How Can Quantum Mechanics Explain the Eigenstates of a Spherical Pendulum?

In summary, the conversation discusses the difficulty in finding the eigenstates of a spherical pendulum under the small angle approximation. The final result is expected to be a combination of a harmonic oscillator in $\theta$ and a free particle in $\phi$. Different attempts, such as using series solutions and angular momentum operators, have been made but have not been successful in finding the desired energy quantizations. A possible solution is suggested, involving the use of spherical harmonics and 3j symbol calculations. The conversation ends with the question of whether the approximate form can yield something related to the harmonic oscillator.
  • #1
LarryC
6
0
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?
 
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  • #2
LarryC said:
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.).
 
  • #3
DrClaude said:
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##.
 

Related to How Can Quantum Mechanics Explain the Eigenstates of a Spherical Pendulum?

1. What is a Quantum Spherical Pendulum?

A Quantum Spherical Pendulum is a theoretical physical system that combines the concepts of a pendulum and quantum mechanics. It consists of a mass attached to a string or rod, which is free to swing in a spherical motion, and is influenced by quantum effects such as superposition and uncertainty.

2. How does a Quantum Spherical Pendulum differ from a classical pendulum?

A classical pendulum follows the laws of classical mechanics, while a Quantum Spherical Pendulum follows the laws of quantum mechanics. This means that the behavior and movement of the pendulum are governed by different principles, leading to distinct differences in their motion and properties.

3. What are the applications of a Quantum Spherical Pendulum?

Currently, there are no practical applications for a Quantum Spherical Pendulum, as it is a theoretical concept. However, it can be used as a model to better understand the behavior of quantum systems and to explore the possibilities of quantum computing.

4. What are some challenges in studying a Quantum Spherical Pendulum?

One of the main challenges in studying a Quantum Spherical Pendulum is the complexity of quantum mechanics and the difficulty in accurately measuring and predicting the behavior of quantum systems. Additionally, the concept of a spherical pendulum is already a difficult system to analyze, and adding the quantum aspect makes it even more complex.

5. Can a Quantum Spherical Pendulum exist in the physical world?

Currently, there is no evidence to suggest that a Quantum Spherical Pendulum can exist in the physical world. It is a theoretical concept that has not been observed or created in any experiments. However, with advancements in quantum technology, it is possible that we may be able to create and observe a Quantum Spherical Pendulum in the future.

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