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

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Is it possible to have transformations that transform the action and the measure while leaving the functional integral invariant?

+ 2 like - 0 dislike
1045 views

Anomalous symmetries are those for which the Lagrangian stays invariant but the measure of the functional integral does not. I wonder if there are transformations that change both the action and the measure, such that the functional integral is left invariant, is this possible?

asked Jan 31, 2016 in Theoretical Physics by Dmitry hand me the Kalashnikov (735 points) [ revision history ]

Any arbitrary change of variables does this, no?

@RyanThorngren: Not quite; see my answer.

1 Answer

+ 3 like - 0 dislike

On a formal level, the functional integral is an infinite-dimensional generalization of a scaled Lebesgue integral in $R^n$. For the latter, any diffeomorphism (bijective $C^1$ transformation) of the vector $x$ over which the integration extends can be compensated by changing the measure by the corresponding absolute value of the Jacobian determinant. In infinite dimensions, $x$ becomes a field, but the transformation law should still be valid.

However, since there is no proper definition of the functional integral (and an infinite-dimensional Lebesue integral doesn't exist) it is not clear to which extent the above reasoning is valid. People use it (just like the functional integral itself) on a heuristic basis.

Note that the property of renormalizability is not preserved under nonlinear field transformations, but renormalizability is needed to give the functional integral a well-defined and unique perturbative meaning. Thus performing a nonlinear tranformation of fields in the context of functional integrals is a hazard that easily leads to inconsistent results.

A proper answer to your question must wait for the day where functional integrals have a meaningful nonperturbative definition.

answered Jan 31, 2016 by Arnold Neumaier (15,787 points) [ no revision ]

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$\hbar$ysicsOverfl$\varnothing$w
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
...