Is the Quantum/Classical Boundary the most important question in Physics?

[DarMM]

Do you believe that there is a measurement problem in quantum mechanics?

People mean many things by that phrase, some split it into two or three separate problems that they then call "small" or "big" and sometimes different groups can mean the same thing but one group views it as a problem and the other views it simply as the way things are, i.e. a property of the theory.

So it depends on what you mean.

 

[A. Neumaier]

What do you mean?

Superselection rules are properties of the observable algebra not states. In fact isn't that the point, that there is more to the physics than just the looking at the states alone can tell you due to restrictions on the observable algebra. If you see Streater's book he calls this a superselection rule.

I also don't understand what you mean by "limitations...affecting one particular state".

Statements about ''the actual limit of material in the universe'' are properties of the state, not the observable algebra. The empty universe and our universe are different states with vastly different amount of material.

On the level of the observable algebra, superselection rules are related to the center of the algebra, and the center does not seem to have any relation to decoherence, as far as I know. That's why I queried you for more information.

The environment induced superselection rules are heavily state-dependent and strictly speaking do not deserve their name.

 

[romsofia]

The "cut" doesn't even need to be brought up when talking about Copenhagen, because Copenhagen doesn't deal with it. Yes, it's mystical that apparently we could make the "cut" appear anywhere in the measurement process, and we get the same results. That's because no where in the formalism does it tell us where to break off, where does the measurement *actually* occur. Apparently, at some point, a break from this quantum system occurs, and a classical outcome becomes reality. So, Copenhagen isn't the interpretation to look at if you care about that. Or in the words of Robert Georch...

But, in any case, no difficulty arises from the failure of the Copenhagen interpretation, in the first experiment, to specify whether the geiger counter was “classical” or “quantum”. The claim of the interpretation is just that:
if you specify to it with sufficient precision what you will do in making a measurement, it will tell you the probability distribution for the results of the measurement.

It does precisely this in every case. It does not have to deal with what “really happens”. It is, in some sense, “consistent”

 

[DarMM]

The environment induced superselection rules are heavily state-dependent and strictly speaking do not deserve their name.

Superselection rules are often defined by the statement that for two sets of states $S_A$ and $S_B$ say, then there is no element of the observable algebra $\mathcal{A}$ that connects these two sets. That's what is going on here. Why is this not a superselection rule then?

 

[A. Neumaier]

Superselection rules are often defined by the statement that for two sets of states $S_A$ and $S_B$ say, then there is no element of the observable algebra $\mathcal{A}$ that connects these two sets. That's what is going on here. Why is this not a superselection rule then?

Because decoherence does not prove this for environment induced superselection rules. At least I haven't seen such a proof, did you?

 

[DarMM]

Because decoherence does not prove this for environment induced superselection rules. At least I haven't seen such a proof, did you?

It shows that the required self-adjoint operator involves systems of $\mathcal{O}\left(10^{10^{18}}\right)$ particles that would have to violate relativity. These don't exist clearly, so this self-adjoint operator isn't an element of the observable algebra.

 

[A. Neumaier]

It shows that the required self-adjoint operator involves systems of $\mathcal{O}\left(10^{10^{18}}\right)$ particles. These don't exist clearly, so this self-adjoint operator isn't an element of the observable algebra.

These don't exist only in a particular state - namely the Heisenberg state specifying our universe. But the observable algebra allows many states in which these exist!

 

[DarMM]

These don't exist only in a particular state - namely the Heisenberg state specifying our universe. But the observable algebra allows many states in which these exist!

It does? The states also require a violation of relativity. How could the observable algebra allow them?

 

[A. Neumaier]

It does?

Yes. A free QFT allows states with arbitrarily many particles.

The states also require a violation of relativity.

This is a different statement, which I do not believe since decoherence is usually discussed in a strictly nonrelativistic framework.

Where precisely is this proved? Page numbers please.

 

[vanhees71]

But in every application there is a cut, so one could just as well say that everything requires a classical treatment. Basically, there is a classical-quantum cut, the topic of this thread.

Where in the description of the Stern-Gerlach experiment and its outcome is the necessity for a cut? You can describe it pretty easily completely by solving the Schrödinger equation for a particle in an inhomogeneous magnetic field of the appropriate kind.

 

[A. Neumaier]

Where in the description of the Stern-Gerlach experiment and its outcome is the necessity for a cut? You can describe it pretty easily completely by solving the Schrödinger equation for a particle in an inhomogeneous magnetic field of the appropriate kind.

