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QM - Measuring Angular Momentum

  • Thread starter Biest
  • Start date
67
0
Hi,

I have a quick homework problem because I am confused. So a wave function with l = 1 in the state:

[tex] |\psi> = \frac{1}{\sqrt{14}} \[ \left( \begin{array}{ccc}
1 \\
2 \\
3i \end{array} \right)\] [/tex]

and i have to find the probability to be in state [tex] \hbar, -\hbar, 0 [/tex] in [tex] L_z [/tex], so i applied [tex] L_z [/tex] so [tex] |\psi> [/tex] and got

[tex] L_z|\psi> = \frac{\hbar}{\sqrt{14}} \[ \left( \begin{array}{ccc}
1 \\
0 \\
-3i \end{array} \right)\] [/tex]

How do i find the probabilities from that?


Thanks,

Biest
 

Answers and Replies

979
1
Always, the probability of finding a state [tex]|\psi\rangle[/tex] in a state [tex]|\phi\rangle[/tex] is [tex]\left| \left<\phi \middle| \psi \right>\right|^2[/tex].
 
67
0
But how do i take the [tex] \hbar, -\hbar 0[/tex] condition into account?
 
18
0
assuming your [tex]|\psi\rangle[/tex] is in the basis of normalised angular momentum eigenstates, then your probabilities just correspond to [tex]\frac{1}{14}[/tex] for [tex]\hbar[/tex], [tex]\frac{2}{7}[/tex] for 0 and [tex]\frac{9}{14}[/tex] for [tex]-\hbar[/tex].

so what happens is that for [tex]|\phi\rangle[/tex] to return you an eigenvalue of say [tex]\hbar[/tex], we need

[tex]|\phi\rangle = \[ \left( \begin{array}{ccc} 1 \\ 0 \\ 0 \end{array} \right) \] [/tex]

and taking [tex]\left| \left<\phi \middle| \psi \right>\right|^2[/tex], we get [tex]\frac{1}{14}[/tex].

hope it helps. QM was never my strong point. haha
 

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