# What is observable in a relativistic quantum field theory?

Steven Weinberg wrote: ''In its mature form, the idea of quantum field theory is that quantum fields are the basic ingredients of the universe, and particles are just bundles of energy and momentum of the fields.'' (see p.2 of his essay, ''What is Quantum Field Theory, and What Did We Think It Is?'' http://arxiv.org/pdf/hep-th/9702027v1)

This thread is intended to discuss questions related to the above quote. Such questions came up in other threads ( https://www.physicsforums.com/showthread.php?t=471125 and https://www.physicsforums.com/showthread.php?t=474537 ), but to focus the discussion, I respond to them here.

More precisely, the question to be discussed is; Which fields (or other observables) represent the operational content of relativistic quantum field theory?

It will take some time till I have prepared the proper background for a full discussion.
For the moment, I just begin by commenting on the following, which assumes the QED electron field Psi(x,t):
we use this operator to construct another abstract object called "current density 4-vector"
$$J^{\mu} (x,t)= \overline{\Psi} (x,t) \gamma^{\mu} \Psi(x,t)$$
This is not an ''abstract object'' but a measurable observable, which persists in the macroscopic (classical) limit and is there identical with the current figuring in the Maxwell equations, describing the current density at time t and position x in ordinary space-time.
Moreover, parameters x are just integration variables. There is absolutely no reason to identify them with physical positions. Parameter t is set to 0, so it has not direct relevance to measured time.
I don't know what you are talking about. The parameter x in the standard Maxwell equations in flat space-time has the standard meaning of the physical position. Quantization doesn't change this.

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This is not an ''abstract object'' but a measurable observable, which persists in the macroscopic (classical) limit and is there identical with the current figuring in the Maxwell equations, describing the current density at time t and position x in ordinary space-time.

I don't know what you are talking about. The parameter x in the standard Maxwell equations in flat space-time has the standard meaning of the physical position. Quantization doesn't change this.

My post that you've quoted has begun with the statement that once we know the full interacting Hamiltonian H = H_0 + V in the Fock space, we are fully equipped to do all kinds of calculations and to obtain all physical information about the system. For each state in the Fock space we can determine its particle content and how the particle observables change with time. We don't need any extra input, like Maxwell equations.

Recall again the logic outlined in Weinberg's book. His goal was to construct the interaction operator V that satisfies the cluster separability and relativistic invariance. He found that this can be achieved if he writes

$$V = \int dx V(x,0)$$

where $$V(x,0)$$ is written as a product of certain operator functions $$\Psi(x,0)$$ and $$A^{\mu} (x,0)$$. Weinberg does not claim that these operator functions have any physical relevance or that they can be directly or indirectly observed. They are just convenient mathematical objects that happen to help to achieve our goal of building V.

It is true that operator functions $$\Psi(x,t)$$ and $$A^{\mu} (x,t)$$ satisfy certain differential equations. It is also true that instead of the above *free* fields we can construct *interacting* fields, which satisfy other differential equations, which happen to resemble Maxwell equations from the middle of 19th century. But, in my opinion, this is just a coincidence and not a sufficient ground to claim that $$\Psi(x,t)$$ and $$A^{\mu} (x,t)$$ have any relevance to the behavior of interacting electrons and photons. Everything we need in order to calculate such a behavior is already contained in the Hamiltonian H = H_0 + V, and any reference to the fields $$\Psi(x,t)$$ and $$A^{\mu} (x,t)$$ is certainly redundant and, quite possibly, wrong.

Eugene.

My post that you've quoted has begun with the statement that once we know the full interacting Hamiltonian H = H_0 + V in the Fock space, we are fully equipped to do all kinds of calculations and to obtain all physical information about the system. For each state in the Fock space we can determine its particle content and how the particle observables change with time. We don't need any extra input, like Maxwell equations.

We are talking past each other. In the process of constructing your Hamiltonian you mentioned the current operator and claimed that it was merely an abstract object (whereas, supposedly, the Hamiltonian you construct is the ''real'' thing). I was responding that (not only the Hamiltonian) but the current operator is an observable with a direct physical meaning. Hence it is no more abstract than your Hamiltonian.

