(a) Following the same procedure that led to (5.73)
\hat{J}_{z}=\hbar \left(\begin{matrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{matrix} \right).
and (5.75),
\hat{J}_{x}=\frac{\hbar }{\sqrt{2} } \left(\begin{matrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right) , \hat{J}_{y}= \frac {\hbar }{\sqrt{2} } \left(\begin{matrix} 0 & -i & 0 \\ i & 0 & -i \\ 0 & i & 0 \end{matrix} \right).
we can verify that for s=\frac{3}{2} we have
S_{z} =\frac{\hbar }{2} \left(\begin{matrix} 3 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -3 \end{matrix} \right), (5.314)
\hat{S}_{-} =\hbar \left(\begin{matrix} 0 & 0 & 0 & 0 \\ \sqrt {3} & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & \sqrt{3} & 0 \end{matrix} \right), \hat{S}_{+} =\hbar \left(\begin{matrix} 0 & \sqrt{3} & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & \sqrt{3} \\ 0 & 0 & 0 & 0 \end{matrix} \right), (5.315)
which, when combined with \hat{S}_{x}=(\hat{S}_{+}+\hat {S}_{-})/2 and \hat{S}_{y}=i(\hat{S}_{-}-\hat{S}_{+}) /2, lead to
\hat{S}_{x}=\frac{\hbar }{2} \left(\begin{matrix} 0 & \sqrt{3} & 0 & 0 \\ \sqrt{3} & 0 & 2 & 0 \\ 0 & 2 & 0 & \sqrt{3} \\ 0 & 0 & \sqrt{3} & 0 \end{matrix} \right) , \hat{S}_{y}=\frac{i\hbar }{2} \left(\begin{matrix} 0 & -\sqrt{3} & 0 & 0 \\ \sqrt{3} & 0 & -2 & 0 \\ 0 & 2 & 0 & -\sqrt{3} \\ 0 & 0 & \sqrt{3} & 0 \end{matrix} \right). (5.316)
Thus, we have
\hat{S}^{2}_{x} =\frac{\hbar ^{2} }{4} \left(\begin{matrix} 3 & 0 & 2\sqrt{3} & 0 \\ 0 & 7 & 0 & 2\sqrt{3} \\ 2\sqrt{3} & 0 & 7 & 0 \\ 0 & 2\sqrt{3} & 0 & 3 \end{matrix} \right),\hat{S}^{2}_{y} =\frac{\hbar ^{2} }{4} \left(\begin{matrix} 3 & 0 & -2\sqrt{3} & 0 \\ 0 & 7 & 0 & -2\sqrt{3} \\ -2\sqrt{3} & 0 & 7 & 0 \\ 0 & -2\sqrt{3} & 0 & 3 \end {matrix} \right). (5.317)
(b) The Hamiltonian is then given by
H=\frac{\varepsilon _{0} }{\hbar ^{2} } (\hat{S}^{2}_{x}-\hat{S}^{2}_{y})-\frac{\varepsilon _{0} }{\hbar}\hat{S}_{z}=\frac{1}{2} \varepsilon _{0}\left(\begin{matrix} -3 & 0 & 2\sqrt{3} & 0 \\ 0 & -1 & 0 & 2\sqrt{3} \\ 2\sqrt{3} & 0 & 1 & 0 \\ 0 & 2\sqrt{3} & 0 & 3 \end{matrix} \right). (5.318)
The diagonalization of this Hamiltonian yields the following energy values:
E_{1} =-\frac{5}{2} \varepsilon _{0}, E_{2} =-\frac{3}{2} \varepsilon _{0}, E_{3} =\frac{3}{2} \varepsilon _{0}, E_{4} =\frac{5}{2} \varepsilon _{0}. (5.319)
The corresponding normalized eigenvectors are given by
|1 〉=\frac{1}{2} \left(\begin{matrix} -\sqrt{3} \\ 0 \\ 1 \\ 0 \end{matrix} \right) , |2 〉=\frac{1}{2} \left(\begin{matrix} 0 \\ -\sqrt{3} \\ 0 \\ 1 \end{matrix} \right), |3 〉=\frac{1}{\sqrt{12} }\left(\begin{matrix} \sqrt{3} \\ 0 \\ 3 \\ 0 \end{matrix} \right), |4 〉=\frac{1}{2} \left(\begin{matrix} 0 \\ 1 \\ 0 \\ \sqrt{3} \end{matrix} \right). (5.320)
None of the energy levels is degenerate.
(c) Since the initial state |\psi _{0} 〉 can be written in terms of the eigenvectors (5.320) as follows:
|\psi _{0} 〉=\left(\begin{matrix} 1 \\ 0 \\ 0 \\ 0 \end{matrix} \right)=-\frac{\sqrt{3}}{2} |1 〉+\frac{1}{2}|3 〉; (5.321)
the eigenfunction at a later time t is given by
|\psi (t) 〉=-\frac{\sqrt{3}}{2} |1 〉e^{-iE_{1}t/\hbar } +\frac{1}{2}|3 〉e^{-iE_{3}t/\hbar }
=-\frac{\sqrt{3}}{4}\left(\begin{matrix} -\sqrt{3} \\ 0 \\ 1 \\ 0 \end{matrix} \right)\exp \left[\frac{5i\varepsilon _{0}t}{2\hbar } \right] +\frac{1}{2\sqrt{12} }\left(\begin{matrix} \sqrt{3} \\ 0 \\ 3 \\ 0 \end{matrix} \right)\exp \left[-\frac{3i\varepsilon _{0}t}{2\hbar } \right] . (5.322)