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Rigorous Quantum Field Theory.

by DarMM
Tags: field, quantum, rigorous, theory
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Fredrik
#55
Nov18-09, 03:07 AM
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Quote Quote by meopemuk View Post
...Weinberg's book. It postulates particles as primary physical objects,...
I think you got a little carried away here. The way I see it, Weinberg is just saying that if you start with QM (as defined by the Dirac-von Neumann axioms), define what a symmetry is, and impose the condition that there's a group of symmetries that's isomorphic to the proper orthochronous Poincaré group (a very natural assumption of you want a special relativistic theory), we're immediately led to the concept of non-interacting particles.

To me this suggests that particles are at best an approximate concept that can be useful in situations where gravity can be neglected and the interactions are weak.
Quote Quote by meopemuk View Post
In my opinion there are actually TWO radically different quantum field theories. Let me call them QFT1 and QFT2. It seems to me that in our discussion we sometimes mix them together, though they should be clearly separated.
I don't know about that. The QFT2 approach teaches us the significance of irreducible representations of (the universal covering group of) the proper orthochronous Poincaré group, and the QFT1 approach teaches us a way to construct those representations.
Bob_for_short
#56
Nov18-09, 04:48 AM
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Quote Quote by meopemuk View Post
...which we have built even before introducing interactions.
Quote Quote by Fredrik View Post
...we're immediately led to the concept of non-interacting particles.
To me this suggests that particles are at best an approximate concept that can be useful in situations where gravity can be neglected and the interactions are weak.
You speak of obtained "non-interacting particles" as if they were non-observable, non-physical, etc. If it were so, then there would not be necessity to talk about Poincaré representations, Lorentz covariance, spin-statistics and all that.

In fact they are well observable and they posses the features furnished by us coming from the experimental data. I see the only reasonable way to understand it without uneasiness: to consider the usual decoupled Dirac and Maxwell equations (free equations) as equations describing subsystems of one compound system, i.e., the equations describing dynamics of separated variables. If so, they stay decoupled if there is no external force or presence of another charge. Their solutions (or variables) get into equations of other charges as "external" ones, i.e., the free equation solutions (or better, variables) are observable in this way. In other words, we have to introduce the interaction between different charges rather than between a charge and its own filed. Such a theory construction is free from self-action and full of physical meaning. It can be constructed in a rigorous way and it will be a rigorous QFT with everything physically meaningful and mathematically well defined from the very beginning. There is no problem with an infinite number of excitation modes here since they carry finite energy and do not contribute to (modify perturbatively) masses and charges.

The other understanding is logically inconsistent and mathematically questionable, IMHO.

We all are looking forward to seeing a rigorous QFT from DarMM.
DrFaustus
#57
Nov18-09, 09:21 AM
P: 90
strangerep -> Rigorous QFT is pretty much a very mathematical topic. One does not care as much about eventual physical end results (those have been widely tested) as much as, say, proving that you can exchange a limit with an integral when doing it leads to a physical result that agrees with experiment. Or understanding if the perturbation series converges or not (it doesn't). Or figuring out the commonalities of physical states (widely believed to be the so called "Hadamard states"). Also, tring to construct representations of the CCR algebra. Or justifying the use of the Gell-Mann & Low formula in perturbation theory - when and for what models one can use it. And it also includes much more technical results like showing that the "relativistic KMS condition is indeed satisfied by the 2-point function of an interacting scalar theory in 2D with polynomial interaction." And so on. Think about mathematicians and what they do and then let them work in QFT. In a sense, that's an accurate picture. And of course, the grand goal of Rigorous QFT is to construct an interacting field theory in 4D. For a better idea of RQFT check out the work of people like Glimm, Jaffe, Buchholz, Borchers, Jost, Wightman, Haag, Wald, Kay, Roberts and a million of others related to them. I think Jaffe has some short descriptive paper about RQFT on his webpage.

And thanks for the reference!

meopemuk -> The creation/annihilation operators for a free and an interacting theory are different. For free theory they are time independent whereas in the interacting case they depend on time. (See book by Srednicki, chapter 5)

Also, the "infinite degrees of freedom" do not refer to the number of particles but to the fact that you have an infinite numer of harmonic oscillators, i.e. one at each point in spacetime.

Renormalization is not an open fundamental issue in QFT - it is mathematically well understood. Why is this so difficult to accept?

