# What are the states in QFT?

#### A. Neumaier

I am very interested in this question rised by Eugene. I would like to reformulate it in a more general form as follows. Basically standard QFT is based on the following two assumptions:

1) A representation of the Poincaré group is defined on the Hilbert space of a relativistic quantum system;

2) All the operators of the Hilbert space are generated by fields, i.e., operator valued distributions defined on Minkowski space-time and transforming covariantly under the above representation.

For example, these two assumptions can be easily recognized in Wightman's axiomatic formulation of QFT. Of course 1 and 2 are different and independent assumptions, and a theory which only satisfies 1 can obviously be developed.

My problem is that I am not completely conviced of the need of assumption 2
If you have the representation, every operator A_0 defines a field satisfying 2) by mean of the construction given in my previous mail. Thus 2) in itself is an empty requirement.

The important missing thing is that there must be such a (distribution-valued) field that is nonzero and satisfies causal commutation rules.

#### bg032

If you have the representation, every operator A_0 defines a field satisfying 2) by mean of the construction given in my previous mail. Thus 2) in itself is an empty requirement.

The important missing thing is that there must be such a (distribution-valued) field that is nonzero and satisfies causal commutation rules.
Ok, I did not mention that the fields are required to be causal (I also did not mention the spectrum condition for the energy-momentum operators P_\mu and the existence, uniqueness and translation-invariance of the vacuum). I will change the post.

However this simply reinforce my question: why do we require the existence of covariant and (microscopically) causal fields? Can we renounce to covariance or to microscopic causality? Note that our experimental evidence of causality is a macroscopic evidence, and I think it is not impossible to built a theory which violates microscopic causality but nevertheless is compatible with our macroscopic evidence of causality.

For example, if I am not wrong your fields A(x):=U(x)A_0U(-x) cannot be causal. In fact A(x) is a well defined operator for every x, and it is well known that a covariant causal field cannot be a well defined operator at every point x (if we also assume the spectral condition for the energy-momentum operators and the translation invariance of the vacuum).

Eventually, it is well known that Wightman axioms are very difficult to satisfy, and actually impossible in gauge field theories (http://arxiv.org/abs/hep-th/0401143).

#### A. Neumaier

why do we require the existence of covariant and (microscopically) causal fields?
Because, as Weinberg showed, these properties are expected to hold for all QFTs that can be constructed from local actions - and this includes all QFTs in current use! Wightman's axioms are not arbitrarily chosen but are the ones that one can derive (on the level of rigor of theoretical physics) from the assumptions that particle physicists rely on all the time, plus the assumption of a mass gap. (The infrared behavior of a massless theory seems to require appropriate modifications of Wightman's axioms.)

Moreover, in 2D and 3D, this expectation is actually rigorously verifiable in many cases.

Third, Weinberg also showed (in Chapters 3 and 4) that these properties seem necessary in order that a theory has a covariant S-matrix and satisfies the cluster decomposition property.

Fourth, given the Wightman axioms, one can deduce a lot of sensible physical properties (e.g., a well-defined scattering theory).

Finally, there is no no-go theorem that would say that the requirements are too strong.

Can we renounce covariance or microscopic causality?
Of course, one can renounce each of Wightman's axiom, but at a high price.

Renouncing the first drops the connection to relativity theory. It would be very difficult to find such a theory whose classical limit reduces in the standard situations to special relativistic mechanics.

Renouncing the second makes the first trivial to satisfy, as I have shown. The remaining axioms are far too little constraining to allow one to draw useful conclusions.

For example, if I am not wrong your fields A(x):=U(x)A_0U(-x) cannot be causal. In fact A(x) is a well defined operator for every x, and it is well known that a covariant causal field cannot be a well defined operator at every point x (if we also assume the spectral condition for the energy-momentum operators and the translation invariance of the vacuum).
The scalar free field Phi(x) is well-defined as a densely defined quadratic form, which is enough for my expression to make sense. (The formal Hamiltonians that Weinberg discusses have no better properties.) If we put A_0:=Phi(0), then A(x)=U(x)A_0U(-x)=Phi(x) satisfies the Wightman axioms

Of course, this is not enough for a mathematically rigorous proof.

