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Angular momentum and Expectation values (Another question)

  1. Nov 15, 2009 #1
    1. The problem statement, all variables and given/known data

    Using the fact that ,[tex]\left\langle \hat{L}_{x}^{2} \right\rangle = \left\langle \hat{L}_{y}^{2} \right\rangle[/tex] show that [tex]\left\langle \hat{L}_{x}^{2} \right\rangle = 1/2 \hbar^{2}(l(l+1)-m^{2}.[/tex]

    3. The attempt at a solution

    [tex]L^{2} \left|l,m\right\rangle = \hbar^{2}l(l+1) \left|l,m\right\rangle[/tex]

    [tex]L_{z} \left|l,m\right\rangle = \hbar m \left|l,m\right\rangle[/tex]

    [tex]\left\langle \hat{L}^{2} \right\rangle = \hbar^{2}l(l+1)[/tex]

    [tex]\left\langle \hat{L}_{z}^{2} \right\rangle = (\hbar m)^{2}[/tex]


    [tex]\left\langle \hat{L}^{2} \right\rangle = \left\langle \hat{L}_{x}^{2} \right\rangle + \left\langle \hat{L}_{y}^{2} \right\rangle + \left\langle \hat{L}_{z}^{2} \right\rangle[/tex]

    [tex]\left\langle \hat{L}_{x}^{2} \right\rangle = 1/2 (\left\langle \hat{L}^{2} \right\rangle - \left\langle \hat{L}_{z}^{2} \right\rangle )[/tex]



    I dont think that this is really showing the solution since i have just stated that

    [tex]\left\langle \hat{L}^{2} \right\rangle = \left\langle \hat{L}_{x}^{2} \right\rangle + \left\langle \hat{L}_{y}^{2} \right\rangle + \left\langle \hat{L}_{z}^{2} \right\rangle[/tex]

    and

    [tex]\left\langle \hat{L}_{z}^{2} \right\rangle = (\hbar m)^{2}[/tex]. Unless it is genrally true that [tex]\left\langle \hat{L}^{2} \right\rangle = \left\langle \hat{L} \right\rangle^{2[/tex]



    What do you think?
     
    Last edited: Nov 15, 2009
  2. jcsd
  3. Nov 16, 2009 #2

    gabbagabbahey

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    In order to calculate any expectation value, you need to know the state of the system. I assume that you are trying to calculate [itex]\langle \hat{L}_x^2\rangle[/itex] in the state [itex]|l,m\rangle[/itex]?



    Well, certainly [itex]\langle\hat{L}^2\rangle=\langle\hat{L}_x^2+\hat{L}_y^2+L_z^2\rangle[/itex]...What is the definition of expectation value?...Do the inner products involved satisfy properties that will allow you to conclude that [itex]\langle\hat{L}_x^2+\hat{L}_y^2+L_z^2\rangle=\langle\hat{L}_x^2\rangle+\langle\hat{L}_y^2\rangle+\langle\hat{L}_z^2\rangle[/itex]?

    Again, appeal to the definition of expectation value...
     
  4. Nov 16, 2009 #3
    Yes


    [tex] \left\langle \hat{L}^{2} \right\rangle = \left\langle l,m\right|
    L^{2} \left|l,m\right\rangle
    [/tex]

    [tex] \left\langle l,m\right|
    L^{2} \left|l,m\right\rangle = \left\langle l,m\right|
    L_{x}^{2}+L_{y}^{2}+L_{z}^{2} \left|l,m\right\rangle
    [/tex]


    [tex] \left\langle l,m\right|
    L^{2} \left|l,m\right\rangle = \left\langle l,m\right|
    L_{x}^{2}\left|l,m\right\rangle + \left\langle l,m\right|L_{y}^{2}\left|l,m\right\rangle +\left\langle l,m\right| L_{z}^{2} \left|l,m\right\rangle
    [/tex]

    [tex]
    \left\langle \hat{L}^{2} \right\rangle = \left\langle \hat{L}_{x}^{2} \right\rangle + \left\langle \hat{L}_{y}^{2} \right\rangle + \left\langle \hat{L}_{z}^{2} \right\rangle
    [/tex]

    yes?

    Still not so sure about this one...

    [tex]
    \left\langle l,m\right| L_{z} \left|l,m\right\rangle = \hbar m
    [/tex]

    [tex]
    L_{z} \left|l,m\right\rangle = \hbar m \left|l,m\right\rangle
    [/tex]

    [tex]
    L_{z}^{2} \left|l,m\right\rangle = (\hbar m)^{2} \left|l,m\right\rangle
    [/tex]

    [tex]
    \left\langle l,m\right| L_{z}^{2} \left|l,m\right\rangle = (\hbar m)^{2}
    [/tex]
     
  5. Nov 16, 2009 #4

    gabbagabbahey

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    Why aren't you sure about this?

    [tex]\langle L_{z}^{2} \rangle=\left\langle l,m\right| L_{z}^{2} \left|l,m\right\rangle=\langle l,m| (\hbar m)^{2} |l,m\rangle=\hbar^2m^2\langle l,m |l,m\rangle[/tex]
     
  6. Nov 16, 2009 #5

    I am not sure how you can infer that [tex]
    \hat{L}_{z}^{2} = (\hbar m)^{2}
    [/tex] from [tex]
    L_{z} \left|l,m\right\rangle = \hbar m \left|l,m\right\rangle
    [/tex]
     
    Last edited: Nov 16, 2009
  7. Nov 16, 2009 #6

    gabbagabbahey

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    You can't!

    However, you can infer that

    [tex]L_z^2|l,m\rangle=L_zL_z|l,m\rangle=L_z(\hbar m|l,m\rangle)=\hbar m L_z|l,m\rangle=\hbar^2 m^2|l,m\rangle[/itex]

    and that's all that's needed.
     
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