What Are the States in Quantum Field Theory?

[cgoakley]

Stueckelberg constructs only the leading terms in a formal power series.

The power series is obtained by expanding the QED equations of motion. If one simply defines one's theory such that the interacting fields are the free fields plus these leading terms then everything will be perfectly finite and rigorous, and the simple scattering amplitudes will be correctly reproduced. The cost will just be that the interacting CCRs will not be the expected ones.

But it is well-known that formal power series never define a function: There are always infinitely many functions whose Taylor expansion agrees with the formal power series.

True, but irrelevant.

What is missing to make his methods rigorous is a recipe for picking the right function in a way that is consistent with the action of some self-adjoint Hamiltonian acting on Fock space (or another space).

This is not in my view the issue. The Hamiltonian for the free fields works just as well for the interacting fields in this approach. The thing that does not exist is a "free Hamiltonian" that time-displaces the interacting field as though it was the free field (and if it did, it would violate Haag's theorem).

 

[A. Neumaier]

The power series is obtained by expanding the QED equations of motion. If one simply defines one's theory such that the interacting fields are the free fields plus these leading terms then everything will be perfectly finite and rigorous, and the simple scattering amplitudes will be correctly reproduced. The cost will just be that the interacting CCRs will not be the expected ones.

But then the commutation relations are not causal, and what you get is not Poincare invariant. It is very easy to construct theories that are well-defined and approximate QED one way or another. It is also easy to construct Poincare invariant theories if you make compromises with causality.

But real QED at the same time
(i) is Poincare invariant,
(ii) is causal, i.e., field commutators at space-like related arguments vanish,
(iii) satisfies the cluster decomposition property,
and no amount of tinkering a la Stueckelberg so far has lead to a consistent and nontrivial 4D field theory with these properties.

 

[cgoakley]

But then the commutation relations are not causal,

Spacelike (anti)commutativity is not guaranteed, unless one introduces higher-order terms, certainly. What this has to do with causality is not so clear, though, as defining what one means by this in the quantum world is a lot harder than in the classical world.

and what you get is not Poincare invariant.

This I absolutely do not get. How and why is this not Poincare invariant?

It is very easy to construct theories that are well-defined and approximate QED one way or another. It is also easy to construct Poincare invariant theories if you make compromises with causality.

Examples?

 

[meopemuk]

As just mentioned, the correct, intended statement was ''The interaction Psi_i(x) is a limit of a sequence of unitary transform of the free Psi(x), hence satisfies the same transformation rules.'' This is indeed a nontrivial statement; for the 1+1d case, see, e.g.,
the paper by Glimm and Jaffe in
Acta Mathematica 125, 1970, 203-267
http://www.springerlink.com/content/t044kq0072712664/
and the two papes that preceded their part III.

Thanks. I'll check it out.

Since this nontrivial statement (which was assumed to be self-evident by Haag) was proven only much later, does it mean that Haag's proof is not complete? Shall we call it Haag-Glimm-Jaffe theorem now? Does it mean that this theorem is rigorously valid only in the 1+1d case? The story of Haag's theorem becomes rather confusing.

Eugene.

 

[A. Neumaier]

Since this nontrivial statement (which was assumed to be self-evident by Haag) was proven only much later, does it mean that Haag's proof is not complete? Shall we call it Haag-Glimm-Jaffe theorem now? Does it mean that this theorem is rigorously valid only in the 1+1d case? The story of Haag's theorem becomes rather confusing.

No. Haag's theorem is a theorem that holds in general when the Wightman axioms are satisfied, while the statement under discussion is a significantly stronger statement that can be proved for the specific theories that were explicitly constructed.

 

[A. Neumaier]

Spacelike (anti)commutativity is not guaranteed, unless one introduces higher-order terms, certainly. What this has to do with causality is not so clear, though, as defining what one means by this in the quantum world is a lot harder than in the classical world.

It means that one can prepare a state with given exact values of a Hermitian fields
everywhere at any particular time, in any Lorentz frame. Violation of the causal commutation rules mean that this is impossible (due to the uncertainty relation for m=noncommuting observables), so that there must be an instantaneous influence of part of the world to other parts of the world that would forbid this.

Therefore, at least for the electromagnetic field which is observable and preparable, the causal commutation rules ar a necessity for a consistent relativistic QFT.

How and why is this not Poincare invariant?

