Evanescent probability current ($J=\frac{\hslash }{m}\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$)

A nonzero probability density inside a barrier does not by itself imply transport. In one dimension, the probability current of a wavefunction is $J=(\hslash /m)\mathrm{Im}({\psi }^{\ast }{\psi }^{\prime })$. A single evanescent wave $C{e}^{-\kappa x}$ has ${\psi }^{\ast }{\psi }^{\prime }=-\kappa |C{|}^2{e}^{-2\kappa x}$, a real quantity, hence $J=0$. This is what happens on the inaccessible side of a potential barrier extending to infinity with $E<V_0$.^[ref]

For the finite-barrier wavefunction $A{e}^{\kappa x}+B{e}^{-\kappa x}$, the cross terms instead giveThe 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 ${\partial }_{x}J=0$ for this stationary state. Matching the wavefunction and its derivative also makes $J$ continuous at both faces, soHere $R=|r{|}^2$ 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.