# Haag's Theorem loophole?

by Riposte
Tags: haag, loophole, theorem
 Sci Advisor P: 1,136 Interacting QFT and SR are a perfect match. Where it goes wrong is with "second quantization" which violates special relativity already at the most elementary level. If one mixes "momentum" π defined in an "internal space" or "unspecified space" with real 4-momentum in Minkowski space then you can expect to get a mess... It goes wrong from the start: The Hamiltonian H, the Lagrangian L and the term pv of a classical relativistic particle transform like: $$\begin{array}{rll} H & ~~\mbox{transforms as} ~~& \gamma \\ pv & ~~\mbox{transforms as} ~~& \beta^2\gamma \\ L & ~~\mbox{transforms as} ~~& 1/\gamma \\ \end{array}$$ The volume of a wave-function transforms like 1/γ due to Lorentz contraction. So, the densities become higher by a factor γ, hence the Hamiltonian density, the Lagrangian density and the density of the pv term transform like. $$\begin{array}{lll} {\cal H} & ~~\mbox{transforms as}~~ & \gamma^2 \\ {\cal {\small PV}} & ~~\mbox{transforms as}~~ & \beta^2\gamma^2 \\ {\cal L} & ~~\mbox{transforms as}~~ & "1" \\ \end{array}$$ The Lagrangian density is Lorentz invariant and the expressions for the densities become quadratic in nature, already in classical relativistic physics: $${\cal H-PV = -L} & ~~~~\mbox{transforms as}~~~~& ~~~~E^2~~-p^2 =~~ m^2 & ~~~~\mbox{which transforms as}~~~~ \gamma^2-\beta^2\gamma^2 = 1$$ Because E transforms like γ, p transforms like βγ and m transforms like "1". This means that we can write familiar expressions like: $$\begin{array}{lllll} {\cal H} & ~~~~\mbox{transforms as}~~~~& \,+\tfrac12E^2 +\tfrac12p^2 +\tfrac12 m^2 ~=~ ~~E^2 & ~~~~\mbox{which transforms as}~~~~& \gamma^2 \\ {\cal L} & ~~~~\mbox{transforms as}~~~~& -\tfrac12E^2 +\tfrac12p^2 -\tfrac12 m^2 ~=~ -m^2 & ~~~~\mbox{which transforms as}~~~~& '1' \end{array}$$ In second quantization this gets messed up by making replacements like $$(mv)^2 \implies \dot{\varphi}^2$$ where mv is the (non-relativistic!) momentum in an "internal" or "unspecified" space. The latter term does not transform like the pv density but like: $$\begin{array}{lll} \dot{\varphi} & ~~~~\mbox{transforms as}~~~~& \gamma \\ \dot{\varphi}^2 & ~~~~\mbox{transforms as}~~~~& \gamma^2 \\ \end{array}$$ and so the standard expression. $${\cal H}-\dot{\varphi}^2 = {\cal -L}$$ already violates special relativity at the most basic level leading to wrong expressions for either the Hamiltonian or Lagrangian densities: $$\begin{array}{lllll} {\cal H} & ~~~~\mbox{transforms (not) as}~~~~& \,+\tfrac12E^2 +\tfrac12p^2 -\tfrac12 m^2 ~=~ p^2 & ~~~~\mbox{which transforms as}~~~~& \beta^2\gamma^2 \\ {\cal L} & ~~~~\mbox{transforms (not) as}~~~~& -\tfrac12E^2 +\tfrac12p^2 +\tfrac12 m^2 ~=~0 & ~~~~\mbox{which transforms as}~~~~& '0' \\ \end{array}$$ The signs of the invariant mass terms are changed compared with the correct versions. The quantities differ by a constant value which generally goes unnoticed but spoils the way they transform. Many textbooks use the form of the Lagrangian density which is zero for all non-interacting eigenstates. Zero is Lorentz invariant but is not correct. Regards, Hans
P: 1,160
 Quote by Hans de Vries ...Many textbook use the form of the Lagrangian density which is zero for all non-interacting eigenstates....
How a Lagrangian can be zero if it contains unknown variables?

Let us note that nobody is interested in calculations (values) of action or Lagrangian expressed via solutions.
P: 1,136
 Quote by Bob_for_short How a Lagrangian can be zero if it contains unknown variables? Let us note that nobody is interested in calculations (values) of action or Lagrangian expressed via solutions.
You would have noticed how the Lagrangian densities, as typically presented,
for the Klein Gordon field or the Dirac field become zero for free particle solutions
IF you would have been simply interested enough to do the calculations....

