# Haag's Theorem and the Poincare Group

• A
Gold Member

## Main Question or Discussion Point

But realistic Hilbert spaces are direct sums of many such irreps
I'll have to think a bit about what you've written, but just to note this is not true due to Haag's theorem.

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I'll have to think a bit about what you've written, but just to note this is not true due to Haag's theorem.
Why would Haag's theorem change things here?

I think it is fair to assume that there is at least one good Hilbert space representation for the realistic observables in Nature, with the appropriate unitaries for symmetries. It may be non-unique, but it will predict expectation values to any required degree of precision. If such a Hilbert space is Poincare-covariant (i.e. if physics on flat spacetime is internally consistent) then that Hilbert space can be decomposed into Poincare irreps. We may not know what these will look like, but there should be some nice ones, like say, the set of states (position wavefunctions) of a single ground-state hydrogen atom. Each irrep will have a definite invariant mass, so any operator that changes the mass of the system, such as a smeared electric field operator, will not be covered by the Moretti/Oppio result.

Gold Member
Why would Haag's theorem change things here?

I think it is fair to assume that there is at least one good Hilbert space representation for the realistic observables in Nature, with the appropriate unitaries for symmetries. It may be non-unique, but it will predict expectation values to any required degree of precision. If such a Hilbert space is Poincare-covariant (i.e. if physics on flat spacetime is internally consistent) then that Hilbert space can be decomposed into Poincare irreps. We may not know what these will look like, but there should be some nice ones, like say, the set of states (position wavefunctions) of a single ground-state hydrogen atom. Each irrep will have a definite invariant mass, so any operator that changes the mass of the system, such as a smeared electric field operator, will not be covered by the Moretti/Oppio result.
I'm not saying it as something that corrects or defends Moretti & Oppio's result, just as an interesting fact in itself. The Hilbert spaces of interacting quantum field theories are not sums of irreps of the Poincaré group with definite invariant masses. The hydrogen atom in QED for example won't belong to a Poincaré irrep of such a form.

dextercioby
I'm not saying it as something that corrects or defends Moretti & Oppio's result, just as an interesting fact in itself. The Hilbert spaces of interacting quantum field theories are not sums of irreps of the Poincaré group with definite invariant masses. The hydrogen atom in QED for example won't belong to a Poincaré irrep of such a form.
Really? This is very surprising to me! Of course we don't really know what the Hilbert space looks like, but the states of ground-state hydrogen are clearly physical and clearly go only to each other under Poincare transformations. The invariant mass is also well defined. Why would this not be an irrep?

BTW, I need to correct what I wrote before:
For instance, for a free real scalar field, some irreps are: the vacuum (for which we will need to invent an extra dimension as its "imaginary partner"); the space of one-particle states; the spaces of N-particle states where all the particles are in the same mode; the spaces of two-particle states that, when expressed at wavefunctions on momentum space, have support only for pairs of momenta with some fixed scalar product; etc.
Actually, on a Hilbert space of two-scalar-particle states, the Pauli-Lubanski vector is nontrivial (except in the case where they share a single "wavefunction"). So we need to decompose the space further, to states that also have fixed values of the Casimir, i.e. fixed squared orbital angular momentum in the rest frame, along with the fixed scalar product of momenta.

Gold Member
Really? This is very surprising to me! Of course we don't really know what the Hilbert space looks like, but the states of ground-state hydrogen are clearly physical and clearly go only to each other under Poincare transformations. The invariant mass is also well defined. Why would this not be an irrep?
This would really take another thread, but you have example constructions of interacting Hilbert spaces from the works of Magnen, Glimm, Jaffe and others.

For the QED case the book of Othmar Steinmann gives a good construction of some aspects of the Hilbert space.

In general Haag's theorem prevents the decomposition which you are referring to.

dextercioby
This would really take another thread, but you have example constructions of interacting Hilbert spaces from the works of Magnen, Glimm, Jaffe and others.

For the QED case the book of Othmar Steinmann gives a good construction of some aspects of the Hilbert space.

In general Haag's theorem prevents the decomposition which you are referring to.
Can you (or someone else) give a one-line general idea of what goes wrong?