The magnetic field and the detector are described in classical terms.

 

[DarMM]

Just to be clear, this is definition of superselection rules given by Wightman in "Superselection Rules: Old and New":

We shall say that a superselection rule operates between subspaces if there are neither spontaneous transitions between their state vectors (i.e. if a selection rule operates between them) and if, in addition to this, there are no measurable quantities with finite matrix elements between their state sectors

In the same paper Wightman also refers to "Superselection rules induced by the environment". So it seems to be considered valid by both Streater and Wightman that there can be superselection rules induced by the environment. Can you perhaps say where they are wrong with this?

 

[DarMM]

Yes. A free QFT allows states with arbitrarily many particles

I don't understand how this is relevant.

This is a different statement, which I do not believe since decoherence is usually discussed in a strictly nonrelativistic framework.

Where precisely is this proved? Page numbers please.

Omnes discusses it on p.269 of his book. He doesn't give it in full detail, but it is easy enough to spell out.

An object of $10^{10^{18}}$ particles would need to be at least several quadrillion light years to not collapse in on itself. At this size it cannot operate fast enough (due to the speed of light) to have full reversible control of the first device, whose thermal fluctuations and so on operate on much smaller time scales.

 

[Morbert]

The magnetic field and the detector are described in classical terms.

But does this imply a classical treatment is required? Omnes states that empirical data are 'best expressed in classical terms' but he still associates them with a quantum mechanical description of the detector.

 

[kith]

However sometimes a state of the form:

$$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|a\rangle + |b\rangle\right)$$​

simply represents classical "ignorance" based probability. Such as when |a⟩|a⟩ and |b⟩|b⟩ belong to different superselection sectors.

If $|a \rangle$ and $|b \rangle$ belong to different superselection sectors, doesn't your definition in post #82 imply that $|\psi \rangle$ can't be prepared in the first place?

 

[DarMM]

If $|a \rangle$ and $|b \rangle$ belong to different superselection sectors, doesn't your definition in post #84 imply that $|\psi \rangle$ can't be prepared in the first place?

Mixed states can be prepared. Post #84 isn't mine, obviously this is just a minor typo, which one of my posts are you referring to?

 

[kith]

Mixed states can be prepared.

Yes but then why did you write down a ket?

I meant post #82 (Thanks, I edited my last post) where you gave the definition of superselection rules.

 

[DarMM]

Yes but then why did you write down a ket?

It's still a vector state, i.e. it is a sum in the Hilbert space of two pure states. It's just that sometimes kets are actually mixed states. That's the "interesting" part about superselection rules.

 

[vanhees71]

Usually a pure state is defined as being represented by a projector, i.e., obeying $\hat{\rho}^2=\hat{\rho}$, i.e., there's a normalized vector, defined up to a phase factor, with $\hat{\rho}=|\psi \rangle \langle \psi|$. There's a one-to-one mapping between Hilbert-space vectors modulo a phase (rays) and the pure states.

In which sense can kets represent mixed states and what has it to do with superselection rules? Don't superselection rules simply say that certain superpositions cannot describe perparable (pure) states? E.g., usually one argues that one cannot have the superposition of states with half-integer and integer spin, because rotations by angles $2 \pi$ should be a symmetry?

 

[kith]

Where in the description of the Stern-Gerlach experiment and its outcome is the necessity for a cut? You can describe it pretty easily completely by solving the Schrödinger equation for a particle in an inhomogeneous magnetic field of the appropriate kind.

The cut is between what you as the user of quantum mechanics consider to be the quantum system and everything else (including the measurement apparatus and yourself).

In a simple description of the Stern-Gerlach experiment, you cut between the spin degree of freedom and everything else. So your quantum mechanical system is a simple two-level system.

In the more sophisticated version you refer to above, you move the cut behind the magnets. So your quantum mechanical system consists of a spin-1/2 particle and the magnet$^1$. Note that the measurement result gets recorded only when part the quantum system interacts with the screen. This is why in this description, the screen alone constitutes the external measurement apparatus which isn't included in the quantum description.

Can the cut be shifted further? In principle, sure. If you start your experiment, isolate and leave the lab, you could take the spin-1/2 particle, the SG apparatus and the whole lab to be the quantum system which evolves unitarily until you break the isolation by opening the lab door.

