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I Decoherence Pure States Into Mixed States

  1. Jul 27, 2017 #1
    According to decoherance.

    Say there is a pure state initially in state:

    |ψ⟩=α|0⟩+β|1⟩

    After decoherance (interaction with environment), the system will transform into the improper mixed state of:

    ρ=|α|2|0⟩⟨0|+|β|2|1⟩⟨1|

    This is the "apparent" collapse that decoherance refers to. With the probabilities of the mixed state being equal to the Born rule probabilities. However, why is this the mixed state that the pure state transforms into? For example, doesn't this assumes that the "measurement" interaction of the environment is for the observable/basis that the pure state was initially expressed in?

    Is the mixed state shown above just one of many possible mixed states that the pure state could have transformed into? Depending on the type of interaction of the system and the environment. Or in other words, which observable was being measured?

    For example, using a standard measurement collapse postulate, I could have collapsed an original wavefunction into many different final states with different probabilities depending on which operator/observable I am measuring. Shouldn't this be reflected in the pure state being decohered into a mixed state? All examples I've seen online have shown exactly what I wrote above.

    Thanks.
     
  2. jcsd
  3. Jul 28, 2017 #2

    Orodruin

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    Yes. It depends on the system's interaction with the environment.
     
  4. Jul 28, 2017 #3
    Yes consider |0><a| for a different basis.
     
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