= Solution
Choose the two real <energy eigenstates> in part (a) and take their coefficients to be in phase initially. A different initial relative phase only shifts the oscillation in time. The normalized <quantum superposition> is
$$
\Psi(x,t)=\frac{\psi_1(x)e^{-iE_1t/\hbar}+\psi_2(x)e^{-iE_2t/\hbar}}{\sqrt2}.
$$
<Orthogonality> removes the cross terms in the <energy> <expectation value>, so
$$
\boxed{\langle E\rangle=\frac{E_1+E_2}{2}=\frac{5\pi^2\hbar^2}{4mL^2}.}
$$
For <position>, the diagonal <matrix elements> equal $L/2$ by reflection symmetry, but the off-diagonal element is nonzero. Using the product-to-sum identity and $\int_0^Lx\cos(j\pi x/L)\,dx=L^2[(-1)^j-1]/(j^2\pi^2)$ gives
$$
x_{12}=\frac2L\int_0^Lx\sin\frac{\pi x}L\sin\frac{2\pi x}L\,dx=-\frac{16L}{9\pi^2}.
$$
Therefore the <equal superposition of the first two box states> has
$$
\boxed{\langle x\rangle=\frac L2-\frac{16L}{9\pi^2}\cos\Omega t,\quad \Omega=\frac{E_2-E_1}{\hbar}=\frac{3\pi^2\hbar}{2mL^2}.}
$$
The <amplitude> is $\boxed{16L/(9\pi^2)\simeq0.1801L}$, and the ordinary <frequency> is $\boxed{f=\Omega/(2\pi)=3h/(8mL^2)}$. The motion of the mean <position> is an interference effect between distinct energies, while their populations, and hence the mean <energy>, remain fixed.
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