Mixed states vs pure states - physical POV

[DrDu]

For example how does it fit with Gleason? It proves, providing you have non contextuality, all states are positive operators of unit trace - assuming the strong principle of superposition any positive operator is a legit state. Mixed states are different, by their very definition, to pure states. The state 1/2 |a><a| + 1/2 |b><b| can never be put in the form |u><u| - its simply not possible by the axioms of linear algebra and what a positive operator of unit trace is. The decomposition is not unique - one can find other mixed states that observationaly are the same - but not a pure state.

Gleason? It says that every subset of one fixed Hilbert space can be projected onto by a density matrix. That's certainly true. However what I want to say is that you can embed this Hilbert space into a larger Hilbert space so that these representations become all vector representations.
Basically this is the content of the GNS theorem:
https://en.wikipedia.org/wiki/Gelfand–Naimark–Segal_construction

When I was first introduced into QFT, I had the feeling that somebody had pulled the carpet from under my feet. You hardly consider states any more, and even less density matrices. Nevertheless the distinction between pure and mixed states is still somewhere hidden and is important. Take for example the ground states of a superconductor. There are several ones with either the number of Cooper pairs fixed or the phase of the condensate fixed. Can you tell which one is pure and which one is mixed?
If not, have a look at
Haag, Rudolf. "The mathematical structure of the Bardeen-Cooper-Schrieffer model." Il Nuovo Cimento (1955-1965) 25.2 (1962): 287-299.

Therefore I think it would make much sense to explain the ideas behind the GNS theorem already in ordinary QM.
I tried to derive a representation of the mixed vector states a la GNS already in an earlier post,
https://www.physicsforums.com/threa...of-entangled-state.920907/page-2#post-5810047

 

[bhobba]

Gleason? It says that every subset of one fixed Hilbert space can be projected onto by a density matrix. That's certainly true.

It says more than that - but I may be getting at least a bit of your drift.

Because of that I would really like to see a discussion about this between you and people more knowledgeable than myself like Vanhees etc.

Just to outline my thinking - suppose you have 1/root2 Ia>|b> + 1/root2 Ib>|a> - its a pure state but if you just observe one system then it acts like a mixed state. Thats my current drift - but I want to see more knowledgeable peoples opinion.

Thanks
Bill

 

[kith]

I think it is operationally ok to define a statistical operator as the outcome of a measurement where the result is not observed. I am only concerned whether we can really say that the system is really in a definite eigenstate but we do not know which one. This may be a philosophical question, but I think it is the original question of the OP.

I'm wondering about this too. In principle, different preparations of the same mixed state are distinguishable if we consider the correlations of the system with the environment.

Let's say Xavier prepares the state \frac{1}{2}|\uparrow_x\rangle \langle \uparrow_x| + \frac{1}{2}|\downarrow_x\rangle \langle \downarrow_x| and Yvonne prepares the state \frac{1}{2}|\uparrow_y\rangle \langle \uparrow_y| + \frac{1}{2}|\downarrow_y\rangle \langle \downarrow_y| by orienting their preparation devices in the respective directions. These states cannot be distinguished by measurements which involve only the system. So on the system level, they are two ways of writing the same state. (Which is reflected by the fact that we can get from one to the other by a change of basis).

But do Bell et al. object if Xavier claims that his system really is either in state |\uparrow_x\rangle or |\downarrow_x\rangle and Yvonne claims that her system really is either in state |\uparrow_y\rangle or |\downarrow_y\rangle? They could justify their way of speaking by noting that if we keep track of the state of the preparation device, we'll find different correlations with the system state in both cases.

 

[kith]

When I was first introduced into QFT, I had the feeling that somebody had pulled the carpet from under my feet. You hardly consider states any more, and even less density matrices. Nevertheless the distinction between pure and mixed states is still somewhere hidden and is important.

That's my feeling still. I took some lectures on particle physics and QFT and learned a bit of it on my own a while ago, but I could never fully reconcile it with my intuitions based on the non-relativistic quantum dynamics of open systems.

