Rigorous Quantum Field Theory.

[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.

 

[Bob_for_short]

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]

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.

What energy can absorb/emit new quanta corresponding to operators A(k)?

I'm not sure what you are asking.

 

[Bob_for_short]

..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.

I see it as a problem with an external source, not a problem in an external filed. If j(x) is a known function, then I have an exact solution.

The old quanta are physical - they exchange with energy-momentum corresponding to experimental data (in case of photons). To what do correspond new quanta, if any?

 

[DarMM]

I see it as a problem with an external source, not a problem in an external filed.

What's the difference?

If j(x) is a known function, then I have an exact solutions.

So do I, so has everybody for the last fifty years. They are written above or are to be found in Reed and Simons books.

The old quanta are physical - they exchange with energy-momentum corresponding to experimental data (in case of photons). To what correspond new quanta, if any?

They are energy eigenstates, the eigenstates of the new Hamiltonian. That's why they diagonalize it, just like regular QM. The old quanta do not diagonalize the new Hamiltonian and so are not energy eigenstates. Hence they do not correspond to the energy measured in experiments. Just like regular QM.

 

[Fredrik]

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}.
...
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.

You said that you're using "QFT2", which means that you start with a Hilbert space \mathcal H of one-particle states. That Hilbert space is the representation space of a massive spin-0 irreducible representation. The Fock space \mathcal F is (I assume) the Fock space constructed from \mathcal H. So you must be using that to define a(k). And then you define A(k)=a(k)+f(k)I where f is a complex-valued function and I is the identity operator. This means that A(k) is just another operator on \mathcal F.

How can any argument you make after that lead to the conclusion that A(k) is an operator on a Hilbert space that's disjoint with \mathcal F?

 

[Bob_for_short]

What's the difference?

An external source (current) emits photons in coherent states. The energy is not defined - there is no an eigenstate but a superposition (coherent states). There is an average energy and its dispersion, etc., since the number of photons is not certain but the wave phase is.

An external filed is understood for a particle, not for photons. It can lead to bound states different from free states. So the difference between these two cases is obvious - the subjects are different.

They are energy eigenstates, the eigenstates of the new Hamiltonian. That's why they diagonalize it, just like regular QM. The old quanta do not diagonalize the new Hamiltonian and so are not energy eigenstates. Hence they do not correspond to the energy measured in experiments. Just like regular QM.

A classical current (antenna) emits uncertain number of photons. It is not an eigentsate and it is known. It does not prevent the emitted photons from being the true quanta with E=hf.

As soon as your, new quanta, correspond to a certain energy (do they?) there is an obvious contradiction with the exact solution. That is why I am asking about them.

EDIT: We can take an exact solution in terms of old operators, make the variable change and express it in terms of the new operators. If the new operators correspond to new eigenstates (let us admit it for instance), then the original solution will be a superposition of new eigenstates too, so the solution is not transformed in a state with a certain energy (a pure eigenstate).

An analogue of such a variable change exists already in the usual QED if you transform the "linearly polarized" c/a operators into "circularly polarized" ones. The original solution remains a coherent state with uncertain energy.

 

[DarMM]

You said that you're using "QFT2", which means that you start with a Hilbert space \mathcal H of one-particle states. That Hilbert space is the representation space of a massive spin-0 irreducible representation. The Fock space \mathcal F is (I assume) the Fock space constructed from \mathcal H. So you must be using that to define a(k). And then you define A(k)=a(k)+f(k)I where f is a complex-valued function and I is the identity operator. This means that A(k) is just another operator on \mathcal F.

How can any argument you make after that lead to the conclusion that A(k) is an operator on a Hilbert space that's disjoint with \mathcal F?

Good question. Basically by staying in the abstract algebra of operators at all times. I have the operators for creating free massive spin-0 bosons, I can then express the "external-source" Hamiltonian as a function of these operators.
Of course I can also express it as a function of operators which diagonalize it.

So I have two different Hamiltonians, one diagonalized by one set of creation and annihilation and another by another set.
However I have not yet passed to a representation. If I do I can use the quantity Z to measure how disjoint the two representations are. If Z is non-zero, then the reps are unitarily equivalent, otherwise they are not.
Essentially A(k) is made a function of a(k), at the abstract operator level, before I pass to reps. I apoligise if this is unclear.

