# Homework Help: Change of Variables Legendre's Equation

1. Mar 4, 2017

### Crush1986

1. The problem statement, all variables and given/known data
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

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

2. Relevant 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 $$\frac{d^2}{d^2 \theta} = -sin [ \frac {d}{d \theta} \frac{d}{du} + \frac{du}{d \theta} \frac {d^2}{du^2}] = -sin \theta [-sin \theta \frac{d^2}{du^2}-sin \theta \frac{d^2}{du^2}] = 2 sin^2 \theta \frac{d^2}{du^2}$$

3. 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.

Last edited: Mar 4, 2017
2. Mar 4, 2017

### Orodruin

Staff Emeritus
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.

3. Mar 4, 2017

### 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?

4. Mar 4, 2017

### kuruman

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

5. Mar 4, 2017

### 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?

6. Mar 4, 2017

### 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.