   PHY.K02UF Molecular and Solid State Physics

## Harmonic oscillator

The Schrödinger equation for a harmonic oscillator is,

$$-\frac{\hbar^2}{2m}\nabla^2\psi(x) +\frac{k^2x^2}{2}\psi(x) = E\psi(x),$$

where $m$ is the mass of the particle and $k$ is the spring constant. The energy levels of the quantum harmonic oscillator are,

$$E_n = \hbar\omega(n+1/2)\qquad n=0,1,2,\cdots$$

and the angular frequency $\omega$ is,

$$\omega = \sqrt{\frac{k}{m}}.$$

The harmonic oscillator wave functions have the form,

$$\psi_n(x) = \frac{1}{\sqrt{2^nn!}}\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\exp\left(-\frac{m\omega x^2}{2\hbar}\right)H_n\left(\sqrt{\frac{m\omega}{\hbar}}x\right)\qquad n=0,1,2,\cdots,$$

where $H_n(x)$ are the Hermite polynomials. The wave function crosses zero $n$ times and decays due to the exponential factor with a characteristic length $\xi = \sqrt{\frac{\hbar}{m\omega}}$. There is a peak in amplitude of the wave function near the classical turning point, $x_n = \sqrt{2E_n/k}$. The wave function can be written in terms of $\xi$,

$$\psi_n(x) = \frac{1}{\pi^{1/4}\sqrt{2^nn!}}\frac{1}{\sqrt{\xi}}\exp\left(-\frac{x^2}{2\xi^2}\right)H_n\left(x/\xi\right)\qquad n=0,1,2,\cdots.$$

Note that in this form it is clear that $\psi_n(x)$ has the units of m-1/2. The code below will calculate the the value of the wave function $\psi_n(x)$.



The harmonic oscillator wave functions are plotted and tabulated below. $n=$ , $m=$ kg, $k=$ N/m, . The red dots are the classical turning points. The parameters for some diatomic molecules can be loaded with these buttons.

 $\psi_n$ [m-1/2] $\psi^2_n$ [m-1] $\omega=$ rad/s $\hbar\omega=$ eV $E_n=$ eV $\xi=$ Å $x_n=$ Å $x$ [Å] $x$ [Å]
 $x$ [m]    $\psi_n(x)$ [m-1/2]    $\psi_n^2(x)$ [1/m]

### Correspondence principle

The energy of a harmonic oscillator is a sum of the kinetic energy and the potential energy,

$$E= \frac{mv^2}{2}+\frac{kx^2}{2}.$$

At the turning points where the particle changes direction, the kinetic energy is zero and the classical turning points for this energy are $x=\sqrt{2E/k}$. For large quantum numbers $n$, the probability of finding a particle at some location $|\psi|^2$ approximates the distribution you would expect for a classical particle which would also be more likely to be found near the classical turning points. This is an example of the correspondence principle.

 $\psi^2_{20}$ [m-1] $x$ [Å]