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Derivation of B.31

  1. Jan 25, 2016 #1
    I was looking at this PDF (http://www.ifi.unicamp.br/~maplima/fi001/2012/aula20b.pdf) showing the derivation of the squared-angular momentum operator. Everything seems okay although I am just slightly lost in how exactly B.31 was derived. Isn't B.31 equal to the dot product of B.24 with itself? Where exactly does the sin in the numerator and denominator in B.31 come from when the theta component of B.24 is simply the partial derivative with respect to theta with no other terms? Shouldn't the theta components dotted with each simply yield the second order partial derivative with respect to theta when considering only this component?
     
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  3. Jan 26, 2016 #2

    blue_leaf77

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    It's indeed just the dot product, but you cannot neglect the fact that the spherical unit vectors are not fixed, therefore differentiating them against some variable will not necessarily equal zero. For example
    $$
    \frac{\hat{\theta}}{\sin \theta} \frac{\partial}{\partial \phi} \cdot \frac{\hat{\theta}}{\sin \theta} \frac{\partial}{\partial \phi} = \frac{\hat{\theta}\cdot \hat{\theta}}{\sin^2 \theta} \frac{\partial^2}{\partial \phi^2} + \frac{\hat{\theta}}{\sin^2 \theta} \cdot \frac{\partial \hat{\theta}}{\partial \phi} \frac{\partial }{\partial \phi}
    $$
     
  4. Jan 26, 2016 #3
    Interesting. What exactly is the value of ## \frac{\partial \hat{\theta}}{\partial \phi} ##? I don't seem to know of an expression that would help simplify this, although I seem to understand what you are saying now. Thank you for pointing out that these unit vectors are not fixed...so easy to forget.
     
  5. Jan 26, 2016 #4

    blue_leaf77

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    Do you know how to express the spherical unit vectors in terms of Cartesian unit vectors? ##\hat{\theta}## is something like
    $$
    \hat{\theta} = A_{\theta x} \hat{x} + A_{\theta y} \hat{y} + A_{\theta z} \hat{z}.
    $$
    Find out how the ##A##'s look like as a function of ##r##, ##\theta##, and ##\phi##. The Cartesian unit vectors have fixed direction, thus they won't be acted upon by any differentiation.
     
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