Bob_for_short -> "Non-interacting" particles are not observable. The very fact that they can be observed means they are interacting. No interaction = no observation. In fact, free particles as described by free fields simply do not exist. At least we don't have any evidence that they do. Mathematical free fields are another story.


Rigorous QFT is a MATHEMATICAL subject. And as such, it has its axioms (the Wightman axioms or, alternatively, the Haag - Kastler axioms) and draws consequences from them. Using various different mathematical techniques. Loosely speaking, one could say that RQFT is the part of functoinal analysis with the Wightman axioms on top. Or the part of Operator Algebras with the Haag-Kastler axioms on top. Loads of theorems and even more hard core maths. And NONE of the popular QFT books is mathematically rigorous. And this includes Peskin&Schroeder, Srednicki and Weinberg as well. If you don't believe me, email them and ask them.
Bob_for_short
#58
Nov18-09, 09:50 AM
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Quote Quote by DrFaustus View Post
...Renormalization is not an open fundamental issue in QFT - it is mathematically well understood. Why is this so difficult to accept?
Having a working QFT without renormalizations and its "ideology" is quite a desirable thing. Why is this so difficult to accept?
Quote Quote by DrFaustus View Post
Bob_for_short -> "Non-interacting" particles are not observable. The very fact that they can be observed means they are interacting. No interaction = no observation. In fact, free particles as described by free fields simply do not exist. At least we don't have any evidence that they do. Mathematical free fields are another story.
A plane wave is a free solution. It is an observable state. We can measure its energy, momentum, spin, etc., whatever it describes - an "elementary" particle or a compound system center of mass motion. Nobody doubts it but you.
(A "self-interacting" particle is probably "self-observable"?)
Rigorous QFT is a MATHEMATICAL subject. And as such, it has its axioms (the Wightman axioms or, alternatively, the Haag - Kastler axioms) and draws consequences from them.
QFT may, of course, be a subject of study of mathematicians, nobody forbids it. But it is first of all a working tool of physicists-theorists. It should proceed from and contain physically meaningful stuff. It is a natural human desire to work with meaningful stuff.

By the way, why they (mathematicians) call the Dirac delta-function a "distribution" rather than a "concentrution"?
meopemuk
#59
Nov18-09, 01:39 PM
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Quote Quote by DrFaustus View Post
meopemuk -> The creation/annihilation operators for a free and an interacting theory are different. For free theory they are time independent whereas in the interacting case they depend on time. (See book by Srednicki, chapter 5)
I wrote about my understanding of creation/annihilation operators in post #44. The time-dependent versions of creation operators are given by

[tex] a^{dag}_{free}(p,t) = \exp(-iH_0t)a^{dag}(p,0) \exp(iH_0t) [/tex]
[tex] a^{dag}_{int}(p,t) = \exp(-iHt)a^{dag}(p,0) \exp(iHt) [/tex]

where [itex]H_0[/itex] and [itex]H[/itex] are the free and interacting Hamiltonians, respectively. At time 0 both sets of creation operators reduce to [itex] a^{dag}(p,0) [/itex], i.e., they coincide. At non-zero times the free and interacting creation operators can be expressed as linear combinations of (products of) each other. Both these sets act in the same Fock space. If you agree with these descriptions, then there is nothing to argue about.

Quote Quote by DrFaustus View Post
Renormalization is not an open fundamental issue in QFT - it is mathematically well understood. Why is this so difficult to accept?
I agree that renormalization works fine as a tool for obtaining accurate S-matrix. However, in the process of renormalization we obtain a totally unacceptable Hamiltonian (with infinite counterterms), so there is no chance to get physical unitary time evolution. One can say: who cares? the time evolution is not measurable in scattering experiments, anyway. But I think that quantum theory without a well-defined Hamiltonian and time evolution cannot be considered complete.
Bob_for_short
#60
Nov18-09, 02:27 PM
P: 1,160
Quote Quote by meopemuk View Post
I agree that renormalization works fine as a tool for obtaining accurate S-matrix.
I disagree. The renormalizations themselves do not provide good S-matrix elements. One is obliged to consider the IR divergences as well and calculate (at least partially) inclusive cross sections. Only then one obtains something reasonable. In fact, all inclusive cross sections describe inelastic processes rather than elastic ones. Exact elastic S-matrix elements are equal identically to zero in QED.

