Graduate Stephen Weinberg on Understanding Quantum Mechanics

[A. Neumaier]

Yngvason's point about Type III systems without pure states

In the light of the example of the previous post, things for algebras of type III_1 are similar in spirit but technically more complex and algebraically more varied. Now there are infinitely many unitarily inequivalent irreducible representations on Hilbert spaces (corresponding to the different superselection sectors of the theory). But in each such representation, the algebra of bounded observables is vanishingly small compared to the algebra of all bounded operators.

Thus what breaks down is the simple equation observable = Hermitian linear operator. Once this equation is broken, the question whether a state is pure becomes dependent on the precise specification of which operators are observables. In gauge theories the situation is further complicated by the fact that the observable algebra has a nontrivial center consisting of charges that in each irreducible representation are represented trivially. Thus a single irreducible representation on a single Hilbert space (corresponding to a single superselection sector) is no longer sufficient to characterize the complete algebra of observables.

 

[rubi]

Something was still fishy about my post. I think I figured it out now:

So let's look at type $III_1$ factors $\mathfrak A$ now. Can there be irreducible vector representations? I think what is going on is the following: At first, the fact that every state on $\mathfrak A$ must be mixed seems to invalidate this. However, what it really means is the following: Let's take the GNS representation $(\mathcal H_\omega,\pi_\omega,\Omega_\omega)$ induced by a state $\omega$ on $\mathfrak A$. Clearly, it must be reducible due to the theorem I stated in the beginning. However, that means that there is an invariant subspace of $\mathcal H_0\subseteq \mathcal H_\omega$ and I can take it to be minimal, i.e. $(\mathcal H_0,\pi_\omega\rvert_{\mathcal H_0})$ is irreducible and I can take any vector $\Psi\in\mathcal H_0$. Then I can define the state $\xi(A)=\left<\Psi,\pi_\omega(A)\Psi\right>_{\mathcal H_0}$ and it clearly defines a vector state in an irreducible representation on $\mathfrak A$. How can this be? The answer is that the data $(\mathcal H_0, \pi_\omega\rvert_{\mathcal H_0},\Psi)$ does not arise as the GNS data of some state on $\mathfrak A$. If I apply the GNS construction to the state $\xi$, I will end up with a reducible representation, because Yngvason tells us that $\xi$ must be a mixed state and our theorem tells us that mixed states produce reducible GNS representations.

The problem is that $\mathcal H_0$ is supposed to be irreducible and hence also cyclic. Thus, by some general theorems about operator algebras, it should be unitarily equivalent to the GNS representation of $\xi$, which means that the GNS representation of $\xi$ would be irreducible and thus $\xi$ would be pure. The solution is: $\mathcal H_\omega$ is reducible, but it doesn't have a minimal invariant subspace. Every invariant subspace of $\mathcal H_\omega$ is again reducible. Type $III_1$ algebras just don't have irreducible representations.

Thus what breaks down is the simple equation observable = Hermitian linear operator.

This was never a postulate. QM requires observables to be self-adjoint operators, but it doesn't require every self-adjoint operator to be an observable. I don't think any quantum theorist believes that the algebra of observables must encompass all bounded operators. What you explained in your post is that by restricting the algebra of observables to a subalgebra, a pure state can become mixed. I think stevendaryl knows this already. However, the issue with type $III_1$ algebras is even more subtle, because in that case, one can't single out the physical states solely by considering the algebra of observables and asking for irreducibility and continuity. Instead, one needs to take into account the dynamics of the theory.

 

[A. Neumaier]

but it doesn't require every self-adjoint operator to be an observable.

I agree that one has to give up the assumption that every bounded self-adjoint operator is an observable. But this has serious consequences for the foundations! Indeed, a test for a pure state is in terms of observables an observation of the orthogonal projector to the subspace spanned by the state. If this is not an observable then it is in principle impossible to make this test. But then Born's rule hangs in the air, and the whole foundations that start with it and derive everything else from it break down completely!

 

[A. Neumaier]

Type III_1 algebras just don't have irreducible representations.

Yes. insisting on irreducibility is a restriction of the scope of QM. It also excludes doing quantum mechanics on phase space - which gives a highly reducible but also highly useful view of quantum mechanics.

 

[rubi]

I agree that one has to give up the assumption that every bounded self-adjoint operator is an observable.

