What Are the Implications of a New Relativistic Quantum Theory?

[dextercioby]

First, Maxwell's theory claims to be something more general than the theory of a single photon.

Maxwell fields are classical objects, photons are concepts from a quantum theory. Maxwell theory couldn't claim to be <more general> than a theory of one photon, because there's no such thing as a theory of one photon. There's a theory of photons, or of photon quantum states.
As per my understading, Maxwell's theory leads through quantization to a theory of massless, free, relativistic spin 1 quantum fields. The Fock space of this theory comprises as many multi-photon states as you want.

Second, I still insist that *any* relativistic theory must possesses 10 generators of the Poincare group with corresponding commutation relations.

OK.

Maxwell's theory is not formulated in such a way.

Of course not, because it's purely classical.

So, it violates the most basic requirements of quantum mechanics and relativity.

Of course it violates the requirement of QM, because it's a classical field theory. Relativity violated? Of course not, since relativity itself was built to explain Maxwell's theory...

EDIT: Apparently your use of <Maxwell theory> is different from mine and gives rise to confusions. For me Maxwell's name cannot be associated to a quantum theory.

 

[A. Neumaier]

First, Maxwell's theory claims to be something more general than the theory of a single photon.

In terms of quantum mechanics, Maxwell's theory is what you get if you restrict the Photon Fock space (or rather Kibble's far bigger space) to the manifold defined by all coherent states. The coherent states are in 1-1 correspondence with the solutions of the maxwell equations, and behave essentially classically, as you can glance from any quantum optics book. Mandel&Wolf show that this very well accounts for classical optics.

Second, I still insist that *any* relativistic theory must possesses 10 generators of the Poincare group with corresponding commutation relations. Maxwell's theory is not formulated in such a way. So, it violates the most basic requirements of quantum mechanics and relativity.

Maxwell's theory possesses 10 such generators, since it is manifestly covariant, and constitutes an irreducible representation of the full Poincare group. Perhaps you can't see this in your favorite QFT book by Weinberg, but he derives it in one of his any spin papers: Phys.Rev. 134 (1964), B882-B896

In contrast, you theory possesses only approximate such generators, to the order that you do your expansion.

 

[meopemuk]

Maxwell's theory is not formulated in such a way.

Of course not, because it's purely classical.

So, it violates the most basic requirements of quantum mechanics and relativity.

Of course it violates the requirement of QM, because it's a classical field theory. Relativity violated? Of course not, since relativity itself was built to explain Maxwell's theory...

A proper relativistic classical theory must be derived as a limit of a relativistic quantum theory. So, the classical theory must keep the most important features of its quantum counterpart. Namely, it must possesses 10 generators of the Poincare group. The quantum commutator should be represented by its classical analog, such as the Poisson bracket. And the 10 classical generators must satisfy the well known Poincare bracket relations. I don't see these features in Maxwell's theory, so I have a reason to doubt that this theory is a classical limit of a more general quantum approach.

Eugene.

 

[dextercioby]

A proper relativistic classical theory must be derived as a limit of a relativistic quantum theory. So, the classical theory must keep the most important features of its quantum counterpart. Namely, it must possesses 10 generators of the Poincare group. The quantum commutator should be represented by its classical analog, such as the Poisson bracket.

Agree so far.

And the 10 classical generators must satisfy the well known Poincare bracket relations. I don't see these features in Maxwell's theory, so I have a reason to doubt that this theory is a classical limit of a more general quantum approach.

Eugene.

Well, in a classical field theory in the Lagrangian formulation (which has the advantage of being manifestly covariant), we have the famous Noether's theorem which gives us the 10 generators of the Poincare Lie algebra in terms of the classical fields, the 1-forms $A_{\mu} (x)$. You must know that...

 

[meopemuk]

Maxwell's theory possesses 10 such generators, since it is manifestly covariant, and constitutes an irreducible representation of the full Poincare group. Perhaps you can't see this in your favorite QFT book by Weinberg, but he derives it in one of his any spin papers: Phys.Rev. 134 (1964), B882-B896

You've possible mistaken this paper for another Weinberg's work

Weinberg, S. Photons and gravitons in perturbation theory: derivation of
Maxwell's and Einstein's equations. Phys. Rev. 138 (1965) B988-B1002

where he discusses Maxwell equations in section VI. This work is the basis of my subsection 8.1.2 and Appendix N. It is true that Weinberg has 10 quantum generators of the Poincare group. It is also true that his fields $$A^{\mu}$$ and $$J^{\mu}$$ satisfy relationships that formally resemble the set of Maxwell equations. However, this is only a superficial resemblance.

