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Bosons and Fermions in a rigorous QFT

by kof9595995
Tags: bosons, fermions, rigorous
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kof9595995
#19
Feb10-12, 07:45 PM
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Quote Quote by DrDu View Post
Is there anybody who believes that a rigorous QED exists?
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
#20
Feb11-12, 03:24 AM
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Quote Quote by kof9595995 View Post
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.
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
#21
Feb11-12, 08:36 AM
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Quote Quote by tom.stoer View Post
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.
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
#22
Feb11-12, 01:50 PM
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Quote Quote by kof9595995 View Post
I'd like the ask the question again, if a rigorous QED doesn't exist, what would be underlying theory of perturbative QED?
Lattice gauge theory or an asymptotically free gauge theory in a partial higgs phase can provide a UV completion of low energy QED.
atyy
#23
Feb11-12, 03:20 PM
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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
#24
Feb11-12, 07:42 PM
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Quote Quote by kof9595995 View Post
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,
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.

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.
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
#25
Feb11-12, 09:09 PM
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Quote Quote by strangerep View Post
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]).
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
#26
Feb12-12, 12:58 PM
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Quote Quote by strangerep View Post
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.
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
#27
Feb12-12, 01:03 PM
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Quote Quote by Physics Monkey View Post
Lattice gauge theory or an asymptotically free gauge theory in a partial higgs phase can provide a UV completion of low energy QED.
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
#28
Feb12-12, 03:50 PM
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Quote Quote by strangerep View Post
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]).
Quote Quote by atyy View Post
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?
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
#29
Feb12-12, 08:41 PM
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Quote Quote by 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.
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
#30
Feb12-12, 08:56 PM
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Quote Quote by kof9595995 View Post
[...] 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, [...]
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
#31
Feb13-12, 03:48 PM
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Quote Quote by strangerep View Post
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... :-)
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
#32
Feb13-12, 04:59 PM
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Quote Quote by strangerep View Post
The field operators in the exact form of Gauss' law correspond to the physical fields, not the free fields.

...

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. ...
This is exactly what I had in mind in #3

Quote Quote by tom.stoer View Post
The above mentioned field equation cannot be quantized directly b/c it has to be gauge fixed. In A=0 at least for the time-indep. constraint (the Gauss law) this equaton is implemented on the physical Hilbert space. This does not require any "free field approach".
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
#33
Feb13-12, 05:03 PM
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Quote Quote by kof9595995 View Post
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.
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
#34
Feb13-12, 05:11 PM
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Quote Quote by tom.stoer View Post
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.
...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.

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.
Yes, yes, and yes.

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.
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.
tom.stoer
#35
Feb13-12, 05:40 PM
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I remember the 'dressing' and the 'gauge fixing by unitary transformations'; both approaches seem to be similar b/c they partially 'solve' some field equations.

The first approach (dressing) changes the operator algebra; it can be solved explicitly in some 1+1 dim. field theories like the Schwinger model (here one formally solves the Dirac equation by exponentiation using a gauge field string with a path ordered product).

The second approach does not change the operator algebra; the Gauss law is solved but b/c a unitary trf. is used, all operator algebras remain unchanged. This may be spoiled by regularization which requires gauge-invariant point splitting (I can only remember the two-dim. case).

In the first case the interaction is "hidden" in the dressed fields; they create the physical Coulomb interaction, but the interaction term itself looks trivial algebraically. In the second case the interaction terms are constructed explicitly and in principle they can be expressed using physical Fock space operators.

In the second case the (A=0 & Coulomb gauge) Hamiltonian contains one piece which shows directly the color-electric Coulomb potential:

[tex]H_C = g^2 \int d^3x \int d^3y \,\text{tr}\,J^{-1}(x)\,\rho(x)\,(-D\partial)^{-1}\,(-\partial^2)\,(-D\partial)^{-1}\,J(y)\,\rho(y)[/tex]

with D = ∂ + gA, A being the gauge-fixed gluon field, J being the Fadeev-Popov determinant J = det(-D∂), ρ = ρ[q] + ρ[A] being the total color charge with quark and gluon contribution (w/o J, D and ρ[A] in HC the usual Coulomb gauge interaction in QED is recovered)
atyy
#36
Feb13-12, 05:47 PM
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@strangerep, BTW, another interesting comment in Haag was that since the Hilbert spaces for each representation of the CCRs are different, presumably the selection of the representation depends on dynamics. He then says that the advantage of the Lagrangian approach is that it makes it easy to choose the dynamics based on symmetries, and then construct the appropriate Hilbert space after that. (Again, I don't have the page reference, but it should be in one of the two sections I mentioned above.)

Also, it's really interesting to me that BCS has this "rigourous treatment" - I'd always taken it to be unrigourous since it's modelling a condensed matter phenomenon where one can definitely take a lattice cut-off so that Haag's theorem won't apply.


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