Therefore I think it would make much sense to explain the ideas behind the GNS theorem already in ordinary QM.

Maybe you could write an Insight about this?

My current understanding of this matter is that every state of a system (be it pure or mixed) can be represented by a pure state in a larger Hilbert space. But the physical significance is unclear to me.

Going in the other direction is straighforward: if we have an entangled state of a large system, we get a mixed state for a subsystem by tracing out over the other degrees of freedom. But given a mixed state without context, enlarging the Hilbert space in order to represent it as a ket vector can be done in many ways and seems kind of arbitrary to me.

 

[stevendaryl]

In the discussion of pure versus mixed, there are a few different concepts that get mixed up (no pun intended):

1. Mixed states reflecting lack of information about a system.
2. Mixed states due to considering one subsystem out of a larger, entangled system.

It's kind of a strange fact that these two situations are described by the same mathematics.

What might be clarifying is to talk about a particular subsystem being mixed or pure, given a particular preparation procedure. There are important, because:

• A subsystem may be mixed (in sense 2 above), even though it is part of a larger, pure system.
• Different preparation procedures can be described by different density matrices, even though there may be no objective, physical difference in the system.

To give an example of the latter, suppose that we have some device that flips a coin, and then produces a spin-up electron if the result is heads, and spin-down if the result is tails. If the device hides the result of the coin flip, then you would describe the situation using the density matrix:

\rho = \frac{1}{2}|up\rangle\langle up| + \frac{1}{2}|down\rangle\langle down|

If you later find out that, in fact, the result was heads, then the density matrix would be changed to \rho =|up\rangle\langle up|. The physical situation with the electron wasn't changed by the knowledge, but the density matrix was.

I'm not exactly sure how to define "subsystem" here, though. It's natural enough to think of splitting the universe into the particle(s) of interest (the electron, maybe) and everything else. But that's actually not quite good enough, because

1. After all, electrons are indistinguishable, so it doesn't literally make any sense to separate the electron of interest from all other electrons.
2. Sometimes, the split isn't really along particle lines. For example, the spin of electron is independent of its location, so an electron might be in a mixed state for location, but a pure state for spin.

Maybe the correct way to talk about subsystems is to use "degrees of freedom".

So for a particular chosen set of degrees of freedom, and for a particular preparation procedure, we can say that:

• The result is a pure state if there is an observable that "acts on" those degrees of freedom that has a definite value (it's always the same, given the same preparation procedure).
• The result is a mixed state if all observables that act on those degrees of freedom have uncertain values (the value is not determined by the preparation procedure)

The state "spin-up in the x-direction", defined by |up_x\rangle = \frac{1}{\sqrt{2}} (|up_z\rangle + |down_z\rangle) is a pure state, because it produces a consistent value of +1 for the observable "spin in the x-direction", which is an observable that acts on the same degrees of freedom, the spin state, as spin in the z-direction.

 

[Lord Jestocost]

Quoting J. Allday (Allday, Jonathan. Quantum Reality, CRC Press):

"In his brilliant textbook on quantum mechanics, Paul Dirac identified the existence of quantum mixed states as the central puzzle of quantum theory. Dirac points out that a quantum mixed state is not some sort of average. It doesn't describe an existence that is a blend of the separate states, which a classical ‘mixed state’ would."

"The distinction between a mixed state and an eigenstate is not an absolute divide. States |U〉 and |D〉 are eigenstates as far as (UP, DOWN) measurements are concerned but mixed states for (LEFT, RIGHT) measurements. Similarly, states |R〉 and |L〉 are eigenstates for (LEFT, RIGHT) measurements but mixed states for (UP, DOWN) measurements."

Edit: I always wonder why this is discussed again and again.

 

[PeterDonis]

Quoting J. Allday (Allday, Jonathan. Quantum Reality, CRC Press):

This quote is probably not very helpful in this discussion, since it is using "mixed state" to mean "superposition".