But free fields indeed satisfy all Wightman axioms rigorously and satisfy
A(x)=U(x)A_0U(-x).

Eventually, it is well known that Wightman axioms are very difficult to satisfy, and actually impossible in gauge field theories (http://arxiv.org/abs/hep-th/0401143).
You over-interpret the paper. There are 2-dimensional gauge theories (e.g., the Schwinger model) satisfying the Wightman axioms. In 4D, there is not a single theorem against the existence of interacting Wightman fields; it is just that we currently lack the mathematical tools to decide either way.

Nothing excludes gauge fields since there is no agreed-upon way how to formulate the requirement of gauge invariance in the Wightman setting. If one formulation can be proved to lead to nonexistence, it only rules out this formulation as good.

#### meopemuk

Interacting fields are introduced only in Chapter 7. Section 7.1 discusses the standard Hamiltonian approach in the instant form and the Schroedinger picture, and introduces in (7.1.28/29) the interacting field operators in the Heisenberg picture. Since in the instant form, space translations are implemented kinematically, these equations imply that
(1) .... A(x) = U(x) A_0 U(-x)
for all Operators A_0=F(Q,P), where - unlike in (3.5.12) - the translations U(x) are the physical (interacting) ones.

...

Finally, from (1) and the fact that the translations are part of an (interacting) unitary representation of the Poincare group, it is not difficult to show that one also gets the relations
(2) .... A(Lambda x) = U(Lambda) A(x) U(Lambda^{-1}),
which prove that the interacting quantum fields are Poincare-covariant with respect to the interacting representation of the Poincare group.
This seems to be your best attempt at the proof so far. I agree with your point (1). Could you please elaborate on your phrase "...it is not difficult to show..."? Your formula (2) doesn't look obvious to me.

Eugene.

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#### bg032

Because, as Weinberg showed, these properties are expected to hold for all QFTs that can be constructed from local actions - and this includes all QFTs in current use! Wightman's axioms are not arbitrarily chosen but are the ones that one can derive (on the level of rigor of theoretical physics) from the assumptions that particle physicists rely on all the time, plus the assumption of a mass gap. (The infrared behavior of a massless theory seems to require appropriate modifications of Wightman's axioms.)

Moreover, in 2D and 3D, this expectation is actually rigorously verifiable in many cases.

Third, Weinberg also showed (in Chapters 3 and 4) that these properties seem necessary in order that a theory has a covariant S-matrix and satisfies the cluster decomposition property.

Fourth, given the Wightman axioms, one can deduce a lot of sensible physical properties (e.g., a well-defined scattering theory).

Finally, there is no no-go theorem that would say that the requirements are too strong.
For me the problem here is that we oscillate between the rigour of Wightman axioms and what you call the "rigor of theoretical physics". I had no problem if the rigour of Weinberg were the same of Wightman. On the other hand, Weinberg makes true physics and obtain empirical predictions, while Whigtman is very abstract. My hope is that one day we will obtain the results of Weinberg with the rigour of Wightman.

Renouncing the first [covariance] drops the connection to relativity theory.
With reference to my two points (1: Hilbert space with a representation of Poincaré group and 2: causal covariant fields), I remark that a connection with relativity is already present in point 1. Points 1 and 2 are two different and independent connections with relativity.

It would be very difficult to find such a theory whose classical limit reduces in the standard situations to special relativistic mechanics.
Mhhh. I think in his book Eugene has given an example, by developing a theory empirically equivalent to QED but not based on covariant causal fields.

You over-interpret the paper.
Quote from http://arxiv.org/abs/hep-th/0401143: [Broken]

"In conclusion, quite generally one can prove that in the quantization of gauge field theories the (correlation functions of the) charged fields cannot satisfy all the quantum mechanical constraints QM1, QM2 and the relativity constraint R1, R2, since locality and positivity are crucially in conflict. Therefore, the general framework discussed in Sect.3.3 has to be modified (modified Wightman axioms)"

Maybe the paper is wrong, but certainly I do not over-interpret it.

