= Solution
Put $\tau=t-T$ and retain the <global phase> from the preceding <Lüders rule> update. Since $P\phi_n=(\hbar n/R)\phi_n$, the same <orthonormal basis> diagonalizes the new <Hamiltonian operator>, with
$$
H_{\mathrm{new}}=\frac{(P-ms)^2}{2m},\qquad
\widetilde E_n=\frac1{2m}\left(\frac{\hbar n}{R}-ms\right)^2.
$$
In particular, $\widetilde E_{\pm2}=E_\star+ms^2/2\mp2\hbar s/R$. Applying <unitary time evolution> from $T$ therefore gives
$$
\boxed{\begin{aligned}
\Psi(t,x)&=-\frac{e^{-iE_\star T/\hbar}}{\sqrt2}
\left(e^{-i\widetilde E_2\tau/\hbar}\phi_2(x)+e^{-i\widetilde E_{-2}\tau/\hbar}\phi_{-2}(x)\right)\\
&=-\frac{e^{-iE_\star T/\hbar}e^{-i(E_\star+ms^2/2)\tau/\hbar}}{\sqrt{\pi R}}
\cos\!\left(\frac{2(x+s\tau)}R\right).
\end{aligned}}
$$
This 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>.
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