Difference between 'Quantum theories'

[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) $$[a(x), a^{\dag}(x')] = 0$$ , 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 $$[\psi(x), \psi^{\dag}(x')] =0$$ 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

$$\mathbf{R}' = U_g \mathbf{R} U_g^{-1}$$

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

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

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

$$\mathbf{R}' = U_g \mathbf{R} U_g^{-1}$$

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

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

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.

 

[StevieTNZ]

Glad I asked because I was definitely confused looking at Wikipedia pages on second quantization, QFT, etc.

 

[meopemuk]

But how does this transformation change the description of the Newton-Wigner annihilation field?

This is easy. Each operator transforms to the new reference frame by the same general formula. For the Newton-Wigner "annihilation field" this formula is

$$a(x) \to U_g a(x) U_g^{-1} = \int dp e^{ipx }U_g a(p) U_g^{-1}$$

Transformation formulas for the momentum-space annihilation operators a(p) can be found in Weinberg's textbook.

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.

So, basically you are saying that any operator expression that depends on arguments (x,t) can be called "quantum field". But I hope you wouldn't claim that all these various expressions can be used in the Weinberg's quote, from which we started our discussion:

''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."

Eugene.

 

[meopemuk]

Glad I asked because I was definitely confused looking at Wikipedia pages on second quantization, QFT, etc.

Wikipedia is terribly bad as a source of information about QFT. My best advise is to learn QFT from Weinberg's textbook. This is not an easy read for a beginner, but, at least, this book presents a clear and correct logic of what follows from what. If you complement the reading by articles from the literature lists in each chapter, then you will make it.

Eugene.

 

[A. Neumaier]

This is easy. Each operator transforms to the new reference frame by the same general formula. For the Newton-Wigner "annihilation field" this formula is

$$a(x) \to U_g a(x) U_g^{-1} = \int dp e{ipx }U_g a(p) U_g^{-1}$$

Transformation formulas for the momentum-space annihilation operators a(p) can be found in Weinberg's textbook.

Yes, messy formulas that don't tell what's going on. Compare my uniform, manifestly covariant formulas with his (5.1.11) and (5.1.12). I don't understand why you prefer the latter.

So, basically you are saying that any operator expression that depends on arguments (x,t) can be called "quantum field".

Yes. that s the standard usage. Just like any vector expression that depends on arguments (x,t) can be called ''vector field''.

But I hope you wouldn't claim that all these various expressions can be used in the Weinberg's quote, from which we started our discussion:

''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."

Insert ''certain'' before ''quantum'', and this problem is resolved. Which quantum fields qualify as ''basic ingredients'' is something that we'd need to discuss further.

I'd like to focus our further discussion by dissecting it into a number of more or less independent parts:

(i) Which fields represent the operational content of quantum field theory. This is directly related to your preceding comment. I'll open a new thread about this, in which I'll summarize the main issues from that part of our discussion here.

(ii) Indistinguishable particles and fields. I'll open a new thread about this, in which I'll summarize the main findings from that part of our discussion here. See
https://www.physicsforums.com/showthread.php?t=474321
https://www.physicsforums.com/showthread.php?t=474293

(iii) What deserves to be called a quantum field. Should this not yet be agreed upon by your remark above, I'd like to have this discussed in the present thread.

(iv) How to display and use Lorentz covariance. It seems to me that you only pay lip service to it by saying that given the transformation laws in Weinberg, everything is said about it and one doesn't need to consider it further. I'd like to have this discussed in the present thread. (By the way, I think that this attitude is part of the reason why you get certain unfounded, crackpottish results in parts of your book on QED - but I won't discuss the latter directly, unless you open a thread on your book in the Independent Research forum.)

 

[A. Neumaier]

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*.

In which sense are your *particles* the true masters? Do you mean to imply with your statement that in reality there exists some discrete physical substance, called particles?

You agreed already that N(x) is not a number operator but a field density, and integrating N(x) over a region Omega of space will never result in an operator with integral spectrum, unless you take Omega=R^3.

