Difference between 'Quantum theories'

[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).

 

[A. Neumaier]

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).

Well, in the Heisenberg picture, your 3D N(x) clearly changes with time, hence should be written N(x,t), or in 4D notation, again N(x); the same holds for a(x). The only well-defined a(x) is the one Weinberg defines - with different notation - in (5.1.4), and it is covariant, as he states in (5.1.6).

If you don't agree, please write down a fully precise definition of your version of a(x), so that we can discuss its properties.

First, I must apologize for not being clear. I am interested only in relativistic quantum fields here.

But this doesn't mean that you can change the traditional terminology to suit your narrow focus. If you fill standard concepts with your private meaning you don't need to be surprised that misunderstandings result.

Second, the covariance and the space-like (anti)commutativity are absolutely essential for the definition of relativistic quantum fields.

Assuming it were so, why then does Weinberg talk about annihilation fields and creation fields? And why does wikipedia in the link given talk about ''composite fields, which are usually nonlocal, are used to model asymptotic bound states''? Both statements refer to relativistic quantum field theory!

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."

It is needed _only_ for those fields used (p.198 top) ''to construct a scalar interaction density that satisfies the'' [properties derived in earlier chapters]. But there are many other fields, with other uses. In particular, these other fields are used to construct the fields that satisfy your (1) and (2). Indeed, to achieve this, Weinberg proceeds ''to combine annihilation and creation fields in linear combinations:'' (5.1.31).

Since the field N(x) is not needed to construct the interaction density, it is not restricted by Weinberg's considerations on p.198.

 

[A. Neumaier]

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).

For purists like you, let me restate in a slightly more precise way what I had expressed before on the level of rigor of Weinberg's book:

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

But you should know that few quantum field theorists feel the necessity to emphasize in their wording that their ''functions'' are distributions only. And since we are discussing Weinberg, it makes little sense to insist on more rigor than his level.

Moreover, there is no assumption in axiomatic QFT that the operators have to act on Fock space.

 

[meopemuk]

If you don't agree, please write down a fully precise definition of your version of a(x), so that we can discuss its properties.

There are two things that look very similar, but have very different physical interpretations. One of them I denote a(x) and call "operator annihilating particle at space point x". The other one I denote \psi(x) and call "particle annihilation field".

Let me start with the former operator a(x). It is defined as the following integral of particle annihilation operators in the momentum representation

a(x) = \int dp \exp(ipx) a(p)

The physical meaning of this operator is that it annihilates the particle at a given point x in physical space. One can show that (1) a(x,t) does *not* transform covariantly under boosts, and that (2) [a(x), a^{\dag}(x&#039;)] = 0 , because Newton-Wigner wave functions of particles localized at x and x' are orthogonal. Due to the property (1), operator a(x) cannot be used in construction of relativistic interaction operators a la Weinberg. I do not apply the word "field" to operator functions a(x).


The particle annihilation field is defined as

\psi(x) = \int \frac{dp}{\omega_p} \exp(ipx) a(p)

Note the presence of the extra factor \omega_p = \sqrt{p^2 + m^2} there. Due to this factor, the field \psi(x,t) *does* transform covariantly under boosts. However, the commutation relation [\psi(x), \psi^{\dag}(x&#039;)] =0 is not valid. In order to achieve this commutator one needs to make a sum "annihilation field + creation field" as discussed in Weinberg's section 5.2. Then one obtains the full *quantum field* \Psi(x,t) which is good to be used as a factor in interacting Hamiltonians. However, it is important to note that \psi(x) can *not* be interpreted as operator annihilating the particle at a space point x. Such interpretation would be in conflict with properties of Newton-Wigner position-space wave functions. This is why I say that quantum fields \Psi(x,t) are just abstract mathematical constructs. Their argument x has nothing to do with real spatial position.

Eugene.

 

[A. Neumaier]

There are two things that look very similar, but have very different physical interpretations. One of them I denote a(x) and call "operator annihilating particle at space point x". The other one I denote \psi(x) and call "particle annihilation field".

Thanks for the clarification. Let me rewrite what you said in covariant notation. For simplicity, I only consider neutral scalar fields, and take hbar=c=1. I use the inner product with signature +---, and write Dp for the appropriately normalized invariant measure on the mass shell. I write a(p) for the annihilation operator with 4-momentum p, scaled such that the smeared annihilators
a(f):=\int Dp f(p) a(p)
satisfy
[a(f),a(g)^*]=\int Dp f(p) g(p)^*
for square integrable test functions f,g on the mass shell.
(This differs from Weinberg's annihilation operators by a factor proportional to sqrt(p_0) but has the advantage of making everything manifestly covariant.) Then we have several kinds of quantum fields:

A. The annihilation field
\psi^+(x) = \int Dp e^{ip\cdot x}a(p)
that annihilates the vacuum. It satisfies covariance but violates causality. The adjoint creation field
\psi^-(x) = \int Dp e^{-ip\cdot x}a(p)^*
creates 1-particle states from the vacuum and also satisfies covariance but violates causality.

