What does Haag's theorem say about the Schrodinger picture?

Suppose there are two interacting fields $$\phi _1$$ and $$\phi_2$$. Let $$\psi [\phi_1, \phi_2]$$ be a functional with the two fields as the input functions and complex numbers as the output, such that the functional integral $$\int |\psi|^2 d[\phi_1]d[\phi_2]$$ is normalizable to 1.

This means that $$\psi$$ can serve as the initial state of the system. The time evolution will be given by the functional Schrodinger equation.

The initial state is a vector in a Hilbert space, the Schrodinger equation is bound to map vectors to vectors. Given all this, I see no way the time evolution isn't part of the Hilbert space. But Haag's theorem seems to say this exact thing: The states of an interacting field theory aren't part of the Hilbert space.

Can someone please explain this? Or if I've got Haag's theorem wrong, please explain what exactly does Haag's theorem say in the Schrodinger picture.

Haag's theorem says that the representation of the CCR of a free field cannot be isomorphic to the representation of the CCR of an interacting field.

So if you start with an interacting field, there's no issue: All the states and operators live in the representation of the interacting field, and everything works normally. We're not saying "time evolution doesn't exist".

What Haag's theorem does is draw into question the rigorous meaning of the usual physical treatment of interacting quantum field theories, where we talk about the theory being the free theory in the asymptotic past/future and write formulae that seem to connect the free and the interacting vacuum (e.g. stuff that looks schematically like $$\lvert 0\rangle = \lim_{T\to\infty} \mathrm{e}^{\mathrm{i}Ht} \lvert \Omega\rangle$$): These formulae that involve both free and interacting states, connected by the action of an operator on "the Hilbert space", make no sense in light of Haag's theorem: The free and interacting states live in different Hilbert spaces, and the field operators that act on the free states are not the same field operators that act on the interacting states - the free and interacting space are different, non-isomorphic representations.

This, in particular, says that "the interaction picture" doesn't exist: In the interaction picture we must assume that the free $$H_0$$ and the interaction part $$H_I$$ are operators on the same space - otherwise nothing we write down in the interaction picture makes any sense. Therefore, this is a problem for the standard derivations of the LSZ formula in canonical quantization, which makes heavy use of the interaction picture.

Haag's theorem doesn't really have anything to say about the Schrödinger or Heisenberg pictures.

• Does Haag's theorem disprove the adiabatic theorem in QFT? In the post, I assumed that the Hilbert space of the interacting theory is that of functionals $\psi [\phi_1 , \phi_2]$. That makes this space simply the direct product of two free field Hilbert spaces. Is Haag's theorem against this? May 2 at 12:15
• @RyderRude Yes, that's exactly what Haag's theorem says isn't true - while you may write both interacting and free states as functionals $\psi[\phi_i]$, Haag's point is that the "$\phi_i$" appearing there in the free and interacting cases are different objects, and there is no "good" mapping (isomorphism) between them. Perhaps your confusion is this: The Hilbert spaces of the free and interacting theory are of course isomorphic as Hilbert spaces - all Hilbert spaces of the same cardinality are. Haag's theorem is about isomorphisms as representations of the CCR. May 2 at 12:19
• Hey, about the CCR part, were you saying this because conjugate momenta of interacting theories may be different, and hence the CCR have to be imposed with different conjugate momenta compared to free theories? Jul 6 at 10:05
• @RyderRude No, it is not the case that "different conjugate momenta" necessarily lead to different representations: Compare to the ordinary QM case, where the Stone-von Neumann theorem guarantees that all representations of $[x_i,p_j] = \mathrm{i}$ are unitarily equivalent - even for theories for which the canonical $p_j$ is are "different". Jul 6 at 10:25
• But even then, if you fix one operator, say $X=x\delta (x-x')$, then $P$ gets severely restricted. In interacting QFT, $X$ remains fixed but the corresponding momentum may change drastically due to the interaction term of the Lagrangian. This should lead to a very different Hilbert space. Is Haag's argument any similar to this? Jul 6 at 10:32