Mixed states vs pure states - physical POV

[SchroedingersLion]

Hey guys,

I am having issues with understanding the physical nature of pure and mixed states. Maybe you can help me out?

1) A pure state - superposition is a state that consists of different states at the same time. It's like having several waves, each one belonging to an Eigenstate of the Hamilton operator, with known phase and amplitude, that interfere with each other, right?

Does that mean, when I have a pure superposition of |0> and |1>, I can find times and positions, where the wave of |0> has minimal deflection and the wave of |1> is at its peak, so the state looks more like a |1> state?

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.
For an ensemble, that means that for a maximal incoherent mixture, half of the particles will be in |1> and the other half in |0>, right?

Does that mean the description of mixed states is only a result of the observer's lack of knowledge, for example: " I know that physically, the particle must be represented by a wave, that corresponds either to |1> or to |0>, but I only know the amplitudes, not the phases, so I have to describe my system as a statistical mixture"

That's it for the beginning. Would be nice, if someone could clear me up on this. And please don't talk too much about density matrices, I know about the mathematical differences in their respective descriptions.

Best wishes
SchroedingersLion

 

[Khashishi]

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.

 

[Khashishi]

Does that mean, when I have a pure superposition of |0> and |1>, I can find times and positions, where the wave of |0> has minimal deflection and the wave of |1> is at its peak, so the state looks more like a |1> state?

Not sure exactly what you are asking. It depends on what you mean by "looks more like a |1> state". If |0> and |1> are eigenstates of the Hamiltonian, then the absolute values of the amplitudes aren't changing in time. So, the whole wavefunction is going to look like a superposition of |0> and |1>. But some particular positions can be influenced by one eigenstate more than the other. For example, for a particle in a box superposition state,
$\psi = a_1 \sin\left(\frac{x \pi}{L}\right) + a_2 \sin\left(\frac{2x \pi}{L}\right)$
The point at L/2 is a node for the "2" state, so it looks like state "1". On the other hand, the point at L/4 is somewhat more strongly influenced by the "2" state than the "1" state, though both contribute.

The complex phases will be rotating with respect to each other.

 

[PeterDonis]

A pure state - superposition is a state that consists of different states at the same time.

Not really, no, unless you want to say that a vector which is the sum of two other vectors "consists of different vectors at the same time". Quantum states form a vector space, so you can always add any number of states together to form another state; it's just adding vectors. And you can choose the basis of the vector space however you like, so you can always make any given state a superposition, or not a superposition, depending on how you choose the basis (either choose the basis so one basis vector is pointing in the same direction as your chosen state vector, or not). This is all just basic linear algebra and is not particular to quantum mechanics.

 

[SchroedingersLion]

Hey guys, thanks for your replies!

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.

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?

Not sure exactly what you are asking. It depends on what you mean by "looks more like a |1> state". If |0> and |1> are eigenstates of the Hamiltonian, then the absolute values of the amplitudes aren't changing in time. So, the whole wavefunction is going to look like a superposition of |0> and |1>. But some particular positions can be influenced by one eigenstate more than the other. For example, for a particle in a box superposition state,
$\psi = a_1 \sin\left(\frac{x \pi}{L}\right) + a_2 \sin\left(\frac{2x \pi}{L}\right)$
The point at L/2 is a node for the "2" state, so it looks like state "1". On the other hand, the point at L/4 is somewhat more strongly influenced by the "2" state than the "1" state, though both contribute.

The complex phases will be rotating with respect to each other.

I think you understood me correctly. Of course, the absolute values do not depend on time, my mistake. What happens if I were to measure the energy at L/2 (in the energy eigenbasis). Would I see the energy of state |1>, since the state "2" wave has a node there?

Not really, no, unless you want to say that a vector which is the sum of two other vectors "consists of different vectors at the same time". Quantum states form a vector space, so you can always add any number of states together to form another state; it's just adding vectors. And you can choose the basis of the vector space however you like, so you can always make any given state a superposition, or not a superposition, depending on how you choose the basis (either choose the basis so one basis vector is pointing in the same direction as your chosen state vector, or not). This is all just basic linear algebra and is not particular to quantum mechanics.

