Interacting Hamiltonian commutes with momentum operator? In Peskin's textbook chapter 7 Radiative Corrections: some formal developments (page 212 second paragraph), he describes two-point functions and chooses states to be eigenstates of the full interacting Hamiltonian $H$. Since the momentum operator $P$ commutes with $H$, we can also choose the states to be eigenstates of $P$.
So my question is how can an interacting Hamiltonian $H$ still commute with the total momentum operator $P$? In my mind, only a free system's Hamiltonian $H$ can commutes with $P$, so is there anything wrong? I am very confused by this part. 
 A: According to Heisenberg's equations of motion,
$$
[H,P]\propto \partial_t P
$$
which equals zero because $P$ is time-independent. This assumes nothing about the nature of $P,H$, (except for the fact that these generate translations). The system need not be free: the only thing you need is HEoM, which is one of the postulates of QM.
But how do we know that $\partial _tP=\partial _t H=0$? that is, how do we know that the generators are time-independent?
Well, for the Hamiltonian its fairly easy to see that $\partial_t H\propto [H,H]=0$, because ever operator commutes with itself. Therefore, $H$ is time independent. In the case of the momentum operator, the easiest way to see that $\partial_t P=0$ is to realise that $P^\mu$ is the Noether charge of $x^\mu\to x^\mu+a^\mu$, that is, $P^\mu$ is the generator of translations. In other words,
$$
P^\mu=\int\mathrm d\boldsymbol x\ T^{0\mu}
$$
where $T^{\mu\nu}$ is the stress-energy tensor (the Noether current). Noether's theorem implies that charges are conserved, that is, $\partial_t P^\mu=0$.
A: The assumption is that momentum is conserved. So total momentum is a constant of motion. The operator for a constant of motion would commute with the Hamiltonian, because, being a constant of motion, its derivative with respect to time vanishes.
