# FeaturedA Stephen Weinberg on Understanding Quantum Mechanics

1. Jan 11, 2017

### rubi

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.

2. Jan 12, 2017

### A. Neumaier

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.

3. Jan 12, 2017

### RockyMarciano

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?

4. Jan 12, 2017

### Auto-Didact

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.

5. Jan 13, 2017

### A. Neumaier

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.

6. Jan 13, 2017

### A. Neumaier

7. Jan 13, 2017

### RockyMarciano

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?

8. Jan 13, 2017

### A. Neumaier

9. Jan 13, 2017

### rubi

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.

10. Jan 13, 2017

### RockyMarciano

11. Jan 13, 2017

### A. Neumaier

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.

12. Jan 13, 2017

### A. Neumaier

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.

13. Jan 13, 2017

### RockyMarciano

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.

14. Jan 13, 2017

### rubi

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 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.

15. Jan 13, 2017

### A. Neumaier

But it is well-known that the Wightman axioms describe only the vacuum sector and fail to describe QED.

16. Jan 13, 2017

### A. Neumaier

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.

17. Jan 13, 2017

### rubi

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.

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.

18. Jan 13, 2017

### A. Neumaier

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.

Last edited: Jan 13, 2017
19. Jan 13, 2017

### rubi

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.

20. Jan 13, 2017

### A. Neumaier

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!

Last edited: Jan 13, 2017