First, Weinberg stresses many times that his goal is to calculate the S-matrix. So, he is not interested in interacting time dynamics of charges and electromagnetic fields (which is the subject of Maxwell's theory). He is right not to go there, because the field Hamiltonian (eqs. (8.10) - (8.14) in my book) is incapable of describing the time evolution even for simplest one-charge states. I've explained that failure in section 10.1 of my book. So, Weinberg's fields are just formal mathematical objects that are not related to anything observed in experiments.

In contrast, you theory possesses only approximate such generators, to the order that you do your expansion.

This is true. Higher perturbation orders require more work.

Eugene.

 

[meopemuk]

Well, in a classical field theory in the Lagrangian formulation (which has the advantage of being manifestly covariant), we have the famous Noether's theorem which gives us the 10 generators of the Poincare Lie algebra in terms of the classical fields, the 1-forms $A_{\mu} (x)$. You must know that...

See my response to Arnold.

Eugene.

 

[dextercioby]

See my response to Arnold.

Eugene.

My post was about the classical theory by itself and not thinking of it as a limit of a quantum theory.

Your statement <A proper relativistic classical theory must be derived as a limit of a relativistic quantum theory> sets conditions on the quantum theory rather than on the classical counterpart. So it's ok.

Further you claim that <the classical theory must keep the most important features of its quantum counterpart>. I think it's again a matter of perhaps wrong wording: the quantum theory must generalize the classical one and could(and should have novel features compared to it. So it's likely that some (if not all) of the <the most important features> of the quantum theory couldn't have any classical analogue. Just think of the H-atom. Also your statement suggests the opposite of the first one: it looks like it sets conditions on the classical theory based on knowledge of the quantum one. But that's wrong, the quantum theory is the issue, the classical one is to be postulated as it's simply a particular case of the specially relativistic dynamics which we already know to be correct.

Futher <Namely, it must possesses 10 generators of the Poincare group>. The classical theory does that, of course.

Next: <The quantum commutator should be represented by its classical analog, such as the Poisson bracket>. It does (actually this is a classical gauge theory, so there's some ambiguity at the classical level in the definition of fields which forces us to use a different simplectic structure on the classical state space than the normal PB). But again, you reverse the logics: to the classical Poisson bracket one must find a proper quantum commutator and not viceversa.

Next: <I don't see these features in Maxwell's theory>. Can you justify your statement ?

 

[meopemuk]

Arnold, dextercioby,

Yes, you've convinced me that Maxwell's theory has a set of 10 generators, e.g., those defined as classical analogs of QED functions $$H, \mathbf{K}, \mathbf{P}, \mathbf{J}$$ of operator fields $$A^{\mu}, J^{\mu}$$ in subsection 8.1.2. So, formally, it is OK as a relativistic theory.

However, I think, it is important to note that this alleged quantum-classical correspondence is only formal. The point is that quantum QED generators in subsection 8.1.2 do not form a viable physical theory. So, its classical limit cannot be viable too. The bad features of QED in 8.1.2 are seen from the fact that the S-matrix computed with the Hamiltonian (8.10) is divergent. Of course, in QED this problem is fixed by adding counterterms, which effectively result in infinite masses and charges. If we were to take a classical limit of QED with counterterms, we wouldn't get the familiar Maxwell's theory.

However, even QED with counterterms is not satisfactory as a quantum theory of interacting charges and photons. As I discuss in section 10.1, the time evolution of states is not acceptable in this approach. The next step should be taken, which is a unitary transformation of the QED Hamiltonian with counterterms to the dressed particle form. Then, finally, we obtain an acceptable quantum theory with a finite Hamiltonian, realistic time evolution of particle states, and experimentally confirmed S-matrix. But the classical limit of this theory is not going to look as Maxwell's theory at all. It looks more like the Darwin-Breit Hamiltonian.

