13.4 Ladder operators and the harmonic oscillator

The quantum harmonic oscillator is the most important solved system in physics — every oscillation, every field mode, every particle in quantum field theory is one. And it can be solved without touching a differential equation, using only the commutator $[\hat x,\hat p]=i\hbar$ and a clever factorisation. The method — ladder operators — is the algebraic heart of the operator formalism and the direct ancestor of the creation and annihilation operators of QFT.

Factorising the Hamiltonian

The oscillator’s energy operator (Hamiltonian) is

\hat H \;=\; \frac{\hat p^2}{2m} + \frac12 m\omega^2 \hat x^2,

a sum of two squares. Classically a sum of squares factors as $(\hat x - i\hat p)(\hat x + i\hat p)$ ; because $\hat x$ and $\hat p$ do not commute, the quantum factorisation leaves a remainder — and that remainder is the whole story. Define the lowering and raising 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).

They are adjoints of each other (hence the dagger) and, unlike $\hat x$ and $\hat p$ , are not self-adjoint — they are not observables but tools.

where

$\hat a$: lowering (annihilation) operator
$\hat a^\dagger$: raising (creation) operator — the adjoint of â
$\hat N = \hat a^\dagger \hat a$: number operator; its eigenvalue n counts the quanta
$\omega$: oscillator angular frequency rad/s

▶ [â, â†] = 1 and Ĥ = ℏω(â†â + ½) Derivation

Compute the commutator using $[\hat x,\hat p]=i\hbar$ and bilinearity:

[\hat a,\hat a^\dagger] = \frac{m\omega}{2\hbar}\left[\hat x + \frac{i\hat p}{m\omega},\; \hat x - \frac{i\hat p}{m\omega}\right] = \frac{m\omega}{2\hbar}\left(\frac{-i}{m\omega}[\hat x,\hat p] + \frac{i}{m\omega}[\hat p,\hat x]\right) = \frac{m\omega}{2\hbar}\cdot\frac{2\hbar}{m\omega} = 1.

So $[\hat a,\hat a^\dagger] = 1$ — the canonical relation, repackaged. Now expand $\hat a^\dagger\hat a$ ; the cross terms are $\tfrac{i}{m\omega}(\hat x\hat p - \hat p\hat x) = \tfrac{i}{m\omega}(i\hbar) = -\tfrac{\hbar}{m\omega}$ , and collecting,

\hat a^\dagger\hat a = \frac{1}{\hbar\omega}\left(\frac{\hat p^2}{2m} + \frac12 m\omega^2\hat x^2\right) - \frac12 = \frac{\hat H}{\hbar\omega} - \frac12.

Rearranged, and writing the number operator $\hat N = \hat a^\dagger\hat a$ :

\boxed{\;\hat H = \hbar\omega\left(\hat a^\dagger\hat a + \tfrac12\right) = \hbar\omega\left(\hat N + \tfrac12\right).\;}

Finding the energy spectrum is now reduced to finding the eigenvalues of $\hat N$ .

Climbing the ladder

The names “raising” and “lowering” come from how $\hat a^\dagger$ and $\hat a$ move between the eigenstates of $\hat N$ . Everything follows from two commutators.

▶ â raises and lowers the number by one Derivation

From $[\hat a,\hat a^\dagger]=1$ one gets $[\hat N,\hat a^\dagger] = \hat a^\dagger$ and $[\hat N,\hat a] = -\hat a$ (a two-line calculation with the product rule). Let $|n\rangle$ be an eigenstate of $\hat N$ with eigenvalue $n$ , $\hat N|n\rangle = n|n\rangle$ . Then

\hat N\big(\hat a^\dagger|n\rangle\big) = \big([\hat N,\hat a^\dagger] + \hat a^\dagger\hat N\big)|n\rangle = \big(\hat a^\dagger + n\hat a^\dagger\big)|n\rangle = (n+1)\,\hat a^\dagger|n\rangle.

So $\hat a^\dagger|n\rangle$ is an eigenstate with eigenvalue $n+1$ : the raising operator climbs one rung. Identically, $\hat a|n\rangle$ has eigenvalue $n-1$ — it descends. The eigenvalues of $\hat N$ therefore come in a ladder spaced by $1$ .

\hat a|0\rangle = 0, \qquad \hat N|0\rangle = 0.

\hat a^\dagger|n\rangle = \sqrt{n+1}\,|n+1\rangle, \qquad \hat a|n\rangle = \sqrt{n}\,|n-1\rangle.

The spectrum

With $\hat N$ ‘s eigenvalues fixed at the non-negative integers, the energies follow from $\hat H = \hbar\omega(\hat N + \tfrac12)$ :

\boxed{\;E_n = \hbar\omega\left(n + \tfrac12\right), \qquad n = 0, 1, 2, \dots\;}

Three features, all obtained without solving a single differential equation:

Equal spacing. Levels are a uniform ladder separated by one quantum $\hbar\omega$ . Adding energy to an oscillator means climbing rungs, one $\hbar\omega$ at a time.
Zero-point energy. The ground state has $E_0 = \tfrac12\hbar\omega \neq 0$ : the oscillator cannot be brought fully to rest, a direct consequence of the uncertainty principle (13.3) — pinning both $\hat x$ and $\hat p$ to zero is forbidden.
Everything from algebra. The spectrum came entirely from $[\hat a,\hat a^\dagger]=1$ and positivity. The eigenfunctions $\psi_n(x)$ (Hermite polynomials times a Gaussian) can be generated afterward by applying $\hat a^\dagger$ to the ground state, but they were never needed to find the energies.

n = 0 · E₀ = 0.5 ℏω · a†|0⟩ = 1 |1⟩ · a|0⟩ = 0

The whole spectrum follows from the algebra [a, a†] = 1 and N = a†a: a† adds one quantum of energy ℏω, a removes one, and a annihilates the ground state. Each mode of a field is an oscillator like this — and a†(k) creating a quantum is what “a particle” means in quantum field theory.

Press $\hat a^\dagger$ to climb and $\hat a$ to descend the ladder; the highlighted rung is the current level $E_n = \hbar\omega(n+\tfrac12)$ , and the curve riding on it is the eigenfunction $\psi_n$ . At the bottom, $\hat a|0\rangle = 0$ — the lowering operator annihilates the ground state, and the ladder can go no lower.

Why this is the doorway to quantum field theory

The importance is out of all proportion to the simplicity. A field — the sound field of Sound 4.8, the electromagnetic field, a crystal lattice — is, after a Fourier decomposition (Ch 7), a collection of independent harmonic oscillators, one per mode. Quantising the field means quantising each of those oscillators with its own $\hat a_k, \hat a_k^\dagger$ . Then $\hat a_k^\dagger$ does not merely “raise a rung” — it creates a quantum of the mode $k$ , a particle: a photon for the electromagnetic field, a phonon for the sound field, an electron or quark for a matter field. The number operator $\hat N_k$ counts how many quanta occupy that mode. “A particle” in quantum field theory is an excitation of a mode’s ladder, and the algebra $[\hat a,\hat a^\dagger]=1$ is the whole grammar. The final lesson makes that step — canonical quantization — explicit.