A Stephen Weinberg on Understanding Quantum Mechanics

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rubi

Of course, one can get and measure projectors of observables whose exponentials are in the observable algebra. For these, Born's rule works and gives probabilities. But this does not alter the fact that the test for being in a particular vector state cannot be carried out since this particular projector is not obtainbable in this way (since it is not in the observable, although it is a bounded operator).
Well, projectors that don't correspond to physical questions about observables are not in the algebra. But is that a problem? All projections of physical observables are available and can be obtained by the functional calculus, so we have the standard Born rule for all physical questions.

A. Neumaier

All projections of physical observables are available and can be obtained by the functional calculus, so we have the standard Born rule for all physical questions.
Yes, so the impact on the foundations is a bit different than what I had initially indicated. The net effect of the III_1 discussion are the first two points of the following list, with a third point not yet discussed:
1. It makes no sense to consider mixed states as being composed of pure states: They are much vector states as any other vector states.
2. Pure states don't exist on the most fundamental level. Vector states do exist but their interpretation is dependent on which representation one is considering.
3. The superposition principle is invalid on the most fundamental level since it fails between vector states from two different representations. Note that the GNS construction produces a different Hilbert space for each state. Fixing one of these Hilbert spaces gives access to states from only one superselection sector.
In particular, it is not even clear what the superposition of a decaying particle state (described by QM) and a cat (described by QFT) should be. If one simply takes the tensor product one has to specify a particular interaction, which is unlikely to be the interaction that would follow from the standard model or an even more fundamental description including gravity. Lacking an acceptable formalization it is moot to ask what happens when the particle decays.

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RockyMarciano

The superposition principle is invalid on the most fundamental level since it fails between vector states from two different representations.
The superposition principle is an essential property of linear theories, are you suggesting that QM is not fundamentally linear in Weinberg's sense or something else altogether?

Auto-Didact

1. It makes no sense to consider mixed states as being composed of pure states: They are much vector states as any other vector states.
2. Pure states don't exist on the most fundamental level. Vector states do exist but their interpretation is dependent on which representation one is considering.
3. The superposition principle is invalid on the most fundamental level since it fails between vector states from two different representations. Note that the GNS construction produces a different Hilbert space for each state. Fixing one of these Hilbert spaces gives access to states from only one superselection sector.
In particular, it is not even clear what the superposition of a decaying particle state (described by QM) and a cat (described by QFT) should be. If one simply takes the tensor product one has to specify a particular interaction, which is unlikely to be the interaction that would follow from the standard model or an even more fundamental description including gravity. Lacking an acceptable formalization it is moot to ask what happens when the particle decays.
Have you read Penrose' The Road To Reality?
Apart from the Bloch sphere description of two state density matrices given there and your particular conclusion, this post somewhat mirrors parts of chapter 29. In fact, as far as I can see, almost all points mentioned in this entire thread as well as many others are discussed in comparable (or perhaps even greater) detail in that same single chapter. In case you haven't read it, it is duly recommended.

A. Neumaier

The superposition principle is an essential property of linear theories, are you suggesting that QM is not fundamentally linear in Weinberg's sense or something else altogether?
I am just drawing conclusions from the structure of quantum field theory. For example, it is impossible to prepare a superposition of a charged and an uncharged electron.

Already on the level of single-particle quantum mechanics, it is impossible to prepare a superposition of a state of spin 1/2 and a state of spin 0, since these transform differently under rotations. This was known for a long time (1950s). Thus those interested could have known long ago that the superposition principle is not universally valid.

• dextercioby

A. Neumaier

Have you read Penrose' The Road To Reality?
Apart from the Bloch sphere description of two state density matrices given there and your particular conclusion, this post somewhat mirrors parts of chapter 29. In fact, as far as I can see, almost all points mentioned in this entire thread as well as many others are discussed in comparable (or perhaps even greater) detail in that same single chapter. In case you haven't read it, it is duly recommended.
I had read it some time ago. Penrose adds his own speculations about the role of quantum gravity, which I don't support. I'll reread his book in due time in the light of your comments.

RockyMarciano

I am just drawing conclusions from the structure of quantum field theory. For example, it is impossible to prepare a superposition of a charged and an uncharged electron.

Already on the level of single-particle quantum mechanics, it is impossible to prepare a superposition of a state of spin 1/2 and a state of spin 0, since these transform differently under rotations. This was known for a long time (1950s). Thus those interested could have known long ago that the superposition principle is not universally valid.
But it can be said that if the structure of QFT makes the principle of superposition invalid that should at least partially solve the "measurement problem", at least in the form of why certain outcomes are measured, since there would not be all those outcomes in superposition to begin with. Is this what you mean when you claim that there is no measurement problem or at least that it is ill-posed in QM?

