Quote by waterfall
It's in another chapter but I learnt it first in the quantum physics forum by Fredrik who says Fock Space in QFT is noninteracting (which of the following is inaccurate, please correct it):
"A Fock space is constructed from the Hilbert space associated with the singleparticle theory. You use the singleparticle space to construct a space of 2particle states, a space of 3particle states, and so on, and then you combine them all into a Hilbert space that contains all the 1particle states, all the 2particle states, and so on. This Hilbert space is called a Fock space. So it's just an algebraic construction. You need nothing more than the Hilbert space from the singleparticle theory to define it, and the singleparticle theory can be defined using a Lagrangian with no products of more than two field components or derivatives of field components.
However, in nonrigorous QFT, I think the idea is just to ignore that the interacting Hilbert space is really a different Hilbert space, and just introduce operators that can take nparticle states to (n+1)particle states for example. In this context, Fock space is, as you put it, "pretending to have interaction when it doesn't really". I really suck at QFT beyond the most basic stuff, so I can't explain it better, and I might even be wrong (about the stuff in this paragraph)."

I must admit this is the first time I hear of a Hilbert space being interacting or not. You may claim that the basis vectors constructed as a direct product of singleparticle kets of arbitrary power are eigenkets of the Hamiltonian of the system
only when the theory is noninteracting, but the space spanned by them is independent of the basis, and, at least in principle, one should be able to diagonalize even the interacting Hamiltonian acting on kets in this Fock space.
In my opinion, the most important sentence in your post is the bolded one. If there are products of more than 2 field operators in the Lagrangian, then this is necessarily an interction.