Hm, but isn't this the general interpretation of a superposition? You have two waves from two states, interfering each other to form a third state?
If I am in a superposition of two eigenstates of H and I measure the energy, I will get either the energy of the first or the energy of the second state, there is no in between, so it makes sense to interpret the superposition as "both states at once" before measuring. Or would it be more accurate to say "with respect to energy, this superposition consists of two energy eigenstates"?

 

[jfizzix]

The difference between a pure, even superposition of $|0\rangle$ and $|1\rangle$:
e.g., $|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$
.. and an even mixture of $|0\rangle$ and $|1\rangle$:
e.g., $\hat{\rho} = \frac{1}{2} |0\rangle\langle 0| + \frac{1}{2}|1\rangle\langle1|$
is that when measuring the right observable of a pure state, one can get a single outcome 100 percent of the time, while the evenly mixed state will yield 50/50 probabilities no matter what observable is measured.

A quantum mixed state can indeed represent a mixture of quantum states. For example, an ensemble of imperfectly prepared particles will have a statistical distribution of quantum states, relative to the precision of the preparation.
However, mixed states also accurately represent the statistics of individual particles, when they happen to be entangled with other particles.
When the joint state of a pair of particles is maximally entangled, the state of either particle by itself is best represented as a maximally mixed state.

 

[Physics Footnotes]

Hm, but isn't this the general interpretation of a superposition? You have two waves from two states, interfering each other to form a third state? If I am in a superposition of two eigenstates of H and I measure the energy, I will get either the energy of the first or the energy of the second state, there is no in between, so it makes sense to interpret the superposition as "both states at once" before measuring. Or would it be more accurate to say "with respect to energy, this superposition consists of two energy eigenstates"?

Suppose you have a system with two energy eigenstates $|a\rangle$ and $|b\rangle$ with eigenvalues $a$ and $b$, respectively. When you perform an energy measurement and get a value of $a$, say, this does not mean that the state of the system was $|a\rangle$. In fact, from an individual measurement you cannot determine what the initial state $|\psi\rangle$ was. At best you can say that the initial state had an overlap with $|a\rangle$ (i.e. $\langle a|\psi\rangle \neq 0$) or, equivalently, has a non-zero $|a\rangle$ component in an energy basis expansion.

If the initial state of the system was $|\psi\rangle=\frac {|a\rangle + |b\rangle} {\sqrt {2}}$, for example, then measurements on this state would give values of $a$ and $b$ 50% of the time each. None of these measurements in and of themselves tell you the initial state of the system.

Also, you should think of the state $|\psi\rangle$ as a state in and of itself, rather than being genuinely composed of other individual states. A spatial displacement, for example, is a vector in and of itself. It is convenient to express it in terms of other vectors, but it is still a new entity with its own identity and properties.

EDITED TO ADD: You might also find it helpful to read this post and this one for a more detailed explanation of these points.

 

[SchroedingersLion]

Good evening,

I have been busy with other courses the last weeks so I completely forgot about this thread. Now I am back and I need to understand this topic in order to learn for another examination in 2 weeks :D

@Phyics Footnotes
Thanks, I've red your posts and I understand. I always used to think about superpositions as if they were "2 states at once", and not an "unique state on itself", since if we have a superposition of energy eigenstates and I were to measure the energy, I would get EITHER the energy of state 1 or state 2, and NOT some strange mixture energy. However, this is not because the superposition is not a unique state, but because it is not an EIGENstate of H.

Now back to a few questions:
If I have a particle source that creates state 1 with a certain probability p and state 2 with 1-p, then I can describe my system as a statistical mixture: p|1><1|+(1-p)|2><2|. So this time, I know for certain that each particle is either in |1> or in |2>.
So I describe every particle in this ensemble as being in a mixed state. Just because I don't KNOW in which pure state my particle is. I know it has to be in a pure state, but I have to describe it as a mixed state, right? Feels a bit strange.

