Difference between 'Quantum theories'

[StevieTNZ]

Hey there

This may sound a bit silly to a lot of 'experts' but, what are the differences between standard qunatum mechanics, quantum field theory, and quantum electrodynamics?

Are they all predicting different aspects of natures workings, or has one superseded another? I get confused because I think standard qm has been done away with, and quantum field theory is what is used now. Perhaps a MAJOR correction is in order? :P

 

[Pengwuino]

They're different fields, but none of them really supersede each other, they just are different areas of mechanics at the quantum level. Standard quantum mechanics is kind of like your Newtonian mechanics of classical mechanics. QED is the electrodynamics of the classical world and field theory, I suppose is the field theory of the classical world heh. Not that that helps I suppose...

 

[Avodyne]

Quantum electrodynamics is a particular example of a quantum field theory.

A quantum field theory is a system whose basic objects are fields, and which is "quantized" by following a standard (or "canonical", which means "according to the laws of the church") procedure for turning a classical system into a quantum one.

I'm not sure what you mean by "standard" quantum mechanics. If you mean wave functions for nonrelativistic particles, this can be understood as a low-energy limit of an appropriate quantum field theory.

 

[alxm]

Yeah, I'm not sure what you would call 'standard quantum mechanics'.

The "original" quantum mechanics would basically be the Schrödinger equation. Now, the S.E. has several deficiencies: It doesn't take into account Special Relativity and it doesn't describe fields. Which doesn't mean you can't take fields into account in the S.E., just that they're classical fields and it doesn't describe the field itself. You can't really say the Schrödinger equation has been 'superseded' with respect to these things, though. The fact that it didn't describe these things was clear from the start. In fact, Einstein himself was there to point out the S.E. didn't obey relativity at the famous 1927 Solvay Conference. The work on remedying this had already started then, notably by another attendee, the young Paul Dirac.

So, in addition to the original quantum mechanics, you have relativistic quantum theory, which takes S.R. into account, and quantum field theory, which deals with fields quantum-mechanically. QED deals specifically with electromagnetic fields. QFT is usually relativistic, but for completeness I guess one should mention the existence of non-relativistic QFT, which sees some use in condensed-matter physics. There's a difference in formalism here; The "original" QM deals mainly with the wave function, whereas QFT/QED use http://en.wikipedia.org/wiki/Propagator" (or Green's functions, in non-relativistic contexts). But you don't need quantized fields and special relativity to describe a lot of things. I'd even dare say that that most things of immediate practical significance don't, since almost all of chemistry and solid-state physics and material science ends up in that category. So the original quantum mechanics with its formalism is quite alive and well.

In my field (quantum chemistry) we don't need quantized fields, and despite some valiant efforts to introduce them (Hello Yngve..), propagator methods don't see much use. We have no need for QED in most cases, because those effects are too small to be relevant. Relativistic effects are relevant in heavy atoms (e.g. causing gold to be yellow and mercury to be a liquid). That gets handled either by applying a pseudopotential to the Schrödinger equation, mimicking the approximate effects, or by using the Dirac equation, which in this context can be viewed as a relativistic version of the Schrödinger equation.

But when the Dirac equation is used in QC, it's to get a relativistic description of the electrons, whereas the field only gets a semi-classical description (the "Coulomb gauge", where the magnetic potential is quantized but the Coulomb interaction isn't). This as opposed to the "Lorenz gauge", which leads to QED. So you use a theory where where it makes sense to do so, rather than what theory is 'best'. There's no point in complicating your description for an insignificant or nonexistent improvement in the result. Also, just because the Schrödinger equation has been around the longest, doesn't mean the formalism and methodology hasn't developed. Newtonian mechanics got reformulated in Lagrangian and Hamiltonian mechanics. The electronic Schrödinger equation has been reformulated in Density Functional Theory, for instance. (Which can be sub-divided further into the Kohn-Sham DFT and 'orbital-free' DFT, and again into chemical methods versus solid-state methods)

 

[A. Neumaier]

what are the differences between standard quantum mechanics, quantum field theory, and quantum electrodynamics?

Quantum mechanics is the theory of that part of physics that can be understood in terms of linear operators on a Hilbert space called observables.

Quantum field theory is the theory of that part of quantum mechanics in which some observables depend on a space-time point. Quantum field theory is the least understood part of quantum mechanics.

Quantum electrodynamics is the quantum field theory in which the basic observables are the electromagnetic field and the electron current. Quantum electrodynamics is the best understood quantum field theory.

 

[A. Neumaier]

We have no need for QED in most cases, because those effects are too small to be relevant.

QED is essential for predicting reactions in laser chemistry.

 

[alxm]

QED is essential for predicting reactions in laser chemistry.

Could you elaborate on that and give an example, please?

QED is necessary for describing some absorption/emission processes (or all of them, if you want enough detail). That's not the same thing as predicting a (photo-)chemical reaction, which is something that starts after the absorption.

There wasn't much QED in the one grad course I took in laser chemistry.

 

[yoda jedi]

Yeah, I'm not sure what you would call 'standard quantum mechanics'.

The "original" quantum mechanics would basically be the Schrödinger equation.

SE (schrödinger equation) can be linear (Standard Quantum Mechanics) or nonlinear (Non Standard Quantum Mechanics)

 

[cgk]

QED is essential for predicting reactions in laser chemistry.

I'm also working in quantum chemistry, and I don't think I've ever seen QED corrections being applied to anything. Admittedly, I rarely read papers on <= 4electron systems where one can still make variational explicitly correlated wave functions, and where one could potentially obtain enough accuracy in the description of the electronic wave functions such that QED corrections might become relevant.

For systems with more electrons (say, up to 10--20), an accuracy of energy differences of about 1 meV or 0.1 kJ/mol is about the maximum one could expect to obtain with the currently available technology. And even that only in favorable circumstances and by combining everything in programs and methodology there currently is. Something like that is very not fun.

 

[A. Neumaier]

Could you elaborate on that and give an example, please?

QED is necessary for describing some absorption/emission processes (or all of them, if you want enough detail). That's not the same thing as predicting a (photo-)chemical reaction, which is something that starts after the absorption.

Enter the following keywords into http://scholar.google.com/ to get lots of references:
laser chemistry QED.

There is also a book called Molecular Quantum Electrodynamics, which describes QED for chemical applications.

 

[meopemuk]

Quantum mechanics is the theory of that part of physics that can be understood in terms of linear operators on a Hilbert space called observables.

Quantum field theory is the theory of that part of quantum mechanics in which some observables depend on a space-time point. Quantum field theory is the least understood part of quantum mechanics.

Quantum electrodynamics is the quantum field theory in which the basic observables are the electromagnetic field and the electron current. Quantum electrodynamics is the best understood quantum field theory.

I would like to suggest slightly different definitions.

The main feature of quantum field theory is that it describes systems where the number of particles can vary. So, it would be appropriate to say "quantum mechanics of systems with varying numbers of particles" instead of "quantum field theory". Otherwise, QFT is not different from *ordinary* quantum mechanics (which usually deals with systems having fixed number of particles). All the usual machinery of quantum mechanics (Hilbert spaces, wave functions, Hermitian operators, etc.) works in QFT just as well as in ordinary quantum
mechanics.

