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 there a general relationship between the conformal weight of a field and its (classical) scaling dimension?

+ 4 like - 0 dislike
3339 views

A field $\phi(z)$ has the conformal weight $h$, if it transforms under $z\rightarrow z_1(z)$ as

$$ \phi(z) = \tilde{\phi}(z_1)\left(\frac{dz_1}{dz}\right)^h $$

The (classical) scaling dimension can be obtained for each field by appearing in the Lagrangian by making use of the constraint that has to be dimensionless, resulting for example in

$$ [\phi] = [A^{\mu}] = 1 $$

for a scalar and a gauge field or

$$ [\Psi_D] = [\Psi_M] = [\chi] = [\eta] = \frac{3}{2} $$

for Dirac, Majorana, and Weyl spinors.

Are these two concepts of scaling dimension and conformal weight somehow related?

asked Jul 17, 2013 in Theoretical Physics by Dilaton (6,240 points) [ revision history ]
@Matthew ok, should this be plain obvious? In this case I am too stupid to see it ... :-/

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Dilaton
I deleted my sarcastic comment and gave a hopefully more helpful answer.

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Matthew
It is probably easier to see this if you use the general (not 2d) formalism for the conformal group.

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Vibert
@Vibert hm I have mostly seen some CFT in 2d so far ... So can you expand a bit what you mean?

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Dilaton
The formalism you use ($z \mapsto z_1(z)$) only works in 2d, where conformal maps are holomorphic maps. If you just have the global conformal group (in $d\neq 2$) operators transform as $\phi(x) \mapsto |\partial x'/\partial x|^{-\Delta} \phi(x).$

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Vibert

2 Answers

+ 5 like - 0 dislike

From "Perturbative quantum field theory" Edward Witten (page 446 in volume 1 of "Quantum fields and strings : A course for mathematicians"):

enter image description here

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user user10001
answered Jul 17, 2013 by user10001 (635 points) [ no revision ]
Thanks for this answer. I always thought that one has to be careful if one looks at mass/energy or lenght dimensions, which would flip the sign too? From other text I have read I thought that for example for a scalar field $\phi$, the enginiearing dimension is what one obtains from the action has to be dimensionless considerations and this dimension coincides with the scalind dimension if there are no quantum effects that would lead to corrections called anomalous dimensions ... Is this wrong?

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Dilaton
@Dilaton Engineering dimension can not be changed by quantum corrections unless you at some step redefine a field by multiplying it with some dimensionful constant. On the other hand scaling dimension is eigenvalue under scale transformations. Since conformal algebra may develop anomaly via quantization so it may or may not be possible to quantize a classical theory while preserving the scaling dimensions of fields.

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user user10001
@Dilaton I re-edited the link. Most of the lectures in the two volumes of the book are freely available here.

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user user10001
Oh yeah thanks, this is even better :-)

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Dilaton
+ 1 like - 0 dislike

They're the same thing; if I have a CFT, then the dimension of a field $[\phi]$ is equal to its conformal dimension. This is because $[\phi]$ is defined to be the behavior of $\phi$ under rigid scalings, which is a special case of a conformal transformation.

Note, however, that one can also define a dimension $[\phi]$ for theories that aren't conformally invariant by promoting the couplings to background fields. For example, one needs to scale all mass parameters by $m\mapsto \lambda^{-1}m$ under $x\mapsto \lambda x$. In general any regulator that you can think of breaks this symmetry, and this leads to RG flows.

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Matthew
answered Jul 17, 2013 by Matthew (320 points) [ no revision ]
Yes thanks, this is quite helpful already. What exactly does promoting the couplings to background fields mean, I am not sure if I understand this correctly.

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Dilaton
If you want, you can just think of this as a symmetry that acts as $m\mapsto \lambda^{-1}m$ and $x\mapsto \lambda x$. I find it kind of strange to think of symmetries acting on numbers, like $m$, so it's convenient to think of temporarily promoting $m$ to a field with no kinetic term, on which the symmetry acts.

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Matthew
P.S. This trick turns out to be surprisingly helpful in many different contexts, including proving nonrenormalization theorems using holomorphy, and also in the (second) proof of the $a$-theorem.

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Matthew
Interesting, could you give some links to these applications?

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Dilaton
Sure, arxiv.org/abs/hep-ph/9309335 and arxiv.org/abs/1112.4538.

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user Matthew

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:
$\varnothing\hbar$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
...