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Is it possible to distinguish between pure and mixed states?

  1. Jan 26, 2015 #1
    Is it possible, in principle, for an experiment to distinguish between an ensemble of pure states and an ensemble of mixed states?
    If so, how?
    In particular, I am thinking of an ensemble of particles whose spin has been measured, one at a time, on the "Vertical" axis. The ensemble consists of an equal number of spin-up particles and spin-down particles. Thus there should be no entanglement. This is the ensemble of pure states.
    The other ensemble has been prepared so that all particles are in a mixed state.
    Would a black hole consisting of one ensemble be different in any way from a black hole formed from the second ensemble?
     
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  3. Jan 26, 2015 #2

    kith

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    You can't distinguish between a pure and a mixed state by using only a single observable (or axis in your example).

    You can if you perform measurements wrt to different axes. For a pure state, you will find an axis where all members of the ensemble are spin up. For a mixed state, you can't find such an axis. For a maximally mixed state, the amount of spin up doesn't depend on the axis at all.
     
  4. Jan 26, 2015 #3
    You're kidding. Why does everyone want to make them all Up? I said it is an ensemble consisting of an equal number of pure Up and pure Down.
    And you can measure them any way you like, I just want to know if you can get different results in SOME conceivable experiment.
     
  5. Jan 26, 2015 #4

    kith

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    Maybe I misunderstood you. So in the first case, you have an ensemble which is formed by mixing two ensembles which correspond to pure states of spin up and spin down wrt to a certain axis. In the second case, you have a ensemble of particles in a mixed state. If this mixed state is maximal (i.e. if the probabilities for spin up and down are equal), you have the same density matrix in both cases and there's no way to distinguish them experimentally.

    If in the second case, the state is not maximally mixed, you can use the method of my previous post to distinguish between the two ensembles.
     
  6. Jan 26, 2015 #5
    According to John Baez, each particle in the pure state is described by a wavefunction, not a density matrix:
    http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/info_loss.html
    "Take a quantum system in a pure state and throw it into a black hole. Wait for some amount of time until the hole has evaporated enough to return to its mass previous to throwing anything in. What we start with is a pure state and a black hole of mass M. What we end up with is a thermal state and a black hole of mass M. We have found a process (apparently) that converts a pure state into a thermal state. But, and here's the kicker, a thermal state is a MIXED state (described quantum mechanically by a density matrix rather than a wave function). In transforming between a mixed state and a pure state, one must throw away information. For instance, in our example we took a state described by a set of eigenvalues and coefficients, a large set of numbers, and transformed it into a state described by temperature, one number. All the other structure of the state was lost in the transformation."

    And the first ensemble has never necessarily been separated into two other ensembles. It might have been a mixed-state ensemble each of whose particles was separately measured on the vertical axis and passed on, in the preparation process.
     
  7. Jan 26, 2015 #6

    kith

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    That's interchangable. The density matrix formalism is more general than the state vector / wavefunction formalism. For a pure state, you can use both but for a mixed state you have to use the former.

    What John Baez writes is in accordance with my first post, so you should try to understand what I wrote. Your posts give off an argumentative vibe to me. I'm happy to assist if you want to learn but I'm not interested in arguing with you.
     
  8. Jan 26, 2015 #7

    bhobba

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    An ensemble of mixed states is an ensemble of pure states.

    A mixed state is of the form Σpi |ui><ui|. Take an ensemble of such states and you end up with the same form.

    Thanks
    Bill
     
  9. Jan 26, 2015 #8
    So are you saying that we can always refer to an ensemble of mixed states as consisting of pure states, which have already been measured, even though the mixed states have not been measured?
    John Baez seems to be saying that the ensemble of pure states has more information than the ensemble of mixed states.
     
  10. Jan 26, 2015 #9
    Does that mean that the information loss that John Baez ascribes to the difference between the density matrix and the wavefunction does not occur, and therefore there is no Information Paradox?
     
