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For the oppositely travelling components, write $u=c_1{e}^{ikx}$ and $v=c_2{e}^{-ikx}$ after removing the common time phase. Then ${\partial }_x\psi =ik(u-v)e^{-i\omega t}$. In $\mathrm{Im}({\psi }^{\ast }{\partial }_x\psi )$ the interference terms contribute no imaginary part, soA standing-wave interference pattern can therefore change density without adding a cross-term to the net current.

For barrier-top scattering, the stationary equation inside the barrier is $-{\hslash }^2{\psi }_2^{\prime \prime }/(2m)=0$. Thus its spatial part is linear: ${\psi }_2=(bx+c)e^{-i\omega t}$. Take incident amplitude one, $k=\sqrt{2mE}/\hslash$ and $\omega =E/\hslash$. Continuity of the spatial wave function and its derivative at the two finite steps givesSolving these four equations yieldsThe equal exterior velocities mean flux ratios are squared amplitude ratios:This is the threshold limit of finite-barrier scattering, not exponential decay with a nonzero interior decay constant.

The interior constants areSince $\mathrm{Im}[(bx+c{)}^{\ast }b]=\mathrm{Im}(c^{\ast }b)=k/(1+k^2{a}^2)$,It is positive and independent of position, matching the transmitted flux. Although the classical kinetic energy inside is zero, the complex linear quantum wave function carries conserved probability current; its phase varies with position.