General normalized state for a state vector

  • #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] ?

Answers and Replies

  • #2
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.
  • #3
can't believe i forgot about something as simple as that. thanks for the answer!

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