The wavefunction \(\psi(L,t)\) is confined to a circle whenever the eigenvalues L of a particle are only nonzero on the points along a circle. When the wavefunction \(\psi(L,t)\) associated with a particle has non-zero values only on points along a circle of radius \(r\), the eigenvalues \(p\) (of the momentum operator \(\hat{P}\)) are quantized—they come in discrete multiples of \(n\frac{ℏ}{r}\) where \(n=1,2,…\) Since the eigenvalues for angular momentum are \(L=pr=nℏ\), it follows that angular momentum is also quantized.


The wavefunction \(\psi\) is a tensor field defined at each point along the circle: at a particular point the value of this field reads “\(\psi\)” independent of the coordinate value used to label that point. Notice that the coordinate values \(x\) and \(x+(n)(2πr)\) label the same point. It follows that


since the value of a scalar field must be the same at a particular point. If the wavefunction is confined to a circle, then this condition that the wavefunction must “come back to itself” applies to any wavefunction corresponding to any state. The eigenfunction \(\psi_p\) (associated with momentum) must satisfy


It is fairly straightforward to show that \(\psi_p\) is always given by \(\psi_p(x,t)=\psi_p(x)=Ae^{ipx/ℏ}\). The momentum eigenvectors \(|\psi_p⟩\) are those special vectors which satisfy the equation


We can rewrite this equation in terms of the wavefunction as

Let’s multiply both sides of Equation # by \(i/ℏ\) to obtain
The solution to Equation # is given by
$$\psi_p (x)=Ae^{ipx/ℏ}.$$
If we substitute this result into Equation # we get
We can now use algebra to determine what values of \(p\) satisfy Equation #:
$$e^{2πrnip/ℏ}=1⇒2πrip/ℏ=(n)2π\text{     (n=1,2,…)}$$