The interesting point is that once we have allowed the number of particles to change in QFT, there is no place to stop. We are forced to treat simultaneously all systems, where this number can be anywhere between zero and infinity, as well as superpositions of such states. Then the Hilbert space of QFT takes the specific form known as the Fock space, where the number of particles is a variable. Defining an interacting relativistic quantum theory in the Fock space is a challenging task. It appears that the easiest way to achieve this goal is to introduce certain abstract operator functions (called quantum fields) in the Fock space and build interaction operators (e.g., in the Hamiltonian) as polynomials of such operator functions. Hence the name "quantum field theory" for this approach. The most coherent explanation of this logic can be found in Weinberg's "The quantum theory of fields" vol. 1.

QED is a specific version of QFT applied to electromagnetic interactions of charged particles and photons.

Eugene.

 

[A. Neumaier]

The main feature of quantum field theory is that it describes systems where the number of particles can vary.

This is one feature but not the main one. The main feature, as the name says, is that it describes systems containing fields. For example, QED features operator-valued versions of the Maxwell equations for the electromagnetic field.

So, it would be appropriate to say "quantum mechanics of systems with varying numbers of particles" instead of "quantum field theory". Otherwise, QFT is not different from *ordinary* quantum mechanics

QFT is different from QM in the same sense as quantum physics is different from physics.
In both cases, the former is a proper subset of the latter.

 

[meopemuk]

This is one feature but not the main one. The main feature, as the name says, is that it describes systems containing fields. For example, QED features operator-valued versions of the Maxwell equations for the electromagnetic field.

Here we disagree. In my opinion fields (both Dirac field for electrons and Maxwell field for photons) are just mathematical tools for describing systems with variable number of particles. E.g., photons can be absorbed and emitted. Fields are not measurable by themselves.

QFT is different from QM in the same sense as quantum physics is different from physics.
In both cases, the former is a proper subset of the latter.

I basically agree with this. Though some people prefer the following terminology:

quantum theory = the most general quantum framework
quantum mechanics = quantum theory applied to systems with fixed number of particles
quantum field theory = quantum theory applied to systems with variable number of particles

Eugene.

 

[Fra]

I basically agree with this. Though some people prefer the following terminology:

quantum theory = the most general quantum framework
quantum mechanics = quantum theory applied to systems with fixed number of particles
quantum field theory = quantum theory applied to systems with variable number of particles

Good point.

With "basic quantum mechanics" I mean the abstractions of QM, state spaces, state spaces, expectations, measurements. Ie. the CORE feature of the measurement theory as it stands.

Wether it's fixed particle count, variable or field are all details within the process of indexing information. This is IMHO partly an open question, and I am not satisfied with the mainstream ways, RG etc. For ME at least, analysing this problem unavoidably connects to the problem of understanding the emergence of the spacetime index, which necessarily also connets to some aspects of gravity. I also think this couples back to the basic QM structure, which needs to be revised or put in a larger context (generalized).

In my opinion fields (both Dirac field for electrons and Maxwell field for photons) are just mathematical tools for describing systems with variable number of particles. E.g., photons can be absorbed and emitted. Fields are not measurable by themselves.

I think that not only fields, but also the notion of particles are all just different ways to index and structure information. Actually the INDEX itself, does carry information in the sense that you can apply the ergodic principle to any index and get an a priori expectation. My opinon is that the exact way of indexing information is something that's not yet understood.

What I consider to elementary (measurable as you say) are simply a set of distinguishable events. I make not interpretation that these corrsepond to "particle counts" at a detector. I just see them as distinguished events. This is the a discrete structure, where the "field" picture is more like an smooth approximation.

To think of "particles" is far too confusing and even more classical. I just think of the "directly observable" as detector counts, sitting in the observers boundary (loosely speaking). The internal structure is then recoded forms of histories of counts, and in this recoding - there is a preferred index. I think changing the index amounts to different forms of datacompression. This is how I picture the selection principle.

/Fredrik

 

[A. Neumaier]

Here we disagree. In my opinion fields (both Dirac field for electrons and Maxwell field for photons) are just mathematical tools for describing systems with variable number of particles. E.g., photons can be absorbed and emitted. Fields are not measurable by themselves.

Of course, fields require detectors to be measurable. But the electromagnetic field is not only the photons - these are only the harmonic excitations of the latter.

some people prefer the following terminology:

quantum theory = the most general quantum framework
quantum mechanics = quantum theory applied to systems with fixed number of particles
quantum field theory = quantum theory applied to systems with variable number of particles

Only a few people like you. Most quantum physicists know why quantum field theory is called quantum _field_ theory.

Even Weinberg, your favorite textbook writer, writes (on p. xii):
''The point of view of this book is that quantum field theory [...] is the only way to reconcile the principles of quantum mechanics [...] with those of relativity.''
This view is incompatible with your above definitions of quantum mechanics and quantum field theory.

 

[cgk]

Enter the following keywords into http://scholar.google.com/ to get lots of references:
laser chemistry QED.

I've looked through the first 10 pages of that, and I found no title indicating any relation to chemical reactions. In fact, most of them don't even deal with molecules.

There is also a book called Molecular Quantum Electrodynamics, which describes QED for chemical applications.

According to the amazon description, this also doesn't deal with reactions.

 

[A. Neumaier]

I've looked through the first 10 pages of that, and I found no title indicating any relation to chemical reactions. In fact, most of them don't even deal with molecules.

According to the amazon description, this also doesn't deal with reactions.

Indeed, I had somehow mixed up molecular spectroscopy with laser chemistry.
''this self-contained, systematic introduction features formal derivations of the quantized field matrix elements for numerous laser-molecule interaction effects: one- and two-photon absorption and emission, Rayleigh and Raman scattering, linear and nonlinear optical processes, the Lamb shift, and much more.''
https://www.amazon.com/dp/0121950808/?tag=pfamazon01-20

http://repository.kulib.kyoto-u.ac.jp/dspace/bitstream/2433/43521/1/JChemPhys_115_3497.pdf
treats chemical reactions on a QED basis, but I must admit, this is one of very few papers.

 

[alxm]

[link]treats chemical reactions on a QED basis, but I must admit, this is one of very few papers.

Even that paper isn't actually treating the reaction using QED; they're treating the reaction using HF/6-311G* (which has an error many orders of magnitude greater than any pure QED effects). They're doing a QED calculation on the results of those calculations to get a spatial decomposition of the electronic energy, something which the authors think gives a more interesting qualitative picture of what's going on during the reaction. There are other, more common, ways of doing that. Frontier-orbital analysis, Bader analysis, a whole bunch of ways that utilize the electrostatic potential, etc.

Using QED to calculate properties which you can use to visualize aspects of a reaction is of course not the same thing as saying a QED treatment is necessary to describe the reaction itself. As cgk said, the maximum accuracy of QC methods is about 0.1 kJ/mol, and purely QED effects are several orders of magnitude smaller than that. Yet, about 1 kJ/mol is about what's considered "chemical accuracy". I just don't see how a purely-QED effect could be energetically large enough to "make-or-break" a chemical reaction at just about any temperature.