 

[vanhees71]

Yes, you even know that it is in either the pure state $|1 \rangle \langle 1|$ or in the pure state $|2 \rangle \langle 2|$, but before measuring it, you don't know in which. The only thing you know that it's in state 1 with probability $p$# and in state 2 with probability $1-p$. That's described by the mixed state
$$\hat{\rho}=p |1 \rangle \langle 1| + (1-p) |2 \rangle \langle 2 |.$$
Any state, whose statistical operator is not a projection operator is called "mixed". A pure state is always represented by a projection operator.

 

[DrDu]

Yes, you even know that it is in either the pure state $|1 \rangle \langle 1|$ or in the pure state $|2 \rangle \langle 2|$, but before measuring it, you don't know in which. The only thing you know that it's in state 1 with probability $p$# and in state 2 with probability $1-p$. That's described by the mixed state
$$\hat{\rho}=p |1 \rangle \langle 1| + (1-p) |2 \rangle \langle 2 |.$$
Any state, whose statistical operator is not a projection operator is called "mixed". A pure state is always represented by a projection operator.

No, generally you don't know that it is either in state 1 or 2, as the decomposition of a mixed state into pure states isn't unique.

 

[Nugatory]

No, generally you don't know that it is either in state 1 or 2, as the decomposition of a mixed state into pure states isn't unique.

In general you don't, but for some preparation procedures you can.

 

[SchroedingersLion]

Yeah I was explicitely talking about a particle source that gives me state |1> or state |2> with known probabilities.

So next step, how is it generally? 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.

 

[Nugatory]

What is the physical nature of a mixed state.

That's the sort of question that has no answer within the mathematical formalism of QM. The state tells you the probability of getting various results from various possible measurements, but that's all the physical nature that's there.

Many people are unsatisfied with this state of affairs, which is why we hear so many questions along the lines of "but what's REALLY happening under the covers?" and why there are so many interpretations of quantum mechanics... But the universe doesn't seem to care much about whether we're satisfied with its rules.

 

[bhobba]

Your question is both easy and hard - I will not touch the hard stuff since it is interpretive and will not lead to an answer in the terms you have posed your question.

First a state is an operator. Forget what beginner and some intermediate texts tell you - they are wrong - but we all have to start somewhere.

Specifically the are positive operators of unit trace. Since this is an I level thread you should know what that means from a course in linear algebra. If not see the following:
http://quantum.phys.cmu.edu/CQT/chaps/cqt03.pdf

Now by definition states of the form |u><u| are called pure. States of the form ∑pi |ui><ui| are called mixed, where since it is a positive operator of unit trace you can easily prove the pi are positive and ∑pi = 1. Straight away you can see they are likely to have a connection to probability and the Born rule shows exactly what that connection is. I will leave that as an exercise for you.

Now, again you can show it, its not hard, all states are either are either pure or mixed. Pure states are easily mapped to a vector space and show why phase makes no difference to states |cu><cu| = |u><u| where c is a complex number of unit length ie a phase factor.

Without telling you they are only a subset of states, beginner texts tell you they are part of a vector space. That is only pure states.

Its not hard - what does it mean physically. Well the Born Rule for all states says - the average of the outcome of an observable O, E(O), is E(O) = Trace(SO) where S is the state the system is in. In fact it to some extent can be proved via Gleason's theorem, which you can look up. Playing with the Born rule will show you all sorts of things - eg if you assume continuity the state of a system after observing it is a pure state and an eigenvector of the observable. This is the so called collapse postulate - which isn't a postulate at all and a simple consequence of the Born Rule.

That all there really is to it theory wise, but what it means - that's much much harder and the part of interpretations.

