How is Hilbert space constructed in interacting theory?

Question

In the free field theory, we an decompose the field with creation and annihilation operators $a^{\dagger}_k$ and $a_k$. $a_k$ acts on some state $|0\rangle$ and outputs $0$. We call that state the vacuum state—a state with no particles. $a^{\dagger}_k$ in contrast acts on the vacuum stays and creates a particle. We can use these operators to build the Hilbert space.

When we turn interactions on, from what I’ve read, the interacting field no longer has a pretty decomposition into plane waves scaled by creation and annihilating operators. So it seems to me that in the interacting theory we can’t even talk about particle states. Yet, we talk about the interacting vacuum state $|\Omega \rangle$. I have two questions:

1. How can the interacting vacuum state even be defined if we do not have creation/annihilation operators for the interacting field? I’ve seen it defined as $H|\Omega \rangle = 0$ where $H$ is the full Hamiltonian, but then without having creation/annihilating operators, how can we even have such a relation (no particle destruction operators in the Hamiltonian)?

2. Similarly, how can we talk about interacting particle states? Particle states in the free theory are created by acting on the vacuum state by the creation operator, but again, there is no creation operator for the interacting field.

Answer 1

The description of the complete Hilbert space of one interacting theory is a very non-trivial problem and it is not known in four spacetime dimensions, although as far as I know there are results in fewer dimensions. The investigation of the proper rigorous construction of such Hilbert space is the subject of the so-called "constructive QFT" and is an active area of research.

Nevertheless, that is not to say that we know nothing about such Hilbert space. In fact, we know two things which answer your two questions:

1. Whatever this Hilbert space is it must carry one unitary representation of the universal cover of the Poincaré group. This is simply because this is the basic requirement of relativisic symmetry. In particular this means that the Poincaré generators $P_\mu$ and $M_{\mu\nu}$ will be realized by Hermitian operators in such Hilbert space.

   In that regard, the vacuum is simply defined to be a Poincaré invariant state. In particular this implies it must be annihilated by the Poincaré generators, so that $P_\mu |\Omega\rangle=M_{\mu\nu}|\Omega\rangle=0$.

2. Whatever this Hilbert space ${\cal H}$ is we assume it has a subspace of in/out scattering states ${\cal H}_{\rm in/out}\subset{\cal H}$. Such subspaces are identified through the use of Moller operators to the corresponding free theory Hilbert spaces (see e.g. this answer of mine for a short description). The characteristic property of such states is that one such state, when observed in the in/out region will be found to be a state described by a certain collection of free particles. In that regard, particles are defined asymptotically: they enter spacetime as free particles, exit as free particles, and we can find the probability amplitudes for one in configuration to evolve to an out configuration, but we never really describe what is going on during the interaction.

Answer 2

I asked this as a graduate student to Tom Kibble at Imperial. At that time I knew of exactly integrable 1+1 dimension field theories. Even then to go from the interacting basis back to the noninteracting basis is non trivial. If you do conventional perturbation theories they do not converge to the exact solutions as the exact results were non perturbative. A challenge to Renormalization theorists. Nevertheless in many cases the quasi particle theories do work as in metals and semiconductors.