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For a normalized sufficiently regular solution of the Schrödinger equation on $[0,L]$ with Dirichlet walls, the instantaneous outward forces on the walls have magnitudesA wall displacement changes the kinetic-energy quadratic form by the boundary term $-P_{j}d{L}_j$, giving the local force. Equivalently, direct integration by parts in the time derivative of the mean particle momentum givesThe force on the particle is the opposite of the net force $-P_0+P_L$ on the walls. For a stationary energy eigenstate each magnitude equals $2E_n/L$. For the equal superposition of the first two box states, writing $C=E_1/L$ and $\Omega =3E_1/\hslash$ gives $P_0=C(5+4\cos \Omega t)$ and $P_L=C(5-4\cos \Omega t)$. Both time averages equal $2⟨E⟩/L$, the classical mean force on one wall at that energy, even though the quantum forces vary continuously in time.

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 isOrthogonality removes the cross terms in the energy expectation value, soFor 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 ${\int }_0^{L}x\cos (j\pi x/L)dx=L^2[(-1{)}^j-1]/(j^2{\pi }^2)$ givesTherefore the equal superposition of the first two box states hasThe amplitude is $\boxed{16L/(9{\pi }^2)\simeq 0.1801L}$, and the ordinary frequency is $\boxed{f=\Omega /(2\pi )=3h/(8m{L}^2)}$. The motion of the mean position is an interference effect between distinct energies, while their populations, and hence the mean energy, remain fixed.