Bosons and Fermions in a rigorous QFT

kof9595995

But here if you creat the Hilbert space using creation operators, it might not be the same with the true Hilbert space, then the best you can get is a perturbation theory.

kof9595995

I'd like the ask the question again, if a rigorous QED doesn't exist, what would be underlying theory of perturbative QED?

tom.stoer

At least in QM tis is not true.

Which states do I miss?

I can construct nearly arbitrary operators from the creation and annihilation operators; look at the method of coherent states, for example.

And please not that up to now we haven't defined the Hilbert space, neither for x and p, nor for the creation and annihilation operators! So if we need to change the entire Hilbert space for some reason we are not forced to do this by introducing creation and annihilation which are nothing else but linear combinations of x and p. There is physics behind it so far.

kof9595995

I kinda get what you mean, but I think in QFT the situation is quite different, and I'm more curious about QFT senario.
In QM, before we define any operator, we set the full Hilbert space as all square-integrable functions, and we define how x and p (so are a and a+) act on this full Hilbert space, then according to the specific Hamiltonian we have, we can construct the subspace using a and a+, though the construction may not be as simple as harmonic oscillator case. I guess this is what you mean.
However in an interacting QFT, the true structure of the full Hilbert space is not clear yet, let alone defining how a and a+ act on the space. The normal treatment is to use the Hilbert space from free theory (a Fock space) and then do things perturbatively. And my point is, if the interaction actually destroys any existence of direct-product structure of the true Hilbert space, then whatever kind of Fock space you use will be just an approximation at best, and if this is the case, I think Fock space does necessarily imply a perturbative approach.

Physics Monkey

Lattice gauge theory or an asymptotically free gauge theory in a partial higgs phase can provide a UV completion of low energy QED.

atyy

Googling "Grassmann Osterwalder-Schrader" produced these papers which may be relevant.

Jaffe, Constructive Quantum Field Theory
"The Euclidean methods also apply to theories with fermions, at least for examples with interactions that are quadratic in the fermions. This is the case for free and for “Yukawa type” interactions, used extensively in physics."

Benfatto, Falco, Mastropietro, Functional Integral Construction of the Thirring model
"Proposed by Thirring half a century ago, the Thirring model is a Quantum Field Theory of a spinor field in a two dimensional space-time, with a self interaction ..."

strangerep

In modern QM, that's become kinda backwards. We start with an algebra of observable quantities which characterize the (class of) system being modelled, and then (try to) construct a unitary representation (i.e., construct a Hilbert space) in which those quantities are represented as operators on the Hilbert space. For many interesting case (like x,p) above, it turns out that a rigged Hilbert space is more convenient than an ordinary Hilbert space. See the early chapters of Ballentine for a more detailed exposition of all this.

That's essentially what renormalization tries to fix. At each order of perturbation, we adjust the space we're working with, either by adjusting the Hamiltonian, or applying a Bogoliubov-like transformation to the basic c/a operators (cf. Shirokov's approach: http://arxiv.org/abs/nucl-th/0102037[/url]).

BTW, the presence of interaction doesn't necessarily "destroy" direct-product structure. It simply means that the Hamiltonian can now mix stuff from different sectors.

atyy

How does this tie in with the view that renormalization, say in QED, just preserves the important low energy terms, and that at high energies non-renormalizable terms or even new degrees of freedom should enter?

kof9595995

But even the perturbative Hamiltonian can "mix stuff from different sectors". Let me elaborate my original point: my starting point is that in a mathematically rigorous QED, the equation of motion in my original post must be satisfied. However the perturbative approach definitely fails to satisfy it, because LHS and RHS are defined on independent sectors, so for example, if I sandwich (i.e. taking rayleigh quotient) both sides using photon states, then LHS is mostly nonzero but RHS is always 0. So to satisfy the equation, the true Hilbert space must somehow take Bosonic and Fermionic sector as one, but I can't proceed with the reasoning because I haven't acquired enough maths.

kof9595995

But as far as I know, an constant issue of Lattice theory is about how to get to the continuum limit, and this isn't very well-established yet. And what I'm trying to argue is, QED is probably some limit of underlying bigger theory, but within this limit we should have a mathematically rigorous framework. Just like classical mechanics and quantum mechanics, the former is a limit of the latter, but still classical mechanics is a mathematically rigorous theory.