 

[A. Neumaier]

As cgk said, the maximum accuracy of QC methods is about 0.1 kJ/mol, and purely QED effects are several orders of magnitude smaller than that. Yet, about 1 kJ/mol is about what's considered "chemical accuracy". I just don't see how a purely-QED effect could be energetically large enough to "make-or-break" a chemical reaction at just about any temperature.

I now realize where my misunderstanding came from.

QED is needed to model the interaction between a laser and a molecule, since this involves both photons and electrons. Thus it is needed to derive from first principles
the conditions under which a laser is able to excite a molecule from its ground state to an excited state - where it moves on a different potential energy surface, which may enables reaction that are otherwise impossible or inefficient.

However, concerning the practice of laser chemistry, one can forget QED once one has
drawn the qualitative conclusion from QED that adding laser light of the right frequency to bridge the energy gap between ground state and excited state has the desired effect.
And one doesn't need QED at all if one simply assumes this statement without invoking QED, but referring instead to experimental practice.

 

[meopemuk]

.
Even Weinberg, your favorite textbook writer, writes (on p. xii):
''The point of view of this book is that quantum field theory [...] is the only way to reconcile the principles of quantum mechanics [...] with those of relativity.''
This view is incompatible with your above definitions of quantum mechanics and quantum field theory.

Weinberg's book is remarkably different from other QFT textbook. Weinberg *does not* start his book like "Here we have this thing called field. Let us apply quantum mechanics to it and see what we get."

Instead, his starting point is: "We have things called particles. We want to calculate their relativistically invariant scattering amplitudes. How can we build the appropriate Hamiltonian?"

Eugene.

 

[A. Neumaier]

Weinberg's book is remarkably different from other QFT textbook. Weinberg *does not* start his book like "Here we have this thing called field. Let us apply quantum mechanics to it and see what we get."

Instead, his starting point is: "We have things called particles. We want to calculate their relativistically invariant scattering amplitudes. How can we build the appropriate Hamiltonian?"

Instead of quoting fictitious lines put into the mouth of Weinberg, you'd read what he actually writes in his preface! His starting point is (quoted verbatim with some omissions indicated explicitly):

''Why another book on quantum field theory? [...] specific examples are frequently used [...] chosen from contemporary particle physics or nuclear physics as well as from quantum electrodynamics. [...] The point of view of this book is that quantum field theory is the way it is because (aside from theories like string theory that have an infinite number of particle types) it is the only way to reconcile the principles of quantum mechanics [...] with those of relativity.''

This quote contains all (two) occurrences of the word ''particle'' in the first five paragraphs, while the word ''field'' appears quite frequently, and my previous quote is there, too.

In the sixth paragraph come his reasons why particles enter the picture: ''The most immediate and certain consequences of relativity and quantum mechanics are the properties of particle states, so here particles come first - they are introduced in Chapter 2''

Indeed, Chapter 1 is exclusively about classical field equations. Particles make their appearance first on p.67 bottom. At this point, the book is already more than 10% on the way. In sharp contrast to your claim, the particles do not appear as starting point but (as promised in the preface) as the most immediate and certain consequences of relativity and quantum mechanics.

From Chapter 4 on he discusses (in the remaining 72% of the book) the less immediate consequences of relativity and quantum mechanics - called quantum field theory.

 

[meopemuk]

Arnold,

it seems that we are reading the same book, but understand it quite differently. Let me recap briefly what I've learned from Weinberg.

Chapter 1: I skip this as it is just an historical introduction.

Chapter 2: Definition of particles as unitary irreducible representations of the Poincare group. No mention of fields.

Chapter 3: Definition of the S-matrix as transition amplitudes between different *particle* states. Requirement of the relativistic invariance of the S-matrix. The important piece is eqs. (3.5.11) - (3.5.18), which suggest that the theory can be made relativistically invariant if the interaction Hamiltonian is represented as an integral of some "density" with a particular transformation law and commutators. Note also that there is no proof that this is the *only* way to build relativistic interactions. Note the phrase "let's try the hypothesis" before (3.5.11). No mention of fields.

Chapter 4. The cluster decomposition principle. Creation and annihilation operators of *particles* are introduced. The important point is that the theory can be made cluster-separable if the Hamiltonian is expressed as a polynomial in c/a operators (4.4.1) with smooth coefficient functions. No mention of fields.

Chapter 5. Here Weinberg tries to build specific examples of Hamiltonians with properties specified in Chapter 3. This is where the fields finally come into play. Weinberg defines free fields as some abstract linear combinations of creation and annihilation operators with postulated properties, like the covariant transformation law (5.1.6) - (5.1.7) and (anti)commutators (5.1.32). He briefly mentions field equations (5.1.34) and then says "...we start with the particles, and derive the fields according to the dictates of Lorentz invariance, with the field equations arising almost accidentally as a byproduct of this construction."

It seems pretty clear to me that Weinberg is not interested in fields by themselves, and he does not assign any physical significance to fields. For him, the only reason to introduce free quantum fields is to have a mathematical tool for constructing relativistically invariant and cluster separable interaction Hamiltonians. Once such an interaction is constructed and expressed in terms of *particle* creation and annihilation operators, we can happily forget about fields. We will have all rules (see Chapter 6) for calculating scattering amplitudes for any configuration of initial *particles*, and this is the only thing we really care about in quantum field theory. *Particle* scattering cross-sections and energies of *particle* bound states are the only numbers that can be reliably calculated in QFT and compared with experiment. Fields play a certain role at intermediate steps of these calculations, but this role is completely technical.

Possibly one can pretend that field is some kind of material substance and try to use QFT to calculate the field strength at given points. But is this information useful? Is there an experiment to verify these numbers? I don't think so. Even the most revered *electromagnetic field* is an approximate concept. As far as I know, light is not a continuous field or wave. If we make light of sufficiently low intensity and look at it with sufficiently high resolution, we will find out that it consists of indivisible and countable particles called photons. So, actually, there are no truly continuous material fields or waves in nature. Everything becomes granular and particle-like if we look at it with a high enough resolution.

Eugene.

 

[A. Neumaier]

it seems that we are reading the same book, but understand it quite differently.

Yes indeed.

It seems pretty clear to me that Weinberg is not interested in fields by themselves, and he does not assign any physical significance to fields.

You completely ignore what Weinberg himself writes as motivation and goal in his preface, and replace it by your subjective conjectures about the interests of Weinberg, though these contradict his statements in the preface.

Once such an interaction is constructed and expressed in terms of *particle* creation and annihilation operators, we can happily forget about fields. We will have all rules (see Chapter 6) for calculating scattering amplitudes for any configuration of initial *particles*, and this is the only thing we really care about in quantum field theory.

Weinberg never forgets the fields. All later chapters are full of them.

*Particle* scattering cross-sections and energies of *particle* bound states are the only numbers that can be reliably calculated in QFT and compared with experiment.

You don't realize how much is done with quantum field theory. People calculate a lot more: Thermodynamic properties of equilibrium states, hydrodynamic equations for flowing fields, kinetic transport equations governing the behavior of semiconductors, etc.. The fact that this is not in Weinberg's book doesn't mean that it is not done. If scattering of particles were the only application of QFT, the latter wouldn't play the fundamental role it plays.