If you want to look into that - it can't be done in a post - here is THE textbook on it:
https://www.amazon.com/dp/3540357734/?tag=pfamazon01-20

Thanks
Bill

 

[DrDu]

In general you don't, but for some preparation procedures you can.

I don't see this. All of the information we have is encoded in the density matrix. If the density matrix does not allow for a unique decomposition into eigenstates (which is basically only possible for pure states) then we also can't say that it is a mixture of specific eigenstates.

 

[DrDu]

Now by definition states of the form |u><u| are called pure. States of the form ∑pi |ui><ui| are called mixed,

No, this is representation dependent. Any state, whether pure or mixed can be represented as |u><u|.
E.g. take the operators of angular momentum for a spin 1/2 particle. You can represent them (neglecting all hbar prefactors) by Pauli matrices $\sigma_i$, but you can also represent them as
$\sigma_i \otimes 1_2$. In the latter case, a mixed state with a times eigenstate 1 of s_z and (1-a) times eigenstate -1 of s_z can be represented as
$| \phi \rangle= a |1\rangle |1 \rangle + (1-a) |-1 \rangle |-1 \rangle$.

Edit: Let me dwell a little further on this:
The representation of the state $\phi$ isn't unique. Apparently, we could also write
$| \phi \rangle = a| 1 \rangle |-1 \rangle +(1-a) | -1 \rangle | 1 \rangle$.
If I call the first representation $|\phi_1\rangle$ and the second one $|\phi_2\rangle$, then any combination $|\tilde{\phi}\rangle= \lambda |\phi_1\rangle + (1-\lambda) |\phi_2 \rangle$ represents the same mixed state. Hence, in contrast to pure states, mixed states can be represented as sums of other states. This also holds for density matrices. Pure states are the extremal points of the convex set formed by all states.

 

[vanhees71]

No, generally you don't know that it is either in state 1 or 2, as the decomposition of a mixed state into pure states isn't unique.

Of course, but I understood the original question such as if somebody is preparing a system either in state $|1 \rangle \langle 1|$ or $|2 \rangle \langle 2 |$ with probabilities $p$ and $1-p$ respectively.

 

[DrDu]

Of course, but I understood the original question such as if somebody is preparing a system either in state $|1 \rangle \langle 1|$ or $|2 \rangle \langle 2 |$ with probabilities $p$ and $1-p$ respectively.

The OP said: "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 can't say why, but with the EPR paradoxon, Bells inequalties and all that stuff in mind, I am reluctant to make statements about a quantum system being in some definite state assuming is only unknown to me.

 

[bhobba]

No, this is representation dependent. Any state, whether pure or mixed can be represented as |u><u|.

As much as I hate contradicting my fellow science adviser and someone who has a doctorate - I much prefer to listen to such - this is wrong.

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.

See page 53 Ballentine - the following is equivalent to being a pure state. If p is the state it is pure iff trace (p^2) =1 - it is mixed otherwise ie if trace (p^2) <1. Nothing can change that - if trace (p^2) <1 it can't be made to be equal to one. Think of the Born rule and take the state as the observable - Trace (p^2) < 1 will give observational outcomes different statistically to trace (p^2) = 1.

Thanks
Bill

 

[vanhees71]

But that's the operational definition of mixed states, leading to the formal description of the states as a statistical operator.

Where are your specific quibbles concerning the EPR paradoxon, Bell's inequalities and the like?

 

[vanhees71]

But that's the operational definition of mixed states, leading to the formal description of the states a

The OP said: "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."

But that's the operational definition of mixed states.

I can't say why, but with the EPR paradoxon, Bells inequalties and all that stuff in mind, I am reluctant to make statements about a quantum system being in some definite state assuming is only unknown to me.

Where are your specific quibbles concerning the EPR paradoxon, Bell's inequalities and the like?

 

[DrDu]

But that's the operational definition of mixed states, leading to the formal description of the states as a statistical operator.

Where are your specific quibbles concerning the EPR paradoxon, Bell's inequalities and the like?

