Rigorous Quantum Field Theory.

[meopemuk]

I don't understand this.
What states are the creation and annihilation operators creating and annihilating then? A two particle state at $$t=\infty$$ is not the same state as a two particle state at $$t=-\infty$$. So two different operators are necessary to create them from the vacuum.

Which time corresponds to the plane wave $N\exp(ipx)$? You cannot assign any time label to it. This is just a single state vector in the Hilbert space. Creation operators create only such plane wave states (or their linear combinations) without any reference to time. You are right when you say: "A two particle state at $$t=\infty$$ is not the same state as a two particle state at $$t=-\infty$$". This is simply because the state vector created at $$t=\infty$$ has evolved through time into another state vector. But both "infinite past" and "infinite future" state vectors can be build by applying (linear combinations of) the same creation operators to the same vacuum vector.

In/Out states form two completely different basis of the Hilbert space which are related to each other through a unitary transformation known as the S-matrix. So if what you were saying is true, then the S-matrix would be the identity and the theory would be trivial.
This isn't even an aspect of rigorous field theory, it's in usual field theory textbooks like Srednicki.
If you object to this aspect of QFT, you are basically saying all theories are free.

I've never understood what's the idea about separation of the in and out states. Suppose that I've shown you a wave function or a vector in the Hilbert space. Would you be able to tell if this is an "in" state or an "out" state? Or something else? They all look pretty similar to me. The difference is that in "in" states particles are moving toward each other (before the collision), while in "out" states particles are moving away from each other (after the collision). But nobody can forbid me to prepare a state in which particles are moving away from each other at time $t = - \infty$.

Excluding some pathological cases, the S-matrix is not trivial as long as there is a non-trivial interaction between particles. So, most interacting theories are not free.

Eugene.

 

[Bob_for_short]

I completely agree with Eugene.

Scattering result is just a redistribution of the initial energy/momentum over another population of states in the same basis. This was meant at the beginning of QED and other QFTs, and this has been finally achieved after renormalizations. The initial and final particles are of the same nature.

 

[DarMM]

Which time corresponds to the plane wave $N\exp(ipx)$? You cannot assign any time label to it. This is just a single state vector in the Hilbert space. Creation operators create only such plane wave states (or their linear combinations) without any reference to time. You are right when you say: "A two particle state at $$t=\infty$$ is not the same state as a two particle state at $$t=-\infty$$". This is simply because the state vector created at $$t=\infty$$ has evolved through time into another state vector. But both "infinite past" and "infinite future" state vectors can be build by applying (linear combinations of) the same creation operators to the same vacuum vector.

Yes, but that is QM where I can explicitly right down wavefunctions. QFT is not so simple. Plus an in state has different expectation values for certain observables than an out state, so their difference can be characterised.

You guys have ideas on scattering theory that I've never really seen anywhere else. I mean the LSZ formalism and its more rigorous version of Haag-Ruelle theory uses the fact that the in and out basis are different. Srednicki uses it in his introductory textbook, as does Weinberg.

I don't really understand what is objectionable about there being different creation and annihilation operators for in and out states. It doesn't imply they live in a different Hilbert space. It's just like having ladder operators for the harmonic oscillators states and having different ladder operators for the anharmonic oscillators states. States of one can still be written as a superposition of the others. It's just a matter of them being different states, so they have different operators to create them. Sure using a combination of in-creation operators you can create an out state and vice-versa, but they are still different operators.

Do you guys object to this? If so why? I must admit I don't fully understand since it is in most QFT books.

 

[Bob_for_short]

We can use plane waves in the In-states and bound states in Out-states if we describe, for example, recombination an other reactions, no problem.

We do not understand how you distinguish In- and Out-electrons if they are free in both states. Similarly for photons.

Question of mine: does this necessity come from renormalizations? Use QED as an example, please. (I am afraid this "difference" of operators has no physical meaning and is explained exclusively with difficulties in constructing a working interaction model.)

