In 1 (see references below), I'm trying to derive how a spinless field transforms under a conformal transformation, specifically eq. (2.41). CFT references/lectures are the most confusing I've seen EVER, even the basic definitions are a mess.

Reading 226464 and 469205 clarified bits and pieces but overall I still can't form a clear picture. @MBolin's answer about how to write conformal transformations based on Zee's book is the most clarifying/natural definition I've seen. On the other hand, @MannyC's clarification on the distinctions between diffeomorphism/Weyl/conformal transformations kind of gave me a big picture, BUT the notations and conventions are in contrast to some references which makes it hard to form a picture.

What I learned is that when people say conformal transformations they really mean a diffeomorphism in the sense that $g'_{\mu\nu}(x') = \frac{\partial x^\rho}{\partial x'^\mu} \frac{\partial x^\sigma}{\partial x'^\nu} g_{\rho\sigma}(x) = \Omega(x)^2 g_{\rho\sigma}(x)$ (Here, I didn't adopt @MBolin's answer since most people don't use that) plus a Weyl transformation $\bar{g}_{\rho\sigma}(x) = \Omega(x)^2 g_{\rho\sigma}(x)$.

First question, can anyone clarify which is which, some people define conformal transformations (which I think is really a Weyl transformation) in the first page of every reference to be $g'_{\mu\nu}(x') = \Omega(x) g_{\mu\nu}(x)$, while others define it to be $g'_{\mu\nu}(x') = \Omega(x)^2 g_{\mu\nu}(x)$. Why aren't they consistent? The extra square factor is confusing me. On top of that @MannyC's answer defined it to be $g'_{\mu\nu}(x') = \Omega(x)^{-2} g_{\mu\nu}(x)$.

Second, going back to my original question about how fields transform. I'll use the convention of prime $\phi'(x')$ for diffeomorphism and a bar $\bar{\phi}(x)$ for Weyl transformation. In some posts, they immediately defined that fields transform as $\phi(x) \rightarrow \bar{\phi'}(x') = \Omega^{-\Delta}(x) \phi(x)$, but I would like to derive this similar to how references go on about it. I think one way to derive this is through the action of a free scalar field with kinetic term only (I've read in Lecture Notes on String Theory that conformal invariance is hard to check even in flat spacetime so we can just check scale invariance and presume that conformal invariance holds),

\begin{align} S' & = \int d^dx' \bar{g}'^{\mu \nu} \partial'_\mu \bar{\phi'}(x') \partial'_\nu \bar{\phi'}(x') = \int d^dx \Omega^d \bar{g}'^{\mu \nu} \frac{\partial x^\rho}{\partial x'^\mu} \frac{\partial x^\sigma}{\partial x'^\nu} \partial_\rho \bar{\phi'}(x') \partial_\sigma \bar{\phi'}(x')\\ & = \int d^dx \Omega^d \Omega^{-2} g'^{\mu \nu} \frac{\partial x^\rho}{\partial x'^\mu} \frac{\partial x^\sigma}{\partial x'^\nu} \partial_\rho \bar{\phi'}(x') \partial_\sigma \bar{\phi'}(x'), \qquad \text{inverse Weyl:}\; \bar{g}^{\rho\sigma}(x) = \Omega(x)^{-2} g^{\rho\sigma}(x) \\ \end{align}

From the last line I'm not sure how to proceed, but the end result should look like,

\begin{equation} \int d^dx \partial_\rho (\Omega^{\Delta} \bar{\phi'}(x') ) \partial^\rho ( \Omega^{\Delta} \bar{\phi'}(x') ) \end{equation}

so that,

\begin{equation} \Omega^{\Delta} \bar{\phi'}(x') = \phi(x) \rightarrow \bar{\phi'}(x') = \Omega^{-\Delta} \phi(x) \end{equation}

Lastly, in 1, the scale factor $\Omega$ was expressed in terms of the Jacobian, i.e. eq. (2.41), but I can't seem to find a way to derive it.


[1]Lectures on Conformal Field Theory by Joshua D. Qualls

[2]Lecture Notes on String Theory by Kevin Zhou


1 Answer 1


Let $g_{\mu \nu}$ be the metric tensor in a $d$-dimensional spacetime. A conformal transformation is a change of coordinates $x\mapsto x'=f(x)$ that leaves the metric tensor invariant up to a scale: \begin{equation} g_{\mu \nu}(x) \mapsto g'_{\mu \nu}(x')=\Omega^2(x)g_{\mu \nu}(x), \end{equation} where we assume $\Omega(x)>0$. Some other definition just write $\Omega$ instead of $\Omega^2$, but it is just a question of taste.

Primary fields are defined based on the action of the generators of the conformal group on them. This is a general and purely geometrical definition, similarly to how you would define normal rotations in 3D.

Definition1. In $d>2$ spacetime dimensions, let $\hat{D}$ be the generator of dilations and let $\hat{K}_{\mu}$ be the generator of special conformal transformations. A conformal primary field $\hat{\phi}^M_{\rho}(x)$, in the $\rho$ representation of the Lorentz group and with conformal dimension $\Delta$, satisfies the following conditions at $x=0$: \begin{equation} \left[\hat{D},\hat{\phi}^M_{\rho}(0)\right]=-i\Delta\hat{\phi}^M_{\rho}(0);\\ \left[\hat{K}_{\mu},\hat{\phi}^M_{\rho}(0)\right]=0. \end{equation}

Now, it turns out that this definition is equivalent to the one you are talking about. This second definition gives the explicit behaviour of the field under a conformal transformation.

Definition2. In $d>2$ spacetime dimensions, a conformal primary field $\hat{\phi}^M_{\rho}(x)$, in the $\rho$ representation of the Lorentz group and with conformal dimension $\Delta$, transforms under a conformal transformation $\eta_{\mu \nu}\mapsto \Omega^2(x)\eta_{\mu \nu}$ as \begin{equation} \hat{\phi'}^M_{\rho}(x')=\Omega^{\Delta}(x)\mathcal{D}{\left[R(x)\right]^M}_{N}\hat{\phi}^N_{\rho}(x) \end{equation} where ${R^{\mu}}_{\nu}(x)=\Omega^{-1}(x)\frac{\partial x^{\mu}}{\partial x'^{\nu}}$ and $\mathcal{D}{\left[R(x)\right]^M}_{N}$ implements the action of $R$ in the $SO(d-1,1)$ representation of $\hat{\phi}^{M}_{\rho}(x)$.

The above equation is valid for any type of fields, so it generalizes the scalar field behaviour you are talking about.

The proof of the equivalence, as well as consistent notations and definitions (as you said, many CFT references are confusing), you find them in this paper: https://arxiv.org/abs/2112.01837.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.