= Solution
For a plane wave $A e^{i(kx-Et/\hbar)}$, the <probability current> is
$$
J=\frac{\hbar k}{m}|A|^2.
$$
The incident amplitude is one and the transmitted amplitude is $T$, so
$$
\boxed{
\frac{J_{\rm tr}}{J_{\rm in}}
=|T|^2
=\operatorname{sech}^2(2ka)
}.
$$
This is the transmission probability for <quantum tunnelling>. In the <stationary state>, $\partial\rho/\partial t=0$, so the <continuity equation> makes the net current independent of position. The reflected current therefore supplies the remainder:
$$
\boxed{\frac{|J_{\rm refl}|}{J_{\rm in}}
=1-\operatorname{sech}^2(2ka)
=\tanh^2(2ka).}
$$
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