Poincare invariance of a quantum field theory is a very nontrivial statement that is not easy to get. Thus as long as no proof is available that a given construction is Poincare invariant (by giving the interacting generators with P_0 and verifying that they satisfy the Lie algebra of Poincare) it is very likely that it is not Poincare invariant. In particular, truncating the Hamiltonian in a field theory generally destroys Poincare invariance, since there is no matching truncation of the other generators that would preserve the Lie algebra. (This can be made more rigorous in terms of cohomology...)

Indeed, Weinberg argues in his book that Poincare invariance of a field theory requires causal commutation rules.

 

[meopemuk]

No. Haag's theorem is a theorem that holds in general when the Wightman axioms are satisfied, while the statement under discussion is a significantly stronger statement that can be proved for the specific theories that were explicitly constructed.

Well, now I am confused even more. Isn't it true that one of Wightman axioms (the one named W2 in the Wikipedia article http://en.wikipedia.org/wiki/Wightman_axioms ) defines exactly the Lorentz transformations of the fields. Then it appears that Glimm and Jaffe have proven one of Wightman axioms?

Eugene.

 

[A. Neumaier]

Well, now I am confused even more. Isn't it true that one of Wightman axioms (the one named W2 in the Wikipedia article http://en.wikipedia.org/wiki/Wightman_axioms ) defines exactly the Lorentz transformations of the fields. Then it appears that Glimm and Jaffe have proven one of Wightman axioms?

What Glimm and Jaffe proved in part iV, based on the results in part iii (my reference),
is that P(Phi)_2 quantum field theories satisfy the Wightman axioms. This was a major achievement at the time since before their work it wasn't known whether any interacting field theory satisfying Wightmans's axiom exist at all.

In case your confusion is about why one should prove an ''axiom'' (which is supposed to be an assumption): This is not more strange than when verifying that the integers form a group under addition - you need to prove for them the axioms of group theory.

 

[cgoakley]

How and why is this not Poincare invariant?

Poincare invariance of a quantum field theory is a very nontrivial statement that is not easy to get. Thus as long as no proof is available that a given construction is Poincare invariant (by giving the interacting generators with P_0 and verifying that they satisfy the Lie algebra of Poincare) it is very likely that it is not Poincare invariant. In particular, truncating the Hamiltonian in a field theory generally destroys Poincare invariance, since there is no matching truncation of the other generators that would preserve the Lie algebra. (This can be made more rigorous in terms of cohomology...)

Indeed, Weinberg argues in his book that Poincare invariance of a field theory requires causal commutation rules.

If you are saying this, then with respect, I do not think that you have understood the approach at all.

The steps are these:

1. Build a free field relativistic quantum field theory

We are not going to argue about this - or are we?

The Hamiltonian and other Poincare generators are perfectly well defined in terms of the free field creation and annihilation operators (Noether's theorem). The Poincare algebra is obeyed and a faithful representation of the Poincare algebra is obtained on the space of physical states.

2. Construct interacting fields as sums of products of free fields, the zeroth order in each case being the free field. As long as the coefficient functions in each multilinear product are chosen correctly (i.e. covariantly, with spacetime co-ordinates only appearing as differences), then the transformation properties of the interacting field under the Poincare group will be the same as for the free fields. The time displacement generator, a.k.a. the Hamiltonian, will be the same for both. There are no "free" and "interacting" Hamiltonians - there is just a Hamiltonian.

3. Following Stueckelberg, the matrix elements for elementary processes can then be read off directly, after applying the interacting fields to the vacuum to create particle states. The free vacuum is the same as the interacting one.

Note: no interaction picture, no time-ordered products. The approach is NOT equivalent to the one given (e.g.) in Weinberg's books.

 

[A. Neumaier]

1. Build a free field relativistic quantum field theory

The Hamiltonian and other Poincare generators are perfectly well defined in terms of the free field creation and annihilation operators (Noether's theorem). The Poincare algebra is obeyed and a faithful representation of the Poincare algebra is obtained on the space of physical states.

2. Construct interacting fields as sums of products of free fields, the zeroth order in each case being the free field. As long as the coefficient functions in each multilinear product are chosen correctly (i.e. covariantly, with spacetime co-ordinates only appearing as differences), then the transformation properties of the interacting field under the Poincare group will be the same as for the free fields. The time displacement generator, a.k.a. the Hamiltonian, will be the same for both. There are no "free" and "interacting" Hamiltonians - there is just a Hamiltonian.

If the Hamiltonian is still the Hamiltonian of the free field, then the dynamics is trivial.
You get exactly the same eigenstates and asymptotic behavior, there are no bound states, the scattering is trivial.

Renaming the field observables is not enough to create a nontrivial dynamics.
Of course one can use the revised field operators to create a mock scattering scenario, but this scenario has nothing to do anymore with Schroedinger equations.