Regards, Hans
P: 1,160
 Quote by Hans de Vries You would have noticed how the Lagrangian densities, as typically presented, for the Klein Gordon field or the Dirac field become zero for free particle solutions IF you would have been simply interested enough to do the calculations....
I do not argue that substituting the solutions in L makes it zero, but first, it is not Lagrangian any more, next, why should I need to perform such a calculation? I need solutions, so I use the Lagrangian to derive the equations. In this case any Lagrangian contains unknown variables and cannot be zero by definition.
P: 278
 Quote by meopemuk I often see this statement, but I am not sure about its validity. In my opinion, this statement goes against basic postulates of quantum theory. Let me explain why I think the Hilbert space used to describe a physical system should be independent on whether the system is interacting or not. Let us first ask why we use Hilbert spaces to describe physical systems (their states and observables) in QM? The answer is given by "quantum logic". This theory tells us that subspaces in the Hilbert space are representatives of "yes-no experiments" or logical "propositions" or experimental "questions". Meets, joins, and orthogonal complements of subspaces represent usual logical operations OR, AND, and NOT. It seems reasonable to assume that the same questions can be asked about interacting and non-interacting system. The logical relationships between these questions should not depend on the interaction as well. Therefore, the same Hilbert space (= logical propositional system) should be applied to both interacting and non-interacting system, if their particle content is the same.
That's an excellent point meopemuk, saying the interacting theory lives in a different Hilbert Space is in truth a very lazy way of saying what is really going on.

As you said all questions that can be asked about the free theory can also be asked about the free theory. These “questions” form the C*-algebra of observables if I may be very formal.
Just like QM, the interacting case and the free case give different answers to these questions. The only difference is that for QFT there exists no unitary transformation to take you from the free answers to the interacting answers, which is the basic content in Haag's theorem. To paraphrase F. Strocchi this is because the interacting and free answers correspond to “two totally seperate worlds”. It would be impossible to prepare a system which gives “free” answers to questions and have it evolve into one which gives “interacting” answers, unlike regular QM where this can be done with the right choice of Hamiltonian.

The reason people usually say they live in different Hilbert spaces is because even the though the Hilbert space is formally identical (all seperable Hilbert spaces are isomorphic), the algebra of observables acts in a different way. So we say they are two different reps of the algebra.
P: 278
 Quote by Haelfix Getting back to the original question, there are a few workarounds that I can remember, and mentioned in a previous thread about this subject. Generically they all relax one or more of the axioms of the theorem 1) If you introduce a volume cutoff, this explicitly breaks the covariance condition and the theorem does not hold 2) Renormalization by formally infinite counterterms (it does so by making such a mess of the mathematics and being so illdefined, that it violates a number of Wightman axioms to start with, so its no suprise that it also evades Haags theorem) 3) Supersymmetry. One of the assumptions of the theorem is that the fields either obey commutation relations, or anticommutation relations, but not both. So strictly speaking it does not apply to SuSY I think 4) If the interacting theory lives in a different Hilbert space (the idea Eugene hates) also strictly speaking bypasses one of the assumptions. Anyway, the point is that perturbative QFT that we all know and measure everyday is fine. But a real, sensible, nonperturbative definition of an interacting field theory is going to be difficult.
I just wanted to say a few things in relation to this specifically renormalization and Haag's theorem.
First of all, in the field theories that have been nonperturbatively constructed the free and interacting theory do live in seperate Hilbert spaces, e.g. $$\phi^{4}$$ in two or three dimensions do not live in Fock space. From this point of view the field theories living in different Hilbert space or being in different reps of the algebra of operators is the reason for Haag's theorem. There can't be a unitary relation between them because they're in different reps.
Renormalization enters in this picture as a way of fixing the mismatch. The free and interacting field theory live in seperate Hilbert spaces so the differences between them involve very singular objects which physicists call counterterms.
P: 1,160
 Quote by DarMM ...The reason people usually say they live in different Hilbert spaces is because even the though the Hilbert space is formally identical (all seperable Hilbert spaces are isomorphic), the algebra of observables acts in a different way. So we say they are two different reps of the algebra.
Now let us have a look at a non-interacting QFT. Its initial state (particle populations) does not change with time. The initial superposition remains intact with time.

