# Relationship between electronic current and electron momentum?

I remembered when reading Laughlin's famous argument for quantum Hall, it implied the actual current should be proportional to the electron's mechanical momentum ($$p-eA/c$$) instead of $$p$$ itself. Why is it the case?

When one deals with quantum problems involving a gauge freedom, all physical observables must be gauge invariant. This seems simple, but this simple claim means that for an electron, the operator $p$ is not observable, as it is not gauge invariant.
The way to proceed is to say that actually, we have been measuring the operator $(p-eA/c)$ all along, and if we are in a situation where $A=0$ the 2 concepts merge. This also means that if we want to assign a velocity operator to the electron, we have to use $v_{op}=\frac{1}m (p-eA/c)$.
The next step is to define a current operator using $v_{op}$ by $j_{op}=\sum _{i}e\delta (x-x_i)v_{op}$, by analogy with the classical $j=\rho v$ . (actually there is an issue with symmetrizing this for hermiticity).