\nonumber
{{\mathbf{j}}}&= -Di{{\mathbf{p}}}j^0+i\sigma (-{{\mathbf{p}}}A^0+\omega {{\mathbf{A}}})\\
\nonumber
&~~~-\sigma\tau_J \omega  (-{{\mathbf{p}}}A^0+\omega {{\mathbf{A}}})-\kappa_B ([{{\mathbf{p}}}\cdot{{\mathbf{A}}}]{{\mathbf{p}}}-{{\mathbf{p}}}^2{{\mathbf{A}}}) \\
\label{eq:constituent-eq-p} 
&~~~+{\cal O}(p^2A, p^2j^0).\end{aligned}$$ Let us solve these equations for the transverse and the longitudinal components of the current. By introducing the transverse component, $j^i_T(p)\equiv P^{ij}_T(p) j^j(p)$, we get the solutions as follows: $$\begin{aligned}
{{\mathbf{j}}}_T(p)&= \left(i\sigma\omega-\sigma\tau_J\omega^2+\kappa_B{{\mathbf{p}}}^2
\right) {{\mathbf{A}}}_T \nonumber \\
&~~~ +{\cal O}(\omega^3{{\mathbf{A}}}_T, \omega{{\mathbf{p}}}^2{{\mathbf{A}}}_T, {{\mathbf{p}}}^4{{\mathbf{A}}}_T),\\
\nonumber
\omega j^0(p)&= -iD{{\mathbf{p}}}^2 j^0
-i{{\mathbf{p}}}^2\sigma A^0 \\
&~~~+ {\cal O}(\omega{{\mathbf{p}}}^2 A^0, {{\mathbf{p}}}^4 A^0, \omega{{\mathbf{p}}}^2 j^0, {{\mathbf{p}}}^4 j^0) \nonumber \\
&~~~+({\text {terms that are proportional to }}{{\mathbf{p}}}\cdot{{\mathbf{A}}}).\end{aligned}$$

By using the linear response theory, the induced current is written as $$\begin{aligned}
\label{eq:linear-response}
j_\mu(p)=-G^R_{\mu\nu}(p)A^\nu(p),\end{aligned}$$ from which we obtain $$\begin{aligned}
\label{eq:hydro-GR-T}
G_{T}(p) &= i\sigma\omega-\sigma\tau_J\omega^2+\kappa_B{{\mathbf{p}}}^2
+{\cal O}(\omega^3, \omega{{\mathbf{p}}}^2, {{\mathbf{p}}}^4),\\
\label{eq:hydro-GR-L}
G^R_{00}(p) &= i\sigma{{\mathbf{p}}}^2\frac{1+{\cal O}(\omega,{{\mathbf{p}}}^2)}{\omega+iD{{\mathbf{p}}}^2+{\cal O}(\omega{{\mathbf{p}}}^2,{{\mathbf{p}}}^4)}.\end{aligned}$$ We note that there is a pole at $\omega=-iD{{\mathbf{p}}}^2$ in the longitudinal channel, which we call the diffusion pole. This is a novel structure that appears only at finite ${{\mathbf{p}}}$. This pole appears as a peak in the spectral function, $$\begin{aligned}
\label{eq:hydro-rho-L}
\rho_{00}(p) &= \sigma{{\mathbf{p}}}^2
\frac{\omega}{\omega^2+(D{{\mathbf{p}}}^2)^2} ,\end{aligned}$$ while at $|{{\mathbf{p}}}|=0$, it reduces to the delta function, $$\begin{aligned}
\label{eq:hydro-rho-L-p=0}
\rho_{00}(\omega,{{\mathbf{0}}}) 
&= \pi\frac{\sigma}{D}\omega\delta(\omega) .\end{aligned}$$

For constructing our sum rules, we, in principle, also need to evaluate the zero temperature part, which should be subtracted 