Possibly one can pretend that field is some kind of material substance and try to use QFT to calculate the field strength at given points. But is this information useful? Is there an experiment to verify these numbers?

Every engineer is able to measure the electromagnetic field at a particular point.

Even the most revered *electromagnetic field* is an approximate concept.

Everything in physics is an approximate concept when applied to reality.

In particular, the particle concept is even more approximate than the field concept, since matter frequently behaves far from particle-like.

 

[meopemuk]

Weinberg never forgets the fields. All later chapters are full of them.

I agree that fields make a useful mathematical concept. They allow to simplify calculations considerably. For example, it would be very difficult (though possible) to calculate loop integrals without using fields and propagators but relying only on interaction expressed as a function of creation/annihilation operators. The mathematical usefulness of fields does not prove that quantum field is a measurable physical concept.

You don't realize how much is done with quantum field theory. People calculate a lot more: Thermodynamic properties of equilibrium states, hydrodynamic equations for flowing fields, kinetic transport equations governing the behavior of semiconductors, etc..

I am not an expert in this, but I always thought that the current in semiconductors can be reliably calculated with usual classical methods. Actually, presently I am working in semiconductor industry where such calculations are performed routinely without any involvement of quantum mechanics or QFT. I can agree that simple quantum effects (e.g., tunneling) may become visible in nanostructures at low temperatures. However, I've never heard that radiative corrections (whose description demands QFT methods) have any visible effect on the behavior of semiconductors. It is very difficult to see such corrections (e.g., the Lamb shift) in super precise measurements of individual atoms. But observation of radiative corrections in such a mess as a working semiconductor device would be truly amazing. So, I remain sceptical.

It would be a different matter if you are talking about condensed matter quantum field theory, which uses approximately continuous fields (such as the phonon field or electron current density). I agree that this heuristic quantum field theory has made a lot of progress in such areas as superconductivity, etc. But I thought that we are discussing fundamental, presumably exact relativistic QFT of elementary particles, which is the subject of Weinberg's book.

If scattering of particles were the only application of QFT, the latter wouldn't play the fundamental role it plays.

I am glad that we agree about that.

Every engineer is able to measure the electromagnetic field at a particular point.

... by placing a test charged *particle* at that point and measuring its acceleration.

In particular, the particle concept is even more approximate than the field concept, since matter frequently behaves far from particle-like.

Like when?

Eugene.

 

[A. Neumaier]

I agree that fields make a useful mathematical concept.

But Weinberg wrote a book about physics. Clearly he thinks that fields make a useful physical concept.

The mathematical usefulness of fields does not prove that quantum field is a measurable physical concept.

The expectation values of the quantum electromagnetic field is routinely measured by many engineers, hence the quantum electromagnetic field is as much a measurable physical concept as the position of a particle, where one commonly - namely except in diffraction experiments - also measures only the expectation value.

I am not an expert in this, but I always thought that the current in semiconductors can be reliably calculated with usual classical methods. Actually, presently I am working in semiconductor industry where such calculations are performed routinely without any involvement of quantum mechanics or QFT.

One doesn't need quantum methods on the engineering level, just as one doesn't need quantum theory to use superconducting solenoids as magnets.

Nevertheless, the classical formulas are all derived from quantum field theory.
The behavior of semiconductors on the atomic level is determined in solid state physics by QED in an external periodic potential with or without impurities. In some cases, nonrelativistic QED (or even coarser approximations) is enough, but not always: http://scholar.google.com lists about 18,800 references for the keywords
semiconductor relativistic

The valence electrons in a metal are so much delocalized that the electrons must be regarded as a quantum fluid (described by a electron density field) rather than as a collection of particles. Moreover, even when talking in the 'particle language, solid state physicists think of the valence electrons as quasi-particles whose very definition is in terms of fields.

I can agree that simple quantum effects (e.g., tunneling) may become visible in nanostructures at low temperatures. However, I've never heard that radiative corrections (whose description demands QFT methods) have any visible effect on the behavior of semiconductors.

The classical Maxwell equations don't fall from heaven as classical equations independent of quantum mechanics but are the macroscopic limit of QED. This is completely independent of either semiconductors or radiative corrections.

It would be a different matter if you are talking about condensed matter quantum field theory, which uses approximately continuous fields (such as the phonon field or electron current density). I agree that this heuristic quantum field theory has made a lot of progress in such areas as superconductivity, etc. But I thought that we are discussing fundamental, presumably exact relativistic QFT of elementary particles, which is the subject of Weinberg's book.

The _only_ difference between condensed matter quantum fields and relativistic quantum fields is that the former have nonlocal interactions and transform under the Galilei group, while the latter have local interactions and transform under the Poincare group. But both are based on the same concept of a quantum field.

Quantum fields are also more basic than particles in condensed matter theory. Indeed, the concept of identical particles has no logical basis on the particle level and must be introduced in an ad hoc way to get agreement with experiment. And it completely destroys the very basis of a reasonable particle concept: the possibility to assign observables (self-adjoint operators) to the position of a particle inside a multiparticle system. There _are_ no such observables.

On the other hand, identical particles are one of the most elementary consequences of the
quantum field concept, and certain smeared fields are both observable in the formal sense of being self-adjoint operators on the Hilbert space of the system _and_ in the sense of measurable expectation values.

Like when?.

Electrons bound in a molecule or a crystal. Chemists think of them in terms of charge density (a field concept), and solid state physicists think of them in terms of a fluid (another field concept). Particles are not experimentally identifiable except under very extreme circumstances.


Finally, the difficulties you have in forcing QED into the Procrustes bed of a pure particle picture
are proof of that the latter is unnatural and incomplete. Indeed, the canonical Fock space electrons that figure in your approach lack the electromagnetic field with which they need to be accompanied as asymptotic, physical particles) in order to exhibit the correct infrared behavior.
You would notice further problems if you were to extend your theory to handle the relaticistic plasma (see, e.g., H.A. Weldon, Phys. Rev. D 26, 1394–1407 (1982), http://llacolen.ciencias.uchile.cl/~vmunoz/download/papers/w82.pdf ). I wonder how you'd do that in your version of QED.

By the way, in your book you give a fairly complete list of publications related to Faddeev's dressing method, which is the basis of your perturbative particle-only approach to QED. Maybe you want to add the following one...

I.Ya. Aref'eva
Renormalized scattering theory for the Lee model
Teoret. Mat. Fiz. 12 (1972) 331-348
Theor. Math. Phys. 12 (1972), 859-872
http://mi.mathnet.ru/eng/tmf2995

Maybe you should send a link to your book to your fellow countryman Faddeev, whom you owe so much, and ask him why he gave up his dressing approach in favor of the usual field theoretic methods that lead him to the today accepted solution of the infrared problem in QED.

 

[meopemuk]

Arnold,

being asked about examples of the field-like behavior in nature your answer was:

Electrons bound in a molecule or a crystal. Chemists think of them in terms of charge density (a field concept), and solid state physicists think of them in terms of a fluid (another field concept).