If Stueckelberg's idea had been the breakthrough that your interpretation claims it is, it would have had far more impact.

 

[cgoakley]

If the Hamiltonian is still the Hamiltonian of the free field, then the dynamics is trivial.
You get exactly the same eigenstates and asymptotic behavior, there are no bound states, the scattering is trivial.

Renaming the field observables is not enough to create a nontrivial dynamics.
Of course one can use the revised field operators to create a mock scattering scenario,

Yes, of course if you put free fields into the scattering calculations, you will get trivial results. That is not what he does. The interacting electron field contains, in higher order, a (free) electron combined with a (free) photon, an electron combined with an electron-positron pair and so on. Similarly, the interacting photon field contains, in higher-order, an electron-positron pair, a photon combined with an electron-positron pair, and so forth. It is the non-zero matrix elements between the higher-order pieces of the interacting fields that enable one to correctly obtain all tree-level scattering amplitudes for QED.

but this scenario has nothing to do anymore with Schroedinger equations.

Why should this be a requirement? Schroedinger equations are very resistant to being made relativistic when there are interactions - this is what Haag's theorem is all about. Stueckelberg's method is relativistic right from the start.

If Stueckelberg's idea had been the breakthrough that your interpretation claims it is, it would have had far more impact.

"Already in 1934 [...] it seemed that a systematic theory could be developed in which these infinities [divergent radiative corrections] are circumvented. At that time nobody attempted to formulate such a theory [...].There was one tragic exception [...], and that was Ernst C.G. Stueckelberg. He wrote several important papers in 1934-38 putting forward a manifestly invariant formulation of field theory. This could have been a perfect basis for developing the ideas of renormalization. Later on, he actually carried out a complete renormalization procedure in papers with D. Rivier, independently of the efforts of other authors. Unfortunately, his writings and his talks were rather obscure,and it was very difficult to understand them or to make use of his methods. He came frequently to Zurich in the years 1934-6, when I was working with Pauli, but we could not follow his way of presentation. Had Pauli and I myself been capable of grasping his ideas, we might well have calculated the Lamb shift and the correction to the magnetic moment of the electron at the time."

Weisskopf's words (in 1981) - not mine.

 

[A. Neumaier]

It is the non-zero matrix elements between the higher-order pieces of the interacting fields that enable one to correctly obtain all tree-level scattering amplitudes for QED.

Tree level is considered trivial. It doesn't even explain the anomalous magnetic moment of the lectron, or the Lamb shift - the successes that made modern QED respectable.

Why should this be a requirement? Schroedinger equations are very resistant to being made relativistic when there are interactions - this is what Haag's theorem is all about. Stueckelberg's method is relativistic right from the start.

The Schroedinger equation is still the thing that makes quantum field theory consistent (at least on a formal level). As you can read in any QFT textbook, it is needed to derive the form of the S-matrix and its unitarity. Therefore everyone (except you) requires the Schroedinger equation, though it is no longer very practical to use it in computations since its use breaks manifest Lorentz covariance.

All relativistic QFTs that have been constructed rigorously have a Hamiltonian generating both the time evolution and relating to the S-matrix in the same way as in simple scattering at external potentials. Therefore, if you can't construct the Hamiltonian as part of a nontrivial representation of the Poincare group, you didn't construct a relativistic quantum field theory according to today's standards. In particular, Stueckelberg didn't construct one.

"Ernst C.G. Stueckelberg. He wrote several important papers in 1934-38 putting forward a manifestly invariant formulation of field theory. This could have been a perfect basis for developing the ideas of renormalization. Later on, he actually carried out a complete renormalization procedure in papers with D. Rivier, independently of the efforts of other authors. Unfortunately, his writings and his talks were rather obscure,and it was very difficult to understand them or to make use of his methods."

Weisskopf's words (in 1981) - not mine.

It is still very difficult to understand them or to make use of his methods. In the 30 years since this revelation, nobody found them useful enough to develop his methods further. They are not useful - they are far less powerful than the real thing, and they are approximate only, violating causality. You may find that this is irrelevant, but the experts know better.

 

[meopemuk]

What Glimm and Jaffe proved in part iV, based on the results in part iii (my reference),
is that P(Phi)_2 quantum field theories satisfy the Wightman axioms. This was a major achievement at the time since before their work it wasn't known whether any interacting field theory satisfying Wightmans's axiom exist at all.