 

[vanhees71]

I don't believe that Dirac made such utterly wrong statements. A pure state is a pure state and a mixed state is a mixed state. A state is represented by a positive semidefinite self-adjoint operator of trace one, the Statistical Operator of the system, $\hat{\rho}$. It represents a pure state if and only if it is a projection operator, i.e., if $\hat{\rho}^2=\hat{\rho}$.

An observable $A$ is represented by a self-adjoing operator $\hat{A}$ with orthonormal eigenstates $|a,\beta \rangle$, where $a$ is a (necessarily real) eigenvalue of $\hat{A}$ and $\beta$ is a set of variables labeling the orthonormal basis of the eigenspace of $\hat{A}$ with eigenvalue $a$. If a system is prepared in a pure or mixed state, represented by $\hat{\rho}$, it contains probabilistic information about the outcome of measurements, i.e., the probability for finding the eigenvalue $a$ (and an observable can take only eigenvalues) of $\hat{A}$ is given by
$$P(a|\hat{\rho})=\sum_{\beta} \langle a,\beta|\hat{\rho}|a,\beta \rangle.$$

 

[Lord Jestocost]

... If a system is prepared in a pure or mixed state...

What is your exact definition of "system"?

 

[stevendaryl]

Quoting J. Allday (Allday, Jonathan. Quantum Reality, CRC Press):

"In his brilliant textbook on quantum mechanics, Paul Dirac identified the existence of quantum mixed states as the central puzzle of quantum theory. Dirac points out that a quantum mixed state is not some sort of average. It doesn't describe an existence that is a blend of the separate states, which a classical ‘mixed state’ would."

I think that your quoted material sounds more like he's talking about superpositions, instead of mixed states. |U\rangle and |D\rangle are not mixed states as far as left-right measurements, but are superpositions.

 

[stevendaryl]

What is your exact definition of "system"?

I think that perhaps a better phrase might be "degrees of freedom". The universe has infinitely many degrees of freedom, but in a measurement, you're focusing on just some tiny number of them, such as the spin of a single particle.

 

[vanhees71]

What is your exact definition of "system"?

Anything you describe with quantum theory from a single elementary particle up to many-body systems of condensed-matter physics, i.e., nearly everything you can treat in theoretical physics.

 

[Lord Jestocost]

...If a system is prepared in a pure or mixed state, represented by ...

Now I see the point, my misunderstanding! I have checked my old German textbook (Blochinzew). The term „mixed state“ is synonymous to the German terms „gemischte Gesamtheit“ or „Gemenge“.

 

[cube137]

Energy eigenstates aren't the only possible choice of eigenstates. A pure state that is a superposition in one choice of basis states can be written as an eigenstate in another choice of basis states. (That's not true for mixed states.) For example, a particle in the spin right state is written as a linear combination of spin up and spin down states, but there isn't really a fundamental difference between spin right and spin up other than choice of coordinate system -- one is not more "mixed" than the other.

A mixed state will be mixed in any complete basis. Mixed state can arise due to incomplete knowledge or due to taking a subsystem of an entangled system. It is not "DEFINITELY" in one of the eigenstates, and a mixed state cannot be unambiguously represented as a statistical combination of pure states. For example, a 50/50 mixture of a spin up and spin down particle is the same as a 50/50 mixture of a spin left and spin right particle. Mathematically, they are the same state, and cannot be represented as a pure state in any basis.

but eigenstates can also be mixed state, is it not? For example.. consider a quantum state with the following terms: "Schrödinger cat exposed to radioactive source and dead" and "Schrödinger cat not exposed to radioactive source and alive". These are eigenstates. Now they are exposed to environmental decohererence, the terms "Schrödinger cat exposed to radioactive source and dead" and "Schrödinger cat not exposed to radioactive source and alive" become mixed states. so mixed states is one of the eigenstates chosen so this contradicts the statement above.. unless something wrong with my analysis?

