# 1D Time independent Schrodinger equation applied to a ring of Radius $R$

In the lectures we have been doing basic examples of Applying the TISE to determine the solutions of simple situations such as the finite and infinite square well and I understand how to find the solutions (eigenstates) in the form u(x).

However if we consider a free particle of mass m which is constrained to move along a circumference of a ring of radius R (keeping it all in 1D) how would I begin to solve the 1D time independent schrodinger equation if V(x) = 0 along the ring and periodic boundary conditions were applied. I am confident that the solution would be of the form $u(x) = Aexp(±ikx)$ but other than that I am stuck.

Any help would be appreciated.

Let's start with the Schrodinger equation in 2D with $V = 0$.

$$H \psi = - {\hbar^2 \nabla^2 \psi \over 2m} = - {\hbar^2 \over 2m} \left({1 \over r} {\partial \over \partial r} \left( r {\partial \psi \over \partial r}\right) + {1 \over r^2} {\partial^2 \psi \over \partial \theta^2}\right).$$

Now if we constrain the particle to move on a circle of radius R, the radial derivative drop out and we're left with

$$H \psi = - {\hbar^2 \over 2m R^2} {\partial^2 \psi \over \partial \theta^2}.$$

The eigenfunctions of this operator are $e^{il \theta}$, where we require $l$ to be an integer in order to satisfy the periodicity condition.

Defining $\psi_l = {1 \over 2 \pi} \int_0^{2 \pi} d\theta\; e^{-il\theta} \psi(\theta)$, applying the azimuthal Fourier transform to both sides, and integrating by parts twice on the left-hand-side yields

$$E_l \psi_l = {\hbar^2 l^2 \over 2 m R^2}\psi_l ~~~\Rightarrow~~~ E_l = {\hbar^2 l^2 \over 2 m R^2}.$$

To find the time-dependent wavefunction, we simply tack on the exponential time-dependence for each of the eigenfunctions,

$$\psi(\theta,t) = \sum_{-\infty}^{\infty} \psi_l(0) e^{il\theta - i E_l t/ \hbar},$$

where $\psi_l(0) = {1 \over 2 \pi} \int_0^{2 \pi} d\theta\; e^{-il\theta} \psi(\theta,0)$. To get the time-independent version, just set $t = 0$.