I didn't quite grasp the essence of Haags Theorem in the the way it is presented (for example on wikipedia), but the issue seems to be that if one wants to represent infinitely degrees of freedom using operators acting on a Hilbert space that satisfy the canonical commutation relations, then one can't find an Unitary transformation between two such representations. (Correct me if any of the things I said is either wrong or "not well formulated". Nitpicking is allowed).

The Theorem is often stated to be the cause for no interaction-picture to exist (which is understandable, changing to an interaction-picture from the Heisenberg-Picture needs a unitary transformation). However, by the same argument I could argue that no time-evolution exists (of which I think that this is an even more fundamental problem). Why has nobody ever pointed this out? Or am I wrong and Haag's theorem doesn't prevent time-evolution?


1 Answer 1


The answer is no, Haag's theorem does not prevent time-evolution in the Heisenberg picture.

Given any representation and any unitary transformation, applying the latter to the former gives a unitarily equivalent representation. And well-defined unitary representations do exist, as long as we're using a well-defined formulation to begin with. In particular, given a well-defined formulation with a Hilbert space, an algebra of local operators, and a Hamiltonian $H$, we can use the unitary time-evolution operator $U(t)=\exp(-iHt)$ with no problems. Haag's theorem says that if we start with the vacuum representation in a free scalar model, then no unitary transformation can give the vacuum representation of an interacting scalar model — at least not when working in infinite volume, etc; some caveats are highlighted below. So the interaction picture doesn't work, at least not in terms of operators acting on state-vectors, again with some caveats highlighted below.

What does Haag's theorem say?

For reference, this is how Haag's theorem is expressed on page 12 in "Haag's theorem in renormalised quantum field theories" (https://arxiv.org/abs/1602.00662):

Haag’s Theorem. If a scalar quantum field is unitarily equivalent to a free scalar quantum field, then, by virtue of the reconstruction theorem, it is also a free field because all vacuum expectation values coincide.

This is how it is expressed more carefully on page 49 in the same paper:

Theorem 11.7 (Haag’s Theorem). Let $\varphi$ and $\varphi_0$ be two Hermitian scalar fields of mass $m\geq 0$ in the sense of the Wightman framework. Suppose the sharp-time limits $\varphi(t,f)$ and $\varphi_0(t,f)$ exist and that at time $t = 0$ these two sharp-time fields form an irreducible set in their respective Hilbert spaces ${\cal H}$ and ${\cal H}_0$. Furthermore let there be an isomorphism $V:{\cal H}_0\rightarrow {\cal H}$ such that at time $t$, $\varphi(t,f)=V\varphi_0(t,f)V^{-1}$. Then $\varphi$ is also a free field of mass $m\geq 0$.

(I assume that $\varphi_0$ denotes a free field.) Notice that these statements of Haag's theorem are specific to scalar fields. As far as I know, the theorem has never been generalized to models that have gauge fields.

Neither of the excerpts shown above says anything that conflicts with time-evolution in the Heisenberg picture. A free scalar field remains free (with the same mass) under time-evolution, and an interacting scalar field remains interacting (with the same mass and coupling constants) under time-evolution. So Haag's theorem does not forbid time-evolution.

However, according to these excerpts, Haag's theorem does imply that a scalar field cannot begin free and then become interacting (or vice versa), so it implies that the interaction picture doesn't work — at least, it doesn't work under the theorem's strict conditions (which I didn't copy here), including strict Poincare symmetry.

Theorem 17.1 in the same paper highlights a related result that the author calls "Haag's theorem for free fields". The theorem says that two models of free scalar fields with different masses cannot be unitarily equivalent to each other.

By the way, Haag's theorem can be considered irrelevant in practice, for two reasons:

  • The only known mathematically well-defined constructions of most${}^{[1]}$ interacting QFTs involve treating space (or spacetime) as a finite lattice, but Haag's theorem relies on Poincare symmetry, or at least on an infinite-volume limit.

  • Well-defined lattice-based constructions don't use the interaction picture anyway.

We don't normally use a lattice formulation explicitly (because it's messy), but it's important in the modern view of renormalization. We can think of the usual ill-defined perturbative calculations as a convenient abbreviation for messy-but-well-defined calculations on a finite lattice. With this perspective, Haag's theorem essentially becomes irrelevant. Similar sentiments were expressed in another post about Haag's theorem.

${}^{[1]}$ In a comment, Abdelmalek Abdesselam pointed out that there are exceptions: "There are models (in 2d and 3d) constructed in the continuum and satisfying Poincare invariance."

  • $\begingroup$ +1, but I would also add a paragraph about how Haag’s theorem actually handicaps naive construction of interacting QFTs and why it isn’t related to tone evolution (to answer op’s original question). $\endgroup$ Nov 7, 2018 at 7:48
  • $\begingroup$ I'm left with some questions in this answer: 1: What has unitary equivalence of 2 models with different masses to do with the existence / nonexistence of a unitary transformation between 2 representations of the field- operators? 2: What exactly does "Unitary Unequivalent mean"? I thought it would mean that one couldn't find a unitary transformation between two representations of the CCR, but this seems to be wrong then? 3: If time-evolution is not forbidden by Haags theorem, isn't that a contradiction to the statement that no unitary transformation can be found? $\endgroup$ Nov 7, 2018 at 22:04
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    $\begingroup$ @Dan Yand: Do I understand right: Haags Theorem doesn't say that unitary equivalence doesn't exist, it only says that unitary equivalence between interaction picture and Heisenberg picture doesn't exist? So my proposed understanding in my original post is wrong? $\endgroup$ Nov 15, 2018 at 11:59
  • $\begingroup$ @DanYand : Since this was the main confusion of me, I'd gladly accept your answer if you could edit this little detail into it (My question is answered now, but comment sections tend to get deleted some times, so it'd be nice if you could prevent this helpful comment of yours to vannish at some point) :) $\endgroup$ Nov 15, 2018 at 14:12
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    $\begingroup$ "The only known mathematically well-defined constructions of most interacting QFTs involve treating space (or spacetime) as a finite lattice" is an erroneous statement. There are models (in 2d and 3d) constructed in the continuum and satisfying Poincare invariance. They are obtained as limits of lattice models. $\endgroup$ Mar 16, 2019 at 12:04

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