How can I derive the quadrupole momentum tensor in terms of dyadic product of wave vector and electric field?

Question

In page 3 of PhysRevB.78.121101 they used an expression of electric quadrupole tensor as

$Q_{ij} = i \alpha_Q(k_i E_j + k_j E_i)$ or $\overleftrightarrow{Q}=i\alpha_Q(\vec{k}\vec{E}+\vec{E}\vec{k})$ , which is different from that derived from vector expansion in Jackson’s book $Q_{ij}=\int_V(3x'_ix'_j-r'^2\delta_{ij})\rho(x')dv'$ or $\overleftrightarrow{Q}=\int_V(3\vec{x}'\vec{x}'-r'^2\overleftrightarrow{I})dv'$. I am wondering how can I derive or understand this expression?

From formula (1.129) in Multipole Theory In Electromagnetism Classical, Quantum, And Symmetry Aspects, With Applications $\vec{D}=\epsilon_0\vec{E}+\vec{P}-\frac{1}{2}\nabla\cdot\overleftrightarrow{Q}$. At weak field approximation we have $\vec{P}=\chi \vec{E}$, I think the electric quadrupole polarizability $\alpha_Q$ can be defined like this. But I fail to prove or derive it.

Answer 1

You can define the quadrupole interaction tensor in electrostatic situations as $\partial_i E_j=\partial_i\partial_j \phi$.In this case, the expression given in Jackson represents the (pure) quadrupolar component of $\phi$. The expression that you find in the paper is for an electromagnetic (plane) wave $E=E^0 exp(-ikr-i\omega t)$ If you calculate the symmetrized derivative $\partial_iE_j+\partial_jE_i$, you get $-i(k_iE_j+K_jE_i)$. So the two definitions are related to each other, but they don't describe the same object.