 

[DarMM]

We do not understand how you distinguish In- and Out-electrons if they are free in both states. Similarly for photons.

Question of mine: does this necessity come from renormalizations? Use QED as an example, please. (I am afraid this "difference" of operators has no physical meaning and is explained exclusively with difficulties in constructing a working interaction model.)

Alright maybe the misunderstanding is the picture being used. Surely you can see that in the Heisenberg picture they must different states. In the Schrodinger picture where states are labeled only by their properties and not their properties at a time, there is of course no real difference because the language of in/out states would not arise.

Surely you can see how, in the Heisenberg picture, in/out states with the same properties are different states and hence would need different creation and annihilation operators?

Now if you guys would prefer I carry out the argument in the Schrodinger picture, that's fine. Although it is a little less natural for QFT. I'm assuming this is the only problem and that you do not disagree with the difference of in/out states in the Heisenberg picture.

 

[Bob_for_short]

In the Heisenberg picture the operators depend on time. It is like momentum dependence on time in CM p(t): in the initial state it is p, in the final one it is p'. Do you speak of this "difference"? It is not a problem in question. At least for me. What is a principal difference in QFT (QED)?

 

[meopemuk]

Yes, but that is QM where I can explicitly right down wavefunctions. QFT is not so simple.

This is exactly the source of our disagreement. I hold the opinion that there is no fundamental difference between QM and QFT. The only difference is that interaction in QFT can change the number of particles, while in QM this number is fixed. If I had Weinberg's book with me right now, I could find a quote there to support this view.

 

[Bob_for_short]

Yes, but that is QM where I can explicitly right down wavefunctions. QFT is not so simple.

It is not necessary to write down the exact wave-functions in QFT. It is sufficient to choose a good initial/final approximations taking into account an essential part of interaction. The interaction reminder can be considered perturbatively, in my humble but not groundless opinion.

 

[meopemuk]

Hi DarMM,

I think I understood which in-out operators you are talking about. Let me explain it in my own words and see whether you'd agree or not.

Let me define a set of creation operators $a^{\dag}_{in}(p, -\infty)$. These operator create from vacuum 1-particle plane waves at time $t = -\infty$. If we apply the interacting time evolution operator to these plane waves, then we'll find that at time $t = +\infty$ they do not look like plane waves at all. This is because scattering has occurred at times around zero. These time-evolved states can be obtained by using time-evolved in-creation operators

$$a^{\dag}_{in}(p, +\infty) = \exp(-iHT) a^{\dag}_{in}(p, -\infty) \exp(iHt)$$

where H is the total Hamiltonian and by T I formaly denoted an infinite time interval between $t = -\infty$ and $t = +\infty$.

As I said, these new creation operators do not create plane wave states at $t = +\infty$. If we still want to create plane wave states at $t = +\infty$, we can define a new set of creation operators $a^{\dag}_{out}(p, +\infty)$.

Both sets of operators $a^{\dag}_{in}(p, -\infty)$ and $a^{\dag}_{out}(p, +\infty)$ create the same plane wave states. However the former set creates them in the infinite past, while the latter set creates them in the infinite future.

Operators $a^{\dag}_{in}(p, +\infty)$ can be expressed as linear combinations of $a^{\dag}_{out}(p, +\infty)$. The expansion coefficients are elements of the S-matrix.

I think I can agree that scattering theory can be built on these principles, though this is not my favorite approach. Returning to our previous discussion, I would like to say that both in- and out-operators live in the same Hilbert space, and I still don't see a good reason why interacting and non-interacting theories must inhabit different Hilbert spaces.

Eugene.

 

[DarMM]

This is exactly the source of our disagreement. I hold the opinion that there is no fundamental difference between QM and QFT. The only difference is that interaction in QFT can change the number of particles, while in QM this number is fixed. If I had Weinberg's book with me right now, I could find a quote there to support this view.

