 Solution

Put $\tau =t-T$ and retain the global phase from the preceding Lüders rule update. Since $P{\varphi }_n=(\hslash n/R){\varphi }_n$, the same orthonormal basis diagonalizes the new Hamiltonian operator, withIn particular, ${\tilde{E}}_{\pm 2}=E_{\star }+m{s}^2/2\mp 2\hslash s/R$. Applying unitary time evolution from $T$ therefore givesThis is rigid translation from a momentum term in the Hamiltonian: the position probability density moves with velocity $-s$. The plus sign in $x+s\tau$ follows directly from the negative sign of $-sP$. At $\tau =0$ the expression agrees exactly with the preceding post-measurement state.