Thus, how do you identify a particle, given a state of the whole system which is not just a single particle state?

What is discrete about the reality underlying QED? The only discrete thing I can see in QED is the spectrum of the object called (for historical reasons) the number operator. But this is an extremely nonlocal object that cannot be observed at all. So why should it be considered more real than the charge density, the current, or the electromagnetic field, which are observed every day?

Finally, why does this discrete reality (if it can be pinned down at all) deserve to carry the name particles?

 

[meopemuk]

In which sense are your *particles* the true masters? Do you mean to imply with your statement that in reality there exists some discrete physical substance, called particles?

Yes, this is exactly what I mean.

integrating N(x) over a region Omega of space will never result in an operator with integral spectrum, unless you take Omega=R^3.

Here I disagree. If we integrate N(x) over any region Omega of space we obtain an Hermitian operator with the spectrum 0,1,2,3,... whose interpretation is "the number of particles in the region Omega".

Thus, how do you identify a particle, given a state of the whole system which is not just a single particle state?

Any QFT state vector can be expanded as a linear superposition of 0-particle, 1-particle, 2-particle etc. states, according to the Fock space structure. So, I can always calculate the probability of finding N particles in any given state.

What is discrete about the reality underlying QED? The only discrete thing I can see in QED is the spectrum of the object called (for historical reasons) the number operator. But this is an extremely nonlocal object that cannot be observed at all.

Scattering amplitudes is the final result of most QFT calculations. When we calculate scattering amplitudes in QFT, we always specify the number of particles (and their observables, like momentum and spin) in both initial and final states. So, discrete particles are always present in their explicit form. I am not sure what you mean by the word "nonlocal", and why the particle number cannot be observed in your opinion?

So why should it be considered more real than the charge density, the current, or the electromagnetic field, which are observed every day?

As I wrote already, if you look at charge and current densities with high enough resolution, you'll see that they are not continuous quantities, but averaged descriptions for a large number of particles - electrons. The same is true for the free electromagnetic field. If the field has high frequency and low intensity, we can easily distinguish discrete particles there - photons. It is true that static electric and magnetic fields are really continuous, but they are not observable in the absence of charged particles. So, one can defend the position that there is nothing but charged particles directly interacting with each other via distance- and velocity-dependent forces.

 

[A. Neumaier]

Yes, this is exactly what I mean.

This is what I feared you mean. But there is no evidence at all for this point of view.

Here I disagree. If we integrate N(x) over any region Omega of space we obtain an Hermitian operator with the spectrum 0,1,2,3,... whose interpretation is "the number of particles in the region Omega".

That you make such claims without getting red in your face shows that your intuition about the behavior of field operators is very poorly developed. You'd make lots of calculations with quantum fields to get some practice...

Please check the action of N(x) (at time t=0) on the 1-particle sector. It is a simple exercise to check that when the support of the state is all of R^3 then the expectation of the integral of N(x) over a nonempty, open and bounded domain is always strictly between 0 and 1. This implies that the eigenvalues also must lie in this interval.

So your number interpretation is a figment of your imagination.

Any QFT state vector can be expanded as a linear superposition of 0-particle, 1-particle, 2-particle etc. states, according to the Fock space structure. So, I can always calculate the probability of finding N particles in any given state.

Please show me how you calculate, given an arbitrary Fock state, the probability of finding a single particle in a fixed bounded region Omega.

Scattering amplitudes is the final result of most QFT calculations. When we calculate scattering amplitudes in QFT, we always specify the number of particles (and their observables, like momentum and spin) in both initial and final states. So, discrete particles are always present in their explicit form.

Although you had often claimed the above, QFT is good for much more than calculating scattering amplitides.

I am not sure what you mean by the word "nonlocal", and why the particle number cannot be observed in your opinion?

The only integral over N(x) with an integral spectrum is the integral over all of R^3, a very nonlocal operator: One needs to gather information at every point x in space to be able to say something about this integral.