Is it possible to verify the Wightman axiom about Lorentz transformations in more realistic theories, such as QED? Perhaps P(Phi)_2 quantum field theories (where, according to Glimm and Jaffe, the axiom is true) are some exceptional pathological cases, where the transformation law becomes simple due to some cancellations? Still, I don't see any *physical* reason to believe in this simple transformation law. I agree, it makes a nice formula, but what is the physics of it?

Eugene.

 

[A. Neumaier]

Is it possible to verify the Wightman axiom about Lorentz transformations in more realistic theories, such as QED?

QED is currently still too hard for mathematical physicists, though they can construct various reasonable approximations to QED, but not one satisfying the Wightman axioms.
The problem is still open. (On the other hand, they can prove that Phi^4 theory exists in dimensions 2 and 3, and is trivial in dimensions >4. The 4D case is borderline and therefore hardest.)

Perhaps P(Phi)_2 quantum field theories (where, according to Glimm and Jaffe, the axiom is true) are some exceptional pathological cases, where the transformation law becomes simple due to some cancellations?

There is nothing pathological for 2D theories. If you can prove the Wightman axioms, you can apply all the nice results that can be derived from these axioms, including the existence of a good scattering theory with a covariant S-matrix.

Still, I don't see any *physical* reason to believe in this simple transformation law. I agree, it makes a nice formula, but what is the physics of it?

The necessity for the Wightman axioms stems from the belief in fundamental physical principles - relativity, causality, the existence of fields and a vacuum, and a separable Hilbert space accommodating all these. These together make the Wightman axioms essentially unescapable.

A transformation law that does not satisfy the commutation rules of the Poincare algebra
(your nice formulas) has no representation of the Poincare group in which H=P_0 (the interacting Hamiltonian) generates the physical time translations.

Anyway, I don't understand why you complain about the transformation laws that you champion yourself in your book, though in perturbation theory, and in the instant form only. This restricted form of the representation has no effect on the transformation law.

 

[meopemuk]

QED is currently still too hard for mathematical physicists, though they can construct various reasonable approximations to QED, but not one satisfying the Wightman axioms.
The problem is still open.

A rigorous proof would be very difficult, I agree. But one can still try to verify the transformation formula in low perturbation orders, I think. In QED we have both the interacting Hamiltonian and the boost operator. So, in principle, we should be able to insert them in the Wightman's transformation formula, make the perturbation expansion and see directly whether this formula holds, at least in low orders.

If this formula does hold, I would be very surprised. If the formula is violated, then I wouldn't blame QED, which is our best physical theory, after all. I would rather say that Wightman's assumption is unrealistic.

A transformation law that does not satisfy the commutation rules of the Poincare algebra
(your nice formulas) has no representation of the Poincare group in which H=P_0 (the interacting Hamiltonian) generates the physical time translations.

I don't quite understand your logic. According to Weinberg, a theory is relativistically invariant if it has a unitary representation of the Poincare group. In other words, if there exist 10 Hermitian operators, satisfying the corresponding Lie algebra commutators. This says nothing about the explicit transformation law of the interacting field. If you can prove that Wightman's formula follows directly from commutation relations of Poincare generators, then I would agree with you. Can you prove that?

Anyway, I don't understand why you complain about the transformation laws that you champion yourself in your book, though in perturbation theory, and in the instant form only. This restricted form of the representation has no effect on the transformation law.

The transformation law in question is postulated for free quantum fields. However, this is not a reflection of any physical principle, like the principle of relativity. According to Weinberg, free fields are intentionally defined in such a way that this transformation law is valid. Then Weinberg proves that if one builds the interacting Hamiltonian and boost operators out of products of such free fields, then one obtains 10 non-trivial Poincare generators with required commutation relations. This is enough to obtain a satisfactory interacting quantum field theory (apart from renormalization, which we do not discuss here). The behavior of the *interacting* field is not relevant in this construction. If I remember correctly, Weinberg does not mention Lorentz transformations of the interacting field anywhere in his book. They are simply not needed for calculations of scattering amplitudes.

In my opinion, free quantum fields (and their covariant transformation laws) have no relationship to any physical object observed in experiments. They are just mathematical entities, which are useful for constructing interaction operators. *Interacting* fields don't have even this limited meaning. A full interacting theory can be constructed without mentioning interacting fields at all. Yes, one can formally build these objects and study their transformation laws. But I don't understand why one should a priori assume some specific form of these transformations?

Eugene.

 

[strangerep]

[...]
The necessity for the Wightman axioms stems from the belief in fundamental physical principles - relativity, causality, the existence of fields and a vacuum, and a separable Hilbert space accommodating all these. These together make the Wightman axioms essentially unescapable.

