Why doesn't the free vacuum transform under Poincare?

[Riposte]

My understanding of Haag's theorem (see link below) is that there is a mismatch between the Hilbert spaces of free and interacting particles. The argument seems to be that we require both the free and interacting vacuum states to be invariant under Poincare transformations. Now since the entire Hilbert space of the free particle is built by acting on the vacuum with creation operators which are not invariant under Poincare transformations, then our only choice for the interacting vacuum is to have it be proportional to the free vacuum. Here we get a contradiction, since the free vacuum cannot be an eigenstate of both the free Hamiltonian and the full Hamiltonian.

http://philsci-archive.pitt.edu/archive/00002673/01/earmanfraserfinalrevd.pdf" )

All that's good and well, except for one thing. Why do we require the free vacuum to be invariant under Poincare transformations? The free fields aren't the physical ones, the interacting ones are. I see no reason why the non-physical free vacuum can't transform non-trivially under a translation or a rotation.

If, instead, we require only that the interacting vacuum be invariant under Poincare transformations, the mismatch between the Hilbert spaces disappears.

 

[strangerep]

[...] Why do we require the free vacuum to be invariant under Poincare transformations? The free fields aren't the physical ones, the interacting ones are. I see no reason why the non-physical free vacuum can't transform non-trivially under a translation or a rotation.

If, instead, we require only that the interacting vacuum be invariant under Poincare transformations, the mismatch between the Hilbert spaces disappears.

Your final sentence above seems like a sweeping statement, imho...

If you abandon the principle that the (free) vacuum is not annihilated
by the (free) Poincare generators, then... how do you construct a Fock
space? What Lie algebra do you start from?? And how do you construct
the interacting Hilbert space?

 

[Bob_for_short]

My understanding of Haag's theorem (see link below) is that there is a mismatch between the Hilbert spaces of free and interacting particles.

There is no such a mismatch in Atomic Physics or non relativistic QM. So the "mismatch" in QFT is due to wrong interaction term, first of all.

The argument seems to be that we require both the free and interacting vacuum states to be invariant under Poincare transformations. Now since the entire Hilbert space of the free particle is built by acting on the vacuum with creation operators which are not invariant under Poincare transformations, then our only choice for the interacting vacuum is to have it be proportional to the free vacuum. Here we get a contradiction, since the free vacuum cannot be an eigenstate of both the free Hamiltonian and the full Hamiltonian.

If the full Hamiltonian "coincides" with the free one in asymptotic, initial and final states, then the interaction transforms the state vectors within the same Hilbert space, so there is no problem with it. Your statement about non-observability of the free states is mostly dictated with infinities that are "hidden" into the "bare" parameters in course of renormalizations, isn't it? This, conceptual problem is due to self-action interaction term. The theory can be reformulated without self-action (only with the true interaction) and in such a formulation there is no "corrections" to the masses and charges. Think of non relativistic QM as an example where the Haag's theorem fails.

 

[Riposte]

If you abandon the principle that the (free) vacuum is not annihilated
by the (free) Poincare generators, then... how do you construct a Fock
space? What Lie algebra do you start from?? And how do you construct
the interacting Hilbert space?

Sorry, I didn't make myself clear. The Fock space would still be constructed in the usual way. The only difference is how the free vacuum transforms under Poincare transformations:

Before, we have U(\Lambda)|0\rangle = |0\rangle
Now, we would some non-trivial transformation U(\Lambda)|0\rangle = \sum c_i |i\rangle
where the coefficients of this transformation are chosen such that the interacting vacuum is invariant.

 

[Riposte]

So the "mismatch" in QFT is due to wrong interaction term, first of all.

Wrong interaction term? What do you mean by this?

If the full Hamiltonian "coincides" with the free one in asymptotic, initial and final states, then the interaction transforms the state vectors within the same Hilbert space, so there is no problem with it.

I agree, if you have interactions which turn adiabatically on and off, then there's no problem. However, that's usually not the case.

Your statement about non-observability of the free states is mostly dictated with infinities that are "hidden" into the "bare" parameters in course of renormalizations, isn't it?

No, what I meant by calling the free vacuum non-physical is that once interactions are added to the theory the eigenstates of the theory (vacuum, 1-particle states, etc) are shifted from the free ones. The free eigenstates no longer correspond to physical states, and therefore there is no need for them to retain the expected behavior under Poincare transformations. It's like watching an electron whiz by and say "Oh look, there's a 1-particle state of QED". No matter how long you keep looking, you'll never see the free 1-particle state of a neutral electron go floating by.

 

[Bob_for_short]

Wrong interaction term? What do you mean by this?

A self-action term.

I agree, if you have interactions which turn adiabatically on and off, then there's no problem. However, that's usually not the case.

You can separate the projectile and the target in the experimental setup. Such a separation is described with non overlapping wave packets.

No, what I meant by calling the free vacuum non-physical is that once interactions are added to the theory the eigenstates of the theory (vacuum, 1-particle states, etc) are shifted from the free ones.

Not always. Sometimes "free" states correspond to the separated variables (elementary modes) of one compound system, so the main part of interaction is already present in them.

 

[meopemuk]

No, what I meant by calling the free vacuum non-physical is that once interactions are added to the theory the eigenstates of the theory (vacuum, 1-particle states, etc) are shifted from the free ones.

Yes, that's exactly the weird property of QFT (in its current formulation) that makes it so different from the ordinary quantum mechanics. It appears that QFT (bare) vacuum "interacts with itself" and QFT (bare) particle also "interacts with itself". This self-interaction is ultimately responsible for ultraviolet divergences.

One approach to deal with this problem is to take this difference between bare and dressed states for granted, consider different Hilbert spaces for non-interacting and interacting theories, etc, etc. Personally, I find this approach cumbersome and not appealing.

There is, however, another line of thought. We can ask ourselves, are we sure that the interaction adopted in QFT (take QED as an example) is the correct one? After all, this interaction (the "minimal" coupling between the photon field and electron current; I think this is what Bob_for_short had in mind when writing about the "wrong interaction term" in QFT.) was "derived" by using rather shaky analogies with classical Maxwell's electrodynamics. The only experimental data supporting this choice of the QED interaction is related to scattering. But it is well-known that there are many (scattering-equivalent) Hamiltonians, which produce exactly the same S-matrix. So, perhaps, we can choose QED interaction in another form, such that 1) the self-interaction is not present anymore; 2) there is no difference between free (bare) and interacting (dressed) states, just as in ordinary quantum mechanics; 3) the S-matrix remains the same (i.e., agreeing with experiment) as in renormalized QED; 4) ultraviolet divergences are not present.

