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Complex probability in QM system?

  1. Sep 10, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider a quantum system described by a basis [tex]\mid 1 \rangle[/tex] and [tex]\mid 2 \rangle[/tex].

    The system is initially in the state: [tex]\psi_i = \frac{i}{\sqrt3} \mid 1 \rangle + \sqrt{\frac{2}{3}} \mid 2 \rangle[/tex].

    (a) Find the probability that the initial system is measured to be in the state: [tex]\psi_f = \frac{1 + i}{\sqrt 3} \mid 1 \rangle + \frac{1}{\sqrt{3}} \mid 2 \rangle[/tex]

    2. Relevant equations

    The basis is assumed to be orthonormal, hence [tex]\langle 1 \mid 1 \rangle = \langle 2 \mid 2 \rangle = 1[/tex]

    Probability is calculated as [tex](\langle \psi_f \mid \psi_i \rangle)^2[/tex]

    3. The attempt at a solution

    Calculating this, I get a complex answer. I'm not sure but I think a probability (a real observable) should be a real number. Is that right?

    The answer I get is [tex]\frac{2 + 2\sqrt{2}}{9}(1+i)[/tex].
     
  2. jcsd
  3. Sep 10, 2009 #2
    I just tried to normalize the two states and found that they are both unit-normal already...

    This is confusing.
     
  4. Sep 10, 2009 #3

    gabbagabbahey

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    Surely you mean:

    [tex]\frac{\langle \psi_i \vert \psi_f \rangle \langle \psi_f \vert \psi_i \rangle}{\langle \psi_i \vert \psi_i \rangle \langle \psi_f \vert \psi_f \rangle}=\vert\langle \psi_i \vert \psi_f \rangle \vert^2[/tex]

    (for normalized [itex]\psi_i[/itex] and [itex]\psi_f[/itex])....right?:wink:
     
  5. Sep 10, 2009 #4
    Thanks for the response ggh.

    Since my states are normalized, the denominator drops to 1. The numerator seems correct. I think you are right that I missed the absolute value signs.

    ...

    After a quick calculation, I see that's where I screwed up. I get a real answer if I take the absolute value before squaring.

    Thanks!
     
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