Consider a Euclidean path integral say in a real scalar field theory. $$ \int d[\phi]\exp(-I[\phi]) $$ In the semiclassical approximation, we consider stationary points of the action and expand around them. Now, consider I want to make a semiclassical expansion of the generating functional $$ Z[J]=\int d[\phi]\exp\bigg(-I[\phi]-\int d^4x\,J\phi\bigg) $$ I have a doubt, should I consider saddles of $I$ of all the sourced action? $$ I_J[\phi]\equiv I[\phi]+\int d^4x\,J\phi $$ Naively I would guess that I gotta take the saddles of the whole exponent, but my biggest concern then is that if I take saddles of the sourced action, the stationary field configurations will, in general, have $J$ dependence, and thus after expanding the action around these stationary points $\phi_s$, taking functional derivatives of $Z$ with respect to $J$ will be very dirty since I will have $J$ dependence in every place I have a $\phi_s$.

So, saddles of the sourced or the unsourced action?

Consider a Euclidean path integral say in a real scalar field theory. $$ \int d[\phi]\exp(-I[\phi]) $$ In the semiclassical approximation, we consider stationary points of the action and expand around them. Now, consider I want to make a semiclassical expansion of the generating functional $$ Z[J]=\int d[\phi]\exp\bigg(-I[\phi]-\int d^4x\,J\phi\bigg) $$ I have a doubt, should I consider saddles of $I$ or those of all the sourced action? $$ I_J[\phi]\equiv I[\phi]+\int d^4x\,J\phi $$ Naively I would guess that I gotta take the saddles of the whole exponent, but my biggest concern then is that if I take saddles of the sourced action, the stationary field configurations will, in general, have $J$ dependence and thus after expanding the action around these stationary points $\phi_s$, taking functional derivatives of $Z$ with respect to $J$ will be very dirty since I will have $J$ dependence in every place I have a $\phi_s$.

So, saddles of the sourced or of the unsourced action?

Consider a Euclidean path integral say in a real scalar field theory. $$ \int d[\phi]\exp(-I[\phi]) $$ In the semiclassical approximation we consider stationary points of the action, and expand around them. Now, consider I want to make a semiclassical expansion of the generating functional $$ Z[J]=\int d[\phi]\exp\bigg(-I[\phi]-\int d^4x\,J\phi\bigg) $$ I have a doubt, should I consider saddles of $I$ of all the sourced action? $$ I_J[\phi]\equiv I[\phi]+\int d^4x\,J\phi $$ Naively i would guess that I gotta take the saddles of the whole exponent, but my biggest concern then is that if I take saddles of the sourced action, the stationary field configurations will in general have $J$ dependence, and thus after expanding the action around these stationary points $\phi_s$, taking functional derivatives of $Z$ with respect to $J$ will be very dirty since I will have $J$ dependence in every place I have a $\phi_s$.

So, saddles of the sourced or the unsourced action?

Consider a Euclidean path integral say in a real scalar field theory. $$ \int d[\phi]\exp(-I[\phi]) $$ In the semiclassical approximation, we consider stationary points of the action and expand around them. Now, consider I want to make a semiclassical expansion of the generating functional $$ Z[J]=\int d[\phi]\exp\bigg(-I[\phi]-\int d^4x\,J\phi\bigg) $$ I have a doubt, should I consider saddles of $I$ of all the sourced action? $$ I_J[\phi]\equiv I[\phi]+\int d^4x\,J\phi $$ Naively I would guess that I gotta take the saddles of the whole exponent, but my biggest concern then is that if I take saddles of the sourced action, the stationary field configurations will, in general, have $J$ dependence, and thus after expanding the action around these stationary points $\phi_s$, taking functional derivatives of $Z$ with respect to $J$ will be very dirty since I will have $J$ dependence in every place I have a $\phi_s$.

So, saddles of the sourced or the unsourced action?