Math Infinitum

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Intuition

Near the bottom of any smooth valley the potential is a parabola: expand it about its minimum and the first term that matters is quadratic. So a molecule’s bond, an atom in a crystal, a mode of an electric circuit and a mode of light all vibrate, for small motions, like a mass on a spring. The quantum harmonic oscillator is therefore the most used model in physics, and it has a remarkable answer: its energies are evenly spaced, one quantum apart, and the lowest is half a quantum above the bottom.

A marble in any bowl, pushed gently, rocks back and forth as if the bowl were a perfect parabola. Only large swings feel the bowl’s true shape.

The potential $\tfrac12m\omega^{2}x^{2}$ and its first five levels, in units of $\hbar\omega$ and of the oscillator length $x_{0}$ . Each level is drawn as far as a classical particle of that energy could go. The levels are $\tfrac12,\tfrac32,\tfrac52,\dots$ : one quantum $\hbar\omega$ apart, and the lowest is not at the bottom.

The oscillator and its scales

A particle of mass $m$ in the potential $\tfrac12m\omega^{2}x^{2}$ . Measured in the units the oscillator itself supplies — a length, a momentum and an energy built from , and — every oscillator is the same problem.

\begin{aligned}\hat{H}&=\frac{\hat{p}^{2}}{2m}+\frac12m\omega^{2}\hat{x}^{2}\\ x_{0}&=\sqrt{\frac{\hbar}{m\omega}},\qquad p_{0}=\frac{\hbar}{x_{0}}=\sqrt{\hbar m\omega}\end{aligned}

What to know first

Near a minimum at $x=a$ , $V(x)\approx V(a)+\tfrac12V''(a)(x-a)^{2}$ : an oscillator with .

The oscillator in its own units

\begin{aligned}&x=x_{0}\xi,\quad x_{0}=\sqrt{\hbar/m\omega}\\ &\Longrightarrow\ \hat{H}=\frac{\hbar\omega}{2}\Bigl(-\frac{d^{2}}{d\xi^{2}}+\xi^{2}\Bigr)\end{aligned}

Write the Hamiltonian in the position representation and change the variable to $\xi=x/x_{0}$ . The kinetic term gains $1/x_{0}^{2}$ and the potential term ; choosing makes both coefficients .

Proof steps

\hat{H}=-\frac{\hbar^{2}}{2m}\frac{d^{2}}{dx^{2}}+\frac12m\omega^{2}x^{2}

The Hamiltonian acting on wavefunctions, with $\hat{p}=-i\hbar\,d/dx$ .

x=x_{0}\xi,\qquad \frac{d^{2}}{dx^{2}}=\frac{1}{x_{0}^{2}}\frac{d^{2}}{d\xi^{2}}

Each derivative with respect to $x$ is $1/x_{0}$ times one with respect to $\xi$ .

\hat{H}=-\frac{\hbar^{2}}{2mx_{0}^{2}}\frac{d^{2}}{d\xi^{2}}+\frac12m\omega^{2}x_{0}^{2}\,\xi^{2}

Substitute both.

x_{0}^{2}=\frac{\hbar}{m\omega}\ \Rightarrow\ \frac{\hbar^{2}}{mx_{0}^{2}}=\hbar\omega=m\omega^{2}x_{0}^{2}

This choice of $x_{0}$ makes the two coefficients equal.

\hat{H}=\frac{\hbar\omega}{2}\Bigl(-\frac{d^{2}}{d\xi^{2}}+\xi^{2}\Bigr)

Factor out $\tfrac12\hbar\omega$ : no constant is left in the bracket.

Applications

Practice

Every Minimum Is a Parabola

Expand any smooth potential about a minimum: the constant does nothing, the linear term vanishes at the minimum, and the quadratic term is the first that matters. Small vibrations about any minimum are therefore harmonic.

V(x)\approx V(a)+\tfrac12V''(a)\,(x-a)^{2}

Try it

Why does the harmonic oscillator describe small vibrations of almost any system?