 

[vanhees71]

A state is not a vector. So it cannot be an eigenvector. The quantum state is represented by a statistical operator, not a Hilbert-space vector. Since the pure states by definition are represented by projection operators as statistical operators there's always a normalized vector (determined up to a phase factor) such that $\hat{\rho}=|\psi \rangle \langle \psi|$. So an equivalent formulation you can find in many textbooks is that a pure state can be also represented by a (unit) ray in Hilbert space.

Sometimes you find also the inaccurate statement that pure states are represented by (normalized) Hilbert-space vectors. Such a book is not a priori bad, but it's inaccurate, and I'd automatically enhance my level of scepticism against it!

 

[SchroedingersLion]

Thanks for the further answers - even though I feel it drifted a bit too far sometimes :D

When I was talking about mixed states as a statisticial mixture:

2) Mixed states are statistical mixtures.
A mixture of |1> and |0> means that the particle is DEFINITELY in one of these two states, but I do only know the corresponding probability.

I clarified (or tried to) what I meant:

Ok, I remember that one has to be careful in interpreting the coefficients in the mixed states sum. In general, you can not say that the system is definitely in one of the eigenstates, BUT: If I have a source that produces known pure states with a known probability, then I have to describe my system as a mixed state, and then I can write it as a sum of these pure states, where the prefactors are the known probabilities, right?

So in general a mixed state can't be thought as a statistical mixture of pure states, BUT if I have a certain experimental setup that creates state |i> with probability p and state |j> with probability (1-p) I have to describe the system as a mixed state. Is this correct?

From your answers, I see that the physical nature of a mixed state is not trivial to understand.
Can someone explain it in terms of coherences? Is it like I tried to explain here:

What is the physical nature of a mixed state. Let's talk about a single particle.
I have red it has something to do with coherence. In a pure superposition, I also know the phases of the waves, since I know the coefficients a,b in a|1>+b|2>.
However, wenn I have a mixture, that can be represented as, let's say, |p|² |1><1| + |q|² |2><2|, I do not know anything about the phase of the complex coefficients p and q.

Regards

 

[Khashishi]

Sometimes you find also the inaccurate statement that pure states are represented by (normalized) Hilbert-space vectors.

Well now, we could get into a philosophical debate about the difference between a state and the representation of a state. If there is an isomorphism between projection matrices and vectors modulo normalization (for pure states), then it doesn't matter which you use. My issue with your statement is that it is very common to represent pure states by vectors of the form $\left|\psi\right>$, and it isn't wrong for pure states because the calculations work out just fine. Density matrices are usually introduced as a construction on top of the vector representation, and then demonstrated to reduce to the vector representation for pure states.

 

[Khashishi]

but eigenstates can also be mixed state, is it not? For example.. consider a quantum state with the following terms: "Schrödinger cat exposed to radioactive source and dead" and "Schrödinger cat not exposed to radioactive source and alive". These are eigenstates. Now they are exposed to environmental decohererence, the terms "Schrödinger cat exposed to radioactive source and dead" and "Schrödinger cat not exposed to radioactive source and alive" become mixed states. so mixed states is one of the eigenstates chosen so this contradicts the statement above.. unless something wrong with my analysis?

The problem with Schrödinger's cat is that a cat is far too big and complicated to be ever be seen in a pure state. It constantly interacts with the environment. "Alive" and "Dead" aren't simple eigenstates like spin up and spin down. They each encompass a huge number of states and aren't precisely defined to the level demanded by physics. Any macroscopic system is always in a mixed state. A pure state is one where we know everything there is to know about each atom and particle that makes up the system. It is simply not possible to know this for anything but the tiniest systems.

 

[vanhees71]

Well now, we could get into a philosophical debate about the difference between a state and the representation of a state. If there is an isomorphism between projection matrices and vectors modulo normalization (for pure states), then it doesn't matter which you use. My issue with your statement is that it is very common to represent pure states by vectors of the form $\left|\psi\right>$, and it isn't wrong for pure states because the calculations work out just fine. Density matrices are usually introduced as a construction on top of the vector representation, and then demonstrated to reduce to the vector representation for pure states.