Yes, but in QFT it is almost impossible to write the wavefunctions down since they are square integrable functions on an infinite dimensional space. Only in a few simple cases can you actually write down what the wavefunction is, which is why I said "explicitly right down wavefunctions". I wouldn't see this as a source of disagreement, since it isn't really related to what we are discussing and surely, for instance, you couldn't write down the wavefunction $$\psi\left[A_{\mu},t\right)$$ for a proton. Yes there is no "fundamental" difference in some sense, but QFT is a lot harder, especially with regard to getting explicit results. Of course there are other ways in which QFT is quite different from QM, the presence of Lorentz invariance and locality imply strong analytic constraints on correlation functions in QFT that aren't there in QM. For instance one obtains crossing symmetry and various other properties of the S-matrix, where as in QM the S-matrix can be much more general. Also one has the spin and statistics theorem, e.t.c.

I'm sure Weinberg would agree, since he used to say that imposing analyticity, locality, e.t.c. and keeping the rules of quantum theory is literally what QFT is. Of course since QFT is a quantum theory it will have a lot in common with nonrelativistic QM, but it is also very different in some ways.

I think I can agree that scattering theory can be built on these principles, though this is not my favorite approach. Returning to our previous discussion, I would like to say that both in- and out-operators live in the same Hilbert space, and I still don't see a good reason why interacting and non-interacting theories must inhabit different Hilbert spaces.

Well the creation and annihilation operators weren't supposed to explain why, they were merely meant to be a step on the way there. However I can see that scattering is probably not the way to go, considering the difficulty in agreeing on the basics of the subject. So instead I'll write a post concerning an exactly soluable model in four dimensions where the presence of different Hilbert spaces becomes obvious, perhaps that will clear things up.

 

[meopemuk]

Yes, but in QFT it is almost impossible to write the wavefunctions down since they are square integrable functions on an infinite dimensional space. Only in a few simple cases can you actually write down what the wavefunction is, which is why I said "explicitly right down wavefunctions". I wouldn't see this as a source of disagreement, since it isn't really related to what we are discussing and surely, for instance, you couldn't write down the wavefunction $$\psi\left[A_{\mu},t\right)$$ for a proton. Yes there is no "fundamental" difference in some sense, but QFT is a lot harder, especially with regard to getting explicit results. Of course there are other ways in which QFT is quite different from QM, the presence of Lorentz invariance and locality imply strong analytic constraints on correlation functions in QFT that aren't there in QM. For instance one obtains crossing symmetry and various other properties of the S-matrix, where as in QM the S-matrix can be much more general. Also one has the spin and statistics theorem, e.t.c.

I'm sure Weinberg would agree, since he used to say that imposing analyticity, locality, e.t.c. and keeping the rules of quantum theory is literally what QFT is. Of course since QFT is a quantum theory it will have a lot in common with nonrelativistic QM, but it is also very different in some ways.

I am glad that you've mentioned Weinberg, because his book presents the only approach to QFT that makes sense to me.

On page 49 he writes:

"First, some good news: quantum field theory is based on the same quantum mechanics that was invented by Schroedinger,..."

I don't think he ever considers the necessity to introduce a special Hilbert space for interacting theory. He also does not question the validity of Hermitian particle number operators. Although, I should admit that he doesn't bother to describe in any detail the Hilbert space he is working in. (the term "Fock space" is not even in the index). However, my guess is that he is doing "quantum mechanics in the Fock space", as I described earlier.

The crucial point of Weinberg's approach is indicated on page 200, where he expalins how it is different from the traditional QFT:

"Traditionally in quantum field theory one begins with such field equations or with the Lagrangian from which they are derived, and then uses them to derive the expansion of the fields in terms of one-particle annihilation and creation operators. In the approach followed here, we start with the particles, and derive the fields according to the dictates of Lorentz invariance, with the field equations arising almost accidentally as a byproduct of this construction."

