Velocity is actually used in quantum mechanics, e.g., when evaluating transition dipole moments and orbital magnetization for solids. People use $\mathbf{r \times v}$ instead of $\mathbf{r \times p}$ to calculate the orbital magnetization, [see PRB 74, 024408], for normal DFT implementations. My question is can I use $\mathbf{r \times p}$ within the PAW sphere.

@ProfM The position operator $\mathbf{r}$ is well defined within the PAW spheres, so there should be no issues considering the matrix elements of the position operator.

So the question is can I use $\mathbf{v=p/}m$ within the PAW sphere?

Both ${\bf p}$ and ${\bf r}$ are always well defined: they're observables.

PRB 74, 024408 defines the "velocity" as ${\bf v}=i[\hat{H},{\bf r}]$ and notes that the value might not be ${\bf p}/2m$ since the Hamiltonian can also include field terms like ${\bf p} \cdot {\bf A}$, spin-orbit interactions or the pseudopotentials. So the answer would be no. But this has nothing to do with your question about the orbital angular momentum!

${\bf p}$ is always ${\bf p}$, and ${\bf L}$ is always defined as ${\bf L} = {\bf r} \times {\bf p}$. It's just a wholly other question whether ${\bf L}$ makes sense in your calculation, since ${\bf L}$ is only a good quantum number in the central field problem.

I see that equation 18 in the paper has the form ${\bf R} \times {\bf v}$ but ${\bf v}$ is not the classical velocity ${\bf p}/m$.