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#### A. Neumaier

For me the problem here is that we oscillate between the rigour of Wightman axioms and what you call the "rigor of theoretical physics". I had no problem if the rigour of Weinberg were the same of Wightman. On the other hand, Weinberg makes true physics and obtain empirical predictions, while Wiggtman is very abstract. My hope is that one day we will obtain the results of Weinberg with the rigour of Wightman.
Since QED and QCD so far have no mathematical definition, there is no choice than to oscillate. Of course, the purpose of the axioms is to characterize empirically relevant theories. Mathematical physicists believe that at least QCD (and other asymptotically free QFTs) admits a rigorous description satisfying the Wightman axioms, though it hasn't been found yet. Opinions on QED are divided.

With reference to my two points (1: Hilbert space with a representation of Poincaré group and 2: causal covariant fields), I remark that a connection with relativity is already present in point 1. Points 1 and 2 are two different and independent connections with relativity.
But this connection is far too weak to conclude a covariant scattering theory and the cluster decomposition principle. The fact that Weinberg's semirigorous derivation ''proves'' the Wightman axioms means that they are needed to characterize local QFTs if they have a mass gap.

Mhhh. I think in his book Eugene has given an example, by developing a theory empirically equivalent to QED but not based on covariant causal fields.
But:

(i) the equivalence is bought by referring to Weinberg's analysis in Chapters 3 and 4, since his construction starts with the field theory and produces a unitarily equivalent theory, which means - though he tries to disown this fact - that the fields Weinberg has are of course also there in his theory (by applying to the Weinberg fields his unitary transform).

(ii) his construction is perturbative only, and hence doesn't yet make sense on a rigorous level. I'd count it as an effective theory on the level of NRQED, but much more awkward to use. In any case, it doesn't count in the present context, where full rigor is the goal.

(iii) he hasn't even been able to match the basic tests of QED, the anomalous magnetic moment of the electron and the Lamb shift.

(iii) he completely ignores the infrared problem.

Quote from http://arxiv.org/abs/hep-th/0401143: [Broken]

"In conclusion, quite generally one can prove that in the quantization of gauge field theories the (correlation functions of the) charged fields cannot satisfy all the quantum mechanical constraints QM1, QM2 and the relativity constraint R1, R2, since locality and positivity are crucially in conflict. Therefore, the general framework discussed in Sect.3.3 has to be modified (modified Wightman axioms)"

Maybe the paper is wrong, but certainly I do not over-interpret it.
I'd have said that you over-interpret the evidence given in the paper.

If one reads section 4 and look at what precedes this statement on p.23, one finds that the author doesn't give a proof. Reference (28) lists two sources,both by the author himself (already not a good sign), and seems to contain the evidence. The Phys Rev paper starts off with ''... standard QFT can be formulated in terms of fields satisfying all the standard axioms (positivity included)'', hence shows that he doesn't work on the rigorous level - since none of the standard QFTs in 4D has been shown rigorously to satisfy these axioms.
I don't have access to the book, but don't expect a higher level of rigor there.

Since nobody understands the IR problem for nonabelian gauge theories, let alone is able to prove anything about them rigorously 9in a positive or negative direction), his arguments are nothing more than plausibility considerations. And his conclusions are not shared by many. (Vienna, where I live, is the host of the Erwin Schroedinger Institute for Mathematical Physics; so I am informed first hand....)

There is even a 1 Million Dollar price for showing that 4D Yang Mills theory (the simplest nonabelian gauge theory) exists in the Wightman sense!

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#### A. Neumaier

This seems to be your best attempt at the proof so far. I agree with your point (1). Could you please elaborate on your phrase "...it is not difficult to show..."? Your formula (2) doesn't look obvious to me.
I didn't claim it is obvious but that it is easy to prove. Expand both sides using the definition (1) and simplify the result using the rules of the group representation and the fact that A_0 is Lorentz invariant. (This is the case for Weinberg's examples; I forgot to assume this as a general condition on A_0.)

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#### meopemuk

I didn't claim it is obvious but that it is easy to prove. Expand both sides using the definition (1) and simplify the result using the rules of the group representation and the fact that A_0 is Lorentz invariant. (This is the case for Weinberg's examples; I forgot to assume this as a general condition on A_0.)
I guess, by "Lorentz invariant" you mean "commutes with the interacting boost operator"? I don't think this condition is true even for the simplest scalar field. Anyway, since this is a very important result (one of Wightman's axioms which allegedly forms the basis for entire rigorous QFT) I would appreciate if you make your proof as detailed as possible. I expect to see some subtle points there, which cannot be resolved by simple handwaving.

