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The Harmonic Oscillator

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) = $\displaystyle - \int_{-\infty}^{\infty} (-kx) dx$ (118)
  = $\displaystyle V_0 + \frac{1}{2} kx^2$  

If we choose the energy scale such that V0 = 0 then V(x) = (1/2)kx2. This potential is also appropriate for describing the interaction of two masses connected by an ideal spring. In this case, we let x be the distance between the masses, and for the mass m we substitute the reduced mass $\mu$. Thus the harmonic oscillator is the simplest model for the vibrational motion of the atoms in a diatomic molecule, if we consider the two atoms as point masses and the bond between them as a spring. The one-dimensional Schrödinger equation becomes

\begin{displaymath}- \frac{\hbar^2}{2 \mu} \frac{d^2\psi}{dx^2} + \frac{1}{2} kx^2 \psi(x) =
E \psi(x)
\end{displaymath} (119)

After some effort, the eigenfunctions are

\begin{displaymath}\psi_n(x) = N_n H_n(\alpha^{1/2} x) e^{-\alpha x^2 / 2} \hspace{1.0cm}
n=0,1,2,\ldots
\end{displaymath} (120)

where Hn is the Hermite polynomial of degree n, and $\alpha$ and Nn are defined by

\begin{displaymath}\alpha = \sqrt{\frac{k \mu}{\hbar^2}} \hspace{1.5cm}
N_n = \frac{1}{\sqrt{2^n n!}} \left( \frac{\alpha}{\pi} \right)^{1/4}
\end{displaymath} (121)

The eigenvalues are

\begin{displaymath}E_n = \hbar \omega (n + 1/2)
\end{displaymath} (122)

with $\omega = \sqrt{k/ \mu}$.


next up previous contents
Next: The Rigid Rotor Up: Some Analytically Soluble Problems Previous: The Particle in a