Stephen Weinberg on Understanding Quantum Mechanics

[DrDu]

I notice that in the paper by Yngvason , he concludes with a warning that this discussion does not resolve any of the foundational issues with QM:

Yngvarson sais :"On the other hand, the framework of LQP does not per se resolve all “riddles” of quantum physics. " The shift from "all" to "any" is yours.

 

[stevendaryl]

Yngvarson sais :"On the other hand, the framework of LQP does not per se resolve all “riddles” of quantum physics. " The shift from "all" to "any" is yours.

My mistake.

 

[A. Neumaier]

Yngvason is defining a "pure" state $\omega$ as one that cannot be written in the form $\omega = \frac{1}{2} \omega_1 + \frac{1}{2} \omega_2$ with $\omega_1 \neq \omega_2$. I don't understand the motivation for this definition.

It says that the state cannot be a mixture of two different states with equal weight. One can easily see that this is true for states $\omega(A)=\psi^*A\psi$ given by a wave vector $\psi$. One can see with a little more work that it is false for any state $\omega(A)=$trace$(\rho A)$ where $\rho$ has rank greater than 1. Thus the two definitions are equivalent for states on algebras of bounded operators, as they are used in Bell-type experiments.

 

[A. Neumaier]

• There are types of systems for which there are no pure states.
• (Therefore) density matrices cannot be interpreted as classical probabilities for being in this or that pure state.

I don't understand the definition or the conclusion, so I need to think about it a little more.

Point 2 follows directly from point 1 since there are no pure states, so talking about them is meaningless.

Even in the case of quantum mechanics of a 2-level system, where operators are matrices, density matrices cannot be interpreted as classical probabilities for being in this or that pure state. Unpolarized light has as density matrix half the identity matrix. One has infinitely many essentially different decompositions of the kind you describe - which one gives the ''correct'' interpretation? None, since the state is symmetric under the helicity $SU(2)$ while none of the decompositions is. Picking out one of them is well-defined only at the moment of measurement. Thus one can only get probabilities for passing an experimental test measuring this or that polarization. Thus your ontology is defective already at the level of single photons in a fixed beam.

 

[stevendaryl]

Point 2 follows directly from point 1 since there are no pure states, so talking about them is meaningless.

Even in the case of quantum mechanics of a 2-level system, where operators are matrices, density matrices cannot be interpreted as classical probabilities for being in this or that pure state. Unpolarized light has as density matrix half the identity matrix. One has infinitely many essentially different decompositions of the kind you describe - which one gives the ''correct'' interpretation? None, since the state is symmetric under the helicity $SU(2)$ while none of the decompositions is. Picking out one of them is well-defined only at the moment of measurement. Thus one can only get probabilities for passing an experimental test measuring this or that polarization. Thus your ontology is defective already at the level of single photons in a fixed beam.

I certainly recognize that. That was actually going to be my next point.

In my opinion, every interpretation of QM amounts to shuffling the defect around, rather than addressing it.

 

[A. Neumaier]

Thus one can only get probabilities for passing an experimental test measuring this or that polarization.

In the notation of Yngvason, testable statements are represented by Hermitian elements $P$ with only eigenvalues 0 and 1, corresponding to the two possible results of disproving or confirming the statement in the experiment. This is equivalent to saying that $P^2=P=P^*$, the condition for an orthogonal projector. The probability for testing a statement $P$ is given by $p=\omega(P)$. This is completely independent of the notion of a pure state. von Neumann's classification is essentially a classification of the possible orthogonal projectors. Pure states exist only when the algebra contains projectors of rank 1. Then $P=\psi\psi^*$ where $\psi$ is a normalized state in the range, and if $\omega$ is also a pure state we get Born's rule.

 

[A. Neumaier]

every interpretation of QM amounts to shuffling the defect around, rather than addressing it.

If one acknowledges that specifying the state (in Yngvason's sense) is having specified the system then at least the defect that I mentioned in my post is completely absent, without introducing another defect anywhere else. Thus it is a superior ontology.

 

[stevendaryl]

If one acknowledges that specifying the state (in Yngvason's sense) is having specified the system then at least the defect that I mentioned in my post is completely absent, without introducing another defect anywhere else. Thus it is a superior ontology.

I suppose, but the issues that Yngvason said are unresolved by using density matrices are the most vexing aspects of the foundations of quantum mechanics, in my opinion.

