# 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?

• $\psi^*\mathbf p\psi$ looks like a density current from a physical point of view, as $\mathbf p = m\mathbf v$ for a single particle of mass $m$. All the other terms wouldn't have such an interpretation – Phoenix87 Sep 29 '17 at 21:27
• @Phoenix87 With more complicated Hamiltonians we don't necessarily have $\mathbf{p} = m \mathbf{v}$, though. Generally $\mathbf{v}$ can be an arbitrarily complicated function of $(\mathbf{x}, \mathbf{p})$. Would the right definition of $\mathbf{J}$ change in this case? – knzhou Sep 29 '17 at 21:51
• this is just an interpretation that leads you to introduce a new quantity that satisfies the continuity equation. As you point out there are many other quantities that satisfy the same property, and that can be considered "probability currents", but they lack a somewhat "clean" relation with the momentum vector in general. – Phoenix87 Sep 29 '17 at 21:58
• Further nitpick: not directly related, but the probability current may have more latitude to it. – Cosmas Zachos Sep 29 '17 at 22:34
• There's a similar indeterminacy issue with the Poynting vector; you can justify one choice over the others but the issue never really goes away. – Emilio Pisanty Sep 30 '17 at 15:32

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!

Firstly, you can derive the continuity equation from Schroedinger's equation ($\hbar = 1$) $$-\frac{1}{2m} \nabla^2 \psi + V(x) \psi = i \partial_t \psi$$ by multiplying it by $\overline{\psi}$ and its complex conjugate by $\psi$ and summing them. With a little manipulation, you get:

$$\partial_t (\psi \overline{\psi}) -\frac{i}{2m} \nabla \cdot \left( \overline{\psi} \nabla \psi - \psi \nabla \overline{\psi} \right) = 0.$$

Since you want local conservation of probability density, you define the current $\vec{j}$ as

$$\vec{j}_0 = \frac{-i}{2m} \left( \overline{\psi} \nabla \psi - \psi \nabla \overline{\psi} \right)$$

So there's nothing "special" about it, it just imposes local conservation of probability density, and there is no physical reason to add another term to it, since the current arises naturally from Schroedinger's equation.

Now, the above current has to be redefined depending on the system you are working on. Supposing there is an external magnetic field ($\vec{B} = \nabla \times \vec{A}$), deriving the current in the same way as before you will get another term in $\vec{j}$, proportional to $\overline{\psi} \vec{A}\psi$.

Now, as said, there is no physical reason to add another term, but there is a special case. In a system where you have an external magnetic field AND you consider its interaction with spin, you define your current with an extra term, proportional to $\nabla \times (\overline{\psi} \vec{S} \psi)$, even though it does not naturally arise try to derive $\vec{j}$ as before (note that $\nabla \cdot \nabla \times \vec{f} = 0$). This is physically justified. In non-relativistic quantum mechanics spin does not arise naturally, but this term exists in the conserved current of the Dirac equation, and survives in the non-relativistic limit. So this is imposed and you get $$\vec{j}= \vec{j}_0 - \frac{q}{m} \vec{A} \overline{\psi} \psi\ + \frac{\mu_s}{s} \nabla \times (\overline{\psi} \vec{S} \psi)$$

P.S.1: Also, remember that Quantum Mechanics is invariant under a global phase transformation $e^{i\alpha}$, and $\vec{j}_0$ is the current associated to this kind of symmetry.

P.S.2: Maybe there are some constants missing, if there are, I apologize.

I think the fundamental reason is that the standard expression is the (standard) Noether current associated with the continuous global gauge invariance $|\psi\rangle \to e^{i \theta} |\psi\rangle$. E.g. the same expression gives the Noether current for a relativistic scalar field with a global $U(1)$ symmetry.

• But... isn't that the OP's very point? Noether currents are routinely "improved" by the addition of immaterial constant or curl terms, as in supersymmetric theories... – Cosmas Zachos Nov 1 '17 at 13:39
• @CosmasZachos True, good point. – tparker Nov 1 '17 at 14:43