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I Actual example of a mixed state not part of entangled state

  1. Jul 23, 2017 #1
    A mixed state is when the system is actually in one state or another, but you just don't know which, and you use probabilities to describe your uncertainty. I'm referring to a mixed state of the entire system. I want an actual example. Can you think of some? Note I wasn't describing mixed state of a subsystem only, obtained by tracing over the rest of the system that aren't measured, with the system as a whole being in a pure state (like in case of decoherence)

    My concern is, how do you know the system is actually in one state even prior to measurement or observation? (ever heard of the Bertelsmann's socks argument in Bell's Theorem?) Mixed state argument is like that of Bertelsmann's socks or properties/state existing even before observation. What's the proof or experimental example of this?
     
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  3. Jul 23, 2017 #2

    Nugatory

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    The preparation procedure determines the initial state; depending on what that procedure is the initial state may be either pure (and therefore a superposition in some basis) or mixed.
    For example:
    1) The beam of particles coming out of one side of a horizontally oriented Stern-Gerlach device has been prepared in the state spin-left, which is a superposition of spin-up and spin-down. A measurement of the vertical spin of these particles will yield spin-up or spin down with equal probability.
    2) If I combine the two beams coming out of the two sides of a vertically oriented Stern-Gerlach device, I will have prepared the output beam in the mixed state "50% spin-up, 50% spin-down". As with the previous case, a measurement of the vertical spin of these particles will yield spin-up or spin-down with equal probability.

    However, the two beams will behave differently if measured on any other axis except the vertical axis. The particles in superposition (case 1 above) will be spin-left on the horizontal axis 100% of the time, and the up/down probability at intermediate angles will be ##\cos^2\frac{\theta}{2}## where ##\theta## is the angle from horizontal. The particles in the mixed state (case 2 above) will be up or down with 50% probability at all angles.

    This is a fairly easy experiment, although in practice we're more likely to do it with polarized photons. Either way, the results are unequivocal.
     
    Last edited: Jul 23, 2017
  4. Jul 23, 2017 #3
    Note there are two kinds of mixed states (of whole systems, let's ignore the mixed state from entangled system for now). The first one is your example where the mixture occurs after or when we measured them. I was referring to another class of mixed state where ''The system is in some pure state, but we don't know which'' or in other words, the pure state has already a result prior to measurement, and we don't know just what it is due to classical ignorance. This is known as proper mixed states. I want some actual example. Your example is not this because before you measure the spin up or spin down.. it doesn't have spin up or spin down before measurements. I'm just perflexed of how pure state can have value before measurement giving rise to this ''The system is in some pure state, but we don't know which".
     
  5. Jul 23, 2017 #4

    Strilanc

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    In the many worlds / universal wavefunction interpretation, all mixed states are actually subsystems of a larger entangled system. A system that was really truly in a probabilistic mix of superpositions is incompatible with many worlds. So the fact that many-worlds is one of the more popular interpretations is a decent hint that there's no such case known; it's an unobservable philosophical distinction.

    However, there is certainly a pragmatic sense in which I can create quantum states that are much closer to the idea of a randomized pure state. It's just that, instead of relying on the inherently random, we rely on the fact that you're not omniscient. I simply go find things whose state has low mutual information w.r.t. the state of your body (your body has a maximum information capacity so really all I have to do is find lots of things). Then I write the states of those things down, run them through a [cryptographic randomness extractor](https://en.wikipedia.org/wiki/Randomness_extractor) to get them nice and mixed, then use what comes out to decide which state to produce. The output state should be unpredictable to you, so you will have to think about it in terms of a mixed state despite it being one specific pure state. (Loophole: what if the state of the entire world is extremely compressible?)
     
  6. Jul 23, 2017 #5
    basically.. can mixed states be produced by unitary evolution from pure states?

    without measurement, how can you put each pure state system is in a different pure state?
     
  7. Jul 23, 2017 #6

    Strilanc

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    You can't produce mixed states with unitary evolution. But you can produce states that factor into two halves, where one half has low mutual information (in a pragmatic sense) with the other half. Call the one half "you" and the other half "some state you want to reason about". By definition you won't know the other half of the state, because that would imply maximum mutual information, and therefore you will have to reason about it as a mixed state.
     
  8. Jul 23, 2017 #7

    Nugatory

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    I must confess that I'm not seeing the distinction you're looking for here. The second preparation procedure I described produces an ensemble of particles each of which is in a proper mixed state, one in which the particle is in a pure state but we don't know which.

    Perhaps if you could write down a density matrix for the mixed state that you're thinking about, so we can see how it differs from the state produced by my hypothetical preparation procedure ? Given that, we can probably come up with a preparation procedure that will produce that state.
     
  9. Jul 24, 2017 #8
    I think you are right as your case 2 sinks in. Pure state has operator in the form |u><u|.
    mixed state is the convex sum of pure states ie ∑ pi |u><u| where ∑ pi = 1.

    Mixed state is still related to pure state and not a classical state which I initially thought so. Some believed that in mixed state, measurement problem is solved, the state you observe is what's there prior to observation, nothing collapses or changes. But its still derived from pure state which still has superposition so the measurement problem still remains (or you believed gone?)
     
