# In the semiclassical approximation, should I expand the generating functional around saddles of the sourced or the unsourced action?

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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?

asked Jun 11, 2017

You get a far better approximation (but of course far less tractable expressions) if you expand around the sourced action.

The thing is that I want to prove that the trivial saddle yields perturbation theory, as it is claimed in many places, but I do not know what saddle to use. In order to show this, do you know if the saddle of the unsourced is enough?@ArnoldNeumaier

For perturbation theory it should be enough, since there only infinitely small sources matter.

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