Change of Variables Legendre's Equation

[Crush1986]

Homework Statement

Determine the spherical harmonics and the eigenvalues of \vec{\hat{L}}^2 by solving the eigenvalue equation \vec{\hat{L}}^2 |\lambda, m \rangle in position space,

[\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}

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

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

Homework Equations

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

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

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

I'm then saying -sin \theta = \frac{du}{d \theta} then using the chain rule I get \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}<br />

The Attempt at a Solution

I'd appreciate any help. When i put this all together I'm getting
2(1-u^2)\frac{d^2 \Theta_{\lambda,0}}{du^2} - u \frac{d \Theta_{\lambda,0}}{du} + \lambda \Theta_{\lambda,0}=0
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.

 

[Orodruin]

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.

 

[Crush1986]

Ok,

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

\frac{d}{d \theta}(sin \theta \frac {d}{d \theta})
= sin \theta \frac{d}{du}(sin^2 \theta \frac{d}{du})
one term is sin^3 \theta \frac{d^2}{du^2}
I think I keep messing up the second term. How do you compute it?

 

[kuruman]

Note that $\sin^2\theta = 1 - u^2$. That should help.

 

[Crush1986]

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 \frac{d}{du} sin^2 \theta come out to?

 

[Crush1986]

I JUST GOT IT! ugh, I don't know why I just didn't see how to do \frac{d}{du} sin^2 \theta

Thanks all.