= Solution
At the midpoint the second sine vanishes, whereas the derivative of the first sine vanishes. Thus the <wavefunction> and its spatial derivative simplify to
$$
\Psi(a/2,t)=\frac{e^{-iE_1t/\hbar}}{\sqrt a},\qquad \partial_x\Psi(a/2,t)=-\frac{2\pi}{a^{3/2}}e^{-iE_2t/\hbar}.
$$
Substitution in the <probability current> gives
$$
\boxed{j(a/2,t)=\frac{2\pi\hbar}{ma^2}\sin\Omega t.}
$$
The <continuity equation>, integrated over the left half, gives $P_{\mathrm L}'(t)=j(0,t)-j(a/2,t)$. The <Dirichlet boundary condition> makes $j(0,t)=0$, and the preceding <probability> formula gives
$$
P_{\mathrm L}'(t)=-\frac{4\Omega}{3\pi}\sin\Omega t=-\frac{2\pi\hbar}{ma^2}\sin\Omega t=-j(a/2,t).
$$
\b[Positive midpoint current removes probability from the left half.] This is precisely the <midpoint current of a two-mode box state>.
\Image[/past-exam-of-the-quantum-theory-course-of-the-university-of-oxford/2014-box-current.png]
{title=Interference between the first two box modes transfers probability between the two halves; midpoint current is minus the derivative of left-half probability}
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