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Homework Help: Change of Variables Legendre's Equation

  1. Mar 4, 2017 #1
    1. The problem statement, all variables and given/known data
    Determine the spherical harmonics and the eigenvalues of [tex] \vec{\hat{L}}^2 [/tex] by solving the eigenvalue equation [tex] \vec{\hat{L}}^2 |\lambda, m \rangle [/tex] in position space,

    [tex] [\frac{1}{sin \theta} \frac{\partial}{\partial \theta} ( sin \theta \frac{\partial}{\partial \theta} ) + \frac{1}{sin^2 \theta} \frac{\partial^2}{\partial \phi^2}] \Theta_{\lambda,m}(\theta) e^{im\phi} [/tex]

    restrict your attention to the m = 0 case. Rewrite the equation in terms of u = cos \theta and show that it becomes

    [tex] (1-u^2) \frac{d^2 \Theta_{\lambda, 0}}{du^2}-2u \frac{d \Theta_{\lambda,m}}{du} + \lambda \Theta_{\lambda,0} = 0 [/tex]

    2. Relevant equations
    distributing the [tex] \frac{\partial}{\partial \theta} [/tex] and recognizing that the double phi derivative turns to zero I am working with the equation [tex] -[cot \theta \frac{\partial}{\partial \theta} + \frac{\partial^2}{\partial \theta^2}] \Theta_{\lambda,0} = \lambda \Theta_{\lambda,0} [/tex]

    so, using [tex] u = cos \theta [/tex] I get that [tex] du = -sin \theta d \theta [/tex], which in turn gives me that [tex] \frac{d}{d \theta} = - sin \theta \frac{d}{du} [/tex]

    So I'm pretty sure I'm messing up when I am substituting in for [tex] \frac{d^2}{d^2 \theta} = -sin \theta \frac{d}{du} (-sin \theta \frac{d}{du}) [/tex]

    I'm then saying [tex] -sin \theta = \frac{du}{d \theta} [/tex] then using the chain rule I get [tex] \frac{d^2}{d^2 \theta} = -sin [ \frac {d}{d \theta} \frac{d}{du} + \frac{du}{d \theta} \frac {d^2}{du^2}] = -sin \theta [-sin \theta \frac{d^2}{du^2}-sin \theta \frac{d^2}{du^2}] = 2 sin^2 \theta \frac{d^2}{du^2}

    3. The attempt at a solution
    I'd appreciate any help. When i put this all together I'm getting
    [tex] 2(1-u^2)\frac{d^2 \Theta_{\lambda,0}}{du^2} - u \frac{d \Theta_{\lambda,0}}{du} + \lambda \Theta_{\lambda,0}=0 [/tex]
    I'm pretty sure it's just me doing something stupid while trying to find what the second derivative with respect to theta equals after the change of variable.
    Last edited: Mar 4, 2017
  2. jcsd
  3. Mar 4, 2017 #2


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    The LaTeX of your most important steps seems screwed upnan it is therefore difficult to tell exactly what you have done.

    Let me just say this though: You are taking the long way around. I suggest not using the derivatives to make it two terms before making the substitution.
  4. Mar 4, 2017 #3

    I've tried this way as well and I always get snagged at the same part.

    [tex] \frac{d}{d \theta}(sin \theta \frac {d}{d \theta}) [/tex]
    [tex] = sin \theta \frac{d}{du}(sin^2 \theta \frac{d}{du}) [/tex]
    one term is [tex] sin^3 \theta \frac{d^2}{du^2} [/tex]
    I think I keep messing up the second term. How do you compute it?
  5. Mar 4, 2017 #4


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    Note that ##\sin^2\theta = 1 - u^2##. That should help.
  6. Mar 4, 2017 #5
    Kuruman, I got that. that term works out perfect. But I'm not getting a -2u for my first derivative term, I only have -u. I'm not applying the chain or product rule correctly when computing what d^2/d(theta)^2 is.

    What does [tex] \frac{d}{du} sin^2 \theta [/tex] come out to?
  7. Mar 4, 2017 #6
    I JUST GOT IT! ugh, I don't know why I just didn't see how to do [tex] \frac{d}{du} sin^2 \theta [/tex]

    Thanks all.
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