What Are the States in Quantum Field Theory?

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In summary: NON-COMPACT nature of the Poincare' group. All FINITE-DIMENSIONAL irreducible representations of any non-compact group are NOT UNITARY. Therefore these IR representations can not be carried by functions on Hilbert space, but by functions on spacetime where non-unitarty does not cause any problem.
  • #141
bg032 said:
Quote from http://arxiv.org/abs/hep-th/0401143:

"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:

A. Neumaier said:
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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  • #142
meopemuk said:
Arnold, let me see if I got your hints right. I can always find a boost parameter [tex] \theta [/tex], such that

[tex]
[\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 [/tex]

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 [tex] \Phi^4 [/tex] theory with the full Hamiltonian [tex] H = H_0 + V [/tex], where

[tex] V = \int d \mathbf{r} \Phi^4 (\mathbf{r}, 0) [/tex]

and constructed interacting field

[tex] \Phi (\mathbf{r}, t) = e^{iHt} \Phi (\mathbf{r}, 0) e^{-iHt} [/tex]

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.
 
  • #143
meopemuk said:
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 [tex] \Phi^4 [/tex] theory with the full Hamiltonian [tex] H = H_0 + V [/tex], where

[tex] V = \int d \mathbf{r} \Phi^4 (\mathbf{r}, 0) [/tex]

and constructed interacting field

[tex] \Phi (\mathbf{r}, t) = e^{iHt} \Phi (\mathbf{r}, 0) e^{-iHt} [/tex]

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.
 
  • #144
meopemuk said:
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 [tex] \Phi^4 [/tex] theory with the full Hamiltonian [tex] H = H_0 + V [/tex], where

[tex] V = \int d \mathbf{r} \Phi^4 (\mathbf{r}, 0) [/tex]

and constructed interacting field

[tex] \Phi (\mathbf{r}, t) = e^{iHt} \Phi (\mathbf{r}, 0) e^{-iHt} [/tex]

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.
 
  • #145
A. Neumaier said:
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 [tex] \Phi^4 [/tex] 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

[tex] \Phi = a^{\dag} + a [/tex]
[tex] 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 [/tex]

The first term in this expansion acts non-trivially on the vacuum, so the action of the full Hamiltonian [tex] H = H_0 + V [/tex] 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.
 
  • #146
meopemuk said:
I think that the claim is that the vacuum of the interacting [tex] \Phi^4 [/tex] 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.
 
  • #147
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:


A. Neumaier said:
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 [tex]V= a^{\dag}a^{\dag}aa [/tex]. 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.
 

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