| ||||
PHY.K02UF Molecular and Solid State Physics | ||||
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 $\psi_n(x)$.
|