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Is the cat alive, dead, both or unknown

  1. Jun 17, 2015 #1
    my girlfriend is wondering if while shroodingers cat was in the box, we should consider it as both unknown rather than both alive and dead. please shed some light on the subject.
     
  2. jcsd
  3. Jun 17, 2015 #2
    Unknown, or more technically, in the superposition of two states.
     
  4. Jun 17, 2015 #3

    Nugatory

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    You'll find a bunch of threads on Schrodinger's cat here if you search. The mainstream answer is also the commonsense one: The cat is either alive or dead, just as a tossed coin is either heads or tails even before we look.

    I should add that that's pretty much always been the standard answer. When Schrodinger posed his famous though experiment almost a century ago, it wasn't because he or anyone else was seriously suggesting that the cat would somehow be "both alive and dead" or "in a superposition of alive and dead". He was pointing out a problem in the then-current understanding of quantum mechanics: nothing in the mathematical formalism clearly explained why the cat would be either dead or alive but not something in between.
     
  5. Jun 17, 2015 #4

    atyy

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    Unknown. It "is" in a superposition of two states, but we don't now what that means. Quantum mechanics is not about what "is", but about what we can predict about what we will observe.
     
  6. Jun 17, 2015 #5
    Unknown, but we DO know the probability of the cat being either alive or dead.
    The thought experiment involves a radioactive atom, which when it decays triggers the mechanism which kills the cat.
    The atom type has a known half life, and after this amount of time has passed the probability of the cat being alive or dead is 1:1.
    If the cat is left in the box for a longer time the probability of it being dead increases and given sufficient time it become a near certainty.
     
  7. Jun 17, 2015 #6

    OCR

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    So never, never, put a dead cat in "the box", at the beginning... such an action would have a probability of

    creating fundamental disturbances in the Force, or at the vary least, a minor "tremor"... :oldwink:
     
  8. Jun 17, 2015 #7

    Nugatory

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    "Unknown" and "in the superposition of two states" are not the same thing.

    If toss a coin, it's either heads or tails but I don't know which until I look at it. That's "unknown".

    A spin-1/2 particle with its spin aligned up along the z axis is not in an unknown state. It's in the up eigenstate of the ##S_z## operator and that's a complete and unambiguous specification of its state that leaves nothing unknown. However, that state is also a superposition of spin-up along the x axis and spin-down along the x axis.
     
  9. Jun 17, 2015 #8

    bhobba

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    Indeed.

    Technically its the difference between a superposition and a mixed state. If you don't know the difference look it up.

    Thanks
    Bill
     
  10. Jun 17, 2015 #9

    Nugatory

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    What Bhobba said.
    You have to use the density matrix formalism to see the difference. Googling for "quantum density matrix" will find some good links - it's unfortunate that this is not covered by some intro textbooks.
     
  11. Jun 17, 2015 #10
    Thanks for the correction.
     
  12. Jun 17, 2015 #11

    atyy

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    But since this is the density matrix before the measurement outcome, it could be an improper mixture, which is a superposition.
     
  13. Jun 17, 2015 #12

    bhobba

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    Improper mixtures are not superpositions from the very definition of a mixture. Its called improper because its physical origin is different to a proper one - not that its not a mixture - which it obviously is.

    Thanks
    Bill
     
  14. Jun 17, 2015 #13

    atyy

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    Improper mixtures are superpositions which is why they are not proper. A superposition refers to a pure state, which is what an improper mixture is.
     
  15. Jun 18, 2015 #14

    bhobba

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    An improper mixture is NOT a superposition.

    Outside the system it remains in superposition - inside it isn't. That is the key difference - see the section 1.2.3:
    http://philsci-archive.pitt.edu/5439/1/Decoherence_Essay_arXiv_version.pdf

    It is removing system B from our control by tracing over the environment that does it.

    Thanks
    Bill
     
  16. Jun 18, 2015 #15

    atyy

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    There is no difference because the entire system is still in a superposition. Everything you do on the improper mixture has a counterpart in the full system.
     
  17. Jun 18, 2015 #16
    Bernard d'Espagnat makes the distinction between an improper mixture v a proper mixture in his book "On Physics and Philosophy". atyy is correct in stating an improper mixture refers to a superposition (pure) state.
     
  18. Jun 18, 2015 #17

    bhobba

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    Of course. But we are talking about inside the system. That's what tracing over the environment does.

    Thanks
    Bill
     
  19. Jun 18, 2015 #18

    bhobba

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    It cant - by the definition of a mixed state. You have traced over the environment to do it.

    Thanks
    Bill
     
  20. Jun 18, 2015 #19

    bhobba

    Staff: Mentor

    Just to expand further on the issue, Lubos has done an excellent post giving the detail - if I was to write something it would basically be what he wrote:
    http://physics.stackexchange.com/qu...ake-the-partial-trace-to-describe-a-subsystem.

    What's going on is you have the whole system A+B - that will be assumed to be a pure state |p><p| - here I have used the operator description of a pure state to avoid confusion later. The mixed state comes about because we are only interested in observations on A. But its entangled with B. Its that entanglement that leads to it being a mixed state because you need to do a partial trace over B due to only A being observed. It is this only observing A when it entangled with B that's the cause ie you are observing inside the system ie the state is (trace over B |p><p|) - which, lo and behold, turns out to be a mixed state - see equation 1.2.3 in the paper I linked to before. Outside the system its a pure state |p><p|, hence a superposition of all sorts of things, and remains in superposition, until, of course, it becomes entangled with something else. But because of the partial trace inside the system, since you are observing only A, its a mixed state.

    Thanks
    Bill
     
    Last edited: Jun 18, 2015
  21. Jun 18, 2015 #20
    This thread is not about the radioactive atom, but about the cat which will be killed when that atom decays. The question is if the special state that the atom is in according to the theory, necessarily affects the cat because we don't know about it. It was Schrödinger's intention to ridicule that idea by means of his cat example.
    - https://en.wikipedia.org/wiki/Schrödinger's_cat#The_thought_experiment
     
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