The reason this is an excellent question is that it leads naturally to the point of view of algebraic field theory. That is, the operators themselves and their algebra are fundamental. Things become clearer at the level of the operators in abstract, before we pass to a rep. This is an important insight in Rigorous Quantum Field Theory.

 

[DarMM]

An external source (current) emits photons in coherent states. The energy is not defined - there is no an eigenstate but a superposition (coherent states). There is an average energy and its dispersion, etc., since the number of photons is not certain but the wave phase is.

I'm not talking about photons. My example above discusses massive spin-0 bosons, not massless spin-1 bosons. The problems you are talking about concern massless particles. I'm aware of these problems, but I am not discussing massless particles, hence there is no need for coherent states.

 

[Bob_for_short]

I'm not talking about photons. My example above discusses massive spin-0 bosons, not massless spin-1 bosons. The problems you are talking about concern massless particles. I'm aware of these problems, but I am not discussing massless particles, hence there is no need for coherent states.

First of all there is no problem in case of the coherent radiation in QED, so I do not impose " a problem" to your model. Next, if your source (current j(x)) is sufficiently energetic (ω>>mc²), your case is not different from a massless case. That is why I have not noticed any particular difference in physics in these both cases.

 

[Bob_for_short]

Rigorous QFT divides into three areas:

Axiomatic Field Theory
...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...

...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. ...

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?"

Here a certain interaction Lagrangian/Hamiltonian is implcitely implied and this fact is crucial. It is not about "the most general" properties of fields but in certain frames imposed with the interaction term. An the the corresponding properties are crucially dependent on this term, on its physics. \phi^{4} ot jA imply a self-action.

 

[DarMM]

First of all there is no problem in case of the coherent radiation in QED, so I do not impose " a problem" to your model. Next, if your source (current j(x)) is sufficiently energetic (ω>>mc²), your case is not different from a massless case. That is why I have not noticed any particular difference in physics in these both cases.

There are some similarities in that both models have particles being created by an external source, but with the presence of a mass gap there is the fundamental difference that there is no coherent states. If the source is strong enough to create particles, then yes particles will be emitted. However there is no soft particles here, so I don't have to treat infrared issues such as coherent states. This is a big difference, the physics of massive and massless particles are not similar even if both are being created by the same mechanism. This is the importance of the mass gap.

 

[DarMM]

Here a certain interaction Lagrangian/Hamiltonian is implcitely implied and this fact is crucial. It is not about "the most general" properties of fields but in certain frames imposed with the interaction term. An the the corresponding properties are crucially dependent on this term, on its physics. \phi^{4} ot jA imply a self-action.

In Axiomatic Quantum Field and Algebraic Quantum Field theory no interaction term is implied or any properties of the interaction, besides that it obey relativity.
In Constructive Field Theory a certain interaction is explicitly chosen, because you are trying to show a specific theory is well-defined. So by showing Yukawa Theory exists a Yukawa interaction is obviously chosen.

In none of the cases is it implict though, it's either not assumed (Axiomatic and Algebraic) or explicitly chosen (Constructive).

 

[Bob_for_short]

There are some similarities in that both models have particles being created by an external source, but with the presence of a mass gap there is the fundamental difference that there is no coherent states. If the source is strong enough to create particles, then yes particles will be emitted. However there is no soft particles here, so I don't have to treat infrared issues such as coherent states. This is a big difference, the physics of massive and massless particles are not similar even if both are being created by the same mechanism. This is the importance of the mass gap.

Fortunately there is no IR problem in QED description of a classical current radiations. The exact solution exists and is physically meaningfull. Your fears are groundless and the solutions are physically very similar. So in both cases we have an exact solution with incertain energy. This is clear and there is no problem with it at all.

As soon as A(k), A(k)⁺ have different commutation relationships, what states do they create? Are their "energy levels" equidistant, etc.?

 

[Bob_for_short]

In Axiomatic Quantum Field and Algebraic Quantum Field theory no interaction term is implied or any properties of the interaction, besides that it obey relativity.

But any vertex implies a certain interaction, doesn't it?

