# Units of momentum eigenstate probability current have an extra density in momentum

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?

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.