Usually they say that IR and UV divergences are of different nature. I agree with it in a very narrow sense: with very massive photons one can obtain non-zero elastic S-matrix elements. But in my opinion both divergences are removed from QED at one stroke with correct description of interaction physics.
strangerep
#61
Nov18-09, 06:36 PM
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Quote Quote by meopemuk View Post
QFT1 is taught in most QFT textbooks (excluding the Weinberg's one). [...]
we build the Fock space by applying creation operators to the vacuum many times, etc. etc. In this approach, it is not clear whether interacting theory lives in the same Hilbert (Fock) space as the non-interacting one. Are the creation/annihilation operators of the two theories different? Do they have different particle number operators? Possibly, it makes sense to build the interacting Hilbert space as a representation space of canonical (anti)commutation relations? Due to the "infinite number of degrees of freedom", it might even happen that mathematically the two Hilbert spaces are different or inequivalent.

On the other hand, QFT2 is a version presented in Weinberg's book. It postulates particles as primary physical objects, and uses quantum mechanics from the beginning (so, no need for "quantization"). First, the Hilbert (Fock) space is build as a direct sum of N-particle spaces. Then creation and annihilation operators are explicitly defined in this Fock space. The next step is to define dynamics (= an unitary representation of the Poincare group) in the Fock space. The non-interacting representation can be constructed trivially (as a direct sum of N-tensor products of irreducible representations). The difficult part is to define an interacting representation of the Poincare group, which satisfies cluster separability and permits changes in the number of particles. This is the place where quantum fields (= certain formal linear combinations of creation and annihilation operators) come handy. We simply notice that if the interaction Hamiltonian (= the generator of time translations) and interacting boost operators are build as integrals of products of fields at the same "space-time points", then all physical conditions are satisfied automatically. In this approach, there is no need to worry about different Hilbert spaces for the non-interacting and interacting theories. [...]
The Hilbert space of "QFT2" is still infinite-dimensional -- so any mathematical issues arising
as a consequence of infinite degrees of freedom still lurk here, just as they lurk in "QFT1".
meopemuk
#62
Nov18-09, 06:44 PM
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Quote Quote by strangerep View Post
The Hilbert space of "QFT2" is still infinite-dimensional -- so any mathematical issues arising
as a consequence of infinite degrees of freedom still lurk here, just as they lurk in "QFT1".
I agree with that, but the Hilbert space of QFT2 is built *before* any interaction is introduced. So, the same Hilbert space is used in both non-interacting and interacting theories.
strangerep
#63
Nov18-09, 07:18 PM
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Quote Quote by meopemuk View Post
[...] the Hilbert space of QFT2 is built *before* any interaction is introduced. So, the same Hilbert space is used in both non-interacting and interacting theories.
Except that the a/c operators in QFT2 are not bona-fide operators, but are really
operator-valued distributions. Hence fields constructed as linear combinations of them
are also operator-valued distributions. Hence QFT2 suffers exactly the same "ill-defined
equal-point multiplication of distributions" mathematical problem as QFT1.
meopemuk
#64
Nov18-09, 07:51 PM
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Quote Quote by strangerep View Post
Except that the a/c operators in QFT2 are not bona-fide operators, but are really
operator-valued distributions. Hence fields constructed as linear combinations of them
are also operator-valued distributions. Hence QFT2 suffers exactly the same "ill-defined
equal-point multiplication of distributions" mathematical problem as QFT1.
Could you give an example in which a product of a/c operators or quantum fields is "ill-defined"?
Bob_for_short
#65
Nov19-09, 04:04 AM
P: 1,160
I think Strangerep speaks of loops in practical calculations, not of something different.

I would say the loop expressions are not "ill-defined" but simply divergent. There are many "cut-off" approaches to make them temporarily finite but they are just infinite as they should be.
DarMM
#66
Nov19-09, 04:39 AM
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Okay, here is a model which is exactly solvable nonperturbatively and is under complete analytic control. Also virtually every aspect of this model is understood mathematically.

Now in the is model, the field does not transform covariantly. The reason I'm using the model is to show that even with this property removed there are still different Hilbert spaces.

Firstly, the model is commonly known as the external field problem. It involves a massive scalar quantum field interacting with an external static field.