 

[DarMM]

Fortunately there is no IR problem in QED description of a classical current radiations. The exact solution exists and is physically meaningfull. Your fears are groundless and the solutions are physically very similar. So in both cases we have an exact solution with incertain energy. This is clear and there is no problem with it at all.

Okay, I'm not talking about QED. I'm talking about the simple model posted in message #66. I have no fears about QED and I'm aware that there aren't any problems with the infrared part of the theory since those problems are solved by coherent states. However what I'm talking about is not QED, so I'm not going to talk about those things. It's not that I think QED has a problem, it is simply that I'm not talking about it and have chosen a model where coherent states are absent. I understand what you are talking about, but it doesn't occur here.

As soon as A(k), A(k)⁺ have different commutation relationships, what states do they create? Are their "energy levels" equidistant, etc.?

They create the physical eigenstates, as I have said.

 

[DarMM]

But any vertex implies a certain interaction, doesn't it?

Sure, but they don't treat field theory using diagrams. Particularly Algebraic Field Theory. They just study consequences of conditions on the field themselves.

 

[Bob_for_short]

Sure, but they don't treat field theory using diagrams. Particularly Algebraic Field Theory. They just study consequences of conditions on the field themselves.

OK, for me everything is clear. My message is that we can construct a QFT in a rigorous way similar to atom-atomic interaction mentioned above, i.e., with everyhting physically meaningful.

Thanks, DarMM, for your posts and replies.

Regards,

Vladimir.

 

[meopemuk]

DarMM, thank you for the good example. It will allow us to discuss a few important issues.

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)}

Please note that your interaction V \propto a^{*} + a is called "bad" in the language of "dressed particle" theory. This theory explicitly forbids using such "bad" interactions. If you had selected a "good" interaction, then you wouldn't get all the problems that you've described below. In particular, your "dressed particles" (eigenstates of the interacting Hamiltonian) would be no different from "bare" particles (eigenstates of the free Hamiltonian).

On the other hand, such "bad" interactions are used very often in QFT (they will appear every time you build interactions as products of quantum fields), and we need to discuss how to make sense of them.

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}.

This is exactly what other people call a "dressing transformation". The new a/c operators A^{*}, A are said to create/annihilate "dressed" particles. One troublesome point is that your operators A^{*}, A do not satisfy usual commutation relations. I think this is unphysical. The fact that there exists a vacuum vector annihilated by all A(k) is not sufficient to declare that A^{*}, A are valid a/c operators. The canonical form of commutation relations is important as well. This form ensures that a/c operators behave as they supposed to do, i.e., change the number of particles in the system by \pm 1. This condition can be satisfied by using unitary "dressing transformation". Apparently, your "dressing transformation" is not unitary. However, this is not such a big deal, and does not affect the issue of "different Hilbert spaces". I think your transformation can be made unitary without much trouble, if needed.

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.

I think we should be careful before claiming that a(k), a^{*}(k) and A(k), A^{*}(k) give us orthogonal Hilbert spaces. I can agree that expansion coefficients of "dressed" states wrt "bare" states are zero. But this does not mean that they belong to different Hilbert spaces. Let me explain what I mean on this simple example:

Consider simple 1-particle quantum mechanics. Eigenstates of the momentum operator are usual plane waves in the position representation. If we want these eigenstates to be normalized, we must multiply them by a normalization factor that is effectively zero (but not exactly zero!). So, it would be tempting to conclude that normalized plane waves are orthogonal to all "normal" states in the 1-particle Hilbert space. Do they belong to some other orthogonal Hilbert space? Some people try to resolve this problem by introducing "rigged Hilbert spaces", "Gelfand triples", etc. Personally, I don't like these ideas. My opinion is that we are using too narrow a definition of the Hilbert space. We should use a broader definition of Hilbert spaces, i.e., such that eigenvectors of unbounded operators (like momentum) can find their place there. This requirement is dictated to us by physics, and our math must follow physical requirements, not the other way around. I am not sure exactly what mathematics should be used for this purpose. Perhaps the "non-standard analysis" of A. Robinson could help, but my math skills are too weak to go there.

My guess is that in a properly defined "broad" or "non-standard" Fock space there should be enough place for both "bare" and "dressed" particles.