[...] the quantum theory must generalize the classical one and could(and should have novel features compared to it. [...] the classical one is to be postulated [...] to the classical Poisson bracket one must find a proper quantum commutator and not viceversa.

I strongly disagree with the idea that first we must postulate a classical theory and then "quantize" it in order to get a quantum-mechanical counterpart. The most general and exact theory of nature must be both quantum and relativistic. So, if we don't want to make mistakes, we must first postulate a self-consistent fully quantum approach with Hilbert space, commutators, and all that. Then, the classical analog should be obtained in the limit $$\hbar \to 0$$. In this limit we may lose some fine quantum features, but, at least, we can be confident that our classical approach has a solid foundation in quantum postulates. If for some reason we find that the classical limit doesn't work or disagrees with experiment, then we should modify our quantum theory and try again.

The idea of "quantization" can possibly work as a heuristic tool for guessing the form of yet unknown quantum theory in the absence of other theoretical options. But I wouldn't consider "quantization" as a rigorous theoretical mechanism.

Eugene.

 

[A. Neumaier]

you've convinced me that Maxwell's theory has a set of 10 generators, e.g., those defined as classical analogs of QED functions $$H, \mathbf{K}, \mathbf{P}, \mathbf{J}$$ of operator fields $$A^{\mu}, J^{\mu}$$ in subsection 8.1.2. So, formally, it is OK as a relativistic theory.

With time, we'll convince you that, formally, all the things you didn't like about field theory and that you claimed before to be actually false or unproved, are OK. But though all of that stuff is empirically validated, you'll always say that these are only theoretical speculations that have no weight compared with those unvalidated practical speculations that figure in your book...

However, I think, it is important to note that this alleged quantum-classical correspondence is only formal.

Your judgment reveals a serious lack of knowledge of the state of the art. This quantum-classical correspondence is extremely well established and backed up by lots of experimental evidence.

For example, the quantum optics book by Mandel and Wolf discusses in Chapters 5-9 classical Maxwell theory and its optical consequences. Chapter 10-20 do the same for the quantum version, starting with standard QED. They find that the quantum theory of coherent fields is essentially identical with that of classical fields, except for corrections of the order of hbar (that give highly interesting nonclassical effects).

 

[meopemuk]

It is possible to make a source of light, which produces single photons one-by-one on demand. If I shine this light on a double-slit, I get a famous picture on the screen, where each photon makes a separate tiny spot, and only after long exposure the interference pattern emerges. I don't see other way to explain this behavior but the quantum mechanical picture in which light particles are described by a probabilistic wave function. The Maxwellian field representation of light is incapable of describing this experiment at all. So, I conclude that electromagnetic field is just a crude approximation, which works only for high-intensity light, where large numbers of photons are present at once and measuring devices are not sensitive enough to distinguish each individual photon.

Eugene.

 

[A. Neumaier]

It is possible to make a source of light, which produces single photons one-by-one on demand. If I shine this light on a double-slit, I get a famous picture on the screen, where each photon makes a separate tiny spot, and only after long exposure the interference pattern emerges. I don't see other way to explain this behavior but the quantum mechanical picture in which light particles are described by a probabilistic wave function. The Maxwellian field representation of light is incapable of describing this experiment at all.

See
http://arnold-neumaier.at/ms/lightslides.pdf
http://arnold-neumaier.at/ms/optslides.pdf
for the standard quantum optics view of this matter. It is all explainable by the field picture - particles entering only in a semiclassical view.

 

[meopemuk]

See
http://arnold-neumaier.at/ms/lightslides.pdf
http://arnold-neumaier.at/ms/optslides.pdf
for the standard quantum optics view of this matter. It is all explainable by the field picture - particles entering only in a semiclassical view.

So, photons are "localized lumps of energy". Do they pass through one slit or through both slits at once in the double-slit experiment?

Eugene.

 

[A. Neumaier]

So, photons are "localized lumps of energy". Do they pass through one slit or through both slits at once in the double-slit experiment?

As any wave, localized or not, through both slits.

Only upon recording the photons, they materialize - at a single spot only.