  11. Jan 26, 2015 #10

    PeterDonis

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    It's an ensemble of pure states, but that's not the same as the entire system, consisting of all the particles in the ensemble, being in a pure state.

    How? You'll need to be more specific as to this preparation process.

    We don't have a theory of quantum gravity so this question can't be answered. (I know you're interested in the black hole information paradox, but as I said in the thread in the relativity forum that spawned this one, we don't have an answer because we don't have a theory of quantum gravity, so there's no point in posing questions that assume there is a known answer on the issue.) This thread should stay focused on the question given in its title.
     
    Last edited: Jan 26, 2015
  12. Jan 26, 2015 #11

    PeterDonis

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    He's using "Up" in a different sense than you are; what he means is that, for a pure state, you will be able to find some direction in space (which will not, in your example, be the direction you are calling "Up") along which you can orient the measuring apparatus so that the measurement will always give the same result. For a mixed state, you can't.
     
  13. Jan 26, 2015 #12
    Do you mean that, on some axis, an Up particle will give the same result as a Down particle? If that were true, then it would be easy to distinguish the two ensembles. But some say they have the same density matrix.
     
  14. Jan 26, 2015 #13

    PeterDonis

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    Do you know the actual technical definitions of these terms? Have you worked through a basic textbook on QM where they are discussed? If not, I would strongly suggest doing so.

    That said, you are missing a basic distinction, which I mentioned in a previous post, between the state of individual constituents of a system, and the state of the system as a whole, including all the constituents. Note carefully that Baez, in what you quoted, says "Take a quantum system and throw it into a black hole". He does not say "Take an ensemble of quantum particles and throw them into a black hole". When he says the system starts out in a pure state, he means the system as a whole; he does not mean each of the individual particles composing the system. The individual particles may have no well-defined "state" at all (this will be the case whenever the state of the system as a whole is not a simple product state, which will be true of most quantum systems).
     
  15. Jan 26, 2015 #14
    "What we start with is a pure state and a black hole of mass M. What we end up with is a thermal state and a black hole of mass M."
    Or, for luminous matter, an oven.
     
  16. Jan 26, 2015 #15

    PeterDonis

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    As I said in the previous thread in the relativity forum: Baez' article is a statement of the problem, not a statement of the solution. You cannot help yourself to his scenario as a preparation process. Anyway, as I said in a previous post, this thread should stay focused on the title question, without bringing in the black hole information paradox. So we need a mundane preparation process for a mixed state that doesn't involve a black hole.

    How would the (thermal) mixed state of the particles when they come out of the oven be described? How will the oven affect their spins?
     
  17. Jan 26, 2015 #16
    I see. So even if you measure each particle in an ensemble, the ensemble is in a indeterminate state, although each of the particles has a known exact property. I presume this holds true for an ensemble of two particles?
     
  18. Jan 26, 2015 #17

    kith

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    No.

    His argument is that the initial state is a pure state and the final state is a statistical mixture of pure states. A pure state is a state of maximal knowledge because you can't obtain more information about your system by performing measurements without forgetting something. If you have a statistical mixture, you can obtain more information by performing a measurement, namely which pure state the system actually is in. So if you start with a pure state and end up with a mixed state, you have no longer a state of maximal knowledge and in this sense you have lost information.

    As PeterDonis writes, the question whether this is a real loss of information or an artifact of the missing quantum gravity description cannot be decided.
     
  19. Jan 26, 2015 #18
    I should think that each one's spin would be indeterminate. Or can we also speak of those spins as being exact also? I have been told here that we cannot speak of a particle having an exact property until it is measured (in a fundamental way, not merely unknown to that particular experimenter). Thus I should think that the measured particles have exact spins and the others... don't.
     
  20. Jan 26, 2015 #19
    So it would not be unreasonable to say that each particle in the thermal mixture emitted by the black hole has an exact pure spin, even before it is measured?
    And neutrinos emitted in Hawking radiation in such a statistical mixture, is each one of them a pure states?
     
