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.Auto-Didact said:
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 said:
Could you please cite the context where I made this claim?RockyMarciano said:
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 said:
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.A. Neumaier said:
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.rubi said:
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).RockyMarciano said:
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.A. Neumaier said:
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.A. Neumaier said:
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.
But it is well-known that the Wightman axioms describe only the vacuum sector and fail to describe QED.rubi said:
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.rubi said:
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.A. Neumaier said:
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.A. Neumaier said:
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.rubi said:
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.A. Neumaier said:
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.rubi said:
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.A. Neumaier said:
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.
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.rubi said:
Anderson-like emergence may help, provided that you know what you are talking about.Bird on a wire said:
Bird on a wire said:
cosmik debris said:
PeterDonis said:
cosmik debris said:
PeterDonis said:
This is the 20th century string theory, where string really means string. But philosophy of 21th century string theory is different. For instance, according to AdS/CFT duality, string theory (on a certain background) is supposed to be equivalent to a field theory (on the boundary of that background).PeterDonis said:
Demystifier said:
Yes, to both questions. But a step backwards is not necessarily bad. Sometimes you need to make a step backwards to continue the walk in the right direction.PeterDonis said:
Demystifier said:
Demystifier said:
martinbn said:Шаг вперёд, два шага назад.
Are you a Russian? I thought no Russian likes Bourbaki style.martinbn said:Шаг вперёд, два шага назад.