As I wrote already, if you look at charge and current densities with high enough resolution, you'll see that they are not continuous quantities,

I cannot see this. Please demonstrate it for me.

 

[meopemuk]

Please check the action of N(x) (at time t=0) on the 1-particle sector. It is a simple exercise to check that when the support of the state is all of R^3 then the expectation of the integral of N(x) over a nonempty, open and bounded domain is always strictly between 0 and 1. This implies that the eigenvalues also must lie in this interval.

So your number interpretation is a figment of your imagination.



Please show me how you calculate, given an arbitrary Fock state, the probability of finding a single particle in a fixed bounded region Omega.

Operator for the number of particles in the volume $$\Omega$$ is

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

Consider an N-particle state vector in the Fock space, with each particle having a definite position

$$\Psi = a^{\dag}(y_1) a^{\dag}(y_2) \ldots a^{\dag}(y_N) |0 \rangle$$...(1)

Then one can prove that

$$N(\Omega) \Psi = M \Psi$$

where M is an integer number, which tells us how many particles lie inside the volume $$\Omega$$. States of the type (1) form a basis in the Fock space. If $$\Psi$$ is a one-particle state

$$\Psi = a^{\dag}(y_1) |0 \rangle$$

then

$$N(\Omega) \Psi = \Psi$$ if $$y_1 \in \Omega$$
$$N(\Omega) \Psi = 0$$ if $$y_1 \not{\in} \Omega$$

For a state that is not well-localized, the expectation value of $$N(\Omega)$$ gives the probability of finding the particle in the volume $$\Omega$$.

Although you had often claimed the above, QFT is good for much more than calculating scattering amplitides.

QFT can also calculate energies of bound states (e.g., Lamb shifts). What else?

I cannot see this. Please demonstrate it for me.

I hope you wouldn't argue that the current in metals is produced by discrete electrons. Continuous charge and current fields are just very rough approximations. The discrete nature of the radiation field was demonstrated by Einstein in his 1905 paper about the photo-electric effect.

 

[A. Neumaier]

$$N(\Omega) = \int_{\Omega} dx a^{\dag}(x)a(x)$$
Consider an N-particle state vector in the Fock space, with each particle having a definite position
$$\Psi = a^{\dag}(y_1) a^{\dag}(y_2) \ldots a^{\dag}(y_N) |0 \rangle$$...(1)

Unfortunately, your proposed state is not a Fock state, not even when you symmetrize it. For it is not normalizable.

Then one can prove that
$$N(\Omega) \Psi = M \Psi$$
where M is an integer number

However, I see that the situation N=1 that I had considered was too simple to decide things since there is nothing yet to symmetrize. Indeed, on the 1-particle sector, N(Omega) is an orthogonal projector with eigenvalues 0 and 1 only. This shows that my previous argument was flawed, and that in an infinite-dimensional Hilbert space one can conclude only that the spectrum is in [0,1], not that the endpoints are excluded.

Upon reexamining the situation, I can see that a precise formulation of your idea is the following:

If Omega is open, the kernel of N(Omega)=integral_Omega dx a(x)^*a(x) is the Fock space F_0(Omega) whose 1-particle space is the Hilbert space of wave functions whose support is disjoint with Omega. And for M>0, we have N(Omega) Psi = M Psi for every Psi in the (non Fock) space F_M(Omega) obtained from F_0(Omega) by applying to its elements M smeared creation operators whose wave function has support in the closure of Omega. This defines a direct decomposition of the full Fock space. Therefore, yes, you are right, N(Omega) has spectrum {0,1,2,...}.

So it is me who is getting red in the face. I apologize, and thank you for having learned something new. (I should have known from Haag's local field theory, but I had never seen discussed that certain spectra are preserved under localization. Maybe DarMM can point to a place in the literature?)

Thus there is _something_ discrete about QFT, namely the existence of number operators localized in some region. Thus we can consistently talk about the expected number of massive particles in a region Omega of space at a given time t (which was zero in the above discussion).