The important point is that pure states are represented by projection operators or, equivalently, rays in Hilbert space. This is not some metaphysical subtlety but clearly observed. One important example is the existence of half-integer spin particles and fermions.

 

[Simon Phoenix]

So in general a mixed state can't be thought as a statistical mixture of pure states, BUT if I have a certain experimental setup that creates state |i> with probability p and state |j> with probability (1-p) I have to describe the system as a mixed state. Is this correct?

Sorry for coming to this late - I've been traveling over Summer with somewhat random internet access (perhaps my connection was in a mixed state of 'off' and 'on').

Anyway, you've had some good answers to your questions so far and I'm not sure I can add very much - and most of what I'm going to write is simply a restatement of what others have said. I tend to be a bit of a windbag so I hope you get something of value from my verbal excess

In general a mixed state can indeed be thought of as a statistical mixture of pure states - but only because it's formally indistinguishable from that. For example, suppose I have the mixed state of a spin-1/2 particle in a mixed state described by $$ | \psi \rangle = \frac 1 2 |z+ \rangle \langle z+ | + \frac 1 2 |z- \rangle \langle z- | $$ then I can interpret that as having been prepared as the pure state $| z+ \rangle$ with probability 1/2 and the pure state $|z- \rangle$ with probability 1/2. So basically having been prepared as an 'up ' or 'down' eigenstate of spin-z with equal probability.

However, the actual preparation procedure might have been to prepare the particle with equal probability in the state $| x+ \rangle$ or $|x- \rangle$, that is, in one of the eigenstates of spin-x chosen at random.

The thing is both preparation procedures lead to the same quantum state. So yes, it can be thought of as a statistical mixture of the pure spin-z eigenstates, but it can also be thought of as a statistical mixture of the spin-x eigenstates. In fact it can be thought of as a statistical mixture of the spin-$\theta$ eigenstates - where $\theta$ is any arbitrary direction.

There's just no way to tell from the formalism the difference between them. So even though the initial preparation procedure might have been to choose eigenstates of spin-z at random, it wouldn't actually be 'wrong' to think of the state as having been prepared by choosing spin-x eigenstates at random. There's no way to experimentally distinguish between the two preparation procedures.

We can set this up operationally by having Clive receive a spin-1/2 particle from Alice and a spin-1/2 particle from Bob (let's imagine the 2 particles delivered in a nice presentation case such that Clive doesn't know which is from Alice and which is from Bob). Alice has chosen the state of her particle to be one of the spin-z eigenstates at random, Bob has chosen the state of his particle to be one of the spin-x eigenstates at random. Clive's job is to tell which particle has come from Bob and which particle has come from Alice. There is nothing Clive can do that would give him a better probability of identifying the particles that is better than guessing.

So in a nutshell - if it helps to think of a density operator as a prepared statistical mixture of particular pure states then it's not 'wrong' to do so - even though the actual preparation might have been very different. You'll get the same answers because there's no way to tell the specific 'preparation procedure' from a given density operator. In general the description of a density operator in terms of pure states is not unique.

To give another example the classic entangled singlet state consists of 2 spin-1/2 particles. Let's suppose I give you just one of these particles. Now the density operator you would use to determine the results of measurements on your particle is mathematically identical to the density operator you'd need if I simply gave you an unentangled spin-1/2 particle prepared in a randomly chosen eigenstate of spin-z (for example). Once again we can think of this operationally - if I gave you a spin-1/2 particle and told you it was (a) either one from an entangled pair or (b) in an eigenstate of spin-z chosen at random, you would not be able to do any better than guessing as to whether it was preparation (a) or (b).

