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QM - Measuring Angular Momentum

  1. Apr 6, 2008 #1
    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
     
  2. jcsd
  3. Apr 6, 2008 #2
    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].
     
  4. Apr 6, 2008 #3
    But how do i take the [tex] \hbar, -\hbar 0[/tex] condition into account?
     
  5. Apr 6, 2008 #4
    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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