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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?

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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).

I dont feel like I should tell the whole technicalities of the story, but the main idea that gauge invariance forces us to redefine concepts like velocity and current, and the replacement of the momentum by the canonical momentum in all of the equtions is the way to do so.

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