A. Neumaier

Is this what you mean when you claim that there is no measurement problem or at least that it is ill-posed in QM?
Could you please cite the context where I made this claim?

rubi

The superposition principle is invalid on the most fundamental level since it fails between vector states from two different representations. Note that the GNS construction produces a different Hilbert space for each state. Fixing one of these Hilbert spaces gives access to states from only one superselection sector.
I don't think you can say this. Taking convex combinations of algebraic states is not the superposition principle (not even for type $I$ algebras). It rather corresponds to how you add density matrices to get another density matrix. If you have fixed one algebraic state $\omega$ and realized it on a Hilbert space $\mathcal H$, you can consider its folium $\mathrm{Fol}(\omega)=\{\xi(A)=\mathrm{Tr}(\rho A):\rho \text{ is a density matrix on }\mathcal H\}$, i.e. the set of all algebraic states that can be realized as density matrices on $\mathcal H$. One can show (Fell's theorem) that this folium is in a certain way dense in the set of all algebraic states, i.e. there is no physical way to measure whether your theory is defined by a state that can't be realized as a density matrix on $\mathcal H$. Thus, physically, it is admissible to restrict the discussion to only one Hilbert space, like in ordinary QM. So the fact that there are states that don't lie in the folium of $\omega$ needn't bother us much.

A. Neumaier

the set of all algebraic states that can be realized as density matrices on $\mathcal H$. One can show (Fell's theorem) that this folium is in a certain way dense in the set of all algebraic states, i.e. there is no physical way to measure whether your theory is defined by a state that can't be realized as a density matrix on $\mathcal H$. Thus, physically, it is admissible to restrict the discussion to only one Hilbert space, like in ordinary QM.
Maybe it is admissible to restrict discussion to the vacuum sector of a quantum field theory, and discuss charged states in these terms. But this is not the way quantum field theory is done.

In addition, on the foundational level, arguments with approximate states are very dangerous since it is virtually impossible to gauge the effect of even minor imperfections in the state on the analysis of system plus its detector since the corresponding multiparticle dynamics is extremely sensitive to details (it is chaotic in the kinetic approximation).

The point is that being dense ''in a certain way'' is completely irrelevant if the certain way is not the topology relevant for the discussion of the observables of interest.

• RockyMarciano

A. Neumaier

Yes, in the context of this thread https://www.physicsforums.com/threads/the-typical-and-the-exceptional-in-physics.885480/page-20 I got the impression(maybe wrong) that you agreed with vanhees71 that there was no measurement problem, I think now that you might be referring to this in the context of QFT and the superposition principle.
Hm, on this page there is not a single answer by myself. Please properly quote my text (cut and paste since the thread is already locked, and refer to the post number).

There is no measurement problem in a practical sense. On the other hand, there are lots of challenging problems in modeling the measurement process in terms of statistical mechanics. So everything boils down to what one precisely means, and for a term like ''measurement problem'' this depends to some extent on the context. Statements taken out of context become ambiguous and easily convey the opposite of what was intended.

RockyMarciano

Hm, on this page there is not a single answer by myself. Please properly quite my text (cut and paste since the thread is already locked, and refer to the post number).

There is no measurement problem in a practical sense. On the other hand, there are lots of challenging problems in modeling the measurement process in terms of statistical mechanics. So everything boils down to what one precisely means, and for a term like ''measurement problem'' this depends to some extent on the context. Statements taken out of context become ambiguous and easily convey the opposite of what was intended.
The thread is very long and probably there is not a specific post where you explicitly say that, it was just the general tone of your position in the discussion, but I think I can give it a more precise context now, and understand better what you mean and agree.

rubi

Maybe it is admissible to restrict discussion to the vacuum sector of a quantum field theory, and discuss charged states in these terms. But this is not the way quantum field theory is done.
If you do QFT in the Wightman sense, you are supposed to work on a fixed Hilbert space and model your states as vectors or density matrices on that space.

The point is that being dense ''in a certain way'' is completely irrelevant if the certain way is not the topology relevant for the discussion of the observables of interest.
The precise statement is that if you have some fixed observable algebra $\mathfrak A$ and a finite (but arbitrarily large) number of observables $A_i\in\mathfrak A$ and a corresponding list of arbitrarily small measurement uncertainties $\epsilon_i>0$, then for any state $\omega$, there exists a state $\xi\in\mathrm{Fol}(\omega_0)$ of some fixed state $\omega_0$ such that $\left|\omega(A_i)-\xi(A_i)\right|\lt\epsilon_i$, i.e. there exists no way to physically distinguish the state $\omega$ from a state in the folium of the state $\omega_0$. Hence, all states of physical relevance can be taken to lie in $\mathrm{Fol}(\omega_0)$ and thus be written as density matrices on a fixed Hilbert space $\mathcal H$. This is clearly the topology of physical interest.

A. Neumaier

If you do QFT in the Wightman sense, you are supposed to work on a fixed Hilbert space and model your states as vectors or density matrices on that space.
But it is well-known that the Wightman axioms describe only the vacuum sector and fail to describe QED.