Weinberg does not claim that these operator functions have any physical relevance or that they can be directly or indirectly observed. They are just convenient mathematical objects that happen to help to achieve our goal
By the same token, the Hamiltonian is also nothing else than a convenient mathematical object that happens to help to achieve our goal of constructing QED.

One needs more definite criteria to demarcate what is ''physical'' from what is ''mere abstract''.
It seems we need first to agree upon these criteria before we can have a fruitful discussion.

we can construct *interacting* fields, which satisfy other differential equations, which happen to resemble Maxwell equations from the middle of 19th century. But, in my opinion, this is just a coincidence
Whereas everyone else regards this as one of the major reasons to consider QED at all.
Just look at the history of the subject!

It doesn't make sense to state mere opinions. To have a fruitful discussion, opinions must be grounded on common, accepted criteria.

In any case, one must be able to derive the macroscopic Maxwell equations in flat space-time from QED and recover the traditional form, which is amply verified by current technology. In the latter, the parameter x has the standard meaning of the physical position. I wonder how this meaning could possibly materialize when it starts off as being a meaningless integration parameter.

So, if you want to have another simple challenge, please derive the macroscopic Maxwell equations in vacuum (certainly something observable) from your particle picture.
I think I can defend the position that experimentalists see only accelerations of test particles (massive and/or charged), so gravitational and electromagnetic fields by themselves are non-observable and redundant. This idea has been developed in Chapters 12 and 13 of the book.
Unfortunately, my expectation to find there a derivation of the Maxwell equations was disappointed. You write a book about QED, and not even recover its macroscopic consequences in the appropriate approximation. A quantum theory that is not able to serve as a foundation for classical electrodynamics as used by every engineer is not worth being called QED.

If you'd try to do it, you'd find that the meaning of x and t in the Maxwell equations and in the quantum current are identical.

Here is a derivation of semiconductor dynamics from QED:
Holger F. Hofmann and Ortwin Hess
Quantum Maxwell-Bloch equations for spatially inhomogeneous semiconductor lasers
Phys. Rev. A 59, 2342–2358 (1999)
http://arxiv.org/pdf/physics/9807011
There is no trace of your idea that x (there called r) is only a formal integration parameter.
It everywhere figures as the position. (One also sees that QED is a dynamical theory, and not only provides scattering information.)

In any case, one must be able to derive the macroscopic Maxwell equations in flat space-time from QED and recover the traditional form, which is amply verified by current technology.

This is a fair point, and I must address it in order to be taken seriously. Here is my explanation:

As I said, the full information about interacting particle dynamics can be obtained from the QED Hamiltonian. However, the traditional form of this Hamiltonian is not suitable, because in addition to correct particle-particle interactions, this Hamiltonian also includes spurious self-interactions of bare particles, which are responsible for divergences among other bad things. This problem can be fixed by changing to the so-called "dressed particle" representation, in which the self-interactions are cancelled. Note that this cancellation is done by means of a unitary transformation, so it does not affect scattering properties of the Hamiltonian, so all usual predictions of QED (scattering amplitudes and energies of bound states) remain unchanged. The resulting charge-charge interaction potential in the 2nd perturbation order can be found in

E. V. Stefanovich "Quantum field theory without infinities", Ann. Phys. 292 (2001), 139; see first equation on page 153.

This potential can be further simplified by taking the (v/c)^2 approximation. It is not difficult to show that in the position representation this leads to the well-known Darwin-Breit potential (see section 83 in Berestetskii, Livgarbagez, Pitaevskii, "Quantum electrodynamics"). If we further ignore spins, then we obtain the Darwin potential. See, e.g., (12.82) in Jackson's book on electrodynamics. It is also well-established that the Darwin potential is sufficient to describe all electromagnetic effect (with the exception of radiation, which requires a higher 3rd order of the perturbation theory; this order can be dealt with in a similar manner, but is more challenging mathematically). There are many examples of that in Chapter 12 of my book.