This strategy (known as the "dressed particle" approach) was first formulated in a beautiful paper

O. W. Greenberg and S. S. Schweber, "Clothed particle operators in simple models of quantum field theory", Nuovo Cim. 8 (1958), 378.

It appears rather promising, as you can check by following multiple references to this work. Within this approach, Haag's theorem does not present any difficulty. See, for example

M.I. Shirokov, "Dressing" and Haag's theorem. http://www.arxiv.org/abs/math-ph/0703021

 

[Bob_for_short]

Yes, that's exactly the weird property of QFT (in its current formulation) that makes it so different from the ordinary quantum mechanics. It appears that QFT (bare) vacuum "interacts with itself" and QFT (bare) particle also "interacts with itself". This self-interaction is ultimately responsible for ultraviolet divergences.

And I would add that the renormalizations remove (perturbatively) the self-action contribution.

Understanding the "free" particles as elementary excitation modes of one compound system makes it unnecessary to add the self-action term. One can write just an interaction term. For example, scattering of charges can be constructed as a potential scattering of compound systems (just like atoms) with inevitable exciting their internal degrees of freedom (photon oscillators).

 

[javierR]

Why do we require the free vacuum to be invariant under Poincare transformations? The free fields aren't the physical ones, the interacting ones are. I see no reason why the non-physical free vacuum can't transform non-trivially under a translation or a rotation.

We need a vacuum state (for the free theory too), which we define to be that state which is annihilated by lowering operators. The Hamiltonian should be a bounded operator in the theory. If your state doesn't satisfy the annihilation property, it cannot be the ground state and you have to continue till you find it or have defined it. So where is yours?

 

[strangerep]

The Fock space would still be constructed in the usual way.

OK, so have the usual (free) a/c operators, i.e., a_k^*,
etc, satisfying the usual CCRs, and these operators constitute an
irreducible set, (meaning that any operator on the Fock space can be
expressed as a polynomial or function of the a/c ops).

That means you still have the usual (free) representation
of the Poincare generators, so there are still operators
U(\Lambda) such that U(\Lambda)|0\rangle =|0\rangle.

Then...

The only difference is how the free vacuum transforms under
Poincare transformations:

Before, we have U(\Lambda)|0\rangle = |0\rangle
Now, we would some non-trivial transformation
U(\Lambda)|0\rangle = \sum c_i |i\rangle
where the coefficients of this transformation are chosen such that the
interacting vacuum is invariant.

I believe this is pretty much the same thing as constructing an "interacting
representation of the Poincare group". Let's call these operators W(\Lambda)
and let's call the interacting vacuum |\Omega\rangle,
where W(\Lambda) |\Omega\rangle = 0,
but W(\Lambda) |0\rangle \ne 0.

But the original (free) a/c ops constitute an irreducible set, so W
must be expressible as function of them (if indeed the free and interacting
Hilbert spaces coincide). One would also expect to find "interacting" a/c operators,
i.e., \alpha_k^*, etc, corresponding to single-particle states in the interacting
theory (which are eigenstates of the interacting Hamiltonian). These must satisfy
\alpha_k |\Omega\rangle = 0.

The task is then to express the \alpha_k ops in terms of the a_k^ ops.
That's pretty much what the "dressed particle" approach involves (also known under the
phrase "unitary dressing transformation"). As well as the references mentioned by
meopemuk, there's also this review article:

Shebeko, Shirokov:
"Unitary Transformations in Quantum Field Theory and Bound States"
Available as nucl-th/0102037

Unfortunately, one generally encounters infinite quantities when
perturbatively constructing the transformation, which must be
"renormalized" in a manner reminiscent of standard renormalization.
One thus encounters a different variation of Haag's theorem in that
the interacting representation cannot live in the Fock space constructed
from the free a/c ops.

 

[meopemuk]

Unfortunately, one generally encounters infinite quantities when
perturbatively constructing the transformation, which must be
"renormalized" in a manner reminiscent of standard renormalization.
One thus encounters a different variation of Haag's theorem in that
the interacting representation cannot live in the Fock space constructed
from the free a/c ops.

Yes, it is true that the perturbative unitary dressing transformation is rather messy. However, there are good reasons to believe that this difficulty is technical rather than fundamental. It is easy to imagine how a full interacting theory can be constructed in a single Fock space, without self-interactions, ultraviolet divergences, and Haag's theorem problems.

First, one can define the usual "Fock space constructed from the free a/c ops". Then one automatically gets the non-interacting representation of the Poincare group there. The next step is to construct the interaction part of the Hamiltonian (the Poincare generator of time translations) as a normally-ordered polynomial in "a/c ops". Of course, in order to be physically admissible, this polynomial must satisfy a few conditions:

1. Each term in the polynomial must have at least two annihilation operators and at least two creation operators. This is necessary to avoid self-interactions in the vacuum and 1-particle states, i.e., to make sure that these states are low-energy eigenvectors of the interacting Hamiltonian.
2. The momentum-dependent coefficient functions in each term must vanish sufficiently rapidly away from the "energy shell". This is necessary to ensure that all loop integrals encountered in S-matrix calculations are finite, so that there are no ultraviolet divergences.
3. In parallel with the above construction of the interacting energy one needs to build the "interacting boost" operator having similar properties, and making sure that commutation relations of the Poincare Lie algebra remain intact. This would guarantee the relativistic invariance of the theory.
4. Finally, one needs to make sure that the S-matrix calculated with the above Hamiltonian is exactly the same as the S-matrix of the renormalized QFT. This can be done by properly adjusting coefficient functions in each perturbation order. Then the new theory is guaranteed to agree with all existing experiments.

I don't have a full mathematical proof that this construction is possible, but there are many indirect indications that there are no fundamental obstacles on this path.

 

[Bob_for_short]

Yes, it is true that the perturbative unitary dressing transformation is rather messy. However, there are good reasons to believe that this difficulty is technical rather than fundamental. It is easy to imagine how a full interacting theory can be constructed in a single Fock space, without self-interactions, ultraviolet divergences, and Haag's theorem problems.

This raises the following question: "How does the dressed electron look like in QED?" considered in another thread.

...
4. Finally, one needs to make sure that the S-matrix calculated with the above Hamiltonian is exactly the same as the S-matrix of the renormalized QFT. This can be done by properly adjusting coefficient functions in each perturbation order. Then the new theory is guaranteed to agree with all existing experiments.

I don't have a full mathematical proof that this construction is possible, but there are many indirect indications that there are no fundamental obstacles on this path.

Apart from UV divergences there are IR ones in QED an in other QFTs with massless (thresholdless) excitations. In fact it is the inclusive cross section that is well comparable with the experiments, not S-matrix itself. Unfortunately, working with the creation/annihilation operators in each perturbative order does not reveal how this problem can be resolved. In order to resolve it one needs a better physical idea and the corresponding mathematical construction for interacting fields. I tried to propose something constructive in my publications which I am not allowed to cite here.

