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A variation of the Bell experiment

  1. Nov 18, 2014 #1

    If we start with a Bell state


    and (after moving the second qbit a significant distance away) apply the interferometer transformation
    |0> -> 0.5(|0>+|1>)
    |1> -> 0.5(|0>-|1>)

    to the first qbit, we get
    which gives equal probability of the first qbit ending up in |0> or |1>

    Lets now start again with the same spatially separated Bell state but first apply the transformation

    |0> -> 0.5(|0>+|1>)
    |1> -> 0.5(|0>+|1>)
    to the second qbit:


    then apply the original (interferometer) transformation to the first qbit:

    Now, the first qbit is in state |0> with 100% (as opposed to 50%) probability as a result of what was done to the second one.

    So...can anyone tell me if I made any false assumptions or stupid math mistakes here?

    Dustin Soodak
  2. jcsd
  3. Nov 18, 2014 #2


    User Avatar
    Science Advisor

    This transformation isn't reversible, so I think you have to measure the state and collapse the wave function before doing the transformation. After collapsing the wave function, the state should be a mixed state, not a pure state, ie. with 50% chance the state is |0>(|0>+|1>), and with 50% chance the state is |1>(|0>+|1>).
  4. Nov 19, 2014 #3
    The way I've always seen it described in this sort of experiment, something counts as a measurement if the information about the state leaks out.
    I originally thought this transformation could be done (using photons as qbits) with an interferometer that has one path length 1/4 wavelength longer than the other. However, I suppose you could then get info about the photon's path information from the difference in travel time.
  5. Nov 19, 2014 #4


    Staff: Mentor

    Cant quite follow your idea here - information about state leaks out?

    In modern times an observation is generally considered to have occurred just after decoherence.

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