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Decoherence and the number of degrees of freedom

  1. Jun 1, 2006 #1

    hellfire

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    When reading in the web about decoherence especially in popular articles I find very often explanations that point out the fact that the enviroment has a large number of degrees of freedom. It is unclear to me in which extent and in which aspects this is relevant for decoherence.

    My understanding is that when one has a system A in a pure state that interacts with another system B, leading to an entangled state between A and B, the reduced density matrix of the subsystem A will always describe a mixed state, provided that the states of the subsystem B are orthonormal.

    If this is correct, the fact that decoherence arises is independent of the number of degrees of freedom of B.
     
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  3. Jun 1, 2006 #2
    My intuitive answer(though haven't worked this out) is:-cancellations won't occur if there aren't large number of degrees of freedom i.e. the off-diagonal elements won't go to zero.
     
    Last edited: Jun 1, 2006
  4. Jun 1, 2006 #3

    vanesch

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    What you write is correct of course, but you also want "this orthogonality to last" through time evolution. When the system B has a few degrees of freedom, the unitary evolution of both together might very well "unentangle" (partially) the two systems again. It is only when the number of degrees of freedom is very large that this is very unlikely to happen.
    (at least, that's how understand this issue).
     
  5. Jun 1, 2006 #4
    Vanesch,if the system remains entangled(say) then what hellfire says is right--right?I thought the following :-If the system with a large number of degrees of freedom happens to unentangle,then what I say in post #2 should take over i.e. cause vanishing of off-diagonal elements of reduced density matrix.But I could be wrong:-my knowledge of decoherence is limited to what I've read in these forums.
     
    Last edited: Jun 1, 2006
  6. Jun 1, 2006 #5

    hellfire

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    If A and B are entangled their base states cannot be separated. In the Schrödinger picture, for example, base states remain stationary, so how can it happen that two entangled systems disentangle with unitary time evolution?
     
  7. Jun 2, 2006 #6

    vanesch

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    If the hamiltonian is H = H1 + H2, where H1 is acting upon system A only, and H2 is acting on system 2 only, then there is no way, in the first place, to go from a product state to an entangled state.

    So even in order to produce your entangled state, you'd need to add a term H12 which makes the two systems interact.
    It is then not excluded that this H12 will "disentangle" them in the same way. In fact, this is unavoidable if you have a very low number of degrees of freedom in a finite volume (a kind of "Poincare recurrence").

    In other words, *if* the two systems can interact so that their states entangle, there must be a term H12, and hence the individual base states of the separate systems are NOT stationary states anymore.
     
  8. Jun 2, 2006 #7

    vanesch

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    Saying that the system is entirely entangled is mathematically equivalent to saying that the off-diagonal matrix elements of the reduced density matrix are 0, if I'm not mistaking.
    But the entire question is: will this REMAIN so ?
    And this is, if my understanding of the issue is correct, where the large number of degrees of freedom comes in.
     
  9. Jun 2, 2006 #8

    hellfire

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    I had suspected that the answer might have been about an interaction term in the Hamiltonian but it was not clear to me.

    Thank you, this completely answers my question.
     
  10. Jun 2, 2006 #9
    I thought the sequence was this:-off -diagonal elements go to zero for a system(with interactions) with a large no. of degrees of freedom easily [tex]\implies [/tex] entangled state follows,where the entangled state remains entangled over time.A small no. of degrees of freedom [tex]\implies [/tex] less chances of the entangled state to form &. even if it formed it is unstable i.e. disentangles easily.
     
  11. Jun 2, 2006 #10

    vanesch

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    Yes, that's also what I thought was the idea...
     
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