← 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


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,


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


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


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


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").