Topic 6 of 6

Harmonic Approximation

The second derivative test says a stable equilibrium is where $V''(x_0) > 0$. The Taylor expansion tells you more: every potential near a stable equilibrium looks like a harmonic oscillator to leading order. This single observation underlies molecular vibrations, crystal phonons, quantum field theory, and perturbation theory.

Key Concepts

Expanding

V(x)

in a Taylor series around equilibrium

x_0

(where

V'(x_0)=0

) gives

V \approx V_0 + \frac{1}{2}V''(x_0)(x-x_0)^2

. The coefficient

k = V''(x_0)

is the effective spring constant.

Anharmonicity

The

x^3

x^4

\ldots

terms in the Taylor expansion are anharmonic corrections. They cause overtone frequencies, thermal expansion, and are responsible for deviations from simple harmonic motion.

Zero-Point Energy

Quantum mechanics gives each harmonic mode an energy

E_0 = \hbar\omega/2

even in the ground state. This zero-point energy is a direct consequence of the harmonic approximation combined with quantum uncertainty.

Phonons

The atoms in a crystal sit near potential energy minima. The harmonic approximation treats their vibrations as independent harmonic oscillators — the quanta of these vibrations are phonons.

Key Equations

Taylor Expansion near Equilibrium

V(x) = V(x_0) + \underbrace{V'(x_0)}_{=\,0}(x-x_0) + \frac{1}{2}V''(x_0)(x-x_0)^2 + \cdots

At equilibrium

V'(x_0) = 0

, leaving the harmonic term as leading correction.

Harmonic Oscillator Frequency

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{V''(x_0)}{m}}

The angular frequency of small oscillations around stable equilibrium.

Zero-Point Energy

E_0 = \frac{1}{2}\hbar\omega

The ground-state energy of a quantum harmonic oscillator. Non-zero even at absolute zero.

Worked Example

Harmonic Approximation for a Cubic Potential

Problem

A particle of mass $m = 1$ kg is in the potential $V(x) = x^3 - 3x$ . Find the equilibrium point(s), the spring constant $k$ , and the angular frequency $\omega$ of small oscillations.

Solution

Find equilibria: $V'(x) = 3x^2 - 3 = 0 \Rightarrow x = \pm 1$ .

Check stability: $V''(x) = 6x$ . At $x=1$ : $V''(1) = 6 > 0$ (stable). At $x=-1$ : $V''(-1) = -6 < 0$ (unstable).

The spring constant at $x=1$ is $k = V''(1) = 6$ N/m.

\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{6}{1}} = \sqrt{6} \approx 2.449 \text{ rad/s}

Answer Stable equilibrium at

x=1

;

k = 6

N/m;

\omega = \sqrt{6} \approx 2.449

rad/s

Practice

Exercises

6 problems

1 of 6

For the potential $V(x) = 1 - \cos(x)$ , expand near $x=0$ to find the effective spring constant $k$ (from $V \approx \frac{1}{2}kx^2$ ). What is $k$ ?

Your answer

2 of 6

For $V(x) = 2x^2 + x^3$ , what is the effective spring constant $k = V''(0)$ at the equilibrium $x = 0$ ?

Your answer

N/m

3 of 6

With $k = 4$ N/m and mass $m = 0.5$ kg, what is the angular frequency $\omega$ of small oscillations (in rad/s)?

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4 of 6

A particle in potential $V(x) = 3x^2 - x^4/4$ at equilibrium $x=0$ . What is the effective spring constant $k = V''(0)$ ?

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5 of 6

The zero-point energy of a quantum harmonic oscillator with $\omega = 4\times 10^{13}$ rad/s and $\hbar = 1.055\times10^{-34}$ J·s is $E_0 = \hbar\omega/2$ . Express $E_0$ in units of $10^{-21}$ J.

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6 of 6

For the Morse potential near its minimum, $V''(r_0) = 8$ N/m and the reduced mass is $\mu = 2$ kg. What is the classical oscillation frequency $\omega$ in rad/s?

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Key Takeaways

Every stable equilibrium ( $V'(x_0)=0$ , $V''(x_0)>0$ ) looks like a harmonic oscillator to leading order, with spring constant $k = V''(x_0)$ .
The angular frequency of small oscillations is $\omega = \sqrt{k/m} = \sqrt{V''(x_0)/m}$ .
This is why diatomic molecules, crystal phonons, and quantum field modes are all treated as harmonic oscillators.
Anharmonic corrections (from $x^3$ , $x^4$ terms) cause frequency shifts, thermal expansion, and decay of phonons.
Quantum mechanically, each harmonic mode has zero-point energy $E_0 = \hbar\omega/2$ , unavoidable by the uncertainty principle.