= Solution
At time $t$ the state is
$$
\chi(t)=\frac12e^{-iE_1t/\hbar}\chi_1
+\frac{\sqrt3}{2}e^{-iE_2t/\hbar}\chi_2.
$$
Direct evaluation of the radial <integral> gives the <matrix element>[matrix elements]
$$
\langle\chi_1|r|\chi_1\rangle=\frac{3a}{2},
\qquad
\langle\chi_2|r|\chi_2\rangle=6a,
$$
and
$$
\begin{aligned}
\langle\chi_1|r|\chi_2\rangle
&=4\pi\int_0^\infty r^3\chi_1(r)\chi_2(r)\,dr\\
&=-\frac{64a}{81\sqrt2}.
\end{aligned}
$$
It follows that
$$
\boxed{
R(t)=\frac{39a}{8}
-\frac{32a}{81}\sqrt{\frac32}
\cos\!\left(\frac{E_2-E_1}{\hbar}t\right)
}.
$$
Thus $R(t)$ oscillates sinusoidally about $39a/8$. Its <angular frequency> is
$$
\boxed{\omega=\frac{E_2-E_1}{\hbar}
=\frac{3mK^2}{8\hbar^3}
=\frac{3K}{8a\hbar}}.
$$
Back to article page
