# Rigorous Quantum Field Theory.

1. ### DarMM

331
Hey everybody, since the previous thread got locked I thought I would open this thread as a place to discuss rigorous issues in quantum field theory, be it on the constructive or axiomatic side of things.

I apologize if one is not supposed to start a discussion with posts from old threads, but I thought it would be a nice way to get the discussion going.

Yes, that is essentially my understanding. The upshot is that at the end you've constructed the perturbative expansion for $$S(g)$$ (the S-matrix in finite volume) in a completely rigorous way. Modern work on the Epstein-Glaser approach tries to take the limit $$g \rightarrow 1$$, to go to infinite volume, although it has proven extremely difficult.

Yes, the Hamiltonian is not a well-defined finite operator on Fock space, but in most QFTs it can be shown to be a well-defined finite operator on another Hilbert space. Do you view this as a problem?

For the two-dimensional models the best reference is:

Segal, I. Notes towards the construction of nonlinear relativistic quantum elds
I: The Hamiltonian in two spacetime dimensions as the generator of a C* -
automorphism group", Proc. Nat. Acad. Sci. U.S.A. 57, 1178-1183.

For the proof that general field theories have signals propogating at the speed of light a good read might be the treatise on algebraic quantum field theory:
Araki, H. Mathematical Theory of Quantum Fields, Oxford, 2000.

For several nonperturbatively constructed field theories there are actual calculations showing this to be the case. If you want references I can provide them.
I understand you are skeptical because there isn't a well-defined finite Hamiltonian on Fock space, however there is one on the correct Hilbert space.

2. ### meopemuk

Yes, I view this as a problem (though this could be a problem of my ignorance).

As far as I understand, when doing routine S-matrix calculation in QFT, we are working in the normal Fock space. We calculate matrix elements of the S-operator with states like $a^{\dag}a^{\dag} |0 \rangle$ where $|0 \rangle$ is the free vacuum and $a^{\dag}$ are creation operators for free particles. These calculations are perfectly successful and agree with experiment.

Now, we seem to agree that renormalized QED does not provide a finite (Fock space) Hamiltonian, that is consistent with the S-matrix above. In order to define a Hamiltonian you are suggesting to go to another Hilbert space with different vacuum and different creation operators $A^{\dag}$. There, presumably, we will be able to calculate the finite time evolution of states and observables. If we can do that, then why don't we extend the time interval to infinity and get the S-matrix as well? If our finite time evolution is unitary and physically reasonable, then the S-matrix should come out OK too. Then why don't we forget about the original Fock space and move our entire formalism in your new Hilbert space? This Hilbert space has all we need: a finite Hamiltonian and a finite and accurate S-matrix. Then we will avoid the confusing necessity of dealing with two Hilbert (Fock) spaces.

Why this one-Hilbert-space description is not possible? Is it because asymptotic particles (at large distances where the interaction can be neglected) live in the non-interacting Fock space, while the same particles at short distances "migrate" to the interacting Hilbert space? How this "migration" occurs? Shouldn't we also consider some "intermediate" Hilbert spaces? Does the Hilbert space of interacting system change with time?

I just can't comprehend the overall picture.

I find it much easier to understand a theory in which entire life of an interacting system of particles is spent in the same fixed Hilbert (or Fock) space, where (just as in ordinary quantum mechanics) we have a fixed Hamiltonian and other operators of observables; the time evolution operator and the S-operator are related to each other by usual QM formulas. Such a theory can easily accomodate interactions which change the number of particles. The only limitation is that interaction should not contain "bad" terms like $aaa + a^{\dag}aa +a^{\dag} a^{\dag} a + a^{\dag}a^{\dag}a^{\dag}$. But the absence of such terms in the Hamiltonian is perfectly justifiable, because nobody has seen a physical process in which, e.g., three particle spontaneously appear out of vacuum. Without these "bad" terms, there is no distinction between "bare" and "dressed" (or "physical" or "interacting") particles. So there is no need to struggle with (at least) two Hilbert spaces.

Eugene.

P.S. Thank you for the references, I will check them out.

Last edited: Nov 10, 2009
3. ### meopemuk

DarMM,

by the way, thanks for opening this new thread. I really appreciate the discussion.

Eugene.

4. ### meopemuk

Bob_for_short,

I see your point: When two charges collide, many soft photons must be emitted, and the probability of scattering without such an emission should be negligibly small. On the other hand, the lowest order QED predicts a sizeable probability (cross section) for no-photons events. I don't know how to answer this question. I hope it is handled satisfactorily, when IR divergences are discussed, but I am not good at those.

Eugene.

5. ### strangerep

2,102
(Sigh.) Bob, you're committing the same impoliteness again: slowly hijacking someone
else's thread off in a tangential direction. DarMM clearly intended this thread to
concentrate on constructive/axiomatic/rigorous QFT.

thread if necessary) so that any followup answers can appear there, with
redirections to your thread in the IR forum. That would be more polite to others.

6. ### strangerep

2,102
Thank you. I was indeed a bit disappointed that, while I slept in my part of the world,
my other thread became derailed enough to get it shut down.

