Quantum Harmonic Oscillator

The quantum harmonic oscillator is one of the most important systems in physics, underpinning everything from molecular vibrations to quantum field theory. Its energy levels are equally spaced, and it is solved elegantly using ladder operators.

Key Concepts

Hamiltonian: Ĥ = p²/2m + mω²x²/2
Energy eigenvalues: Eₙ = (n+½)ℏω, n=0,1,2,...
Zero-point energy: E₀ = ℏω/2
Raising operator: â†|n⟩ = √(n+1)|n+1⟩
Lowering operator: â|n⟩ = √n|n-1⟩

Key Equations

Energy levels

E_n = \left(n+\frac{1}{2}\right)\hbar\omega

Raising operator

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

Lowering operator

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

Commutator

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

Worked Example

Example Problem

Problem

A diatomic molecule has vibrational frequency ν = 9.0×10¹³ Hz. Find the zero-point energy in eV.

Solution

E₀ = ℏω/2 = hν/2 = (6.626×10⁻³⁴ × 9×10¹³)/2 = 2.98×10⁻²⁰ J = 0.186 eV.

Practice

Exercises

7 problems

1 of 7

A quantum harmonic oscillator has ω = 2.0×10¹⁴ rad/s. Observe the parabolic potential and Gaussian ground state. Find the zero-point energy E₀ = ℏω/2 in eV.

The Gaussian ground state ψ₀ (green) oscillates inside the parabolic potential V(x) = ½mω²x². The yellow dashed line shows the zero-point energy E₀. Enter E₀ = ℏω/2 in eV.

E₀ = eV

2 of 7

Using the energy ladder diagram for the same oscillator (ω = 2.0×10¹⁴ rad/s), find E₃ in eV (n=3 state).

The ladder shows equally-spaced levels Eₙ = (n + ½)ℏω separated by ℏω = 0.1317 eV. The green level (n=3) is highlighted. An animated photon marks a downward transition. Enter E₃ in eV.

E₃ = eV

3 of 7

Find the energy spacing ΔE = Eₙ₊₁ - Eₙ for ω = 2.0×10¹⁴ rad/s in eV.