= Solution
Use the <orthonormal> <energy eigenstates> of the <infinite square well>,
$$
\phi_n(x)=\sqrt{\frac2a}\sin\frac{n\pi x}{a},\qquad E_n=n^2E_1,\qquad E_1=\frac{\pi^2\hbar^2}{2ma^2}.
$$
The identity $\sin u\cos u=\tfrac12\sin2u$ resolves the initial <wavefunction> into $\Psi(0)=(2\phi_1+\phi_2)/\sqrt5$. Each <energy eigenstate> acquires its own <dynamical phase>, so
$$
\boxed{\Psi(x,t_0)=\frac{2\phi_1(x)e^{-iE_1t_0/\hbar}+\phi_2(x)e^{-4iE_1t_0/\hbar}}{\sqrt5}}
$$
inside the well, and the <wavefunction> vanishes outside. The <energy> probabilities are $4/5$ and $1/5$ at every time. Consequently
$$
\boxed{\langle H\rangle_0=\langle H\rangle_{t_0}=\frac85E_1=\frac{4\pi^2\hbar^2}{5ma^2}.}
$$
For the left half, each diagonal <integral> $\int_0^{a/2}|\phi_n|^2dx$ is $1/2$, while
$$
\int_0^{a/2}\phi_1\phi_2\,dx=\frac4{3\pi}.
$$
The <interference> term therefore yields
$$
\boxed{P_{\mathrm{left}}(t_0)=\frac12+\frac{16}{15\pi}\cos\frac{3E_1t_0}{\hbar}.}
$$
This <probability> lies between approximately $0.1605$ and $0.8395$, with <period> $2\pi\hbar/(3E_1)$. A conserved <energy> distribution is compatible with time-dependent position probabilities because the relative <quantum phase> keeps evolving.
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