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B Why is state transition probability symmetric?

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  1. Mar 31, 2017 #1
    Restricting to finite dimensional QP, suppose a system is in a state S1, an experiment is done, and state S2 is one of the eigenstates (assume all eigenvalues are distinct). The probability that the system transitions from S1 to S2 is p = Trace( S1*S2), using state operator notation. On the other hand, if a system is in state S2, a different experiment is done, and state S1 is one of the eigenstates,, the probability that the system transitions from S2 to S1 is, again, p, due to the symmetry of the Trace inner product. Is there a physical rationale why these two state transition probabilties are the same?
     
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  3. Apr 2, 2017 #2

    mfb

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    It is not directly the probability, it is just proportional to it.

    This is the time-symmetry of (nonrelativistic) quantum mechanics.
     
  4. Apr 2, 2017 #3
    Time symmetry is an interesting perspective. The difficulty I have with this, though, is that the two directions of state transition require two different experiments.
    BTW, it is directly probability, as the p's add up to 1, owing to the states having trace 1.
     
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