Solution
= Solution
Expanding the <scalar product> gives $j_\alpha=[nQ^2/(2k_BT)]\sum_j f\,r_{ij,\alpha}\sum_\beta r_{ij,\beta}\mathcal E_\beta$. Comparing coefficients of the independently variable field components yields
$$
\boxed{\sigma_{\alpha\beta}=\frac{nQ^2}{2k_BT}\sum_{j=nn(i)}f(\omega_0,\Delta E_{ij},T)\,r_{ij,\alpha}r_{ij,\beta}}.
$$
The <tensor> is symmetric and <positive semidefinite> because its contraction with any vector twice is a nonnegative sum of squares.