# Representations of Lorentz group in interacting QFT

In QFT, we obtain a representation of the Lorentz group by defining a set of unitary operators whose action on (spinless) free particle states is given by $$U(\Lambda) |k \rangle = |\Lambda k \rangle$$ (and similarly for multiparticle states).

Physically, if two observers $O_1$ and $O_2$ are related by a Lorentz transformation $\Lambda$ and a free particle appears to observer $O_1$ to be in state $|k \rangle$, then it will appear to observer $O_2$ to be in state $|\Lambda k \rangle$. This transformation property extends to arbitrary elements of the Hilbert space, even elements that are not single particle momentum eigenstates: if $O_1$ sees a state $|s\rangle$ then $O_2$ sees a state $U(\Lambda)|s \rangle$.

My question is the following: If both observers are situated in a highly interacting region (i.e. a region of space where particles do not behave as free particles), and observer $O_1$ sees a state $|s \rangle$, what does observer $O_2$ see? Do we now need a new representation of the Lorentz group to answer this question? If so, how can such a representation be obtained if I know the Hamiltonian for the full interacting theory?

My guess is that we do need a new representation. I am basing this on the fact that, if we consider the full Poincare group, then it is clear that the free particle representation does not provide us with the correct transformation between observers, since it fails to give the correct time-translation.

Yes you would need a new representation. The reason is the Lorentz group is connected with the group of translations and as you point out the time-translation is different. If $K^i$ is the generator of boosts in the $i$ direction and $P^j$ the generator of space translations (i.e. the linear momentum) $$[K^i,P^j]=iH\delta_{ij}$$ So if the interacting Hamiltonian is different, the $K^i$ operators would need to be different as well.