When we have a physical interaction, the superposition coefficients become time-dependent and the populations change. After scattering the populations reach their new constant values and the theory becomes non-interacting again. So we stay within the same Hilbert space.

When your calculations give infinite corrections to the superposition coefficients, it is not because of "too strong interaction at short distances" but because of too poor initial approximation and wrong interaction term.

It is not correct to extrapolate the Haag's theorem obtained for badly guessed theories to good theories with physical interactions.
P: 1,475
 Quote by Bob_for_short It is not correct to extrapolate the Haag's theorem obtained for badly guessed theories to good theories with physical interactions.
That's a rather sweeping statement. Which formulation of Haag's theorem
did you have in mind? The usual Hall-Wightman form doesn't mention
specific interacting theories. Here is a statement of the theorem in
H-W form (from Shirokov's math-ph/0703021, p7):

 Suppose we have two field theories. One is a free theory described by a set of free fields $A_0(x)$ acting in the Hilbert space $\mathcal{H}_0$. The other is described by an irreducible set of fields $A(x)$. Further, let us assume that the following conditions are satisfied: 1) $A(x)$ is an operator in $\mathcal{H}$ which carries a unitary representation of translations and rotations $$U(\mathbf{a},R) A(x) U^{\dag}(\mathbf{a},R) ~=~ A(Rx + \mathbf{a})$$ and 1') Lorentz transformations $$U(\Lambda) A(x) U^{\dag}(\Lambda) ~=~ A(\Lambda x) ~.$$ (These relationships are written for the particular case of a scalar field.) 2) There is a unique invariant state $U \Omega ~=~ \Omega$ in $\mathcal{H}/itex]. 3) There exists a unitary operator [itex]V$, from $\mathcal{H}_0$ to $\mathcal{H}$, such that at a time instant $t$ we have $$A(\mathbf{x}, t) ~=~ V(t)A_0(\mathbf{x}, t) V^{\dag}(t)$$ 4) The spectrum of energies is bounded from below. Then $A(x)$ is a free field.
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The same paper (pp5-6) also gives an alternate formulation
of the theorem as follows:

 Lemma. Suppose we have a Euclidean (i.e., translational and rotational) invariant field theory written in terms of creation-annihilation operators $a^{\dag}_{\mathbf{p}}, a_{\mathbf{p}}$. Suppose also that a Fock representation of these operators in the Hilbert space $\mathcal{H}_0$ with the no-particle vector $\Omega_0$ is given. Then, in $\mathcal{H}_0$ there is a unique normalizable eigenstate of the total momentum operator $\mathbf{P}$, and this state coincides with $\Omega_0$. Proof: (See the paper.) Theorem. Suppose that conditions of the Lemma are satisfied and there is a unique normalizable eigenstate of the full Hamiltonian $H$ with lowest energy, i.e., the vacuum vector $\Omega$. Then $\Omega$ must coincide with $\Omega_0$. Proof. Since $[H, P_j]= 0$, $\Omega$ must be a common eigenvector of $H$ and $\mathbf{P}$. However there is only one normalizable eigenstate of $\mathbf{P}$ in $\mathcal{H}_0$, and this eigenstate coincides with $\Omega_0$. Thus $\Omega = \Omega_0$. (End of proof.) In fact, in all local theories $\Omega$ does not coincide with the no-particle vector of ”bare” creation-annihilation operators which diagonalize $H_0$. This means such theories violate some assumptions of the theorem.
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Neither formulation relies on any specific details of the
interaction term in the Hamiltonian (except perhaps locality).

To evade Haag's theorem, one must either state which of its input assumptions should be abandoned or modified, or demonstrate that the maths in the existing proofs are somehow erroneous. E.g., Shirokov goes on to show that condition (1') is not valid in his "dressed field" approach.
P: 1,742
 Quote by strangerep To evade Haag's theorem, one must either state which of its input assumptions should be abandoned or modified, or demonstrate that the maths in the existing proofs are somehow erroneous. E.g., Shirokov goes on to show that condition (1') is not valid in his "dressed field" approach.
Hi strangerep,

I agree with Shirokov on this point. As I wrote earlier, condition (1') does not follow from any deep physical principle, so it should be abandoned.

