19.3: The Harmonic Oscillator
We now consider a particle subject to a restoring force \(F = -kx\), as might arise for a mass-spring system obeying Hooke’s Law. The potential is then:
\[ V(x) = - \int_{-\infty}^{\infty} (-kx) dx = V_0 + \dfrac{1}{2} kx^2. \label{20.3.1} \]
If we choose the energy scale such that \(V_0 = 0\) then: \(V(x) = \dfrac{1}{2}kx^2\), and the TISEq looks:
\[ - \dfrac{\hbar^2}{2 \mu} \dfrac{d^2\psi}{dx^2} + \dfrac{1}{2} kx^2 \psi(x) = E \psi(x) \label{20.3.2} \]
After some effort, the eigenfunctions are:
\[ \psi_n(x) = N_n H_n(\alpha^{1/2} x) e^{-\alpha x^2 / 2} \quad n=0,1,2,\ldots,\infty, \label{20.3.3} \]
where \(H_n\) is the Hermite polynomial of degree \(n\), and \(\alpha\) and \(N_n\) are defined by
\[ \alpha = \sqrt{\dfrac{k \mu}{\hbar^2}} \hspace{1.5cm} N_n = \dfrac{1}{\sqrt{2^n n!}} \left( \dfrac{\alpha}{\pi} \right)^{1/4}. \label{20.3.4} \]
The eigenvalues are:
\[ E_n = \hbar \omega \left(n + \dfrac{1}{2} \right), \label{20.3.5} \]
with \(\omega = \sqrt{k/ \mu}\). Notice how, once again, the eigenfunctions and eigenvalues are not continuous. In this case, however, the first eigenvalue corresponds to \(n=0\), but because of the \(\dfrac{1}{2}\) factor in Equation \ref{20.3.5}, the lowest energy state is, once again, not zero. In other words, the two masses of a quantum harmonic oscillator are always in motion. The frequencies at which they vibrate do not form a continuous spectrum. That is, the vibration frequency cannot take any value that we can think of, but only those given by Equation \ref{20.3.5}. The lowest possible energy (the ZPE) will be \(E_0 = \dfrac{1}{2} \hbar \omega\).