atyy

I took a look at Haag's "Local Quantum Physics". He describes both views of renormalization, but doesn't give any link between them. The view you describe is related to the "Algebriac QFT" formalism, while the latter Wilsonian view is related to the "Constructive QFT" formalism, which typically constructs a statistical field theory, and checks that it satisfies the Osterwalder-Schrader conditions in order to make it a Minkowski spacetime QFT.

strangerep

Which sections of Haag? There's some stuff in the early sections, but also in the later ch VIII.
(I should probably refresh my memory of Haag a bit more thoroughly, else I risk talking out of my rear... :-)

strangerep

This is an example of how the space of states in the free theory does not satisfactorily span the space of physical states in the full physical (interacting) theory.

The field operators in the exact form of Gauss' law correspond to the physical fields, not the free fields. Some approaches to QFT try to construct physical field operators perturbatively in terms of (increasingly-complicated) products of the free field operators. Among other things, this procedure must ensure that the new field operators still correspond to suitable Poincare unirreps with the physically correct spin, etc.

Do you recall the earlier thread where I talked about dressing the asymptotic electron states using coherent photon operators? After such dressing has been applied, the commutation relations are a bit different. E.g., the commutator between an electron operator and the electric field operator is no longer zero. Instead, it gives the usual Coulomb field of a charged electron. (This is all at low momenta, since the main task there was to deal with IR divergences.) A similar construction (of Dirac) also shows how to banish some parts of the unphysical EM gauge freedom.

At least one of the rigorous QFT results that I know proceeds via a related process of dressing transformations applied to the basic operators:

J. Glimm,
"Boson Fields with the [itex]:Φ^4[/itex]: Interaction in Three Dimensions",
http://projecteuclid.org/DPubS?servi...cmp/1103840981

Warning: very few people can safely read that on an empty stomach...

atyy

In the second edition, the first sort of renormalization is in section 2.4, while the second type is in the section starting p323 "Algebraic Approach versus Euclidean Quantum Field Theory". I think that in his language "renormalization" always means the first type, because when discussing the second type he says something like it does away with renormalization (can't remember the exact words, I read it in the library, and am getting the section references from a search on Amazon).

tom.stoer

This is exactly what I had in mind in #3

It is interesting that in the very end the physical fields can be decomposed using physical creation and annihilation operators spanning a physical Fock space. Now I understand where the confusion comes from; people seem to confuse Fock space with free particle Fock space plus perturbation theory. This (implicit) assumption is wrong of course.

A Fock space is nothing else but a direct sum of tensor products of copies of a single-particle Hilbert space H.

But it is not necessary that the Hilbert space H is spanned by free fields; any single particle Hilbert space with the correct creation and annihilation operator algebra is sufficient. So if it's possible to construct a suitable transformation from free to 'physical' or 'dressed' fields and a physical Hilbert space then the latter one can be decomposed into physical Fock states. This has been done in QCD in orer to study confinement in the canonical formulation (the problem with the construction of the physical Hilbert space is of course always the same: complete gauge fixing, taming Gribov ambiguities etc.; anyway - these problems do by no means spoil the Fock space approach using physical fields)

Physics Monkey

I would disagree about the continuum limit. We understand very well that the lattice theory flows to the continuum theory plus irrelevant operators. The essential properties of the low energy theory are there. For example, the heat capacity in lattice qed at low temperatures in the deconfined phase is proportional to [itex] T^3 [/itex] just as you would get for free photons. Similarly, the long distance decay of gauge invariant correlation functions is exactly what you would expect for free photons. Lattice theories of qcd are also quite advanced, including fairly good numbers for hadron masses, although treating fermions dynamically is always troublesome because of the sign problem.

I also think it is interesting to note that classical electromagnetism is also not a rigorous theory, at least when thinking about point charges. Indeed, classical fluid dynamics and general relativity are also not known to be free of singularities.

strangerep

...except that they might not satisfy exactly the same CCRs/CARs as in the free case.
DarMM once mentioned something about this, but we never got to hear the full story.

Yes, yes, and yes.

It's still not clear to me what the "correct" c/a operator algebra should be. E.g., physical electron operators probably shouldn't commute with physical photon operators like they do in the free theory. Not sure about this, though. I recall a theorem (in Barut?) about how a very large class of algebras can be expressed in terms of operators from a Heisenberg algebra.