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

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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
@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

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

This post imported from StackExchange Physics at 2014-03-09 16:14 (UCT), posted by SE-user user10001
answered Jul 17, 2013 by (635 points)
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
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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 (320 points)
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

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