Eugene.

 

[meopemuk]

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.

I would like to avoid "putting words in strangerep's mouth" again. If you are right and the mentioned "ill-definiteness" is related to loops (as we discussed already in the post #53 in https://www.physicsforums.com/showthread.php?t=348911&page=4), then I have a few comments.

If our interaction Hamiltonian is constructed as a product of quantum fields, then when we calculate the S-matrix in perturbation theory we must evaluate products of such products. These terms lead inevitably to the appearance of non-zero (in most cases even infinite) loop contributions to vacuum->vacuum and 1-particle->1-particle scattering amplitudes. This "ill-definiteness" is exactly what is cured by the renormalization prescription. You may not like this "cooking recipe", but it works well as far as practical applications are concerned.

There is however a different way to deal with the "ill-definiteness" of the products of fields. Just make sure that products of fields (and associated "bad" terms) never appear in your interactions. If your interaction is built from "good" terms only, then there is no need for renormalization, and there are no divergences.

Eugene.

 

[Bob_for_short]

The equations of motion are:
\left( \Box + m^{2} \right)\phi\left(x\right) = gj\left(x\right) ...(1)

Now I'm actually going to start from what meopemuk calls "QFT2". The Hamiltonian of the free theory is given by:
H_0 = \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)} ...(2)

So far, so good.

I did not verify what Hamiltonian corresponds to the equation (1). Is (2) a modification of (1) or just the total Hamiltonian?

If (2) does not correspond to (1), then (2) is a new problem with its own dynamics different from what I discussed above. Is it indeed?

 

[Bob_for_short]

I would like to avoid "putting words in strangerep's mouth" again.

I guess I am right. "Ill-defined" things occur in practical calculations and make them impossible. Otherwise nobody would care.

If you are right and the mentioned "ill-definiteness" is related to loops (as we discussed already in the post #53 in https://www.physicsforums.com/showthread.php?t=348911&page=4), then I have a few comments.

If our interaction Hamiltonian is constructed as a product of quantum fields, then when we calculate the S-matrix in perturbation theory we must evaluate products of such products. These terms lead inevitably to the appearance of non-zero (in most cases even infinite) loop contributions to vacuum->vacuum and 1-particle->1-particle scattering amplitudes. This "ill-definiteness" is exactly what is cured by the renormalization prescription. You may not like this "cooking recipe", but it works well as far as practical applications are concerned.

I call "ill defined" expressions divergent, I am not shy. Renormalizations are discarding perturbative corrections to the masses and charge. (I am not shy to call the things as they are.) In most cases in QFT this prescription does not work.

There is however a different way to deal with the "ill-definiteness" of the products of fields. Just make sure that products of fields (and associated "bad" terms) never appear in your interactions. If your interaction is built from "good" terms only, then there is no need for renormalization, and there are no divergences.

I know that. Imagine, in my Hamiltonian they come in such combinations that the vacuum and one-particle (electronium) states are stable.

 

[DarMM]

Please note that your interaction V \propto a^{*} + a is called "bad" in the language of "dressed particle" theory.

Why is it bad?

In particular, your "dressed particles" (eigenstates of the interacting Hamiltonian) would be no different from "bare" particles (eigenstates of the free Hamiltonian).

I don't understand this, even in regular quantum mechanics the eigenstates of the interacting Hamiltonian are different to the eigenstates of the free Hamiltonian. I thought this would be easily agreed on.

One troublesome point is that your operators A^{*}, A do not satisfy usual commutation relations. I think this is unphysical.
The fact that there exists a vacuum vector annihilated by all A(k) is not sufficient to declare that A^{*}, A are valid a/c operators. The canonical form of commutation relations is important as well. This form ensures that a/c operators behave as they supposed to do, i.e., change the number of particles in the system by \pm 1.

It doesn't matter. You can prove these operators have a Fock representation, hence there is a Hilbert space where they move the number of particles by \pm 1

I think we should be careful before claiming that a(k), a^{*}(k) and A(k), A^{*}(k) give us orthogonal Hilbert spaces. I can agree that expansion coefficients of "dressed" states wrt "bare" states are zero. But this does not mean that they belong to different Hilbert spaces. Let me explain what I mean on this simple example:

They are orthogonal to every single free state, so at the very list the Hilbert space is a direct sum,
\mathcal{F}_{I} \oplus \mathcal{F}. However you can show that such a thing is not true using representation theory, so they truly do live in different Hilbert spaces.