Eugene.

#### A. Neumaier

I guess, by "Lorentz invariant" you mean "commutes with the interacting boost operator"? I don't think this condition is true even for the simplest scalar field. Anyway, since this is a very important result (one of Wightman's axioms which allegedly forms the basis for entire rigorous QFT) I would appreciate if you make your proof as detailed as possible.
I am not willing to spoon-feed you. Please pay for my effort with your effort. Let us proceed at least according to the rules for homework help in this forum.

Thus please present your evidence that, for a free scalar field (which is the simplest), the operator
$$A_0:=\Phi(x)|x=0$$
is not Lorentz invariant, i.e., does not commute with the free boost operators, by giving a supporting calculation.

I expect to see some subtle points there, which cannot be resolved by simple handwaving.
The only subtle point (that Weinberg consistently ignores and that is the only obstacle for making everything rigorous in a simple way) is that all manipulations are formal rather than rigorously justified. But all the manipulations in your book are of this kind, too, so that you shouldn't demand here more rigor. That would be unreasonable to expect, since a rigorous interactive QFT in 4D is worth a million of dollars.

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#### meopemuk

I am not willing to spoon-feed you. Please pay for my effort with your effort. Let us proceed at least according to the rules for homework help in this forum.

Thus please present your evidence that, for a free scalar field (which is the simplest), the operator
$$A_0:=\Phi(x)|x=0$$
is not Lorentz invariant, i.e., does not commute with the free boost operators, by giving a supporting calculation.

Fair enough. I do agree that $$\Phi(0)$$ commutes with the *free* boost operators $$\mathbf{K}_0$$. Here is my proof:

First, I define the free field by usual formula (e.g. (5.2.11) in Weinberg)

$$\Phi(\mathbf{r}, t) = \int \frac{d \mathbf{p}}{\sqrt{\omega_{\mathbf{p}}}}\left(a(\mathbf{p}) e^{i \mathbf{pr} - i \omega_{\mathbf{p}}t} + a^{\dag}(\mathbf{p}) e^{-i \mathbf{pr} + i \omega_{\mathbf{p}}t} \right)$$

Then I am going to show that the field in the origin $$(\mathbf{r}=0, t=0)$$ commutes with the free boost operator $$\mathbf{K}_0$$. Actually, it is sufficient to provide the proof for the negative frequency part of the field only

$$\Phi^-(0, 0) = \int \frac{d \mathbf{p}}{\sqrt{\omega_{\mathbf{p}}}}a(\mathbf{p})$$

I choose some non-trivial Lorentz transformation $$\Lambda$$, which is represented in the Hilbert space by the unitary operator $$U_0 (\Lambda) = \exp(i \mathbf{K}_0 \vec{\theta})$$. Then I use the transformation law for the annihilation operator $$a(\mathbf{p})$$ as in Weinberg's (5.1.11)

$$\exp(i \mathbf{K}_0 \vec{\theta}) a(\mathbf{p}) \exp(-i \mathbf{K}_0 \vec{\theta}) = \sqrt{\frac{\omega_{\Lambda \mathbf{p}}}{\omega_{\mathbf{p}}}}a(\Lambda \mathbf{p})$$

So, the proof goes like this

$$U_0 (\Lambda) \Phi^-(0, 0) U_0^{-1} (\Lambda)= \int \frac{d \mathbf{p}}{\sqrt{\omega_{\mathbf{p}}}}\exp(i \mathbf{K}_0 \vec{\theta}) a(\mathbf{p}) \exp(-i \mathbf{K}_0 \vec{\theta})$$

$$= \int \frac{d \mathbf{p}}{\sqrt{\omega_{\mathbf{p}}}} \sqrt{\frac{\omega_{\Lambda \mathbf{p}}}{\omega_{\mathbf{p}}}} a(\Lambda \mathbf{p})$$

$$= \int \frac{d (\Lambda \mathbf{p})}{\omega_{\Lambda \mathbf{p}}} \sqrt{\omega_{\Lambda \mathbf{p}}} a(\Lambda \mathbf{p} )$$

$$= \int \frac{d \mathbf{p}}{\sqrt{\omega_{ \mathbf{p}}}} a( \mathbf{p}) = \Phi^-(0, 0)$$

