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?


closed as off-topic by sammy gerbil, Aaron Stevens, Sebastian Riese, Emilio Pisanty, stafusa Sep 30 '18 at 0:15

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  • $\begingroup$ Consider doing separation of variables before expanding everything to first/second/whatever order. Write down the separated equations and then do an approximation. $\endgroup$ – Gabriel Golfetti Sep 29 '18 at 22:13
  • $\begingroup$ That's an interesting suggestion, Gabriel. I've actually tried it and it doesn't seem to produce anything cheerful. I guess it doesn't matter whether you take the approximation before or after the differentiation operations as long as no divergent series shows up. $\endgroup$ – LarryC Sep 29 '18 at 22:26
  • $\begingroup$ You know where you should end up, which is a quantum harmonic oscillator in $\theta$ and a free particle in $\phi$. Do you know what the equation for the qho oscillator looks like in polar coordinates? $\endgroup$ – bRost03 Sep 29 '18 at 23:45
  • $\begingroup$ Your question was frozen as I was writing out an answer for you. To solve this, use operator algebra. The entire first term is just $\frac{L^2}{2ml^2}$, where $L$ is the angular momentum operator. and the second term can be recast as $\frac{1}{2}mgl\theta^2$ so then your equation looks just like the quantum harmonic oscillator and you can use the techniques associated with that to solve this problem. Good luck! $\endgroup$ – bRost03 Sep 30 '18 at 0:23