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

[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.

 

[DarMM]

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

Pure and Mixed refer to a state's entropy though. Thus a state is pure if it has vanishing entropy.

In many (most) cases of course we represent pure states with vector states. Thus the two are often, but not always, the same.

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.

It's a vector state, not a pure state as its entropy does not vanish.

 

[A. Neumaier]

Pure and Mixed refer to a state's entropy though. Thus a state is pure if it has vanishing entropy.

In many (most) cases of course we represent pure states with vector states. Thus the two are often, but not always, the same.It's a vector state, not a pure state as its entropy does not vanish.

With which definition of entropy? Where is this discussed?

 

[DarMM]

With which definition of entropy? Where is this discussed?

So first of all one cannot use the usual formula:
$$S\left[\rho\right] = -Tr\left(\rho\ln\rho\right)$$
to compute the entropy in the most general cases. This will give the wrong answer in many cases, or be impossible to apply in some cases. An example being the case of superselection sectors where indeed a state such as:
$$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|a\rangle + |b\rangle\right)$$
with $|a\rangle$ and $|b\rangle$ being elements of different superselection sectors. This is actually a mixed state, but of course the usual formula will give $S\left[\rho\right] = 0$ which is incorrect.

It is for this (and other cases) that Araki developed the more general method of computing entropy in the following papers:
H. Araki,Relative entropy of states of von Neumann algebras I, Publ. RIMS Kyoto Univ. 11, 809-833 (1976)
H. Araki,Relative entropy of states of von Neumann algebras II, Publ. RIMS Kyoto Univ. 13, 173-192 (1977)

Although I should say these papers use modular theory and other laborious methods from the theory of C*-algebras. A good introduction to modular theory (in my opinion) is:
Bratteli, O., Robinson, D.W. (1979): Operator algebras and quantum statistical mechanics I
(Springer, New York, Berlin, Heidelberg)
Bratteli, O., Robinson, D.W. (1981): Operator algebras and quantum statistical mechanics
II, (Springer, New York, Berlin, Heidelberg)

The original papers by Takesaki are harder to read.

Another good guide to all of this is:
Ohya M, Petz D. 1993 Quantum entropy and its use. Berlin, Germany: Springer.

You have the ultimate development of entropy here as related to Connes cocycles.

 

[vanhees71]

I'm really very puzzled.

It thought in a model with superselection rules the above superposition is simply not describing a physical preparable state. If it were preparable mathematically it'd by definition represent a pure state (or rather in the usual sense of $\hat{\rho}=|\psi \rangle \langle \psi|$ representing this pure state to be pedantic).

Whether von Neumann entropy describes the entropy measure for a given physical situation is of course a question to be decided for the specific case, but at least it's the standard definition of the missing information, given the statistical operator relative to complete information, defined as given by preparation in (any) pure state.

 

[DarMM]

It thought in a model with superselection rules the above superposition is simply not describing a physical preparable state. If it were preparable mathematically it'd by definition represent a pure state (or rather in the usual sense of $\hat{\rho}=|\psi \rangle \langle \psi|$ representing this pure state to be pedantic).

It is preparable, it's just that it describes a mixed state not a pure state. That's the whole point, just because a state can be represented by a vector state does not mean it is a pure state.

You don't typically see this in QM in practice, but it is the case.

 

[DarMM]

If it helps the actual connection between pure states and vector states is that in an irreducible representation of the observable algebra the vector states are pure. What's happening with superselection is that one is dealing with vector states in a reducible representation. Thus they are vector states, but can be mixed.

 

[Morbert]

It thought in a model with superselection rules the above superposition is simply not describing a physical preparable state. If it were preparable mathematically it'd by definition represent a pure state (or rather in the usual sense of $\hat{\rho}=|\psi \rangle \langle \psi|$ representing this pure state to be pedantic).

I think it's the case that the state is not physical only insofar as there is no observable that is a projector onto the state. But it is still physical as a preparation.

 

[DarMM]

I think it's the case that the state not physical only insofar as there is no observable that is a projector onto the state. But it is still physical as a preparation.

Precisely. It's as physical as any other mixed state, i.e. describes a preparation.

 

[kith]

Just a quick question: are there kets which correspond to partially mixed states?

 

[DarMM]

Just a quick question: are there kets which correspond to partially mixed states?

In the GNS construction you can form a vector state corresponding to any mixed state.