 

[A. Neumaier]

Suppose now that we constructed a relativistic interacting theory in which the interacting Hamiltonian is *not* a product of fields. For example, it can be $$V= a^{\dag}a^{\dag}aa$$.

This still is a product of nonrelativistic local fields. I guess you mean: not (a linear combination of) products of relativistic, local fields.

Then [...] two important conditions of Haag's theorem will not be satisfied, and we will not be able to prove that the Fock space is excluded.

Yes.

As a result of this exercise we will obtain a non-trivial interacting theory in the Fock space. "Dressed particle" theories are exactly of this form. Their only problem is that interacting fields are non-covariant and non-commuting. Could you please explain why you think that this is an important problem? Is there any measurable property that proves the impossibility of non-covariant and non-commuting interacting fields?

First, you now have infinitely many interacting terms. So additional problems about well-definedness and self-adjointness arise.

Second, observed physics is Poincare invariant, to very high accuracy. Your Poincare representation is only approximate at any order.

Third, lack of covariance makes all calculations (especially those of higher order) much more messy. This is the main reason, I think, why very few people today work with noncovariant methods.

Finally, you do everything at zero temperature. However, most application of QED where the time evolution is important take place at finite temperature. There everything is different again - in place of the asymptotic particles at T=0 one now has only effective particles, which are different.

 

[meopemuk]

As any wave, localized or not, through both slits.

Only upon recording the photons, they materialize - at a single spot only.

For the double-slit experiment with visible light the distance between two slits can be macroscopic, e.g. 0.1 millimeter, or something like that. This means that the photon "lump" should be no smaller than this size. So, you are saying that the energy lump associated with a single visible-light photon can be as big as 0.1 millimeter or so? And that all this volume is filled with a time-changing electromagnetic field? Do I understand your model correcly?

Eugene.

 

[A. Neumaier]

For the double-slit experiment with visible light the distance between two slits can be macroscopic, e.g. 0.1 millimeter, or something like that. This means that the photon "lump" should be no smaller than this size. So, you are saying that the energy lump associated with a single visible-light photon can be as big as 0.1 millimeter or so? And that all this volume is filled with a time-changing electromagnetic field?

Of course. Photons can be very delocalized. This is precisely what happens in a double slit experiment. Each photon in a laser beam has the same shape as the classical field by which this beam is described.

My views about what photons are turned by almost 180 degrees after I had begun to talk to the experimentalists in Zeilinger's group (who moved to Vienna in 1999)...

 

[meopemuk]

This still is a product of nonrelativistic local fields. I guess you mean: not (a linear combination of) products of relativistic, local fields.

In our example we've considered a $$\Phi^4$$ theory. So, I am saying that $$V = a^{\dag} a^{\dag}aa$$ cannot be expressed as a linear combination of integrals

$$\int d\mathbf{r} \Phi^n(\mathbf{r}, 0)$$

First, you now have infinitely many interacting terms. So additional problems about well-definedness and self-adjointness arise.

What do you mean by "many interaction terms"? My interaction has only one term $$V = a^{\dag} a^{\dag}aa$$.

Second, observed physics is Poincare invariant, to very high accuracy. Your Poincare representation is only approximate at any order.

So, you are saying that it is impossible to formulate a Poincare-invariant theory with interaction $$V = a^{\dag} a^{\dag}aa$$? Can you prove that?

You are making the claim that Haag's theorem does not leave any chance for using Fock space in relativistic quantum theories. So, it is your job to prove that interaction $$V = a^{\dag} a^{\dag}aa$$ violates relativity.

Third, lack of covariance makes all calculations (especially those of higher order) much more messy. This is the main reason, I think, why very few people today work with noncovariant methods.

Messy or not messy is a matter of taste and convenience. This is not a scientific argument.

Finally, you do everything at zero temperature. However, most application of QED where the time evolution is important take place at finite temperature. There everything is different again - in place of the asymptotic particles at T=0 one now has only effective particles, which are different.

I am interested in what happens when two electrons (or other particles) move close to each other. I think it is important to understand this simplest event first. Only then we can switch to systems with many particles, non-zero temperature, etc.

Eugene.

 

[A. Neumaier]

What do you mean by "many interaction terms"? My interaction has only one term $$V = a^{\dag} a^{\dag}aa$$.

