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?


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 $\frac{particles}{m^2 s}$.

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