Translation of generators when derive representations of Poincare group on fields As stated in section 4.2.1 of Di Francesco's book on conformal field theory. In order to find out representation of Poincare group on fields, we can start by studying the subgroup that leaves the point $x=0$ invariant, that is, the Lorentz group. Then we have $L_{\mu\nu}\Phi(0)=S_{\mu\nu}\Phi(0)$, here $L_{\mu\nu}$ is generator of Lorentz transformation, $S_{\mu\nu}$ is the spin operator associated with the field $\Phi(0)$. Next we translate the generator $L_{\mu\nu}$ to a nonzero value of $x$:
$e^{ix\cdot P}L_{\mu\nu}e^{-ix\cdot P}=S_{\mu\nu}-x_\mu P_\nu+x_\nu P_\mu$
Where $P$ is generator of translation. This allows us to write the action of generators
$L_{\mu\nu}\Phi(x)=i(x_\mu \partial_\nu-x_\nu \partial_\mu)\Phi(x)+S_{\mu\nu}\Phi(x)$
Here is my question. $L_{\mu\nu}$ is not a field. It is an operator act on fields.  What is the meaning of translation of $L_{\mu\nu}$?  Why can we translate generator $L_{\mu\nu}$ in this way?
 A: For any unitary operator $U$, use the abbreviation
$$
 \sigma_U X\equiv U^{-1}XU.
$$
Now, let $G$ be a group of unitary transformations $\sigma_U$ that act on field operators $X$ in this way. If $\sigma_U\in G$ and $\sigma_V\in G$, then $\sigma_U\sigma_V\sigma_U^{-1}\in G$. This follows just from the fact that unitary operators are invertible and can be multiplied. Notice that
\begin{align*}
 \sigma_U\sigma_V\sigma_U^{-1}X 
  &= \sigma_U\sigma_V(UXU^{-1}) \\
  &= \sigma_U(V^{-1}UXU^{-1}V) \\
  &= U^{-1}V^{-1}UXU^{-1}VU \\
  &=\sigma_{U^{-1}VU}X.
\end{align*}
Therefore, if we can transform a field $X$ as $\sigma_UX$ and as $\sigma_VX$, we can also transform it as $\sigma_{U^{-1}VU}X$.
That's what the authors are doing in section 4.2.1. They are using this general fact to take the results they already derived about the transformations $V$ that fix the point $x=0$ and extended those results to transformations $U^{-1}VU$ that fix any other point $x\neq 0$, with $U$ the translation from $x=0$ to $x\neq 0$. They're using the generator of $V$ instead of $V$ itself, but it's the same idea.
