Ladder Operators and the Number States

The analytic solution of the harmonic oscillator passes through Hermite's differential equation: guess a Gaussian asymptotic form, expand the remainder in a power series, and demand that the series terminate. The energies $E_n=(n+\frac{1}{2})\hslash \omega$ emerge as the termination condition. Dirac's algebraic method reaches the same spectrum without solving any differential equation. It rewrites the Hamiltonian as a product of two first-order operators, and from a single commutator every energy, every eigenstate, and every matrix element of $\hat{x}$ and $\hat{p}$ follows.¹

Factoring the Hamiltonian

The one-dimensional oscillator Hamiltonian is

Classically $p^2+(m\omega x{)}^2$ factors as $(m\omega x-ip)(m\omega x+ip)$. The operators $\hat{x}$ and $\hat{p}$ do not commute, so the quantum factorization leaves a residue, and that residue is the entire physics. Define the dimensionless lowering and raising operators

They are adjoints of each other because $\hat{x}$ and $\hat{p}$ are Hermitian. Multiplying them out, with $\hat{x}\hat{p}-\hat{p}\hat{x}=i\hslash$,

The oscillator Hamiltonian is therefore

Reversing the product order and using $[\hat{x},\hat{p}]=i\hslash$ once more gives $\hat{a}{\hat{a}}^{†}=\hat{H}/\hslash \omega +\frac{1}{2}$. Subtracting the two expressions leaves the one relation that drives everything:

The number operator

Define the number operator $\hat{N}={\hat{a}}^{†}\hat{a}$, so that $\hat{H}=\hslash \omega (\hat{N}+\frac{1}{2})$. It is Hermitian and positive: for any state $|\psi ⟩$, $⟨\psi |\hat{N}|\psi ⟩=\Vert \hat{a}|\psi ⟩{\Vert }^2\ge 0$. Since $\hat{H}$ and $\hat{N}$ differ only by a scale and a shift, they share every eigenstate; label them by the eigenvalue $\nu$,

The commutators of $\hat{N}$ with the ladder operators follow from $[\hat{a},{\hat{a}}^{†}]=1$:

These two relations are the ladder. Apply $\hat{N}$ to the state ${\hat{a}}^{†}|\nu ⟩$:

So ${\hat{a}}^{†}|\nu ⟩$ is an eigenstate of $\hat{N}$ with eigenvalue $\nu +1$: the raising operator climbs one rung. The identical calculation with $[\hat{N},\hat{a}]=-\hat{a}$ shows $\hat{a}|\nu ⟩$ carries eigenvalue $\nu -1$. The operators shift the energy by exactly one quantum $\hslash \omega$ in either direction.

The vacuum and the integer spectrum

Lowering cannot continue forever. Because $\hat{N}$ is positive, no eigenvalue can be negative, yet each application of $\hat{a}$ lowers $\nu$ by one. The only escape from producing a negative eigenvalue is that the descending chain terminates: there is a state $|0⟩$ with

This state has $\hat{N}|0⟩={\hat{a}}^{†}\hat{a}|0⟩=0$, so its eigenvalue is $\nu =0$ exactly. Suppose instead the smallest surviving eigenvalue were some ${\nu }_0$ not equal to zero with $\hat{a}|{\nu }_0⟩\ne 0$. Then $\hat{a}|{\nu }_0⟩$ would carry eigenvalue ${\nu }_0-1$, and repeating the descent would eventually cross zero into negative eigenvalues, contradicting positivity. The chain can stop only on the state annihilated by $\hat{a}$, which sits at $\nu =0$. Every eigenvalue is reached from it by integer steps of the raising operator, so

The full spectrum, ground-state energy included, comes from the commutator and the positivity of $\hat{N}$ alone. No boundary condition, no Hermite polynomial, no series termination.

Normalization and the tower

The raised and lowered states are eigenstates, but the ladder relations do not fix their length. Compute the norms. Using ${\hat{a}}^{†}\hat{a}=\hat{N}$ and $\hat{a}{\hat{a}}^{†}=\hat{N}+1$,

Choosing the phase so the coefficients are real and positive gives the standard normalization,

The factor $\sqrt{n}$ in the lowering formula reproduces $\hat{a}|0⟩=0$ automatically. Applying the raising operator $n$ times to the vacuum and dividing by the accumulated norms builds every state:

The $\sqrt{n!}$ is the product $\sqrt{1}\cdot \sqrt{2}\cdots \sqrt{n}$ of the raising norms collected on the way up. Orthonormality follows because the states belong to distinct eigenvalues of the Hermitian operator $\hat{N}$.

