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Dickfore
#7
Feb5-12, 05:41 AM
P: 3,014
Quote Quote by waterfall View Post
It's in another chapter but I learnt it first in the quantum physics forum by Fredrik who says Fock Space in QFT is non-interacting (which of the following is inaccurate, please correct it):

"A Fock space is constructed from the Hilbert space associated with the single-particle theory. You use the single-particle space to construct a space of 2-particle states, a space of 3-particle states, and so on, and then you combine them all into a Hilbert space that contains all the 1-particle states, all the 2-particle 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 single-particle theory to define it, and the single-particle theory can be defined using a Lagrangian with no products of more than two field components or derivatives of field components.

However, in non-rigorous 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 n-particle 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 single-particle kets of arbitrary power are eigenkets of the Hamiltonian of the system only when the theory is non-interacting, 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.