Is there a limit how far you can shift the cut even in principle? Yes. If you are doing experiments, you can't include yourself in the description of these experiments.

______
$^1$ Actually, a semiclassical approximation is used, so the magnet isn't treated completely quantum mechanically.

 

[DarMM]

In which sense can kets represent mixed states and what has it to do with superselection rules?

I'd have to go into the representation theory of the observable algebra. Basically in more advanced contexts pure states and kets aren't the same thing, superselection being one aspect of this.
However I might cover this in a separate thread or Insight.

 

[kith]

It's still a vector state, i.e. it is a sum in the Hilbert space of two pure states.

Yes, mathematically, this state may be possible. But if we consider what states can be actually prepared, this state seems unphysical to me. If we prepare an arbitrary initial state by measuring a complete set of commuting observables of our system, there neither exists a Hamiltonian which leads to your state by unitary time evolution nor do open system dynamics lead to it.

 

[DarMM]

Yes, mathematically, this state may be possible. But if we consider what states can be actually prepared, this state seems unphysical to me. If we prepare an arbitrary initial state by measuring a complete set of commuting observables of our system, there neither exists a Hamiltonian which leads to your state by unitary time evolution nor do open system dynamics lead to it.

Do you mean evolution from an initial pure state under unitary dynamics does not lead to this state? Of course pure states evolve into pure states under unitary dynamics and this state is not pure.

That's the point of superselection rules, that not every sum of two pure states is another pure state.

All of this is immediate from the claim that the state is mixed despite it being a ket. What aspect are disagreeing with or pointing out?

 

[A. Neumaier]

Just to be clear, this is definition of superselection rules given by Wightman in "Superselection Rules: Old and New":

In the same paper Wightman also refers to "Superselection rules induced by the environment". So it seems to be considered valid by both Streater and Wightman that there can be superselection rules induced by the environment. Can you perhaps say where they are wrong with this?

I need to reread the paper to see what they claim with which justification.

It's still a vector state, i.e. it is a sum in the Hilbert space of two pure states. It's just that sometimes kets are actually mixed states. That's the "interesting" part about superselection rules.

Can you explain how a vector state is a mixed state? I understand how a mixed state can be pure in a bigger Hilbert space constructed by the GNS construction, but what you say is the opposite...

I'd have to go into the representation theory of the observable algebra. Basically in more advanced contexts pure states and kets aren't the same thing, superselection being one aspect of this.
However I might cover this in a separate thread or Insight.

I am looking forward to this. For now, can you just give a reference?

 

[DarMM]

Can you explain how a vector state is a mixed state? I understand how a mixed state can be pure in a bigger Hilbert space constructed by the GNS construction, but what you say is the opposite...
...
For now, can you just give a reference?

I think you may have terminology confused. A mixed state is any state which, as an algebraic state, is the sum of another two states, i.e. for all observables:
$$\omega\left(A\right) = c_{1}\omega_{1}\left(A\right) + c_{2}\omega_{2}\left(A\right)$$
A pure being one for which this is not true.
This is equivalent to having vanishing (pure) or nonvanishing (mixed) Von Neumann Entropy.

The GNS construction then allows one to express any state, pure or mixed, as a ket/vector state.
See Robert M. Wald "Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics" p.83:

Note that the GNS construction always expresses any state (pure or mixed) as a vector state

Thus a vector state can be pure or mixed. However a pure state (entropy is zero) is never mixed (entropy non-zero).

 

[vanhees71]

In the SGE you cannot locate the cut between the spin of the atom and everything else. The measurement of the spin component is based on a full entanglement between the atom's spin and its position by its dynamical evolution in presence of the magnetic field. Of course, here we use the semiclassical approximation in treating the magnetic field as classical. I'm not sure, whether a full QED description is feasible.

Even in the usual S-matrix formalism you'd describe the magnetic field as classical background field (it's an interesting question, whether there's a fully relativistic description of the SGE, which I'm not so sure about since the properties of spin are more complicated in the relativistic theory than in the non-relativistic one to begin with).

The other thing with the preparability is also easy: If there's by some principle (like a SSR) no Hamiltonian exists that can prepare the state, then this state is indeed simply unphysical, because it cannot be observed. Isn't this the deeper reason for an SSR (like the spin or charge superselection rule) to begin with?

Another example that comes to my mind are neutrinos. You can't prepare mass eigenstates because there's no Hamiltonian available to do this. Consequently there are no neutrinos that can be interpreted as free particles (asymptotic free single-particle Fock states).