Consider simple 1-particle quantum mechanics. Eigenstates of the momentum operator are usual plane waves in the position representation. If we want these eigenstates to be normalized, we must multiply them by a normalization factor that is effectively zero (but not exactly zero!). So, it would be tempting to conclude that normalized plane waves are orthogonal to all "normal" states in the 1-particle Hilbert space. Do they belong to some other orthogonal Hilbert space? Some people try to resolve this problem by introducing "rigged Hilbert spaces", "Gelfand triples", etc. Personally, I don't like these ideas. My opinion is that we are using too narrow a definition of the Hilbert space. We should use a broader definition of Hilbert spaces, i.e., such that eigenvectors of unbounded operators (like momentum) can find their place there. This requirement is dictated to us by physics, and our math must follow physical requirements, not the other way around. I am not sure exactly what mathematics should be used for this purpose. Perhaps the "non-standard analysis" of A. Robinson could help, but my math skills are too weak to go there.

My guess is that in a properly defined "broad" or "non-standard" Fock space there should be enough place for both "bare" and "dressed" particles.

Can I ask why Fock space is so important? In another thread you said that relativistic covariance could be given up to ensure that the interacting theory remained in Fock space. In this model we do give that property up and yet we still have a different Hilbert space. Now you want to change the concept of Hilbert space as used in QM, just so that we remain in Fock space? Why? It's just a Hilbert space, why change QM and relativity to avoid this fact?
In my opinion your doing the reverse of what you claim, giving up physics we know (no lorentz covariance, superluminal signaling) in order to keep a mathematical structure (Fock space).

 

[Bob_for_short]

And my questions about the new spectrum and the total Hamiltonian, please!

 

[meopemuk]

Please note that your interaction V \propto a^{*} + a is called "bad" in the language of "dressed particle" theory.

Why is it bad?

By definition, "good" operators in the normally-ordered form (all annihilation operators on the right, all creation operators on the left) must have at least two creation operators and at least two annihilation operators. The simplest example of a "good" operator is a^*a^*aa. Their importance stems from the fact that they yield zero when acting on any 1-particle state or on the vacuum. Another important property is that the product or commutator of any number of "good" operators is also "good".

There is also a class of operators that I call "renorm". They are either a^*a or multiplication by a constant. The free Hamiltonian is one example of a "renorm" operator.

In simple theories that we are discussing, all other operators are "bad". Your operator V \propto a^{*} + a is "bad" in this classification. Its unpleasant property is that (normally ordered) products of such operators contain "renorm" terms. If V happens to be in the interaction Hamiltonian, then the S-operator expansion S \propto 1 + V + VV + VVV +... contains "renorm" terms (in addition to the first "1"), which signify the presence of self-interaction and self-scattering in the vacuum and 1-particle states. This leads immediately to the necessity of renormalization and other unpleasant effects.

On the other hand if interaction V contained only "good" terms, then all multiple products of V would yield zero when acting on the vacuum and 1-particle states, which agrees with the intuitive understanding that vacuum and single particle cannot scatter off themselves. There is no need for renormalization if V is "good".

I don't understand this, even in regular quantum mechanics the eigenstates of the interacting Hamiltonian are different to the eigenstates of the free Hamiltonian. I thought this would be easily agreed on.

Yes, generally the interacting and free Hamiltonians have different eigenstates. However, the important issue is about the vacuum and 1-particle eigenstates. In your Hamiltonian H_0 +a^* + a, the interaction acts non-trivially on the free vacuum and 1-particle states. Therefore, 0-particle and 1-particle eigenstates of this Hamiltonian are different from 0-particle and 1-particle eigenstates of the free Hamiltonian. Your "dressed" particles are different from "bare" particles. This is why the (mass) renormalization is needed.