On page 191 he explains why he needs to consider fields at all. The physical requirements of Poincare invariance and cluster separability are key here. Somewhat earlier he showed that interacting theory can be made Poincare-invariant if the interaction operator is represented as a space integral of "interaction density", which is a scalar with respect to the non-interacting representation of the Poincare group. The easiest way to construct such a scalar is to make a product of N "quantum fields". In Weinberg's interpretation, quantum fields appear as auxiliary technical constructs, whose only role is to ensure the Poincare invariance of interaction. Another benefit is that building interactions as products of quantum fields ensures cluster separability (or, as Weinberg calls it - the cluster decomposition principle). The obvious downside of such field theories is the inevitable need for renormalization: even single particle experiences unphysical self-interaction and self-scattering.

Note that Weinberg cannot prove that building interactions from fields is the only possible way to provide Poincare invariance, cluster separability, and changing number of particles. As an example of a failed alternative approach he mentions Bakamjian-Thomas theory. However, he doesn't tell that there are recent developments of this theory (see works by W.N. Polyzou), which show progress in both cluster separability and particle-number-changing interactions.

Weinberg doesn't mention yet another alternative - the "dressed particle" approach - which can easily accommodate Poincare invariance, cluster separability, and particle-number-changing interactions. Its main advantage is that renormalization is never needed. I consider this as a major sign of physical/mathematical consistency. The main disadvantage was mentioned by strangerep in one of recent threads: Since the "dressed particle" approach does not use fields, it does not accommodate the idea of "gauge invariance", so, unlike in QFT, there is no (relatively) simple and regular way to construct interaction Hamiltonians that immediately match observations. Currently we must either fit "dressed" Hamiltonians to experiment or use complicated tricks, like "unitary dressing". Though, this may be not a high price for getting rid of renormalization and divergences.

Eugene.

 

[Bob_for_short]

... the "dressed particle" approach - which can easily accommodate Poincare invariance, cluster separability, and particle-number-changing interactions. Its main advantage is that renormalization is never needed. I consider this as a major sign of physical/mathematical consistency. The main disadvantage was mentioned by strangerep in one of recent threads: Since the "dressed particle" approach does not use fields, it does not accommodate the idea of "gauge invariance", so, unlike in QFT, there is no (relatively) simple and regular way to construct interaction Hamiltonians that immediately match observations.

I do not remember mentioning such a "disadvantage" by Strangerep. I don not consider this as a "disadvantage". On the contrary, gauge-invariant formulation that uses tensions (transverse polarizations) involves only physical degrees of freedom that can carry and exchange the energy-momentum with charges.

IMHO there is a simple way to construct interaction Hamiltonians in the "dressed" particle approach but I am not sure of its unambiguity. Anyway, the final word is after experiment. There may be many trial Hamiltonians to test against the experimental data.

 

[meopemuk]

I do not remember mentioning such a "disadvantage" by Strangerep.

Post #66 in https://www.physicsforums.com/showthread.php?t=348911&page=5

He didn't mention local gauge invariance used as a tool for formulating interacting QFT Hamiltonians, but I guess he meant exactly this, because AFAIK currently there is no other means to obtain realistic interactions.

Eugene.

 

[strangerep]

Post #66 in https://www.physicsforums.com/showthread.php?t=348911&page=5

He didn't mention local gauge invariance used as a tool for formulating interacting QFT Hamiltonians, but I guess he meant exactly this, because AFAIK currently there is no other means to obtain realistic interactions.

No, I wasn't thinking about local gauge invariance in that post.
(You're probably on safer ground if you don't put words in my mouth. :-)

But let us not derail DarMM's thread off onto such tangents. I'm very interested to
discuss the exactly solvable 4D model he promises to write a post about...