 

[kith]

All of this is immediate from the claim that the state is mixed despite it being a ket. What aspect are disagreeing with or pointing out?

I agree that your ket formally corresponds to a completely mixed state and that we can use it to calculate the correct probabilities.

In order to consider it physical, I'd like to see a description where this state ket emerges from a well-prepared pure state (which necessarily belongs to only one of the superselection sectors) by a time evolution law. I don't see how this could be possible except for the ad hoc way where the density matrix is used in the calculation and only certain special mixed state density matrices are identified with kets by hand later on.

 

[DarMM]

In order to consider it physical, I'd like to see this state emerge from a well-prepared pure state (which necessarily belongs to only one of the superselection sectors) by a time evolution law

Why are you requiring this? All mixed states don't result from unitary time evolution of a pure state.

 

[kith]

Why are you requiring this? All mixed states don't result from unitary time evolution of a pure state.

I don't require unitary time evolution.

/edit: I slightly edited the wording of my last post.

 

[A. Neumaier]

I think you may have terminology confused. A mixed state is any state which, as an algebraic state, is the sum of another two states, i.e. for all observables:
$$\omega\left(A\right) = c_{1}\omega_{1}\left(A\right) + c_{2}\omega_{2}\left(A\right)$$
A pure being one for which this is not true.
This is equivalent to having vanishing (pure) or nonvanishing (mixed) Von Neumann Entropy.

The GNS construction then allows one to express any state, pure or mixed, as a ket/vector state.
See Robert M. Wald "Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics" p.83:Thus a vector state can be pure or mixed. However a pure state (entropy is zero) is never mixed (entropy non-zero).

I know all this. But you were referring to the addition of kets, not of states (linear functionals). A sum of kets is always a pure state (in the Hilbert space of the ket). My question was how it is possible that you refer to it as a mixed state. Or was this only a slip of the pen?

 

[DarMM]

I know all this. But you were referring to the addition of kets, not of states (linear functionals). A sum of kets is always a pure state (in the Hilbert space of the ket). My question was how it is possible that you refer to it as a mixed state. Or was this only a slip of the pen?

First of all you said this:

I understand how a mixed state can be pure in a bigger Hilbert space constructed by the GNS construction

That's not true right? A mixed state may be a vector state in the GNS Hilbert space, but not a pure state.

Secondly a sum of kets is not a pure state precisely in the case where both belong to different superselection sectors. Unless you are have a certain precise definition of Hilbert space in mind.

 

[DarMM]

I don't require unitary time evolution.

/edit: I slightly edited the wording of my last post.

Well a very dumb example, but it could be fleshed out with a more realistic one, but imagine a "trap" of some kind that can only hold a single particle. Either spin-1/2 or spin-1. It's in contact with a reservoir of spin-1/2 and spin-1 particles with which it can exchange particles. The trap starts off with a single spin-1/2 particle. Its state after a finite time would be described by such a state.

 

[DarMM]

The cut is between what you as the user of quantum mechanics consider to be the quantum system and everything else (including the measurement apparatus and yourself).

Just to note the "cut" as such is present in any non-Kolomogorvian probability theory not just quantum mechanics. For instance PR boxes and the "Nearly quantum theory" of Barnum et al. This is because in all such theories you need some system external to the modeled one to select out the Boolean algebra of events.

So in a sense if one wishes to remove the cut, that is have a theory without the cut, you need to somehow restore classical probability.

 

[A. Neumaier]

A mixed state may be a vector state in the GNS Hilbert space, but not a pure state.

A vector state in the GNS Hilbert space is a pure state in the Hilbert space sense, but not in the $C^*$ algebra sense.

Secondly a sum of kets is not a pure state precisely in the case where both belong to different superselection sectors. Unless you are have a certain precise definition of Hilbert space in mind.

It is a pure state in the Hilbert space defined by the direct sum of the superselection sectors. What I don't understand is in which sense it can be a mixed state.

 

[kith]

Well a very dumb example, but it could be fleshed out with a more realistic one, but imagine a "trap" of some kind that can only hold a single particle. Either spin-1/2 or spin-1. It's in contact with a reservoir of spin-1/2 and spin-1 particles with which it can exchange particles. The trap starts off with a single spin-1/2 particle. Its state after a finite time would be described by such a state.

I like the example. It will take me probably a few days to answer, though.