More in line with the case here you can just change the coefficients in front of the kets

 

[vanhees71]

It is preparable, it's just that it describes a mixed state not a pure state. That's the whole point, just because a state can be represented by a vector state does not mean it is a pure state.

You don't typically see this in QM in practice, but it is the case.

Then we have different definitions of pure states.

 

[DarMM]

Then we have different definitions of pure states.

Possibly. The definition I'm using is typical in rigorous QFT, quantum foundations and quantum information. It reduces to your definition (i.e. a pure state is a ket) in the case when the kets are an irreducible representation of the observable algebra.

However due to things like the GNS construction your definition isn't very useful in the case of QFT in curved spacetimes and situations like that, since one can take any statistical operator and recast it as a vector state in another representation.

This is a problem because if we take a statistical operator then it has $S > 0$, however in the GNS representation, according to the usual definition of entropy, we would have $S = 0$. Even though it's really the same state. Thus we need a representation independent notion of purity and entropy.

 

[Elias1960]

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.

This seems relevant to the discussion we had about generalizations of probability theory.

Because it suggests that in all those non-Kolomogorvian probability theories all one needs is to add some external system which selects the "Boolean algebra of events", and for the combined system we have a Kolmogorovian probability theory, with the elementary events defined as the external animal which defines the Boolean algebra of events together with the corresponding events. Thus, simply the straightforward generalization of the Kochen-Specker construction for quantum theory.

 

[DarMM]

This seems relevant to the discussion we had about generalizations of probability theory.

Because it suggests that in all those non-Kolomogorvian probability theories all one needs is to add some external system which selects the "Boolean algebra of events", and for the combined system we have a Kolmogorovian probability theory, with the elementary events defined as the external animal which defines the Boolean algebra of events together with the corresponding events. Thus, simply the straightforward generalization of the Kochen-Specker construction for quantum theory.

There is a significant difference though. One selects the Boolean subalgebra from a fundamentally non-Boolean structure, but that choice is not given a probability. The Kochen-Specker construction has the choice itself as an event with a probability in a larger Boolean algebra.

 

[Elias1960]

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.

Sorry, but this makes no sense. Kets are pure states. Point. For a mathematician, the only reasonable reaction to an attempt to name them mixed states is a facepalm.

Usual superselection rules give only that they cannot be prepared. But that they cannot be prepared does not justify to give them false names.

The same holds for those weaker notions of superselection rules where one can prepare such states, like the state of a Schrödinger cat. Here we obviously have no operator to measure this pure state. All measurements we have give the same result as some large number of other pure as well as mixed states. But this also does not give a permission to name a pure state a mixed state. These are simply indistinguishable but nonetheless different states.

I would not object to phrases like "pure states become indistinguishable from mixed states", but to name a pure state a mixed state because of this is positivism going insane.

(Of course, the whole idea of considering $C^*$ algebras as something of some fundamental importance smacks of positivism. So, I would not really be surprised if this is indeed standard language in that community. In this case, I would characterize this as an example of an absurd consequence of the remaining influence of positivism in physics.)

 

[DarMM]

Sorry, but this makes no sense. Kets are pure states. Point. For a mathematician, the only reasonable reaction to an attempt to name them mixed states is a facepalm.

Nobody is naming kets as mixed states. Mixed states can be represented as vector states, such as via the GNS construction. That's a mathematical fact.

You basically don't accept the whole area of quantum probability or its nomenclature, so you're simply never going to accept any of this. All I can say is that most experts do. The work of Huzhiro Araki and others in defining a representation independent notion of pure and mixed states and entropy might be a "face palm" for you and the definition of pure and mixed state in rigorous QM and quantum information might be dumb to you. However there is little I can say on that point.

 

[Elias1960]

Nobody is naming kets as mixed states.

Ok, nobody has written this (emphasis mine):

An example being the case of superselection sectors where indeed a state such as:
$$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|a\rangle + |b\rangle\right)$$
with $|a\rangle$ and $|b\rangle$ being elements of different superselection sectors. This is actually a mixed state, ...

Mixed states can be represented as vector states, such as via the GNS construction. That's a mathematical fact.

Already better, but also sloppy language. The GNS construction is a larger space, the mixed states are projections into a smaller space. Ok, I'm not a native speaker, but "represented as" sounds problematic and misleading to me, "represented using a" or "represented by" would be ok. Because the "represented as" suggests some map object $\to$ representation, while "represented using a" suggests representation $\to$ object.

Again, just to clarify, I do not object against the mathematics, I have objections here against the misleading language used.

