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How to prepare pure state?

  1. Jul 9, 2009 #1

    KFC

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    After reading some introudction of quantum states (pure and mixed), I know that states produced statistically are called mixed state. My question is: since there always exists stochastic process, so in practical system, how can we produce "pure" state?
     
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  3. Jul 9, 2009 #2

    JK423

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    Very cold hydrogen atoms should all be in the ground state. Thats a pure state. All of your atoms are in the same state
     
  4. Jul 9, 2009 #3

    KFC

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    But why only ground state is pure state?
     
  5. Jul 9, 2009 #4

    Pythagorean

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    It's not the only pure state, and it doesn't necessarily have to be quantum particles. A pure state is an abstract idea that depends on your definitions and your detection scheme.

    If you send a beam of polarized light through a filter that's aligned with the beam's direction, you'll get the whole beam on the other side. You measure the same beam intensity on both sides, because the beam is a pure state.

    However, if you throw a mixture of polarizations into the apparatus, only the portion of the beam that corresponds to the test apparatus' polarizations will make it through. You will have lost all the information about the rest of the beam. The intensity measured coming out of the apparatus will be less than what went in. This beam was a mixed state.

    (note, the beam coming out of the apparatus is of course a pure state now based on our detection scheme, but the beam going in was mixed. The apparatus basically filtered out the pure state and threw the rest away.)
     
    Last edited: Jul 9, 2009
  6. Jul 9, 2009 #5

    KFC

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    For the second example, can I say the beam after the polarizer (i.e. the part only contains particles have same polarization) is pure state?

     
  7. Jul 9, 2009 #6

    Pythagorean

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    yes, I mentioned this in my last paragraph
     
  8. Jul 9, 2009 #7

    KFC

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    I still quite confuse about the difference of mixed state and pure state. Since in the textbook, there is a criterion to determine if the state is pure or mixed. For pure state, the trace of square of density matrix is one while for mixed state it is less than 1. But if I have my initial state go through a blackbox and how do I know if the outcome is pure or mixed? What I am asking is, how can I find out the density matrix experimentally?
     
  9. Jul 9, 2009 #8

    Pythagorean

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    In my experiments (I'm not doing quantum, but I use pure states and density matrices) what we measure experimentally is the state vector.

    if your state vector is A, then the density matrix is the outer product of A with itself:

    S = |A><A|

    This doesn't tell you whether it's pure or mixed though. That depends on how you define your system, and how your apparatus measures things.
     
  10. Jul 9, 2009 #9

    Dale

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    It sounds like "pure state" just means "eigenstate". Is there some subtlety I am missing?
     
  11. Jul 9, 2009 #10

    JK423

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    We have a pure state if all quantum systems are described by the same state vector |Ψ>. This |Ψ> could be anything, from an eigenstate to a superposition of eigenstates etc
     
  12. Jul 10, 2009 #11

    Fredrik

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    If you're familiar with wavefunctions/state vectors, but not density operators*, then every state you've ever come across is a pure state. There's an obvious bijection between state vectors (e.g. [itex]|\psi\rangle[/itex]) and projection operators of the form [itex]|\psi\rangle\langle\psi|[/itex]. We can use more general projection operators to represent the situation when the system is known to be in a specific state, but it's not known which one. For example, you have just performed a measurement, but you're unaware of the result.

    A projection operator of the form

    [tex]\rho=\sum_i w_i|\psi^{(i)}\rangle\langle\psi^{(i)}|[/tex]

    with [itex]\sum_i w_i=1[/itex] represents an ensemble of systems such that a fraction wi of them are in state [itex]|\psi^{(i)}\rangle[/itex]. This is called a "mixed state". A "pure state" is the special case with all of the wi=0, except one of them, which is =1.

    *) The term "density matrix" seems to be more common, even though it really only makes sense to use that for the matrix representation of the density operator. Ballentine uses the term "statistical operator".
     
  13. Jul 14, 2009 #12
    I would add to this discussion that sometimes one idealized notion cover another. Polarizers without experimental uncertainties, hundred percent efficiency photon detectors, temperature of zero kelvin and plane waves are some examples. In real experimental setup I see as very, very difficult to make sure we have a pure state in hand. We may appeal to FAPP (for all practical purpose) notion to save the day, but we must pay attention to the degree of compromise these idealized assumptions require to make our theoretical approach consistent.

    best regards,

    DaTario
     
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