 

[meopemuk]

This raises the following question: "How does the dressed electron look like in QED?" considered in another thread.

In the approach I am describing here the electron is just a point (structureless) particle characterized by measured values of mass, spin, and charge. Nothing fancy.

 

[Bob_for_short]

According to my experience, any interaction smears quantum mechanically the charge. Probably your interacting (dressed) electron is still point-like because you have not completed the dressing, isn't it?

 

[meopemuk]

According to my experience, any interaction smears quantum mechanically the charge. Probably your interacting (dressed) electron is still point-like because you have not completed the dressing, isn't it?

The point-like character of particles and the spread of their wave functions are two separate issues. In QM, the stationary wave function of a free electron is a plane wave, i.e., it is completely delocalized. However, this does not prevent us from talking about the point-like electron.

 

[meopemuk]

Apart from UV divergences there are IR ones in QED an in other QFTs with massless (thresholdless) excitations.

You are right that infrared divergences is a difficult problem. I don't think it has been studied in the "dressed particle" approach. However, this problem has been successfully solved within the standard renormalized QED. I believe that a similar solution could be found in the "dressed particle" approach as well.

 

[Bob_for_short]

The point-like character of particles and the spread of their wave functions are two separate issues. In QM, the stationary wave function of a free electron is a plane wave, i.e., it is completely delocalized. However, this does not prevent us from talking about the point-like electron.

Yes, it prevents. We speak of de Broglie waves, probability amplitude instead of point-like particle. And I speak of the electron coupled to the quantized EMF. Its charge should be smeared, just like in an atom: the compound system has a center of inertia and relative motion wave functions. Both are smeared quantum mechanically. Such a smearing is different from a classical one but it is sufficient to eliminate divergences.

 

[meopemuk]

Yes, it prevents. We speak of de Broglie waves, probability amplitude instead of point-like particle. And I speak of the electron coupled to the quantized EMF. Its charge should be smeared, just like in an atom: the compound system has a center of inertia and relative motion wave functions. Both are smeared quantum mechanically. Such a smearing is different from a classical one but it is sufficient to eliminate divergences.

It appears that we use different definitions of "point-like" or "localizable" particles. In my terminology, a particle is "point-like" if there is a well-defined position operator whose eigenfunctions are Dirac delta functions. The particle can be always prepared in such localized states. According to this definition, the electron is "localizable". However, this does not mean that only delta-functions are allowed as descriptions of electron's states. All kinds of delocalized functions (e.g., plane waves or atomic orbitals) are permissible as well. In most potentials, electron's stationary wave functions are pretty much delocalized, but the electron itself is considered a point-like particle.

 

[Bob_for_short]

It appears that we use different definitions of "point-like" or "localizable" particles. In my terminology, a particle is "point-like" if there is a well-defined position operator whose eigenfunctions are Dirac delta functions. The particle can be always prepared in such localized states. According to this definition, the electron is "localizable". However, this does not mean that only delta-functions are allowed as descriptions of electron's states. All kinds of delocalized functions (e.g., plane waves or atomic orbitals) are permissible as well. In most potentials, electron's stationary wave functions are pretty much delocalized, but the electron itself is considered a point-like particle.

So what is electron de-localization due to dressing in the "dressed" particle approach? Is it different from a "free" electron de-localization and in what respect?

 

[meopemuk]

So what is electron de-localization due to dressing in the "dressed" particle approach? Is it different from a "free" electron de-localization and in what respect?

I am not sure I understand the question, but let me try to explain how I understand the issue of electron localization in QFT.

In the standard textbook QFT, the "bare" electron is a normal point-like particle. As I said above, its wave function can be either localized (delta function) or delocalized (plane wave), but the important point is that one can easily define the position operator corresponding to the single electron, so the "bare" electron is "localizable", just as in ordinary quantum mechanics.

When the interaction is turned on in QFT, the "bare" electron states are no longer eigenstates of the Hamiltonian. So, they cannot correspond to states of "physical" or "dressed" electrons. One can try to build states of such "physical" electrons as linear combinations of "bare" particle states. Such linear combinations must involve contributions of infinite number of "bare" states like (1 electron) + (1 electron + 1 photon) + (1 electron + 1 electron-positron pair) + ... This is a rather complicated wave function. I don't think it is possible to write it down in a complete form. I am also not sure how one can talk about the "localizability" of such states.

The above was the situation in the standard QFT. The difference between "bare" and "dressed" electrons (and the difficulty in explicit definition of "physical" states) was determined by the fact that QFT interaction operator had a nontrivial action on the "bare" 1-particle states (and on the "bare" vacuum states).

Now, the "dressed particle" approach says: The interaction operator chosen in QFT is not good. The correct interaction between "bare" particles must have a trivial action (i.e., yield zero) on the vacuum and 1-particle states. If such a correct interaction is chosen, then "bare" states remain eigenstates of the full interacting Hamiltonian, and there is no difference between "bare" and "physical" particles. In this case the localizability of "physical" particles is not different from the localizability of "bare" particles discussed above. "Physical" electron is a point-like particle. One can easily write down a position operator whose eigenvectors represent localized states of the "physical" electron in the "dressed particle" approach.

So, in answering your question, I can say that there is no "de-localization due to dressing in the "dressed" particle approach", or, at least, I don't understand what you mean by that.

 

[Bob_for_short]

I am not sure I understand the question, but let me try to explain how I understand the issue of electron localization in QFT.

In the standard textbook QFT, the "bare" electron is a normal point-like particle. As I said above, its wave function can be either localized (delta function) or delocalized (plane wave), but the important point is that one can easily define the position operator corresponding to the single electron, so the "bare" electron is "localizable", just as in ordinary quantum mechanics.

As soon as we describe the electron with help of a wave function or coordinate operators rather than just coordinates, it is quantum mechanical with its inevitable smearing. I could understand if you would invoke the r-dependence of the interaction potential to say: "See, it depends on distance or position, thus the electron is point-like", but even then the QM smearing is involved in calculations. The simplest example is a charge form-factor: is is determined with the wave function.

Now, the "dressed particle" approach says: The interaction operator chosen in QFT is not good. The correct interaction between "bare" particles must have a trivial action (i.e., yield zero) on the vacuum and 1-particle states. If such a correct interaction is chosen, then "bare" states remain eigenstates of the full interacting Hamiltonian, and there is no difference between "bare" and "physical" particles. In this case the localizability of "physical" particles is not different from the localizability of "bare" particles discussed above. "Physical" electron is a point-like particle. One can easily write down a position operator whose eigenvectors represent localized states of the "physical" electron in the "dressed particle" approach.