The equations of motion are:
[tex]\left( \Box + m^{2} \right)\phi\left(x\right) = gj\left(x\right)[/tex]

Now I'm actually going to start from what meopemuk calls "QFT2". The Hamiltonian of the free theory is given by:
[tex]H = \int{dk \omega(k)a^{*}(k)a(k)}[/tex]
Where [tex]a^{*}(k)[/tex],[tex]a(k)[/tex] are the creation and annihilation operators for the Fock space particles.

In order for this to describe the local interactions with an external source, I would modify the Hamiltonian to be:
[tex]H = \int{dk \omega(k)a^{*}(k)a(k)} + \frac{g}{(2\pi)^{3/2}}\int{dk\frac{\left[a^{*}(-k) + a(k)\right]}{\sqrt{2}\omega(k)^{1/2}}\tilde{j}(k)}[/tex]

So far, so good.

Now the normal mode creation and annihilation operators for this Hamiltonian are:
[tex]A(k) = a(k) + \frac{g}{(2\pi)^{3/2}}\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}[/tex]
A short calculation will show you that these operators have different commutation relations to the usual commutation relations.

They bring the Hamiltonian into the form:
[tex]H = \int{dk E(k)A^{*}(k)A(k)}[/tex],
where [tex]E(k)[/tex] is a function describing the eigenspectrum of the full Hamiltonian.

Now, if you use Rayleigh-Schrödinger perturbation theory you obtain the interacting ground state as a superposition of free states:
[tex]\Omega = Z^{1/2} \sum^{\infty}_{n = 0} \frac{1}{n!} - \left(\frac{g}{(2\pi)^{3/2}}\int{\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}a^{*}(k)}\right)^{n} \Psi_{0}[/tex].
Where [tex]\Psi_{0}[/tex] is the free vacuum.
Also [tex]Z = exp\left[\int{\frac{g^{2}}{(2\pi)^{3}}\frac{|\tilde{j}(k)|^{2}}{\sqrt{2}\omega(k )^{3}}}\right][/tex]

Now for a field weak enough that:
[tex]\frac{\tilde{j}(k)}{\omega(k)^{3/2}} \in L^{2}(\mathbb{R}^{3})[/tex]
then everything is fine. I'll call this condition (1).

However if this condition is violated, by a strong external field, then we have some problems.

First of all [tex]A(k), A^{*}(k)[/tex] are just creation and annihilation operators. They have a different commutation relations, but essentially I can still use them to create a Fock basis, since I can prove the exists a Hilbert space with a state annihilated by all [tex]A(k)[/tex]. Now this constructed Fock space always exists, no problem. Let's call this Fock space [tex]\mathcal{F}_{I}[/tex].

However, if condition (1) is violated something interesting happens. [tex]Z[/tex] vanishes. Now the expansion for [tex]\Omega[/tex] is a sum of terms expressing the overlap of [tex]\Omega[/tex] with free states. If [tex]Z=0[/tex], then [tex]\Omega[/tex] has no overlap with and hence is orthogonal to all free states. This can be shown for any interacting state.
So every single state in [tex]\mathcal{F}_{I}[/tex] is completely orthognal to all states in [tex]\mathcal{F}[/tex], the free Fock space. Hence the two Hilbert spaces are disjoint.

So the Fock space for [tex]a(k),a^{*}(k)[/tex] is not the same Hilbert space as the Fock space for [tex]A(k), A^{*}(k)[/tex]. They are still both Fock spaces, however [tex]A(k), A^{*}(k)[/tex] has a different algebra, so it's the Fock representation of a new algebra. If one wanted to still use the [tex]a(k),a^{*}(k)[/tex] and their algebra, you would need to use a non-Fock rep in order to be in the correct Hilbert space.

This is what a meant by my previous comment:
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.


I hope this post helps.
DarMM
#67
Nov19-09, 04:49 AM
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P: 303
I also want to say that in [tex]\mathcal{F}_{I}[/tex], the Hamiltonian and S-matrix are both finite. So one has unitary evolution in this space.