It then follows that

$$[\mathbf{K}_0, \Phi^-(0, 0)] = 0$$

Similarly, we can prove

$$[\mathbf{K}_0, \Phi^+(0, 0)] = [\mathbf{K}_0, \Phi(0, 0)] = 0$$

This much I understand. However, how are we going to calculate the commutator with the *interacting* boost operator? If I understand correctly, your claim is that

$$[\mathbf{K}, \Phi(0, 0)] = [\mathbf{K}_0 + \mathbf{W}, \Phi(0, 0)] = 0$$

regardless of the interacting part $$\mathbf{W}$$. Can you prove that? Now it is your turn.

Eugene.

#### A. Neumaier

Fair enough. I do agree that $$\Phi(0)$$ commutes with the *free* boost operators $$\mathbf{K}_0$$.

However, how are we going to calculate the commutator with the *interacting* boost operator? If I understand correctly, your claim is that

$$[\mathbf{K}, \Phi(0, 0)] = [\mathbf{K}_0 + \mathbf{W}, \Phi(0, 0)] = 0$$

regardless of the interacting part $$\mathbf{W}$$.
Not regardless of the interacting part, but only if the interacting part is constructed canonically from a local action without derivative interaction. This covers Phi^4 theory and QED.

In view of what you proved already, it is enough to show that W commutes with Phi(0,0). Now W is defined in (3.5.17), and H(x,0) for a real scalar field is, according to (7.1.35), a function of Q=Phi(x,0), namely the normally ordered product g:Phi(x,0)^4: . Thus it suffices to verify that [Phi(x,0),Phi(0,0)]=0, which is straightforward given your formula for Phi(r,t).

The same kind of arguments also work for QED, except that one needs to be slightly more careful for the fermion fields.

#### meopemuk

Hi Arnold,

Thanks, I see your point. In the $$\Phi^4(x)$$ theory $$\mathbf{W} = \int d \mathbf{r} \mathbf{r} \Phi^4(\mathbf{r}, 0)$$

so

$$[\mathbf{W}, \Phi(0,0)] = \int d \mathbf{r} \mathbf{r} [\Phi^4(\mathbf{r}, 0), \Phi(0,0)] = 0$$

due to zero free field commutators at spacelike separations, in particular

$$[\Phi (\mathbf{r}, t), \Phi(0,0)] = 0$$.....................(1)

if $$(\mathbf{r}, t)$$ is a spacelike 4-vector.

Then the covariant transformation law for the interacting field $$\Phi_i(x)$$ can be proven as follows

$$\Phi_i(0,0,z, t) = e^{iP_{0z} z} e^{-i Ht} \Phi(0,0,0,0) e^{i Ht} e^{-iP_{0z} z}$$

and

$$e^{i K_z \theta} \Phi_i(x) e^{ -i K_z \theta }= e^{i K_z \theta }e^{i P_{0z} z} e^{-i Ht} \Phi(0,0) e^{i Ht} e^{-i P_{0z}z } e^{ -i K_z \theta }$$

$$= e^{i K_z \theta } e^{i P_{0z} z} e^{-i Ht} e^{-i K_z \theta } e^{i K_z \theta }\Phi(0,0) e^{-i K_z \theta } e^{i K_z \theta }e^{i Ht} e^{-i P_{0z}z } e^{ -i K_z \theta }$$

$$= e^{i K_z \theta } e^{i P_{0z} z} e^{-i Ht} e^{-i K_z \theta } \Phi(0,0) e^{i K_z \theta }e^{i Ht} e^{-i P_{0z}z } e^{ -i K_z \theta }$$

$$= e^{i P_{0z} (z \cosh \theta - ct \sinh \theta) } e^{-i H (t cosh \theta - (z/c) \sinh \theta)} \Phi(0,0) e^{i H (t cosh \theta - (z/c) \sinh \theta)} e^{-i P_{0z}(z \cosh \theta - ct \sinh \theta) }$$

$$= \Phi_i(0,0, z \cosh \theta - ct \sinh \theta, t cosh \theta - (z/c) \sinh \theta) = \Phi_i (\Lambda x)$$

