Probability current

We have $\psi'=ikAe^{ikx}$, so

$$\psi^*\psi'=A^*e^{-ikx}\cdot ikAe^{ikx}=ik|A|^2, \qquad j=\frac{\hbar}{m}\operatorname{Im}(ik|A|^2)=\frac{\hbar k}{m}|A|^2 .$$

With $p=\hbar k$, we recognise $j=v\,\rho$, where $\rho=|A|^2$ is the probability density and $v=p/m$ the classical velocity. This is the usual relation between flux, density, and velocity in a fluid: the plane wave describes a uniform probability flux moving towards increasing $x$ at velocity $v$.

We have $\psi'=ik(Ae^{ikx}-Be^{-ikx})$, and

$$\psi^*\psi'=ik\bigl(A^*e^{-ikx}+B^*e^{ikx}\bigr)\bigl(Ae^{ikx}-Be^{-ikx}\bigr) =ik\Bigl(|A|^2-|B|^2+B^*Ae^{2ikx}-A^*Be^{-2ikx}\Bigr).$$

Let $z=B^*Ae^{2ikx}$. The two cross terms are $z-z^*=2i\operatorname{Im}z$, which is purely imaginary. Multiplication by $ik$ gives $-2k\operatorname{Im}z$, which is real and therefore does not contribute to the imaginary part. Thus

$$j=\frac{\hbar k}{m}\bigl(|A|^2-|B|^2\bigr).$$

The total current is the difference between the current of the wave propagating to the right and that of the wave propagating to the left: the two waves transport probability independently of each other. For the standing wave $C\sin(kx)=\frac{C}{2i}(e^{ikx}-e^{-ikx})$, we have $|A|=|B|=|C|/2$, and the current vanishes: the two opposing fluxes cancel. This can also be seen directly: $\psi^*\psi'=|C|^2k\sin(kx)\cos(kx)$ is real.

We have $\psi^*\psi'=e^{-i\theta}f\cdot e^{i\theta}f'=ff'$, which is real: its imaginary part vanishes, and $j=0$. The constant phase $\theta$ plays no role. The evanescent wave $Ae^{-\kappa x}$ has this form, with $f(x)=|A|e^{-\kappa x}$ and $\theta=\arg A$. It therefore carries no current: this is why an evanescent wave beyond an infinite potential step corresponds to no transmission.

We have $\psi'=\kappa(Ae^{\kappa x}-Be^{-\kappa x})$, and

$$\psi^*\psi'=\kappa\bigl(A^*e^{\kappa x}+B^*e^{-\kappa x}\bigr)\bigl(Ae^{\kappa x}-Be^{-\kappa x}\bigr) =\kappa\Bigl(|A|^2e^{2\kappa x}-|B|^2e^{-2\kappa x}+B^*A-A^*B\Bigr).$$

The first two terms are real. The last two can be written $B^*A-(B^*A)^*=2i\operatorname{Im}(AB^*)$. Hence

$$j=\frac{2\hbar\kappa}{m}\operatorname{Im}(AB^*).$$

As required, this current is independent of $x$. It is non-zero if and only if $A$ and $B$ are both non-zero and do not have the same phase modulo $\pi$. A single real exponential carries no current, but a combination of the two exponentials with coefficients of different phases does. This is exactly what occurs inside a potential barrier of finite width: the matching conditions impose such a combination, and the current through the barrier equals the transmitted current.

Differentiating the product gives

$$\psi'=\Bigl(f'+\frac{i}{\hbar}fS'\Bigr)e^{iS/\hbar}, \qquad \psi^*\psi'=ff'+\frac{i}{\hbar}f^2S' .$$

The imaginary part is $f^2S'/\hbar$, and therefore

$$j=\frac{\hbar}{m}\cdot\frac{f^2S'}{\hbar}=f^2\,\frac{S'}{m}=\rho\,v .$$

The current depends only on the phase gradient: it is the phase of the wave function that encodes the local velocity of the probability flow. For the plane wave, $f=|A|$ and $S=\hbar kx+\hbar\arg A$, so $S'=\hbar k$ and $v=\hbar k/m$, in agreement with the first question.