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Qubits probability

  1. Apr 15, 2008 #1
    1. The problem statement, all variables and given/known data

    3. The attempt at a solution

    I know to calculate the pr we square the number in front of the state.

    e.g) for state psi 00, the pr is (1/root(2))^2

    But for state psi 01, do I include the imaginary number part? That is, do I calculate the square of (1+i/2root(2))?

  2. jcsd
  3. Apr 15, 2008 #2


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    The probability is the square of the magnitude of the coefficient (if the wavefunction is normalized). So the formula is that the probability of measuring E_{01} for example is

    [tex] \vert C_{01} \vert^2 = C_{01} \times C_{01}^* [/tex]

    which is obviousy a rel number.
    This should be familiar to you. Of course, when the coefficient is real, this becomes simply the ordinary square of the coefficient.
  4. Apr 16, 2008 #3
    Ah so its the complex conjugate. Makes sense!
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