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

Question

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.

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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)?

Answer 1

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.

Answer 2

There is a simple to see the action of momentum operator on this state. Write your state as creation operators acting on the vacuum. Then use commutation relations to arrive at the answer. You will see that the state is a momentum eigenstate with momentum, p_1+p_2.

Cheers!