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xmW
5:08 PM
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.
 
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A: Evaluate the matrix elements of the orbital angular momentum within the PAW sphere

Susi Lehtola Velocity is not really used in quantum mechanics, since it is the momentum that is the canonical variable. Leave velocity to classical physics. The momentum operator ${\bf p}$ makes sense for whatever Hamiltonian. It will just only share eigenstates with the Hamiltonian in cases where $\hat{H}=\...

 
xmW
@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 ${\bf p}/m$.
 

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