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Vector calculus: angular momentum operator in spherical coordinates

  1. Feb 23, 2014 #1
    Note: physics conventions, [itex]\theta[/itex] is measured from z-axis
    We have a vector operator

    [itex]\vec{L} = -i \vec{r} \times \vec{\nabla} = -i\left(\hat{\phi} \frac{\partial}{\partial \theta} - \hat{\theta} \frac{1}{\sin\theta} \frac{\partial}{\partial \phi} \right)[/itex]
    And apparently
    [itex]\vec{L}\cdot\vec{L}= -\left(\frac{\partial^2}{\partial \theta^2} + \cot \theta\frac{\partial}{\partial \theta}+ \frac{1}{\sin^2\theta} \frac{\partial^2}{\partial \phi^2} \right)[/itex]
    I am wondering about a way to obtain the second expression (the [itex]\cot \theta[/itex] term in particular) from the first expression without taking the circuitous route followed in my references. I realize the unit vectors aren't constant.
    [itex]\frac{\partial \boldsymbol{\hat{\theta}}} {\partial \phi} = \cos \theta\boldsymbol{\hat \phi}[/itex]
    [itex]\frac{\partial \boldsymbol{\hat{\phi}}} {\partial \theta} = 0[/itex]

    The basis vectors are orthonormal however, and for the norm-squared of a regular vector in spherical coordinates, we can just square each of the components and add.
     
  2. jcsd
  3. Mar 3, 2014 #2

    joshmccraney

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    while i'm not quite sure what you're asking for (not a physicist), i can say the del operator is not explicitly defined in spherical coordinates. thus, if you are looking for a way to literally take [itex]\vec{L}\cdot \vec{L}[/itex] perhaps the only way is the "circuitous" route from your references.

    maybe this helps a little?
     
  4. Mar 3, 2014 #3

    joshmccraney

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    while i'm not quite sure what you're asking for (not a physicist), i can say the del operator is not explicitly defined in spherical coordinates. thus, if you are looking for a way to literally take [itex]\vec{L}\cdot \vec{L}[/itex] perhaps the only way is the "circuitous" route from your references.

    maybe this helps a little?
     
  5. Mar 3, 2014 #4

    joshmccraney

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    Gold Member

    while i'm not quite sure what you're asking for (not a physicist), i can say the del operator is not explicitly defined in spherical coordinates. thus, if you are looking for a way to literally take [itex]\vec{L}\cdot \vec{L}[/itex] perhaps the only way is the "circuitous" route from your references.

    maybe this helps a little?
     
  6. Mar 3, 2014 #5

    vanhees71

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    The gradient operator is defined in spherical coordinates (except along the polar axis, where spherical coordinates are singular).

    To evaluate [itex]\hat{\vec{L}}^2[/itex], it's easier to go back to Cartesian coordinates first.
    [tex]\hat{\vec{L}}^2 \psi =-\epsilon_{ijk}x_j \partial_k (\epsilon_{ilm} x_l \partial_m) \psi.[/tex]
    First we use the product rule and then to get
    [tex]\epsilon_{ijk} \epsilon_{ilm}=\delta_{jl} \delta_{km}-\delta_{jm} \delta_{kl}[/tex]
    [tex]\hat{\vec{L}}^2 \psi =-(delta_{jl} \delta_{km}-\delta_{jm} \delta_{kl}) x_j (\delta_{kl} \partial_m+x_j x_l \partial_k \partial_m)=-(r^2 \Delta - 2 (\vec{x} \cdot \vec{\nabla}) \psi - x_j (\vec{x}\cdot \vec{\nabla}) \partial_j \psi).[/tex]
    The latter expression is not nice, but we can use
    [tex](\vec{x} \cdot \vec{\nabla})(\vec{x} \cdot \vec{\nabla})=x_j \partial_j (x_k \partial_k \psi) = (\vec{x} \cdot \vec{\nabla}) \psi + x_j (\vec{x} \cdot \vec{\nabla}) \partial_j \psi[/tex]
    to finally write
    [tex]\hat{\vec{L}}^2 \psi = - (r^2 \Delta - (\vec{x} \cdot \vec{\nabla}) \psi - (\vec{x} \cdot \vec{\nabla})(\vec{x} \cdot \vec{\nabla}) \psi.[/tex]
    Now you use the Laplacian in spherical coordinates and
    [tex]\vec{x} \cdot \vec{ \nabla} \psi = r \hat{r} \cdot \vec{\nabla} \psi = r \partial_r \psi[/tex]
    to finally get
    [tex]\hat{\vec{L}}^2 \psi=-\frac{1}{\sin \vartheta} \left [\frac{\partial}{\partial \vartheta} \left (\sin \vartheta \frac{\partial \psi}{\partial \vartheta} \right ) + \frac{1}{\sin \vartheta}\frac{\partial^2 \psi}{\partial \varphi^2} \right ].[/tex]
     
  7. Mar 3, 2014 #6
    I guess what I'm wondering is if there is a way without going back to Cartesian coordinates.
     
  8. Mar 4, 2014 #7

    vanhees71

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    It's a bit tricky in non-Cartesian coordinates as soon as vector (or higher-rank tensor) fields are involved, and [itex]\hat{\vec{L}} \psi[/itex] already is a vector field. That's why it is much simpler to first derive a scalar expression in Cartesian coordinates, which you already now how to express in spherical coordinates.
     
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