left\{\hat{A},\hat{B} \right\}, however, does form a CSCO since the opertators \left\{\hat{A}\right\} and \left\{\hat{B}\right\} commute. The set of eigenstates that are common to \left\{\hat{A},\hat{B} \right\} are given by
|a_{2},b_{1}〉=\frac{1}{\sqrt{2}} \left(\begin{matrix} 0 \\ i \\ 1 \end{matrix} \right), |a_{1},b_{2}〉=\frac{1}{\sqrt {2}} \left(\begin{matrix} 0 \\ -i \\ 1 \end{matrix} \right), |a_{3},b_{3}〉= \left(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right). (3.211)