On the other hand, if I have a Hamiltonian with a "good" interaction, for example H_0 + a^*a^*aa, then its 0-particle and 1-particle eigenvectors are exactly the same as the free vacuum and free particles, respectively. There is no need for the (mass) renormalization. Both free and interacting theories live in the same Fock space. 0-particle and 1-particle sectors of the two theories are exactly the same. On the other hand, 2-particle, 3-particle, etc. sectors of the two theories are quite different. In the interacting theory, 2-particles are allowed to have a non-trivial scattering, form bound states, etc. This is consistent with the intuitive understanding that interactions can occur only if there are 2 or more particles. By choosing "good" interactions only we eliminate unphysical self-interactions in the vacuum and 1-particle states. At the same time we avoid unphysical renormalizations and divergences.

Can I ask why Fock space is so important? In another thread you said that relativistic covariance could be given up to ensure that the interacting theory remained in Fock space. In this model we do give that property up and yet we still have a different Hilbert space. Now you want to change the concept of Hilbert space as used in QM, just so that we remain in Fock space? Why? It's just a Hilbert space, why change QM and relativity to avoid this fact?
In my opinion your doing the reverse of what you claim, giving up physics we know (no lorentz covariance, superluminal signaling) in order to keep a mathematical structure (Fock space).

I like Fock space, because it is a natural consequence of two physically transparent statements:

(1) Particle numbers are valid physical observables (hence there are corresponding Hermitian operators);
(2) Numbers of particles of different types are compatible observables (hence their operators commute)

Please note that I make a distinction between "Lorentz covariance" and "relativistic invariance":

"Lorentz covariance" is the assumption that certain quantities (like space-time coordinates of events or components of quantum fields) have simple (e.g., linear) transformation properties with respect to boosts.

"Relativistic invariance" is the requirement that the theory is invariant with respect to the Poincare group.

In my opinion, the relativistic invariance is the most important physical law. It can be never compromised. On the other hand, Lorentz covariance cannot be derived from this law directly (some additional dubious assumptions are needed for this "derivation"). So, Lorentz covariance is just an approximate property, which makes sense for non-interacting or weakly-interacting systems only. Superluminal propagation of interactions is definitely in conflict with Lorentz covariance, but it is fully consistent with relativistic invariance. The conflict with causality is resolved by the fact that boost transformations of particle observables must be non-linear and interaction-dependent. This follows from the interaction-dependence of the total boost operators, as explained in Weinberg's book.

Note also that I don't want to modify quantum mechanics in any significant way. I just want to broaden the definiton of the Hilbert space, so that eigenvectors of unbounded operators (like momentum) can live there.

Eugene.

 

[strangerep]

Hmmm. So once again, while I slept, an avalanche of posts in this thread has
overwhelmed me. Although, curiously, a couple of people seemed to be
whispering in my dreams, debating about what I "really" meant but not actually
asking me. Quite bizarre. Anyway, you guys are probably all in bed by now,
so I can have a some peace to read carefully and respond... :-)

First, DarMM's posting of the "external field" example...

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)} +<br /> \frac{g}{(2\pi)^{3/2}}\int{dk\frac{\left[a^{*}(-k) +<br /> a(k)\right]}{\sqrt{2}\omega(k)^{1/2}}\tilde{j}(k)}

So far, so good.

I'm guessing that \tilde{j}(k)} is a c-number, right? I.e., it commutes with everything?
(I'll proceed on this assumption, but if it's wrong, please tell me what commutation relations it satisfies.)

Now the normal mode creation and annihilation operators for this Hamiltonian
are:
A(k) = a(k) +<br /> \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.

OK, so we want to diagonalize the full Hamiltonian in terms of new a/c ops
A^{*}(k),\,A(k), and in this case it's fairly easy to guess what
they are in terms of the free a/c ops. Let me re-write the 2nd-last formula
above in a simpler form:
<br /> A(k) ~=~ a(k) ~+~ z(k)<br />
where (hopefully) the definition of my z(k) is obvious.

Now, you said above that the A(k) don't satisfy the usual commutation relations.
I don't understand this. If z(k) commutes with the free a/c ops a(k), etc,
then the A(k) do satisfy the canonical commutation relations, afaict.
Or did I miss something?

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!} -<br /> \left(\frac{g}{(2\pi)^{3/2}}\int{\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}a^{*}(k)}\right)^{n}<br /> \Psi_{0}.