Although the Darwin potential is rather accurate, it is fundamentally different from Maxwell's theory in many aspects. First, the Darwin interaction between charges is instantaneous instead of Maxwellian (Lienard-Wiechert) retarded potentials. Second, the Darwin's approach does not involve the idea of momentum and energy contained in electromagnetic fields. So, it doesn't suffer from usual paradoxes associated with the infinite energy of the field of a point charged particle, "hidden momentum", Trouton-Noble, etc. Interestingly, even the Aharonov-Bohm effect can be explained without the use of electromagnetic potentials.

So, you are right, you will not find a derivation of Maxwell's theory there. You will find a theory of directly interacting charges, whose experimental predictions are not worse than in the standard field-based approach.

Of course, you may argue that the non-retarded form of the Darwin-Breit potential is an artefact of its approximate low-order character. We can discuss this point later, if you like. The main point of this post was to show that classical electromagnetic theory follows (in its basic features) directly from the QED Hamiltonian H without any need to consider quantum fields and their arguments (x,t) as observable quantities. In principle, the same approach can be extended to higher perturbation orders and to higher powers of (v/c), so as to incorporate the radiation emission/absorption effects.

Eugene.

strangerep
we use this operator to construct another abstract object called "current
density 4-vector"
$$J^\mu(x,t) ~=~ \overline{\Psi}(x,t) \gamma^\mu \Psi(x,t)$$

[...]

Restricting to such bilinears loses the electron's spinorial properties.
In particular, that a rotation through 360deg multiplies the (single)
fermion state by -1.

In Penrose's "Spinors & Spacetime vol-1", pp46-47, he mentions that
spinorial effects have been observed in a split beam of single neutrons.
(IIRC, it involves subjecting the split parts to a relative rotation
and observing what happens when the parts are recombined.)

Such experiments seem sufficient to show that spinors are
physically significant and necessary in their own right.
Bilinear combinations alone are not enough.

This is a fair point, and I must address it in order to be taken seriously. Here is my explanation:

As I said, the full information about interacting particle dynamics can be obtained from the QED Hamiltonian. [...]
So, you are right, you will not find a derivation of Maxwell's theory there. You will find a theory of directly interacting charges, whose experimental predictions are not worse than in the standard field-based approach. [...]
The main point of this post was to show that classical electromagnetic theory follows (in its basic features) directly from the QED Hamiltonian H without any need to consider quantum fields and their arguments (x,t) as observable quantities.

The most basic features of classical electrodynamics are given by the Maxwell equations,
from which everything else follows.You showed that a few phenomenological items valid in classical electromagnetic theory follow. But this is a far cry from showing that the classical theory (in either the form taken as basic in a course on classical electrodynamics, or the form used by engineers) can be derived from the quantum version without using quantum fields.

For the classical limit of QED in the functional integral formulation, see, e.g.,
Phys. Rev. D 12, 1733–1738 (1975).

Restricting to such bilinears loses the electron's spinorial properties.

Such experiments seem sufficient to show that spinors are physically significant and necessary in their own right.Bilinear combinations alone are not enough.
Yes. The true dynamical variable is the Wigner transform of the spinor field; it is a 4x4 matrix dependent on a 4-position x and a 4-momentum p. It generalizes the phase space density that figures in the Boltzmann equation.

I mentioned the current only as the prime example of an obviously physical field, denigrated by meopemuk to only an abstract object.

DrDu
Restricting to such bilinears loses the electron's spinorial properties.
In particular, that a rotation through 360deg multiplies the (single)
fermion state by -1.

In Penrose's "Spinors & Spacetime vol-1", pp46-47, he mentions that
spinorial effects have been observed in a split beam of single neutrons.
(IIRC, it involves subjecting the split parts to a relative rotation
and observing what happens when the parts are recombined.)

Such experiments seem sufficient to show that spinors are
physically significant and necessary in their own right.
Bilinear combinations alone are not enough.

As far as I understand it, the statistics can be reconstructed from the quasi-local observable operators (which contain a like number of field creators and anihilators) and the observed charge superselection sectors, see e.g.:
http://arxiv.org/PS_cache/hep-th/pdf/9910/9910243v2.pdf
or A.S. Wightman, Superselection rules, old and new, Il Nuovo Cimento 110, (1995) pp. 751-769 who also discusses the above mentioned neutron experiment.