Indeed, the electron charge density in a molecule or crystal is continuous. This is because the charge density is basicaly the square of a continuous wave function. However, the continuity of the wave function does not mean the continuity of measurements. The wave function of one electron can be continuous, but in precise measurements we always see a single discrete electron, rather than a spread-out field. If our measurements are not precise (e.g., the scanning tunneling microscopy or X-ray diffraction), then we can see an averaged continuous charge density. But we should be aware that this continuity is just an artefact of the imprecise measurement technique, and not a fundamental feature.

Regarding "charge fluid" and continuous charge and current densities used in solid state QFT methods. They are just crude approximations. There is a finite countable number of electrons in any crystal. Their representation in terms of a continuous "field" can be useful in some applications, but this is not a precise concept, for sure.

The _only_ difference between condensed matter quantum fields and relativistic quantum fields is that the former have nonlocal interactions and transform under the Galilei group, while the latter have local interactions and transform under the Poincare group. But both are based on the same concept of a quantum field.

Yes, on the formal mathematical level there is no much difference between these two kinds of fields. However, on the physical level they are very different. Condensed matter fields (phonons, charge density, etc.) are formulated from the beginning as approximate concepts. We know that in reality crystals are formed by discrete countable nuclei and electrons. Their behavior can be assumed to be continuous and field-like only in the large-scale large-number limit.

On the other hand, relativistic fields of elementary particles are supposed to be *exactly* continuous. They are supposed to be basic non-reducible ingredients of nature. They are supposed to be even more fundamental than particles.

Indeed, the concept of identical particles has no logical basis on the particle level and must be introduced in an ad hoc way to get agreement with experiment. And it completely destroys the very basis of a reasonable particle concept: the possibility to assign observables (self-adjoint operators) to the position of a particle inside a multiparticle system. There _are_ no such observables.

Sorry, I don't understand this. Why are you saying that it is impossible to define the position observable for one particle in a system? I think, it can be done rather easily.

Finally, the difficulties you have in forcing QED into the Procrustes bed of a pure particle picture
are proof of that the latter is unnatural and incomplete. Indeed, the canonical Fock space electrons that figure in your approach lack the electromagnetic field with which they need to be accompanied as asymptotic, physical particles) in order to exhibit the correct infrared behavior.

I agree that infrared problem does not have a satisfactory solution in the present day dressed particle approach. There are so many things that remain to be done!

You would notice further problems if you were to extend your theory to handle the relaticistic plasma (see, e.g., H.A. Weldon, Phys. Rev. D 26, 1394–1407 (1982), http://llacolen.ciencias.uchile.cl/~vmunoz/download/papers/w82.pdf ). I wonder how you'd do that in your version of QED.

I don't know much about plasma. I prefer to think only about simplest physical systems. In most cases, 2-particle systems are the most complicated things that I can handle.

By the way, in your book you give a fairly complete list of publications related to Faddeev's dressing method, which is the basis of your perturbative particle-only approach to QED. Maybe you want to add the following one...

I.Ya. Aref'eva
Renormalized scattering theory for the Lee model
Teoret. Mat. Fiz. 12 (1972) 331-348
Theor. Math. Phys. 12 (1972), 859-872
http://mi.mathnet.ru/eng/tmf2995

Thank you for the reference. I've added it to the book's draft.

Regards.
Eugene.

 

[A. Neumaier]

Indeed, the electron charge density in a molecule or crystal is continuous. This is because the charge density is basically the square of a continuous wave function.

No. The modulus square of an N-particle wave function is a function of 3N arguments, while the charge density of an N-electron system is a function of 3 arguments only. It is precisely the quantum field expectation value rho(x) := <e psi^*(x) psi(x)>.

The wave function of one electron can be continuous, but in precise measurements we always see a single discrete electron, rather than a spread-out field. If our measurements are not precise (e.g., the scanning tunneling microscopy or X-ray diffraction), then we can see an averaged continuous charge density. But we should be aware that this continuity is just an artefact of the imprecise measurement technique, and not a fundamental feature.

No Measurements of bound electrons never measure a single, discrete electron but the expectation value of the electron density field. This measurement need not be more inaccurate than the measurement of the electron position of a single, free electron in a cathode ray.
To call the latter the real thing and the former an artifact is very subjective.

Regarding "charge fluid" and continuous charge and current densities used in solid state QFT methods. They are just crude approximations. There is a finite countable number of electrons in any crystal. Their representation in terms of a continuous "field" can be useful in some applications, but this is not a precise concept, for sure.

No. There is a finite, uncountable number of electrons in any crystal. It is impossible to count them, except very roughly - so according to your own arguments, it should be just an artifact.

Yes, on the formal mathematical level there is no much difference between these two kinds of fields. However, on the physical level they are very different. Condensed matter fields (phonons, charge density, etc.) are formulated from the beginning as approximate concepts. We know that in reality crystals are formed by discrete countable nuclei and electrons. Their behavior can be assumed to be continuous and field-like only in the large-scale large-number limit.

Condensed matter physics often treats a crystal as an electron field in an external periodic potential - at th same level as QED.

On the other hand, relativistic fields of elementary particles are supposed to be *exactly* continuous. They are supposed to be basic non-reducible ingredients of nature. They are supposed to be even more fundamental than particles.

Yes, but that doesn't make the latter less physical than the former. on the contrary...

Why are you saying that it is impossible to define the position observable for one particle in a system? I think, it can be done rather easily.

Then please tell me how the operator corresponding to the position in x-direction of a single nonrelativistic electron in an N-electron system acts as a Hermitian operator on the N-particle sector of the electron Fock space. The only position operator one can define is that of the center of mass of all electrons together.

I agree that infrared problem does not have a satisfactory solution in the present day dressed particle approach. There are so many things that remain to be done!

The point was that your electrons are the wrong things to start with since they don't carry the electromagnetic field that they possesses in reality (and in QED). To account for the latter you need to revise your whole basis.

I don't know much about plasma. I prefer to think only about simplest physical systems. In most cases, 2-particle systems are the most complicated things that I can handle.

That's also the reason why you stick to particles rather than fields. They are the simplest physical systems. Fields are infinitely more complex.

I on the other hand, think of all the systems that a theory at the QED level must be able to cater for. This gives a much more complex scenario with many constraints that immediately rule out too simplistic modes of explanations such as yours.

 

[meopemuk]

Then please tell me how the operator corresponding to the position in x-direction of a single nonrelativistic electron in an N-electron system acts as a Hermitian operator on the N-particle sector of the electron Fock space. The only position operator one can define is that of the center of mass of all electrons together.

Consider a 2-electron system. Its Hilbert space (or the 2-electron sector of the Fock space) is the antisymmetrized tensor product of two 1-electron spaces. H_2 = H_1 \otimes_{as} H_1. The tensor product construction basically defines mappings from subspaces in each H_1 to subspaces in H_2. These mappings also map each operator in H_1 to some operator in H_2. Each 1-electron space H_1 carries an unitary irreducible representation of the Poincare group. The Newton-Wigner construction allows us to define the position operator in each one-particle space H_1. The tensor product mapping mentioned above can be applied to position operators in H_1 in order to define single particle position operators in the two-particle space H_2.

