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 \begin{equation} U(\Lambda) |k \rangle = |\Lambda k \rangle \end{equation} (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.

However even in the interacting theory if you have a single stable particle it will transform nicely like a free particle under the interacting representation of the Poincare group. The catch is that the stable particles may be different from the free theory you started with. For instance if you have a single electron alone in the universe that state will transform like a particle. But also a single hydrogen atom in its ground state (in a theory in which it is stable) looks just as much like a single particle as far representations of the Poincare group is concerned.

Weinberg's QFT textbook (Volume I) has a good discussion of both these points. Chapter 2 goes over the representations of the Poincare group, and parts of Chapter 3 talk about the problem of the interacting boost generators.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.