The most basic features of classical electrodynamics are given by the Maxwell equations,
from which everything else follows.

QED is a quantum theory. In any quantum theory the Hamiltonian H is supposed to contain all dynamical information. This means that if we succeeded in solving the Schroedinger equation

$$i d |\Psi \rangle /dt = H |\Psi \rangle$$

we should be able to reconstruct the time dependence of *all* relevant physical observables. The Fock space of QED is a multiparticle Hilbert space, where, in addition, the number of particles is allowed to change. So, the relevant observables are the number of particles and properties (position, momentum, spin, etc.) of each particle in the system. These basic features must be preserved in the classical limit as well. So, the classical analog of QED should have the form of a Hamiltonian theory, where particle observables are the canonical variables.

Now, you are saying that this picture is wrong or incomplete. If I understand correctly, you are saying that we need to introduce fields (the electric current and electromagnetic potentials) and solve Maxwell equations in both quantum and classical limits. (Or only in the classical limit?) Could you please clarify your position and explain what is the relationship between the Hamiltonian formalism that I've indicated above and the formalism of Maxwell equations. This is a multiple choice question:

1) Both approaches are equivalent
2) The Hamiltonian approach is more general and Maxwell equations are just an approximation.
4) Maxwell's approach is more general. Shroedinger and Hamilton equations are approximations.
3) Fields and Maxwell equations provide additional physical info that is not present in the particle-based Hamiltonian approach.

Eugene.

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QED is a quantum theory. In any quantum theory the Hamiltonian H is supposed to contain all dynamical information. This means that if we succeeded in solving the Schroedinger equation
$$i d |\Psi \rangle /dt = H |\Psi \rangle$$
we should be able to reconstruct the time dependence of *all* relevant physical observables.
True.
The Fock space of QED is a multiparticle Hilbert space, where, in addition, the number of particles is allowed to change.
The state space of QED is not a Fock space. It looks like that only in perturbation theory.
So, the relevant observables are the number of particles and properties (position, momentum, spin, etc.) of each particle in the system.
This is your claim to be proved, not a fact that could be assumed.

The standard view is that the relevant observables are the (local) electromagnetic field F(x) and the (microlocal) electron Wigner matrix field W(x,p), obtained by a Wigner transform from the spinor bilinears.
These basic features must be preserved in the classical limit as well.
Indeed, the classical limit gives what is called a quantum kinetic equation, relating F(x) and W(x,p). See, e.g., http://arxiv.org/pdf/1007.1099 . It is the QED analogue of the Klein-Gorden equation, the classical limit of a free scalar quantum field theory.
So, the classical analog of QED should have the form of a Hamiltonian theory, where particle observables are the canonical variables.
This only holds under your unproved assumption, which conflicts with the standard view.
Now, you are saying that this picture is wrong or incomplete. If I understand correctly, you are saying that we need to introduce fields (the electric current and electromagnetic potentials) and solve Maxwell equations in both quantum and classical limits.
I was saying that every engineer uses the traditional Maxwell equations, and that any theory that deserves the name QED must be able to reproduce it in the appropriate approximation, no matter which starting point it takes. If you can derive it from your Hamiltonian particle theory, fine - but we'll discuss that part in your newly created thread https://www.physicsforums.com/showthread.php?t=474666

This thread, however, is for the discussion of what it means in a relativistic quantum field theory to be ''observable'' - this is independent of the choice of foundations. Rather, it is a matter of relating theoretical entities to experimental options.

The state space of QED is not a Fock space.

I strongly disagree. In any QFT we have a unique vacumm state and a set of particle creation and annihilation operators. Using those, we can always divide the full Hilbert space of the theory into n-particle sectors and reproduce the Fock space structure.

It is a different matter that bare particles of QED are fictitious objects, and their Fock space does not have a good physical interpretation. So, one needs to consider the physical vacuum, physical (or dressed) particles and associated physical Fock space. Then the particle interpretation and the Fock space picture are alive and well in QED.

Eugene.

