Particle in a Ring
There is a potential of zero where a particle lies on a ring of radius, (r), therefore,
\(H = - \frac{\hbar^{2}}{2m} \frac{d^{2}}{dx^{2}}\) for all coordinates.
Accordingly, the Hamiltonian of the particle of the ring would be as follows, where the coordinate system is now represented in terms of \(r\) and \(\phi\):
\[H = - \frac{\hbar^{2}}{2m} \left(\frac{d^{2}}{dx^{2}} + \frac{d^{2}}{dy^{2}}\right)\] \[= - \frac{\hbar^{2}}{2mr^{2}} \frac{d^{2}}{d\phi^{2}}\] \[= - \frac{\hbar^{2}}{2I} \frac{d^{2}}{d\phi^{2}}\]where (I =) moment of inertia.
The blue squiggle line is a representation of the wavefunction.
Show that (e^{i m_{l} \phi} = \cos(m_{l} \phi) + i \sin(m_{l} \phi)) is an eigenfunction of the Hamiltonian.
\[H \Psi = E \Psi\] \[H \Psi = - \frac{\hbar^{2}}{2I} \frac{d^{2}}{d\phi^{2}} \left(e^{i m_{l} \phi}\right)\] \[H \Psi = - \frac{\hbar^{2}}{2I} \frac{d}{d\phi} \left(i m_{l} e^{i m_{l} \phi}\right)\] \[H \Psi = - \frac{\hbar^{2}}{2I} \left(- m_{l}^{2} e^{i m_{l} \phi}\right)\] \[H \Psi = \frac{\hbar^{2} m_{l}^{2}}{2I} e^{i m_{l} \phi}\] \[= \frac{\hbar^{2} m_{l}^{2}}{2I} \Psi\]
Where \(E = \frac{\hbar^{2} m_{l}^{2}}{2I}\) for a particle in a ring. At this time, there is no quantization because there are no restrictions or boundaries. A restriction must be set to achieve quantization: the wave must meet smoothly and continuously, or the wavefunction must start and end in the same place as seen in the wavefunction figure above. Therefore for a ring where the phase (\phi) ranges from (0) to (2\pi), a wavefunction at a given angle (\phi) must be equal to the wavefunction at (\phi + 2\pi).
\[\Psi(\phi) = \Psi(\phi + 2\pi)\]If the wavefunction is defined as \(\Psi = e^{i m_{l} \phi}\) then how does this new restriction (\(\Psi(\phi) = \Psi(\phi + 2\pi)\)) lead to quantization? Note: the wavefunction is now in spherical coordinates.
Plug \(\Psi = e^{i m_{l} \phi}\) into \(\Psi(\phi) = \Psi(\phi + 2\pi)\)
\[e^{i m_{l} \phi} = e^{i m_{l} (\phi + 2\pi)}\] \[= e^{i m_{l} \phi} \cdot e^{2\pi i m_{l}}\] \[1 = e^{2\pi i m_{l}}\] \[1= (e^{i \pi})^{2 m_{l}}\] \[1= (-1)^{2 m_{l}}\]Therefore:
\[m_{l} = ..., -2, -1, 0, 1, 2, ...\]In order for \(1= (-1)^{2 m_{l}}\) to equal each other then (m_{l}) must equal any whole number. Because of this restriction we get our (m_{l}) which is a quantum number and demonstrates the quantization of the particle on a ring.