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.

 

[bhobba]

But that's the operational definition of mixed states.

Indeed.

But if that was the only way to get mixed states all our interpretation issues would be over.

Its of zero importance of course to the Ensemble interpretation - it does't care how the state was formed, but the fact is states formed by that operational definition are not formed the same way as mixed states from decoherence. One in such cases can't say for sure it was in some state from the mixture prior to observation and therein lies the whole argument about if decoherence solves the observation problem or not.

Thanks
Bill

 

[vanhees71]

I don't understand your problems with this interpretation of the mixed state. Usually it's quoted as "the measurement problem", i.e., if you have given a system in some pure state $\hat{R}_0=|\psi \rangle \langle \psi|$ and you measure an observable (for which $|\psi \rangle$ is not an eigenstate), then (assuming you have made an ideal von Neumann filter measurement), before taking note of the outcome, you have to associate the statistical operator
$$\hat{R}_1=\sum_{n} |\langle n \rangle \rangle n| \hat{R}_0|n \rangle \langle n|,$$
where $|n \rangle$ denote a complete orthonormal set of eigenstates of the measured observable, and this is a mixed state. There cannot be a unitary time evolution from $\hat{R}_0$ to $\hat{R}_1$, and that's taken as the "measurement problem" by proponents of the collapse conjecture, where they take this "transition" as a physical process rather than an adaption of the probabilistic description after gaining information (i.e., in this case the knowledge that somebody has done a von Neumann filter measurement on the system, but not taking note of the measured result).

 

[vanhees71]

Indeed.

But if that was the only way to get mixed states all our interpretation issues would be over.

Its of zero importance of course to the Ensemble interpretation - it does't care how the state was formed, but the fact is states formed by that operational definition are not formed the same way as mixed states from decoherence. One in such cases can't say for sure it was in some state from the mixture prior to observation and therein lies the whole argument about if decoherence solves the observation problem or not.

Thanks
Bill

Yes, and that's why I always say that the ensemble interpretation solves all the apparent paradoxa concerning the socalled "measurement problem".

 

[bhobba]

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 think that needs elaboration. Any pure state is an eigenstate of some observable.

As far as EPR goes the particles are entangled and a bit of math shows for such systems if you observe one particle due to that entanglement its not in a pure state - it in fact acts as if its in a mixed state. You can view it as being in one state or the other just prior to measurement. But before measurement who knows. That's what separates QM from the red and green paper example and Bertelmann's socks. It shows in the different statistical properties such states have to ordinary correlations.

Thanks
Bill

 

[bhobba]

Yes, and that's why I always say that the ensemble interpretation solves all the apparent paradoxa concerning the socalled "measurement problem".

Trouble is, with me, after many many years of struggling with what QM means you are preaching to the converted.

Thanks
Bill

 

[stevendaryl]

Yes, and that's why I always say that the ensemble interpretation solves all the apparent paradoxa concerning the so called "measurement problem".

I don't see how the ensemble interpretation addresses the measurement problem. It seems like a non sequitur to me. At best, maybe, it's a matter of just saying that you don't care about the problem. But you don't need an ensemble if you're willing to dismiss the problem. Copenhagen works fine in practice.

 

[vanhees71]

Copenhagen without collapse is fine, and the statistical (or ensemble) interpretation just takes the states as what they are, probabilities for what to expect when measuring an observable given the (pure or mixed) state of the system. Updating the state after a measurement accoring to its outcome is just updating ones probabilistic description, nothing "in the real world". In a way, you are right, it's just not caring about a "problem" that you have, because you insist on sticking to a classical picture of nature.

 

[Physics Footnotes]

Yes, and that's why I always say that the ensemble interpretation solves all the apparent paradoxa concerning the socalled "measurement problem".

Invoking the so-called 'ensemble interpretation' doesn't solve the measurement problem; it simply expresses your optimism for the direction in which you hope the solution will be found.

 

[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.