(a) According to Postulate 2 of Chapter 3, the results of the measurements are given by the eigenvalues of the measured quantity. Here the eigenvalues of \hat{J}_{x}, which are obtained by diagonalizing the matrix J_{x} are j_{x}=-\hbar ,0, and \hbar; the respective (normalized) eigenstates are
|-1〉=\frac{1}{2}\left(\begin{matrix} -1 \\ \sqrt{2} \\ -1 \end {matrix} \right) , |0〉=\frac{1}{\sqrt{2} } \left(\begin{matrix} -1 \\ 0 \\ 1 \end{matrix} \right), |1〉=\frac{1}{2}\left(\begin{matrix} 1 \\ \sqrt{2} \\ 1 \end{matrix} \right). (5.260)
(b) If the system is in the state j_{x}=-\hbar, its eigenstate is given by |-1〉 . In this case 〈\hat{J}_{z} 〉 and 〈\hat{J}^{2}_{z}〉 are given by
〈-1|\hat{J}_{z}|-1 〉=\frac{\hbar }{4} \left(\begin{matrix} -1 & \sqrt{2} & -1\end{matrix} \right) \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{matrix} \right)\left(\begin{matrix} -1 \\ \sqrt{2} \\ -1 \end{matrix} \right)=0, (5.261)
〈-1|\hat{J}^{2}_{z}|-1 〉=\frac{\hbar ^{2} }{4} \left(\begin {matrix} -1 & \sqrt{2} & -1\end{matrix} \right) \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix} \right)\left(\begin {matrix} -1 \\ \sqrt{2} \\ -1 \end{matrix} \right)=\frac{\hbar ^{2} }{2}. (5.262)
Thus, the uncertainty \Delta J_{z} is given by
\Delta J_{z}=\sqrt{〈-1|\hat{J}^{2}_{z}|-1 〉-〈-1|\hat{J}_{z}|-1 〉^{2} } =\sqrt{\frac{\hbar ^{2} }{2}} =\frac{\hbar }{\sqrt{2} } . (5.263)
(c) Following the same procedure in (b), we have
〈-1|\hat{J}_{y}|-1 〉=\frac{\hbar }{4\sqrt{2} } \left(\begin {matrix} -1 & \sqrt{2} & -1\end{matrix} \right)\left(\begin{matrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{matrix} \right)\left(\begin{matrix} -1 \\ \sqrt{2} \\ -1 \end{matrix} \right)=0, (5.264)
〈-1|\hat{J}^{2}_{y}|-1 〉=\frac{\hbar ^{2} }{8}\left(\begin {matrix} -1 & \sqrt{2} & -1\end{matrix} \right)\left(\begin{matrix} 1 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 1 \end{matrix} \right)\left(\begin {matrix} -1 \\ \sqrt{2} \\ -1 \end{matrix} \right)= \frac{\hbar ^{2} }{2}; (5.265)
hence
\Delta J_{y}=\sqrt{〈-1|\hat{J}^{2}_{y}|-1 〉-〈-1|\hat{J}_{y}|-1 〉^{2} } =\frac{\hbar }{\sqrt{2} } . (5.266)
(d) We can express |\psi 〉=\frac{1}{\sqrt{14} } \left(\begin {matrix} -\sqrt{3} \\ 2\sqrt{2} \\ \sqrt{3} \end{matrix} \right) in terms of the eigenstates (5.260) as
\frac{1}{\sqrt{14} }\left(\begin{matrix} -\sqrt{3} \\ 2\sqrt{2} \\ \sqrt{3} \end{matrix} \right) =\sqrt{\frac{2}{7} }\frac{1}{2} \left (\begin{matrix} -1 \\ \sqrt{2} \\ -1 \end{matrix} \right)+\sqrt{\frac{3}{7} } \frac{1}{\sqrt{2} } \left(\begin{matrix} -1 \\ 0 \\ 1 \end{matrix} \right)+\sqrt{\frac{2}{7} }\frac{1}{2}\left(\begin{matrix} 1 \\ \sqrt{2} \\ 1 \end{matrix} \right) (5.267)
or
|\psi 〉=\sqrt{\frac{2}{7} }|-1〉+\sqrt{\frac{3}{7} }|0〉+\sqrt {\frac{2}{7} }|1〉. (5.268)
A measurement of \hat{J}_{x} on a system initially in the state (5.268) yields a value J_{x}=-\hbar with probability
P_{-1} =\left|〈-1|\psi 〉\right| ^{2} =\left|\sqrt{\frac{2}{7} }〈-1|-1〉+\sqrt{\frac{3}{7}}〈-1|0〉+\sqrt{\frac{2}{7} }〈-1|1〉 \right| ^{2} =\frac{2}{7} , (5.269)
since 〈-1|0〉=〈-1|1〉=0 and 〈-1|-1〉=1, and the values J_{x}=0 and J_{x}=\hbar with the
respective probabilities
P_{0} =\left|〈0|\psi 〉\right| ^{2} =\left|\sqrt{\frac{3}{7}}〈0|0〉\right| ^{2} =\frac{3}{7}, P_{1} =\left|〈1|\psi 〉\right| ^{2} =\left|\sqrt{\frac{2}{7}}〈1|1〉\right| ^{2} =\frac{2}{7}. (5.270)