There's one more: the belief that multi-particle physics takes place in a common
Minkowski spacetime -- the truth/falsehood of that belief is what this discussion
centers on. In Haag's original paper, he says (in effect) that any other choice is
"unnatural", but gives no further justification. OTOH, the fact that 2-particle
non-relativistic QM requires a tensor product space (i.e., does not work properly
if the particles are assumed to occupy a common position space) gives me reason
to doubt whether Haag's "natural" choice is indeed physically correct.

 

[A. Neumaier]

There's one more: the belief that multi-particle physics takes place in a common
Minkowski spacetime -- the truth/falsehood of that belief is what this discussion
centers on. In Haag's original paper, he says (in effect) that any other choice is
"unnatural", but gives no further justification. OTOH, the fact that 2-particle
non-relativistic QM requires a tensor product space (i.e., does not work properly
if the particles are assumed to occupy a common position space) gives me reason
to doubt whether Haag's "natural" choice is indeed physically correct.

Minkowski space-time is not to be confused with configuration space. The configuration space of a field theory is i(n the absence of gauge invariance and assuming asymptotic completeness) the disjoint union of a (particle content dependent) number of copies of R^3N for N=0,1,2,... This accounts sufficiently for your tensor products.

On the other hand, Minkowski space is the space of arguments of the fields, and this is just 4D for any traditional field theory (excluding Kaluza-Klein, strings, etc.).

 

[A. Neumaier]

But one can still try to verify the transformation formula in low perturbation orders, I think. In QED we have both the interacting Hamiltonian and the boost operator. So, in principle, we should be able to insert them in the Wightman's transformation formula, make the perturbation expansion and see directly whether this formula holds, at least in low orders.

If this formula does hold, I would be very surprised.

So be surprised! Weinberg does so to all orders in Chapter 3.3 of Volume 1.
If the formula were violated, it would have been the end of QED or relativity.

According to Weinberg, a theory is relativistically invariant if it has a unitary representation of the Poincare group. In other words, if there exist 10 Hermitian operators, satisfying the corresponding Lie algebra commutators.

If this were the only condition then even a nonrelativistic field theory would be relativistically invariant. For one can define on nonrelativistic Fock space over R^3 the free relativistic fields. But they are unrelated to all the other structure of the nonrelativistic field theory, and hence meaningless.

But Weinberg is not so stupid to make meaningless definitions.

The theory is relativistic if the physical creation operators (that create physical particles from the vacuum and satisfy causal commutation relations) generate n-point vacuum expectation values that are Poincare covariant. But this is just what the Wightman axioms require.

Weinberg proves this condition in a heuristic fashion in Section 3.3 (there for the S-matrix, which, in view of the the LSZ-formula p.430 proves it for the time-ordered expectation values, which is only little weaker than the Wightman axioms. Using closed-time-path integrals, one can extend the argument to contour-ordered expectation values, which include the Wightman functions. of course, this ''proof'' is only in perturbation theory, and not a mathematical proof but only one according to the usual standards of theoretical physics.

This says nothing about the explicit transformation law of the interacting field. If you can prove that Wightman's formula follows directly from commutation relations of Poincare generators, then I would agree with you. Can you prove that?

Since (as I can see from your book) you accept the usual standards of theoretical physics as sufficient for proofs, you should now agree.

The transformation law in question is postulated for free quantum fields.

This is completely irrelevant. The free operators are only the scaffolding of the building.What counts is the transformation law for the interacting fields.

Weinberg proves that if one builds the interacting Hamiltonian and boost operators out of products of such free fields, then one obtains 10 non-trivial Poincare generators with required commutation relations. This is enough to obtain a satisfactory interacting quantum field theory (apart from renormalization, which we do not discuss here). The behavior of the *interacting* field is not relevant in this construction.

Of course it is, since the interacting field is defined using the interacting representation of the Poincare group.

If I remember correctly, Weinberg does not mention Lorentz transformations of the interacting field anywhere in his book. They are simply not needed for calculations of scattering amplitudes.

You don't remember correctly. The nontrivial ones are constructed in Section 3.3.; see formulas (3.3.18), (3.3.20), and more explicitly (3.5.17) and (7.4.20). They are not needed for the calculations, but they are essential for the proof of covariance of the S-matrix.

In my opinion, [...] *Interacting* fields don't have even this limited meaning. A full interacting theory can be constructed without mentioning interacting fields at all.

The challence is not to construct some interacting theory but to construct one that has the physically verifiable properties - giving a Lorentz invariant and cluster separable scattering matrix. To verify this you need all the stuff you despise.

