(a) A measurement of A yields any of the eigenvalues of A which are given by a_{1}=-1,a_{2}=0,a_{3}=1; the respective (normalized) eigenstates are
|a_{1}〉=\frac{1}{2}\left(\begin{matrix} -1 \\ \sqrt{2} \\ -1 \end{matrix} \right) , |a_{2}〉=\frac{1}{\sqrt{2}}\left(\begin{matrix} -1 \\ 0 \\ 1 \end{matrix} \right) , |a_{3}〉=\frac{1}{2}\left(\begin{matrix} 1 \\ \sqrt{2} \\ 1 \end{matrix} \right) . (3.190)
The probability of obtaining a_{1}=-1 is
P(-1)=\frac{\left|〈a_{1}|\psi(t)〉\right| ^{2}}{〈\psi(t)|\psi(t)〉} = \frac{1}{6}\left|\frac{1}{2}\left(\begin{matrix} -1 & \sqrt{2} & -1\end {matrix} \right)\left(\begin{matrix} -1 \\ 2 \\ 1 \end{matrix} \right) \right| ^{2}=\frac{1}{3}. (3.191)
where we have used the fact that 〈\psi(t)|\psi(t)〉=\left(\begin {matrix} -1 & 2 & 1\end{matrix} \right)\left(\begin{matrix} -1 \\ 2 \\ 1 \end{matrix} \right)=6.
(b) A measurement of B yields a value which is equal to any of the eigenvalues of B: b_{1}=-1,b_{2}=0, and b_{3}=1 ; their corresponding eigenvectors are
|b_{1}〉=\left(\begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right), |b_{2}〉=\left(\begin{matrix} 0 \\ 1 \\ 0 \end {matrix} \right), |b_{3}〉=\left(\begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right). (3.192)
Since the system was in the state |\psi(t)〉, the probability of obtaining the value b_{2}=0 for B is
P(b_{2})=\frac{\left|〈b_{2}|\psi(t)〉\right| ^{2}}{〈\psi(t)| \psi(t)〉}=\frac{1}{6}\left|\frac{1}{2}\left(\begin{matrix} 0 & 1 & 0\end{matrix} \right)\left(\begin{matrix} -1 \\ 2 \\ 1 \end{matrix} \right) \right| ^{2}=\frac{2}{3}. (3.193)
We deal now with the measurement of the other observable, A. The observables A and B do not have common eigenstates, since they do not commute. After measuring B (the result is b_{2}=0), the system is left, according to Postulate 3, in a state |\phi〉 which can be found by projecting |\psi(t)〉 onto |b_{2}〉:
|\phi〉= |b_{2}〉〈b_{2}|\psi(t)〉=\left(\begin{matrix} 0 \\ 1 \\ 0 \end{matrix} \right)\left(\begin{matrix} 0 & 1 & 0\end{matrix} \right) \left(\begin{matrix} -1 \\ 2 \\ 1 \end{matrix} \right)=\left (\begin {matrix} 0 \\ 2 \\ 0 \end{matrix} \right). (3.194)
The probability of finding 1 when we measure A is given by
P(a_{3})=\frac{\left|〈a_{3}|\phi〉\right| ^{2} }{〈\phi |\phi〉}= \frac{1}{4}\left|\frac{1}{2}\left(\begin{matrix} 1 & \sqrt{2} & 1\end {matrix} \right)\left(\begin{matrix} 0 \\ 2 \\ 0 \end{matrix} \right) \right| ^{2}=\frac{1}{2}, (3.195)
since 〈\phi |\phi〉=4. In summary, when measuring B then A, the probability of finding a value of 0 for B and 1 for A is given by the product of the probabilities (3.193) and (3.195):
P(b_{2},a_{3})=P(b_{2})P(a_{3})=\frac{2}{3}\frac{1}{2}=\frac {1}{3}. (3.196)
(c) Next we measure A first then B. Since the system is in the state |\psi(t)〉, the probability of measuring a_{3}=1 for A is given by
P^{′}(a_{3})=\frac{\left|〈a_{3}|\psi(t)〉\right| ^{2}}{〈\psi(t) |\psi(t)〉}=\frac{1}{6}\left|\frac{1}{2}\left(\begin{matrix} 1 & \sqrt{2} & 1\end{matrix} \right)\left(\begin{matrix} -1 \\ 2 \\ 1 \end{matrix} \right) \right| ^{2}=\frac{1}{3}, (3.197)
where we have used the expression (3.190) for |a_{3}〉.
We then proceed to the measurement of B. The state of the system just after measuring A (with a value a_{3}=1 ) is given by a projection of |\psi(t)〉 onto |a_{3}〉:
|\chi 〉=|a_{3}〉〈a_{3}|\psi(t)〉=\frac{1}{4}\left(\begin{matrix} 1 \\ \sqrt{2} \\ 1 \end{matrix} \right)\left(\begin{matrix} 1 & \sqrt{2} & 1\end{matrix} \right)\left(\begin{matrix} -1 \\ 2 \\ 1 \end{matrix} \right)=\frac{\sqrt{2}}{2}\left(\begin{matrix} 1 \\ \sqrt{2} \\ 1 \end{matrix} \right). (3.198)
So the probability of finding a value of b_{2}=0 when measuring B is given by
P^{′}(b_{2})=\frac{\left|〈b_{2}|\chi〉\right| ^{2}}{〈\chi|\chi〉} =\frac{1}{2}\left|\frac{\sqrt{2}}{2}\left(\begin{matrix} 0 & 1 & 0\end {matrix} \right)\left(\begin{matrix} 1 \\ \sqrt{2} \\ 1 \end{matrix} \right) \right| ^{2}=\frac{1}{2}, (3.199)
Since 〈\chi|\chi〉=2
So when measuring A then B, the probability of finding a value of 1 for A and 0 for B is given by the product of the probabilities (3.199) and (3.197):
P(a_{3},b_{2})=P^{′}(a_{3})P^{′}(b_{2})=\frac{1}{3}\frac{1}{2}=\frac{1}{6}. (3.200)
(d) The probabilities P(b_{2},a_{3}) and P(a_{3}, b_{2}), as shown in (3.196) and (3.200), are different. This is expected, since A and B do not commute. The result of the successive measurements of A and B therefore depends on the order in which they are carried out. The probability of obtaining 0 for B then 1 for A is equal to \frac{1}{3}. On the other hand, the probability of obtaining 1 for A then 0 for B is equal to \frac{1}{6}. However, if the observables A and B commute, the result of the
measurements will not depend on the order in which they are carried out (this idea is illustrated in the following solved problem).
(e) As stated in the text, any operator with non-degenerate eigenvalues constitutes, all by itself, a CSCO. Hence each of \left\{\hat{A} \right\} and \left\{\hat{B}\right\} forms a CSCO, since their eigenvalues are not degenerate. However, the set \left\{\hat{A} ,\hat{B}\right\} does not form a CSCO since the opertators \left\{\hat{A} \right\} and \left\{\hat{B}\right\} do not commute.