 

[DarMM]

Ok, nobody has written this (emphasis mine):

Yes that state is a mixed state. However that's not the same as saying vector states and mixed states are the same thing. Just in this case a mixed state is being represented as a vector state.

Already better, but also sloppy language

I don't see how. In the GNS construction a mixed state is represented as a vector state.
I don't get this "as/using a/by" thing. "Represented as" is a fairly common phrase.

 

[DarMM]

An old colleague who was reading this thread has kindly pointed out to me that the literature has seven (!) different definitions of a superselection rule, with some equivalent, others stronger/weaker. I'll do a bit of reading and report back hopefully if I emerge from the morass!

 

[Elias1960]

Yes that state is a mixed state. However that's not the same as saying vector states and mixed states are the same thing. Just in this case a mixed state is being represented as a vector state.
I don't get this "as/using a/by" thing. "Represented as" is a fairly common phrase.

I have to give up. There seems to be no chance. Possibly a culture thing: Mathematicians are taught to care even about minor differences, and assuming such a cultural background among physicists seems wrong.

 

[DarMM]

Mathematicians are taught to care even about minor differences, and assuming such a cultural background among physicists seems wrong.

Plenty of mathematicians write "represented as". I can give you some examples. I just don't think the difference between the English phrases "represented as" or "represented using a" are what you are claiming since nobody else seems to comment on or adhere to your distinction.

I don't really think this is a physicist vs mathematician cultural distinction, just that English isn't as fine grained as you are imagining here.

 

[Morbert]

I have to give up. There seems to be no chance. Possibly a culture thing: Mathematicians are taught to care even about minor differences, and assuming such a cultural background among physicists seems wrong.

I'm not sure I see the controversy. I don't normally work with these kinds of equations but:

We have the ket $|\psi\rangle = \frac{1}{\sqrt{2}}|a\rangle+\frac{1}{\sqrt{2}}|b\rangle$ mentioned previously. Thus, we have a quantum state $\omega_\psi$ such that

$$\omega_\psi(A) = \frac{1}{2}\omega_a(A)+\frac{1}{2}\omega_b(A)+ \frac{1}{2}\langle a,Ab\rangle+\frac{1}{2}\langle b,Aa\rangle$$

for all observables $A$. If $|a\rangle$ and $|b\rangle$ are subject to superselection rules such that $\langle a,Ab\rangle = \langle b,Aa\rangle = 0$ for all observables $A$, then the above reduces to the mixed state

$$\omega_\psi(A) = \frac{1}{2}\omega_a(A)+\frac{1}{2}\omega_b(A)$$

But $\omega_\psi$ is still a vector state, since $\psi\in\mathcal{H},||\psi||=1$.

Are you perhaps discriminiating $|\psi\rangle$ from $\omega_\psi$?

[edit] - made notation more consistent

 

[Elias1960]

Are you perhaps discriminiating $|\psi\rangle$ from $\omega_\psi$?

Yes. This is my point, they are different but indistinguishable because we can distinguish them only if we have operators which we don't have because of the superselection rule. The superselection rule reduces our ability to construct resp. identify by measurement particular states. But such restriction of human abilities to distinguish different objects does not make them identical.

 

[DarMM]

Yes. This is my point, they are different but indistinguishable because we can distinguish them only if we have operators which we don't have because of the superselection rule. The superselection rule reduces our ability to construct resp. identify by measurement particular states. But such restriction of human abilities to distinguish different objects does not make them identical.

Well the typical terminology in the most general applications of quantum theory, such as in algebraic field theory and quantum information, is that $|\psi\rangle$ is a vector state which represents $\omega_{\psi}$. However it is to $\omega_{\psi}$ that the mixed/pure distinction refers, since regardless of the representation $\omega$ has finite entropy and all the other characteristics of a mixed state. Thus it is a mixed state represented by a vector state in this particular representation.

 

[A. Neumaier]

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.

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

Pure and Mixed refer to a state's entropy though. Thus a state is pure if it has vanishing entropy.

In many (most) cases of course we represent pure states with vector states. Thus the two are often, but not always, the same.

It's a vector state, not a pure state as its entropy does not vanish.

It thought in a model with superselection rules the above superposition is simply not describing a physical preparable state. If it were preparable mathematically it'd by definition represent a pure state (or rather in the usual sense of $\hat{\rho}=|\psi \rangle \langle \psi|$ representing this pure state to be pedantic).

