In quantum mechanics, position coordinates x,y,z of a particle are replaced with position operators and the components of the momentum vector are replaced with the operators . The most important property of these operators is the commutation relation between a coordinate and the corresponding momentum component:
In the position representation, the momentum operator has the form of a differential operator
and the operator is just a multiplication of the wavefunction with the variable x.
Can these results be extended to coordinate systems that are not Cartesian? Consider a plane polar coordinate system where the position is given as the pair , with
The “momentum” corresponding to the variable is obviously the angular momentum operator
But what about the radial momentum operator? It would have to be
in the position representation, but is it an acceptable quantum operator?
Actually, it is not. An operator corresponding to a measurable quantity has to be Hermitian, and it’s easy to see that if we have a wavefunction similar to a hydrogen atom ground state
where a is a real constant, the state is an eigenstate of with an imaginary eigenvalue . Therefore it’s not possible to use the radial momentum operator as an observable.
Another fun way to see this is to consider a “radial translation operator”, which is generated by the radial momentum:
and operate with it on a rotation symmetric function . The result is
However, because a radius coordinate always has to be a nonnegative number, we have irreversibly lost all information about the values of the function for here, which means that this radial translation operator does not have an inverse, and therefore can’t be unitary as would be expected if were Hermitian!