Well, we don't need to give it up, because we have never assumed it in the first place. It just accidently happens to be the case in some situations.

But this has serious consequences for the foundations! Indeed, a test for a pure state is in terms of observables an observation of the orthogonal projector to the subspace spanned by the state. If this is not an observable then it is in principle impossible to make this test. But then Born's rule hangs in the air, and the whole foundations that start with it and derive everything else from it break down completely!

What you're saying is that not every projector in a representation of a type $III_1$ factor corresponds to a physical proposition. That's of course true. However, all physical propositions still have associated projectors and their probabilities can still be calculated by the Born rule. The ordinary quantum formalism needs no modification in order to work with type $III_1$ factors.

Maybe the interpretational consequences that you want to point out are that density matrices aren't a different type of statistical mixture than vector states, contrary to what many people believe. But I think you don't need type $III_1$ factors in order to make that point. The algebraic formalisms shows that the particular realization of a state on some Hilbert space doesn't have any physical content, because there is no way to detect the Hilbert space. The lack of irreducible representations just tells us that there is also no mathematically preferred realization.

 

[A. Neumaier]

What you're saying is that not every projector in a representation of a type $III_1$ factor corresponds to a physical proposition. That's of course true. However, all physical propositions still have associated projectors and their probabilities can still be calculated by the Born rule.

My main point here was that testing for being in a pure state is impossible, since these are no longer physical propositions. So one cannot decide whether a system is in a pure state. So assuming it is a metaphysical act. One can dispense with it without any loss of reality content.

Maybe the interpretational consequences that you want to point out are that density matrices aren't a different type of statistical mixture than vector states, contrary to what many people believe. But I think you don't need type $III_1$ factors in order to make that point.

Well, I had argued for the objectivity of mixed states as indivisible things long before I knew these facts about III_1. The latter only make it unavoidable. But the real reason why working with densities (or corresponding general states in AQFT) is much preferable is that it makes the closeness to classical reasoning much more conspicuous. On the level of densities, the quantum classical correspondence is very direct in essentially every respect, strongly facilitating understanding. My thermal interpretation is the result of this.

 

[rubi]

My main point here was that testing for being in a pure state is impossible, since these are no longer physical propositions. So one cannot decide whether a system is in a pure state. So assuming it is a metaphysical act. One can dispense with it without any loss of reality content.

Well, after fixing a representation $(\mathcal H,\pi)$, a proposition $P$ will still be represented as a projector $\pi(P)$, because the representation properties imply $\pi(P)^2=\pi(P)=\pi(P)^*$. Hence, it will still project onto a subspace of the Hilbert space consisting of those vector states $\psi$, for which $\pi(P)\psi=\psi$. So we could still say that $\omega(P)$ computes the probability for measuring the system to be in the subspace $\pi(P)\mathcal H$ of vector states (if we really want to use this bad Copenhagen terminology). It will just be the case that no such $\psi$ defines a pure algebraic state. So in order to be rigorous, we just have to replace every occurence of the word "pure" by "vector" or not mention the those words in the first place. I don't think that this changes the interpretation of the Born rule.

 

[A. Neumaier]

, we just have to replace every occurence of the word "pure" by "vector" or not mention the those words in the first place.

That's not quite sufficient since the projector to the vector state is not in the observable algebra!

 

[rubi]

That's not quite sufficient since the projector to the vector state is not in the observable algebra!

Well, observables in the von Neumann algebra are encoded as one-parameter groups of normal elements $W(s)$, which are supposed to be interpreted as $e^{i s X}$ for some self-adjoint (possibly unbounded) operator $X$. If we have fixed a (strongly continuous) representation $(\mathcal H,\pi)$, we can define $\hat X = -i \left.\frac{\mathrm d}{\mathrm d s}\right|_{s=0} \pi(W(s))$ and compute its projectors $\hat P_B = \chi_B(\hat X)$ for Borel sets $B$. This is always how we are supposed to obtain physical observables in the $C^*$ algebraic setting and it works also for type $III_1$ algebras.

 

[A. Neumaier]

compute its projectors

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).

 

[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.

 

[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.

 

[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.

 

[RockyMarciano]

Could you please cite the context where I made this claim?

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]

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.

 

[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.

 

[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.

 

[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.

 

[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.

 

[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.

 

[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!