- #1

LarryC

- 6

- 0

$$

-\frac{\hbar^2}{2ml^2}\bigg[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\bigg(\sin\theta\frac{\partial\psi}{\partial\theta}\bigg) + \frac{1}{\sin^2\theta}\frac{\partial^2\psi}{\partial\phi^2} \bigg] + mgl(1-\cos\theta)\psi(\theta,\phi) = E\psi(\theta,\phi)

$$

Using $\sin\theta \approx \theta$ and $\cos\theta\approx 1-\theta^2/2$ leads me to

$$

-\frac{\hbar^2}{2ml^2}\bigg(\frac{\theta}{\Theta}\frac{d\Theta}{d\theta} + \frac{\theta^2}{\Theta}\frac{d^2\Theta}{d\theta^2} + \frac{1}{\Phi}\frac{d^2\Phi}{d\phi^2} \bigg) + \frac{1}{2}mgl\theta^4 = E\theta^2

$$

Here I've already used the ansatz $\psi(\theta,\phi)=\Theta(\theta)\Phi(\phi)$. Of course I can throw away the $\theta^4$ term, but any further simplifications with $\theta^2$ terms would also eliminate the energy, which is what I want. I've also tried to solve the $\Theta(\theta)$ equation with series solutions, and the result seems weird and cannot give my any energy quantizations.

Another attempt is to write the entire kinetic energy term in terms of angular momentum operators, which gives

$$

H=\frac{1}{2ml^2}\bigg(L_\theta^2 + \frac{L_\phi^2}{\sin^2\theta} \bigg) + mgl(1-\cos\theta)

$$

I was hoping to solve this with raising and lowering operators, but that $1/\sin^2\theta$ term is really a pain in the ass. I have no idea of finding a suitable ladder operator that satisfies $[H,\hat{a}] = c\hat{a}$.

Any ideas?