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?
 A: 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:
First way
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 ($\rho \equiv \psi\psi^*$), $$\begin{split} \frac{\partial }{\partial t} \psi \psi^* &= \frac{1}{i\hbar} \left( - \frac{\hbar^2}{2m} \nabla^2 \psi + V\psi\right)\psi^* - \frac{1}{i\hbar} \left( - \frac{\hbar^2}{2m} \nabla^2 \psi^* + V\psi^*\right) \psi\\
 &= - \frac{\hbar}{2mi} \vec{\nabla}\cdot \left( \psi^* \vec{\nabla} \psi - \psi \vec{\nabla} \psi^* \right) = - \vec{\nabla} \cdot \vec{J}.\end{split}$$
Second way
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.
A: An intuitive approach, which is less formal, is the conservation of charge:
\begin{equation}
\nabla \cdot\vec{j} + \partial_t \rho = 0
\end{equation}
we know the "charge density" of a wave function is $|\psi|^2$, and we are left to figure out the correct $\vec{j}$ to conserve charge. This is, of course, the Noether current, but intuitively this is what the current is saying.
A: The way I would approach this problem is to integrate the probability distribution,
$P(r,t)=|\psi(r,t)|^2$ over all space and see how this varies with time (take a time derivative). So using the bra ket formalism we have
$$ 
\frac{\partial}{\partial t} \langle\psi|\psi\rangle
$$
Which can be expanded as 
$$
\langle\frac{\partial \psi}{\partial t}|\psi\rangle + \langle\psi|\frac{\partial \psi}{\partial t}\rangle
$$
The re-express $\frac{\partial \psi}{\partial t}$ and its complex conjugate $\frac{\partial \psi *}{\partial t}$ using the time dependent schrodinger equation in 3D.
ie.
$$
\frac{\partial}{\partial t} \psi(\mathbf{r},t) = -\frac{i}{ \hbar} \left [ \frac{-\hbar^2}{2m}\nabla^2 + V(\mathbf{r},t)\right ] \psi(\mathbf{r},t)
$$
The result should follow using this method.
You may also find the divergence theorem useful in completing the argument. 
$$
\iiint_\tau\left(\mathbf{\nabla}\cdot\mathbf{F}\right)\,dV=\iint_{\partial \tau} \mathbf{F}\cdot\mathbf{dS} .
$$ 
I hope this helps, it should at least give you some idea of the motivation behind the definition of $\mathbf{j}$. 
