Checkpoint: Oscillator Energies

This checkpoint consolidates the core results of the module. Before attempting the exercise, review the three key facts established in earlier lessons.

The energy formula

The discrete energy levels of the quantum harmonic oscillator are

This follows from diagonalising the Hamiltonian $\hat{H} = \hbar\omega(\hat{N} + \tfrac{1}{2})$, where $\hat{N} = \hat{a}^\dagger \hat{a}$ is the number operator with non-negative integer eigenvalues.

Equal spacing

The energy difference between any two consecutive levels is constant:

More generally, the gap between levels $n$ and $m$ (with $n > m$) is

This is a direct consequence of the uniform ladder: each rung is exactly $\hbar\omega$ above the previous one. The result simplifies calculations considerably — you do not need to evaluate the half-integer offsets individually when computing a gap, because they cancel.

Zero-point energy

Even the lowest level, $n = 0$, carries a non-zero energy

This zero-point energy cannot be removed; it reflects the Heisenberg uncertainty principle, which forbids the particle from simultaneously having zero position uncertainty and zero momentum uncertainty. A perfectly stationary particle at the bottom of the well would violate $\Delta x\,\Delta p \geq \hbar/2$.

Connecting to physical units

In spectroscopy and molecular physics, the product $\hbar\omega$ is the natural energy unit. For a diatomic molecule with a bond vibrational frequency near $\nu \approx 10^{13}\,\text{Hz}$, the spacing works out to

a value in the mid-infrared range. Excitation of the $n = 0 \to n = 1$ transition absorbs a photon of exactly that energy — this is the basis of infrared molecular spectroscopy.

Try it

This is a numerical exercise — your code should return a number. A harmonic oscillator has $\hbar\omega = 0.25\,\text{eV}$. What is the energy difference $E_5 - E_2$ in eV?

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