= Solution
The <boundary force in an infinite square well> is obtained from the kinetic <momentum> flux at each wall. With $C_1=\hbar^2\pi^2/(2mL^3)=E_1/L$, the derivatives of the state in part (c) are
$$
\Psi'(0,t)=\frac{\pi}{L^{3/2}}(e^{-iE_1t/\hbar}+2e^{-iE_2t/\hbar}),\quad \Psi'(L,t)=\frac{\pi}{L^{3/2}}(-e^{-iE_1t/\hbar}+2e^{-iE_2t/\hbar}).
$$
The outward force magnitudes are $P_j=(\hbar^2/2m)|\Psi'(x_j,t)|^2$. Taking positive force to mean rightwards, the forces exerted by the particle on the walls are
$$
\boxed{F_{\mathrm{left}}=-C_1(5+4\cos\Omega t),\qquad F_{\mathrm{right}}=C_1(5-4\cos\Omega t).}
$$
Both point outwards at all times, but their magnitudes oscillate out of phase. This force formula follows directly by differentiating the mean particle <momentum> and integrating the interior <Schrödinger equation> by parts: $d\langle p\rangle/dt=(\hbar^2/2m)(|\Psi'(0)|^2-|\Psi'(L)|^2)$. As a check, the net force on the particle is $8C_1\cos\Omega t=m\,d^2\langle x\rangle/dt^2$, the opposite of the net force on the two walls.
A classical particle of <energy> $E=\langle E\rangle$ transfers <momentum> $2mv$ in each collision with a particular wall, and returns to that wall after $2L/v$. Its mean outward force on that wall is
$$
\boxed{\overline P_{\mathrm{classical}}=\frac{2mv}{2L/v}=\frac{2E}{L}=5C_1=\overline P_{\mathrm{quantum}}.}
$$
Thus the time averages agree separately at each wall, although the classical force consists of collision impulses and the quantum expectation varies continuously. Constant mean <energy> alone does not make the instantaneous quantum wall forces constant; their unequal instantaneous values supply the oscillation of $\langle x\rangle$.
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