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In general, the orbital magnetic dipole moment operator is defined as: $$- {e\over2} \mathbf{r \times v} \tag{1}$$ $\mathbf{v=p}/m$ is true only for simple Hamiltonian like $-1/2\ \nabla^2+V$, but not for normal DFT implementations with semi-local or nonlocal pseudopotentials. For the PAW method (projector augmented wave), e.g., implemented in vasp, if I would evaluate the matrix elements of the orbital magnetic dipole moment operator within the PAW spheres, i.e., $\langle \phi_i|\mathbf{r \times v}|\phi_j\rangle$, where $\phi$ is the all-electron partial wave, then can I use the linear momentum representation?

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2 Answers 2

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  1. Velocity is not really used in quantum mechanics, since it is the momentum that is the canonical variable. Leave velocity to classical physics.
  2. The momentum operator ${\bf p}$ makes sense for whatever Hamiltonian. It will just only share eigenstates with the Hamiltonian in cases where $\hat{H}=\hat{H}({\bf p})$, e.g. the free particle $\hat{H}={\bf p}^2/2m$.
  3. The concept of the angular momentum operator $\hat{\bf L}$ originates from the central field problem, where the potential is purely radial: $V({\bf r})=V(r)$. In this case you can show that in classical physics ${\bf L}={\bf r} \times {\bf p}$ is a constant of motion. However, it turns out that also in quantum mechanics $\hat{H}$ and $\hat{\bf L}$ are compatible operators, i.e. $[\hat{H},\hat{\bf L}^2]=0$ and $[\hat{H},\hat{\bf L}_z]=0$, meaning that the eigenstates of the system are eigenstates of both the Hamiltonian and the angular momentum operator.
  4. You can evaluate matrix elements of ${\bf L}={\bf r} \times {\bf p}$ within the quantum mechanical sense. There are two obvious ways to do this. If you are in coordinate space then ${\bf r}={\bf r}$ and ${\bf p}=-i\hbar \nabla_{\bf r}$. However, you can also evaluate in momentum space: ${\bf p}={\bf p}$ and ${\bf r}=i\hbar \nabla_{\bf p}$.
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    $\begingroup$ +1 Nice overview. However, my guess is that xmW has in mind the issues that arise in periodic boundary conditions with the operator $\hat{\mathbf{r}}$, and the question is whether this is also an issue if only considering the PAW spheres. But I may have misunderstood the question. $\endgroup$
    – ProfM
    Commented Sep 10, 2020 at 10:41
  • $\begingroup$ 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. $\endgroup$ Commented Sep 10, 2020 at 11:23
  • $\begingroup$ @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. $\endgroup$ Commented Sep 10, 2020 at 11:29
  • $\begingroup$ So the question is can I use $\mathbf{v=p/}m$ within the PAW sphere? $\endgroup$ Commented Sep 10, 2020 at 11:40
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    $\begingroup$ ${\bf p}$ 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. $\endgroup$ Commented Sep 10, 2020 at 14:03
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I think we can safely use $\mathbf{v=p}/m$ to evaluate the orbital magnetic dipole moment $-e/2m(\mathbf{r\times p})$ within the PAW sphere, since within the PAW sphere the all-electron wave functions are subjected to the Hamiltonian $H=\mathbf{p}^2/2m + V$[Phys. Rev. B 50, 17953 (1994)].

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