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
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Does symmetry breaking in the course of the RG flow contradict the c- or a-theorem respectively?

+ 3 like - 0 dislike
2021 views

The c-theorem says, that for 2d QFTs there exists a function  $C(g_i,\mu)$ of the coupling constants and the energy scale, which monotonically decreases in the course of the RG flow. It measures roughly speaking the number of degrees of freedom of the EFT valid at the scale $\nu$, and means that the RG flow has to be reversible.

A higher dimensional (in particular 4d) generalization is the a-theorem, where the coefficient $a$ of the Weyl anomaly takes the role of the function $C$.

How are these two theorems compatible with the breaking of symmetries in the course of the RG flow, which as I understand it rather increases the number of degrees of freedom? For example thinking about starting from a GUT at the unification scale,  the number of coupling constants certainly increases when flowing to the low-energy observable scales... ?

A more general question would be under what conditions can the c- or a-theorems not be applied, for example because the number of relevant quantities needed to describe the EFT when flowing to lower energy-scales increases? Or should such cases not occur ...?

asked Jun 22, 2014 in Theoretical Physics by Dilaton (6,240 points) [ no revision ]

1 Answer

+ 6 like - 0 dislike

Spontaneous breaking of a (global or gauge) symmetry does not increase the number of degrees of freedom, it preserves it. When a spontaneous symmetry breaking happens, the system goes from a unstable vacuum to a true stable vacuum. This implies that what are the physical excitations around the vacuum change but it is simply a reorganization of the same degrees of freedom. This is obvious with a Lagrangian at the classical level: if a symmetry is broken by some expectation values of the fields then to understand the theory after the symmetry breaking, we have to expand the fields around these expectation values: it is just the same theory expanded around a different point so the number of degrees of freedom are the same.

In general, the reorganization of the degrees of freedom is non-trivial (example: Higgs mechanism for spontanously broken gauge symmetries, "absorption of the Goldstone bosons by the broken gauge bosons") and will deeply affect the RG flow: for example, in the Higgs mechanism, some gauge bosons become massive and so are integrated out in the IR whereas they would survive as massless particles in the unbroken theory.

The GUT example is not correct: if we start with a big gauge group which is broken by a Higgs mechanism to several small gauge groups, the number of degrees of freedom remains the same as argued before (they are more groups but they are smaller) and after pushing the RG flow and integrating the massive gauge bosons, the number of degrees of freedom decreases.

In fact, the a- and c-theorems apply without the assumption of no symmetry breaking. It is one of the greatest interest of this results. More precisely: in general, it is very difficult to know if a spontaneous symmetry breaking occurs in a strongly coupled QFT whereas this question is of central importance to understand the IR behavior of the theory. But the a-,c- type theorems give us some tools: one can say, let us assume that such symmetry is (or not) spontaneously broken then determine the IR theory with this assumption and compare it with the UV theory: if this violates the general a-,c-type inequalities, then this means that in fact the symmetry is not (or is) broken.

answered Jun 24, 2014 by 40227 (5,140 points) [ no revision ]

Thanks (again) for these very nica answer and the valuable clarifications :-)
 

I am a little mystified about bound-states in this picture. There can't possibly be a bound on the number of bound states, and you can surely make an effective theory where the bound states are considered elementary. Perhaps the catch is that the bound states must be considered unbound in order to have a relativistic theory (the binding energy is less than the mass energy)? I don't know.

Ron Maimon: you are right. In fact, I have hesitated for a few days before giving this answer because I was thinking: what about the tower of hadrons in QCD...?(which is exactly your point) I think that the notion of "number of degrees of freedom" is not well-defined in general. The quantities appearing in the a-,c-theorems are rather kind of "number of massless degrees of freedom" and coincide with the "number of degrees of freedom" only for theories with only massless degrees of freedom, such as the CFTs and so in particular for the CFT UV and IR limits of a given QFT (the fact that the "a" and "c" quantities are only sensible to massless states is clear from the fact that they are some anomalies). 

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
...