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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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 $\hat{H}=\... @ProfM The position operator$\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\$.