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Deterministic systems and disproving the schrodinger's cat theory

  1. Aug 23, 2004 #1
    about deterministic systems and disproving the schrodinger's cat theory.

    if your system state is determined by a quantum wave function then the theory states that your wave function collapses as soon as it is observed... if you set up your system so that after a set time period based on the state of the system, the system say releases a particle or doesn't release a particle. the act of the particle being observed will collapse the system, but the trick is if the system doesn't release a particle then according the theory the system will still be in a quantum super position (and half of the time it will have released a particle) and it will there for either defy causality or have collapsed by some other means.

    Is this correct logic, or am I missing something?
    Last edited: Aug 23, 2004
  2. jcsd
  3. Aug 23, 2004 #2
    I do not understand sufficiently to answer.
  4. Aug 23, 2004 #3
    The original theory or my post?
  5. Aug 23, 2004 #4


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    What do you mean by "release a particle"? In Heisenberg's argument, the particle was released (emitted) by a radioactive nucleus and so you have no control over whether a particle is released or not. Heisenberg's point was that quantum changes (emitting a particle) CAN have macro consequences.
  6. Aug 23, 2004 #5


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    May I give you a friendly advise that you first do a search on the "Schrodinger Cat" on this forum? It will save you a lot of "grief" figuring out if what you are asking has already been answered numerous times. It will also help you to understand this further because it appears that you are utterly confused of what a "Schrodinger Cat" is.

    [Hint: it ISN'T a "theory", it is a thought experiment to illustrate the superposition principle of Quantum Mechanics. So you are looking at the tail end of an animal and confusing that it IS the whole animal]

  7. Aug 23, 2004 #6


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    No, this is not correct logic. Suppose that we're talking about an atom in an excited state. Let's call the excited state |e> and the ground state |g>. The system starts out in the state |e> and after a while, it has evolved into a superposition, something like |e> + |g>. This means that there is a 50% probability that a measurement will find that the atom has decayed to the ground state, and a 50% probability that a measurement will find that it is still in the excited state.

    "the system doesn't release a particle" means that the state is |e>.

    "the system will still be in a quantum superposition" means that the state is |e> + |g>.

    These two statements contradict each other.
  8. Aug 24, 2004 #7
    Makes sense, thanks.
  9. Aug 24, 2004 #8
    I think it was the post. Maybe it was just the lack of capital letters. I'm sure you'll be understanding this soon enough. Welcome to the boards BTW, we'll be happy to discuss these things with you.
  10. Aug 26, 2004 #9
    Was thinking 'can't have conflicting super positions which would cause a super position of the super position colapsing' but the word guess floated around in my mind looking for somthing to attach to.
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