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Probability of an entanglement state

  1. Feb 17, 2014 #1
    given two system is entangled, |A>=(|[itex]0_{A}[/itex]>+|[itex]1_{A}[/itex]>)/√2, |B>=(|[itex]0_B{}[/itex]>+|[itex]1_B{}[/itex]>)/√2. entangle state |AB>what is the probability to find |[itex]0_A{}[/itex][itex]0_B{}[/itex]> and |[itex]1_A{}[/itex][itex]1_B{}[/itex]>. are there still 1/2 just like normal inner product formulation?
     
    Last edited: Feb 18, 2014
  2. jcsd
  3. Feb 18, 2014 #2

    bhobba

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    You need to rephrase that - states with two particles have terms of the form |a>|b>.

    Thanks
    Bill
     
  4. Feb 18, 2014 #3

    bhobba

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    Well your rephrased question would be |U> = (|0a>|0b> + |1a>|1b>)/root 2. If you have an observation |0a>|0b> or |1a>|1b> then one or the other will be detected with probability 1/2. This is directly from the Born rule trace formula eg trace (|U><U| |1a>|1b>) = 1/2.

    Thanks
    Bill
     
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