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Entangled particles and measurement

  1. Oct 10, 2009 #1


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    We know that if, for example, two particles are entangled in a state like: |Ψ>=a |0>|0> + b |1>|1>, then measuring an observable on only one of the particles it makes the interference dissapear.
    But isnt it impossible to measure both particles simultaneously in order to maintain the interference?? There should be an infinitsimal time difference!
    That leads me to the conclusion that if two particles get entangled then no interferences will ever appear. Would you agree?

    Also, the above can explain for example why in the double slit experiment interference is lost when we try to specify the electron`s position (which slit). When the photon scatters from the electron, their interaction makes them somehow get entangled. So, when the electron hit the wall (position measurement) the interference is lost just because it`s entangled to the photon that scattered!
    If the scattered photon is detected first (measurement), then its' state collapses and so does the electron`s. That means, no interference again.
    Is the above description any close to reality?

    And one last question! I mentioned above that in the photon-electron scattering the particles somehow get entangled. In general, when particles interact with each other, they get entangled??
  2. jcsd
  3. Oct 11, 2009 #2


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    A couple of points. First, interference disappears for entangled photons if the possibility merely exists to determine which-slit information. It only "returns" (this is a bit too complex to get into in a paragraph) if you erase that possibility. Second, as you suggest, there is a time difference when measuring Alice and Bob pretty much no matter how you set things up.

    So essentially, your conclusion is correct. Groups of entangled photons NEVER show an interference pattern directly.
  4. Oct 12, 2009 #3


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    Thanks a lot DrChinese!
    If there are any other comments on the rest of the OP, they`d be welcome!
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