I strongly disagree. In any QFT we have a unique vacumm state and a set of particle creation and annihilation operators.
Who constructed these?
Using those, we can always divide the full Hilbert space of the theory into n-particle sectors and reproduce the Fock space structure.
DarMM had already pointed out in other threads that there are interacting QFT's in 3D (e.g. Phi^4_3) where the Hilbert space has been rigorously constructed and does not carry a Fock representation. Thus your claim is provably false at least for _some_ QFT.

To save your claim you would have to argue why dimension 4 is so different from dimension 3 that one can _know_ in the former case that the representation must be Fock.

It is a different matter that bare particles of QED are fictitious objects, and their Fock space does not have a good physical interpretation. So, one needs to consider the physical vacuum, physical (or dressed) particles and associated physical Fock space. Then the particle interpretation and the Fock space picture are alive and well in QED.
In Phi^4_3, the _physical_ multiparticle states do not have the structure of a Fock space.

So what is the additional property that QED has from which you can conclude that its multiparticle states form a Fock space?

meopemuk said:
In any QFT we have a unique vacuum state and a set of particle creation and annihilation operators.
Who constructed these?

S. Weinberg in section 4.2.

In my opinion, the Fock space and c/a operators do not require any *derivation*. They are simply postulated from the beginning based on our knowledge (which you seem to deny) that all physical systems are just collections of discrete interacting particles. So, we build the most general Hilbert space, that can accomodate states of such systems. This Hilbert space is a direct sum of 0-particle, 1-particle, 2-particle, etc. sectors, which is called the Fock space.

After this Hilbert space is constructed, we define an unitary representation of the Poincare group there. That is it. The theory is completed. Now we can simply run calculations and get all desired results and compare them with experiment. This is the logic, which I picked up from Weinberg's textbook, and I like it.

I understand that this point of view on QFT is different from yours. You claim that "fields" are the most fundamental objects. The fields interact with each other, satisfy some non-linear field equations... Particles and the Fock space structure should be "derived" from fields through some kind of "quantization" procedure, and if this procedure fails, then you say that the Fock space does not exist. (If I misrepresented your views, then please correct me.) So, you and I are talking about two different theories and put different meaning into the same words. No wonder it is so difficult to understand each other.

Cheers.
Eugene.

In Phi^4_3, the _physical_ multiparticle states do not have the structure of a Fock space.

My rule of thumb is the following: If _physical_ (or _dressed_) 0-particle and 1-particle states are different from _bare_ 0-particle and 1-particle states, then this is a bad theory, meaning that interaction is not chosen properly. A correct interaction operator should not affect the definitions of the vacuum and 1-particle states.

In this sense QED is a bad theory as well. The need for renormalization is just a part of this problem. Luckily, the bad features of QED can be fixed by means of the *dressing transformation*. Perhaps, Phi^4_3 is so bad that it cannot be fixed even by dressing. This is just my wild guess since I haven't studied Phi^4_3 closely.

Eugene.

S. Weinberg in section 4.2.
He constructed them for noninteracting theories, not for QED.
This Hilbert space is a direct sum of 0-particle, 1-particle, 2-particle, etc. sectors, which is called the Fock space.

After this Hilbert space is constructed, we define a unitary representation of the Poincare group there. That is it. The theory is completed.
In principle yes. But the tricky point is the construction of a unitary representation of the Poincare group defining a nontrivial interaction satisfying the cluster decomposition principle. Haag's theorem says that this cannot be done in Fock space. You'd need to find a loophole in Haag's theorem.

Note that you derive in Section 8.1.2 and Appendix N only a formal representation of the Poincare algebra, not a unitary representation of the Poincare group. For to get the latter, you must show that the generators are self-adjoint operators on some Hilbert space. But because the fields are distributions only, your operators (borrowed from Weinberg) are only formal strings of symbols, not well-defined operators on a dense subspace of your Fock space.

It is here that things go wrong with Fock space. It is known from Haag's theorem that Fock space cannot work, but in 4D it is not yet known how to fix that.
I understand that this point of view on QFT is different from yours.
No. If your construction would really give a unitary representation of the Poincare group on a QED Fock space, I wouldn't object to your above line of reasoning.