I on the other hand, think of all the systems that a theory at the QED level must be able to cater for. This gives a much more complex scenario with many constraints that immediately rule out too simplistic modes of explanations such as yours.

Can you give an example of such a "much more complex scenario", where precise experimental data exists, which was shown to be impossible to explain in terms of particles and their interactions?

 

[meopemuk]

The point was that your electrons are the wrong things to start with since they don't carry the electromagnetic field that they possesses in reality (and in QED). To account for the latter you need to revise your whole basis.

I believe that the source of infrared difficulties is rather different. Photons have zero mass, so infinite number of "soft photons" can be created in interactions of charged particles. Therefore, the usual definition of the scattering operator does not apply, which makes the fitting of the dressed particle Hamiltonian rather difficult. The mathematics is complicated, but this doesn't mean that the particle-based worldview should be abandoned.

Eugene.

 

[A. Neumaier]

Consider a 2-electron system. Its Hilbert space (or the 2-electron sector of the Fock space) is the antisymmetrized tensor product of two 1-electron spaces. H_2 = H_1 \otimes_{as} H_1. The tensor product construction basically defines mappings from subspaces in each H_1 to subspaces in H_2. These mappings also map each operator in H_1 to some operator in H_2. Each 1-electron space H_1 carries an unitary irreducible representation of the Poincare group. The Newton-Wigner construction allows us to define the position operator in each one-particle space H_1. The tensor product mapping mentioned above can be applied to position operators in H_1 in order to define single particle position operators in the two-particle space H_2.

Please give a bit more details. I assumed that the electron is treated nonrelativistically. Hence if the position representation is used, the single particle Newton-Wigner operator is simply the multiplication by x. So please conclude your discussion by an example giving explicitly the antisymmetric 2-electron wave function that is the image of the antisymmetric wave function phi(x_1)psi(x_2)-psi(x_1)phi(x_2) under the operator defining the z-position of particle 1.

By the way, the Latex symbol for the antisymmetrized tensor product is \wedge.

Can you give an example of such a "much more complex scenario", where precise experimental data exists, which was shown to be impossible to explain in terms of particles and their interactions?

I already gave many, and you responded with that you only consider the simplest case of 2-particle systems.

As you mention in your other post, the IR problem already requires that you regard the physical electron as an infinite-particle system composed of one of your electrons and an infinite number of soft photons. If the particle view of QED can only do 2-particle systems easily, it is vastly inferior to the field view, which has no difficulties handling coherent states with infinitely many soft photons. But these states do not live in your Fock space anymore!

 

[A. Neumaier]

It seems pretty clear to me that Weinberg is not interested in fields by themselves, and he does not assign any physical significance to fields. For him, the only reason to introduce free quantum fields is to have a mathematical tool for constructing relativistically invariant and cluster separable interaction Hamiltonians.

On p.2 of his essay, ''What is Quantum Field Theory, and What Did We Think It Is?'' http://arxiv.org/pdf/hep-th/9702027v1, Weinberg wrote (two years after his book appeared):

''In its mature form, the idea of quantum field theory is that quantum fields are the basic ingredients of the universe, and particles are just bundles of energy and momentum of the fields. In a relativistic theory the wave function is a functional of these fields, not a function of particle coordinates. Quantum field theory hence led to a more unified view of nature than the old dualistic interpretation in terms of both fields and particles.''

Just the opposite of what you put into his mouth and mind. I think you should retract your statement.

 

[meopemuk]

Please give a bit more details. I assumed that the electron is treated nonrelativistically. Hence if the position representation is used, the single particle Newton-Wigner operator is simply the multiplication by x. So please conclude your discussion by an example giving explicitly the antisymmetric 2-electron wave function that is the image of the antisymmetric wave function phi(x_1)psi(x_2)-psi(x_1)phi(x_2) under the operator defining the z-position of particle 1.

I see your point. You wanted to say that if I simply multiplied this function by z_1, I would get a non-antisymmetric function, which is not permitted in the 2-electron system. OK, then I need to anti-symmetrize the multiplication result. Thus obtained position operator is not really the position operator of particle 1. It has a bit different meaning. Its spectral projections correspond to experimental questions like "is it true that there is one particle (either particle 1 or particle 2, I don't care) at position z?" This is consistent with the indistinguishability of the two particles.

I already gave many, and you responded with that you only consider the simplest case of 2-particle systems.

I knew that you would not forgive me this little joke. But seriously, I don't see a good reason for abandoning particles in your examples. Plasma is just a collection of oppositely-charged particles. Isn't it? Perhaps, some field-like techniques are useful in plasma calculations, but this doesn't prove that it would be hopeless to use the particle-only picture there.

As you mention in your other post, the IR problem already requires that you regard the physical electron as an infinite-particle system composed of one of your electrons and an infinite number of soft photons. If the particle view of QED can only do 2-particle systems easily, it is vastly inferior to the field view, which has no difficulties handling coherent states with infinitely many soft photons. But these states do not live in your Fock space anymore!

"Vastly inferior" does not mean "physically wrong". I can agree that in some cases the field-like formalism provides a convenient calculation technique. But this cannot disprove the fact that in precise experiments we always see individual particles. I agree that it is almost impossible to count soft photons or radio photons experimentally, because their energy is just too low to trigger any response in detectors. However, photons of visible light and higher energies can be counted easily by photomultipliers. The visible light is not a continuous field, no matter how you prepare its state. This gives me confidence to say that lower-frequency radiation is not a continuous field as well.

Eugene.

 

[meopemuk]

On p.2 of his essay, ''What is Quantum Field Theory, and What Did We Think It Is?'' http://arxiv.org/pdf/hep-th/9702027v1, Weinberg wrote (two years after his book appeared):

''In its mature form, the idea of quantum field theory is that quantum fields are the basic ingredients of the universe, and particles are just bundles of energy and momentum of the fields. In a relativistic theory the wave function is a functional of these fields, not a function of particle coordinates. Quantum field theory hence led to a more unified view of nature than the old dualistic interpretation in terms of both fields and particles.''

Just the opposite of what you put into his mouth and mind. I think you should retract your statement.

Well, even such giants as Weinberg can be inconsistent in their writings. By the way, I am not advocating a "dualistic interpretation in terms of both fields and particles." I am for the particles-only view.

Eugene.

 

[meopemuk]

meopemuk: I think I need more than, saying the man is inconsistent to justify an apparent 180 deg turn. Do you have some evidence that this was just an inconsistency, or is your view the rarer beast?

When I read Weinberg's textbook, I see a consistent effort to present quantum field theory from the point of view of particles interacting with each other. I don't have direct quotes, but the whole logic of his treatise suggests that he considers particles as primary physical objects, while quantum fields are being introduced as formal technical devices, which help to establish a relativistically invariant and cluster-separable operator of interaction between *particles*.

I think, Weinberg's book is the best text written about quantum field theory ever. Before reading that book I struggled to understand QFT for many years. After the book all pieces fell in their places and the entire logic of QFT became transparent.

Now, Weinberg's quote mentioned by Arnold sounds very disturbing to me. But I would prefer to ignore that quote and rather focus on the beautiful physics revealed by the textbook.

