2.9 The harmonic oscillator

V(x)=\frac12 m\omega^2x^2\tag{2.9.1}

Why a parabola

Expand any potential $V(x)$ about a minimum at $x_0$ :

V(x)=V(x_0)+\underbrace{V'(x_0)}_{=0}(x-x_0)+\frac12V''(x_0)(x-x_0)^2+\cdots\tag{2.9.2}

The linear term vanishes at a minimum, and the constant can be absorbed into the zero of energy. The leading surviving term is always quadratic. Set $m\omega^2\equiv V''(x_0)$ and you have a harmonic oscillator.

So any time you study “small vibrations” you are doing harmonic-oscillator physics: the vibrational spectra of diatomic molecules, phonons in a crystal, every mode of the electromagnetic field, LC circuits, the approximate description of a superconducting qubit… the list goes on.

The ladder-operator method

Grinding through the differential equation (series solution plus Hermite polynomials) certainly works, but the algebraic route is faster and far more illuminating.

Derivation: from a commutator to the entire spectrumadvanced~12 min

Step 1: factorise the Hamiltonian.

\hat H=\frac{\hat p^2}{2m}+\frac12m\omega^2\hat x^2\tag{2.9.3}

If $\hat p$ and $\hat x$ were ordinary numbers we could write $p^2+(m\omega x)^2=(m\omega x-\ii p)(m\omega x+\ii p)$ . Operators do not commute, so the factorisation is slightly off — off by exactly the commutator. Define

\hat a_\pm=\frac{1}{\sqrt{2\hbar m\omega}}\left(\mp\ii\hat p+m\omega\hat x\right)\tag{2.9.4}

Step 2: compute the commutator. Using $[\hat x,\hat p]=\ii\hbar$ :

\hat a_-\hat a_+=\frac{1}{2\hbar m\omega}\left(\hat p^2+(m\omega\hat x)^2-\ii m\omega[\hat x,\hat p]\right) =\frac{\hat H}{\hbar\omega}+\frac12\tag{2.9.5}

Similarly $\hat a_+\hat a_-=\frac{\hat H}{\hbar\omega}-\frac12$ . Subtracting,

[\hat a_-,\hat a_+]=1\tag{2.9.6}

and

\hat H=\hbar\omega\left(\hat a_+\hat a_-+\frac12\right)\tag{2.9.7}

Step 3: show that they raise and lower the energy. Suppose $\hat H\psi=E\psi$ and consider $\hat a_+\psi$ :

\hat H(\hat a_+\psi)=\hbar\omega\left(\hat a_+\hat a_-+\tfrac12\right)\hat a_+\psi\tag{2.9.8}

Use $\hat a_-\hat a_+=\hat a_+\hat a_-+1$ to move $\hat a_+$ all the way left:

=\hbar\omega\,\hat a_+\left(\hat a_-\hat a_++\tfrac12\right)\psi =\hat a_+\left[\hbar\omega\left(\hat a_+\hat a_-+\tfrac12\right)+\hbar\omega\right]\psi =(E+\hbar\omega)(\hat a_+\psi)\tag{2.9.9}

So $\hat a_+\psi$ is also an eigenstate, with energy higher by $\hbar\omega$ . Likewise $\hat a_-\psi$ has energy lower by $\hbar\omega$ . Hence “ladder operators”.

Step 4: the ladder must have a bottom rung. Applying $\hat a_-$ repeatedly would lower the energy indefinitely. But the energy is bounded below:

E=\langle\hat H\rangle=\frac{\langle\hat p^2\rangle}{2m}+\frac12m\omega^2\langle\hat x^2\rangle>0\tag{2.9.10}

(both terms are expectation values of non-negative operators, and they cannot vanish together). The only escape from the contradiction is a ground state $\psi_0$ satisfying

\hat a_-\psi_0=0\tag{2.9.11}

Step 5: solve for the ground state. This is a first-order equation, far easier than the original second-order one:

\frac{1}{\sqrt{2\hbar m\omega}}\left(\hbar\frac{\dd}{\dd x}+m\omega x\right)\psi_0=0 \;\Longrightarrow\; \frac{\dd\psi_0}{\psi_0}=-\frac{m\omega}{\hbar}x\,\dd x\tag{2.9.12}

\psi_0(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\ee^{-m\omega x^2/2\hbar}\tag{2.9.13}

(the prefactor comes from normalisation). The ground state is a Gaussian — a fact that becomes important again in the next section.

