Basis for F. Mandl's interpretation of the amplitude of a plane wave I'm going through Mandl's Quantum Mechanics and I'm having trouble understanding some of the moves he makes when discussing the finite potential barrier.
He begins by interpreting the plane wave $Ae^{ikx}$ as a beam "of particles of uniform density", saying that $|A|^2$ represents the particles per unit volume. I don't see how this necessarily follows.
Everywhere else I read of the plane wave with the finite potential barrier, $A^2$ is generally set to 1 to represent a single particle, and the reflected and transmitted beams' amplitudes' squares add to 1 (which makes sense). It would make sense, based on this, that if $A^2$ represented particle density, then the other two amplitudes would square and add to $A^2$.
However, using this interpretation, for which I don't necessarily understand the justification, he claims that we can define the variable $j_I = \frac{\hbar k}{m} |A^2|$ as the current density ("particles per unit area normal to the direction of travel per unit time"). I'm not ready for this step. Do I have enough of a foundation to understand it if I thought about it at length? Is this $j_I$ representing a concept I'll see later in general? How can I breach into an understanding of this subject?
 A: I've done some thinking about this.
If we integrate $\psi^{*}\psi$ over a given volume $V$ we get $\int{A^2 e^{-ikx}e^{ikx}dv}$ which we know equals the probability $P$ of finding a particle in a given volume. The integral simplifies to $\int{A^2 dV} = A^2 V = P_{V}$. Since the plane wave is non-normalisable, we may get a $P > 1$, representing the number of particles in that volume. For the purposes of this argument, we can just interpret $P_{V}$ as the number of particles in $V$, and if we divide both sides of the equation $A^2 V = P_{V}$ by $V$ then we get that $A^2 = \frac{P_V}{V}$ which is more sensibly interpreted as the expected particles per unit volume.
Knowing now that $A^2$ clearly represents particle density (because of the fact that the plane wave is periodic, and thus we can expect generally uniform particle density represented only by $A^2$ and not dependent on position), it is clear that $j_I$ represents current density, as $\frac{\hbar k}{m}$ represents velocity (in $\frac{m}{s}$) and so $j_I = \frac{\hbar k}{m} A^2$ is in the units $\frac{particles}{m^2 s}$.
