The units of probability current should be $m^{-2} s^{-1}$, a rate per area, basically. However, on page 530 of Shankar's Quantum mechanics book, we are speaking in terms of the momentum eigenstate in the position basis: $\lvert {\mathbf p} \rangle= (2 \pi \hbar)^{-3/2} e^{i \mathbf p \cdot \mathbf r /\hbar} $. We get the probability current from the formula:

$$ \mathbf j = \frac{\hbar}{2mi} \left [ \psi^* \nabla \psi - \psi \nabla \psi^* \right ] \\=\frac{1}{2m} \left [ \psi^* \hat p \psi - \psi \hat p \psi^* \right ] \\ = \frac{\hbar k}{m} |\psi |^2\\ = \frac{\hbar k }{m} \left (\frac{1}{2 \pi \hbar }\right )^3$$

The units of the term in parens is $p^{-3} m^{-3}$ where $p$ is units of momentum. First of all, that should have units of $1/m^3$ because it's a probability density, and when you combine it with the left term which has units of velocity, you get something with units $p^{-3} m^{-2} s^{-1}$, basically, I have a momentum density that seems to be spuriously showing up. What gives?


1 Answer 1


There is a certain degree of arbitraryness in normalizing momentum eigenstates, i.e., in the prefactor of the plane wave (strictly speaking, momentum eigenstates cannot be normalized).

For calculating the current density, another normalization than the one you have quoted would normalize to a big volume $V$, such that $$ \left|\left.\mathbf{p} \right>\right. = \frac{1}{\sqrt{V}}e^{i\mathbf{p}\cdot\mathbf{r}/\hbar}. $$ This normalizes the wave function within $V$ and results in the correct units of the current density. We keep in mind here, that in the end, we are interested in the limit where $V\rightarrow\infty$. With this normalization, the current density becomes $$ \mathbf{j} = \frac{\hbar\mathbf{k}}{m}\frac{1}{V}.$$ If you consider periodic boundary conditions, as often done when considering infinite systems, you find the volume of a particular $\mathbf{k}$-state to be $d^3k = (2\pi)^3/V$ (note that here we made the transition $V\rightarrow\infty$ by going to the differential $d^3k$), such that you can rewrite your current density as $$ d\mathbf{j} = \frac{\hbar\mathbf{k}}{m}\frac{d^3k}{(2\pi)^3}.$$ You see that this is an infinitesimal contribution to the current density caused by a single momentum eigenstate. A proper current density needs some sort of integration over a range of momentum states.


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