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Quantum measurement projective?

  1. Apr 22, 2008 #1

    neu

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    1. The problem statement, all variables and given/known data

    Have two Greenberger-Horne-Zeilinger (GHZ) states of qubits A, B, C and
    D, E, F as follows:

    [tex] \mid GHZ \rangle_{ABC} = \frac{1}{\sqrt{2}} \left( \mid 0 \rangle_{A}\mid 0 \rangle_{B}\mid 0 \rangle_{C} + \mid 1 \rangle_{A}\mid 1 \rangle_{B}\mid 1 \rangle_{C}\right)[/tex]

    and

    [tex] \mid GHZ \rangle_{DEF} = \frac{1}{\sqrt{2}} \left( \mid 0 \rangle_{D}\mid 0 \rangle_{E}\mid 0 \rangle_{F} + \mid 1 \rangle_{D}\mid 1 \rangle_{E}\mid 1 \rangle_{F}\right)[/tex]

    If you perform a measurement in the Bell basis on the qubits A and D, and obtain
    the outcome: [tex] \mid \Psi^{+} \rangle_{AD} = \frac{1}{\sqrt{2}} \left(\mid 0 \rangle_{A}\mid 1 \rangle_{D} + \mid 1 \rangle_{A}\mid 0 \rangle_{D}\right) [/tex]

    Write down the state to which qubits B, C, D and F are projected?

    Attempted Solution

    Total state is : [tex] \mid GHZ \rangle_{ABCDEF}= \mid GHZ \rangle_{ABC} \mid GHZ \rangle_{DEF} [/tex]

    Projector operator for measurement of A and D is [tex] \mid Bell \rangle \langle Bell \mid_{AD} [/tex]

    so [tex] \mid Bell \rangle \langle Bell \mid_{AD} \mid GHZ \rangle_{ABCDEF} = \mid Bell \rangle_{AD} \mid \Psi^{+} \rangle_{AD} \mid \Psi^{+} \rangle_{BCEF} [/tex]; is this right?

    Here I get confused. I think the method is simply:

    [tex] \langle \Psi^{+} \mid_{AD}\mid GHZ \rangle_{ABCDEF} = \mid \Psi^{+} \rangle_{BCEF} [/tex]

    But if so, why?
     
    Last edited: Apr 22, 2008
  2. jcsd
  3. Apr 23, 2008 #2

    neu

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    one bump then I'll give up.

    I'm still stuck on this one, no takers?
     
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