Eugene.

 

[Fredrik]

I just read that Weinberg article (actually a transcript from a lecture). It contains another interesting quote on the topic of particles vs. fields:

The hope of S-matrix theory was that, by using the principles of unitarity, analyticity, Lorentz invariance and other symmetries, it would be possible to calculate the S-matrix, and you would never have to think about a quantum field. In a way, this hope reflected a kind of positivistic puritanism: we can’t measure the field of a pion or a nucleon, so we shouldn’t talk about it, while we do measure S-matrix elements, so this is what we should stick to as ingredients of our theories.
...
By the mid-1960’s it was clear that S-matrix theory had failed in dealing with the one problem it had tried hardest to solve, that of pion–pion scattering.
...
From a practical point of view, this was the greatest defeat of S-matrix theory. The irony here is that the S-matrix philosophy is not that far from the modern philosophy of effective field theories, that what you should do is just write down the most general S-matrix that satisfies basic principles. But the practical way to implement S-matrix theory is to use an effective quantum field theory — instead of deriving analyticity properties from Feynman diagrams, we use the Feynman diagrams themselves. So here’s another answer to the question of what quantum field theory is: it is S-matrix theory, made practical.

 

[meopemuk]

I just read that Weinberg article (actually a transcript from a lecture). It contains another interesting quote on the topic of particles vs. fields:

Fredrik,

I think that the problem with the pure S-matrix theory was that it denied the existence of underlying dynamics and assumed that scattering cross-sections are the only truly measurable things. In particular, this theory refused to consider such things as the Hamiltonian. This is too radical and too restrictive, in my opinion.

One important (yet not well appreciated) point in Weinberg's book is that any theory that strives to be both quantum and relativistic must be formulated as a unitary representation of the Poincare group in a Hilbert space. The Hamiltonian is an important part of this construction.

Eugene.

 

[A. Neumaier]

I see your point. You wanted to say that if I simply multiplied this function by z_1, I would get a non-antisymmetric function, which is not permitted in the 2-electron system. OK, then I need to anti-symmetrize the multiplication result.

Please do it and check what you get, instead of speculating about its meaning! Its meaning is precisely what I had claimed before: the position of the center of mass of the 2-electron system. And for an N-electron system you get the same conclusion.

But this cannot disprove the fact that in precise experiments we always see individual particles.

We see individual particles only when they directly fall into our eyes. But this never constitutes a precise experiment.

Particles we can see through an electron microscope, say, appear there as lumps with slightly fuzzy boundaries, described by a field density.

In the most precise experiments, we hear sounds, read pointers or digital numbers, or see ionization tracks that are only indirectly related to the particles - whose existence we infer.

 

[A. Neumaier]

Weinberg's quote mentioned by Arnold sounds very disturbing to me. But I would prefer to ignore that quote

That was the whole point of my efforts: to disturb your belief in that you understand things well enough to be sure of your assessment.

If you ignore it, you waste a precious opportunity.

Think of what you wouldn't have learned if you had ignored my disturbing comments about the causal commutation relations or the position operator of a single electron inside an N-electron system! Not everybody takes the effort it takes to conquer your self-assuredness, just to let you see what others can see.

 

[meopemuk]

Please do it and check what you get, instead of speculating about its meaning! Its meaning is precisely what I had claimed before: the position of the center of mass of the 2-electron system. And for an N-electron system you get the same conclusion.

You have nailed me down here. Very good! I agree that using antisymmetrization is not the way to go. If I do that, I get

x_1 \phi(x_1) \psi(x_2)- x_1 \psi(x_1) \phi(x_2) - (x_2 \phi(x_2) \psi(x_1)- x_2 \psi(x_2) \phi(x_1))
= (x_1 + x_2)(\phi(x_1) \psi(x_2)- \psi(x_1) \phi(x_2) )

which is not what I want.

But I can still use the observable called "the number of electrons at point x" and defined as

N(x) = \int dx a^{\dag}(x) a(x)

where a^{\dag}(x), a(x) are the creation and annihilation operators at position x.

We see individual particles only when they directly fall into our eyes. But this never constitutes a precise experiment.

Particles we can see through an electron microscope, say, appear there as lumps with slightly fuzzy boundaries, described by a field density.

In the most precise experiments, we hear sounds, read pointers or digital numbers, or see ionization tracks that are only indirectly related to the particles - whose existence we infer.

However imprecise or indirect, experiments still show us that *everything* is made of discrete, countable and indivisible lumps of matter, which I call "particles". You still haven't provided an example of a physical system for which the particle representation is not applicable.

Eugene.

 

[meopemuk]

That was the whole point of my efforts: to disturb your belief in that you understand things well enough to be sure of your assessment.

If you ignore it, you waste a precious opportunity.

Think of what you wouldn't have learned if you had ignored my disturbing comments about the causal commutation relations or the position operator of a single electron inside an N-electron system! Not everybody takes the effort it takes to conquer your self-assuredness, just to let you see what others can see.

I am greatful for your efforts from which I've learned a lot. For example, I've removed the section about Haag's theorem from the book's draft as a result of our discussions.

Thanks.
Eugene.

 

[A. Neumaier]

You have nailed me down here. Very good! I agree that using antisymmetrization is not the way to go. If I do that, I get

x_1 \phi(x_1) \psi(x_2)- x_1 \psi(x_1) \phi(x_2) - (x_2 \phi(x_2) \psi(x_1)- x_2 \psi(x_2) \phi(x_1))
= (x_1 + x_2)(\phi(x_1) \psi(x_2)- \psi(x_1) \phi(x_2) )

which is not what I want.

But I can still use the observable called "the number of electrons at point x" and defined as

N(x) = \int dx a^{\dag}(x) a(x)

where a^{\dag}(x), a(x) are the creation and annihilation operators at position x.

This is still not what you want, since the right hand side is independent of x. What you want is
N(x)=a^*(x)a(x) - which is a field operator, not a particle operator, describing the electron density, as I had claimed.

However imprecise or indirect, experiments still show us that *everything* is made of discrete, countable and indivisible lumps of matter, which I call "particles". You still haven't provided an example of a physical system for which the particle representation is not applicable.

Certainly experiments haven't shown that the gravitational field is made of discrete, countable and indivisible lumps of matter.

The electromagnetic fields in a conventional vacuum (_not_ the vacuum state in a Fock space) aren't made of matter either, by the very definition of a vacuum.

So, if you want to have another simple challenge, please derive the macroscopic Maxwell equations in vacuum (certainly something observable) from your particle picture.

 

[A. Neumaier]

I am greatful for your efforts from which I've learned a lot. For example, I've removed the section about Haag's theorem from the book's draft as a result of our discussions.

Fine. If you don't waste the opportunities given by my disturbing posts, you'll need to modify some more of your book. After these changes it will be a much better (and even publishable) book since it then agrees on the most important things with the main stream.

 

[meopemuk]

This is still not what you want, since the right hand side is independent of x. What you want is
N(x)=a^*(x)a(x) - which is a field operator, not a particle operator, describing the electron density, as I had claimed.