My rule of thumb is the following: If _physical_ (or _dressed_) 0-particle and 1-particle states are different from _bare_ 0-particle and 1-particle states, then this is a bad theory, meaning that interaction is not chosen properly.
''The ﬁrst principle is that you must not fool yourself - and you are the easiest person to fool.'' (Richard Feynman)

The study of lower-dimensional theories suggests instead that the Hilbert space is not chosen properly.
A correct interaction operator should not affect the definitions of the vacuum and 1-particle states.

In this sense QED is a bad theory as well. The need for renormalization is just a part of this problem. Luckily, the bad features of QED can be fixed by means of the *dressing transformation*. Perhaps, Phi^4_3 is so bad that it cannot be fixed even by dressing. This is just my wild guess since I haven't studied Phi^4_3 closely.
All the stuff about perturbative dressing can be applied to any renormalizable QFT.
There is nothing special about QED_4 in this respect.

First, we use this operator to construct another abstract object called "current density 4-vector"

$$J^{\mu} (x,t)= \overline{\Psi} (x,t) \gamma^{\mu} \Psi(x,t)$$

Then we take the product with another abstract "photon field" to form the "interaction Hamiltonian density"

$$V(x,t) = J^{\mu} (x,t) A_{\mu} (x,t)$$

Eugene.

i agree with you, why one should naïvely take operators to represent physical quantities.

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Note that you derive in Section 8.1.2 and Appendix N only a formal representation of the Poincare algebra, not a unitary representation of the Poincare group. For to get the latter, you must show that the generators are self-adjoint operators on some Hilbert space. But because the fields are distributions only, your operators (borrowed from Weinberg) are only formal strings of symbols, not well-defined operators on a dense subspace of your Fock space.

I would like to stay at the same rigor level as in Weinberg's book. It seems that he is not particularly concerned about "dense subspaces", so I am not concerned either. For me it is sufficient if 10 Hermitian operators are found, which satisfy Poincare commutation relations. Then I would say that we have a relativistic quantum theory.

Eugene.

The study of lower-dimensional theories suggests instead that the Hilbert space is not chosen properly.

I am not sure how one can make a wrong choice here. If we accept that world is made of particles, then the Fock space is the only natural choice for a Hilbert space accommodating systems with any/variable number of particles.

The lower-dimensional theories you are talking about use a completely different field-based logic. This is why they have so much trouble with the Fock space.

Eugene.

DarMM
Gold Member
My rule of thumb is the following: If _physical_ (or _dressed_) 0-particle and 1-particle states are different from _bare_ 0-particle and 1-particle states, then this is a bad theory, meaning that interaction is not chosen properly. A correct interaction operator should not affect the definitions of the vacuum and 1-particle states.
James Glimm has constructed $$\phi^{4}_{3}$$ in a non-Fock Hilbert space. This construction directly gives the Hilbert space of the physical particles, no mention of "bare states".

DarMM
Gold Member
I am not sure how one can make a wrong choice here. If we accept that world is made of particles, then the Fock space is the only natural choice for a Hilbert space accommodating systems with any/variable number of particles.
Fock space is not the only natural choice. Take two elements of $$\mathcal{H}_{1}$$, the single particle Hilbert space, whose support overlaps in space. In Fock space their tensor product will simply be a two particle state, which you can compute easily. In an interacting theory however, the resulting state could have an overlap with four particle states for example, an indication of 2 -> 4 scattering processes. This is a structure that cannot be replicated in Fock space.

Fock space is not the only natural choice. Take two elements of $$\mathcal{H}_{1}$$, the single particle Hilbert space, whose support overlaps in space. In Fock space their tensor product will simply be a two particle state, which you can compute easily. In an interacting theory however, the resulting state could have an overlap with four particle states for example, an indication of 2 -> 4 scattering processes. This is a structure that cannot be replicated in Fock space.

If I understand correctly, this means that we cannot prepare pure 2-particle states. Such states will always have some admixture of 4-particle states. Is it true that such an admixture will be present even if the two particles are infinitely separated? Or this approach says that 2particle/4particle coefficients depend on the distance between the particles. This would be extremely weird.

If your description of the non-Fock space is correct, then we would not be able to define such common things as 2particle->2particle or 2particle->4particle S-matrix coefficients, because it is impossible to say exactly how many particles the system has. Are there any experimental confirmations of such unusual predictions?

