Probability current (Integral in all space) So , when we take the integral in all space of the probability current j (as defined in the first relationship here https://en.wikipedia.org/wiki/Probability_current) in non relativistic quantum mechanics , the result is zero? If yes, that's because of the normalization of probability and the fact that the particle can be anywhere in space and there are no special points in space that favor probability as a notion?
Can you provide me of an intuitive explanation of the probability current in order to understand the result?
 A: As Al Brown suggested, the current operator is equivalent to
\begin{align}
  \vec j(\vec r, t) &= \frac{1}{2} \left\{\psi^*(\vec r, t) \frac{\hbar}{m i}\mathbf \nabla \psi(\vec r, t)  - \psi(\vec r, t) \frac{\hbar}{m i}\mathbf \nabla \psi^*(\vec r, t)  \right\}\\
&= \frac{1}{2} \left\{\psi^*(\vec r, t) \mathbf v \psi(\vec r, t)  - \psi(\vec r, t) \mathbf v \psi^*(\vec r, t)  \right\}
\end{align}
And the volume integral
\begin{align}
 \iint d^3r \vec j(\vec r, t) &= \iiint d^3r \frac{1}{2} \left\{\psi^*(\vec r, t) \mathbf v \psi(\vec r, t)  - \psi(\vec r, t) \mathbf v \psi^*(\vec r, t)  \right\},\\
&= \frac{1}{2}\left\{ \langle\vec v\rangle - \langle \vec v\rangle^* \right\},\\
&=0. 
\end{align}
Since the operator $\mathbf v$ is hermitian, the expectation value of $ \langle \vec v\rangle$ is a real number, therefore $\langle \vec v\rangle = \langle \vec v\rangle^*$.
Note that the vanished surface integral of current doesn't imply the vanished volume integral of current, as deriving from the continuity equation $\frac{\partial \rho}{\partial t} + \vec \nabla\cdot \vec j = 0$.
A: The probability density at a particular location is essentially the chance something will happen, or will be found at, or is located at, that location in the next infinitesimal amount of time $dt$ (except, loosely, that times $\tfrac{1}{dt}$ to make it per unit time). Just think of density as the chance it will appear there or be there or is there per unit time. Or even just the chance it will appear in the next second. This can vary throughout space. The reason it’s a density is technically we need a finite volume of space for finite chance of it to be in that region. We cannot have a finite chance of something being at a point, which is infinitely small (with $(x,y,z) \in \mathbb{R}^3$). This is applying the normal probability density function (pdf $f(\cdot)$) to three dimensional space.
If we integrate this density we get one. It is somewhere. There are exceptions if it is used for an event, a decay for example. In that case it usually has a constant magnitude, less than one.
Current is simply how this density changes through time. The flow of probability from one place to another. The increase in chance at a point is $\frac{\partial \rho}{\partial t}$ which is the influx of probability current:
$$ \frac{\partial \rho}{\partial t}=- \nabla \cdot \vec{j}$$
These flows mustn’t necessarily add up to zero. In fact we can look at velocity as the volume integral $\int j dv $ over all of space if probability density is of particle position and change in expected location is velocity.
But we definitely can and almost always do favor certain locations, ie higher densities
It is often assumed, but not needed for the above, that the probability density cannot change in a discontinuous manner. (We can’t have it have a 0.01 per second that it is in a certain volume of space and the next instant it is 0.03.).
A: The continuity equation is $$\frac{\partial \rho}{\partial t} + {\bf \nabla}\cdot{\bf j}=0\tag1$$ where $\rho=\Psi^*\Psi$ is the probability density and $\bf j$ is the current density.
Using the divergence theorem and integrating over a volume $V$ bounded by a surface $S$ gives $$\frac{d}{dt}\int_{V}{\Psi^*\Psi dV} =- \oint_{S}{{\bf j}\cdot{\bf dS}}\tag2$$
This means that probability is conserved. And note that $$\frac{d}{dt}\int_{\text{all space}}{\Psi^*\Psi d^3r} = \oint{{\bf j}\cdot {\bf dS}}=0\tag3$$
which I think is what you were getting at.
The value of the integral is zero because $\Psi$ and consequently $\bf j$ goes to zero at infinity, and the equality must hold for all space. This is a statement of the conservation of probability and  consequence of the condition $$\int_{\text{all space}}{\Psi^*\Psi d^3r}=1\tag4$$ which is the normalization condition, and equivalent to the statement that the probability of finding the particle anywhere in space must equal one.
To get an intuitive idea, recall that in quantum theory, particles never have exactly defined positions in space, but instead are more or less smeared out in space following a particular probability distribution given by $\rho$ which tells us the likelihood that a particle will be in a  given location.
For a quantum particle, this distribution "flows" like a fluid. This is completely analogous to charge and current.  Probability current behaves like any other current, having varying density at separate points in space corresponding to the likelihood of finding the particle at these points in space, and naturally, the particle must be located somewhere in space which is reflected in the equations (3) and (4).