 

[DrFaustus]

meopemuk -> There is one crucial difference between QM and QFT, which is related to the question of Hilbert spaces, the number of degrees of freedom. In particular, I'm talking about the Stone - Von Neumann theorem (see Wald - QFT in Curved Spacetime and Black Hole Thermodynamics, or simply Wikipedia). In short, the theorem states that in the case of finitely many degrees of freedom, all the representations of Weyl relations, i.e. of the canonical commutation relations, are unitarlity equivalent. And this is the reason why for QM you don't need to consider any other Hilbert space than the usual $$L^2(R)$$. But this is not so in QFT, when the theorem breaks down. And there are, in fact, infinitely many representations of the CCR relations that are unitarily inequivalent (!), hence yielding different physical predictions. And one has to use physics to pick the appropriate Hilbert space. In other words, having different Hilbert spaces is ordinary in QFT. This is why you cannot draw consequences about QFT starting from QM. At best you can seek guidance, but you cannot, in general, just adopt arguments from QM to QFT.

An immediate example of the necessity of a different Hilbert space comes from QFT at finite temperature. A KMS state (or finite temperature state) is in no sense an excitation of the vacuum, and hence does not inhabit Fock space. An example of a KMS state is quark gluon plasma at some temperature. You cannot describe this state in any way by using states of Fock space. And this is only the tip of the iceberg. I think I've read somewhere that multiparticle states are a set of measure zero in the set of all the (physically relevant) states. And this brings me to the next point.

Weinberg does like his particles indeed. And sure in Minkowski spacetime, while dealing with particle scattering it leads to a consistent theory, this is not so when considering a generic Lorentzian spacetime. The very concept of particle is not at all clear. Even by staying in Minkowski space, and attempting to quantize a field in Rindler space (a wedge of Minkowski space), one ends up with a different particle interpretation. (See again book by Wald - chapter 5.1, The Unruh effect in flat space - or S.Fulling - Nonuniqueness of canonical field quantization in Riemannian spacetime, Phys.Rev.D 7, (1973)). The particle aspects of QFT are just a part of what QFT is. Sure, you can start with particles and eventually arrive to fields. But it is conceptually more elegant to start from a generic concept, the field, that has validity beyond the particle aspect.

When you talk about the dressed particle approach, what I'd like to know is how does such an approach cope with field theory at finite temperature. Or in condensed matter. Or on curved spaces. (In particluar I'd like to know about the latter as on a generic curved background you will not even be able to write down a mode expansion. And if you define some creation and annihilation operators you will still need to tell me why is that specific choice the physically correct one, given that there is no preferred vacuum state. Actually, there is no vacuum state at all.) Oh, and I don't want to turn this into a discussion about the dressed particle approach... or any other approach for that matter!

My bottom line is basically the following. One, you cannot draw parallels between QM and QFT because there's a world of difference between the two when it comes to the formal, mathematical aspects of the theories. Two, there is MUCH MORE to QFT than just particle scattering in Minkowski space. Finite temperature states, finite density states, QFT on curved spaces are just some examples. And any approach at modifying QFT will have to be able to deal with at least these scenarios, as QFT does. And each of these requires a different Hilbert space.

(I know this is not an argument that free theory andinteracting theory live in different Hilbert spaces, but just a suggestion that one should not be too surprised about it. Waiting for DarMM's post...)

 

[Bob_for_short]

...in QFT...there are, in fact, infinitely many representations of the CCR relations that are unitarily inequivalent (!), hence yielding different physical predictions. And one has to use physics to pick the appropriate Hilbert space.

No problem. Physicists are obliged to choose the right models all the time. That's their job.

I do not know why you appeal to the "quark gluon plasma at some temperature". As if this were an exactly solvable system with some "impossibilities" demonstrated. We speak first of all of QED. As well, a curved space is off-topic. Do not complicate the things. As far as I know, Eugene prefers a flat space-time and so do I.

...free and interacting theories live in different Hilbert spaces...

OK, this statement would make sense if we started from free (bare) particles and finished with interacting (dressed) ones. But who prevents us from starting from interacting (dressed) particles and finish with them in the same Hilbert space? I cannot get rid of feeling that it is all about aggravation brought up with the interaction neglected in the zeroth approximation. As soon as nobody could demonstrate so far the difference between In- and Out-electrons, such statements serve just to cover the renormalization prescription.

