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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?

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  • $\begingroup$ @Matthew ok, should this be plain obvious? In this case I am too stupid to see it ... :-/ $\endgroup$
    – Dilaton
    Commented Jul 17, 2013 at 19:08
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    $\begingroup$ I deleted my sarcastic comment and gave a hopefully more helpful answer. $\endgroup$
    – Matthew
    Commented Jul 17, 2013 at 19:13
  • $\begingroup$ It is probably easier to see this if you use the general (not 2d) formalism for the conformal group. $\endgroup$
    – Vibert
    Commented Jul 17, 2013 at 19:28
  • $\begingroup$ @Vibert hm I have mostly seen some CFT in 2d so far ... So can you expand a bit what you mean? $\endgroup$
    – Dilaton
    Commented Jul 17, 2013 at 19:34
  • $\begingroup$ 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 $\phi(x) \mapsto |\partial x'/\partial x|^{-\Delta} \phi(x).$ $\endgroup$
    – Vibert
    Commented Jul 17, 2013 at 19:44

2 Answers 2

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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"):

enter image description here

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  • $\begingroup$ 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? $\endgroup$
    – Dilaton
    Commented Jul 17, 2013 at 22:11
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    $\begingroup$ @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. $\endgroup$
    – user10001
    Commented Jul 17, 2013 at 22:32
  • $\begingroup$ @Dilaton I re-edited the link. Most of the lectures in the two volumes of the book are freely available here. $\endgroup$
    – user10001
    Commented Jul 17, 2013 at 23:02
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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.

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  • $\begingroup$ 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. $\endgroup$
    – Dilaton
    Commented Jul 17, 2013 at 19:19
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    $\begingroup$ 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. $\endgroup$
    – Matthew
    Commented Jul 17, 2013 at 19:24
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    $\begingroup$ 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. $\endgroup$
    – Matthew
    Commented Jul 17, 2013 at 19:27
  • $\begingroup$ Interesting, could you give some links to these applications? $\endgroup$
    – Dilaton
    Commented Jul 17, 2013 at 19:32
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    $\begingroup$ Sure, arxiv.org/abs/hep-ph/9309335 and arxiv.org/abs/1112.4538. $\endgroup$
    – Matthew
    Commented Jul 17, 2013 at 19:35

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