= Solution
For a one-dimensional <wavefunction>, the <probability density> and <probability current> are
$$
\rho=|\psi|^2,\qquad j=\frac{\hbar}{2mi}\left(\psi^*\psi_x-\psi\psi_x^*\right)=\frac\hbar m\operatorname{Im}(\psi^*\psi_x).
$$
The <Time-dependent Schrödinger equation> with a real potential is $i\hbar\psi_t=-\hbar^2\psi_{xx}/(2m)+V\psi$. Multiply it and its <complex conjugate> by $\psi^*$ and $\psi$, respectively. Subtracting cancels the real-potential terms and gives
$$
\rho_t=\frac{i\hbar}{2m}(\psi^*\psi_{xx}-\psi\psi_{xx}^*)=-j_x.
$$
This derives the <continuity equation> $\rho_t+j_x=0$.
For $k>0$, the specified free-region <wavefunction> combines a right-moving incoming wave of unit amplitude and a left-moving reflected wave with complex reflection amplitude $R$. Applying the free Schrödinger operator to either plane wave gives $E=\hbar^2k^2/(2m)$. Its time factor has modulus one, so
$$
\rho=1+|R|^2+2\operatorname{Re}(Re^{-2ikx}),\qquad \rho_t=0,\qquad j=\frac{\hbar k}{m}(1-|R|^2).
$$
The density includes the stationary interference pattern; $|R|^2$, rather than $R$, is the reflection probability.
For the <finite square well>, let $k=\sqrt{2mE}/\hbar$ outside and $q=\sqrt{2m(E+V_0)}/\hbar$ inside. Write the stationary parts as $e^{ikx}+Re^{-ikx}$ to the left, $Ae^{iqx}+Be^{-iqx}$ inside, and $Te^{ikx}$ to the right. At both finite potential steps, the <wavefunction> and its first spatial <derivative> are continuous. Integrating the stationary <Schrödinger equation> across a shrinking step proves derivative continuity; a jump of the <wavefunction> would create an unmatched distributional derivative and is excluded.
Propagate the boundary values at $a$ back through the well. The values at zero are
$$
\chi(0)=Te^{ika}\left(\cos qa-i\frac kq\sin qa\right),\qquad
\chi'(0)=Te^{ika}\left(q\sin qa+ik\cos qa\right).
$$
Matching them to $1+R$ and $ik(1-R)$, and adding the two equations after division of the second by $ik$, gives
$$
2=Te^{ika}\left[2\cos qa-i\left(\frac kq+\frac qk\right)\sin qa\right].
$$
Since the exterior wave numbers agree, the <transmission probability> is the current ratio $|T|^2$, so
$$
|T|^2=\left[1+\frac{(q^2-k^2)^2}{4k^2q^2}\sin^2qa\right]^{-1}.
$$
When $V_0=3E$, $q=2k=\sqrt{8mE}/\hbar$, yielding
$$
\boxed{|T|^2=\left[1+\frac9{16}\sin^2\frac{a\sqrt{8mE}}{\hbar}\right]^{-1}.}
$$
Classically an incoming particle of positive energy crosses the attractive well and emerges with its original energy, so transmission is one. The quantum result agrees exactly when $qa$ is an integer multiple of $\pi$:
$$
\boxed{a=\frac{n\pi\hbar}{\sqrt{8mE}},\qquad n=1,2,\ldots.}
$$
These are widths with <resonant transmission through a square well>; the degenerate width zero also has unit transmission.
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