= Solution
The initial <quantum state> is the <equal superposition of the first two box states>, $(\psi_1+\psi_2)/\sqrt2$. The <orthonormality> of the two <energy eigenstates> makes its squared <norm> equal to one. Their separate <stationary state> phases give
$$
\boxed{\Psi(x,t)=\frac1{\sqrt a}\left[\sin\frac{\pi x}{a}e^{-iE_1t/\hbar}+\sin\frac{2\pi x}{a}e^{-iE_2t/\hbar}\right].}
$$
Write $\Omega=(E_2-E_1)/\hbar=3\pi^2\hbar/(2ma^2)$. The <probability density> is
$$
\rho(x,t)=\frac1a\left[\sin^2\frac{\pi x}a+\sin^2\frac{2\pi x}a+2\sin\frac{\pi x}a\sin\frac{2\pi x}a\cos\Omega t\right].
$$
Each squared sine integrates to $a/4$ on the left half. For the <quantum interference> term, the product-to-sum identity gives
$$
\int_0^{a/2}\sin\frac{\pi x}a\sin\frac{2\pi x}a\,dx=\frac12\int_0^{a/2}\left(\cos\frac{\pi x}a-\cos\frac{3\pi x}a\right)dx=\frac{2a}{3\pi}.
$$
Hence the <Born rule> gives
$$
\boxed{P_{\mathrm L}(t)=\int_0^{a/2}\rho(x,t)\,dx=\frac12+\frac4{3\pi}\cos\Omega t.}
$$
The <quantum interference> moves probability between the halves even though each component separately is a <stationary state>.
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