# Finite conformal transformations of fields

I want to work out the finite change of a Field in Conformal Field Theory. In Di Francesco's Conformal Field Theory he states "In principle we can derive [it] from the [local generators at x=0]" but then he only states that for spinless field we have \begin{align*} \Phi(x)\to \Phi^\prime(x^\prime) = \left|\frac{\partial x^\prime}{ \partial x}\right| ^{-\Delta/d} \Phi(x) \end{align*} I dont see how he derives this. How is the Jacobian is connected with the generators? And what would I need to do, to derive it for a field with spin ($$S_{\mu\nu}\neq 0$$)? My guess is that I need to look at something like \begin{align*} \Phi^\prime(x^\prime) = exp(-ia^\mu P_\mu) exp( -i \alpha D)exp( -i M^{\mu\nu}L_{\mu\nu})exp(-i b^\mu K_\mu)\Phi(x) \end{align*} but then I dont know how these operators should be ordered, since they dont commute and I dont really know how to find the Jacobian and the factor $$\Delta/d$$ from this expression.

I know that a similar question (Finite conformal transformations of fields from infinitesimal) was already asked, but there is no clear answer on how to derive it.

First of all, the equation you write for $$\Phi'(x')$$ is not correct. The reason is that in principle you are allowed to use the same generator multiple times in the product since the order matters (incidentally, since $$\Phi$$ is a scalar primary, the action of $$D$$ and $$K$$ would vanish). An analogy is the Euler angles where you can parametrize a rotation as a product, e.g., $$XYX$$. Another possible way to parametrize the transformation is by having all generators in the same exponent $$\Phi'(x') = \exp\left(-ia^\mu P_\mu -i \alpha D - iM^{\mu\nu}L_{\mu\nu} -i b^\mu K_\mu\right)\,\Phi(x)\,.$$ The proof of that statement can be done by looking at all transformations separately. For $$P$$ and $$L$$, which belong to the Poincaré group, we have, after $$x \mapsto x' = \Lambda \cdot x + a$$ $$\Phi'(x') = \Phi(x)\,.$$ For $$D$$ we have, after $$x \mapsto x' = \lambda x$$ $$\Phi'(x') = \lambda^{-\Delta} \,\Phi(x)\,.$$ Now $$K$$ is a bit complicated, but we can study the action of inversions $$I: x^\mu \mapsto {x'}^\mu = \frac{x^\mu}{x^2}\,.$$ It is easy to show that a special conformal transformation can be obtained by composing an inversion, a translation and another inversion. Namely $$K_\mu = I P_\mu I$$. So let's check inversions $$\Phi'(x') = (x^{2})^\Delta\Phi(x)\,.$$ Now you can check that every single transformation that we did satisfies $$\Phi'(x') = \left|\det\,\middle(\frac{\partial x'}{\partial x}\middle)\right|^{-\Delta/d}\Phi(x)\,,$$ simply because Poincaré transformations have determinant $$1$$, $$x\mapsto \lambda x$$ has determinant $$\lambda^d$$ and inversions have determinant $$x^{-2d}$$.
One thing that might disappoint you is the use of inversions, because we cannot derive their action directly from the algebra (as they are not connected to the identity). In that case one would have to prove the result using $$K$$, and this becomes a bit cumbersome in terms of computations.
• Thanks for the answer. But I still dont really understand how you get $\lambda^{-\Delta}\Phi(x)$ from $exp(-i\alpha D)$. Could you elaborate on that ? Since we would have $exp(- i x^\nu\partial_\nu + \Delta)\Phi(x)$ How is this computed to be $\lambda^{-\Delta}$ ? – T. Die Mar 18 at 0:07
• First, I implicitly defined $\alpha = \log \lambda$, sorry about that. Second, the $x^\mu\partial_\mu$ part is what changes $x \to \lambda x$ inside $\Phi$ and the $\Delta$ is what gives the $\lambda^\Delta$ in front. – MannyC Mar 18 at 0:17
• Okay, I got everything now, besides how you come up with the $\Delta$ factor in $(x^2)^\Delta$ for the inversion. In my calculation the $\Delta$ only appears if $\alpha \neq 0$. And If I only look at the SCT/Inversion I would need to set $\alpha$ to zero? But really thank you, your answers are great ! – T. Die Mar 18 at 7:02
• You are right, I didn't motivate that well. You can see that $\Delta$ shows up if you act with a special conformal transformation $\exp(-i b^\mu K_\mu)$. The reason is the following: $\Phi$ at a point $x$ can be seen as $\exp(-i x^\mu P_\mu) \Phi(0)$. So you'd have to get $K_\mu$ past the $P_\mu$ in the exponential and there is a $D$ in their commutator. Then $K_\mu$ on $\Phi(0)$ is zero. – MannyC Mar 18 at 9:22
• I am still not getting it. If I have $iD \Phi(x) = (\Delta+ x\cdot \partial) \Phi(x)$ then I get $\Phi'(x') = e^{i\alpha D} \Phi(x) = \lambda^\Delta \Phi(\lambda x)$, where $\alpha = \log \lambda$ and $x' = \lambda x$. This is far from $\Phi'(x') = \lambda^{-\Delta} \Phi(x)$. What is wrong here? – jkb1603 Mar 19 at 15:32