← Quantum Mechanics

The Simple Harmonic Oscillator

Sunday, February 27, 2022

What is an oscillator?

A classical oscillator is an object of mass mm attached to a spring of force constant kk. The spring exerts a restoring force F=kxF=-kx on the object, where xx is the displacement from the equilibrium position.

Harmonic oscillators have an angular frequency ω0=k/m\omega_0=\sqrt{k/m} and period T=2πm/kT=2\pi\sqrt{m/k}. Their amplitude is found by their maximum displacement, x0x_0, and their maximum kinetic energies occur at the turning points x=±x0x=\pm x_0. Therefore, the motion is confined to x0<x<+x0-x_0\lt x\lt +x_0.

The Quantum Mechanical Version of a Harmonic Oscillator

Although no natural example of a one-dimensional quantum oscillator exist, there are plenty of systems that behave approximately like oscillators. For instance, a vibrating diatomic molecule.

A force F=kxF=-kx has potential energy U=12kx2U=\frac{1}{2}kx^2, meaning

22md2ψdx2+12kx2ψ=Eψ-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+\frac{1}{2}kx^2\psi=E\psi

There are no boundaries in this situation, but we know that as x+x\rightarrow +\infty and xx\rightarrow -\infty, ψ0\psi\rightarrow 0. The simplest function that satisfies this condition is ψ(x)=Aeax2\psi\left(x\right)=Ae^{-ax^2}, where the constant aa and energy EE can be found by finding the first and second derivatives of the wave function:

dψdx=2ax(Aeax2)d2ψdx2=2a(Aeax2)2ax(2ax)Aeax2=(2a+4a2x2)Aeax2\frac{d\psi}{dx}=-2ax\left(Ae^{-ax^2}\right)\newline \frac{d^2\psi}{dx^2}=-2a\left(Ae^{-ax^2}\right)-2ax\left(-2ax\right)Ae^{-ax^2}=\left(-2a+4a^2x^2\right)Ae^{-ax^2}

Plugging this into the Schrödinger equation from above,

2am2a22mx2+12kx2=E\frac{\hbar^2a}{m}-\frac{2a^2\hbar^2}{m}x^2+\frac{1}{2}kx^2=E

Instead of solving for xx, we are trying to make this true for all values of xx by finding constants that make that true. For this to be true,

2a22m+12k=02am=E-\frac{2a^2\hbar^2}{m}+\frac{1}{2}k=0\newline \frac{\hbar^2a}{m}=E

Which results in

a=km2E=12k/ma=\frac{\sqrt{km}}{2\hbar}\newline E=\frac{1}{2}\hbar\sqrt{k/m}

Using the equation for the angular frequency of a classical harmonic oscillator, E=12ω0E=\frac{1}{2}\hbar\omega_0.

The coefficient AA can be found using the normalization condition, and the result for the ground state wave function is A=(mω0/π)1/4A=\left(m\omega_0/\hbar\pi\right)^{1/4}. Therefore, the complete wave function is

ψ(x)=(mω0π)1/4e(km/2)x2\psi\left(x\right)=\left(\frac{m\omega_0}{\hbar\pi}\right)^{1/4}e^{-\left(\sqrt{km}/2\hbar\right)x^2}

Note that the wave function penetrates into the forbidden region (beyond x=±x0x=\pm x_0), whereas the classical oscillator does not.

Energies

The solution above only works for the ground-state. The more general solution is ψn(x)=Afn(x)eax2\psi_n\left(x\right)=Af_n\left(x\right)e^{-ax^2}, where fn(x)f_n\left(x\right) is a polynomial in which the highest power of xx is xnx_n. The energies are

En=(n+12)ω0        n=0,1,2,...E_n=\left(n+\frac{1}{2}\right)\hbar\omega_0~~~~~~~~n=0,1,2,...

Probability distributions

Below are a few examples of what probability densities look like for harmonic oscillators:

Harmonic oscillator

Uncertainties

The resulting uncertainties for this situation are as follows:

Δx=/2mω0Δp=ω0m/2\Delta x=\sqrt{\hbar/2m\omega_0}\newline\Delta p=\sqrt{\hbar\omega_0m/2}

And the product of the uncertainties is ΔxΔp=/2\Delta x\Delta p=\hbar/2, meaning the uncertainty is at a minimum (called "compact").