Gold Member
Can you (or someone else) give a one-line general idea of what goes wrong?
There's a variety of effects that lead to it. The fact that interacting Hilbert spaces are not Fock, that the interactions in gauge theories prevent the formation of asymptotic Hilbert spaces that permit Wigner reps (infraparticle problem), at the most extreme end you'd have that there is no such thing as a pure state for a system confined to a region of spacetime, e.g. a hydrogen atom.

dextercioby
stevendaryl
Staff Emeritus
There's a variety of effects that lead to it. The fact that interacting Hilbert spaces are not Fock, that the interactions in gauge theories prevent the formation of asymptotic Hilbert spaces that permit Wigner reps (infraparticle problem), at the most extreme end you'd have that there is no such thing as a pure state for a system confined to a region of spacetime, e.g. a hydrogen atom.
Well, the specific question was why can't a hydrogen atom be considered an "elementary particle". In the sense of irreducible representations, I think it's almost by definition that a composite object can't have an irreducible representation.

Gold Member
Well, the specific question was why can't a hydrogen atom be considered an "elementary particle". In the sense of irreducible representations, I think it's almost by definition that a composite object can't have an irreducible representation.
I understood the question as being more generally about sums of (tensor products) of irreps, hydrogen isn't even that.

dextercioby
A. Neumaier
2019 Award
Why would Haag's theorem change things here?

I think it is fair to assume that there is at least one good Hilbert space representation for the realistic observables in Nature, with the appropriate unitaries for symmetries. It may be non-unique, but it will predict expectation values to any required degree of precision. If such a Hilbert space is Poincare-covariant (i.e. if physics on flat spacetime is internally consistent) then that Hilbert space can be decomposed into Poincare irreps. We may not know what these will look like, but there should be some nice ones, like say, the set of states (position wavefunctions) of a single ground-state hydrogen atom. Each irrep will have a definite invariant mass, so any operator that changes the mass of the system, such as a smeared electric field operator, will not be covered by the Moretti/Oppio result.
The Hilbert spaces of interacting quantum field theories are not sums of irreps of the Poincaré group with definite invariant masses.
In general Haag's theorem prevents the decomposition which you are referring to.
Can you (or someone else) give a one-line general idea of what goes wrong?
What goes wrong is that the irreps of the Poincare group are labelled among others by a continuous mass parameter. Therefore one has to take into account direct integrals of irreps rather than just direct sums. This is reflected by the Kallen-Lehmann formula, which contains an (in general Stieljes) integral over the mass.

However, Haag's theorem only forbids a Fock space structure of interacting theories, and says nothing about the representation theory of the Poincare group involved. Indeed, the Hilbert space of an interacting quantum field theory is always a direct integral over unitary irreducible representations of nonnegative mass, integrated over a complicated set of labels - labeling in addition to mass and spin/helicity the different copies of a representation of fixed mass and spin/helicity.

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dextercioby, king vitamin and maline
Gold Member
However, Haag's theorem only forbids a Fock space structure of interacting theories, and says nothing about the representation theory of the Poincare group involved.
Does it not tell you it can't be sums of tensor products of irreps?

A. Neumaier
2019 Award
Does it not tell you it can't be sums of tensor products of irreps?
No. Where is a formulation of the theorem that would say so? It only says that it cannot be the Fock sum of tensor products, and indeed it is something more complicated.

The unitary representation theory of the Poincare group is completely known; any unitary representation is a direct integral of irreducible ones. Since the Wightman axioms require a unitary representation, there must be such a decomposition.

dextercioby and maline
Gold Member
Oh that's interesting, what's an example of a Hilbert space formed from a sum of irreps that's not the Fock sum?

A. Neumaier
2019 Award
Oh that's interesting, what's an example of a Hilbert space formed from a sum of irreps that's not the Fock sum?
Take the 1-particle space of a quasifree scalar field theory. It is not a Fock space and the Poincare group acts reducibly (but as a direct integral).

dextercioby
Gold Member
Take the 1-particle space of a quasifree scalar field theory. It is not a Fock space and the Poincare group acts reducibly (but as a direct integral).
Yes, I'm not saying Haag's theorem forbids integrals of irreps. As you said the representation theory of the Poincaré group is known, but what's an example of a sum of irreps that's not Fock? Since it forbids Fock sums, unless there is an example of a sum that is not Fock that means it forbids sums.

Unless you are taking sums to include integerals.

A. Neumaier
2019 Award
what's an example of a sum of irreps that's not Fock?
Well, take the sum of nonsymmetric tensor products, or the sum of symmetric tensor products up to some degree, or the tensor product of a Fock space with a 1-particle space, etc. Fock space is a very special situation, defined by the presence of annihilation and creation operators satisfying CCR/CAR, which is absent in general.