But there is another piece relevant to Haag's theorem. This is the assumption about canonical commutators of interacting fields. In particular, this condition claims that in analogy with (1) we should have

$$0 = [\Phi_i (\mathbf{r}, t), \Phi(0,0) ] = [e^{iP_{0z} z} e^{-i Ht} \Phi(0,0,0,0) e^{i Ht} e^{-iP_{0z} z}, \Phi(0,0, 0, 0) ]$$

Can we prove this condition in the $$\Phi^4(x)$$ theory?

Eugene.

#### A. Neumaier

Excellent. So you now have a proof that interacting fields transform covariantly under the interacting Poincare representation.

But there is another piece relevant to Haag's theorem. This is the assumption about canonical commutators of interacting fields. In particular, this condition claims that in analogy with (1) we should have

$$0 = [\Phi_i (\mathbf{r}, t), \Phi(0,0) ] = [e^{iP_{0z} z} e^{-i Ht} \Phi(0,0,0,0) e^{i Ht} e^{-iP_{0z} z}, \Phi(0,0, 0, 0) ]$$

Can we prove this condition in the $$\Phi^4(x)$$ theory?
Of course. We do not even need to do an explicit calculation.

There is a Lorentz transformation that moves (r,t) to (r',0) for some r', because of the space-like assumption. Since Lorentz transformation fix (0,0), the statement you want follows from the transformation properties of the fields that you had proved already.

Thus the Wightman axioms relating to relativity and causality are verified on the formal level.

The existence of the vacuum and the spectral boundedness assumptions are necessary in order to be able to interpret the theory (excluding for example the covariant field theory of Horwitz and Piron), but cannot be proved that easily, not even on the formal level, since these properties emerge only in the renormalized limit. So a lot of technicalities would need to be considered, which is beyond what I am prepared to discuss in the forum.

#### meopemuk

There is a Lorentz transformation that moves (r,t) to (r',0) for some r', because of the space-like assumption. Since Lorentz transformation fix (0,0), the statement you want follows from the transformation properties of the fields that you had proved already.

Thus the Wightman axioms relating to relativity and causality are verified on the formal level.

Arnold, let me see if I got your hints right. I can always find a boost parameter $$\theta$$, such that

$$[\Phi_i (\mathbf{r}, t), \Phi(0,0) ] = [e^{i\mathbf{K} \theta} \Phi_i (\mathbf{r}', 0) e^{-i\mathbf{K} \theta}, \Phi(0,0) ] = e^{i\mathbf{K} \theta} [\Phi (\mathbf{r}', 0) , \Phi(0,0) ] e^{-i\mathbf{K} \theta} = 0$$

I need some more time to think this through. But it looks like I've been proven wrong. Thank you for your patience.

Eugene.

#### A. Neumaier

Arnold, let me see if I got your hints right. I can always find a boost parameter $$\theta$$, such that

$$[\Phi_i (\mathbf{r}, t), \Phi(0,0) ] = [e^{i\mathbf{K} \theta} \Phi_i (\mathbf{r}', 0) e^{-i\mathbf{K} \theta}, \Phi(0,0) ] = e^{i\mathbf{K} \theta} [\Phi (\mathbf{r}', 0) , \Phi(0,0) ] e^{-i\mathbf{K} \theta} = 0$$
Yes. By Weinberg's table on p.66, the restricted Lorentz groups has six orbits on R^4:
the open future cone,
the open past cone,
the future light cone,
the past light cone,
the complement of the closed, 2-sided causal cone,
and the zero point.
In particular, since (r,t) is outside the causal cone, it can be mapped by a Lorentz transformation to any other point outside the causal cone, and hence to a point of the form (r',0).

