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Decoherence And What happens in reality

  1. Mar 12, 2013 #1
    I know that the theoretical decoherence mechanism makes the density mattrix diagonal, but my question is if that happens in real life. I mean, lets take the hidrogen atom (lets work in the QM framework And not in the QFT framework in order to have multiple stationary states) And suppose that the electron is in a superposition of energy/stationary states. In this case, does in real life the electron, through the interaction with the environment, spontaneously "choose" one of those stationary states or it remains in a superposition of energy states?

    Thanks
     
    Last edited: Mar 13, 2013
  2. jcsd
  3. Apr 2, 2013 #2
    I think we do not know, there are theories but we lack experimental proof.
     
  4. Apr 2, 2013 #3

    f95toli

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    I don't think a hydrogen atom is a good example. An artifical qubit would be a better system to study since it then clear what the "classical" state would be (this is never obvious for an atom).

    In the latter case then yes, that is pretty much what happens. Due to decoherence, a system that is initially in a superposition of states will end up permanently in ONE of those states.

    There are experiments where people have traced out the "trajectory" of a qubit on a Bloch sphere, and this can be used as a nice illustration of this.
     
  5. Apr 2, 2013 #4
    The superposition you mention is a superposition of Energy states, right?

    Thanks
     
  6. Apr 2, 2013 #5

    kith

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    I'm not sure if decoherence is possible if we have strictly stationary states in the system. Usually, the interaction with the environment causes both decoherence and relaxation. The time scale for decoherence is just much much shorter for most environments.

    The final state of the system is a mixed state. This means that the probability to find the system in any of the states of the initial superposition remains unchanged. I don't think that we should say that it is in only one of them, see also below.

    I tend to disagree. Consider a Bell state of two entangled qubits. We wouldn't think of a single qubit as having one definite state here. But the reduced density matrix is the same as for a qubit which has undergone decoherence. I don't think that it should matter wether the qubit is entangled with another qubit or with the environment.
     
  7. Apr 2, 2013 #6

    bhobba

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    Yea - but just to elaborate - decoherence transforms a superposition into an improper mixed state. The difference here is that mixed states are usually interpreted as a system where a randomly selected state is given for observation. If that was the case for the mixed states of decoherence then low and behold the measurement problem would be solved, you are observing a reality that exists prior to observation and everything is sweet. But it was not physically prepared that way so you can't do that - bummer - thats why its called improper - Schlosshauer carefully explains it in his book on Decoherence. But wait - it is both observationally and mathematically exactly the same so what you can do is assume it is like that - no observation or mathematical analysis can prove you wrong. This is the simple assumption I make. Other ways of using it exist eg Decoherent Histories and Many Worlds. Many worlds is particularly neat - nothing happens - the mixed state simply keeps evolving but each state of the ensemble is considered a separate world. You can read about Decoherent Histories here:
    http://quantum.phys.cmu.edu/CHS/histories.html

    Thanks
    Bill
     
    Last edited: Apr 2, 2013
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