Remark: The classical self-response

The background-field prescription drops from (eq:fluct) the $u$-independent terms, whose retarded response would be ${{\Delta }_{\mathrm{R}}}^{\mathrm{gr}}(-\frac{1}{2}{\sigma }_{\epsilon }{G}_{\mathrm{lin}}[h_{\mathrm{cl}}])$ at order ${\kappa }^0$ and ${{\Delta }_{\mathrm{R}}}^{\mathrm{gr}}(-\frac{1}{2}{\sigma }_{\epsilon }(G_{\mathrm{lin}}[k_{\mathrm{cl}}]+{\mathcal{G}}_2[h_{\mathrm{cl}},h_{\mathrm{cl}}]))$ at order $\kappa$. Including them would make the quantum fluctuation carry a classical, field-independent shift, and under $h_{\mathrm{cl}}\mapsto h_{\mathrm{cl}}+K\xi$ the second would change by ${{\Delta }_{\mathrm{R}}}^{\mathrm{gr}}(-{\mathcal{Q}}^{(1)}[K\xi ]h_{\mathrm{cl}})={\mathcal{L}}_{\xi }{h}_{\mathrm{cl}}-{{\Delta }_{\mathrm{R}}}^{\mathrm{gr}}{\mathcal{L}}_{\xi }{P}_{\mathrm{gr}}{h}_{\mathrm{cl}}$ up to the gauge-fixing part, where $P_{\mathrm{gr}}{h}_{\mathrm{cl}}\ne 0$ off shell is a c-number: the frozen two-jet law for $k_{\mathrm{cl}}$ would acquire a second copy of ${\mathcal{L}}_{\xi }{h}_{\mathrm{cl}}$. The prescription keeps $k_{\mathrm{cl}}$ off shell and arbitrary, as in X-A. The classical self-response is named and not opened.