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Pure, proper mixed, and improper mixed states in laymen's terms

  1. Jan 26, 2014 #1
    This is how one poster tried to explain it to me but for people who have only taken a basic physics course in college it leaves alot wanting.

    "If a system is in a pure state, and you know what the pure state is, then your knowledge of the system is complete, and all uncertainty is quantum. If we take a state to apply to an ensemble, this means that every member of the ensemble has been identically prepared and is in the same state."
    I interpret this as a controlled experiment in a lab where all values are accounted for and a photon is fired which has collapsed the particle. Here there is no uncertianty about actual collapse because the enviorment has been purposflley taken out of the equation yet collapse occurs.

    "A proper mixed state means that you do not know exactly what the quantum state is, but only what the state is with some probability, so uncertainty is due to intrinsic quantum uncertainty, as well as your ignorance of the state. In an ensemble, this means that not all members of the ensemble have been identically prepared."
    This is also conducted in a controlled lab where the enviorment has been taken out of the equation and a photon is not fired at the particle and superposition is apparent.

    "An improper mixed state comes about when the entire system is in a pure state, but you restrict yourself to observing a subsystem. The improper mixed state describes the behaviour of the subsystem."
    I assume the subsystem is the enviorment and when the enviorment is not isolated in a vaccum it acts like a pure state where superposition isnt present and the enviorment has apparently collapsed but since the enviorment has not been taken out of this equation in fact it is the core of an improper mixed state it is impossible to know if collapse has actually occured because the variables are not controlled. It is assumed the enviorment has collapsed the system by interacting with itself but it is uncertian because the variables are not controlled.


    Im sure Im wrong about alot of this so thats why Im asking for help in figuring this out. Any help would be much aprecciated.
     
  2. jcsd
  3. Jan 26, 2014 #2

    atyy

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    In a pure state, one has many systems all prepared in the same pure state, and there is in principle, a measurement that when performed on each system will give exactly the same result.

    In a proper mixed state, each system is in a different pure state, so there is no measurement that will give exactly the same result when performed on each system.

    In an improper mixed state, each system is in the same pure state, so there is in principle a measurement that when performed on each system will give exactly the same result. The improper mixed state is the state of a subsystem, and the uncertainty is due to different subsystems being entangled, and to performing a measurement on a subsystem rather than the whole system.
     
  4. Jan 26, 2014 #3
    By system you mean the meusuring device is a system, the photon is a system and the particle is also a system. Is this correct. What is a system in this case.

    What is a subsystem exactly and what is the whole system.
     
  5. Jan 26, 2014 #4

    atyy

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    No I just mean the quantum system. If the quantum system has two spins, then each spin would be a subsystem. I consider the measuring apparatus classical.
     
  6. Jan 26, 2014 #5

    bhobba

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    Mate you have run into one of the problems of physics.

    Some things require the full technical detail to explain, and that includes the entire mathematical machinery.

    Using math its a doodle to explain those things - without it rather difficult.

    Here is the mathematical explanation.

    By definition a quantum state is a positive operator of unit trace. By definition a pure state is such an operator of the the form |u><u|. A mixed state is the convex sum of pure states ie ∑pi |bi><bi| where ∑pi =1. It can be shown all states are either pure or mixed. For filtering type observations (ie ones where the system is not destroyed by the observation) the outcome is a pure state and is in fact the so called collapse postulate - but that's a bit of a misnomer because it can be derived from other considerations such as physical continuity. If you observe a system whose state is already in one those possible outcomes then nothing happens, no collapse - nothing - QM is easy. One way to physically get a mixed state is to randomly present pure states to be observed, the pi representing the probability that state has been selected for observation. Suppose the |bi><bi| are the outcomes of your observation. For such a situation measurement problem solved - the state you observe is whats there prior to observation, nothing collapses or changes, and everything is sweet in quantum land. Such states are called proper mixed states. However what decoherence does is transform a superposition into a mixed state. Its mathematically and observationally exactly the same as a proper mixed state, but because it wasn't prepared the same way as a proper state so is called an improper state.

    This is the crux and the essence of decoherence and the measurement problem. If it was a proper mixed state measurement problem solved. But it isn't - it wasn't prepared the same way. However its impossible to tell the difference. It APPARENTLY has solved the measurement problem - it hasn't really - but without any kind of contradiction you can assume it's a proper mixed state and collapse has already occurred. It hasn't, but it looks like it has, smells like it has, and in every way acts like it has. It is the basis of the ignorance ensemble interpretation I hold to.

