In quantum mechanics, position coordinates x,y,z of a particle are replaced with position operators \large \hat{x},\hat{y},\hat{z} and the components of the momentum vector \mathbf{p} are replaced with the operators \large \hat{p_x},\hat{p_y},\hat{p_x}. The most important property of these operators is the commutation relation between a coordinate and the corresponding momentum component:

\large [\hat{x},\hat{p_x}] = i\hbar

In the position representation, the momentum operator has the form of a differential operator

\large \hat{p_x} = -i\hbar \frac{\partial}{\partial x}

and the \hat{x} 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 \large (r,\theta), with

\large x = r\cos \theta
\large y = r\sin \theta

The “momentum” corresponding to the variable \large \theta is obviously the angular momentum operator

\large \hat{L}_z = -i\hbar \frac{\partial}{\partial \theta}.

But what about the radial momentum operator? It would have to be

\large \hat{p}_r = -i\hbar \frac{\partial}{\partial r}

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

\large \psi (r) = Ae^{-ar}

where a is a real constant, the state is an eigenstate of \hat{p}_r with an imaginary eigenvalue \large ia\hbar. 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:

\large \hat{T}_r (\Delta r) = \exp \left(\frac{i\Delta r \hat{p}_r}{\hbar}\right)

and operate with it on a rotation symmetric function \psi (r). The result is

\large \hat{T}_r (\Delta r)\psi (r) = \psi (r + \Delta r)

However, because a radius coordinate always has to be a nonnegative number, we have irreversibly lost all information about the values of the function \large \psi (r) for \large r<\Delta r here, which means that this radial translation operator does not have an inverse, and therefore can’t be unitary as would be expected if \large \hat{p}_r were Hermitian!