Something looks wrong with that expression.
Should the minus sign be inside the parentheses?

Where \Psi_{0} is the free vacuum.
Also Z =<br /> 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:
<br /> \frac{\tilde{j}(k)}{\omega(k)^{3/2}} \in L^{2}(\mathbb{R}^{3})<br /> ~~~~~~~~(1)<br />

then everything is fine. [...]

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

However, if condition (1) is violated something interesting happens.
Z vanishes.

Did you forget a minus sign in your definition of Z above? It looks like
it goes to infinity rather than zero when condition (1) is violated.

(Or did you perhaps mean that \Omega becomes non-normalizable when Z\to\infty?)

- - - - - - - - - - - - - - - - -

I'm also guessing that (after correcting any errors) the example is really
just the well-known so-called "field displacement" transformation, i.e.,
<br /> A^*(k) ~:=~ a^*(k) ~+~ g\,\bar{z}(k) ~~;~~~~~~<br /> A(k) ~:=~ a(k) ~+~ g\,z(k)<br />
which alternatively can be expressed as a formally-unitary transformation
<br /> A(k) ~:=~ U[z] \, a(k) \, U^{-1}[z]<br />
where U is of the form
<br /> U[z] ~:=~ exp\int\!\!dk\Big( \bar{z}(k)a(k) - z(k)a^*(k) \Big)<br />
(where possibly I might have a sign wrong.)
This U[z] is essentially equivalent to the operator you used to go from the free vacuum \Psi_0 to the interacting vacuum \Omega.

The alert reader may have noticed that the form of U[z] is exactly the
same as that which generates ordinary (Glauber) coherent states in the
inf-dof case. I might say more about that later, depending on what else
Bob_for_short wrote (and if he doesn't badger me about it).

The point is that states generated by U[z] acting on the free vacuum \Psi_0 are only in the free Fock space if z(k) is square-integrable.
(I have some more detailed latex notes on this calculation that I could possibly post if anyone cares. :-)

For mathematical literature on this model:
Reed, M. and Simon, B. Methods of Modern Mathematical Physics, Vols. II-III

If you have those volumes handy, could you possibly give a more precise reference?

Rigorous QFT divides into three areas
[...]

Thanks. I see now that I'm mostly interested in algebraic-constructive
stuff. (I.e., constructing under the more general algebraic umbrella.)

And... hmm... I've run out of time, and can't do any more posts today. :-(

(Eugene, I know there's some posts aimed in my general direction that I haven't
answered yet. I'll try again tomorrow. :-)

 

[Bob_for_short]

...What's the difference?...

In your case (an external or classical current/source) your Lagrangian density contains the term jφ. In case of an external filed V(x) it would contain φ²V and V would get into the field equation as a potential.

I think your H is just H_tot corresponding to the original equation. There is no modification in the problem but "modification" of H₀ to get the original, full equation.

I verified, A and A⁺ have the same CCR so the "new" excitation spectrum is the same. Although in terms of A the Hamiltonian is diagonalized, the problem solution remains a superposition of different eigenstates, i.e., a state without a certain energy (a la coherent states). The particluar for massive φ spectrum \\w(k) (dispersion law with or without gap) is not important.

---> Strangerep. I did not mean to offend you or answer for you. We were chatting on-line and I expressed what I thought following good sense in order to advance in discussion. I think my answer was reasonable (although I did not mention vertices). If you had been on-line, you would have answered yourself, I guess. A live chat needs quick responses. I am sorry, Strangerep, forgive me. (Consider it as my politeness - I did not want to wake you up and bother with a minor question.)

As to our problems with practical calculations in QED and QFT, the example of DarMM shows that there is no mathematical and conceptual problems in case of a know source j(x). (I hope we all agree that it is not a "free" case.)

The problems arise when we couple the unknown current j(x) with unknown filed φ (or A in QED via jA). It is precisely here where the self-action is introduced.
I take the advantage to show how one can proceed in a more physical way. In case of a know current j_ext the charge motion is determined with a strong external field so the charge acceleration in j_ext can be expressed via the external force F_ext. So instead of j_ext we can substitute its expression via the external force. Then the original equation reads as excitation of quanta due to the external force: it is the external force that stands in the right-hand side of the original equation for the quanta being excited. Then it is quite natural to suppose that this charge is a part of oscillators, - perturbing a part of oscillator excites the oscillator, like I described in my publications. Then a self-consistent theory is built quite straightforwardly in terms of compound systems without self-action - it is a theory of interacting compound systems.