 

[meopemuk]

But one can still try to verify the transformation formula in low perturbation orders, I think. In QED we have both the interacting Hamiltonian and the boost operator. So, in principle, we should be able to insert them in the Wightman's transformation formula, make the perturbation expansion and see directly whether this formula holds, at least in low orders.

If this formula does hold, I would be very surprised.

So be surprised! Weinberg does so to all orders in Chapter 3.3 of Volume 1.
If the formula were violated, it would have been the end of QED or relativity.

The theory is relativistic if the physical creation operators (that create physical particles from the vacuum and satisfy causal commutation relations) generate n-point vacuum expectation values that are Poincare covariant. But this is just what the Wightman axioms require.

Weinberg proves this condition in a heuristic fashion in Section 3.3 (there for the S-matrix, which, in view of the the LSZ-formula p.430 proves it for the time-ordered expectation values, which is only little weaker than the Wightman axioms. Using closed-time-path integrals, one can extend the argument to contour-ordered expectation values, which include the Wightman functions. of course, this ''proof'' is only in perturbation theory, and not a mathematical proof but only one according to the usual standards of theoretical physics.

Arnold, I don't see any explicit proof of the covariant transformation law for interacting fields in the places you mentioned. Section 3.3 is titled "Symmetries of the S-matrix". There is not a word there about fields and their transformations. Actually, the whole concept of a *free* quantum field is introduced much later in chapter 5. On page 430 Weinberg discusses the pole structure of the S-matrix. Again, not a word about field transformations. It seems that you are reading something between Weinberg's lines. I would like to see a more explicit proof.

If I remember correctly, Weinberg does not mention Lorentz transformations of the interacting field anywhere in his book. They are simply not needed for calculations of scattering amplitudes.

You don't remember correctly. The nontrivial ones are constructed in Section 3.3.; see formulas (3.3.18), (3.3.20), and more explicitly (3.5.17) and (7.4.20). They are not needed for the calculations, but they are essential for the proof of covariance of the S-matrix.

(3.3.18) - general formulas for Poincare generators (space-time translations and rotations) in any instant form interacting relativistic dynamics.

(3.3.20) - general formula for the boost generator in the instant form

(3.5.17) - expression of the boost interaction as an integral of the Hamiltonian density

(7.4.20) - another expression for the boost operator in a field theory.

The *interacting* quantum field and its transformations are not mentioned there.

Eugene.

 

[A. Neumaier]

I don't see any explicit proof of the covariant transformation law for interacting fields in the places you mentioned. Section 3.3 is titled "Symmetries of the S-matrix". There is not a word there about fields and their transformations.

This is because the transformations are more fundamental and must be present in any relativistic quantum theory, whether with or without fields. (Indeed, as long as one works in the Schroedinger representation, one can completely dispense with the fields; but they are nevertheless there, as shown by my construction below.)

Later he simply take this for granted and specializes it to quantum fields. This specialization is done first tentatively in Section 3.5 (see the middle of p.144), and further justified in Chapter 4 (see p.169); later it is assumed without further ado.

Note that the first few chapters are in the Schroedinger picture. The translation to the Heisenberg picture is as follows: For an arbitrary observable A_0 in the Schroedinger picture, the corresponding quantum field A(x) satisfies
A(x) = U(x) A_0 U(-x) ... (1)
A(Lambda x) = U(Lambda) A(x) U(Lambda^{-1}) ... (2)
where the translations U(x) and the Lorentz transforms U(Lambda) are the physical (interacting) ones. This transformation law appears for the special case of the interaction part of the energy density field in (3.5.11), but is a general property of the Heisenberg picture. (Proof: Take (1) as the definition of the field, and deduce (2) from (1) and the properties of arbitrary unitary representations of the Poincare group. The proof doesn't depend on whether the representation is given in the instant form or any other form.)

It seems that you are reading something between Weinberg's lines. I would like to see a more explicit proof.

I read the whole book and fully understand at least the first eight chapters. This is enough to read between the lines. I hope that the details above fill in what you missed.

The *interacting* quantum field and its transformations are not mentioned there.

I am surprised that you can't see them: On p. 144f, H_{curly}(x,t), the interaction part of the energy density is an interacting field since for free fields, it vanishes identically. Later chapters specialize the expression for H_{curly}(x,t) to those corresponding to Lagrangian field theories, expressing it in terms of the corresponding free field operators.
Section 3.5 discusses the needed properties of H_{curly}(x,t) for creating a good interacting representation of the Poincare group, resulting in the requirement of causal commutation rules (with caveats in the footnote for contact terms; cf. p. 277ff).

