# Bound states of the Schrödinger equation on $S^1\times \mathbb{R}^3$

I wana study the $\textbf{bound states}$ of a particle in the space $\mathbb{R}^3\times S^1$ which is under the effect of a time independent potential $V(\vec{r})$ but is free on $S^1$. So, I begin by writing the Schrödinger equation

$i\hbar\frac{\partial}{\partial{}t}\psi=\left(\frac{-\hbar^2}{2m}[\nabla^2+\frac{\partial}{R^2\partial\theta^2}]+V(\vec{r}\right)\psi$

and proceed to solve by separation of variables (the coordinate $\theta$ is an angle used to parameterize a point of $S^1$, and $R$ is the radius of the space)

$\psi(\vec{r},\theta;t)=\phi(\vec{r})\beta(\theta)N(t)$

This way I get three equations, the first is just the time dependent one, just the same we get solving the Schrödinger equation in $\mathbb{R}^3$

$i{}N'(t)=\omega{}N(t)$

whose solution is $N(t)=C_1e^{-i\omega{}t}$

The second equation is the one corresponding to $\theta$

$-\frac{\hbar}{2mR^2}\beta''(\theta)=\omega_1\beta(\theta)$

where $\omega_1$ is just a constant that appears after separating the variables. The solutions are $\beta(\theta)=C_2e^{i\theta{}R\sqrt{\frac{2m\omega_1}{\hbar}}}+C_3e^{-i\theta{}R\sqrt{\frac{2m\omega_1}{\hbar}}}$

The second equation is analogous to the typical time independent Schrödinger equation but this time the $\omega$ is not the same as in $N(t)$ because of the $\theta$ part.

$\frac{-\hbar^2}{2m}\nabla^2\phi+V(\vec{r})\phi=\hbar(\omega-\omega_1)\phi$

Having arrived here I am unsure how to proceed. I want to study the $\textbf{bound states}$, so, do I have to consider the solutions with $\omega$ negative or the ones with $(\omega-\omega_1)$ negative, since this is the constant appearing in the equation who has the potential?

Furthermore, what boundary conditions should I apply to $\beta(\theta)$?

Writing the derivation in detail would have taken too much space so I fear that some things I have written may not be too clear, let me know.

I would appreciate any hint. Thanks!

-
You have the constraint of periodicity for the $\beta$ function : $\beta(\theta)= \beta(\theta + 2\pi)$. With your equation $\frac{\hbar}{2mR^2}\beta''(\theta)=\omega_1\beta(\theta)$, it is only possible if $\omega_1 <0$ (to have imaginary exponents in the exponentials). The periodicity constraint implies that $R \sqrt{\frac{-2m\omega_1}{\hbar}}$ is an integer. –  Trimok Dec 14 '13 at 17:57
after posting the question I have found this pagehttp://en.wikipedia.org/wiki/Particle_in_a_ring, and we can have solutions with positive $\omega$, right? actually omega needs to be always positive (maybe you just confused the sign and you meant that)thanks for the answer anyway! –  silvrfück Dec 14 '13 at 18:10
I trusted you for the equation : $\frac{\hbar}{2mR^2}\beta''(\theta)=\omega_1\beta(\theta)$... With this equation, $\omega_1$ has to be negative –  Trimok Dec 14 '13 at 18:22
hey you are right, the equation is supposed to have a minus, my bad! –  silvrfück Dec 14 '13 at 18:24