# Evaluate the orbital magnetic dipole moment within the PAW sphere

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?

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 $${\bf r}=i\hbar \nabla_{\bf p}$$.
• +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. – ProfM Sep 10 '20 at 10:41
• 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. – Xiaoming Wang Sep 10 '20 at 11:23
• @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. – Xiaoming Wang Sep 10 '20 at 11:29
• So the question is can I use $\mathbf{v=p/}m$ within the PAW sphere? – Xiaoming Wang Sep 10 '20 at 11:40
• ${\bf p}$ 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. – Susi Lehtola Sep 10 '20 at 14:03
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)].