This is the reason why Chapter 5 bothers to construct free fields, since it is with their help that this condition can be satisfied if the interaction is represented as a sum of integrals of local products of free fields.

 

[bg032]

Note that the first few chapters are in the Schroedinger picture. The translation to the Heisenberg picture is as follows: For an arbitrary observable A_0 in the Schroedinger picture, the corresponding quantum field A(x) satisfies
A(x) = U(x) A_0 U(-x) ... (1)
A(Lambda x) = U(Lambda) A(x) U(Lambda^{-1}) ... (2)
where the translations U(x) and the Lorentz transforms U(Lambda) are the physical (interacting) ones. This transformation law appears for the special case of the interaction part of the energy density field in (3.5.11), but is a general property of the Heisenberg picture.

I do not understand. In 3.5.12 (Weinberg) the interaction energy density field H(x) is required to transform covariantly according to the non-interacting representation U_0.

 

[A. Neumaier]

I do not understand. In 3.5.12 (Weinberg) the interaction energy density field H(x) is required to transform covariantly according to the non-interacting representation U_0.

Yes, you are right. Thanks for pointing out my mistake.

Indeed, the section discusses the S-matrix in terms of the asymptotic fields, which are free by definition - one field for every bound state (see p.110 after (3.1.10).) Thus the situation is a bit more complicated than I had described before.

I'll correct my description later, after having figured out how to describe things more properly (which is not so easy since no textbook discusses this in simple terms).

 

[bg032]

... (which is not so easy since no textbook discusses this in simple terms).

I agree!

 

[A. Neumaier]

Yes, you are right. Thanks for pointing out my mistake.

Indeed, the section discusses the S-matrix in terms of the asymptotic fields, which are free by definition - one field for every bound state (see p.110 after (3.1.10).) Thus the situation is a bit more complicated than I had described before.

I'll correct my description later, after having figured out how to describe things more properly (which is not so easy since no textbook discusses this in simple terms).

Ignoring for simplicity infrared issues in case of massless particles,
the situation is the following:

In order to be able to talk about the S-matrix, one needs to have asymptotic 1-particle states whose tensor product describes the possible input to a scattering event. Clearly, we can prepare independently beams of any kind of free physical particles (elementary or bound states) in the theory and bring them to a collision. I'll call these particles asymptotic particles.

Thus, in QED, we can prepare photons, electrons, and positrons, which are the only asymptotic particles of the theory. In QCD, we can prepare mesons and baryons, but not quarks or gluons as - due to confinement -the latter are not asymptotic particles.

Weinberg now assumes (on p.110) that the unperturbed Hamiltonian describes the free motion of all these asymptotic particles, with their observable quantum numbers (mass, spin, charges). Asymptotic in-states are therefore elements of a Fock space generated from the asymptotic vacuum by means of free creation operators, one for each asymptotic particle species. These creation operators define the free quantum fields introduced on p.144 and used in the remainder of the chapter and in Chapter 4.

According to (3.1.8), the interaction is defined as the difference of the actual Hamiltonian and this free Hamiltonian.

There is, however, a difficulty that Weinberg does not directly discuss in the book: The asymptotic particles need not correspond one to one to the bare particles in which the Hamiltonian is derived from an action. This is most obvious in case of QCD, where the action involves quarks and gluons only, while the asymptotic particles are mesons and baryons. The assumptions break down, and perturbation theory is meaningless - a nonperturbative approach is called for, about which Weinberg is silent in Volume 1. He only says that the bound state problem is poorly solved in QFT (p.560), though with some trickery he is able to consider bound states for QED in an external field (needed to get the Lamb shift).

This breakdown of perturbation theory is the formal reason why low energy predictions from QCD are very hard - it is part of the unsolved confinement problem of QCD. (The derivation of effective actions for mesons on baryons from QCD is still a web of guesswork, with few hard results and much input of phenomenology in addition to intuition derived from QCD proper.)

Even in case of QED (and all other field theories without bound states), the problem remains that the masses of the asymptotic particles don't match the corresponding coefficients of the action from which the Hamiltonian is derived (the so-called bare masses and charges) - rather they are complicated functions of these, determined only as part of the solution process. The simplest instance of this is the anharmonic oscillator,
which can be viewed as a 1+0-dimensional quantum field theory. Here the mass corresponds to the difference between the first two eigenvalues, and this difference changes as a function of the interaction strength.