I think it is a good, productive way to focus on particular items arising in another thread.

What exactly do you mean here by "most QFTs"? Those that rigorously exist, or
other stuff like $\phi^4$, QED, etc (in 4D)?

I see on Amazon that a much cheaper paperback version was released recently (July 2009).
I have ordered it, although the book is apparently for "mathematics graduates".
I'm hoping I now know enough topology and functional analysis to get by.

17,346
Staff Emeritus
Strangrep is absolutely right. If this thread is hijacked too, it too will be closed. Please stick to the subject.

8. ### DarMM

331
I mean the ones which have either been completely rigorously constructed, or at least constructed far enough to have their Hamiltonian defined.
$\phi^4$ in 4d is one such theory, the only question left is if it is trivial or not, not if it exists. If it is not trivial, then it can be shown to have a Hamiltonian defined on another Hilbert space.
Virtually all theories in 2d and 3d have been shown to have this property.

Believe it or not the book is not that difficult, especially considering it is a translation of a Japanese original. Dr. Carow-Watamura must be congratulated for an excellent translation. Araki introduces the terminology when relevant and the appendicies are very well written. The only difficult mathematics in the book is operator algebras, but the book basically doubles as an intro to operator algebra so this doesn't matter too much.

For the causal properties of QFTs you should look at sections four and six.

9. ### Avodyne

1,270
$\phi^4$ in 4d has been proved to be trivial: http://arxiv.org/abs/0808.0082

10. ### DarMM

331
This is not a proof in the sense that I mean. For instance it is difficult to analytically control the renormalization group in a mathematically rigorous way. Basically most treatments ignore something which is called the large field problem, which is what makes the renormalization group so difficult to treat rigorously. The large field problem is essentially that the renormalization group, even after a very short flow will produce a nonpolynomial action whose behaviour at very large fields $$\phi$$ is not known and hence may not be integrable and the flow will halt. Papers such as the above ignore this problem.

There is nothing wrong with such ways of demonstrating the triviality of $$\phi^{4}$$, Luscher and Weisz and Wolff have all given excellent numerical demonstrations. However it still does not constitute a mathematical proof of triviality.

11. ### DarMM

331
Yes, that is correct.

Indeed, however I shall stick to QFT in general rather than just QED in my answers below.

Yes, another Hilbert space which is a different representation of the creation/annhilation operator algebra or canonical commutation relations, which ever you prefer.

It does indeed. We can take the time interval to infinity and obtain a well-defined S-matrix which agrees with the one from usual caculations. References on this available if you wish.

Now we come to the main issue. In the correct Hilbert space everything is defined, the Hamiltonian, the S-matrix, finite time evolution, e.t.c. The problem is that without very high powered mathematics, particularly the representation theory of operator algebras and functional integration, you cannot explicitly construct the correct Hilbert space. Fock space is the only rep which is easy to find/construct. Constructive Quantum Field Theory is basically the theory of constructing the interacting Hilbert space.
So once we find the correct Hilbert space everything can be done in one single space, however the kilometer high wall of advanced mathematics standing in the way of finding this Hilbert space basically forces this to be a concern only of mathematical physics.
Also you can show that the perturbative expansions of the S-matrix on the true Hilbert space can be written using Fock space expectation values, or to put it another way Fock space can simulate the real Hilbert space well enough for perturbative calculations. So standard perturbation theory has a mathematical backing.

By the way if anybody thinks this is too canonical, all this can be translated in path integral language:
Free Hilbert space is different to interacting Hilbert space
becomes
Free field measure on field path space is mutually singular with respect to the interacting measure.

Hopefully my answer above has explained things. It is possible, just prohibitively difficult. The Clay mathematics institute Yang-Mills problem is basically asking you to construct/find the correct Hilbert space for 4D Yang-Mills theory.

12. ### meopemuk

If you don't mind I would like to learn a bit more about the interacting Hilbert space. Let me know where my logic fails.

Let us assume that we have constructed this Hilbert space somehow. If it represent physics correctly, then it should have some specific properties. For example, it should have a particle interpretation, i.e., we can identify 0-particle, 1-particle, 2-particle, etc. states there. It seems natural that N-particle states are orthogonal to M-particle states. Therefore the Hilbert space can be divided into orthogonal N-particle sectors. Also, there must be an unitary representation of the Poincare group. I presume that both 0-particle and 1-particle sectors are invariant with respect to this representation. This is because in experiments we don't see spontaneous creation of particles from the vacuum and 1-particle states (suppose that we are talking about stable particles only). Furthermore, based on this structure, we should be able to define particle creation and annihilation operators, which change the number of particles by +1 or -1. Finally, the Hamiltonian (and other operators) should be expressable as a polynomial in these creation and annihilation operators.

In other words, we get all relationships the same as in the usual Fock space. (These relationships are forced upon us by physics. If some of them fails, there should be observable consequences.) Where is the difference? Why I am not allowed to do the same stuff in my favorite Fock space? Why can't I say that the interacting Hiolbert space and the Fock space are simply equivalent?

Eugene.