Then my argumentation about non point-like electron is applicable again.

 

[meopemuk]

Then my argumentation about non point-like electron is applicable again.

It looks like we are arguing about terminology rather than substance. Of course the electron's wave function can take any shape from point-like (delta function) to completely delocalized (plane wave). So, one can say that the electron is not a point particle. How is it related to the original posting?

 

[Bob_for_short]

... So, one can say that the electron is not a point particle. How is it related to the original posting?

Directly. A real, observable electron and the quantized EMF form a compound system described quantum mechanically in the same way as an atom: with the center of inertia and relative coordinates. Such a dressed electron (I call it "electronium") is observable (not bare) and a potential interaction of such electrons is accompanied with radiation as naturally as atom excitation in atomic collisions. In this formulation there are no non-physical (non-observable) particles, vacuums, etc., and the Haag's theorem fails: the interaction does not bring problems with the Hilbert space.

 

[meopemuk]

Directly. A real, observable electron and the quantized EMF form a compound system described quantum mechanically in the same way as an atom: with the center of inertia and relative coordinates. Such a dressed electron (I call it "electronium") is observable (not bare) and a potential interaction of such electrons is accompanied with radiation as naturally as atom excitation in atomic collisions. In this formulation there are no non-physical (non-observable) particles, vacuums, etc., and the Haag's theorem fails: the interaction does not bring problems with the Hilbert space.

We already discussed your proposal several times. Few important yet unresolved issues remain. What is the Hamiltonian? How it was derived? Is it relativistically invariant? Can you calculate and compare with experiment scattering amplitudes for some simple collisions?

 

[Bob_for_short]

We already discussed your proposal several times. Few important yet unresolved issues remain. What is the Hamiltonian? How it was derived? Is it relativistically invariant? Can you calculate and compare with experiment scattering amplitudes for some simple collisions?

Yes, I can but these are questions to my results that belong to the Independent Research section. I cannot answer them here.

 

[DarMM]

All that's good and well, except for one thing. Why do we require the free vacuum to be invariant under Poincare transformations? The free fields aren't the physical ones, the interacting ones are. I see no reason why the non-physical free vacuum can't transform non-trivially under a translation or a rotation.
If, instead, we require only that the interacting vacuum be invariant under Poincare transformations, the mismatch between the Hilbert spaces disappears.

There are two things to consider here.
1)It is easy to prove that the free vacuum is invariant under Poincare transformation, so there isn't really anyway around the fact that it transforms in the required way.
2)For any field theory that has actually been rigorously constructed nonperturbatively the result is that they live in a different Hilbert space than the free theory.

Haag's theorem isn't really something to be got around. Perturbatively there is no problem with doing Fock space calculations to approximate the interacting theory. All it says is that nonperturbatively you'll require heavier mathematical machinery if you want to do things rigorously.

 

[Bob_for_short]

...Perturbatively there is no problem with doing Fock space calculations to approximate the interacting theory.

Except for statements like free particles are not observable, they have infinite parameters, etc.
It is like developing a function in Taylor series with saying that the zeroth order approximation is non observable. => There are severe problems in perturbative calculations in actual QFTs.

 

[DarMM]

Except for statements like free particles are not observable, they have infinite parameters, etc.

I don't understand how this is a problem. If the system is not free then you shouldn't observe free particles.

It is like developing a function in Taylor series with saying that the zeroth order approximation is non observable. => There are severe problems in perturbative calculations in actual QFTs.

I don't see how. The expansion makes sense once you know its rigorous justification and it agrees fantastically with experiment. So I don't think there are severe problems.

 

[Bob_for_short]

It don't understand how this is a problem. If the system is not free then you shouldn't observe free particles.

That is the problem: you do not understand.

Let us take the following experiment: an atom as a target and another atom as a fast projectile. We study scattering backwards, at large angles. We can write mechanical or quantum mechanical equations, whatever, containing all involved particles: electrons and nuclei. Then we solve these equations. Scattering backward occurs due to nucleus-nucleus (Coulomb or not) interaction because the target electrons cannot cause such a projectile deviation (i.e., at large angles). So we are tempted to neglect the electron-nucleus interaction in the zeroth approximation. At this stage we still have two different possibilities:

1) to take exactly into account a certain part of the electron-nucleus interaction in order to correctly represent the experimental initial and final conditions (neutral atoms in some states). The rest can be considered perturbatively.

2) But we can also take all electron-nucleus interactions perturbatively, so we do not have neutral atoms in the initial and final states in our theory. Nevertheless such a theory give good prediction for scattering at large angles (the Rutherford formula, for example). Then we try to take into account the remaining electron-nucleus interactions, in particular, "build" atomic wave functions perturbatively. Naturally here we encounter mathematical difficulties: divergences, etc. In order to overcome these difficulties we invent the notion of bare initial particles with infinite masses, etc. It is evident that such an invention has nothing to do with reality but is solely due to our incapability to correctly build the initial and final states (atoms) perturbatively.

On the other hand, we can use a better initial approximation (1) for our problem and obtain a correct cross section of atom-atomic scattering at large angles. In the latter approach there is no need in "bare" particle ideology at all. The atoms are not "free" but are interacting in our theory, yet there are no bare (non observable) particles with bizarre properties to cancel bizarre perturbative corrections. The initial and final states of (interacting = free) atoms are observable and the atom-atomic interaction can be considered perturbatively (remember, our projectile is fast). In this case there is no problem like the Haag's theorem.

Do you understand now where the difference comes from?

 

[DarMM]

Do you understand now where the difference comes from?

I must admit that I do not fully understand. I my mind, this is how things work:
A quantum field theory may (basically) be characterised by its set of Green's functions. These contain all the information in the theory. Now let's say I want to study particle scattering, the information on this is contained within the S-matrix. The S-matrix itself, by the LSZ formalism can be calculated as the on-shell poles of the Green's functions.

So basically all I need to do is calculate the Green's functions and I have the scattering predictions of the theory. Now since QFT is too difficult to do calculations with, one usually needs an expansion to calculate the Green's functions. The expansion we take is the one around a theory we can actually solve, namely a free field theory.

However because the interacting theories are singular with respect to the free theory we also need terms with infinite coefficients in the expansion. This is not a problem, it is simply a fact. QFTs are highly singular mathematical objects and unlike non-relativistic quantum mechanics two theories can be so mutually singular as to live in different Hilbert spaces, so your expansion must take this into account, hence the counterterms.

This has nothing really to do with Haag's theorem which is related to nonperturbative infrared divergences, not perturbative ultraviolet ones that show up in usual calculations.