This is the origin of the need for renormalization. Renormalization is a technique for parameterizing the bare parameters as a function of the observable parameters (or parameters related to these in a fairly insensitive fashion). For my view on this, see
Renormalization without infinities - an elementary tutorial
http://arnold-neumaier.at/ms/ren.pdf

An additional problem in QFTs of dimension 1+d (d>0) is that perturbation theory is infinitely sensitive to changes in the bare parameters, leading to divergent integrals in second-order perturbation theory. Fortunately, renormalization cures this defect automatically, at the cost of making the bare parameters tend to infinity in a particular, fairly well-understood fashion. This was the breakthrough that earned Feynman, Tomonaga and Schwinger the Nobel prize. But the computations become quite technical...

Returning to Weinberg, it is fortunate that (because of the LSZ formula) the formal S-matrix contains essentially the same information as more rigorous approaches that work with the Wightman axioms. Therefore his derivations in Chapter 3 and 4 remain plausible (though not at the level of a mathematical proof) even in the face of the above difficulties. The main insight from Chapter 3.5 is the need for the causal commutation rules for the interaction density to get Lorentz invariance (which is not dependent on a particular representation of it in terms of the asymptotic Fock space), and from Chapter 4 hints for the particular structure of the interaction from the cluster decomposition principle.

The result is that one should represent the interaction as a Lorentz invariant scalar in terms of integrals over products of local field operators satisfying causal commutation relations and carrying an irreducible representation of the Poincare group. Chapter 5 describes the possibilities for the free part.

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. Moreover, (7,1,27) defines the form of the free Lagrangian in terms of the physical parameters. As in the Hamiltonian case discussed in Chapter 3, the interaction is defined as the difference V=L_0-L where L is the full action (with bare parameters). The fact that bare and physical parameters are generally different leads to the observation that the so defined interaction automatically has counterterms (for QED, this is done on p.473).

Section 7.2 then reviews the construction of a Hamiltonian from the Lagrangian. Sections 7.3 and 7.4 verify that there is a unitary representation of the Poincare group in which P_0 is the interacting Hamiltonian defined in Section 7.2. The most important commutation relations (those needed to derive the Lorentz invariance of the S-matrix in Section 3.3) are verified on p.p. 316-317.

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.

At various points in the developments of Chapter 3 and 7, Weinberg points out problems due to singularities at equal times, which may complicate matters (but not in Phi^4 theory or QED). These must be resolved on the basis of more detailed investigations involving the cohomology of the representations, and lead (sometimes) to anomalies, a quite advanced subject that doesn't alter the basic correctness of his analysis (on the level of rigor customary for theoretical physics) and the importance of his conclusions.

 

[bg032]

As far as I know, the proof of Haag's theorem uses the assumption that *interacting* fields have specific transformation laws under the *interacting* representation of the Lorentz group. Namely, the transformation is assumed to be of the form

\Psi_i(x) \to \sum_j D_{ij}(\Lambda^{-1}) \Psi_j(\Lambda x)

Can somebody explain me the reason for this formula? In particular, it is interesting why there are no traces of interaction there?

Thanks.
Eugene.

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 assumprions:

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 causal fields, i.e., operator (valued distributions) defined on Minkowski space-time, transforming covariantly under the above representation and satisfying causal commutation rules*.

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.

(*) Added after a remark of Neumaier

 

[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.

 

[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:

"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.

 

[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:

"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!

 

[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.)

 

[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.

 

[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.

Good. That's already half the answer to the full problem.

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]

Thanks, I see your point.

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

<br /> [\Phi_i (\mathbf{r}, t), \Phi(0,0) ] = [e^{i\mathbf{K} \theta} \Phi_i (\mathbf{r}&#039;, 0) e^{-i\mathbf{K} \theta}, \Phi(0,0) ] = e^{i\mathbf{K} \theta} [\Phi (\mathbf{r}&#039;, 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

<br /> [\Phi_i (\mathbf{r}, t), \Phi(0,0) ] = [e^{i\mathbf{K} \theta} \Phi_i (\mathbf{r}&#039;, 0) e^{-i\mathbf{K} \theta}, \Phi(0,0) ] = e^{i\mathbf{K} \theta} [\Phi (\mathbf{r}&#039;, 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:

"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.

 

[meopemuk]

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

<br /> [\Phi_i (\mathbf{r}, t), \Phi(0,0) ] = [e^{i\mathbf{K} \theta} \Phi_i (\mathbf{r}&#039;, 0) e^{-i\mathbf{K} \theta}, \Phi(0,0) ] = e^{i\mathbf{K} \theta} [\Phi (\mathbf{r}&#039;, 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.

 

[A. Neumaier]

I answered in https://www.physicsforums.com/showthread.php?p=3164679