# Electric quadrupole operator for periodic systems

Considering one component of the electric quadrupole operator $$\hat{r}_x \hat{r}_y$$, I'm wondering if the following equalities hold: $$\langle u_{n\mathbf{k}} | \hat{r}_x \hat{r}_y |u_{m\mathbf{k}} \rangle = -i\langle \partial_{k_x} u_{n\mathbf{k}} | \hat{r}_y |u_{m\mathbf{k}} \rangle = \langle \partial_{k_x} u_{n\mathbf{k}} |\partial_{k_y} u_{m\mathbf{k}} \rangle$$ where $$u_{n\mathbf{k}}=e^{-i\mathbf{k \cdot r}}\psi_{n\mathbf{k}}$$ is the cell-periodic part of the Bloch function with $$n$$ and $$\mathbf{k}$$ the band index and crystal momentum.

To my understanding, the (approximate) substitution to transform the dipole from real to k-space is: $$\mathbf{r}\to i\exp(i\mathbf{k}\cdot \mathbf{r})\nabla_k\exp(-i\mathbf{k}\cdot \mathbf{r})\tag{1}\label{1}$$ Applying this same identity for the quadrupole operator, you should get something like:

$$\langle\psi_{n\mathbf{k}}|\mathbf{r}\mathbf{r}|\psi_{m\mathbf{k}}\rangle=-\langle u_{n\mathbf{k}}|\nabla_{\mathbf{k}}\nabla_{\mathbf{k}}|u_{m\mathbf{k}}\rangle\tag{2}\label{2}$$

where the main differences between our formulas are the sign of the right hand expression and the use of $$\psi_{n\mathbf{k}}$$ in the integrals on the lefthand side.

As discussed in , this substitution is only valid for insulating systems in weak electric fields, as there should be an additional term in Eq \eqref{1} related to induced current within the bands of the material.

### References:

1. Kudin, K. N.; Scuseria, G. E. An efficient finite field approach for calculating static electric polarizabilities of periodic systems. J. Chem. Phys. 2000, 113 (18), 7779–7785. DOI: 10.1063/1.1315999.
• Thanks. Are you missing a minus sign in the second exponential in Eq. (1)? For Bloch states, the left hand side expression of Eq. (2) should be equal to the lhs expression of my equation, since $e^{i\mathbf{k \cdot r}}$ commute with $\mathbf{r}$, right? Jan 16, 2022 at 7:06
• @XiaomingWang the missing minus sign was a typo. For eq 2, $\exp(ik\cdot r)$ won't commute with $r$ because it's not just a function of $r$, but also of $k$, which doesn't commute with $r$.
– Tyberius
Jan 16, 2022 at 14:42
• Well, I think $\psi_{n\mathbf{k}}$ is a function of $\mathbf{r}$ only, while both $n$ and $\mathbf{k}$ are parameters, am I wrong? Jan 16, 2022 at 17:00
• @XiaomingWang don't think $k$ is just a parameter in this case, though we write it that way because in practice we discretize over k-space. If it was just a parameter, which value of $k$ are we putting in the substitution in equation 1? Wouldn't the right half of equation 2 be zero since the $u_{nk}$ would have no dependence $k$ dependence and it's derivative would thus be zero?
– Tyberius
Jan 16, 2022 at 18:54
• Thanks, I got you. But why k doesn't commute with r? Jan 16, 2022 at 20:44