quantum mechanics - Intuition/derivation behind the probability current definition

The definition is: $${\bf{j}} = \frac{\hbar}{2mi} (\psi^* \nabla \psi - \psi \nabla \psi^*)$$ However: Where ever I have looked, the above "pops out of nowhere".

I was wondering how can I obtain some intuition about this, and/or, can it be derived from some related definition/s?

Answer

I know of two ways to derive this: the first is to take the time derivative of $|\psi|^2$, then use the Schrödinger and the continuity equations, and the second is to start with the Schrödinger lagrangian and find the Noether current. Indeed:

The Schrödinger equation is $$\left( - \frac{\hbar^2}{2m} \nabla^2 + V(\vec{r}) \right) \psi(\vec{r},t) = i\hbar \frac{\partial \psi}{\partial t}(\vec{r},t)$$ and the continuity equation $$\vec{\nabla}\cdot \vec{J} + \frac{\partial \rho}{\partial t} = 0.$$ We have ($\rho \equiv \psi\psi^*$), $$\begin{split} \frac{\partial }{\partial t} \psi \psi^* &= \frac{1}{i\hbar} \left( - \frac{\hbar^2}{2m} \nabla^2 \psi + V\psi\right)\psi^* - \frac{1}{i\hbar} \left( - \frac{\hbar^2}{2m} \nabla^2 \psi^* + V\psi^*\right) \psi\\ &= - \frac{\hbar}{2mi} \vec{\nabla}\cdot \left( \psi^* \vec{\nabla} \psi - \psi \vec{\nabla} \psi^* \right) = - \vec{\nabla} \cdot \vec{J}.\end{split}$$

In the following $k$ denotes spatial and $\mu$ spacetime indices. The lagrangian is $$\mathcal{L} \equiv \frac{\hbar}{2m} \vec{\nabla} \psi^* \cdot \vec{\nabla} \psi - i \frac{\partial \psi}{\partial t} \psi^* + V \psi\psi^*.$$ This has a global U(1) symmetry, $\psi \to \psi e^{i a}$, hence $\delta \psi = i\psi$, and Noether's theorem yields: $$\begin{split} J^0 &= \frac{\partial \mathcal{L}} { \partial \partial_0\psi} \delta \psi = \psi\psi^* =\rho,\\ J^k &= \frac{\partial \mathcal{L}} {\partial \partial_k \psi} \delta \psi + \frac{\partial \mathcal{L}} {\partial \partial_k \psi^*} \delta \psi^* = \frac{\hbar}{2mi} ( \psi^* \partial^k \psi - \psi \partial^k \psi^*), \end{split}$$ and from $\partial_\mu J^\mu = 0$ the continuity equation follows.