B. The Heisenberg field (as it is commonly called)
\psi(x) = \psi^+(x) + \psi^-(x)
that figures in the interaction density. It satisfies covariance and causality, hence can be used to define an interaction density.

C. For each future-pointing velocity 4-vector u with u^2=1, a Newton-Wigner field
a_u(x) = \int Dp \sqrt{u\cdot p} e^{ip\cdot x}a(p)
and its adjoint. They are frame-dependent and violate covariance but satisfy a CCR of the form
[a_u(x),a_u(y)^*]=0~~~~~~~ (u\cdot x=u\cdot y,~~ x\ne y).
An observer moving along the world line x(s) with velocity u(s)=\dotx(s) has at each moment s its private time coordinate t(s)=u(s)\cdot x(s), its private 3-space defined by the hyperplane u(s)\cdot x=t(s), and its private Newton-Wigner field a(s)=a_{u(s)} (dependence on x suppressed). Because of the CCR, the latter can in principle be prepared and measured independently at each point of the private 3-space.

D. For every sufficiently nice kernel K(p,q) the composite field
N_K(x) = \int Dp Dq K(p,q) e^{i(q-p)\cdot x}a(p)^*a(q)
It transforms covariantly iff K(p,q) is Lorentz invariant, and its commutation properties can be worked out; I'll do this another time. Some of these fields - which ones we'll have to discuss - deserve to be called mass density field, energy density field, etc..

According to the traditional terminology, all these fields deserve to be called quantum fields, since they are operator-valued distributions.

''In physics, a field is a physical quantity associated to each point of spacetime. [...] a field can be either a classical field or a quantum field, depending on whether it is characterized by numbers or quantum operators respectively.'' (http://en.wikipedia.org/wiki/Quantum_field )

 

[meopemuk]

''In physics, a field is a physical quantity associated to each point of spacetime. [...] a field can be either a classical field or a quantum field, depending on whether it is characterized by numbers or quantum operators respectively.'' (http://en.wikipedia.org/wiki/Quantum_field )

I would like to stress three major points:

1. In relativistic quantum mechanics, position space must be defined according to Newton-Wigner.

2. Annihilation quantum field \psi^+(x,t) does annihilate a single particle. However, this particle's wavefunction is *not* localized in the Newton-Wigner position space. Therefore, the field argument x should *not* be interpreted as spatial coordinate. Quantum field \psi^+(x,t) is just a formal mathematical object, which does not need to be interpreted or compared with experiment. So, in your wikipedia quote the words "physical quantity" and "point of spacetime" should not be applied to the fields constructed in Weinberg's Chapter 5.

3. On the other hand, the operator function a(x) has a clear interpretation as operator annihilating a particle localized exactly at point x in the physical space. One can also form the particle number operator for the point x as N(x) = a*(x)a(x). So, these operators are definitely associated with "points of space". However, their usefulness does not indicate that in reality there exists some continuous physical substance, called field. These operators are humble mathematical servants to the true masters - *particles*.

Eugene.

 

[A. Neumaier]

1. In relativistic quantum mechanics, position space must be defined according to Newton-Wigner.

In a covariant description of relativistic QM, there are many Newton-Wigner operators and hence many position spaces - namely one for each distinguished time direction.
You declare just one of them to be the right one - against the established consensus that
relativistic physics must look the same in every frame of reference. My quantum field operators (C) have this property and precisely encode the Newton-Wigner information that you were writing in a preferred frame of reference.

2. Annihilation quantum field \psi^+(x,t) does annihilate a single particle. However, this particle's wavefunction is *not* localized in the Newton-Wigner position space. Therefore, the field argument x should *not* be interpreted as spatial coordinate. Quantum field \psi^+(x,t) is just a formal mathematical object, which does not need to be interpreted or compared with experiment. So, in your wikipedia quote the words "physical quantity" and "point of spacetime" should not be applied to the fields constructed in Weinberg's Chapter 5.

You are the only one that requires an arbitrary quantum field to be localized.

Newton-Wigner position is pure space, not space-time, and hence has no connection with
the terminology in wikipedia. Whereas the four kinds of fields I constructed have a space-time parameter and qualify as quantum field by the standard definition.

3. On the other hand, the operator function a(x) has a clear interpretation as operator annihilating a particle localized exactly at point x in the physical space. One can also form the particle number operator for the point x as N(x) = a*(x)a(x). So, these operators are definitely associated with "points of space".

Quantum fields are definitely associated with space-time as a whole, not with an observer-dependent 3-space slice of space-time.