Of course, I was sloppy. I actually wanted to write the operator for the number of particles in the volume \Omega as

N(\Omega) = \int_{\Omega} dx a^{\dag}(x) a(x)

but ended up with something meaningless.

Now I see that we have used different terminologies all the time. This probably explains some misunderstandings. I would never call N(x)=a^*(x)a(x) a field operator. In my definition, quantum field is a very specific linear combination of particle creation and annihilation operators as defined in Weinberg's Chapter 5. Quantum field must have (1) covariant transformation rules and (2) (anti)commute at spacelike separations. In my notation x is eigenvalue of the Newton-Wigner position operator. So, I doubt that N(x)=a^*(x)a(x) satisfies the above two conditions.

Certainly experiments haven't shown that the gravitational field is made of discrete, countable and indivisible lumps of matter.

The electromagnetic fields in a conventional vacuum (_not_ the vacuum state in a Fock space) aren't made of matter either, by the very definition of a vacuum.

So, if you want to have another simple challenge, please derive the macroscopic Maxwell equations in vacuum (certainly something observable) from your particle picture.

I think I can defend the position that experimentalists see only accelerations of test particles (massive and/or charged), so gravitational and electromagnetic fields by themselves are non-observable and redundant. This idea has been developed in Chapters 12 and 13 of the book.

 

[meopemuk]

Fine. If you don't waste the opportunities given by my disturbing posts, you'll need to modify some more of your book. After these changes it will be a much better (and even publishable) book since it then agrees on the most important things with the main stream.

I would be happy to discuss with you which modifications you think are appropriate. There are only two problems: (1) I am rather stubborn, (2) moderators don't like discussions of unpublished stuff, so we will earn penalty points rather quickly. Perhaps by e-mail?

Eugene.

 

[A. Neumaier]

I would never call N(x)=a^*(x)a(x) a field operator. In my definition, quantum field is a very specific linear combination of particle creation and annihilation operators as defined in Weinberg's Chapter 5. Quantum field must have (1) covariant transformation rules and (2) (anti)commute at spacelike separations. In my notation x is eigenvalue of the Newton-Wigner position operator. So, I doubt that N(x)=a^*(x)a(x) satisfies the above two conditions.

But your doubt is unfounded. You can easily check it by assuming (1) and (2) for a(x) and a^*(x), and working out what it implies for N(x). N(x) commutes at spacelike separations, no matter which statistics the field a(x) has.

And you don't need to invoke the Newton-Wigner position operator to do that. The latter is needed only if you want to go from an unspecified representation of the Poincare group to the positin representation. But we are already working in the position representation.

I think I can defend the position that experimentalists see only accelerations of test particles (massive and/or charged), so gravitational and electromagnetic fields by themselves are non-observable and redundant. This idea has been developed in Chapters 12 and 13 of the book.

I'll look at your defense as given in the book - but not now.

 

[A. Neumaier]

I would be happy to discuss with you which modifications you think are appropriate. There are only two problems: (1) I am rather stubborn, (2) moderators don't like discussions of unpublished stuff, so we will earn penalty points rather quickly.

(1) I know that, but seem to have found a way to conquer your corresponding defenses.

(2) We can make it a discussion about published stuff, since the relevant statements in your book are based on misunderstandings of what is already in the published literature.

For example, you could open a new thread on the forms of relativistic dynamics, questioning some of the statements in Chapter B1 of my FAQ. I'd then try to convince you that all forms are equivalent, and the instant form is not as privileged as you claim it is. This is something everyone can learn from, so the moderators have no reason to penalize it.

 

[meopemuk]

But your doubt is unfounded. You can easily check it by assuming (1) and (2) for a(x) and a^*(x), and working out what it implies for N(x). N(x) commutes at spacelike separations, no matter which statistics the field a(x) has.

And you don't need to invoke the Newton-Wigner position operator to do that. The latter is needed only if you want to go from an unspecified representation of the Poincare group to the positin representation. But we are already working in the position representation.

I don't agree with you here. I would appreciate if you can provide a proof.

So, a(x) and a^*(x) are defined as annihilation and creation operators for a particle at point x (an eigenvalue of the Newton-Wigner position operator). If I understand correctly, you are saying that a(x) and a^*(x) have covariant transformation laws (as in eq. (5.1.6) - (5.1.7) in Weinberg) and they commute at spacelike separations (as in (5.1.32)). Can you prove these statements?

Eugene.

 

[A. Neumaier]

I don't agree with you here. I would appreciate if you can provide a proof.

You were right to disagree. Since I wrote my post quickly and from memory, I made a mistake; I was confusing a(x) and a(p).

So, a(x) and a^*(x) are defined as annihilation and creation operators for a particle at point x (an eigenvalue of the Newton-Wigner position operator). If I understand correctly, you are saying that a(x) and a^*(x) have covariant transformation laws (as in eq. (5.1.6) - (5.1.7) in Weinberg) and they commute at spacelike separations (as in (5.1.32)). Can you prove these statements?

Let me state what I really claim:

a(x) and a^*(x) have covariant transformation laws, and therefore so does N(x). They do not (anti)commute at spacelike separations, but only have an exponential falloff. In the customary language http://en.wikipedia.org/wiki/Composite_field , they are nonlocal, covariant fields.

Of course, this conflicts with your requirement that ''Quantum field must [...] (anti)commute at spacelike separations.'' But this is your personal requirement - it has nothing to do with the conventional terminology. Indeed, you can see just before (5.1.4) that Weinberg calls both
a(x) and a^*(x) [in his notation psi^+ (x) and psi^-(x)] fields - he talks about ''annihilation fields'' and ''creation fields''.

Quantum fields do not even need to be covariant - only the fields in relativistic quantum field theory are. In general, the only requirement for a quantum field is that it is an operator depending on a space-time argument.

 

[meopemuk]

a(x) and a^*(x) have covariant transformation laws

I don't think so. It is known that the Newton-Wigner position operator does not transform covariantly under boosts. So, I wouldn't expect covariant transformations of the corresponding operators a(x) and a^*(x).

Quantum fields do not even need to be covariant - only the fields in relativistic quantum field theory are. In general, the only requirement for a quantum field is that it is an operator depending on a space-time argument.

First, I must apologize for not being clear. I am interested only in relativistic quantum fields here. Second, the covariance and the space-like (anti)commutativity are absolutely essential for the definition of relativistic quantum fields. Only if these two conditions are satisfied, one can build interacting Hamiltonian density with properties (5.1.2) and (5.1.3) as products of fields (5.1.9). Note also the sentence at the bottom of page 198: "The point of view taken here is that Eq. (5.1.32) [the space-like (anti)commutativity] is needed for the Lorentz invariance of the S-matrix, without any ancillary assumptions about measurability or causality."

Eugene.

 

[dextercioby]

In general, the only requirement for a quantum field is that it is an operator depending on a space-time argument.

I strongly disagree with this, knowing that, in axiomatical QFT, the fields are both operators on the Fock space and distributions, so it's not \psi (x), a(x), a^{\dagger} (x), N(x), but rather \psi (f), a(f), a^{\dagger} (f), N(f), where typically f\in S\left(\mathbb{R}^4\right) (the so-called <smeatring> of fields).