For mathematical literature on this model:
Reed, M. and Simon, B. Methods of Modern Mathematical Physics, Vols. II-III, New York: Academic Press.
Wightman's article in Partial Differential Equations edited by D. Spencer, Symposium in Pure Mathematics (American Mathematical Society, Providence), Vol. 23.
Bob_for_short
#68
Nov19-09, 05:00 AM
P: 1,160
Thank you, DarMM, for this example. Before reading it, I would like to know if it is very different from "radiation of classical current", i.e., from coherent states (if m=0 and the field is the quantized EMF)? Or it is akin to the coherent states?
DarMM
#69
Nov19-09, 05:12 AM
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Quote Quote by strangerep View Post
What's not clear to me is whether rigorous QFT is really all about taming this uncountable
infinity of inequivalent representations, or something else entirely.
Rigorous QFT divides into three areas:

Axiomatic Field Theory
This is basically, as Dr. Faustus said, Functional analysis with the Wightman axioms on top. One tries to understand what type of mathematical object quantum fields are and what conditions they should obey, either in canonical or path integral form. This has been accomplished by Wightman, Osterwalder, Schrader, Frohlich, Nelson, Symanzik and others. Given these conditions (axioms), you then try to figure out properties of the quantum fields, such as:
Analyticity of the vertex functions.
Poles in correlation functions.
The connection between spin and statistics.
e.t.c.

Basically the study of what mathematical objects fields are and the consequences of this.

Algebraic Field Theory
Here one is being very general and simply concentrates on the properties of local quantum theories in Minkowski or other spacetimes, not specifically requiring there to be fields involved. This area can be seen as working out the general physical consequences of quantum theory and relativity. For instance this area provides the simplest treatment of Bell's inequalities in general spacetimes. It is this area that really tries to understand the uncountable infinity of inequivalent representations, since it is trying to understand the very general implications of quantum theory and relativity.

You could see Axiomatic field theory as a subset of Algebraic field theory, which focuses on Minkowski spacetime and assumes the theory is a field theory and is non-thermal, which leads to more detailed properties. However because of fields being involved, the mathematics and the focus of the two areas tend to be quite different.

For instance an algebraic field theory question might be "What are the general characteristics of thermal states that seperate them from pure states? What are the different representations in which they live? Can we characterise them?"
Axiomatic Field Theory would ask "Where are the poles in the three-point vertex function? What physical information is contained in these poles?"

Constructive Field Theory
This area essentially tries to prove that the field theories that physicists work with actually belong to the catagories above. For instance in Axiomatic Field Theory you have the Wightman axioms, but how do you know there exists any mathematical object that actually satisfies them? Constructive field theory builds (constructs) such objects by nonperturbatively controlling actual quantum field theories.

So constructive field theory would ask something like:
"Does [tex]\phi^{4}[/tex] in two dimensions exist as a well-defined mathematical entity?, If it does exist, does it satisfy the axioms from axiomatic field?"
DarMM
#70
Nov19-09, 05:14 AM
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Quote Quote by Bob_for_short View Post
Thank you, DarMM, for this example. Before reading it, I would like to know if it is very different from "radiation of classical current", i.e., from coherent states (if m=0 and the field is the quantized EMF)? Or it is akin to the coherent states?
If I had choosen the scalar field to be massless there would be coherent states. I haven't choosen this specifically to avoid dealing with that issue. So the issues dealt with are quite different.
Bob_for_short
#71
Nov19-09, 05:23 AM
P: 1,160
Quote Quote by DarMM View Post
If I had choosen the scalar field to be massless there would be coherent states. I haven't choosen this specifically to avoid dealing with that issue. So the issues dealt with are quite different.
No, there is no any issue in case of quantized EMF radiated with a classical current (source).

In your case there is just a threshold for massive quanta but the external current can be sufficiently powerful to radiate even massive quanta, so I see it as a quite akin problem.

What energy can absorb/emit new quanta corresponding to operators A(k)? (You missed dk in two integrals but it is not essential.)
DarMM
#72
Nov19-09, 05:29 AM
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Quote Quote by Bob_for_short View Post
No, there is no any issue in case of quantized EMF radiated with a classical current (source).

In your case there is just a threshold for massive quanta but the external current can be sufficiently powerful to radiate even massive quanta, so I see it as a quite akin problem.
Okay, but it isn't an akin problem. Having a mass gap to the emitted quanta fundamentally changes things. If I had massless quanta there would be two features, change of Hilbert space and coherent states. Since the quanta are massive there are no coherent states. A problem with no coherent states is not akin to a problem with coherent states.
Unless you are saying they are akin simply because in both the external field can create quanta, however that's no different than saying they are both external field problems. The main focus here is the change in Hilbert space, not coherent states or anything like them, because they are not present.

Quote Quote by Bob_for_short View Post
What energy can absorb/emit new quanta corresponding to operators A(k)?
I'm not sure what you are asking.


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