  10. Jul 24, 2017 #9

    DrDu

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    Note that also a mixed state can be expressed as |u><u| if you chose your Hilbert space appropriately.
     
  11. Jul 24, 2017 #10

    vanhees71

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    Hm, by definition a projector as the statistical operator is a pure state, or what do you mean with "if you chose your Hilbert space appropriately"?
     
  12. Jul 24, 2017 #11

    DrDu

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    I mean that e.g. with the GNS construction, you can express any state as a vector state, whether pure or mixed.
     
  13. Jul 24, 2017 #12
    This thread stems from the statements of one of our science advisers where he wrote numerous times in numerous threads that in proper mixed states, "the state you observed is what's there prior to observation, nothing collapses or changes, everything is sweet in quantum land" see https://www.physicsforums.com/threa...-states-in-laymens-terms.734987/#post-4642357

    Is Bill right? What he seems to be implying is that in proper mixed states which are ensembles of pure states.. the state is there prior to observation or measurement.. this is wrong.. because the state only occurs after measurement or collapse (or one outcome occurs). A proper mixed state already has born rule applied.. but Bill is implying proper mixed states are there before observation or measurement. Isn't this incorrect? If he is incorrect, then all the threads he wrote about it must be altered. It disturbed my minds for many days. (dear Bill, please defend or clarify your incredibly ambiguous statements). Thank you.
     
  14. Jul 24, 2017 #13

    Nugatory

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    Yes.
    Whatever you man by that, it is incorrect. Perhaps you are misunderstanding that a superposition is always a pure state?

    A mixed state is one that may be any of several different pure states with various probabilities; and depending on your choice of basis these states may themselves be superpositions. If you perform a measurement on a particle whose state is a mixture of ##|\psi_1\rangle## and ##|\psi_2\rangle##, the probability of a given result will be what you get by applying the Born rule to ##|\psi_1\rangle## times the weight of of ##|\psi_1\rangle## in the mixture, plus what you get by applying the Born rule to ##|\psi_2\rangle## times the weight of ##|\psi_2\rangle## in the mixture.... this should feel like just a routine exercise in conditional probability.

    A good exercise would be to write down the density matrix for the two states we prepared in post #2 above, in both the left-right and up-down bases. In one basis, the states that are being mixed are superpositions and in the other they are not.
     
  15. Jul 24, 2017 #14
    It's simple. Born rule applied is collapse. Bill argued no collapses needed in mixed states and everything sweet in quantum land. I don't think it's sweet.
     
  16. Jul 24, 2017 #15
    To elaborate. Mixed states can't be derived from the unitary evolution in pure states. You need born rule applied which is equal to collapse. Hence mixed states only occur after it is measured when born rule is applied. Bill was emphasizing that even before measurement, the mixed states already have definite outcome.
     
  17. Jul 24, 2017 #16
    here's a little math... the density matrix can be decomposed into the form p = ∑ pi |u><u|... After this decomposition, it can mean "The system is in one of the pure states |u>, but we don't know which." However, in one way or another, i assume the pure state |u> has to collapse first to get out of the superposition? yes.. I assume pure state means it is in superposition, hence not a thing in our world yet. To appear in our world, it has to collapse first. Unless you are saying something concrete and already existing can be a pure state?
     
  18. Jul 25, 2017 #17
    After an agonizing day of reading. I think I finally understood. Pure state is simply when there is interference(s) of the possibilities (or probabilities).. while Mixed state is just ensemble of the pure states when you don't know which possibilities of the pure states actually manifesting so it's classical ignorance.. But note that even in mixed state, there is still a pure state inside with interference of the possibilities *within* that pure state. Remember at the end of the day, the proper mixed state still has one pure state selected classically (agree?) So I don't agree with Bill that measurement problem solved for proper mixed state. He being into ensemble interpretation ignores the dynamics but just think wave function only makes sense for ensembles or when you have many copies of the object. And he thinks the interference of possibilities is just tool of probability while we Copenhagenists (or Bohmians or MWI lovers) think interferences of possibilities really occur in actual. Whatever, I think measurement problems also involves knowing the nature of the interferences of possibilities (or probabilities) right? Because Bill considers measurement problem as only pertaining to why the improper mixed states in decoherence becomes proper and this is very limiting.
     
  19. Jul 25, 2017 #18

    Nugatory

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    You will want to be careful with that definition, as it risks misleading you into thinking that there are pure states that are superpositions and pure states that are not superpositions.

    A better definition might be: A pure state is one that can be represented in some basis by a density matrix which is completely zero except for a single element on the main diagonal that is equal to one.
     
  20. Jul 25, 2017 #19

    bhobba

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    That's basically the difference between impure mixed states and pure one - in pure ones you are sure, for some reason or another it definitely is, in impure ones from decohence you aren't. But there is no way, observationally to tell the difference. That is the exact reason decoherence does not solve the measurement problem.
     
  21. Jul 25, 2017 #20

    bhobba

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    Or in other language it can be shown any state is of the form ∑pi|ui><ui|. If it only has one element in that sum ie is <u|u> its pure and u can be mapped to a Hilbert space. The pi are the probabilities you will get |ui><ui| if you did an observation.

    Now you can view it as though prior to observation it was in state |ui><ui|, but only on some cases can you be sure of that. In decoherence you can't.

    Thanks
    Bill
     
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