I am reading this review paper "Introduction to Generalized Global Symmetries in QFT and Particle Physics". In equation (2.43)-(2.47), the paper tried to prove that when $$U_g(\Sigma_2)=exp\{\frac{i\lambda}{g^2}\oint_{\Sigma_2} *F\},$$ $$U_g(\Sigma_2)=\exp\{\frac{i\lambda}{g^2}\oint_{\Sigma_2} *F\},\tag{2.41}$$ the operator correspond to the 1-form $U(1)_E$ symmetry, acts on a Wilson line $W(q,\gamma)=e^{iq\int_\gamma A}$,$$W(q,\gamma)=e^{iq\int_\gamma A},\tag{2.42}$$ the result is
$$ \langle U_g(\Sigma)W(q,\gamma)\rangle=e^{iq\lambda\ Link(\Sigma_2,\gamma)}\langle W(q,\gamma)U_g(\Sigma')\rangle. $$$$ \langle U_g(\Sigma)W(q,\gamma)\rangle=e^{iq\lambda\ Link(\Sigma_2,\gamma)}\langle W(q,\gamma)U_g(\Sigma')\rangle.\tag{2.43} $$
The technique the paper used is that, consider inserting the Wilson loop in path integral
$$ \int[DA] e^{iS}W(q,\gamma)=\int [DA] e^{iq\int_\gamma A+iS}=\int [DA] e^{iq\int_{M_4}\delta^3(\gamma)A+\frac{1}{2g^2}F\wedge*F} $$$$ \int[DA] e^{iS}W(q,\gamma)=\int [DA] e^{iq\int_\gamma A+iS}=\int [DA] e^{iq\int_{M_4}\delta^3(\gamma)A+\frac{1}{2g^2}F\wedge*F}\tag{2.44} $$
the equation of motion change to
$$ d*F=qg^2\delta^3(\gamma). $$$$ d*F=qg^2\delta^3(\gamma).\tag{2.46} $$
My question is, why does inserting an operator in path integral change the equation of motion? If inserting operator really does change the e.o.m, then it seems $\langle U_g(\Sigma)W(q,\gamma)\rangle$ further change the e.o.m. because we have one additional operator ($U_g$) inserted.
My question is, why does inserting an operator in path integral change the equation of motion?
If inserting operator really does change the e.o.m, then it seems $\langle U_g(\Sigma)W(q,\gamma)\rangle$ further change the e.o.m. because we have one additional operator ($U_g$) inserted.