 

[A. Neumaier]

the issues that Yngvason said are unresolved by using density matrices are the most vexing aspects of the foundations of quantum mechanics

Maybe, but they are more likely to be solved based on a physically correct basis (general states) rather than based on one (pure states) that is already known to be only of limited validity.

 

[rubi]

Let me clarify this issue.

An (algebraic) state on a $C^*$-algebra $\mathfrak A$ is a linear functional $\omega:\mathfrak A\rightarrow\mathbb C$ with $\omega(a^*a)\geq 0$ and $\omega(1)=1$. One example of a $C^*$-algebra is the set of bounded operators $\mathcal B(\mathcal H)$ on a Hilbert space $\mathcal H$ and an example of a state is the functional $\omega(A)=\left<\Omega,A\Omega\right>$, where $\Omega\in\mathcal H$ is any vector with $\lVert\Omega\rVert=1$. Another example would be $\omega(A)=\mathrm{Tr}(\rho A)$, where $\rho:\mathcal H\rightarrow\mathcal H$ is any positive trace-class operator with $\mathrm{Tr}(\rho)=1$. An algebraic state $\omega$ on $\mathfrak A$ is said to be mixed, if there exist $\omega_1,\omega_2 \neq \omega$ and $\lambda\in(0,1)$ such that $\omega=\lambda\omega_1+(1-\lambda)\omega_2$. Otherwise, it is said to be pure.

Theorem (Gelfand, Naimark, Segal): Given a $C^*$-algebra $\mathfrak A$ and a state $\omega$ on $\mathfrak A$, there exists a Hilbert space $\mathcal H$, a representation $\pi:\mathfrak A\rightarrow\mathcal B(\mathcal H)$ of $\mathfrak A$ and a vector $\Omega\in\mathcal H$, such that for all $A\in\mathfrak A$, we have
$$\omega(A)=\left<\Omega,\pi(A)\Omega\right> \text{.}$$

Now, what is the relevance to quantum theory?
1. The algebraic terminology of pure and mixed states matches the QM terminology in the following sense: If $\omega_1(A)=\left<\psi,A\psi\right>$ and $\omega_2(A)=\left<\phi,A\phi\right>$ are vector states on $\mathcal B(\mathcal H)$, then $\omega=p_1\omega_1+p_2\omega_2$ with probabilities $p_1$, $p_2$ is given by $\omega(A)=\mathrm{Tr}(\rho A)$ with $\rho=p_1\left|\psi\right>\left<\psi\right|+p_2\left|\phi\right>\left<\phi\right|$. In that sense, the algebraic terminology agrees with the standard terminology.
2. Every algebraic state $\omega$ can be realized as a vector state $\omega(A)=\left<\Omega,\pi(A)\Omega\right>$ by the GNS construction, even a state $\omega(A)=\mathrm{Tr}(\rho\pi(A))$ defined by a density matrix. The confusion between Arnold and stevendaryl is due to a disagreement about terminology. While Arnold refers to states as "pure", if they are pure algebraic states (which is what AQFT people do), stevendaryl refers to states are pure, if they are defined by vectors in a Hilbert space (which is common in QM). Of course, Arnold's terminology is more restrictive, because every state can be realized as a vector state in some Hilbert space by the GNS theorem and thus the distinction between vector states ("pure states" in stevendaryl's terminology) and non-vector states is completely unphysical. Nevertheless, in standard QM, it has practical relevance, because in that case we have the Stone-von-Neumann theorem, which (pretty much) singles out the standard Schrödinger representation and we usually like to represent our states in that particular representation. Of course, in the case of QFT, no such uniqueness result is available and hence, it is unreasonable to distinguish between vector states and non-vector states.

 

[Auto-Didact]

After carefully reading Yngvason's paper, one can conclude: its a cogent summary of mathematical objects in contemporary AQFT and gives possible distinctions w.r.t. similar objects in standard non-relativistic QM. It is however not concerned per se with interpretational issues, i.e. what Weinberg et al. are directly concerned with. However, Yngvason does state in a footnote on p.18

24 See [12] for important steps in this direction and [49] for a thorough analysis of foundational issues of QM

giving 2 references w.r.t interpretations, specifically:

[12] P. Blanchard, R. Olkiewicz: “Decoherence induced transition from quantum to classical
dynamics”, Rev. Math. Phys. 15, 217–244 (2003).

[49] J. Fröhlich, B. Schubnel: “Quantum Probability Theory and the Foundations of Quantum mechanics”, arXiv:1310.1484v1 [quant-ph].