I don't see what all this talk about "bare" particles is about. The difference between the "bare" particles, by which I assume you mean the free theory Fock space particles and the "real" particles, by which I assume you mean the asymptotically free particles in the real Hilbert space doesn't really show up in scattering calculations due to the whole LSZ amputation of the external legs. Where it will show up is in finite time calculations.

 

[Bob_for_short]

...A quantum field theory may (basically) be characterised by its set of Green's functions. These contain all the information in the theory.

One cannot hide oneself behind Green's functions. If one knows the exact solutions, one can build the exact Green's function and vice versa.

...However because the interacting theories are singular with respect to the free theory we also need terms with infinite coefficients in the expansion.

It is only so in theories with self-action. Renormalizations remove (not always though) the singular self-action contribution. So the net result belongs to another theory - with an interaction without self-action, just like in non relativistic QM. Remember, the interaction term is our guess and it can be improved. And renormalization is also our guess or try, it does not work automatically.

I don't see what all this talk about "bare" particles is about. The difference between the "bare" particles, by which I assume you mean the free theory Fock space particles and the "real" particles, by which I assume you mean the asymptotically free particles in the real Hilbert space doesn't really show up in scattering calculations due to the whole LSZ amputation of the external legs. Where it will show up is in finite time calculations.

No, the self-action gives (unnecessary) corrections to the particle constants involved into the Green's functions, so "the whole LSZ" approach is as plagued with these problems as any other (standard) QFT. These problems, including the Haag's theorem, are eliminated with a better choice of the initial approximation, just like in the atom-atomic scattering problem outlined in the previous post of mine (post #29).

 

[DarMM]

One cannot hide oneself behind Green's functions. If one knows the exact solutions, one can build the exact Green's function and vice versa.

Maybe I didn't make myself clear. I was only mentioning this as a set up to my discussion.

It is only so in theories with self-action. Renormalizations remove (not always though) the singular self-action contribution. So the net result belongs to another theory - with an interaction without self-action, just like in non relativistic QM. Remember, the interaction term is our guess and it can be improved. And renormalization is also our guess or try, it does not work automatically.

No theory has "self-action" contributions. That only results from a Feynman diagram type expression of the Wick contractions in perturbation theory. The full nonperturbative theory has no "self-actions". That is to say I can represent an interacting Green's function as a sum of convolutions of free Green's functions. These convolutions if interpreted literally would look like a particle interacting with itself, but why would you literally interpret a term in the perturbative expansion. Also renormalization is a guess as you say, however it has now been proven to work. Also a lot of non-relativistic QM systems also involve corrections to coupling constants, they're just not infinite like in QFT. In QM you will also have "bare" and "physical" coupling constants.

No, the self-action gives (unnecessary) corrections to the particle constants involved into the Green's functions, so "the whole LSZ" approach is as plagued with these problems as any other (standard) QFT. These problems, including the Haag's theorem, are eliminated with a better choice of the initial approximation, just like in the atom-atomic scattering problem outlined in the previous post of mine (post #29).

These things are not "problems", they are facts. The interacting theory lives in a different Hilbert space so the perturbative expansion is going to be very, very singular. I don't really see Haag's theorem as something to be gotten around, it makes perfect sense that free and interacting particles live in different reps because otherwise it would be possible for a real electron to evolve into a fictitious particle from a world with no photons, i.e. a free electron.

 

[Bob_for_short]

I have already expressed myself on this subject and I do not want to repeat my arguments.

 

[meopemuk]

The interacting theory lives in a different Hilbert space...

I often see this statement, but I am not sure about its validity. In my opinion, this statement goes against basic postulates of quantum theory. Let me explain why I think the Hilbert space used to describe a physical system should be independent on whether the system is interacting or not.

Let us first ask why we use Hilbert spaces to describe physical systems (their states and observables) in QM? The answer is given by "quantum logic". This theory tells us that subspaces in the Hilbert space are representatives of "yes-no experiments" or logical "propositions" or experimental "questions". Meets, joins, and orthogonal complements of subspaces represent usual logical operations OR, AND, and NOT. It seems reasonable to assume that the same questions can be asked about interacting and non-interacting system. The logical relationships between these questions should not depend on the interaction as well. Therefore, the same Hilbert space (= logical propositional system) should be applied to both interacting and non-interacting system, if their particle content is the same.

So, interaction has no any effect on the Hilbert space structure. Interaction only changes the representation of the Poincare group (the group of transformations between inertial observers) acting in this Hilbert space. The non-interacting system is described by one representation of the Poincare group. The interacting system is described by a different representation. In particular, this means that generators of the two representations (e.g., the Hamiltonians) are different.

 

[Haelfix]

Getting back to the original question, there are a few workarounds that I can remember, and mentioned in a previous thread about this subject. Generically they all relax one or more of the axioms of the theorem

1) If you introduce a volume cutoff, this explicitly breaks the covariance condition and the theorem does not hold
2) Renormalization by formally infinite counterterms (it does so by making such a mess of
the mathematics and being so illdefined, that it violates a number of Wightman axioms to start with, so its no surprise that it also evades Haags theorem)
3) Supersymmetry. One of the assumptions of the theorem is that the fields either obey commutation relations, or anticommutation relations, but not both. So strictly speaking it does not apply to SuSY I think
4) If the interacting theory lives in a different Hilbert space (the idea Eugene hates) also strictly speaking bypasses one of the assumptions.

Anyway, the point is that perturbative QFT that we all know and measure everyday is fine. But a real, sensible, nonperturbative definition of an interacting field theory is going to be difficult.

 

[meopemuk]

breaks the covariance condition

There are many different ways to formulate Haag's theorem. In one formulation one proves that "the interacting field does not transform covariantly", meaning that the field transformation law does not have the required form (for simplicity I consider the scalar field)

\phi(x) \to \phi(\Lambda(x+a))......(1)

where \Lambda, a are Poincare transformation parameters. This is usually regarded as a serious contradiction.

However I can not see what the problem is. OK, the field does not transform by simple formula (1). So what? Let it transform as it wish.

In any case, the violation of (1) does not mean a violation of relativistic invariance. As we know (see Weinberg's vol. 1) the relativistic invariance of a quantum theory is guaranteed by the existence of a unitary representation of the Poincare group in the Hilbert space of the system. Such an unitary representation can coexist happily with interacting fields transforming non-covariantly. Where is the alleged contradiction? Who said that interacting fields must transform covariantly?

 

[Hans de Vries]

Interacting QFT and SR are a perfect match.

Where it goes wrong is with "second quantization" which violates special relativity
already at the most elementary level. If one mixes "momentum" π defined in an
"internal space" or "unspecified space" with real 4-momentum in Minkowski space
then you can expect to get a mess...

