Topic

The Harmonic Oscillator

Quantum Mechanics Classical Mechanics Foundations Examples

Why the Harmonic Oscillator Matters

The harmonic oscillator is the most important system in physics. This is not an exaggeration — it appears everywhere:

Vibrations of atoms in molecules and crystals
Electromagnetic field modes (photons)
Phonons in solids
The foundation of quantum field theory
Any system near a stable equilibrium

The reason is mathematical: any smooth potential near a minimum looks quadratic. If $V(x)$ has a minimum at $x_0$ , Taylor expansion gives:

$V(x) \approx V(x_0) + \frac{1}{2}V’‘(x_0)(x - x_0)^2 + \ldots$

The first derivative vanishes (it’s a minimum), so the leading term is quadratic — a harmonic oscillator!

The Classical Harmonic Oscillator

Definition

A harmonic oscillator is a system with a restoring force proportional to displacement:

$F = -kx$

where $k > 0$ is the spring constant.

Equation of Motion

Newton’s second law gives:

$m\ddot{x} = -kx$

Or:

$\ddot{x} + \omega^2 x = 0$

where $\omega = \sqrt{k/m}$ is the angular frequency.

Solution

The general solution is:

$x(t) = A\cos(\omega t) + B\sin(\omega t) = C\cos(\omega t + \phi)$

where $C = \sqrt{A^2 + B^2}$ is the amplitude and $\phi$ is the phase.

Energy

The potential energy is:

$V(x) = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2$

The kinetic energy is:

$T = \frac{1}{2}m\dot{x}^2$

The total energy:

$E = T + V = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\omega^2 x^2$

This is conserved and can take any positive value.

Example 1: Mass on a Spring

Setup

A mass $m = 0.5$ kg is attached to a spring with $k = 200$ N/m. It is displaced 0.1 m from equilibrium and released.

Calculation

Angular frequency: $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{200}{0.5}} = \sqrt{400} = 20 \text{ rad/s}$

Period: $T = \frac{2\pi}{\omega} = \frac{2\pi}{20} = 0.314 \text{ s}$

Frequency: $f = \frac{1}{T} = \frac{\omega}{2\pi} = 3.18 \text{ Hz}$

Position (with initial conditions $x(0) = 0.1$ m, $\dot{x}(0) = 0$ ): $x(t) = 0.1 \cos(20t) \text{ m}$

Velocity: $\dot{x}(t) = -2 \sin(20t) \text{ m/s}$

Maximum velocity: $v_{\max} = \omega A = 20 \times 0.1 = 2$ m/s

Total energy: $E = \frac{1}{2}kA^2 = \frac{1}{2}(200)(0.1)^2 = 1 \text{ J}$

Example 2: Simple Pendulum (Small Angles)

Setup

A pendulum of length $L$ swings through small angles $\theta$ .

Derivation

The tangential restoring force is:

$F = -mg\sin\theta \approx -mg\theta$

for small $\theta$ . With arc length $s = L\theta$ :

$m\ddot{s} = -mg\frac{s}{L}$

$\ddot{s} + \frac{g}{L}s = 0$

This is a harmonic oscillator with:

$\omega = \sqrt{\frac{g}{L}}, \quad T = 2\pi\sqrt{\frac{L}{g}}$

Calculation

For $L = 1$ m and $g = 9.8$ m/s²:

$T = 2\pi\sqrt{\frac{1}{9.8}} = 2\pi \times 0.319 = 2.01 \text{ s}$

This is the famous “seconds pendulum” — about 2 seconds per complete swing.

The Quantum Harmonic Oscillator

The Hamiltonian

In quantum mechanics, the Hamiltonian is:

$\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}m\omega^2\hat{x}^2$

We need to solve the eigenvalue problem:

$\hat{H}|\psi\rangle = E|\psi\rangle$

Method 1: Differential Equation

In position representation, $\hat{p} = -i\hbar \frac{d}{dx}$ , so:

$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2} + \frac{1}{2}m\omega^2 x^2 \psi = E\psi$

This is solved by the Hermite functions:

$\psi_n(x) = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4} \frac{1}{\sqrt{2^n n!}} H_n\left(\sqrt{\frac{m\omega}{\hbar}}x\right) e^{-\frac{m\omega x^2}{2\hbar}}$

where $H_n$ are the Hermite polynomials.

Method 2: Ladder Operators (Algebraic)

Define the annihilation and creation operators:

$\hat{a} = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} + \frac{i\hat{p}}{m\omega}\right)$

$\hat{a}^\dagger = \sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x} - \frac{i\hat{p}}{m\omega}\right)$

These satisfy:

$[\hat{a}, \hat{a}^\dagger] = 1$

The Hamiltonian becomes:

$\hat{H} = \hbar\omega\left(\hat{a}^\dagger\hat{a} + \frac{1}{2}\right) = \hbar\omega\left(\hat{n} + \frac{1}{2}\right)$

where $\hat{n} = \hat{a}^\dagger\hat{a}$ is the number operator.

Energy Levels

The eigenvalues of $\hat{n}$ are $n = 0, 1, 2, \ldots$

Therefore, the energy levels are:

$E_n = \hbar\omega\left(n + \frac{1}{2}\right), \quad n = 0, 1, 2, \ldots$

Key features:

Energy is quantized (discrete values only)
Levels are equally spaced by $\hbar\omega$
There is zero-point energy: $E_0 = \frac{1}{2}\hbar\omega \neq 0$

Ladder Operator Action

$\hat{a}|n\rangle = \sqrt{n}|n-1\rangle$

$\hat{a}^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$

The ground state satisfies $\hat{a}|0\rangle = 0$ .

