quantum mechanics - How do we choose the standard probability current?

In quantum mechanics, the probability current is defined as $$\mathbf{J} \propto \text{Im}(\psi^* \nabla \psi)$$ and satisfies the continuity equation $$\nabla \cdot \mathbf{J} = - \frac{\partial \rho}{\partial t}$$ where $\rho$ is the probability density. However, the continuity equation remains true if we add to $\mathbf{J}$ any divergence-free function. For example, the following expressions all satisfy the continuity equation: $$\mathbf{J}_1 = \mathbf{J} + \nabla \times (\rho \mathbf{J}), \quad \mathbf{J}_2 = \mathbf{J} + \langle \mathbf{p} \rangle, \quad \mathbf{J}_3 = \mathbf{J} - \langle \mathbf{J} \rangle.$$ However, since I've never heard anybody mention any alternate probability currents, the definition $\mathbf{J}$ must be special in some way. What is special about it? Does it satisfy some nice properties besides the continuity equation?

Answer

Well, here is a property it satisfies. I'm sure the Bohm-Aharonov advection mavens have neat geometric names about such things, but I'm drawing a blank for the moment. Irrotational flow?

Nevertheless, peremptorily normalizing, for simplicity, $$ \mathbf{J}\equiv \psi^* \nabla \psi - \psi \nabla \psi^*= - \nabla \rho + 2 \psi^* \nabla \psi = \nabla \rho - 2 \psi \nabla \psi ^*, $$ you see that $$ \rho ~\nabla \times \mathbf{J}=2\rho (\nabla \psi^* \times \nabla \psi) = \nabla \rho \times \mathbf{J}~, $$ so $$ \nabla \times \mathbf{J} -\frac{\nabla \rho}{\rho} \times \mathbf{J}=0=\rho~~\nabla\times (\mathbf{J}/\rho)~. $$ At a minimum, this might exclude your 2nd and 3rd options, $\mathbf{J}_2,~\mathbf{J}_3$.

Current over density might serve to define some sort of effective velocity of probability flow (Dirac, Landau-Lifschitz), so then with vanishing vorticity.

$\mathbf{J} /\rho$ being irrotational, it may be thought of as a potential flow (a gradient of the phase of the wavefunction), which goes under the name of the Madelung quantum Euler equations' formulation.

Your first option $\mathbf{J}_1$ is dimensionally inconsistent, anyway, so it needs a dimensionfull constant in front of the extra curl term. But there is no good reason the power of the density needs to be one, as you have chosen. If you considered $\nabla \times (\rho^n \mathbf{J})$ instead, then you see the irrotational condition automatically projects out the n = -1 case, but not all other ones.

• (One may always reinstate the normalization $-i\hbar/2m$ in the current, but why?) For broader contexts, see WP.

A simplest exploration is afforded by a stationary system, such as an atom, where the current is divergenceless, $ \nabla \cdot \mathbf{J}=0$, as the only complex part of the wavefunction is the azimuthal exponential, so only the φ component of it survives in polar coordinates: uniform rotational flow for a stationary system!