= Solution
\b[Single travelling wave.] Differentiation of the <plane wave> gives $\partial_x\psi=ik\psi$, so its <probability current> is
$$
J=\operatorname{Re}\left(\psi^*\frac{\hbar}{im}\partial_x\psi\right)
=\boxed{\frac{\hbar k}{m}|A|^2}.
$$
This is <probability density> times the free-particle velocity $\hbar k/m$.
\b[Counterpropagating waves.] Remove the common time phase and set $z=B^*Ae^{2ikx}$. Direct multiplication gives
$$
\psi^*\partial_x\psi
=ik\left(|A|^2-|B|^2+z-z^*\right).
$$
Consequently
$$
J=\frac{\hbar k}{m}\operatorname{Re}\left(|A|^2-|B|^2+z-z^*\right)
=\boxed{\frac{\hbar k}{m}\left(|A|^2-|B|^2\right)}.
$$
The <interference> term $z-z^*$ is purely imaginary in the last real-part operation. The <probability density> can have <interference fringes> even though this net <probability current> is spatially constant.
\b[Scattering amplitudes.] Locate the step at $x=0$ and first take $E>V$. With $k_1=\sqrt{2mE}/\hbar$ and $k_2=\sqrt{2m(E-V)}/\hbar$, use
$$
\psi_1=e^{ik_1x}+r_a e^{-ik_1x},\qquad
\psi_2=t_a e^{ik_2x}.
$$
The common factor $e^{-iEt/\hbar}$ has been suppressed. A finite step with the same mass on both sides requires continuity of $\psi$ and $\psi'$, giving $1+r_a=t_a$ and $k_1(1-r_a)=k_2t_a$. Hence
$$
\boxed{r_a=\frac{k_1-k_2}{k_1+k_2}
=\frac{\sqrt E-\sqrt{E-V}}{\sqrt E+\sqrt{E-V}}},
\qquad
\boxed{t_a=\frac{2k_1}{k_1+k_2}}.
$$
If $0<E<V$, choose $k_2=i\kappa$, $\kappa=\sqrt{2m(V-E)}/\hbar$, so the transmitted wave decays. The same amplitude formulae continue to hold, but the semi-infinite forbidden region carries no outgoing <probability current>: $R=1$ and $T=0$.
\b[Flux balance.] In the propagating case,
$$
R=|r_a|^2,\qquad
T=\frac{k_2}{k_1}|t_a|^2
=\frac{4k_1k_2}{(k_1+k_2)^2},\qquad R+T=1.
$$
In particular,
$$
J_1=\frac{\hbar k_1}{m}(1-R)
=\frac{\hbar k_2}{m}|t_a|^2=J_2.
$$
The velocity ratio in $T$ is essential: the squared transmission amplitude alone is not the <transmission probability> when the two velocities differ.
\b[Numerical transmission.] The mass cancels from the ratio, and the <energies> can remain in the same <electronvolt> units:
$$
\boxed{T=\frac{4\sqrt3}{(\sqrt3+1)^2}=0.92820}.
$$
Thus about $92.8\%$ is transmitted and $7.18\%$ reflected, despite the incident <energy> exceeding the step.
Back to article page
