(a) A measurement of the energy yields the values E_{1}=-5 ,E_{2} =3,E_{3}=5; the respective (orthonormal) eigenvectors of these values are
|\phi _{1} 〉=\frac{1}{\sqrt{2}} \left(\begin{matrix} 0 \\ -1 \\ 1 \end{matrix} \right), |\phi _{2} 〉=\left(\begin{matrix} 1 \\ 0 \\ 0 \end {matrix} \right), |\phi _{3} 〉=\frac{1}{\sqrt{2}} \left(\begin {matrix} 0 \\ 1 \\ 1 \end{matrix} \right). (3.217)
The probabilities of finding the values E_{1}=-5,E_{2} =3,E_{3} =5 are given by
P(E_{1})=\left|〈\phi _{1}|\psi (0) 〉\right| ^{2}= \left|\frac{1}{5 \sqrt{2}}\left(\begin{matrix} 0 & -1 & 1\end{matrix} \right) \left (\begin {matrix} 3 \\ 0 \\ 4 \end{matrix} \right) \right| ^{2} =\frac{8}{25}, (3.218)
P(E_{2})=\left|〈\phi _{2}|\psi (0) 〉\right| ^{2}= \left|\frac{1}{5}\left(\begin{matrix} 1 & 0 & 0\end{matrix} \right) \left(\begin {matrix} 3 \\ 0 \\ 4 \end{matrix} \right) \right| ^{2} =\frac{9}{25}, (3.219)
P(E_{3})=\left|〈\phi _{3}|\psi (0) 〉\right| ^{2}= \left|\frac{1}{5\sqrt{2}}\left(\begin{matrix} 0 & 1 & 1\end{matrix} \right) \left (\begin{matrix} 3 \\ 0 \\ 4 \end{matrix} \right) \right| ^{2} =\frac{8}{25}. (3.220)
(b) To find |\psi (t)〉 we need to expand |\psi (0)〉 in terms of the eigenvectors (3.217):
|\psi (0)〉=\frac{1}{5}\left(\begin{matrix} 3 \\ 0 \\ 4 \end {matrix} \right)=\frac{2\sqrt{2} }{5}|\phi _{1} 〉+\frac{3}{5}|\phi _{2} 〉+\frac{2\sqrt{2} }{5} |\phi _{3} 〉; (3.221)
hence
|\psi (t)〉=\frac{2\sqrt{2} }{5} e^{-iE_{1}t} |\phi _{1} 〉+\frac{3}{5}e^{-iE_{2}t} |\phi _{2} 〉+\frac{2\sqrt{2} }{5}e^{-iE_{3}t} |\phi _{3} 〉=\frac{1}{5}\left(\begin{matrix} 3e^{-3it} \\ -4i\sin 5t \\ 4\cos 5t \end{matrix} \right). (3.222)
(c)We can calculate the energy at time t = 0 in three quite different ways. The first method uses the bra-ket notation. Since 〈\psi (0)|\psi (0)〉=1,〈\phi _{n})|\phi _{m}〉=\delta _{nm} and since \hat{H}|\phi _{n}〉=E_{n}|\phi _{n}〉, we have
E(0)=〈\psi (0)|\hat{H}|\psi (0)〉=\frac{8}{25}〈\phi _{1}|\hat {H}|\phi _{1}〉+\frac{9}{25}〈\phi _{2}|\hat{H}|\phi _{2}〉+\frac{8}{25} 〈\phi _{3}|\hat{H}|\phi _{3}〉
=\frac{8}{25}(-5)+\frac{9}{25}(3)+\frac{8}{25}(5)=\frac{27}{25}. (3.223)
The second method uses matrix algebra:
E(0)=〈\psi (0)|\hat{H}|\psi (0)〉=\frac{1}{25}\left(\begin {matrix} 3 & 0 & 4 \end{matrix} \right)\left(\begin{matrix} 3 & 0 & 0 \\ 0 & 0 & 5 \\ 0 & 5 & 0 \end{matrix} \right)\left(\begin{matrix} 3 \\ 0 \\ 4 \end{matrix} \right)=\frac{27}{25}. (3.224)
The third method uses the probabilities:
E(0)=\sum\limits_{n=1}^{2}P(E_{n})E_{n}=\frac{8}{25}(-5)+\frac{9}{25}(3)+\frac{8}{25}(5)=\frac{27}{25}. (3.225)
The energy at a time t is
E(t)=〈\psi (t)|\hat{H}|\psi (t)〉=\frac{8}{25}e^{iE_{1}t}e^{-iE_{1}t}〈\phi _{1}|\hat{H}|\phi _{1}〉+\frac{9}{25}e^{iE_{2}t} e^{-iE_{2}t}〈\phi _{2}|\hat{H}|\phi _{2}〉
+\frac{8}{25}e^{iE_{3}t}e^{-iE_{3}t} 〈\phi _{3}|\hat{H}|\phi _{3}〉=\frac{8}{25}(-5)+\frac{9}{25}(3)+\frac{8}{25}(5)=\frac{27}{25} =E(0). (2.226)
As expected, E(t)=E(0) since d\left\{\hat{H} \right\}/dt=0.
(d) Since none of the eigenvalues of \hat{H} is degenerate, the eigenvectors |\phi _{1} 〉,|\phi _{2} 〉,|\phi _{3} 〉 form a compete (orthonormal) basis. Thus \left\{\hat {H}\right\} forms a complete set of commuting operators.