Energy Levels Eₙ = ħω(n+½)

The quantum harmonic oscillator potential

A particle of mass $m$ in a parabolic potential $V(x) = \frac{1}{2}m\omega^2 x^2$ is the quantum harmonic oscillator. The angular frequency $\omega$ characterises how steeply the potential rises; every restoring-force system near a stable equilibrium reduces to this form in the small-displacement limit.

The time-independent Schrödinger equation for this system is

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

Solving it — either by the power-series (Frobenius) method or, more elegantly, with raising and lowering operators — forces the energy to take only discrete values.

Deriving the energy spectrum

The algebraic (ladder-operator) approach introduces two operators

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

whose commutator is $[\hat{a},\hat{a}^\dagger] = 1$ . The Hamiltonian rewrites as

\hat{H} = \hbar\omega\!\left(\hat{a}^\dagger\hat{a} + \tfrac{1}{2}\right) = \hbar\omega\!\left(\hat{N} + \tfrac{1}{2}\right),

where $\hat{N} = \hat{a}^\dagger\hat{a}$ is the number operator. Its eigenstates $|n\rangle$ satisfy $\hat{N}|n\rangle = n|n\rangle$ with $n = 0, 1, 2, \ldots$

Reading off the eigenvalues of $\hat{H}$ :

\boxed{E_n = \hbar\omega\!\left(n + \tfrac{1}{2}\right), \quad n = 0, 1, 2, \ldots}

The ladder operators act as $\hat{a}^\dagger|n\rangle = \sqrt{n+1}\,|n+1\rangle$ (raising by one quantum) and $\hat{a}|n\rangle = \sqrt{n}\,|n-1\rangle$ (lowering), with $\hat{a}|0\rangle = 0$ establishing the ground state floor.

Reading the formula

The factor $\frac{1}{2}\hbar\omega$ that persists even at $n = 0$ is the zero-point energy — a direct consequence of the uncertainty principle (covered in the next lesson). Above the ground state each successive level is spaced by exactly $\hbar\omega$ , so the spectrum is a uniform ladder:

| $n$ | $E_n$ | |-----|--------| | 0 | $\tfrac{1}{2}\hbar\omega$ | | 1 | $\tfrac{3}{2}\hbar\omega$ | | 2 | $\tfrac{5}{2}\hbar\omega$ | | 3 | $\tfrac{7}{2}\hbar\omega$ |

Equal spacing is a hallmark of the harmonic oscillator and is exploited heavily in quantum optics, where $|n\rangle$ are called Fock states (or photon-number states).

Physical scale: a diatomic molecule

For a diatomic molecule such as HCl the bond acts like a spring. Infrared spectroscopy shows absorption lines near $\hbar\omega \approx 0.36\,\text{eV}$ for HCl, consistent with lines measured near $\tilde{\nu} \approx 2886\,\text{cm}^{-1}$ . The vibrational transition from $n=0$ to $n=1$ absorbs a photon of energy exactly $\hbar\omega$ , as the equal-spacing rule predicts.

Try it

This is a numerical exercise — your code should return a number. For a vibrational mode with $\hbar\omega = 0.1\,\text{eV}$ , compute the energy of the $n = 3$ level using $E_n = \hbar\omega(n + \frac{1}{2})$ .

Run your code to see the quantum state.