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Qunatum Angular Momentum

  1. Feb 4, 2008 #1
    [SOLVED] Qunatum Angular Momentum

    1. The problem statement, all variables and given/known data
    Particle is in state

    [tex]\psi=A(x+y+2z)e^{-\alpha r}[/tex]

    A and alpha are real constants.

    a) Normalize angular part of wave function.
    b) Find [tex]<\vec{L}^{2}> , <L_{z}>[/tex]
    c) Find probability of finding [tex]L_{z}=+\hbar[/tex].

    2. Relevant equations

    [tex]{L}^{2}= \hbar^{2}l(l+1)|lm>[/tex]
    [tex]L_{z}=m \hbar|lm>[/tex]

    3. The attempt at a solution
    I have found a) part



    Since [tex]Y^{m}_{l}=|lm>[/tex]

    Using [tex]L_{z}=m \hbar|lm>[/tex]

    <Lz>= 1/(4*3)*2<1-1|Lz|1-1> + 1/(4*3)*2<11|Lz|11> + 4/6<10|Lz|10> = 0

    To find [tex]<\vec{L}^{2}>[/tex] I would apply operator of L^2 to angular part of wave function, just like I have done for Lz.

    [tex]{L}^{2}= \hbar^{2}l(l+1)|lm>[/tex]

    Is this is the way to find expectation values ?


  2. jcsd
  3. Feb 4, 2008 #2


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    All correct!

    You do realize that the value [itex]\langle\vec{L}^{2}\rangle[/itex] is immediately obvious, right?
  4. Feb 5, 2008 #3
    No, can you explain ?
    I can see that l=1 , and m={-hbar, 0 , hbar}, but why is [itex]\langle\vec{L}^{2}\rangle[/itex] obvious ?
  5. Feb 5, 2008 #4


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    You only have states with [tex] l = 1[/tex] , then what can you say about the expactation value?
  6. Feb 5, 2008 #5
    Possible values for measurment of L are L=hbar*sqrt(l(l+1)) so L^2=hbar^2*(l(l+1)) .
    Because I have l=1
    L^2=2*(hbar)^2 , but I only have states with l=1 so it must be also
    <L^2>=L^2=2*(hbar)^2 right ?
    Last edited: Feb 5, 2008
  7. Feb 5, 2008 #6


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    Right. Your state is an eigenstate of [itex]\vec{L}^{2}[/itex] with eigenvalue [itex]2\hbar^2[/itex]. So the expectation value (for a normalized state) is the same as the eigenvalue.
  8. Feb 5, 2008 #7
    ok, thanks for hints and replies.
    Last edited: Feb 5, 2008
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