Derivation of the full generator of the Lorentz transformations Let us study the subgroup of the Poincare group that leaves the point $x=0$ invariant, that is the Lorentz group. The action of an infinitesimal Lorentz transformation on a field $\Phi(0)$ is $L_{\mu \nu}\Phi(0) = S_{\mu \nu}\Phi(0)$.  By use of the commutation relations of the Poincare group, we translate the generator $L_{\mu \nu}$ to a nonzero value of $x$: $$e^{ix^{\rho}P_{\rho}} L_{\mu \nu} e^{-ix^{\sigma}P_{\sigma}} = S_{\mu \nu} - x_{\mu}P_{\nu} + x_{\nu}P_{\mu}\,\,\,\,\,\,\,\,\,(1),$$ where the RHS is computed using the Baker-Campbell-Hausdorff formula to the second term.  Then we can write the action of the generators $$P_{\mu} \Phi(x) = -i\partial_{\mu}\Phi(x) \,\,\,\,\text{and}\,\,\,\,L_{\mu \nu}\Phi(x) = i(x_{\mu}\partial_{\nu} - x_{\nu}\partial_{\mu})\Phi(x) + S_{\mu \nu}\Phi(x)\,\,\,\,\,(2)$$ 
Could someone explain the contents of Eqn (1) and how (2) is deduced from it?  I think there is an unitary transformation on the LHS of Eqn (1), but I don't understand what the equation is doing and why the contents are what they are in the exponents of the exp terms.
 A: The LHS describes the transformation of $L_{\mu \nu}$ under "space-time translations". This is the natural generalization of how time evolution works in the Heisenberg picture. So instead of acting by just the generator of time translation $e^{-\frac{i H t}{\hbar}}$, you also have the $p_i$ in the exponent, which are the generators of space translations. You start with the action of the subgroup which leaves $x=0$ invariant, this is just the spin part of the angular momentum. Your total angular momentum operator is the sum of the orbital angular momentum and spin angular momentum which is evident from your equation 2. 
You get equation 2 by just Taylor expanding $\Phi$ about 0, and then comparing with equation 1. (Here you are comparing the Heisenberg and Schrödinger pictures)
$$L_{\mu \nu}\omega^{\mu \nu}\Phi(x)=S_{\mu \nu} \omega^{\mu \nu} \Phi(0)+ x_{\mu} \omega^{\mu \nu}\partial_{\nu}\Phi(0)$$
where $\omega_{\mu \nu}$ are the boost and rotation paramaters, and I have considered the infinitesimal lorentz transformation $x^{\rho}=0 + \omega^{\rho}_{\nu} x^{\nu}$ and noting $\omega_{\mu \nu}$ is antisymmetric you get the correct actions. 
THe infinitesimal version of equation 1 is just the commutator of the field with $L_{\mu \nu}$. This is a common feature of actions in QFT, the infinitesimal lie algebra version acts through the commutator which on exponentiating gives the conjugation action defined above. 
A: I am still not entirely sure what your question is, but I will try to explain the left-hand side of equation (1) in general. More information can be found in "Lie algebras in particle physics" by Georgi.
A representation of matrix Lie group can in be written as:
$$
D(g) = e^{i \alpha_a X_a}
$$
where $X_a$ denotes a generator and $\alpha_a$ denotes the group parameter. The representation acts on some vector space:
$$
|i \rangle \to |i' \rangle = e^{i \alpha_a X_a} |i \rangle
$$
Now, let $O$ denote an operator giving the ket $O|i\rangle$. Clearly, this will transform as:
$$
O|i \rangle \to O' | i' \rangle = e^{i \alpha_a X_a} O |i \rangle = e^{i \alpha_a X_a} O e^{-i \alpha_a X_a} e^{i \alpha_a X_a} |i \rangle = e^{i \alpha_a X_a} O e^{-i \alpha_a X_a} |i' \rangle
$$
Thus, we see from above that the operator transforms as:
$$
O' = e^{i \alpha_a X_a} O e^{-i \alpha_a X_a}
$$
which corresponds to the left-hand side of equation (1) in your question. In other words, the left-hand side shows how the operator $L_{\mu \nu}$ transforms under the 4-momentum generator (i.e. space and time translations).
A: Since $\Phi(x)=(e^{x\cdot\partial}\Phi)(0)=(e^{ix\cdot P}\Phi)(0)$, which can be get from Taylor series, we have
$$
(L_{\mu\nu}\Phi)(x)=(e^{ix\cdot P}L_{\mu\nu}\Phi)(0)=(e^{ix\cdot P}L_{\mu\nu}e^{-ix\cdot P}e^{ix\cdot P}\Phi)(0)
$$
With $e^{ix\cdot P}L_{\mu\nu}e^{-ix\cdot P}=L_{\mu\nu}-x_\mu P_\nu+x_\nu P_\mu$ and $(L_{\mu\nu}\Phi)(0)=S_{\mu\nu}\Phi(0)$, we can see
$$
\begin{align}
(L_{\mu\nu}\Phi)(x)&=[(L_{\mu\nu}-x_\mu P_\nu+x_\nu P_\mu)(e^{ix\cdot P}\Phi)](0)\\
&=S_{\mu\nu}[(e^{ix\cdot P}\Phi)(0)]+[e^{ix\cdot P}(-x_\mu P_\nu+x_\nu P_\mu)\Phi](0)\\
&=S_{\mu\nu}\Phi(x)+(-x_\mu P_\nu+x_\nu P_\mu)\Phi(x)\\
&=(S_{\mu\nu}+ix_\mu \partial_\nu-ix_\nu \partial_\mu)\Phi(x)
\end{align}
$$