We have the ket $|\psi\rangle = \frac{1}{\sqrt{2}}|a\rangle+\frac{1}{\sqrt{2}}|b\rangle$ mentioned previously. Thus, we have a quantum state $\omega_\psi$ such that $$\omega_\psi(A) = \frac{1}{2}\omega_a(A)+\frac{1}{2}\omega_b(A)+ \frac{1}{2}\langle a,Ab\rangle+\frac{1}{2}\langle b,Aa\rangle$$ for all observables $A$. If $|a\rangle$ and $|b\rangle$ are subject to superselection rules such that $\langle a,Ab\rangle = \langle b,Aa\rangle = 0$ for all observables $A$, then the above reduces to the mixed state $$\omega_\psi(A) = \frac{1}{2}\omega_a(A)+\frac{1}{2}\omega_b(A)$$ But $\omega_\psi$ is still a vector state, since $\psi\in\mathcal{H},||\psi||=1$.

Well the typical terminology in the most general applications of quantum theory, such as in algebraic field theory and quantum information, is that $|\psi\rangle$ is a vector state which represents $\omega_{\psi}$. However it is to $\omega_{\psi}$ that the mixed/pure distinction refers, since regardless of the representation $\omega$ has finite entropy and all the other characteristics of a mixed state. Thus it is a mixed state represented by a vector state in this particular representation.

The misunderstandings in this thread related to the above are due to the fact that the notion of a state and the distinction between pure and mixed states are both ambiguous because both depend on the algebra on which they are considered. That's why we have been talking past each other.
A good open source general discussion is in Section 6.3 of the paper

• N.P. Landsman, Between classical and quantum, Handbook of the Philosophy of Science 2 (2006), 417-553.

The essence is here:

In standard quantum physics (i.e., that not using the $C^*$ algebra terminology), a state is a density operator (i.e., a Hermitian positive semidefinite trace 1 operator) $\rho$ on a Hilbert space $H$, and it is pure when it has rank one and mixed otherwise. In the pure case (only) it can be represented by a state vector $\psi$ as $\rho=\psi\psi^*$, uniquely up to a phase. Hence state vectors always represent pure states.

On the other hand, algebraic quantum physics uses the $C^*$ algebra terminology. There a state is a positive linear functional $\omega$ on a $C^*$ algebra $A$ (called the observable algebra) satisfying $\omega(1)=1$, and the state is pure iff it is not decomposable into a convex combination of two different states. This is a generalization of the notion from standard quantum physics, which is the special case where the observable algebra is the algebra $B(H)$ of bounded linear operators on a Hilbert space $H$ and the density operator $\rho$ is identified with the positive linear functional $\omega$ defined by $$\omega(A):=Tr~ \rho A.~~~~~~~~(1)$$ This defines a 1-1 correspondence, making the identification possible. With this identification the notion of pure and mixed states agree.

One says that there are superselection rules when the observable algebra has a center consisting not only of the multiples of the identity. When working in a fixed Hilbert space this implies that $A$ is a proper *-subalgebra of $B(H)$. In this case, (1) no longer defines a 1-1 correspondence between density operators and positive linear functionals, and the notions both of state and of the pure/mixed distinction differ algebraically.

This is easiest to see if one takes the Hilbert space as $C^n$ and the observable algebra to be the algebra of diagonal matrices. Then $\omega(A)=\omega(Diag(A))$, where $Diag(A)$ is the matrix obtained from $A$ by setting the off-diagonal elements to zero. The pure states are now only those whose density matrix have a single 1 and otherwise zeros on the diagonal. In particular, most pure states become mixed.

The nonuniqueness also implies that the Schrödinger equation becomes meaningless; only the von Neumann equation for the dynamics of the density operator remains meaningful, and only for Hamiltonians belonging to the observable algebra. In particular, this dynamics does not lead outside of a superselection sector; if the initial state is pure (and hence belongs to a single superselection sector, as it is often assumed), the state remains pure and lies for all times in the same superselection sector.

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.

This can work only in a setting where the initial state of the system of trap plus particle is already mixed. The reduced dynamics of the particle alone is no longer Hamiltonian. Thus it is no surprise that the initially pure state becomes mixed.

 

[Quantum Alchemy]

Quantum theory has a classical-quantum cut. Hence the use of quantum theory itself means the assumption of a classical "something".

No, it just means the appearance of something classical. A Heisenberg cut is subjective to the Observer. As I've shown earlier, observation can strengthen or weaken interference.