It goes wrong from the start:

The Hamiltonian H, the Lagrangian L and the term pv of a classical relativistic
particle transform like:<br /> \begin{array}{rll}<br /> H &amp; ~~\mbox{transforms as} ~~&amp; \gamma \\<br /> pv &amp; ~~\mbox{transforms as} ~~&amp; \beta^2\gamma \\<br /> L &amp; ~~\mbox{transforms as} ~~&amp; 1/\gamma \\<br /> \end{array}<br />

The volume of a wave-function transforms like 1/γ due to Lorentz contraction.
So, the densities become higher by a factor γ, hence the Hamiltonian density,
the Lagrangian density and the density of the pv term transform like.<br /> \begin{array}{lll}<br /> {\cal H} &amp; ~~\mbox{transforms as}~~ &amp; \gamma^2 \\<br /> {\cal {\small PV}} &amp; ~~\mbox{transforms as}~~ &amp; \beta^2\gamma^2 \\<br /> {\cal L} &amp; ~~\mbox{transforms as}~~ &amp; &quot;1&quot; \\<br /> \end{array}<br />The Lagrangian density is Lorentz invariant and the expressions for the densities
become quadratic in nature, already in classical relativistic physics:<br /> {\cal H-PV = -L} &amp; ~~~~\mbox{transforms as}~~~~&amp; ~~~~E^2~~-p^2 =~~ m^2 <br /> &amp; ~~~~\mbox{which transforms as}~~~~ \gamma^2-\beta^2\gamma^2 = 1<br />Because E transforms like γ, p transforms like βγ and m transforms like "1".
This means that we can write familiar expressions like:<br /> \begin{array}{lllll}<br /> {\cal H} &amp; ~~~~\mbox{transforms as}~~~~&amp; \,+\tfrac12E^2 +\tfrac12p^2 +\tfrac12 m^2 ~=~ ~~E^2 <br /> &amp; ~~~~\mbox{which transforms as}~~~~&amp; \gamma^2 \\<br /> {\cal L} &amp; ~~~~\mbox{transforms as}~~~~&amp; -\tfrac12E^2 +\tfrac12p^2 -\tfrac12 m^2 ~=~ -m^2 <br /> &amp; ~~~~\mbox{which transforms as}~~~~&amp; &#039;1&#039;<br /> \end{array}<br />In second quantization this gets messed up by making replacements like(mv)^2 \implies \dot{\varphi}^2where mv is the (non-relativistic!) momentum in an "internal" or "unspecified"
space. The latter term does not transform like the pv density but like:<br /> \begin{array}{lll}<br /> \dot{\varphi} &amp; ~~~~\mbox{transforms as}~~~~&amp; \gamma \\<br /> \dot{\varphi}^2 &amp; ~~~~\mbox{transforms as}~~~~&amp; \gamma^2 \\<br /> \end{array}<br />and so the standard expression.<br /> {\cal H}-\dot{\varphi}^2 = {\cal -L} <br /> already violates special relativity at the most basic level leading to wrong
expressions for either the Hamiltonian or Lagrangian densities:<br /> \begin{array}{lllll}<br /> {\cal H} &amp; ~~~~\mbox{transforms (not) as}~~~~&amp; \,+\tfrac12E^2 +\tfrac12p^2 -\tfrac12 m^2 ~=~ p^2<br /> &amp; ~~~~\mbox{which transforms as}~~~~&amp; \beta^2\gamma^2 \\<br /> {\cal L} &amp; ~~~~\mbox{transforms (not) as}~~~~&amp; -\tfrac12E^2 +\tfrac12p^2 +\tfrac12 m^2 ~=~0 <br /> &amp; ~~~~\mbox{which transforms as}~~~~&amp; &#039;0&#039; \\<br /> \end{array}<br />The signs of the invariant mass terms are changed compared with the
correct versions. The quantities differ by a constant value which generally
goes unnoticed but spoils the way they transform.

Many textbooks use the form of the Lagrangian density which is zero for
all non-interacting eigenstates. Zero is Lorentz invariant but is not correct.Regards, Hans

 

[Bob_for_short]

...Many textbook use the form of the Lagrangian density which is zero for all non-interacting eigenstates...

How a Lagrangian can be zero if it contains unknown variables?

Let us note that nobody is interested in calculations (values) of action or Lagrangian expressed via solutions.

 

[Hans de Vries]

How a Lagrangian can be zero if it contains unknown variables?

Let us note that nobody is interested in calculations (values) of action or Lagrangian expressed via solutions.

You would have noticed how the Lagrangian densities, as typically presented,
for the Klein Gordon field or the Dirac field become zero for free particle solutions
IF you would have been simply interested enough to do the calculations...Regards, Hans

 

[Bob_for_short]

You would have noticed how the Lagrangian densities, as typically presented, for the Klein Gordon field or the Dirac field become zero for free particle solutions IF you would have been simply interested enough to do the calculations...

I do not argue that substituting the solutions in L makes it zero, but first, it is not Lagrangian any more, next, why should I need to perform such a calculation? I need solutions, so I use the Lagrangian to derive the equations. In this case any Lagrangian contains unknown variables and cannot be zero by definition.

 

[DarMM]

I often see this statement, but I am not sure about its validity. In my opinion, this statement goes against basic postulates of quantum theory. Let me explain why I think the Hilbert space used to describe a physical system should be independent on whether the system is interacting or not.

Let us first ask why we use Hilbert spaces to describe physical systems (their states and observables) in QM? The answer is given by "quantum logic". This theory tells us that subspaces in the Hilbert space are representatives of "yes-no experiments" or logical "propositions" or experimental "questions". Meets, joins, and orthogonal complements of subspaces represent usual logical operations OR, AND, and NOT. It seems reasonable to assume that the same questions can be asked about interacting and non-interacting system. The logical relationships between these questions should not depend on the interaction as well. Therefore, the same Hilbert space (= logical propositional system) should be applied to both interacting and non-interacting system, if their particle content is the same.

That's an excellent point meopemuk, saying the interacting theory lives in a different Hilbert Space is in truth a very lazy way of saying what is really going on.

As you said all questions that can be asked about the free theory can also be asked about the free theory. These “questions” form the C*-algebra of observables if I may be very formal.
Just like QM, the interacting case and the free case give different answers to these questions. The only difference is that for QFT there exists no unitary transformation to take you from the free answers to the interacting answers, which is the basic content in Haag's theorem. To paraphrase F. Strocchi this is because the interacting and free answers correspond to “two totally separate worlds”. It would be impossible to prepare a system which gives “free” answers to questions and have it evolve into one which gives “interacting” answers, unlike regular QM where this can be done with the right choice of Hamiltonian.

