• 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.


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


(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 ...

  Field redefinitions and new counterterms

+ 6 like - 0 dislike

My question was motivated by my attempt to answer this question. Suppose we are given an action and we make a change of variables such that the theory is non-renormalizable. Does the new theory then require an infinite number of counterterms?

As an explicit example lets consider the situation brought up in the linked question (though I change notation for my convenience). We start with the Lagrangian, $${\cal L}= \frac{1}{2}\partial^\mu\phi_0\partial_\mu\phi_0-\frac{1}{2}m^2\phi_0^2$$

Then we make the substitution, If I make $\phi_0=\phi+\frac{\lambda}{M} \phi^2$ such that $\lambda$ is dimensionless and $M$ is some mass scale. Then the Lagrangian is $${\cal L}= \frac{1}{2}\partial^\mu\phi\partial_\mu\phi-\frac{1}{2}m^2\phi^2+2\frac{\lambda}{M}\phi\partial^\mu\phi\partial_\mu\phi-\frac{\lambda}{M} m^2\phi^3 + 2\frac{\lambda^2}{M^2}\phi^2\partial^\mu\phi\partial_\mu\phi-\frac{1}{2}\frac{\lambda^2 }{M^2}m^2\phi^4$$

Now originally we could have found all the counterterms with calculating a few simple diagrams. On the one hand I'd think that since we still have a single coupling, $\lambda$, we should still have the same number of counterterms in the new theory. On the other hand I've learned that operators get renormalized, and not couplings, so since we have more operators we also need more counterterms. How many counterterms does this new theory actually need?

This post imported from StackExchange Physics at 2014-06-19 11:27 (UCT), posted by SE-user JeffDror
asked Jun 17, 2014 in Theoretical Physics by JeffDror (650 points) [ no revision ]
retagged Jun 19, 2014
Haven't you forget a Jacobian with this change of variable? I think that they usually do the job to make the theory after change of variable ok.

This post imported from StackExchange Physics at 2014-06-19 11:27 (UCT), posted by SE-user Adam

Even without Lagrangian, it is clear that the new $\phi$ is expressed via the old $\phi_0$ and $\lambda$ in a complicated way. If you want to develop a series in powers of $\lambda$, try it first in this variable change to make sure it is sensible. Only one root out of two makes it possible.

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:
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