Apart from "observing," or detecting, the electrons, the detector had no effect on the current. Yet the scientists found that the very presence of the detector-"observer" near one of the openings caused changes in the interference pattern of the electron waves passing through the openings of the barrier. In fact, this effect was dependent on the "amount" of the observation: when the "observer's" capacity to detect electrons increased, in other words, when the level of the observation went up, the interference weakened; in contrast, when its capacity to detect electrons was reduced, in other words, when the observation slackened, the interference increased.

https://www.sciencedaily.com/releases/1998/02/980227055013.htm

So based on things like this, I see wave function collapse as the reduction of interference. This doesn't mean the probable states collapse. Without interference, these probable states appear as classical to the observer.

 

[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?

It is only an informal notion, not a mathematical one. This is revealed by the fact that Section 6, which discusses this, has no significant formulas. Indeed, in the final paragraph of his paper Wightman writes:

Even the definition of what one means by a superselection rule for a system induced by its environment has not been made sufficiently precise.

Maybe assuming a thermodynamic limit for the environment can make this mathematically precise, but of course the latter is an approximation only.

 

[DarMM]

One says that there are superselection rules when the observable algebra has a center consisting not only of the multiples of the identity

It is only an informal notion, not a mathematical one.

Note that there seem to be several definitions of a superselection in the literature, which may also be a source of talking past each other. Some authors permit superselections to simply correspond to a non-trivial commutant, allowing the center to be trivial. Even weaker, some authors, such as R.F. Streater permit it to correspond to the absence of an interference operator from the algebra of measurables. I use "measurable" here to specify a difference from the abstract C*-algebra of observables, where by we may be excluding observables for reasons beyond the quantum kinematical.

So if $\mathcal{M}$ is the Von-Neumann algebra of abstract observables constructed from the basic degrees of freedom, the different notions of superselection in the literature are in decreasing strength:

1. $\mathcal{Z}\left(\mathcal{M}\right)$ non-trivial and $\mathcal{Z}\left(\mathcal{M}\right) \subset \mathcal{M}$
2. $\mathcal{Z}\left(\mathcal{M}\right)$ non-trivial
3. Simply that $\mathcal{M}^{'}$ is non-trivial. Which may permit $\mathcal{Z}\left(\mathcal{M}\right)$ to be trivial. There are many different definitions of superselection in the literature that all reduce to this, or when made mathematically precise become this.
4. The absence of the appropriate interference observable from the set of physically realizable observables.

Only the fourth, weakest notion, is the one commonly invoked for the Schrodinger cat paradox in order to view the uncertainty between macroscopic states as being due to ignorance. An example would be Streater in section 4.6 of his text "Lost Causes in and beyond Theoretical Physics" where he says on p.51:

This justifies the assumption that two states of a system differing in the value of a classical variable are separated by a superselection rule.

Thus he uses this weakest notion of superselection to argue for classical ignorance amongst macroscopic degrees of freedom. As you said this is not superselection in the quantum kinematical sense. Although note that some authors permit a weaker notion of superselection that is still kinematical (3 in the list above).

You've meant Definition 1 or perhaps also 2, where as I have meant the weakest: 4.

 

[DarMM]

The misunderstandings in this thread...

Indeed, @vanhees71 essentially spotted this in #116. It's also the differences between the notions of entropy.

 

[kith]

Just a quick question: are there kets which correspond to partially mixed states?

In the GNS construction you can form a vector state corresponding to any mixed state.

More in line with the case here you can just change the coefficients in front of the kets.

I have thought about it some more and your posts helped me to fix an error in my thinking.

In decoherence, we start with a superposition and evolve to a maximally mixed state. During this, the off-diagonal elements of the density matrix decay. In your example, we also end up in a maximally mixed state but we start with a basis state. What bothered me somehow was that the off-diagonal elements remain zero at all times but now I see that that's not a problem at all.

There's also been the related misunderstanding that when I asked about "partially mixed states", I had in mind a density matrix with fixed populations but didn't spell this out. Keeping this in mind, the answer to my question above is "no, these states are forbidden". Until now, I have viewed superselection rules from the viewpoint of the Hilbert space / state vectors / density matrices but if we start with the algebra of observables it seems very natural that certain density matrices simply don't occur as valid states.

Thanks for your help! I see that my knowledge has become a bit rusty in parts.

 

[DarMM]

Thanks for your help! I see that my knowledge has become a bit rusty.

No problem! It's very abstract stuff, I get mixed up myself sometimes!