## The problematic radial momentum operator

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!