Position and momentum as ladder combinations

Inverting the definitions of $\hat{a}$ and ${\hat{a}}^{†}$ expresses the physical operators in the number basis:

Their matrix elements read off from the ladder action. Since $\hat{a}|n⟩$ lands on $|n-1⟩$ and ${\hat{a}}^{†}|n⟩$ on $|n+1⟩$, position connects only neighboring states:

In matrix form $\hat{x}$ is a tridiagonal band with zero diagonal, nonzero only one step off the main diagonal. The same holds for $\hat{p}$. This is the operator content of the $\Delta n=\pm 1$selection rule: the dipole operator $\hat{x}$ has no matrix element between states differing by more than one rung, so an oscillator emits or absorbs only the single quantum $\hslash \omega$.

Each nonzero cell shows the integer $k$ whose root $\sqrt{k}$ is the matrix element; the common prefactor $\sqrt{\hslash /2m\omega }$ is suppressed to expose the integer pattern.

The vacuum wavefunction recovered

The algebra never invoked the position representation, yet it must agree with it. The vacuum condition $\hat{a}|0⟩=0$ is a first-order differential equation once $\hat{a}$ is written with $\hat{p}=-i\hslash \mathbf{d}/\mathbf{d}x$:

Separating and integrating,

with the prefactor fixed by $\int |{\psi }_0{|}^2\mathbf{d}x=1$. This is the Gaussian ground state of the analytic treatment. Every excited state then follows by applying the differential form of ${\hat{a}}^{†}$, which manufactures exactly the Hermite polynomials: acting with $\hat{x}-(i/m\omega )\hat{p}$ on a Gaussian brings down a factor of $x$ minus a derivative, the Rodrigues operation that generates $H_n$.

Two consequences fall out of this example. The kinetic and potential energies split the total evenly,

the quantum virial theorem for a quadratic potential. And the uncertainty product is

saturated at $n=0$. The Gaussian vacuum is the minimum-uncertainty state, and the $\frac{1}{2}\hslash \omega$ ground-state energy is precisely the energy of that minimum-uncertainty compromise between kinetic and potential terms.

The ladder operators in the Heisenberg picture

The same commutator that fixed the spectrum also fixes the dynamics. In the Heisenberg picture an operator evolves by $\dot{\hat{A}}=(i/\hslash )[\hat{H},\hat{A}]$. For the lowering operator, using $\hat{H}=\hslash \omega ({\hat{a}}^{†}\hat{a}+\frac{1}{2})$ and $[{\hat{a}}^{†}\hat{a},\hat{a}]=-\hat{a}$,

and by conjugation ${\hat{a}}^{†}(t)={\hat{a}}^{†}(0)e^{+i\omega t}$. The lowering operator rotates in the complex plane at the classical frequency, unforced and undamped. Substituting into $\hat{x}=\sqrt{\hslash /2m\omega }(\hat{a}+{\hat{a}}^{†})$ reproduces the classical solution as an operator identity:

The operators obey Newton's equation $\ddot{\hat{x}}=-{\omega }^2\hat{x}$ exactly, with no expectation value taken. Whether a given state oscillates visibly is a separate question: a stationary state $|n⟩$ has $⟨\hat{x}⟩=0$ at all times because $\hat{a}$ and ${\hat{a}}^{†}$ connect it only to orthogonal neighbors. Producing a moving $⟨\hat{x}(t)⟩$ requires a superposition of adjacent number states, and the superposition that follows the classical trajectory most faithfully is the eigenstate of $\hat{a}$ constructed in the next lesson.

The algebraic and analytic routes to the oscillator agree on every result but differ in what they compute directly.

Why the algebra generalizes

The construction used only three inputs: a Hamiltonian written as $\hslash \omega (\hat{N}+\frac{1}{2})$ with $\hat{N}={\hat{a}}^{†}\hat{a}$, the commutator $[\hat{a},{\hat{a}}^{†}]=1$, and the positivity of $\hat{N}$. Any system whose excitations obey these relations has the same evenly spaced spectrum and the same ladder of states. Quantized field modes, phonons in a crystal, and photons in a cavity are each a collection of such oscillators, and ${\hat{a}}^{†}$, $\hat{a}$ become the creation and annihilation operators that add or remove one quantum of excitation. The number state $|n⟩$ is then read as $n$ quanta present, and the vacuum $|0⟩$ as the state with none. The coherent states of the next lesson are the superpositions of number states that behave most like a classical oscillation, and they are built directly from $\hat{a}$ and its adjoint.