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Prove angular momentum operator identity

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data

    Using the operator identity:
    [tex]
    \hat{L}^2=\hat{L}_-\hat{L}_+ +\hat{L}_z^2 + \hbar\hat{L}_z
    [/tex] show explicitly:
    [tex]
    \hat{L}^2 = -\hbar^2 \left[
    \frac{1}{\sin^2\theta} \frac{\partial^2}{\partial\phi^2} +
    \frac{1}{\sin\theta} \frac{\partial}{\partial\theta}
    \left(\sin\theta\frac{\partial}{\partial\theta}\right)
    \right]
    [/tex](Note: all L are operators, i.e. L(hat))

    2. Relevant equations
    [tex]
    \hat{L}_\pm = \hbar e^{\pm i\phi}\left(\pm \frac{\partial}{\partial\theta} + i\cot\theta\frac{\partial}{\partial\phi}\right) \\
    \hat{L}_z = -i\hbar \frac{\partial}{\partial\phi}
    [/tex]

    3. The attempt at a solution

    \begin{align*}
    \hat{L}^2 &= \hat{L}_-\hat{L}_+ + \hat{L}_z^2 + \hbar \hat{L}_z \\
    &= \hbar e^{-i\phi}\left(- \frac{\partial}{\partial\theta} + i\cot\theta\frac{\partial}{\partial\phi}\right) \times \hbar e^{+i\phi}\left(+ \frac{\partial}{\partial\theta} + i\cot\theta\frac{\partial}{\partial\phi}\right) + \left(-i\hbar \frac{\partial}{\partial\phi}\right)^2 + \hbar \left(-i\hbar \frac{\partial}{\partial\phi}\right) \\
    &= \hbar^2\left[\left(- \frac{\partial}{\partial\theta} + i\cot\theta\frac{\partial}{\partial\phi}\right)\left( \frac{\partial}{\partial\theta} + i\cot\theta\frac{\partial}{\partial\phi}\right)\right] -\hbar^2\frac{\partial^2}{\partial\phi^2} - i\hbar^2\frac{\partial}{\partial\phi} \\
    &= -\hbar^2\left[\left( \frac{\partial}{\partial\theta}\right)^2 + \left(\cot\theta\right)^2 + \left(\frac{\partial}{\partial\phi}\right)^2 + i\frac{\partial}{\partial\phi}\right] \\
    &= -\hbar^2\left[\left( \frac{\partial}{\partial\theta}\right)^2 + \left(\frac{\partial}{\partial\phi}\right)^2\left(\cot^2\theta+1\right) + i\frac{\partial}{\partial\phi}\right] \\
    &= -\hbar^2\left[\left( \frac{\partial}{\partial\theta}\right)^2 + \left(\frac{\partial}{\partial\phi}\right)^2\left(\frac{1}{\sin^2\theta}\right) + i\frac{\partial}{\partial\phi}\right]
    \end{align*}

    not sure how to procced from here, it's close to the required form but I do not know how to deal with the [itex] i\frac{\partial}{\partial\phi} [/itex] term or I might have made mistakes....

    Hope someone can help, thanks
     
    Last edited by a moderator: Apr 4, 2013
  2. jcsd
  3. Apr 4, 2013 #2

    DrClaude

    User Avatar

    Staff: Mentor

    Hi Tom, welcome to PF!

    In this step, you seem to have forgotten that you are dealing with operators. For instance, you can't just cancel out ##e^{-i\phi}## and ##e^{+i\phi}## because there is a ##∂/∂\phi## inbetween them.
     
  4. Apr 4, 2013 #3
    Ok, I have no idea if what I'm doing is right, but would really appreciate some guidance here:
    \begin{align*}
    &= \hbar e^{-i\phi} \left(-\frac{\partial}{\partial\theta} + i\cot\theta\frac{\partial}{\partial\phi}\right)
    \hbar e^{+i\phi} \left(+\frac{\partial}{\partial\theta} + i\cot\theta\frac{\partial}{\partial\phi}\right) +
    \left(-i\hbar\frac{\partial}{\partial\phi}\right)^2 -
    i\hbar \frac{\partial}{\partial\phi} \\
    &= \hbar \left(e^{-i\phi}\left(-\frac{\partial}{\partial\theta}\right) + ie^{-i\phi}\cot\theta\frac{\partial}{\partial\phi}\right) \hbar \left(e^{+i\phi}\left(\frac{\partial}{\partial\theta}\right) + ie^{+i\phi}\cot\theta\frac{\partial}{\partial\phi}\right) +\left(-i\hbar \frac{\partial}{\partial\phi}\right)^2 -i\hbar \frac{\partial}{\partial\phi} \\
    &= -\hbar^2 \left[e^{-2i\phi} \left(\frac{\partial}{\partial\theta}\right)^2 - e^{-2i\phi}\cot\theta\frac{\partial}{\partial\theta} - \cot\theta\frac{\partial}{\partial\theta} + \cot^2\theta + \left(\frac{\partial}{\partial\phi}\right)^2 + i\frac{\partial}{\partial\phi}\right] \\
    &= -\hbar^2 \left[e^{-2i\phi} \left(\frac{\partial}{\partial\theta}\right)^2 + e^{-2i\phi}\csc^2\theta + \csc^2\theta + \cot^2\theta +\left(\frac{\partial}{\partial\phi}\right)^2 + i\frac{\partial}{\partial\phi}\right]
    \end{align*}
    I think I'm doing something where as this doesn't seem to be going anyway.....
     
    Last edited by a moderator: Apr 4, 2013
  5. Apr 4, 2013 #4

    DrClaude

    User Avatar

    Staff: Mentor

    Start by considering only ##\hat{L}_- \hat{L}_+##:
    $$
    \hat{L}_- \hat{L}_+ = \hbar^2 e^{-i \phi} \left( - \frac{\partial}{\partial \theta} + i \cot \theta \frac{\partial}{\partial \phi} \right) e^{i \phi} \left( \frac{\partial}{\partial \theta} + i \cot \theta \frac{\partial}{\partial \phi} \right)
    $$
    If you distribute ##\hat{L}_-## on ##\hat{L}_+##,
    $$
    \frac{\hat{L}_- \hat{L}_+}{\hbar^2} = - e^{-i \phi} \frac{\partial}{\partial \theta} e^{i \phi} \left( \frac{\partial}{\partial \theta} + i \cot \theta \frac{\partial}{\partial \phi}\right) + e^{-i \phi} i \cot \theta \frac{\partial}{\partial \phi} \left[ e^{i \phi} \left( \frac{\partial}{\partial \theta} + i \cot \theta \frac{\partial}{\partial \phi} \right) \right]
    $$
    Then
    $$
    - e^{-i \phi} \frac{\partial}{\partial \theta} \left( e^{i \phi} \frac{\partial}{\partial \theta} \right) = - \frac{\partial^2}{\partial \theta^2}
    $$
    and
    $$
    - e^{-i \phi} \frac{\partial}{\partial \theta} \left( e^{i \phi} i \cot \theta \frac{\partial}{\partial \phi}\right) = i \csc^2 \theta \frac{\partial}{\partial \phi} - i \cot \theta \frac{\partial}{\partial \theta} \frac{\partial}{\partial \phi}
    $$
    and so on.

    It is a bit tedious, but just remember the "rule"
    $$
    \frac{\partial}{\partial x} f(x) g(y) = \frac{df}{dx} g(y) + f(x) g(y) \frac{\partial}{\partial x}
    $$
     
  6. Apr 4, 2013 #5
    I see, thank you very much
     
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