Can I give you a better insight into the physical meaning of the density operator? Not really, I'm afraid. It's a great mathematical tool and very important but I confess I don't have a really good way to describe what it is in a 'physical' sense because it's not really all that clear to me either

 

[bhobba]

Can I give you a better insight into the physical meaning of the density operator? Not really, I'm afraid. It's a great mathematical tool and very important but I confess I don't have a really good way to describe what it is in a 'physical' sense because it's not really all that clear to me either

The rock bottom basis of the Born Rule and the existence of states is really non-contextuality (plus a few other things like the principle of strong superposition that most people pretty much assume without even being told its an assumption - and I think there may be others like that as well - non-contextuality is by far the main one) as shown by Gleason. For me states, including mixed ones, are simply a mathematical artifact of that assumption. Definitionally they are the equivalence classes of preparation procedures of all the different outcomes of the Born rule - a preparation procedure belongs to the same class if it gives exactly the same result from the Born Rule when all possible operators are applied using it. I don't know how physical you would call that - but it seems to be the modern view.

Thanks
Bill

 

[stevendaryl]

The rock bottom basis of the Born Rule and the existence of states is really non-contextuality (plus a few other things like the principle of strong superposition that most people pretty much assume without even being told its an assumption - and I think there may be others like that as well - non-contextuality is by far the main one) as shown by Gleason. For me states, including mixed ones, are simply a mathematical artifact of that assumption. Definitionally they are the equivalence classes of preparation procedures of all the different outcomes of the Born rule - a preparation procedure belongs to the same class if it gives exactly the same result from the Born Rule when all possible operators are applied using it. I don't know how physical you would call that - but it seems to be the modern view.

My bugaboo about standard quantum mechanics is the role that a macroscopic/microscopic partition of the universe seems built-in to the very framework. If you describe the Born rule in terms of preparation procedures and measurement results, the two ends--the starting point and the ending point of an experiment---are macroscopic, while the middle part--the transition from initial state to final state--is treated microscopically, using smooth evolution (whether you use density matrices or wave functions). My feeling is that if there is nothing additional going on at the two macroscopic ends, then there should be a description of the whole shebang in microscopic terms, and the macroscopic description (in terms of preparation procedures and measurements) should be derivable, in a similar way that thermodynamics can be derived from Newton's laws plus statistics.

I'm not really asking for any response, just commenting that I personally don't find clarifying QM in terms of preparation procedures to be particularly enlightening. It's mathematically an elegant thing to do, but it seems like it's elegant at the cost of shuffling the hard parts out of the picture. What's left can be made elegant.

 

[vanhees71]

What you want, of course, exists for decades: It's called quantum statistics aka many-body theory and is as old as quantum theory itself.

 

[stevendaryl]

In general a mixed state can indeed be thought of as a statistical mixture of pure states - but only because it's formally indistinguishable from that. For example, suppose I have the mixed state of a spin-1/2 particle in a mixed state described by $$ | \psi \rangle = \frac 1 2 |z+ \rangle \langle z+ | + \frac 1 2 |z- \rangle \langle z- | $$ then I can interpret that as having been prepared as the pure state $| z+ \rangle$ with probability 1/2 and the pure state $|z- \rangle$ with probability 1/2. So basically having been prepared as an 'up ' or 'down' eigenstate of spin-z with equal probability.

That's the miracle of quantum mechanics that both makes it hard to nail down an interpretation and practically unnecessary.

• Start with a pure state of some small system.
• Let it evolve according to Schrödinger's equation, and eventually it will start to interact with larger systems--measurement devices, the environment, observers, etc.
• At this point, the original system is no longer describable by pure state, since it's become entangled with other systems. The best you can do is to trace out the degrees of freedom due to the parts of the universe that are unobservable or too complicated to analyze in detail. The result is a mixed state for the original system.
• But a mixed state can be interpreted as a statistical mixture, which we all understand from classical probability.
• So we can pretend that the system is in this or that state with a certain probability.

So somewhere along the way, we went from talking about superpositions, which are quantum-mechanical and a little mysterious, to mixtures, which can be understood in classical terms. Exactly how that transition took place is irrelevant when it comes to making predictions about the results of experiments.

 

[stevendaryl]

What you want, of course, exists for decades: It's called quantum statistics aka many-body theory and is as old as quantum theory itself.

No. That's not true. Quantum statistical mechanics does not answer the problems. It doesn't solve the measurement problem, but instead simply applies statistics to the ad hoc interpretation.

