Mixed states vs pure states - physical POV

[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: "Schrodinger cat exposed to radioactive source and dead" and "Schrodinger cat not exposed to radioactive source and alive". These are eigenstates. Now they are exposed to environmental decohererence, the terms "Schrodinger cat exposed to radioactive source and dead" and "Schrodinger 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: "Schrodinger cat exposed to radioactive source and dead" and "Schrodinger cat not exposed to radioactive source and alive". These are eigenstates. Now they are exposed to environmental decohererence, the terms "Schrodinger cat exposed to radioactive source and dead" and "Schrodinger 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 Schrodinger'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 Schrodinger'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.

 

[stevendaryl]

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"

Absolutely, I can't prove that, and I don't believe it. I believe that there should be a non-schizophrenic formulation of quantum theory, but the current formulation is schizophrenic.

If so, then you'd be right in saying that there is a division of the world in microscopic and macroscopic behavior.

I'm saying that the formalism has this division. I'm not saying that the division is inherent in nature. I believe it's not.

 

[vanhees71]

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.

Physics is defined as describing preparation procedures, measurements, quantitative observations of Nature. All theories of physics are right about this way to investigate what's objectively going on. If you want something else, it's not physics!

 

[vanhees71]

Absolutely, I can't prove that, and I don't believe it. I believe that there should be a non-schizophrenic formulation of quantum theory, but the current formulation is schizophrenic.

This is your very individual idea about what the natural sciences should provide. Nature doesn't care about our feelings and expectations how she should behaves. She is just is she is.

 

[stevendaryl]

This is your very individual idea about what the natural sciences should provide. Nature doesn't care about our feelings and expectations how she should behaves. She is just is she is.

My complaint is not in how nature behaves, but the schizophrenic way that we describe it. I don't believe that nature is schizophrenic.

 

[stevendaryl]

Physics is defined as describing preparation procedures, measurements, quantitative observations of Nature. All theories of physics are right about this way to investigate what's objectively going on. If you want something else, it's not physics!

You're confusing your own philosophy of physics with physics itself.

 

[vanhees71]

It's not my philosophy of physics. It's what physicists do in all kinds of physics labs and theory institutes around the world.