A question about the Weinberg QFT Vol.1 Section2.4 I'm self-studying Weinberg QFT, and I'm confused about the connection between the momentum operator and the generator of translations
In Section 2.4, Weinberg shows the Lie algebra of Poincare Group,
\begin{align}
i[J^{\mu\nu} ,J^{\rho\sigma}] & =\eta^{\nu\rho}J^{\mu\sigma}-\eta^{\mu\rho}J^{\nu\sigma}-\eta^{\sigma\mu}J^{\rho\nu}+\eta^{\sigma\nu}J^{\rho\mu}\\
i[P^\mu , J^{\rho\sigma}] &=\eta^{\mu\rho}P^\sigma-\eta^{\mu\sigma}P^{\rho}\\
i[P^\mu, P^\nu] &=0
\end{align}
For now, we just know $P^\mu$ and $J^{\rho\sigma}$ are some Hermitian operators, but next, Weinberg said the $P^0$ is the energy operator $H$, and the commutation rules can show that ${P^1,P^2,P^3}$ are the components of the momentum operators.
But Why? Why can we just take the $P^0$ as the operator $H$? Or do we just define the $P^0$ as $H$? But if we use this definition, how can we make sure   that ${P^1,P^2,P^3}$ are the components of the momentum operator？I know we can use the commutation rules to define the angular momentum in Quantum Mechanics, Is this also for the same reason?
If not, then how can we use the Poincare Group or space-time symmetry to derive the momentum operator $-i\hbar\frac{\partial}{\partial x^i}$?
 A: I will try to clarify your thoughts, but if something is not clear, please do not hesitate to comment.
In Section 2.4, Weinberg is considering the infinitesimal Poincare transformations
$$x^{\mu}\rightarrow x^{\mu}+\epsilon^{\mu}$$
$$x^{\mu}\rightarrow \Lambda^{\mu}_{\nu}x^{\nu}=(\delta^{\mu}_{\nu}+\omega^{\mu}_{\nu})x^{\nu}$$
which correspond to a spacetime (as opposed to simply space) translation and a set of rotations and boosts. These transformations form a group, called the Poincare group. It's generators are found by considering group elements near the identity. Those generators are the hermitian operators you have mentioned (i.e. the set $\{P^{\mu},J^{\mu\nu}\}$, which are in total 4+6 generators).
This set of 10 generators contains a generator that commutes with all the generators $\{P^1,P^2,P^3,J^{23},J^{31},J^{12}\}$. This generator is $P^0$. Since $P^0$ is the generator of time translation, we call this one the Hamiltonian. Since every other generators in the aforementioned set commutes with $P^0$, then they correspond to conserved quantities because $i\frac{d\mathcal{O}}{dt}=[\mathcal{O},H]$.
The conserved quantities (/generators) associated with space translations are the components of the momentum operator, whereas the conserved quantites (/generators) associated with rotations are the angular momentum operators. Those are used to label physical states. The remaining generators correspond to boosts...
So, to recap, we take $P^0$ to be $H$ if the system is invariant under time translations. And if this is the case, then the system will be also invariant under space translations and rotations and hence we identify the operators that correspond to those transformations as the momentum and angular momentum operators respectively...
I hope this helps.