So you are talking about a harmonic oscillator?

Or do you mean $$V = \int dx a^*(x) a^*(x)a(x)a(x)$$? This is already a linear combination of field products, not only one product.

So, you are saying that it is impossible to formulate a Poincare-invariant theory with interaction $$V = a^{\dag} a^{\dag}aa$$? Can you prove that?

No. What I was saying is spelled out in post #74.

What you though I was saying is probably possible in the style of Bakamjian, at least when one allows an additional energy shift. But then you don't have cluster separability.

You are making the claim that Haag's theorem does not leave any chance for using Fock space in relativistic quantum theories. So, it is your job to prove that interaction $$V = a^{\dag} a^{\dag}aa$$ violates relativity.

No. I am only claiming that Haag's theorem says what it says. This is in the literature, freeing me from any further obligation.

If you create a theory that violates the assumptions of Haag's theorem, you are not bound to its conclusions.

Messy or not messy is a matter of taste and convenience. This is not a scientific argument.

Of course it is. Kepler's theory was in his time not more accurate than Ptolemy's. But it was far less messy, and made things so much easier that it was adopted as the standard.

I am interested in what happens when two electrons (or other particles) move close to each other.

Whereas I an interested to explain standard QFT to those who want to understand it.

I think it is important to understand this simplest event first. Only then we can switch to systems with many particles, non-zero temperature, etc.

2-electron systems are useless approximations, once one has interactions terms changing particle number.

Once one wants to insist on a dynamical view (rather than the asymptotic scattering view), one needs to prepare the particles at finite time, hence they'll never be exact 2-particle states (which make sense only asymptotically). But even if they were exact for one moment, they'd lose that property the very next moment.

Thus you are chasing a chimera. I prefer to chase the insights of the community of high energy physicists, who all work in the quantum field paradigm.

 

[meopemuk]

So you are talking about a harmonic oscillator?

Or do you mean $$V = \int dx a^*(x) a^*(x)a(x)a(x)$$?

No, I meant a 2-particle interaction whose full expression involves momentum-space a/c operators

$$V = \int d\mathbf{p} d\mathbf{q} d\mathbf{k} V(\mathbf{p}, \mathbf{q}, \mathbf{k}) a^{\dag}(\mathbf{p-k}) a^{\dag}(\mathbf{q+k}) a(\mathbf{p}) a(\mathbf{q})$$

Here $$V(\mathbf{p}, \mathbf{q}, \mathbf{k})$$ is a numerical coefficient function, which can be selected in such a way that interaction is translationally and rotationally invariant. I also claim that one can find a corresponding interacting boost operator, so that entire theory becomes relativistically invariant.

My other claim is that in this theory interacting quantum fields constructed by usual formulas do *not* transform covariantly. So, the main condition of Haag's theorem is not satisfied, the theorem is not applicable, and we are permitted to work in the Fock space.
More details on this example can be found in

H. Kita, "A non-trivial example of a relativistic quantum theory of particles without divergence difficulties", Progr. Theor. Phys., 35 (1966), 934.

If you create a theory that violates the assumptions of Haag's theorem, you are not bound to its conclusions.

Great! I am glad that we agree about that.

Eugene.

 

[meopemuk]

Of course. Photons can be very delocalized. This is precisely what happens in a double slit experiment. Each photon in a laser beam has the same shape as the classical field by which this beam is described.

OK, so we have a macroscopic lump of electromagnetic energy, which falls on the double slit and interferes with itself according to Maxwell equations. Then this macroscopic lump reaches the photographic plate and suddenly collapses to a microscopic point, whose size is comparable to the size of a grain of photo-emulsion. The photon energy, that was previously spread up in a macroscopic lump now gets released within a group of few atoms.

How does Maxwell equation explain this collapse?

Eugene.

 

[A. Neumaier]

No, I meant
$$V = \int d\mathbf{p} d\mathbf{q} d\mathbf{k} V(\mathbf{p}, \mathbf{q}, \mathbf{k}) a^{\dag}(\mathbf{p-k}) a^{\dag}(\mathbf{q+k}) a(\mathbf{p}) a(\mathbf{q})$$

OK, not a product but an integral over a product. It is difficult to understand you if you shorten formulas to an extent that they become unrecognizable.