The reason people usually say they live in different Hilbert spaces is because even the though the Hilbert space is formally identical (all seperable Hilbert spaces are isomorphic), the algebra of observables acts in a different way. So we say they are two different reps of the algebra.

 

[DarMM]

Getting back to the original question, there are a few workarounds that I can remember, and mentioned in a previous thread about this subject. Generically they all relax one or more of the axioms of the theorem

1) If you introduce a volume cutoff, this explicitly breaks the covariance condition and the theorem does not hold
2) Renormalization by formally infinite counterterms (it does so by making such a mess of
the mathematics and being so illdefined, that it violates a number of Wightman axioms to start with, so its no surprise that it also evades Haags theorem)
3) Supersymmetry. One of the assumptions of the theorem is that the fields either obey commutation relations, or anticommutation relations, but not both. So strictly speaking it does not apply to SuSY I think
4) If the interacting theory lives in a different Hilbert space (the idea Eugene hates) also strictly speaking bypasses one of the assumptions.

Anyway, the point is that perturbative QFT that we all know and measure everyday is fine. But a real, sensible, nonperturbative definition of an interacting field theory is going to be difficult.

I just wanted to say a few things in relation to this specifically renormalization and Haag's theorem.
First of all, in the field theories that have been nonperturbatively constructed the free and interacting theory do live in separate Hilbert spaces, e.g. \phi^{4} in two or three dimensions do not live in Fock space. From this point of view the field theories living in different Hilbert space or being in different reps of the algebra of operators is the reason for Haag's theorem. There can't be a unitary relation between them because they're in different reps.
Renormalization enters in this picture as a way of fixing the mismatch. The free and interacting field theory live in separate Hilbert spaces so the differences between them involve very singular objects which physicists call counterterms.

 

[Bob_for_short]

...The reason people usually say they live in different Hilbert spaces is because even the though the Hilbert space is formally identical (all seperable Hilbert spaces are isomorphic), the algebra of observables acts in a different way. So we say they are two different reps of the algebra.

Now let us have a look at a non-interacting QFT. Its initial state (particle populations) does not change with time. The initial superposition remains intact with time.

When we have a physical interaction, the superposition coefficients become time-dependent and the populations change. After scattering the populations reach their new constant values and the theory becomes non-interacting again. So we stay within the same Hilbert space.

When your calculations give infinite corrections to the superposition coefficients, it is not because of "too strong interaction at short distances" but because of too poor initial approximation and wrong interaction term.

It is not correct to extrapolate the Haag's theorem obtained for badly guessed theories to good theories with physical interactions.

 

[strangerep]

It is not correct to extrapolate the Haag's theorem obtained for badly guessed theories to good theories with physical interactions.

That's a rather sweeping statement. Which formulation of Haag's theorem
did you have in mind? The usual Hall-Wightman form doesn't mention
specific interacting theories. Here is a statement of the theorem in
H-W form (from Shirokov's math-ph/0703021, p7):

Suppose we have two field theories. One is a free theory described by a
set of free fields A_0(x) acting in the Hilbert space \mathcal{H}_0.
The other is described by an irreducible set of fields A(x). Further, let us
assume that the following conditions are satisfied:

1) A(x) is an operator in \mathcal{H} which
carries a unitary representation of translations and rotations

<br /> U(\mathbf{a},R) A(x) U^{\dag}(\mathbf{a},R) ~=~ A(Rx + \mathbf{a})<br />

and

1') Lorentz transformations

<br /> U(\Lambda) A(x) U^{\dag}(\Lambda) ~=~ A(\Lambda x) ~.<br />

(These relationships are written for the particular case
of a scalar field.)

2) There is a unique invariant state U \Omega ~=~ \Omega in \mathcal{H}/itex].<br /> <br /> 3) There exists a unitary operator V, from \mathcal{H}_0 to<br /> \mathcal{H}, such that at a time instant t we have<br /> <br /> &lt;br /&gt; A(\mathbf{x}, t) ~=~ V(t)A_0(\mathbf{x}, t) V^{\dag}(t)&lt;br /&gt;<br /> <br /> 4) The spectrum of energies is bounded from below.<br /> <br /> Then A(x) is a free field.<br />

<br /> <br /> ------------------------------------------<br /> <br /> The same paper (pp5-6) also gives an alternate formulation<br /> of the theorem as follows:<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> <b>Lemma.</b> Suppose we have a Euclidean (i.e., translational and<br /> rotational) invariant field theory written in terms of creation-annihilation<br /> operators a^{\dag}_{\mathbf{p}}, a_{\mathbf{p}}. Suppose also that a Fock<br /> representation of these operators in the Hilbert space \mathcal{H}_0 with the<br /> no-particle vector \Omega_0 is given. Then, in \mathcal{H}_0 there is a<br /> unique normalizable eigenstate of the total momentum operator \mathbf{P},<br /> and this state coincides with \Omega_0.<br /> <br /> Proof: (See the paper.)<br /> <br /> <b>Theorem.</b> Suppose that conditions of the Lemma are satisfied and<br /> there is a unique normalizable eigenstate of the full Hamiltonian<br /> H with lowest energy, i.e., the vacuum vector \Omega.<br /> Then \Omega must coincide with \Omega_0.<br /> <br /> Proof. Since [H, P_j]= 0, \Omega must be a common<br /> eigenvector of H and \mathbf{P}. However there is<br /> only one normalizable eigenstate of \mathbf{P} in<br /> \mathcal{H}_0, and this eigenstate coincides with<br /> \Omega_0. Thus \Omega = \Omega_0. (End of proof.)<br /> <br /> In fact, in all local theories \Omega does not coincide with the no-particle<br /> vector of ”bare” creation-annihilation operators which diagonalize H_0.<br /> This means such theories violate some assumptions of the theorem. </div> </div> </blockquote><br /> ------------------------------------------<br /> <br /> Neither formulation relies on any specific details of the<br /> interaction term in the Hamiltonian (except perhaps locality).<br /> <br /> To evade Haag&#039;s theorem, one must either state which of its input assumptions should be abandoned or modified, or demonstrate that the maths in the existing proofs are somehow erroneous. E.g., Shirokov goes on to show that condition (1&#039;) is not valid in his &quot;dressed field&quot; approach.

 

[meopemuk]

To evade Haag's theorem, one must either state which of its
input assumptions should be abandoned or modified, or demonstrate
that the maths in the existing proofs are somehow erroneous.
E.g., Shirokov goes on to show that condition (1') is not valid in his
"dressed field" approach.

