What Are the States in Quantum Field Theory?

[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