All states can be built from the ground state:

$|n\rangle = \frac{(\hat{a}^\dagger)^n}{\sqrt{n!}}|0\rangle$

Example 3: Quantum Energy Levels

Setup

A particle of mass $m = 10^{-26}$ kg oscillates with frequency $\nu = 10^{13}$ Hz (typical for molecular vibrations).

Calculation

Angular frequency: $\omega = 2\pi\nu = 2\pi \times 10^{13} = 6.28 \times 10^{13} \text{ rad/s}$

Energy quantum: $\hbar\omega = (1.055 \times 10^{-34})(6.28 \times 10^{13}) = 6.63 \times 10^{-21} \text{ J}$

In electron volts: $\hbar\omega = \frac{6.63 \times 10^{-21}}{1.6 \times 10^{-19}} = 0.041 \text{ eV}$

Energy levels:

$n$	$E_n$ (eV)
0	0.021
1	0.062
2	0.103
3	0.144

Zero-point energy: $E_0 = \frac{1}{2}\hbar\omega = 0.021 \text{ eV}$

Even at absolute zero, the oscillator has this minimum energy — a purely quantum effect!

Example 4: Ground State Wave Function

The Ground State

From $\hat{a}|0\rangle = 0$ in position representation:

$\left(\hat{x} + \frac{i\hat{p}}{m\omega}\right)\psi_0 = 0$

$\left(x + \frac{\hbar}{m\omega}\frac{d}{dx}\right)\psi_0 = 0$

$\frac{d\psi_0}{dx} = -\frac{m\omega}{\hbar}x\psi_0$

Solution

$\psi_0(x) = A \exp\left(-\frac{m\omega x^2}{2\hbar}\right)$

Normalizing:

$A = \left(\frac{m\omega}{\pi\hbar}\right)^{1/4}$

Properties

The ground state is a Gaussian centered at the origin.

Width (standard deviation): $\Delta x = \sqrt{\langle x^2 \rangle} = \sqrt{\frac{\hbar}{2m\omega}}$

Momentum uncertainty: $\Delta p = \sqrt{\frac{\hbar m\omega}{2}}$

Uncertainty product: $\Delta x \cdot \Delta p = \frac{\hbar}{2}$

The ground state saturates the uncertainty principle — it’s a minimum uncertainty state!

Example 5: Transition Frequencies

Photon Emission

When the oscillator transitions from level $n$ to level $n-1$ , it emits a photon with energy:

$E_\gamma = E_n - E_{n-1} = \hbar\omega$

All transitions emit the same frequency! This is unique to the harmonic oscillator (equally spaced levels).

Selection Rule

The matrix element for electric dipole transitions:

$\langle m | \hat{x} | n \rangle = \sqrt{\frac{\hbar}{2m\omega}}\left(\sqrt{n}\delta_{m,n-1} + \sqrt{n+1}\delta_{m,n+1}\right)$

This gives the selection rule: $\Delta n = \pm 1$

Only transitions between adjacent levels are allowed (for electric dipole radiation).

The Classical Limit

Correspondence Principle

For large $n$ , quantum mechanics should approach classical behavior.

Classical energy: Any $E > 0$ is allowed

Quantum energy: $E_n = \hbar\omega(n + 1/2)$

For large $n$ : $\frac{E_{n+1} - E_n}{E_n} = \frac{\hbar\omega}{\hbar\omega(n + 1/2)} \approx \frac{1}{n} \to 0$

The energy levels become so closely spaced that they appear continuous — the classical limit is recovered.

Probability Density

Classically, the particle spends more time near the turning points (where it moves slowly).

The classical probability density is:

$P_{\text{classical}}(x) = \frac{1}{\pi\sqrt{A^2 - x^2}}$

For large $n$ , the quantum probability density $|\psi_n(x)|^2$ oscillates rapidly and its average approaches the classical distribution.

Connection to Quantum Field Theory

In QFT, the electromagnetic field is treated as a collection of harmonic oscillators — one for each mode (frequency and direction).

The photon is a quantum of excitation:

$|0\rangle$ = vacuum (zero photons)
$|1\rangle = \hat{a}^\dagger|0\rangle$ = one photon
$|n\rangle$ = n photons

The creation operator $\hat{a}^\dagger$ creates a photon. The annihilation operator $\hat{a}$ destroys a photon.

This is why the harmonic oscillator is the foundation of quantum field theory!

Summary

Aspect	Classical	Quantum
Energy	Any $E \geq 0$	$E_n = \hbar\omega(n + 1/2)$
Ground state	$E = 0$ (at rest)	$E_0 = \frac{1}{2}\hbar\omega$ (zero-point)
Position	Definite trajectory	Probability distribution
Minimum uncertainty	N/A	$\Delta x \Delta p = \hbar/2$
Transitions	Continuous	Discrete, $\Delta n = \pm 1$

Key Formulas

Quantity	Formula
Angular frequency	$\omega = \sqrt{k/m}$
Energy levels	$E_n = \hbar\omega(n + 1/2)$
Ladder operators	$[\hat{a}, \hat{a}^\dagger] = 1$
Ground state	$\psi_0 \propto e^{-m\omega x^2/2\hbar}$
Number operator	$\hat{n} = \hat{a}^\dagger\hat{a}$

References

D. J. Griffiths, Introduction to Quantum Mechanics, 3rd ed. (Cambridge University Press, 2018), Chapter 2
J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 2nd ed. (Addison-Wesley, 2011), Chapter 2
L. D. Landau and E. M. Lifshitz, Quantum Mechanics: Non-Relativistic Theory, 3rd ed. (Pergamon, 1977), §23
R. Shankar, Principles of Quantum Mechanics, 2nd ed. (Springer, 1994), Chapter 7
C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Vol. 1 (Wiley, 1977), Chapter V