I also claim that one can find a corresponding interacting boost operator, so that entire theory becomes relativistically invariant.

My other claim is that in this theory interacting quantum fields constructed by usual formulas do *not* transform covariantly.

The quantum fields of interest are those constructed from
$$\phi(0)=\int dp (a(p) +a^*(p))$$
by conjugating with the representation of the Poincare group you claim exists. Since the stabilizer of zero is the Lorentz group, the translations gives a Hermitian field phi(x) which transforms covariantly. But the field is not local, and hence not of interest for particle physics. One can construct plenty of similar field with Bakamjian's construction.

So, the main condition of Haag's theorem is not satisfied, the theorem is not applicable, and we are permitted to work in the Fock space.

Yes, but you get something lacking cluster separation.

More details on this example can be found in

H. Kita, "A non-trivial example of a relativistic quantum theory of particles without divergence difficulties", Progr. Theor. Phys., 35 (1966), 934.

Probably this is the reason why this pseudo-breakthrough only generated 7 references in over 50 years.

 

[A. Neumaier]

OK, so we have a macroscopic lump of electromagnetic energy, which falls on the double slit and interferes with itself according to Maxwell equations. Then this macroscopic lump reaches the photographic plate and suddenly collapses to a microscopic point, whose size is comparable to the size of a grain of photo-emulsion. The photon energy, that was previously spread up in a macroscopic lump now gets released within a group of few atoms.

How does Maxwell equation explain this collapse?

The Maxwell equations are valid in vacuum and need not explain their failure in the presence of a detector.

The behavior of the detector in the presence of the incident classical electromagnetic field is fully explained by the detector's quantum structure. See Chapter 9 of the quantum optics book by Mandel & Wolf.

That the photon ''suddenly collapses to a microscopic point, whose size is comparable to the size of a grain of photo-emulsion'' is pure fantasy. The photon is absorbed by the detector, and doesn't survive as a localized photon.

 

[meopemuk]

The Maxwell equations are valid in vacuum and need not explain their failure in the presence of a detector.

The behavior of the detector in the presence of the incident classical electromagnetic field is fully explained by the detector's quantum structure. See Chapter 9 of the quantum optics book by Mandel & Wolf.

That the photon ''suddenly collapses to a microscopic point, whose size is comparable to the size of a grain of photo-emulsion'' is pure fantasy. The photon is absorbed by the detector, and doesn't survive as a localized photon.

So, is it correct to say that the macroscopically delocalized lump of the photon's EM field is absorbed by the entire photographic plate, and then all this absorbed energy gets channeled somehow to a single grain of photoemulsion?

If this model is not correct, then how should I visualize the interaction between the photon and the photographic plate? Do I need to take the quantum structure of the photographic plate into account?

Eugene.

 

[A. Neumaier]

So, is it correct to say that the macroscopically delocalized lump of the photon's EM field is absorbed by the entire photographic plate

I answered in the thread https://www.physicsforums.com/showthread.php?p=3165407 where this part of the discussion belongs.

 

[meopemuk]

OK, not a product but an integral over a product. It is difficult to understand you if you shorten formulas to an extent that they become unrecognizable.

Sorry about that.

The quantum fields of interest are those constructed from
$$\phi(0)=\int dp (a(p) +a^*(p))$$
by conjugating with the representation of the Poincare group you claim exists. Since the stabilizer of zero is the Lorentz group, the translations gives a Hermitian field phi(x) which transforms covariantly.

You've probably meant

$$\Phi(\mathbf{0}, 0) = \int \frac{d \mathbf{p}}{\sqrt{\omega_{\mathbf{p}}}}\left(a(\mathbf{p}) + a^{\dag}(\mathbf{p}) \right)$$

We went throgh this exercise in posts #135, #137, #139 of the thread
https://www.physicsforums.com/showthread.php?t=388556&page=9 and we found there that covariant transformation law and space-like commutativity can be proven if the interaction is written in the form

$$\int d \mathbf{r} \Phi^n(\mathbf{r}, 0)$$

My interaction does not have this form, so I don't see how you're going to prove the covariance and commutativity?

But the field is not local, and hence not of interest for particle physics.

