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 ...

  Nonlinear field redefinitions

+ 7 like - 0 dislike
3827 views

I'm trying to understand why nonlinear field redefinitions are valid. They seem very peculiar to me since they take a renormalization QFT to a non-renormalizable QFT. Aren't there any subtlities associated with such a transformation? In particular is it still valid to impose the renormalizability requirement on the original Lagrangian (prior to the nonlinear transformation) if you will be making such a transformation? 


To be concrete let me present the context in which this came up. I'm watching lectures on Effective Field Theory. When deriving different representations of a nonlinear sigma model the professor starts out with a linear sigma model:
\begin{align} 
{\cal L} _\sigma & = \frac{1}{4} \mbox{Tr} \left( \partial _\mu \pi \partial ^\mu \pi \right)  + \frac{ \mu ^2 }{ 4 } \mbox{Tr} \left( \pi ^\dagger \pi \right) - \frac{ \lambda }{ 4 !} \left( \mbox{Tr} \left( \pi ^\dagger \pi \right) \right)  ^2   
\end{align} 
with $ \pi = \sigma + i {\vec \tau} \cdot {\vec \pi} $ (${\vec \tau} $ are the Pauli matrices). He then performs a few field redefinitions one of which being,
\begin{align} 
& S = \sqrt{  (\tilde{\sigma}  +v )  ^2 + {\vec \pi} ^2 } - v \\ 
& \vec{ \phi } = \frac{ v \vec{ \pi } }{ ( \tilde{\sigma} + v ) ^2 + \vec{ \pi } ^2 } 
\end{align} 
This (if I understand correctly) leads to an infinite number of new terms in the Lagrangian and the Lagrangian ceases to be renormalizable. Why is this valid, or is it only true in the low energy regime?

asked Apr 6, 2014 in Theoretical Physics by JeffDror (650 points) [ no revision ]

It's a low-energy regime thing--- the theory is expanded around the vacuum value, and the low-energy excitations are along the tangent space.

2 Answers

+ 4 like - 0 dislike

The S-matrix is field-redefinition invariant, provided that the new operator and the old operator have non-trivial overlap on one-particle states. In other words, if $\phi$ created a one particle state, then when the new operator acts on the vacuum, it can create other stuff too, but it needs to also create some of that same one particle state. Individual off-shell correlators are not invariant under this field redefinition, but on-shell S-matrix elements are, essentially because of LSZ reduction. I believe this is described in Weinberg Chapter 10.

answered Apr 7, 2014 by DanielK (40 points) [ no revision ]

Thanks, I'll take a look at Weinberg. If I understand correctly, then the redefinition should hold at all energies (since the field redefinition shouldn't effect whether or not the one-particle states overlap). Does this contradict RonMaimon's view in the comment, that it only works at low energies?

A proper field redefinition works at all energies, but this redefinition is not sensible, it doesn't have a single valued inverse except locally, near a single vacuum, for small fluctuations. The resulting theory is superficially non-renormalizable (I say "superficially" because if you do a proper field redefinition, it doesn't do anything to renormalizability), so that you can only practically use the new action to extract the low energy theory, by taking low energy limit, so that the interactions scale to zero at large distances. Sorry for being confusing.

Ahh, I see. Thanks, that makes sense!

+ 1 like - 0 dislike

Renormalization is only a condition for _perturbative_ well-definedness. Nonlinear transformations change the meaning of the perturbative expansion, hence may make a renormalizable theory nonrenormalizable and vice versa. In the (apart from its mathematically unresolved definition nonperturbative) path integral formulation, nonlinear transformations are very natural, generalizing nonlinear transformations of a finite-dimensional integral $\int_{R^n} e^{i S(x)} dx$ via the substitution rule.

Note that upon discretizing a field theory to a finite number of space-time points reduces the path integral to such an integral. In infinite dimensions (i.e., for a full QFT) subtle things must be taken account of to get valid results, and this is an art, not a science, as so far nobody has found a logically sound definition of the path integral for a nonquadratic action.

answered Apr 12, 2014 by Arnold Neumaier (15,787 points) [ no revision ]
Most voted comments show all comments

Hi Arnold, I wonder how you got to know the stuff on renormalization and S-matrix, which don't seem to be any standard textbook material. Does it come from years of gathering bits and pieces or is there some book talking about all these?

Much of it comes from years of gathering bits and pieces. I know many things that _should_ be in textbooks but I had to learn them the hard way. Perhaps I'll write one day my own textbook on QFT. For the moment, I can offer only my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html , which, however, is still silent on many things, due to lack of time.

I've always wondered why you chose to make your FAQ with text documents. This has the disadvantage of not having LaTeX, advanced formatting, etc...  

I'd love to read your book on these topics.

I started the FAQ around 1996 when internet support was still poor and communication was mainly through the sci.physics mailing list. The FAQ is book-size, hence quite difficult to convert, and it is impossible to do it automatically. Every now and then, when I update single entries, I convert them to html, which is already nontrivial.

In principle, should there be a desire for having a FAQ section within PO, I could imagine having the FAQ integrated in suitably modified form into such a FAQ section.

Most recent comments show all comments

I'm not sure I'm following you. 

What do you understand by perturbative renormalizability? To my understanding, it is not enough if you can extract or predict an infinite number of observables encoded in the S-Matrix from a finite number of observations (up to Landau-pole-like problems and non-convergence of the perturbative series). Therefore, if the S-Matrix is invariant, both theories (or both descriptions of the theory) are perturbatively renormalizable.

1. Renormalizability means that the perturbation series can be made finite to arbitrary order by finitely many counterterms. This definition makes sense only in perturbation theory, thus it is perturbative renormalizability. Changing the field changes the perturbation series and hence the applicability of the criterion. (Even linear changes change the series, but renormalization takes advantage of this.)

2. The whole setting has nothing to do with numbers of observations.  

3. The perturbative expansion (LSZ) does not define the S-matrix. It only defines (an asymptotic expansion of) that part of the S-matrix for which the elementary fields correspond to single-particle bound states. The asymptotic expansion may have arbitrarily many function interpretations, or none.

4. There are 2D non-free theories (e.g., https://en.wikipedia.org/wiki/Sine-Gordon_equation#Quantum_version) that can be nonlinearly transformed into a free theory. The perturbative expansion of the non-free version is nontrivial but spurious. The S-matrix is trivial.

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$\hbar\varnothing$sicsOverflow
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
...