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Thread: Quantum Mechanics - Angular Momentum part 2

  1. #1
    Senior Member bugatti79's Avatar
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    Quantum Mechanics - Angular Momentum part 2

    Consider

    $\displaystyle
    [\hat L^2, \hat L_x]=[\hat L_x^2+\hat L_y^2+\hat L_z^2, \hat L_x]=[\hat L_x^2,\hat L_x]+[\hat L_y^2,\hat L_x]+[\hat L_z^2,\hat L_x]
    $

    $\displaystyle
    =\hat L_x [\hat L_x,\hat L_x]+ [\hat L_x,\hat L_x] \hat L_x +\hat L_y [\hat L_y,\hat L_x]+ [\hat L_y,\hat L_x] \hat L_y+\hat L_z [\hat L_z,\hat L_x]+ [\hat L_z,\hat L_x] \hat L_z
    $

    $\displaystyle
    = i \hbar (- \hat L_y \hat L_z -\hat L_z \hat L_y +\hat L_z \hat L_y +\hat L_y \hat L_z)=0
    $

    On the second line I am wondering how the first two terms go to 0 and how the third term

    $\displaystyle
    \hat L_y [\hat L_y,\hat L_x]=- i \hbar \hat L_z
    $ etc?
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  2. #2
    A Plied Mathematician
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    Any operator commutes with itself, which is another way of saying its commutator is zero. As for your second question, note that

    $\displaystyle \left[\hat{L}_{x},\hat{L}_{y}\right]=i\hbar\hat{L}_{z},$

    $\displaystyle \left[\hat{L}_{y},\hat{L}_{z}\right]=i\hbar\hat{L}_{x},$

    $\displaystyle \left[\hat{L}_{z},\hat{L}_{x}\right]=i\hbar\hat{L}_{y}.$

    Also, note that, in general, $\displaystyle \left[\hat{A},\hat{B}\right]=-\left[\hat{B},\hat{A}\right].$

    Hence, you have

    $\displaystyle \hat{L}_{y}\left[\hat{L}_{y},\hat{L}_{x}\right]=\hat{L}_{y}\left(-i\hbar\hat{L}_{z}\right)=-i\hbar\hat{L}_{y}\hat{L}_{z}.$

    Does that make sense?
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