1
$\begingroup$

Wikipedia derives the eigenstates and eigenenergies of a particle on a ring by arguing that since the particle's position only depends on $\theta$

$ \nabla^2 = \frac{1}{R^2}\frac{\partial^2}{\partial \theta^2} $

and thus, with periodic boundary conditions, the Schrodinger equation yields $\psi(\theta) = \frac{1}{\sqrt{2 \pi}} e^{in\theta}$

I'm curious how this generalizes to ellipses. On one hand, the particle is still constrained to a closed ring, and so it's position can only depend on one parameter, but on the other hand the particle is no longer constrained to a single radius. How would derivation change?

$\endgroup$
3
$\begingroup$

The canonical quantization procedure suggests that the shape of the closed curve largely doesn't matter. Briefly: suppose we have a classical particle confined to a closed loop that's given by a function $\gamma: [0, 1] \to \mathbb{R}^3$, with $\gamma(0) = \gamma(1)$. We can use the arclength $s$ as a generalized coordinate in the Lagrangian, in which case the Lagrangian is simply $$ L = \frac{1}{2} m \dot{s}^2. $$ The corresponding Hamiltonian is then $$ H = \frac{p_s^2}{2 m}, $$ where $p_s \equiv \partial L/\partial \dot{s}$. The Poisson brackets between our coordinates are then $\{s,s\}=\{p_s,p_s\} = 0$ and $\{s,p_s\} = 1$.

To pass from the classical to the quantum system, we "promote" the Hamiltonian coordinates to operators, and "promote" the Poisson brackets to commutators. In other words, we want to find operators that satisfy the algebra $$ [\hat{s}, \hat{s} ] = [\hat{p}_s, \hat{p}_s] = 0, \qquad [\hat{s}, \hat{p}_s ] = i \hbar. $$ But in terms of the wavefunctions $\psi(s)$, it's not hard to show that the operators $$ \hat{s} \psi = s \psi, \qquad \hat{p}_s \psi = -i \hbar \frac{d}{ds}\psi $$ satisfy this relationship. (You might recognize these as the standard position and momentum operators for a particle moving on a straight line, with $x \to s$.)

The time-independent Schrödinger equation is then $$ E \psi = \hat{H} \psi = \frac{\hat{p}_s^2}{2m} \psi = - \frac{\hbar^2}{2 m} \frac{d^2 \psi}{ds^2}. $$ for which the solutions are of the form $\psi(s) \propto e^{\pm i \sqrt{2 m E}s/\hbar}.$ If the total circumference of the curve is $S$, then the requirements that $\psi(0) = \psi(S)$ and $\psi'(0) = \psi'(S)$ imply that we must have $$ \frac{\sqrt{2 m E}S}{\hbar} = 2 \pi n \quad \Rightarrow \quad E = \frac{h^2 n^2}{2 m S^2}. $$ Note that these are the same energy eigenvalues as the particle on a ring, since the circumference of the ring is $2 \pi R$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for the answer. Introducing a generalized coordinate is a great way to look at it. $\endgroup$ – user2944352 May 28 '19 at 8:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.