2
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

Since we want to deal with path integral by using the stationary phase approximation, you will need to take the stationary points of the sourced action: these regions contribute coherently to the path integral (cause the phase changes very little over a small region surrounding each stationary point) giving the main contribution to the value of the quantity you compute, like in a constructive interference.

As you noticed this will be obtained for configurations depending on the source in a messy way. This dependence however is precisely what you are interested in: it will bring to the properties of the generating functional (if you derive it with respect with J you get expectation values of powers of $\phi$), even if in general this is way easier to see without making the actual computation of the partition function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.