We first rewrite the wave vector in Dirac notation:
| ψ ⟩ = 1 5 | 1 , − 1 ⟩ + 3 5 | 1 , 0 ⟩ + 1 / 5 | 1 , 1 ⟩ {\displaystyle |\psi \rangle ={\frac {1}{\sqrt {5}}}|1,-1\rangle +{\sqrt {\frac {3}{5}}}|1,0\rangle +1/{\sqrt {5}}|1,1\rangle }
We see that the possible results for a measurement of L ^ z {\displaystyle {\hat {L}}_{z}} are − ℏ , {\displaystyle -\hbar ,} 0 , {\displaystyle 0,\!} and ℏ . {\displaystyle \hbar .}
The probablity for obtaining − ℏ {\displaystyle -\hbar } is
P ( − ℏ ) = | ⟨ 1 , − 1 | ψ ⟩ | 2 = | 1 5 ⟨ 1 , − 1 | 1 , − 1 ⟩ + 3 5 ⟨ 1 − 1 | 1 , 0 ⟩ + 1 5 ⟨ 1 , − 1 | 1 , 1 ⟩ | 2 = 1 5 . {\displaystyle P(-\hbar )=|\langle 1,-1|\psi \rangle |^{2}=\left|{\frac {1}{\sqrt {5}}}\langle 1,-1|1,-1\rangle +{\sqrt {\frac {3}{5}}}\langle 1-1|1,0\rangle +{\frac {1}{\sqrt {5}}}\langle 1,-1|1,1\rangle \right|^{2}={\tfrac {1}{5}}.}
Similarly, the probablites of obtaining 0 {\displaystyle 0\!} and ℏ {\displaystyle \hbar } are
P ( 0 ) = | ⟨ 1 , 0 | ψ ⟩ | 2 = 3 5 {\displaystyle P(0)=|\langle 1,0|\psi \rangle |^{2}={\tfrac {3}{5}}}
and
P ( ℏ ) = | ⟨ 1 , 1 | ψ ⟩ | 2 = 1 5 . {\displaystyle P(\hbar )=|\langle 1,1|\psi \rangle |^{2}={\tfrac {1}{5}}.}
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