In this connection I would like to mention the work

H. Kita, "A non-trivial example of a relativistic quantum theory of particles without divergence difficulties", Progr. Theor. Phys., 35 (1966), 934.

in which a perfectly valid and relativistically invariant interacting QFT model is constructed, "interacting fields" are explicitly calculated, and it is demonstrated that the condition (1') does not hold.
P: 278
 Quote by meopemuk Hi strangerep, I agree with Shirokov on this point. As I wrote earlier, condition (1') does not follow from any deep physical principle, so it should be abandoned. In this connection I would like to mention the work H. Kita, "A non-trivial example of a relativistic quantum theory of particles without divergence difficulties", Progr. Theor. Phys., 35 (1966), 934. in which a perfectly valid and relativistically invariant interacting QFT model is constructed, "interacting fields" are explicitly calculated, and it is demonstrated that the condition (1') does not hold.
Two things to note are:
(a) All models which have been constructed nonperturbatively do satisfy the covariant transformation law. For instance in all scalar and Yukawa theories in two and three dimensions, the Gross-Neveu model in two and three dimensions and also QED and the Higgs model in two dimensions, the interacting field transforms covariantly.
(b) Kita's model breaks another Wightman axiom besides the covariant transformation law axiom, so it may not have all the properties associated with a QFT. Although the model is still very interesting. Similar models also appear in Algebraic quantum field theory, where the "field" is not covariant, in fact it can even be nonlocal. However the problem is that only theories which satisfy the Wightman axioms have been proven to have all the healthy properties normally associated with a QFT.
P: 1,160
 Quote by strangerep ...That's a rather sweeping statement. ... Neither formulation relies on any specific details of the interaction term in the Hamiltonian (except perhaps locality). To evade Haag's theorem, one must either state which of its input assumptions should be abandoned or modified, or demonstrate that the maths in the existing proofs are somehow erroneous. E.g., Shirokov goes on to show that condition (1') is not valid in his "dressed field" approach.
Concerning Wightman's axioms, maybe it was too early to axiomatize something? The theory is not ready yet, but one hurries up to axiomaize or impose his vision to it.

Concerning Haag, it is funny: you take a filed A(x,t) and it happens to be free whatever interaction you use. Maybe Haag himself should have pointed out where his proof was interaction-dependent?

Here I would like to draw you attention to another reading of free fields. For the sake of sloppiness, I will take a Dirac field ψ and the quantized electromagnetic filed Atr and I will write the free equations in a lazy way:

(γ∂+m)ψ = 0, ∂2Atr = 0

These are two independent equations that in the standard QED are "coupled" with an "interaction" term jA. As soon as jA includes the self-action and non-linearity, one obtains problems with vacuum instability and other self-action provoked rubbish. This is a dead end in the theory development. There are too few things to axiomatize in it.

Now let us look at the free equations as at equations of the center of inertia of a compound system and internal motion equations. They should be independent - they describe independent degrees of freedom of one compound system. So there in no need to couple them if there is no external force acting on the charge.

An external force acting on a charge makes two independent (=additive) works: it changes the CI energy-momentum and it pumps the internal degrees of freedom (oscillations). In presence of an external filed Aext the equations are modified with the external filed contributions. The equation system is then essentially linearized. One can be sure that the physical solutions exist.

In this description the main problem is to express the charge coordinates via CI coordinates and relative ones. This expression should be dictated with experimental facts first of all rather than with axioms of somebody's. Then the theory becomes a routine theory of compound systems without any mathematical and conceptual problems. After having been developed and verified experimentally, it might serve as a model for axiomatization although I personally am against axiomatization. Only experiment can give an idea how other theories (strong, weak, gravity forces) could be constructed. I mean, for example, what quasi-particles exist in these compound systems and how they are excited.

