# Quantum Spherical Pendulum

I have trouble with finding the eigenstates of a spherical pendulum (length $l$, mass $m$) under the small angle approximation. My intuition is that the final result should be some sort of combinations of a harmonic oscillator in $\theta$ and a free particle in $\phi$, but it's not obvious to see this from the Schrodinger equation:
$$-\frac{\hbar^2}{2ml^2}\bigg[\frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\bigg(\sin\theta\frac{\partial\psi}{\partial\theta}\bigg) + \frac{1}{\sin^2\theta}\frac{\partial^2\psi}{\partial\phi^2} \bigg] + mgl(1-\cos\theta)\psi(\theta,\phi) = E\psi(\theta,\phi)$$
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?

## Answers and Replies

DrClaude
Mentor
Another attempt is to write the entire kinetic energy term in terms of angular momentum operators, which gives
$$H=\frac{1}{2ml^2}\bigg(L_\theta^2 + \frac{L_\phi^2}{\sin^2\theta} \bigg) + mgl(1-\cos\theta)$$
More simply, you have
$$H=\frac{1}{2ml^2} L^2 + mgl(1-\cos\theta)$$
The eigenfunctions of ##L^2## are the spherical harmonics, and
$$Y_1^0(\theta,\phi) = \sqrt{\frac{3}{4\pi}} \cos \theta$$
so
$$mgl(1-\cos\theta) = mgl \left[ 1 - \sqrt{\frac{4\pi}{3}} Y_0^1 \right]$$
The coupling between the spherical harmonics due to this term is then the integral of a product of three spherical harmonics, which can be expressed in terms of 3j symbol and easily calculated (check out the books on angular momentum by Zare or by Varshalovich et al.).

More simply, you have
$$H=\frac{1}{2ml^2} L^2 + mgl(1-\cos\theta)$$
The eigenfunctions of ##L^2## are the spherical harmonics, and
$$Y_1^0(\theta,\phi) = \sqrt{\frac{3}{4\pi}} \cos \theta$$
so
$$mgl(1-\cos\theta) = mgl \left[ 1 - \sqrt{\frac{4\pi}{3}} Y_0^1 \right]$$
The coupling between the spherical harmonics due to this term is then the integral of a product of three spherical harmonics, which can be expressed in terms of 3j symbol and easily calculated (check out the books on angular momentum by Zare or by Varshalovich et al.).

Thanks for the great reference! I am just wondering if I can get something related to the harmonic oscillator from out of the approximate form
$$\bigg[\frac{L^2}{2ml^2}+\frac{1}{2}mgl\theta^2\bigg]\psi = E\psi$$
This looks like a harmonic oscillator, but ##L## contains both ##\theta## and ##\phi## and I don't know how to commute it with ##\theta##.