Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

206 submissions , 164 unreviewed
5,103 questions , 2,249 unanswered
5,355 answers , 22,797 comments
1,470 users with positive rep
820 active unimported users
More ...

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

+ 1 like - 0 dislike
1198 views

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 in Theoretical Physics by Dmitry hand me the Kalashnikov (735 points) [ no revision ]

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.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\varnothing$ysicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...