Consider a physical system whose Hamiltonian H and initial state |\psi _{0}〉 are given by
H=\varepsilon \left(\begin{matrix} 0 & i & 0 \\ -i & 0 & 0 \\ 0 & 0 & -1 \end{matrix} \right), |\psi _{0}〉=\frac{1}{\sqrt{5}}\left(\begin{matrix} 1-i \\ 1-i \\ 1 \end{matrix} \right),
where ε has the dimensions of energy.
(a) What values will we obtain when measuring the energy and with what probabilities?
(b) Calculate 〈\hat{H}〉, the expectation value of the Hamiltonian.
(a) The results of the energy measurement are given by the eigenvalues of H. A diagonalization of H yields a nondegenerate eigenenergy E_{1}=\varepsilon and a doubly degenerate value E_{2} =E_{3}=-\varepsilon whose respective eigenvectors are given by
|\phi _{1}〉=\frac{1}{\sqrt{2}}\left(\begin{matrix} 1 \\ -i \\ 0 \end{matrix} \right), |\phi _{2}〉=\frac{1}{\sqrt{2}} \left(\begin{matrix} -i \\ 1 \\ 0 \end{matrix} \right), |\phi _{3}〉=\left(\begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right); (3.177)
these eigenvectors are orthogonal since H is Hermitian. Note that the initial state |\psi _{0}〉 can be written in terms of |\phi _{1}〉,|\phi _{2}〉, and |\phi _{3}〉 as follows:
|\psi _{0}〉=\frac{1}{\sqrt{5}}\left(\begin{matrix} 1-i \\ 1-i \\ 1 \end{matrix} \right)=\sqrt{\frac{2}{5}}|\phi _{1}〉+\sqrt {\frac{2}{5}}|\phi _{2}〉+\frac{1}{\sqrt{5}}|\phi _{3}〉. (3.178)
Since |\phi _{1}〉,|\phi _{2}〉, and |\phi _{3}〉 are orthonormal, the probability of measuring E_{1}= \varepsilon is given by
P_{1}(E_{1})=\left|〈\phi _{1}|\psi _{0}〉\right| ^{2}=\left |\sqrt{\frac{2}{5}}〈\phi _{1}|\phi _{1}〉\right| ^{2}=\frac{2}{5} . (3.179)
Now, since the other eigenvalue is doubly degenerate, E_{2} =E_{3}=-\varepsilon, the probability of measuring -ε can be obtained from (3.3):
P_{2}(E_{2})=\left|〈\phi _{2}|\psi _{0}〉\right| ^{2}=\left|〈\phi _{3}|\psi _{0}〉\right| ^{2}=\frac{2}{5}+\frac{1}{5}=\frac{3}{5}. (3.180)
(b) From (3.179) and (3.180), we have
〈\hat{H}〉=P_{1}E_{1}+P_{2}E_{2}=\frac{2}{5}\varepsilon -\frac{3}{5}\varepsilon =-\frac{1}{5}\varepsilon. (3.181)
We can obtain the same result by calculating the expectation value of \hat{H} with respect to |\psi _{0}〉.
Since 〈\psi _{0}|\psi _{0}〉=1, we have 〈\hat{H}〉=〈\psi _{0}|\hat{H}|\psi _{0}〉/〈\psi _{0}|\psi _{0}〉=〈\psi _{0}|\hat {H}|\psi _{0}〉:
〈\hat{H}〉=〈\psi _{0}|\hat{H}|\psi _{0}〉=\frac{\varepsilon }{5}\left(\begin{matrix} 1+i & 1+i & 1\end{matrix} \right)\left(\begin {matrix} 0 & i & 0 \\ -i & 0 & 0 \\ 0 & 0 & -1 \end{matrix} \right)\left (\begin{matrix} 1-i \\ 1-i \\ 1 \end{matrix} \right)=-\frac{1}{5} \varepsilon. (3.182)