 

[Bob_for_short]

By definition, "good" operators in the normally-ordered form (all annihilation operators on the right, all creation operators on the left) must have at least two creation operators and at least two annihilation operators. The simplest example of a "good" operator is a^*a^*aa. Their importance stems from the fact that they yield zero when acting on any 1-particle state or on the vacuum. Another important property is that the product or commutator of any number of "good" operators is also "good".

Just in this example it is clearly seen that there is nothing bad or non-physical it the solution expressed via certain combinations of "bad" operators. Many-photon states, like coherent ones, are natural and accompany most of scattering processes.

 

[meopemuk]

Just in this example it is clearly seen that there is nothing bad or non-physical it the solution expressed via certain combinations of "bad" operators. Many-photon states, like coherent ones, are natural and accompany most of scattering processes.

Could you please be more specific what you mean by "solution"?

My point is that "bad" operators should not be present in the interaction Hamiltonian.

By the way, "bad" operators are not present in the S-operator of any traditional QFT. This is because the sets of particles created and annihilated by "bad" operators have different energies. (For example, in the case of a^*(p), the energy of annihilated particles is 0 and the energy of created particles is E_p). Therefore, these terms are always "killed" by the presence of energy delta function.

However, the absence of "bad" terms in the S-operator does not forbid formation of many-photon states in scattering. For example, the "good" operator responsible for the emission of two photons in a collision of two electrons is a^*a^*c^*c^*aa (electron operators are denoted by a; photon operators are denoted by c).

Eugene.

 

[Bob_for_short]

Could you please be more specific what you mean by "solution"?

My point is that "bad" operators should not be present in the interaction Hamiltonian.

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)}

\Psi = Z^{1/2} \sum^{\infty}_{n = 0} \frac{1}{n!}\left(\frac{-g}{(2\pi)^{3/2}}\int{dk\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}a^{*}(k)}\right)^{n} \Psi_{0} = Z^{1/2} \exp\left(\frac{-g}{(2\pi)^{3/2}}\int{dk\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}a^{*}(k)}\right)\Psi_{0}.

Where \Psi_{0} is the free vacuum.

The total Hamiltonian contains the c/a operators in a "bad" way, the exact evolution operator does not leave the vacuum and one-particle state stable, the exact solution can be expressed in Glauber's form explicitly containing a "bad" combination. In fact, such a solution gives a Poisson probability distribution determined with one parameter - the average number of quanta (photons in QED).

 

[meopemuk]

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)}

\Psi = Z^{1/2} \sum^{\infty}_{n = 0} \frac{1}{n!}\left(\frac{-g}{(2\pi)^{3/2}}\int{dk\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}a^{*}(k)}\right)^{n} \Psi_{0} = Z^{1/2} \exp\left(\frac{-g}{(2\pi)^{3/2}}\int{dk\frac{\tilde{j}(k)}{\sqrt{2}\omega(k)^{3/2}}a^{*}(k)}\right)\Psi_{0}.

Where \Psi_{0} is the free vacuum.

The total Hamiltonian contains the c/a operators in a "bad" way, the exact evolution operator does not leave the vacuum an one-particle state stable, the exact solution can be expressed in Glauber's form explicitly containing a "bad" combination. In fact, such a solution is a Poisson probability distribution determined with one parameter - the average number of photons.

Please correct me if I misunderstood. Your \Psi is the lowest-energy eigenvector of the interacting Hamiltonian. This is the "dressed" vacuum vector, which is different from the "free" vacuum vector \Psi_{0}. Taken literally, your expression means that in the "dressed" (or "physical") vacuum there is a non-zero chance to find one or more photons. However, this prediction disagrees with experiments. If I place a photon detector in absolute vacuum I will never see photons there. This is exactly the reason why I think that "bad" terms should not be present in interactions and that "physical" vacuum should be a zero-particle state.

Eugene.