Look at the way classical statistical mechanics works. You start with laws of motion for particles. Those laws are non-statistical. Then you consider the larger problem of many, many, maybe 10^20 or so, particles all obeying those laws, but with different initial conditions. This problem is completely intractable using the original laws of motion, but to get macroscopic properties, which is good enough for many purposes, you can treat the collection statistically, and it all becomes manageable again.

The problem with QM is that the "laws of motion" themselves are only understood in terms of macroscopic quantities--measurements and preparation procedures. So we don't have the microscopic form of the laws of physics. That's what's weird.

 

[stevendaryl]

No. That's not true. Quantum statistical mechanics does not answer the problems. It doesn't solve the measurement problem, but instead simply applies statistics to the ad hoc interpretation.

Quantum statistical mechanics is describing a different problem, which is that you have many, many identical systems. That is not what is going on in say an EPR experiment. In that case, it isn't that we have many, many twin pairs. We have a single twin-pair. What's macroscopic are:

• The device that produces the twin pairs
• The device that measures the spin (or polarization)
• The observers

 

[stevendaryl]

The problem with QM is that the "laws of motion" themselves are only understood in terms of macroscopic quantities--measurements and preparation procedures. So we don't have the microscopic form of the laws of physics. That's what's weird.

So the laws themselves are described in macroscopic terms (probabilities for measurement outcomes given certain preparation procedures). Then you propose to eliminate the dependency on macroscopic phenomena by describing those macroscopic phenomena as many, many microscopic objects obeying the same laws. But since the laws themselves refer to macroscopic phenomena, you haven't actually eliminated the reference to macroscopic phenemena.

It seems circular to me, but maybe it's actually some kind of bootstrapping, or recursive understanding of the universe. The microscopic is understood in terms of macroscopic, which is understood in terms of microscopic, etc. Maybe there is some mathematical sense in which this is a convergent process.

 

[vanhees71]

Well, can you prove that the measurement devices used to measure the polarization of photons or the spin of particles etc. etc. are not described by quantum statistics of macroscopic "very-many-body systems"? If so, then you'd be right in saying that there is a division of the world in microscopic and macroscopic behavior.

I don't see the necessity for such a schizophrenic worldview at all, and there's also nothing in QT which tells you where the microscopic world should end and the macroscopic should begin. The more refined our preparation methods become the more success quantum experimenters have in preparing larger and larger (partially even macroscopic) systems in specific quantum states and showing, as expected, all the expectations related with them (e.g., double-slit experiments with buckyball molecules, entangled phonon states of diamonds (even at room temperature!), Bose-Einstein condensates of trapped gases, etc. etc.).

 

[Mentz114]

Quantum statistical mechanics is describing a different problem, which is that you have many, many identical systems. That is not what is going on in say an EPR experiment. In that case, it isn't that we have many, many twin pairs. We have a single twin-pair. What's macroscopic are:

• The device that produces the twin pairs
• The device that measures the spin (or polarization)
• The observers

I don't think this is entirely accurate. In a typical QM experiment the apparatus can be described as a series of concentic spheres, with the coldest, most-shielded part in the centre surrounded by ever more 'classical' apparatus
There was a recent paper which describes an experiment that mimics 'Maxwells demon' in which the experiment can be driven from outside the central area by laser or MW pulses shining into the ultra-cold region where the QM laws of motion rule and information extracted by pulses coming out. So the apparatus has quantum parts.
Furthermore what is happening in there is predictable and well-understood.

The paper is here https://arxiv.org/abs/1702.05161

The figure on page 10 is interesting.

 

[stevendaryl]

I don't see the necessity for such a schizophrenic worldview at all

I'm calling your approach schizophrenic. If the only meaning of the quantum state is to give probabilities for measurement outcomes (or other macroscopic quantities such as mean values) in terms of preparation procedures, then that's necessarily schizophrenic. A non-schizophrenic theory would not mention macroscopic quantities such as measurements or preparation procedures in the fundamental laws, but would be able to derive whatever is being claimed about those things.