    That's the math. Explaining it without that I simply can't do. The jig is up - you need to study the math.

    I recommend - Ballentine - Quantum Mechanics - A Modern Development.

    Thanks
    Bill
     
  7. Jan 27, 2014 #6
    so a pure state there is no collapse whereas mixed states do exhibt collapse. Is this correct.
     
  8. Jan 27, 2014 #7

    bhobba

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    No.

    Pure states are the states you read about in the populist publications and beginning textbooks. They are the states the principle of superposition applies to and implies they form a vector space. But states are more general than that. First, strictly speaking they are not elements of a vector space - they are the projection operators formed from those elements ie if |u> is the state in the vector space sense the actual state is the operator |u><u|. This is also the reason why multiplying the state by a phase factor does nothing - it gives exactly the same operator.

    However states are more general than this and also includes convex sums of such states ie operators of the form ∑pi |bi><bi|. These are the mixed states.

    What decoherence does is transform a superposition into a mixed state, and that state LOOKS like collapse has occurred. Decoherence explains APPARENT collapse - not actual collapse.

    I realize the above is probably gobbly-gook because you don't have the required physical and mathematical background.

    However the jig is up here - I simply cant explain it any other way.

    If you want to understand it you must learn the technical detail.

    Thanks
    Bill
     
  9. Jan 27, 2014 #8
    ok thank you anways
     
  10. Jan 27, 2014 #9

    atyy

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    Let's consider a system of one spin.

    A pure state means that we have prepared many copies of the system of one spin, and each copy of the system is in the same pure state. For example, every copy of the single spin is pointing up.

    A proper mixed state means that we have prepared many copies of the system of one spin, and each copy of the system is in a different pure state. For example, some copies of the single spin are pointing up and some are pointing down.

    To illustrate an improper mixture, we have to consider a system of two spins. The system is in a pure state, which means we have many copies of the system of two spins, and each copy of the system of two spins is in the same pure state. If the pure state is one in which the two spins are entangled, then the state of a subsystem of one spin is an improper mixed state. We call the state of the subsystem a mixed state, because the subsystem in every copy behaves as if it is a proper mixed state. However, it is an improper mixed state, because if we consider the system of two spins, it is pure. So while we cannot distinguish between the proper and improper mixed state by looking at a subsystem of one spin, we can if we look at the entire system of two spins.
     
    Last edited: Jan 27, 2014
  11. Jan 27, 2014 #10
    is this even possible to look at the whole system , is there evidence that it is infact noticbley diffrent when you look at the whole system or is this just a theory
     
  12. Jan 27, 2014 #11

    atyy

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    In simple cases such as a system of two spins, it is possible to look at the whole system.

    In complex cases, when the system includes a complex environment, it is just a theory. This practical inability to look at the whole system in such cases is why decoherence is said to be "apparent collapse".

    In real collapse, the system evolves from a pure state to a proper mixed state.

    In decoherence, the system consisting of environment + experiment is in a pure state and does not collapse. Here the experiment is a subsystem. Because we can only examine the experiment and not the whole system, the experiment through getting entangled with the environment will evolve from a pure state into an improper mixed state. Since the improper mixed state looks like a proper mixed state that results from collapse as long as we don't look at the whole system, decoherence is said to be apparent collapse.
     
    Last edited: Jan 27, 2014
  13. Jan 27, 2014 #12
    so how do you explain the wolrd of classical physics I mean the real world if decoherance doesnt cause actual collapse why cant I walk through walls.
     
  14. Jan 27, 2014 #13

    atyy

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    In textbook quantum mechanics with collapse, quantum mechanics does not explain the world of classical physics. When we do quantum mechanics, we always divide the universe into classical and quantum realms. Basically, you and your measuring apparatus are classical, and the measurement results you get when the apparatus probes a quantum system are also classical. Quantum mechanics assumes that a classical world exists, and provides a way to calculate the probabilties for various classical results that you measure. Collapse is what happens to the quantum system when the classical measuring apparatus interacts with the quantum system and you get a classical result.

    There are interpretations without collapse, such as de Broglie-Bohm theory and many-worlds. You can look at those for an explanation of why the classical world exists. But the standard textbook interpretation with collapse doesn't explain the classical world, and assumes its existence.
     
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