Step 6: the ground-state energy. From $\hat H=\hbar\omega(\hat a_+\hat a_-+\frac12)$ with $\hat a_-\psi_0=0$ :

E_0=\frac12\hbar\omega\tag{2.9.14}

Step 7: the whole spectrum. Climb from the bottom:

\psi_n=\frac{1}{\sqrt{n!}}\left(\hat a_+\right)^n\psi_0,\qquad E_n=\left(n+\frac12\right)\hbar\omega,\quad n=0,1,2,\dots\tag{2.9.15}

The entire solution used only the commutation relation $[\hat x,\hat p]=\ii\hbar$ and never once actually solved the second-order differential equation.

The picture

Equally spaced levels. This is unique to the oscillator. The infinite well’s spacing grows with $n$ , hydrogen’s shrinks with $n$ ; only the oscillator keeps a constant $\hbar\omega$ .

Consequence: the oscillator’s spectrum is a single line (every adjacent transition has the same frequency). Real molecules show several vibrational lines, which is precisely the evidence that the well is not exactly parabolic — anharmonicity is a rich source of information in molecular spectroscopy.

Zero-point energy. $E_0=\hbar\omega/2\ne0$ . Even at absolute zero the oscillator still jitters. This is no theoretical ornament: helium stays liquid at atmospheric pressure all the way down because the zero-point motion is violent enough to prevent freezing, and the Casimir force comes from the zero-point energy of electromagnetic field modes.

The mathematics

E_n=\left(n+\tfrac12\right)\hbar\omega\tag{2.9.16}

\psi_n(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\frac{1}{\sqrt{2^nn!}}H_n(\xi)\,\ee^{-\xi^2/2}\tag{2.9.17}

with $\xi=\sqrt{m\omega/\hbar}\,x$ and $H_n$ the Hermite polynomials:

H_0=1,\quad H_1=2\xi,\quad H_2=4\xi^2-2,\quad H_3=8\xi^3-12\xi\tag{2.9.18}

The recurrence $H_{n+1}=2\xi H_n-2nH_{n-1}$ is what this site’s simulations use to evaluate them (see appendix C).

Zero-point energy cannot simply be defined away

Someone will object: the zero of energy is arbitrary anyway, so why not just subtract $\hbar\omega/2$ ?

For a single isolated oscillator, fine. But changes in the zero-point energy are observable:

The Casimir effect: changing the separation of two metal plates changes which field modes are allowed, hence the total zero-point energy, producing a measurable force (verified precisely in 1997).
Isotope effects: $\omega\propto1/\sqrt{\mu}$ , so deuterium has a lower zero-point energy than hydrogen, giving C–H and C–D bonds different dissociation energies and reaction rates differing by factors of several. This is a routine tool in chemistry.

What is physically meaningful is always an energy difference, and the zero-point energy varies with the parameters — so it cannot be defined away.

Take a look

The one-dimensional harmonic oscillator

Equally spaced levels, zero-point energy, tails reaching into the classically forbidden region, and the one quantum state that really does behave like a classical particle. Units: ħ = m = 1.

Eₙ = (n+½)ħω: 0.500
Level spacing ħω: 1.00
Classical turning point ±A: ±1.00
Zero-point energy E₀: 0.500

Angular frequency ω1.00

A steeper well → larger spacing ħω and a narrower wavefunction

Quantum number n0

nodes = n = 0

View

Show ψ

Try this

In the "level ladder" view, push n up to 8: the levels stay exactly ħω apart. This is unique to the oscillator — look back at the infinite well, where the spacing grows as you climb.
Turn ω down to 0.4 and note that the ground-state energy E₀ = ħω/2 shrinks but never reaches zero. Making E₀ = 0 would require the particle to sit exactly at x = 0 with p = 0, and ΔxΔp ≥ ħ/2 forbids it.
Switch to "single eigenstate" with ψ shown and look at the n = 0 curve beyond the two dashed lines (the classical turning points): it has not gone to zero. The particle has a definite probability of being where classical energy conservation forbids — the very same root as tunnelling.
Show |ψ|², turn on "compare with classical", and take n from 0 to 20: the envelope of the quantum oscillation hugs the classical curve ever more closely (the ends are favoured because a classical particle moves slowest at the turning points).
Switch to "coherent state" and press play: the packet oscillates back and forth as a whole, keeps its shape, and its period is exactly the classical T = 2π/ω. Schrödinger found this most classical of quantum states back in 1926; it is also the theoretical description of laser light.