• RockyMarciano

A. Neumaier

This is clearly the topology of physical interest.
No. The Hilbert space topology is not the topology relevant for discussion of the behavior of coarse-grained expectation values. For the latter one needs a topology relevant for discussing the dynamics of the hierarchy of quantum BBGKY equations.

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rubi

But it is well-known that the Wightman axioms describe only the vacuum sector and fail to describe QED.
No, it's not known that QED isn't a Wightman theory. There are only heuristic arguments. And Fell's theorem shows that if the Wightman axioms fail to describe QED, then any other state $\omega$ on the QED observable algebra will also fail to do so.

No. The Hilbert space topology is not the topology relevant for discussion of the behavior of coarse-grained expectation values. For the latter one needs a topology relevant for discussing the dynamics of the hierarchy of quantum BBGKY equations.
I have not mentioned the Hilbert space topology at all. My post was about a topology on the space of algebraic states and it of course also applies to algebras on which a dynamics is defined. A theorem that tells us that we cannot physically distinguish (i.e. by measurements), whether we need to leave the folium is clearly physically relevant.

• atyy and Auto-Didact

A. Neumaier

No, it's not known that QED isn't a Wightman theory.
It certainly does not describe the charged sectors. It cannot, since these are in different superselection sectors. It can only give local approximations to the states there (in a finite region), but these are practically useless. In bounded regions, all QFTs are III_1 algebras, and one loses their physical differences. Thus the description in unbounded regions is essential. This also applies to the second argument of your post.

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• RockyMarciano

rubi

In bounded regions, all QFTs are III_1 algebras, and one loses their physical differences.
This doesn't follow. Of course, there is only one type $III_1$ factor, but it is of crucial importance for physical theories to identify how the quantum fields are realized as operators in this algebra. This is what is difficult about constructive QFT.

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A. Neumaier

it is of crucial importance for physical theories to identify how the quantum fields are realized as operators in this algebra. This is what is difficult about constructive QFT.
Having this realization in one of these algebras is not enough for constructive QFT. One needs to have it in all, in a compatible way, which is equivalent to have it on unbounded domains.

And having a construction of the vacuum sector of QED in Wightman's sense would not yet be a construction of QED unless one also has constructed the charged sectors that are not described by the Wightman axioms. While they are probably determined by the vacuum sector they are not given by it. One needs a description of the 1-electron states (renormalized, dressed by their electromagnetic field), and these don't exist in the vacuum sector!

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• RockyMarciano

rubi

Having this realization in one of these algebras is not enough for constructive QFT. One needs to have it in all, in a compatible way, which is equivalent to have it on unbounded domains.
Nevertheless, you can't specify a QFT by telling me the von Neumann classification of its observable algebra. (And one can even argue that one can measure observables only in bounded regions anyway, so the algebras of interest form a type $III_1$ factor.) A QFT consists of a concrete observable algebra, a concret state on it and a concrete representation of the symmetries as *-automorphisms. Specifying concretel examples is what is difficult.

And having a construction of the vacuum sector of QED in Wightman's sense would not yet be a construction of QED unless one also has constructed the charged sectors that are not described by the Wightman axioms. While they are probably determined by the vacuum sector they are not given by it. One needs a description of the 1-electron states (renormalized, dressed by their electromagnetic field), and these don't exist in the vacuum sector!
Well, Fell's theorem tells us that you cannot physically distinguish any algebraic state from a density matrix in the folium of a/the vacuum state. Hence, it is enough to construct a Wightman QFT and specify states in terms of density matrices. Of course, however, nobody stops you from working in the algebraic framework if you want to.

A. Neumaier

A QFT consists of a concrete observable algebra, a concrete state on it and a concrete representation of the symmetries as *-automorphisms. Specifying concrete examples is what is difficult.
Sure, but it must be specified on all bounded subsets, and hence everywhere, including the unbounded sets. And the approximations to the charged states guaranteed by Fell's theorem are awkward to work with; they are not needed for mathematical existence but for physical useability.

Bird on a wire

I am fascinated by this discussion and a somewhat out my depth. Nevertheless, a question: Can emergence as defined by Anderson help ? The quantum states collapse. A single world is what we have, even allowing multiple sensory pictures or for multiple human perceptions of reality.

Demystifier

2018 Award
Can emergence as defined by Anderson help ? The quantum states collapse.
Anderson-like emergence may help, provided that you know what you are talking about. • bhobba

Bird on a wire

Thanks for the reply -I was hoping for something less condescending. There is a lot of critique by better physicists than us that emergence is developing from being the preserve of biologists and philosophers to something that is necessary to explain basic physical phenomena. If, as I do, you need to explain quantum effects to high school students, then the Copenhagen agreement is pretty weak and the standard model of matter sounds like black magic science. Anderson may not be an easy read for someone with a biochemical background, but McGlaughlin explains it well; by always looking for answers by going smaller and smaller, physics is losing the plot.

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"Stephen Weinberg on Understanding Quantum Mechanics"

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