Compatible Observables in QM vs OQS In elementary QM courses we always consider that components of momentum vector form a complete set of commuting observables.
I am confused whether this is an input to our theory or whether we somehow derive this? As far as I can recall we take this as an intuition that the components of momentum vector will form a complete set because momentum measured in one direction cannot affect momentum measured in another direction. On the other hand, since both momentum and position are promoted to operators (as compared to classical mechanics) we "expect" them to defy the intuition from classical mechanics.
If so, then I wonder what modern quantum theory says about entanglement in such scenarios. Say I prepare a state with two entangled electrons such that one of them goes in x direction and the other goes in y direction. Do the components of momentum operator still form a complete set of commuting variables? How?
 A: Good question. The supposition that the components of four vector momentum form a complete set of commuting observables is taken as an input.
The trivial commutation relations for four momentum vector components is equivalent (after a Fourier transform) to the relation $$[\hat{x_{a}},\hat{x_{b}}]=0.$$ That is equivalent to say that the underlying spacetime geometry is taken to be commutative. A reasonable hypothesis in light of the fact that we haven't observed (yet) spacetime "fuzziness".
The point is that something like $$[\hat{x_{a}},\hat{x_{b}}]= i\hbar\delta_{ab}$$ is perfectly fine, at least in principle and in the non-relativistic setup.
It's worth to say that maybe noncommutative geometry appear in some relevant physical situations(example), but nothing too relevant for elementary purposes.
Completeness of the set of four momentum operators is also an input because classically any system has momentum, so it is reasonable to expect that quantum mechanically any state can be labbeled with a reasonable expectation value for the momentum operator.
