Quantum mechanics: Probability current density in terms of velocity vs. in terms of continuity equation For simplicity, consider a one-electron system. Some sources tell that the probability current density can be written in terms of the velocity operator $\mathbf{v} = -i[\mathbf{r}, H]$ as
$$ \mathbf{j} = \frac{1}{2}(\psi^\ast \mathbf{v} \psi + (\mathbf{v}\psi)^\ast \psi). \quad\quad(1)$$
I try to reconcile this with the continuity equation. Using the Schrödinger equation, one can obtain
$$ \frac{\partial \rho}{\partial t} = \frac{\partial}{\partial t}(\psi^\ast \psi) = -i[\psi^\ast H\psi -(H\psi)^\ast \psi] \stackrel{!}{=}-\nabla \cdot \mathbf{j}\quad\quad(2)$$
Combining these two equations would mean that
$$ \frac{1}{2} \nabla \cdot (\psi^\ast[\mathbf{r}, H]\psi + ([\mathbf{r},H]\psi)^\ast \psi)= -[\psi^\ast H \psi - (H\psi)^\ast \psi].\quad\quad(3)$$
I have tried a lot to show explicitly that both sides of this equation are identical but have failed so far. I would be happy if someone knows how to do this, or can hint me to some additional assumption that is needed for this equation to be true.
Note that for specific forms of the Hamiltonian, e.g. a nonrelativistic electron in an electrostatic external potential (where only the momentum operator contributes to the velocity) or also including a static magnetic field (where also the vector potential contributes to the velocity), I am able to show the equality of both sides of Eq. (3). But I am interested in a general proof.
 A: I normally approach this from a different angle, but I hope that it may give some insight. It seems like you want to show that the Schrödinger equation leads to a meaningful probability current, but I would start with probability and show that it requires unitary time evolution satisfying the conditions of Stone's theorem, and from there derive the Schrödinger equation. Then it must be clear that the Schrödinger equation is a statement of probability and must give a conserved probability current.
For a normalised state $|f\rangle$, the probability density $\rho(x) $at time $t$ is $$ \rho(x)=\langle f| x \rangle \langle x| f \rangle = |f(x)|^2 , $$ where $$\int d^3x \rho(x) = \int d^3x\langle f| x \rangle \langle x| f \rangle = \int d^3x\langle f| f \rangle = 1.$$ The evolution operator is constrained by unitarity to ensure that probability is conserved. Consequently, it obeys a conservation law with the form $$ \frac{\partial}{\partial t}\rho + \nabla . \mathbf{j}=0 $$ where $\mathbf{j}$ is the probability current. The expected flow of probability must correspond to a flow of mass, that is it relates to momentum, $$\mathbf{P}= - \int d^3x |x\rangle i\nabla \langle x|.$$ The Hermitian operator with the required property is the current density operator, $$\mathbf{J}(x)= - \frac{i}{2m} |x\rangle (\nabla  -  \overleftarrow \nabla) \langle x|$$ where the arrow means that differentiation acts to the left. I have used this notation to emphasise that $\mathbf J$ is a density at a point, which can get lost otherwise. This form is also used in some sources, and I think it should be clear that for the particular Hamiltonians you have considered it is equivalent to the form which you gave. I am not clear whether it is identically the same. If it is not, that may be the source of the problem. The form you have appears to have been taken from Ehrenfest's theorem, but as we are not dealing with an observable I am not sure if it is correctly applied.
