 Half revival in an infinite square well

For the infinite square well of length $a$, the energy eigenvalues are $E_n=n^2{E}_1$, with $E_1={\pi }^2{\hslash }^2/(2m{a}^2)$. At $t_{\ast }=\pi \hslash /E_1$, the dynamical phase is $e^{-i{E}_n{t}_{\ast }/\hslash }=(-1{)}^n$. Since the midpoint parity operator has eigenvalues $(-1{)}^{n+1}$ in the same complete orthonormal basis, the time-evolution operator is exactly $U(t_{\ast })=-R_a$ on every state. At $2t_{\ast }$ every phase is one, giving a full revival.

Thus a state becomes its reflected copy, up to a global phase, at the half revival. Its position probability density is reflected, while its mean momentum changes sign because $P{R}_a=-R_{a}P$. For an initially reflection-symmetric probability density, the initial and half-revival position distributions agree even when their probability currents point in opposite directions.