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  Nonlinear field redefinitions

+ 7 like - 0 dislike
750 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 (12,640 points) [ no revision ]
Most voted comments show all comments

Ok. I undo the downvote until I have time to think about what you are saying.

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://www.mat.univie.ac.at/~neum/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.

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.

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