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General normalized state for a state vector

  1. Aug 20, 2011 #1
    A particle is in a state for which the probabilities are [tex]P(L_{z} = 1) = \frac{1}{4}, P(L_{z} = 0) = \frac{1}{2}, and P(L_{z} = -1) = \frac{1}{4}[/tex]. Show that the most general normalized state with this property is:
    [tex]|\phi> = \frac{e^{i \delta_{1}}}{2}|L_{z} = 1> + \frac{e^{i \delta_{2}}}{2^{\frac{1}{2}}}|L_{z} = 0> + \frac{e^{i \delta_{3}}}{2}|L_{z} = -1>.[/tex]

    i know to calculate [tex]P(L_{z} = 1)[/tex] it is done by [tex] |< L_{z} = 1 |\phi> |^2 [/tex] and assuming i'm in an orthonormal eigenbasis i get that [tex] < L_{z} = 1 |\phi> = \frac{e^{i \delta_{1}}}{2} [/tex] and when its squared i get [tex] \frac{e^{2i \delta_{1}}}{2} [/tex] which should equal to [tex] \frac{1}{2} [/tex] and this is the part i am having trouble with. how do i show that [tex] e^{2i \delta_{1}} = 1 [/tex] ?
     
  2. jcsd
  3. Aug 20, 2011 #2

    fzero

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    To compute the probability, you don't just compute the square of the amplitude. You compute the modulus of the complex number (which is always real) and then square that.
     
  4. Aug 21, 2011 #3
    can't believe i forgot about something as simple as that. thanks for the answer!
     
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