13. ### DarMM

331
Remember in an interacting theory a one particle state could decay into two particles, so there is no reason to expect them to be orthogonal. Of course you mention later that we are discussing an interacting theory with only stable particles, so this doesn't matter too much.
Yes this is the same, although I think the one-particle space is covariant.
This is where you run into a problem. There is no state which is annihilated by all annihilation operators. There is still a vacuum, i.e. a state with no energy or momentum which is invariant under Poincaré transformations, but there is no "lowest state on the ladder" for creation and annihilation operators.
You can define creation and annihilation operators for which this isn't a problem, but they have a different algebra. So you can say it two ways:
The interacting Hilbert space is a non-Fock representation of the usual creation and annihilation operators
or
It is a Fock representation of unusual creation and annihilation operators.

Last edited: Nov 13, 2009
14. ### meopemuk

This is what I don't understand.

Suppose that I have divided my Hilbert space into N-particle sectors. For each sector I have a basis of vectors with defined particle momenta. For example, in the 1-particle sector the basis is $|p1\rangle$. In the 2-particle sector the basis is $|p1, p2 \rangle$ etc. Then I can simply define my creation and annihilation operators by going through all basis vectors and postulating how the operators act on these vectors. For example, the annihilation operator $a(p1)$ is *defined* by the following relationships

$$a(p1) |0 \rangle = 0$$
$$a(p1) |p1\rangle = |0 \rangle$$
$$a(p1) |p2\rangle = 0$$
$$a(p1) |p1, p2\rangle = |p1 \rangle$$
...........

The action on vectors not belonging to the basis is extended by linearity. Furthermore, I can simply check that thus defined operators satisfy usual (anti)commutation relations. Where is my mistake?

Eugene.

15. ### Avodyne

1,270
You cannot do this, because the "number of particles" is not a well-defined concept. Even in a theory like $\varphi^4$ with a single type of particle that is stable, you can have pair creation, e.g. a process in which 2 high-energy particles scatter into 4.

Consider (for simplicity) states with zero 3-momentum. What are the allowed energies? Answer: E=0, E=m, and E>=2m. The E=0 state is the vacuum (no particles), the E=m state is one particle (at rest, because we have assumed zero 3-momentum), and the states with E>=2m form the multi-particle continuum. Consider, for example, a state with E=4m. This could be 4 particles at rest, or it could be 2 particles with equal and opposite 3-momenta such that their energies add up to 4m. But there is no reason for these states to be orthogonal, because we know one can scatter into the other. So what are the right basis states in the multiparticle continuum? This is the hard part of the problem.

But the theories that make sense perturbatively and have physical particles like those in the corrsponding free theory ($\varphi^4$, QED) are almost certainly trivial, which almost certainly means that they can't be rigorously constructed (at least IMO).

16. ### meopemuk

Pair creation does not mean that the particle number operators do not exist. It simply means that these operators do not commute with the total Hamiltonian, i.e., particle numbers are not conserved.

Yes, I agree with this.

Here we disagree. In experiments the number of particles behaves as any any other observable. So, there should exist an Hermitian operator corresponding to this observable. Different eigenvalues of this operator (i.e., different numbers of particles) must correspond to orthogonal eigen-subspaces.

The scattering matrix element between two orthogonal states is generally non-zero. So, two states can be orthogonal and still one can scatter into the other. So, your explanation is not convincing.

From physical requirements, N-particle sectors should be constructed as tensor products of N 1-particle subspaces. This also determines the basis in such sectors. Physics does not leave much frredom in the construction of N-particle states.

Eugene.

17. ### Dmitry67

So, how many particles proton consists of?
I've heard about something called "scale problem" where the number of particles in QCD is different on different scale.

+Unruh effect, where the number of particles depend on frame.

18. ### meopemuk

In ordinary QM the number of particles is regarded as a valid observable, i.e., the corresponding operator N is Hermitian (with real eigenvalues 0,1,2,3,...) and its eigensubspaces are orthogonal. The (expectation value of the) number of particles can depend on time (e.g., in decays). This simply means that the operator N does not commute with the Hamiltonian. The proton (which is an eigenstate of the interacting Hamiltonian) may be not an eigenstate of N. This is not unusual. The operator N may also not commute with generators of boosts, so the number of particles could be frame-dependent as well. This is not unusual also.

If I understand Avodyne correctly, then in rigorous 2D QFT models the particle number operator either does not exist or is not Hermitian. This seems rather strange and unphysical to me. I would like to learn more about it.

19. ### DarMM

331
I'm not sure what is meant here, in Quantum Mechanics isn't the total number of particles fixed?

Let's start here. Given the Hilbert space of a relativistic quantum field theory, how would you go about dividing it into N-particle sectors? That is, how would you build the number operator?

20. ### Bob_for_short

Any particular atomic state ψn can be considered as a different "particle" in a narrow sense. After scattering one has a superposition of all ψn' allowed by the conservation laws. One can say (in a narrow sense) that in atom-atomic scattering the "particles" transform into each other. If the transferred energy is sufficient for ionization, then the number of different possible final n' is infinite.

Last edited: Nov 14, 2009