I think this, not yet explored direction is worth developing and practising.
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 Sci Advisor P: 1,128 The theorem says that if $A_0(x)$ is a free field of mass $m_0$, and $V(t)$ is a unitary operator, then $A(x) = V(t)A_0(x) V^{\dag}(t)$ is also a free field of mass $m_0$. But suppose we consider adding an extra $A_0^2$ term to the Hamiltonian to shift the mass, and take this extra term as an interaction that we treat perturbatively. We could then construct the usual Dyson series for $V(t)$, and this should transform $A_0(x)$ with mass $m_0$ into $A(x)$ with mass $m$. According to the theorem, though, such an operator does not exist! So what goes wrong? Consider the vacuum state of the original hamiltonian. Since the field is free, we can describe it in terms of noninteracting momentum modes, each of which is a harmonic oscillator. The ground state wave function of each mode is a gaussian in the Fourier-transformed field mode, with a width controlled by the energy $E=({\bf k}^2+m_0^2)^{1/2}$, and the ground state wave functional of the field theory is the product of these individual mode wave functions. Of course, after the mass is shifted, the ground state wave functional looks the same, except $m_0$ is replaced by $m$. Now, the point is that the inner product of these two grounds states is zero. This is because the mode wave functions are not the same (one has $m_0$ in it and the other $m$), so the inner product of the two normalized mode wave functions (with the same three-momentum $\bf k$) is less than one. Now we have to take the product of an infinite number of numbers, each less than one, and so we get zero. The same argument shows that the ground state with the shifted mass is orthogonal to every state in the Fock space of the original field. So there is no unitary map from one set to the other. But this is clearly a function of having used an infinite number of modes. If we provide infrared and ultraviolet cutoffs, so that the number of modes is finite, the inner products will be finite, and the unitary operator will exist. Lorentz invariance requires an infinite number of modes, and that's why it's a condition of the theorem. I believe it is not a coincidence that the only general method we know for nonperturbative regularization of QFT is the lattice, which of course breaks Lorentz symmetry. Lorentz symmetry then emerges only in the limit in which the regulator is removed.
P: 1,160
 Quote by Avodyne ...I believe it is not a coincidence that the only general method we know for nonperturbative regularization of QFT is the lattice, which of course breaks Lorentz symmetry. Lorentz symmetry then emerges only in the limit in which the regulator is removed.
By saying "QFT" you imply a certain interaction term, that is the problem, not Lorentz invariance.
 Sci Advisor P: 1,636 Thats sort of the point, there is no obstruction for doing QFT 'as done by physicists', say treated as a lattice in the Wilsonian framework. The problem is the existence of the 'thing' we are actually approximating, which may or may not exist mathematically in general, even if we have a handful of explicit examples of fully nonperturbative models (usually in 2d) Its wonderfully miraculous that the whole business works at all. I think a lot of physicists became a lot less worried about Haags theorem after Wilsons work on effective field theory, where it became clear that in general there might need to be a UV completion for most phenomenological field theories of interest.
P: 1,475
 Quote by meopemuk [...] I would like to mention the work H. Kita, "A non-trivial example of a relativistic quantum theory of particles without divergence difficulties", Progr. Theor. Phys., 35 (1966), 934. in which a perfectly valid and relativistically invariant interacting QFT model is constructed, "interacting fields" are explicitly calculated, and it is demonstrated that the condition (1') does not hold.
"Perfectly valid" is perhaps not quite accurate, since Kita's theory
is constructed merely perturbatively, and only to rather low order.
AFAICT, he doesn't address convergence questions thoroughly and
rigorously (beyond some wishful thinking that mollifier functions
will save the day at all orders). Hence I don't think he has
a nonperturbatively valid theory.
P: 1,475
 Quote by DarMM (b) Kita's model breaks another Wightman axiom besides the covariant transformation law axiom,
Which other axiom does it break?
P: 1,475
 Quote by Bob_for_short Concerning Wightman's axioms, maybe it was too early to axiomatize something? The theory is not ready yet, but one hurries up to axiomatize or impose his vision to it.
I see "axiomatize" as merely a synonym for "let's use clear and honest mathematics",
which seems like a good thing at any stage. (Of course, choosing particular axioms
must be a physically-motivated endeavour).

 Concerning Haag, it is funny: you take a filed A(x,t) and it happens to be free whatever interaction you use.
I believe the point is that an interacting theory cannot share a Hilbert space with
a free theory.

 Maybe Haag himself should have pointed out where his proof was interaction-dependent?
IIRC, Haag's original paper basically just pointed out that all known physically-relevant
interactions contained products of creation operators which necessarily lead outside
the free Hilbert space.

 This expression should be dictated with experimental facts first of all rather than with axioms of somebody's.
As I said above, those axioms are merely someone's attempt to construct a
theory using honest mathematics. It might of course be physically incorrect.
But even if we take a phenomenlogical approach (starting from experimental facts),
one still needs to use honest mathematics in whatever theory one constructs.

 [...] I personally am against axiomatization. [...]
Perhaps you should re-phrase that statement. It sounds like you're
against honest mathematics, which I'm sure is not what you mean.
P: 1,742
 Quote by strangerep Kita's theory is constructed merely perturbatively, and only to rather low order.
The same can be said about our most accurate theory - QED. I am pretty sure that Shirokov's condition (1') does not apply in QED. Does it make QED invalid in your opinion?

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