Sign error in representation of angular momentum on fields Consider a field $\phi$ that belongs to a representation of the Poincaré algebra. We assume that under translations $\phi\mapsto\phi'$, where $\phi'(x')=\phi(x)$. This fixes the action of the momentum operator to $P_\mu=-i\partial_\mu$. Now, let us further assume that the representation of angular momentum at $x=0$ is given by a matrix $L_{\mu\nu}\phi(0)=S_{\mu\nu}\phi(0)$. Furthermore, we have the formula (see https://arxiv.org/abs/1511.04074 eqn (2.34))
$$e^{ix\cdot P}L_{\mu\nu}e^{-ix\cdot P}=L_{\mu\nu}-x_{\mu}P_\nu+x_{\nu}P_\mu.$$
Then we can apply $\phi(x)$ to both sides. On the one hand, we can use a Taylor series at $x=0$ to see that
$$\phi(x)=e^{x^\mu\partial_\mu}\phi(0)=e^{ix^\mu P_\mu}\phi(0).$$
Thus, the left hand side is $S_{\mu\nu}\phi(x)$. Then, using the representation of momentum found above, we conclude that
$$L_{\mu\nu}=S_{\mu\nu}-i(x_\mu\partial_\nu-x_\nu\partial_\mu).$$
But, according to (2.34) in the reference above (and every other reference I've looked at), the sign of the second term I obtained with this procedure is wrong. I've had this same sign error when computing the other operators of a conformal field theory so I must be missing something conceptually. I would appreciate if someone can help me point out my mistake.
 A: You are misreading your reference, complacently ignoring that risk by failing to label his objects suitably, even though he says, in words, he is translating the generator.
By using the 5th commutation relation in (2.27), you see that
$$
L_{\mu\nu}(x)\equiv  e^{ix\cdot P}L_{\mu\nu}(0)e^{-ix\cdot P}=L_{\mu\nu}(0)-x_{\mu}P_\nu+x_{\nu}P_\mu,$$
so that
$$
L_{\mu\nu}(x) \phi(x)=  e^{ix\cdot P} \bigl (L_{\mu\nu}(0) \phi(0)\bigr),
$$
as dictated by the similarity isomorphism of the translation.
It is then evident that
$$L_{\mu\nu}(x) =S_{\mu\nu} - x_{\mu}P_\nu+x_{\nu}P_\mu = S_{\mu\nu}+i(x_\mu\partial_\nu-x_\nu\partial_\mu).\tag{2.34}$$
A: I just realized that Cosmas Zachos wrote an answer as I was writing this. I think we effectively said the same thing in a different notation. Thanks Cosmas!

I believe I've found the source of my confusion. This is all a notation problem. The thing is that $L_{\mu\nu}$ is a differential operator. It acts on a field $\phi$, not on its value at a point $\phi(x)$. Accordingly, the notation $L_{\mu\nu}\phi(x)$ really stands for $(L_{\mu\nu}\phi)(x)$. Using this more explicit notation, we have
$$(L_{\mu\nu}\phi)(a)=(e^{ia\cdot P}L_{\mu\nu}\phi)(0)=((L_{\mu\nu}-a_{\mu}P_\nu+a_\nu P_\mu)e^{ia\cdot P}\phi)(0)$$
Now, we know that $(L_{\mu\nu}\phi)(0)=S_{\mu\nu}\phi(0)$. In particular, we can apply this to the right hand side to get $(L_{\mu\nu}e^{ia\cdot P}\phi)(0)=(S_{\mu\nu}e^{ia\cdot P}\phi)(0)=S_{\mu\nu}\phi(a).$ For the other terms we can just use the fact that momenta commute with one another. Then $(a_{\mu}P_\nu e^{ia\cdot P}\phi)(0)=(e^{ia\cdot P}a_{\mu}P_\nu\phi)(0)=-ia_{\mu}(e^{ia\cdot P}\partial_\nu\phi)(0)=-ia_\mu(\partial_\nu\phi)(a).$ This ends up giving use the correct relation
$$(L_{\mu\nu}\phi)(a)=S_{\mu\nu}\phi(a)+ia_\mu(\partial_\nu\phi)(a)-ia_\nu(\partial_\mu\phi)(a),$$
or
$$L_{\mu\nu}=S_{\mu\nu}+i(a_\mu\partial_\nu-a_\nu\partial_\mu)$$

Ok, the above is just simply wrong. The trick is to note precisely that $L_{\mu\nu}$ is not a differential operator in spacetime, but rather a differential operator on the space of fields, as remarked in my answer here Primary fields in di Francesco's CFT reverses the commutation relations, so that they should actually be
$$e^{ix\cdot P}L_{\mu\nu}e^{-ix\cdot P}=L_{\mu\nu}+x_\mu P_\nu-x_\nu P_\mu.$$ We can then use the procedure I mentioned when asking the question to obtain the correct equation that
$$L_{\mu\nu}=S_{\mu\nu}+x_\nu P_\mu-x_\mu P_\nu$$
