Rigorous Quantum Field Theory.

meopemuk

This depends very much on your starting philosophical position. If you believe (as many people do) that world is basically made of continuous fields, and particles (that we observe) are just some "excitations" of these fields, then you immediately have infinite number of degrees of freedom and all problems associated with them. On the other hand, one can assume that our world is made of discrete countable particles, and quantum fields are just formal mathematical objects, then I'm not sure that "infinite number of degrees of freedom" and "inequivalent representations of CCR" are useful ideas.


I decline your invitation to go into such complex matters. I would like to understand most basic QFT examples, such as QED. I think there are enough open fundamental questions in QED. Renormalization being one of them. Moreover, I don't think that "QFT on curved spaces" is an example of a successful theory, that must be replicated. So far, we don't have a consistent theory of quantum gravity. Perhaps this is because we are looking for such theory in wrong places ("continuous fields", "curved spaces")? Perhaps "quantum theory of systems with variable number of particles" can suggest some other places to look at?

strangerep

FYI, the "unitary dressing transformations" are reminiscent of the Bogoliubov transformations
used in condensed matter theory to find a more physically-suitable Hilbert space. I.e., the
dressing transformations do indeed map between unitarily inequivalent reps in general.
(Shebeko & Shirokov talk a bit more about this in Appendix B of nucl-th/0102037.)

Such "improper" unitary transformations, moving between inequivalent Hilbert spaces,
do indeed seem useful in many areas of physics, including those you mentioned.

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.

meopemuk

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.

QFT1 is taught in most QFT textbooks (excluding the Weinberg's one). This approach begins with postulating continuous fields as primary physical objects. Then we postulate a field Lagrangian, derive the equation of motion, and expand its solutions into plane waves. Then we "quantize" this theory by postulating certain (anti)commutation relations between fields and their conjugated momenta. In the non-interacting case we find out that (anti)commutation relations between coefficients in the plane-wave expansion allow us to interpret them as annihilation and creation operators. Then 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. We have only one Fock space, which we have built even before introducing interactions. This space is not affected by our choice of interactions at all. The particle interpretation (creation/annihilation operators, particle number operators) is not affected by the interaction as well.

Remarkably, both QFT1 and QFT2 approaches lead to the same Feynman rules, so they are equivalent as far as comparison with scattering experiments is concerned. However, these are two completely different philosophies. It does matter which philosophy you choose if you want to go beyond traditional QFT, e.g., if you want to resolve the problem of renormalization and divergences. My choice is the particle-based philosophy QFT2.

Eugene.

Fredrik

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.
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

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

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

Having a working QFT without renormalizations and its "ideology" is quite a desirable thing. Why is this so difficult to accept?
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"?)
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

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.

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

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

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

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

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

Could you give an example in which a product of a/c operators or quantum fields is "ill-defined"?

Bob_for_short

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

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

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

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?