# How is particle in a ring is in bound state?

Particle in a potential well forms a bound state which gives rise to discrete energy levels. But states for particle in a ring also have discrete energy levels even where there is no potential well. In what aspect hence the system is 'bound'? Or is it that it's not necessary for a potential well to exist for system to be bound?

If we immagine a unidimensional infinite potential well, the confined particle do not have access to the region where the potetial assumes the value $+\infty$. So the wave describing the particle must be null on the walls, because the squared modulus of the wave function represents the probability of finding the particle in a given point: you can not find the particle there. Since not every wave nullifies in those points, only the waves with an appropriate amount of periods are allowed, and from this comes the discrete values of energy allowed.
Describing a particle on a ring, instead, requires another condition. Suppose $\psi(\theta)$ is the wavefunction describing the particle, where $\theta$ is the angle the paramterizes the ring. $\psi(\theta)$ allows us to calculate the probability of finding the particle at every angle: but one can immagine that the following condition must be satisfied: $$\psi(\theta)=\psi(\theta+2\pi)$$
Note: this condition can be made more weak, because one could require that the probability must be the same after a complete lap, not the wavefunction. This condition can be expressed as follows: $$|\psi(\theta)|^2=|\psi(\theta+2\pi)|^2$$ And its solutions can lead to strange results: $$\psi(\theta)=e^{i\alpha}\psi(\theta+2\pi)\qquad \alpha\in \mathbb{R}$$ It turns out that different values of $\alpha$ describe different types of particles, in particular bosons ($\alpha =0$), fermions ($\alpha =\pi$), and anyons ($\alpha \neq 0,\pi$).