= Solution
For
$$
T=\frac{2ma^2}{\pi\hbar},
$$
the phase of the $n$th stationary state is
$$
e^{-iE_nT/\hbar}=e^{-in^2\pi}=(-1)^n.
$$
Reflection in the midpoint gives
$$
\chi_n(a-x)=(-1)^{n+1}\chi_n(x),
$$
so
$$
\boxed{\psi(x,T)=-f(a-x)}.
$$
The minus sign is a global <quantum phase> and therefore does not change the physical state. The relative even--odd phases first acquire this reflection pattern at $T$. Applying it twice gives
$$
\boxed{\psi(x,2T)=f(x)}.
$$
If $f$ vanishes on the right half of the well, then $\psi(x,T)$ vanishes on the left half. The <Born rule> consequently gives zero probability of finding the particle in $0\leq x\leq a/2$ at time $T$.
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