Given that the wavefunction is $\psi=A \mathrm{e}^{\mathrm{i}(k z-\omega t)}+B \mathrm{e}^{-\mathrm{i}(k z+\omega t)}$, where $A$ and $B$ are constants, show that the probability current density is
$$
\mathbf{J}=v\left(|A|^{2}-|B|^{2}\right) \hat{\mathbf{z}}
$$
where $v=\hbar k / m .$ Interpret the result physically.