# How does the momentum operator act on a multi-particle state?

My issue is about the proper development of the action of the momentum operator $P^{\mu}$ - the generator of spacetime translations - on multi-particle states. I'm a little clueless on this, so I'm going to build up my question without developing it at all. Hopefully somebody can fill in the holes.

Consider a scalar quantum field theory with underlying spacetime symmetry group being the orthochronous restricted Poincare transformations. A scalar field operator $\phi (x)$ can be generated by the (unitary) translation operator $T(x)= \exp (iP^{\mu}x_{\mu})$ and the operator at some other point (say, the origin):

$$\phi(x)=\exp(iP^{\mu}x_{\mu})\phi(0)\exp (-iP^{\mu}x_{\mu})$$

The generator(s) of space-transformations, $P^{\mu}$, act on single-particle states $|p\rangle$ as follows:

$$P^{\mu}|p\rangle=p^{\mu}|p\rangle$$

My question is, how does the operator $P^{\mu}$ act on multi-particle states? For example, how does the 4-momentum operator act on a two-particle state:

$$P^{\mu}|k_1,k_2\rangle = ?$$

Is a two-particle state simply not an eigenstate of the 4-momentum operator? Or maybe it's an eigenstate when the two are equal? Or maybe we need to define 4-momentum operators that act on different $n$-particle Hilbert spaces ((anti-)symmetrized appropriately)?

I think you're overthinking. The total momentum of a state with particles of momentum $k_1$ and $k_2$ is just $k_1 + k_2$, so $$P^\mu |k_1, k_2 \rangle = (k_1 + k_2)^\mu |k_1, k_2 \rangle.$$ If you wish, you can show this more explicitly by thinking of momentum as the generator of translations; both particles are translated, so you pick up phase factors from both, so the momenta of the two just sum.