  21. Jan 26, 2015 #20

    kith

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    It doesn't lead to contradictions but given only the density matrix there are many inequivalent ways to assign pure states.
     
  22. Jan 26, 2015 #21

    Nugatory

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    So the preparation procedure for the members of that ensemble is to pass them through a vertically oriented SG apparatus? So far, so good.
    This preparation procedure produces a mixed state. Half the particles are aligned up along the measurement axis and half are aligned down, and if we pick any one of them it's 50/50 which we'll get.
    (Did you mean to say "superposition of up and down states" instead of "entanglement"? "Entanglement" makes no sense in this context.)
    Prepared how? You've described the preparation procedure you used for the first ensemble. What's the procedure you're using for this second ensemble? We need to know that before we can tell you how measurements on it will turn out.
     
  23. Jan 26, 2015 #22
    The (possibly alien, in a former scenario) scientist who measured the spins knows which ones are Up and which ones are Down. You are saying that the other scientist just doesn't have that knowledge; but the information exists. Is that, experimentally, exactly the same as an ensemble of particles of totally indeterminate spin?
    Is this just semantics? It seems more fundamental than that.
     
  24. Jan 26, 2015 #23
    I wanted to exclude the suggestion made earlier that entanglement might provide a way to distinguish the two ensembles. Entanglement made sense then, I am pleased that it no longer makes sense for this scenario.
     
  25. Jan 26, 2015 #24

    bhobba

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    Ok - lets start from scratch.

    By definition a state is a positive operator of unit trace. A pure state, by definition is a state of the form |u><u|. A mixed state, by definition, is a state of the form ∑pi |ui><ui| ie the convex sum of pure states. It can be shown all states are either mixed or pure. It easy to see a mixed state is an ensemble of pure states where the pi is the probability of a random element of the ensemble being the pure state |ui><ui|. Now if you take an ensemble of mixed states obviously you end up with another mixed state. If pk is the probability of mixed state k = ∑pik |uik><uik| in that ensemble the probability of state |uik><uik| in the ensemble is pk*pik so the overall mixed state is Σpk*pik |uik><uik| where the sum is over i and k.

    However your question in the title is if its possible to tell the difference between mixed and pure states. Of course it is - in theory from a large ensemble of systems prepared that same way to be in the same pure state, by observing each system to see if its in that state, providing you get a yes on each observation and the ensemble is very large you can say what the state is. If it was a mixed state you would not get a yes on each observation. Or rather its a vanishingly small probability of that being the case - but that probability can be made as small as desired because the ensemble can be made as large as desired.

    Thanks
    Bill
     
    Last edited: Jan 26, 2015
  26. Jan 26, 2015 #25

    PeterDonis

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    That's not at all what I said. Once again, have you actually looked at the technical definitions of these terms in a textbook? The questions you are asking indicate that you do not understand the terms you are using very well.

    If you measure each particle in an ensemble, then the ensemble's state is just a product state, the product of the measured states of all the particles. This is a perfectly determinate state. (I am assuming that we are only interested in the particles' spins, not any other properties. I am also assuming that the particles are distinguishable, which just means we are able to keep track of which particle is which. This isn't really a good assumption in general, but for this discussion I don't think we want to bring in the added complication of quantum statistics.)

    Why? Suppose each particle's spin was measured before the particle was put in the oven. Does the oven change the particles' spins? (Assume there is no magnetic field or anything like that in the oven.)

    But if you prepare a particle in a certain state (as in your scenario, where you specified that half the particles were prepared as Up and the other half as Down), then you have measured it. Preparation and measurement are the same thing. As long as you don't subject the particle to some interaction that affects the prepared state (for example, measuring its spin in a direction different from Up-Down), it will stay in that state.

    It obviously can't if you specify a scenario in which there is no possibility of entanglement. But that just means you picked a special scenario; it doesn't mean entanglement is not relevant to the general question you are asking.
     
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