I don't think there is a proof that particle physics can be explained only in terms of local quantum fields. In my opinion, there is some historically motivated prejudice, but not a proof.

Yes, but you get something lacking cluster separation.

The cluster separability of the "dressed particle" approach is proven in section 10.2. Interaction of the type presented above is cluster-separable if the coefficient function $$V(\mathbf{p}, \mathbf{q}, \mathbf{k})$$ is smooth, i.e., non-singular. This is not difficult to achieve.

Eugene.

 

[A. Neumaier]

You've probably meant
$$\Phi(\mathbf{0}, 0) = \int \frac{d \mathbf{p}}{\sqrt{\omega_{\mathbf{p}}}}\left(a(\mathbf{p}) + a^{\dag}(\mathbf{p}) \right)$$

Well, as I had said in our other discussion, I normalize the annihilator fields differently in order to get rid of the sqrt factors. (Anyway, it doesn't matter in this case, since no matter which p-dependent factor we use, we don't get a causal field.)

My interaction does not have this form, so I don't see how you're going to prove the covariance and commutativity?

This is precisely the point, with my recipe to construct a covariant field you don't get causality. With your recipe to construct a causal field you don't get covariance. But one needs both. Thus either field is useless. And the lack of citations of Kita's paper is empirical proof of that - only 7 people in over 50 years found it useful enough to merit a citation.

The cluster separability of the "dressed particle" approach is proven in section 10.2. Interaction of the type presented above is cluster-separable if the coefficient function $$V(\mathbf{p}, \mathbf{q}, \mathbf{k})$$ is smooth, i.e., non-singular. This is not difficult to achieve.

Please give the page; I didn't see it. I only found a passing reference on p.356, which referred to p.271, which establishes cluster separability only in a preferred frame of reference. You declare this to be enough on p.428 - but very few people will follow you in that. You discard the basic insight of 20th century physics. There are no experimental hints that Nature has preferred frames.

While searching for information, I came across the following unrelated remark on p. 426: ''We can record time even if we do not measure anything, even if there is no physical system to observe.'' I wonder how you record time without observing (i.e., measuring) a clock. If there is no physical system to observe, there is in particular no clock and no observer.

And on p.427; ''the formally quantum nature of clocks and rulers does not play any role in experimental physics.'' But they play a big role if the times and distances become short enough. Accurate experiments become impossible (or extremely difficult) when these are so short that the quantum nature starts playing a role.

 

[meopemuk]

And the lack of citations of Kita's paper is empirical proof of that - only 7 people in over 50 years found it useful enough to merit a citation.

If you want to make fun of Kita's paper, I can help you. One of these 7 citations is in my book, two of them are by my friend Shirokov in Dubna, and three are Kita's self-citations. So, we are not talking about mainstream here. You are right.

Eugene.

 

[meopemuk]

This is precisely the point, with my recipe to construct a covariant field you don't get causality. With your recipe to construct a causal field you don't get covariance. But one needs both. Thus either field is useless.

I agree that the interacting field built in this manner is useless. It is useless not because it is non-covariant or non-commuting, but because actual calculations of the S-matrix, bound states, time evolution, etc. do not involve this field at all. All we need for practical calculations is the Hamiltonian, and there is no need to worry about the properties of the interacting field.

Please give the page; I didn't see it. I only found a passing reference on p.356, which referred to p.271, which establishes cluster separability only in a preferred frame of reference. You declare this to be enough on p.428 - but very few people will follow you in that. You discard the basic insight of 20th century physics. There are no experimental hints that Nature has preferred frames.

I agree that the point about cluster separability is not well emphasized. I will think how to rewrite section 10.2 to make this point more clear. But the idea is very simple. Interaction is guaranteed to be cluster-separable if the coefficient function is smooth. This is Statement 7.7 on page 271. This is the same condition as in Weinberg's sections 4.3-4.4.