#### A. Neumaier

Quote from http://arxiv.org/abs/hep-th/0401143: [Broken]

"In conclusion, quite generally one can prove that in the quantization of gauge field theories the (correlation functions of the) charged fields cannot satisfy all the quantum mechanical constraints QM1, QM2 and the relativity constraint R1, R2, since locality and positivity are crucially in conflict. Therefore, the general framework discussed in Sect.3.3 has to be modified (modified Wightman axioms)"

Maybe the paper is wrong, but certainly I do not over-interpret it.
I continue the discussion of these matters in another thread:

Strocchi [...] puts them into the framework of axiomatic field theory (where the completely different notation and terminology makes things look very different). This results in theorem that precisely specify the assumptions that go into the conclusions.

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#### meopemuk

Arnold, let me see if I got your hints right. I can always find a boost parameter $$\theta$$, such that

$$[\Phi_i (\mathbf{r}, t), \Phi(0,0) ] = [e^{i\mathbf{K} \theta} \Phi_i (\mathbf{r}', 0) e^{-i\mathbf{K} \theta}, \Phi(0,0) ] = e^{i\mathbf{K} \theta} [\Phi (\mathbf{r}', 0) , \Phi(0,0) ] e^{-i\mathbf{K} \theta} = 0$$

I need some more time to think this through. But it looks like I've been proven wrong. Thank you for your patience.

Eugene.
Arnold, I would like to return to this discussion, if you don't mind. I think we came close to understanding Haag theorem on this example, but haven't completed the job. In my understanding we've achieved the following: We took a $$\Phi^4$$ theory with the full Hamiltonian $$H = H_0 + V$$, where

$$V = \int d \mathbf{r} \Phi^4 (\mathbf{r}, 0)$$

and constructed interacting field

$$\Phi (\mathbf{r}, t) = e^{iHt} \Phi (\mathbf{r}, 0) e^{-iHt}$$

We have shown that this field (1) commutes with itself at space-like separations, (2) transforms covariantly under interacting boost operator. So far I don't see any problem. Could you please indicate what's wrong in this picture and why we need to abandon the Fock space here? I have my ideas about that, but I would like to know your opinion first.

Thanks.
Eugene.

#### A. Neumaier

Arnold, I would like to return to this discussion, if you don't mind. I think we came close to understanding Haag theorem on this example, but haven't completed the job. In my understanding we've achieved the following: We took a $$\Phi^4$$ theory with the full Hamiltonian $$H = H_0 + V$$, where

$$V = \int d \mathbf{r} \Phi^4 (\mathbf{r}, 0)$$

and constructed interacting field

$$\Phi (\mathbf{r}, t) = e^{iHt} \Phi (\mathbf{r}, 0) e^{-iHt}$$

We have shown that this field (1) commutes with itself at space-like separations, (2) transforms covariantly under interacting boost operator. So far I don't see any problem. Could you please indicate what's wrong in this picture and why we need to abandon the Fock space here?
The reasoning was on Weinberg's level of rigor - i.e., pretending that Phi(r) is densely defined operator of Fock space, and that a formally Hermitian Hamiltonian H is automatically self-adjoint, so that e^{itH} makes sense, and that therefore no renormalization is needed. It is a pretense since, in fact, the bare Phi and H = H_0+V don't have these properties.

Renormalization removes this pretense (in <4D rigorously, in 4D only in a semi-rigorous, perturbative fashion) and turns H into a self-adjoint operator in spite of the fact that fields are only operator-valued distributions, by changing both the Hilbert space and the way the limit that removes the cutoff is taken.

Haag's theorem is a fully rigorous mathematical theorem, without any pretense, and tells that if renormalization results in a covariant theory with a physical energy spectrum then the representation of the equal-time CCR cannot be a Fock representation.

#### A. Neumaier

we came close to understanding Haag theorem on this example, but haven't completed the job. In my understanding we've achieved the following: We took a $$\Phi^4$$ theory with the full Hamiltonian $$H = H_0 + V$$, where

$$V = \int d \mathbf{r} \Phi^4 (\mathbf{r}, 0)$$

and constructed interacting field

$$\Phi (\mathbf{r}, t) = e^{iHt} \Phi (\mathbf{r}, 0) e^{-iHt}$$

We have shown that this field (1) commutes with itself at space-like separations, (2) transforms covariantly under interacting boost operator. So far I don't see any problem. Could you please indicate what's wrong in this picture and why we need to abandon the Fock space here?
The reasoning was on Weinberg's level of rigor - i.e., pretending that Phi(r) is densely defined operator of Fock space, and that a formally Hermitian Hamiltonian H is automatically self-adjoint, so that e^{itH} makes sense, and that therefore no renormalization is needed. It is a pretense since, in fact, the bare Phi and H = H_0+V don't have these properties.

