In Lectures on CFT by J. D. Qualls, there appears the following statement without proof (page 30):
The scaling dimension $\Delta$ of a field is defined by the action of a scale transformation on the field $\Phi$ according to $$ \Phi(\lambda x) = \lambda^{-\Delta}\Phi(x) \tag1 $$
$\lambda$ is defined in such a way that upon dilatations spacetime coordinates transform as $x\rightarrow x^\prime = \lambda x$.
According to other posts in the Physics site, like this, it's been stated in the comments to the accepted answer that this can be obtained via the action of the dilatation generator $D = -i(\Delta + x^\mu\partial_\mu)$. I'm trying to figure this out but I must be doing something wrong. My attemp is the following:
$$ \Phi(\lambda x) = \Phi(x + (\lambda - 1)x) = \exp\{ -(\lambda-1)x^\mu\partial^\prime_\mu \}\Phi(\lambda x)|_{\lambda = 1} $$ $$ = \exp\{ -(1-\lambda^{-1})(x^\mu \partial_\mu) \}\Phi(\lambda x) |_{\lambda = 1} \tag2 $$ where $\partial^\prime_\mu = \partial/\partial(\lambda x^\mu)$. Evaluation on $\lambda = 1$ only affects the field.
And I'm stuck. Maybe another way would be to start with the usual transformation of the field with exponentiation of the dilatation generator, this is, the transformed field at the same point $x$ is given by $$ \Phi^\prime(x) = e^{-i\epsilon D}\Phi(x)e^{i\epsilon D^\dagger}\tag3 $$
but this wouldn't relate $\Phi(\lambda x)$ to $\Phi(x)$ but rather $\Phi^\prime(x)$ to $\Phi(x)$.
Could you clarify the way to obtain both $\Phi^\prime(x)$ and $\Phi(x^\prime = \lambda x)$?
Comment: Reading ''Conformal field theory'' by P. Di Francesco et al., in their eq (2.121) (page 38), it appears that the our eq (1) above should be in fact $$ \Phi^\prime(\lambda x) = \lambda^{-\Delta}\Phi(x) \tag4 $$ Hence, may Qualls have an errata there?