The latter is a comprehensive review of the field, while the former a more specific environmentally induced decoherence proposal for open quantum systems. It seems that this environmental decoherence proposal is Yngvason's preferred interpretation given his AQFT understanding. In other words, taking density operators (or matrices) as more primary ontologic concepts implies embracing environmental decoherence, which is after all a purely FAPP pragmatic (as opposed to fundamental) philosophy as John Bell said.

 

[DrDu]

rubi, I think at least in ordinary QM vector representations of mixed states are reducible, while those of pure states are irreducible. Is this true for all C* algebras?

 

[rubi]

rubi, I think at least in ordinary QM vector representations of mixed states are reducible, while those of pure states are irreducible. Is this true for all C* algebras?

This is a little bit subtle. What is true is the following: The vector representation $(\mathcal H_\omega,\pi_\omega,\Omega_\omega)$ given by the GNS construction of a state $\omega$ is irreducible if and only if $\omega$ is pure. However, that doesn't mean that all vector representations are irreducible. For example, we can take the irreducible GNS representation $(\mathcal H_\omega,\pi_\omega,\Omega_\omega)$ of a pure state $\omega$ and specify the following data: $\mathcal H=\mathcal H_\omega\oplus\mathcal H_\omega$, $\pi=\pi_\omega\oplus\pi_\omega$ and $\Omega=\frac{1}{\sqrt{2}}\Omega_\omega\oplus\Omega_\omega$. Define a state $\xi(A)=\left<\Omega,\pi(A),\Omega\right>_{\mathcal H}$. We then find $$\xi(A) = \frac{1}{2}\left<\Omega_\omega\oplus\Omega_\omega,\pi_\omega(A)\Omega_\omega\oplus\pi_\omega(A)\Omega_\omega\right>_{\mathcal H}=\frac{1}{2}\left<\Omega_\omega,\pi_\omega(A)\Omega_\omega\right>_{\mathcal H_\omega}+\frac{1}{2}\left<\Omega_\omega,\pi_\omega(A)\Omega_\omega\right>_{\mathcal H_\omega}=\frac{1}{2}\omega(A)+\frac{1}{2}\omega(A)=\omega(A) \text{.}$$ Thus $\xi=\omega$. However, $(\mathcal H,\pi)$ is clearly reducible, because $\mathcal H\oplus 0$ is an invariant subspace. So we have found a reducible vector representation of a pure state $\omega$. This is of course possible, because only the GNS representation of $\omega$ needs to be irreducible. Our representation $(\mathcal H,\pi,\Omega)$ is not the GNS representation of $\omega$. So even in QM, vector representations needn't be irreducible.

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.

(However, I still find it fishy. The states should form a convex set and as such, it should have extremal points, which should correspond to pure states. I need to think about this more.)

 

[A. Neumaier]

Every algebraic state can be realized as a vector state by the GNS construction, even a state defined by a density matrix.

But this is not the way stevendaryl uses the terms since for him, mixed states are composed of pure states, in a fixed Hilbert space, while the GNS construction produces a different Hilbert space for each (pure or mixed) state!

 

[A. Neumaier]

The states should form a convex set and as such, it should have extremal points

This does not follow. The plane is a convex set without extremal points.

 

[A. Neumaier]

it clearly defines a vector state in an irreducible representation

A vector state in an irreducible representation of the observable algebra $\cal A$ in a Hilbert space $\cal H$ can still be mixed. It is guaranteed to be pure only relative to the algebra of all bounded operators on $\cal H$. But this algebra is far bigger than the III_1 algebra $\cal A$, and contains lots of operators that have no interpretation as observables. This is the essential difference to the case of type I algebras.

 

[stevendaryl]

A request: If you feel you understand Yngvason's point about Type III systems without pure states, could you write up a post about it, aimed at people who understand the basics of Hilbert spaces, but not the technical points of $C^*$ algebras, and all that? Someone mentioned a sort-of accessible example involving an infinite array of spin-1/2 particles, or something like that.

 

[A. Neumaier]

example involving an infinite array of spin-1/2 particles

This example is given in equations (27) and (29) in Yngvason's paper. The GNS construction is fully described in Wikipedia. It produces a large Hilbert space with a huge algebra of bounded operators, most of which are (typically) not in the algebra of observables one starts with. In particular, this happens when one starts with a mixed state of the algebra of bounded operators of a small Hilbert space. (For example, one obtains from an $n$-dimensional physical Hilbert space in most cases an $n^2$-dimensional nonphysical Hilbert space.)