In building the "dressed particle" version of QED I start from the usual QED Hamiltonian, whose interaction V satisfies the above separability condition already. (Well, strictly speaking, this condition is satisfied only if we ignore the singularity associated with the zero photon mass. But I don't want to open this can of worms again here.) I obtain the dressed particle interaction by applying an unitary transformation $$e^{i\Phi}$$ to the QED Hamiltonian. As I claim in Theorem 10.2, the Hermitian operator $$\Phi$$ must be smooth (=separable) in order to preserve the S-matrix. So, I pay a special attention to make sure that $$\Phi$$ satisfies this condition. Then, when I do dressing transformation order-by-order in section 10.2, I obtain new interaction terms in the form of multiple commutators involving operators V and $$\Phi$$. Both of these operators are smooth (=separable), so by theorem 7.11 all their commutators are smooth (=separable) too. This proves that the dressed particle Hamiltonian is separable in each perturbation order.

You are right that these calculations have been done in a single (but arbitrary) reference frame. If we want to obtain the Hamiltonian in a moving frame, we would need to apply the boost generator to it. Due to the principle of relativity, the Hamiltonian in the moving frame will have the same expansion coefficients with respect to to a/c operators in the moving frame as the coefficients of the rest-frame Hamiltonian with respect to rest-frame a/c operators. So, if the rest-frame Hamiltonian is separable, then the moving-frame Hamiltonian is separable too.

I am not sure where you found any evidence that I adhere to the preferred frame idea. I am 100% behind the principle of relativity. All inertial frames are equal. It is true that specific calculations can be done more easily in a specific frame. And that's what I do. But this should not be a problem, since there exist well-defined rules about how to translate our descriptions between different frames.

Eugene.

 

[meopemuk]

While searching for information, I came across the following unrelated remark on p. 426: ''We can record time even if we do not measure anything, even if there is no physical system to observe.'' I wonder how you record time without observing (i.e., measuring) a clock. If there is no physical system to observe, there is in particular no clock and no observer.

I've tried to explain my views on measurements (and on time measurements, in particular) on pages xxix and xxx that refer to Figure 1. In this figure there is a clear separation between the measuring device and the observed physical system. The measuring device and the clock are not parts of the physical system. They are parts of the experimental equipment, which every observer must have. The clock plays a specific role in a sense that it records something (time), which has no relevance to the observed physical system. Time is a quantity that exists by itself, without any connection to the observed system. And time can be recorded even if there is no system to observe, i.e., if the space between the preparation device and the measuring apparatus in Fig. 1 is empty. This allows me to say in subsection 11.3.4 that there cannot be a "time observable"

Of course, we may decide to treat our clock as a physical system, i.e., put it between the preparation device and the measuring apparatus in Fig. 1. Then we can find all kinds of quantum uncertainties associated with the clock. E.g., we can find that the leads cannot have certain velocities and positions simultaneously. But then, in order to keep time labels of our measurements, we would need to choose some other (reference) clock in our laboratory. The readings of this reference clock will be considered as true classical time labels.

So, any laboratory has a clock associated with it. The readings of this clock are postulated to be exact and classical time labels that are attached to all measurements that we do in the laboratory.

It might also happen that our reference clock behaves irregularly due to quantum fluctuations. This simply means that we've chosen a bad instrument to serve as a clock. We would need to replace it with some other (e.g., more massive) device whose ticks are more regular and predictable.

This is kind of philosophical point, and I am still struggling to formulate it in a coherent fashion.

And on p.427; ''the formally quantum nature of clocks and rulers does not play any role in experimental physics.'' But they play a big role if the times and distances become short enough. Accurate experiments become impossible (or extremely difficult) when these are so short that the quantum nature starts playing a role.

I agree that this is not a well-formulated sentence. I am going to replace it with the following: "So, for theoretical purposes, it is reasonable to assume the availability of ideal clocks and rulers, whose performance is not affected by quantum effects."

Eugene.

 

[A. Neumaier]

If you want to make fun of Kita's paper, I can help you. One of these 7 citations is in my book, two of them are by my friend Shirokov in Dubna, and three are Kita's self-citations. So, we are not talking about mainstream here.

I had no intention to be funny here. ''useful'' is a sociological term with a different meaning of ''mainstream''. Mainstream is the dominant view on a subject; useful is something if people start using it once its exist.

I wouldn't regard Polyzou's work on covariant few-particle systems as mainstream. But it has about 200 citations, and looking at some of them, one finds that is useful to some extent. On the other hand, something that hasn't been used in many years can hardly be called useful.