Renormalization removes this pretense (in <4D rigorously, in 4D only in a semi-rigorous, perturbative fashion) and turns H into a self-adjoint operator in spite of the fact that fields are only operator-valued distributions, by changing both the Hilbert space and the way the limit that removes the cutoff is taken.

Haag's theorem is a fully rigorous mathematical theorem, without any pretense, and tells that if renormalization results in a covariant theory with a physical energy spectrum then the representation of the equal-time CCR cannot be a Fock representation.

#### meopemuk

The reasoning was on Weinberg's level of rigor - i.e., pretending that Phi(r) is densely defined operator of Fock space, and that a formally Hermitian Hamiltonian H is automatically self-adjoint, so that e^{itH} makes sense, and that therefore no renormalization is needed. It is a pretense since, in fact, the bare Phi and H = H_0+V don't have these properties.

Renormalization removes this pretense (in <4D rigorously, in 4D only in a semi-rigorous, perturbative fashion) and turns H into a self-adjoint operator in spite of the fact that fields are only operator-valued distributions, by changing both the Hilbert space and the way the limit that removes the cutoff is taken.

Haag's theorem is a fully rigorous mathematical theorem, without any pretense, and tells that if renormalization results in a covariant theory with a physical energy spectrum then the representation of the equal-time CCR cannot be a Fock representation.

In my opinion, the alleged problem is less subtle than the difference between Hermitian and self-adjoint operators. I think that the claim is that the vacuum of the interacting $$\Phi^4$$ theory is othrogonal to any non-interacting Fock state. This statement becomes plausible if we understand that the non-interacting Fock vacuum cannot be an eigenstate of the interacting Hamiltonian. The field and interaction operator can be formally written in a normally-ordered form as

$$\Phi = a^{\dag} + a$$
$$V = \Phi^4 = (a^{\dag} + a)^4 = a^{\dag}a^{\dag}a^{\dag}a^{\dag} + a^{\dag}a^{\dag}a^{\dag}a + a^{\dag}a^{\dag}aa + a^{\dag}aaa + \ldots$$

The first term in this expansion acts non-trivially on the vacuum, so the action of the full Hamiltonian $$H = H_0 + V$$ is non-trivial as well. This indicates that interacting vacuum is different from the free vacuum.

Let me know if you agree with this line of reasoning before I continue.

Eugene.

#### A. Neumaier

I think that the claim is that the vacuum of the interacting $$\Phi^4$$ theory is orthogonal to any non-interacting Fock state. This statement becomes plausible if we understand that the non-interacting Fock vacuum cannot be an eigenstate of the interacting Hamiltonian.
The latter is obvious and has nothing to do with Haag's theorem.

#### meopemuk

Arnold, Let me try a different argument. When we discussed covariance and commutativity of interacting fields, you've noticed correctly that our proof works only for interactions constructed as products of fields. Here is your quote:

Not regardless of the interacting part, but only if the interacting part is constructed canonically from a local action without derivative interaction. This covers Phi^4 theory and QED.
Suppose now that we constructed a relativistic interacting theory in which the interacting Hamiltonian is *not* a product of fields. For example, it can be $$V= a^{\dag}a^{\dag}aa$$. Then the above proofs will not be valid. Interacting fields will not be covariant and they will not commute at space-like separations. Then two important conditions of Haag's theorem will not be satisfied, and we will not be able to prove that the Fock space is excluded.

As a result of this exercise we will obtain a non-trivial interacting theory in the Fock space. "Dressed particle" theories are exactly of this form. Their only problem is that interacting fields are non-covariant and non-commuting. Could you please explain why you think that this is an important problem? Is there any measurable property that proves the impossibility of non-covariant and non-commuting interacting fields?

Eugene.