 

[meopemuk]

My bottom line is basically the following. One, you cannot draw parallels between QM and QFT because there's a world of difference between the two when it comes to the formal, mathematical aspects of the theories.

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.

Two, there is MUCH MORE to QFT than just particle scattering in Minkowski space. Finite temperature states, finite density states, QFT on curved spaces are just some examples. And any approach at modifying QFT will have to be able to deal with at least these scenarios, as QFT does. And each of these requires a different Hilbert space.

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]

An immediate example of the necessity of a different Hilbert space comes from QFT at finite temperature. A KMS state (or finite temperature state) is in no sense an excitation of the vacuum, and hence does not inhabit Fock space.

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]

...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.

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]

...which we have built even before introducing interactions.

...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]

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]

...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?

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]

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

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

where $H_0$ and $H$ are the free and interacting Hamiltonians, respectively. At time 0 both sets of creation operators reduce to $a^{dag}(p,0)$, 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.

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]

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]

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]

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]

[...] 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]

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]

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:
$$\left( \Box + m^{2} \right)\phi\left(x\right) = gj\left(x\right)$$

Now I'm actually going to start from what meopemuk calls "QFT2". The Hamiltonian of the free theory is given by:
$$H = \int{dk \omega(k)a^{*}(k)a(k)}$$
Where $$a^{*}(k)$$,$$a(k)$$ 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:
$$H = \int{dk \omega(k)a^{*}(k)a(k)} + \frac{g}{(2\pi)^{3/2}}\int{dk\frac{\left[a^{*}(-k) + a(k)\right]}{\sqrt{2}\omega(k)^{1/2}}\tilde{j}(k)}$$

So far, so good.

Now the normal mode creation and annihilation operators for this Hamiltonian are:
$$A(k) = a(k) + \frac{g}{(2\pi)^{3/2}}\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}$$
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:
$$H = \int{dk E(k)A^{*}(k)A(k)}$$,
where $$E(k)$$ 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:
$$\Omega = Z^{1/2} \sum^{\infty}_{n = 0} \frac{1}{n!} - \left(\frac{g}{(2\pi)^{3/2}}\int{\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}a^{*}(k)}\right)^{n} \Psi_{0}$$.
Where $$\Psi_{0}$$ is the free vacuum.
Also $$Z = exp\left[\int{\frac{g^{2}}{(2\pi)^{3}}\frac{|\tilde{j}(k)|^{2}}{\sqrt{2}\omega(k)^{3}}}\right]$$

Now for a field weak enough that:
$$\frac{\tilde{j}(k)}{\omega(k)^{3/2}} \in L^{2}(\mathbb{R}^{3})$$
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 $$A(k), A^{*}(k)$$ 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 $$A(k)$$. Now this constructed Fock space always exists, no problem. Let's call this Fock space $$\mathcal{F}_{I}$$.

However, if condition (1) is violated something interesting happens. $$Z$$ vanishes. Now the expansion for $$\Omega$$ is a sum of terms expressing the overlap of $$\Omega$$ with free states. If $$Z=0$$, then $$\Omega$$ 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 $$\mathcal{F}_{I}$$ is completely orthognal to all states in $$\mathcal{F}$$, the free Fock space. Hence the two Hilbert spaces are disjoint.

So the Fock space for $$a(k),a^{*}(k)$$ is not the same Hilbert space as the Fock space for $$A(k), A^{*}(k)$$. They are still both Fock spaces, however $$A(k), A^{*}(k)$$ has a different algebra, so it's the Fock representation of a new algebra. If one wanted to still use the $$a(k),a^{*}(k)$$ 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 $$\mathcal{F}_{I}$$, 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?

 

[DarMM]

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 separate 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 $$\phi^{4}$$ in two dimensions exist as a well-defined mathematical entity?, If it does exist, does it satisfy the axioms from axiomatic field?"

 

[DarMM]

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.