 

[rubi]

But this is not the way stevendaryl uses the terms since for him, mixed states are composed of pure states, in a fixed Hilbert space, while the GNS construction produces a different Hilbert space for each state!

Yes, stevendaryl uses a different terminology. There are even algebraic states that can't even be realized as density matrix states, if the Hilbert space is fixed (although the density matrix states are in some sense dense in the space of algebraic states). The concept of algebraic states is much better for theory building if the Hilbert space isn't known yet.

This does not follow. The plane is a convex set without extremal points.

Of course, you're right. I'm stupid. The state space may be non-compact.

The point is that a vector state in an irreducible representation of the observable algebra $\cal A$ in a Hilbert space $\cal H$ can still be mixed. It is guaranteed to be pure only relative to the algebra of all bounded operators on $\cal H$. But this algebra is far bigger than the III_1 algebra $\cal A$, and contains lots of operators that have no interpretation as observables. This is the essential difference to the case of type I algebras.

Yes, I acknowledged this in my post. Even an irreducible vector state needn't be a pure algebraic state. $\mathfrak A$ was a type $III_1$ algebra. But the fact that you're dealing with type $III_1$ algebras doesn't mean that you need to drop the standard QM formalism of vector states in irreducible representations. In fact, the pure/mixed decomposition isn't the most interesting one. People are more interested in theories with unique vacuum states and in order to get that, one should rather be looking at the ergodic decomposition with respect to the time evolution.

 

[A. Neumaier]

could you write up a post about it, aimed at people who understand the basics of Hilbert spaces, but not the technical points of C^* algebras, and all that?

The GNS construction is fully described in Wikipedia. It produces a large Hilbert space with a huge algebra of bounded operators, most of which are (typically) not in the algebra of observables one starts with.

Let me illustrate what happens in a C^*-algebra-free way (hence without GNS):

We consider an explicit representation of mixed states by a state vector in the simplest case of an $n$ level system. Here the Hilbert space $\cal H$ consists of all complex vectors of size $n$, and the associated observable algebra $\cal A$ consists of all $n\times n$ matrices. (This is the simplest example of a $C^*$-algebra, but we do not need that.) $\cal A$ has a unitary representation on another Hilbert space $\cal\hat H$, consisting of all $n\times n$ matrices with inner product $<\Phi|\Psi>:=$tr $\Phi^*\Psi$. The action of $A\in\cal A$ on $\Psi$ is simply given by multiplication $A\Psi$.

Every state of $\cal A$ is representable as a vector state on $\cal\hat H$. Indeed, the linearity of the state $\omega$ implies that we can write $\omega(A)=$tr$\rho A$ with some $\rho\in\cal A$, and the state properties then imply that $\rho$ is Hermitian positive semidefinite with trace 1. Thus it is a (typically mixed) state in the traditional sense. It has a Hermitian positive semidefinite square root $\hat\rho\in\cal\hat H.$ By construction, $\hat\rho\hat\rho^*=\hat\rho^2=\rho$. In the inner product of $\cal\hat H$, we have for any $A\in\cal A$ the relation $<\hat\rho|A|\hat\rho>=$tr $\hat\rho^* A\hat\rho=$tr$\hat\rho \hat\rho^*A=$tr $\rho A=\omega(A)$. Since this implies for $A=1$ that $\hat\rho$ is a vector of norm 1 in $\cal\hat H,$ we have represented the state $\omega$ as a vector state on $\cal\hat H$.

Thus $\omega$ is seemingly represented as a pure state on $\cal\hat H$. But of course it is still the same mixed state that it was initially! This seeming paradox finds its explanation in the fact that there is a subtle difference between the state $\omega$ (defined on $\cal A$) and the state $\hat\omega$ on $\cal\hat H$ defined on the algebra $\cal \hat A$ of linear operators $\hat A$ on $\cal\hat H$ by $\hat\omega(\hat A):=<\hat\rho|\hat A|\hat\rho>$. The precise mathematical relation between these two states is that the state $\omega$ is the restriction of $\hat\omega$ to the observable algebra $\cal A$. That the state $\omega$ can be written as a mixture of pure states applies only to observables from $\cal A$. On the other hand, $\cal \hat A$ contains many linear operators (e.g., the generator of the Lindblad equations considered in Weinberg's paper) that do not belong to $\cal A$, hence have no physical meaning as observables. The decomposition of $\omega$ as a mixture does not extend to all these other linear operators. Thus there is no conflict with $\hat\omega$ being pure.

 

[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.