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In QM, can the trajectory in the phase-space split?

  1. Oct 28, 2014 #1

    ORF

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    Hello.

    I was told that "in QM the information doesn't disappear". Does it mean that, in QM, the trajectory in the phase-space can't split? [The typical example is a vertical simple pendulum, with the mass above the spin-point; in classical mechanics you can't know if the pendulum will fall one side or another.]

    If this question is already answered in this forum, just tell me, and I will delete this thread.

    Thank you for your time :)

    Greetings
    PS: My mother language is not English, so I'll be glad if you correct any mistake.
     
  2. jcsd
  3. Oct 28, 2014 #2

    atyy

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    In the classical case you bring up, a small change in initial condition leads to a large change in final condition.

    In the quantum case, I believe this cannot happen, because the time evolution of the quantum state is linear. This only refers to the time evolution of the state between measurements, and not to measurement outcomes, which are random and whose probabilities are given by the Born rule. However, a quick google brings up this interesting discussion:

    http://arxiv.org/abs/quant-ph/9802047
    http://dx.doi.org/10.1103/PhysRevA.45.R555
    Quantum state sensitivity to initial conditions
    Gonzalo Garcia de Polavieja
    (Submitted on 17 Feb 1998)
    The different time-dependent distances of two arbitrarily close quantum or classical-statistical states to a third fixed state are shown to imply an experimentally relevant n
    otion of state sensitivity to initial conditions. A quantitative classification scheme of quantum states by their sensitivity and instability in state space is given that reduces to the one performed by classical-mechanical Lyapunov exponents in the classical limit.
     
    Last edited: Oct 28, 2014
  4. Oct 29, 2014 #3

    ORF

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    Hello

    Thank you for your interest (I have not access to the Physical Review A; I looked for similar papers, but with no results). Well, I suppose the time evolution of the state will depend on the Hamiltonian that you consider :)

    Let's going to remake the question: can we build a Hamiltonian which cause a no-linear time evolution?

    Greetings.
     
  5. Oct 29, 2014 #4

    atyy

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    There are two links - I think the arXiv link should be free?

    The time evolution (between measurements) in quantum mechanics is always linear, no matter what the Hamiltonian is.
     
  6. Oct 29, 2014 #5

    ORF

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    Yes, the arXiv link is free, thank you for it :)

    Ouch, such a lapse of memory. Thank you, you are right: in the "Schrödinger picture", the operators are constant, and the states evolve.

    But in the "Dirac picture" (or "interaction picture"), operators and states have a time dependence :)

    Thank you for your interest.

    Greetings
     
  7. Oct 29, 2014 #6

    atyy

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    I found some more discussions about the relationship between quantum systems and classical systems that are sensitively dependent on the initial condition.

    http://arxiv.org/abs/math-ph/0503032v2
    Quantum Chaos: Spectral Analysis of Floquet Operators
    James Matthew McCaw

    http://xxx.lanl.gov/abs/quant-ph/9906092
    Continuous Quantum Measurement and the Emergence of Classical Chaos
    Tanmoy Bhattacharya, Salman Habib, Kurt Jacobs


    The thesis by James McCaw has an interesting overview of the literature. One things he mentions is that in the Bohmian formulation, particles do have trajectories, so one can ask whether Bohmian particles exhibit sensitive dependence on their initial conditions.
     
  8. Oct 30, 2014 #7

    ORF

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    Hello

    I only gave the McCaw's work a glance, but it looks a very good starting point (I didn't know anything about quantum chaos when I did the question).

    The question seemed to me simple, but finding an answer is harder than I thought :)

    Thank you for the links, I am going to read them right now.

    Greetings!
     
  9. Oct 30, 2014 #8

    naima

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    I am interested too.
    Let us take a quadripartite system. It is in an initial pure state say 0110>
    each particle has a complete local information. There is an unitary evolution operator U(t) which act on the state. Local information may disappear. But ## U^{-1}## will give it back
    where is encoded the information at time t?
     
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