Note: physics conventions, [itex]\theta[/itex] is measured from z-axis(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Vector calculus: angular momentum operator in spherical coordinates

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