# How do fields transform under special conformal transformations?

The fields transform under finite conformal transformations as$$^1$$ $$\Phi^a(x') \mapsto {\Phi^a}'(x) = \Omega(x')^\Delta\,D(R(x'))^{\phantom{b}a}_b \,\Phi^b(x')\,. \tag{1}\label{main}$$ as given in equation $$(55)$$ of $$[1]$$. Let's break it down:

• $$\Delta$$ is the conformal dimension of $$\Phi$$.
• $$\Omega$$ is the conformal factor of the transformation.
• $$D$$ is the spin representation of $$\Phi$$.
• $$R$$ is the rotation Jacobian of the transformation

So let's compute these things. The spin and the conformal dimension $$\Delta$$ are given. The first thing we have to look at is the Jacobian. $$\frac{\partial x^{\prime \mu}}{\partial x^\nu} = \Omega(x') R^\mu_{\phantom{\mu}\nu}(x')\,.$$ This implicitly defines both $$\Omega$$ and $$R$$ and it is not ambiguous because we require $$R \in \mathrm{SO}(d)$$, namely $$R^{\mu}_{\phantom{\mu}\nu} \,\eta^{\nu\rho}\,R^{\lambda}_{\phantom{\lambda}\rho} \,\eta_{\lambda\kappa}= R^{\mu}_{\phantom{\mu}\kappa}\,.$$ You can immediately see that for the Poincaré subgroup of the conformal group $$\Omega(x')= 1$$, whereas for dilatations $$\Omega(x') = \lambda$$ and for special conformal transformations $$\Omega(x') = \frac{1}{1+2(b\cdot x') + b^2 {x'}^2}\,.\tag{2}\label{omega}$$ This can be proven with a bit of algebra. Using your definition of SCT $${x'}^\mu = \frac{x^\mu - b^\mu x^2}{1+-2(b\cdot x) + b^2 {x}^2}\,,$$ one can check $$\frac{\partial x^{\prime \mu}}{\partial x^\rho} \eta_{\mu\nu} \frac{\partial x^{\prime \mu}}{\partial x^\lambda} = \frac{\eta_{\rho\lambda}}{(1-2(b\cdot x) + b^2 x^2)^2}\,.$$ That means that the Jacobian is an orthogonal matrix up to a factor, which is the square root of whatever multiplies $$\eta_{\rho\lambda}$$. Then we have to re-express that as a function of $$x'$$. After some algebra again one finds that it suffices to change the sign to the term linear in $$b$$.

Finally, how does one compute $$R$$? Well, it's just the Jacobian divided by $$\Omega$$. In the case of special conformal transformations one has (there might be mistakes, redo it for safety) $$R^{\mu}_{\phantom{\mu}\nu} = \delta^\mu_\nu + \frac{2 b_\nu x^\mu - 2 b^\mu (b_\nu x^2+ x_\nu - 2 (b\cdot x) x_\nu) -2b^2 x^\mu x_\nu }{1-2b\cdot x +b^2 x^2}\,,$$ which, as before, needs to be expressed in terms of $$x'$$.

If you are interested in $$\Phi$$ scalar then $$D(R) = 1$$ and you can just plug \eqref{omega} into \eqref{main} to obtain the transformation. If you want to consider also spinning $$\Phi$$ then it's not much harder.

For spin $$\ell=1$$ the $$D$$ is just the identity, namely $$D(R)^{\phantom{\nu}\mu}_\nu = R^{\phantom{\nu}\mu}_\nu\,.$$ For higher spins one just has to take the product $$D(R)^{\phantom{\nu_1\cdots \nu_\ell}\mu_1\cdots \mu_\ell}_{\nu_1\cdots \nu_\ell} = R^{\phantom{\nu_1}\mu_1}_{\nu_1}\cdots R^{\phantom{\nu_\ell}\mu_\ell}_{\nu_\ell}\,.$$ Again, by plugging these definitions in \eqref{main} you obtain the desired result.

$$\;[1]\;\;$$TASI Lectures on the Conformal Bootstrap, David Simmons-Duffin, 1602.07982

$$\;{}^1\;\;$$The way the transformations are written in the lecture notes linked above differs a bit from Di Francesco Mathieu Sénéchal. The difference is that Di Francesco et al. make an active transformation $$x \to x'$$ with $$\Phi(x) \mapsto \Phi'(x') = \mathcal{F}(\Phi(x))\,,$$ while David Simmons Duffin makes essentially the inverse transformation $$x' \to x$$ $$\Phi(x') \mapsto \Phi'(x) = \mathcal{F}^{-1}(\Phi(x'))\,.$$ That is why in the above discussion the indices of $$R^\mu_{\phantom{\mu}\nu}$$ get swapped when passed inside $$D$$ as $$D(R) = R^{\phantom{\nu}\mu}_{\nu}$$. And that's also why we get a factor $$\lambda^\Delta$$ rather than $$\lambda^{-\Delta}$$ as Di Francesco et al. This is all consistent as long as it is clear what one is doing.