General normalized state for a state vector

[demonelite123]

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

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

 

[fzero]

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.

 

[demonelite123]

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