= Evanescent probability current
{title2=$J=\frac{\hbar}{m}\operatorname{Im}(\psi^*\psi')$}
A nonzero <probability density> inside a barrier does not by itself imply transport. In one dimension, the <probability current> of a <wavefunction> is $J=(\hbar/m)\operatorname{Im}(\psi^*\psi')$. A single <evanescent wave> $Ce^{-\kappa x}$ has $\psi^*\psi'=-\kappa|C|^2e^{-2\kappa x}$, a real quantity, hence $J=0$. This is what happens on the inaccessible side of a <potential barrier> extending to infinity with $E<V_0$.<Quantum Physics I (8.04), Spring 2016/Lecture 6>[p. 5, equations (3.4)-(3.6): probability continuity equation and probability current]{ref}
For the finite-barrier <wavefunction> $Ae^{\kappa x}+Be^{-\kappa x}$, the cross terms instead give
$$
J=\frac{2\hbar\kappa}{m}\operatorname{Im}(B^*A).
$$
The relative <quantum phase> of the two <probability amplitudes> permits a nonzero, spatially constant <probability current>, even though neither <exponential function> separately carries it. The <probability continuity equation> gives $\partial_xJ=0$ for this <stationary state>. Matching the <wavefunction> and its <derivative> also makes $J$ continuous at both faces, so
$$
J_{\mathrm{inc}}=\frac{\hbar k}{m},\qquad
J_{\mathrm{ref}}=-\frac{\hbar k}{m}|r|^2,\qquad
J_{\mathrm{tr}}=\frac{\hbar k}{m}|t|^2,\qquad
R+T=1.
$$
Here $R=|r|^2$ is the <quantum reflection probability>, and $T=J_{\mathrm{tr}}/J_{\mathrm{inc}}$ is the <transmission probability>. The net exterior <probability current> equals the interior <probability current>; it is not lost as the <wavefunction> decreases. A physical incident <wave packet> supplies finite total probability, while the <stationary state> describes the <energy> component used to compute its transmitted flux.
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