# Haag's Theorem loophole?

1. Sep 3, 2009

### 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" [Broken])

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.

Last edited by a moderator: May 4, 2017
2. Sep 3, 2009

### strangerep

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?

3. Sep 3, 2009

### Bob_for_short

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.
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.

Last edited: Sep 3, 2009
4. Sep 3, 2009

### Riposte

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.

5. Sep 3, 2009

### Riposte

Wrong interaction term? What do you mean by this?

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

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.

6. Sep 3, 2009

### Bob_for_short

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

7. Sep 3, 2009

### meopemuk

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

8. Sep 3, 2009

### Bob_for_short

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).

9. Sep 3, 2009

### javierR

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?

10. Sep 3, 2009

### strangerep

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....

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.

11. Sep 4, 2009

### meopemuk

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.

12. Sep 4, 2009

### Bob_for_short

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

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.

13. Sep 4, 2009

### meopemuk

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.

14. Sep 4, 2009

### 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?

15. Sep 4, 2009

### meopemuk

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.

16. Sep 4, 2009

### meopemuk

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.

17. Sep 4, 2009

### Bob_for_short

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.

18. Sep 4, 2009

### meopemuk

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.

19. Sep 4, 2009

### Bob_for_short

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?

20. Sep 4, 2009

### meopemuk

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.

21. Sep 5, 2009

### Bob_for_short

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

22. Sep 5, 2009

### meopemuk

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?

Last edited: Sep 5, 2009
23. Sep 5, 2009

### Bob_for_short

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.

24. Sep 5, 2009

### meopemuk

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?

25. Sep 5, 2009

### Bob_for_short

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