= Solution
Let $\Pi_\star=|\phi_2\rangle\langle\phi_2|+|\phi_{-2}\rangle\langle\phi_{-2}|$ be the <orthogonal projection> for $E_\star$. The <Lüders rule> gives the normalized <post-measurement state>
$$
\boxed{\begin{aligned}
\Psi(T^+,x)&=\frac{\Pi_\star\Psi(T^-,x)}{\sqrt{1/3}}\\
&=-\frac{e^{-iE_\star T/\hbar}}{\sqrt2}\bigl(\phi_2(x)+\phi_{-2}(x)\bigr)
=-\frac{e^{-iE_\star T/\hbar}}{\sqrt{\pi R}}\cos\!\left(\frac{2x}{R}\right).
\end{aligned}}
$$
The factor $-e^{-iE_\star T/\hbar}$ is a <global phase>, so $\cos(2x/R)/\sqrt{\pi R}$ is an equivalent representative. The <projective measurement> preserves the <quantum coherence> between the two <momentum eigenstates>; it does not select one of them. This matters for the subsequent position <probabilities>.
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