Equal superposition of the first two box states

For an infinite square well on $[0,L]$, choose ${\varphi }_n=\sqrt{2/L}\sin (n\pi x/L)$ and $E_n=n^2{E}_1$, $E_1={\pi }^2{\hslash }^2/(2m{L}^2)$. The state $\psi =({\varphi }_1{e}^{-i{E}_{1}t/\hslash }+{\varphi }_2{e}^{-i{E}_{2}t/\hslash })/\sqrt{2}$ has constant expectation value $⟨E⟩=5E_1/2$, butIndeed, the diagonal position matrix elements are $L/2$, whereas the off-diagonal element isA different initial relative phase shifts the oscillation in time. The oscillation comes from quantum interference, while the constant mean energy follows from the fixed populations of the two energy eigenstates.

For the equal superposition of the first two box states with zero initial relative quantum phase, put $\Omega =(E_2-E_1)/\hslash$. The probability in the left half is $P_{\mathrm{L}}=1/2+4\cos (\Omega t)/(3\pi )$. The continuity equation and zero wall probability current implyDirect substitution of $\psi (L/2,t)=L^{-1/2}{e}^{-i{E}_{1}t/\hslash }$ and ${\partial }_x\psi (L/2,t)=-2\pi L^{-3/2}{e}^{-i{E}_{2}t/\hslash }$ into the probability current verifies the sign. Positive current means probability leaves the left half. The quantum interference controls transport even though each component has stationary probability density.