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
fzero
Science Advisor
Homework Helper
Gold Member
3,119
289
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!
 

Related Threads on General normalized state for a state vector

Replies
2
Views
4K
  • Last Post
Replies
1
Views
8K
  • Last Post
Replies
1
Views
915
Replies
6
Views
2K
Replies
1
Views
642
Replies
2
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
16
Views
3K
Replies
1
Views
606
Replies
5
Views
1K
Top