But of course, this is a nonrelativistic situation, and it is the relativistic treatment that is relevant to the foundations of quantum field theory. That something nontrivial happens is that the above can be replicated in the relativistic situation only for massive particles since one needs a Newton-Wigner position operator to bring the Fock space in a representation where the annihilator fields satisfy the nonrelativistic spatial CCR. In particular, nothing of the above discussion applies to massless particles; so QED is not covered since photons are massless.

In any case, the problem of which operators are associated to observable reality gets nontrivial in relativistic field theory and must be discussed carefully. This is a situation that I haven't yet fully analyzed, so I need more time to prepare for this discussion, which may take a few days. When I am ready, I'll start a new thread.

QFT can also calculate energies of bound states (e.g., Lamb shifts). What else?

I had answered this already:

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.

Indeed, _all_ of the many applications of quantum field theory to macroscopic matters are done with little or no reference to scattering processes. Look at any book on nonequilibrium statistical mechanics. Note that there is little difference in principle between relativistic and nonrelativistic treatments - in both cases one uses the same field theory. The nonrelativistic case differs only in that
- antiparticles and particle-number changing processes are neglected, and
- interactions are nonlocal.
Nothing else.

I hope you wouldn't argue that the current in metals is produced by discrete electrons. Continuous charge and current fields are just very rough approximations. The discrete nature of the radiation field was demonstrated by Einstein in his 1905 paper about the photo-electric effect.

I answered this one earlier today ago in a separate thread,
https://www.physicsforums.com/showthread.php?p=3147920
and ask you to reply there concerning this aspect.

See also my previous post #37 in the present thread.

 

[meopemuk]

Unfortunately, your proposed state is not a Fock state, not even when you symmetrize it. For it is not normalizable.

Sorry, I don't get it. I thought that this is the standard QFT method for building symmetrized and normalized N-particle states. See, for example, (4.2.2) in Weinberg's book. The only difference in my formula is that I am using position representation instead of Weinberg's momentum representation.

That something nontrivial happens is that the above can be replicated in the relativistic situation only for massive particles since one needs a Newton-Wigner position operator to bring the Fock space in a representation where the annihilator fields satisfy the nonrelativistic spatial CCR.

This is true. A satisfactory position operator for photons has not been constructed yet. Perhaps, it cannot be defined at all. Perhaps, photons cannot be exactly localized in the position space. I don't have a certain opinion about that. However, they can be perfectly localized and counted in the momentum space. So, they still look like discrete particles.

Indeed, _all_ of the many applications of quantum field theory to macroscopic matters are done with little or no reference to scattering processes. Look at any book on nonequilibrium statistical mechanics. Note that there is little difference in principle between relativistic and nonrelativistic treatments - in both cases one uses the same field theory. The nonrelativistic case differs only in that
- antiparticles and particle-number changing processes are neglected, and
- interactions are nonlocal.
Nothing else.

Yes, QFT applications to condensed matter processes use many of the same techniques as the fundamental relativistic QFT of elementary particles. However, at the fundamental level these two frameworks are quite different. Condensed matter QFT is inherently approximate as it ignores the discrete atomistic structure of matter. For example, the continuous phonon field is not a good approximation at distances comparable with interatomic separations in the crystal. On the other hand, relativistic QFT is supposed to be exact at all distances.

Of course, there is also an "effective field theory" school of thought in relativistic QFT, which claims that relativistic fields are not exact concepts, and that there is some yet unknown space-time granularity, which plays the role similar to the "crystal lattice" for fundamental particle fields. However, in my opinion, this is a wild hypothesis without any experimental support.

The field approximation in condensed matter is justified, because we deal with systems containing a huge number of particles. So, using the particle-based picture would be mathematically hopeless. In the fundamental relativistic QFT, we usually focus on processes involving small number of particles, which can be treated separately.

Another difference between the condensed matter and fundamental field theories is that the former does not use the concept of (Galilean) relativity, because there is always one distinguished frame of reference - the one connected with the underlying crystal lattice.

Eugene.