This is an interactive simulation (harmonic-oscillator) and needs JavaScript to run. The derivations and conclusions in the text do not depend on it, so feel free to read on.

Each of the three views answers one question:

Level ladder: is the spacing really constant? (Drag $n$ and watch the gaps between the lines.)
Single eigenstate: how much wavefunction lies beyond the classical turning points? (Look at the tails outside the two dashed lines.)
Coherent state: is there any quantum state that really moves like a classical particle? (Press play.)

Tails: leaking into the classically forbidden region

A classical particle with energy $E_n$ can only reach the turning point $x_n=\sqrt{2E_n}/\omega\sqrt{m}$ . Beyond that the quantum wavefunction decays as $\ee^{-\xi^2/2}$ , but it is not zero.

Compute it: the probability of finding the ground state in the forbidden regionadvanced~5 min

For the ground state $E_0=\hbar\omega/2$ , the turning point follows from $\frac12m\omega^2x_0^2=\frac12\hbar\omega$ :

x_0=\sqrt{\frac{\hbar}{m\omega}}\quad\Longleftrightarrow\quad \xi_0=1\tag{2.9.19}

(in the dimensionless coordinate $\xi=\sqrt{m\omega/\hbar}\,x$ the turning points sit exactly at $\xi=\pm1$ ).

P_{\text{forbidden}}=2\int_1^\infty\frac{1}{\sqrt\pi}\ee^{-\xi^2}\dd\xi =1-\operatorname{erf}(1)=1-0.8427=0.157\tag{2.9.20}

About 15.7%. That is not a small number — a ground-state particle has roughly a one-in-six chance of being found where it classically could never be.

What about excited states? The fraction shrinks as $n$ grows (about 5% at $n=10$ ), because the classically allowed region widens while the decay length of the tail stays put. Once again, the quantum effect weakens in relative terms as $n$ grows — the correspondence principle.

"The particle borrows energy to get out there" is the wrong story

The common explanation runs: the particle “borrows” an energy $\Delta E$ from the vacuum, sneaks out for a time $\Delta t\sim\hbar/\Delta E$ , and pays it back.

That invokes the energy–time uncertainty relation, but $\Delta t$ in quantum mechanics is not an observable corresponding to any operator; the meaning of that relation is quite different from $\sigma_x\sigma_p\ge\hbar/2$ , and the analogy does not hold.

The cleaner statement: finding the particle in the forbidden region cannot be distinguished from “measuring a definite negative kinetic energy” in any operational sense. The moment you localise the particle inside the forbidden region to within $\Delta x\lesssim1/\kappa$ , the uncertainty principle hands it a momentum spread $\Delta p\gtrsim\hbar\kappa$ , whose kinetic energy is enough to cover the potential deficit. In other words, the contradictory scenario of “measuring negative kinetic energy” simply cannot be set up experimentally.

Key formulas

Ladder operators
$\hat a_\pm=\frac{1}{\sqrt{2\hbar m\omega}}\left(\mp\ii\hat p+m\omega\hat x\right),\qquad[\hat a_-,\hat a_+]=1$
The whole structure hides in this one commutator
Hamiltonian
$\hat H=\hbar\omega\left(\hat a_+\hat a_-+\tfrac12\right)$
â₊â₋ is the number operator n̂
Energy levels
$E_n=\left(n+\tfrac12\right)\hbar\omega,\qquad n=0,1,2,\dots$
Equally spaced; the zero-point energy ħω/2 cannot be removed
Ground state
$\psi_0(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4}\ee^{-m\omega x^2/2\hbar}$
A Gaussian — and also a minimum-uncertainty state
General eigenstate
$\psi_n=\frac{1}{\sqrt{n!}}(\hat a_+)^n\psi_0\propto H_n(\xi)\,\ee^{-\xi^2/2}$
ξ = √(mω/ħ)·x, with Hₙ the Hermite polynomials

Self-check4 questions

What comes next

So far everything has been a bound state: the particle is trapped and the spectrum is discrete. The next section lets go — set the particle free and see what happens. The answer is a little surprising.

The harmonic oscillator

Recommended first

After this section you should be able to

Why a parabola

The ladder-operator method

The picture

The mathematics

Take a look

The one-dimensional harmonic oscillator

Tails: leaking into the classically forbidden region

Key formulas

Self-check4 questions

What comes next

Why a parabola

The ladder-operator method

◑The picture

∑The mathematics

Take a look

Tails: leaking into the classically forbidden region

∎Key formulas

What comes next

The picture

The mathematics

Key formulas