Is the probability current an observable? Is the probability current in Quantum Mechanics an observable? If so, how can it me measured (directly or indirectly)?
 A: The probability current can be expressed in terms of an operator. Furthermore the electric current is charge times probability current, so measuring the probability current for a charged particle is as simple as measuring the electrical current and dividing by the charge.
The following Hermitian operator is the current operator
$$\hat{j}(r)=\frac{1}{2m}[|r\rangle\langle r|\hat{p}+\hat{p}|r\rangle\langle r|]$$
so that the probability current in state $|\psi\rangle$ is the usual expression
$$\langle\psi|\hat{j}(r)|\psi\rangle=\frac{1}{2m}[\psi^*(r)\{-i\hbar\nabla\psi(r)\}+\psi(r)\{-i\hbar\nabla\psi(r)\}^*]=\frac{\hbar}{2mi}[\psi^*\nabla\psi-\psi\nabla\psi^*]$$
When there are gauge fields around ones needs to be more careful about the momentum but the idea is still the same. For more details see this article.
A: The probability current is just the EM current divided by the charge. It is the charge-current that is the observable. The proportionality is no longer true in relativistic quantum mechanics, as the example of the Klein-Gordon equation shows. 
A: No. It is not an observable. An observable is something that can be measured in a single measurement, not just inferred statistically from a series of measurements over time. Probability currents never fit that description.
It is important to distinguish between your question "is a probability current observable" for which the clear answer is "no", and the quite similar question that is closely related:
"Is a probability current real (i.e. something that has a physical existence)?" 
The second question is up for debate and it isn't easy to think of experiments that determine if something that is not observable is real or not.
