= Solution
Distinct <eigenvalues> of the self-adjoint <angular momentum operator> imply $\langle\psi_1,\psi_2\rangle=0$. The given combination is not normalized: its squared norm is $(\cos^2\omega t+3\sin^2\omega t)/4$, which is always positive. Therefore its properly normalized <expectation value> is
$$
\langle L_1\rangle_\psi=\frac{\langle\psi,L_1\psi\rangle}{\langle\psi,\psi\rangle}=\hbar\frac{\cos^2\omega t-3\sin^2\omega t}{\cos^2\omega t+3\sin^2\omega t}.
$$
It vanishes when $\tan^2\omega t=1/3$. The first positive solution has $\omega t=\pi/6$, so
$$
\boxed{t_0=\frac{\pi}{6\omega}.}
$$
Normalization changes the value of the <expectation value> away from zero, but not its zero times.
Back to article page