Hi strangerep,

I agree with Shirokov on this point. As I wrote earlier, condition (1') does not follow from any deep physical principle, so it should be abandoned.

In this connection I would like to mention the work

H. Kita, "A non-trivial example of a relativistic quantum theory of particles without divergence difficulties", Progr. Theor. Phys., 35 (1966), 934.

in which a perfectly valid and relativistically invariant interacting QFT model is constructed, "interacting fields" are explicitly calculated, and it is demonstrated that the condition (1') does not hold.

 

[DarMM]

Hi strangerep,

I agree with Shirokov on this point. As I wrote earlier, condition (1') does not follow from any deep physical principle, so it should be abandoned.

In this connection I would like to mention the work

H. Kita, "A non-trivial example of a relativistic quantum theory of particles without divergence difficulties", Progr. Theor. Phys., 35 (1966), 934.

in which a perfectly valid and relativistically invariant interacting QFT model is constructed, "interacting fields" are explicitly calculated, and it is demonstrated that the condition (1') does not hold.

Two things to note are:
(a) All models which have been constructed nonperturbatively do satisfy the covariant transformation law. For instance in all scalar and Yukawa theories in two and three dimensions, the Gross-Neveu model in two and three dimensions and also QED and the Higgs model in two dimensions, the interacting field transforms covariantly.
(b) Kita's model breaks another Wightman axiom besides the covariant transformation law axiom, so it may not have all the properties associated with a QFT. Although the model is still very interesting. Similar models also appear in Algebraic quantum field theory, where the "field" is not covariant, in fact it can even be nonlocal. However the problem is that only theories which satisfy the Wightman axioms have been proven to have all the healthy properties normally associated with a QFT.

 

[Bob_for_short]

...That's a rather sweeping statement. ...
Neither formulation relies on any specific details of the interaction term in the Hamiltonian (except perhaps locality).

To evade Haag's theorem, one must either state which of its input assumptions should be abandoned or modified, or demonstrate that the maths in the existing proofs are somehow erroneous. E.g., Shirokov goes on to show that condition (1') is not valid in his "dressed field" approach.

Concerning Wightman's axioms, maybe it was too early to axiomatize something? The theory is not ready yet, but one hurries up to axiomaize or impose his vision to it.

Concerning Haag, it is funny: you take a filed A(x,t) and it happens to be free whatever interaction you use. Maybe Haag himself should have pointed out where his proof was interaction-dependent?

Here I would like to draw you attention to another reading of free fields. For the sake of sloppiness, I will take a Dirac field ψ and the quantized electromagnetic filed A_tr and I will write the free equations in a lazy way:

(γ∂+m)ψ = 0, ∂²A_tr = 0

These are two independent equations that in the standard QED are "coupled" with an "interaction" term jA. As soon as jA includes the self-action and non-linearity, one obtains problems with vacuum instability and other self-action provoked rubbish. This is a dead end in the theory development. There are too few things to axiomatize in it.

Now let us look at the free equations as at equations of the center of inertia of a compound system and internal motion equations. They should be independent - they describe independent degrees of freedom of one compound system. So there in no need to couple them if there is no external force acting on the charge.

An external force acting on a charge makes two independent (=additive) works: it changes the CI energy-momentum and it pumps the internal degrees of freedom (oscillations). In presence of an external filed A_ext the equations are modified with the external filed contributions. The equation system is then essentially linearized. One can be sure that the physical solutions exist.

In this description the main problem is to express the charge coordinates via CI coordinates and relative ones. This expression should be dictated with experimental facts first of all rather than with axioms of somebody's. Then the theory becomes a routine theory of compound systems without any mathematical and conceptual problems. After having been developed and verified experimentally, it might serve as a model for axiomatization although I personally am against axiomatization. Only experiment can give an idea how other theories (strong, weak, gravity forces) could be constructed. I mean, for example, what quasi-particles exist in these compound systems and how they are excited.

I think this, not yet explored direction is worth developing and practising.
.

 

[Avodyne]

The theorem says that if A_0(x) is a free field of mass m_0, and V(t) is a unitary operator, then A(x) = V(t)A_0(x) V^{\dag}(t) is also a free field of mass m_0.

But suppose we consider adding an extra A_0^2 term to the Hamiltonian to shift the mass, and take this extra term as an interaction that we treat perturbatively. We could then construct the usual Dyson series for V(t), and this should transform A_0(x) with mass m_0 into A(x) with mass m. According to the theorem, though, such an operator does not exist! So what goes wrong?

Consider the vacuum state of the original hamiltonian. Since the field is free, we can describe it in terms of noninteracting momentum modes, each of which is a harmonic oscillator. The ground state wave function of each mode is a gaussian in the Fourier-transformed field mode, with a width controlled by the energy E=({\bf k}^2+m_0^2)^{1/2}, and the ground state wave functional of the field theory is the product of these individual mode wave functions. Of course, after the mass is shifted, the ground state wave functional looks the same, except m_0 is replaced by m.

Now, the point is that the inner product of these two grounds states is zero. This is because the mode wave functions are not the same (one has m_0 in it and the other m), so the inner product of the two normalized mode wave functions (with the same three-momentum \bf k) is less than one. Now we have to take the product of an infinite number of numbers, each less than one, and so we get zero. The same argument shows that the ground state with the shifted mass is orthogonal to every state in the Fock space of the original field. So there is no unitary map from one set to the other.

But this is clearly a function of having used an infinite number of modes. If we provide infrared and ultraviolet cutoffs, so that the number of modes is finite, the inner products will be finite, and the unitary operator will exist. Lorentz invariance requires an infinite number of modes, and that's why it's a condition of the theorem.

I believe it is not a coincidence that the only general method we know for nonperturbative regularization of QFT is the lattice, which of course breaks Lorentz symmetry. Lorentz symmetry then emerges only in the limit in which the regulator is removed.

 

[Bob_for_short]

...I believe it is not a coincidence that the only general method we know for nonperturbative regularization of QFT is the lattice, which of course breaks Lorentz symmetry. Lorentz symmetry then emerges only in the limit in which the regulator is removed.

By saying "QFT" you imply a certain interaction term, that is the problem, not Lorentz invariance.

 

[Haelfix]

Thats sort of the point, there is no obstruction for doing QFT 'as done by physicists', say treated as a lattice in the Wilsonian framework. The problem is the existence of the 'thing' we are actually approximating, which may or may not exist mathematically in general, even if we have a handful of explicit examples of fully nonperturbative models (usually in 2d)

Its wonderfully miraculous that the whole business works at all.

I think a lot of physicists became a lot less worried about Haags theorem after Wilsons work on effective field theory, where it became clear that in general there might need to be a UV completion for most phenomenological field theories of interest.