Eugene.

DarMM
Gold Member
If I understand correctly, this means that we cannot prepare pure 2-particle states. Such states will always have some admixture of 4-particle states. Is it true that such an admixture will be present even if the two particles are infinitely separated? Or this approach says that 2particle/4particle coefficients depend on the distance between the particles. This would be extremely weird.
It will not occur when they are infinitely separated, that is the reason for mentioning that their spatial supports overlap. It has nothing to do with preparation of particle states, just the mathematical structure of the theory. A tensor product of one-particle states is automatically a two particle state in Fock space, in an interacting theory it is not.

If your description of the non-Fock space is correct, then we would not be able to define such common things as 2particle->2particle or 2particle->4particle S-matrix coefficients, because it is impossible to say exactly how many particles the system has. Are there any experimental confirmations of such unusual predictions?
The non-Fock spaces do not make such predictions. The S-matrix is described as a unitary mapping of asymptotic in states to asymptotic out states. These states have a Fock structure. So all n-particle to m-particle processes are well-defined. Scattering deals with widely separated particles, so the issues mentioned above do not effect the labelling of scattering states.

I would like to stay at the same rigor level as in Weinberg's book. It seems that he is not particularly concerned about "dense subspaces", so I am not concerned either. For me it is sufficient if 10 Hermitian operators are found, which satisfy Poincare commutation relations. Then I would say that we have a relativistic quantum theory.
(since this has nothing to do with the question posed in the title of the thread).

It will not occur when they are infinitely separated, that is the reason for mentioning that their spatial supports overlap. It has nothing to do with preparation of particle states, just the mathematical structure of the theory. A tensor product of one-particle states is automatically a two particle state in Fock space, in an interacting theory it is not.

The non-Fock spaces do not make such predictions. The S-matrix is described as a unitary mapping of asymptotic in states to asymptotic out states. These states have a Fock structure. So all n-particle to m-particle processes are well-defined. Scattering deals with widely separated particles, so the issues mentioned above do not effect the labelling of scattering states.

DarMM, Arnold,

let me summarize our discussion as I understand it.

There are two distinct ways to look at relativistic quantum theory of systems with variable number of particles (aka QFT).

In one approach (advocated by you) the Hilbert space has a non-Fock structure. The subspace of a 2-particle system is not necessarily the tensor product of 1-particle subspaces. The subspaces of n particles are not necessarily orthogonal to subspaces of m particles. All these non-Fock features appear only when particles are close to each other, i.e., when they interact. In the limit of widely separated particles the Fock-like structure of the Hilbert space gets restored, which allows you to define asymptotic n-particle states. All these unusual features result from the desire to keep the idea of interacting quantum fields, which satisfy covariant transformation laws and (anti)commute at space-like separations. Basically, it is assumed that fields and their postulated properties are paramount and untouchable. If these assumptions result in some weird particle properties, then so be it. Particles are secondary, anyway.

A different approach (that I read between the lines of Weinberg's textbook) is that particles are the primary objects. 2-particle Hilbert space is always a tensor product of 1-particle spaces, whether patricles interact or not. Independent on interaction, one can always define and prepare pure n-particle states, which are always orthogonal to m-particle states. This leads to the Fock structure of the total Hilbert space, and this structure is independent on interactions. Interactions are introduced by specifying an unitary representation of the Poincare group in this Fock space. As I understand, this approach is fully capable to explain scattering, decays, and the entire range of processes in which particles can be created or annihilated. In this approach, *free* quantum fields are used simply as abstract building blocks for the interaction Hamiltonian. Field equations are not given any physical interpretation. Field commutators are not associated with causality or other physical conditions. *Interacting* fields do not play any role here. Their field equations and commutators are not used in calculations at all. All physical properties can be computed if we know the Hamiltonian and its expression in terms of particle a/c operators.

As far as I know, there is no experiment that can directly say, which of these two worldviews is correct and which is wrong. One can propose various philosophical arguments in favor or against both of these two approaches. So, it is too early to claim victory of one approach over the other. Possibly, the best we can do is (1) recognize the differences (2) agree to coexist peacefully.

Eugene.