A number operator must have a spectrum consisting precisely of {0,1,2,3,...}.
Therefore, your N(x) is _not_ a particle number operator, but a spatial field density.

 

[meopemuk]

In a covariant description of relativistic QM, there are many Newton-Wigner operators and hence many position spaces - namely one for each distinguished time direction.
You declare just one of them to be the right one - against the established consensus that
relativistic physics must look the same in every frame of reference. My quantum field operators (C) have this property and precisely encode the Newton-Wigner information that you were writing in a preferred frame of reference.

Your requirement that "relativistic physics must look the same in every frame of reference" is already encoded in the Poincare group representation U_g, which is the basis of all relativistic physics. No extra efforts are needed to ensure that.

I can write the Newton-Wigner position operator for one observer as (for a spinless massive particle, to make it simple)

\mathbf{R} = -(1/2)(\mathbf{K}H^{-1} + H^{-1} \mathbf{K})...(1)

where H is the Hamiltonian and \mathbf{K} is the boost operator. For any other observer that is related to the original one by the group element g, the Newton-Wigner operator is obtained by standard formula

\mathbf{R}&#039; = U_g \mathbf{R} U_g^{-1}

The new operator \mathbf{R}&#039; has the same dependence on the transformed H' and \mathbf{K}&#039; as (1)

\mathbf{R}&#039; = -(1/2)(\mathbf{K}&#039;(H&#039;)^{-1} + (H&#039;)^{-1} \mathbf{K}&#039;).

You are the only one that requires an arbitrary quantum field to be localized.

Perhaps, I've misunderstood you. I thought that you interpret \psi^+(x,0) as an operator annihilating the particle at point x at time 0. Now, you are saying that this is not the case. If so, then what is your physical interpretation of the operator \psi^+(x,0)?

In my opinion, this operator has no clear physical interpretation, and such an interpretation is not necessary to develop the QFT formalism.

A number operator must have a spectrum consisting precisely of {0,1,2,3,...}.
Therefore, your N(x) is _not_ a particle number operator, but a spatial field density.

Agreed.

Eugene.

 

[A. Neumaier]

I can write the Newton-Wigner position operator for one observer as (for a spinless massive particle, to make it simple)

\mathbf{R} = -(1/2)(\mathbf{K}H^{-1} + H^{-1} \mathbf{K})...(1)

where H is the Hamiltonian and \mathbf{K} is the boost operator. For any other observer that is related to the original one by the group element g, the Newton-Wigner operator is obtained by standard formula

\mathbf{R}&#039; = U_g \mathbf{R} U_g^{-1}

The new operator \mathbf{R}&#039; has the same dependence on the transformed H' and \mathbf{K}&#039; as (1)

\mathbf{R}&#039; = -(1/2)(\mathbf{K}&#039;(H&#039;)^{-1} + (H&#039;)^{-1} \mathbf{K}&#039;).

But how does this transformation change the description of the Newton-Wigner annihilation field? This remains obscure in your style of writing things while it is obvious in my manifestly covariant form. Keeping manifest covariance whenever possible usually makes working with formulas a lot simpler.

Perhaps, I've misunderstood you. I thought that you interpret \psi^+(x,0) as an operator annihilating the particle at point x at time 0. Now, you are saying that this is not the case. If so, then what is your physical interpretation of the operator \psi^+(x,0)?

In my opinion, this operator has no clear physical interpretation, and such an interpretation is not necessary to develop the QFT formalism.

I was doing neither. I mainly tried teaching you that all these expressions fully deserve to be called quantum fields, as they are called by everyone except you.

Physical interpretation is a different matter, and in terms of interpretation, there is far less agreement in the literature, so I claim much less. To discuss interpretation issues, we need to clarify what it means to assign the label ''physical'' or ''measurable'' to a field. But this is far from trivial, and cannot be done by a simple declaration.

So let us first agree to use the standard terminology regarding quantum fields, as specified e.g., by the wikipedia quote.

Then we can see whether we can reach an agreement about what the various fields could mean. I don't have a set mind about the latter, since I had never had an opportunity to discuss measurement in a quantum field context. But I have very high standards about the properties that need to be satisfied before a particular claim is justified.

 

[StevieTNZ]

So... QED and QFT still contain superposition of states, 'collapse of the wavefunction', entanglement, etc. I.e. All the implications of QM?

 

[meopemuk]

So... QED and QFT still contain superposition of states, 'collapse of the wavefunction', entanglement, etc. I.e. All the implications of QM?

Yes, this is exactly true. QFT and QED are just particular cases of the general formalism of quantum mechanics. Unfortunately, many QFT textbooks don't emphasize this fact and make some readers believe that QFT is a kind of "next step" beyond quantum mechanics. At least, this was the impression I got in my early studies of QFT, when I read about "second quantization" and stuff. I hope you wouldn't make the same mistake.

Eugene.