Gold Member
Are those reps possible for a quantum field theory?

I would have thought the only possible summed rep for a field theory is the Fock one, which is then eliminated by Haag's theorem.

A. Neumaier
2019 Award
Are those reps possible for a quantum field theory?

I would have thought the only possible summed rep for a field theory is the Fock one, which is then eliminated by Haag's theorem.
Well, not these (you were talking in more general terms to which my examples applied) but others.

The Hilbert space of a Wightman field theory decomposes into a complicated direct sum of direct integrals of irreducibles - not into a direct sum of tensor products of irreducibles. I guess this is what @maline had in mind but expressed in an informal fashion.

I know only a small part of the structure of this decomposition; but I wouldn't talk about this in public. (Send me an email if interested.)

DarMM
Well, the specific question was why can't a hydrogen atom be considered an "elementary particle". In the sense of irreducible representations, I think it's almost by definition that a composite object can't have an irreducible representation.
No, the definition of a composite object is that its field doesn't appear in the Lagrangian.
An irreducible representation of a group is, practically speaking, a Hilbert subspace that is closed under the group's action and that has no proper subspaces that are similarly closed. The representation is then the set of group operators, as confined to this subspace. The single particle states of a stable particle are special in that they form an irrep with no continuous parameter labels (such as the scalar product of momenta for the two-free-scalar-particle case I mentioned above), and so they contain normalizabe states that live inside the full Hilbert space, The vacuum is also a trivial irrep on its own. The multi-particle irreps, OTOH, must be indexed continuously by various scalar values, and a state with a fixed value of an observable with a continuous spectrum will not be normalizable within the full Hilbert space, just like plane waves or "position states" in nonrelativistic QM. Then the decomposition of the full Hilbert space into such "subspaces" will be an direct integral rather than a direct sum, as Arnold pointed out.
But the single particles need not be elementary. All that is required is an object that can be specified without continuous parameters - as bound states usually are - and that is stable under the full Hamiltonian. It can be an electron, a proton, a stable nucleus, the ground state of an atom, or even a macroscopic perfect crystal of iron.
Does it not tell you it can't be sums of tensor products of irreps?
I am not discussing tensor products of irreps, but only single irreps. Even in the free case, it makes sense to decompose the Fock space in this way, labelling each irrep according to the scalars that appear in it, such as total mass, particle number, scalar products of momenta, and spin (squared). In the interacting case, the irreps will look very different and ugly, because we need to use the full Hamiltonian for time translations. But some nice, stable cases, like the ground-state hydrogen atom, should still appear.

A. Neumaier
2019 Award
the single particles need not be elementary. All that is required is an object that can be specified without continuous parameters - as bound states usually are - and that is stable under the full Hamiltonian. It can be an electron,
It cannot be a QED electron, since these are infraparticles and have a continuous mass spectrum.

Gold Member
I am not discussing tensor products of irreps, but only single irreps.
My response with Haag's theorem was originally in response to this:
But realistic Hilbert spaces are direct sums of many such irreps
Can you tell me what I'm misunderstanding. I took it to be referring to the whole of the QFT's Hilbert space which by Haag's theorem cannot be just a direct sum, but must involve integrals over irreps, although the integral might involve point masses.

Were you in fact referring to direct sums of integrals of irreps as @A. Neumaier said.

Gold Member
But some nice, stable cases, like the ground-state hydrogen atom, should still appear.
Ground state hydrogen will involve electrons which are infraparticles, so it's a least not clear to me that it's an irrep as opposed to an integral over irreps.

stevendaryl
Staff Emeritus
No, the definition of a composite object is that its field doesn't appear in the Lagrangian.
An irreducible representation of a group is, practically speaking, a Hilbert subspace that is closed under the group's action and that has no proper subspaces that are similarly closed.
Thank you. I was misremembering what I learned years ago. When in SU(3) one writes ##3 \otimes \bar{3} = 8 \oplus 1##, the ##3, \bar{3}, 8, 1## are irreducible, but ##3 \otimes \bar{3}## is not. Is that right?

maline
Gold Member
No, the definition of a composite object is that its field doesn't appear in the Lagrangian.
Just to say it is difficult to maintain this definition of composite when it comes to gauge theories like QCD. There none of the states in the Hilbert space have a field present in the Lagrangian. Using your definition all states in QCD are composite.

A. Neumaier