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The boundary force in an infinite square well is obtained from the kinetic momentum flux at each wall. With $C_1={\hslash }^2{\pi }^2/(2m{L}^3)=E_1/L$, the derivatives of the state in part (c) areThe outward force magnitudes are $P_j=({\hslash }^2/2m)|{\Psi }^{\prime }(x_j,t){|}^2$. Taking positive force to mean rightwards, the forces exerted by the particle on the walls areBoth 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: $d⟨p⟩/dt=({\hslash }^2/2m)(|{\Psi }^{\prime }(0){|}^2-|{\Psi }^{\prime }(L){|}^2)$. As a check, the net force on the particle is $8C_1\cos \Omega t=m{d}^2⟨x⟩/d{t}^2$, the opposite of the net force on the two walls.

A classical particle of energy $E=⟨E⟩$ 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 isThus 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 $⟨x⟩$.

Write $k=\pi /a$, $A=\sqrt{2/a}$ and ${\Psi }_0=A{e}^{ikx}\sin kx$. Its derivative isIntegration by parts with the Dirichlet boundary conditions gives the energy expectationIt is constant in time, since the time-independent Hamiltonian operator has fixed energy measurement probabilities: $⟨H{⟩}_t={\sum }_n{E}_n|c_n{|}^2$.

For the midpoint parity operator, $R{\Psi }_0=-A{e}^{-ikx}\sin kx$. Hence, again putting $u=kx$,Indeed the imaginary part integrates to zero by reflection and ${\int }_0^{\pi }\cos 2u{\sin }^{2}udu=-\pi /4$. This expectation value is also constant in time, because the bounded self-adjoint operator $R$ commutes with $H$ and hence with its time-evolution operator. Equivalently, $⟨R{⟩}_t={\sum }_n(-1{)}^{n+1}|c_n{|}^2$.

For the momentum operator, direct evaluation givesThe real sine-cosine integral vanishes. This expectation value is not constant in time: the infinite square well walls reverse motion. One can prove the change exactly without approximating the wavefunction. Atthe energy eigenstate expansion and $R{\psi }_n=(-1{)}^{n+1}{\psi }_n$ give $\Psi (t_{\ast })=-R{\Psi }_0$. Differentiation shows $PR=-RP$, soThis half revival in an infinite square well is a concrete counterexample to momentum conservation. The sign from $-R$ is only a global phase.

The apparent interior identity $[H,P]=0$ cannot justify conservation here: the momentum operator does not preserve the Dirichlet boundary conditions of the Hamiltonian operator's operator domain. For sufficiently regular states, retaining the boundary terms in Ehrenfest theorem instead gives the boundary force in an infinite square well,In the strictly confined interval, $-i\hslash {\partial }_x$ with both endpoint values zero is symmetric but not itself a self-adjoint operator. The requested integral is the conventional mean momentum; it also equals the full-line momentum mean of the zero-extended state, as measured upon releasing the walls. This domain distinction explains why the formal interior commutator misses the wall force.